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Finding range of a log function with variable base How can I find the range of $\log_{x-7}(x-5)$?
Here is my attempt.
Let $$y=\log_{x-7}(x-5)$$
Let $z=x-7$, $y=\log_z(z+2)$
then $y$ satisfies
$$z^y-z-2=0, z>0, z \ne 1.$$
$y$ be written as $\frac{\ln (x-5)}{\ln (x-7)}$.
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The domain of the function is $x>7$ and $x \ne 8$. We note that $\ln (x-5)>0$. $\ln(x-7)$ takes negative value when $x<8$ and it takes positive value when $x>8$.
Let's differentiate it,
$$f'(x) = \frac{\frac{\ln (x-7)}{x-5}-\frac{\ln (x-5)}{x-7}}{(\ln (x-7))^2}=\frac{(x-7)\ln (x-7) - (x-5) \ln (x-5)}{(x-5)(x-7)(\ln (x-7))^2}<0$$
Now let's examine $\lim_{x \to \infty}\frac{\ln (x-5)}{\ln(x-7)}=\lim_{x \to \infty} \frac{x-7}{x-5}=1$. Furthermore, it is clear that $1$ can't be attained.
$$\lim_{x\to 8^+}\frac{\ln (x-5)}{\ln (x-7)}=\infty$$
$$\lim_{x\to 8^-}\frac{\ln (x-5)}{\ln (x-7)}=-\infty$$
$$\lim_{x \to 7^+}\frac{\ln (x-5)}{\ln (x-7)}=0$$
Hence the range is $(-\infty, 0) \cup (1, \infty)$.
|
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"url": "https://math.stackexchange.com/questions/4257909",
"timestamp": "2023-03-29T00:00:00",
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|
Proving $\frac{2\sin x+\sin 2x}{2\sin x-\sin 2x}=\csc^2x+2\csc x \cot x+\cot^2x$
Prove
$$\dfrac{2\sin x+\sin 2x}{2\sin x-\sin 2x}=\csc^2x+2\csc x \cot x+\cot^2x$$
Proving right hand side to left hand side:
$$\begin{align}\csc^2x+2\csc x \cot x+\cot^2x
&= \frac{1}{\sin^2x}+\dfrac{2\cos x}{\sin^2x}+\dfrac{\cos^2x}{\sin^2x} \tag1 \\[0.8em]
&=\frac{\cos^2x+2\cos x+1}{\sin^2x} \tag2 \\[0.8em]
&=\frac{1-\sin^2x+1+\dfrac{\sin 2x}{\sin x}}{\sin^2x} \tag3 \\[0.8em]
&=\frac{\;\dfrac{2\sin x-\sin^3x+\sin 2x}{\sin x}\;}{\sin^2x} \tag4 \\[0.8em]
&=\frac{2\sin x-\sin^3x+\sin2 x}{\sin^3x} \tag5
\end{align}$$
I could not prove further to the left hand side from here. I would need help. Thank you in advance.
|
The following relations would be used in the answer:$$1-\cos^2x = \sin^2x$$ $$\sin 2x=2\sin x\cos x$$
Taking LHS,
$$\frac{2\sin x+\sin 2x}{2\sin x-\sin 2x} = \frac{2\sin x(1+\cos x)}{2\sin x(1-\cos x)}$$
Cancelling $\sin 2x$
$$=\frac{1+\cos x}{1-\cos x}$$
Multiplying $1+\cos x$ to numerator and denominator
$$=\frac{(1+\cos x)^2}{1-\cos^2x}$$
which is equivalent to $$=\frac{(1+\cos^2x+2\cos x)}{\sin^2x}$$
Divinding each term by denominator, we get LHS as $$=\csc^2x+2\csc x\cot x+\cot^2x$$
which is the RHS.
Hence, Proved.
|
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"timestamp": "2023-03-29T00:00:00",
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Roots of An Equation - Finding Equations in other degrees of roots (Link: https://pastpapers.papacambridge.com/viewer/caie/cambridge-advanced-as-and-a-level-mathematics-9231-2021-may-june-9231s21gtpdf-9231s21ms11pdf-9231s21ms12pdf-9231s21ms13pdf-9231s21ms21pdf-9231s21ms22pdf-9231s21ms23pdf-9231s21ms31pdf-9231s21ms32pdf-9231s21ms33pdf-9231s21ms41pdf-9231s21ms42pdf-9231s21ms43pdf-9231s21qp11pdf)
This is from a paper, so I'll give a summary of the question first.
$x^{4}-2x^{3}-1=0$
has roots $\alpha ,\beta ,\gamma , \delta$
a) Find equation with $\alpha^{3} ,\beta^{3} ,\gamma^{3} , \delta^{3}$
Answer $y = x^{3}$
Therefore, $y^{4}-8y^{3}-12y^{2} - 6y - 1 = 0$
b) Find the value of $1/\alpha ^{3}+ 1/\beta ^{3} + 1/\gamma ^{3} + 1/\delta ^{3}$
Answer -6
*** c) Find the value of $\alpha ^{4}+ \beta ^{4} + \gamma ^{4} + \delta ^{4}$ ***
Now the mark scheme says
$\alpha ^{4}+ \beta ^{4} + \gamma ^{4} + \delta ^{4} = 2(\alpha ^{3}+\beta ^{3}+\gamma^{3}+\delta ^{3}) + 4$ which is = 20. The mark scheme also says Uses original equation
Can someone explain to me how they got this equation for question C? Thanks in advance.
|
(From dxiv)
Original Equation $x^4 - 2x^3 - 1 = 0$
Therefore $x^4 = 2x^3 +1 $
Using each of the roots instead of x, we get
$\alpha^4 = 2\alpha^3 +1$
.....
.....
$\delta^4 = 2\delta^3 +1$
Add these all together to get $\alpha^4+\beta^4+\gamma^4+\delta^4 = 2(\alpha^3+\beta^3+\gamma^3+\delta^3) + 4$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
$10$ circles ($2$ large of radius $R$, $6$ small of radius $r$ and 2 small of radius $t$) are enclosed in a square. How we find $r$ in terms of $t$? Let us embed $2$ large intersecting circles of radius $R$ into a square as depicted by the figure below. These two circles are highlighted green. Into these $2$ circles we embedd $6$ smaller ones of equal radius $r$ (highlighted orange).
Finally we have $2$ circles of the same radius $t$, which both touch the large (green) circles and the square.
How we can find a formula that calculates the radius $r$ of the $6$ small circles when inputting the radius $t$?
One idea to proceed might be to define a distance $d$ from the center of one of the large (green) circles to the point at which the large radius $R$ is touching one of the small (orange) circles. Then at least we would get the Pythagorean triangle equation $r^2+d^2=(R − r)^2$ as a possibly useful starting point:
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Let the centres of the green circles be $O_1$ (lower) and $O_2$ (upper).
Let $P$ be the midpoint of $O_1O_2$ which is also the centre of the square.
Let the uppermost two orange circles above the line $O_1O_2$ have centres $C_1$ (lower) and $C_2$ (upper).
Let $Q$ be the point where the circle centre $C_2$ touches the line $O_1O_2$.
Now if we write $O_1P=d$, then $O_2P=d$, and since $\triangle O_1C_1P\equiv\triangle O_2C_2Q$, then $O_2Q=d\implies O_1Q=3d$.
Applying Pythagoras to $\triangle O_1C_1P\implies (R-r)^2=d^2+r^2\implies d^2=R^2-2Rr$
Applying Pythagoras to $\triangle O_1C_2Q\implies (R+r)^2=9d^2+r^2\implies 9d^2=R^2+2Rr$
Solving these gives $$8R^2=20Rr\implies R=\frac{5r}{2}$$
From this we can get $$d=\frac15R^2\implies d=\frac{R}{\sqrt{5}}\implies d=\frac{r\sqrt{5}}{2}$$
Now let $B$ be the centre of the lower blue circle of radius $t$, and let $BP=y$.
Applying Pythagoras in $\triangle PC_1B\implies (R+t)^2=d^2+y^2$
$$\implies(R+t)^2-\frac{R^2}{5}=y^2$$
$$\implies \frac45R^2+2Rt+t^2=y^2$$
Hence, $$y^2=5r^2+5rt+t^2$$
Now consider the lengths of the diagonals of the square: we have $$2(y+t\sqrt{2})=2d+2R\sqrt{2}$$
$$\implies y+t\sqrt{2}=d+\frac52r\sqrt{2}=r\left(\frac{\sqrt{5}+5\sqrt{2}}{2}\right)$$
Substituting for $y$ we now have
$$5r^2+5rt+t^2=\left[r\left(\frac{\sqrt{5}+5\sqrt{2}}{2}\right)-t\sqrt{2}\right]^2$$
So for a given value of $t$ you can obtain $r$ by solving a quadratic equation:
$$r^2\left(\frac{35+10\sqrt{10}}{4}\right)-rt(15+\sqrt{10})+t^2=0$$
|
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|
Prove that $\sqrt1+\sqrt2+\sqrt3+....+\sqrt n
$Q.$ Prove that $$\underbrace{\sqrt1+\sqrt2+\sqrt3+....+\sqrt n}_{\alpha}<n\sqrt{\underbrace{\frac{n(n+1)}{2}}_\beta}$$
Basically we need to prove that $$\alpha^2<n^2\beta$$
And $$\alpha^2=\beta+2(\sqrt{1.2}+\sqrt{2.3}+\sqrt{3.4}+......+\sqrt{(n-1)n})$$
using AM-GM on $(n-1)$ and $n$ $$\frac{2n-1}{2}>\sqrt{(n-1)(n)}$$
So , $$\alpha^2<\beta+\sum_{i=2}^{n}(2n-1)$$ $$\alpha^2<\beta-1+\sum_{i=1}^{n}2n-1$$ $$\alpha^2<\beta-1-n+n(n+1)$$ $$\alpha^2<\beta+(n+1)(n-1)$$
Now I'm not able to solve further . Can you give some hints ?
|
Here's what I did , $$\alpha^2=(\sqrt{1}+\sqrt{2}+\sqrt{3}+......+\sqrt{n})^2\Rightarrow$$ $$\color{red}{\sqrt1.\sqrt1}+\sqrt1.\sqrt2+\sqrt1.\sqrt3+.....+\sqrt1.\sqrt n\\\sqrt2.\sqrt1+\color{red}{\sqrt2.\sqrt2}+\sqrt2.\sqrt3+.....+\sqrt2.\sqrt n\\\sqrt3.\sqrt1+\sqrt3.\sqrt2+\color{red}{\sqrt3.\sqrt3}+.....+\sqrt3.\sqrt n\\.\\.\\.\\\sqrt n.\sqrt1+\sqrt n.\sqrt2+\sqrt n.\sqrt3+.....+\color{red}{\sqrt n.\sqrt
n} $$
Now apply AM-GM on each term , $$\alpha^2<\color{red}{(1+2+3+......+n)}+2\left(\frac{1+2}{2}+\frac{1+3}{2}+\frac{1+4}{2}+.......\right)$$OR $$\alpha^2<(1+2+3+.....+n)+(n-1)(1+2+3+......+n)$$ $$\alpha^2<n(1+2+3+.....+n)$$
since , $$n^2(1+2+3+.....+n)>n(1+2+3+......+n)$$ then $$(\sqrt1+\sqrt2+\sqrt3+......+\sqrt n)^2<n^2(1+2+3+......+n)$$
Hence
$$\sqrt1+\sqrt2+\sqrt3+.......+\sqrt n<n\sqrt{\frac{n(n+1)}{2}}$$
|
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Solve the following limit $\lim _{n\to \infty }\left( e^{\sqrt{n+1} -\sqrt{n}} -1\right) \times \sqrt{n}$ can someone try to solve this limit? $$\lim _{n\to \infty }\left( e^{\sqrt{n+1} -\sqrt{n}} -1\right) \times \sqrt{n}$$
I tried to solve it like $\lim _{n\to \infty }\left( e^{\sqrt{n+1} -\sqrt{n}} -1\right) \times \lim _{n\to \infty }\sqrt{n}$ but it always tends to 0, but in wolfram the result is $\frac{1}{2}$ so Im a bit confused, thanks!
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Note that $$ \sqrt{n+1} - \sqrt{n} = \frac{1}{\sqrt{n+1} +\sqrt{n}} $$
Let us denote this expression $x_n$. We have $\lim_{n\to\infty} x_n = 0$. Knowing that $\lim_{x\to0} \frac{e^x-1}{x} =1$, we get
\begin{align} & \lim_{n\to\infty} (e^{\sqrt{n+1} - \sqrt{n}} - 1)\sqrt{n} = \\
&=\lim_{n\to\infty} \frac{e^{x_n} - 1}{x_n} x_n \sqrt{n} = \\
&=\lim_{n\to\infty} \frac{e^{x_n} - 1}{x_n} \frac{\sqrt{n}}{\sqrt{n+1} +\sqrt{n}} = \\
&=\lim_{n\to\infty} \frac{e^{x_n} - 1}{x_n} \cdot \frac{1}{\sqrt{1+\frac{1}{n}}+1} = \\
&=\lim_{x\to 0} \frac{e^{x} - 1}{x} \cdot \lim_{n\to\infty} \frac{1}{\sqrt{1+\frac{1}{n}}+1} = \\
&= 1 \cdot \frac12 = \frac12 \end{align}
|
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f(xy)=f(x) * f(y) Let f be a function such that $f(mn) = f(m) f(n)$ for every positive integers m and n. If $f(1), f(2)$ and $f(3)$ are positive integers, $f(1) < f(2),$ and $f(24) = 54$, then $f(18)$ equals ?
Process:-
I attempted to solve this question using 2 approaches , but couldn't reach to the answer using approach 2
Approach 1:-
$f(24) = f(2^3 \cdot 3) = f(2^3) \cdot f(3) = [f(2)]^3 \cdot f(3) = 54$
Similarly writing $f(18)$ as $f (2 \cdot 3^2) = f(2) \cdot [f(3)]^2$
Also $f(1 \cdot 2) = f(1) \cdot f(2) > f(1)\Rightarrow f(1) \cdot [f(2) - 1] > 0,$ now as $f(1)$ is a positive integer we get $f (2) > 1$ , also $f(2) < 4$ , so we are left with $f(2) = 2$ or $f(2) = 3$
by using $f(2) = 2,$ we don't get $f(3) $ as an integer , and $f(2) = 3$ we get $f(3) = 2$
so finally putting $f(2)=3$ and $f(3)=2$ in the expression of $f(18)$ we get the value as $3 \cdot 4 = 12$
My 2nd approach :-
Using the result that $f(x y) = f(x) \cdot f(y)$ gives us a function of the form $f(x) = x ^ t$ , where $x, y$ are positive integers and t is a real number { I am not sure if I am using the condition correctly in this step , please correct me if wrong }
so $24 ^ t = 54$ $\Rightarrow t = \frac{\log (54)}{ \log (24) }$
and now we have to find $f(18)$ so it would be $18 ^ t = 18 ^{\frac{\log (54)}{ \log (24)}} = 37.63 $
where exactly am I going wrong in my 2nd approach , also please clarify when does $f(xy) = f(x) f(y)$ gives us a function of the form $x ^ t.$
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With $f(2)=3$ and $f(3)=2$ the form of $f(x)=x^t$ is not the correct form.
Note that $2^t=3 \implies t=\log_2 3$ while $3^t=2 \implies t=\log_3 2$
We do not necessarily get a closed form for this function.
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Probability all number will appear when rolling k n-sided dice If k balanced n-sided dice are rolled, what is the probability that each of the n different numbers will appear at least once?
A case of this was discussed here, but I’m not sure how to extend this. Specifically, I’m not sure how the to calculate the numbers that can repeat term in the accepted answer.
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(Please allow me to use $n$ in place of your $k$, and $m$ in place of your $n$)
So we have $n$ fair $m$-face dice.
If you consider the dies to be distinct, by color or by launching them in sequence, then
the space of events is given by $m^n$ equiprobable words (strings , m-tuples) of length $n$
formed out of the alphabet $\{1, 2, \ldots, m \}$.
Let's consider the development of
$$
\begin{array}{l}
\left( {x_{\,1} + x_{\,2} + \cdots + x_{\,m} } \right)^n = \\
= \cdots + x_{\,j_{\,1} } x_{\,j_{\,2} } \cdots x_{\,j_{\,n} } +
\cdots \quad \left| {\;j_i \in \left\{ {1, \ldots ,m} \right\}} \right.\quad = \\
= \sum\limits_{\left\{ {\begin{array}{*{20}c}
{0\, \le \,k_{\,j} \,\left( { \le \,n} \right)} \\
{k_{\,1} + k_{\,2} + \, \cdots + k_{\,m} \, = \,n} \\
\end{array}} \right.\;} {\left( \begin{array}{c}
n \\ k_{\,1} ,\,k_{\,2} ,\, \cdots ,\,k_{\,m} \\
\end{array} \right)x_{\,1} ^{k_{\,1} } x_{\,2} ^{k_{\,2} } \cdots x_{\,m} ^{k_{\,m} } } \\
\end{array}
$$
where
$$x_{\,j} ^{k_{\,j} } $$
accounts for the $j$th face (character) repeated $k_j$ times, and where putting the $x$'s at 1 we get
$$
\begin{array}{l}
\left( {\underbrace {1 + 1 + \cdots + 1}_m} \right)^n = m^n = \\
= \sum\limits_{\left\{ {\begin{array}{*{20}c}
{0\, \le \,k_{\,j} \,\left( { \le \,n} \right)} \\
{k_{\,1} + k_{\,2} + \, \cdots + k_{\,m} \, = \,n} \\
\end{array}} \right.\;} {\left( \begin{array}{c}
n \\ k_{\,1} ,\,k_{\,2} ,\, \cdots ,\,k_{\,m} \\
\end{array} \right)1^{k_{\,1} } 1^{k_{\,2} } \cdots 1^{k_{\,m} } } = \\
= \sum\limits_{\left\{ {\begin{array}{*{20}c}
{0\, \le \,k_{\,j} \,\left( { \le \,n} \right)} \\
{k_{\,1} + k_{\,2} + \, \cdots + k_{\,m} \, = \,n} \\
\end{array}} \right.\;} {\left( \begin{array}{c}
n \\ k_{\,1} ,\,k_{\,2} ,\, \cdots ,\,k_{\,m} \\
\end{array} \right)} \\
\end{array}
$$
Out of these we want to number the cases in which $k_1, k_2 , \ldots, k_m$ are at least one, i.e.
$$
N\left( {n,m} \right) = \sum\limits_{\left\{ {\begin{array}{*{20}c}
{1\, \le \,k_{\,j} \,\left( { \le \,n} \right)} \\
{k_{\,1} + k_{\,2} + \, \cdots + k_{\,m} \, = \,n} \\
\end{array}} \right.\;} {\left( \begin{array}{c}
n \\ k_{\,1} ,\,k_{\,2} ,\, \cdots ,\,k_{\,m} \\
\end{array} \right)}
$$
There are $\binom{m}{m-l} = \binom{m}{l}$ ways to choose $m-l$ characters not appearing and leaving $l$ to appear at least once
so it shall be
$$
\begin{array}{l}
m^n = \sum\limits_{\left\{ {\begin{array}{*{20}c}
{0\, \le \,k_{\,j} \,\left( { \le \,n} \right)} \\
{k_{\,1} + k_{\,2} + \, \cdots + k_{\,m} \, = \,n} \\
\end{array}} \right.\;} {\left( \begin{array}{c}
n \\ k_{\,1} ,\,k_{\,2} ,\, \cdots ,\,k_{\,m} \\
\end{array} \right)} = \\
= \sum\limits_{\left( {0\, \le } \right)\,l\,\left( { \le \,m} \right)} {\left( \begin{array}{c}
m \\ l \\
\end{array} \right)\sum\limits_{\left\{ {\begin{array}{*{20}c}
{1\, \le \,c_{\,j} \,\left( { \le \,n} \right)} \\
{c_{\,1} + \,c_{\,2} + \cdots \, + c_{\,l} \, = \,n} \\
\end{array}} \right.} {\;\left( \begin{array}{c}
n \\ c_{\,1} ,\,c_{\,2} , \cdots \,,c_{\,l} \, \\
\end{array} \right)} } \\
\end{array}
$$
But also it is, from the definition of the Stirling N. of 2nd kind
$$
\;m^{\,n} = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left\{ \begin{array}{c}
n \\ k \\ \end{array} \right\}\,m^{\,\underline {\,k\,} } }
= \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {k!\left\{ \begin{array}{c}
n \\ k \\ \end{array} \right\}\,\left( \begin{array}{c} m \\ k \\ \end{array} \right)}
$$
and therefore
$$
N\left( {n,m} \right) = \sum\limits_{\left\{ {\begin{array}{*{20}c}
{1\, \le \,k_{\,j} \,\left( { \le \,n} \right)} \\
{k_{\,1} + k_{\,2} + \, \cdots + k_{\,m} \, = \,n} \\
\end{array}} \right.\;} {\left( \begin{array}{c}
n \\ k_{\,1} ,\,k_{\,2} ,\, \cdots ,\,k_{\,m} \\
\end{array} \right)} = m!\left\{ \begin{array}{c}
n \\ m \\
\end{array} \right\}
$$
|
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Eliminating $r$ from $6\tan(r+x)=3\tan(r+y)=2\tan(r+z)$ Here is the question:
If $$\boldsymbol{6\tan (r+x)=3\tan(r+y)=2\tan(r+z)}$$
show that
$$\boldsymbol{3\sin^2(x-y)+5\sin^2(y-z)-2\sin^2(z-x)=0}$$
It seems easy but the natural approach goes off the rails.
What I have so far,
Take the two equalities and cross multiply,
$$2\sin(r+x)\cos(r+y)=\sin(r+y)\cos(r+x)$$
$$3\sin(r+y)\cos(r+z)=2\sin (r+z)\cos(r+y)$$
then use trigonometric addition formula,
$$2\sin(2r+x+y)+2\sin(x-y)=\sin (2r+x+y)+\sin(y-x)$$
$$3\sin(2r+y+z)+3\sin(y-z)=2\sin(2r+y+z)+2\sin(z-y)$$
And get
$$\sin(2r+x+y)=3\sin(y-x)$$
$$\sin(2r+y+z)=5\sin(z-y)$$
Now use the sum formula for sine and solve the linear system giving,
$$\sin 2r \sin(x-z)=3\sin(y-x)\sin(y+z)-5\sin(z-y)\sin(x+y)$$
$$\cos 2r \sin (x-z) =3\sin(y-x)\cos(y+z)-5\sin(z-y)\cos(x+y)$$
If we square them and add we eliminate $r$. and get
$$\sin^2 (x-z)=9\sin^2(y-x)+25\sin^2(z-y)$$
$$-30\sin(y-x)\sin(z-y)\cos(x-z)$$
But how to go from here ? This doesnt look anything like the desired formula.
Can there be an error in the original problem ? Am I making an error ? Or missing something ?
|
Just as you derive,
$$\sin(2r+x+y)=3\sin(y-x)\tag1$$
$$\sin(2r+y+z)=5\sin(z-y)\tag2$$
you can get, $$\sin(2r+z+x)=-2\sin(z-x)\tag3$$
From (1), can you get, $$\begin{align}\sin(2r+x+y)\sin(y-x)&=3\sin^2(y-x)\\ \frac12\left[\cos(2r+2x)-\cos(2r+2y)\right]&=3\sin^2(y-x)\tag{1´}\end{align}$$?
From (2), can you get, $$\begin{align} \sin(2r+y+z)\sin(z-y)&=5\sin^2(z-y)\\ \frac12\left[\cos(2r+2y)-\cos(2r+2z)\right]&=5\sin^2(z-y)\tag{2´}\end{align}$$?
Similarly, from (3), can you get, $$\begin{align} \sin(2r+z+x)\sin(x-z)&=-2\sin^2(z-x)\\ \frac12\left[\cos(2r+2z)-\cos(2r+2x)\right]&=-2\sin^2(z-x)\tag{3´}\end{align}$$?
Observe what happens when you add, $(1´)$, $(2´)$ and $(3´)$.
Additionally, you can solve this more general problem:
If $$\small \boldsymbol{a\tan (r+x)=b\tan(r+y)=c\tan(r+z)}$$
show that $$\small \boldsymbol{\left(\frac{a+b}{a-b}\right)\sin^2(x-y)+\left(\frac{b+c}{b-c}\right)\sin^2(y-z)+\left(\frac{c+a}{c-a}\right)\sin^2(z-x)=0}.$$
and enjoy! : )
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4277603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
the value of $\lim_{x\to0}\frac{(x^2+1) \ln(x^2+1)-\sin(x^2)}{x\sin(x^2)}$ I want to compute this limit
$$\lim_{x\to0}\frac{(x^2+1) \ln(x^2+1)-\sin(x^2)}{x\sin(x^2)}.$$
I tried to apply Hopital rule, but I cannot compute it.
|
Make life easier writing
$$\frac{(x^2+1) \ln(x^2+1)-\sin(x^2)}{x\sin(x^2)}=x\frac{(x^2+1) \ln(x^2+1)-\sin(x^2)}{x^2\sin(x^2)}$$ Now, let $t=x^2$ to make
$$\frac{(x^2+1) \ln(x^2+1)-\sin(x^2)}{x^2\sin(x^2)}=\frac{(t+1) \ln(t+1)-\sin(t)}{t\sin(t)}$$ Using equivalent $\ln(t+1)\sim t$ and $\sin(t)\sim t$
$$\frac{(t+1) \ln(t+1)-\sin(t)}{t\sin(t)}\sim \frac{(t+1)t-t }{t^2 }=1$$
So, the expression is "similar" to $x$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4277769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Determining the period of $ \frac{\sin(2x)}{\cos(3x)}$ I would like to compute the period of this function which is a fraction of two trigonometric functions. $$ \frac{\sin(2x)}{\cos(3x)}$$
Is there a theorem for this? what trick to use to easily find the period?
I started by reducing the fraction but I'm stuck on the rest.
For example, let $T$ be the period to be calculated: $$\frac{\sin(2x)}{\cos(3x)} =\frac{\sin(2x + 2 T)}{\cos(3x+3T)} = \frac{\sin(2x) \cos(2T)+\sin(2T) \cos(2x)}{\cos(3x) \cos(3T)-\sin(3T)\sin(3x)}$$
Thanks for your help.
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Suppose the period is $p$, and suppose the domain of the function is suitably defined, then for all values of $x$ in the domain, we must have $$\frac{\sin 2x}{\cos 3x}=\frac{\sin(2(x+p))}{\cos(3(x+p))}$$
$$\implies \sin 2x\cos(3x+3p)=\cos 3x\sin(2x+2p)$$
If we set $x=0$, we have $$\sin(2p)=0\implies 2p=k\pi.k\in\mathbb{Z}$$
On the other hand, noting that $\sin 2x=\cos 3x$ when $x=\frac{\pi}{10}$, amongst other possible values, if we set $x=\frac{\pi}{10}$, we get $$\cos\left(\frac{3\pi}{10}+3p\right)=\sin\left(\frac{\pi}{5}+2p\right)=\cos\left(\frac{3\pi}{10}-2p\right)$$
$$\implies\frac{3\pi}{10}+3p=\pm\left(\frac{3\pi}{10}-2p\right)+n\cdot2\pi,n\in\mathbb{Z}$$
From this we get $p=n\cdot2\pi$ or $p=\frac{3\pi}{5}+n\cdot2\pi$
In order to satisfy this and the previous result for $p$ we have to choose $k$ and $n$ such that both equations $p=\frac{k\pi}{2}$ and $p=n\cdot2\pi$ are satisfied and $p$ has minimum value, so we choose $k=4$ and $n=1$.
Therefore the period is $2\pi$.
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
If $\frac{(1+x)^2}{1+x^2}=\frac{13}{37}$, then find the value of $\frac{(1+x)^3}{1+x^3}$
Let $x$ be a real number such that $\frac{(1+x)^2}{1+x^2}=\frac{13}{37}$. What is the value of $\frac{(1+x)^3}{1+x^3}$?
Of course, one way is that to solve for $x$ from the quadratic $37(1+x)^2=13(1+x^2)$ which gives the value $x=\frac{-37\pm\sqrt{793}}{24}$. Thus we can compute $\frac{(1+x)^3}{1+x^3}=\frac{13}{49}$.
But I am trying to find the result from the equation $\frac{(1+x)^2}{1+x^2}=\frac{13}{37}$ without solving for $x$ using some algebra. Here are my attempts to do that:
We have $$\begin{align} & \frac{(1+x)^2}{1+x^2}=\frac{13}{37}\\ &\implies \frac{(1+x)^2}{(1+x)^2-2x}=\frac{13}{37}\\ &\implies \frac{(1+x)^3}{(1+x)^3-2x(1+x)}=\frac{13}{37}\\ &\implies \frac{(1+x)^3}{1+x^3+3x(2+x)-2x(1+x)}=\frac{13}{37}\\ &\implies \frac{(1+x)^3}{1+x^3+x(4+x)}=\frac{13}{37}
\end{align}$$
I can't seem to proceed from here.
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Given
$$\frac{(1+x)^2}{1+x^2}=a,$$
we have
$$\frac{(1+x)^3}{1+x^3}=\frac{(1+x)^3}{(1+x)(1-x+x^2)}=\frac{(1+x)^2}{1-x+x^2}.$$
Now,
$$\frac{1-x+x^2}{(1+x)^2}=\frac{\frac32(1+x^2)-\frac12(1+x)^2}{(1+x)^2}=\frac{3}{2a}-\frac12,$$
so the answer is
$$\frac1{\frac{3}{2a}-\frac12}=\frac{2a}{3-a}.$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4282424",
"timestamp": "2023-03-29T00:00:00",
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|
Approximate solution to a transcendental equation I'm working on a physics problem and stumbled upon the following equation:
$$h=\frac{2n\pi+\arctan\left(\frac{c}{b}\right)}{b}$$
where $n \in \mathbb{Z}$, $c \in [0,20]$ and $h \in \mathbb{R}^+$. This equation has to be solved for $b$, which can be done numerically ofcourse, but if possible, I would love to have a quite accurate expression in the given interval. Ive already did some work and obtained the following fair approximation for $b$:
$$b\approx\frac{(1+2n)\pi c}{2+ch}$$ However, this equation breaks down at low $c$. Thus, I was wondering if more accurate (analytical) expressions are possible.
|
We deal with the case $c > 0, h > 0, n\ge 1$.
The equation is written as
$$b = \frac{2n\pi}{h} + \frac{\arctan\frac{c}{b}}{h}.$$
Clearly, the equation has a unique real solution $b$. More precisely,
this solution is located in $ (\frac{2n\pi}{h}, \frac{2n\pi + \pi/2}{h})$.
Using the famous Lagrange inversion theorem, the solution is given by
$$b = \frac{2n\pi}{h} + \sum_{k=1}^\infty \frac{1}{k! h^k}
\frac{\partial^{k - 1}}{\partial x^{k - 1}}\left[\left(\arctan\frac{c}{x}\right)^k\right]\Bigg\vert_{x = \frac{2n\pi}{h} }.$$
An approximation is
\begin{align*}
b &\approx \frac{2n\pi}{h} + \sum_{k=1}^2 \frac{1}{k! h^k}
\frac{\partial^{k - 1}}{\partial x^{k - 1}}\left[\left(\arctan\frac{c}{x}\right)^k\right]\Bigg\vert_{x = \frac{2n\pi}{h} }\\
&= \frac{2n\pi}{h} + \frac{1}{h}\arctan \frac{ch}{2n\pi}
- \frac{c}{4\pi^2 n^2 + c^2h^2}\arctan \frac{ch}{2n\pi}.
\end{align*}
Examples:
(1) $c = 1/2, h = 1, n = 2$,
the solution is $\approx 12.60601344$,
and the approximation is $12.60601266$.
(2) $c = 1/100, h = 1/5, n = 10$,
the solution is $\approx 314.1594245$,
and the approximation is $314.1594246$.
(3) $c = 1/10, h = 100, n = 10$,
the solution is $\approx 0.6298929625$,
and the approximation is $0.6298929436$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Integration operator appears inside an integration operator as I try $~\mathcal{L}\left[x \cdot \sin^{}\left(x\right) \right]\left(s\right)~$ $$ A:=\mathcal{L}\left[x \cdot \sin^{}\left(x\right) \right]\left(s\right) \tag{1} $$
$$=\lim_{\beta\to\infty}\int_{0}^{\beta}\left(x\cdot\sin^{}\left(x\right)\right)\cdot\exp\left(-sx\right)\,dx$$
$$ = \lim_{\beta\to\infty}\int_{0}^{\beta}\left(x\right)\cdot \underbrace{\left( \sin^{}\left(x\right) \exp\left(-sx\right) \right)}_{\text{This part is to be integrated} } \,dx $$
$$ = \lim_{ \beta \to \infty} \left\{ \left[ x \cdot \left( \int_{ }^{ } \sin^{}\left(x\right) \exp\left(-sx\right) \,dx \right) \right]_{0}^{\beta} - \int_{0 }^{\beta } \left( \int_{ }^{ } \sin^{}\left(x\right) \exp\left(-sx\right) \,dx \right) \,dx \right\} $$
$$ \int_{ }^{ } \sin^{}\left(x\right) \exp\left(-sx\right) \,dx = - \frac{ s \cdot \exp\left(-sx\right) }{ \left( s^2+1 \right) } \left( \sin^{}\left(x\right) + \frac{1}{s}\cos^{}\left(x\right) \right) +\text{const} \tag{2} $$
Should I have written the derivation of the above equation?
$$ A= \lim_{ \beta \to \infty} \left\{ \left[ -\frac{ x \cdot s \cdot \exp\left(-sx\right) }{ \left( s^2+1 \right) } \left( \sin^{}\left(x\right) + \frac{1}{s}\cos^{}\left(x\right) \right) \right]_{0}^{\beta} - \int_{0 }^{\beta } \left( \int_{ }^{ } \sin^{}\left(x\right) \exp\left(-sx\right) \,dx \right) \,dx \right\} $$
$$ = \lim_{ \beta \to \infty} \left\{ \underbrace{\left[ -\frac{ x \cdot s }{ \left( s^2+1 \right) e^{sx} } \left( \sin^{}\left(x\right) + \frac{1}{s}\cos^{}\left(x\right) \right) \right]_{0}^{\beta}}_{\text{About}~\beta~ \text{,it converges to }~0 } - \int_{0 }^{\beta } \left( \int_{ }^{ } \sin^{}\left(x\right) \exp\left(-sx\right) \,dx \right) \,dx \right\} $$
$$ = -\lim_{ \beta \to \infty} \underbrace{\int_{0 }^{\beta } \left( \int_{ }^{ } \sin^{}\left(x\right) \exp\left(-sx\right) \,dx \right) \,dx}_\text{What can be done at here?} $$
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Just proceed calculations using equations of laplace transformations of $~ \sin^{}\left(x_{}\right) ~~,~~ \cos^{}\left(x\right) ~$
$$ A=\mathcal{L}\left[x \cdot \sin^{}\left(x\right) \right]\left(s\right) \tag{1} $$
$$ = -\lim_{ \beta \to \infty} \int_{0 }^{\beta } \left( \int_{ }^{ } \sin^{}\left(x\right) \exp\left(-sx\right) \,dx \right) \,dx $$
$$ = -\lim_{ \beta \to \infty} \int_{0 }^{\beta } \left(- \frac{ s }{ s^2+1 } \right) \frac{ 1 }{ e^{sx} } \left( \sin^{}\left(x\right) + \frac{1}{ s } \cos^{}\left(x\right) \right) \,dx $$
$$ = \frac{ s }{ s^2+1 } \lim_{ \beta \to \infty} \int_{ 0}^{ \beta} \frac{1}{ e^{sx} } \left( \sin^{}\left(x\right) + \frac{1}{ s } \cos^{}\left(x\right) \right) \,dx $$
$$ = \frac{ s }{ s^2+1 } \lim_{ \beta \to \infty} \int_{ 0}^{\beta } \left( \sin^{}\left(x\right) \exp\left(-sx\right) + \frac{1}{ s } \cos^{}\left(x\right) \exp\left(-sx\right) \right) \,dx $$
$$ = \frac{ s }{ s^2+1 } \left( \mathcal{L}\left[\sin^{}\left(x\right) \right]\left(s\right) + \frac{1}{ s } \mathcal{L}\left[\cos^{}\left(x\right) \right]\left(s\right)\right) $$
$$ = \frac{ s }{ s^2+ 1 } \left( \frac{ 1 }{ s^2+1 } + \frac{1}{ s } \frac{ s }{ s^2+1 } \right) $$
$$ = \frac{ s }{ s^2+ 1 } \left( \frac{ 2 }{ s^{2}+1 } \right) $$
$$ = \frac{ 2s }{ \left( s^{2}+1 \right) ^2 } $$
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"url": "https://math.stackexchange.com/questions/4293189",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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|
Quadratic Recurrences is there any way to solve 'simple' recurrences like $a_0=2$, $a_{n+1}=(2a_n-1)^2$ or even $a_0=1$, $a_{n+1}=(2a_n+1)^2$ ? Any help much welcomed !
Regards, Knud
|
Let us denote $b_n = 2a_n -1$, then
$$b_0 = 3$$
$$b_{n+1} = 2b_n^2 -1$$
We seek $x$ satisfying $b_0 = \text{cosh}(x) := \frac{e^x+e^{-x}}{2}$ then
$$x = \text{arcosh}(3) = \ln(3+2\sqrt{2})=2\ln(\sqrt{2}+1)$$
Now, it is easy to prove that $b_n = \text{cosh}(2^nx)$, indeed
$$b_{n+1} = 2\left(\frac{e^{2^n x} +e^{-2^n x}}{2} \right)^2 -1 =\frac{e^{2^{n+1} x} +e^{-2^{n+1} x}}{2} = \text{cosh}(2^{n+1}x)$$
Conclusion:
$$a_n =\frac{\text{cosh}(2^{n+1}\ln(\sqrt{2}+1)) +1}{2} $$
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{
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"url": "https://math.stackexchange.com/questions/4293944",
"timestamp": "2023-03-29T00:00:00",
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|
$x_1 \leq 4, x_1+x_2 \leq 13, x_1+x_2+x_3 \leq 29, x_1+...+x_4 \leq 54, x_1+...+x_5 \leq 90$. Find the maximum of ...
$x_1 \leq 4, x_1+x_2 \leq 13, x_1+x_2+x_3 \leq 29, x_1+...+x_4 \leq 54, x_1+...+x_5 \leq 90 \text{ for } x_1, ..., x_5 \in R_0^+. \\ \ \\ \text{Find the maximum of } \sqrt{x_1}+\sqrt{x_2}+\sqrt{x_3}+\sqrt{x_4}+\sqrt{x_5}.$
Of course, the answer will be $20(x_1=4, x_2=9, x_3=16, x_4=25, x_5=36.)$
My attempt:
\begin{align}
&\text{let } \sum_k \ x_l = x_l+x_{l+1}+x_{l+2}+...+x_{k+l-1}. \ (x_{5n+m}=x_m) \\
&\bigg(\sum_5 \sqrt{x_1} \bigg)^2 \leq 5\bigg(\sum_5 x_1\bigg) \leq 450. \\
&\therefore \bigg( \sum_5 \sqrt{x_1} \bigg) \leq 15\sqrt{2}. \\
\ \\
&\bigg(\sum_4 \sqrt{x_1} \bigg)^2 \leq 4\bigg(\sum_4 x_1 \bigg) \leq 216. \\
&\therefore \bigg( \sum_4 \sqrt{x_1} \bigg) \leq 6\sqrt{6}. \\
\ \\
&\bigg(\sum_3 \sqrt{x_1} \bigg)^2 \leq 3\bigg(\sum_3 x_1 \bigg) \leq 87. \\
&\therefore \bigg(\sum_3 \sqrt{x_1} \bigg) \leq \sqrt{87}. \\
\ \\
&\bigg(\sum_2 \sqrt{x_1} \bigg)^2 \leq 2\bigg(\sum_2 x_1 \bigg) \leq 26. \\
&\therefore \bigg(\sum_2 \sqrt{x_1} \bigg) \leq \sqrt{26}. \\
\ \\
&\bigg(\sqrt{x_1} \bigg)^2 \leq 4. \\
&\therefore \sqrt{x_1} \leq 2. \\
\end{align}
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$$x_1 \leq 4 \tag {1}$$ $$x_1+x_2 \leq 13\tag{2}$$ $$x_1+x_2+x_3 \leq 29\tag{3}$$ $$x_1+...+x_4 \leq 54\tag {4}$$ $$x_1+...+x_5 \leq 90\tag 5$$
now $$(5)\times 10+(4)\times 2+(3)\times 3+(2)\times 5+(1)\times 10 \implies$$ $$10x_5+12x_4+15x_3+20x_2+30x_1\le 1200$$ also by C-S $${\left(10x_5+12x_4+15x_3+20x_2+30x_1 \right)}{\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{15}+\frac{1}{20}+\frac{1}{30}\right)}\ge {\left(\sum \sqrt{x_i}\right)}^2$$ can you end it now?
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve: $\frac{1}{\sin(\pi/n)}=\frac{1}{\sin(2\pi/n)}+\frac{1}{\sin(3\pi/n)}$
The positve integer satisfying value of $n: n>3$ satisfying the equation: $\dfrac{1}{\sin(\pi/n)}=\dfrac{1}{\sin(2\pi/n)}+\dfrac{1}{\sin(3\pi/n)}$ is
This question is from a Practice Book for CBSE Term$-1$ Maths Preparation Class $10$.
What I tried:
$ \dfrac{1}{\sin ({\pi / n})}-\dfrac{1}{\sin({3\pi / n})}=\dfrac{1}{\sin({2 \pi / n})} $
$ \dfrac{\sin({3\pi / n})-\sin({\pi / n})}{\sin({\pi / n})\times \sin({3\pi / n})}=\dfrac{1}{\sin ({2 \pi / n})} $
$\dfrac{2\cos({2 \pi / n})\sin({\pi / n})}{\sin({\pi / n})\sin({3\pi / n})}=\dfrac{1}{\sin ({2 \pi / n})}$
$ 2\cos ({2 \pi / n})\sin({2 \pi / n})=\sin({3\pi / n}) $
$ \sin (4\pi/n)=\sin ({3\pi / n}) $
How to continue this solution? Also this solution is getting pretty complex? Is there any other way to solve this?
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Using the formula
$$\sin A-\sin B=2\sin\left(\frac{A-B}{2}\right)\cos\left(\frac{A+B}{2}\right)$$
we have
\begin{align*}
\sin \dfrac{4\pi}{n}&=\sin \dfrac{3\pi}{n}\\
\sin \dfrac{4\pi}{n}-\sin \dfrac{3\pi}{n}&=0\\
2\sin\left(\frac{\pi}{2n}\right)\cos\left(\frac{7\pi}{2n}\right)&=0
\end{align*}
then
$$
\begin{array}{rcl}
\sin\left(\dfrac{\pi}{2n}\right)&=&0\\[2mm]
\dfrac{\pi}{2n}&\in&\{k\pi:k\in\mathbb Z_{\neq 0}\}\\
n&\in&\left\{\dfrac{1}{2k}:k\in\mathbb Z_{\neq 0}\right\}
\end{array}\quad\text{ or }\quad
\begin{array}{rcl}
\cos\left(\dfrac{7\pi}{2n}\right)&=&0\\[2mm]
\dfrac{\pi}{2n}&\in&\left\{(2k+1)\frac{\pi}{2}:k\in\mathbb Z\right\}\\
n&\in&\left\{\dfrac{7}{2k+1}:k\in\mathbb Z\right\}
\end{array}
$$
Thus
$$n\in\left\{\dfrac{1}{2k}:k\in\mathbb Z_{\neq 0}\right\}\cup\left\{\dfrac{7}{2k+1}:k\in\mathbb Z\right\}$$
|
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|
Find all integer solutions of $(a+b+3)^2+2ab=3ab(a+2)(b+2)$ Here is another number theory problem that I am not able to do
Find all integer solutions of $$(a+b+3)^2+2ab=3ab(a+2)(b+2)$$
My attempt$:$
On expanding we get,
$$b^2+4ab+6b+a^2+6a+9=3a^2b^2+6a^2b+6ab^2+12ab$$ or $$-6ab^2+b^2-8ab+6b+a^2+6a+9=3a^2b^2+6a^2b$$ or $$-6ab^2+b^2-8ab+6b+a^2+6a+9-3a^2b^2-6a^2b=0$$ or $$\left(1-3b^2-6b\right)a^2+\left(-6b^2-8b+6\right)a+b^2+6b+9=0$$ or $$a=\frac{-\left(-6b^2-8b+6\right)\pm \sqrt{\left(-6b^2-8b+6\right)^2-4\left(1-3b^2-6b\right)\left(b^2+6b+9\right)}}{2\left(1-3b^2-6b\right)}$$ or $$a=\frac{-\left(-6b^2-8b+6\right)\pm \:4\sqrt{3}\left(b+1\right)\sqrt{b\left(b+2\right)}}{2\left(1-3b^2-6b\right)};\quad \:b\ne \:-\frac{3+2\sqrt{3}}{3},\:b\ne \frac{2\sqrt{3}-3}{3}$$ and after further solving $$a=\frac{3b^2+4b-3+2\sqrt{3}b\sqrt{b\left(b+2\right)}+2\sqrt{3}\sqrt{b\left(b+2\right)}}{1-3b^2-6b}$$ and $$a=\frac{3b^2+4b-3-2\sqrt{3}b\sqrt{b\left(b+2\right)}-2\sqrt{3}\sqrt{b\left(b+2\right)}}{1-3b^2-6b}$$ where $b\ne \:-\frac{3+2\sqrt{3}}{3},\:b\ne \frac{2\sqrt{3}-3}{3}$
So, now I can assign some integer values to $a$ and check if that value of $a$ is giving some integer values of $b$ or not. Like if $a=0$ then $b=-3$. But I think it is merely hit and trial because this thing can also apply to the original equation. Any help will be greatly appreciated.
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Let x=a+1, y = b+1, so
$(x+y+1)^2 + 2(x-1)(y-1) = 3(x^2-1)(y^2-1)$.
The right hand side is quite obviously larger when |x|, |y| >= 3. Finding solutions for x=-2, -1, 0, 1, 2 are easy to find.
|
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|
An integral identity with a complex parameter Conjecture: For $z\in\mathbb{C}\setminus[-1,1]$, we have
$$
\int_{0}^{\pi}\frac{dx}{|z-\cos x|}=\frac{\pi}{\sqrt{|z^{2}-1}|}.
$$
I know it is true for $z\in\mathbb{R}\setminus[-1,1]$ but not in general.
|
First note that the function
$$f(z) = \frac{dx}{|z-\cos x|}$$
is an even function. Hence,
$$
I = \int_{0}^{\pi}\frac{dx}{|z-\cos x|} = \frac{1}{2}\int_{-\pi}^{\pi}\frac{dx}{|z-\cos x|}
$$
If you do the following substitution:
$$ \cos x = \frac{w+w^{-1}}{2} = \frac{w^2+1}{2w}$$
$$dx = \frac{dw}{wi}$$
$$ I = \frac{1}{2}\int_{-\pi}^{\pi}\frac{dx}{|z-\cos x|} = \frac{1}{i} \oint_{|w|=1} \frac{1}{|w^2+2wz+1|} dw $$
your integral is transformed in a contour integral round the unit complex circle
If $z\in \mathbb{R}$, the function
$$ g(w) = \frac{1}{|w^2+2wz+1|} = \frac{1}{|w+\sqrt{z^2+1}+z||w-\sqrt{z^2+1}+z|}$$ has two singularities:
$$ w_{1} = -\sqrt{z^2-1}-z$$
$$ w_{2} = \sqrt{z^2-1}-z$$
To apply the residue theorem, one of the singularites has to be inside the unit circle:
$$|w_{1}|<1 \Longrightarrow z<-1 $$
$$|w_{2}|<1 \Longrightarrow z>1 $$
Then if $z<- 1$ then
$$I = \frac{1}{i} \oint_{|w|=1} \frac{1}{|w^2+2wz+1|} dw = 2\pi \operatorname{Res}(g,w_{1}) = 2\pi \lim_{w \to -\sqrt{z^2-1}-z} \frac{w+\sqrt{z^2+1}+z}{|w+\sqrt{z^2+1}+z||w-\sqrt{z^2+1}+z|} = \frac{2\pi}{|-2\sqrt{z^2+1}|} = \frac{\pi}{|\sqrt{z^2+1}|}$$
If $z>1$ then
$$I = \frac{1}{i} \oint_{|w|=1} \frac{1}{|w^2+2wz+1|} dw = 2\pi \operatorname{Res}(g,w_{2}) = 2\pi \lim_{w \to -\sqrt{z^2-1}+z} \frac{w-\sqrt{z^2+1}+z}{|w+\sqrt{z^2+1}+z||w-\sqrt{z^2+1}+z|} = \frac{2\pi}{|2\sqrt{z^2+1}|} = \frac{\pi}{|\sqrt{z^2+1}|}$$
Note that we have used the following property to calculate these limits:
If $w=x+iy$
$$ \lim_{x\to a, y\to 0} \frac{w}{|w|} = \lim_{x\to a, y\to 0} \frac{x+iy}{\sqrt{x^2+y^2}} = \lim_{x \to a} \frac{x}{|x^2|} = \lim _{x \to a} \frac{x}{|x|} = \operatorname{sign}(a)$$
Hence we can conclude
$$
\boxed{\int_{0}^{\pi}\frac{dx}{|z-\cos x|}=\frac{\pi}{|\sqrt{z^{2}-1}|} \quad z\in \mathbb{R}\backslash[-1,1] } $$
The statement is not true in general
Consider $z=i$,
The left hand side is
$$\int_{0}^{\pi} \frac{1}{|i-\cos x|} dx = 2\int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{1+\cos^2 x}} dx = 2\int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{2-\sin^2 x}}dx = \frac{2}{\sqrt{2}} \int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{1-\frac{1}{2}\sin^2 x}} dx = \sqrt{2}K\left(\sqrt{\frac{1}{2}}\right)\approx 2.62 $$
The integral converges and the solution is the complete elliptic integral of the first kind. However, the right hand side of the formula
$$\frac{\pi}{|\sqrt{i^{2}-1}|} =\frac{\pi}{|\sqrt{-2}|} = \frac{\pi}{\sqrt{2}} \approx 2.22$$
So
$$
\int_{0}^{\pi}\frac{dx}{|i-\cos x|}\neq \frac{\pi}{|\sqrt{i^{2}-1}|} $$
|
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Is it possible to prove the formula to factorize $(a^n + b^n)$ using the formula that holds for $(a^n - b^n) $? ( with $n\geq 3$) Is it possible to prove the factorization of $a^3 + b^3$ such that $a^3 + b^3= (a+b) ( a^2 - ab + b^2)$ using the formula for the factorization of
$ (a^n - b^n) = (a-b) ( a^{n-1} + a^{n-2}b^1
+ ... + ... + a^1 b^{n-2} + b^{n-1}) $ ,
and , in particular the instance of this formula with $n=3$, I mean :
$ a^3 - b^3 = (a-b) ( a^2 + ab + b^2) ?$
The problem I see is this : certainly, $a^n + b^n = a^n - (-b^n)$. But , in order to apply the formula that holds for the difference , shouldn't we have also : $ a^n - (-b)^n$?
If I'm right in asserting this necessary condition, the question arises : is it the case in general that
$a^n - (-b^n)=a^n - (-b)^n$ ?
More generally, how to prove the " sum of $n^{\mathbb {th}}$ powers" formula?
|
Just let $c=-b$. Then $a^3+b^3=a^3-c^3=(a-c)(a^2+ac+c^2)=(a+b)(a^2-ab+b^2)$.
Note that this only works because $3$ is odd.
|
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show that $\lim\limits_{n\to\infty} \frac{a_1+\cdots + a_n}n = 1$
Let $(a_n)$ be a bounded sequence of positive real numbers such that $\lim\limits_{n\to\infty} \sqrt[n]{a_1\cdots a_n} = 1.$ Prove that $\lim\limits_{n\to\infty} \dfrac{a_1+\cdots + a_n}n = 1.$
By the AM-GM inequality, $ \sqrt[n]{a_1\cdots a_n} \leq \dfrac{a_1+\cdots + a_n}n $ for all $n$ so taking limits on both sides yields $\lim\limits_{n\to\infty} \frac{a_1+\cdots + a_n}n \ge \lim\limits_{n\to\infty} \sqrt[n]{a_1\cdots a_n} = 1.$ Thus it suffices to show that $\lim\limits_{n\to\infty} \dfrac{a_1+\cdots + a_n}n \leq 1.$ Also, if a sequence of real numbers $(a_n)$ satisfies $\lim\limits_{n\to\infty} a_n = L,$ then $\lim\limits_{n\to\infty} \dfrac{a_1+\cdots + a_n}n = L,$ so if we could show that $\lim\limits_{n\to\infty} a_n = 1,$ we would be done. Suppose without loss of generality that $L = \lim\limits_{n\to\infty} a_n > 1$ (the case where $L < 1$ is similar). Let $N > 1$ be such that $a_n > \frac{1+L}2$ for all $n\ge N.$ Then for $n \ge N, \sqrt[n]{a_1\cdots a_n} = \sqrt[n]{a_1\cdots a_{N-1}}\sqrt[n]{a_N \cdots a_n} > R_n \sqrt[n]{(\frac{1+L}2)^{n-N+1}}$, where $R_n = \sqrt[n]{a_1\cdots a_{N-1}}.$ Hence taking the limit of both sides (for $n\ge N$) yields $\lim\limits_{n\to\infty} \sqrt[n]{a_1\cdots a_n}\ge \lim\limits_{n\to\infty} R_n (\dfrac{1+L}2)^{(n-N+1)/n} = \frac{1+L}2 > 1,$ a contradiction. Hence $\lim\limits_{n\to\infty} a_n = 1,$ so the result follows.
Is there something missing, or is the above proof correct?
|
This is false. Consider the sequence
$$a_{j} = \begin{cases}2 && \text{if }j\text{ is odd}\\
\frac{1}{2} && \text{if }j \text{ is even} \end{cases}$$
we have for all $n$
$$1 \leq (a_{1}...a_{n})^{\frac{1}{n}} \leq 2^{\frac{1}{n}}$$
Thus
$$\lim_{n \rightarrow \infty}(a_{1}...a_{n})^{\frac{1}{n}} = 1$$
But
$$\lim_{n \rightarrow \infty}\frac{\sum_{j=1}^{2n}a_{j}}{2n} = \lim_{n \rightarrow \infty}\frac{2n+\frac{1}{2}n}{2n} = \frac{5}{4}.$$
|
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|
prove the sign of $1+2x-e^x$ around 0 I want to prove that the sign of $1+2x-e^x$ around $0$ is negative if $x\to 0^-$ and positive if $x\to 0^+$. If I replace $e^x$ by its equivalent function (the tangent at 0) I conclude that the sign of this expression depends on the sign of $x$ as I have $1+2x - 1-x $.
I need this step to compute the limit of $\frac{1+2x+e^x}{1+2x-e^x}$ when $x\to 0$
|
You should already know from the definition that:
$$
e^x=\lim_{n\to\infty}(1+\frac xn)^n=1+x+\frac{x^2}{2!}+...+\frac{x^k}{k!}+...
$$
Obviously for all $x>0$: $e^x>1+x$. To consider negative $x$'s we write:
$$
e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}(1-\frac x4)+...+\frac{x^{2m+1}}{(2m+1)!}(1-\frac x{2m+2})+...
$$
Thus for small negative $x$, all the terms starting on the $\frac{x^3}{3!}(1-\frac x4) $ term are negative and we conclude:
$$
e^x<1+x+\frac{x^2}2
$$
for small negative $x$.
From the above we get that $1+2x-e^x>1+2x-(1+x)=x>0$ for small positive $x$ and $1+2x-e^x<1+2x-(1+x+\dfrac{x^2}{2})=x-\dfrac{x^2}{2}<0 $ for small negative $x$.
|
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|
Is this method of showing that $\frac{2n^2}{n^3+3} \rightarrow 0$ correct? Given
$$
a_n = \frac{2n^2}{n^3+3}
$$
I want to show that
$$
a_n \rightarrow 0
$$
My solution proceeds as follows.
\begin{align}
\left\lvert \frac{2n^2}{n^3+3} - 0 \right\rvert &< \epsilon \\
\frac{2n^2}{n^3+3} &< \epsilon
\end{align}
since
\begin{align}
\frac{2n^2}{n^3+3} &< \frac{2n^2}{n^3} \\
&= \frac{2}{n}
\end{align}
then
\begin{align}
\frac{2}{n} < \epsilon \implies n > \frac{2}{\epsilon}
\end{align}
Therefore, let $N = \frac{2}{\epsilon}$. Fix $\epsilon > 0$ such that $\forall n > N$,
\begin{align}
\left\lvert \frac{2n^2}{n^3+3} - 0 \right\rvert < \frac{2}{n} < \frac{2}{N} = \epsilon
\end{align}
I am relying on the fact that since $\lvert a_n - 0 \rvert$ is upper-bounded by $\frac{2}{n}$, then the $N$ that I find for $\frac{2}{n}$ will be greater than the $N$ that I find for $\lvert a_n - 0 \rvert$.
Is all of this correct? If so, is there a name for this kind of method/reasoning?
|
Two points:
*
*You set $N:=\tfrac{2}{\varepsilon}$. But what is $\varepsilon$? Also $N$ need not be an integer. That's fine, but the symbol $N$ does suggest an integer value.
*Next you 'fix $\varepsilon$ such that $\forall n>N$...'.
But to prove that the limit equals $0$, you want to show that $$(\forall\varepsilon>0)(\exists N)\left(n>N\quad\Longrightarrow\quad \left|\tfrac{2n^2}{n^3+3}-0\right|<\varepsilon\right).$$
So instead, you want to start off from something like 'Let $\varepsilon>0$ be given...', and then give an appropriate value of $N$, and show that the inequality does indeed hold.
Your first few steps are a good way to determine such an $N$ given some $\varepsilon>0$. You can follow these steps in reverse to show that given $\varepsilon>0$, the desired inequality holds for all $n>N$ for the $N$ that you found.
For example, my answer to the question
Show that $\lim_{n\to\infty}\frac{2n^2}{n^3+3}=0$.
would simply be:
Given any $\varepsilon>0$, for all $n>\frac{2}{\varepsilon}$ we have
\begin{align}
\left\lvert \frac{2n^2}{n^3+3} - 0 \right\rvert <\left\lvert\frac{2n^2}{n^3}\right\rvert=\frac{2}{n}<\epsilon.
\end{align}
|
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|
show this $(f(x))''+2\ge 0$,for any real numbers, where $f(x)=\frac{x^2-1}{x^{2n}-1}$ When I did a question today, I turned to the following question
let $n$ be postive integer,and $x\in R$, let $f(x)=\dfrac{x^2-1}{x^{2n}-1}$,show this
$$(f(x))''+2\ge 0 \tag{1}$$
For example $n=2$
then
$$(f(x))''+2=\dfrac{2x^2(x^4+3x^2+6)}{(x^2+1)^3}\ge 0$$
$n=3$,then
$$(f(x))''+2=\dfrac{2x^4(x^8+3x^6+6x^4+17x^2+15)}{(x^4+x^2+1)^3}\ge 0$$
and when $n=4$,see links1
$n=5$ see links2,
$n=6$ see links3
always hold,so How to prove for $(1)$,maybe this usefull:
\begin{align*}(f(x))''+2&=(f(x)+x^2)''=\left(\dfrac{x^{2n+2}-1}{x^{2n}-1}\right)''\\
&=\dfrac{x^{2n-2}(4n^2(x^2-1)(x^{2n}+1)-2n(3x^2+1)(x^{2n}-1)+2x^2(x^{2n}-1)^2)}{(x^{2n}-1)^3}\end{align*}see links4
|
As stated above in a comment, we can write
$$f(x)=\frac{1}{1+x^2...+x^{2n-2}}=\frac{1}{g(x)}$$
which yields that
$$f''(x)+2=\frac{2g'^2-gg''+2g^3}{g^3}$$
so we only have to decide the sign of the numerator, given that $g(x)>0$ for all x. Since $f(x)=f(-x)$ it also follows that the second derivative is even, which allows us to only examine $x\geq 0$. It seems that the coefficients of the numerator are explicitly positive, so let us see how we can prove this. First, write out the expressions for $g'^2 , gg''$:
$$2g'^2=\left(\sum_{r=0}^{n-1}\sum_{l=0}^r+\sum_{r=n}^{2n-2}\sum_{\ell=r-n+1}^{r}\right)8\ell(r-\ell)x^{2r-2}$$
$$g''g=\left(\sum_{r=0}^{n-1}\sum_{l=0}^r+\sum_{r=n}^{2n-2}\sum_{\ell=r-n+1}^{r}\right)2\ell(2\ell-1)x^{2r-2}$$
Subtracting these two we see that
$$2g'^2-gg''=-\sum_{r=0}^{n-1}x ^{2r-2}r(1+r)+\sum_{r=n}^{2n-2}[4(r-n)^2+6r-5n+1]x^{2r-2}$$
Note that in the 2nd sum all the coefficients are positive and in the 1st sum negative. We will prove that the expansion of $2g^3$ exactly cancels out the $n-1$ smallest powers (note that $r=0$ does not contribute but is included for convenience).
After some manipulations we see that
$$g^3=\sum_{r=0}^{3n-3}x^{2r}\sum_{s=\max\{0,r-n+1\}}^{r}\sum_{m=\max\{0,s-n+1\}}^{s}1$$
Here, all coefficients are manifestly positive as well. Now for $0\leq r\leq n-2$ we immediately see that after performing the sum and a change of index:
$$2g^3=\sum_{r=0}^{n-2}x^{2r}\sum_{s=0}^r\sum_{m=0}^s 1+...=\sum_{r=0}^{n-1}x^{2r-2}r(1+r)+\sum_{r=n-1}^{3n-3}A_{nr}x^{2r}~, ~A_{nr}>0$$
which finally shows that the first $n-1$ coefficients are cancelled and the remaining expression is a polynomial in $x^2$ of degree $3(n-1)$ with positive coefficients and therefore is positive for all $x>0$. At $x=0$ evidently, $f''(0)+2=0$ and hence $f''(x)+2\geq 0$ for all $x$, with a zero of order $2n-2$ at the origin.
|
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|
Number of real solutions of $\begin{array}{r} {\left[\frac{2 x+1}{3}\right]+\left[\frac{4 x+5}{6}\right]} =\frac{3 x-1}{2} \end{array}$ Solve for $x \in \mathbb{R}$ $$\begin{array}{r}
{\left[\frac{2 x+1}{3}\right]+\left[\frac{4 x+5}{6}\right]}
=\frac{3 x-1}{2}
\end{array}$$ where $[x]$ denotes greatest integer less than or equal to $x$.
My try:
Letting $a=\frac{2x+1}{3}$ we get
$$\left[a\right]+\left[a+\frac{1}{2}\right]=\frac{9 a-5}{4} \tag{1}$$
Now knowing that:
$$a-1<[a]\leq a$$ and $$a-\frac{1}{2}<\left[a+\frac{1}{2}\right]\leq a+\frac{1}{2}$$
Adding both the above inequalities and using $(1)$ we get
$$2a-\frac{3}{2}<\frac{9a-5}{4}\leq 2a+\frac{1}{2}$$
So we get $$a \in (-1, 7]$$
Any help here?
|
Thanks for hermites identity that @mymolecules showed...
$\left[\frac{\mathrm{2}{x}+\mathrm{1}}{\mathrm{3}}\right]+\left[\frac{\mathrm{4}{x}+\mathrm{5}}{\mathrm{6}}\right]=\frac{\mathrm{3}{x}−\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\: \\ $
$\left[\frac{\mathrm{2}{x}+\mathrm{1}}{\mathrm{3}}\right]+\left[\frac{\mathrm{2}{x}+\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{2}}\right]=\frac{\mathrm{3}{x}−\mathrm{1}}{\mathrm{2}} \\ $
$\left[{p}\right]=\left[\frac{\mathrm{4}{x}+\mathrm{2}}{\mathrm{3}}\right]=\frac{\mathrm{3}{x}−\mathrm{1}}{\mathrm{2}}={n}\in{Z}\:\:\:\:\:\:\:\:\: \\ $
${x}=\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{3}}\Rightarrow{p}=\frac{\mathrm{8}{n}+\mathrm{10}}{\mathrm{9}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $
$\frac{\mathrm{8}{n}+\mathrm{10}}{\mathrm{9}}−{n}=\frac{\mathrm{10}−{n}}{\mathrm{9}}=\left\{{p}\right\}\in\left[\mathrm{0},\mathrm{1}\right)\:\:\: \\ $
${n}\in\left\{\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6},\mathrm{7},\mathrm{8},\mathrm{9},\mathrm{10}\right\}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $
${x}\in\left\{\frac{\mathrm{5}}{\mathrm{3}},\frac{\mathrm{7}}{\mathrm{3}},\frac{\mathrm{9}}{\mathrm{3}},\frac{\mathrm{11}}{\mathrm{3}},\frac{\mathrm{13}}{\mathrm{3}},\frac{\mathrm{15}}{\mathrm{3}},\frac{\mathrm{17}}{\mathrm{3}},\frac{\mathrm{19}}{\mathrm{3}},\frac{\mathrm{21}}{\mathrm{3}}\right\} \\ $
|
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|
Ways of Changing 85/90 dollars The number of ways of changing 85 dollars with 1,5,10 and 20 dollar bills should be the coefficient of $x^{85}$ in $(1+x+x^2+\ldots+x^{85})(1+x^5+\ldots+x^{85})(1+x^{10}+\ldots+x^{80})(1+x^{20}+\ldots+x^{80})$, that is, $\frac{(1-x^{81})^2}{(1-x)(1-x^5)}\cdot\frac{(1-x^{86})^2}{(1-x^{10})(1-x^{20})}$, however if we want to change 90 dollars with 5,10,20 and 50 bills we can just say its the coefficient of $x^{90}$ in $\frac{1}{(1-x^5)(1-x^{10})(1-x^{20})(1-x^{50})}$. Why can we "eliminate" everything in the numerator in that case?
|
First thing, your formula for finite geometric sums is incorrect. It should be
$$1+x^a+x^{2a}+x^{3a}+x^{4a}+\ldots +x^{na}=\frac{x^{(n+1)a}-1}{x-1}$$
So what you actually should have is that in the $85$ case, it is the coefficient of $x^{85}$ in
$$\frac{(1-x^{86})(1-x^{90})(1-x^{90})(1-x^{100})}{(1-x)(1-x^5)(1-x^{10})(1-x^{20})}$$
The reason why you can omit all the terms of the numerator (aside from $1$) is from considering the expansion of $(1-x^{86})(1-x^{90})(1-x^{90})(1-x^{100})$. This will of course contain the term $1$. The rest of the terms, however, will be powers of $x$ that are greater than $x^{85}$. When multiplied with the series expansion of the denominator, none of these other terms will affect the coefficient of $x^{85}$ because they are already bigger than $x^{85}$.
Another reason to see this is from a similar reasoning. While you are trying to compute the coefficient of $x^{85}$ in the finite polynomial
$$(1+x+x^2+\ldots+x^{85})(1+x^5+\ldots+x^{85})(1+x^{10}+\ldots+x^{80})(1+x^{20}+\ldots+x^{80})$$
What's the difference between the coefficient of $x^{85}$ in the original polynomial and
$$(1+x+x^2+\ldots+x^{85}+x^{86})(1+x^5+\ldots+x^{85})(1+x^{10}+\ldots+x^{80})(1+x^{20}+\ldots+x^{80})$$
You will see that there is none, and we can keep on adding powers higher than $x^{85}$ because they do not affect the coefficient of $x^{85}$ in the expansion. Hence, this is the exact same as computing the coefficient of $x^{85}$ in the expansion of
$$1+x+x^2+\ldots)(1+x^5+\ldots)(1+x^{10}+\ldots)(1+x^{20}+\ldots)$$
$$=\frac{1}{(1-x)(1-x^5)(1-x^{10})(1-x^{20})}$$
|
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Simplify $\sqrt{\frac{\sqrt[4]{x^3}-8}{\sqrt[4]{x}-2}+2\sqrt[4]{x}}\left(\frac{\sqrt[4]{x^3}+8}{\sqrt[4]{x}+2}-\sqrt{x}\right)$ Simplify $$\sqrt{\dfrac{\sqrt[4]{x^3}-8}{\sqrt[4]{x}-2}+2\sqrt[4]{x}}\left(\dfrac{\sqrt[4]{x^3}+8}{\sqrt[4]{x}+2}-\sqrt{x}\right)$$
Is it a good idea to simplify the square root with the common denominator $\sqrt[4]{x}-2$. I tried it and it seemed useless at the end. Any hints would be appreciated. Thank you!
|
Start with setting $\sqrt[4]{x} = u$ (also mentioned by $\textit{Achilles hui}$ in the comments).
We get
$$L= \sqrt{\frac{u^3-8}{u-2}+2u}\left(\frac{u^3+8}{u+2}-u^2\right)$$
Using the identities for sum and differences of cubes we have
$$L=\sqrt{(u^2+2u+4)+2u}((u^2-2u+4)-u^2)$$
$$L=(u+2)\cdot ( -2(u-2))=-2(u^2-4)=-2(\sqrt{x}-4)$$
|
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How to find the solution of $x^2\equiv 25\pmod{32}$? I am trying to find square root of $57$ modulo $32\times 49$. For that I need to find the solutions of $x^2\equiv 57\pmod{32}$ and $x^2\equiv 57\pmod{49}$ which are $x^2\equiv 25\pmod{32}$ and $x^2\equiv 8 \pmod{49}$. Now I could find the solution of the second one as follows: $x_1=1$ is a solution to $x^2\equiv 8 \pmod{7}$. Let $x_2=x_1+7k=1+7k$ be a solution of
$x^2\equiv 8 \pmod{49}$, so
$$\begin{align*}
(1+7k)^2\equiv 8 \pmod{49}&\implies 1+14k\equiv 8\pmod{49} \\
&\implies 14k\equiv 7 \pmod{49}\\
&\implies 2k\equiv 1 \pmod{7},
\end{align*}$$ thus we can find $k$ as $4$, so $1+7k=29$ and hence $x_2=29$ is a solution of $x^2\equiv 8 \pmod{49}$.
Now for the first equation, $x_1=1$ is a solution of $x^2\equiv 25\pmod{8}$. Now I take $x_2=x_1+8k=1+8k$ to be a solution of $x^2\equiv 25 \pmod{16}$, then I try to proceed in a similar way. But I get stuck because it boils down to finding $k$ such that $2k\equiv 1\pmod{2}$ which has no solution. My question is how to find a solution for $x^2\equiv 25\pmod{32}$?
|
Well, the easiest way to find a solution of $x^2\equiv 25\pmod{32}$ is to observe that $25$ is a perfect square, so $5^2=25$ which means that certainly $5^2\equiv 25\pmod{32}$ as well. If it holds in the good old integers, it certainly holds mod whatever.
You quickly earn another solution for free: $(-5)^2 \equiv 27^2 \equiv 25\pmod{32}$ as well. As you’ve observed, a simple Hensel-style lifting doesn’t work mod powers of $2$, but you can easily check that adding (or subtracting) $16$ to each of those solutions yields another one. Thus $11$ and $21$ are two other residues mod $32$ whose squares are congruent to $25$. Indeed, $(a+16)^2 \equiv a^2+32a+16^2 \equiv a^2\pmod{32}$.
|
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|
Solid of revolution problem I would like to check with you guys if this exercise is right. The problem is:
"Let be $R$ be the region bounded by the graph of $y=2x$", $y=\frac{x^2}{4}$ and $y=2$. Find the volume of the solid generated by the rotation of the region $R$ about the $y$-axis. Make a sketch of the region. Show your calculations."
This is the graph I made:
This is what I did:
Since we are integrating with respect to $y$ I solve for $x$ both equations. $y=\frac{x^2}{4}$ becomes $x=2\sqrt{y}$ and $y=2x$ becomes $x=\frac{1}{2}y$. So for the volume:
$$V=\pi\int_{0}^{2} ((2\sqrt{y})^2-(\frac{1}{2}y)^2) \,dy$$
$$V=\pi\int_{0}^{2} (4y-\frac{1}{4}y^2) \,dy$$
$$V=\pi(4\cdot\frac{y^2}{2}-\frac{1}{4}\cdot\frac{y^3}{3})\bigg|_{0}^2$$
$$V=\pi(2y^2-\frac{1}{12}y^3)\bigg|_{0}^2$$
$$V=\pi[(2(2)^2-\frac{1}{12}(2)^3)-(2(0)^2-\frac{1}{12}(0)^3)]$$
$$V=\pi[(2(4)-\frac{1}{12}(8)]$$
$$V=\pi[(8-\frac{2}{3}]$$
$$V=\pi[(\frac{24}{3}-\frac{2}{3}]$$
$$V=\pi[\frac{22}{3}]$$
$$V=\frac{22}{3}\pi$$
Therefore the volumen of the solid of revolution is $\frac{22}{3}\pi$ cubic units.
Thanks in advance for the help.
|
Yes, this is exactly correct.
You can also verify your solution by using the method of cylindrical shells. For instance, the volume of the parabolic region including the cone is just $$\int_{x=0}^{2 \sqrt{2}} 2\pi x \left(2 - \frac{x^2}{4}\right) \, dx = \left[2x^2 - \frac{x^4}{8}\right]_{x=0}^{2\sqrt{2}} \pi = 8\pi.$$ Then subtract out the volume of the cone, which has a circular base of radius $1$ and height $2$, thus volume $\frac{1}{3}\pi r^2 h = \frac{2}{3}\pi$. This gives the desired volume $$\left(8 - \frac{2}{3}\right) \pi = \frac{22}{3}\pi.$$ We could also have computed the parabolic region's volume via disks, which is
$$\int_{y=0}^2 4\pi y \, dy = 8\pi.$$
|
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|
Angles satisfying $(\cos\theta,\sin\theta)=(\frac{\sqrt a }{\sqrt{a+b}},\frac{\sqrt b}{\sqrt{a+b}})$ for integer $a$, $b$ If we list the commonly used angles in the first quadrant $\theta_0=0$, $\theta_1=\dfrac{\pi}{6}$, $\theta_2=\dfrac{\pi}{4}$, $\theta_3=\dfrac{\pi}{3}$, and $\theta_4=\dfrac{\pi}{2}$,
then $$(\cos \theta_i, \sin \theta_i)=\left(\dfrac{\sqrt{4-i}}{2},\dfrac{\sqrt{i}}{2}\right) \qquad\mbox{for}\qquad i=0,\dots,4.$$
The angles $\dfrac{\pi}{6}, \dfrac{\pi}{4},\dfrac{\pi}{3} $ are obtained by bisecting or trisecting the first quadrant.
Question: Are there special constructions for angles $\theta$ such that
$$(\cos \theta, \sin \theta)=\left(\dfrac{\sqrt{a}}{\sqrt{a+b}},\dfrac{\sqrt{b}}{\sqrt{a+b}}\right)$$
where $a,b$, and $\sqrt{a+b}\ $ are positive integers? Or specifically, is there a rational number $r=r(a,b)$ such that $\theta=r\pi$?
Note that if $a,b$ are nonnegative integers such that a+b=4, then we are describing the angles $\theta_0,\dots,\theta_4$.
|
Regarding the second question, there are no other possible values of $\theta\in[0,\pi/2]$. Squaring, we have
$$\cos^2\theta=\frac{a}{a+b}\Leftrightarrow\frac{1+\cos 2\theta}{2}=\frac{a}{a+b}\Leftrightarrow\cos 2\theta=\frac{a-b}{a+b} $$
This means that $\cos2\theta$ is rational and $2\theta$ is a rational multiple of $\pi$ (since $\theta$ is a rational multiple of $\pi$). By Niven's theorem, $\cos2\theta=0,\pm\frac 12$ or $\pm 1$, which gives us $\theta\in\left\{0,\frac{\pi}{6},
\frac{\pi}{4},\frac{\pi}{3},\frac{\pi}{2}\right\}$.
|
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|
Epsilon - delta proof $\lim_{x \to \frac{\pi}{4}} \tan(x)=1$ I need to prove the limit using the definition of limit $$\lim_{x\to c}f(x)=L \leftrightarrow \forall \epsilon >0 \hspace{0.5 cm} \exists \delta >0 : 0<|x-c|<\delta \rightarrow |f(x)-L|<\epsilon $$
My attempt
$$|\tan(x)-1|<\epsilon\\ -\epsilon <\tan(x)-1 < \epsilon\\ \tan^{-1}(-\epsilon+1)<x<\tan^{-1}(\epsilon +1)\\ \tan^{-1}(-\epsilon+1)- \frac{\pi}{4}<x-\frac{\pi}{4}<\tan^{-1}(\epsilon +1)-\frac{\pi}{4}$$
I don't know if with this I can give an adequate value for $\delta$. Any advice on how to solve it would be very helpful, thank you.
|
Here is a more explicit way to do it, without using arctan of epsilon.
Since $\tan(x)$ has an asymptote at $x=\frac{\pi}{2}$, we should stipulate at the outset a maximum value of delta which is less than the distance from the point we are working at and the asymptote (this distance is $\frac{\pi}{4}$), for instance $\delta<\frac{\pi}{8}$. So in what follows we will consider only $x \in (\frac{\pi}{4}-\frac{\pi}{8},\frac{\pi}{4}+\frac{\pi}{8})=(\frac{\pi}{8},\frac{3\pi}{8})$. Now, if we suppose that $\epsilon$ is arbitrary and $0<|x-\frac{\pi}{4}|<\delta$ holds, we have
$$ |\tan(x)-1| = \left|\frac{\sin(x)}{\cos(x)}-\frac{\sin \left(\frac{\pi}{4} \right)}{\cos \left(\frac{\pi}{4} \right)} \right| = \left|\frac{\sin(x)\cos \left(\frac{\pi}{4} \right)-\sin \left(\frac{\pi}{4} \right) \cos(x)}{\cos(x)\cos \left(\frac{\pi}{4} \right)} \right|=\frac{|\sin\left(x-\frac{\pi}{4} \right)|}{|\cos(x)|\cos \left(\frac{\pi}{4} \right)}$$
Now in the numerator we can use the inequality $|\sin \left(x-\frac{\pi}{4} \right)| \leq |x-\frac{\pi}{4}|$ while to bound the denominator we notice that cosine is decreasing on the interval $(\frac{\pi}{8},\frac{3\pi}{8})$ and so $\frac{1}{|\cos(x)|} <\frac{1}{\cos (\frac{3\pi}{8})}$. These two inequalities allow us to write
$$ \frac{|\sin\left(x-\frac{\pi}{4} \right)|}{|\cos(x)|\cos \left(\frac{\pi}{4} \right)} < \frac{|x-\frac{\pi}{4}|}{\cos \left(\frac{3\pi}{8} \right) \cos \left(\frac{\pi}{4} \right)}<\frac{\delta}{\cos \left(\frac{3\pi}{8} \right) \cos \left(\frac{\pi}{4} \right)} \leq \epsilon $$
If we chose $\delta = \min \{\frac{\pi}{8}, \cos \left(\frac{3\pi}{8} \right)\cos \left(\frac{\pi}{4} \right) \epsilon \}$.
|
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|
how to solve this linear system of three equations using Cramer's rule? I have a 3-by-3 matrix,
A=$\left [ \begin{matrix}
1 & 2 & 3 \\
1 & 0 & 1 \\
1 & 1 & -1\\
\end{matrix} \right]$
the known terms are (-6, 2, -5), at the right of "=" symbol.
(1) I've calculated the determinant,
(2) I've used Cramer's rule to find x, y, and z.
but the result isn't correct (the right solutions is (x, y,z) = (1, -5, 1)).
(1) determinant of A,
I used row operations: row 3 <-> row 2. I swapped the rows. (therefore the determinant is $-\det A$).
(2) row 2 <-- $row 2 - row 1$ and I got:
A=$\left [ \begin{matrix}
1 & 2 & 3 \\
0 & -1 & -4 \\
1 & 1 & -1\\
\end{matrix} \right]$
then, I used Laplace in the first column, and I got:
$-\det A$=$\left [ \begin{matrix}
-1 & -4 \\
0 & 1 \\
\end{matrix} \right]$ + $\left [ \begin{matrix}
2 & 3 \\
-1 & -4 \\
\end{matrix} \right]$,
doing algebra here, I get: -1-(-8+3) = -6 but it was -detA, and therefore detA = 6.
I used Cramer's rule, therefore I put the known terms in first column, then in second, and so on.
x = A=$\left [ \begin{matrix}
-6 & 2 & 3 \\
2 & 0 & 1 \\
-5 & 1 & -1\\
\end{matrix} \right] $ (this matrix divided by the determinant of the original matrix)
x is equal to 3, it should've been equal to 1.
y = A=$\left [ \begin{matrix}
1 & -6 & 3 \\
0 & 2 & 1 \\
1 & -5 & -1\\
\end{matrix} \right] $ (this matrix divided by the determinant of the original matrix)
-2-6-6-5 = -14-5 = -19 it should have been equal to -5. (Sarrus' rule)
z = A=$\left [ \begin{matrix}
1 & 2 & -6 \\
0 & 0 & 2 \\
1 & 1 & -5\\
\end{matrix} \right] $ (this matrix divided by the determinant of the original matrix)
4-2 = 2. it should've been equal to 1. (Sarrus' Rule).
|
The determinant of the matrix representing $x$ is, expanding down the middle column (which you definitely can do, no row operations needed!):
$$\det\begin{pmatrix}-6&2&3\\2&0&1\\-5&1&-1\end{pmatrix}=(-2)\cdot\det\begin{pmatrix}2&1\\-5&-1\end{pmatrix}+0+(-1)\cdot\det\begin{pmatrix}-6&3\\2&1\end{pmatrix}$$
Which is: $$(-2)(-2-(-5))+(-1)(-6-6)=-2\cdot3+12=6$$Cramer tells that $x$ is then this determinant divided by the original one, namely $6/6=1$. When you’re saying: “this matrix divided by the determinant of the original matrix” this is wrong. That would imply $x$ is a matrix divided by a scalar, i.e. another matrix. You just take the determinants of both, and find their ratios.
|
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|
Geometry problem with triangle Given a triangle $ABC$ with side lengths $a$, $b$, and $c$. There exists a unique point $P$ such that
$$d^2 = AB^2+AP^2+BP^2 = AC^2+AP^2+CP^2=BC^2+BP^2+CP^2.$$
Question: How to express $d^2$ in terms of $a$, $b$, and $c$, and the circumradius $R$ of the triangle $ABC$?
What I tried: $R$ can be obtained using Heron's formula: $R=\frac{abc}{4S}$, with $S=\sqrt{s(s-a)(s-b)(s-c)}$, $s=(a+b+c)/2$.
I tried to estimate $d^2$ for some specific values.
*
*For example, when $a=b=c$, then $d^2=5R^2=5a^2/3$.
*In the extreme cases of $a=b+c$, $b=a+c$, or $c=a+b$ (with $a>0$, $b>0$, $c>0$), we have $d^2=32R^2$.
*And if $ABC$ is a right triangle (e.g., if $a^2+b^2=c^2$), we have $d^2=8R^2=a^2+b^2+c^2$.
I have no idea how to come to a general expression for $d^2$ in terms of $a$, $b$, $c$, and $R^2$ and I would be curious to see solutions.
|
$\begin{array}{} AB=c & BC=a & AC=b \\ AP=f & BP=g & CP=h \end{array}$
$\begin{array}{} d^2=c^2+f^2+g^2 & (1) \\ d^2=b^2+f^2 +h^2 & (2) \\ d^2=a^2+g^2+h^2 & (3) \ \end{array}$
$\begin{array}{} A=(p,q) & B=(a,0) & C=(0,0) \end{array}$
where:
$\begin{array}{} p=\frac{a^2+b^2-c^2}{2a} & q=\frac{2S}{a} & S=\sqrt{s(s-a)(s-b)(s-c)} & s=\frac{a+b+c}{2}\end{array}$
$\begin{array}{} f^2=(x-p)^2+(y-q)^2 \\ g^2=(x-a)^2+y^2 \\ h^2=x^2+y^2 \\ \end{array}$
$\begin{array}{} (1)(3) & c^2+f^2-a^2-h^2=0 & ⇒ & f^2=a^2+h^2-c^2 \\ (2)(3) & b^2+f^2-a^2-g^2=0 & ⇒ & f^2=a^2+g^2-b^2 \\ & ⇒ a^2+h^2-c^2=a^2+g^2-b^2 & ⇒ & h^2-c^2-g^2+b^2=0 \\ \end{array}$
$\begin{array}{} x^2+y^2-c^2-[(x-a)^2+y^2]+b^2=0 \\ -c^2+2ax-a^2+b^2=0 \\ x_{P}=\frac{a^2-b^2+c^2}{2a} \end{array}$
$f^2=a^2+h^2-c^2$
$(x-p)^2+(y-q)^2=a^2+x^2+y^2-c^2$
$y=\frac{-1}{2q}(2px-p^2-q^2+a^2-c^2)$
replace $p$, $q$ and $S^2$
$y_{P}=\frac{1}{8·aS}(b^4-2b^2c^2+c^4-3a^4+2a^2c^2+2a^2b^2)$
P was identified as L (de Longchamps point).
The distance from the vertex of a triangle to the de Longchamps point is:
$\begin{array}{} AL^2=f^2=16R^2-2(b^2+c^2)-a^2 \\ BL^2=g^2=16R^2-2(c^2+a^2)-b^2 \\ CL^2=h^2=16R^2-2(a^2+b^2)-c^2 \end{array}$
$\begin{array}{} d^2=c^2+f^2+g^2 \\ d^2=c^2+16R^2-2(b^2+c^2)-a^2+16R^2-2(c^2+a^2)-b^2 \\ d^2=32R^2-3(a^2+b^2+c^2) \end{array}$
$\begin{array}{} d^2=c^2+f^2+g^2 \\ d^2=c^2+16R^2-2(b^2+c^2)-a^2+16R^2-2(c^2+a^2)-b^2 \\ d^2=32R^2-3(a^2+b^2+c^2) \end{array}$
|
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|
For real numbers $z$ and $w$, $|(1+z)(1+w)-1| \leq (1+|z|)(1+|w|)-1$. I have written an attempted proof of the theorem on the title, and I need help verifying it.
I have used the following theorems to proof the theorem on the title.
Theorem 5.14)a) Let $x$ be a real number. $-|x| \leq x \leq |x|$.
Theorem 5.14)b) Let $a \geq 0$. $|x| \leq a$ if and only if $-a \leq x \leq a$.
Theorem 5.14)c) Let $x$ and $y$ be real numbers. $|x+y| \leq |x| + |y|$ (The Triangle Inequality).
For real numbers $z$ and $w$, $|(1+z)(1+w)-1| \leq (1+|z|)(1+|w|)-1$.
Proof. From Theorem 5.14)a), $(1+z) \leq |(1+z)|$ and $(1+w) \leq |(1+w)|$. Multiplying the $(1+z) \leq |(1+z)|$ by $|(1+w)|$, one obtains
\begin{align}
(1+z)|(1+w)| \leq |(1+z)||(1+w)|
\end{align}
Since $(1+w) \leq |(1+w)|$, it follows that
\begin{align}
(1+z)(1+w) \leq |(1+z)||(1+w)|
\end{align}
Observe that from The Triangle Inequality,
\begin{align}
|(1+z)| \leq 1 + |z|\\
|(1+w)| \leq 1 + |w|
\end{align}
Since $|(1+z)|(1 + |w|) \leq (1 + |z|)(1 + |w|)$ and $|(1+w)| \leq (1 + |w|)$, it follows that
\begin{align}
(1+z)(1+w) -1 \leq |(1+z)||(1+w)| -1 \leq (1 + |z|)(1 + |w|) -1
\end{align}
It is clear that $(1 + |z|)(1 + |w|) -1 \geq 0$.
In case that $(1+z)(1+w) -1 \geq 0$, $|(1+z)(1+w) -1| = (1+z)(1+w) -1$.
Hence,
\begin{align}
-[(1 + |z|)(1 + |w|) -1] \leq 0 \leq (1+z)(1+w) -1 \leq (1 + |z|)(1 + |w|) -1
\end{align}
From Theorem 5.14)b),
\begin{align}
|(1+z)(1+w) -1| \leq (1 + |z|)(1 + |w|) -1
\end{align}
establishing the result for this case.
On the other hand, in case that $(1+z)(1+w) -1 < 0$, $|(1+z)(1+w) -1| = -[(1+z)(1+w) -1]$.
Hence,
\begin{align}
-[(1 + |z|)(1 + |w|) -1] \leq (1+z)(1+w) -1 < 0
\end{align}
Since $(1 + |z|)(1 + |w|) -1 \geq 0$,
\begin{align}
-[(1 + |z|)(1 + |w|) -1] \leq (1+z)(1+w) -1 \leq (1 + |z|)(1 + |w|) -1
\end{align}
Multiplying the inequalities by $-1$, one obtains
\begin{align}
(1 + |z|)(1 + |w|) -1 \geq -[(1+z)(1+w) -1] \geq -[(1 + |z|)(1 + |w|) -1]
\end{align}
From Theorem 5.14)b),
\begin{align}
|(1+z)(1+w) -1| \leq (1 + |z|)(1 + |w|) -1
\end{align}
establishing the result for this case.
Because the result for all cases of $(1+z)(1+w) -1$ have been established, it is the case that $|(1+z)(1+w) -1| \leq (1 + |z|)(1 + |w|) -1$.
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To answer the solution-verification part of the question, this step is wrong.
one obtains
\begin{align}
(1+z)|(1+w)| \leq |(1+z)||(1+w)|
\end{align}
Since $(1+w) \leq |(1+w)|$, it follows that
\begin{align}
(1+z)(1+w) \leq |(1+z)||(1+w)|
\end{align}
The above is of the form $\, a \cdot |b| \le c \implies a \cdot b \le c\,$, which does not hold true in general. For example, if $\,a=b=-2, \,c=1\,$ then $\,a \cdot |b| = -4 \le 1 = c\,$, but $\,a \cdot b = 4 \gt 1 = c\,$. The implication does hold true for $\,a \ge 0\,$, but here $\,a = 1 + z\,$ which is not necessarily positive.
The correct way to derive the inequality is to use that $\,|a \cdot b| = |a| \cdot |b|\,$, then:
$$
(1+z)\cdot (1+w) \;\leq\; |(1+z)\cdot(1+w)| \;=\; |1+z|\cdot|1+w|
$$
|
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|
How many four-digit positive integers are there that contain the digit $3$ and are divisible by $5$?
How many four-digit positive integers are there that contain the digit $3$ and are divisible by $5$?
The answer is:
the number of four-digit integers that are divisible by $5\;-\;$ the number of four-digit integers that are divisible by $5$ and not contain the digit $3$
So, $9\cdot10\cdot10\cdot2-8\cdot 9\cdot 9\cdot 2=1800-1296=504\tag{*}$
I know that, but when I tried to solve this problem with an other way I got a different result.
Four digit integers : $\overline{xyzw}$
Suppose $\overline{xyzw}$ contain at least one digit equal $3$
So, $x = \{1, \ldots ,9\}, y = 3, z = \{0, \ldots, 9\}, w = \{0,5\}$
or $x = \{1, \ldots, 9\}, y = \{0, \ldots, 9\}, z = 3, w = \{0,5\}$
or $x = 3, y=\{0, \ldots, 9\}, z = \{0, \ldots, 9\}, w = \{0,5\}$
The number of all $\overline{xyzw}$ that must be divisible by $5$ and contain $3$ is: $9\cdot 10\cdot 2+9\cdot 10\cdot 2+10\cdot 10\cdot 2=180+180+200=560\tag{**}$
but (*) contradicts (**), so where is the mistake?
|
You are over counting a few cases. For example, $1330$ is counted in the first case when $x\in\{1,…,9\},y=3,z\in\{0,…,9\},w\in\{0,5\}$ and it is counted again in the second case when $x\in\{1,…,9\},y\in\{0,…,9\},z=3,w\in\{0,5\}$
|
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|
Prove inequality using a given inequality Question: It is given that $$a^r(a-b)(a-c)+b^r(b-c)(b-a)+c^r(c-a)(c-b)\geq 0$$ for positive integers $a,b,c,r$. We are supposed to show that $$\frac{1}{a^5}+\frac{1}{b^5}+\frac{1}{c^5}+\frac{a+b+c}{a^2b^2c^2}\geq \frac{b^2+c^2}{a^3b^2c^2}+\frac{c^2+a^2}{a^2b^3c^2}+\frac{a^2+b^2}{a^2b^2c^3}.$$
Attempt: I tried to multiply $a^5b^5c^5$ throughout in an effort to "reverse engineer" and hopefully, see the answer. What I got was $$b^5c^5+a^5c^5+a^5b^5+a^3b^3c^3(a+b+c)\geq a^2b^3c^3(b^2+c^2)+a^3b^2c^3(c^2+a^2)+a^3b^3c^2(a^2+b^2).$$ Then I tried to let $x=bc$, $y=ac$ and $z=ab$ so that I have $x^5+y^5+z^5+\dots\geq\dots$ and hopefully this would provide an expression that is identical to letting $r=5$. Unfortunately, I didn't succeed.
Any hints or suggestions would be much appreciated!
|
Alternative (brute force/buffalo way) solution: Your inequality is upon full expansion (using symmetric sum notation) $$\frac{\frac{\sum_{\text{sym}} a^5 b^5}2+\frac{\sum_{\text{sym}}a^4 b^3 c^3}2-\sum_{\text{sym}}a^5 b^3 c^2}{a^5b^5c^5}\ge 0.$$
Since $a,b,c>0$, you just need to prove $$\frac{\sum_{\text{sym}} a^5 b^5}2+\frac{\sum_{\text{sym}}a^4 b^3 c^3}2-\sum_{\text{sym}}a^5 b^3 c^2\ge 0$$ or $$\sum_{\text{sym}} a^5 b^5+\sum_{\text{sym}}a^4 b^3 c^3- 2 \sum_{\text{sym}}a^5 b^3 c^2\ge 0.$$
Based on how the exercise is posed, they seem to want you to find some smart substitution to turn this into Schur's inequality. A more brute force approach is this: Assume without loss of generality that $a\le b$ and $a\le c$. We may then write $b=a+x, c=a+y$ for some $x,y\ge 0$. Then the last inequality becomes upon full expansion $$8 a^8 x^2\color{red}-8 a^8 x y+8 a^8 y^2+8 a^7 x^3+20 a^7 x^2 y+20 a^7 x y^2+8 a^7 y^3+2 a^6 x^4+24 a^6 x^3 y+118 a^6 x^2 y^2+24 a^6 x y^3+2 a^6 y^4+6 a^5 x^4 y+140 a^5 x^3 y^2+140 a^5 x^2 y^3+6 a^5 x y^4+66 a^4 x^4 y^2+178 a^4 x^3 y^3+66 a^4 x^2 y^4+12 a^3 x^5 y^2+92 a^3 x^4 y^3+92 a^3 x^3 y^4+12 a^3 x^2 y^5+18 a^2 x^5 y^3+50 a^2 x^4 y^4+18 a^2 x^3 y^5+10 a x^5 y^4+10 a x^4 y^5+2 x^5 y^5\ge 0$$ which is true since $$8a^8 x^2+8a^8 y^2 \ge 16 a^8 xy$$ by AM-GM.
|
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|
Partial differential equation - integral surface I have to find the integral surface of the PDE
$$x(y^2+z)p - y(x^2+z)q = (x^2-y^2)z$$
containing the straight line $x+y=0, z=1$.
From the auxiliary equations, I can obtain $xyz=C_{1}$ for some constant $C_{1}$. To proceed further, I need to find another relation involving $x,y,z$ or at least one of them. However, I am having difficulty finding this. Would someone be able to help me determine the other relation?
|
$x(y^2+z)p - y(x^2+z)q = (x^2-y^2)z$ is ambiguous because the symbols are not defined. I suppose that the PDE is :
$$x(y^2+z)\frac{\partial z}{\partial x} - y(x^2+z)\frac{\partial z}{\partial y} = (x^2-y^2)z(x,y)\tag 1$$
The Charpit-Lagrange characteristic ODEs are :
$$\frac{dx}{x(y^2+z)}=\frac{dy}{-y(x^2+z)}=\frac{dz}{(x^2-y^2)z}$$
Search for a first characteristic equation :
$$\frac{ydx+xdy}{y(x(y^2+z))+x(-y(x^2+z))}=\frac{dz}{(x^2-y^2)z}=\frac{d(xy)}{xy(y^2-x^2)}$$
$$\frac{dz}{z}=-\frac{d(xy)}{xy}$$
$$xyz=c_1$$
Seach for a second characteristic equation :
$$\frac{xdx+ydy}{x(x(y^2+z))+y(-y(x^2+z))}=\frac{dz}{(x^2-y^2)z}=\frac{d(x^2+y^2)/2}{(x^2-y^2)z}$$
$$dz=d(x^2+y^2)/2$$
$$z-\frac12 (x^2+y^2)=c_2$$
The general solution of the PDE on implicit form $c_2=F(c_1)$ is :
$$z-\frac12 (x^2+y^2)=F(xyz)\tag 2$$
$F$ is an arbitrary function (to be determined according to the boundary condition).
Condition : $z=1$ on $y=-x$
$$1-\frac12 (x^2+(-x)^2)=F(x(-x))$$
$$F(-x^2)=1-x^2$$
Let $X=-x^2$
$$F(X)=1+X$$
Now the function $F$ is determined. We put it into the general solution Eq.$(2)$ where $X=xyz$
$$z-\frac12 (x^2+y^2)=1+xyz$$
$$z(x,y)=\frac{1+\frac12 (x^2+y^2)}{1-xy}$$
|
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|
Composed series expansion understanding Recently I'm studying series expansion and there is something I don't understand but it's hard to tell what... For example, imagine I need to calculate the series expansion of $\sqrt{1+\sin(x)}$ at $ a=0$.
I want to do it as fast as possible, so my idea is to have the series expansion of $\sqrt{1+x}$ and then plug the series expansion of $\sin(x)$ in it. Since when $x = 0$, $\sin(x) = 0$ too...
So for $\sqrt{1+x}$ I find $1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 + O(x^4)$ for example.
For $\sin(x)$ it's $x-\frac{x^3}{3!}+O(x^5)$ Which are both correct but when I replace the $x$ of the first expansion by the $\sin(x)$'s one I find the result don't change from $1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 + O(x^4)$ while wolfram tells me the $x^3$ term should be $\frac{-1}{48}$ And it's where it doesn't make sense to me, why would it be the only term to change? Computers are always right so I do believe it's a misunderstanding but where could it be? Here's the thing I have for $x^3$ : $\frac{3}{8}\cdot\frac{1}{3!}\cdot\frac{1}{(1+x^3-\frac{x^3}{3!})^{5/2}}\cdot x^3 = \frac{1}{16}\cdot\frac{1}{(1+x^3-\frac{x^3}{3!})^{5/2}}\cdot x^3$ and for me the last term is equal to $1$ since $x$ is close to $0$.
Thanks in advance
|
Your approach is correct. Until you've got it mastered I'd use two different variables: so
$$
\sqrt{1+y} = 1 + \frac{1}{2}y - \frac{1}{8}y^2 + \frac{1}{16}y^3 + O(y^4)
$$
Then you want to substitute
$$
y = x - \frac{1}{3!}x^3 + O(x^5)
$$
So
$$
\sqrt{1+y} = 1 + \frac{1}{2}\left(x - \frac{1}{3!}x^3 + O(x^5)\right) - \frac{1}{8}\left(x + O(x^3)\right)^2 + \frac{1}{16}\left(x + O(x^3)\right)^3 + O(x^4)
$$
Note that there's no need to expand the third and fourth terms beyond $O(x)$ because they're squared/cubed and will give terms $O(x^4)$. This is why it's only the cubic term that changes.
So the extra term in $x^3$ you're looking for comes from collecting all terms in $x^3$ above.
|
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|
Find all positive integers $(a,b)$ such that $\displaystyle\frac{a^{2^{b}}+b^{2^{a}}+11}{2^{a}+2^{b}}$ is an integer
Find all positive integers $(a,b)$ such that
$\displaystyle\frac{a^{2^{b}}+b^{2^{a}}+11}{2^{a}+2^{b}}$ is an
integer.
The machine/code says that $a=2$ and $b=3$ are suitable up to symmetry. And I bet that these are the only solutions when $a≤b$. If $a=0$, then $1+2^b≤b+11$ implies $b≤3$ and the only solution is $b=1$. In a similar way $a=1$ implies $b=0$. Since the formula is symmetrical, we can assume $a,b>1$.
$a$ and $b$ cannot have the same parity, because the numerator will be odd and the denominator will be even, which is impossible. Suppose $a≥2$ is even and $b≥3$ is odd. The pair $(a,b)=(2,3)$ is a solution giving $\displaystyle\frac{a^{2^{b}}+b^{2^{a}}+11}{2^{a}+2^{b}}=29$. For $a=2$, there is no other solution for $b≤25$. For $a=4$ and $a=6$, there is no solution for $b≤25$.
But I don't think this reasoning will be enough for a complete proof...
|
As noted by the comments one of $a,b$, say $b$ must be $2$ and the parity of $a$ and $b$ must differ. So let us assume that $a$ is odd.
We first make the following claim:
Claim 1: Let $c$ and $a'$ be nonnegative integers. Then for some nonegative integer $d\le a'$ the equation holds:
$$2^c \equiv_{2^{a'}+1} \pm 2^d.$$
Proof: Induction on $c$. Clearly true for $c \le a'$. Now
$2^{a'+1} \equiv_{2^{a'}+1} -1$ so
$$2^{c} \quad = \quad 2^{c-a'}2^d \quad \equiv_{2^{a'}+1} \quad -1 \times 2^{c-a'} \pmod{(2^{a'}+1)}.$$ However, by the induction hypothesis, there is an integer $d' \le a'$ such that the equation $$2^{c-a'} \equiv_{2^{a'}+1} \pm 2^{d'}$$ holds, so from this Claim 1 follows. $\surd$
Claim 2: For an integer $a>31$ the strict inequality $$a^4+11<2^{a-3}$$ holds.
Proof: Induction on $a$. Check directly for $a=31$, and then note that $(a+1)^4+11 < 2×(a^4+11)$, whereas $2^{(a+1)-3} = 2×2^{a-3}$. So informally, the RHS of the inequality, already larger than LHS, at least doubles when $a$ increases by $1$, whereas the LHS does not increase by such a large factor. $\surd$
We now use this to show there is no solution $(a,b)$ with $a$ odd and $a>31$, and with $b=2$. To this end, with $b=2$ and $a$ odd, the denominator becomes $4 \times 2^{a-2}+1$, and the numerator becomes $a^4+2^{2^a}+11$. Thus, it suffices to show
that $2^{a-2}+1$ cannot divide $11+a^4+2^{2^{a}}$ for $a > 31$.
Now, by the Claim 1, $$2^{2^{a}} \equiv_{2^{a-2}+1} \pm 2^d$$
for some nonegative integer $d \le a-2$. We consider 2 cases:
Case 1: $ \ 2^{2^{a}} \equiv_{2^{a-2}+1} -2^d$ for some nonnegative integer $d$. Then for $2^{a-2}+1$ to divide $11+a^4+2^{2^{a}}$, the equation
$a^4+11-2^d \equiv_{2^{a-2}+1} 0$ must hold. So from this the following holds:
For there to be an integer $a>31$ such that $2^{a-2}+1$ divides $11+a^4+2^{2^{a}}$ for some nonnegative integer $d$, there has to be such an $a,d$ for which either the equation (a) $a^4+11-2^d=0$ or the
equation (b) $a^4+11-2^d =n \times (2^{a-2}+1)$ holds.
However, there is no such $a,d$ that satisfies either of these equations (a), (b). [Indeed, the equation $a^4+11-2^d=0$ does not hold for any nonnegative integers $a,d$. [Indeed, checking mod 16, this cannot hold $d \ge 4$. And by exhaustive search this cannot hold for $d \le 4$ either.] And as the
strict inequality $a^4+11< 2^{a-2}$ holds for $a>31$ by Claim 2, the
equation $a^4+11-2^d =n \times (2^{a-2}+1)$ cannot hold for any nonnegative integer $d$ and any integer $a \ge 31$ either.]
Case 2: $ \ 2^{2^{a}} \equiv_{2^{a-2}+1} + 2^d$ for some nonnegative
integer $d$. Then we can assume that $d<a-2$ otherwise revert to Case 1. But then $|2^{a-2}-2^d| \ge 2^{a-3}$, so the only way that
$a^4+11+2^d$ is a multiple of $2^{a-2}+1$ is if $a^4+11$ is at least $|2^{a-2}-2^d|$ $\ge$
$2^{a-3}$, and this is impossible for $a \ge 31$ by Claim 2.
So we have indeed shown that there are no solutions $(a,b)$ with $a >31$ and $b=2$. So we can then reasonably finish the original problem simply by checking directly each of the $16$ solutions $(2k+1,2)$; $k=0,1,2,\ldots, 15$ [or we can trust Servaes' calculations in the Comments.]
|
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|
Where did I go wrong in finding the sum of $1+2k+3k^2+...+nk^{n-1}$ using Abel's formula The version of Abel's formula I'm using is $a_1b_1+a_2b_2+...+a_nb_n=(a_1-a_2)B_1+(a_2-a_3)B_2+...+(a_{n-1}-a_n)B_{n-1}+a_nB_n$ where $B_1=b_1$, $B_2=b_1+b_2$, $B_3=b_1+b_2+b_3$...$B_n=b_1+b_2+b_3+...+b_n$ and $a_n$ and $b_n$ are two sets of real number sequences.
Letting $a_1=1, a_2=2, a_3=3$ etc. and $b_1=1, b_2=k, b_3=k^2$ etc. the sum becomes $(1-2)(1)+(2-3)(1+k)+(3-4)(1+k+k^2)+...+(n-1-n)(1+k+k^2+...+k^{n-1})+n(1+k+k^2+...+k^n)$
Moving the last term to the front, it's equal to $n\left(\frac{k^n-1}{k-1}\right)-(1+(1+k)+(1+k+k^2)+...+(1+k+k^2+...+k^{n-1}))$
$=n\left(\frac{k^n-1}{k-1}\right)-\left(1+\frac{k^2-1}{k-1}+\frac{k^3-1}{k-1}+...+\frac{k^n-1}{k-1}\right)$
$=n\left(\frac{k^n-1}{k-1}\right)-\frac{1}{k-1}(k-1+k^2-1+k^3-1+...+k^n-1)$
$=n\left(\frac{k^n-1}{k-1}\right)-\frac{1}{k-1}(k+k^2+k^3+...+k^n+n(-1))$
$=n\left(\frac{k^n-1}{k-1}\right)-\frac{1}{k-1}(\frac{k(k^n-1)}{k-1}-n)$
$=\frac{nk^n}{k-1}-\frac{k(k^n-1)}{(k-1)^2}$
However, I seem to have an extra multiple of $k$, the answer should be $=\frac{nk^n}{k-1}-\frac{(k^n-1)}{(k-1)^2}$ (I've doubled checked my result against $1+2(2)+3(2)^2+4(2)^3+5(2)^4$ and it doesn't get the correct answer) but I can't spot where I went wrong.
|
Hint:
With the correct expression $b_n=k^{n-1}$ instead of $b_n=k^n$ you will obtain the desired result.
Observe also that you wrongly simplified the last term $a_nB_n$ in your expression (these two errors have eliminated each other to give correctly the first term in the final expression).
|
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|
$1^2 - 2^2 + 3^2 - 4^2 + \dots + 1999^2$ using closed form formula for sum of squares of first $n$ natural numbers The question is simple, the solution, not so much
Q.Find the sum of the given expression $1^2- 2^2 + 3^2 - 4^2 + \dots + 1999^2$
My idea is we know $1^2 + 2^2 + 3^2 + 4^2 + \dots + n^2 = \frac{n(n + 1)(2n + 1)}{6}$
So for $n=1999$ I get the sum as $2,66,46,67,000$
From this I need to subtract the squares of the even terms twice because subtracting once leaves with only the sum of the squares of the odd nos.
I observed something :
$2^2 + 4^2 + 6^2 + \dots + 1998^2 = (2 \cdot 1)^2 + (2 \cdot 2)^2 + \dots + (2 \cdot 999)^2$
Therefore to obtain the sum of the square of the even terms, I can take $4$ as common and use the aforementioned formula for $n=999$ and multiply it by $4$.
therefore sum of square of even terms = $1,33,13,34,000$
I need to subtract this sum twice to get the answer, because subtracting once simply leaves me with the sum of the squares of the odd numbers.
The answer is now $1999000$, which still doesn't match the answer key. Can someone explain where I am going wrong ?
|
There is a slight mistake
From this I need to subtract the sum of squares of even terms.
No. You have to subtract twice of above because
$1^2-2^2+3^2+\cdots+1999^2=(1^2+2^2+3^2+\cdots+1999^2)-2(2^2+4^2+6^2+1998^2)$
|
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|
What is the transformation matrix of the linear mapping? $ f:M_{2.2}(\mathbb{R}) \rightarrow \mathbb{R^3}, $$
\begin{pmatrix}
a & b \\
c & d
\end{pmatrix}
\mapsto
\begin{pmatrix}
2a - 4b \\
-6d \\
8a-16b+2d
\end{pmatrix}\beta =\big
(\begin{pmatrix}
1 & 0 \\
0 & 0
\end{pmatrix}
,\begin{pmatrix}
0 & 1 \\
0 & 0
\end{pmatrix}
\big),
\begin{pmatrix}
0 & 0 \\
1 & 0
\end{pmatrix}
\big),
\begin{pmatrix}
0 & 0 \\
0 & 1
\end{pmatrix}
\big)$
$\beta'= ( e_{2}, 2e_{1}, -e_{3})$ of $\mathbb{R^{3}}$ (
these are the standard unit vectors of $\mathbb{R^{3}}$)
What is the transformation matrix $M_{\beta'}^{\beta}(f)= ?$
Well my thought is that
$f(\left(
\begin{array}{c}
0\\
1\\
0
\end{array}
\right)) =\left(
\begin{array}{c}
-4\\
0\\
-16
\end{array}
\right) = \square \left(
\begin{array}{c}
1\\
0\\
8
\end{array}
\right) +$$\square \left(
\begin{array}{c}
-4\\
0\\
2
\end{array}
\right) +$
$\square \left(
\begin{array}{c}
0\\
0\\
0
\end{array}
\right) +$
$\square \left(
\begin{array}{c}
0\\
-6\\
2
\end{array}
\right)$
$f(\left(
\begin{array}{c}
2\\
0\\
0
\end{array}
\right)) =\left(
\begin{array}{c}
4\\
0\\
16
\end{array}
\right) = \square \left(
\begin{array}{c}
1\\
0\\
8
\end{array}
\right) +$$\square \left(
\begin{array}{c}
-4\\
0\\
2
\end{array}
\right) +$
$\square \left(
\begin{array}{c}
0\\
0\\
0
\end{array}
\right) +$
$\square \left(
\begin{array}{c}
0\\
-6\\
2
\end{array}
\right) $
$f(\left(
\begin{array}{c}
0\\
0\\
-1
\end{array}
\right)) =\left(
\begin{array}{c}
0\\
6\\
-2
\end{array}
\right) = \square \left(
\begin{array}{c}
1\\
0\\
8
\end{array}
\right) +$$\square \left(
\begin{array}{c}
-4\\
0\\
2
\end{array}
\right) +$
$\square \left(
\begin{array}{c}
0\\
0\\
0
\end{array}
\right) +$
$\square \left(
\begin{array}{c}
0\\
-6\\
2
\end{array}
\right) $
And from the linear combinations of the $squares$ I can fill the transformation matrix in the columns and in this way I will get the transformation matrix. Is my idea correct? And how does the transformation matrix look like?
Thank you in advance
|
Your idea is not correct, since the domain of $f$ is $\Bbb R^{2\times2}$; so it makes no sense to talk about, say$$f\left(\begin{bmatrix}0\\1\\0\end{bmatrix}\right).$$
Note that\begin{align}f\left(\begin{bmatrix}1&0\\0&0\end{bmatrix}\right)&=\begin{bmatrix}2\\0\\8\end{bmatrix}\\&=0\times e_2+1\times(2e_1)-8(-e_3).\end{align}Therefore, the entries of the first column of the matrix that you're after are $0$, $1$, and $-8$.
Can you take it from here?
|
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|
A die is rolled once. Call the result N. Then, the die is rolled N times, and those rolls which are equal to or greater than N are summed
A die is rolled once. Call the result N. Then, the die is rolled N times, and those rolls which are
equal to or greater than N are summed (other rolls are not summed). What is the distribution of the
resulting sum? What is the expected value of the sum?
The probability of a sum of $k$ is the coefficeint on $x^k$ in the polynomial,
$$\frac{1}{6}(\frac{1}{6}(x + x^2 + x^3 + x^4 + x^5 + x^6))+ \frac{1}{6}(\frac{1}{6}(1 + x^2 + x^3 + x^4 + x^5 + x^6))^2+\frac{1}{6}(\frac{1}{6}(2 + x^3 + x^4 + x^5 + x^6))^3+ \frac{1}{6}(\frac{1}{6}(3 + x^4 + x^5 + x^6))^4 + \frac{1}{6}(\frac{1}{6}(4 + x^5 + x^6))^5+ \frac{1}{6}(\frac{1}{6}(5 + x^6))^6$$
Why do we have the constants $1 ,2 3, 4,5$ in these terms? Why dont we just have the $x$'s?
|
I will expand slightly on my and Jaap's comments.
Denote by $S$ the random variable of the resulting sum. By the law of total probability, you can find its distribution by
$$ P(S=s) = \sum_{n=1}^6 P(S=s|N=n) P(N=n) = \frac16 \sum_{n=1}^6 P(S=s|N=n). $$
As the OP described, the probabilities $P(S=s|N=n)$ can be found via generating functions. I will shortly describe the idea in the following and highlight the importance of including the constants within the terms. Define the generating functions
$$f_n(x) = \left( \frac{n-1}6 + \sum_{i=n}^6 \frac{x^i}{6} \right)^n. $$
Then, $P(S=s|N=n)$ is the coefficient of $x^s$ in $f_n(x)$. This is because, conditioned on $N=n$, the variable $S$ is the sum of $n$ indendepent variables, which each assume the value $0$ with probability $\frac{n-1}{6}$ and assume each of the values $n,n+1,\dots,6$ with probability $\frac16$. Hence, the coefficient $x^s$ in $f_n(x)$ comprises the sum of all events in which the sum is $S=s$. Crucially, in the computation of the probability, we need to take into account those events, where some of the rolls might be $0$. Consider for example, $N=2$ and $S=5$. The sum $S=5$ can be obtained by the events $0+5$, $5+0$, $2+3$, or $3+2$. Therefore, it is necessary to include the constants $\frac{n-1}{6}$ in the generating functions as those accound for the case of rolling $0$.
Putting everything together, we see that the probability of $P(S=s)$ is obtained through the formula presented by the OP.
|
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|
Question abut the ambiguity of a maths problem. I am trying to prove by induction that:
$$1^3-2^3 +\cdots+n^3=(1+2+\cdots+n)^2 $$
This was a problem from a practice worksheet, but I don't understand how to interpret the LHS.
Is the following correct:
$$1^3-2^3+3^3-4^3+5^3-6^3+\cdots+n^3$$
Or this:
$$1^3-2^3+3^3+4^3+5^3+6^3+\cdots+n^3$$
I have a feeling it is the former. If this is the case, I presumably have to consider odd and even cases for n right?
Thanks
|
The true formula is that
$1^3+2^3+...+n^3=(1+2+...+n)^2$,
and you can prove this by induction. Assume that
$1^3+2^3+\cdots +(n-1)^3=(1+2+\cdots +(n-1))^2$,
then
$1^3+2^3+\cdots +n^3=(1+2+...+n-1)^2+n^3=\frac{n^2(n-1)^2}{4}+n^3=(1+2+\cdots n)^2.$
|
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|
Calculate residue of pole with order 5 I am trying to evaluate the residue of $\frac{(z^6-1)^2}{z^5(2z^4-5z^2+2)}$ at z=0.
Is there a way to do this without having to do the long derivative calculation?
I know we have the formula $Res(f,0)=\lim_{z\to0}\frac{1}{(5-1)!}\frac{\mathrm{d}^{5-1}f}{\mathrm{d}z^{5-1}}z^5f(z)$. But finding the 5th derivative of this function would take a long time and be pretty messy. My first try was to find the residues at the other poles and at infinity since all the residues must sum to zero, but when I tried to do this the residue at infinity ended up being the same as the residue at zero.
I've also considered finding the Laurent expansion but the Partial Fraction Decomposition is also long.
|
For practical calculations, I have never used that derivative formula to calculate any residue. What is useful is learning how to perform a limited Laurent expansion. One of the most useful "tricks" is the simple geometric series formula $\frac{1}{1-\zeta}=\sum_{n=0}^{\infty}\zeta^n$ provided $|\zeta|<1$. You should also be slightly comfortable with big Oh notation to keep track of the order of the various terms (I mean this is just an efficient tracker of how many terms you need to actually care about).
So:
\begin{align}
\frac{(z^6-1)^2}{z^5(2z^4-5z^2+2)}&=\frac{1}{2z^5}\cdot(z^6-1)^2\cdot\frac{1}{1-\left(\frac{5}{2}z^2-z^4\right)}
\end{align}
Notice how I have the $\frac{1}{z^5}$ term. Since I want the residue (i.e coefficient of $\frac{1}{z}$), I should expand terms atleast up to order $4$, meaning I should keep explicitly terms involving $z^4$; I can ignore $z^5$ or higher terms. The third term is precisely of the form $\frac{1}{1-\zeta}$ I mentioned previously, so we shall use this now:
\begin{align}
\frac{(z^6-1)^2}{z^5(2z^4-5z^2+2)}&=\frac{1}{2z^5}\cdot \left[1+\mathcal{O}(z^6)\right]\cdot \left[1+\left(\frac{5}{2}z^2-z^4\right)+
\left(\frac{5}{2}z^2-z^4\right)^2+
\mathcal{O}\left(\left(\frac{5}{2}z^2-z^4\right)^3\right)\right]\\
&=\frac{1}{2z^5}\cdot \left[1+\mathcal{O}(z^6)\right]\cdot \left[1+
\left(\frac{5}{2}z^2-z^4\right) + \frac{25z^4}{4}+\mathcal{O}(z^6)+\mathcal{O}(z^6)\right]\\
&=\frac{1}{2z^5}\cdot \left[1+\mathcal{O}(z^6)\right]\cdot \left[1+\frac{5}{2}z^2+\frac{21}{4}z^4+\mathcal{O}(z^6)\right]\\
&=\frac{1}{2z^5}\cdot \left[1+\frac{5}{2}z^2+\frac{21}{4}z^4+\mathcal{O}(z^6)\right]
\end{align}
I leave the final simplification to you to determine the coefficient of $\frac{1}{z}$, and hence the residue.
Note for example that in the first line, $(z^6-1)^2=z^{12}-2z^6+1$, but I simply shortened this to $1+\mathcal{O}(z^6)$, because that's more than enough (remember we only need to keep track of expansions up to $z^4$ term; anything higher will not contribute to the residue). Similarly, in the second step rather than fully expanding out $\left(\frac{5}{2}z^2-z^4\right)^2$, I write this briefly as $\frac{25}{4}z^4+\mathcal{O}(z^6)$, since that's more than enough. And so on.
|
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"url": "https://math.stackexchange.com/questions/4342553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
}
|
Volume of a container with unknown side and known surface area We have to make a closed container. All side surfaces (walls, ceiling and bottom) must be rectangles and stand perpendicular to each other. One of the sides should be $3 \,m$ long. The surface area of the container shall be $32 \,m^2$. Let one of the other two sides be $x$ meters long. Show that the volume is given by the formula: $$V = \frac{48x-9x^2}{x + 3}$$
|
Unless we let $x=y$ the formula does not have enough information to identify volume. If, however, we let $x=y$, then we can show that the formula works, with the restriction that $x=-8$ or $x=2$.
$$
2(3\cdot x)+2(3\cdot y) + 2(x\cdot y) = 32 \\
2 x y + 6 x + 6 y = 32\\
x = \dfrac{16 - 3 y}{y + 3},
y=x\implies x,y\in\big\{-8,2\big\}\\
$$
\begin{align*}
x=y\in\big\{-8,2\big\}\implies V=3\cdot (-8)^2=192\\
V = 3\cdot (2)^2=12
\end{align*}
$$
V = \frac{48x-9x^2}{x + 3}, x=-8 \implies V=192\\
V = \frac{48x-9x^2}{x + 3}, x=2 \implies V=12
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4342680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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|
Find parameters $a,b$ such that $x^6-2 x^5+2 x^4+2 x^3-x^2-2 x+1-\left(x^3-x^2+a x+b\right)^2>0$ The probrem is to prove that
$$x^6-2 x^5+2 x^4+2 x^3-x^2-2 x+1>0.$$
(the minimum value is about 0.02, tested by wolframalpha.)
I use sos(sum of squares) method, my idea is to reduce the degree of the polynomial gradually.
First I need to find $a,b$ so that
$$x^6-2 x^5+2 x^4+2 x^3-x^2-2 x+1-\left(x^3-x^2+a x+b\right)^2>0.$$
It's equivalent to
$$(1-2 a) x^4+(2 a-2 b+2)x^3 +\left(-a^2+2 b-1\right)x^2 +(-2 a b-2)x -b^2+1>0,$$ or
$$
\left( (1-2\!\:a)\!\:x^4+(2+2\!\:a-2\!\:b)\!\:x^3+(-1-a^2+2\!\:b)\!\:x^2+(-2-2\!\:a\!\:b)\!\:x+1-b^2 \right) _{\min}>0.$$
Let
$$f(x)=
(1-2\!\:a)\!\:x^4+(2+2\!\:a-2\!\:b)\!\:x^3+(-1-a^2+2\!\:b)\!\:x^2+(-2-2\!\:a\!\:b)\!\:x+1-b^2$$
and $f'(x)=0$, I get
$$
-2-2\!\:a\!\:b+2\!\:(-1-a^2+2\!\:b)\!\:x+3\!\:(2+2\!\:a-2\!\:b)\!\:x^2+4\!\:(1-2\!\:a)\!\:x^3=0.
$$ However, it's difficult and ugly to solve the equation. And it's impossible to solve the extreme point when the degree of the polynomial increases, for example, prove that $$x^{12}-x^{11}+x^8-x^7+x^6+x^3-2 x+1>0,$$ we need to find 5 parameters $a,b,c,d,e$ such that $$
x^{12}-x^{11}+x^8-x^7+x^6+x^3-2x+1-\left( x^6-\frac{x^5}{2}+ax^4+bx^3+cx^2+dx+e \right) ^2>0.
$$
How can I solve the problem? Is there any general method?
|
Let $p(x)=x^6-2x^5+2x^4+2x^3-x^2-2x+1$. A possible way to show $p(x)>0$ for all $x\in\mathbb{R}$ follows. Consider 3 cases.
Case 1: $x<-1$. Let $r(x)=p(x-1)$. Then $r(x)=x^6-8x^5+27x^4-46x^3+40x^2-18x+5$. Clearly, $r(x)>0$ for $x<0$, so $p(x)=r(x+1)>0$ for $x<-1$.
Case 2: $x>\frac 12$. Let $q(x)=p\left(x+\frac 12\right)$. Then $q(x)=x^6+x^5+\frac34x^4+\frac72x^3+\frac{55}{16}x^2-\frac{15}{16}x+\frac{5}{64}$. We have $\frac{55}{16}x^2-\frac{15}{16}x+\frac{5}{64}>0$ (check by completing the square or computing the discriminant) so $q(x)>0$ for $x>0$ which implies that $p(x)=q\left(x-\frac 12\right)>0$ for $x>\frac 12$.
Case 3: $-1\le x\le \frac 12$. Write $p(x)=x^4(x^2-2x+2)+(2x-1)(x^2-1)$. We have $x^4(x^2-2x+2)=x^4((x-1)^2+1)\ge 0$ with equality only if $x=0$ and $(2x-1)(x^2-1)\ge 0$ for $-1\le x\le \frac 12$ with equality only if $x=\frac 12$ or $x=-1$, so $p(x)>0$ in this case as well.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4342792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Convergence of the series with $a_n= \frac{1}{\sqrt{1} \cdot n^2} + \frac{1}{\sqrt{2} \cdot (n-1)^2} +\cdot\cdot\cdot + \frac{1}{\sqrt{n} \cdot 1^2}$ I need to determine the convergence of the series whose general term is given by:
$$a_n= \frac{1}{\sqrt{1} \cdot n^2} + \frac{1}{\sqrt{2} \cdot (n-1)^2} +\cdot\cdot\cdot + \frac{1}{\sqrt{n} \cdot 1^2}$$
Observation
*
*$\frac{1}{n^\frac{3}{2}} \lt a_n \lt \sum_{k=1}^n \frac{1}{k^2 }$
*$a_n$ is monotonic decreasing sequence and converges to zero.
This series is bigger than a convergent and smaller than a divergent series by the second observation, if I have not incorrectly done. But its of no help here to determine the convergence. Kindly help, thanks in advance.
|
Let $N \in \mathbb{N}$. One has
\begin{align*}
\left( \sum_{k=1}^N \frac{1}{\sqrt{k}}\right)\left( \sum_{j=1}^N \frac{1}{j^2}\right) &= \sum_{k=1}^N \sum_{j=1}^N \frac{1}{j^2\sqrt{k}}\\
&\leq \sum_{n=2}^{2N} \sum_{j+k=n} \frac{1}{j^2\sqrt{k}}\\
&= \sum_{n=1}^{2N-1} \sum_{j+k=n+1} \frac{1}{j^2\sqrt{k}}\\
&= \sum_{n=1}^{2N-1} \sum_{k=1}^{n} \frac{1}{\sqrt{k}(n+1-k)^2}\\
\end{align*}
So you get that
$$\sum_{n=1}^{2N-1} a_n \geq \left( \sum_{k=1}^N \frac{1}{\sqrt{k}}\right)\left( \sum_{j=1}^N \frac{1}{j^2}\right) $$
and because $\displaystyle{\sum \frac{1}{\sqrt{k}}}$ is divergent and $\displaystyle{\sum \frac{1}{j^2}}$ is convergent, you get that
$$\boxed{\text{the series }\sum a_n \text{ diverges}.}$$
Edit : This gives a precise estimation of the partial sum of the series $\sum a_n$, but as @PeterSzilas noticed, the divergence can be obtained much more directly by noticing that
$$a_n \geq \frac{1}{\sqrt{n}}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4344372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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|
How far can I go with the integral $\int \frac{\sin ^{n} x \cos ^{n} x}{1-\sin x \cos x} d x, $ where $n\in N$? Latest Edit
By the aid of my recent post, a closed form for its definite integral is obtained as below:
$$
\int_{0}^{\frac{\pi}{2}} \sin ^{k} \theta d \theta= \frac{\sqrt{\pi} \Gamma\left(\frac{k+1}{2}\right)}{k \Gamma\left(\frac{k}{2}\right)}
$$
Hence \begin{aligned}
\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{n} x \cos ^{n} x}{1-\sin x \cos x} d x&=\frac{2}{\sqrt{3}}\left[\tan ^{-1}\left(\frac{2 \tan x-1}{\sqrt{3}}\right)-x\right]_{0}^{\frac{\pi}{2}} -\sum_{k=1}^{n-1} \frac{1}{2^{k}} \cdot \frac{\sqrt{\pi} \Gamma\left(\frac{k+1}{2}\right)}{k \Gamma\left(\frac{k}{2}\right)}\\&=\frac{\pi}{3\sqrt3}- \sum_{k=1}^{n-1} \frac{1}{2^{k}} \cdot \frac{\sqrt{\pi} \Gamma\left(\frac{k+1}{2}\right)}{k \Gamma\left(\frac{k}{2}\right)}
\end{aligned}
In my answer, I have found the integral
$$\int \frac{d x}{1-\sin x \cos x} =\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \tan x-1}{\sqrt{3}}\right)+C_0
$$
Next,
$$
\begin{aligned}
& \int \frac{\sin x \cos x}{1-\sin x \cos x} d x \\
=& \int \frac{d x}{1-\sin x \cos x}-\int 1 d x \\
=& \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \tan x-1}{\sqrt{3}}\right)-x+C_{1}
\end{aligned}
$$
and
$$
\begin{aligned}
& \int \frac{\sin ^{2} x \cos ^{2} x}{1-\sin x \cos x} d x \\
=& \int \frac{1-\left(1-\sin ^{2} x \cos ^{2} x\right)}{1-\sin x \cos x} d x
\\
=& \int \frac{d x}{1-\sin x \cos x}-\int(1+\sin x \cos x) d x \\
=& \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \tan x-1}{\sqrt{3}}\right)-x+\frac{\cos 2 x}{4}+C_2
\end{aligned}
$$
Now I want to go further, $$
\begin{aligned}
& \int \frac{\sin ^{3} x \cos ^{3} x}{1-\sin x \cos x} d x \\
=& \int \frac{1-\left(1-\sin ^{3} x \cos ^{3} x\right)}{1-\sin x \cos x} d x \\
=& \int \frac{d x}{1-\sin x \cos x}-\int\left(1+\sin x \cos x+\sin ^{2} x \cos ^{2} x\right) d x \\
=& \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \tan x-1}{\sqrt{3}}\right)-x+\frac{\cos 2 x}{4}-\int \frac{\sin ^{2} 2 x}{4} d x \\
=& \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \tan x-1}{\sqrt{3}}\right)-x+\frac{\cos 2 x}{4}-\frac{1}{4}\int\frac{1-\cos 4 x}{2} d x\\
=& \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \tan x-1}{\sqrt{3}}\right)-x+\frac{\cos 2 x}{4}-\frac{1}{8}\left(x-\frac{\sin 4 x}{4}\right) +C\\=& \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \tan x-1}{\sqrt{3}}\right)-\frac{9}{8} x+\frac{\cos 2 x}{4}+\frac{\sin 4 x}{32} +C_3
\end{aligned}
$$
Then I discovered that the integral
$$
I(n)=\int \frac{\sin ^{n} x \cos ^{n} x}{1-\sin x \cos x} d x
$$
has a telescoping series
$$I(k+1)-I(k)=-\int \sin ^{k} x \cos ^{k} x d x$$
Hence $$
I(n)-I(1)=-\sum_{k=1}^{n-1} \frac{1}{2^{k}} \int\sin ^{k}(2 x) d x
$$
We can conclude that
$$
I(n)=\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \tan x-1}{\sqrt{3}}\right)-x-\sum_{k=1}^{n-1} \frac{1}{2^{k}}\int\sin ^{k}(2 x) d x
$$
Then I was stuck with the last sum.
My question is whether we can find a closed form for the last sum.
|
Let’s find a more general result and simplify with standard functions. The first step is a geometric series and the double angle formula:
$$\int \frac{\sin ^{n} x \cos ^{n} x}{1-\sin x \cos x} d x = \int \frac{\frac1{2^n}\sin ^{n}(2 x) }{1-\frac12\sin(2x)} d x\mathop=^{\big|\frac12\sin(2x)\big|<1} 2^{-n}\int \sin ^{n}(2 x)\sum_{m=0}^\infty \left(\frac12\sin(2x)\right)^m=2^{-n}\sum_{m=0}^\infty 2^{-m}\int\sin^{m+n}(2x)dx$$
The $(1m+1n)$ causes a coefficient of $\frac12$ in the resulting Hypergeometric/ Incomplete Beta function argument which cannot be used for a closed form since the coefficient should be a natural number
I tried with a few substitutions, but could never get natural number coefficients times the sum index showing that there is no closed with the Kampé de Fériet function, a double hypergeometric series function.
$$\,_2\text F_1(a,b;b+1;z)=bz^{-b} \text B_z(b,1-a)\implies 2^{-n}\sum_{m=0}^\infty 2^{-m}\int\sin^{m+n}(2x)dx = 2^{-n}\sum_{m=0}^\infty 2^{-m}\frac{\text{sgn}(\cos(2x))\sin^{m+n+1}(2x)}{2(m+n+1)}\,_2\text F_1\left(\frac12,\frac{m+n+1}2,\frac{m+n+1}2+1,\sin^2(2x)\right)= 2^{-n}\sum_{m=0}^\infty 2^{-m}\frac{\text{sgn}(\cos(2x))\sin^{m+n+1}(2x)}{2(m+n+1)}\left[\frac{m+n+1}2 \sin^2(2x)^ {-\frac{m+n+1}2}\text B_{\sin^2(2x)}\left(\frac{m+n+1}2,\frac12 \right)\right]$$
Therefore:
$$\boxed{\int \frac{\sin ^{n} x \cos ^{n} x}{1-\sin x \cos x} d x = C+2^{-n-2}\sum_{m=0}^\infty \frac{\text{sgn}(\sin(2x))^{m+n+1}\text{sgn}(\cos(2x))}{2^m} \text B_{\sin^2(2x)}\left(\frac{m+n+1}2,\frac12 \right) , |\sin(2x)|<2 }$$
Which again cannot be a closed form, but the integral can take a few complex values of $x$ with the Incomplete Beta function becoming a square root times a polynomial for certain values. Please correct me and give me feedback!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4344864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
}
|
How to grasp the relationship between Laurent series depending on the region they are developed at in trying to understand how to set up the Laurent series for a fractional expression, I have the given function
\begin{equation}
\frac{1}{(z-2)}-\frac{1}{(z-1)}
\end{equation}
The Laurent series of each of the two fractions are evaluated over the three intervals:
a) On the ring $|z|<1$ the series are:
\begin{equation}
\frac{1}{(z-1)}=-\frac{1}{(1-z)}=\sum_{n=0}^{\infty}(-1)^nz^n
\end{equation}
\begin{equation}
\frac{1}{(z-2)}=-\frac{1}{2}\frac{1}{(1-z/2)}=\sum_{n=0}^{\infty}(-1)^n\frac{z^n}{2^{n+1}}
\end{equation}
b) On the annulus $1<|z|<2$
\begin{equation}
\frac{1}{(z-1)}=\frac{1}{z}\frac{1}{(1-1/z)}=\sum_{n=0}^{\infty}\frac{1}{z^{n+1}}
\end{equation}
\begin{equation}
\frac{1}{(z-2)}=-\frac{1}{2}\frac{1}{(1-z/2)}=\sum_{n=0}^{\infty}(-1)^n\frac{z^n}{2^{n+1}}
\end{equation}
c) On the open region $|z|>2$
\begin{equation}
\frac{1}{(z-1)}=\frac{1}{z}\frac{1}{(1-1/z)}=\sum_{n=0}^{\infty}\frac{1}{z^{n+1}}
\end{equation}
\begin{equation}
\frac{1}{(z-2)}=\frac{1}{z}\frac{1}{(1-2/z)}=\sum_{n=0}^{\infty}\frac{2^n}{z^{n+1}}
\end{equation}
If one looks closely, whenever the fractions are evaluated over an open disk, a factor of $\frac{1}{z}$ is extracted from the fraction, so that one obtains the form $\sum_{n=0}^\infty \frac{a^n}{z^{n+1}}$. Then whenever the fractions are evaluated over a closed disk, one seeks a solution in the form of $\sum_{n=0}^\infty \frac{z^n}{a^{n(+1)}}$, with alternating sign. When one considers an annulus squeezed between the two other regions, then the fraction with the pole close to the upper bound is evaluated as for the closed ring, while the fraction with the pole whose value is closer to the lower bound, is rather "inverted" such that one seeks a solution by extracting the factor $\frac{1}{z}$ from the denominator. There is an evident pattern here, but can it be generalized in some understandable formula or explanation?
Thanks
|
The function
\begin{align*}
&f:\mathbb{C}\setminus\{1,2\}\to\mathbb{C}\\
&f(z)=\frac{1}{z-2}-\frac{1}{z-1}
\end{align*}
has simple poles at $z=1$ and $z=2$. Let's assume we want a Laurent expansion of $f$ at $z=0$. We distinguish three regions of convergence
\begin{align*}
\color{blue}{D_1:}&\color{blue}{\quad 0\leq |z|<1}\\
\color{blue}{D_2:}&\quad \color{blue}{1<|z|<2}\\
\color{blue}{D_3:}&\quad \color{blue}{ |z|>2}
\end{align*}
We recall if a function $g$ admits a power series expansion
\begin{align*}
g(z)=\sum_{n=1}^{\infty}a_nz^n
\end{align*}
with positive convergence radius $R$ the series
converges for $|z|<R$ and diverges for $|z|>R$ whereas no general statement at the boundary of the disc can be given. This implies that
\begin{align*}
g\left(\frac{1}{z}\right)=\sum_{n=1}^{\infty}a_n\frac{1}{z^n}
\end{align*}
converges for $\left|\frac{1}{z}\right|<R$ i.e. converges for $|z|>R$. With this in mind we can analyse the regions $D_1, D_2$ and $D_3$.
*
*The first region $D_1$ is a disc with center $z=0$, radius $1$ and the pole at $z=1$ at the boundary of the disc. It admits for both fractions a representation as power series.
*The region $D_2$ is an annulus containing all points outside the closure of $D_1$ and the closure of $D_3$. It admits for the fraction with pole at $z=1$ a representation as principal part of a Laurent series and for the fraction with pole at $z=2$ a power series.
*The region $D_3$ contains all points outside the disc with center $z=0$ and radius $2$. It admits for both fractions a representation as principal part of a Laurent series.
Region $D_1: 0\leq |z|<1$
We obtain
\begin{align*}
\color{blue}{f(z)}&=\frac{1}{z-2}-\frac{1}{z-1}\\
&=-\frac{1}{2}\cdot\frac{1}{1-\frac{z}{2}}+\frac{1}{1-z}\\
&=-\frac{1}{2}\sum_{n=0}^{\infty}\frac{1}{2^n}z^n+\sum_{n=0}^{\infty}z^n\\
&\,\,\color{blue}{=\sum_{n=0}^{\infty}\left(1-\frac{1}{2^{n+1}}\right)z^n}
\end{align*}
Region $D_2: 1<|z|<2$
We obtain
\begin{align*}
\color{blue}{f(z)}&=\frac{1}{z-2}-\frac{1}{z-1}\\
&=-\frac{1}{2}\cdot\frac{1}{1-\frac{z}{2}}-\frac{1}{z}\cdot\frac{1}{1-\frac{1}{z}}\\
&=-\frac{1}{2}\sum_{n=0}^{\infty}\frac{1}{2^n}z^n-\sum_{n=0}^{\infty}\frac{1}{z^{n+1}}\\
&\,\,\color{blue}{=-\sum_{n=0}^{\infty}\frac{1}{2^{n+1}}z^n-\sum_{n=1}^{\infty}\frac{1}{z^n}}
\end{align*}
Region $D_3: |z|>2$
We obtain
\begin{align*}
\color{blue}{f(z)}&=\frac{1}{z-2}-\frac{1}{z-1}\\
&=\frac{1}{z}\cdot\frac{1}{1-\frac{2}{z}}-\frac{1}{z}\cdot\frac{1}{1-\frac{1}{z}}\\
&=\sum_{n=0}^{\infty}2^n\frac{1}{z^{n+1}}-\sum_{n=0}^{\infty}\frac{1}{z^{n+1}}\\
&\,\,\color{blue}{=\sum_{n=1}^{\infty}\left(2^{n-1}-1\right)\frac{1}{z^n}}
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4348500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
How do we handle the integral $I_{4}=\int\left(\frac{\cos \theta}{1+\sin ^{2} \theta}\right)^{4} d \theta$? After struggling with $I_2,I_3$ in the post, I dare to tackle $I_4$ now.
We first rewrite the integral
$$I_{4}=\int\left(\frac{\cos \theta}{1+\sin ^{2} \theta}\right)^{4} d \theta =\int \frac{\sec ^{2} \theta}{\left(\sec ^{2} \theta+\tan ^{2} \theta\right)^{4}} d(\tan \theta) $$
Letting $x=\tan \theta$ converts $$
\begin{aligned}I_4 &=\int \frac{1+x^{2}}{\left(2 x^{2}+1\right)^{4}} d t \\
&=\frac{1}{2}\left[\int \frac{d t}{\left(2 x^{2}+1\right)^{3}}+\int \frac{d t}{\left(2 x^{2}+1\right)^{4}}\right]
\end{aligned}
$$
By my post, we have an elegant reduction formula:
$$
\begin{aligned}
J_{n} &=\int \frac{d x}{\left(a x^{2}+b\right)^{n}}, \text { where } n \geqslant 2 \\
&=\frac{x}{2 b(n-1)\left(a x^{2}+b\right)^{n-1}}+\frac{2 n-3}{2 b(n-1)} J_{n-1}
\end{aligned}
$$
When $a=2$ and $b=1,$ $$
J_{n}=\frac{x}{2(n-1)\left(2 x^{2}+1\right)^{n-1}}+\frac{2 n-3}{2(n-1)} J_{n-1}
$$
Hence
$$
I_{4}=\frac{1}{2}\left(J_{3}+J_{4}\right)
$$
Let’s start with $$J_1= \int \frac{d x}{2 x^{2}+1}=\frac{1}{\sqrt{2}} \tan ^{-1}(\sqrt{2} x)+C_1 $$
$$
J_{2}=\frac{x}{2\left(2 x^{2}+1\right)}+\frac{1}{2} J_{1}=\frac{x}{2\left(2 x^{2}+1\right)}+\frac{1}{2 \sqrt{2}} \tan ^{-1}(\sqrt{2} x)+C_2
$$
$$
\begin{aligned}
J_{3} &=\frac{x}{4\left(2 x^{2}+1\right)^{2}}+\frac{3}{4} J_{2} \\
&=\frac{x}{4\left(2 x^{2}+1\right)^{2}}+\frac{3 x}{8\left(2 x^{2}+1\right)}+\frac{3}{8 \sqrt{2}} \tan ^{-1}(\sqrt{2} x)+C_3
\end{aligned}
$$
$$
J_{4}=\frac{x}{6\left(2 x^{2}+1\right)^{3}}+\frac{5 x}{24\left(2 x^{2}+1\right)^{2}}+\frac{5 x}{16\left(2 x^{2}+1\right)}+\frac{5}{16 \sqrt{2}} \tan ^{-1}(\sqrt2 x)+C_4
$$
Now we can conclude that $$
\begin{array}{c}
\displaystyle I_{4}=\frac{\tan \theta}{96}\left[\frac{8}{\left(2 \tan ^{2} \theta+1\right)^{3}}+\frac{22}{\left(2 \tan ^{2} \theta+1\right)^{2}}+\frac{33}{2 \tan ^{2} \theta+1}\right] +\frac{11}{32 \sqrt{2}} \tan ^{-1}(\sqrt{2} \tan \theta)+C
\end{array}
$$
Please let me know if there is any mistake.
Can we go further?
Wish you all a happy and healthy New Year 2022!
|
In general, for any even power, substitute $\tan^2\theta = \frac12\tan^2 x$ to transform the integral to a manageable one
$$I_{2m}= \int\left(\frac{\cos \theta}{1+\sin ^{2} \theta}\right)^{2m} d \theta
= \frac1{2^{m-\frac12}}\int \cos^{2m}x\>(1+\cos^2x)^{m-1}dx
$$
which, for $m=2$, reduces to
\begin{align}
I_4
= &\>\frac1{2^{3/2}}\int \cos^{4}x\>(1+\cos^2x)\>dx\\
= &\>\frac{1}{2^{5/2}}\left(\frac13\cos^5x +\frac{11}{12}\cos^3x +\frac{11}{8}\cos x \right)\sin x
+ \frac{11}{2^{11/2}}x+C
\end{align}
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4349163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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|
Density function of average of max and min of uniform random variables Suppose $X_1, \ldots, X_n \sim U(\theta - \frac{1}{2}, \theta + \frac{1}{2})$. I want to find the density of $Z = \frac{X_{(1)} + X_{(n)}}{2}$. My strategy is to use the transformation $F(x, y) = (\frac{x + y}{2}, y)$ and the joint density of $X_{(1)}$ and $X_{(n)}$, which is given by
$$
f(x,y) = n (n-1) (y - x)^{n-2}
$$
if $\theta - \frac{1}{2} < x < y < \theta + \frac{1}{2}$. Note that $F^{-1}(z, y) = (2z - y, y)$. By the transformation law, the joint density of $Z$ and $X_{(n)}$ is given by
$$
g(z,y) = f(F^{-1}(z, y)) |J F^{-1}(z, y) | = n(n-1)(y - (2z - y))^{n-2} \cdot 2 = 2^{n-1} n (n-1) (y - z)
$$
if $\theta - \frac{1}{2} < 2z - y < y < \theta + \frac{1}{2}$. Then I marginalize over $y$ to get the density of $Z$:
$$
g(z) = \int_z^{\theta + \frac{1}{2}} g(z,y) dy = n 2^{n-1}(\theta + \frac{1}{2} - z)^{n-1}
$$
for $\theta - \frac{1}{2} < z < \theta + \frac{1}{2}$. But this is not a density function (doesn't integrate to $1$). The culprit is the factor of $2^{n-1}$. Can someone point to where I went wrong? Much thanks
|
The mistake is in the integral to find density function of $Z$.
$ \displaystyle \theta - \frac{1}{2} < 2z - y < y < \theta + \frac{1}{2}$ would mean
For $\theta - \frac 12 \lt z \lt \theta, z \lt y \lt 2z + \frac 12 - \theta$
And for $\theta \lt z \lt \theta + \frac 12, z \lt y \lt \theta + \frac 12$
Integrating you get,
$ \displaystyle g(z) = n \cdot 2^{n-1} \cdot \left(\frac 12 + z - \theta\right)^{n-1} ~, \theta - \frac 12 \lt z \lt \theta$
$ \displaystyle g(z) = n \cdot 2^{n-1} \cdot \left(\frac 12 + \theta - z \right)^{n-1} ~, \theta \lt z \lt \theta + \frac 12$
|
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|
Area of a circle passing through two vertices of a parallelogram touching one edge. For reference:
Let $ABCD$ be a parallelogram, $AB = 6, BC= 10$ and $AC = 14$ ; traces a circle passing through the vertices $C$ and $D$ with $AD$ being tangent and this circle is $BC$ a secant. Calculate the area of the circle. (Answer: $12\pi$)
My progress:
$AD^2 =AI\cdot AC \implies 10^2=AI\cdot14$
$ \implies AI = \dfrac{50}{7} ,\ IC = \dfrac{48}{7}$
In $\triangle ABC$
$14^2=6^2+10^2-2\cdot6\cdot10\cdot\cos \angle B\implies
\cos \angle B =-\dfrac{1}{2}\therefore \angle B =120^o $
....I can't see many options????
|
The center $(x,y)$ of the circle has obvious $x=10$; the ordinate $y$ is given by the intersection of lines $x=10$ and the perpendicular to the segment $DC$ at its midpoint (and this ordinate is clearly the radius also by the tangency).
$► \cos(\angle{ABC})=-\dfrac12\Rightarrow \angle{BAD}=60^{\circ}$.
It follows:
$►$ Coordinates $D=(10,0),C=(13,3\sqrt3)$, midpoint of $DC=(11.5,1.5\sqrt3)$
$►$ Line $DC$ has pente $\sqrt3$ then the perpend. has equation $2y-3\sqrt3=-\dfrac{1}{\sqrt3}(2x-23)$ which gives for $x=10$ the value $y=2\sqrt3$.
$►$ Finally the area is $\pi r^2=12\pi$.
|
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|
Identity involving double sum with binomials (Edit: This question has now been answered in MathOverflow, here)
In the course of a calculation, I have met the following complicated identity. Let $A$ and $a$ be positive integers. Then I believe that
$$ \sum_{B\ge A,b\ge a} (-1)^{a+b}{b\choose a} {B-1\choose A-1}\frac{(B-1)!b!}{(B+b)(x-b)^{(B+b)}}=\frac{(A-1)!a!}{(A+a)(x-A+1)^{(A+a)}},$$
where $x$ is some variable and $(x)^{(a)}=x(x+1)\cdots(x+a-1)$ is the rising factorial.
This is what I would like to prove. It is like a generalization of the identity $$\sum_{a=0}^\infty \frac{1}{x^a}=\frac{x}{x-1}.$$
Notice a kind of miracle: the left hand side in principle has poles at all integer values of $x$, while the right hand side only has poles at integers smaller than $A-1$.
Following the criticism in the answer by Paul Sinclair, I think the meaning of the equality is a bit unclear. I have in mind a large $x$ expansion. Take the case $A=a=1$, for example. If I sum $B$ and $b$ both from 1 to 2 I get
$$ \frac{1}{2}\frac{x^2-3x-8}{x(x-2)(x^2-1)}$$
for the left hand side, which is
$$ \frac{1}{2x^2}- \frac{1}{2x^3}+O\left(\frac{1}{x^4}\right)$$
for large $x$. If I sum $B$ and $b$ up to larger values, more terms in the large $x$ expansion of the left hand side agree with
$$\frac{1}{2x^2}-\frac{1}{2x^3}+\frac{1}{2x^4}-\frac{1}{2x^5}+\cdots,$$
which is the large $x$ series of $\frac{1}{2x(x+1)}$, the right hand side.
(I think the double sum on the left hand side is convergent only for large enough $x$, and it is in this regime that it agrees with the right hand side. So finding different residues at $x=0$ does not invalidate the identity.)
|
In the case $A = 1, a = 1$, and letting $n = B + b -1$, this reduces to
$$-\sum_{n=1}^\infty\sum_{b=1}^n (-1)^bb\dfrac{(n-b)!b!}{(n+1)(x-b)^{(n+1)}} = \dfrac 1{2x(x+1)}$$
But that cannot be. The residue at $0$ can be found by multiplying both sides by $x$ and taking the limit as $x \to 0$. The right-hand side easily gives $1$. The left-hand is as little harder. $x-b \to -b$ and since $b \le n$
$$(x-b)^{(n+1)} = (x-b)(x-b+1)\cdots(x-1)x(x+1)\cdots(x+n-b)$$
So $$\lim_{x\to 0} \frac 1x (x-b)^{(n+1)} = (-b)(-b+1)\cdots(-1)(1)\cdots(n-b) = (-1)^bb!(n-b)!$$
Which means the left-hand side reduces to $$-\sum_{n=1}^\infty\sum_{b=1}^n \frac{b}{n+1}= -\sum_{n=1}^\infty\frac n2=-\infty$$
It seems clear tome that even if $x$ is only near $0$, this series is still going to diverge. Thus it cannot be equal to the much better behaved right-hand side.
Either you've made a mistake in your calculations that led you to this conclusion, or else in pulling this particular "equality" out as the cause your larger calculation is true.
|
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|
Solving $(x+1)\sqrt{2(x^2 + 1)} + \sqrt{6x^2 + 18x +12} = \frac{3x^2 + 7x + 10}{2}$ Today, I came across this problem.
$$(x+1)\sqrt{2(x^2 + 1)} + \sqrt{6x^2 + 18x +12} = \dfrac{3x^2 + 7x + 10}{2}$$
We are asked to find the possible values of $x$ satisfying this equation.
The first thought which came to my mind is to use some kind of factorisation. I tried for like an hour but all in vain.
Then, I tried to solve the problem by multiplying both sides by 2 and then squaring both sides. The equation became too complicated.
By using some hit and trial, I get to know that $x = 1$ satisfies the equation. But what about complex solutions. So this method is also of no use.
I am sure this question has to be solved using some special equality which I'm unaware of. I want a method so that, I could obtain all the possible values. Can anyone help me or just give some hints?
|
Square the equation rearrange & square again. This will allow you to remove the radicals & leave you with a polynomial.
\begin{eqnarray*}
(3x^2+7x+10)^2-8(x+1)^2(x^2+1)-4(6x^2+18x+12)=8(x+1)\sqrt{2(x^2+1)(6x^2+18x+12)} \\
\end{eqnarray*}
\begin{eqnarray*}
((3x^2+7x+10)^2-8(x+1)^2(x^2+1)-4(6x^2+18x+12))^2-128(x+1)^2(x^2+1)(6x^2+18x+12)=0.
\end{eqnarray*}
Now I would resort to CAS ...
So $x=1$ is the only real solution as $x^6+54x^5+153x^4+104x^3 -72x^2+400$ is always positive. The complex roots will be jolly unpleasant & unilluming.
|
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|
Solve the system $3^x-2^{y^2}=77$, $3^{\frac{x}{3}}-2^{\frac{y^2}{2}}=7$ in $\mathbb{R}$ I had to solve the similar system $3^x-2^{y^2}=77,\; 3^{\frac{x}{2}}-2^{\frac{y^2}{2}}=7$ before, which can be solved like this:
$$3^{\frac{x}{2}}-2^{\frac{y^2}{2}}=7 \implies 2^{\frac{y^2}{2}}=3^{\frac{x}{2}}-7$$
$$2^{\frac{y^2}{2}}=3^{\frac{x}{2}}-7 \implies 2^{y^2}=(3^{\frac{x}{2}}-7)^2$$
$$2^{y^2}=(3^{\frac{x}{2}}-7)^2 \implies 2^{y^2}=3^{x}-14\cdot3^{\frac{x}{2}}+49$$
So putting that in the first equation:
$$3^x-2^{y^2}=77 \implies 3^x-3^{x}+14\cdot3^{\frac{x}{2}}-49=77$$
$$\mathbf{3^x-3^{x}+14\cdot3^{\frac{x}{2}}-49=77 \implies 14\cdot3^{\frac{x}{2}}=126}$$
$$14\cdot3^{\frac{x}{2}}=126 \implies 3^{\frac{x}{2}}=9$$
$$3^{\frac{x}{2}}=9 \implies x = 4$$
And putting the value of $x$ in the second equation we get $y$:
$$3^{\frac{x}{2}}-2^{\frac{y^2}{2}}=7 \implies 3^{2}-2^{\frac{y^2}{2}}=7$$
$$3^{2}-2^{\frac{y^2}{2}}=7 \implies 2^{\frac{y^2}{2}}=2$$
$$2^{\frac{y^2}{2}}=2 \implies y^2 = 2$$
$$y^2 = 2 \implies y = \sqrt{2}$$
So I tried solving the system in the question title with the same strategy until the step in bold, which gave me $3^x - 3^{\frac{2x}{3}}+14\cdot3^{\frac{x}{3}}=126$, and I have no idea how to progress further.
|
Using the hyperbolic method when there is only one real root as for$$u^3-u^2+14u-126=0$$ the solution is not ugly at all. It write
$$u=\frac{1}{3} \left(1+2 \sqrt{41} \sinh \left(\frac{1}{3} \sinh
^{-1}\left(\frac{1639}{41 \sqrt{41}}\right)\right)\right)$$ while using Cardano method, it write
$$\frac{1}{3} \left(1-\frac{41}{\sqrt[3]{1639+3 \sqrt{306138}}}+\sqrt[3]{1639+3
\sqrt{306138}}\right)$$
Which one do you prefer ?
|
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|
On Sylvester’s criterion for $2 \times 2$ matrices In my electrical engineering text, I have an expression of the form:
$$E_m(x_1, x_2) = \frac{1}{2}(A_{11}x_{1}^2+2Mx_1x_2+A_{22}x_{2}^2) > 0$$
where $$A = \begin{bmatrix}
A_{11} & M \\
M & A_{22}
\end{bmatrix}$$
and so the expression can be written in matrix form as
$$E_m(x_1, x_2) = \frac{1}{2}\begin{bmatrix}
x_1 & x_2
\end{bmatrix}\begin{bmatrix}
A_{11} & M \\
M & A_{22}
\end{bmatrix} \begin{bmatrix}
x_1 \\ x_2
\end{bmatrix} >0 $$
The claim is that $E_m(x_1, x_2) >0$ if and only if $A_{11}>0$, $A_{22} >0$, and $M^2 < A_{11}A_{22}$ .
But I'm having a hard time seeing why? $x_1$ and $x_2$ are arbitrary and can be positive or negative independently. Can someone show me how we know these sign constraints about the elements of $A$?
|
(In this answer, I assume that $x_1$ and $x_2$ are not zero simultaneously.)
Let us first exclude the case where both $A_{11}$ and $A_{22}$ are both zero because $M x_1 x_2 >0$ cannot hold for arbitrary values of $x_1$ and $x_2$.
Without loss of generality, since the expression is symmetric with respect to $x_1$ and $x_2$, let us assume that $A_{11} \neq 0$.
For $x_2=0$ one can easily see that$$E_m(x_1,0)=\frac{1}{2}(A_{11}x_1^2 + 2M x_10+ A_{22}0)=\frac{1}{2}A_{11}x_1^2 >0$$holds if and only if $A_{11} >0$. Similarly, for the case $A_{22} \neq 0$ one concludes that $A_{22} >0$.
So, let us assume that $x_2 \neq 0$. By factoring out $\frac{1}{x_2^2}$, we have$$E_m(x_1,x_2)= \frac{1}{2x_2^2}\left (A_{11} \left ( \frac{x_1}{x_2} \right )^2 + 2M \left ( \frac{x_1}{x_2} \right ) + A_{22} \right ).$$Since $\frac{1}{2x_2^2}>0$, the above expression is greater than zero if and only if$$A_{11} \left ( \frac{x_1}{x_2} \right )^2 + 2M \left ( \frac{x_1}{x_2} \right ) + A_{22} >0.$$Since $x_1$ and $x_2$ are arbitrary, the above relation holds if and only if$$A_{11}x^2+2Mx + A_{22} >0,$$where $x=\frac{x_1}{x_2}$ is any arbitrary real number.
Now, one can easily conclude (see "Addendum") the last inequality holds if and only if
\begin{gather}(2M)^2 < 4A_{11}A_{22}, \quad A_{11} >0 \\ \Rightarrow \quad M^2 < A_{11}A_{22}, \quad A_{11}>0.
\end{gather}
Considering all possible cases, we conclude that$$E_m(x_1, x_2) = \frac{1}{2}(A_{11}x_1^2+ 2Mx_1x_2+A_{22}x_2^2) > 0$$if and only if$$A_{11}>0, \quad A_{22}>0, \quad M^2< A_{11}A_{22}.$$
Addendum
Proposition. The inequality $ax^2+bx+c>0$, where $a \neq 0$, holds if and only if$$b^2<4ac, \qquad a>0.$$
Proof.
The following fact can be easily obtained:
\begin{align}
& b^2<4ac \\
\Leftrightarrow \quad & \frac{4ac-b^2}{4a^2} >0, \qquad (a\neq 0) \\
\Leftrightarrow \quad & \left ( x+ \frac{b}{2a} \right ) ^2 + \frac{c}{a} - \frac{b^2}{4a^2}>0 \\
\Leftrightarrow \quad & x^2 + \frac{b}{a}x + \frac{c}{a} > 0 \\
\Leftrightarrow \quad & ax^2+bx+c >0 \qquad (a>0).
\end{align}
|
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|
Why is $\left ( 1+\frac{1}{k} \right )^kIn Artin's Gamma function (page 20), he says in footnote
(..) we consider the elementary inequalities
$$
\left ( 1+\frac{1}{k} \right )^k<e<\left ( 1+\frac{1}{k} \right )^{k+1}
$$
for $k=1,2,\dots, n-1$.
Where does the inequalities come from? From what I know is that
$$
e=\lim_{n\to\infty}\left ( 1+\frac{1}{n} \right )^n.
$$
Is it because that $n\mapsto (1+1/n)^n$ is increasing?
|
Partial answer.
Let $t_k = \left( 1 + \frac{1}{k} \right) ^k.$
"Borrowing" from Rudin's PMA, in Theorem $3.31$ he gives the nice formula, due to the Binomial theorem:
$$t_k = 1 + 1 + \frac{1}{2!}\left( 1 - \frac{1}{k} \right) + \frac{1}{3!}\left( 1 - \frac{1}{k} \right) \left( 1 - \frac{2}{k} \right) + \ldots$$
$$ + \frac{1}{k!}\left( 1 - \frac{1}{k} \right) \left( 1 - \frac{2}{k} \right) \ldots \left( 1 - \frac{k-1}{k} \right)$$
Hence,
\begin{align} t_{k+1} = 1 + 1 + \frac{1}{2!}\left( 1 - \frac{1}{k+1} \right) + \frac{1}{3!}\left( 1 - \frac{1}{k+1} \right) \left( 1 - \frac{2}{k+1} \right) + \ldots\\
\\
\\ + \frac{1}{k!}\left( 1 - \frac{1}{k+1} \right) \left( 1 - \frac{2}{k+1} \right) \ldots \left( 1 - \frac{k-1}{k+1} \right) \\
\\
\\+ \frac{1}{(k+1)!}\left( 1 - \frac{1}{k+1} \right) \left( 1 - \frac{2}{k+1} \right) \ldots \left( 1 - \frac{k}{k+1} \right)\\
\\
\\ > 1 + 1 + \frac{1}{2!}\left( 1 - \frac{1}{k+1} \right) + \frac{1}{3!}\left( 1 - \frac{1}{k+1} \right) \left( 1 - \frac{2}{k+1} \right) + \ldots \\
\\
\\ + \frac{1}{k!}\left( 1 - \frac{1}{k+1} \right) \left( 1 - \frac{2}{k+1} \right) \ldots \left( 1 - \frac{k-1}{k+1} \right)\\
\\
\\ > t_k.\\
\end{align}
This proves that $\ t_k\ $ is strictly increasing. Since we know that $
\displaystyle\lim_{k\to\infty}\left( 1+\frac{1}{k} \right )^k=e,$ we have the left half of the inequality.
I want to prove the right half of the inequality in the same way, although this is trickier, I think. I'll try again later.
|
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|
Evaluating $\frac{a^3+b^3+c^3-3abc}{a^4+b^4+c^4-4abc}$, given $a+b+c=a^2+b^2+c^2=1$ How do I solve this problem?
Says so:
If $$a+b+c=a^2+b^2+c^2=1$$
find the value of
$$M=\frac{a^3+b^3+c^3-3abc}{a^4+b^4+c^4-4abc}$$
edit:
I know these identities
$$1)(a + b + c)^2 = a^2 + + c^2 + b^2+2 a b + + 2 a c + 2 b c$$
$$2)(a + b + c)^3 =3 a^2 b+3 a^2 c+a^3+3 a b^2+6 a b c+3 a c^2+3 b^2 c+b^3+3 b c^2+c^3$$
$$3)(a + b + c)^4=6 a^2 b^2+12 a^2 b c+4 a^3 b+6 a^2 c^2+4 a^3 c+a^4+12 a b^2 c+4 a b^3+12 a b c^2+4 a c^3+6 b^2 c^2+4 b^3 c+b^4+4 b c^3+c^4$$
|
Expanding on the comments, given that $\,a+b+c=1\,$ and $\,ab+bc+ca=0\,$:
$$
\begin{align}
p(x) = (x-a)(x-b)(x-c) &= x^3 - (a+b+c)x^2 + (ab+bc+ca)x - abc
\\ &= x^3 - x^2 - abc
\end{align}
$$
Since $\,p(a)=p(b)=p(c)=0\,$:
$$
\begin{cases}
a^3 = a^2 + abc
\\ b^3 = b^2 + abc
\\ c^3 = c^2 + abc
\end{cases}
$$
Adding the three equalities:
$$
a^3+b^3+c^3=a^2+b^2+c^2 + 3 abc = 1 + 3 abc \;\;\iff\;\; a^3+b^3+c^3 - 3abc = 1
$$
Multiplying the same equalities by $a,b,c$, respectively, and adding up:
$$
a^4+b^4+c^4 = a^3+b^3+c^3 + abc(a+b+c) = 1 + 4abc \;\;\iff\;\; a^4+b^4+c^4-4abc = 1
$$
Note: first identity is equivalent to Vieta's formulas, the others Newton's identities in disguise.
|
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|
Find all solutions to $\frac{a+b}{b} = \frac{a}{a+b}$ I want to find all solutions to $\frac{a+b}{b} = \frac{a}{a+b}$. I have plugged this into a calculator which tells us that if we solve for $a$, without loss of generality, we obtain
$$a = - \frac{b}{2} \pm \frac{\sqrt{3}}{2}i$$
and vice versa. I am not sure how to show this, however. I have tried a couple of different things. For example,
Method 1:
$$\frac{a}{a+b} = \frac{a+b}{b} \implies (a+b)^2=ab \implies a+b = \pm \sqrt{ab}$$
Method 2:
$$\frac{a}{a+b} \cdot \frac{a-b}{a-b}= \frac{a+b}{b} \cdot \frac{a-b}{a-b} \implies \frac{a(a-b)}{a^2-b^2}=\frac{a^2-b^2}{b(a-b)} \implies \big( a^2-b^2 \big)^2 = ab(a-b)^2$$
But so far neither of these methods are shedding any light for me. Any advice? Thanks in advance.
|
To begin with, let us assume that $b\neq 0$ and $a\neq -b$.
Now we can multiply both sides by $b(a+b)$ in order to obtain
\begin{align*}
(a + b)^{2} = ab & \Longleftrightarrow a^{2} + 2ab + b^{2} = ab\\\\
& \Longleftrightarrow a^{2} + ab + b^{2} = 0\\\\
& \Longleftrightarrow \left(a + \frac{b}{2}\right)^{2} + \frac{3b^{2}}{4} = 0\\\\
& \Longleftrightarrow \left(a + \frac{b}{2}\right)^{2} = -\frac{3b^{2}}{4}\\\\
& \Longleftrightarrow a + \frac{b}{2} = \pm\frac{bi\sqrt{3}}{2}\\\\
& \Longleftrightarrow a = -\frac{b \pm bi\sqrt{3}}{2}
\end{align*}
Hopefully this helps !
|
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|
Prove $\sum_{k=1}^{n-1} k \cos\frac{3 \pi k}{n} \sin\frac{\pi k}{n}=\frac n4\csc\frac{2\pi}n$ I would like to evaluate the sum
$$\sum_{k=1}^{n-1} k \cos\frac{3 \pi k}{n} \sin\frac{\pi k}{n}
$$
which is said to reduce to the simple close-form $\frac n4\csc\frac{2\pi}n$. I have verified it numerically for a large number of $n$’s. However, I have struggled to prove it by using familiar trigonometric identities. The index term $k$ in front of the sequence is problematic. I am not sure how to get around it. Appreciate any hint or proof.
|
To get rid of the $k$, I use the following identity:
$$\sum_{k=1}^{n-1} kz^k=z\frac{d}{dz}\sum_{k=0}^{n-1}z^k=z\frac{d}{dz}\frac{1-z^n}{1-z}=\frac{(n-1)z^{n+1}-nz^n+z}{(1-z)^2}$$
Thus using $z=e^{\frac {4i\pi}n}$,
$$\begin{split}
\sum_{k=1}^{n-1} k \cos\frac{3 \pi k}{n} \sin\frac{\pi k}{n} &= \frac 1 2\sum_{k=1}^{n-1} k \left(\sin \left (\frac {4\pi k}{n}\right) -\sin \left (\frac {2\pi k}{n}\right)\right)\\
&= \frac 1 2 \Im \left [ \sum_{k=1}^{n-1} k e^{\frac {4i\pi k}{n}} -\sum_{k=1}^{n-1} k e^{\frac {2i\pi k}{n}}\right]\\
&= \frac 1 2 \Im\left[ \frac{(n-1)e^{\frac {4i\pi}{n}}-n+e^{\frac {4i\pi}{n}}}{(1-e^{\frac {4i\pi}{n}})^2} - \frac{(n-1)e^{\frac {2i\pi}{n}}-n+e^{\frac {2i\pi}{n}}}{(1-e^{\frac {2i\pi}{n}})^2}\right]\\
&= \frac 1 2 \Im\left[ \frac{(n-1)e^{\frac {4i\pi}{n}}-n+e^{\frac {4i\pi}{n}}}{(1-e^{\frac {4i\pi}{n}})^2} - \frac{(n-1)e^{\frac {2i\pi}{n}}-n+e^{\frac {2i\pi}{n}}}{(1-e^{\frac {4i\pi}{n}})^2}(1+e^{\frac {2i\pi}{n}})^2\right]\\
&= \frac 1 2 \Im\left[ \frac{ne^{\frac {2i\pi}{n}}(1-e^{\frac {4i\pi}{n}})}{(1-e^{\frac {4i\pi}{n}})^2} \right]\\
&= \frac 1 2 \Im\left[ \frac{ne^{\frac {2i\pi}{n}}}{(1-e^{\frac {4i\pi}{n}})} \right]\\
&= \frac n 2 \Im\left[ \frac{1}{-2i\sin\left(\frac {4\pi}{n}\right)} \right]\\
&= \frac n 4 \csc \left(\frac {4i\pi}{n}\right)
\end{split}$$
|
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|
Solving $\overline{z}\cdot|z|\cdot z^5=8\sqrt{2}\left(-\sin\frac{\pi}{5}+i\cos\frac{\pi}{5}\right)^8$ I have this equation to solve:
$$\overline{z}\cdot|z|\cdot z^5=8\sqrt{2}\left(-\sin\frac{\pi}{5}+i\cos\frac{\pi}{5}\right)^8$$
Since $\overline{z}\cdot z = |z|^2$ and utilizing the de Moivre's formula this can be simplified to:
$$|z|^3z^4=8\sqrt{2}\left(\cos\frac{8\pi}{5}+i\sin\frac{8\pi}{5}\right)$$
$$|z|^7(\cos{4\alpha} + i\sin{4\alpha})=8\sqrt{2}\left(\cos\frac{8\pi}{5}+i\sin\frac{8\pi}{5}\right)$$
From here I thought just to compare $|z|^7 = 8\sqrt{2}$ and the sine, cosine part.
Is this the correct way to go about it or could it be done using some simpler method?
|
From the beginning:
Since $i^8 = 1$, we can multiply by $i^8$ to get $(- \cos(\pi/5) - i \sin (\pi/5))^8$ which is the same as $(\cos \pi/5 + i \sin \pi/5)^8$.
So now you have:
$$|z|^3z^4=8\sqrt{2} e^{8i \pi/5}$$
which means that $z$ has modulus $(8 \sqrt{2})^{1/7} = \sqrt{2}$. Now since $|z|^3$ is real, you have:
$$\arg z = \frac{1}{4} \arg z^4 = \frac{1}{4} \arg |z|^3z^4.$$
|
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|
Using Lagrange Mult. to find maximum of $2x + y^2$ with constraints: $x^2 + y^2 + z^2 = 2$ and $x + y^2 + z = 0$ I was trying to maximize the function $2x + y^2$ with the following constraints: $x^2 + y^2 + z^2 = 2$ and $x + y^2 + z = 0$ using Lagrange multipliers. I already did all the work of proving that the maximum does exist and showing that we can apply the technique of Lagrange multiplies. After that I arrived at the following systems of equations:
$$\begin{cases}2 = 2\alpha x + \beta \\
2y = 2\alpha y + 2 \beta y \\
0 = 2\alpha z + \beta \\
x^2 +y^2 + z^2 =2 \\
x + y^2 + z = 0\end{cases}$$
Where $\alpha$ and $\beta$ are the Lagrange multipliers. The problem is that I'm having a lot of trouble solving this system for $(x,y,z)$.
My plan was trying to write $x,y,z$ in terms of $\alpha,\beta$ and then using the equations $x^2 + y^2 + z^2 = 2$ and $x + y^2 + z = 0$ find the values for $\alpha $ and $\beta$, but I wasn't able to find an expression for $y$ in terms of $\alpha$ and $\beta$.
How can this system be solved?
|
Using the two constraint-multiplier equations you wrote,
$$ 2 \ - \ 2\alpha x \ - \ \beta \ \ = \ \ 0 \ \ \ , \ \ \ 2y \ - \ 2\alpha y \ - \ 2 \beta y \ \ = \ \ 0 \ \ \ , \ \ \ 0 \ - \ 2\alpha z \ - \ \beta \ \ = \ \ 0 \ \ , $$
we could eliminate $ \ \beta \ $ between the first and third equations to write $ \ 2·(1 - \alpha·x) \ = \ -2·\alpha·z $ $ \Rightarrow \ 1 - \alpha·(x - z) \ = \ 0 \ \Rightarrow \ \alpha \ = \ \frac{1}{x \ - \ z} \ \ . $ The second equation is factorable as $ \ y·(1 - \alpha - \beta) \ = \ 0 \ \ . $
The first case from this equation is the easier one to deal with: with $ \ y \ = \ 0 \ \ , $ the constraint equation $ \ x + y^2 + z \ = \ 0 \ \ $ requires that $ \ z \ = \ -x \ \ . $ The constraint sphere then produces
$$ x^2 \ + \ 0^2 \ + \ (-x)^2 \ \ = \ \ 2·x^2 \ \ = \ \ 2 \ \ , $$
giving us the extremal points $ \ ( \ \pm 1 \ , \ 0 \ , \ \mp 1 \ ) \ \ $ and the function values $ \ \mathbf{f( \ 1 \ , \ 0 \ , \ -1 \ ) } \ = \ 2·1 + 0 \ \mathbf{= \ +2} \ \ $ and $ \ \ \mathbf{f( \ -1 \ , \ 0 \ , \ 1 \ )} \ = \ 2·(-1) + 0 \ \mathbf{= \ -2} \ \ . $
With $ \ y \ \neq \ 0 \ \ , $ we have $ \ 1 - \alpha - \beta \ = \ 1 - \frac{1}{x \ - \ z} - \beta \ = \ 0 \ \Rightarrow \ \beta \ = \ 1 - \frac{1}{x \ - \ z} \ \ . $ As the third equation gave us $ \ \beta \ = \ -2\alpha z \ \ , $ we obtain $ \ \frac{x \ - \ z \ - \ 1}{x \ - \ z} \ = \ -\frac{2·z}{x \ - \ z} \ \Rightarrow \ x - z \ = \ 1 - 2z \ \Rightarrow \ z \ = \ 1 - x \ \ . $ Inserting this into the constraint parabolic-cylinder equation yields
$$ x \ + \ y^2 \ + \ ( \ 1 \ - \ x \ ) \ \ = \ \ 0 \ \ \Rightarrow \ \ y^2 \ \ = \ \ -1 \ \ . $$
As this is impermissible for coordinate values, the only two extrema are those found above. We can see this in the graph below: the constraint sphere in orange and the constraint skewed-parabolic cylinder $ \ x + y^2 + z \ = \ 0 \ \ $ [in yellow] intersect on a three-dimensional curve. However, the level-surfaces $ \ f(x,y) \ = \ x + y^2 \ = \ c \ $ are (upright) parabolic cylinders which are only tangent to this curve when the vertex of the parabolic cross-section lies at $ \ x \ = \ -1 \ , \ y \ = \ 0 \ \ [ \ c \ = \ -2 \ ] \ \ $ or $ \ x \ = \ +1 \ , \ y \ = \ 0 \ \ [ \ c \ = \ +2 \ ] \ \ . $
|
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|
Why is $ab$ likely to have more divisors than $(a-b)(a+b)$? Consider the two numbers $ab$ and $(a-b)(a+b), \gcd(a,b) = 1, 1 \le b < a$. On an average, which of these two numbers has more distinct prime factors? All the prime factors of $a$ and $b$ divide $ab$ and similarly all the prime factors of $a-b$ and $a+b$ divide $(a-b)(a+b)$. So one number does not seem to have an obvious advantage over the other. However if we look at the data than we see that $ab$ dominates.
Let $f(x)$ be the average number of distinct prime factors in all such $ab, a \le x$ and
$g(x)$ be the average number of distinct prime factors in all such $(a-b)(a+b), a \le x$.
Update: Experimental data for the first $6.1 \times 10^{9}$ pairs of $(a,b)$ shows that $f(x) - g(x) \sim 0.30318$. Instead of distinct prime factors, if we count the number of divisors than $f(x) - g(x) \sim 0.848$.
Question: Why is $ab$ likely to have more distinct prime factors or divisors than $(a-b)(a+b)$ and what is the limiting value of $f(x) - g(x)$?
|
On one hand,
\begin{align*}
\sum_{a,b\le x} \omega(ab) &= \sum_{a,b\le x} \sum_{\substack{p\le x \\ p\mid ab}} 1 = \sum_{p\le x} \sum_{\substack{a,b\le x \\ p\mid ab}} 1 \\
&= \sum_{p\le x} \biggl( \sum_{\substack{a,b\le x \\ p\mid a}} 1 + \sum_{\substack{a,b\le x \\ p\mid b}} 1 - \sum_{\substack{a,b\le x \\ p\mid a,\, p\mid b}} 1 \biggr) \\
&= \sum_{p\le x} \biggl( \bigl( \tfrac xp+O(1) \bigr)(x+O(1)) + (x+O(1))\bigl( \tfrac xp+O(1) \bigr) - \bigl( \tfrac xp+O(1) \bigr)^2 \biggr) \\
&= 2x^2 \sum_{p\le x} \tfrac1p - x^2 \sum_{p\le x} \tfrac1{p^2} + O\biggl( x \sum_{p\le x} \bigl( 1+\tfrac1p \bigr) \bigg) = 2x^2 \sum_{p\le x} \tfrac1p - x^2 \sum_p \tfrac1{p^2} + o(x^2).
\end{align*}
On the other hand, $p\mid(a+b)$ and $p\mid(a-b)$ simultaneously if and only if either $p\mid a$ and $p\mid b$, or $p=2$ and $a$ and $b$ are both odd. Therefore
\begin{align*}
\sum_{a,b\le x} \omega\bigl( (a+b)(a-b) \bigr) &= \sum_{a,b\le x} \sum_{\substack{p\le x \\ p\mid (a+b)(a-b)}} 1 = \sum_{p\le x} \sum_{\substack{a,b\le x \\ p\mid (a+b)(a-b)}} 1 \\
&= \sum_{p\le x} \biggl( \sum_{\substack{a,b\le x \\ p\mid (a+b)}} 1 + \sum_{\substack{a,b\le x \\ p\mid (a-b)}} 1 - \sum_{\substack{a,b\le x \\ p\mid (a+b),\, p\mid (a-b)}} 1 \biggr) \\
&= \sum_{p\le x} \biggl( \sum_{a\le x} \sum_{\substack{b\le x \\ b\equiv-a\,(\mathop{\rm mod}\,p)}} 1 + \sum_{a\le x} \sum_{\substack{b\le x \\ b\equiv a\,(\mathop{\rm mod}\,p)}} 1 - \sum_{\substack{a,b\le x \\ p\mid a,\, p\mid b}} 1 \biggr) \\
&\qquad{}- \sum_{\substack{a,b\le x \\ a,b \text{ both odd}}} 1 \\
&= \sum_{p\le x} \biggl( (x+O(1))\bigl( \tfrac xp+O(1) \bigr) + (x+O(1))\bigl( \tfrac xp+O(1) \bigr) - \bigl( \tfrac xp+O(1) \bigr)^2 \biggr) \\
&\qquad{}- \bigl( \tfrac x2+O(1) \bigr)^2 \\
&= 2x^2 \sum_{p\le x} \tfrac1p - x^2 \sum_p \tfrac1{p^2} - \tfrac{x^2}4 + o(x^2).
\end{align*}
From these two asymptotic formulas it follows that the difference of the two sums is asymptotic to $\frac{x^2}4$, so that the average difference in the number of distinct prime factors is asymptotically $\frac14$.
Heuristically (in hindsight), the difference is entirely caused by the prime $2$: since $a$ and $b$ are even or odd independently, there is a $\frac34$ chance that $p=2$ will contribute to $\omega(ab)$; but since $a+b$ and $a-b$ are both even or both odd, there is only a $\frac12$ chance that $p=2$ will contribute to $\omega\bigl( (a+b)(a-b) \bigr)$.
|
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|
Finding the number of complex numbers satisfying $|z|=\text{max}\{|z-1|,|z+1|\}$
Find the number of complex numbers satisfying $|z|=\text{max}\{|z-1|,|z+1|\}$
My Attempt: Let $z=x+iy$, so, $$\sqrt{x^2+y^2}=\text{max}\{\sqrt{(x-1)^2+y^2},\sqrt{(x+1)^2+y^2}\}$$
Case I: $\sqrt{x^2+y^2}=\sqrt{(x-1)^2+y^2}\implies \pm x=x-1\implies x=\frac12$
Case II: $\sqrt{x^2+y^2}=\sqrt{(x+1)^2+y^2}\implies \pm x=x+1\implies x=-\frac12$
Does that mean if the complex numbers are lying on two lines, either $x=\frac12$ or $x=-\frac12$ then they would satisfy the required equation?
Therefore, the required number of complex numbers is infinite?
But the answer given is zero.
Can we solve this graphically?
|
It follows from
$$
|z-1|^2 = (x-1)^2 + y^2 = x^2 + y^2 - 2x + 1 \\
|z+1|^2 = (x-1)^2 + y^2 = x^2 + y^2 + 2x + 1
$$
that
$$
\max(|z-1|, |z+1|)^2 = x^2 + y^2 + 2|x| + 1 > x^2 + y^2 = |z|^2
$$
which shows that the equation has no solution.
Geometrically:
$$
\max(|z-1|, |z+1|) = |z+1|
$$
means that the distance from $z$ to $1$ is less than or equal to the distance from $z$ to $-1$, that are exactly the points in the closed right half-plane. But points in the right half-plane are closer to the origin than to the point $-1$, so that
$$
\max(|z-1|, |z+1|) = |z+1| > |z|
$$
In the same way you can show that the equation has no solution in the (closed) left half-plane.
With respect to your approach: It is correct that $|z| = |z-1|$ if and only if $x=1/2$ (case I). But for those $z$ is
$$
|z-1| < |z+1|
$$
so that does not give any solution. Similar in case II.
|
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|
$ab+ac+bc=2\quad ,\quad\min(10a^2+10b^2+c^2)=?$
If $ab+ac+bc=2$, find minimum value of $10a^2+10b^2+c^2$
$1)3\qquad\qquad2)4\qquad\qquad3)8\qquad\qquad4)10$
I used AM-GM inequality for three variables:
$$ab+ac+bc\ge3(abc)^{\frac23}\quad\Rightarrow\quad 2\ge3(abc)^{\frac23}$$
$$10a^2+10b^2+c^2\ge 3(10abc)^{\frac23}$$
Hence $10a^2+b^2+c^2\ge2\times 10^{\frac23}$. But it is not in the options.
|
Hint: $$10a^2+10b^2+c^2-4(ab+bc+ca)=\frac{2}{5}(5a-b-c)^2+\frac{3}{5}(4b-c)^2$$
|
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|
How do I prove: $xyz(x-2)(y-2)(z-2)\leq\left(1+\frac{2(xy+yz+xz)}{3}\right)^3,\ x,y,z>0$? I have a question which askes to prove:
$\displaystyle \tag*{} xyz(x-2)(y-2)(z-2)\leq\left(1+\frac{2(xy+yz+xz)}{3}\right)^3,\ x,y,z>0$?
My approach: I tried AM-GM inequality, didn't work well. So I tried substituting $2(xy+yz+xz)=(x+y+z)^2-x^2-y^2-z^2$, I couldn't proceed further. I even tried expanding the LHS, it's tedious. Any help would be appreciated, thanks.
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We want $x+y+z$ largest when $xyz,xy+yz+zx$ fixed. By 3.83 in the Mathematical Inequalities (Volume 1) by Vasile Cîrtoaje, we can assume $x=y\le z$. Now we rewrite the inequality as $x=y$:
$$ x^2z(x-2)^2(z-2)\leq\left(1+\frac{2(x^2+2xz)}{3}\right)^3$$
We must have $z\ge 2$ otherwise left is smaller than zero, and right is larger than zero.
Now expand everything, we have
$$RHS-LHS=\frac{64x^3z^3+(69x^4+108x^3+36x^2)z^2+(48x^5+54x^4-72x^3+216x^2+108x)z+8x^6+36x^4+54x^2+27}{27}$$
Since $z\ge 2$ so $64x^3z^3\ge72x^3z$. All others are positive, so this is proved.
Remark: I suspect
$$xyz(x-2)(y-2)(z-2)\leq\left(1+\frac{(xy+yz+xz)}{3}\right)^3$$
|
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|
Evaluate $\int_0^\infty\frac{x^2\ln x}{x^4+x^2+1}dx$ by the residue theorem The result should be $\frac{\pi^2}{12}$.
Edit: I have tried to reproduce the image, but limitations of MathJax required some reformatting. Here is the original image.
$$
\begin{align}
\int_0^\infty\frac{x^2\ln x\,\mathrm{d}x}{x^4+x^2+1}&=\int_0^\infty\frac{x^2\ln x}{\left(x^2+x+1\right)\left(x^2-x+1\right)}\\
&\text{poles}\left\{\begin{array}{}
\boxed{\textstyle\frac12+\frac{i\sqrt3}2}&\raise{5pt}{\text{lies inside}\\\text{contour}}\\
\frac12-\frac{i\sqrt3}2\\
\boxed{\textstyle-\frac12+\frac{i\sqrt3}2}&\raise{5pt}{\text{lies inside}\\\text{contour}}\\
-\frac12-\frac{i\sqrt3}2\\
\end{array}\right.
\end{align}
$$
$$
\begin{align}
\int_{-\infty}^0\frac{x^2\ln x\,\mathrm{d}x}{x^4+x^2+1}+\int_0^\infty\frac{x^2\ln x\,\mathrm{d}x}{x^4+x^2+1}&=2\pi i\sum\text{Res}\\
\int_0^\infty\frac{x^2\ln(-x)\,\mathrm{d}x}{x^4+x^2+1}+\int_0^\infty\frac{x^2\ln(x)\,\mathrm{d}x}{x^4+x^2+1}&=\qquad"\\
\int_0^\infty\frac{x^2(i\pi+\ln x)\,\mathrm{d}x}{x^4+x^2+1}+\int_0^\infty\frac{x^2\ln(x)\,\mathrm{d}x}{x^4+x^2+1}&=\qquad"\\
\int_0^\infty\frac{x^2(i\pi)\,\mathrm{d}x}{x^4+x^2+1}+2\int_0^\infty\frac{x^2\ln(x)\,\mathrm{d}x}{x^4+x^2+1}&=\qquad"\\
\end{align}
$$
We need to find the residue
Then we can compare RHS & LHS side
|
You can use Feynman's method to evaluate.
Let
$$I(a)=\int_0^\infty\frac{x^a}{x^4+x^2+1}dx.$$
Then
\begin{eqnarray}
I(a)&=&\int_0^1\frac{x^a+x^{2-a}}{x^4+x^2+1}dx\\
&=&\int_0^1\frac{(1-x^2)(x^a+x^{2-a})}{1-x^6}dx\\
&=&\int_0^1\sum_{n=0}^\infty(1-x^2)(x^a+x^{2-a})x^{6n}dx\\
&=&\sum_{n=0}^\infty\int_0^1(1-x^2)(x^a+x^{2-a})x^{6n}dx\\
&=&\sum_{n=0}^\infty\int_0^1(x^{6n+a}+x^{6n+2-a}-x^{6n+2+a}-x^{6n+4-a})dx\\
&=&\sum_{n=0}^\infty\bigg(\frac{1}{6n+a+1}+\frac{1}{6n+3-a}-\frac{1}{6n+3+a}-\frac{1}{6n+5-a}\bigg)
\end{eqnarray}
and hence
\begin{eqnarray}
I'(2)&=&\sum_{n=0}^\infty\bigg(-\frac{2}{(6n+3)^2}+\frac{1}{(6n+1)^2}+\frac{1}{(6n+5)^2}\bigg)\\
&=&-\frac{2}{9}\sum_{n=0}^\infty\frac{1}{(2n+1)^2}+\sum_{n=0}^\infty\bigg(\frac{1}{(6n+1)^2}+\frac{1}{(6n+5)^2}\bigg)\\
&=&-\frac{2}{9}\sum_{n=0}^\infty\frac{1}{(2n+1)^2}+\sum_{n=-\infty}^\infty\frac{1}{(6n+1)^2}\\
&=&-\frac{2}{9}\frac{\pi^2}{8}+\frac{\pi^2}{9}\\
&=&\frac{\pi^2}{12}.
\end{eqnarray}
Here
$$ \sum_{n=0}^\infty\frac{1}{(2n+1)^2}=\frac{\pi^2}{8},\sum_{n=-\infty}^\infty\frac{1}{(6n+1)^2}=\frac{\pi^2}{9} $$
are used.
|
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|
Seeing if a polynomial is irreducible over a field extension of $\Bbb Q$. Exercise. Determine $m_{i\sqrt[3]{2}}$ over $\Bbb Q(i)$, where $m_{i\sqrt[3]{2}}$ represents the minimal polynomial of ${i\sqrt[3]{2}}$.
My attempt. We know that $m_i = x^2 +1$ over $\Bbb Q$ (this is obvious) and thus $\deg(i) = \deg(m_i) = 2$ over $\Bbb Q$. Given this, we can say that
\begin{equation*}
\Bbb Q(i) = <1,i>_\Bbb Q = \{a + b i : a,b \in \Bbb Q\}
\end{equation*}
Now, obviously $i\sqrt[3]{2} \notin \Bbb Q(i)$ since $\sqrt[3]{2} \notin \Bbb Q$ (the only chance would be taking $a=0$ and $b = \sqrt[3]{2}$). So, $\deg(i\sqrt[3]{2}) > 1$ over $\Bbb Q(i)$.Further more,
\begin{equation*}
(i\sqrt[3]{2})^3 = -i*2 = (-2) i
\end{equation*}
And so $i\sqrt[3]{2}$ is a root of the polynomial $f = x^3 + 2i$ over $\Bbb Q(i)$. Since $f$ is monic, all we have to do now is to prove that $f$ is irreducible over $\Bbb Q(i)$. Since $f$ has $\deg = 3$ and $\Bbb Q(i)$ is a field extension (and thus a field) all we have to is check if $f$ has roots in $\Bbb Q(i)$. Roots of $f$ are the following:
\begin{equation*}
x=\frac{\sqrt[3]{2}\sqrt{3}}{2}-i\frac{\sqrt[3]{2}}{2},\:x=\sqrt[3]{2}i,\:x=-\frac{\sqrt[3]{2}\sqrt{3}}{2}-i\frac{\sqrt[3]{2}}{2}
\end{equation*}
None of this is in $\Bbb Q(i)$ and so $f$ is irreducible, proving that $ m_{i\sqrt[3]{2}} = x^3 + 2i$ over $\Bbb Q(i)$.
My doubts. First, I would obviously like to know if what I did is right. Other than this, I would like to know how to compute the complex roots of $f$ manually (I used symbolab software to find the roots).
Thanks for any help in advance!
|
What you've done is correct. There are a few ways to do it by hand, without using a computer.
The most elementary way is to take a general element $a + bi \in \mathbb{Q}(i)$ and show that it can't be a root of $f(x)$. That is, let $a, b \in \mathbb{Q}$ and suppose that $f(a + bi) = 0$. Then
$$
(a + bi)^3 = -2i,
$$
which implies
$$
(a^3 - 3ab^2) + (3a^2b - b^3)i = -2i.
$$
Comparing real parts, we get $a^3 - 3ab^2 = 0$, so $a=0$ or $a^2 = 3b^2$. If $a \neq 0$, then $a^2 = 3b^2$, so $b\neq 0$ and $(a/b)^2 = 3$, which famously is impossible, since $a/b \in \mathbb{Q}$. Therefore, $a = 0$, and comparing imaginary parts gives $b^3 = 2$, which is also impossible.
If you want to actually find the complex roots, just solve the equations $a^3 = 3ab^2$ and $3a^2b - b^3 = -2$ over $\mathbb{R}$. Assuming that $a$ is nonzero (the $a=0$ case is easier), we have $a^2 = 3b^2$, implying that $9b^3 - b^3 = -2$, so $8b^3 = -2$ and therefore $b^3 = - 1/4$. We get $b = - \frac{\sqrt[3]{2}}{2}$, and then you can get $a$ from $a^2 = 3b^2$.
A slightly less elementary way would be to put $z = re^{i\theta}$ for $r>0$ and $\theta \in [0, 2\pi)$, and suppose that $f(z) = 0$. Then $r^3e^{i3\theta} = -2i$, which gives $r^3 = 2$ and $3\theta \equiv \frac{3\pi}{2} \pmod{2\pi}$. This implies $r = \sqrt[3]{2}$ and $\theta = \frac{\pi}{2}, \frac{7\pi}{6}, \frac{11\pi}{6}$, from which it is easy to compute the real and imaginary parts of the roots.
|
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Find the $\lim_\limits{n\to\infty}{\frac{\sqrt{n^2+1}+\sqrt{n}}{(n^4+1)^{1/4}-\sqrt{n}}}$ How to solve the limits without using L-hospital law, like using rationalisation
L-hospital method is taking too long
The final answer I got was
$$\lim_\limits{n\to\infty}{\frac{\sqrt{n^2+1}+\sqrt{n}}{(n^4+1)^{1/4}-\sqrt{n}}} = 1$$
|
$$\lim_\limits{n\to\infty}{\frac{\sqrt{n^2+1}+\sqrt{n}}{(n^4+1)^{1/4}-\sqrt{n}}} = \lim_\limits{n\to\infty}{\frac{(\sqrt{n^2+1}+\sqrt{n})((n^4+1)^{1/4}+\sqrt{n})}{((n^4+1)^{1/4}-\sqrt{n})((n^4+1)^{1/4}+\sqrt{n})}} = \lim_\limits{n\to\infty}{\frac{(\sqrt{n^2+1}+\sqrt{n})((n^4+1)^{1/4}+\sqrt{n})}{(n^{4}+1)^{\frac{1}{2}}-n}} = \lim_\limits{n\to\infty}{\frac{(\sqrt{n^2+1}+\sqrt{n})((n^4+1)^{1/4}+\sqrt{n})((n^{4}+1)^\frac{1}{2}+n)}{((n^{4}+1)^{\frac{1}{2}}-n)((n^{4}+1)^\frac{1}{2}+n)}} = \lim_\limits{n\to\infty}{\frac{(\sqrt{n^2+1}+\sqrt{n})((n^4+1)^{1/4}+\sqrt{n})((n^{4}+1)^\frac{1}{2}+n)}{n^{4}+1 - n^2}}.$$Can you take it from there? Now it is merely a matter of comparing the coefficients of the highest powers of n in the numerator and the denominator.
|
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How can I show that $\underset{\left(x,y\right)\rightarrow\left(0,0\right)}{\lim}{\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}}$ exists, using limit def? I am trying to solve an exercise to show that this function $$f(x,y)=\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}$$ has a limit as $(x,y)$ approaches $(0,0)$:
$$\underset{\left(x,y\right)\rightarrow\left(0,0\right)}{\lim}{\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}}$$
To do so, I have to use the limit precise definition:
$$\forall\epsilon>0\ \exists\delta>0∶\ \left|\left(x,y\right)-\left(a,b\right)\right|<\delta\ {\Rightarrow}\left|f\left(x,y\right)-L\right|<\ \epsilon$$
I've found out that the the potential limit $L=2$ by evaluating the limits across the x-axis, y-axis, arbitrary line and x^2 parabola and y^2 parabola.
I have plugged all the necessary values and ended up with this:
$$\forall\epsilon>0\ \exists\delta>0∶\ \sqrt{x^2+y^2}<\delta\ {\Rightarrow}\left|\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}-2\right|<\ \epsilon$$
Now, I am in trouble to construct inequalities that lead to prove that the following is true:
$$\left|\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}-2\right|<\sqrt{x^2+y^2}\ $$
Can someone help me out here?
|
Let $r^2=x^2+y^2$
$|\frac{r^2}{\sqrt{r^2+1}-1}-2|<\epsilon$
$2-\epsilon<\sqrt{r^2+1}+1<2+\epsilon$ by rationalizing the denominator, cancelling terms and pushing $2$ to the outsides of the compound inequality.
$ \sqrt{r^2+1}\le1+\frac{r^2}{2} \implies -\epsilon< r^2/2<\epsilon$ by Taylor's Theorem, then cancelling, then swapping $x$ and $y$ back in.
$x^2+y^2<2\epsilon$. So let $\delta=\epsilon$ for a simpler, tighter bound.
|
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|
What's the ratio between the segments $\frac{AF.BG}{FG}$ in the figure below? For reference: In the figure below the trapezoid has height $13$ and is inscribed in a circle of radius $15$. Point $E$ is on the minor arc determined by $A$ and $B$, the points $F$ and $G$ intersect with $ED$, $E$C and $AB$. Then the ratio between the segments $\frac{AF \times BG}{FG}$ is?
My progress:
I'm lost in this question...Perhaps using some of the segment $AB$ ratio with the diameter $DC$ ??
$\triangle EFG\sim HJC: \frac{JC}{FG} = \frac{JH}{FE}=\frac{HC}{EG}\\
\triangle BJG \sim \triangle HJC: \frac{JC}{JG}=\frac{JH}{BJ}=\frac{HC}{BG}\\
\triangle KID \sim \triangle KAF: \frac{JC}{FK}=\frac{JH}{AF}=\frac{HC}{AK}$
...???
|
Notice that points $H$ and $I$ in this answer are different from those in the original post.
Let $\frac{AF\cdot BG}{FG}=R$. First of all, $$AB=AF+FG+BG=2\sqrt{15^2-13^2}=4\sqrt{14}$$
Power of a point gives us
\begin{align*}
AF\cdot FB&=AF\cdot\left(4\sqrt{14}-AF\right)=EF\cdot DF&&(1)\\
BG\cdot AG&=BG\cdot\left(4\sqrt{14}-BG\right)=EG\cdot GC&&(2)
\end{align*}
and it follows that
\begin{align*}
\frac{(1)\cdot(2)}{FG^2}\implies R\left(4\sqrt{14}+R\right)=\frac{EF}{FG}\cdot \frac{EG}{FG}\cdot DF\cdot GC
\end{align*}
Now, with $\triangle EFG\sim\triangle HDF$ we have $\frac{EF}{FG}=\frac{HD}{DF}$. Similarly, with $\triangle EFG\sim\triangle IGC$ we have $\frac{EG}{FG}=\frac{IC}{GC}$. Therefore, $$R\left(4\sqrt{14}+R\right)=HD\cdot IC$$
Since $\triangle HDF\sim\triangle IGC$, it follows that $$HD\cdot IC=HF\cdot IG=13^2=169$$ and we finally have the quadratic equation $$R^2+4\sqrt{14}R-169=0$$ whose solutions are $$R=-2\sqrt{14}\pm15$$
We'll take $\color{blue}{R=-2\sqrt{14}+15}$ since $R>0$.
Maybe there are solutions that are more straightforward or concise, but I'm just gonna put my answer out here. I believe $R=-2\sqrt{14}+15$ is the correct answer since I've verified it on GeoGebra. Some lengthy brute-force simplifications are skipped in this answer.
|
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|
Confusions in Holder's Inequality Holder's Inequality states that for nonnegative real numbers $a_1,...,a_n$ and $b_1,...,b_n$ we have $$\left(\sum_{i=1}^na_i\right)^p\left(\sum_{i=1}^nb_i\right)^q\ge \left(\sum_{i=1}\sqrt[p+q]{a_i^pb_i^q}\right)^{p+q}$$
Where $p$ and $q$ are positive real numbers.
Here is my problem : $a,b,c$ are positive reals, prove that $$(a^5-a^2+3)(b^5-b^2+3)(c^5-c^2+3)\ge (a+b+c)^3$$
But for $x>0$ we have $x^5-x^2+3\ge x^3+2$ so we only need to prove $$(a^3+2)(b^3+2)(c^3+2)\ge (a+b+c)^3$$
But when I read the solution, it says ''From Holder's Inequality, it follows '' $$(a^3+2)(b^3+2)(c^3+2)\ge (a+b+c)^3$$
But I don't see that.
|
I would think Holder's inequality implies
$$(a^3+2)(b^3+2)(c^3+2)\geq \left((a^3b^3c^3)^{1/3}+ (2^3)^{1/3} \right)^3=(abc+2)^3$$
So it would be good to know how your last inequality follows from Holder's.
|
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|
Markov Chain Contest Problem Five cards labeled 1, 3, 5, 7, 9 are laid in a row in that order, forming the five-digit number 13579 when
read from left to right. A swap consists of picking two distinct cards, and then swapping them. After
three swaps, the cards form a new five-digit number n when read from left to right. Compute the
expected value of n.
I have shown my attempt to solve the problem in an answer. Could someone vet the solution.
|
Here's a slightly simpler way of calculating the exact expected value of $\ n\ $. After the OP's correction to my original calculation (in which I made an arithmetical error), it gives the same answer as his.
After $3$ swaps, there will be a probability $\ p\ \ \big($which turns out to be $\ \frac{3}{10}\ \big)
$ that the digit in any given place will be the same as it was at the start, and a probability of $ \frac{1-p}{4}\ \left(=\frac{7}{40}\right)\ $ that it will be any one of the other $4$ digits. The expected value of $\ n\ $ is therefore
\begin{align}
10^4\Big(&p\cdot1+\frac{(1-p)(25-1)}{4}\Big)+10^3\left(p\cdot3+\frac{(1-p)(25-3)}{4}\right)\\
&+10^2\Big(p\cdot5+\frac{(1-p)(25-5)}{4}\Big)+10\left(p\cdot7+\frac{(1-p)(25-7)}{4}\right)\\
&\hspace{1.5em}+p\cdot9+\frac{(1-p)(25-9)}{4}\\
&\hspace{3em}=\left(\frac{5p-1}{4}\right)13579+\left(\frac{25(1-p)}{4}\right)11111\\
&\hspace{3em}=\frac{13579+35\cdot11111}{8}\\
&\hspace{3em}=50308
\end{align}
You can calculate the value of $\ p\ $ by treating the presence or absence of the original digit in any place as a two-state Markov chain with transition matrix
$$
\pmatrix{\frac{3}{5}&\frac{2}{5}\\\frac{1}{10}&\frac{9}{10}}\ ,
$$
since the probability is $\ \frac{1}{10}\ $ that a single swap will return any given digit back to its original place once it has been swapped out of it. The initial state of the chain is $1$, and there are $4$ sequences of states that end with with the digit originally in the place being back there after $3$ swaps. Those sequences and their probabilities are:
$$
\begin{array}{cc}
\text{state sequence}&\text{probability}\\
1111&\left(\frac{3}{5}\right)^3=\frac{27}{125}\\
1121&\left(\frac{3}{5}\right)\left(\frac{2}{5}\right)\left(\frac{1}{10}\right)=\frac{3}{125}\\
1211&\left(\frac{2}{5}\right)\left(\frac{1}{10}\right)\left(\frac{3}{5}\right)=\frac{3}{125}\\
1221&\ \left(\frac{2}{5}\right)\left(\frac{9}{10}\right)\left(\frac{1}{10}\right)=\frac{9}{250}\ ,
\end{array}
$$
and the sum of these probabilities is $\ \frac{3}{10}\ $, as stated above.
|
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Prove that the sequence $t_n$ defined by $\dfrac{1}{n}(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}})$ converges determine if the following sequence converges
$t_n=\dfrac{1}{n}(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}})$
my solution:
$\text{ using AM-GM inequality, }\space\space t_n\ge (1\cdot2\cdot 3\cdots n)^{\dfrac{-1}{2n}}$
$\implies (1\cdot2\cdot 3\cdots n)^{\dfrac{-1}{2n}}\le t_n\le \dfrac{1}{n}(1+\cdots+1)=\dfrac{n}{n}=1$
$(1\cdot2\cdot 3\cdots n)^{\dfrac{1}{n}}=(1)^{\frac{1}{n}}\cdot (2)^{\frac{1}{n}}\cdot (3)^{\frac{1}{n}}\cdots\cdot (n)^{\frac{1}{n}}\stackrel{n\to\infty}{\to}1$
$\implies (1\cdot2\cdot 3\cdots n)^{\dfrac{-1}{2n}}\to1$
Hence, by sandwich theorem $t_n\to1$
Is this correct?
|
There is a Riemann sum in disguise.
Note, however, that the function to integrate is unbounded. It still works with the improper integral, as is explained here: Convergence of Riemann sums for improper integrals.
$$t_n=\frac{1}{n}\sum_{k=1}^n\frac{1}{\sqrt k}=\frac{1}{n}\sum_{k=1}^n\frac{1/\sqrt{n}}{\sqrt {k/n}}=\frac{1}{n\sqrt n}\sum_{k=1}^n\frac{1}{\sqrt {k/n}}$$
Hence,
$$\sqrt{n}\,t_n=\frac{1}{n}\sum_{k=1}^n\frac{1}{\sqrt {k/n}}\to\int_0^1\frac{1}{\sqrt x}\mathrm dx=2$$
And
$$t_n\sim \frac{2}{\sqrt n}$$
|
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|
solve the recurrence relation $a_n = 7a_{n-1} - 16a_{n-2} +12a_{n-3} + n4^n $ Solve the recurrence relation
$$a_n = 7a_{n-1} - 16a_{n-2} +12a_{n-3} + n4^n $$
With initial conditions:
$a_0 = −2$
$a_1 = 0$
$a_2 = 5$
I solve the homogenous part like that:
$$a_n - 7a_{n-1} + 16a_{n-2} -12a_{n-3} = 0 $$
$$ p(r) = r^3 -7r^2 + 16r - 12 $$
$$ p(r) = (r-2)^2(r-3) $$
But I am stucked with the non-homogenous part.
For $$ g(n) = 4^n $$ We can use $$ q(r) = A4^n $$ But for $$ g(n) = n4^n $$
What we should do?
|
You can find a particular solution using the $Z$ transform:
$$
Z[a_n-7a_{n-1}+16a_{n-2}-12a_{n-3}]=Z[n 4^n]
$$
then
$$
\left(1-\frac 7z+\frac {16}{z^2}-\frac{12}{z^3}\right)Z[a_n]=\frac{4z}{(z-4)^2}+7a_{-1}-16 \left(a_{-2}+\frac{a_{-1}}{z}\right)+12\left(a_{-3}+\frac{a_{-2}}{z}+\frac{a_{-1}}{z^2}\right)
$$
now taking the initial values $a_{-1}= a_{-2}=a_{-3}=0$ (a particular solution) we have
$$
Z[a_n] = \frac{4 z^4}{(z-4)^2 \left(z-2\right)^2(z-3)}
$$
and now inverting we have a particular solution
$$
a_{n}^p = 4 \left(2^n \left(2^{n+2} (n-5)-n-7\right)+3^{n+3}\right)
$$
|
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|
Solving $6^{50-x} = 2^{50}$ I'm being asked to solve an equation for $x$, giving the answer in the form below: $$a \log_{b} c$$
One of them is: $$6^{50-x} = 2^{50}$$
So I started by taking the $\log_6$ of both sides to give:
$$\log_{6}6^{50-x}=\log_{6}2^{50} \tag1$$
Leading to:
$$\begin{align}
50-x &=\log_{6}2^{50} \tag2 \\
50-x &=50\log_{6}2 \tag3 \\
-x &=-50+50\log_{6}2 \tag4
\end{align}$$
or finally:$$x=50-50\log_{6}2 \tag5$$
I've been given the answer of: $$x=50\log_{6}3 \tag6$$
How do I get from my answer to the correct form, or have I made a mistake along the way?
Thanks,
|
Your last line:
$x= 50 -50\log_6 2$;
$x=50-50\log_6 ((2 \cdot 3)/ 3));$
Can you finish?
Option:
$(2 \cdot 3)^{50}=2^{50}6^x ;$
$2^{50}3^{50}=2^{50}6^x;$
$6^x=3^{50};$
$x=50 \log_6 3.$
|
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Use induction to prove $ S(n) = \sum_{k=1}^n \sum_{i=1}^k 2^{n-k} = 2^{n+1} - (n+2)$ Use induction to prove $$ S(n) = \sum_{k=1}^n \sum_{i=1}^k 2^{n-k} = 2^{n+1} - (n+2)$$
So far I worked out that you can convert the two summations into one:
$$\sum_{k=1}^n k\cdot 2^{n-k} $$
My problem comes when I try to write $S(n+1)$ as the sum of $S(n)$ plus the 'last term'. I realised that finding $n+1^{th}$ term isn't as simple as I am used to. Here is an example:
$$S(3) = 1 \cdot 2^2 + 2 \cdot 2^1 + 3 \cdot 2^0 = 11$$
$$S(4) = 1 \cdot 2^3 + 2 \cdot 2^2 + 3 \cdot 2^1 + 4 \cdot 2^0 = 26$$
Ideally I would like to write $$\sum_{k=1}^n k\cdot 2^{n-k} = 1 \cdot 2^{n-1} + 2 \cdot 2^{n-2} + ... + n \cdot 2^{n-n}$$
Then $$\sum_{k=1}^{n+1} k\cdot 2^{n+1-k} = 1 \cdot 2^{n+1-1} + 2 \cdot 2^{n+1-2} + ... + (n+1) \cdot 2^{n+1-n-1}$$
Then substitute from the induction hypothesis that $$\sum_{k=1}^n k\cdot 2^{n-k} = 2^{n+1} - (n+2)$$
But now I'm totally stuck and don't know how to proceed. Any help would be most appreciated!
|
$$S(n+1)=\sum_{k=1}^{n+1} k \, 2^{n+1-k} = 2 \sum_{k=1}^n k \, 2^{n-k} + (n+1) \\ \stackrel{\text{I.H.}}{=} 2\left(2^{n+1}-n-2\right) + n + 1 = 2^{n+2} - n - 3$$
which is what you would expect.
|
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|
Is $ T: \mathbb{R^3} -> \mathbb{R^2} : T(x,y,z) = (x+1, y+z)$ a linear transformation? Need help for my proof So I have
$$ T: \mathbb{R^3} \rightarrow\mathbb{R^2} $$
$$ T(x,y,z) = (x+1, y+z),$$
and I know that it is indeed a linear transformation if I show both cases:
$$ \forall v \in T : F(a+b) = F(a)+F(b),$$
and
$$ \forall k \in K: F(k \cdot a) = k \cdot F(a).$$
So I just start and I would like to know if what I'm doing is correct, I as am a little bit unsure if I'm not doing any mistakes:
$$ T(a+b)= \begin{pmatrix}
x_1+1+x_2+1\\
(y_1+z_1)+(y_2+z_2)
\end{pmatrix}$$
$$ T(a)= \begin{pmatrix}
x_1+1\\
(y_1+z_1)
\end{pmatrix}$$
$$ T(b)= \begin{pmatrix}
x_2+1\\
(y_2+z_2)
\end{pmatrix}$$
For $T(a) + T(b)$ follows:
$$ \begin{pmatrix}
x_1+1 + x_2+1\\
(y_1+z_1)+(y_2+z_2)
\end{pmatrix} = T(a+b)$$
Now for the second case, the skalar: Let $k = 2$
$$ F(k \cdot a) = \begin{pmatrix}
2 \cdot (x_1 + 1)\\
2 \cdot(y_1+z_1)
\end{pmatrix} $$
now finally:
$$ k \cdot F( a) = 2 \cdot \begin{pmatrix}
x_1+1\\
(y_1+z_1)
\end{pmatrix} =\begin{pmatrix}
x_1+1\\
(y_1+z_1)
\end{pmatrix} + \begin{pmatrix}
x_1+1\\
(y_1+z_1)
\end{pmatrix} = F(k \cdot a) $$
therefore
$$ T(x,y,z) = T(x+1, y+z),$$
is indeed a linear transformation.
I am correct or did I do something wrong? I am not sure about the skalar multiplication part but would appreciate it a lot if somebody could guide me if I'm on the right path.
Thanks in advance!
Greetings!
|
Any linear transformation can be written in the form $$T(\overline{v})=M \overline{v}$$ where $M$ is a matrix of transformation. The transformation cannot be written in this form because
$$T(x,y,z)= \begin{pmatrix} x+1 \\ x+z
\end{pmatrix} =\begin{pmatrix} 1 & 0 & 0 \\
1 & 0 & 1
\end{pmatrix} \begin{pmatrix} x \\ y \\ z
\end{pmatrix} + \begin{pmatrix} 1 \\ 0
\end{pmatrix} =M\overline{v}+\overline{c}$$
Hence, it is not a linear transformation.
|
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|
Proving a Rational Expression is Surjective I need to prove that the function $f:(-1,1)\rightarrow\mathbb{R}$ defined by $f(x) = \frac{x}{x^2 - 1}$ is surjective.
My work.
$b = \frac{a}{a^2 - 1} \iff b(a^2 -1)=a \iff ba^2 - b - a=0$
From here I did a few cases:
Case 1) $b=0$. Then $a=0$.
Using the quadratic formula: $a = \frac{1\pm \sqrt{1+4b^2}}{2b}$.
Case 2) $b \gt0$. Then $a = \frac{1+ \sqrt{1+4b^2}}{2b}$ $\notin (-1,1)$ $\forall$ b $\in R$.
Case 3) $b \gt0$. Then $a = \frac{1- \sqrt{1+4b^2}}{2b}$ $\in (-1,1)$ $\forall$ b $\in R$.
Using Case 1 and Case 3, f is subjective. Is this correct? I cannot use Calculus.
|
It is correct. But you should justify the assertions that $\frac{1+\sqrt{1+4b^2}}{2b}\notin(-1,1)$ and that $\frac{1-\sqrt{1+4b^2}}{2b}\in(-1,1)$. This follows easily from the fact that$$\frac{1+\sqrt{1+4b^2}}{2b}\times\frac{1-\sqrt{1+4b^2}}{2b}=-1.$$
|
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|
Proving $\sum_{n=1}^\infty (-1)^n\tan\left(\frac{n+1}{n^2}\right)$ converges according to the Leibniz test I need to prove the following series converges:
$$\sum_{n=1}^\infty (-1)^n\tan\left(\frac{n+1}{n^2}\right)$$
I'm trying to understand a solution that uses Leibniz's test.
I understand why $\tan\left(\frac{n+1}{n^2}\right)\to0$,
but cannot figure out how to formally prove it monotonically decreases and is positive for every $n\geq2$.
Thank you!
|
HINT
To begin with, notice the $\tan$ function is strictly increasing.
Moreover, its argument is decreasing, as the following reasoning proves:
\begin{align*}
a_{n + 1} - a_{n} & = \frac{(n + 1) + 1}{(n + 1)^{2}} - \frac{n + 1}{n^{2}}\\\\
& = \frac{n^{2}(n + 1) + n^{2} - (n + 1)^{3}}{n^{2}(n + 1)^{2}}\\\\
& = \frac{n^{3} + 2n^{2} - n^{3} - 3n^{2} - 3n - 1}{n^{2}(n + 1)^{2}}\\\\
& = -\left(\frac{n^{2} + 3n + 1}{n^{2}(n + 1)^{2}}\right) < 0
\end{align*}
Consequently, one has that:
\begin{align*}
a_{n + 1} < a_{n} \Rightarrow \tan(a_{n + 1}) < \tan(a_{n})
\end{align*}
whence it can be concluded that $\tan(a_{n})$ is strictly decreasing.
Based on the previous results, can you take it from here?
|
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Assume $\sum \frac{1}{a_n} < \infty$, How about $\sum \frac{1}{a_n - 1}$? Problem :
Given sequence $\left\{a_n\right\}$ s. t. $\forall n \in \mathbb{N},a_n(a_n-1)\neq 0$.
If $\displaystyle\sum_{n=1}^\infty\frac{1}{a_n}$ converge, then how about $\displaystyle\sum_{n=1}^\infty\frac{1}{a_n-1}$?
Rewrite $$\frac{1}{a_n-1} = \frac{1}{a_n}\times \frac{1}{1-\frac{1}{a_n}}$$
Since $\displaystyle\sum_{n=1}^\infty\frac{1}{a_n}$ converge, we know that $\displaystyle\frac{1}{a_n}\to 0$.
From $\displaystyle\frac{1}{1-x}\approx1+x$, $\quad\displaystyle\frac{1}{a_n-1} = \frac{1}{a_n}\times \frac{1}{1-\frac{1}{a_n}}\approx\frac{1}{a_n}\times\left(1+\frac{1}{a_n}\right)=\frac{1}{a_n}+\frac{1}{a_n^2}$
And we know both $\displaystyle\sum\frac{1}{a_n},\quad\sum\frac{1}{a_n^2}$ converge, finally $\displaystyle\sum\frac{1}{a_n-1}$ also converge.
Is this proof valid? if not, I wonder to know the counterexample for this one.
Thank you.
|
For the same example given by Lorago (assuming $n>1$), notice that $\dfrac{1}{a_n(a_n-1)}$ would be positive for sufficiently large $n$ (in fact positive for every $n>1$). Specifically, $$\dfrac{1}{a_n(a_n-1)} = \dfrac{1}{n+(-1)^{n+1}\sqrt{n}}.$$
Thus $\displaystyle\sum \dfrac{1}{a_n(a_n-1)} $(considering from the stage when it turns positive) is divergent by limit comparison test (compare it with $\displaystyle\sum \dfrac{1}{n})$.
Notice also that
$$\sum\left( \dfrac{1}{a_n-1} -\dfrac{1}{a_n}\right)= \sum \dfrac{1}{a_n(a_n-1)}.$$
Hence $\displaystyle\sum\dfrac{1}{a_n-1} $ must be divergent.
|
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show this $\left(\left(\frac{p-1}{3}\right)!\right)^3+1\not\equiv 0\pmod p$ let $p\equiv 1\pmod 3$ is prime number,show that
$$\left(\left(\dfrac{p-1}{3}\right)!\right)^3+1\not\equiv 0\pmod p$$
I know this well
Show that $[(\frac{p-1}{2})!]^2+1\equiv0 \pmod p$
and I try use same idea to
$$-1\equiv (p-1)!=\left(\dfrac{p-1}{3}\right)!\cdot\left(\dfrac{p+2}{3}\cdot\dfrac{p+5}{3}\cdot\dfrac{p+p-2}{3}\right)\left(\cdot\dfrac{p+p+1}{3}\cdot\dfrac{p+p+4}{3}\cdots\dfrac{p+p+p-3}{3}\right)\equiv \left(\dfrac{p-1}{3}\right)!\cdot\left(\dfrac{2}{3}\cdot\dfrac{5}{3}\cdots\dfrac{p-2}{3}\right)\left(\dfrac{1}{3}\cdot\dfrac{4}{3}\cdots\dfrac{p-3}{3}\right)\pmod p$$
|
Let $X$ be the elliptic curve $u^3+v^3+w^3= 0$, which has the Weierstrass form $y^2 = x^3 - 432$. Now one can confirm the following facts (not necessarily easily).
*
*If $p \equiv 1 \bmod 3$, the number of points on $X$ has the form $1 + p - a_p$, where
$|a_p| < 2 \sqrt{p}$ (Weil bounds).
*If $p \equiv 1 \bmod 3$, then there is a congruence
$$a_p \equiv \frac{(p-1)!}{((p-1)/3)!^3} \bmod p,$$
Hence your requirement (given Wilson's theorem $(p-1)! \equiv -1 \bmod p$) is that $a_p \equiv 1 \bmod 3$, and thus $a_p = 1$ by the Hasse bounds. This comes by noting that one can count points modulo $p$ by evaluating the sum
$$\sum 1 - (u^3 + v^3 + z^3)^{p-1},$$
because this expression equals $1$ if $u^3+v^3+z^3=0$ and equals zero otherwise. Explicitly, if there are $(p+1-a_p)$ projective points, there are $1 + (p-1)(p+1-a_p)$ affine points in $[u,v,w]$, and this is $a_p \bmod p$. Hence
$$a_p \equiv - \sum (u^3 + v^3 + z^3)^{p-1} \equiv - \sum u^{3i} v^{3j} z^{3k} (p-1)!/(i! j! k!).$$
But the the power sums $\sum u^{3i} v^{3j} z^{3k}$ vanish unless $3i$, $3j$, and $3k$ are all positive and divisible by $p-1$, so the only term contributing a non-zero sum is the coefficient of $(uvz)^{p-1}$.
*There is a congruence $a_p \equiv 2 \bmod 3$ for primes $p \equiv 1 \bmod 3$. More on this below. (One proof: the elliptic curve $E$ has a rational $3$-torsion point, so $3$ divides $1+p-a_p$ and hence $a_p \equiv 2 \bmod 3$ when $p \equiv 1 \bmod 3$.)
These facts combined imply that the congruence never holds.
An additional fact is that any $p \equiv 1 \bmod 3$ can be written as
$$p = N(a + \omega b) = a^2 - a b + b^2.$$
Here $\omega^3 = 1$ and $a + \omega b$ is unique up to multiplication by powers of $-1$ and $\omega$,
so there are six possible choices. But there is a unique choice which is congruent to $1 \bmod 3$ in the ring of Eisenstein integers, and for that choice, one has
$$a_p = \mathrm{Tr}(a + \omega b) = 2a - b,$$ and hence
$$\frac{1}{((p-1)/3)!^3} \equiv b - 2 a \bmod p.$$
This is a special case of the theory of complex multiplication, although known in these cases by Gauss via the relation to Jacobi sums.
More concretely, this implies that $3|b$ and $a \equiv 1 \bmod 3$, which certainly implies that $2a-b \equiv 2 \bmod 3$.
As an alternative (to see a related example where there are solutions), consider instead the congruence:
$$((p-1)/3)!^3 \equiv 1 \bmod p,$$
this is the same as asking that $a_p = -1$, or that $2a-b=-1$ and since $(a,b) \equiv (1,0) \bmod 3$, that $a = 1+3n$, or $p=7 + 27 n + 27 n^2$ for some integer $n$ (positive or negative). The first few primes with this property are
$$p = 7, 61, 331, 547, 195, 2347, \ldots $$
For example $61 = N(4 + 9 \omega)$ and $8 - 9 = -1$. But also $20! \equiv 47 \bmod 61$ and $47^3 \equiv 1 \bmod 61$.
These will be exactly the primes with $((p-1)/3)!^3 \equiv 1 \bmod p$. Presumably there are infinitely many such $p$.
|
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Prove that $\dfrac{nm^2-n+1}{2mn-2n+1} \notin\mathbb{Z}$ when $m \geq 2$ Given: $n$ and $m$ are positive integers with
$m≥2$, show that:
$$\frac {nm²-n+1}{2mn - 2n+1}\notin \mathbb Z$$
My attempt:
$n, m \in \mathbb{Z}$
$nm^2 \in \mathbb{Z}$
$nm^2 - n \in \mathbb{Z}$
$2mn - 2n \in \mathbb{Z}$
So $\dfrac{nm^2 - n }{2mn - 2n+1}$ may or may not $\in \mathbb{Z}$
But $\dfrac{1}{2mn - 2n + 1} \in \mathbb{Z}$ only when
$$
\begin{aligned}
2mn - 2n +1&= 1 \\
2mn&= 2n \\
m&= 1
\end{aligned}
$$
Given $m \geq 2$, $\dfrac{1}{2mn - 2n + c} \notin \mathbb{Z}$
So $\dfrac{nm^2 - n + 1}{2mn - 2n+1} \notin \mathbb{Z}$
|
If $\frac{2nm^2-2n+2}{2mn-2n+1}$ is not an integer, then the original fraction $\frac{nm^2-n+1}{2mn-2n+1}$ cannot be an integer either. So we show that $\frac{2nm^2-2n+2}{2mn-2n+1}$ is not an integer.
\begin{align*}
\frac{2nm^2-2n+2}{2mn-2n+1} &= m + \frac{2mn-2n-m+2}{2mn-2n+1} \\[6pt]
&=m + 1 + \frac{1-m}{2mn-2n+1} \\[6pt] &= m+1 - \frac{m-1}{2n(m-1)+1}.
\end{align*}
As $m+1$ is an integer, it is sufficient to show that the last term $\frac{m-1}{2n(m-1)+1}$ is not an integer.
Indeed, since $n>0$ and $m\geq 2$, we have
$$ 0<\frac{m-1}{2n(m-1)+1} \leq \frac{m-1}{2(m-1)+1}< \frac{m-1}{2(m-1)}=\frac{1}{2}$$
so $\frac{m-1}{2n(m-1)+1}\in\left(0,\frac{1}{2}\right)$ and in particular not an integer, as desired.
|
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Calculating $\lim _{x\rightarrow 0^{+}}\left( \dfrac{1+x}{2+x}\right) ^{\dfrac{1-\sqrt{x}}{1-x}}$ Maybe someone could explain the following to me:
So I want to calculate $$\lim _{x\rightarrow 0^{+}}\left( \dfrac{1+x}{2+x}\right) ^{\dfrac{1-\sqrt{x}}{1-x}}$$
It is pretty much obvious that $$\lim _{x\rightarrow 0^{+}}\left( \dfrac{1+x}{2+x}\right) ^{\dfrac{1-\sqrt{x}}{1-x}}=\dfrac{1}{2}$$
The thing that I don't understand is that if we use the known limit
$\lim _{x\rightarrow 0}\left( 1+x\right) ^{\dfrac{1}{x}}=e$, then:
$$\begin{aligned}\lim _{x\rightarrow 0^{+}}\left( \dfrac{1+x}{2+x}\right) ^{\dfrac{1-\sqrt{x}}{1-x}}
=\\
\lim _{x\rightarrow 0^{+}}\left( \dfrac{2+x-1}{2+x}\right) ^{\dfrac{1-\sqrt{x}}{1-x}}=\\
\lim _{x\rightarrow 0^{+}}\left( \dfrac{2+x}{2+x}-\dfrac{1}{2+x}\right) ^{\dfrac{1-\sqrt{x}}{1-x}}
=\\
\lim _{x\rightarrow 0^{+}}\left( 1+\dfrac{1}{-x-2}\right) ^{\dfrac{1-\sqrt{x}}{1-x}}\end{aligned}$$
And from here we will eventually get $$\lim _{x\rightarrow 0^{+}}\left( \dfrac{1+x}{2+x}\right) ^{\dfrac{1-\sqrt{x}}{1-x}}=\dfrac{1}{\sqrt{e}}$$
What am I missing? Thanks!
|
You didn't correctly use the known limit though, because one cannot here.
To be able to use the result $\lim_{z \rightarrow 0^+}(1+z)^{\frac{1}{z}} = e$ to evaluate $\lim_{x \rightarrow 0^+} \left(1+\frac{1}{-x-2}\right)^{\frac{1-\sqrt{x}}{1-x}}$, you would need both [among other things]
*
*the equation $\lim_{x \rightarrow 0^+} \frac{1}{-x-2} = 0$ to be true, [because put informally $\frac{1}{-x-2}$ is playing the role of $z$]
*and also to be true the equation $\lim_{x \rightarrow 0^+} \frac{1-\sqrt{x}}{1-x} = \infty$.
Neither of these is true though. As $x$ goes to $0$, $\frac{1}{-x-2}$ goes to $-\frac{1}{2}$ instead of $0$, and $\frac{1-\sqrt{x}}{1-x}$ goes to $1$, not $\infty$.
|
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Rhombus rotates in a circle. Two vertexes of the rhombus is on the circle, rotate the rhombus $ABCD$ clockwise around the point A to the rhombus $AB'C'D'$, where the point $B'$ falls on the circle, link $B'D,C'C$.
If $B'D:CC'=4:3$, then value of $tan\angle BAD$ will be?
It seems $B',B,D'$ or $D,D',C'$ are in a straight line on the image, but I can't proof it.By the way, there are also many things that seem right but hard to proof.It confuses me a lot.
|
Let $\angle AOD=\angle DOC=\alpha, \angle EOC=\beta, \angle ODC=\gamma$
We have that $\sin \alpha: \sin \beta=4:3$ or $3\sin \alpha=4\sin \beta$, also $\sin(2\alpha+\beta)=0$Thus,
$$\sin 2\alpha\cos \beta+\cos2 \alpha\sin\beta=0$$
$$2\sin \alpha\cos \alpha \cos \beta+\frac{3}{4}\cos2 \alpha\sin\alpha=0$$
$$2\cos \alpha \cos \beta+\frac{3}{4}\cos2 \alpha=0$$
Using that: $9\sin^2\alpha=16\sin^2\beta \implies \cos \beta=\frac{\sqrt{7+9\cos^2\alpha}}{4}$
$$\cos \alpha \frac{\sqrt{7+9\cos^2\alpha}}{2}+\frac{3}{4}(2\cos^2 \alpha-1)=0$$Thus, we have the equation to find $x=\cos \alpha$: $$x\sqrt{7+9x^2}+3x^2=\frac{3}{2}$$
$$x^2(7+9x^2)=\frac{9}{4}-9x^2+9x^4$$
$$16x^2=\frac{9}{4} \implies x=\frac{3}{8}$$ Next, notice that we need to find $\tan BAD=\tan (\pi-2\gamma)=-\tan 2\gamma$ and $\gamma=\frac{\pi}{2}-\frac{\alpha}{2} \implies \tan BAD=-\tan(\pi-\alpha)=\tan \alpha=\frac{\sqrt{55}}{3}$
|
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|
$\frac{1}{a+b}+\frac{1}{a+c}=\frac{3}{a+b+c}$ in a triangle Find the angle $\alpha$ of a triangle with sides $a,b$ and $c$ for which the equality $$\dfrac{1}{a+b}+\dfrac{1}{a+c}=\dfrac{3}{a+b+c}$$ holds.
My idea is to use the law of cosines: $$\cos\alpha=\dfrac{b^2+c^2-a^2}{2bc}$$ after simplifying the given equality and plug in something, but this does not seem to be easy.
$$\dfrac{1}{a+b}+\dfrac{1}{a+c}=\dfrac{3}{a+b+c}\\\dfrac{2a+b+c}{(a+b)(a+c)}=\dfrac{3}{a+b+c}\\(2a+b+c)(a+b+c)=3(a+b)(a+c)$$ Am I missing something?
|
You are doing great, expand the last line, you get $b^2+c^2=a^2+bc$, so the $\cos \alpha = 1/2$.
|
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How many ten digit numbers have the sum of their digits equal to $4$? How many ten digit numbers have the sum of their digits equal to $4$?
Well, my solution goes like this:
The first digit of the $10$ digit number cannot be $0$.
It can be $1,2,3$ or $4$ only .
If the first digit is $1$ then we can have $3$ other digits as $1$ so as the sum of digit is $4$ . This can be done in $9\choose3$ ways . Now when the 1st digit is $1$ then we can have two digits $2$ and $1$ as another case such as sum of digit remains $4$. So, this can be done in $9\choose 2$ ways . Now, another case can be the one when 1st digit is $4$ but another digit is $3$ . This can be done in $9\choose 1$ ways . So the number of ways when 1st digit is $1$ is $9\choose3$$+$$9\choose2$$+$$9\choose1$ways.
Now, if the 1st digit is $2$ then the other two digits can be $1$ each . This can done in $9\choose2$ways . If the 1st digit is $2$ the other digit can be $2$ . This can be done in $9\choose1$ways.The total of ways this can be done is $9\choose2$$+$$9\choose1$ ways. If the 1st digit is $3$ then another digit among those $9$ digits must be $1$.this can be done in $9\choose1$ways. If the 1st digit is $4$ then all the other digits are zero . This can be done in $1$ way only. So, the total number of ways in which the sum of digits can be $4$ in a $10$ - digit number is $9\choose3$$+$$9\choose2$$+$$9\choose1$$+$$9\choose2$$+$$9\choose1$$+$ $9\choose1$$+$$1$ ways$ =184$ways
However the answer in the book is given as: $1+2$$9\choose 1$$+$$9\choose1$$+$$9\choose2$$3!$$/2!$$+$$9\choose 3$$=$$220$ways
Where is the mistake? Where is the problem occuring?
|
As Mark Saving pointed out in the comments, the error you made was not taking into account the order of the digits $1$ and $2$ in numbers in the case in which the leading digit is $1$ and the other nonzero digits are $1$ and $2$. In that case, there are $9$ ways to place the second $1$ and eight ways to place the $2$. With that correction, you would have obtained $220$ rather than $184$ since $P(9, 2) - \binom{9}{2} = 9 \cdot 8 - 36 = 36$.
Here is another approach to the problem. Let $x_i$ be the $i$th digit. Then
$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 + x_8 + x_9 + x_{10} = 4 \tag{1}$$
Since the leading digit cannot be zero, the integer $x_1 \geq 1$. Each of the remaining variables represents a nonnegative integer.
Let $x_1' = x_1 - 1$. Then $x_1'$ is also a nonnegative integer. Substituting $x_1' + 1$ for $x_1$ in equation 1 yields
\begin{align*}
x_1' + 1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 + x_8 + x_9 + x_{10} & = 4\\
x_1' + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 + x_8 + x_9 + x_{10} & = 3 \tag{2}
\end{align*}
Equation 2 is an equation in the nonnegative integers. A particular solution of equation 2 corresponds to the placement of $10 - 1 = 9$ addition signs in a row of $3$ ones. For instance,
$$+ + 1 + + + + 1 1 + + +$$
corresponds to the solution $x_1' = x_2 = 0$, $x_3 = 1$, $x_4 = x_5 = x_6 = 0$, $x_7 = 2$, $x_8 = x_9 = x_{10} = 0$ (in which case, the original number was $10,010,002,000$ since $x_1 = x_1' + 1$). The number of solutions of equation 2 is the number of ways we can insert $10 - 1 = 9$ addition signs in a row of $3$ ones, which is
$$\binom{3 + 10 - 1}{10 - 1} = \binom{12}{9} = 220$$
since we must choose which nine of the twelve positions required for three ones and nine addition signs will be filled with addition signs.
|
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|
Find $x,y,z$ satisfying $x(y+z-x)=68-2x^2$, $y(z+x-y)=102-2y^2$, $z(x+y-z)=119-2z^2$ Solve for $x,y,z$:
$$x(y+z-x)=68-2x^2$$
$$y(z+x-y)=102-2y^2$$
$$z(x+y-z)=119-2z^2$$
After some manipulation, I obtain
$$xy+xz=68-x^2$$
$$yz+xy=102-y^2$$
$$xz+yz=119-z^2$$
After combining equations, I get
$$y=\frac{-51-x^2+z^2}{x-z}$$
This seems too tedious. Is there a simpler way?
|
If you add
$$xy+xz=68-x^2$$
$$yz+xy=102-y^2$$
$$xz+yz=119-z^2$$
you get $(x+y+z)^2=17^2$, and so $x+y+z=\pm 17$. If you substitute $x+y=\pm 17-z$ in the third of your equation, you get $\pm 17 z=119$, and so $z=7$ or $z=-7$; now it is easy to get the two solution $x=4$, $y=6$, $z=7$ and $x=-4$, $y=-6$, $z=-7$.
|
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Find the sum of all the numbers less than $10000$ that can be formed using the digits $1, 2, 3$ and $4$ without repetition. I was not able to solve this question on my own and then when I looked up on the web, I found that there is a formula somewhat related to this problem which states that
If all the possible $n$-digit numbers using $n$ distinct digits are
formed, the sum of all the numbers so formed is equal to $(n-1)! \times \text{Sum
of all the digits} \times \text{$111$... n times}$.
And using this formula, I can find out the sum of all the $4$ digit numbers that are formed using $1, 2, 3$ and $4$ which turns out to be $6660$. But how can we extend the application of this formula for all the $3$ digits or $2$ digits numbers that can be formed using $1, 2, 3$ and $4$. Please help !!!
Thanks in advance !!!
|
Let's do some case work.
One-digit numbers: There are four such numbers: $1, 2, 3, 4$. They add to $10$.
Two-digit numbers: Since repetition is not permitted, there are four ways to choose the tens digit and three ways to choose the units digit. Hence, there are $4 \cdot 3 = 12$ such numbers. By symmetry, each of the digits $1, 2, 3, 4$ must appear in each position $12/4 = 3$ times. Hence, the sum of the two-digit numbers which can be formed from the digits $1, 2, 3, 4$ without repetition is
$$\frac{1}{4} \cdot 4 \cdot 3 \cdot (1 + 2 + 3 + 4)(10 + 1) = 330$$
Three-digit numbers: Since repetition is not permitted, for each of the four choices for the hundreds digit, there are three choices for the tens digit, and two choices for the units digit. Hence, there are $4 \cdot 3 \cdot 2 = 24$ such numbers. By symmetry, each of the digits $1, 2, 3, 4$ must appear in each position $24/4 = 6$ times. Hence, the sum of the three-digit numbers which can be formed from the digits $1, 2, 3, 4$ without repetition is
$$\frac{1}{4} \cdot 4 \cdot 3 \cdot 2 \cdot (1 + 2 + 3 + 4)(100 + 10 + 1) = 6660$$
Four-digit numbers: Since repetition is not permitted, for each of the four choices for the thousands digit, there are three choices for the hundreds digit, two choices for the tens digit, and one choice for the units digit. Hence, there are $4 \cdot 3 \cdot 2 \cdot 1 = 24$ such numbers. By symmetry, each of the digits $1, 2, 3, 4$ must appear in each position $24/4 = 6$ times. Hence, the sum of the four-digit numbers which can be formed from the digits $1, 2, 3, 4$ without repetition is
$$\frac{1}{4} \cdot 4 \cdot 3 \cdot 2 \cdot 1 \cdot (1 + 2 + 3 + 4)(1000 + 100 + 10 + 1) = 66~660$$
Total: Since the four cases are mutually exclusive and exhaustive, the sum of all the numbers that can be formed using the digits $1, 2, 3,$ and $4$ without repetition is
$$10 + 330 + 6660 + 66660 = 73660$$
|
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Is there any closed form for $\frac{d^{2 n}( \cot z)}{d z^{2 n}}\big|_{z=\frac{\pi}{4}}$? Latest Edit
Thanks to Mr Ali Shadhar who gave a beautiful closed form of the derivative which finish the problem as
$$\boxed{S_n = \frac { \pi ^ { 2 n + 1 } } { 4 ^ { n + 1 } ( 2 n ) ! } | E _ { 2 n } | }, $$
where $E_{2n}$ is an even Euler Number.
In the post, I had found the sum
$$
\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 k+1)^{3}}= \frac{\pi^{3}}{32},
$$
and want to investigate it in a more general manner, $$
S_{n}=\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 k+1)^{2 n+1}}
$$
where $n\in N.$
$$
\begin{aligned}
S_{n}&= \lim _{N \rightarrow \infty} \sum_{k=0}^{N} \frac{1}{(4k+1)^{2 n+1}}-\sum_{k=0}^{N} \frac{1}{(4 k+3)^{2 n+1}} \\
&=\frac{1}{4^{2 n+1}} \lim _{N \rightarrow \infty} \left[\sum_{k=0}^{N} \frac{1}{\left(k+\frac{1}{4}\right)^{2 n+1}}-\sum_{k=0}^{N} \frac{1}{\left(k+\frac{3}{4}\right)^{2 n+1}}\right] \\
&=\frac{1}{4^{2 n+1}} \lim _{N \rightarrow \infty} \left[\sum_{k=0}^{N} \frac{1}{\left(k+\frac{1}{4}\right)^{2 n+1}}+\sum_{k=0}^{N} \frac{1}{\left(-k-\frac{3}{4}\right)^{2 n+1}}\right] \\
&=\frac{1}{4^{2 n+1}} \lim _{N \rightarrow \infty} \left[\sum_{k=0}^{N} \frac{1}{\left(k+\frac{1}{4}\right)^{2 n+1}}+\sum_{k=-N}^{-1} \frac{1}{\left(k+\frac{1}{4}\right)^{2 n+1}}\right] \\
&=\frac{1}{4^{2 n+1}}\left[\lim _{N \rightarrow \infty} \sum_{k=-N}^{N} \frac{1}{\left(k+\frac{1}{4}\right)^{2 n+1}}\right]
\end{aligned}
$$
Using the Theorem:
$$(*):\pi \cot (\pi z)=\lim _{N \rightarrow \infty} \sum_{k=-N}^{N} \frac{1}{k+z} ,\quad \forall z \not \in Z.$$
Differentiating (*) w.r.t. $z$ by $2 n$ times yields
$$
\begin{aligned}
& \lim _{N \rightarrow \infty} \sum_{k=-N}^{N} \frac{(-1)^{2 n}(2 n) !}{(k+z)^{2 n+1}}=\frac{d^{2 n}}{d z^{2 n}}[\pi \cot (\pi z)] \\
\Rightarrow & \lim _{N \rightarrow \infty} \sum_{k=-N}^{N} \frac{1}{(k+z)^{2 n+1}}=\frac{\pi}{(2 n) !} \frac{d^{2 n}}{d z^{2 n}}[\cot (\pi z)]
\end{aligned}
$$
Now we can conclude that
$$\boxed{\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 k+1)^{2n+1}}=\left.\frac{\pi^{2n+1}}{4^{2 n+1}(2 n) !} \frac{d^{2 n}}{d z^{2 n}}[\cot z]\right|_{z=\frac{\pi}{4}}}$$
My Question:
Is there any closed form for $\displaystyle \left.\frac{d^{2 n} (\cot z)}{d z^{2 n}}\right|_{z=\frac{\pi}{4}}$?
|
This does not answer the question asked in title.
If the goal is to investigate
$$S_{n}=\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 k+1)^{2 n+1}}=\frac 1{4^{2n+1}}\Bigg[\sum_{k=0}^{\infty} \frac 1 {\left(k+\frac{1}{4}\right)^{2 n+1} } -\sum_{k=0}^{\infty}\frac 1 {\left(k+\frac{3}{4}\right)^{2 n+1} }\Bigg]$$ it could be simple to use directly
$$\sum_{k=0}^{\infty} \frac1 {{(k+a)^m}}=\zeta (m,a)$$ where appears Hurwitz zeta function (which is the generalization of Riemann zeta function).
Then
$$S_n=\frac 1{4^{2n+1}} \left(\zeta \left(2 n+1,\frac{1}{4}\right)-\zeta \left(2
n+1,\frac{3}{4}\right)\right)$$
|
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|
How to calculate $\lim\limits_{n\rightarrow \infty }(n-\sum^n_{k=1}\cos\frac{2k}{n\sqrt{n}}) $? I tried this:
$\lim\limits_{n\rightarrow \infty }(n-\sum^n_{k=1}\cos\frac{2k}{n\sqrt{n}}) $=
= $ \lim\limits_{n\rightarrow \infty }(1-\cos\frac{2}{n\sqrt{n}}) $ + $ \lim\limits_{n\rightarrow \infty }(1-\cos\frac{4}{n\sqrt{n}}) $ + ... + $ \lim\limits_{n\rightarrow \infty }(1-\cos\frac{2n}{n\sqrt{n}}) $ =
= $ \lim\limits_{n\rightarrow \infty }\frac{(1-\cos\frac{2}{n\sqrt{n}})(1+\cos\frac{2}{n\sqrt{n}})}{1+\cos\frac{2}{n\sqrt{n}}} $ + $ \lim\limits_{n\rightarrow \infty }\frac{(1-\cos\frac{4}{n\sqrt{n}})(1+\cos\frac{4}{n\sqrt{n}})}{1+\cos\frac{4}{n\sqrt{n}}} $ + ...+ $ \lim\limits_{n\rightarrow \infty }\frac{(1-\cos\frac{2n}{n\sqrt{n}})(1+\cos\frac{2n}{n\sqrt{n}})}{1+\cos\frac{2n}{n\sqrt{n}}} $ =
= $ \lim\limits_{n\rightarrow \infty }\frac{\sin^2\frac{2}{n\sqrt{n}}}{1+\cos\frac{2}{n\sqrt{n}}} $ + $ \lim\limits_{n\rightarrow \infty }\frac{\sin^2\frac{4}{n\sqrt{n}}}{1+\cos\frac{4}{n\sqrt{n}}} $ + ... + $ \lim\limits_{n\rightarrow \infty }\frac{\sin^2\frac{2n}{n\sqrt{n}}}{1+\cos\frac{2n}{n\sqrt{n}}} $ = ...
, but from here I don't know what to do.
|
Call the required limit $L$. Use the Taylor expansion
$$\cos t=1-t^2/2+O(t^4)$$ to deduce that
$$L=\lim_n \sum_{k=1}^n [2(k/n)^2 n^{-1}+O(n^{-2})] =\int_0^1 2x^2 \,dx=2/3 \,,$$
since the sum is a Riemann sum.
|
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|
Question from isi previous years (a) Show that $\left(\begin{array}{l}n \\ k\end{array}\right)=\sum_{m=k}^{n}\left(\begin{array}{c}m-1 \\ k-1\end{array}\right)$.
(b) Prove that
$$
\left(\begin{array}{l}
n \\
1
\end{array}\right)-\frac{1}{2}\left(\begin{array}{l}
n \\
2
\end{array}\right)+\frac{1}{3}\left(\begin{array}{l}
n \\
3
\end{array}\right)-\cdots+(-1)^{n-1} \frac{1}{n}\left(\begin{array}{l}
n \\
n
\end{array}\right)=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}
$$
I was thinking mathematical induction for the second part that is,,,
Let
$$
\begin{aligned}
P(n):\left(\begin{array}{l}
n \\
1
\end{array}\right)-\left(\frac{1}{2}\right)\left(\begin{array}{l}
n \\
2
\end{array}\right) &+\frac{1}{3}\left(\begin{array}{c}
n \\
3
\end{array}\right) \cdots+(-1)^{n-1} \frac{1}{n}\left(\begin{array}{l}
n \\
n
\end{array}\right) \\
&=1+\frac{1}{2}+\cdots \frac{1}{n}
\end{aligned}
$$
$P(1):\left(\begin{array}{l}1 \\ 1\end{array}\right)=1$.
$P(1)$ true.
$$
\begin{aligned}
P(2):\left(\begin{array}{l}
2 \\
1
\end{array}\right)-\frac{1}{2}\left(\begin{array}{l}
2 \\
2
\end{array}\right) \\
=& 2-\frac{1}{2}=1+\frac{1}{2}
\end{aligned}
$$
$P(2)$ is also true.
Let $p(k)$ tque $\Rightarrow\left(\begin{array}{l}k \\ 1\end{array}\right)-\frac{1}{2}\left(\begin{array}{l}k \\ 2\end{array}\right)+\frac{1}{3}\left(\begin{array}{l}k \\ 3\end{array}\right) \cdots+(-1)^{k+\frac{1}{k}}$ Now we try to show $p(k+1)$ will be true $1+\frac{1}{2}+\cdots \frac{1}{k}$ $P(k+1)=\left(\begin{array}{c}k+1 \\ 1\end{array}\right)-\frac{1}{2}\left(\begin{array}{c}k+1 \\ 2\end{array}\right)+\cdots(-1)^{k} \frac{1}{k+1}\left(\begin{array}{l}k+1 \\ k+1\end{array}\right)$
I cannot argue from here,,, please help me for both part. Thank you.
|
Sketch of a proof using Mathematical Induction:
Recall the identity (well-known, easy to prove)
$$\binom{n + 1}{k} = \binom{n}{k} + \binom{n}{k - 1}. \tag{1}$$
For (a):
Using (1), we have
$$\sum_{m=k}^{n + 1} \binom{m-1}{k-1} = \sum_{m=k}^{n} \binom{m-1}{k-1} +
\binom{n}{k-1} = \binom{n}{k} + \binom{n}{k-1} = \binom{n+1}{k}.$$
We are done.
$\phantom{2}$
For (b):
Using (1), we have
\begin{align*}
&\sum_{k=1}^{n+1} \frac{(-1)^{k - 1}}{k}\binom{n+1}{k}\\
=\, & \sum_{k=1}^{n} \frac{(-1)^{k - 1}}{k}\binom{n+1}{k} + \frac{(-1)^n}{n + 1} \\
=\, & \sum_{k=1}^{n} \frac{(-1)^{k - 1}}{k}\binom{n}{k}
+ \sum_{k=1}^{n} \frac{(-1)^{k - 1}}{k}\binom{n}{k-1}
+ \frac{(-1)^n}{n + 1} \\
=\,& \sum_{k=1}^{n} \frac{(-1)^{k - 1}}{k}\binom{n}{k}
+ \frac{1}{n + 1}\left(\sum_{k=1}^{n} \frac{(-1)^{k - 1}(n + 1)}{k}\binom{n}{k-1}
+ (-1)^n\right)\\
=\,& \sum_{k=1}^{n} \frac{(-1)^{k - 1}}{k}\binom{n}{k} + \frac{1}{n + 1}\left(\sum_{k=1}^{n} (-1)^{k - 1}\binom{n + 1}{k}
+ (-1)^n\right)\\
=\,& \sum_{k=1}^{n} \frac{(-1)^{k - 1}}{k}\binom{n}{k} + \frac{1}{n + 1}\left(1 - \sum_{k=0}^{n + 1} (-1)^{k}\binom{n + 1}{k}\right)\\
=\,& \sum_{k=1}^{n} \frac{(-1)^{k - 1}}{k}\binom{n}{k} + \frac{1}{n + 1}
\end{align*}
where we have used the binomial theorem (letting $x = 1$)
$$(1 - x)^{n + 1} = \sum_{k=0}^{n + 1} (-x)^{k}\binom{n + 1}{k}.$$
We are done.
|
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|
Cross covariance and trace identity This may be a simple answer but I can't find any proof. I know that the following identities are true
$$ E\left \{ \left ( \mathbf{x} - E\left ( \mathbf{x} \right ) \right )^T\mathbf{Q}\left ( \mathbf{x} - E\left ( \mathbf{x} \right ) \right ) \right \}=tr\left ( \mathbf{Q}cov\left ( \mathbf{x} \right ) \right )$$
$$ E\left \{ \left ( \mathbf{A}\left ( \mathbf{x} - E\left ( \mathbf{x} \right ) \right ) \right )^T\mathbf{Q}\left ( \mathbf{A}\left ( \mathbf{x} - E\left ( \mathbf{x} \right ) \right ) \right ) \right \}=tr\left ( \mathbf{Q}\mathbf{A}cov\left ( \mathbf{x} \right )\mathbf{A}^T \right )=tr\left ( \mathbf{A}^T\mathbf{Q}\mathbf{A}cov\left ( \mathbf{x} \right ) \right ) $$
where $\mathbf{Q}$ is a diagonal matrix, $\mathbf{x}$ is a random column vector, $\mathbf{A}$ is a matrix of appropriate dimensions and $cov$ is the covariance. For the case of two random vectors, are these identities true?
$$ E\left \{ \left ( \mathbf{x} - E\left ( \mathbf{x} \right ) \right )^T\mathbf{Q}\left ( \mathbf{y} - E\left ( \mathbf{y} \right ) \right ) \right \}=tr\left ( \mathbf{Q}cov\left ( \mathbf{x,y} \right ) \right )$$
$$ E\left \{ \left ( \mathbf{A}\left ( \mathbf{x} - E\left ( \mathbf{x} \right ) \right ) \right )^T\mathbf{Q}\left ( \mathbf{B}\left ( \mathbf{y} - E\left ( \mathbf{y} \right ) \right ) \right ) \right \}=tr\left ( \mathbf{Q}\mathbf{A}cov\left ( \mathbf{x,y} \right )\mathbf{B}^T \right )=tr\left ( \mathbf{A}^T\mathbf{Q}\mathbf{B}cov\left ( \mathbf{x,y} \right ) \right ) $$
Thanks
|
The answer to both is yes.
For the first one, let $\mathbf{c}=\mathbf{x}-E\left ( \mathbf{x} \right)$, $\mathbf{d}=\mathbf{y}-E\left ( \mathbf{y} \right)$ and $Q=\text{diag}(q_1,\ldots,q_n)$, then you can easily check
$$ E\left[ \mathbf{c}^TQ\mathbf{d}\right]=E\left[\sum_i q_i\mathbf{c}_i \mathbf{d}_i\right]=\sum_i q_iE\left[\mathbf{c}_i \mathbf{d}_i\right].$$
Since $\left(cov(\mathbf{x},\mathbf{y})\right)_{ij}=\left(E\left[\mathbf{c}\mathbf{d}^T\right]\right)_{ij}=E\left[\mathbf{c}_i\mathbf{d}_j\right]$, the $i$-th diagonal element of $Qcov\left ( \mathbf{x},\mathbf{y} \right )$ is $q_iE\left[\mathbf{c}_i\mathbf{d}_i\right]$ and thus the first equality follows.
For the second equality, an important fact to keep in mind is that $E\left[A\mathbf{x}\right]=AE\left[\mathbf{x}\right]$. So $A\mathbf{c}=A\left( \mathbf{x} - E\left ( \mathbf{x} \right )\right )=\left ( A\mathbf{x} - E\left ( A\mathbf{x} \right )\right )$ and we can apply the first equation to obtain
$$ E\left [ \left( A\left ( \mathbf{x} - E\left ( \mathbf{x} \right ) \right )\right )^T\mathbf{Q}\left( B\left ( \mathbf{y} - E\left ( \mathbf{y} \right ) \right )\right ) \right ]=tr\left ( \mathbf{Q}cov\left ( \mathbf{Ax},\mathbf{By} \right ) \right ). $$
To complete the proof, note that
$$ cov(A\mathbf{x},B\mathbf{y})= E\left[ A\mathbf{c}\left(B\mathbf{d}\right)^T\right]=E\left[ A\mathbf{c}\mathbf{d}^TB^T\right]=A E\left[ \mathbf{c}\mathbf{d}^T\right]B^T=A cov(\mathbf{x},\mathbf{y})B^T $$
|
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|
Tschirnhaus transformation for cubic equation - "Galois Theory" by Stewart, 3rd edition Consider the following cubic equation:
$$x^3 + ax^2 + bx + c = 0$$
where $a, b, c \in \mathbb{C}$.
Suppose $x \in \mathbb{C}$ satisfies this equation.
Define $y := x + \frac{a}{3}$, so that $x = y - \frac{a}{3}$.
We now substitute this in the original equation.
Stewart's Galois Theory 3rd edition (and also other sources apparently) says that, after substituting:
The equation becomes
$$y^3 + py + q = 0$$
where
$$
\begin{align}
p &= \frac{a^2 - 2a^3 + 3b}{3} \\
q &= \frac{2a^3 - 9ab + 27c}{27}
\end{align}
$$
I have tried to check this in detail myself. We have:
$$
\begin{aligned}
\left(y - \frac{a}{3}\right)^3 &= y^3 - ay^2 + \frac{a^2}{3}y - \frac{a^3}{27} \\
a\left(y - \frac{a}{3}\right)^2 = a\left(y^2 - \frac{2a}{3}y + \frac{a^2}{9} \right) &= ay^2 - \frac{2a^2}{3}y + \frac{a^3}{9} \\
b\left(y - \frac{a}{3}\right) &= by - \frac{ab}{3}\\
c &= c
\end{aligned}
$$
So, summing up everything, we get the equation:
$$y^3 + \left(-\frac{a^2}{3} + b\right) y + \left( \frac{2a^3}{27} - \frac{ab}{3} + c \right) = 0$$
So I'm getting $p = \frac{-a^2 + 3b}{3}$.
Where am I going wrong? I must be making some silly mistake during the algebraic manipulations.
|
Nevermind, I checked the 4th edition of the book and there the author writes $p = \frac{-a^2 + 3b}{3}$, so it must have been a mistake in the 3rd edition that has now been corrected.
I'm leaving this thread up so that other people who encounter this same issue while reading the 3rd edition can resolve it quickly.
|
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|
How to solve this eliptic equation? I want to find a solution to
$$ \frac{x_1}{x_2 + x_3} +\frac{x_2}{x_1 + x_3} + \frac{x_3}{x_1+x_2} = 4 $$
for $x_1,x_2,x_3>0$, and $x_1+x_2+x_3=1$.
We have
$$\frac{x_1}{1-x_1}+\frac{x_2}{1-x_2}+\frac{1-x_1-x_2}{x_1+x_2}=4.$$
Let $\dfrac{x_1}{1-x_1}=s,\dfrac{x_2}{1-x_2}=t,$then $x_1=\dfrac{s}{1+s},x_2=\dfrac{t}{1+t}.$
We then have $$s+t+\dfrac{1-st}{s+t+2st}=4,$$
and $s=x+y,t=x-y,$then$$\frac{4 x^3+3 x^2-4 x y^2+y^2+1}{2 \left(x^2+x-y^2\right)}=4$$
$$y^2=\frac{4 x^3-5 x^2-8 x+1}{4 x-9}$$
which eventually simplifies to solving $$y^2 = x^3 + 121 x^2 + 1144 x + 2704.$$
How can we solve this? I am unfamiliar with eliptic curves, even after reading some basic material, it seems each equation is tackled differently. Is there a routine method for equations like this?
|
First notice that x=0,y=52 is a solution. Take a tangent to the curve at this point and it will necessarily intersect at another rational point. By successively taking intersections of the curve with either tangents of rational points or the lines passing through two different rational points you can find more rational points.
|
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|
Solving the system $a^3 + 15ab^2 = 9$, $\;\frac 35 a^2b + b^3 = \frac 45$ I have problem solving the following system of two cubic equations.
$$\begin{cases}
a^3 + 15ab^2 = 9 \\
\frac 35 a^2b + b^3 = \frac 45
\end{cases}
$$
I don't have any idea how to approach this kind of problem.
I'm looking for solutions included in the set of real numbers (no complex solutions).
Is there any general solution to the problem?
|
Let's convert this to a nice single-variable cubic. Let $a = kb$. Then we have
$$
b^3(k^3 + 15k) = 9
\\ b^3(3k^2 + 5) = 4
$$
Now multiply top by $4$, bottom by $9$, and subtract to get
$$
b^3(4k^3 - 27k^2 + 60k - 45) = 0.
$$
We definitely don't have $b = 0$, so it's the cubic in $k$ that must be $0$. The discriminant is negative, so there's only the one real root, and it's clearly positive. By the rational roots theorem, if $k$ is rational, its numerator will divide $45$ and its denominator with divide $4$. We can then do some divisibility checks to rule out some of the possibilities. Since all coefficients except $k^3$ are divisible by $3$, $k$'s numerator must be divisible by $3$. But $k$'s numerator can't be divisible by $9$, because $4k^3 -27k^2 + 60k$'s numerator is divisible by $27$ but $45$ isn't. So the numerator is either $3$ or $15$. At this point we can just try all 6 remaining possibilities and find $k= 3$ works. So $k = 3$, and we have
$$
b^3(27+45) = 9\Longrightarrow b^3 = \frac{9}{72}\Longrightarrow b = \frac{1}{2}.
$$
So the solution is
$$
(a,b) = (3/2,1/2)
$$
|
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|
Does this proof of the binomial expansion (a+b)^2 work? I was rereading Terence Tao's Analysis 1 and found this question in the section:
Exercise $2.3.4.$ Prove the identity $(a + b)^2 = a^2 + 2ab + b^2$ for all natural
numbers a, b.
Prior to this we already have proved:
$1.a\cdot b=b\cdot a\\2.a\cdot b=0\implies a=0\lor b=0\\3.a\cdot (b\cdot c)=(a\cdot b)\cdot c\\4. a\cdot (b+c)=a\cdot b+a\cdot c\\$
So I wrote this down:
$(x+y)^2\\=(x+y)(x+y)\\=x(x+y)+y(x+y)\\=x^2+xy+xy+y^2\\=x^2+2xy+y^2$
Is this a valid proof?
|
Basically correct, but here are two remarks:
*
*From $(x + y)(x + y)$ to $x(x + y) + y(x + y)$ you need the result $(a + b)\cdot c = a \cdot c + b \cdot c$ which can be proved from 1. 4. but you haven't done it.
*From $x(x + y) + y(x + y)$ you can only get $(x^2 + xy) + (yx + y^2)$. In order to be able to write it as $x^2 + xy + xy + y^2$, you need to know that $+$ is associative, namely $(a + b) + c = a + (b + c)$, so that order of addition doesn't matter.
|
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|
How to Evaluate $\int_1^4 (\frac{1}{2t}+i)^2 dt$ Question:
$$\int \limits _1^4\left (\frac{1}{2t}+i \right )^2\,dt.$$
How can I solve this? I believe its an indefinite integral and I can probably expand it by using $(a+b)^2=a^2+2ab+b^2$ to get something like$$\int \frac{1}{4t^2}+\frac{2i}{2t}+i^2\,dt$$but I'm not even sure if that's right.
|
We have
\begin{align*}\int \limits _1^4\left (\frac{1}{2t}+i\right )^2\,dt & =\int \limits _1^4\left (\frac{i}{t}+\frac{1}{4t^2}-1\right )\,dt \\
& =i\int \limits _1^4\frac{1}{t}\,dt+\frac{1}{4}\int \limits _1^4\frac{1}{t^2}\,dt-\int \limits _1^41\,dt \\
& =\left .i\ln |t|-t-\frac{1}{4t}\right |_1^4 \\
& =\frac{16i\ln 4-65}{16}+\frac{5}{4} \\
& =\frac{16i\ln 4-45}{16}.
\end{align*}
|
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|
Doubts on asymptotic criterion for $\sum_{n=1}^{\infty}a_n=\sum_{n=1}^{\infty}n^{a}\tan^{-1}\bigg(\frac{1}{n^a}\bigg)-e^{1/n}$ with $a>0$. I have to valuate the character of the following series:
$$\sum_{n=1}^{\infty}a_n=\sum_{n=1}^{\infty}n^{a}\tan^{-1}\bigg(\frac{1}{n^a}\bigg)-e^{1/n}$$
with $a>0$.
I have thought that definitely the sequence $a_n$ is made by constant signed terms. So I can apply the asymptotic criterion to study the series.
$=n^{a}\tan^{-1}\big(\frac{1}{n^a}\big)-e^{1/n}\\
=n^{a}\big(\frac{1}{n^a}-\frac{1}{3n^{3a}}+o(\frac{1}{n^{3a+1}})\big)-1-\frac{1}{n}+o(\frac{1}{n})\\
=1-\frac{1}{3n^{2a}}+o(\frac{1}{n^{2a+1}})-1-\frac{1}{n}+o(\frac{1}{n})\\
=\frac{1}{3n^{2a}}+o(\frac{1}{n^{2a+1}})-\frac{1}{n}+o(\frac{1}{n})\\
\color{red}{=\frac{1}{3n^{2a}}-\frac{1}{n}+o(\frac{1}{n})}$
The red passage is right?
I have thought that since $2a+1>1$ then I can put the $o(\frac{1}{n^{2a+1}})$ into $o(\frac{1}{n})$.
Now:
if $a\leq 1/2$ then I have $\frac{1}{3n^{2a}}-\frac{1}{n}+o(\frac{1}{n})=\frac{1}{3n^{2a}}+o(\frac{1}{n})$, so the corresponding series diverges and also the original one.
if $a>1/2$ then $\frac{1}{3n^{2a}}-\frac{1}{n}+o(\frac{1}{n})=-\frac{1}{n}+o(\frac{1}{n})$ and again the series diverges.
My overall attempt is right?
Edit : I understand there are some problems in the final part with the little $o$. Can someone help me
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First note that
$$
\arctan x = x - \frac{{x^3 }}{3} + \mathcal{O}(x^5 )\quad \text{ and }\quad e^x = 1 + x + \mathcal{O}(x^2 )
$$
as $x\to 0$. Thus,
\begin{align*}
a_n &= n^a \left( {\frac{1}{{n^a }} - \frac{1}{{3n^{3a} }} + \mathcal{O}\!\left( {\frac{1}{{n^{5a} }}} \right)} \right) - \left( {1 + \frac{1}{n} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right)} \right) \\ & = - \frac{1}{{3n^{2a} }} - \frac{1}{n} + \mathcal{O}\!\left( {\frac{1}{{n^{4a} }}} \right) + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right)
\end{align*}
as $n\to +\infty$. Now if $a = \frac{1}{2} + \varepsilon > \frac{1}{2}$, then it is readily seen that
$$
a_n = - \frac{1}{n} + \mathcal{O}\!\left( {\frac{1}{{n^{1 + 2\varepsilon } }}} \right) + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right) \sim - \frac{1}{n}
$$
as $n\to +\infty$, whence the series diverges by the limit comparison test (using the fact that the harmonic series diverges). If $
a = \frac{1}{2}$, we find
$$
a_n = - \frac{4}{{3n}} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right) \sim - \frac{4}{{3n}}
$$
as $n\to +\infty$, whence the series diverges by the limit comparison test. If $
0 < a < \frac{1}{2}$, then the first term will dominate, i.e.,
$$
a_n \sim - \frac{1}{{3n^{2a} }}
$$
as $n\to +\infty$, and the series diverges by the limit comparison test and the $p$-series test.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/4453199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
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