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If $(z-2)/(z+2)$ is purely imaginary, find $z$ that satisfies this condition?
If $(z-2)/(z+2)$ is purely imaginary, then what is the value of $z$ that satisfies this condition?
I am troubled solving this equation for a long time. I tried to put $z = a + i b$ and then I got $(a-2+ib)(a+2+ib)$ is imaginary.
Not sure how to go now? Thank youvery much.
| Putting $z=a+ib$, with $z\neq-2$, is a correct way to start. Then, you get \begin{align}\frac{z-2}{z+2}&=\frac{z-2}{(a+2)+ib}=\frac{(a-2)+ib}{(a+2)+ib}\cdot\frac{(a+2)-ib}{(a+2)-ib}\\[0.2cm]&=\frac{(a-2)(a+2)+b^2+ib(a+2-a+2)}{(a+2)^2+b^2}=\underbrace{\frac{(a-2)(a+2)+b^2}{(a+2)^2+b^2}}_{=\text{ real part }=0}+i\frac{4b}{(a+2)^2+b^2}\end{align} So, you get the equation $$(a-2)(a+2)+b^2=0\iff a^2+b^2=2^2 \iff |z|=2$$ which gives a set of possible solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
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The maximum possible value of $x^2+y^2-4x-6y$ subject to the condition $|x+y|+|x-y|$ The maximum possible value of
$x^2+y^2-4x-6y$
subject to the condition $|x+y|+|x-y|$=4
My workout...
Now if we add 13 to the equation we get
$x^2+y^2-4x-6y+13-13$
or,$x^2+y^2-4x-6y+4+9-13$
or,$(x-2)^2+(y-3)^2-13$
Are there any methods other than function.
| Hint: your objective function $x^2+y^2-4x-6y$ is convex, so its maximum is attained at an extreme point of your convex feasible region.
| {
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How many real, rational and complex solutions has this system of equations? Let the system $\left\{\begin{aligned}
a+b+c &= 3\\
a^2+b^2+c^2 &=5\\
a^3+b^3+c^3 & =12
\end{aligned}\right. $
How many real, rational and complex solutions has it?
I read System of three variables of simultaneous equations and found $ e_1 = 3, e_2 = 2, e_3 = 1 $. Then I do not understand the reason but I get the polynomial $ t ^ 3-3t ^ 2 + 2t-1 $. The question is how many , not what are the solutions, so at this point I do not know what to do ... or if I'm in the right way. Can you help me please?
| Note
$$ 3^2=(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca=5+2(ab+bc+ca). $$
from this, one has
$$ ab+bc+ca=2. $$
Also
\begin{eqnarray}
3^3&=&(a+b+c)^3=a^3+b^3+c^3+3a^2b+3ab^2+3a^2c+3ac^2+3b^2c+3bc^2+6abc\\
&=&12+3ab(a+b+c)+3bc(a+b+c)+3ca(a+b+c)-3abc\\
&=&12+3(ab+bc+ca)(a+b+c)-3abc\\
&=&30-3abc
\end{eqnarray}
from which, one has
$$ abc=1. $$
Thus $a,b$ and $c$ are the roots of the following equation
$$ t^3-3t^2+2t-1=0. \tag{1} $$
So the coefficients of (1) are
$$ a_3=1,a_2=-3,a_1=2,a_0=-1 $$
and hence the discriminant is
$$ \Delta=18a_3a_2a_1a_0-4a_2^3a_0+a_2^2a_1^2-4a_3a_1^3-27a_3^2a_0^2=-23<0$$
and hence (1) has one real root and two complex roots.
| {
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"timestamp": "2023-03-29T00:00:00",
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Parametric equation of the intersection between $x^2+y^2+z^2=6$ and $x+y+z=0$ I'm trying to find the parametric equation for the curve of intersection between $x^2+y^2+z^2=6$ and $x+y+z=0$. By substitution of $z=-x-y$, I see that $x^2+y^2+z^2=6$ becomes $\frac{(x+y)^2}{3}=1$, but where should I go from here?
| $z=(-x-y)$ leads to
$$ x^2+y^2+(x+y)^2 = 6 $$
that is equivalent to $x^2+xy+y^2=3$, i.e. the equation of an ellipse in the $xy$-plane. Such ellipse goes through $(1,1)$, hence by considering the intersections between such ellipse and the lines through $(1,1)$, i.e. by solving $x^2+xy+y^2=3$ and $y=m(x-1)+1$, we get that
$$ (x,y)=\left(\frac{m^2-2m-2}{m^2+m+1},\frac{-2m^2-2m+1}{m^2+m+1}\right) $$
for $m\in\overline{\mathbb{R}}$ is a parametrization of the ellipse in the $xy$-plane and
$$ (x,y,z)=\left(\frac{m^2-2m-2}{m^2+m+1},\frac{-2m^2-2m+1}{m^2+m+1},\frac{m^2+4m+1}{m^2+m+1}\right) $$
is a parametrization of the wanted section (circle) in the Euclidean space.
| {
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"timestamp": "2023-03-29T00:00:00",
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Addition of two piecewise function Suppose I have function
$$y(x)=\begin{cases}
x+1\qquad & 0\leq x\leq1 \\
2-x\qquad & 1<x \leq2 \\
0\qquad & \mathrm{elsewhere}
\end{cases}$$
I need to find the function $g(x)=y(x+2)+2y(x+1)$
$$y(x+2)=\begin{cases}
x+3\qquad &-2\leq x\leq-1 \\
-x\qquad & -1<x \leq0 \\
0\qquad &\mathrm{elsewhere}
\end{cases}$$
$$2y(x+1)=\begin{cases}
2x+4\qquad & -1\leq x\leq0\\
2-2x\qquad & 0<x \leq1 \\
0\qquad &\mathrm{elsewhere}
\end{cases}$$
Now,
$$g(x)=y(x+2)+2y(x+1)=\begin{cases}
x+3\qquad & -2\leq x\leq -1 \\
x+4\qquad & -1<x \leq0 \\
2-2x\qquad &0<x\leq1\\
0\qquad &\mathrm{elsewhere}
\end{cases}$$
Now the result looks Okay,
But the problem is the value of $y(x+2)$ at $x=-1$ is $2$ and $2y(x+2)$ at $x=-1$ is also $2$ so if I add both then $y(x+2)+2y(x+1)$ at $x=-1$ will be $2+2=4$,but the value of $g(x)$ at $x=-1$ is $-1+3=2$
| \begin{eqnarray}
g_1(x)=y(x+2)=\begin{cases}
x+3&\text{ for }-2\le x\le -1\\
-x&\text{ for }-1< x\le0\\
0&\text{ otherwise}
\end{cases}
\end{eqnarray}
and
\begin{eqnarray}
g_2(x)=2y(x+1)=\begin{cases}
2x+4&\text{ for }-1\le x\le0\\
2-2x&\text{ for }0< x\le1\\
0&\text{ otherwise}
\end{cases}
\end{eqnarray}
Giving the result
\begin{eqnarray}
g(x)=\begin{cases}
x+3&\text{ for }-2\le x<-1\\
3x+7&\text{ for }x=-1\\
x+4&\text{ for }-1< x\le0\\
2-2x&\text{ for }0< x\le1\\
0&\text{ otherwise }
\end{cases}
\end{eqnarray}
Now we see that $g_1(-1)=2,\,g_2(-1)=2$ and $g(-1)=4$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Factored $z^4 + 4$ into $(z^2 - 2i)(z^2 + 2i)$. How to factor complex polynomials such as $z^2 - 2i$ and $z^2 + 2i$? I was wondering how to factor complex polynomials such as $z^2 - 2i$ and $z^2 + 2i$?
I originally had $z^4 + 4$, which I factored to $(z^2 - 2i)(z^2 + 2i)$ by substituting $X = z^2$ and using the quadratic formula. However, I'm not sure if there exists any method to factor $z^2 - 2i$ and $z^2 + 2i$?
I would greatly appreciate it if people could please explain how one would go about this.
| Solving $$z^2-2i=0$$
we have $$z^2=2i=2\exp\left( \frac{i\pi}{2}\right)$$
$$z=\pm\sqrt2\exp\left( \frac{i\pi}{4}\right)=\pm\sqrt{2}\left(\cos\left( \frac{\pi}{4}\right)+i\sin\left(\frac{\pi}{4} \right)\right)=\pm(1+i)$$
Hence, $$z^2-2i=(z-(1+i))(z+(1+i))$$
Try the same trick on $z^2+2i$.
| {
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Epsilon-Delta proof for a limit of a function $\lim\limits_{x \to 2} \frac{x^2+4}{x+2}=2$
I understand the structure of the epsilon delta proof, but I need help with the scratchwork/setup.
Using |$\frac{x^2+4}{x+2}-2$| $<\epsilon$ , you can factor and you're left with |$x-4$|$< \epsilon$. What I'm stuck on is solving for $x-a$ or $x-2$.
Can I do something with the Triangle Inequality to say |$x-2-2$| $\le$ |$x-2$|$ \;+\; 2$ $< \epsilon$
Any help would be appreciated!
| $$\left|\frac{x^2+4}{x+2} -2 \right|=\left|\frac{x^2-2x}{x+2} \right|=\frac{|x||x-2|}{|x+2|}$$
WLOG, we can assume that $\delta < 1$,
Hence if $|x-2| < \delta$, then $2-\delta < x < 2+\delta$ which implies that $x$ is between $1$ and $3$ while $x+2$ is in between $3$ and $5$.
Hence, $$\left|\frac{x^2+4}{x+2} -2 \right|\leq \frac{3|x-2|}{3}=|x-2|$$
Remark about your triangle inequality approach:
what if $\epsilon < 2$.
| {
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Simplifying $\frac{x}{\sqrt[3]{x+1}-1}$
Simplify the following fraction:
$$\frac{x}{\sqrt[3]{x+1}-1}$$
How should I approach this? unlike $(a-b)(a+b)=a^2+b^2$?
| Depending what you mean by simplify: presumably, get rid of the denominator.
Let $y = \sqrt[3]{x+1}$ then
$$
\frac{x}{\sqrt[3]{x+1}-1} = \frac{y^3 -1}{y-1} = y^2 + y +1 = (x+1)^{\frac23} + \sqrt[3]{x+1} + 1
$$
| {
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"url": "https://math.stackexchange.com/questions/2489098",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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$x+y+z=3$, prove the inequality For $x,y,z>0$ and $x+y+z=3$ prove that $\frac{x^3}{(y+2z)^2}+\frac{y^3}{(z+2x)^2}+\frac{z^3}{(x+2y)^2}\ge \frac{1}{3}$.
QM, AM, GM, HM suggested ;)
| Use the inequality $\dfrac{a^3}{b^2}\geq \dfrac{2a^2}{3b} - \dfrac{a}{9}$ and then use $\dfrac{a^2}{b}\geq \dfrac{2a}{3}-\dfrac{b}{9}$, so $$\sum_{x,y,z}\dfrac{x^3}{(y+2z)^2}\geq\sum_{x,y,z}(\dfrac{x}{3} - \dfrac{2(y+2z)}{27})=\dfrac{3(x+y+z)}{27}=\dfrac{1}{3}.$$
The inequalities used are trivial AM-GM.
| {
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Simplify: $\frac{\sqrt{\sqrt[4]{27}+\sqrt{\sqrt{3}-1}}-\sqrt{\sqrt[4]{27}-\sqrt{\sqrt{3}-1}}}{\sqrt{\sqrt[4]{27}-\sqrt{2\sqrt{3}+1}}}$ I am doing a pretty hard problem:
$$\frac{\sqrt{\sqrt[4]{27}+\sqrt{\sqrt{3}-1}}-\sqrt{\sqrt[4]{27}-\sqrt{\sqrt{3}-1}}}{\sqrt{\sqrt[4]{27}-\sqrt{2\sqrt{3}+1}}}$$
So it is a pretty long and complicated problem. I got stuck though. My idea was to turn $\sqrt[4]{27} =\sqrt[4]{3}\sqrt{3}$ and since I cant make the second part easier with Langranges formula (it doesn't apply to this) I made it $\sqrt[4]{3}-1$.
I seemed happy that I was getting somewhere and I thought that I had it but later on I just got stuck primarily by the 1's that I don't know what to do with.
| Let $$x=\frac{\sqrt{\sqrt[4]{27}+\sqrt{\sqrt{3}-1}}-\sqrt{\sqrt[4]{27}-\sqrt{\sqrt{3}-1}}}{\sqrt{\sqrt[4]{27}-\sqrt{2\sqrt{3}+1}}}$$
The square of numerator is $$2\sqrt[4]{27} -2\sqrt{\sqrt{27}-(\sqrt{3}-1)}=2\sqrt[4]{27}-2\sqrt{2\sqrt{3}+1}$$
The square of denominator is $\sqrt[4]{27}-\sqrt{2\sqrt{3}+1}$.
Hence $x^2 = 2$. Also note that $x>0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that a function $f(x,y)$ is bijective Show that a function is bijective
$ f: \mathbb{R}_{+} \times \mathbb{R}_{+} \rightarrow \mathbb{R}_{+} \times \mathbb{R}, f(x,y) := (x+y; \frac{1}{x} - \frac{1}{y}) $
I know I have to show that this function is injection and surjection.
My attempts:
*
*Injection
$ f(x,y)=f(a,b) \Rightarrow (x,y)=(a,b) $
so $(x+y; \frac{1}{x} - \frac{1}{y})= (a+b; \frac{1}{a} - \frac{1}{b} ) \Rightarrow (x,y)=(a,b)$
$x+y=a+b$ and $\frac{1}{x} - \frac{1}{y} = \frac{1}{a} - \frac{1}{b} $
I write this as $(x-a)=(b-y)$ and $\frac{1}{x} - \frac{1}{a} = \frac{1}{y}-\frac{1}{b} $ which leads to:
$xa=-yb$
That doesn't make much sense consider that i had to prove that $x=a$ and $b=y$
*Surjection
Following the definition
For every $ (a,b) \in \mathbb{R}_{+} \times \mathbb{R}$ there is $ (x,y) \in \mathbb{R}_{+} \times \mathbb{R}_{+} $ such that $f(x,y) = (a,b)$
I tried to extract $y$ from $x+y=a$ and put it in $ \frac{1}{x}-\frac{1}{y}=b$
This gives me
$ y= \frac{ba-2 \pm \sqrt{4+a^2b^2}}{2b}$
so that doesn't look good...
| Injectivity: You have shown that $x-a=b-y$ and $\frac{a-x}{ax}=\frac{b-y}{by}$. If $x-a$ and $b-y$ is not zero, then you have $ax=-by$ (as you got). But this is impossible since $a,b,x,y>0$. Hence, the only possibility is $x-a=b-y=0$. That is $(x,y)=(a,b)$.
Surjectivity: Your approach is fine, but you should separate cases $b>0$, $b=0$ or $b<0$.
$b=0$ is quite easy. You can do it yourself.
$b>0$. Then only $y=\frac{ba-2 + \sqrt{4+a^2b^2}}{2b}$ is positive. And it's easy to show that $y<a$, so $x=a-y>0$.
$b<0$. Both roots are positive. You have to choose $y=\frac{ba-2 - \sqrt{4+a^2b^2}}{2b}$ so that $x=a-y>0$.
| {
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Simplifying an algebraic fraction The image below is from an online algebra quiz I did recently for uni. I got this question wrong and it indicated what the correct answer was, however I cannot understand how they got to this answer.
Question from my algebra quiz
This is what I came up with when working out the question:
$\frac{(3x^3-2x^2y+3xy^2-5y^3)-(2x^3+2x^2y-3xy^2-2y^3)}{x - y}$
$=\frac{x^3-4x^2y+6xy^2-3y^3}{x-y}$
$=\frac{x^2(x-4y)+3y^2(2x-y)}{x-y}$
Then from here I realised I couldn't simplify any further and so answered 'None of the above'.
Should my answer have been marked correct or is there a way to get the answer that has been provided in the yellow box in the picture?
Any help would be appreciated.
| \begin{eqnarray*}
x^3-4x^2y+6xy^2-3y^3 &=& x^3-y^3 -2y(y^2-3xy+2x^2)\\
&=& (x-y)(x^2+xy+y^2)-2y(y-x)(y-2x)\\
&=& (x-y)(x^2+xy+y^2+2y^2-4xy)\\
&=& (x-y)(x^2-3xy+3y^2)
\end{eqnarray*}
So we have
$$...=\frac{x^3-4x^2y+6xy^2-3y^3}{x-y} ={(x-y)(x^2-3xy+3y^2)\over x-y}=x^2-3xy+3y^2$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving $a \sin\theta + b \cos\theta = c$ Could someone help me with the steps for solving the below equation $$a \sin\theta + b \cos\theta = c$$
I know that the solution is $$\theta = \tan^{-1} \frac{c}{^+_-\sqrt{a^2 + b^2 - c^2}} - \tan^{-1} \frac{a}{b} $$
I just can't figure out the right steps to arrive at this solution.
| Well, we have:
$$\text{a}\cdot\sin\left(x\right)+\text{b}\cdot\cos\left(x\right)=\text{c}\tag1$$
Substitute $\text{y}=\tan\left(\frac{x}{2}\right)$, so $\sin\left(x\right)=\frac{2\cdot\text{y}}{1+\text{y}^2}$ and $\cos\left(x\right)=\frac{1-\text{y}^2}{1+\text{y}^2}$:
$$\text{y}^2-\frac{\text{b}-\text{c}}{\text{b}+c}-\frac{2\cdot\text{a}\cdot\text{y}}{\text{b}+\text{c}}=0\tag2$$
Solving for $\text{y}$, gives:
$$\text{y}=\pm\sqrt{\frac{\text{a}^2}{\text{b}+\text{c}}+\frac{\text{b}-\text{c}}{\text{b}+\text{c}}}+\frac{\text{a}}{\text{b}+\text{c}}\tag3$$
| {
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$a_1 = a_2 = 1$ and $a_n = \frac{1}{2} \cdot (a_{n-1} + \frac{2}{a_{n-2}})$. Prove that $1 \le a_n \le 2: \forall n \in \mathbb{N} $
Let $a_n$ be a sequence satisfying $a_1 = a_2 = 1$ and $a_n = \frac{1}{2} \cdot (a_{n-1} + \frac{2}{a_{n-2}})$. Prove that $1 \le a_n \le 2: \forall n \in \mathbb{N} $
Attempt at solution using strong induction:
Base cases: $n = 1$ and $n = 2 \implies a_1 = a_2 = 1 \implies 1\le 1 \le 2$
Inductive assumption (strong induction): Assume that for all $m\in \mathbb{N}$ such that $1\le m \le k$, where $k\in \mathbb{N}$, the condition $1\le a_m \le 2$ holds True.
Show that $m = k+1$ holds true
$a_{k+1} = \frac{1}{2} \cdot (a_k + \frac{2}{a_{k-1}})$
I know that $a_k$ and $a_{k-1}$ are satisfying $ 1\le a_m\le2$ but I am not sure how to use that to prove that $a_{k+1}$ holds true for the condition.
| We have $a_k \geq 1$ and $a_{k-1} \leq 2$ ; which gives $ \frac{2}{a_{k-1}} \geq 1$. So
\begin{eqnarray*}
a_{k+1} = \frac{1}{2} \left( a_k +\frac{2}{a_{k-1}} \right) \geq \frac{1}{2} \left( 1+1 \right)=1.
\end{eqnarray*}
We have $a_k \leq 2$ and $a_{k-1} \geq 1$ ; which gives $ \frac{2}{a_{k-1}} \leq 2$. So
\begin{eqnarray*}
a_{k+1} = \frac{1}{2} \left( a_k +\frac{2}{a_{k-1}} \right) \leq \frac{1}{2} \left( 2+2 \right)=2.
\end{eqnarray*}
| {
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Prove that $\sum\limits_{\mathrm{cyc}} \frac{a^2 + a b + 2 a + b^2 + 3}{a^2 + a b - 2 a + b^2 + 3} \le 6$ for $a, b, c \ge 1$
I have already post this inequality but I add this condition $a,b,c\geq 1$
$$6\geq \frac{a^2 + a b + 2 a + b^2 + 3}{a^2 + a b - 2 a + b^2 + 3}+\frac{c^2 + c b + 2 b + b^2 + 3}{c^2 + c b - 2 b + b^2 + 3}+\frac{a^2 + a c + 2 c + c^2 + 3}{a^2 + a c - 2 c + c^2 + 3}$$
I try many classical inequalities but without success...
My try :
We study the following function :
$$f(x)=\frac{x^2+x+2x+3+1}{x^2+x-2x+3+1}+\frac{x^2+x+2+3+1}{x^2+x-2+3+1}+2$$
It's decreasing and the maximum is reached for $x=1$ and is equal to 6.
Now we put :
$$x= \frac{(\sqrt{4 a^3 + 4 a^2 b - 15 a^2 + 4 a b^2 + 12 a} - a)}{2 a}$$
We get :
$$f(x)=\frac{(a^2+b^2+ab+3+2a)}{(a^2+b^2+ab+3-2a)}+2+\frac{(a^2 + \sqrt{a (4 a b + a (4 a - 15) + 4 b^2 + 12)} + ab-a + b^2 + 3)}{(a^2 - \sqrt{a (4 a b + a (4 a - 15) + 4 b^2 + 12)} + a b + a + b^2 + 3)} $$
So if $a\geq b \geq c$ and if we suppose:$\quad$ $max(\frac{a^2 + a b + 2 a + b^2 + 3}{a^2 + a b - 2 a + b^2 + 3};\frac{c^2 + c b + 2 b + b^2 + 3}{c^2 + c b - 2 b + b^2 + 3})=\frac{a^2 + a b + 2 a + b^2 + 3}{a^2 + a b - 2 a + b^2 + 3}$ it remains to prove that we have :
$$\frac{(a^2 + \sqrt{a (4 a b + a (4 a - 15) + 4 b^2 + 12)} + ab-a + b^2 + 3)}{(a^2 - \sqrt{a (4 a b + a (4 a - 15) + 4 b^2 + 12)} + a b + a + b^2 + 3)}\geq \frac{a^2 + a b + 2 a + b^2 + 3}{a^2 + a b - 2 a + b^2 + 3}$$
Wich is true for $b\geq2$ because we have with this condition :
$$\sqrt{a (4 a b + a (4 a - 15) + 4 b^2 + 12)}-a\geq 2a$$
Edit : I have the end of my proof but it's ugly...Now we know that the inequality is true .
Have you easier ?
Thanks a lot .
| The given inequality can be transformed into
$$\sum \frac {a^2+b^2+ab+2a+3}{a^2+b^2+ab-2a+3}\le 6$$
Which changes to
$$\sum 1+\frac {4a}{a^2+b^2+ab-2a+3}\le 6$$
$$\Rightarrow \sum \frac {4a}{a^2+b^2+ab-2a+3}\le 3$$
Hence it suffices to show that
$$\sum \frac {a}{a^2+b^2+ab-2a+3}\le \frac{3}{4}$$
Hence by AM GM we have LHS as
$$\sum \frac {a}{3(ab+1)-2a}$$
Again by AM GM LHS becomes
$$\sum \frac {a}{6\sqrt {ab}-2a}=\sum \frac {1}{6\sqrt {\frac {b}{a}}-2}$$
Hence it suffices to show that
$$\sum \frac {1}{3\sqrt {\frac {b}{a}}-1}\le \frac {3}{2}$$
That means
$$\frac {1}{3\sqrt {\frac {b}{a}}-1}+ \frac {1}{3\sqrt {\frac {c}{b}}-1}+ \frac {1}{3\sqrt {\frac {c}{a}}-1}\le \frac {3}{2}$$
I hope you can take it from here.
| {
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"url": "https://math.stackexchange.com/questions/2507591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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Why is the dimension of the null space of this matrix 1? Consider this problem from wikipedia:
$$A = \begin{bmatrix}
5 & 4 & 2 & 1 \\
0 & 1 & -1 & -1 \\
-1 & -1 & 3 & 0 \\
1 & 1 & -1 & 2 \end{bmatrix}$$
From the wikipedia link: "Including multiplicity, the eigenvalues of A are λ = 1, 2, 4, 4. The dimension of the eigenspace corresponding to the eigenvalue 4 is 1 (and not 2), so A is not diagonalizable"
The corresponding $A - \lambda I$ matrix for $\lambda = 4$ is
$$A - 4I = \begin{bmatrix}
1 & 4 & 2 & 1 \\
0 & -3 & -1 & -1 \\
-1 & -1 & -1 & 0 \\
1 & 1 & -1 & -2 \end{bmatrix}$$
I see that $dim(\mathbb{R}^4) - rank(A - 4I) = 4 - 3 = 1$, so the dimension of the eigenspace corresonding to $\lambda = 4$ is $1$ and the link is right.
However, when I go and solve it manually by hand, I get:
$$(A - 4I)x = \begin{bmatrix}
1 & 4 & 2 & 1 \\
0 & -3 & -1 & -1 \\
-1 & -1 & -1 & 0 \\
1 & 1 & -1 & -2 \end{bmatrix}x = 0 \iff \begin{bmatrix}
1 & 0 & 0 & -1 & |\space 0\\
0 & 1 & 0 & 0 & |\space 0\\
0 & 0 & 1 & 1 & |\space 0\\
0 & 0 & 0 & 0 & |\space 0\end{bmatrix}$$
So the corresponding equations are:
$$x_1 - x_4 = 0\\ x_2 = 0 \\ x_3 + x_4 = 0$$
and we have:
$$x = x_4\begin{bmatrix}
1 \\
0 \\
-1\\
1\end{bmatrix} + x_2\begin{bmatrix}
0 \\
1 \\
0\\
0\end{bmatrix}$$
From this, it looks like the dimension of the eigenspace corresponding to $\lambda = 4$ is 2? Where did I make a mistake?
EDIT: I don't doubt that the rank of $A - 4I$ is 3, I'm confused as to how the row reducing method yields what seems like a 2 dimensional null space. I didn't show my work for the row reductions because it was done automatically in Matlab with "rref()" function, so I doubt Matlab made a row reduction mistake.
| Let's do the Gaussian elimination slowly:
\begin{align}
A - 4I =
\begin{bmatrix}
1 & 4 & 2 & 1 \\
0 & -3 & -1 & -1 \\
-1 & -1 & -1 & 0 \\
1 & 1 & -1 & -2
\end{bmatrix}
&\to
\begin{bmatrix}
1 & 4 & 2 & 1 \\
0 & -3 & -1 & -1 \\
0 & 3 & 1 & 1 \\
0 & -3 & -3 & -3
\end{bmatrix}
&&\begin{aligned} R_3&\gets R_3+R_1\\R_4&\gets R_4-R_1\end{aligned}
\\[6px]
&\to
\begin{bmatrix}
1 & 4 & 2 & 1 \\
0 & -3 & -1 & -1 \\
0 & 0 & 0 & 0 \\
0 & 0 & -2 & -2
\end{bmatrix}
&&\begin{aligned}R_3&\gets R_3=R_2 \\ R_4&\gets R_4-R_2\end{aligned}
\end{align}
So the rank is indeed $3$. Which is the same rank you get from the RREF.
If we want to get the RREF, the next step is
\begin{align}
&\to
\begin{bmatrix}
1 & 4 & 2 & 1 \\
0 & 1 & 1/3 & 1/3 \\
0 & 0 & 1 & 1 \\
0 & 0 & 0 & 0
\end{bmatrix}
&&\begin{aligned}R_3&\leftrightarrow R_4 \\ R_2&\gets -\tfrac{1}{3}R_2 \\ R_3&\gets -\tfrac{1}{2}R_3\end{aligned}
\\[6px]
&\to
\begin{bmatrix}
1 & 4 & 0 & -1 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 1 \\
0 & 0 & 0 & 0
\end{bmatrix}
&&\begin{aligned}R_2&\gets R_2-\tfrac{1}{3}R_3 \\ R_1&\gets R_1-2R_2\end{aligned}
\\[6px]
&\to
\begin{bmatrix}
1 & 0 & 0 & -1 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 1 \\
0 & 0 & 0 & 0
\end{bmatrix}
&& R_1\gets R_1-4R_2
\end{align}
This matrix again has rank $3$.
A basis of the eigenspace is given by the vector
$$
\begin{bmatrix}1\\0\\-1\\1\end{bmatrix}
$$
| {
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"url": "https://math.stackexchange.com/questions/2508344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
How to calculate $\int_0^\pi x\,\cos^4x\, dx$ Assume that $$\int_{0}^\pi x\,f(\sin(x))dx=\frac{\pi}2 \int_{0}^\pi f(\sin(x))dx$$
and use it to calculate $$\int_{0}^\pi x\,\cos^{4}(x)\, dx$$
Can anyone help me with that? I proved the identity but I am stuck with the rest.
| With different approach. First note:
\begin{align}
I=\int_0^\pi \cos^4(x) dx = 2 \int_0^{\pi/2} \cos^4(x)dx
\end{align}
Substitute $z=\tan(x)$ so that $\cos^4(x)= \frac{1}{(1+z^2)^2}$ and $dx=\frac{dz}{1+z^2}$. So we get:
\begin{align}
I=2\int^\infty_0 \frac{1}{(1+z^2)^3}dz = \int_{-\infty}^\infty \frac{1}{(1+z^2)^3} dz
\end{align}
Use a semi circle contour in the upperhalf plane and by the Residue Theorem we get:
\begin{align}
I= 2\pi i\text{Res}_{z=i} \frac{1}{(1+z^2)^3} = 2\pi i \cdot \left( -i\frac{3}{16} \right) = \frac{3\pi}{8}
\end{align}
Finally use the formula you stated and get:
\begin{align}
\int^\pi_0 x\cos^4(x)dx = \frac{\pi}{2} I =\frac{3\pi^2}{16}
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Prove $ ( \frac{\sin ^{2}\theta}{2}+\frac{2}{\cos ^{2}\theta} )^{1/4} + ( \frac{\cos ^{2}\theta}{2}+\frac{2}{\sin ^{2}\theta} )^{1/4}\ge 68^{1/4}$
Let $0<\theta<\pi/2$. Prove that $$\left ( \frac{\sin ^{2}\theta}{2}+\frac{2}{\cos ^{2}\theta} \right )^{1/4}+\left ( \frac{\cos ^{2}\theta}{2}+\frac{2}{\sin ^{2}\theta} \right )^{1/4}\geqslant (68)^{1/4}$$
and find when the equality case holds.
This is a competition math problem. The material used should only cover up to pre-calculus.
So I quickly found out that equality holds when both of the $\sin^2(\theta)$ and $\cos^2(\theta)$ equals to 1/2, but I am not sure how to prove that this equality is true. I tried to substitute for variables and also use trig identities but just can't find out a way to do this. Thank you guys for helping me.
| Clearly, we only need to prove the case when $\theta \in (0, \pi/4]$.
Let
\begin{align*}
A &= \frac{4}{17}\left(\frac{\sin^2 \theta}{2} + \frac{2}{\cos^2\theta}\right), \\
B &= \frac{4}{17}\left(\frac{\cos^2 \theta}{2} + \frac{2}{\sin^2\theta}\right).
\end{align*}
It suffices to prove that
$$A^{1/4} + B^{1/4} \ge 2$$
or
$$B^{1/4} - 1 \ge 1 - A^{1/4}$$
or
$$\frac{B - 1}{(B^{1/4} + 1)(B^{1/2} + 1)} \ge \frac{1 - A}{(1 + A^{1/4})(1 + A^{1/2})}$$
or
$$\frac{B - 1}{B^{3/4} + B^{1/2} + B^{1/4} + 1}
\ge \frac{1 - A}{1 + A^{1/4} + A^{1/2} + A^{3/4}}.$$
It is easy to prove that
$B \ge 1 \ge A$.
It suffices to prove that
$$\frac{B - 1}{B + B + B + 1}
\ge \frac{1 - A}{1 + A + A + A}$$
or
$$3AB - A - B - 1 \ge 0$$
or (clearing the denominators)
$$(3\sin^4\theta - 3\sin^2\theta + 56)(1 - 2\sin^2\theta)^2 \ge 0$$
which is true, with equality if $\sin^2\theta = 1/2$.
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2513277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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area of points in a square closer to the center than the edge I'm hoping someone can confirm my solution to the following problem or perhaps provide better alternatives. Cheers.
A shape is defined by consisting of all the points closer to the center than the edge of a 2 inch square. Find the area of the shape.
Use symmetry is simplify the integral. Placing the center at the origin, we can find $\frac{1}{8}$ of the area by considering an angle sweep of $\frac{\pi}{4}$ radians. With that condition we know: the left and upper bounds are given by $y=x$ and the lower bound is the $x$-axis. We need to find the right bound.
Find the equation that defines the right bound or the rounded edge from the $x$-axis to the intersection of the line $y=x$.
Let $r$ be the distance from the center, the origin, to a point the edge of the shape $(x,y)$. Then
\begin{align}
r&=\sqrt{x^2+y^2}\\
\text{and for the first $\frac{\pi}{4}$ radians $r$ is restricted to}
r&=1-x\\
\text{$\therefore$}
\sqrt{x^2+y^2}&=1-x\\
x^2+y^2&=(1-x)^2\\
y^2&=1-2x+x^2-x^2\\
y&=\sqrt{1-2x}
\end{align}
Find intersection of $y=x$ and $y=\sqrt{1-2x}$
\begin{align}
x&=\sqrt{1-2x}\\
x^2&=\sqrt{1-2x}^2\\
0&=1-2x-x^2\\
0&=x^2+2x-1\\
1&=x^2+2x\\
1+1&=x^2+2x+1\\
2&=(x+1)^2\\
\sqrt{2}&=x+1\\
\sqrt{2}-1&=x
\end{align}
Find the $x$-intercept of $\sqrt{1-2x}$
\begin{align}
0&=\sqrt{1-2x}\\
0&=1-2x\\
-1&=-2x\\
\frac{1}{2}&=x
\end{align}
Now find the area under the curve $y=x$ from $0$ to $\sqrt{2}-1$ and the area under the curve $y=\sqrt{1-2x}$ from $\sqrt{2}-1$ to $\frac{1}{2}$
\begin{align}
\frac{1}{8}A&=\int_0^{\sqrt{2}-1}xdx+\int_{\sqrt{2}-1}^{\frac{1}{2}}\sqrt{(1-2x)}dx\\
A&=8\bigg(\int_0^{\sqrt{2}-1}xdx+\int_{\sqrt{2}-1}^{\frac{1}{2}}\sqrt{(1-2x)}dx\bigg)\\
A&\approx 0.876in^2
\end{align}
| Since our square is $[{-1},1]^2$ the arc you considered is not characterized by $\sqrt{x^2+y^2}=2-x$, but by $\sqrt{x^2+y^2}=1-x$. It is a parabolic arc given by $$x={1\over2}(1-y^2)\qquad\bigl(0\leq y\leq\sqrt{2}-1\bigr)\ .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2515367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If polynomial is divisible by quadratic, find values of $a$ and $b$ Equation is $z^4+(a+b)z^3+4az^2+(a+b+32)z+45$ which is divisible by $z^2+6z+9$.
We're meant to find values of $a$ and $b$ and then solve the equation itself!
I figured $z^2+6z+9=(z+3)^2$, but was wondering if this automatically implies that it is a double root(?) If so, would differentiating the equation and do the normal simultaneous equations thing suitable for this question?
| Let $p(z)=z^4+(a+b)z^3+4az^2+(a+b+32)z+45$. Then\begin{multline}p(z)=p\bigl((z+3)-3\bigr)=(z+3)^4+(a+b-12)(z+3)^3+\\+(-5a-9b+54)(z+3)^2+(4a+28b-76)(z+3)+6a-30b+30.\end{multline}On the other hand, $z^2+6z+9=(z+3)^2$. Therefore, $p(z)$ is a multiple of $z^2+6z+9$ if and only if$$\left\{\begin{array}{l}4a+28b-76=0\\6a-30b+30=0.\end{array}\right.$$This means that $a=5$ and that $b=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2519646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Probably that an $80\%$-truthful person actually rolled a $6$ A person, $A$, speaks the truth $4$ out of $5$ times. The person throws a die and reports that he obtained a $6$. What is the probability that he actually rolled a $6$?
I know there is a similar question like this
but my doubts are different from it and also I want to identify and solve total probability theorem questions so I posted a side doubt also.
In my attempt, I defined the events
\begin{align*}
E_1&: \text{The person tells the truth.} \\
E_2&: \text{The person lies.} \\
E_3&: \text{The person reports that the die landed on a 6.}
\end{align*}
I noted that $P(E_1)=\frac{4}{5}$, $P(E_2)=\frac{1}{5}$, $P(E_3|E_1)=6^{-1}$ and
$P(E_3|E_2)=0$ and obtained
\begin{align*}
P(E_3) = \frac{4}{5} \cdot \frac{1}{6} + \frac{1}{5} \cdot 0 = \frac{2}{15}.
\end{align*}
However, the correct answer is, $\frac{4}{9}$. What did I do wrong?
Side doubt: Even though the first experiment (truth and lying) is different from the second experiment, can we still apply total probability theorem? In my book the dependent experiment lies inside the sample space associated with the mutually and exhaustive events.
| Let $D_{6}$ denote the event that the die lands on a $6$ and $R_{6}$ denote the event that the person reports that it landed on a $6$. You know that
\begin{align*}
P(R_{6}|D_{6}) &= 0.8 \\
P(D_{6}) &= 6^{-1}
\end{align*}
In order to obtain the correct answer, you also need to assume that $P(R_{6}|D_{6}^{c})=0.2$, that is, if the person lies and doesn't obtain a $6$, then he says he obtained a $6$.
Since you wish to determine $P(D_{6}|R_{6})$, which is an inverse probability of $P(R_{6}|D_{6})$, a strategy that usually works is applying Bayes' theorem.
\begin{align*}
P(D_{6}|R_{6})
&= \frac{P(D_{6})P(R_{6}|D_{6})}
{P(D_{6})P(R_{6}|D_{6})+P(D_{6}^c)P(R_{6}|D_{6}^c)} \\
&= \frac{6^{-1} \cdot 0.8}{6^{-1} \cdot 0.8 + 5 \cdot 6^{-1} \cdot 0.2}
= \frac{4}{9}
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 2
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How many sets of digits are there so each digit would belong to the corresponding set? Given nine sets:
$X_1 \{2,3,9\}, X_2 \{1, 2, 3, 5, 6, 7, 8, 9\}, X_3 \{3, 9\}, X_4 \{1, 2, 3, 7, 9\}, X_5 \{1, 2, 3, 4, 5, 7, 9\}, X_6 \{2, 3, 6, 7\}, X_7 \{1, 6, 7, 9\} , X_8 \{1, 3, 4, 6, 7, 8, 9\}, X_9 \{2, 9\} $
How many sets of different digits $(a_1, a_2, a_3, a_4, a_5, a_6, a_7, a_8, a_9 )$ are there so $a_i \in x_i $ ?
Have tried using combinatorical formules, seems like there are too many variants.
| Consider only $X_1$, $X_3$, and $X_9$. $a_1$ can be 2, 3, or 9. If $a_1=2$, then $a_9=9$ so $a_3=3$. If $a_1=3$, then $a_3=9$ so $a_2=2$. Finally if If $a_1=9$, then $a_3=3$ and $a_9=2$.
The possibilities for $(a_1, a_3, a_9)$ are $(2,3,9)$, $(3,9,2)$, or $(9,3,2)$.
Since 2, 3, 9 are "used up" you can repeat the process with $X_4$, $X_6$, and $X_7$.
The possibilities for $(a_4, a_6, a_7)$ are $(1,6,7)$, $(1,7,6)$, or $(7,6,1)$.
Now 1, 6, 7 are also "used up". Last considering $X_2$, $X_5$, and $X_6$ gives the possibilities for $(a_2, a_5, a_8)$ are $(5,4,8)$ or $(8,5,4)$.
The total number is $3\times 3 \times 2 = 18$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Locating the third vertex of an equilateral triangle Two vertices of an equilateral triangle are at $A=(10,-4)$ and $B=(0,6)$.
How can one locate the third vertex?
Maybe someone could give me the easy way please.
My attempt:
*
*Find the average of $M=(x , y)$ of $A$ and $B$, which is $(10+0)/2 = 5$, $(-4+6)/2 = 1$. So M(5,1)$.
*Find the equation of the perpendicular bisector. Slope of $AB = 10/-10 = -1$, Slope $m$ of the perpendicular bisector $= +1$
$$ \Rightarrow y-1 = 1*(x-5)\Rightarrow y = x-4 .$$
*There are $2$ vertices, $C$ and $D$, both on the line $y = x-4$. The altitude of the triangle is $10$. Find the $2$ points on $y = x-4$ which is $10$ units distance from $M(5,1)$:
$$d^2 = (\text{difference of } x)^2 + (\text{difference of } y)^2 \Rightarrow 100 = (x-5)^2 + (y-1)^2$$
Sub for $y = x-4$:
$$ 100 = (x-5)^2 + (x-5)^2\Rightarrow (x-5)^2 = 50 \Rightarrow x-5 = \pm\sqrt{50}$$
Thus $x = 5 + \sqrt{50}$, $y = 1 + \sqrt{50}$ and so the Vertices are $C$: $x = 5 - \sqrt{50}$, $y = 1 - \sqrt{50}$ --> Vertex $D$
| We know that multiplying with $\varepsilon ={1\over 2}+{\sqrt{3}\over 2}$ is rotation around $0$ for $60^{\circ}$ in counterclockwise direction and with $-\varepsilon$ around $0$ for $60^{\circ}$ in clockwise direction.
Let in complex plain $z=10-4i$ correspond to the point $(10,-4)$ and $w=6i$ to the point $(0,6)$. If $v$ is third point of this triangle then $v-z = \varepsilon(w-z)$ (if we rotate $w$ around $z$ for $60^{\circ}$ in counterclockwise direction) or $v-z = -\varepsilon(w-z)$ (if we rotate $w$ around $z$ for $60^{\circ}$ in clockwise direction). In first case we have
$$v= 10-4i+\varepsilon(-10+10i) = ...$$ and in second case we have $$ v= 10-4i-\varepsilon(-10+10i) = ...$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2523286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the probability that the first head will appear on the even numbered tosses Question
$\text{Consider a coin with probability R to be heads. What is the probability}$
$\text{that the first head will appear on the even numbered tosses?}$
My Approach
let the required Probability$=P$.
Hence we can write our eauation as-:
$$P=(1-R) \times R+(1-R) \times (1-R) \times (1-R) \times P $$
$$P(1-(1-R) \times (1-R) \times (1-R))=(1-R) \times R $$
$$P=\frac{(1-R) \times R }{(1-(1-R) \times (1-R) \times (1-R)}$$
Am i correct?
Answer is given as-:
$$P=\frac{(1 - R)}{(2 - R)}$$
| The probability that the first head will appear on the second toss is $(1 - R)R$.
The probability that the first head will appear on the fourth toss is $(1 - R)^3R$.
The probability that the first head will appear on the sixth toss is $(1 - R)^5R$.
In general, the probability that the first head will appear on the $2k$th toss is $(1 - R)^{2k - 1}R$.
Hence, the desired probability is
$$p = \sum_{k = 1}^{\infty} (1 - R)^{2k - 1}R = (1 - R)R\sum_{k = 1}^{\infty} (1 - R)^{2k}$$
which is a geometric series with common ratio $(1 - R)^2$. If $R < 1$, we obtain
\begin{align*}
p & = (1 - R)R \cdot \frac{1}{1 - (1 - R)^2}\\
& = \frac{(1 - R)R}{1 - (1 - 2R + R^2)}\\
& = \frac{(1 - R)R}{2R - R^2}\\
& = \frac{(1 - R)R}{R(2 - R)}\\
& = \frac{1 - R}{2 - R} && \text{provided that $R \neq 0$}
\end{align*}
If $R = 1$, then heads will be obtained on the first toss, so $p = 0$. If $R = 0$, then heads will never be obtained, so again $p = 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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$\frac{1}{x+1} + \frac{2}{x^2+1} + \frac{4}{x^4+1} + \cdots $ up to $n$ terms in terms of $x$ and $n$. There's a series which I can't seem to find a way to sum. Any help would be highly appreciated. It goes as follows $$\frac{1}{x+1} + \frac{2}{x^2+1} + \frac{4}{x^4+1} + \cdots $$ up to $n+1$ terms. The sum is to be expressed in terms of $x$ and $n$. I tried setting a formula for the $n$-th term and setting it into a difference but ran into a dead end.
| By induction, it is pretty clear that
$$ x^{2^{n+1}}-1 = (x-1)\prod_{k=0}^{n} \left(x^{2^k}+1\right) \tag{A}$$
and by applying $x\cdot\frac{d}{dx}\log(\cdot)$ to both sides we have:
$$\frac{2^{n+1}x^{2^{n+1}} }{x^{2^{n+1}}-1}=\frac{x}{x-1}+\sum_{k=0}^{n}\frac{2^k x^{2^k}}{x^{2^k}+1} \tag{B}$$
and by replacing $x$ with $\frac{1}{z}$:
$$\frac{2^{n+1}}{1-z^{2^{n+1}}}=\frac{1}{1-z}+\sum_{k=0}^{n}\frac{2^k}{z^{2^k}+1}. \tag{C}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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} |
Solving $3^x = 2^y + 1$ with $x,y \in \mathbb{N}^2$ I consider the following equation
$$ 3^x = 1 + 2^y \tag{$\star$} $$
with $(x,y) \in \mathbb{N}^2$ and $y \geq 3$.
I would like to show that :
$$ 3^x \equiv 1 \; [2^y] \; \Leftrightarrow \; 2^{y-2} \mid x. $$
I assume that $3^x \equiv 1 \; [2^y]$. Writing
$$ 3^x = \sum_{k=0}^{x} C_{x}^{k} 2^k = 2^x + x 2^{x-1} + \ldots + 2x + 1 $$
I have that $3^x \equiv 1 \; [2^y]$ is equivalent to :
$$ 2^x + x 2^{x-1} + \ldots + 2x \equiv 0 \; [2^y]. $$
This is also equivalent to :
$$ 2^{x-1} + x 2^{x-2} + \ldots + x \equiv 0 \; [2^{y-1}]. $$
Then, I am unsure about how to proceed. I would like to assume that $x > y$ and then, it would follow that $2^{x-1} \equiv 0 \; [2^{y-1}]$. Therefore, I would end up with :
$$ x \big( 2^{x-2} + \ldots + 1 ) \equiv 0 \; [2^{y-1}]. $$
Since $\operatorname{gcd}\big( 2^{y-1}, 2^{x-2} + \ldots + 1 \big) = 1$, it follows that $2^{y-1} \mid x$. But this is not the expected result. Where did I make a mistake ?
|
I would like to show that :
$$ 3^x \equiv 1 \; [2^y] \; \Leftrightarrow \; 2^{y-2} \mid x. $$
To show this, I would use that, for every pair of positive integers $(a,s)$ where $s$ is odd, there exists an odd integer $t$ such that
$$3^{2^as}-1=2^{a+2}t\tag1$$
(the proof are written at the end of the answer)
Proof for $2^{y-2}\mid x\implies 3^x\equiv 1\pmod{2^y}$
If $2^{y-2}\mid x$, then, from $(1)$, $3^{x}-1$ is divisible at least by $2^y$. $\quad\blacksquare$
Proof for $3^x\equiv 1\pmod{2^y}\implies 2^{y-2}\mid x$
It is easy to see that $4\not\mid 3^{\text{odd}}-1$.
The claim follows from this and $(1)$. $\quad\blacksquare$
Finally, let us prove $(1)$ by induction on $a$.
For $a=1$, we have $3^{2s}-1\equiv 1-1\equiv 0\pmod{2^3}$. Also, writing $s=2u+1$, we have $$3^{2s}-1=9^{2u+1}-1=9\cdot 81^u-1\equiv 9-1\equiv 8\pmod{2^4}.$$
Suppose that $(1)$ holds for some $a$.
Then, we have
$$3^{2^{a+1}s}-1=(3^{2^as})^2-1=(2^{a+2}t+1)^2-1=2^{2a+4}t^2+2^{a+3}t=2^{a+3}t'$$
where $t'$ is odd. $\quad\blacksquare$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2528986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Computing limit of $\sqrt{n^2+n}-\sqrt[4]{n^4+1}$ I have tried to solve this using conjugate multiplication, but I got stuck after factoring out $n^2$.
$\begin{align}
\lim_{n\rightarrow\infty}\dfrac{n^2+n-\sqrt{n^4+1}}{\sqrt{n^2+n}+\sqrt[4]{n^4+1}}
&=\lim_{n\rightarrow\infty}\dfrac{n(1+\dfrac{1}{n}-\sqrt{1+\dfrac{1}{n^4}})}{\sqrt{1+\dfrac{1}{n}}+\sqrt[4]{1+\dfrac{1}{n^4}}}\\&
=\lim_{n\rightarrow\infty}\dfrac{n+1-n\sqrt{1+\dfrac{1}{n^4}}}{\sqrt{1+\dfrac{1}{n}}+\sqrt[4]{1+\dfrac{1}{n^4}}}
\end{align}$
Given that $\dfrac{1}{n}$ tends to $0$ (so denominator is 2), can I reduce $n$ and $-n\sqrt{1+\dfrac{1}{n^4}}$ and say that the limit is $\dfrac{1}{2}$?
I mean $\dfrac{1}{n^4}$ tends to $0$, so $\sqrt{1+\dfrac{1}{n^4}}$ tends to $1$ and in this case $n-n\sqrt{1+\dfrac{1}{n^4}}$ can be simplified to $n-n$.
Solution given in my book uses conjugate multiplication twice to get rid of all the roots in nominator, but I am curious if my answer is correct or my teacher will tell me that simplifying the way I did it is incorrect.
| it must be $$\frac{(n^2+n)^2-(n^4+1)}{(\sqrt{n^2+n}+\sqrt[4]{n^4+1})(n^2+n+\sqrt{n^4+1})}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2531365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
} |
Determine all solutions of the congruence $y^2≡5x^3\pmod7$ in integers $x$, $y$. Determine all solutions of the congruence $y^2≡5x^3\pmod7$ in integers $x$, $y$.
I learned about primitive roots and the theory of indices.
By trial, I check that $3$ is a primitive root of $7$ and $\operatorname{ind}_35≡5\pmod 7$.
| With $p=7$, you have that $6|p-1$, which means there are only a limited number of squares and cubes mod $7$. Whatever $y$ is, $y^2$ can only be $0$, $1$, $4$, or $2$.
And $x^3$ can only be congruent to $0$, $1$, or $6$. So $5x^3$ can only be $0$, $5$, or $2$.
So with the given congruence, for the left side to match the right side, either both sides are $0$ or both sides are $2$.
If both sides are $0$, this implies $x\equiv y\equiv 0$.
If both sides are $2$, then $y$ could be either square root of $2$ (which are $3$ and $4$) and $x$ could be any cube root of $2/5\equiv6$ (which are $3$, $5$, and $6$.)
All together that makes $2\cdot3+1=7$ solution pairs.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2531819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to generate number in ascending order using the expression $a^2\cdot b^3$ where $a$ and $b$ are distinct primes. How to generate numbers in ascending order using the expression $a^2\cdot b^3$ where $a$ and $b$ are distinct primes.
Here are a couple of examples :
$$\begin{array}{c|ccc}
\text{No}&a&b&a^2\cdot b^3\\
\hline
1&3&2&72\\
2&2&3&108\\
3&5&2&200\\
4&7&2&392\\
5&2&5&500\\
6&5&3&675\\
7&11&2&968\\
8&3&5&1125\\
9&7&3&1323\\
10&13&2&1352\\
\end{array}$$
Is there a way I can stop at $1000$th number without pre computing say $100000$ numbers, and then sorting and arriving at $1000$th number ?
| I got Matlab to calculate numbers up to $10^8$, or a hundred billion.
The $N$th number, over that range, was about $9N^{2.5}/\ln N$. The $1000$th number was $42797187 = 3^31259^2$.
The largest prime involved in the first thousand was $2311$, which is the $344$th prime.
According to my formula, the biggest prime $p$, which will be with $2^3p^2\approx 9N^{2.5}/\ln N$, is roughly $p=N^{1.25}/\sqrt{\ln N}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2531956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
How to differentiate a power series? I have a task where I need to write the power series:
$\sum\limits_{n=0}^\infty\frac{(-1)^n}{(2n)!}x^n$
differentiated 2 times where $x=0$.
This is what I have done so far:
$f'(x) = \sum\limits_{n=1}^\infty n\frac{(-1)^n}{(2n)!}x^{n-1} \\
f''(x) = \sum\limits_{n=1}^\infty n(n-1)\frac{(-1)^n}{(2n)!}x^{n-2} = \sum\limits_{n=1}^\infty\frac{((-1)^n (n - 1) n x^{n - 2})}{((2 n)!)}$
However I'm not quite sure how to proceed.
Update:
So I found the solution by replacing x with 0 which meant that the only term that wouldn't be undefined was 2 which resulted in the sollution:
$ f''(0) = \frac{(-1)^2(2-1)2}{(2*2)!} = \frac{2}{4!} = \frac{2}{4*3*2*1} = \frac{1}{4*3} = \frac{1}{12}$
| \begin{eqnarray*}
\sum\limits_{n=1}^\infty\frac{(-1)^n n(n - 1) x^{n - 2}}{(2 n)!} \mid _{x=0}
\end{eqnarray*}
Note that the first term is zero, the third & higher terms are all zero, so only the second term contributes & gives the value $\color{red}{1/12}$.
Another way to figure is the function is $y=\cos(\sqrt{x})$ now differntiate this twice (chain rule and quotient)
\begin{eqnarray*}
\frac{d^2y}{dx^2} =\frac{ \sin(\sqrt{x})-\sqrt{x} \cos(\sqrt{x})}{4 x^{3/2}}
\end{eqnarray*}
Now use $\sin(\sqrt{x})=\sqrt{x}-x^{3/2}/6 +\cdots$ and $ \sqrt{x} \cos( \sqrt{x} ) = \sqrt{x} -x^{3/2}/2$ and again we get $\color{red}{1/12}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2532992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How do I rationalize the following fraction: $\frac{1}{9\sqrt[3]{9}-3\sqrt[3]{3}-27}$? As the title says I need to rationalize the fraction: $\frac{1}{9\sqrt[3]{9}-3\sqrt[3]{3}-27}$. I wrote the denominator as: $\sqrt[3]{9^4}-\sqrt[3]{9^2}-3^3$ but I do not know what to do after. Can you help me?
| We have
$(x+y+z)(x^2+y^2+z^2-xy-xz-yz)=x^3+y^3+z^3-3xyz$
With $x=9\sqrt[3]{9},y=-3\sqrt[3]{3},z=-27$ all terms on the right side are rational, try it. So multiply the given numerator and denominator by $x^2+y^2+z^2-xy-xz-yz$ with $x,y,z$ as rendered above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2535817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Determine the centroid of the region $ D \ $ within the circle of radius $ \ 2 \ $
Determine the centroid of the region $ D \ $ within the circle of radius $ \ 2 \ $ as follows :
Where $ \ AB \ $ is the line joining $ \ A (-\sqrt 2, -\sqrt 2) \ $ and $ \ (\sqrt 2, \sqrt 2 ) \ $
Step 1: Use the formula $ \ \bar x=\frac{1}{A} \iint_D xdA , \ \ \bar y=\frac{1}{A} \iint_D y dA \ $, where $ \ A=\iint_D dxdy \ $ is the area of the shadded region
Step 2: Use the formula $ \bar x=\frac{1}{2A} \int_C x^2 dy \, \ \ \bar y=\frac{1}{2A} \int_C y^2 dx$ , where $C$ is the boundary of $D$.
Show that both case gives same centroids
Answer:
The equation of the line $ AB \ $ is given by
$ y=- \sqrt 2 \ $
$$ A=\iint_D dxdy = 2 \int_{y=-\sqrt 2}^{2} \int_{0}^{\sqrt{4-y^2}} dxdy = \int_{-\sqrt 2}^{2} \sqrt{4-y^2} dy=1+ \frac{3 \pi}{2} =5.7124$$
Thus
$$\bar x=\frac{1}{A} \int_{-\sqrt 2}^{2} \int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}} x dxdy =0 \qquad \bar y= \frac{1}{A} \int_{-\sqrt 2}^{2} \int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}} y dxdy =0.1650. $$
Now using formula in step 2, we get
$$\bar y= \frac{1}{2A} \int_C y^2 dx = \frac{1}{2A} \int_{-2}^{2} (4-x^2) dx = \frac{10.677}{11.4248}=0.93, $$
which is not equal to above $\bar y=0.1650$.
So I am making mistake somewhere.
I think , I am making mistake when integrating along the curve $ \ C \ $ which consists of the line segments $ AB \ $
Help me out.
| Note that
$$|D|=\iint_D dxdy= \int_{y=-\sqrt 2}^{2} \int_{x=-\sqrt{4-y^2}}^{x=\sqrt{4-y^2}} dxdy =2\int_{-\sqrt 2}^{2} \sqrt{4-y^2} dy =2+ 3 \pi.$$
Therefore
$$\bar y=\frac{1}{|D|}\iint_D ydxdy=\frac{1}{A}\int_{y=-\sqrt 2}^{2} y\int_{x=-\sqrt{4-y^2}}^{x=\sqrt{4-y^2}} dxdy\\=\frac{2}{|D|}\int_{-\sqrt 2}^{2} y\sqrt{4-y^2} dy=\frac{(4/3)\sqrt{2}}{2+ 3 \pi}\approx 0.16504.$$
Hence your first evaluation is correct.
On the other hand, the curve $C$ is the union of the segment $C_1$ and the circular arc $C_2$.
For $C_1$: $(x,y)=(t,-\sqrt{2})$ with $t\in [-\sqrt{2},\sqrt{2}]$ then
$$\int_{C_1} y^2dx=\int_{t=-\sqrt{2}}^{\sqrt{2}}(-\sqrt{2})^2(t)'dt=4\sqrt{2}.$$
For $C_2$: $(x,y)=(2\cos(t),2\sin(t))$ with $t\in [-\pi/4,5\pi/4]$ then
$$\int_{C_2} y^2dx=\int_{t=-\pi/4}^{5\pi/4}(2\sin(t))^2(2\cos(t))'dt=
8\int_{t=-\pi/4}^{5\pi/4}(1-\cos^2(t))^2\cos(t)'dt=?$$
Can you take it from here and correct the second evaluation?
P.S. Note that, it should be $\bar y=-\frac{1}{2|D|} \int_C y^2 dx$ where $C$ is oriented counter-clockwise.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding function $f(x)$ which satisfy given functional equation
Find all function $f:\mathbb{R}-\{0,1\}$ in $$f(x)+2f\left(\frac{1}{x}\right)+3f\left(\frac{x}{x-1}\right)=x$$
Attempt: put $\displaystyle x = \frac{1}{x}$, then $$f\left(\frac{1}{x}\right)+2f(x)+3f\left(\frac{1}{1-x}\right) = \frac{1}{x}$$
could some help me how to solve it , thanks
| Just as you replaced $x$ by $\frac{1}{x}$, you can replace $x$ by $1-x$. Then you can replace $x$ by $\frac{1}{x}$ in this new equation.
Repeating this process, I believe you will obtain 6 independent equations in the variables $f(z)$ where $z$ takes the values $$x,\frac{1}{x},\frac{1}{1-x},\frac{x}{x-1},\frac{x-1}{x},1-x.$$
These can be solved for $f(x)$.
Carrying out this process gives $$24f(x)=-3x+\frac{3}{x}+\frac{1}{1-x}+\frac{7x}{x-1}+\frac{x-1}{x}-5(1-x)$$$$f(x)=\frac{2x^3+x^2+5x-2}{24x(x-1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2536550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
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Finding number of circles and area of polygon formed by their centres. Let $S(x,y)=0$ represent a circle with radius $\frac{3}{\sqrt 2}$ such that $S(\lambda -3, \lambda) =0$ has equal roots and $S(\mu, 7- \mu )=0$ also has equal roots then find
1) number of such circles
2) number of circles whose centre lie in first quadrant
3) area of polygon formed by joining all possible centres.
My approach: I wrote $S=0$ in the general form of circle with unknowns g, f, and c. Using the condition of radius I got one equation in these variables. Then substituted the given points in equation of circle to get quadratic equations in $\lambda$ and $\mu$. Then I set the discriminant of these equations equal to 0 to obtain two more equations. But the equations obtained appear to be impossible to solve because of so many of variables in 2nd degree. Can anyone please provide me some hints to deal with this question in an easy way.
| The following is just painful computation from where you left off. I would be very interested to see if there is a slicker way to do this problem; in particular, why did the magic cancelations happen?
The general equation is $$S(x,y) = (x-a)^2 + (y-b)^2 - 9/4.$$
Then
\begin{align}
S(\lambda - 3, \lambda)
&= (\lambda - (a+3))^2 + (\lambda - b)^2 - 9/4
= 2 \lambda^2 - 2(a+b+3)\lambda + (a+3)^2 + b^2 - 9/4
\\
S(\mu, 7-\mu)
&= (\mu - a)^2 + (7-\mu-b)^2 - 9/4
= 2\mu^2 -2(a-b+7)\mu + a^2 + (b-7)^2 - 9/4
\end{align}
Setting the discriminants to zero yields
\begin{align}
(a+b+3)^2 - 2(a+3)^2 - 2b^2 + 9/4=0
\\
(a-b+7)^2 - 2a^2 - 2(b-7)^2 + 9/4=0
\end{align}
If we write the left-hand sides as quadratics in $b$, then we have
\begin{align}
b^2 -(2a+6)b+(a^2 +6a+27/4) &= 0\\
b^2 +(2a-14)b+(a^2 -14a+187/4) &= 0
\end{align}
If you use the quadratic formula to solve for $b$, you get magic cancelations that lead to
\begin{align}
b &= \frac{1}{2} [2a+6 \pm \sqrt{(2a+6)^2-4(a^2 + 6a+27/4)}]
= a + 3 \pm 3/2
\\
b &= \frac{1}{2}[-2a + 14 \pm \sqrt{(2a-14)^2 + 4(a^2 - 14a + 187/4)}] = -a +7 \pm 3/2
\end{align}
Thus, the solutions to "$S(\lambda-3,\lambda)$ has repeated roots" are the parallel lines $b=a+9/2$ and $b=a+3/2$,
while the solutions to "$S(\mu, 7-\mu)$ has repeated roots" are the parallel lines $b=-a+11/2$ and $b=-a+17/2$.
Thus, there are four solutions, and it should be easy to finish the problem from here.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Convergence of $ \sum_{k=1}^{\infty} \frac {3^k}{5^k + 1}$
Show the convergence of the following series:
$$\sum_{k=1}^{\infty}\frac {3^k}{5^k + 1}$$
*
*a) Show the monotony of the partial sums
*b) estimate upwards
*c) remember the geometric series (I do not know how to use that here.)
The following is what I have done so far:
To show by induction: $a_{k+1} < a_k \forall k \in \mathbb N_0$
Induction start: $n=1$
$a_2= \frac{3^2}{5^2+1}=\frac{9}{25+1}=\frac{9}{26}=\frac{18}{52} < \frac{26}{52}=\frac{1}{2}=\frac{3}{6}=\frac{3^1}{5^1+1}=a_1$
Induction step:
$$\begin{align}
a_{k+1}&<a_k \\
\equiv \frac{3^{k+1}}{5^{k+1}+1} &< \frac{3^k}{5^k+1} \\
\equiv \frac{3^{k+1}}{5^{k+1}} &< \frac{3^k}{5^k} \\
\equiv \frac{3^{k+2}}{5^{k+1}} &< \frac{3^{k+1}}{5^k} \\
\equiv \frac{3^{k+2}}{5^{k+2}} &< \frac{3^{k+1}}{5^{k+1}} \\
\equiv \frac{3^{k+2}}{5^{k+2}+1} &< \frac{3^k+1}{5^{k+1}+1} \\
\equiv a_{k+2} &< a_{k+1} \\
\end{align}$$
To Show: $|a_k|= a_k$
$$
\begin{align}
|a_k| &= |\frac{3^k}{5^k+1}| \\
&= \frac{|3^k|}{|5^k+1|} \\
&= \frac{3^{|k|}}{5^{|k|}+|1|} \\
&= \frac{3^k}{5^k+1} \\
&= a_k
\end{align}$$
Because of the induction I can conclude that the sequence $\lim_{k \to \infty}a_k$ becomes smaller and smaller.
And because of $|a_k|=a_k$ are all values $\forall k \in \mathbb N$ positive.
$\Rightarrow $ The sequence $a_k$ is monotically decreasing.
$\Rightarrow $ The series $\sum_{k=1}^{\infty}a_k$ is monotically increasing.
$$
\begin{align}
a_k = \frac{3^k}{5^k+1} &< \frac{3^k}{5^k} \\
&< \frac{3^k}{3^k} \\
&= 1 \\
\end{align}$$
$$\lim_{k \to \infty} 1 = 1$$
And thus: $\exists N \in \mathbb N$, such that
$$|a_k| \le 1, \forall \quad k \ge N$$
Using the direct comparison test there can be concluded that $\sum_{k=1}^{\infty}{a_k}$ converges.
Question: Is my proof correct?
| For a) you have to show, that $\sum_{k=0}^n a_k<\sum_{k=0}^{n+1} a_k$, which is kinda clear, since $a_k>0$ for every $k\in\mathbb{N}$. Then stipulate $\frac{3^k}{5^k+1}<\frac{3^k}{5^k}$ and use the geometric series. With the comparision test we get that the series converges.
But this can be also done immediatly. We kinda do not need the step a).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Index Manipulation of Summation in Real Cosine Transform Fast Computation Consider $X_k = \sum_{n=0}^{N-1}x_{n}\cos\frac{(2n+1)k\pi}{2N}$ for $k=0,1,2,...,N-1$. Now we define $g_{n} = x_{n} + x_{N-1-n}$ and $h_{n} = \frac{x_{n}-x_{N-1-n}}{2\cos\frac{(2n+1)k\pi}{2N}}$ for $\{x_{n}\}_{n=0}^{N-1}$ and $\{g_{n}\}_{n=0}^{\frac{N}{2}-1}$, $\{h_{n}\}_{n=0}^{\frac{N}{2}-1}$. Next, we also define $G_{k} = \sum_{n=0}^{\frac{N}{2}-1}g_{n}\cos\frac{(2n+1)k\pi}{N}$ and $H_{k}= \sum_{n=0}^{\frac{N}{2}-1}h_{n}\cos\frac{(2n+1)k\pi}{N}$. Note that $X_{k}$ is in the context of being the real cosine transform of $x_{n}$.
Show that $G_{k}=X_{2k}$ and $H_{k}+H_{k+1}=X_{2k+1}$!
So, here is my attempt so far:
Observe that
$$X_{2k}= \sum_{n=0}^{N-1}x_{n}\cos\frac{(2n+1)k\pi}{N}$$
$$X_{2k}= \sum_{n=0}^{\frac{N}{2}-1}x_{n}\cos\frac{(2n+1)k\pi}{N}+ \sum_{n=\frac{N}{2}}^{N-1}x_{n}\cos\frac{(2n+1)k\pi}{N}$$
$$X_{2k}= \sum_{n=0}^{\frac{N}{2}-1}x_{n}\cos\frac{(2n+1)k\pi}{N}+ \sum_{i=0}^{\frac{N}{2}-1}x_{i+\frac{N}{2}}\cos\big(\frac{(2i+1)k\pi}{N} + \frac{k\pi}{2}\big) $$
and I do not know what to do from here on since the form does not match.
Then, I also have tried the second part but I find myself to not be able to do further as well.
Observe that
$$H_{k}+H_{k+1}= \sum_{n=0}^{\frac{N}{2}-1}h_{n}\cos\frac{(2n+1)k\pi}{N} + \sum_{n=0}^{\frac{N}{2}-1}h_{n}\cos\frac{(2n+1)(k+1)\pi}{N} $$
$$H_{k}+H_{k+1}= \sum_{n=0}^{\frac{N}{2}-1}h_{n}\bigg[\cos\frac{(2n+1)k\pi}{N} + \cos\frac{(2n+1)(k+1)\pi}{N}\bigg]$$
$$H_{k}+H_{k+1}= \sum_{n=0}^{\frac{N}{2}-1}h_{n}2\cos\frac{(2n+1)(2k+1)\pi}{2N}\cos\frac{(2n+1)\pi}{2N}$$
$$H_{k}+H_{k+1}= \sum_{n=0}^{\frac{N}{2}-1}\frac{x_{n}-x_{N-1-n}}{2\cos\frac{(2n+1)k\pi}{2N}}2\cos\frac{(2n+1)(2k+1)\pi}{2N}\cos\frac{(2n+1)\pi}{2N}$$
$$H_{k}+H_{k+1}= \sum_{n=0}^{\frac{N}{2}-1}(x_{n}-x_{N-1-n})\cos\frac{(2n+1)(2k+1)\pi}{2N}$$
$$H_{k}+H_{k+1}= \sum_{n=0}^{\frac{N}{2}-1}x_{n}\cos\frac{(2n+1)(2k+1)\pi}{2N}-\sum_{n=0}^{\frac{N}{2}-1}x_{N-1-n}\cos\frac{(2n+1)(2k+1)\pi}{2N}$$
Again, I do not know what I have to do at this point since I cannot manipulate the index of the summation to get the result I want.
Any help is much appreciated! Thank you!
| Here is the first part. We want to show equality $G_k=X_{2k}$ with
\begin{align*}
G_k&=\sum_{n=0}^{\frac{N}{2}-1}\left(x_n+x_{N-1+n}\right)\cos\frac{(2n+1)k\pi}{N}\\
X_{2k}&=\sum_{n=0}^{N-1}x_n\cos\frac{(2n+1)k\pi}{N}
\end{align*}
We obtain
\begin{align*}
\color{blue}{G_k}&\color{blue}{=\sum_{n=0}^{\frac{N}{2}-1}\left(x_n+x_{N-1+n}\right)\cos\frac{(2n+1)k\pi}{N}}\\
&=\underbrace{\sum_{n=0}^{\frac{N}{2}-1}x_n\cos\frac{(2n+1)k\pi}{N}}_{=: A}+\sum_{n=0}^{\frac{N}{2}-1}x_{N-1+n}\cos\frac{(2n+1)k\pi}{N}\\
&=A+\sum_{n=0}^{\frac{N}{2}-1}x_{N-1+\left(\frac{N}{2}-1-n\right)}\cos\frac{\left(2\left(\frac{N}{2}-1-n\right)+1\right)k\pi}{N}\tag{1}\\
&=A+\sum_{n=0}^{\frac{N}{2}-1}x_{\frac{N}{2}+n}\cos\frac{\left(N-1-2n\right)k\pi}{N}\\
&=A+\sum_{n=\frac{N}{2}}^{N-1}x_{\frac{N}{2}-\left(n+\frac{N}{2}\right)}\frac{\cos\left(N-1-2\left(n-\frac{N}{2}\right)\right)k\pi}{N}\tag{2}\\
&=A+\sum_{n=\frac{N}{2}}^{N-1}x_n\frac{\cos\left(2N-1-2n\right)k\pi}{N}\\
&=\sum_{n=0}^{\frac{N}{2}-1}x_n\cos\frac{(2n+1)k\pi}{N}+\sum_{n=\frac{N}{2}}^{N-1}x_n\cos\frac{(2n+1)k\pi}{N}\tag{3}\\
&\color{blue}{=\sum_{n=0}^{N-1}x_n\cos\frac{(2n+1)k\pi}{N}=X_{2k}}
\end{align*}
and the claim follows.
Comment:
*
*In (1) we change the order of summation of the right-hand sum: $n\rightarrow \frac{N}{2}-1-n$.
*In (2) we shift the index to start with $n=\frac{N}{2}$.
*In (3) we use that $\cos$ is an even function and periodic with period $2\pi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2540434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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$\frac{(2n − 1)!!}{(2n)!!} \leq \frac{1}{\sqrt{2n+1}}$ by induction? Let $(2n)!!$ be the product of all positive even integers less than or equal to $2n$. Let $(2n − 1)!!$ be the product of all odd positive integers less than or equal to $(2n − 1)$. Prove that
$$\frac{(2n − 1)!!}{(2n)!!} \leq \frac{1}{\sqrt{2n+1}}.$$
I am up to the inductive step, where am I stuck. I can't find a way to substitute in from the assumption because of the square root.
| $$\frac{(2n-1)!!}{(2n)!!}=\frac{2n-1}{2n}\cdot \frac{(2n-3)!!}{(2n-2)!!}\le \frac{2n-1}{2n\sqrt{2(n-1)+1}}=\frac{2n-1}{2n\sqrt{2n-1}}= \frac{\sqrt{2n-1}}{2n}$$
Now to show that $\frac{\sqrt{2n-1}}{2n}\le \frac{1}{\sqrt{2n+1}}$ we show that $\frac{1}{\sqrt{2n+1}}-\frac{\sqrt{2n-1}}{2n}\ge 0$:
$$\frac{1}{\sqrt{2n+1}}-\frac{\sqrt{2n-1}}{2n}=\frac{2n-\sqrt{2n-1}\sqrt{2n+1}}{2n\sqrt{2n+1}}=\frac{2n-\sqrt{4n^2-1}}{2n\sqrt{2n+1}}\ge \frac{2n-\sqrt{4n^2}}{2n\sqrt{2n+1}}=0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find $\sum_{n=1}^{\infty}\frac{\sin(nx)}{2n+1}$? My attempt was : $f(t) = \sum_{n=1}^{\infty}\frac{\sin(nx)}{2n+1}t^{2n+1}$, where $|t|\le 1$. So we can find $f'(t) = \displaystyle \Im\sum_{n=0}^{\infty}(e^{ix}t^2)^{n} = \Im\frac{1}{1-e^{ix}t^2} = \frac{t^2\sin(x)}{t^4-2\cos(x)t^2+1}$. So $f(t) = \displaystyle \int\frac{t^2\sin(x)}{t^4-2\cos(x)t^2+1}dt = \int\frac{t^2 \Im e^{ix}}{(t^2-e^{ix})(t^{2}-e^{-ix})}dt$.
Now if we decompose it and integrate, we will get $\displaystyle \frac{\pi-x}{8}t = \frac{\pi - x}{8}$(when $t = 1$), am I right?
| Let $z=\rho e^{ix}$ with $\rho\in(0,1)$. Then
$$\sum_{n\geq 0}\frac{\rho^n\sin(nx)}{2n+1}=\text{Im}\sum_{n\geq 0}\frac{z^n}{2n+1} =\text{Im}\left(\frac{\text{arctanh}\sqrt{z}}{\sqrt{z}}\right)$$
and by considering the limit as $\rho\to 1^-$ we get
$$ \sum_{n\geq 0}\frac{\sin(nx)}{2n+1}=\frac{\pi}{4}\cos\left(\frac{x}{2}\right)-\frac{1}{2}\sin\left(\frac{x}{2}\right)\log\cot\left(\frac{x}{4}\right) $$
for any $x\in(0,2\pi)$.
| {
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"url": "https://math.stackexchange.com/questions/2541236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Generating fuctions: $\sum_{k=0}^\infty \binom{2k}{k}\binom{n}{k}\left(-\frac{1}{4}\right)^k=2^{-2n}\binom{2n}{n}$ I'm trying to solve one of my combinatorics exercise but I struggle a bit.
Is the equality correct for all the $n\ge 0$?
$$\sum_{k=0}^\infty \binom{2k}{k}\binom{n}{k}\left(-\frac{1}{4}\right)^k=2^{-2n}\binom{2n}{n}$$
First of all:$$\sum_{n=0}^\infty \left(2^{-2n}\binom{2n}{n}\right)x^n=\sum_{n=0}^\infty \left( \frac{x}{4}\right)^n\binom{2n}{n}=\frac{1}{\sqrt{1-x}}$$
Now
$$\sum_{n=0}^\infty \left(\sum_{k=0}^\infty \binom{2k}{k}\binom{n}{k}\left(-\frac{1}{4}\right)^k \right)x^n=\sum_{k=0}^\infty \left(-\frac{1}{4}\right)^k\sum_{n=k}^\infty \binom{2k}{k}\binom{n}{k} x^n=[n-k=m]=\sum_{k=0}^\infty \left(-\frac{1}{4}\right)^k\sum_{m=0}^\infty \binom{2k}{k}\binom{m+k}{m} x^{m+k}=\sum_{k=0}^\infty \left(-\frac{x}{4}\right)^k\binom{2k}{k}\sum_{m=0}^\infty \binom{m+k}{m} x^{m}$$
and here I don't know what to do next. Can anyone help me? Thanks in advice!
| This follows directly (more or less) from Vandermonde's identity:
\begin{align*}
\sum_{k=0}^n \binom{n}{k}\binom{-1/2}{k} &= \binom{n-1/2}{n} \\
&= \frac{(n-1/2)(n-3/2)\cdots(n-1/2-(n-1))}{n!}\\
&= \frac{(2n-1)(2n-3)\cdots (1)}{2^nn!} \\
&= 2^{-2n}\binom{2n}{n}.
\end{align*}
But also,
$$\binom{-1/2}{k} = \frac{(-1/2)(-1/2-1)\cdots(-1/2-(k-1)}{k!}
= (-1)^k\frac{1\cdot 3\cdot 5\cdots (2k-1)}{2^kk!}
= (-1)^k 2^{-2k}\binom{2k}{k}$$
and therefore
$$\sum_{k=0}^n \binom{n}{k}\binom{-1/2}{k} = \sum_{k=0}^n \binom{n}{k}(-1)^k2^{-2k}\binom{2k}{k}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the real solutions for the system: $ x^3+y^3=1$,$x^2y+2xy^2+y^3=2.$
Find the real solutions for the system:
$$\left\{
\begin{array}{l}
x^3+y^3=1\\
x^2y+2xy^2+y^3=2\\
\end{array}
\right.
$$
From a book with exercises for math contests. The solutions provided are: $(x,y)=(\dfrac{1}{\sqrt[3]{2}},\dfrac{1}{\sqrt[3]{2}})$ and $(\dfrac{1}{3^{\frac{2}{3}}},\dfrac{2}{3^{\frac{2}{3}}})$.
Working with the expressions I could find that an equivalent system is
$$\left\{
\begin{array}{l}
(x+y)(x^2-xy+y^2)=1\\
y(x+y)^2=2\\
\end{array}
\right.
$$
Developing these expressions I got stuck.
Hints and answers are appreciated. Sorry if this is a duplicate.
| Well, the second equation is a quadratic equation in $x$:
$$\text{a}\cdot x^2+\text{b}\cdot x+\text{c}=0\space\Longleftrightarrow\space x=\frac{-\text{b}\pm\sqrt{\text{b}^2-4\cdot\text{a}\cdot\text{c}}}{2\cdot\text{a}}\tag1$$
Now, in your example $\text{a}=\text{y},\text{b}=2\cdot\text{y}^2,\text{c}=\text{y}^3-2$, so we get:
$$x=\frac{-2\cdot\text{y}^2\pm\sqrt{\left(2\cdot\text{y}^2\right)^2-4\cdot\text{y}\cdot\left(\text{y}^3-2\right)}}{2\cdot\text{y}}=\pm\frac{\sqrt{2}}{\sqrt{\text{y}}}-\text{y}\tag2$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is it true that $\sum_{k=0}^m\binom{n-k}k$ outputs the $(n+1)$th Fibonacci number, where $m=\frac{n-1}2$ for odd $n$ and $m=\frac n2$ for even $n$?
Does $$\sum_{k=0}^m\binom{n-k}k=F_{n+1}$$ where $m=\left\{\begin{matrix}
\frac{n-1}{2}, \text{for odd} \,n\\
\frac n2, \text{for even} \,n
\end{matrix}\right.$ hold for all positive integers $n$?
Attempt:
I have not yet found a counterexample, so I will attempt to prove it.
$$\text{LHS} =\binom n0 + \binom{n-1}1+\binom{n-2}2+...+\left\{\begin{matrix}
\binom{1+(n-1)/2}{(n-1)/2}, \text{for odd} \,n\\
\binom{n/2}{n/2}, \text{for even} \,n
\end{matrix}\right.$$
Now using the identity that $\binom nk + \binom n{k+1}=\binom {n+1}{k+1}$, where $k$ is a positive integer, I find that $$\binom{n-1}1=\binom n1 - 1, \\ \binom {n-2}2=\binom n2-2\binom n1+3,\\ \binom {n-3}{3} =\binom n3 - 3\binom n2 + 6 \binom n1 - 10, \\ ...$$ This pattern suggests that the coefficients of $\binom{n-4}4$ will be square numbers, those of $\binom{n-5}5$ will be pentagonal numbers, etc. However, I cannot see a way to link these results to any Fibonacci identity.
Edit: @Jack D'Aurizio♢ has provided a very succinct proof to this, but is there a more algebraic method to show the equality?
| Here is more of an algebraic solution through generating functions. We have
\begin{align}
\sum_{n=0}^{\infty}\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n-k}{k}x^n
&=\sum_{k=0}^{\infty}\sum_{n=k}^{\infty}\binom{n}{k}x^{n+k+1}\\
&=\sum_{k=0}^{\infty}x^{2k+1}\sum_{n=k}^{\infty}\binom{n}{k}x^{n-k}\\
&=\sum_{k=0}^{\infty}x^{2k+1}\sum_{n=0}^{\infty}\binom{n+k}{k}x^{n}\\
&=\sum_{k=0}^{\infty}x^{2k+1}\frac{1}{(1-x)^{k+1}}\\
&=\frac{x}{1-x}\sum_{k=0}^{\infty}\left(\frac{x^2}{1-x}\right)^k\\
&=\frac{x}{1-x}\frac{1}{1-\frac{x^2}{1-x}}\\
&=\frac{x}{1-x-x^2}
\end{align}
where on the last line we arrived at the well known generating function for fibonacci numbers. So by equality of coefficients it follows
$$F_{n+1} = \sum_{k=0}^{\lfloor (n)/2\rfloor}\binom{n-k}{k}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Proving trigonometric identity: $ \frac{\sin (y+x)}{\sin (y-x)} = \frac{\tan y + \tan x}{\tan y - \tan x} $
Prove that
$$ \frac{\sin (y+x)}{\sin (y-x)} = \frac{\tan y + \tan x}{\tan y - \tan x} $$
This is my working -
$$\text{LHS} = \frac{\sin (y+x)}{\sin (y-x)} = \frac{\sin y \cos x + \cos y \sin x}{\sin y \cos x - \cos y \sin x} $$
I used the compound angle relations formula to do this step.
However , now I'm stuck.
I checked out the answer and the answer basically carried on my step by dividing it by
$ \frac{\cos x \cos y}{\cos x \cos y} $
If I'm not wrong , we cannot add in our own expression right? Because the question is asking to prove left is equal to right. Adding in our own expression to the left hand side will be not answering the question, I feel .
| Consider fractions that are equivalent to $2/3$.
$$\frac{2}{3}\cdot\frac{2}{2} = \frac{4}{6}$$
$$\frac{2}{3}\cdot\frac{3}{3} = \frac{6}{9}$$
$$etc.$$
Conclusion ... you're allowed to multiply by one.
Similarly, with the expression $\displaystyle\frac{\sin y\cos x+\cos y \sin x}{\sin y \cos x−\cos y \sin x}$, you're allowed to multiply by $\displaystyle\frac{\frac{1}{\cos y \cos x}}{\frac{1}{\cos y \cos x}}$ because $$\displaystyle\frac{\frac{1}{\cos y \cos x}}{\frac{1}{\cos y \cos x}} = 1$$
And just to make things mentally more clear, I may convert the original expression to a fraction of fractions as follows
$$\displaystyle\frac{\frac{\sin y\cos x+\cos y \sin x}{1}}{\frac{\sin y \cos x−\cos y \sin x}{1}}$$ before multiplying.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2547140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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What is the reduction formula for $\int x\cos^nx\,dx$ such that $n \in \mathbb{Z}^+ $? I am pretty bad at deducing reduction formulas but I tried and I reached
$$\frac{1}{2}I_{n-2} + \frac{1}{2}(x\sin x\cos^{n-1}x+\sin^{n-1}x+\frac{1}{n} \cos^nx)$$
Which I tried testing for $n = 2$ and comparing my answer to the integral-calculator's, the calculator's answer was: $$\frac{2x(\sin(2x)+x)+\cos(2x)}{8} + C$$
And mine was: $$\frac{16x^2 + 16x\sin2x + 8\cos2x + 32\sin x}{64}$$
I can tell that I got a couple of things right, like the existence of a $\sin2x$ multiplied by a $x^1$, the existence of an $x^2$ free of any trigonometric functions, and a $\cos2x$ free of any $x$ all three with matching coefficients to the actual answer.
And then I have that annoying extra $sinx$ which I can't really trace.
I know that comparing answers the way I am doing is kinda wrong, but I just couldn't help myself but see the similarities and feel close to the correct answer.
EDIT: here's the process as requested, I'll be skipping some parts because I need to sleep for uni.
$$\int x cos^nx\ dx = \int x\cos^{n-2}x\cos^2x\ dx = \int x\cos^{n-2}(1-\sin^2x)dx$$
then solving that using the fact that $\sin^2x = \frac{1}{2}(1-\cos2x)$ I reached, $$\frac{1}{2} I_{n-2} + \frac{1}{2}\int x\cos^{n-2}\cos2x\ dx$$
Then, solving $\int x\cos^{n-2}\cos2x\ dx$ by parts, $u = x\cos^{n-2}x$ and $dv = \cos2x\ dx$
$$\int x\cos^{n-2}\cos2x\ dx = \frac{x\cos^{n-2}x\sin2x}{2}-\frac{1}{2}\int (-(n-1)x\sin^{n-3}x+\cos^{n-2}x)\sin2x\ dx$$
then, solving that last integral I got: $-2\sin^{n-1}x-\frac{2\cos^nx}{n}$
plugging in and simplifying the $\sin2x$ using $\sin2x = 2\sin x\cos x$ I got
$$I_n = \frac{1}{2}I_{n-2} + \frac{1}{2}(x\sin x\cos^{n-1}x+\sin^{n-1}x+\frac{1}{n}\cos^nx)$$
| The first step is correct. You made a mistake during integration by parts
$$ v = \frac{\sin 2x}{2} = \sin x\cos x, \quad du = \big( \cos^{n-2}x - x(n-2)\cos^{n-3}x\sin x \big)\ dx $$
Then
$$ \begin{align} \int x\cos^{n-2}x\sin 2x\ dx &= x\cos^{n-1}x\sin x - \int \cos^{n-1}x\sin x\ dx \\ &\quad + (n-2)\int x \cos^{n-2}x \sin^2 x\ dx \\
&= x\cos^{n-1}x\sin x + \frac{\cos^n x}{n} + (n-2)I_{n-2} - (n-2)I_n \end{align} $$
Altogether
$$ \begin{align}
I_n &= \frac{1}{2}I_{n-2} + \frac{1}{2}\int x\cos^{n-2}x \sin 2x \ dx \\
&= \frac{n-1}{2}I_{n-2} - \frac{n-2}{2}I_n + \frac{1}{2}x\cos^{n-1}x\sin x + \frac{\cos^n x}{2n}
\end{align} $$
which gives
$$ I_n = \frac{n-1}{n}I_{n-2} + \frac{1}{n}x\cos^{n-1}x\sin x + \frac{\cos^n x}{n^2} $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is $\operatorname{ord}_{22}(5^6)$?
Find $\operatorname{ord}_{22}(5^6)$.
So, basically we want to find:$$\operatorname*{arg\,min}_k 5^{6k} \equiv 1\pmod {22}$$
I found that $5^5 \equiv 1\pmod {22}$ so I know that for $k=5$ we have $5^{6k} \equiv 1 \pmod {22}$. Therefore, $\text{ord}(5^6) \le 5$.
I guess that I could proceed with the somewhat tedious checking:
*
*$k=4$: $(5^6)^4 \equiv (5^4)^6 \equiv 5^4 \equiv 9 \pmod {22}$
*$k=3$ $(5^6)^3 \equiv (5^3)^6 \equiv 5^3 \equiv 15 \pmod {22}$
*$k=2$ $(5^6)^2 \equiv (5^2)^6 \equiv 5^2 \equiv 3 \pmod {22}$
*$k=1$ $(5^6)^1 \equiv (5^1)^6 \equiv 5 \pmod {22}$
So $\text{ord}_{22}(5^6) = 5$.
Questions:
*
*Is that what I'm expected to do? Or is there a simpler way? (I had to use a calculator or to tediously calculate it myself)
*Why can we get rid of the exponent when we do modular arithmetic? (I actually used it along the proof) i.e. $$x \equiv a \pmod m \iff x^b \equiv a \pmod m$$
| You have found that $5^5 \equiv 1 \pmod {22}$. So as $5^6 \equiv 5 \pmod {22}$. Then we must have $\text{ord}(5^6) = \text{ord}(5)$. But from the first equation we have that $\text{ord}(5) \mid 5$ and as $5$ is prime and $5^1 \not \equiv 1 \pmod {22}$ we must have that $\text{ord}(5^6) = \text{ord}(5) = 5$
Note that we used the fact that if $a^{k} \equiv 1\pmod {p}$, then $\text{ord}_p(a) \mid k$. This can be easily proven by the Division Algorithm. Let $k = \text{ord}_p(a)\cdot m + r$, where $0 \le r < \text{ord}_p(a)$. Then:
$$1 \equiv a^k \equiv (a^{\text{ord}_p(a)})^m \cdot a^r \equiv a^r \pmod {p} $$
So $r=0$, as $\text{ord}_p(a)$ is the smallest positive integer $t$ s.t. $a^t \equiv 1 \pmod {p}$ by definition
| {
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"url": "https://math.stackexchange.com/questions/2548808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integral question (Using Fundamental Theorem) Let's say I have the function
$$ x^2 = \int_{\tan(x)}^{y(x)}\frac{1}{\sqrt{2+t^2}}\,\mathrm dt $$
can I replace the y in the upper limit of the integral with $ x^2 $ ? If not, any steps on how to proceed? How could I find $ y'(0) $ ?
| By the fundamental theorem of calculus, we have
$$ F(x) := \int_{a}^{x} \frac{1}{\sqrt{2+t^2}} \,\mathrm{d}t
\implies F'(x) = \frac{1}{\sqrt{2+t^2}}. $$
Splitting the original integral, we obtain
$$ x^2 = \int_{\tan(x)}^{y(x)} \frac{1}{\sqrt{2+t^2}} \,\mathrm{d}t
= \int_{a}^{y(x)} \frac{1}{\sqrt{2+t^2}} \,\mathrm{d}t - \int_{a}^{\tan(x)} \frac{1}{\sqrt{2+t^2}} \,\mathrm{d}t
= F(y(x)) - F(\tan(x)). $$
Differentiating with respect to $x$ on both sides (making use of the chain rule), we get
\begin{align} 2x
&= \frac{\mathrm{d}}{\mathrm{d}x} F(y(x)) - \frac{\mathrm{d}}{\mathrm{d}x} F(\tan(x)) \\
&= F'(y(x)) y'(x) - F'(\tan(x)) \frac{\mathrm{d}}{\mathrm{d}x} \tan(x) \\
&= \frac{1}{\sqrt{2+y(x)^2}} y'(x) - \frac{1}{\sqrt{2+\tan(x)^2}} \sec(x)^2.
\end{align}
Note that this only makes sense if $y$ is differentiable.
EDIT: To address the edited version of the original question, substitute $x=0$ into the last displayed equation in order to obtain
\begin{align} &0 = \frac{1}{\sqrt{2+y(0)^2}} y'(0) - \frac{\sec(0)^2}{\sqrt{2+\tan(0)^2}} = \frac{1}{\sqrt{2+y(0)^2}} y'(0) - \frac{1}{\sqrt{2}} \\
&\qquad\implies \frac{1}{\sqrt{2+y(0)^2}} y'(0) = \frac{1}{\sqrt{2}} \\
&\qquad\implies y'(0) = \frac{\sqrt{2+y(0)^2}}{\sqrt{2}} = \sqrt{1 + \frac{1}{2} y(0)^2}.
\end{align}
EDIT: While GTonyJacobs has already addressed this, I would feel remiss in not completing the exercise:
The original integrand is strictly positive, from which it follows that
$$ 0 = F(0) = \int_{\tan(0)}^{y(0)} \frac{1}{\sqrt{2+t^2}} \,\mathrm{d} t
= \int_{0}^{y(0)} \frac{1}{\sqrt{2+t^2}} \,\mathrm{d} t \iff y(0) = 0, $$
as the integral of a positive function over a set of non-zero measure must be positive, and every non-degenerate interval has positive measure (this can fairly easily be rephrased in terms of the the Riemann integral, but I am feeling lazy). But we have already determined that
$$ y'(0) = \frac{\sqrt{2+y(0)^2}}{\sqrt{2}} = \sqrt{1 + \frac{1}{2} y(0)^2}, $$
and so
$$ y'(0) = \sqrt{1+\frac{1}{2}\cdot 0^2} = 1. $$
| {
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"answer_id": 1
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What is the real part of $ e^{e^{i \theta}} ?$ How to find the real part of the complex number (in Euler's form) $ z = e^{e^{i \theta } } $ ?
I got confused on how to proceed.
I am a beginner to complex numbers.
| For Euler's formula:
$z=x+iy=r(cos\theta+isin\theta)$ (It is easy part.)
$cos\theta=1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}-\frac{\theta^6}{6!}+...$
$sin\theta=\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-\frac{\theta^7}{7!}...$| Then
$cos\theta+isin\theta=1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}-\frac{\theta^6}{6!}+...+i(\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-\frac{\theta^7}{7!}...)$$\hspace{56pt}= 1+i\theta+(\frac{(i\theta)^2}{2!}+(\frac{(i\theta)^3}{3!}+(\frac{(i\theta)^4}{4!}...$ $\hspace{56pt}= e^{i\theta}$. (Because $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$)Hence $$z=e^{e^{i\theta}}=e^{cos\theta +isin\theta}=e^{cos\theta}[\cos (\sin\theta) +i\sin(\sin \theta)]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2550387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding GCD of two polynomials I am finding the GCD ($x^{24}-1, x^{15}-1$) using Euclidean Algorithm. So far I have
EDIT:
$x^{24}-1=x^9(x^{15}-1)$ w remainder $x^9-1$, then $x^{15}=x^6(x^9-1)$ with the remainder $x^6-1$, then $x^9-1=x^3(x^6-1)$ with remainder $X^3-1$, then $x^6-1=x^3(x^3-1)$ with remainder $x^3-1$. Then obviously $x^3-1=1(x^3-1)$ hence our GCD is $x^3-1$.
Thanks everyone for dealing with my dummy moment!
| $$x^6-1 = x^3(x^3 -1 ) + x^3 -1$$
$$x^3-1 = 1(x^3-1) + 0$$
$$\gcd{(x^{24}-1, x^{15}-1)}=\gcd{(x^3-1,0)}$$
Therefore, $\gcd{(x^{24}-1, x^{15}-1)}=x^3-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2553019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculate $\lim\limits_{n\rightarrow \infty} \int\limits_0^\infty \frac{n^2 \sin(x/n)}{n^3x + x(1 + x^3)} \, d x$ In preparation for finals, I am trying to calculate $$\lim_{n\rightarrow \infty} \int\limits_0^\infty \frac{n^2 \sin(x/n)}{n^3x + x(1 + x^3)} \, d x$$ with proof.
Here is my approach/what I have done so far:
If we can find a dominating function, we have $$\lim_{n\rightarrow \infty} \int\limits_0^\infty \frac{n^2 \sin(x/n)}{n^3x + x(1 + x^3)} \, d x = \int\limits_0^\infty \lim_{n\rightarrow \infty} \frac{n^2 \sin(x/n)}{n^3x + x(1 + x^3)} \, d x$$ by the Dominated Convergence Theorem. If we let $f_{n} = \frac{n^2 \sin(x/n)}{n^3x + x(1 + x^3)}$, then $f_{n}(x)$ converges to $0$ for all $x > 0$, which implies the limit is equal to 0 because the Dominated Convergence Theorem only requires a.e. convergence (so not having convergence at $x = 0$ is no issue). Operating under the assumption that dominating function exists, is this correct?
As far as finding a dominating function is concerned, we have
$$
|f_{n}| = \left| \frac{n^2 \sin(x/n)}{n^3x + x(1 + x^3)} \right| = \frac{|n^2 \sin(n/x)|}{n^3x + x(1 + x^3)} \leq \frac{n^2}{n^3x + x(1 + x^3)},
$$
which is where I get stuck. The two directions that seemed the most clear from here was to either
$$
\frac{n^2}{n^3x + x(1 + x^3)} \leq\frac{n^2}{n^3x} = \frac{1}{x} \quad \text{or} \quad \frac{n^2}{n^3x + x(1 + x^3)} \leq \frac{n^2}{x(1 + x^3)}.
$$
The former is not integrable and I cannot seem to grapple with the $n^2$ on the latter and sufficiently bound it. So my main question is how can I bound $|f_{n}|$?
| Use $|\sin(x/n)|\leq|x/n|=x/n$ for $x>0$, $\left|\dfrac{n^{2}\sin(x/n)}{n^{3}x+x(1+x^{3})}\right|\leq\dfrac{nx}{n^{3}x+x(1+x^{3})}\stackrel{\sqrt{ab}\leq \frac{1}{2}(a+b)}{\leq}\dfrac{nx}{2(n^{3}x^{2}(1+x^{3}))^{1/2}}=\dfrac{1}{2n^{1/2}(1+x^{3})}\leq\dfrac{1}{2(1+x^{3})}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2553167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Coefficient of $x^3$ in $(1-2x+3x^2-4x^3)^{1/2}$ The question is to find out the coefficient of $x^3$ in the expansion of $(1-2x+3x^2-4x^3)^{1/2}$
I tried using multinomial theorem but here the exponent is a fraction and I couldn't get how to proceed.Any ideas?
| Say $$(1-2x+3x^2-4x^3)^{1/2} =a+bx+cx^2+dx^3...$$
then
$$1-2x+3x^2-4x^3 =(a+bx+cx^2+dx^3...)^2$$
but $$(a+bx+cx^2+dx^3...)^2 = a^2+2abx+(2ac+b^2)x^2+2(ad+bc)x^3+...$$
So $a=\pm 1$.
If $a=1$ then $b=-1$ and $c=1$ and $d=-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2555399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Find the sum of the series of $\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\frac{1}{7\cdot 9}+...$ Find the sum of the series
$$\frac1{1\cdot3}+\frac1{3\cdot5}+\frac1{5\cdot7}+\frac1{7\cdot9}+\frac1{9\cdot11}+\cdots$$
My attempt solution:
$$\frac13\cdot\left(1+\frac15\right)+\frac17\cdot\left(\frac15+\frac19\right)+\frac1{11}\cdot\left(\frac19+\frac1{13}\right)+\cdots$$
$$=\frac13\cdot\left(\frac65\right)+\frac17\cdot \left(\frac{14}{45}\right)+\frac1{11}\cdot\left(\frac{22}{117}\right)+\cdots$$
$$=2\cdot\left(\left(\frac15\right)+\left(\frac1{45}\right)+\left(\frac1{117}\right)+\cdots\right)$$
$$=2\cdot\left(\left(\frac15\right)+\left(\frac1{5\cdot9}\right)+\left(\frac1{9\cdot13}\right)+\cdots\right)$$
It is here that I am stuck. The answer should be $\frac12$ but I don't see how to get it. Any suggestions?
Also, a bit more generally, are there good books (preferably with solutions) to sharpen my series skills?
| This is a classic telescoping series.
$$\frac1{n\cdot(n+2)}=\frac12\left(\frac1n-\frac1{n+2}\right)$$
Thus
$$\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\frac{1}{7\cdot 9}+\frac{1}{9\cdot 11}+\cdots$$
$$=\frac12\left(\frac11-\frac13+\frac13-\frac15+\frac15-\frac17+\frac17-\frac19+\frac19-\frac1{11}+\cdots\right)$$
$$=\frac12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2556569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Solve the Integral: $\int{\sqrt{8y-x^2}dx}$ I am trying to solve the following integral:
$$\int{\sqrt{8y-x^2}}dx$$
For which I don't understand clearly how to work with root values.
What I've tried:
$$
\int{\sqrt{8y-x^2}dx} = \int{(8y-x^2)^{\frac{1}{2}}dx} \\
= \frac{(8y-x^2)^{\frac{3}{2}}}{\frac{3}{2}}\int{(8y-x^2)dx} \\
= \frac{(8y-x^2)^{\frac{3}{2}}}{\frac{3}{2}}(-\frac{x^3}{3}) \\
= -\frac{2}{3}(8y-x^2)^{\frac{3}{2}}(\frac{x^3}{3}) \\
= -\frac{2}{9}(8y-x^2)^{\frac{3}{2}}x^3
$$
Is that correct?
| Note that if we use the substitution, $x=\sqrt{8y}\sin u$, then the integrand can be easily reduced to:
$$I=\int \sqrt{8y-x^2} dx = \sqrt{8y} \int \cos u \sqrt{8y-8y\sin^2 u} \,du=8y \int \cos^2 u \,du$$
I hope you can take it from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2559844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
The parabola $x^2=12y$ rolls without slipping around the parabola $x^2=-12y$ The parabola $x^2=12y$ rolls without slipping around the parabola $x^2=-12y$ then find the locus of focus of rolling parabola and also find the locus of vertex of rolling parabola
|
\begin{align}
\text{Static parabola: }f(x)&=-\tfrac1{12}x^2
,\\
f'(x)&=-\tfrac1{6}x
.
\end{align}
Tangent line at $x=x_t$:
\begin{align}
f_t(x,x_t)&=
f'(x_t)\,(x-x_t)+f(x_t)
\\
&=
-\tfrac1{6}x_t\,(x-x_t)-\tfrac1{12}x_t^2
\\
&=
-\tfrac16\,x_t\,x+\tfrac1{12}\,x_t^2
.
\end{align}
\begin{align}
|OD|&=\tfrac12x_t \quad\text{(why?)}
,\\
\tan\theta&=\tfrac16\,x_t
,\\
\sin\theta&=
\frac{\tan\theta}{\sqrt{1+\tan^2\theta}}
=\frac{x_t}{\sqrt{x_t^2+36}}
,\\
\cos\theta&=
\frac{6}{\sqrt{x_t^2+36}}
,\\
|OO'|&=2|OD|\sin\theta
=\frac{x_t^2}{\sqrt{x_t^2+36}}
,\\
O'_x&=|OO'|\,\cos(90^\circ-\theta)=|OB|\,\sin\theta
=\frac{x_t^3}{x_t^2+36}
,\\
O'_y&=|OO'|\,\sin(90^\circ-\theta)=|OB|\,\cos\theta
=
\frac{6x_t^2}{x_t^2+36}
.\\
\end{align}
Hence, the locus of vertex of rolling parabola
is defined parametrically as
\begin{align}
O'(x_t)&=
\left(\frac{t^3}{t^2+36}, \frac{6t^2}{t^2+36}\right)
,\quad t\in\mathbb{R}
.
\end{align}
This can be used as a base to derive the locus of the rolling focus.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2562290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Trying to derive expected value of triangularly distributed random variable I'm having trouble deriving the expected value of a triangularly distributed random variable with lower bound $a$ upper bound $b$ and mode $c$ for the case when the distribution is symmetric about the mode. The expected value in this case is known to be the mode $c$. The density function of the triangular distribution can be found here. In the symmetric case, this reduces to
$$ P(x) =
\begin{cases}
\frac{x-a}{\beta^2} & a \leq x < c \\[8pt]
\frac{-(x-b)}{\beta^2} & c \leq x \leq b
\end{cases}
$$
where the parameter $\beta$ is defined by
$$a = c-\beta$$
$$b = c + \beta$$
The expected value integral is then
$$E(x)=1/\beta^2\left(\int_a^cx(x-a)\:dx+\int_c^b -x(x-b)\:dx\right)$$
Which evaluates to
$$E(x)=1/\beta^2\left(\left[\frac{x^3}{3}-\frac{ax^2}{2}\right]^{x=c}_{x=a}-\left[\frac{x^3}{3}-\frac{bx^2}{2}\right]^{x=b}_{x=c}\right)$$
$$E(x)=1/\beta^2\left(\frac{c^3}{3}-\frac{ac^2}{2}+\frac{a^3}{6}-\left(-\frac{b^3}{6} - \frac{c^3}{3}+\frac{bc^2}{2}\right)\right)$$
$$E(x)=1/\beta^2\left(\frac{2c^3}{3}+\frac{a^3}{6}+\frac{b^3}{6}-\frac{ac^2}{2} -\frac{bc^2}{2}\right)$$
$$E(x)=\frac{1}{6\beta^2}\left(4c^3+a^3+b^3-3c^2(a+b)\right)$$
$$E(x)=\frac{1}{6\beta^2}\left(-2c^3+a^3+b^3\right)$$
Here is where I throw my hands in the air and turn to MSE. How the heck is this going to reduce to $c$??
| UPDATE
Finally you got
$$E(x)=\frac{1}{6\beta^2}\left(-2c^3+a^3+b^3\right)$$
but notice that since $a=c-\beta$ and $b=c+\beta$, we have
$$
\begin{split}
a^3+b^3-2c^3
&=(c-\beta)^3 + (c+\beta)^3-2c^3\\
&= \left(c^3-3c^2\beta+3c\beta^2-\beta^3\right)
+ \left(c^3+3c^2\beta+3c\beta^2+\beta^3\right)
- 2c^3\\
&= 3c\beta^2+3c\beta^2 \\
&= 6c\beta^2
\end{split}
$$
Thus
$$
E(x)=\frac{-2c^3+a^3+b^3}{6\beta^2} = \frac{6c\beta^2}{6\beta^2} = c,
$$
as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2563529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Use Viete's relations to prove the roots of the equation $x^3+ax+b=0$ satisfy $(x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2=-4a^3-27b^2$ Use Viete's relations to prove that the roots $x_1$, $x_2$, and $x_3$ of the equation $x^3+ax+b=0$ satisfy the identity $(x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2=-4a^3-27b^2$.
I know that viete's relations state that the roots $x_1$, $x_2$, and $x_3$ of the equation $x^3-px^2+qx-r=0$ have the property $p=x_1+x_2+x_3$, $q=x_1x_2+x_1x_3+x_2x_3$ and $r=x_1x_2x_3$.
My question is whether or not there is a way to do this without multiplying out $(x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2$ and showing that it factors into $4(x_1+x_2+x_3)^3+27(x_1x_2x_3)$ because that algebra involved in that looks like it will be nasty.
| If $b=0$, then the roots are $0,\pm\sqrt{-a}$ from which
$$(x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2=-4a^3$$
follows.
In the following, $b\not=0$.
We have, by Vieta's formulas,
$$x_1+x_2+x_3=0,\quad x_1x_2+x_2x_3+x_3x_1=a,\quad x_1x_2x_3=-b$$
So, we can have $$\begin{align}x_3(x_1-x_2)^2&=x_3((x_1+x_2)^2-4x_1x_2)\\\\&=x_3(-x_3)^2-4x_1x_2x_3\\\\&=x_3^3-4(-b)\\\\&=(-ax_3-b)-4(-b)\\\\&=-ax_3+3b\end{align}$$
Similarly, we get
$$x_1(x_2-x_3)^2=-ax_1+3b,\qquad x_2(x_3-x_1)^2=-ax_2+3b$$
It follows from these that
$$\begin{align}&x_1x_2x_3(x_1-x_2)^2(x_2-x_3)^2(x_3-x_1)^2\\\\&=x_3(x_1-x_2)^2x_1(x_2-x_3)^2x_2(x_3-x_1)^2\\\\&=(-ax_3+3b)(-ax_1+3b)(-ax_2+3b)\\\\&=-a^3x_1x_2x_3+3a^2b(x_1x_2+x_2x_3+x_3x_1)-9ab^2(x_1+x_2+x_3)+27b^3\\\\&=-a^3(-b)+3a^2b\cdot a-9ab^2\cdot 0+27b^3\end{align}$$
Dividing the both sides by $x_1x_2x_3=-b\not=0$ gives
$$(x_1-x_2)^2(x_2-x_3)^2(x_3-x_1)^2=-4a^3-27b^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2564158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Domain of $y=\arcsin\left({2x\sqrt{1-x^{2}}}\right)$ How do you find the domain of the function $y=\arcsin\left({2x\sqrt{1-x^{2}}}\right)$
I know that the domain of $\arcsin$ function is $[-1,1]$
So, $-1\le{2x\sqrt{1-x^{2}}}\le1$ probably?
or maybe $0\le{2x\sqrt{1-x^{2}}}\le1$ , since $\sqrt{1-x^{2}}\ge0$ ?
EDIT: So many people have answered that the domain would be $[-1,1]$ but my book says that its $[\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}}].$
Can anyone explain how are those restrictions made in the given formulas?
| You should see the domain is at first restricted by the square root $\sqrt{1-x^2}$ so that we must have $|x|\le 1$. Also note that $g(x) = 2x\sqrt{1-x^2}$ is an odd function, so we can just analyse for $x> 0$.
$$g'(x) = 2 \sqrt{1-x^2} -\frac{2x^2}{\sqrt{1-x^2}} = \frac{2-4x^2}{\sqrt{1-x^2}}$$
Giving maxima at $x = \tfrac{1}{\sqrt{2}}$. So $g(\tfrac{1}{\sqrt{2}}) = 1$ and $g(0) = 0$. So we have shown that $0 \le g(x) \le 1$ for $x \in [0,1]$. Similarly we show $0 \ge g(x) \ge -1$ for $x \in [-1,0]$
So required domain is $[-1,1]$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2565221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Convergence and divergence of an infinite series The series is
$$1 + \frac{1}{2}.\frac{x^2}{4} + \frac{1\cdot3\cdot5}{2\cdot4\cdot6}.\frac{x^4}{8} + \frac{1\cdot3\cdot5\cdot7\cdot9}{2\cdot4\cdot6\cdot8\cdot10}.\frac{x^6}{12}+... , x\gt0$$
I just stuck over the nth term finding and once I get nth term than I can do different series test but here I am unable to find the nth term of the given series please help me out of this.
The question is different as it contains x terms and its nth term will be totally different from marked as duplicate question
| Hint 1: Since $(4n-2)^2\gt(4n-1)(4n-3)$, we have
$$
\begin{align}
\frac{1\cdot3\cdots(4n-3)}{2\cdot4\cdots(4n-2)}
&\le\sqrt{\frac{1\cdot3\cdots(4n-3)}{3\cdot5\cdots(4n-1)}\cdot\frac{1\cdot3\cdots(4n-3)}{1\cdot3\cdots(4n-3)}}\\
&=\sqrt{\frac1{4n-1}}
\end{align}
$$
Hint 2: Since $(4n-3)^2\gt(4n-2)(4n-4)$, we have
$$
\begin{align}
\frac12\cdot\frac{3\cdot5\cdots(4n-3)}{4\cdot6\cdots(4n-2)}
&\ge\frac12\sqrt{\frac{4\cdot6\cdots(4n-2)}{4\cdot6\cdots(4n-2)}\cdot\frac{2\cdot4\cdots(4n-4)}{4\cdot6\cdots(4n-2)}}\\
&=\frac12\sqrt{\frac1{2n-1}}
\end{align}
$$
Thus,
$$
\frac12\sqrt{\frac1{2n-1}}\,\frac{x^{2n}}{4n}\le\frac{1\cdot3\cdots(4n-3)}{2\cdot4\cdots(4n-2)}\frac{x^{2n}}{4n}\le\sqrt{\frac1{4n-1}}\,\frac{x^{2n}}{4n}
$$
and therefore, since $4n-1\ge3n$ for $n\ge1$, we have
$$
\bbox[5px,border:2px solid #C0A000]{\frac{x^{2n}}{8\sqrt2\,n^{3/2}}\le\frac{1\cdot3\cdots(4n-3)}{2\cdot4\cdots(4n-2)}\frac{x^{2n}}{4n}\le\frac{x^{2n}}{4\sqrt3\,n^{3/2}}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2565939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
How do you go about factoring $x^3-2x-4$? How do you factor $x^3-2x-4$?
To me, this polynomial seems unfactorable.
But I check my textbook answer key, and the answer is $(x-2)(x^2+2x+2)$.
So I got to solve backward:
$$x(x^2-2)-4$$
And add some terms and subtract them later,
$$x(x^2+2x+2-2x-4)-4=0$$
$$x(x^2+2x+2)-4-2x^2-4x=0$$
$$x(x^2+2x+2)-2(x^2+2x+2)=0$$
But I would have thought of this way to factor if I didn't look at the answer key.
| It is better to go with Rational root theorem as suggested. I just want to write another way to see but clearly will not always work out that nicely.
\begin{equation}
x(x^2-2)-4
\end{equation}
instead of $x^2-2$, write $x^2-4$(because it is $(x-2)(x+2)$) and see if it is possible to factor:
\begin{equation}
x(x^2-4) + 2x -4 = x(x-2)(x+2) +2(x-2) = (x-2)(x^2+2x+2)
\end{equation}
There is a guess work here, but for example if you start with $x(x^2-2)-1$, you might wanna consider $x^2-1$ instead of $x^2-4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2567246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
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} |
What is the probability of getting at least one pair in Poker? Problem:
A 5-card hand is dealt from a perfectly shuffled deck of playing cards. What is the probability of each
of the hand has at least two cards with the same rank.
Answer:
By a rank, I mean a card like a $2$ or a king. The set of all poker hands is ${52 \choose 5}$. Let $p$ be the
probability we seek. I note that there are $13$ ranks of cards and for each rank there are $4$ cards.
\begin{eqnarray*}
p &=& \frac{ 13{ 4 \choose 2 }{ 50 \choose 3 } } { { 52 \choose 5 } } \\
{ 4 \choose 2} &=& \frac{4(3)(2)}{2} = 12 \\
{50 \choose 3} &=& \frac{50(49)(48)}{3(2)} = 25(49)(16) \\
%
{52 \choose 5} &=& \frac{52(51)(50)(49)(48)}{5(4)(3)(2)} = \frac{52(51)(10)(49)(48)}{4(3)(2)} \\
&=& \frac{52(51)(10)(49)(12)}{3(2)} = 52(51)(10)(49)(2) \\
p &=& \frac{ 13(12)(25)(49)(16) } { 52(51)(10)(49)(2) } = \frac{ 13(12)(25)(49)(8) } { 52(51)(10)(49) } \\ &=& \frac{ 13(6)(25)(8) } { 52(51)(5) } = \frac{ 13(3)(25)(8) } { 26(51)(5) } \\ &=& \frac{ 13(3)(5)(8) } { 26(51) } \\ &=& \frac{ 1560 } {1586 } \\
\end{eqnarray*}
I am fairly sure that my answer is wrong but I do not understand where I went wrong. I am hoping somebody can tell me where I went wrong.
Thanks,
Bob
| The denominator is correct. So the numerator must be wrong.
The probability of not getting a pair is:
$$
\dfrac{48}{51} \dfrac{44}{50}\dfrac{40}{49}\dfrac{36}{48}\approx 0.507
$$
It's complement is $0.493$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2568067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Sums of $5$th and $7$th powers of natural numbers: $\sum\limits_{i=1}^n i^5+i^7=2\left( \sum\limits_{i=1}^ni\right)^4$? Consider the following:
$$(1^5+2^5)+(1^7+2^7)=2(1+2)^4$$
$$(1^5+2^5+3^5)+(1^7+2^7+3^7)=2(1+2+3)^4$$
$$(1^5+2^5+3^5+4^5)+(1^7+2^7+3^7+4^7)=2(1+2+3+4)^4$$
In General is it true for further increase i.e.,
Is
$$\sum_{i=1}^n i^5+i^7=2\left( \sum_{i=1}^ni\right)^4$$ true $\forall $ $n \in \mathbb{N}$
| The formula is already true for $n=1,2,....,5$ and we know that
$$ \sum_{i=1}^ni= \frac{n\left(n+1\right)}{2}$$
Assume
$$\sum_{i=1}^n i^5+i^7=2\left( \sum_{i=1}^ni\right)^4 =\frac{n^4\left(n+1\right)^4}{8}$$
then,
$$\begin{align}\sum_{i=1}^{n+1} i^5+i^7&=\sum_{i=1}^{n} i^5+i^7 +(n+1)^5 +(n+1)^7\\&=\color{blue}{\frac{n^4\left(n+1\right)^4}{8}}+(n+1)^5 +(n+1)^7
\\&=\color{blue}{\frac{n^4\left(n+1\right)^4}{8}}+(n+1)^4\left[n+1 +(n+1)^3 \right]
\\&=(n+1)^4\left( \frac{n^4}{8} +n+1 +\color{red}{n^3+3n^2+3n+1}\right)
\\&=(n+1)^4\left( \frac{n^4}{8}+ n^3+3n^2+4n+2\right)
\\&=\frac{(n+1)^4}{8}\left( n^4+ \color{blue}{4}\cdot\color{red}{2}\cdot n^3+\color{blue}{6}\cdot\color{red}{2^2}\cdot n^2+\color{blue}{4}\cdot\color{red}{2^3}\cdot n+\color{red}{2^4}\right)
\\&=\color{blue}{\frac{\left(n+1\right)^4\left(n+2\right)^4}{8}=2\left( \sum_{i=1}^{n+1}i\right)^4}\end{align}$$
which prove that the formula is true
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2568157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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} |
How can I calculate $\lim_{x \to 0}\frac {\cos x- \sqrt {\cos 2x}×\sqrt[3] {\cos 3x}}{x^2}$ without L'Hôpital's rule? How can I calculate following limit without L'Hôpital's rule
$$\lim_{x \to 0}\frac {\cos x- \sqrt {\cos 2x}×\sqrt[3] {\cos 3x}}{x^2}$$
I tried L'Hôpital's rule and I found the result $2$.
| $$\cos x=1-\frac{x^2}2+O(x^4),$$
$$\cos nx=1-\frac{n^2x^2}2+O(x^4),$$
$$(\cos nx)^{1/n}=1-\frac{nx^2}2+O(x^4),$$
$$(\cos 2x)^{1/2}(\cos 3x)^{1/3}
=\left(1-\frac{2x^2}{2}+O(x^4)\right)\left(1-\frac{3x^2}{2}+O(x^4)\right)
=1-\frac{5x^2}{2}+O(x^4)$$
and so
$$\cos x-(\cos 2x)^{1/2}(\cos 3x)^{1/3}=2x^2+O(x^4)$$
so the limit you seek is $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2568920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
How to compute $\sum^k_{i=0}{k\choose i}\frac{\prod_{j=0}^{i-1}(x+jm)\prod_{j=0}^{k-i-1}(y+jm)}{\prod_{j=0}^{k-1}(x+y+jm)}$? This basic course on probability and statistics is the first course where I feel like a total idiot... especially since I've already forgotten much from the basic courses at discrete mathematics and analysis. Sigh...
The task from a former test:
A HDD contains $x+y$ infected programs; $x$ are infected by malware $X$ and $y$ are infected by malware $Y$. The user runs random programs. Each time a program infected by malware $X$ is run $m$ uninfected programs become infected by malware $X$; same for malware $Y$. Compute the probability that when the $k$th infected program is run it is infected by malware $X$.
Let $X_1=1$ iff the first infected program is infected by $X$, similary let's define $Y_1$, $X_2$, etc.
We have:
$\operatorname{P}(X_1)=\frac{x}{x+y}$
$\operatorname{P}(Y_1)=\frac{y}{x+y}$
$\operatorname{P}(X_2)=\operatorname{P}(X_1)\operatorname{P}(X_2|X_1)+\operatorname{P}(Y_1)\operatorname{P}(X_2|Y_1)=\frac{x}{x+y}\frac{x+m}{x+m+y}+\frac{y}{x+y}\frac{x}{x+y+m}$
$P(Y_2)=\operatorname{P}(X_1)\operatorname{P}(Y_2|X_1)+\operatorname{P}(Y_1)\operatorname{P}(Y_2|Y_1)=\frac{x}{x+y}\frac{y}{x+m+y}+\frac{y}{x+y}\frac{y+m}{x+y+m}$
$\operatorname{P}(X_3)=\frac{x}{x+y}\frac{x+m}{x+m+y}\frac{x+2m}{x+2m+y}+\frac{y}{x+y}\frac{x}{x+y+m}\frac{x+m}{x+m+y+m}+\frac{x}{x+y}\frac{y}{x+m+y}\frac{x+m}{x+m+y+m}+\frac{y}{y+m}\frac{y+m}{x+y+m}+\frac{x}{x+y+2m}$
More generally:
$\operatorname{P}(X_k)=\sum^k_{i=0}{k\choose i}\frac{\prod_{j=0}^{i-1}(x+jm)\prod_{j=0}^{k-i-1}(y+jm)}{\prod_{j=0}^{k-1}(x+y+jm)}$
This is madness for me. That's not even binomial theorem, albeit it's a little bit similar.
Am I doing something wrong? Or how am I supposed to get a closed formula from this sum?
| You start off good:
$\operatorname{P}(X_1)=\frac{x}{x+y}$
$\operatorname{P}(Y_1)=\frac{y}{x+y}$
$\operatorname{P}(X_2)=\operatorname{P}(X_1)\operatorname{P}(X_2|X_1)+\operatorname{P}(Y_1)\operatorname{P}(X_2|Y_1)=\frac{x}{x+y}\frac{x+m}{x+m+y}+\frac{y}{x+y}\frac{x}{x+y+m}$
But you can simplify here:
$$\frac{x}{x+y}\frac{x+m}{x+m+y}+\frac{y}{x+y}\frac{x}{x+y+m} = $$
$$\frac{1}{x+y}\left(x\cdot\frac{x+m}{x+y+m}+y\cdot\frac{x}{x+y+m} \right)= $$
$$\frac{1}{x+y}\left(x\cdot\frac{x+m}{x+y+m}+x\cdot\frac{y}{x+y+m} \right) =$$
$$\frac{x}{x+y}\left(\frac{x+y+m}{x+y+m}\right) = $$
$$\frac{x}{x+y}$$
And thus by induction, $X_1 = X_k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2573436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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} |
find the range of the function : $y=(3\sin 2x-4\cos 2x)^2-5$ find the range of the function :
$$y=(3\sin 2x-4\cos 2x)^2-5$$
My try :
$$y=9\sin^22x+16\cos^22x-24\sin 2x\cos 2x-5\\y=9+7\cos^22x-12\sin4x-5$$
now what do I do؟
| For real $x,$ $$(a\cos2x-b\sin2x)^2\ge0$$ the equality occurs if $$a\cos2x-b\sin2x=0\iff\dfrac{\cos2x}b=\dfrac{\sin2x}a=\pm\sqrt{\dfrac1{b^2+a^2}}$$
Using Brahmagupta-Fibonacci Identity,
$$(a\cos2x-b\sin2x)^2+(b\cos2x+a\sin2x)^2=(a^2+b^2)(?)$$
$$(a\cos2x-b\sin2x)^2=a^2+b^2-(b\cos2x+a\sin2x)^2\le a^2+b^2$$
the equality occurs if $$b\cos2x+a\sin2x=0\iff\dfrac{\cos2x}a=\dfrac{\sin2x}{-b}=\pm\sqrt{\dfrac1{b^2+a^2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2573749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find $x$ given $\sqrt{x+14-8\sqrt{x-2}}$ + $\sqrt{x+23-10\sqrt{x-2}} = 3$ If $\sqrt{x+14-8\sqrt{x-2}}$ + $\sqrt{x+23-10\sqrt{x-2}} = 3$, then what is the value of $x$?
Is there an easy way to solve such equation, instead of squaring on both sides and replacing $\sqrt{x-2}$ with a different variable?
| $$\sqrt{x+14-8\sqrt{x-2}} +\sqrt{x+23-10\sqrt{x-2}} = 3$$
$$\sqrt{(\sqrt{x-2}-4})^2+\sqrt{(\sqrt{x-2}-5)^2}=3$$
Then
$$|\sqrt{x-2}-4|+|\sqrt{x-2}-5|=3$$
Let $\sqrt{x-2}\ge5 (x\ge27)$. Then
$$\sqrt{x-2}-4+\sqrt{x-2}-5=3$$
$$\sqrt{x-2}=6$$
Let $\sqrt{x-2}\le4 (2\le x\le18)$. Then
$$4-\sqrt{x-2}-\sqrt{x-2}+5=3$$
$$\sqrt{x-2}=3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2574220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Invertible 4x4 matrix $$
\begin{pmatrix}
5 & 6 & 6 & 8 \\
2 & 2 & 2 & 8 \\
6 & 6 & 2 & 8 \\
2 & 3 & 6 & 7 \\
\end{pmatrix}
$$
Is this matrix invertible? I would like to show that it is invertible but first I should find the det(Matrix) which should not be equal to zero. To find the determinant, maybe the best idea is to use row operations and find an upper triangular of zeroes and then multiply the numbers on the diagonal to get the determinant. I have been doing some row operations and get this:
$$
\begin{pmatrix}
5 & 6 & 6 & 8 \\
0 & -1 & -4 & 1 \\
0 & 0 & 2 & 6 \\
-1 & 0 & 0 & -12 \\
\end{pmatrix}
$$
I just need to get rid of the -1 on the last row. But I am stuck. Thank you for your assistance.
| A combination of Gaussian elimination and a well-known trick readily gives the answer.
$$\det
\begin{pmatrix}
5 & 6 & 6 & 8 \\
2 & 2 & 2 & 8 \\
6 & 6 & 2 & 8 \\
2 & 3 & 6 & 7 \\
\end{pmatrix}
=\det\begin{pmatrix}
5 & 6 & 6 & 2 \\
2 & 2 & 2 & 6 \\
6 & 6 & 2 & 2 \\
2 & 3 & 6 & 4 \\
\end{pmatrix}
=4\det\begin{pmatrix}
5 & 6 & 6 & 2 \\
1 & 1 & 1 & 3 \\
3 & 3 & 1 & 1 \\
2 & 3 & 6 & 4 \\
\end{pmatrix}
\\[0.5cm]=4\det\begin{pmatrix}
5 & 6 & 6 & 2 \\
1 & 1 & 1 & 3 \\
2 & 2 & 0 & -2 \\
2 & 3 & 6 & 4 \\
\end{pmatrix}=8\det\begin{pmatrix}
5 & 6 & 6 & 2 \\
1 & 1 & 1 & 3 \\
1 & 1 & 0 & -1 \\
2 & 3 & 6 & 4 \\
\end{pmatrix}
$$
and the last determinant is an odd integer, hence the original matrix is invertible.
$$ \det\begin{pmatrix}
5 & 6 & 6 & 2 \\
1 & 1 & 1 & 3 \\
1 & 1 & 0 & -1 \\
2 & 3 & 6 & 4 \\
\end{pmatrix}\equiv
\det\begin{pmatrix}
1 & 0 & 0 & 0 \\
1 & 1 & 1 & 1 \\
1 & 1 & 0 & 1 \\
0 & 1 & 0 & 0 \\
\end{pmatrix}\equiv 1\pmod{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2574905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit of $h(x)$ as $x \to \infty$ Let $h(x) = \sqrt{x + \sqrt{x}} - \sqrt{x}$ , find $\lim_{x \to \infty } h(x)$ . I've tried substitution , multiplying by conjugate and l'hospital's rule but didn't work .
| Multiplying and dividing by the "conjugate" $\sqrt{x + \sqrt{x}} + \sqrt{x}$ is a good strategy!
Note that for $x>0$,
$$ \sqrt{x + \sqrt{x}} - \sqrt{x}=\frac{ (x + \sqrt{x}) - x}{ \sqrt{x + \sqrt{x}} + \sqrt{x}}=\frac{\sqrt{x}}{\sqrt{x} \sqrt{1 + \frac{1}{\sqrt{x}}} + \sqrt{x}}=\frac{ 1}{ \sqrt{1 + \frac{1}{\sqrt{x}}} + 1}.$$
Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving whether the series $\sum_{n=1}^\infty \frac{(-1)^n}{n-(-1)^n}$ converges. I've updated my proof to be complete now, edited for proof-verification!
We know that for the partial sums with even an uneven terms, the following holds:
$S_{2N}=\sum_{n=1}^{2N} \frac{(-1)^n}{n-(-1)^n} = -\frac{1}{2} + \frac{1}{1} -\frac{1}{4} +\frac{1}{3} -\frac{1}{6} +\frac{1}{5}-\dots+\frac{1}{2N-1}$
$= \frac{1}{2\times1} +\frac{1}{4\times3} + \frac{1}{6\times5}+\dots+\frac{1}{2N(2N-1)} = \sum_{n=1}^{2N} \frac{1}{2n(2n-1)}$
We may rewrite the series in pairs as we know it will have an even amount of terms.
$S_{2N+1} = \sum_{n=1}^{2N+1} \frac{(-1)^n}{n-(-1)^n} = -\frac{1}{2} + \frac{1}{1} -\frac{1}{4} +\frac{1}{3}-\dots+\frac{1}{2N-1} -\frac{1}{2N+2}$
$=\frac{1}{2\times1} +\frac{1}{4\times3} + \frac{1}{6\times5}+\dots+\frac{1}{2N(2N-1)} - \frac{1}{2N+2} = \sum_{n=1}^{2N} \frac{1}{2n(2n-1)}-\frac{1}{2N+2}$
As $n\in\mathbb{N}$, we know that $n\geq1$ so:
$n\geq1 \iff 3n\geq3 \iff 3n^2\geq3n \iff 3n^2-3n\geq0 \iff 4n^2-2n \geq n^2+n$
So: $2n(2n-1)\geq n(n+1) \iff \frac{1}{2n(2n-1)}\leq \frac{1}{n(n+1)}$ for all $n\geq1$.
As the series of the latter sequence converges, we can conclude, by the comparison, test that the series $\sum_{n=1}^{2N} \frac{1}{2n(2n-1)}$ converges.
Suppose it converges to $s$, then we know $^{\lim S_{2N}}_{N\to\infty} = s$ and thus $\lim_{N\to\infty}[S_{2N+1}] = s - (\lim_{N\to\infty}[\frac{1}{2N+2}]) = s-0 = s.$ As the partial sums ending with even and uneven terms both converge to the same limit, the series $\sum_{n=1}^\infty \frac{(-1)^n}{n-(-1)^n}$ converges. $\tag*{$\Box$}$
| $$s =\sum_{n=1}^\infty \frac{(-1)^n}{n-(-1)^n} = $$
$$ -\frac{1}{2} + \frac{1}{1} - \frac{1}{4} + \frac{1}{3} - \frac{1}{6} + \frac{1}{5} - \cdots $$
$$\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+\cdots\right) - \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots\right)$$
$$ \sum_{k =1}^\infty \left(\frac{1}{2k-1}\right) - \sum_{k =1}^\infty \left(\frac{1}{2k}\right)$$
$$ \sum_{k =1}^\infty \left(\frac{1}{2k-1} - \frac{1}{2k}\right)$$
$$\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}$$
$$s = \ln(2)$$
And the series converges to the natural logarithm.
See the special case of the Mercator series
$$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots$$
Setting $x=1$ in the Mercator series yields the Alternating harmonic series
$$\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}=\ln(2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2575967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
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Find the limit of $\log\Big(\frac{(n^2+n)!}{n^2!n^{2n}}\Big)$ I have to prove that
$$\Big[\log\Big(1+\frac 1{n^2}\Big)+ \log\Big(1+\frac 2{n^2}\Big)+...+ \log\Big(1+\frac n{n^2}\Big)\Big] \to \frac 12$$
I know that $$\log\Big(1+\frac 1{n^2}\Big)+ \log\Big(1+\frac 2{n^2}\Big)+...+ \log\Big(1+\frac n{n^2}\Big)= \\ =\log\Big(\frac {n^2+1}{n^2}\Big)+ \log\Big(\frac {n^2+2}{n^2}\Big)+...+ \log\Big(\frac {n^2+n}{n^2}\Big)=\\ =\log\Big(n^2+1\Big)+ \log\Big(n^2+2\Big)+...+ \log\Big(n^2+n\Big)-2n\log n$$
Doing some manipulation we get:
$$\lim_{n \to +\infty}\log\Big(\frac{(n^2+n)!}{n^2!n^{2n}}\Big)=\frac 12$$
How can I prove this limit?
| Via Stirling's formula,
$$
\begin{align}
\frac{(n^2+n)!}{n^2!n^{2n}} &= \frac{\sqrt{2\pi(n^2+n)}\left(\frac{n^2+n}{e}\right)^{n^2+n}}{\sqrt{2\pi n^2}\left(\frac{n^2}{e}\right)^{n^2}n^{2n}}(1+o(1)) \\
&= e^{-n}\left(1+\frac{1}{n}\right)^{n^2+n}(1+o(1))
\end{align}
$$
Taking logarithms,
$$
\begin{align}
\log\frac{(n^2+n)!}{n^2!n^{2n}} &= -n+(n^2+n)\log{\left(1+\frac{1}{n}\right)}+\log(1+o(1)) \\
&=-n+(n^2+n)\left(\frac{1}{n}-\frac{1}{2n^2}+o\left(\frac{1}{n^2}\right)\right)+o(1) \\
&=\frac{1}{2}+o(1).
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2576784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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Convergence/absolute convergence of $\sum_{n=1}^\infty \left(\sin \frac{1}{2n} - \sin \frac{1}{2n+1}\right)$
Does the following sum converge? Does it converge absolutely?
$$\sum_{n=1}^\infty \left(\sin \frac{1}{2n} - \sin
\frac{1}{2n+1}\right)$$
I promise this is the last one for today:
Using Simpson's rules:
$$\sum_{n=1}^\infty \left(\sin \frac{1}{2n} - \sin
\frac{1}{2n+1}\right) = \sum_{n=1}^\infty 2\cos\frac{4n+1}{4n² + 2n}\sin\frac{1}{8n² + 4n}$$
Now, $$\left|2\cos\frac{4n+1}{4n² + 2n}\sin\frac{1}{8n² + 4n}\right| \leq \frac{2}{8n² + 4n}$$
hence by the comparison test, the series converges absolutely, and hence it also converges. Is this correct?
| It's simpler still with equivalents:
$$\sin \frac{1}{2n} - \sin \frac{1}{2n+1}=2\sin\frac1{4n(2n+1)}\underbrace{\cos\frac{4n+1}{4n(2n+1)}}_{\substack{\downarrow\\\textstyle1}}l\sim_\infty2\,\frac1{4n(2n+1)}\cdot 1\sim_\infty\frac1{4n^2},$$
which converges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2580209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
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Find: $\lim_{x\to\infty} \frac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}}.$
Find: $\displaystyle\lim_{x\to\infty} \dfrac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}}.$
Question from a book on preparation for math contests. All the tricks I know to solve this limit are not working. Wolfram Alpha struggled to find $1$ as the solution, but the solution process presented is not understandable. The answer is $1$.
Hints and solutions are appreciated. Sorry if this is a duplicate.
| If you factor out a $\sqrt{x}$ term from the denominator one has
\begin{align*}
\lim_{x \to \infty} \frac{\sqrt{x}}{\sqrt{x + \sqrt{x + \sqrt{x}}}} &= \lim_{x \to \infty} \frac{\sqrt{x}}{\sqrt{x} \sqrt{1 + \frac{1}{x} \sqrt{x + \sqrt{x}}}}\\
&= \lim_{x \to \infty} \frac{1}{\sqrt{1 + \sqrt{\frac{1}{x} + \frac{1}{x^{3/2}}}}}\\
&= 1.
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Is there an infinite number of numbers like $1600$? My reputation is at this moment at $1600$.
I did some experimenting with $1600$ and obtained the following:
Evidently, it is a perfect square $1600=40^2$
Also, it is a hypothenuse of a Pythagorean integer-triple triangle $1600=40^2=32^2+24^2$.
Also, it can be written as the sum of four non-zero squares $1600=20^2+20^2+20^2+20^2$
So, $1600$ is a perfect square, a hypothenuse of a Pythagorean integer-triple (so can be written as a sum of two non-zero squares), and a sum of four non-zero squares.
Is there an infinite number of numbers like $1600$? Can you find some more?
Edit 1: It is also a sum of $4$ non-zero positive cubes: $1600=8^3+8^3+8^3+4^3$
Edit 2: It is also a sum of powers from $1$ to $4$, as we see $1600=7^1+9^2+6^3+6^4$
| Yes, there are an infinite number of them. Given your result for $1600$ we can say that $1600k^2$ is another one. It is a square, it is $(32k)^2+(24k)^2$ and it is $4\cdot (20k)^2$. We can use the parameterization of Pythagorean triples to get others. If you choose $m,n$ relatively prime and of opposite parity, they generate a primitive Pythagorean triple $m^2-n^2, 2mn, m^2+n^2$, so if we choose $m,n$ as legs of a Pythagorean triangle the hypotenuse of that triangle will be a square. As an example, let $m=4,n=3$, which gives the triangle $7,24,25$. The number $25$ is a square, the sum of two squares, and the sum of four squares as $16+4+4+1$. Again you can multiply it by any square.
With the edit, we can still say that $1600k^6$ is a solution by the same reasoning as above plus $1600k^6=3\cdot(8k^2)^3+(4k^2)^3$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Maximum value of 'a' for Given Condition Let $f(x) $ be a positive differentiable function on $[0,a]$ such that $f(0)=1$ and $f(a)=3^{\frac{1}{4}}$. If $f'(x)\ge(f(x))^3 +(f(x))^{-1} $,what is the maximum value of a?
$ a) \frac{\pi}{12} b) \frac{\pi}{36}c) \frac{\pi}{24} d) \frac{\pi}{48}$
From the given conditions i was able to make out that the function is strictly increasing in the interval,but that is about all. And i have no idea how to bring $\pi$ into the mix
I know these kind of questions arent too popular here on SE, but i think this question might be good enough
| On the one hand, because$$
f'(x) \geqslant (f(x))^3 + \frac{1}{f(x)}, \quad \forall 0 < x < a
$$
then $f(x) \neq 0$ for $0 < x < a$. Since $f$ is continuous on $[0, a]$ and $f(0) > 0$, then $f(x) > 0$ for $0 < x < a$. Now,\begin{align*}\def\d{\mathrm{d}}
a &= \int_0^a \,\d x \leqslant \int_0^a \frac{f'(x)}{(f(x))^3 + \frac{1}{f(x)}} \,\d x = \int_0^a \frac{f(x)}{(f(x))^4 + 1} f'(x) \,\d x\\
&= \frac{1}{2} \int_{(f(0))^2}^{(f(a))^2} \frac{\d ((f(x))^2)}{((f(x))^2)^2 + 1} = \frac{1}{2} \int_{(f(0))^2}^{(f(a))^2} \frac{\d u}{u^2 + 1}\\
&= \frac{1}{2} \arctan u \Biggr|_{(f(0))^2}^{(f(a))^2} = \frac{1}{2} (\arctan((f(a))^2) - \arctan((f(0))^2))\\
&= \frac{1}{2} (\arctan \sqrt{3} - \arctan 1) = \frac{1}{2} \left(\frac{π}{3} - \frac{π}{4}\right) = \frac{π}{24}.
\end{align*}
On the other hand, if $\displaystyle f(x) = \sqrt{\tan\left(2x + \frac{π}{4}\right)} \, \left(x \in \left[0, \frac{π}{8}\right]\right)$, then\begin{gather*}
2f(x) f'(x) = ((f(x))^2)' = \left(\tan\left(2x + \frac{π}{4}\right)\right)' = \frac{2}{\displaystyle\cos^2\left(2x + \frac{π}{4}\right)},\\
(f(x))^4 + 1 = \left(\tan\left(2x + \frac{π}{4}\right)\right)^2 + 1 = \frac{1}{\displaystyle\cos^2\left(2x + \frac{π}{4}\right)}.
\end{gather*}
Therefore,$$
f(x) f'(x) = (f(x))^4 + 1 \Longrightarrow f'(x) = (f(x))^3 + \frac{1}{f(x)}.
$$
Note that $\displaystyle f(0) = \sqrt{\tan \frac{π}{4}} = 1$, $\displaystyle f\left(\frac{π}{24}\right) = \sqrt{\tan \frac{π}{3}} = 3^{\frac{1}{4}}$, so $\displaystyle a = \frac{π}{24}$ is the maximum.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is it true that $n = 2t^2-2$ for even $t$ is a congruent number? This post asks for $m$ such that the simultaneous Pythagorean triples,
$$a^2+m^2b^2 = c^2\\b^2+c^2 = d^2\tag1$$
have solutions. Will Jagy found an infinite family given by,
$$m = 2t^2-2 = 0, 6, 16, 30, 48, 70, 96, 126, \dots$$
where,
$$\begin{aligned}
a &= -1 + 9 t^2 - 12 t^4 + 4 t^6\\ b &= -2 t + 4 t^3\\c &= -1 + t^2 - 4 t^4 + 4 t^6\\ d &= 1 + t^2 - 4 t^4 + 4 t^6
\end{aligned}$$
The values $m=6,30,70$ were faintly familiar, as I had posted about congruent numbers before. (See this and this.) A number $n$ is congruent if there is a solution to the simultaneous,
$$p^2 + nq^2 = r^2\\
p^2 - nq^2 = s^2\tag2$$
Q1: Is it true that an infinite family of congruent numbers is given by,
$$n = 2(2v)^2-2 = 6,30,70,126,\dots$$
P.S. A003273 gives a list of congruent numbers $N<10000$ and all $n$ of that form are there.
Q2: If indeed true, what is the connection between systems $(1)$ and $(2)$?
| (This is a partial answer.)
After some persistence and effort, I managed to find a partial answer. It can be proven that
$$n = 2t^2-2$$
is a congruent number for infinitely many $t$ (odd or even).
Proof: If,
$$n = 2(v^2\pm3)^2-2$$
then,
$$p^2+nq^2=r^2\\p^2-nq^2 = s^2$$
has the simple solution,
$$\begin{aligned}p &=v^4\pm4v^2+8
\\q &=2v\end{aligned}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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we need to compute determinant of matrix. If$$
A=\begin{pmatrix}-2 & 9 & -1 \\ 8 & 0 & 7\end{pmatrix}, \quad B=\begin{pmatrix}1 & 5 \\ 6 & 3 \\ 2 & 4\end{pmatrix},
$$
find $|AB|$ without finding $AB$.
| By the Cauchy-Binet formula, we have
\begin{align*}
det(AB) &= \begin{vmatrix}
-2 & 9\\
8 & 0
\end{vmatrix}\begin{vmatrix}
1 & 5\\
6 & 3
\end{vmatrix} + \begin{vmatrix}
9 & -1\\
0 & 7
\end{vmatrix}\begin{vmatrix}
6 & 3\\
2 & 4
\end{vmatrix} + \begin{vmatrix}
-2 & -1\\
8 & 7
\end{vmatrix}\begin{vmatrix}
1 & 2\\
5 & 4
\end{vmatrix}\\
\end{align*}
I think you could continue from here.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Prove that two polynomials are constants if $P(x^2+x+1)=Q(x^2-x+1)$
Let $P$ and $Q$ be two polynomials with complex coefficients. It is known that $P(x^2+x+1)=Q(x^2-x+1)$. How can I prove that $P$ and $Q$ are constants?
I evaluated at various points to deduce $Q(3) = P(1) = Q(1) = P(3)$ but I don't know how to conclude from that.
| Set $x=0$ and you get $P(1)=Q(1)$
With $x=1-0=1$ you get $P(3)=Q(1)$ and $x=-1-0=-1$ gives $P(1)=Q(3)$
With $x=-1-1=-2$ you get $P(3)=Q(7)$ and with $x=1-(-1)=2$ there is $Q(3)=P(7)$
Notice that $x^2+x+1=(-1-x)^2+(-1-x)+1$ and $x^2-x+1=(1-x)^2-(1-x)+1$.
So you can use vieta jumping with the functional relation to show (for $n\in \mathbb Z$)$$\dots P(7)=Q(3)=P(1)=Q(1)=P(3)=Q(7)= \dots$$
But a polynomial $P$ which takes the same value $c$ at an infinite number of points is constant, since then $P-c$ has infinitely many roots hence $P-c = 0$.
To be more explicit $$P(n^2+n+1)=Q(n^2-n+1)=Q((1-n)^2-(1-n)+1)=P((1-n)^2+(1-n)+1)=$$$$[=P(n^2-3n-3)]=P((-1-(1-n))^2+(-1-(1-n))+1)=P((n-2)^2+(n-2)+1)$$ is enough.
| {
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"url": "https://math.stackexchange.com/questions/2590434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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How many complex roots? Find the number of solutions, counting multiplicity, in the domain $\{z \in \mathbb{C} : 1 < |z| <2\}$ to the equation
$$z^9+z^5-8z^3+2z+1=0$$
Notice that along $e^{i \theta}$ where $\theta \in [0,2\pi]$ we have $|8z^3| $dominating the polynomial. Thus, by Rouche's theorem, there are 3 zeroes within that contour. Again, notice that $|z^9|$ dominates on $2e^{i \theta}$ again with $\theta \in [0, 2\pi]$ which gives 9 zeroes. Thus, the annulus contains 6 zeroes. Is this correct?
| The original question was how to calculate the number of roots, now the question has changed to confirming that the number the OP got was correct.
Sketch:
Use Rouche's theorem to count the number of roots within the disk $|z|<2$ and then use Rouche's theorem to count the number of roots within the disk $|z|<1$. A reverse triangle inequality will take care of the disk $|z|=1$ to show that there's no root there.
Details:
For the disk $|z|<2$, write your polynomial as $z^9+(z^5-8z^3+2z+1)$. Using the triangle inequality, we see on $|z|=2$, $|z^9|=512$ while $|z^5-8z^3+2z+1|\leq 32+64+4+1=101<512$, so we can apply Rouche's theorem. In particular, $z^9+(z^5-8z^3+2z+1)$ and $z^9$ have the same number of roots in the disk $|z|<2$. Since $z^9$ has $9$ roots, so does $z^9+(z^5-8z^3+2z+1)$.
For the disk $|z|<1$, write your polynomial as $-8z^3+(z^9+z^5+2z+1)$. On $|z|=1$, $|-8z^3|=8$, while, by the triangle inequality, $|z^9+z^5+2z+1|\leq 1+1+2+1=5<8$. Therefore, we can apply Rouche's theorem. In particular, $-8z^3+(z^9+z^5+2z+1)$ and $-8z^3$ have the same number of roots in the disk $|z|<1$. Since $-8z^3$ has $3$ roots, so does $-8z^3+(z^9+z^5+2z+1)$.
Therefore, there are $6$ roots in the annulus $1\leq |z|<2$. Now, to deal with $|z|=1$, on this circle, $|-8z^3+(z^9+z^5+2z+1)|\geq |-8z^3|-|z^9+z^5+2z+1|\geq 8-5=3$. Therefore, there are no roots on $|z|=1$, and so there are $6$ roots in the annulus $1<|z|<2$.
A quick check with WolframAlpha confirms this computation.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Infinite geometric sum (asking for insight on an easier solution) Let the sequence $F$ be defined as: $F_1=F_2=1$ and $F_n=2F_{n-1}+F_{n-2}$, for $n>2$. Evaluate $\sum_{n=1}^{\infty}\frac{F_n}{10^n}$.
The obvious solution involves solving for the explicit formula for $F$ (using the standard linear recurrence technique): $F_n=(\frac{\sqrt{2}-1}{2})(1+\sqrt{2})^n-(\frac{\sqrt{2}+1}{2})(1-\sqrt{2})^n$. Then we can just split the sum into two infinite geometric series. The computation is annoying, but nonetheless straightforward.
I was wondering if there is any easier solution than solving for an explicit formula for $F$. Maybe there is one that only needs the recursive definition? Any comments are appreciated.
| \begin{align*}
\sum_{n=1}^\infty\frac{F_n}{10^n}&=\frac{1}{10}+\frac{1}{100}+\sum_{n=3}^\infty\frac{F_n}{10^n}\\
&=0.11+\sum_{n=3}^\infty\frac{2F_{n-1}+F_{n-2}}{10^n}\\
&=0.11+\frac{2}{10}\left(\sum_{n=1}^\infty\frac{F_n}{10^n}-\frac{1}{10}\right)+\frac{1}{100}\sum_{n=1}^\infty\frac{F_n}{10^n}\\
0.79\sum_{n=1}^\infty\frac{F_n}{10^n}&=0.09\\
\sum_{n=1}^\infty\frac{F_n}{10^n}&=\frac{9}{79}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2594563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 1
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Factoring the polynomial $3(x^2 - 1)^3 + 7(x^2 - 1)^2 +4x^2 - 4$ I'm trying to factor the following polynomial:
$$3(x^2 - 1)^3 + 7(x^2 - 1)^2 +4x^2 - 4$$
What I've done:
$$3(x^2 -1)^3 + 7(x^2-1)^2 + 4(x^2 -1)$$
Then I set $p=x^2 -1$ so the polynomial is:
$$3p^3 + 7p^2 + 4p$$
Therefore: $$p(3p^2 + 7p + 4)$$
I apply Cross Multiplication Method: $$p(p+3)(p+4)$$
I substitute $p$ with $x^2-1$:
$$(x^2-1)(x^2-1+3)(x^2-1+4)$$
$$(x-1)(x+1)(x^2-2)(x^2-3)$$
I don't know if I've done something wrong or if I have to proceed further and how. The result has to be: $x^2(3x^2+1)(x+1)(x-1)$. Can you help me? Thanks.
| It's $$(x^2-1)(3(x^2-1)^2+7(x^2-1)+4)=(x^2-1)(3(x^2-1)+4)(x^2-1+1)=$$
$$=x^2(x^2-1)(3x^2+1)=x^2(x-1)(x+1)(3x^2+1).$$
By your way.
It should be $$3p^2+7p+4=(p+1)(3p+4).$$
| {
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"url": "https://math.stackexchange.com/questions/2595704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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linear algebra- how to find the representation matrix of a linear transformation when given the base of the origin and its transformations as the title states i want to find the matrix representing some linear transformations $T:V\rightarrow W$ but all i know is the base for $V$is $ (b_1,b_2,b_3)$ and $T(b_1),T(b_2),T(b_3)$
i have done this for the vectors
but in a convoluted way and not in the short and beautiful way my teacher showed me can anyone help me understand if i am taking any redundant steps or if i am missing the point all together
thank you for your time
here is my solution:
$\left[\begin{matrix}x\\
y\\
z
\end{matrix}\right]=\lambda_{1}\left[\begin{matrix}1\\
3\\
1
\end{matrix}\right]+\lambda_{2}\left[\begin{matrix}1\\
2\\
-1
\end{matrix}\right]+\lambda_{3}\left[\begin{matrix}1\\
0\\
5
\end{matrix}\right]\\ T\left(\left[\begin{matrix}x\\
y\\
z
\end{matrix}\right]\right)=T\left(\lambda_{1}
\left[\begin{matrix}1\\
3\\
1
\end{matrix}\right]+\lambda_{2}\left[\begin{matrix}1\\
2\\
-1
\end{matrix}\right]+\lambda_{3}\left[\begin{matrix}1\\
0\\
5
\end{matrix}\right]\right)\\ T\left(\left[\begin{matrix}x\\
y\\
z
\end{matrix}\right]\right)=T\left(\lambda_{1}\left[\begin{matrix}1\\
3\\
1
\end{matrix}\right]\right)+T\left(\lambda_{2}\left[\begin{matrix}1\\
2\\
-1
\end{matrix}\right]\right)+T\left(\lambda_{3}\left[\begin{matrix}1\\
0\\
5
\end{matrix}\right]\right)\\ T\left(\left[\begin{matrix}x\\
y\\
z
\end{matrix}\right]\right)=\lambda_{1}T\left(\left[\begin{matrix}1\\
3\\
1
\end{matrix}\right]\right)+\lambda_{2}T\left(\left[\begin{matrix}1\\
2\\
-1
\end{matrix}\right]\right)+\lambda_{3}T\left(\left[\begin{matrix}1\\
0\\
5
\end{matrix}\right]\right)\\ T\left(\left[\begin{matrix}x\\
y\\
z
\end{matrix}\right]\right)=\lambda_{1}\left[\begin{matrix}1\\
-1\\
0
\end{matrix}\right]+\lambda_{2}\left[\begin{matrix}0\\
1\\
1
\end{matrix}\right]+\lambda_{3}\left[\begin{matrix}2\\
-5\\
-3
\end{matrix}\right]\\ \left[\begin{matrix}1 & 0 & 0 & | & -x+\frac{3y}{5}+\frac{z}{5}\\
0 & 1 & 0 & | & \frac{3x}{2}-\frac{2y}{5}-\frac{3z}{10}\\
0 & 0 & 1 & | & \frac{x}{2}-\frac{y}{5}+\frac{z}{10}
\end{matrix}\right]\\ T\left(\left[\begin{matrix}x\\
y\\
z
\end{matrix}\right]\right)=\left(-x+\frac{3y}{5}+\frac{z}{5}\right)\left[\begin{matrix}1\\
-1\\
0
\end{matrix}\right]+\left(\frac{3x}{2}-\frac{2y}{5}-\frac{3z}{10}\right)\left[\begin{matrix}0\\
1\\
1
\end{matrix}\right]+\left(\frac{x}{2}-\frac{y}{5}+\frac{z}{10}\right)\left[\begin{matrix}2\\
-5\\
-3
\end{matrix}\right]\\ \begin{cases}
\left(-x+\frac{3y}{5}+\frac{z}{5}\right)+2\left(\frac{x}{2}-\frac{y}{5}+\frac{z}{10}\right)=x'\\
-\left(-x+\frac{3y}{5}+\frac{z}{5}\right)+\left(\frac{3x}{2}-\frac{2y}{5}-\frac{3z}{10}\right)-5\left(\frac{x}{2}-\frac{y}{5}+\frac{z}{10}\right)=y'\\
\left(\frac{3x}{2}-\frac{2y}{5}-\frac{3z}{10}\right)-3\left(\frac{x}{2}-\frac{y}{5}+\frac{z}{10}\right)=z'
\end{cases}\\ \begin{cases}
\frac{y}{5}+\frac{2z}{5}=x'\\
-z=y'\\
\frac{y}{5}-\frac{6z}{10}=z'
\end{cases}\\ \left[\begin{matrix}0 & \frac{1}{5} & \frac{2}{5}\\
0 & 0 & -1\\
0 & \frac{1}{5} & \frac{-6}{10}
\end{matrix}\right]$
| Your method is correct and the preferable method in this case.
Usually you first try inspection, so you try to find out what happens to the standard basis vectors $e_1,e_2,e_3$, so you try write $e_1$ as a linear combination of $b_1,b_2$ and $b_3$, and then you can directly calculate $T(e_1)$, which is the first column of the matrix, $T(e_2)$ the second and $T(e_3)$ the third.
If inspection doesn't work (because it's too difficult, which is the case here: at least I cannot see directly how to write $e_1$ as a linear combination of $b_1,b_2,b_3$), then you would have to solve three $Ax=b$ equations, so in that case do what you have done: solving these three equations at the same time by taking them together in a matrix and finding the inverse, etc.
| {
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How do I integrate these functions? I deleted my last question, since some of you wanted me to rewrite the question properly. I feel sorry for inconvenience, but please understand that this is the first time I use $\texttt{MathJax}$.
Up to now, I tried every method I know in integration, like substituition, partial fractions, uv-method, etc. But seems like nothing works. I would appreciate to have your help. Thanks.
$$
\int_{0}^{1}\frac{\mathrm{d}x}{\left(x + 1\right)
\left[x^{2}\left(1 - x\right)\right]^{1/3}}\,,
\qquad\qquad\int_{0}^{1}\frac{\mathrm{d}x}{\left(x^{2} + 1\right)
\left[x^{2}\left(1 - x\right)\right]^{1/3}}
$$
| There are lots of different methods to calculate your integrals,
the simplest way is partial fraction expansion [wikipedia]
(https://en.wikipedia.org/wiki/Partial_fraction_decomposition):
$
\frac{1}{(x+1)\left( x^{2}\left( 1-x\right) \right) ^{\frac{1}{3}}}=\frac{x^{%
\frac{4}{3}}}{2\left( 1-x\right) ^{\frac{1}{3}}}+\frac{\left( 1-x\right) ^{%
\frac{2}{3}}}{x^{\frac{2}{3}}}+\frac{\left( 1-x\right) ^{\frac{2}{3}}x^{%
\frac{4}{3}}}{2\left( 1+x\right) }$
$\frac{1}{(x^{2}+1)\left( x^{2}\left( 1-x\right) \right) ^{\frac{1}{3}}}=%
\frac{x^{\frac{4}{3}}}{2\left( 1-x\right) ^{\frac{1}{3}}}+\frac{\left(
1-x\right) ^{\frac{2}{3}}}{x^{\frac{2}{3}}}+\left( 1-x\right) ^{\frac{2}{3}%
}x^{\frac{1}{3}}-\frac{\left( 1-x\right) ^{\frac{2}{3}}x^{\frac{4}{3}}}{%
2\left( 1+x^{2}\right) }-\frac{\left( 1-x\right) ^{\frac{2}{3}}x^{\frac{7}{3}%
}}{2\left( 1+x^{2}\right) }$
or just the following transformation:
$\int_{0}^{1}\frac{1}{(x+1)\left( x^{2}\left( 1-x\right) \right) ^{\frac{1}{3}%
}}dx=\int_{0}^{1}\frac{x^{-\frac{2}{3}}}{(x+1)\left( 1-x\right) ^{\frac{1}{3}%
}}dx$
$\int_{0}^{1}\frac{1}{(x^{2}+1)\left( x^{2}\left( 1-x\right) \right) ^{\frac{1%
}{3}}}dx=\int_{0}^{1}\frac{x^{-\frac{2}{3}}}{(x^{2}+1)\left( 1-x\right) ^{%
\frac{1}{3}}}dx$
The following antiderivatives appear:
$\int \frac{x^{\alpha }}{\left( 1-x\right) ^{\beta }}=B\left( x,1+\alpha
,1-\beta \right) $
where B is the incomplete Beta -
Function.
$\int \frac{\left( 1-x\right) ^{\alpha }x^{\beta }}{\left( 1+x\right) }=\frac{%
x^{\alpha +1}}{\alpha +1}F_{1}\left( \alpha +1;\beta ,1;\alpha
+2;x,-x\right)$
where $F_{1}$ is the Appell hypergeometric function
AppellF1.
Moreover using the partial fraction decompositions [Dieckmann]
(http://www-elsa.physik.uni-bonn.de/dieckman/InfProd/InfProd.html):
$\frac{1}{\left( 1+x^{2}\right) }=\frac{i}{2\left( i+x\right) }-\frac{i}{%
2\left( x-i\right) }$
the last antiderivative terms are calculated by:
$\int \frac{\left( 1-x\right) ^{\alpha }x^{\beta }}{\left( 1+x^{2}\right) }=%
\frac{1}{2}\frac{x^{\alpha +1}}{\alpha +1}F_{1}\left( \alpha +1;\beta
,1;\alpha +2;x,i~x\right) +\frac{1}{2}\frac{x^{\alpha +1}}{\alpha +1}%
F_{1}\left( \alpha +1;\beta ,1;\alpha +2;x,-i~x\right) $
In special cases, the AppellF1 function can be reduced :
$F_{1}\left( a;b_{1},b_{1};c;x,-x\right) =~_{3}F_{2}\left( \frac{a}{2}+\frac{1%
}{2},\frac{a}{2},b_{1};\frac{c}{2}+\frac{1}{2},\frac{c}{2};x^{2}\right) $
to a hypergeometric function
(http://functions.wolfram.com/HypergeometricFunctions/AppellF1/03/05/0002/)
Put in all the parameters and integration limits finally leads to
the mentioned results [marty cohen] (https://math.stackexchange.com/posts/2611282/edit).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 7,
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Recurrent equation system How to solve this recurrent equation system?
$$\begin{cases}a_{n+1}=a_n+2b_n\\
b_{n+1}=2a_n+b_n\end{cases}$$
The possible solutions should be
\begin{cases}a_n=\frac{1}{2}3^n+\frac{1}{2}(-1)^n\\
b_n=\frac{1}{2}3^n-\frac{1}{2}(-1)^n\end{cases}
| Let $X_n =(a_n,b_n)$ with $X_0=(1,0)=(a_0,b_0)$
\begin{cases}a_{n+1}=a_n+2b_n\\
b_{n+1}=2a_n+b_n\end{cases} Then it is equivalent $$X_{n+1} = AX_n \Longleftrightarrow X_n =A^nX_0$$
where, $$A= \begin{pmatrix}1&2\\ 2&1\end{pmatrix}$$
Then prove by induction that,
$$A^n= \begin{pmatrix}\frac{1}{2}3^n+\frac{1}{2}(-1)^n &\frac{1}{2}3^n-\frac{1}{2}(-1)^n\\\frac{1}{2}3^n-\frac{1}{2}(-1)^n&\frac{1}{2}3^n+\frac{1}{2}(-1)^n\end{pmatrix}$$
Therefore you will get the desired result.
$$X_n= \begin{pmatrix}a_n\\b_n\end{pmatrix}= \begin{pmatrix}\frac{1}{2}3^n+\frac{1}{2}(-1)^n &\frac{1}{2}3^n-\frac{1}{2}(-1)^n\\\frac{1}{2}3^n-\frac{1}{2}(-1)^n&\frac{1}{2}3^n+\frac{1}{2}(-1)^n\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}\\=\begin{pmatrix}\frac{1}{2}3^n+\frac{1}{2}(-1)^n\\
\frac{1}{2}3^n-\frac{1}{2}(-1)^n\end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2597018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\cos3A + \cos3B + \cos3C = 1$ in a triangle, find one of its length I would like to solve the following problem.
In $\triangle ABC, AC = 10, BC = 13$. If $\cos3A + \cos3B + \cos3C = 1$, compute the length of $AB$.
I thought that I could apply the Law of Cosines. Using the fact that $A+B+C=\pi$, I attempted to build the equation up from there.
What I got was that $$\cos3A+\cos3B-\cos(3A+3B)=1$$
Expanding, I got $$\cos3A+\cos3B-\cos3A\cos3B-\sin3A\sin3B=1$$
Now, it's possible that I'd be able to factor it somehow by rewriting it all in terms of cosine and arrive at the answer, but is there a better way to solve the problem? Thanks!
| $\mathbf {Hint...}$
$$\cos {3A}+\cos {3B}+ \cos{3C}=1$$
$$\Rightarrow 4 \sin {\frac {3C}{2}} .\sin {\frac{3B}{2}} .\sin{\frac{3A}{2}}=0$$
Hence the largest angle of triangle is $\frac{2\pi}{3}$ which can be either angle $C$ or angle $A$. By applying cosine rule in each of these cases we get the value of $AB$ as $\sqrt {399}$ or $\sqrt {94}-5$ respectively.
Note:
$$\cos {3A}+\cos {3B}+ \cos{3C}=1$$
$$\Rightarrow -2\cos {\frac {3(A-B)}{2}}\sin {\frac {3C}{2}}- 2\left(\sin {\frac {3C}{2}}\right)^2=0$$
$$=\sin\frac{3C}{2}\left(\cos\frac{3(A-B)}{2}+ \sin\frac{3C}{2}\right)=0$$
$$\sin\frac{3C}{2}\left(\cos\frac{3(A-B)}{2}-\cos\frac{3(A+B)}{2}\right)=0$$
$$\sin\frac{3C}{2}\sin\frac{3B}{2}\sin\frac{3A}{2}=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2597796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Prove $\frac{(m^2 + 2)(m-1)^2(m+1)^2}{(m^2-1)^{3/2}}=(m^2+2) \sqrt{m^2-1}$ Im trying to get from this expression into:
$$\frac{(m^2 + 2)(m-1)^2(m+1)^2}{(m^2-1)^{3/2}}$$
this expression:
$$(m^2+2) \sqrt{m^2-1}$$
someone know how to do it?
i tried it for hours and can't get from the first expression into the second expression.
please explain to me step by step. I'm newbie.
| In order to obtain the second expression you have to know that:
$$a^2b^2=(ab)^2\implies(m-1)^2(m+1)^2=((m+1)(m+1))^2=(m^2-1)^2\tag{1}$$
$$\frac{a^x}{a^y}=a^{(x-y)}\tag{2}$$
Hence
$$\frac{(m^2 + 2)(m-1)^2(m+1)^2}{(m^2-1)^{3/2}}\overbrace{=}^{(1)}\frac{(m^2 + 2)(m^2-1)^2}{(m^2-1)^{3/2}}\overbrace{=}^{(2)}\\=(m^2+2)(m^2-1)^{2-\frac 32}=\\=(m^2+2)(m^2-1)^{\frac 12}=\color{red}{(m^2+2)\sqrt{(m^2-1)}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2598537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
$2^{x-3} + \frac {15}{2^{3-x}} = 256$ $$2^{x-3} + \frac {15}{2^{3-x}} = 256$$
*
*Find the unknown $x$.
My attempt:
We know that $x^y . x^b = x^{y+b}$.
$$2^x . 2^{-3} + 15. 2^{-3+x} = 2^8$$
and
$$2^x . 2^{-3} + 15. 2^{-3} . 2^x = 2^8$$
From here, we get
$$2^x + 15 = 2^8$$
However, I'm stuck at here and waiting for your kindest helps.
Thank you.
| Hint:
As $2^{-y}=\dfrac1{2^y}$
$$\dfrac1{2^{3-x}}=2^{-(3-x)}=?$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculate $\int \frac{1}{x^2+x+1} \, dx$ $$ \int \frac{1}{x^2+x+1}\, dx = \int \frac{1}{(x+\frac 1 2)^{2} + \frac 3 4}\, dx $$
Substitute $x+\frac 1 2 = u$, $dx = du$:
$$\int \frac 4 3 \frac{1}{\left(\frac{2u}{\sqrt 3}\right)^2+1} \, du = \frac 4 3 \int \frac{1}{\left(\frac{2u}{\sqrt 3}\right)^2+1} \, du$$
Substitute: $s = \frac{2u}{\sqrt 3}$, $du = \frac{\sqrt{3}}{2}ds$
$$\frac 4 3 \frac{\sqrt 3}{2}\int \frac{1}{s^2+1} \, ds$$
After multiplying and substituting back, we get the solution:
$$\frac{2}{\sqrt 3}\arctan\left(\frac{2x+1}{\sqrt 3}\right)$$
In a book I have however, this integral is evaluated as $\sqrt 3 \arctan\left(\frac{2x+1}{\sqrt 3}\right)$.
Which solution is right, and if the book's, how did the authors arrived to it?
| Your book is wrong. If you differentiate the answer of the book we get $$\frac{3}{2(x^2+x+1)}$$ which is not the same as the integrand.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2601014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the angle between the two tangents drawn from the point $(1,2)$ to the ellipse $x^2+2y^2=3$.
Find the angle between the two tangents drawn from the point $(1,2)$ to the ellipse $x^2+2y^2=3$.
The given ellipse is $\dfrac{x^2}{3}+\dfrac{y^2}{\frac{3}{2}}=1$
Any point on the ellipse is given by $(a\cos \theta,b\sin \theta)$ where $a=\sqrt 3,b=\frac{\sqrt 3}{\sqrt 2}$.
Now slope of the tangent to the curve at $(a\cos \theta,b\sin \theta)$ is $\dfrac{-a\cos \theta}{2b\sin \theta}$.
Hence we have $\dfrac{b\sin \theta- 2}{a\cos \theta -1}=\dfrac{-a\cos \theta}{2b\sin \theta}$.
On simplifying we get $4b\sin \theta +a\cos\theta =3$
If we can find the value of $\theta $ from above then we can find the two points on the ellipse where the tangents touch them but I am unable to solve them.
Please help to solve it.
Any hints will be helpful
| Let $\alpha$ be the angle for which $\cos\alpha = \frac a{\sqrt{a^2+16 b^2}}$ and $\sin \alpha=\frac {4b}{\sqrt{a^2+16 b^2}}$.
Then $\theta-\alpha$ can be solved from $cos(\theta-\alpha)=\sin \alpha\sin \theta+\cos \alpha \cos \theta=\frac 3{\sqrt {a^2+16 b^2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2602561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find value of $\lambda$ $S$ is a circle having center at $(0, a)$ and radius $b\lt a$. A variable circle centred at $(\alpha, 0)$ and touching the circle $S$ meets $X$ axis at $ M$ and $N$. A point $P=(0,\pm \lambda\sqrt {a^2-b^2})$ on the $y$ axis such that angle $MNP$ is a constant for any choice of $\alpha$ then find $\lambda$.
My work:
I wrote down the equation of circle $S$ and I have tried using the cosine rule in the two triangles having the same constant angle $MNP$. I also tried using basic algebra and the Stewart's theorem in some of the triangles I could do work in. But this was of no use. Moreover so many of variables are confusing me a lot. Any help is greatly appreciated.
| There is no $\lambda$ such that $\angle MNP$ stays constant when $\alpha$ varies, since $\tan(\angle MNP)$ is a ratio of two segments, one of which is constant $\lambda \sqrt{a^2-b^2}\,$, while the other one clearly varies with $\alpha$.
The more interesting question, however, is whether such a $\lambda$ exists so that $\color{red}{\angle MPN}$ stays constant.
Let $c$ be the radius of the second circle, then the tangency condition gives:
$$\alpha^2 + a^2=(b+c)^2 \quad \iff \quad c = \sqrt{\alpha^2+a^2}-b$$
It follows that $M,N = (\alpha \pm c, 0)=\left(\alpha+b \pm \sqrt{\alpha^2+a^2}, 0\right)\,$, then with $p=\lambda\sqrt{a^2-b^2}\,$:
$$\require{cancel}
\begin{align}
\tan(\angle MPN) &= \tan(\angle OPN - \angle OPM) = \frac{\tan(\angle OPN)-\tan(\angle OPM)}{1+\tan(\angle OPN)\tan(\angle OPM)} \\[5px]
&= \frac{\dfrac{\cancel{\alpha}+c}{p}-\dfrac{\cancel{\alpha}-c}{p}}{1+ \dfrac{\alpha+c}{p} \cdot \dfrac{\alpha-c}{p}} = \frac{2pc}{p^2+\alpha^2-c^2} \\[5px]
&= \frac{2\lambda\sqrt{a^2-b^2}(\sqrt{\alpha^2+a^2}-b)}{\lambda^2(a^2-b^2)+ \cancel{\alpha^2}-(\cancel{\alpha^2}+a^2)+2b\sqrt{\alpha^2+a^2}-b^2} \\[5px]
&= \frac{2\lambda\sqrt{a^2-b^2}(\sqrt{\alpha^2+a^2}-b)}{(\lambda^2-1)a^2-(\lambda^2+1)b^2+2b\sqrt{\alpha^2+a^2}} \\[5px]
&= \frac{2\lambda\sqrt{a^2-b^2}}{2b} \,\cdot\, \frac{\sqrt{\alpha^2+a^2}\color{blue}{- b}}{\sqrt{\alpha^2+a^2}\color{blue}{-\dfrac{(1-\lambda^2)a^2+(1+\lambda^2)b^2}{2b}}}
\end{align}
$$
The latter does in fact not depend on $\alpha$ iff the $\color{blue}{\style{font-family:inherit}{\text{blue}}}$ terms are equal $\iff \lambda=\pm1\,$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all the triangles $ABC$ for which the perpendicular line to AB halves a line segment Let $AA_1$ and $BB_1$ be the bisectors of angles in triangle $ABC$. The bisectors intercept at the point $I$. How do I find all the triangles for which the perpendicular line from $I$ to $AB$ halves the line segment $A_{1}B_1$?
Recently I discovered the following:
If $AA_1$ and $BB_1$ are bisectors of sharp angles in a right triangle $ABC$ intersect at the point $I$, then the perpendicular line from $I$ to $AB$ halves the line segment $A_1B_1$.
How do I prove the converse? Are there any other triangles that satisfy the aforementioned condition?
| The answer is as follows :
$$\text{$\triangle{ABC}$ is either a right triangle with $\angle C=90^\circ$ or an isosceles triangle with $|\overline{CA}|=|\overline{CB}|$}$$
We may suppose that $A(-1,0),B(1,0),C(c,d)$ where $c\ge 0$ and $d\gt 0$.
The equation of the line $AC,BC$ is $dx-(c+1)y+d=0,dx-(c-1)y-d=0$ respectively.
For a point $(x,y)$ on the line $AA_1$, we have
$$\small\frac{|dx-(c+1)y+d|}{\sqrt{d^2+(c+1)^2}}=|y|\implies \frac{dx-(c+1)y+d}{\sqrt{d^2+(c+1)^2}}=\pm y\ \rightarrow\ y=\frac{dx+d}{c+1+t}$$
which is the equation of $AA_1$ where $t=\sqrt{d^2+(c+1)^2}$ since the slope of $AA_1$ is positive.
For a point $(x,y)$ on the line $BB_1$, we have
$$\small\frac{|dx-(c-1)y-d|}{\sqrt{d^2+(c-1)^2}}=|y|\implies \frac{dx-(c-1)y-d}{\sqrt{d^2+(c-1)^2}}=\pm y\ \rightarrow\ y=\frac{dx-d}{c-1-s}$$
which is the equation of $BB_1$ where $s=\sqrt{d^2+(c-1)^2}$ since the slope of $BB_1$ is negative.
It follows that
$$ I\left(\frac{2c+t-s}{2+t+s},\frac{2d(c-1)}{(c-1-s)(2+t+s)}\right),\quad A_1\left(\frac{2c+t}{t+2},\frac{2d}{t+2}\right),\quad B_1\left(\frac{2c-s}{s+2},\frac{2d}{s+2}\right)$$
Therefore, we have
$$\begin{align}&\text{the perpendicular line from $I$ to $AB$ halves the line segment $A_{1}B_1$}\\\\&\iff \frac 12\left(\frac{2c+t}{t+2}+\frac{2c-s}{s+2}\right)=\frac{2c+t-s}{2+t+s}\\\\&\iff \frac{2c+t}{t+2}+\frac{2c-s}{s+2}=\frac{4c+2t-2s}{2+t+s}\\\\&\iff (2c+t)(s+2)(2+t+s)+(2c-s)(t+2)(2+t+s)=(4c+2t-2s)(t+2)(s+2)\\\\&\iff (c^2+d^2-1)(2c-s+t)=0\\\\&\iff c^2+d^2=1\quad\text{or}\quad c=0\end{align}$$
where $$2c-s+t=0\iff (2c+t)^2=s^2\iff 4c\left(c+1+\sqrt{d^2+(c+1)^2}\right)=0\iff c=0$$
from which the conclusion written at the top follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2603299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Intersection points of $2^x$ and $x^2$ I am trying to find the intersection points of $2^x$ and $x^2$.
To do this, I set the two functions equal to each other: $2^x = x^2$. It seems intuitive to take the log base two of both sides of the equations, and I got: $x = \log_2{x^2}$.
I'm not entirely sure how to proceed from here. Can someone please help me?
| All three solutions can be found using an analytic approach based on the Lambert W function.
Starting with the equation $2^x = x^2$, taking the square root of both sides gives
$$x = \pm 2^{x/2}.$$
Rewriting this equation in the form for the defining equation for the Lambert W function, namely
$$\text{W} (x) e^{\text{W}(x)} = x,$$
where $\text{W}(x)$ denotes the Lambert W function, we have
\begin{align*}
x &= \pm 2^{x/2}\\
x &= \pm e^{\frac{x}{2} \ln 2}\\
\Rightarrow x e^{-\frac{x}{2} \ln 2} &= \pm 1\\
-\frac{x}{2} \ln 2 e^{-\frac{x}{2} \ln 2} & = \mp \frac{\ln 2}{2},
\end{align*}
which is now exactly in the form for the defining equation for the Lambert W function and yields
$$-\frac{x}{2} \ln 2 = \text{W}_\nu \left (\mp \frac{\ln 2}{2} \right ),$$
or
$$x = - \frac{2}{\ln 2} \text{W}_\nu \left (\mp \frac{\ln 2}{2} \right ). \tag1$$
Here $\nu$ denote the branches for the Lambert W function. For branches that are real we have $\nu = 0$ and corresponds to the principal branch for the Lambert W function while $\nu = -1$ corresponds to the secondary real branch.
We now make use of the following simplification rule for the Lambert W function
$$-\ln x = \begin{cases}
\text{W}_0 \left (-\dfrac{\ln x}{x} \right ), \quad 0 < x \leqslant e,\\[2ex]
\text{W}_{-1} \left (-\dfrac{\ln x}{x} \right ), \quad x \geqslant e,
\end{cases}$$
and consider the negative and positive cases found in the argument of the Lambert W function in (1) separately.
Negative case
Here
$$x = -\frac{2}{\ln 2} \text{W}_\nu \left (-\frac{\ln 2}{2} \right ) = \begin{cases}
-\dfrac{2}{\ln 2} \text{W}_0 \left (-\dfrac{\ln 2}{2} \right )\\
-\dfrac{2}{\ln 2} \text{W}_{-1} \left (-\dfrac{\ln 2}{2} \right )
\end{cases}$$
Now on application of the above simplification rule we have
$$\text{W}_0 \left (-\frac{\ln 2}{2} \right ) = -\ln 2 = 2,$$
since $x = 2$ lies between $0 < x \leqslant e$.
Also
$$\text{W}_{-1} \left (-\frac{\ln 2}{2} \right ) = \text{W}_{-1} \left (-\frac{2 \ln 2}{2^2} \right ) = \text{W}_{-1} \left (- \frac{\ln 4}{4} \right ) = -\ln 4 = -2 \ln 2,$$
since $x = 4$ is greater than $e$. Thus
\begin{align*}
x &= \begin{cases}
-\dfrac{2}{\ln 2} \cdot - \ln 2\\[2ex]
-\dfrac{2}{\ln 2} \cdot - 2\ln 2
\end{cases}\\
\end{align*}
or
$$x = 2, 4.$$
Positive case
Here
$$x = -\frac{2}{\ln 2} \text{W}_\nu \left (\frac{\ln 2}{2} \right ) = -\frac{2}{\ln 2} \text{W}_0 \left (\frac{\ln 2}{2} \right ).$$
Note that as the argument to the Lambert W function is positive, for a real solution, only the principal branch is selected.
So to summarise, the three real solutions to the equation $2^x = x^2$ are:
$$x = 2, 4, -\frac{2}{\ln 2} \text{W}_0 \left (\frac{\ln 2}{2} \right )$$
| {
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"url": "https://math.stackexchange.com/questions/2605114",
"timestamp": "2023-03-29T00:00:00",
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Calculating the probability of a function being real when it has random variables I'm trying to calculate the probability of
$ax^2+bx+c$ root being a real number when the variables $a,$ $b,$ and $c$ values are all randomized by throwing a standard die.
I got to the point where I can get the probability by calculating the chance of $b^2-4ac>0$, but I'm not sure how I can conveniently carry on from here and my attempts at doing it by hand (finding every possible real occurrence and calculating them by the total outcomes) have failed me.
In other words, the values of $a$, $b$ and $c$ are within $\{1,2,3,4,5,6\}$ and each of the three variables is randomly picked from that list with no special weighting (so $\frac 16$ chance to get any of the $6$ values).
| We need $b^2 - 4ac \ge 0$.
If $b= 6$ then $ac \le 9$. If $a =1$ then $c = 1...6$. If $a=2$ then $c=1...4$. If $c=3; a= 1..3$ if $a=4; c=1,2$ if $a=5,6;c =1$. There are $6+4+3+2+1+1 = 17$ ways to do this.
If $b = 5$ then $ac \le 6$. If $a=1$ then $c = 1...6$ if $a= 2$ then $b=1..3$ and if $a = 3; c=1,2$ and if $a \ge 4; c=1$. So $6+3+2+3 = 14$ ways to do this.
If $b = 4$ then $ac \le 4$ So if $a= 1,2,3,4$ then $c \le 4,2,1,1$ so there are $4+2+1+1 = 8$ ways to do that
If $b =3$ then $ac \le 2$ So if $a= 1,2$ then $c \le 2,1$ so there are $2+1=3$ wways to do that.
If $b = 2$ then $ac \le 1$ so $a=c=1$. $1$ way to do that.
$b$ can not be $1$ because $4ac \ge 4 > 1^2$.
So there are $17+14+8+3+1 = 43$ ways to do this.
So the probability is $\frac {43}{6^3} \approx 20\%$ (very slightly less.)
| {
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"timestamp": "2023-03-29T00:00:00",
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Identify $\mathbb{Z}[x]/(x^2-3,2x+4)$. I need help to identify $\mathbb{Z}[x]/(x^2-3,2x+4)$.
I've been solving such problems in an approach like:
$$
2(x^2-3)=2x^2-6, x(2x+4)=2x^2+4x \\
(2x^2+4x)-(2x^2-6)=4x+6, 2(2x+4)=4x+8 \\
(4x+8)-(4x+6)=2
$$
What shall I do next, please? Thank you.
Simon
| we have $2=2(x^2-3)+(2-x)(2x+4)$ and then $(x^2-3,2x+4,2)=(2,x^2-3)$
$\mathbb Z[x]/(x^2-3,2x+4) \cong\mathbb Z[x]/(2,x^2-3)\cong \mathbb Z_2[x]/(x^2-1) \cong \mathbb Z_2[x]/(x-1)^2 \cong \mathbb Z_2[x]/(x)^2 $
| {
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"timestamp": "2023-03-29T00:00:00",
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If $f\left(x\right)=$ $\sum_{n=0}^{\infty}$ $\frac{x^{3n}}{\left(3n\right)!}$ then prove $f''\left(x\right)+f'\left(x\right)+f\left(x\right)=e^{x}$
QuestionIf $f\left(x\right)=$$\sum_{n=0}^{\infty}$$\frac{x^{3n}}{\left(3n\right)!}$
then prove that
$f''\left(x\right)+f'\left(x\right)+f\left(x\right)=e^{x}$
My Approach$f\left(x\right)=$$\sum_{n=0}^{\infty}$$\frac{x^{3n}}{\left(3n\right)!}$
$f'\left(x\right)=$$\sum_{n=1}^{\infty}$$\frac{x^{3n-1}}{\left(3n-1\right)!}$
=$\sum_{n=0}^{\infty}$$\frac{x^{3n+2}}{\left(3n+2\right)!}$
$f''\left(x\right)=$$\sum_{n=2}^{\infty}$$\frac{x^{3n-2}}{\left(3n-2\right)!}$
=$\sum_{n=0}^{\infty}$$\frac{x^{3n+4}}{\left(3n+4\right)!}$
$f''\left(x\right)+f'\left(x\right)+f\left(x\right)=\sum_{n=0}^{\infty}\frac{x^{3n}}{\left(3n\right)!}+\sum_{n=2}^{\infty}\frac{x^{3n+2}}{\left(3n+2\right)!}+\sum_{n=0}^{\infty}\frac{x^{3n+4}}{\left(3n+4\right)!}$
Now i am unable to prove it to be equal to $e$$^{x}$
| Your solution is generally ok, but for $f''$ we should rewrite it like below:
$f''=(f')'=(\sum_{n=0}^{\infty}\frac{x^{3n+2}}{\left(3n+2\right)!})'=\sum_{n=0}^{\infty}\frac{x^{3n+1}}{\left(3n+1\right)!}$
So finally, we will have:
$f\left(x\right)+f'\left(x\right)+f''\left(x\right)=\sum_{n=0}^{\infty}\frac{x^{3n}}{\left(3n\right)!}+\frac{x^{3n+2}}{\left(3n+2\right)!}+\frac{x^{3n+1}}{\left(3n+1\right)!}=\sum_{k=0}^{\infty}\frac{x^{k}}{k!}=e^x$
Note that as we had all $3n, 3n+1$ and $ 3n+2$, we could merge those 3 sigmas.
| {
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"timestamp": "2023-03-29T00:00:00",
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Computing the series$\sum\limits_{k=1}^{\infty}{\frac{3^{k-1}+(2i)^k}{5^k}} $ Show convergence of
$\begin{align}
\sum_{k=1}^{\infty}{\frac{3^{k-1}+(2i)^k}{5^k}} &= \sum_{k=1}^{\infty}{\frac{3^{k-1}}{5^k}}+ \sum_{k=1}^{\infty}{\frac{(2i)^k}{5^k}} \\
&= \sum_{k=1}^{\infty}{\frac{1}{3} \cdot \left( \frac{3}{5} \right) ^k} + \sum_{k=1}^{\infty}{ \left( \frac{2i}{5} \right )^k} \\
&= \frac{1}{3} \cdot \sum_{k=1}^{\infty}{ \left( \frac{3}{5} \right) ^k} + \sum_{k=1}^{\infty}{ \left( \frac{2i}{5} \right )^k} \\
\end{align}$
The first part of the sum converges because it is the geometric series with $q= \frac{3}{5}, 1> \left| \frac{3}{5} \right|$.
$$\sum_{n=1}^{\infty}{\left(\frac{2i}{5} \right)^k}$$
Question: Why does that series diverge (WolframAlpha)?
I mean, if $q=\frac{2i}{5}, |q|<1$ then it should be the geometric series and thus converge?
| Wolfram Alpha says it diverges because you typed in $n=1$ to infinity instead of $k=1$ to infinity. Correct the typo and it works.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2613310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Evaluating an integral2 Evaluate the following integral
$$\int\frac{dx}{\tan^2x +\csc x}$$
I tried the sub $$u=\tan{\frac{x}{2}}$$ to be
$$\int\frac{2u(1-u^2)^2}{(1+u^2)(u^6-u^4+3u^2+1)}du$$
Then i used the partial fractions to get
$$\int\frac{-2u}{1+u^2}+\frac{x^5-x^3+2x}{x^6-x^4+3x^2+1}$$
The first integrand is easy but the second i could not solve it ?
| Try rewriting
$$\begin{align}
\frac{x^5 - x^3 + 2x}{x^6 - x^4 + 3x^2 + 1} &= \frac16 \frac{(6x^5 - 4x^3 + 6x) + (-2x^3 + 6x)}{x^6 - x^4 + 3x^2 + 1} \\
&= \frac16 \ \frac{6x^5 - 4x^3 + 6x}{x^6 - x^4 + 3x^2 + 1} - \frac16 \frac{(x^2-3)2x}{x^6 - x^4 + 3x^2 + 1}
\end{align} $$
The second term can be reduced to a cubic
$$ \int \frac{t-3}{t^3 - t^2 + 3t + 1}dt $$
The denominator doesn't have rational roots, but can be factored as
$$ t^3 - t^2 + 3t + 1 = (t+0.2956)\big((t-0.6478)^2+1.7214^2\big) $$
The roots are given by Wolfram here (with exact forms). You can then use partial fractions again
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2615297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate limit containing $\sum{n^6}$ Evaluate:
$$\lim_{n\to\infty}{\frac{1^6+2^6+3^6+\ldots+n^6}{(1^2+2^2+3^2+\ldots+n^2)(1^3+2^3+3^3+\ldots+n^3)}}$$
I can solve the denominator as:
$$\frac{n(n+1)(2n+1)}{6}\cdot\frac{n^2(n+1)^2}{4}$$
$$n^7\cdot\frac{(1+\frac{1}{n})(2+\frac{1}{n})}{6}\cdot\frac{(1+\frac{1}{n})}{4}$$
$$=\frac{n^7}{12}$$
How can I reduce the numerator?
| Note that
$$\lim_{n\to\infty}{\frac{1^6+2^6+3^6...+n^6}{(1^2+2^2+3^2...+n^2)(1^3+2^3+3^3...+n^3)}}=\lim_{n\to\infty} \frac{12\sum_1^n k^6}{n^7}=\frac{12}{7}$$
indeed by Stolz-Cesaro
$$\lim_{n\to\infty} \frac{12\sum_1^n k^6}{n^7}=\lim_{n\to\infty} \frac{12(n+1)^6}{(n+1)^7-n^7}=\lim_{n\to\infty} \frac{12(n+1)^6}{7n^6+...+1}=\frac{12}{7}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2616313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Real ordered pair $(a,b)$ in equation
If $$a^2+5b^2+2b=6a+2ab-10$$
then all real ordered pair of $(a,b)$ is?
Try: I am trying to convert it into $$(..)^2+(..)^2=0$$ or $$(..)^2+(..)^2+(..)^2=0.$$
So $$(a-b)^2+(2a-0.75)^2+10-(0.75)^2-6a=0.$$
Could some help me to solve it? Thanks.
| The equation can be written as
$$(2b-1)^2+ a^2+b^2+9-2ab-6a+6b=0$$
$$\Rightarrow (2b-1)^2+ (-a)^2+(b)^2+(3)^2+ (2)(-a)(b)+(2)(3)(-a) +(2)(3)(b)=0$$
$$\Rightarrow (2b-1)^2+(a-b-3)^2=0$$
Hence we get $(a, b) = (\frac {7}{2} , \frac {1}{2})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2621439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How to solve the integral $\int \frac{4+12x}{({1-2x-3x^2})^\frac{2}{3}}\,dx$ $$\int \frac{4+12x}{({1-2x-3x^2})^\frac{1}{3}}\,dx$$
So first I tried setting $u = 1-2x-3x^2, du = -2-6x\,dx$
Which gives
$$-2\int u^\frac{-1}{3}du = \frac{-6 u^\frac{2}{3}}{2} =-3u^\frac{2}{3}$$
And ultimately:
$$-3(1-2x-3x^2)^\frac{2}{3}$$
For some odd reason when I take the derivative I'm getting something completely different, what's wrong with my substitution?
| Let's take the derivative:
$$g'(x)=(-3(1-2x-3x^2)^\frac{2}{3})'=-3 \frac 2 3 (1-2x-3x^2)^{\frac {-1}3}(-6x-2)$$
$$g'(x)=2 (1-2x-3x^2)^{\frac {-1}3}(6x+2)=\frac {12x+4}{(1-2x-3x^2)^{\frac {1}3}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2622216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Prove that there are no integers $x, y, z$ such that $x+y+z=0$ and $1/x+1/y+1/z=0$
Prove that there are no integers $x, y, z$ such that $x+y+z=0$ and $\frac1x+\frac1y+\frac1z=0$
My thinking was that since the numbers are integers, then there can't be $2$ negative values that cancel out the positive or $2$ positive numbers to cancel the negative. One integer would have to cancel out the second, and the third wouldn't make the sum zero. Am I right?
| Another proof using symmetric polynomials/Vieta's formulas. This generalizes to any field.
Let $\sigma_1=x+y+z$, $\sigma_2=xy+yz+xz$, and $\sigma_3=xyz$ be the elementary symmetric polynomials. In order to make sense of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$, we can assume that $xyz\ne 0$. Then
$\sigma_3(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})=\sigma_2$, so $\sigma_2=0$.
Thus we know $\sigma_2=\sigma_3=0$. Now by Vieta's formulas, $x$, $y$, and $z$ are the roots of the cubic polynomial $t^3-\sigma_1t^2+\sigma_2t-\sigma_3=t^3-xyz$. Hence $x^3-xyz=0$, and similarly for $y$ and $z$, so we get
$$x^3=y^3=z^3=xyz.$$
Dividing this equation by $x^3$, we get
$$\newcommand{\of}[1]{\left({#1}\right)}1=\of{\frac{y}{x}}^3=\of{\frac{z}{x}}^3=\frac{y}{x}\frac{z}{x}.$$
Now there are two cases. The first case is that $y/x=z/x=1$, in which case $x=y=z$. Together with $x\ne 0$ and $x+y+z=3x=0$, we see that $3=0$, so this case can only happen if the characteristic of the field is 3. In that case, $x=y=z=c$ for any nonzero constant $c$ gives a solution. In particular this can't happen for $\Bbb{Q}$, so there are no integral solutions to the equation in this case.
The other case is that $y/x=\omega$ where $\omega\ne 1$, but $\omega^3=1$. This can only happen when the characteristic of $K$ is not three, since in characteristic three $(t^3-1)=(t-1)^3$, so in characteristic three, the only cube root of unity is 1. (Note that $\omega^3=1$ and $\omega\ne 1$, together imply that $\omega$ satisfies the polynomial $1+t+t^2$.)
Then for any nonzero constant $c$, $x=c$, $y=\omega c$, $z=\omega^2 c$ gives a valid solution and these are the only solutions, since $x+y+z=c+\omega c+\omega^2 c= c(1+\omega+\omega^2)=0$, and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z} = \frac{1}{c}+\frac{1}{c\omega}+\frac{1}{c\omega^2}=\frac{1}{c}\of{1+\omega^2+\omega}=0$. However, once again, there are no primitive cube roots of one in $\Bbb{Q}$, so this case cannot yield any integral solutions.
Note however that there are solutions over $\Bbb{C}$ (namely $1,\omega,\omega^2$, where $\omega=e^{2\pi i/3}$).
To summarize, we have a complete description of solutions to these equations for any field $K$.
The form of the solutions depends on the characteristic of $K$. If the characteristic is 3, the solutions are of the form $x=y=z=c\ne 0$ for any nonzero constant $c$.
Otherwise, there are solutions in $K$ if and only if the equation $1+t+t^2$ has a solution $\omega$ in $K$, in which case the solutions are precisely of the form $x=c$, $y=c\omega$, $z=c\omega^2$ for any nonzero constant $c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2622967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 9,
"answer_id": 2
} |
How do I interpret this sum? So if the sum of $n$ integers $\ge 1$ equal $\frac{n(n+1)}2$. Then my book goes on and says $1 + 2 + 3 +\ldots + 2n = \frac{2n(2n + 1) }2$.
I'm confused about what $1 + 2 + 3 + \ldots +2n$ means. If the sequence is $1, 2, 3, 4$ then where does $2n$ have to do with the $n$th number?
| Let $n = 4$ and $m = 2n = 8$.
Then $1 + 2 + 3 + 4 + 5 +6+7+8 = 36$.
$1+2+3+4+5+6+7+8 = \frac {8(8+1)}{2}= \frac {2*4(2*4 + 1)}2 = 36$.
$1 + 2 + 3 + ....... + m = \frac {m(m+1)}{2}$ and if $m = 8$ then $1 + 2 + 3 +....... + 8 = \frac {8(8+1)}{2} = 36$.
And $1+2+3+...... + 2n -1 + 2n = \frac {2n(2n+1)}2$ and if $n= 4$ then $1 + 2+ 3 + ...... + 8 = \frac {(2*4)(2*4 + 1) } 2 = 36$.
It's nothing profound or difficult.
$1 + 2 + 3 +.......... + whatever = \frac {whatever(whatever + 1)}2$.
And $whatever$ can be .... whatever you want.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2623017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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} |
is there away to give $= \sqrt[3]{5 + 10i } + \sqrt[3]{5- 10i}$ question is there away to give
$$= \sqrt[3]{5 + 10i }
+ \sqrt[3]{5- 10i}$$
in just reals?
I am thinkting maybe use some clever algebra raise it to like 3 and magically imaginary terms dissapear. Or expressed the complex numbers in polar form but idk
Background (might delete if this is not really the question)
Let $$ m(x)=x^3-15x-10 $$
Find the roots of $m(x)$ using the cubic formula and show that they are all real
Using Howell complex approach
Lets change variables $$ \begin{aligned}
&z^3-15z-10
\\=&z^3+0z^2-15z-10
\end{aligned}$$
so, general form is denoted as
$$ z^3+a_2z^2+a_1z+a_0=0$$
so $$\begin{aligned}
a_2&=0
\\a_1&=-15
\\a_0&=-10
\end{aligned}$$
we get by substituting that $z=x-a_2/3$. $a_2=0$ its already depressed cubic but so theses steps are unnecesary for this one problem but will use them for other problems
$$ x^3+bx+c$$
where $$\begin{aligned}
b&=a_1-\frac{a_2^2}{3}
=-15-\frac{0^2}{3}=-15
\\c&=\frac{-a_1a_2}{3}+\frac{2}{27}a_2^3+a_0
=\frac{-(-15)(0)}{3}+\frac{2}{27}(0)^3+-10=-10
\end{aligned} $$
our depressed cubic is of $$ x^3-15x-10 $$
Ferro-Tartaglia Formula
$$ x=\sqrt[3]{\frac{-c}{2} + \sqrt{\frac{c^2}{4} +\frac{b^3}{27}}}\
+ \sqrt[3]{\frac{-c}{2} - \sqrt{\frac{c^2}{4} -\frac{b^3}{27}}}\ $$
in our case $$
\begin{aligned}
x&=\sqrt[3]{\frac{10}{2} + \sqrt{\frac{100}{4} +\frac{(-15)^3}{27}}}\
+ \sqrt[3]{\frac{10}{2} - \sqrt{\frac{100}{4} +\frac{(-15)^3}{27}}}\
\\&= \sqrt[3]{5 + \sqrt{25 -125}}
+ \sqrt[3]{5- \sqrt{25 -125}}
\\ &= \sqrt[3]{5 + 10i }
+ \sqrt[3]{5- 10i}
\end{aligned}
$$
| Since
$$
(5+10i)^{1/3}=\sqrt5\,e^{i\arctan(2)/3}
$$
and
$$
(5-10i)^{1/3}=\sqrt5\,e^{-i\arctan(2)/3}
$$
Thus,
$$
\bbox[5px,border:2px solid #C0A000]{(5+10i)^{1/3}+(5-10i)^{1/3}=2\sqrt5\cos\left(\frac{\arctan(2)}3\right)}
$$
Another Approach to the Original Question
Let $\alpha=2\sqrt5$. Then, using $\cos(3x)=4\cos^3(x)-3\cos(x)$, we get
$$
\begin{align}
10
&=\frac{\alpha^3}4\cos(3x)\\
&=\alpha^3\cos^3(x)-\frac34\alpha^3\cos(x)\\[5pt]
&=\underbrace{\alpha^3\cos^3(x)}_{z^3}-15\,\underbrace{\ \alpha\cos(x)\ }_z
\end{align}
$$
Thus, with $\cos(3x)=\frac{40}{\alpha^3}=\frac1{\sqrt5}$, we get that $z=2\sqrt5\cos(x)$ is a root. This gives $3$ real roots
$$
z_1=2\sqrt5\cos\left(\frac13\arccos\left(\frac1{\sqrt5}\right)\right)
$$
$$
z_2=2\sqrt5\cos\left(\frac{2\pi}3+\frac13\arccos\left(\frac1{\sqrt5}\right)\right)
$$
$$
z_3=2\sqrt5\cos\left(\frac{4\pi}3+\frac13\arccos\left(\frac1{\sqrt5}\right)\right)
$$
Noting that $\arccos\left(\frac1{\sqrt5}\right)=\arctan(2)$, we get that $z_1=(5+10i)^{1/3}+(5-10i)^{1/3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2623575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Coefficients of the stirling's series expansion for the factorial. Knowing the Stirling's approximation for the Gamma function (factorial) for integers:
$$\Gamma(n+1)=n!\approx \sqrt{2\pi n}n^ne^{-n}\bigg(1+\frac{a_1}{n}+\frac{a_2}{n^2}+\cdots\bigg)$$
Using the above approximation one can write:
$$(n+1)!=\sqrt{2\pi(n+1)}(n+1)^{n+1}e^{-(n+1)}\bigg(1+\frac{a_1}{n+1}+\frac{a_2}{(n+1)^2}+\cdots\bigg)$$
We know that following recursion holds:
$$(n+1)!=(n+1)n!$$
One can rewrite this:
$$(n+1)!=(n+1)\sqrt{2\pi n}n^ne^{-n}\bigg(1+\frac{a_1}{n}+\frac{a_2}{n^2}+\cdots\bigg)$$
All this comes from: https://www.csie.ntu.edu.tw/~b89089/link/gammaFunction.pdf (Page 8-9)
Then the author gives this expansion to calculate the $a_k$ coefficients when $n$ becomes large.
Comparing these two expressions for $(n+1)!$ gives
$$1+\frac{a_1}{n}+\frac{a_2}{n^2}+\cdots=\bigg(1+\frac{1}{n}\bigg)^{n+1/2}e^{-1}\bigg(1+\frac{a_1}{n+1}+\frac{a_2}{(n+1)^2}+\cdots\bigg)$$
Then he says, that after "classical series expansion" this equals:
$$1+\frac{a_1}{n}+\frac{a_2-a_1+\frac{1}{12}}{n^2}+\frac{\frac{13}{12}a_1-2a_2+a_3+\frac{1}{12}}{n^3}+\cdots$$
I don't understand how he got there. Only thing that came to my mind was, that as $n\to\infty$ $\big(1+\frac{1}{n}\big)^n$ goes to $e$ but then we are left with $\big(1+\frac{1}{n}\big)^{1/2}$. When i expand this into binomial series, i get:
$$\bigg(1+\frac{1}{2n}-\frac{1}{8n^2}+\frac{1}{16n^3}\cdots\bigg)\cdot\bigg(1+\frac{a_1}{n+1}+\frac{a_2}{(n+1)^2}+\cdots\bigg)$$
And I'm stuck here.
Or is there any other elementary way how to compute the coefficients $a_k$ of the stirling's series expansion for factorial/Gamma function?
|
We expand the series up to terms of $\frac{1}{n^3}$. We obtain for $\left|\frac{1}{n}\right|<1$
\begin{align*}
\color{blue}{\left(1+\frac{1}{n}\right)^{n+\frac{1}{2}}e^{-1}}
&=e^{-1}e^{\left(n+\frac{1}{2}\right)\ln\left(1+\frac{1}{n}\right)}\\
&=\exp(-1)\exp\left[\left(n+\frac{1}{2}\right)\left(\frac{1}{n}-\frac{1}{2n^2}+\frac{1}{3n^3}+\cdots\right)\right]\\
&=\exp(-1)\exp\left(1+\frac{1}{12n^2}-\frac{1}{12n^3}+\cdots\right)\\
&=\exp\left(\frac{1}{12n^2}-\frac{1}{12n^3}+\cdots\right)\\
&\color{blue}{=1+\frac{1}{12n^2}-\frac{1}{12n^3}+\cdots}\tag{1}
\end{align*}
and applying the binomial series expansion
we obtain
\begin{align*}
&\color{blue}{1+\frac{a_1}{n+1}+\frac{a_2}{(n+1)^2}+\frac{a_3}{(n+1)^3}+\cdots}\\
&\qquad =1+\frac{a_1}{n\left(1+\frac{1}{n}\right)}+\frac{a_2}{n^2\left(1+\frac{1}{n}\right)^2}
+\frac{a_3}{n^3\left(1+\frac{1}{n}\right)^3}+\cdots\\
&\qquad =1+\frac{a_1}{n}\left(1-\frac{1}{n}+\frac{1}{n^2}\cdots\right)
+\frac{a_2}{n^2}\left(1+\binom{-2}{1}\frac{1}{n}+\cdots\right)
+\frac{a_3}{n^3}\left(1+\cdots\right)\\
&\qquad\color{blue}{=1+a_1\cdot\frac{1}{n}+\left(-a_1+a_2\right)\frac{1}{n^2}
+\left(a_1-2a_2+a_3\right)\frac{1}{n^3}+\cdots}\tag{2}\\
\end{align*}
Putting (1) and (2) together
we obtain
\begin{align*}
&\left(1+\frac{1}{12n^2}-\frac{1}{12n^3}+\cdots\right)
\left(1+a_1\cdot\frac{1}{n}+\left(-a_1+a_2\right)\frac{1}{n^2}
+\left(a_1-2a_2+a_3\right)\frac{1}{n^3}+\cdots\right)\\
&\quad\color{blue}{ =1+a_1\cdot\frac{1}{n}+\left(\frac{1}{12}-a_1+a_2\right)\frac{1}{n^2}
+\left(-\frac{1}{12}+\frac{13}{12}a_1-2a_2+a_3\right)\frac{1}{n^3}+\cdots}
\end{align*}
and the claim follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2623910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
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