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Proving $\sqrt{3+\sqrt{13+4\sqrt{3}}} = 1+\sqrt{3}$. How would I show $\sqrt{3+\sqrt{13+4\sqrt{3}}} = 1+\sqrt{3}$?
I tried starting from the LHS, and rationalising and what-not but I can't get the result...
Also curious to how they got the LHS expression from considering the right.
| We have $$\sqrt{3+\sqrt{13+4\sqrt{3}}} = 1+\sqrt{3};$$ that is,
$$3+\sqrt{13+4\sqrt{3}}=4+2\sqrt{3},$$ I.e.,
$$\sqrt{13+4\sqrt{3}}=1+2\sqrt{3},$$ which is
$$13+4\sqrt{3}=13+4\sqrt{3}.$$ Now just work backwards.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2405667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
$\frac{1}{2^n}\binom{n}{n}+\frac{1}{2^{n+1}}\binom{n+1}{n}+...+\frac{1}{2^{2n}}\binom{2n}{n}=1$: short proof? The identity $\frac{1}{2^n}\binom{n}{n}+\frac{1}{2^{n+1}}\binom{n+1}{n}+...+\frac{1}{2^{2n}}\binom{2n}{n}=1$ arises from a question on probability in my textbook. A proof by induction on $n$, which exploits the fact that $\binom{a}{b}+\binom{a}{b+1}=\binom{a+1}{b+1}$, is straightforward but not enlightening.
Is it possible to find any very clever approaches? Via a combinatorial or probabilistic interpretation, for instance?
| Suppose you are flipping a coin until you get $n+1$ heads or $n+1$ tails. For $k=0,\dots,n$, what is the probability that you are done after exactly $n+1+k$ flips?
If the last flip was a tails, this means that in the former $n+k$ flips there were exactly $n$ tails, and this happens with probability $\frac{1}{2}\cdot\frac{1}{2^{n+k}}\binom{n+k}{n}$ (the first $\frac{1}{2}$ is due to assuming the last flip is tails). Same for heads, so the probability of finishing after exactly $n+1+k$ flips is $\frac{1}{2^{n+k}}\binom{n+k}{n}$.
Now just note that the number of flips always ends up between $n+1$ and $n+1+n$ (by pigeonhole principle).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2406516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 1
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Simplify $\frac1{\sqrt{x^2+1}}-\frac{x^2}{(x^2+1)^{3/2}}$ I want to know why
$$\frac1{\sqrt{x^2+1}} - \frac{x^2}{(x^2+1)^{3/2}}$$
can be simplified into
$$\frac1{(x^2+1)^{3/2}}$$
I tried to simplify by rewriting radicals and fractions. I was hoping to see a clever trick (e.g. adding a clever zero, multiplying by a clever one? Quadratic completion?)
\begin{align}
\frac{1}{\sqrt{x^2+1}} - \frac{x^2}{(x^2+1)^{3/2}} & = \\
& = (x^2+1)^{-1/2} -x^2*(x^2+1)^{-3/2} \\
& = (x^2+1)^{-1/2} * ( 1 - x^2 *(x^2+1)^{-1}) \\
& = ...
\end{align}
To give a bit more context, I was calculating the derivative of $\frac{x}{\sqrt{x^2+1}}$ in order to use newtons method for approximating the roots.
| $\sqrt x$ is just a shorthand for $x^{1/2}$. Hence we can multiply the two halves of the first fraction in the first term by $x^2+1$:
$$\frac1{\sqrt{x^2+1}}-\frac{x^2}{(x^2+1)^{3/2}}=\frac{x^2+1}{(x^2+1)^{3/2}}-\frac{x^2}{(x^2+1)^{3/2}}$$
and the target expression follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2407653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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In expansion of $(1+x+x^{2}+x^{3}+....+x^{27})(1+x+x^{2}+x^{3}+....x^{14})^{2}$ . Find the coefficient of $x^{28}$ I am not able to apply binomial theorem here
$(1+x+x^{2}+x^{3}+....+x^{27})(1+x+x^{2}+x^{3}+....x^{14})^{2}$
Please help me to find the coefficient of$ x^{28}$
Any help will be appreciated.
| Express is as a product of three factors $F_1,F_2,F_3$:
$$(1+x+x^{2}+x^{3}+....+x^{27})(1+x+x^{2}+x^{3}+....x^{14})^{2}=(1+\cdots +x^{27})(1+\cdots +x^{14})(1+\cdots +x^{14}).$$
Make up a table of powers of $x$:
$$\begin{array}{c|c|lcr}
m& n & \text{$F_1$} & \text{$F_2$} & \text{$F_3$} \\
\hline
& 1 & 14 & 14 & 0 \\
& \downarrow & \downarrow & \downarrow & \vdots \\
1 & 14 & 27 & 1 & 0 \\
& 1 & 13 & 14 & 1 \\
& \downarrow & \downarrow & \downarrow & \vdots \\
1 &15 & 27 & 0 & 1 \\
& 1 & 12 & 14 & 2 \\
&\downarrow & \downarrow & \downarrow & \vdots \\
2 &15 & 26 & 0 & 2 \\
&1 & 11 & 14 & 3 \\
&\downarrow & \downarrow & \downarrow & \vdots \\
3 & 15 & 25 & 0 & 3 \\
& 1 & 10 & 14 & 4 \\
& \downarrow & \downarrow & \downarrow & \vdots \\
4 & 15 & 24 & 0 & 4 \\
& 1 & 9 & 14 & 5 \\
& \downarrow & \downarrow & \downarrow & \vdots \\
5 & 15 & 23 & 0 & 5 \\
& \cdots & \cdots & \cdots & \cdots \\
& 1 & 0 & 14 & 14 \\
& \downarrow & \downarrow & \downarrow & \vdots \\
14 & 15 & 14 & 0 & 14 \\
\end{array}$$
Hence:
$$1\cdot 14 + 14\cdot 15=224.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2408611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Is the double integral equal to the area? I have to compute the double integral $\int_0^{\sqrt{\frac{\pi}{2}}}\int_x^{\sqrt{\frac{\pi}{2}}} 2\sin(y^2)dydx$.
I drawed the region $0\leq x\leq \sqrt{\frac{\pi}{2}}$ and $x\leq y\leq \sqrt{\frac{\pi}{2}}$. This is a triangle with height and base equal to $\sqrt{\frac{\pi}{2}}$, right?
The double integral is equal to the are of the triangle, or not?
So, is the double integral equal to $\frac{1}{2}\cdot \sqrt{\frac{\pi}{2}}\cdot \sqrt{\frac{\pi}{2}}=\frac{\pi}{4}$ ?
| Hint. Note that the double integral over that triangle can be written as iterated integrals in two ways.
\begin{align*}
\int_{x=0}^{\sqrt{\frac{\pi}{2}}}\left(\int_{y=x}^{\sqrt{\frac{\pi}{2}}} 2\sin(y^2)dy\right)dx
&=\int_{y=0}^{\sqrt{\frac{\pi}{2}}}\left(\int_{x=0}^{y} 2\sin(y^2)dx\right)dy\\
&=\int_{y=0}^{\sqrt{\frac{\pi}{2}}}2\sin(y^2)\left(\int_{x=0}^{y} dx\right)dy\\
&=\int_{y=0}^{\sqrt{\frac{\pi}{2}}}2y\sin(y^2)dy.
\end{align*}
Can you take it from here?
P.S. The final result is different from what you have found. It is not the area of the triangle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2411927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve in positive integers, $ x^6+x^3y^3-y^6+3xy(x^2-y^2)^2=1$ Solve in positive integers, $$ x^6+x^3y^3-y^6+3xy(x^2-y^2)^2=1$$
My attempt :
$ x^3y^3+x^6-y^6+3xy(x^2-y^2)^2$
$=x^3y^3+3x^2y^2(x^2-y^2)+3xy(x^2-y^2)^2+(x^2-y^2)^3$
$=(xy+(x^2-y^2))^3$
$=(x^2+xy-y^2)^3 = 1$
so $x^2+xy-y^2 = 1$
Please suggest how to proceed.
Thank you, AmateurMathGuy and lhf.
Please check my work on Induction for Fibonacci sequence.
$$t_{k+2}-3t_{k+1}+t_{k}=0$$
$$y_{k+2}-3y_{k+1}+y_{k}=0$$
$(t_1,y_1)=(3,1)\to(x_1,y_1)=(1,1)$
$(t_2,y_2)=(7,3)\to(x_2,y_2)=(2,3)$
$(t_3,y_3)=(18,8)\to(x_3,y_3)=(5,8)$
$(t_4,y_4)=(47,21)\to(x_4,y_4)=(13,21)$
Since $1, 2, 3, 5, 8, 13, 21$ are in Fibonacci sequence, we predict that $(x_n,y_n)=(F_{2n-1},F_{2n})$
and will prove by Induction.
$(x_n,y_n)=(F_{2n-1},F_{2n})$ is true for $n=1, 2, 3, 4$
Suppose that $(x_k,y_k)=(F_{2k-1},F_{2k})$ is true,
Since $y_{k+1}-3y_k+y_{k-1}=0$, so $y_{k+1}=3y_k-y_{k-1}=3F_{2k}-F_{2k-2}$
$=3F_{2k}-(F_{2k}-F_{2k-1})=2F_{2k}+F_{2k-1}=F_{2k}+F_{2k+1}=F_{2k+2}$
$x_{k+1}=\frac{t_{k+1}-y_{k+1}}{2}=\frac{(3t_k-t_k)-3y_{k-1}-y_{k-1}}{2}$
$=3\left(\frac{t_k-y_k}{2}\right)-\left(\frac{t_k-1-y_{k-1}}{2}\right)=3x_k-x_{k-1}$
Similarly, $x_{k+1}=F_{2k+1}$, so $(x_n,y_n)=(F_{2n-1},F_{2n})$
Answer : $(x_n,y_n)=(F_{2n-1},F_{2n})$
| Hint:
You should keep the cube
$$(x^2+xy-y^2)^3 = 1$$
$$(x^2+xy-y^2)^3 -1^3 = 0$$
You should factorize with the formula $a^3-b^3= (a-b) (a² + ab + b²))$
Edit:
Then solve the equations.But as pointed by Alex, $ a^2+a=-1$ has no solution in $ Z $, so only $a=1$ need to be solved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2412268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving/disproving $∀n ∈ \text{positive integers}$, $\left\lceil{\frac{4n^2+1}{n^2}}\right \rceil = 5$? Is my proof getting anywhere for proving/disproving $∀n ∈ \text{postive integers}$, $$\left\lceil{\frac{4n^2+1}{n^2}}\right \rceil = 5\text{?}$$
$$\left\lceil{\frac{4n^2}{n^2}}+\frac{1}{n^2}\right \rceil = 5$$
$$\left\lceil{\frac{1}{n^2}}\right \rceil = 5 - 4$$
Therefore, for $n \in \mathbb R$, where $n \neq 0$
$$4 < \left\lceil{\frac{4n^2}{n^2}}+\frac{1}{n^2}\right \rceil \le 5$$
$$4 < \left\lceil{4}+\frac{1}{n^2}\right \rceil \le 5$$
$$0 < \left\lceil\frac{1}{n^2}\right \rceil \le 1$$
I feel kind of without purpose once I hit point. Should I be scrapping the inequality and solving for $n$ instead?
| $$\left\lceil \frac{4n^2+1}{n^2} \right\rceil=5\tag{1}$$
means
$$4 < \frac{4n^2+1}{n^2} \le 5\tag{2}$$
so try to prove (2). This should be less error prone that transforming expressions with ceilings in them.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2414453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Sequence : $a_k a_{k+2} +1 = a^2_{k+1}$ Determine all integers $b$ for which there exists a sequence $a_0, a_1, a_2,\ldots $ of rational numbers
satisfying $a_0=0, a_2=b$ and $a_k a_{k+2} +1 = a^2_{k+1}$ for $k=0, 1, 2, \ldots$
that contains at least one nonintegral term.
My attempt :
$a_0=0, a_2=b$, so $a_0a_2+1=a^2_1$ then $a_1=\pm1$
$a_1a_3+1=a^2_2$ so $a_3=\pm(b^2-1)$
We have $a_0, a_1, a_2, a_3 = 0, \;1, \;b, \;b^2-1$ or $\;0, \;-1, \;b, \;-b^2+1$ respectively.
Since $a_k a_{k+2} +1 = a^2_{k+1}$ and $a_{k-1} a_{k+1} +1 = a^2_k$
so $a^2_k - a_{k-1} a_{k+1} = a^2_{k+1} - a_k a_{k+2}$
then $a^2_k + a_k a_{k+2} = a^2_{k+1} + a_{k-1} a_{k+1}$
$a_k(a_k + a_{k+2}) = a_{k+1}(a_{k+1} + a_{k-1})$
we have $\frac{a_k \;+ \;a_{k+2}}{a_{k+1}} = \frac{a_{k+1} \;\; + \;a_{k-1}}{a_k}$, for $a_k, a_{k+1} \not=0$, $\forall k\in \mathbb{N}$
that is $\frac{a_k \;+ \;a_{k+2}}{a_{k+1}} = \frac{a_0 \;+\; a_2}{a_1} = b$ or $-b$
Thus, $a_{k+2} = b\cdot a_{k+1}-a_k $ or $a_{k+2} = -b\cdot a_{k+1}-a_k $
For $a_{k+2} = b\cdot a_{k+1}-a_k $ , we have $x^2-bx=1 = 0$
and $a_n = \frac{1}{\sqrt{b^2-4}}\left[\left(\frac{b+\sqrt{b^2-4}}{2}\right)^n - \left(\frac{b-\sqrt{b^2-4}}{2}\right)^n\right]$, so $b \not= \pm 2$
Since $a_l a_{l+2} +1 = a^2_{l+1}$ , if there exists $l$ such that $a_l = 0$, then $a^2_{l+1} = 1$ and $a_{l+1} = \pm1$
that is, we can always take $a_{l+2} \not\in \mathbb{Z}$ which make the sequence contain nonintegral term.
Hence, to be the integer sequence, only $a_0$ can be equal to $0$.
If $a_2=b=0$ or $a_3=b^2-1=0$, i.e. $b=\pm 1$, the sequence can contain nonintegral term.
For $b \leq -3$ and $b \geq 3$,
since $a_n = \frac{1}{\sqrt{b^2-4}}\left[\left(\frac{b+\sqrt{b^2-4}}{2}\right)^n - \left(\frac{b-\sqrt{b^2-4}}{2}\right)^n\right]$
if $b$ is even, all terms in the sequence will be integers.
| Notice that $a_1 = \pm 1$. For simplicity, let us write $c = b a_1$. Then consider a sequence $(f_n)$ that solves the recurrence relation
$$ f_{n+2} = c f_{n+1} - f_n, \qquad f_0 = a_0, \quad f_1 = a_1. $$
If we consider matrices
$$ A_n = \begin{pmatrix} f_{n+2} & f_{n+1} \\ f_{n+1} & f_n \end{pmatrix}, \qquad
P = \begin{pmatrix} c & -1 \\ 1 & 0 \end{pmatrix}, $$
then it is easy to check that
$$ A_n = P A_{n-1} = \cdots = P^n A_0. $$
So by taking determinant, we obtain
$$ f_{n+2}f_n - f_{n+1}^2 = \det (A_n) = \det(P)^n \det(A_0) = -1. $$
Then it follows from strong induction that $f_n = a_n$ whenever $a_k \neq 0$ for all $1 \leq k \leq n-2$.
If $a_m = 0$ for some $m$, then this forces that $a_{m + 1}^2 = 1$ and we have freedom to choose any value for $a_{m+2}$. And only in such case we can expect non-integral terms for $(a_n)$. So the problem boils down to determine the values of $b$ for which $f_n = 0$ for some $n \geq 2$.
To this end, we appeal to the general theory to compute the general term of $(f_n)$ and proceed our investigation.
*
*If $|b| = |c| \neq 2$, then we have
$$ f_n = \frac{a_1 (\alpha^n - \beta^n)}{\sqrt{c^2 - 4}}
\qquad \text{where} \quad
\alpha = \frac{c + \sqrt{c^2 - 4}}{2}, \quad
\beta = \frac{c - \sqrt{c^2 - 4}}{2}. $$
In particular, if $|b| = |c| \geq 3$ then $\alpha$ and $\beta$ are real numbers such that $|\beta| \neq |\alpha|$, hence it follows that $f_n \neq 0$ for all $n \geq 1$.
On the other hand, if $|c| = 1$ then $\alpha^6 = \beta^6 = 1$ and hence $f_6 = 0$. The case $|c| = 0$ is much easier to investigate, as $\alpha^2 = \beta^2$ and hence $f_2 = 0$.
*If $|b| = |c| = 2$, then
$$ f_n = n \left(\frac{c}{2}\right)^{n+1}. $$
So it follows that $f_n \neq 0$ for all $n \geq 1$.
Therefore, there exists a sequence $(a_n)$ with non-integral term if and only if $b \in \{-1, 0, 1\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2414946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integration of $\int_{0}^{\infty}\frac{dw}{1+\left ( \frac{w}{B} \right )^4}$ with the help of signal properties.
Find the integral
$$\int_{0}^{\infty}\frac{dw}{1+\left ( \frac{w}{B} \right )^4}$$
where $B$ is a constant.
This integration i tried by normal method that gives the result $\frac{\pi B}{2\sqrt2}$ but it goes so lengthy like this
is there any signal properties that i use to solve it in simple steps?
| Just want to share a "smart" method.
(Too long for comment)
Here is a fast way to evaluate the indefinite integral.
Sub $ t=Bx$,
Then,
\begin{align*}
\int \frac 1{1+x^4}\,dx&=\frac 12\left(\int \frac {1+x^2}{1+x^4}\,dx+\int \frac {1-x^2}{1+x^4}\,dx\right)\\
&=\frac 12\left(\int \frac {\color{red}{\left(\frac 1 {x^2}+1\right)dx}}{\frac 1{x^2}+x^2}+\int \frac {\color{blue}{\left(\frac 1 {x^2}-1\right)dx}}{\frac 1{x^2}+x^2}\right)\\
&=\frac 12\left(\int \frac {\color{red}{d\left(x-\frac 1 x\right)}}{\left(x-\frac 1 x\right)^2+2}+\int \frac {\color{blue}{-d\left(x+\frac 1 x\right)}}{\left(x+\frac 1 x\right)^2-2}\right)\qquad\quad\\
&= \frac 12\left(\frac 1{\sqrt2} \tan^{-1}\left(\frac {x-\frac 1 x} {\sqrt2}\right)+\frac 1{\sqrt2} \tanh^{-1}\left(\frac {x+\frac 1 x} {\sqrt2}\right)\right).
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2419814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Laurent Series - What am I doing wrong? I want to obtain the Laurent series around the origin of:
$$ f(z) = \frac{1}{z^2 \sinh z}$$
My plan is to obtain first the laurent series of: $f(z) = \frac{1}{\sinh z}$ and then divide the terms by $z^2$
So I know that: $\frac{1}{\sinh z} = \ln(z + \sqrt{1+z^2}) = \ln(\frac{z}{\sqrt{1+z^2}} + 1) +\ln(\sqrt{1+z^2}) = \ln(\frac{z}{\sqrt{1+z^2}} + 1) +\frac{1}{2}\ln(1+z^2)$
using
$ \ln(z+1) = \sum^{\infty}_{n=1} (-1)^{n-1} \frac{z^n}{n}$
I tried to use this series and write:
$\frac{1}{\sinh z} = \sum^{\infty}_{n=1} \frac{(-1)^{n-1}}{n} \frac{z^n}{(\sqrt{1+z^2})^{n/2}}+ \frac{1}{2}\sum^{\infty}_{n=1} \frac{(-1)^{n-1}}{n} \frac{z^2n}{2}$
but by doing this I dont obtain the right answer, which is:
| It's the inverse of the hyperbolic sine that has that expression (yet another reason to write the inverse as $\arg\sinh{z}$). $1/\sinh{z}=\operatorname{csch}{z}$ is
$$ \frac{2}{e^z-e^{-z}} = 2\left( 2z+\frac{2}{3!}z^3 + \frac{2}{5!}z^5 + \dotsb \right)^{-1}, $$
and then take out a factor of $z$ and use the binomial expansion of $(1+w)^{-1}$.
| {
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"url": "https://math.stackexchange.com/questions/2420286",
"timestamp": "2023-03-29T00:00:00",
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How to prove that $\lim_{n \to \infty} \frac{\sqrt {n^2 +2}}{4n+1}=\frac14$?
How to prove that $\lim_{n \to \infty} \frac{\sqrt {n^2 +2}}{4n+1}=\frac14$?
I started my proof with Suppose $\epsilon > 0$ and $m>?$ because I plan to do scratch work and fill in.
I started with our conergence definition, i.e. $\lvert a_n - L \rvert < \epsilon$
So $\lvert \frac{\sqrt {n^2 +2}}{4n+1} - \frac {1}{4} \rvert$ simplifies to $\frac {4\sqrt {n^2 +2} -4n-1}{16n+4}$
Now $\frac {4\sqrt {n^2 +2} -4n-1}{16n+4} < \epsilon$ is
simplified to $\frac {4\sqrt {n^2 +2}}{16n} < \epsilon$ Then I would square everything to remove the square root and simplify fractions but I end up with $n> \sqrt{\frac{1}{8(\epsilon^2-\frac{1}{16}}}$
We can't assume $\epsilon > \frac{1}{4}$ so somewhere I went wrong. Any help would be appreciated.
| $\lim_{n \to \infty} \frac{\sqrt {n^2 +2}}{4n+1}=\frac14
$
$\begin{array}\\
\dfrac{\sqrt {n^2 +2}}{4n+1}
&\gt \dfrac{n}{4n+1}\\
&= \dfrac{n+1/4-1/4}{4n+1}\\
&=\dfrac14- \dfrac{1}{4(4n+1)}\\
\end{array}
$
so
$\dfrac{\sqrt {n^2 +2}}{4n+1}-\dfrac14
\gt - \dfrac{1}{4(4n+1)}
$.
Since
$(n+1/n)^2
=n^2+2+1/n^2
\gt n^2+2
$,
$\sqrt{n^2+2}
\lt n+1/n
$
so
$\begin{array}\\
\dfrac{\sqrt {n^2 +2}}{4n+1}-\dfrac14
&\lt \dfrac{n+1/n}{4n+1}-\dfrac14\\
&= \dfrac{4n+4/n-(4n+1)}{4(4n+1)}\\
&= \dfrac{4/n-1}{4(4n+1)}\\
&= \dfrac{1}{n(4n+1)}-\dfrac{1}{4(4n+1)}\\
&\lt 0
\qquad\text{for } n > 4\\
\end{array}
$
Therefore,
for $n > 4$,
$|\dfrac{\sqrt {n^2 +2}}{4n+1}-\dfrac14|
\lt \dfrac{1}{4(4n+1)}
$.
By choosing
$\dfrac{1}{4(4n+1)}
\lt \epsilon
$,
or
$n
\gt \dfrac14(\dfrac1{4\epsilon}-1)
$,
the difference is less then $\epsilon$.
Two notes:
Choosing
$n \gt \dfrac1{16\epsilon}$
is sufficient.
From the above,
$0
\lt \dfrac{\sqrt {n^2 +2}}{4n+1}-\dfrac14+\dfrac{1}{4(4n+1)}
\lt \dfrac{1}{n(4n+1)}
$.
| {
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} |
Find all triples $(a, b, c) \in \mathbb{R} : ab + c = bc + a = ca + b = 4$ Any help on this question would be great. Ive found that (1,1,3), (3,1,1), (1,3,1) but am confused on how to find more anymore (if there are any).Any help would be appreciated.
| If $a b+c=4$ then $a=\frac{4-c}{b}$ assuming that $b\not=0$ and then $b c+a =4$ which is $b c+\frac{4-c}{b}=4$ so $c(b-\frac{1}{b})+\frac{4}{b} = 4$ so $c= \frac{4-\frac{4}{b}} { b-\frac{1}{b}} = \frac{4}{b+1}$ assuming that $b\not -1,1$ and to the final equation we get that $c a+b=4$ which is $c(\frac{4-c}{b})+b=4$ which is $\frac{4}{b+1}(\frac{4-\frac{4}{b+1}}{b})+b=4$ and solving for $b$ we get that $b=3,\frac{1}{2}(-1+\sqrt{17}),\frac{1}{2}(-1-\sqrt{17})$ so $c=1,\frac{1}{2}(-1+\sqrt{17}),\frac{1}{2}(-1-\sqrt{17})$ and $a=1,\frac{1}{2}(-1+\sqrt{17}),\frac{1}{2}(-1-\sqrt{17})$ so the only solutions are :
$(1,3,1)$ and all its permutations and $(\frac{1}{2}(-1+\sqrt{17}),\frac{1}{2}(-1+\sqrt{17}),\frac{1}{2}(-1+\sqrt{17})),(\frac{1}{2}(-1-\sqrt{17}),\frac{1}{2}(-1-\sqrt{17}),\frac{1}{2}(-1-\sqrt{17}))$
| {
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"timestamp": "2023-03-29T00:00:00",
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Probability that a point is closer to a side than a diagonal So I have a rectangle in which a point is randomly chosen. One side is $a$ and the other is $b=a\sqrt3$. I am supposed to find the probability that a point is closer to a side than to the closest diagonal.
I have found the probability that the point is closer to $a$ (0.71) and to $b$ (0.24). Now I was wondering how I can put these two probabilities together to form the asked-for probability. Thanks
| The red areas are of points closer to a side of the rectangle. Its area is the sum of the triangles, two acutangle and two obtusangle.
The two acutangle are each $1/3$ of the equilateral triangle having side $1$ and half diagonals. A diagonal is $\sqrt{3+1}=2$ so half diagonal is $1$ (that's why they are equilateral).
So the area of the two acutangle is $2\times \dfrac{1}{3} \cdot\dfrac{\sqrt 3}{4}=\dfrac{\sqrt 3}{6}$
As their sides are the angle bisector of the $30°$ angle the two obtusangle have the vertex that is the incentre (centre of the inscribed circle) of the bigger triangle formed by the diagonals whose area is $\dfrac{\sqrt 3}{4}$ and whose perimeter is $2+\sqrt 3$.
The radius of the inscribed circle is $h=\dfrac{area}{half\,perimeter}=\dfrac{\frac{\sqrt 3}{4}}{\frac{2+\sqrt 3}{2}}=\dfrac{\sqrt{3}}{2 \left(2+\sqrt{3}\right)}$
The two obtusangle triangle have an area $\sqrt{3}\times \dfrac{\sqrt{3}}{2 \left(2+\sqrt{3}\right)}=\dfrac{3}{2 \left(2+\sqrt{3}\right)}$
Therefore the red area is $\dfrac{\sqrt 3}{6}+\dfrac{3}{2 \left(2+\sqrt{3}\right)}=\dfrac{1}{3} \left(9-4 \sqrt{3}\right)$
The area of the rectangle is $\sqrt 3$ so the probability of getting a point closer to the sides of the rectangle is $p=\dfrac{\frac{1}{3} \left(9-4 \sqrt{3}\right)}{\sqrt 3}=\dfrac{1}{3} \left(3 \sqrt{3}-4\right)\approx 0.3987$
| {
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Prove $\lim_{x\to 2} \sqrt{x^2+5} = 3$ As stated in the title, I need to prove that $\lim_{x\to 2} \sqrt{x^2+5} = 3$ using only the precise definition of a limit.
For any given $\varepsilon \gt 0$, there exists a $\delta = $
Such that $0 \lt \lvert x-2 \rvert \lt \delta \Rightarrow \lvert \sqrt{x^2+5} \rvert \lt \varepsilon $
I've attempted to convert $\sqrt{x^2+5}$ into the following:
$\sqrt{x^2+5} = \sqrt{(x^2-4)+9}\\
\sqrt{x^2+5} = \sqrt{(x^2-4x+4)+4x +1} = \sqrt{(x-2)^2+4(x-2)+9} $
I have a hunch that I am heading in the wrong direction. Can I get some advice on what I might be doing wrong?
| You're doing fine! So, in order to have$$\left|\sqrt{(x-2)^2+4(x-2)+9}-3\right|<\varepsilon,$$all you need is to have\begin{multline}\left|\sqrt{(x-2)^2+4(x-2)+9}-3\right|.\left|\sqrt{(x-2)^2+4(x-2)+9}+3\right|<\\<\varepsilon.\left|\sqrt{(x-2)^2+4(x-2)+9}+3\right|,\end{multline}or, in other words,$$\bigl|(x-2)^2+4(x-2)\bigr|<\varepsilon.\left|\sqrt{(x-2)^2+4(x-2)+9}+3\right|.$$In order for this to happen, all you need is that $\bigl|(x-2)^2+4(x-2)\bigr|<3\varepsilon$. So, pick $\delta$ such that $\delta\leqslant\frac38\varepsilon$ and also $\delta\leqslant\sqrt{\frac32\varepsilon}$. Then, if $|x-2|<\delta$,$$\bigl|(x-2)^2+4(x-2)\bigr|\leqslant|x-2|^2+4|x-2|\leqslant\frac32\varepsilon+\frac32\varepsilon=3\varepsilon.$$
| {
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Finding a general formula for two squares whose sum = 2 Presumably, a formula for u & v that generates all the possible u & v such that $u^2 + v^2 = 2$.
With the limitation that the problem must be solved without using roots or fractional powers assuming that u & v are only required to be real numbers.
I'm not exactly sure where to start.
| As $(1-t^2)^2+(2t)^2=(1+t^2)^2$, we can parametrize $a^2+b^2=1$ by letting $a=\frac{1-t^2}{1+t^2}$, $b=\frac{2t}{1+t^2}$.
Now to get to $u^2+v^2=2$ without introducing $\sqrt 2$, just take $u=a+b$, $v=a-b$ (which makes $u^2+v^2=a^2+2ab+b^2+a^2-2ab+b^2=2(a^2+b^)=2$), so take
$$\tag1 u = \frac{1-t^2+2t}{1+t^2},\qquad v = \frac{1-t^2-2t}{1+t^2}.$$
Note that as $|t|$ runs from $0$ to $\infty$, $a$ rund from $1$ (inclusinve) to $-1$ (exclusive). And negating $t$ keeps $a$ while negating $b$. Thus $(a,b)$ above really runs through all of $S^1$ except the single point $(-1,0)$. Accordingly, $(1)$ parametrizes all points with $u^2+v^2=2$ with the single exception $(-1,-1)$. In order to really catch all pairs, you can either accept $t=\infty$ as parameter value, or use the double angle formulas
$\sin 2x=2\sin x\cos x$, $\cos2x=\cos^2x-\sin^2x$ to run around the circle (almost) twice. That is, we'd take $c=2ab$, $d=b^2-a^2$ and then $u=c+d$, $v=c-d$.
| {
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How to determine the $\lim_{n\to \infty} \frac{1+2^2+\ldots+n^n}{n^n}=1$. I stuck to do this,
$$\lim_{n\to \infty} \frac{1+2^2+\ldots+n^n}{n^n}=1.$$
The only thing I have observed is $$ 1\le \lim_{n\to \infty} \frac{1+2^2+\ldots+n^n}{n^n}$$ I am unable to get its upper estimate so that I can apply Sandwich's lemma.
| $$\lim_{n \rightarrow \infty} \frac{1^1 + 2^2 + \dots + (n-1)^{n-1} + n^n}{n^n} = \lim_{n \rightarrow \infty} \left(\frac{1^1}{n^n} + \frac{2^2}{n^n} + \dots + \frac{(n-1)^{n-1}}{n^n} + 1 \right) $$
Let's examine that second to last term:
$$
\begin{align}
\lim_{n \rightarrow \infty} \frac{(n-1)^{n-1}}{n^n} &= \lim_{n \rightarrow \infty} \frac{1}{n}\frac{(n-1)^{n-1}}{n^{n-1}} \\
&= \lim_{n \rightarrow \infty} \frac{1}{n} \left( \frac{n-1}{n} \right)^{n-1}< \lim_{n \rightarrow \infty} \frac{1}{n} \cdot 1\\
&= 0
\end{align}$$
The second largest term is zero, none of the terms are negative, the sum is 1:
$$\lim_{n \rightarrow \infty} \frac{\sum\limits_{m=1}^{n} m^m}{n^n} = 1$$
| {
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"timestamp": "2023-03-29T00:00:00",
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find the total number of possible way to reach to a particular sum suppose you have given a sum like : 5.
we have to find the total number of possible way to reach to 5.
for example
1 + 1 + 1 + 1 + 1 = 5
2 + 1 + 1 + 1 = 5
1 + 2 + 1 + 1 = 5
1 + 1 + 2 + 1 = 5
1 + 1 + 1 + 2 = 5
2 + 2 + 1 = 5
2 + 1 + 2 = 5
1 + 2 + 2 = 5
1 + 1 + 3 = 5
1 + 3 + 1 = 5
3 + 1 + 1 = 5
2 + 3 = 5
3 + 2 = 5
1 + 4 = 5
4 + 1 = 5
5 = 5
above is the total number of way to reach to a particular sum, which is nothing but : 16
| By Stars and Bars the number of ordered $k-$tuples of positive integers that sum to $n$ is $\binom {n-1}{k-1}$. It follows that your answer is $$\sum_{k=1}^n\binom {n-1}{k-1}=\sum_{k=0}^{n-1}\binom {n-1}k=2^{n-1}$$
Direct Proof: List $n$ $*'s$ with blanks between them. There are $n-1$ blanks. your decompositions are entirely determined by an arbitrary subset of those $n-1$ blanks.
| {
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Prove that $\mathbb{Q}$ is dense. Let $a$ and $b$ be the real numbers with $a<b$. Prove that there are integers $m$ and $n\neq 0$ so that
\begin{align*}
a<\frac{m}{n}<b.
\end{align*}
Is my attempt correct?
Proof
Let $x=(b-a)>0$ and $y=1>0$, then there is a positive integer $n$ such that
\begin{align*}
n(b-a)>1\implies nb>na+1 &&\text{By the Archiemedean Property}.
\end{align*}
If $na> 0$ and therefore bounded below, there is a least integer $m$ such that $na<m$ by the Well-Ordering Principle. This means the below is true
\begin{align*}
m-1<na<m
\end{align*}
Then, we have
\begin{align*}
na<m<na+1<nb\implies na<m<nb
\end{align*}
Note that if $na=0$, the above is still true as its bounded both above and below, so there is a least integer $m$ such that $m-1<na<m$ by the Well-Ordering Principle.
If $na < 0$ and therefore bounded above, there is a greatest integer $m$ such that $na>m$ by the Well-Ordering Principle. This means the below is true
\begin{align*}
m<na<m+1<na+1<nb
\end{align*}
Let $m+1=m'$, we have
\begin{align*}
na<m'<nb
\end{align*}
So for any $x$, the below holds:
\begin{align*}
na<m<nb\implies a<\frac{m}{n}<b &&\text{because $n>0$.}
\end{align*}
| I have included some minor corrections in red.
If $na > 0$, there is a least integer $m$ such that $na < m$. This means the below is true
$$m-1 \color{red}{\leq} na < m$$
\begin{align*}
na<m \color{red}{\leq}na+1<nb\implies na<m<nb
\end{align*}
Similar for $na=0$.
If $na < 0$ and therefore bounded above, there is a greatest integer $m$ such that $na \color{red}{\geq}m$ by the Well-Ordering Principle. This means the below is true
\begin{align*}
m\color{red}{\leq}na<m+1\color{red}{\leq}na+1<nb
\end{align*}
Let $m+1=m'$, we have
\begin{align*}
na<m'<nb
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the following nonlinear system of equations:$ xy+xz+yz=12 , xyz=2+x+y+z$
Solve the following system of equations in $\Bbb R^+$:
$$
\left\{
\begin{array}{l}
xy+yz+xz=12 \\
xyz=2+x+y+z\\
\end{array}
\right.
$$
I did as follows.
First equation yields to: $(x+y+z)^2-x^2-y^2-z^2=24$. Now, using second equation we get: $x^2y^2z^2-4xyz=20+x^2+y^2+z^2$ . Here I stopped...
| I will show that the only positive solution is indeed $(2,2,2).$ $$\dfrac{xyz}{xy+yz+zx} = \dfrac{1}{6}+\dfrac{x+y+z}{12} = \dfrac{\sqrt{xy+yz+zx}}{12\sqrt{3}}+\dfrac{x+y+z}{12}\geq\dfrac{xyz}{4(xy+yz+zx)}+\dfrac{3xyz}{4(xy+yz+zx)} = \dfrac{xyz}{xy+yz+zx}$$, by AM-GM inequalities provided that all of them are non-negative.
Therefore, either the problem intended to ask for only non-negative solutions, or it has infinitely many solutions indicated by the answer above.
| {
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"timestamp": "2023-03-29T00:00:00",
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Classification Of Conic Section
Classify $$x^2-4xy+y^2+8x+2y-5=0$$
So the eigenvalues are $\lambda_{1}=3,\lambda_{2}=-1$ so the eigenvectors are $v_{1}=\begin{pmatrix} \frac{1}{\sqrt{2}}\\-\frac{1}{\sqrt{2}}\end{pmatrix}$ and $v_{2}=\begin{pmatrix} \frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}\end{pmatrix}$
So We have $$\begin{pmatrix} x &y \end{pmatrix}\begin{pmatrix} 3 &0\\ 0&-1\end{pmatrix}\begin{pmatrix} x\\y \end{pmatrix}+\begin{pmatrix} 8 &2 \end{pmatrix}\begin{pmatrix} \frac{1}{\sqrt{2}} &\frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\end{pmatrix}\begin{pmatrix} x\\y \end{pmatrix}-5=0$$
$$3x^2-y^2+\frac{6}{\sqrt{2}}x+\frac{10}{\sqrt{2}}y-5=0$$
$$3(x+\frac{1}{\sqrt{2}})^2-\frac{3}{2}-(y-\frac{5}{\sqrt{2}})+\frac{25}{2}-5=0$$
$$3x''^2-y''^2=-6$$
$$-\frac{1}{2}x''^2+\frac{1}{6}y''^2=1$$ Which is?
But in the answer they got
$$\frac{1}{6}x''^2-\frac{1}{2}y''^2=1$$ which is hyperbola as it is in the form of $\frac{x''^2}{a^2}-\frac{y''^2}{b^2}=1$
| You don't have to find the reduced equations to classify the conic:
*
*$\;\begin{vmatrix}1&-2&4\\-2&1&1\\4&1&-5\end{vmatrix}\ne 0$, so the conic is non-degenerate.
*The quadratic form $x^2-4xy+y^2$ has signature $(1,1)$:
$$x^2-4xy+y^2=(x-2y)^2-3y^2.$$
As a conclusion, the conic is a hyperbola.
For an ellipse, the signature would be $(2,0)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the inequality satisfying the given conditions. (Multi-choice-multi-correct question):
If the roots of the equation $ax^2+bx+c=0$, $a\ne0$ are imaginary and $a+c<b$,($a,b,c$ are real numbers) then:
*
*$a+4c< 2b$
*$a+b+c<0$
*$4a+c<2b$
*$4a+c< 2b$ if $a<0$ and $4a+c>2b$ if $a>0$
My attempt:
$b^2-4ac<0 \implies a>0 \land c>0 $ or $a<0 \land c<0$
From condition 1 established above, (a>0,c>0) and given condition $a+c<b$
$4a+c+2b>0$ but this isn't the right answer. Where have I gone wrong? Any hints?
| Roots are imaginary numbers $\implies b^2- 4ac < 0\implies b^2 < 4ac \le(a+c)^2\implies b^2 - (a+c)^2 < 0\implies (b-a-c)(b+a+c) < 0$
Since $b-a-c > 0$, it must be true that $a+b+c < 0$.
Thus $2)$ is true.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving convergence of series $\sum_{k=1}^n {\sqrt{k^3+1}-\sqrt {k^3-1}}$ How to prove convergence of the following?
$$\sum_{k=1}^n {\sqrt{k^3+1}-\sqrt {k^3-1}}$$
Thanks!
| $\sqrt{k^3+1}-\sqrt {k^3-1}=\frac{(\sqrt{k^3+1}-\sqrt {k^3-1})(\sqrt{k^3+1}+\sqrt {k^3-1})}{(\sqrt{k^3+1}+\sqrt {k^3-1})}=\frac{2}{\sqrt{k^3+1}+\sqrt {k^3-1}}\leq \frac{2}{\sqrt{k^3+1}}\leq \frac{2}{\sqrt{k^3}}=\frac{2}{k^{1+\frac{1}{2}}}$
Then apply comparison test.
| {
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$\{a,b,c\}\subset \Bbb R$, $a\not =b$ and $a^2(b+c)=b^2(a+c)=2010$. Find $c^2(a+b)$
Consider that $\{a,b,c\}\subset \Bbb R$, with $a\not =b$, and it is known that $a^2(b+c)=b^2(a+c)=2010$.
Find $c^2(a+b)$.
A question from the third phase of OBM 2010 (Brazilian Math Olympiad). Sorry if it is a duplicate. My developments are leading to some complicated expressions... probably a wrong approach. Hints and answers are welcomed.
| We have
\begin{eqnarray*}
a^2b +a^2 c =2010 \\
a b^2+b^2 c =2010.
\end{eqnarray*}
Subtract these equations ( and use $ a \neq b$) we have $ab+bc+ca=0$ so
\begin{eqnarray*}
c= - \frac{ab}{a+b}. \\
\end{eqnarray*}
Now multiply the first equation by $b^2$ and the second by $a^2$ and subtract ( and again use $ a \neq b$), we have
\begin{eqnarray*}
a^2b^2 =2010 (a+b). \\
\end{eqnarray*}
Now
\begin{eqnarray*}
c^2 (a+b)= \frac{a^2b^2}{a+b}= \color{blue}{2010}. \\
\end{eqnarray*}
| {
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Limit as $x$ approaches infinity of a ratio of exponential functions. The problem appears as follows:
Use algebra to solve the following:
$$\lim_{x\to \infty} \frac{2^{3x+2}}{3^{x+3}}$$
The result is infinity which makes intuitive sense but I can’t get it to yield algebraically.
Attempt:
$$\frac{2^{3x+2}}{3^{x+3}} = \frac{2^{3x}2^2}{3^{x}3^3}$$
I’ve decided against continuing writing out my attempt because writing on an iPad, in Latex, is no fun.
The end-result I produce when taking the limit (as $x$ approaches infinity) of the above expression is $\inf / \inf = 1$.
(Do I have to demonstrate that $2^3x$ grows faster than $3^x$ for large $x$?)
| Hint: $\displaystyle\;\frac{2^{3x+2}}{3^{x+3}}=\frac{2^2}{3^3} \cdot \frac{\left(2^3\right)^x}{3^x}=\frac{4}{27} \cdot \left(\frac{8}{3}\right)^x \gt \frac{4}{27} \cdot 2^x\,$.
| {
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To prove $\lim_{x\to 0}\frac{e^x-e^{x\cos x}}{x-\sin x}=3$ I came across this question to prove the given limit
$$\lim_{x\to 0}\frac{e^x-e^{x\cos x}}{x-\sin x}=3$$
First I tried using LHospital's rule directly.
Then I tried using expansion of $e^x$ and then using LHospital's rule but I am getting stuck.
| I solved it using Taylor expansion:
First we calculate some ugly derivatives:
$$(e^{x\cos(x)})' = (\cos(x) - x\sin(x))e^{x\cos(x)},$$
$$(e^{x\cos(x)})'' = [(-2\sin(x) - x\cos(x))+(\cos(x)-x\sin(x))^2] e^{x\cos(x)}$$
and
$$(e^{x\cos(x)})''' = [(-3\cos(x) + x\sin(x))+(\cos(x)-x\sin(x))^3+2(\cos(x)-x\sin(x))(-2\sin(x)-x\cos(x)) ]e^{x\cos(x)}.$$
From that we have that the first, second and third derivatives of $e^{x\cos(x)}$ at $x=0$ are $1$ ,$1$ and $-2$. Therefore we have
$$e^{x\cos(x)} = 1+x+\frac{1}{2!}x^2-\frac{2}{3!}x^3+o(x^3).$$
Using the Taylor's expansion of $e^x$ and $\sin(x)$ we get:
$$e^x - e^{x\cos(x)} = -\frac{-3}{3!}x^3+o(x^3)$$
and
$$x-\sin(x) = -\frac{1}{3!}x^3+o(x^3).$$
Therefore we obtain
$$\lim_{x\to 0}\frac{e^x - e^{x\cos(x)}}{x-\sin(x) } = \lim_{x\to 0}\frac{-\frac{-3}{3!}x^3+o(x^3)}{-\frac{1}{3!}x^3+o(x^3)} = 3.$$
| {
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Help clarifying the steps to find the derivative of $y=(3x+1)^3(2x+5)^{-4}$ For the problem $y=(3x+1)^3(2x+5)^{-4}$ do I use the chain, quotient and product rules? If so how do I know what parts to break up and where the rules apply? For instance would I consider $f(x)$ to be $(3x+1)^3$ and $g(x)$ to be $(2x+5)^{-4}$? If so do I then need to take the product rule of $f(x)$ and the quotient of $g(x)$ and then combine them? Or do I just use one rule?
| Instead of the "quotient rule" you could write the function as $y= (3x+ 1)^3(2x+ 5)^{-4}$ and use the product rule: $y'= [(3x+ 1)^3]'(2x+ 5)^{-4}+ (3x+ 1)^3[(2x+ 5)^{-3}]'$. Now, using the "chain rule", $[(3x+ 1)^3]'= 3(3x+ 1)^2(3)= 9(3x+ 1)^2$ and $[(2x+ 5)^{-4}]'= (-4)(2x+ 5)^{-5}(2)= -8(2x+ 5)^{-5}$ so that $y'= 9(3x+ 1)^2(2x+ 5)^{-4}- 8(3x+ 1)^3(2x+ 5)^{-5}$. We can then write that in terms of fractions again: $\frac{9(3x+ 1)^2}{(2x+ 5)^4}-\frac{8(3x+ 1)^3}{(2x+ 5)^5}$. We need to multiply both numerator and denominator of the first fraction by 2x+ 5 to get denominator $(2x+ 5)^5$ for both: $\frac{9(3x+ 1)^2(2x+ 4)}{(2x+ 5)^5}- \frac{8(3x+ 1)^3}{(2x+ 5)^5}= \frac{9(3x+ 1)^2(2x+ 5)- 8(3x+1)^3}{(2x+ 5)^5}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof by induction (Combinatorics) This is my first question on here so please bear with me, thank you.
Prove that for all positive integers $n$
$$2(1+2+...+n)^4 = (1^5 + 2^5 +...+ n^5) + (1^7 + 2^7 +...+ n^7)$$
After establishing the base case, I proceeded to the induction:
$2(1+2+...+(n+1))^4 = (1^5 + 2^5 +...+ (n+1)^5) + (1^7 + 2^7 +...+ (n+1)^7)$
$2(1+2+...+(n+1))^4 = (1^5 + 2^5 +...+ n^5)+(n+1)^5 + (1^7 + 2^7 +...+n^7)+(n+1)^7$
$2(1+2+...+(n+1))^4 = 2(1+2+...+n)^4 + (n+1)^5 + (n+1)^7$
$2(1+2+...+(n+1))^4 - 2(1+2+...+n)^4 = (n+1)^5 + (n+1)^7$
$2[(1+2+...+(n+1))^4 - (1+2+...+n)^4] = (n+1)^5 + (n+1)^7$
And from here I've hit a wall and can't figure out how to continue. I tried substituting $(1+...+n) = [n(n+1)]/2$ and $(n+1)=[2(1+...+n)]/n$ but to no avail.
| Note that
$$1+2+\cdots+n =\frac{n(n+1)}{2},$$ and $$1+2+\cdots+n+1 =\frac{(n+1)(n+2)}{2}.$$
Therefore, the left-hand side of the last equation is
$$\frac{(n+1)^4}{8}\left((n+2)^4 -n^4\right)=\frac{(n+1)^4}{8}\left((n+1+1)^4 -(n+1-1)^4\right).$$
But $$(a+1)^4-(a-1)^4 = ((a+1)^2)^2-((a-1)^2)^2\\=((a+1)^2-(a-1)^2)((a+1)^2+(a-1)^2) \\=(a+1+a-1)(a+1-a+1)(a^2+1+2a+a^2+1-2a)\\=8a(a^2+1).$$
Using the above relationship for $a=n+1$, $$\frac{(n+1)^4}{8}\left((n+2)^4 -n^4\right)=(n+1)^4(n+1)((n+1)^2+1) = (n+1)^5+(n+1)^7.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Evaluating limits without L'Hopital's I need to evaluate without using L'Hopital's Rule. They say you don't appreciate something until you lose it. Turns out to be true.
*
*$\lim_{x \to 0} \frac{\sqrt[3]{1+x}-1}{x}$
*$\lim_{x \to 0} \frac{\cos 3x - \cos x}{x^2}$
I have tried to "rationalise" Q1 by using the identity $a^3-b^3=(a-b)(a^2+ab+b)$ but I still end up with a case of $\frac{0}{0}$.
For Q2, I tried to apply the formula and express $\cos 3x-\cos x =2\cos 2x \cos x$ which didn't help.
I would really be grateful for some advice on how I can proceed with the evaluating.
| *
*We use the binomial theorem: $$\frac{\sqrt[3]{1+x}-1}{x}=\frac{(1+x)^{1/3}-1}{x}=\frac{(1+\frac1{3}x+\frac1{9}x^2+\ldots)-1}{x}=\frac{(\frac1{3}x+\frac1{9}x^2+\ldots)}{x}=(\frac1{3}+\frac1{9}x+\ldots)\implies\lim_{x\to0}\frac{\sqrt[3]{1+x}-1}{x}=\frac1{3}$$
2.Again, we use series expansion for $\cos$ function. We get $$\frac{\cos3x-\cos x}{x^2}=\frac{(1-\frac{9x^2}{2}+\frac{81x^4}{4}-\ldots)-(1-\frac{x^2}{2}+\frac{x^4}{4}-\ldots)}{x^2}=\frac{\frac{-9x^2}{2}+\frac{x^2}{2}+\frac{81x^4}{4}-\frac{x^4}{4}+\ldots}{x^2}=-4+20x^2+\ldots\implies\lim_{x\to0}\frac{\cos3x-\cos x}{x^2}=-4$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculate the line integral $I(a,b)=\int_{x^2+y^2=R^2}\limits\ln\frac{1}{\sqrt{(x-a)^2+(y-b)^2}} ds\quad(a^2+b^2\ne R^2).$ Calculate the line integral $$I(a,b)=\int_{x^2+y^2=R^2}\limits\ln\frac{1}{\sqrt{(x-a)^2+(y-b)^2}} ds\quad(a^2+b^2\ne R^2).$$
The parametrized integral path can be given as
$$x=R\cos t,y=R\sin t,t\in[0,2\pi].$$
Then I get into trouble when computing the integral
$$-\frac{1}{2}\int_{0}^{2\pi}R\ln(a^2+b^2+R^2-2aR\cos t-2bR\sin t) dt.$$
Should I take the derivative with respect to $R$ first? Or apply integration by parts?
Update: I find the offered answer is
$$I(a,b)=-2\pi R\ln\max\{R,\sqrt{a^2+b^2}\}$$
and I have examined its correctness.
| Hint : $$W=C\sin(t)+D\cos(t)=(\sqrt{C^2+D^2})(\frac{C}{\sqrt{C^2+D^2}}\sin(t)+\frac{D}{\sqrt{C^2+D^2}}\cos(t))$$
$$W=\sqrt{C^2+D^2}\sin(t+\phi)=E\sin(t+\phi)$$
with $\cos(\phi)=\frac{C}{\sqrt{C^2+D^2}}$ and $ \sin(\phi)=\frac{D}{\sqrt{C^2+D^2}}$
so $$I=\int_0^{2\pi} \ln(A+C\sin(t)+B\cos(t))\text{d}t=\int_0^{2\pi} \ln(A+E\sin(t+\phi))\text{d}t$$
here a formal calculation software can complete the calculation
| {
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For each $a \in \Bbb{Z}$ work out $\gcd(3^{16} \cdot 2a + 10, 3^{17} \cdot a + 66)$ The problem is the following: For each $a \in \Bbb{Z}$ work out $\gcd(3^{16} \cdot 2a + 10, 3^{17} \cdot a + 66)$
This is what I have at the moment:
Let's call $d = \gcd(3^{16} \cdot 2a + 10, 3^{17} \cdot a + 66)$
Then $d \ \vert \ 3^{16} \cdot 2a + 10 \ \land d \ \vert \ 3^{17} \cdot a + 66$
$\Rightarrow d \ \vert \ (3^{16} \cdot 2a + 10) \cdot 3 \ \land \ d \ \vert \ (3^{17} \cdot a + 66) \cdot 2$
$\Rightarrow d \ \vert \ 3^{16} \cdot 6a + 30 \ \land \ d \ \vert \ 3^{16} \cdot 6a + 132$
$\Rightarrow d \ \vert \ 3^{16} \cdot 6a + 132 - (3^{16} \cdot 6a + 30) \ = \ 102 \ =
\ 2 \cdot 3 \cdot 17$
Also, $d \ \vert \ (3^{16} \cdot 2a + 10) \cdot 33 \ \land \ d \ \vert \ (3^{17} \cdot a + 66) \cdot 5$
$\Rightarrow d \ \vert \ 3^{17} \cdot 17a$ (almost with the same method as before)
So I get $d \ \vert \ 102 \ \land d \ \vert \ 3^{17} \cdot 17a$
After this, I can't see how to continue.
| From $d|102$ you have $d\in\{1,2,3,17,34,51,102\}$ and because $d|3^{16}2a+10$ you have $d\notin \{3,51,102\}$. Clearly $2|d$ iff $2|a$. And with $17$ you have a lot of different subcases to deal with. If $17|a$ then $d\ne 17$...
| {
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Three dice are rolled simultaneously. In how many different ways can the sum of the numbers appearing on the top faces of the dice be $9$?
Three dice are rolled simultaneously. In how many different ways can the sum of the numbers appearing on the top faces of the dice be $9$?
What I did:
I know that the maximum value on the dice can be $6$. So, restricting the values of dice, $6-x+6-y+6-z=9$, $x+y+z=9$, $n=9$ and $r=3$.
Applying partition again:
$$\binom{9+3-1}{3-1}=\binom{11}2=55$$
But the answer is $25$. Please, someone explain this one.
This is a gmat exam question.
| In these types of problems, it is not too hard to just consider case by case. Let's list out all the possibilities and how many ways to reorganize them:
$$1,2,6 \rightarrow 3!=6 \text{ ways}$$
$$1,3,5 \rightarrow 3!=6 \text{ ways}$$
$$1,4,4 \rightarrow 3!/2!=3 \text{ ways}$$
$$2,2,5 \rightarrow 3!/2!=3 \text{ ways}$$
$$2,3,4 \rightarrow 3!=6 \text{ ways}$$
$$3,3,3 \rightarrow 3!/3!=1 \text{ way}$$
So in total there are $25$ ways to get a sum of $9$. If you want the probability, just take this over the total number of possibilities and you get $25/6^3 = 25/216$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2441654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Approximate $e$ with $4$-digit chopping and $5$-degree Maclaurin-Taylor polynomial
Number $e$ is defined as
$$ e=\sum^\infty_{n=0} \frac{1}{n!}.$$
Use four-digit chopping arithmatic to compute the following
*
*Approximations to $e$
$$ e\approx \sum^5_{n=0} \frac{1}{n!}$$
Attempt:
\begin{align}
\sum^5_{n=0} \frac{1}{n!}
&= \frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!} \\ &= \frac{1}{1}+\frac{1}{2}+\frac{1}{3 \cdot 2 \cdot 1}+\frac{1}{4 \cdot 3 \cdot 2 \cdot 1}+\frac{1}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} \\
&= 1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}+\frac{1}{120}
\end{align}
to decimals
$$=1+.5000+0.16666666666+.04166666667+.0083333333333333$$
and I am not sure how $4$-digit chopping works.
I think that doing $4$-digit chopping is
$$=1+.5000+0.1666+.04166+.008333=1.716593=1.716$$
I am getting that it is wrong according to webwork wich would never ever be wrong.
| That is quite easy: you forgot the $1/ 0!$ in your calculations. Then you get something like $\mathrm e \approx2.716$. Also rounding in the end of your calculation might be the best.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2442401",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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reduce a rational expression to lowest terms $$\frac{x^2 + 4x + 3}{x^2 - 2x - 3}$$
I'm coming up with $\frac{x + 3}{ x - 3}$ however it seems wrong.
$$x^2 + 4x + 3 = (x + 3) ( x + 1)$$
$$x^2 - 2x - 3 = (x - 3 )(x + 1) $$
cancel out $x + 1$ and left with $\frac{x + 3}{x - 3}$
| It's correct,
$$\require{cancel} \frac{x^2 + 4x + 3} {x^2 - 2x - 3} = \frac{(x+3)\cancel{(x+1)}} {(x-3) \cancel{(x+1)}}$$
Provided $x \neq 3$, $x \neq -1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How do I solve $\int\frac{\sqrt{x}}{\sqrt{x}-3}\,dx?$ $$
\int \frac{\sqrt{x}}{\sqrt{x}-3}dx
$$
What is the most dead simple way to do this?
My professor showed us a trick for problems like this which I was able to use for the following simple example:
$$
\int \frac{1}{1+\sqrt{2x}}dx
$$
Substituting:
$u-1=\sqrt{x}$
being used to create
$\int\frac{u-1}{u}$
which simplifies to the answer which is:
$1+\sqrt{2x}-\ln|1+\sqrt{2x}|+C$
Can I use a similar process for the first problem?
| With $\sqrt x = u$
Let $\sqrt x = u$, then we have $x = u^2$ and $\mathrm{d}x=2u\,\mathrm{d}u$.
\begin{align}
\int\frac{u}{u-3}\times 2u\,\mathrm{d}u
&= 2\int\frac{u^2}{u-3}\,\mathrm{d}u\\
&= 2\int\frac{u^2-9+9}{u-3}\,\mathrm{d}u\\
&= 2\int\frac{u^2-9}{u-3}\,\mathrm{d}u + 18\int\frac{1}{u-3}\,\mathrm{d}u\\
&= 2\int(u+3)\,\mathrm{d}u + 18\int\frac{1}{u-3}\,\mathrm{d}u\\
&= u^2 + 6 u+18\ln(u-3)+C_1\\
&= x+6\sqrt x +18\ln (\sqrt x -3) +C_1
\end{align}
With $\sqrt x - 3 = t$
Let $\sqrt x -3 = t$, then we have $x = (t+3)^2$ and $\mathrm{d}x=2(t+3)\,\mathrm{d}t$.
\begin{align}
\int\frac{t+3}{t}\times 2(t+3)\,\mathrm{d}t
&= 2\int\frac{(t+3)^2}{t}\,\mathrm{d}t \\
&= 2\int\frac{t^2+6t+9}{t}\,\mathrm{d}t \\
&= \int\left(2t+12+\frac{18}{t}\right)\,\mathrm{d}t \\
&= t^2 + 12 t + 18 \ln t +C_2\\
&= (\sqrt x-3)^2 +12(\sqrt x -2) + 18\ln(\sqrt x- 3) +C_2\\
&= x +6\sqrt x +18\ln(\sqrt x -3) + C_2-15
\end{align}
Here $C_1=C_2-15$. You can choose either way and the difference is only in the constant.
With $\frac{\sqrt x}{\sqrt x -3}=v$
It is intentionally left for the readers as an exercise.
| {
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If α and β are the roots of the equation $3x^2+5x+4=0$, find the value of $α^3+β^3$ If α and β are the roots of the equation $3x^2+5x+4=0$, find the value of $α^3+β^3$
How can I factorize the expression to use the rule of sum and product of roots?
The answer is $\frac{55}{27}$
| expanding $$\frac{1}{216} \left(-5-i
\sqrt{23}\right)^3+\frac{1}{216} \left(-5+i
\sqrt{23}\right)^3$$ we obtain $$\frac{55}{27}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to compute symmetrical determinant I'm learning of determinants and am trying to find a trick to compute this one
\begin{pmatrix}
2 & 1 & 1 & 1 & 1\\
1 & 3 & 1 & 1 & 1\\
1 & 1 & 4 & 1 & 1\\
1 & 1 & 1 & 5 & 1\\
1 & 1 & 1 & 1 & 6
\end{pmatrix}
I expanded it out and got $349$ but I feel there must be some trick to easily compute it.
| Use the following rules:
*
*Adding a multiple of one row to another row does not change the determinant.
*If $B$ is obtained by multiplying a row of $A$ by a constant $c$ then $\det B=c \det A$
Start by subtracting the first row from each of the other rows, we get $$A=\begin{pmatrix}
2 & 1 & 1 & 1 & 1\\
-1 & 2 & 0 & 0 & 0\\
-1& 0 & 3 & 0 & 0\\
-1 & 0 & 0 & 4 & 0\\
-1 & 0 & 0 & 0 & 5
\end{pmatrix}$$ which has the same determinant as your matrix. Even just doing this makes the determinant much easier to calculate but we can go further. Divide the second, third, fourth and fifth rows by their corresponding diagonal term, we get $$B=\begin{pmatrix}
2 & 1 & 1 & 1 & 1\\
-1/2 & 1 & 0 & 0 & 0\\
-1/3& 0 & 1 & 0 & 0\\
-1/4 & 0 & 0 & 1 & 0\\
-1/5 & 0 & 0 & 0 & 1
\end{pmatrix}$$ which by rule #$2$, has determinant $\det B =\frac 1{120}\det A$. Finally, subtract each of the other rows from the first row, we get
$$B'=\begin{pmatrix}
147/60 & 0 & 0 & 0 & 0\\
-1/2 & 1 & 0 & 0 & 0\\
-1/3& 0 & 1 & 0 & 0\\
-1/4 & 0 & 0 & 1 & 0\\
-1/5 & 0 & 0 & 0 & 1
\end{pmatrix}$$ where $\det B'=\det B$ by rule #$1$. This is a lower triangular matrix so the determinant is simply the product of the diagonal elements, i.e. $\det B' = 147/60$. Therefore $$\det A = 120 \det B=120 \det B' =120 \cdot \frac{147}{60}=394.$$ This is generally the way to go if you're calculating determinants of large matrices by hand, just be sure to keep track of each time you multiply a row by something so you can get back to the original determinant. Also, the symmetric property makes row-reduction easier but you can do this procedure for any matrix.
| {
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The $n$th prime number is the smallest difference of products of distinct previous primes. For $n\geq 2$, the smallest positive prime number formed by multiplying $2, 3, \dots, p_n$ and up to $1$ subtraction / addition
$$
2 + 3 = 5 \\
2\cdot 5 - 3 = 7 \\
3 \cdot 7 - 2\cdot 5 = 11 \\
5\cdot 11 - 2\cdot 3 \cdot 7 = 13 \\
7 \cdot 13 \cdot 2 - 3 \cdot5 \cdot 11 = 17\\
\vdots
$$
is always $p_{n+1}$. Has this been proven?
| You stopped your examples one step too early. The smallest number $>1$ obtainable from $2,3,\ldots,17$ is
$$2\cdot 5\cdot 7\cdot 11-3\cdot 13\cdot 17=107.$$
(I.e., $663$ and $770$ are the factors of $N:=2\cdot 3\cdot 5\cdot \ldots\cdot 17$ that are closest to $\sqrt N$).
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all n for which $2^n + 3^n$ is divisible by $7$? I am trying to do this with mods. I know that:
$2^{3k+1} \equiv 1 \pmod 7$, $2^{3k+2} \equiv 4 \pmod 7$, $2^{3k} \equiv 6 \pmod 7$ and $3^{3k+1} \equiv 3 \pmod 7$, $3^{3k+2} \equiv 2 \pmod 7$, $3^{3k} \equiv 6 \pmod 7$, so I thought that the answer would be when $n$ is a multiple of $3$ since then $2^{3k} + 3^{3k} \equiv 1 + 6 \pmod 7$, but it doesn't work for $n = 6$
| Powers of $2$ modulo $7$ are $1,2,4,1,2,4,\cdots$ repeating every $3$ and powers of $3$ modulo $7$ are $1,3,2,6,4,5,1,\cdots$ repeating every $6$
Adding these sequence gives $n \equiv 3 \pmod{6}$ so $7$ divides $2^n+3^n$ if and only if $n=3+6m$ (for $ m \in \mathbb{N}_0$).
| {
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Fill out a group table with 6 elements Let $G=\{0,1,2,3,4,5\}$ be a group whose table is partially shown below:
\begin{array}{ c| c | c | c | c |c|c|}
* & 0& 1 & 2 & 3& 4 & 5\\
\hline
0 & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
1 & 1 & 2 & 0 & 4 & & \\
\hline
2 & 2 & & & & & \\
\hline
3 & 3 & 5 & & & & 1\\
\hline
4 & 4 & & & & & \\
\hline
5 & 5 & & & & & \\
\hline
\end{array}
Complete the table.
Needed to use Inverses, cancellation and No 1 element can repeat per row/col, but got the wrong table.
\begin{array}{ c| c | c | c | c |c|c|}
* & 0& 1 & 2 & 3& 4 & 5\\
\hline
0 & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
1 & 1 & 2 & 0 & 4 & 5 & 3 \\
\hline
2 & 2 & 0 & 1 & 5 & 3 & 4 \\
\hline
3 & 3 & 5 & 4 & 2 & 0 & 1\\
\hline
4 & 4 & 3 & 5 & 0 & 1 & 2 \\
\hline
5 & 5 & 4 & 3 & 1 & 2& 0\\
\hline
\end{array}
What is the correct table?
| Suppose you had got this far and needed to fill out the rest:
\begin{array}{ c| c | c | c | c |c|c|}
* & 0& 1 & 2 & 3& 4 & 5\\
\hline
0 & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
1 & 1 & 2 & 0 & 4 & 5 & 3 \\
\hline
2 & 2 & 0 & 1 & 5 & 3 & 4 \\
\hline
3 & 3 & 5 & 4 & ? \\
\hline
4 & 4 & 3 & 5 \\
\hline
5 & 5 & 4 & 3 \\
\hline
\end{array}
One way to proceed would be to use the property of associativity, which says that $(ab)c = a(bc)$. Let $a = 3^2$, and consider $3^3$: since $(3^2)3 = 3(3^2)$, we have $a\cdot3=3\cdot a$. In other words, $3$ and $a$ commute, and since $a \in \{0,1,2\}$, we see that $a=0$.
It's important to understand that the group multiplication table has more properties than just containing a permutation of the elements in each row and column; otherwise you merely have a Latin square.
| {
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Solve $2\log_bx + 2\log_b(1-x) = 4$ I need to solve $$2\log_bx + 2\log_b(1-x) = 4.$$
I have found two ways to solve the problem. The first (and easiest) way is to divide through by $2$:
$$\log_bx + \log_b(1-x) = 2.$$ Then, combine the left side: $$\log_b[x(1-x)] = 2,$$ and convert to the equivalent exponential form, $$b^2 = x(1-x) \;\;\implies\;\; x^2 - x + b^2 = 0,$$ which by the quadratic equation, we get $$x = {1\over2}\left(1\pm\sqrt{1 - 4b^2}\right).$$
My question is: How do I know that neither solution is extraneous? These are the solutions in the back of the textbook, but I am left questioning when these are the solutions.
The alternative solution involves some more clever thinking:
$$\begin{align}2\log_bx + 2\log_b(1-x) &= 4\\\log_bx^2(1-x)^2 &= 4\\ b^4 &= x^2(x^2 - 2x + 1)\\ 0&=x^4 - 2x^3 + x^2 - b^4\\ &=x^2(x - 1)^2 - b^4\\ & = [x(x-1)]^2 - (b^2)^2 \\ &= [x(x-1) + b^2][x(x-1) - b^2],\end{align}$$ which implies that $$x = {1\over2}\left(1 \pm\sqrt{1 + 4b^2}\right) \;\;\;\text{or} \;\;\; {1\over2}\left(1 \pm\sqrt{1 - 4b^2}\right),$$
which is even worse, because now there are $4$ solutions to check.
The most I know is that since $b$ is a logarithmic base, $b>0$ and $b\ne 1$. But what happens when $b$ is something like $2$? Then you end up with a complex number under the root (of the solutions from the easier way), and that doesn't necessarily make sense if I'm trying to solve the logarithmic equation over the reals.
| You are correct to transform $2\log_b x + 2\log_b (1-x) = 4$ into $x^2-x+b^2=0$.
These are the restrictions implicated by the original equation:
*
*$b$ is a positive number not equal to $0$ nor $1$
*$x>0 \iff 0<x$
*$1-x>0 \iff 1>x \iff x<1$
I personally felt that it was easiest to wrap my mind around this using a graph with sliders. This shows us that your first equation for the roots is in fact valid.
Going off of that, $x=\dfrac12\left( 1 \pm \sqrt{1-4b^2} \right)$ is defined only when
*
*$1-4b^2\ge 0 \iff 1\ge 4b^2 \iff \frac14 \ge b^2 \iff \frac12\ge|b|\iff |b|\le \frac12$
Now, using all of the aforementioned restrictions (and the graph if you wish), your first solution
$$\bbox[yellow,5px]{x=\dfrac12\left( 1 \pm \sqrt{1-4b^2} \right)}$$
is correct, and the roots exist only when
$$\bbox[yellow,5px]{0\lt b\le\frac12}$$
(Note that if $b=1/2$, the two roots are the same.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2458109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $(a+2b+\frac{2}{a+1})(b+2a+\frac{2}{b+1})\geqslant16$ if $ ab\geqslant1$ $a$, $b$ are positive real numbers.
Show that $$\left(a+2b+\frac{2}{a+1}\right)\left(b+2a+\frac{2}{b+1}\right)\geqslant16$$
if
$$ ab\geqslant1$$
Should I use AM-GM inequality? If yes, where?
| Expand to get:
$$ab + 2a^2 + \frac{2a}{b+1} + 2b^2 + 4ab + \frac{4b}{b+1} + \frac{2b}{a+1} + \frac{4a}{a+1} + \frac{4}{(a+1)(b+1)} \ge 9 + \frac{2a^2 + 2a + 4ab + 4b + 2b^2 + 2b + 4ab + 4a + 4}{(a+1)(b+1)} \ge 9 + 7 = 16$$
The last inequality follows as:
$$2a^2 + 2a + 4ab + 4b + 2b^2 + 2b + 4ab + 4a + 4 \ge 7(a+1)(b+1)$$
$$2a^2 + 2b^2 + ab \ge a+ b + 3$$
Now it's enough to prove that $2a^2 + 2b^2 \ge a + b + 2$. From the condition we have by AM-GM $a+b \ge 2\sqrt{ab} \ge 2$. Hence by using the quadratic mean inequality:
$$2a^2 + 2b^2 \ge (a+b)^2 \ge 2(a+b) \ge a+b+2$$
Hence the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2461260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solve system of equations: $\sqrt2 \sin x = \sin y, \sqrt2\cos x = \sqrt5\cos y$ $\sqrt2 \sin x = \sin y, \sqrt2\cos x = \sqrt5\cos y$
I tried to use tangent half-angle substitution, tried to sum these two equations and get this
$\sin(x+ \pi/4) = \sqrt3 / \sqrt2 \sin(y+\tan^{-1}(\sqrt5))$
I stuck.
Any Help is appreciated!
| Square both equations, then add them up. $2 \sin^2 x = \sin^2 y$ and $2 \cos^2 x = 5 \cos^2 y$, so $2 = \sin^2 y + 5 \cos^2 y$, and hence $1 = 4 \cos^2 y$. From here, $\cos y = \pm \frac1{2}$, and you should substitute back in to check if both work. You'll also get multiple values of $y$ from this, which you need to check with the $\sin$ relations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2464781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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For what pair of integers $(a,b)$ is $3^a + 7^b$ a perfect square. Problem: For what pair of positive integers $(a,b)$ is $3^a + 7^b$ a perfect square.
First obviously $(1,0)$ works since $4$ is a perfect square, $(0,0)$ does not work, and $(0,1)$ does not work, so we can exclude cases where $a$ or $b$ are zero for the remainder of this post.
I have a few observations made but not much room for a full solution.
First, since powers of an odd number are odd, and the sum of two odd numbers is even, so the base of the square must be an even number.
Secondly, the last digit of the powers of $3$ are $\{1,3,7,9 \}$ , whereas the last digits of the powers of $7$ are $\{7,9,3,1 \}$.
From here I am not sure how to proceed and any hints much appreciated. I'm not sure if there a finite amount of pairs or not either.
| The other case is $1 + 2 \cdot 7^x = 3^y,$ or
$$ 3^y - 1 = 2 \cdot 7^x. $$
Assume $x \geq 1.$ Then both sides are divisible by $7,$ giving
$$ 3^y \equiv 1 \pmod 7, $$
$$ y \equiv 0 \pmod 6. $$ Then $3^y - 1$ is divisible by
$$ 3^6 - 1 = 728 = 8 \cdot 7 \cdot 13. $$
However, then $2 \cdot 7^x$ is divisible by $13,$ which is a contradiction.
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2465086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
} |
Find coordinates of area's center of mass. So we have area $D:{(x^{2}+y^{2})^{2}=2a^{2}xy, x>0, y>0}$. $x= r \cdot cos(\varphi)$ and $y= r \cdot sin(\varphi)$
I want to find the center of mass by formula $y_{c} = \frac{M_{ox}}{m}$, where $M_{ox}=\iint_{D}(\gamma \cdot y \cdot dxdy)$ gamma is density and $m=\iint_{D}(\gamma \cdot dxdy)$.
First of all I decided to use polar coordinates. So $m=\frac{a^{2}}{2}$. and I have problem finding $M_{ox}=\int_{0}^{\pi/2}d(\varphi)\int_{0}^{a \cdot \sqrt{sin(2\varphi)}}r^{2} \cdot sin(\varphi)d(\varphi)$
Help me please to find that integral. Or maybe you have better solutions for this task.
| \begin{align}
M_x
&= \int_0^{\pi/2}d\theta\int_0^{a\sqrt{\sin2\theta}}\gamma r^2\sin\theta\,dr \\
&= \gamma\int_0^{\pi/2}d\theta \dfrac13r^3\sin\theta\Big|_0^{a\sqrt{\sin2\theta}} \\
&= \gamma a^3\dfrac{2\sqrt{2}}{3}\int_0^{\pi/2}\sin^{5/2}\theta\cos^{3/2}\theta d\theta \\
&= \gamma a^3\dfrac{\sqrt{2}}{3}2\int_0^{\pi/2}\sin^{2\frac74-1}\theta\cos^{2\frac54-1}\theta d\theta \\
&= \gamma a^3\dfrac{\sqrt{2}}{3}\beta\left(\dfrac74,\dfrac54\right) \\
&= \gamma a^3\dfrac{\sqrt{2}}{3}\dfrac{\Gamma\left(\dfrac74\right)\Gamma\left(\dfrac54\right)}{\Gamma(3)} \\
&= \gamma a^3\dfrac{\sqrt{2}}{6}\dfrac34\Gamma\left(1-\dfrac14\right)\Gamma\left(1+\dfrac14\right) \\
&= \gamma a^3\dfrac{\sqrt{2}}{6}\dfrac34\dfrac{\pi\dfrac14}{\sin\pi\dfrac14} \\
&= \color{blue}{\dfrac{1}{16}\gamma a^3\pi}
\end{align}
where $\beta(x,y)$ is Beta function and $\Gamma\left(1-x\right)\Gamma\left(1+x\right)=\dfrac{\pi x}{\sin\pi x}$ is Euler's reflection formula.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2466624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find area of triangle ABD Let ABC be a right-angled triangle $B=90$. Let in a triangle pick a point $D$ inside the triangele such that $AD= 20, DC=15, DB=10$ and $AB=2BC$. What is the area of triangle $ABD$?
Thanks!
| Let $BC=a$ and $AB=2a$.
With $B$ as origin let the co-ordinates of $D$ be $(x,y)$. The required area of triangle $ABD$ is $ax$.
Applying Pythagoras' Theorem $$10^2=x^2+y^2 \text{ ... (1)}$$ $$15^2=(a-x)^2+y^2 \text{ ... (2)}$$ $$20^2=x^2+(2a-y)^2 \text{ ... (3)}$$
Subtract (1) from (2) and (3) in turn : $$125=a(a-2x)\implies 2ax=a^2-125$$ $$300=2a(2a-2y) \implies 2ay=2a^2-150$$
Substitute into (1) : $$100(2a)^2=(2ax)^2+(2ay)^2=(a^2-125)^2+(2a^2-150)^2$$ $$400a^2=a^4-250a^2+125^2+4a^4-600a^2+150^2$$ $$0=5a^4-1250a^2+38125$$ $$0=a^4-250a^2+7625$$ $$a^2=\frac12(250\pm \sqrt{250^2-4.7625})=125\pm 40\sqrt{5}$$
We reject the smaller value which corresponds to $D$ outside of the triangle.
$$2ax=a^2-125=40\sqrt{5}$$ The area of triangle $ABD$ is $$ax=20\sqrt{5} \approx 44.72$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2467383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving inhomogeneous heat equation Solve the inhomogeneous heat equation $u_t=c^2u_{xx} +\sin(5\pi x)$ for all $0 < x < 1, t > 0$ subject to homogeneous Dirichlet boundary conditions $u(t,0)=u(t,1)=0$ and initial condition $u(0,x) =4\sin(3\pi x)+9\sin(7\pi x)$.
Solved it as
$$v(x,t)=\sum_{n=1}^{\infty}\bigg[\frac{\int_{0}^{1}[4\sin(3\pi x)+9\sin(7\pi x)]\cdot\sin(n\pi x)dx}{\int_{0}^{1}\sin^2(n\pi x)dx}\cdot e^{-n^2\pi^2c^2t}\cdot\int_{0}^{t}\frac{\int_{0}^{1}\sin(5\pi x)\cdot\sin(n\pi x)dx}{\int_{0}^{1}\sin^2(n\pi x)dx}\cdot e^{n^2\pi^2c^2t}dx\cdot\sin(n\pi x)\bigg]$$
| In order to solve non-homogeneous heat equations, assume that $w(x,t)=u(x,t)+\phi(x)$, where $\phi$ is a function yet to be determined.
So, we've $u_{xx} = w_{xx} - \phi''$ and $u_{t} = w_{t}$. If we substitute this in to our equation, then we have to solve the new problem
$\begin{cases}
w_t - c^2 w_{xx} - c^2 \phi'' - \sin{5 \pi x} = 0\\
w(0,t) - \phi(0) = 0\\
w(1,t) = \phi(1) = 0\\
w(x,0) - \phi(x) = 4\sin{3\pi x} + 9\sin{7 \pi x}
\end{cases}
$
So we really want to choose $\phi$ so that $-c^2\phi^2 - \sin{5 \pi x} = 0$, the choice of $\phi = \frac{1}{25 \pi^2 c^2}\sin{5 \pi x}$ works. And so we want to solve the new equation:
$\begin{cases}
w_t - c^2 w_{xx} = 0\\
w(0,t) = 0\\
w(1,t) = 0\\
w(x,0) = 4\sin{3\pi x} +\frac{1}{25 \pi^2 c^2}\sin{5 \pi x} + 9\sin{7 \pi x}
\end{cases}
$
At this point, this should be easy to do, separation of variables yields that the resulting corresponding Fourier series is a sine series, and we can easily match what the coefficients according to our initial condition. We have
$$w(x,t) = \sum_{n=1}^{\infty} b_n e^{-(cn\pi)^2t}\sin{n\pi x}$$
Matching coefficients, we see $b_3 = 4$, $b_5 = \frac{1}{25 \pi^2 c^2}$, and $b_7 = 9$, and $b_n = 0$ otherwise.
Thus we have
$$w(x, t) = 4 e^{-(3c\pi)^2t}\sin{3\pi x} + \frac{1}{25 \pi^2 c^2} e^{-(5c\pi)^2t}\sin{5\pi x} + 9 e^{-(7n\pi)^2t}\sin{7\pi x}$$
Now, remember that $u(x, t) = w(x, t) - \phi(x)$, the full solution is
$$u(x, t) = 4 e^{-(3c\pi)^2t}\sin{3\pi x} + \frac{1}{25 \pi^2 c^2} e^{-(5c\pi)^2t}\sin{5\pi x} + 9 e^{-(7n\pi)^2t}\sin{7\pi x} - \frac{1}{25 \pi^2 c^2}\sin{5 \pi x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2469422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Semifactorial Identity I was wondering if anyone had any insight on how to prove the following identity:
For all $m \in \mathbb{N}$ $$ \frac{1}{2m-1} + \frac{2m-2}{(2m-1)(2m-3)} + \frac{(2m-2)(2m-4)}{(2m-1)(2m-3)(2m-5)} + \cdots + \frac{(2m-2)!!}{(2m-1)!!} = 1$$
I attempted to rewrite and simplify the left hand side as:
$$ \sum_{k = 1}^m \frac{\frac{(2m-2)!!}{(2m-2k)!!}}{\frac{(2m-1)!!}{(2m-2k-1)!!}} = \sum_{k = 1}^m \frac{(2m-2)!! (2m-2k-1)!!}{(2m-2k)!! (2m-1)!!} = \frac{\left[2^{m-1} (m-1)! \right]^2}{(2m-1)!} \sum_{k=1}^m \frac{(2m-2k-1)!!}{(2m-2k)!!} $$
But I do not seem to be getting anywhere. Does anyone see how to prove the statement?
| Starting from
$$\sum_{k=1}^m \frac{(2m-2)!!/(2m-2k)!!}{(2m-1)!!/(2m-2k-1)!!}$$
and using
$$(2n)!! = 2^n n!
\quad\text{and}\quad
(2n-1)!! = \frac{(2n)!}{2^n n!}$$
we get for the sum
$$\sum_{k=1}^m \frac{2^{m-1} (m-1)!}{2^{m-k} (m-k)!}
\frac{(2m-2k)! / 2^{m-k} / (m-k)!}{(2m)! / 2^{m} / m!}
\\ = \frac{(m-1)! m!}{(2m)!}
\sum_{k=1}^m {2m-2k\choose m-k} 2^{2k-1}
\\ = - \frac{(m-1)! m!}{(2m)!} {2m\choose m} \frac{1}{2}
+ \frac{(m-1)! m!}{(2m)!}
\sum_{k=0}^m {2m-2k\choose m-k} 2^{2k-1}
\\ = -\frac{1}{2m}
+ \frac{(m-1)! m!}{(2m)!}
\sum_{k=0}^m {2m-2k\choose m-k} 2^{2k-1}
\\ = -\frac{1}{2m}
+ \frac{1}{2m} {2m\choose m}^{-1}
\sum_{k=0}^m {2m-2k\choose m-k} 2^{2k}.$$
Continuing with the sum term yields
$$\sum_{k=0}^m [z^{m-k}] (1+z)^{2m-2k} 2^{2k}
= [z^m] (1+z)^{2m}
\sum_{k=0}^m z^k (1+z)^{-2k} 2^{2k}.$$
We may extend $k$ to infinity because when $k\gt m$ there is no
contribution to $[z^m]$. We obtain for the coefficient extractor
$$[z^m] (1+z)^{2m}
\sum_{k\ge 0} z^k (1+z)^{-2k} 2^{2k}
= [z^m] (1+z)^{2m} \frac{1}{1-4z/(1+z)^2}
\\ = [z^m] (1+z)^{2m+2} \frac{1}{(1-z)^2}.$$
This is
$$\sum_{q=0}^m {2m+2\choose q} (m+1-q)
= (m+1) \sum_{q=0}^m {2m+2\choose q}
- \sum_{q=0}^m {2m+2\choose q} q.$$
The first piece yields
$$(m+1) \frac{1}{2} \left(2^{2m+2} - {2m+2\choose m+1} \right)$$
and the second
$$\sum_{q=1}^m {2m+2\choose q} q
= (2m+2) \sum_{q=1}^m {2m+1\choose q-1}
= (2m+2) \sum_{q=0}^{m-1} {2m+1\choose q}
\\ = (2m+2)\frac{1}{2} \left(2^{2m+1} - 2 {2m+1\choose m}\right).$$
Subtract to obtain
$$(2m+2) {2m+1\choose m} - \frac{1}{2} (m+1) {2m+2\choose m+1}
\\ = (2m+2) {2m+1\choose m}
- \frac{1}{2} (m+1) \frac{2m+2}{m+1} {2m+1\choose m}
\\ = (m+1) {2m+1\choose m} = (m+1) \frac{2m+1}{m+1} {2m\choose m}
= (2m+1) {2m\choose m}.$$
Returning to the main computation we thus have
$$-\frac{1}{2m} + \frac{1}{2m} (2m+1) = 1$$
as desired.
| {
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"url": "https://math.stackexchange.com/questions/2470085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How do we minimize $\left(x+a+b\right)\left(x+a-b\right)\left(x-a+b\right)\left(x-a-b\right)$? Find the minimum value of the following function, where a and b are real
numbers.
\begin{align} f(x)&=\left(x+a+b\right)\left(x+a-b\right)\left(x-a+b\right)\left(x-a-b\right) \end{align}
Note: The solution should not contain any calculus methods.
So I conjugated the parenthesizes two by two, and I got:
\begin{align} f(x)&=\left(x^2-(a+b)^2\right)\left(x^2-(a-b)^2\right) \end{align}
Can somebody help me to take this problem further?
| Just another solution in the same spirit as lab bhattacharjee's answer.
$$ f(x)=\left(x+a+b\right)\left(x+a-b\right)\left(x-a+b\right)\left(x-a-b\right) $$
$$f(x)=x^4-2\left( a^2+1\right) x^2+(a^2-1)^2$$ Complete the square
$$f(x)=\left(x^2-(a^2+1)\right)^2-(a^2+1)^2+(a^2-1)^2=\left(x^2-(a^2+1)\right)^2-4a^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2472757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Proving that $\frac{ab}{c^3}+\frac{bc}{a^3}+\frac{ca}{b^3}> \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
Prove that $\dfrac{ab}{c^3}+\dfrac{bc}{a^3}+\dfrac{ca}{b^3}> \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$, where $a,b,c$ are different positive real numbers.
First, I tried to simplify the proof statement but I got an even more complicated: $$a^4b^4+b^4c^4+a^4c^4> a^2b^3c^3+b^2c^3a^3+a^3b^3c^2$$
Then I used Power mean inequality on $\dfrac{1}{a},\dfrac{1}{b},\dfrac {1}{c}$ but that wasn't useful here.
Finally, I tried to solve it using AM-HM inequality but couldn't.
What would be an efficient method to solve this problem? Please provide only a hint and not the entire solution since I wish to solve it myself.
| \begin{eqnarray*}
a^4(b^2-c^2)^2+b^4(a^2-c^2)^2+c^4(b^2-a^2)^2+a^2b^2c^2((a-b)^2+(c-b)^2+(a-c)^2) \geq 0.
\end{eqnarray*}
Rearrange to
\begin{eqnarray*}
a^4b^4+b^4c^4+c^4a^4 \geq a^2b^2c^2(ab+bc+ca).
\end{eqnarray*}
Now divide by $a^3b^3c^3$ and we have
\begin{eqnarray*}
\frac{ab}{c^3}+\frac{bc}{a^3}+\frac{ca}{ b^3} \geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2473942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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Identity with sums of binomials Choose any natural numbers $a \ge 1$ and $b \ge 1$. Then prove that the identity holds:
\begin{equation*}
\frac{a b }{(a+b)^2} {\Large(}\tbinom{a+b}{a} {\Large)}^2 = \\
\sum_{n=1}^{a}\sum_{m=1}^{b} {\Huge[} {\Large(}\tbinom{n+m-2}{n-1} {\Large)}^2 - \frac{ m (a-n)(b-m) }{n(a-n+b-m)^2} {\Large(}\tbinom{n+m-1}{n-1} {\Large)}^2 {\Huge]} {\Large(}\tbinom{a-n+b-m}{a-n} {\Large)}^2
\end{equation*}
In here, the fraction $\frac{ (a-n)(b-m) }{(a-n+b-m)^2}$ is understood to be zero for $a=n$ and $b=m$.
Alternative:
Show that the following equation is in fact an identity for arbitrary $x,y$ with $x+y<1$.
\begin{equation*}
1 = {\Huge[} 1 + \sum_{n=1}^{\infty}\sum_{m=1}^{\infty} {x^n y^m} \frac{2 m n }{(m+n)^2} {\Large(}\tbinom{n+m}{n} {\Large)}^2 {\Huge]}^2 - 4 x y {\Huge[} \sum_{n=0}^{\infty}\sum_{m=0}^{\infty} {x^n y^m} {\Large(}\tbinom{n+m}{n} {\Large)}^2 {\Huge]}^2
\end{equation*}
Indeed, the top identity is - after some manipulation - an equation for the coefficient of $x^a y^b$ in the latter identity.
My effort: I tried to simplify the binomials further but had no success in summing up. A computer program verifies (up to reasonable numbers $a$ and $b$) that indeed equality holds.
| The answer can be derived from two answers to this question: There, the topic is that the partial derivatives of
$f(x,y)=(1+x^2+y^2-2x-2y-2xy)^{-\frac{1}{2}}$
satisfy
\begin{equation*}
\frac{\partial ^{m+n}}{\partial x^{m} \partial y^{n}} f (0,0)= (n+m)! \tbinom{n+m}{n}
\end{equation*}
Now this relation is shown directly (second answer to the said question), and then, by a Taylor expansion of $f(x,y)$, also the following equation holds:
\begin{equation*}
1 = (1+x^2+y^2-2x-2y-2xy) \cdot {\Huge[} \sum_{n=0}^{\infty}\sum_{m=0}^{\infty} {x^n y^m} {\Large(}\tbinom{n+m}{n} {\Large)}^2 {\Huge]}^2
\end{equation*}
This equation, after applying $1+x^2+y^2-2x-2y-2xy = (1-x-y)^2 - 4xy$, equals
\begin{equation*}
1 = {\Huge[} 1 + \sum_{n=1}^{\infty}\sum_{m=1}^{\infty} {x^n y^m} \frac{2 m n }{(m+n)^2} {\Large(}\tbinom{n+m}{n} {\Large)}^2 {\Huge]}^2 - 4 x y {\Huge[} \sum_{n=0}^{\infty}\sum_{m=0}^{\infty} {x^n y^m} {\Large(}\tbinom{n+m}{n} {\Large)}^2 {\Huge]}^2
\end{equation*}
which establishes the identity required here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2474963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving equivalence of norms in $\mathbb{R}^2$
Let $\lVert \cdot \rVert_*:\mathbb{R}^2 \to \mathbb{R}, (x,y) \rightarrow \sqrt{x^2+2xy+3y^2}$ be a norm.
How can I find to constants $k,K \in \mathbb{R}^{>0}$ so that the following equivalence is given:$$k\lVert (x,y) \rVert_2 \leq \Vert (x,y) \rVert_* \leq K\Vert (x,y) \rVert_2.$$
My idea:
$$
k\lVert (x,y) \rVert_2 \leq \Vert (x,y) \rVert_* \leq K\Vert (x,y) \rVert_2
\\
\Leftrightarrow
k\sqrt{|x|^2+|y|^2} \leq \sqrt{x^2+2xy+3y^2} \leq K\sqrt{|x|^2+|y|^2}
$$
Now how can I find $k,K$? The problem is these constants must be minimal.
Can someone give me a tip?
Thx
| Note that
$$x^2 + 2xy + 3y^2 = \begin{pmatrix} x \\ y \end{pmatrix}^T \begin{pmatrix} 1 & 1 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}.$$
The eigenvalues of the matrix have sum 4 and product 2, so they are both stricly positive; let us call them $\lambda_1$ and $\lambda_2$. Since the matrix is symmetric, there exists $Q$ orthogonal such that:
$$\begin{pmatrix} 1 & 1 \\ 1 & 3 \end{pmatrix} = Q^T \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix} Q.$$
Now let us note: $$\begin{pmatrix} q_1 \\ q_2 \end{pmatrix}= Q \begin{pmatrix} x \\ y \end{pmatrix}.$$ Then you can see that $\| \mathbf q \|_2 = \| \mathbf x\|_2$, and
$$x^2 + 2xy + 3y^2 = \begin{pmatrix} q_1 \\ q_2 \end{pmatrix}^T \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix} \begin{pmatrix} q_1 \\ q_2 \end{pmatrix} = \lambda_1 q_1^2 + \lambda_2 q_2^2,$$
from where it is easy to conclude.
Another way to proceed, adapted specifically to the example, is to notice that, on the one hand,
$$ x^2 + 2xy + 3y^2 \leq 2 x^2 + 4y^2 \leq 4(x^2 + y^2),$$
and on the other:
$$ x^2 + 2xy + 3y^2 = \left(\sqrt{a}x + \frac{y}{\sqrt{a}}\right)^2 + (1-a)x^2 + \left(3-\frac{1}{a}\right)y^2 \geq \min\left(1-a,3-\frac{1}{a}\right) (x^2 + y^2) \quad \forall a \in \mathbb R_+.$$
Any $a$ such that $1/3 < a < 1$ will guarantee that the $\min$ is strictly positive.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integrate $\arctan{\sqrt{\frac{1+x}{1-x}}}$ I use partial integration by letting $f(x)=1$ and $g(x)=\arctan{\sqrt{\frac{1+x}{1-x}}}.$ Using the formula:
$$\int f(x)g(x)dx=F(x)g(x)-\int F(x)g'(x)dx,$$
I get
$$\int1\cdot\arctan{\sqrt{\frac{1+x}{1-x}}}dx=x\arctan{\sqrt{\frac{1+x}{1-x}}}-\int\underbrace{x\left(\arctan{\sqrt{\frac{1+x}{1-x}}}\right)'}_{=D}dx.$$
So, the integrand $D$ remains to simplify:
$$D=x\cdot\frac{1}{1+\frac{1+x}{1-x}}\cdot\frac{1}{2\sqrt{\frac{1+x}{1-x}}}\cdot\frac{1}{(x-1)^2} \quad \quad (1).$$
Setting $a=\frac{1+x}{1-x}$ for notation's sake I get
$$D=x\cdot\frac{1}{1+a}\cdot\frac{1}{2\sqrt{a}}\cdot\frac{1}{(x-1)^2}=\frac{x}{(2\sqrt{a}+2a\sqrt{a})(x^2-2x+1)},$$
and I get nowhere. Any tips on how to move on from $(1)?$
NOTE: I don't want other suggestions to solutions, I need help to sort out the arithmetic to the above from equation (1).
| Hint: make the substitution $x = \cos \theta$ with $0 \le \theta \le \pi$. Then $1 + x = 2 \cos^2(\theta/2)$, $1 - x = 2 \sin^2(\theta/2)$, so $\sqrt{\frac{1+x}{1-x}} = |\cot(\theta/2)|$. But by the choice of range of $\theta$, $\cot(\theta/2)$ is positive. So $\arctan \sqrt{\frac{1+x}{1-x}} = \frac{\pi}{2} - \frac{\theta}{2}$. (Note this if you restrict to working with real numbers only this will work only for $x \in [-1,1)$; however, for $x > 1$ or $x < -1$, the square root is pure imaginary anyway. If you want to integrate it anyway, you might need other arguments, for example $x = \cosh \theta$ for $x > 1$.)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving limits without limit theorems I have to prove the following limit without using any limit theorems. I can only do so by using the Archimedean Proprety and the definition of a limit.
I have to prove:
$$\lim_{n\to\infty} \frac{n^2 - 2}{3n^2 + 1} = \frac13$$
| The Archimedean Property can be used to show that
$\tag 1 \displaystyle \lim_{n \to \infty}\frac{1}{n^2} = 0$
It is easy to show that
$\tag 2\frac{n^2 - 2}{3n^2 + 1} \lt \frac{1}{3} \text{ for all } n \ge 0$
The following statements are all equivalent for any $0 \lt \varepsilon \lt \frac{7}{3}$ when $n \gt 0$:
$\tag 3 \frac{1}{3} - \varepsilon \lt \frac{n^2 - 2}{3n^2 + 1}$
$\tag , n^2 + \frac{1}{3} - 3 n^2 \varepsilon - \varepsilon \lt n^2 - 2$
$\tag , \frac{7}{3} - \varepsilon \lt 3 n^2 \varepsilon$
$\tag 4 \frac{1}{n^2} \lt \frac{3 \varepsilon}{\frac{7}{3} - \varepsilon}$
By (1) we can find an $N$ so that (4) is true when $n \ge N$. But then both (2) and (3) show that
$\tag 5 \lim_{n\to\infty} \frac{n^2 - 2}{3n^2 + 1} = \frac13$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Prove that $\binom{ap}{p} \equiv a \pmod{ap}$
Let $N = ap$ be a composite positive integer, where $a > 1$ is a positive integer and $p$ is a prime. Prove that $$\binom{ap}{p} \equiv a \pmod{ap}.$$
We have $$\binom{ap}{p} = \dfrac{ap(ap-1) \cdots (ap-(p-1))}{p!} = \dfrac{a(ap-1)(ap-2) \cdots (ap-(p-1))}{(p-1)!}.$$ If $p$ is the smallest prime factor of $ap$, then $(p-1)!$ has an inverse modulo $ap$. Also, $(ap-1)(ap-2) \cdots (ap-(p-1)) \equiv (-1)^{p-1} (p-1)!$. If $p = 2$, then $$\binom{ap}{p} \equiv a \cdot \dfrac{-(p-1)!}{(p-1)!} \equiv -a \equiv a \pmod{ap}.$$ Otherwise, if $p > 2$ is odd then $$\binom{ap}{p} \equiv a \cdot \dfrac{(p-1)!}{(p-1)!} \equiv a \pmod{ap}.$$ How do we prove it in the case that $p$ is not the smallest prime factor of $ap$?
| using Lucas'Theorem, suppose $a\equiv k \pmod p $, we will have:
$\binom{ap}{p} \equiv\binom{k}{1} =k \equiv a \pmod p $
We also have
$\binom{ap}{p} = \dfrac{a(ap-1)(ap-2) \cdots (ap-(p-1))}{(p-1)!} = a\binom{ap-1}{p-1} \equiv 0 \pmod a$
Let $\binom{ap}{p}=Zp+a$, so $a|Z$, so
$Zp+a=Kap+a \equiv a \pmod {ap} $
| {
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"url": "https://math.stackexchange.com/questions/2486493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Rationalize denominator: $\frac{9\sqrt{2}}{1+\sqrt{2}-2\sqrt{5}-\sqrt{10}}$ $$\frac{9\sqrt{2}}{1+\sqrt{2}-2\sqrt{5}-\sqrt{10}}$$
I simplified it by making it:
$$\frac{9\sqrt{2}}{(1-\sqrt{10})(1+\sqrt{2})}$$
What do I do next? I am confused what to do because it is multiplication. All of my previous examples were either use square binomial or square subtraction. Here I cant think of anything to get rid of these square roots.
| $$\frac{9\sqrt{2}(1+\sqrt{10})(1-\sqrt{2})}{(1-\sqrt{10})(1+\sqrt{2})(1+\sqrt{10})(1-\sqrt{2})}=$$
$$\frac{9\sqrt{2}(1+\sqrt{10})(1-\sqrt{2})}{(1-10)(1-2)}=$$
$$\sqrt{2}(1+\sqrt{10})(1-\sqrt{2})=...$$
| {
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"url": "https://math.stackexchange.com/questions/2488268",
"timestamp": "2023-03-29T00:00:00",
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$f(x)=\sqrt{\dfrac{x-\sqrt {20}}{x-3}}$, $g(x)=\sqrt{x^2-4x-12}$, find the domain of $f(g(x))$
$f(x)=\sqrt{\dfrac{x-\sqrt {20}}{x-3}}$, $g(x)=\sqrt{x^2-4x-12}$, find the domain of $f(g(x))$.
For some reason, I cannot get the correct answer. Here is what I tried.
$D_g:$
$$x^2-4x-12\ge0$$
$$(x-6)(x+2)\ge0$$
$$\boxed{(-\infty,-2] \cup [6,\infty)}$$
$D_f:$
$$\dfrac{x-\sqrt{20}}{x-3}\ge0$$
$$\boxed{(-\infty, \sqrt {20}] \cup (3,\infty)}$$
$D_{g(f)}:$
$$\sqrt{x^2-4x-12}\le \sqrt{20}$$
$$[-4,8]$$
$$\text{or}$$
$$\sqrt{x^2-4x-12}>3$$
$$(-\infty, -3) \cup (7,\infty)$$
$$\boxed{(-\infty, -3)\cup (-3,-2) \cup (7,\infty)}$$
So then my final answer was
$$\color{blue}{\boxed{(-\infty, -3)\cup (-3,-2) \cup [6,7) \cup (7,\infty)}}$$
What did I do wrong?
Correct Answer:
$x\le-4\ \text{or}\ -3<x\le-2\ \text{or}\ 6\le x<7\ \text{or}\ x\ge8$
| Here is the complete solution:
The domain of $g(x)$:
$$x\in (-\infty, -2]\cup[6,+\infty).$$
The domain of $f(x)$:
$$\ 1)\begin{cases}\sqrt{x^2-4x-12}\le \sqrt{20} \\ \sqrt{x^2-4x-12}<3\end{cases} \text{or} \ \ 2)\begin{cases}\sqrt{x^2-4x-12}\ge \sqrt{20} \\ \sqrt{x^2-4x-12}>3\end{cases} \Rightarrow$$
$$1)\ \begin{cases} x\in [-4, 8] \\ x\in (-3, 7) \end{cases} \text{or} \ \ 2)\begin{cases}x\in (-\infty, -4]\cup[8,+\infty) \\ x\in (-\infty, -3)\cup(7,+\infty) \end{cases} \Rightarrow$$
$$1)\ x\in (-3, 7) \ \ \text{or} \ \ 2) \ x\in (-\infty, -4]\cup[8,+\infty).$$
Finally, the domain of $f(g(x))$:
$$1)\ \begin{cases} x\in (-\infty, -2]\cup[6,+\infty)\\ x\in (-3, 7) \end{cases} \text{or} \ \ 2)\begin{cases}x\in (-\infty, -2]\cup[6,+\infty) \\ x\in (-\infty, -4]\cup[8,+\infty) \end{cases} \Rightarrow$$
$$1)\ x\in (-3, -2]\cup[6,7) \ \ \text{or} \ \ 2) \ x\in (-\infty, -4]\cup[8,+\infty) \Rightarrow \\ x\in (-\infty, -4]\cup(-3, -2]\cup[6,7)\cup[8,+\infty).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve a complex number equation I am start learning complex number in my university. I don't know a lot about complex number but I know most of the facts and rules about them.
I have the equation:
$$z^{10} + (z - 1)^{10} = 0$$
It has $10$ roots and they are $z_1$, $\overline{z_1}$, $z_2$, $\overline{z_2}$,..., $z_5$, $\overline{z_5}$. We want to find value of:
$$\sum_{i=1}^5 {\frac{1}{z_i \overline{z_i}}}$$
another method for simplification:
$\frac{1}{z_k} = 1 - e^{i\frac{\pi(2k-1)}{10}}$
we get modulus from both side:
$\frac{1}{|z_k|}=|1-\cos(\frac{\pi(2k-1)}{10}) - i\sin(\frac{\pi(2k-1)}{10}) =$
$1 - (1 - 2\sin(\frac{\pi(2k-1)}{20})^{2}-2i\sin(\frac{\pi(2k-1)}{20})\cos(\frac{\pi(2k-1)}{20})|=$
$2|\sin(\frac{\pi(2k-1)}{20})||\sin(\frac{\pi(2k-1)}{20})-i\cos(\frac{\pi(2k-1)}{20})| =$
for $1 \le k \le 5$:
$\frac{1}{|z_k|} = 2\sin(\frac{\pi(2k-1)}{20})$
Is it possible to help me solve this? Thanks everyone.
|
We will show
$$\sum_{k=1}^5 {\frac{1}{z_k \overline{z_k}}}=10.$$
Note that $z^{10} + (z - 1)^{10} = 0$ is equivalent to
$$\left(1-\frac{1}{z}\right)^{10}=-1=e^{-i\pi}.$$
Hence for $k=1,2,3,4,5$,
$$1-\frac{1}{z_k}=e^{i\pi(2k-1)/10}$$
which implies
$$\frac{1}{z_k \overline{z_k}}=(1-e^{i\pi(2k-1)/10})(1-e^{-i\pi(2k-1)/10})=2-2\cos\left(\frac{\pi(2k-1)}{10}\right).$$
Therefore
$$\sum_{k=1}^5 {\frac{1}{z_k \overline{z_k}}}=\sum_{k=1}^5\left(2-2\cos\left(\frac{\pi(2k-1)}{10}\right)\right)=10-2\sum_{k=1}^5\cos\left(\frac{\pi(2k-1)}{10}\right).$$
So it remains to show that the following sum is zero
$$\sum_{k=1}^5\cos\left(\frac{\pi(2k-1)}{10}\right)=\cos\left(\frac{\pi}{10}\right)+\cos\left(\frac{3\pi}{10}\right)+\cos\left(\frac{5\pi}{10}\right)+\cos\left(\frac{7\pi}{10}\right)+\cos\left(\frac{9\pi}{10}\right)$$
which holds by the symmetry $\cos(\pi-x)=-\cos(x)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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three distinct prime factors $x$, $y$ and $10x+y$, where $x$ and $y$ are each less than 10. What is the largest possible value of $m$? The number $m$ is a three-digit positive integer and is the product of the three distinct prime factors $x$, $y$ and $10x+y$, where $x$ and $y$ are each less than 10. What is the largest possible value of $m$?
Any hints are greatly appreciated.
| $\{x,y\} \subset \{2,3,5,7\}$
$10x + 2$ and $10x + 5$ are not prime. $5 + 7 = 12 = 3*4;2+7 = 9=3*3 $ so $57,75,27,72$ are not prime. So possible contenders are.
$(x,y, 10x+y): = (7,3,73),(3,7,37),(5,3,53), (2,3,23)$.
$7*3*73 > 21*70 > 1400$ and not three digits.
Clearly $2< 3, 3 < 5 < 7$ and $23< 37< 53$ so $(2,3,23)$ is out.
$7*37$ and $5*53$ and similar in size so it will be either $3*7*37$ or $5*3*53$ depending upon whether $7*37 = 210 + 49$ is more or less than $5*53 = 250 +15$.
| {
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"url": "https://math.stackexchange.com/questions/2494175",
"timestamp": "2023-03-29T00:00:00",
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Find: $\lim\limits_{n\to \infty}\arcsin \frac {1}{\sqrt{n^2+1}}+\arcsin \frac {1}{\sqrt{n^2+2}}+...+\arcsin \frac {1}{\sqrt{n^2+n}}$ Let
$$x_n=\arcsin \frac {1}{\sqrt{n^2+1}}+\arcsin \frac {1}{\sqrt{n^2+2}}+...+\arcsin \frac {1}{\sqrt{n^2+n}}$$
I would like to find $\lim\limits_{n\to \infty}x_n$
I attempted to use Riemann rectangle formula but it does not work.
Any Hint?
| for any $k\in\mathbb{N}$ we have
$$\arcsin \frac {1}{\sqrt{n^2+1}}\sim \frac{1}{n},\;n\to\infty$$
Indeed
$$\lim_{n\to \infty } \, \frac{ \arcsin \frac{1}{\sqrt{k+n^2}}}{\frac{1}{n}}=^{(*)}\lim_{n\to \infty } \, \frac{-\frac{n}{\sqrt{(k+n^2)^3} \sqrt{1-\frac{1}{k+n^2}}}}{-\frac{1}{n^2}}=\\=\lim_{n\to \infty } \,\frac{n^3}{\sqrt{k^3+3 k^2 n^2+3 k n^4+n^6}\sqrt{1-\frac{1}{k+n^2}}}=\\=\lim_{n\to \infty } \,\frac{n^3}{n^3\sqrt{\frac{k^3}{n^6}+\frac{3k^2}{n^4}+\frac{3k}{n^2}+1}\sqrt{1-\frac{1}{k+n^2}}}=1$$
$(*)$ L'Hopital rule
Thus $$\sum_{k=1}^n\arcsin \frac {1}{\sqrt{n^2+k}} \sim n\cdot \frac{1}{n}=1$$
Therefore $$\color{red}{\lim_{n\to\infty}\sum_{k=1}^n\arcsin \frac {1}{\sqrt{n^2+k}} =1}$$
Hope this helps
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve summation of $\sum_{j=0}^{n-2}2^j (n-j)$ Question
While Solving a recursive equation , i am stuck at this summation and unable to move forward.Summation is
$$\sum_{j=0}^{n-2}2^j (n-j)$$
My Approach
$$\sum_{j=0}^{n-2}2^j (n-j) = \sum_{j=0}^{n-2}2^j \times n-\sum_{j=0}^{n-2}
2^{j} \times j$$
$$=n \times (2^{n-1}-1)-\sum_{j=0}^{n-2}
2^{j} \times j$$
I am unable to move forward , please help me out!
| $$\sum_{j=1}^{n-2}j\;2^j=2+2\times 2^2+3\times 2^3+\cdots+(n-2)2^{n-2}$$$$=[2+2^2+2^3+\cdots+2^{n-2}]+[2^2+2^3+\cdots+2^{n-2}]+[2^3+\cdots+2^{n-2}]+\cdots+2^{n-2}$$$$=2(2^{n-2}-1)+2^2(2^{n-3}-1)+2^3(2^{n-4}-1)+\cdots+2^{n-2}(2^{n-(n-1)}-1)$$ $$=(2^{n-1}-2)+(2^{n-1}-2^2)+(2^{n-1}-2^3)+\cdots+(2^{n-1}-2^{n-2})$$ $$=(n-2)2^{n-1}-(2+2^2+\cdots+2^{n-2})=(n-2)2^{n-1}-2(2^{n-2}-1)$$ $$=(n-2)2^{n-1}-2^{n-1}+2=(n-3)2^{n-1}+2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2497799",
"timestamp": "2023-03-29T00:00:00",
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Probability that exactly three cards in at least one of the suits in an eight card hand from a standard deck. so my solution was to...
a) find the number of hands where exactly one suit has exactly 3 cards
b) find the number of hands where exactly two suits have exactly 3 cards
c) add a) and b) then divide by c(52,8).
so starting with b) because it's slightly easier...
c(4,1) way to pick a suit and c(13,3) ways to choose 3 cards in said suit...
c(3,1) way to pick rest of suits and c(13,3) ways to pick cards in that suit...
2 cards left among 26, so just c(26,2).
so answer for b) would be - > 12*c(13,3)^2*c(26,2)
then going with a)
c(4,1) way to pick a suit and c(13,3) ways to choose 3 cards in said suit...
c(39,5) ways to pick rest of cards but must consider that we want to exclude possibility of having another 3 in another deck...
so what i did was count the ways in which we could have 3 cards in a suit among the 3 suits and 5 cards we need...
so c(3,1) ways of choosing a suit along with c(13,3) ways of selecting 3, and c(26,2) ways of selecting rest of the suits...
so c(39,5) - 3*c(13,3)*c(26,2) is my answer for a).
add them and divide by c(52,8) but i get a probability that is way too high...
| The numerator:
${13\choose 3}{13\choose 5}{13\choose 0}{13\choose 0}{4\choose 1,1,2} + {13\choose 3}{13\choose 4}{13\choose 1}{13\choose 0}{4\choose 1,1,1,1} + {13\choose 3}{13\choose 3}{13\choose 2}{13\choose 0}{4\choose 2,1,1,1} + {13\choose 3}{13\choose 3}{13\choose 1}{13\choose 1}{4\choose 2,2}+{13\choose 3}{13\choose 2}{13\choose 2}{13\choose 1}{4\choose 1,2,1} = 499163808$
$\frac {499163808}{{52\choose 8}} = 66.3\%$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding $\cos(A+C)$ given that $A,B,C$ are angles in an acute triangle. Given that $A,B,C$ be angles in an acute triangle.
If $(5+4\cos A)(5-4\cos B)=9$ and $(13-12\cos B)(13-12\cos C)=25$
find $\cos(A+C)$.
I know $A+B+C=180^\circ$ and $\cos B=-\cos(A+C)$ and what next?
| Thanks for the challenging problem! I enjoyed working through it.
The tangent half-angle substitution works really well here. Let $a = \tan ({A\over 2})$, $b = \tan({B\over2})$, $c = \tan({C\over2})$ then $a,b,c > 0$ and
$$ \cos A = \frac{1-a^2}{1+a^2}, \quad \cos B = \frac{1-b^2}{1+b^2}, \quad \cos C = \frac{1-c^2}{1+c^2} $$
Substituting these into the given equations and simplifying we (eventually) get $$ a = 3b, \quad b = {1\over 5c}$$
Using the triangle relationship we also have
$$ \tan \left( {A\over2} + {B\over2} \right) = \tan \left(90^\circ-{C\over2}\right) = \cot \left({C\over2}\right) $$
$$ \frac{a+b}{1-ab} = \frac{1}{c} $$
Thus
$$ \frac{4b}{1-3b^2} = 5b $$
$$ b^2 = {1\over 15}$$
Back-substituting gives
$$ \cos B = -\cos(A+C) = {7\over 8} $$
| {
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Find Probability that even numbered face occurs odd number of times A Die is tossed $2n+1$ times. Find Probability that even numbered face occurs odd number of times
First i assumed let $a,b,c,d,e,f$ be number of times $1$ occurred, $2$ occurred and so on $6$ occurred respectively. Then we have
$$a+b+c+d+e+f=2n+1$$ is a linear equation where variables are non negative integers.
So number of non negative integer solutions of above is $\binom{2n+6}{5}$
But $b$, $d$ and $f$ should be odd. So let $b=2x+1$, $d=2y+1$ and $f=2z+1$ where $x,y,z$ are non negative integers.
so we get
$$a+2x+c+2y+e+2z=2n-2$$ which is possible in the following two cases
Case $1.$ $a,c,e$ are all even so let $a=2m+1$, $c=2q+1$ and $e=2r+1$ we get
$$m+x+q+y+r+z=n-1$$ so number of solutions is $\binom{n+4}{5}$
Case $2.$ Exactly two of $a,c,e$ are odd and other is even which can be possible in 3 ways.
So let $a=2m+1$, $c=2q+1$, $e=2r$ we get
$$m+x+q+y+r+z=n-2$$
Number of solutions is $3 \times \binom{n+3}{5}$
Hence Probability is
$$P(A)=\frac{\binom{n+4}{5}+3\binom{n+3}{5}}{\binom{2n+6}{5}}$$
Is my approach fine?
| First of all, it should be noted that $0+3+0+1+0+1=5$ and $1+1+1+1+0+1=5$ are not equally likely. There are $\frac{5!}{3!}$ ways to get the first and $5!$ ways to get the second.
So counting these is not going to be a good approach.
Letting $P(x_2,x_4,x_6)=(3+x_2+x_4+x_6)^{2n+1}$, you then let:
$$P_2(x_2,x_4,x_6)=\frac{1}{2}\left(P(x_2,x_4,x_6)-P(-x_2,x_4,x_6)\right),$$ this gives us the terms where $x_2$ has odd power.
Then $P_4(x_2,x_4,x_6)=\frac{1}{2}\left(P_2(x_2,x_4,x_6)-P_2(x_2,-x_4,x_6)\right)$ has the terms where $x_2$ and $x_4$ have odd powers.
Finally, $P_6(x_2,x_4,x_6)=\frac{1}{2}\left(P_4(x_2,x_4,x_6)-P_4(x_2,x_4,-x_6)\right)$ is the terms where all of $x_2,x_4,x_6$ have odd exponents.
Then the probability you want is:
$$\begin{align}\frac{P_6(1,1,1)}{6^{2n+1}}&=\frac{6^{2n+1}-3\cdot 4^{2n+1}+3\cdot 2^{2n+1}}{8\cdot 6^{2n+1}}\end{align}$$
Another approach is to use Markov processes, with states $0$ to $3$, which is the number of $2,4,6$ which have occurred an odd number of times. Start in state $0$ and you get a transition matrix:
$$A=\begin{pmatrix}\frac{1}{2}&\frac{1}{2}&0&0\\
\frac{1}{6}&\frac{1}{2}&\frac{1}{3}&0\\
0&\frac{1}{3}&\frac{1}{2}&\frac{1}{6}\\
0&0&\frac{1}{2}&\frac{1}{2}
\end{pmatrix}$$
And the probability you want is $$\begin{pmatrix}0&0&0&1\end{pmatrix}A^{2n+1}\begin{pmatrix}1\\0\\0\\0\end{pmatrix}$$
Diagonalizing gives eigenvalues of $A$ equal to $0,\frac{1}{3},\frac{2}{3},1$, and the same probability as above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2502305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How prove this $\sum_{k=2}^{n}\frac{1}{3^k-1}<\frac{1}{5}$ show that $$\sum_{k=2}^{n}\frac{1}{3^k-1}<\dfrac{1}{5}\tag1$$
I try to use this well known: if $a>b>0,c>0$,then we have
$$\dfrac{b}{a}<\dfrac{b+c}{a+c}$$
$$\dfrac{1}{3^k-1}<\dfrac{1+1}{3^k-1+1}=\dfrac{2}{3^k}$$
so we have
$$\sum_{k=2}^{n}\dfrac{1}{3^k-1}<\sum_{k=2}^{n}\dfrac{2}{3^k}=\dfrac{1}{3}$$but this is big than $\dfrac{1}{5}$,so how to prove inequality (1)
| $$\sum_{k\geq 2}\frac{1}{3^k-1}=-\frac{1}{2}+\sum_{k\geq 1}\sum_{m\geq 1}\frac{1}{3^{km}}=-\frac{1}{2}+\sum_{n\geq 1}\frac{d(n)}{3^n}\leq -\frac{1}{2}+\sum_{n=1}^{5}\frac{d(n)}{3^n}+\sum_{n\geq 6}\frac{n}{3^n}$$
leads to the sharper inequality
$$ \mathcal{T}=\sum_{k\geq 2}\frac{1}{3^k-1}\leq \frac{61}{324}.$$
We may actually devise an acceleration method for the series
$$\mathcal{S}=\sum_{n\geq 1}\frac{1}{3^n-1}=\sum_{n\geq 1}\frac{1}{3^n}+\sum_{n\geq 1}\frac{1}{3^n(3^n-1)}=\frac{1}{2}+\mathcal{T}$$
since
$$ \frac{1}{9}\mathcal{T} = \sum_{n\geq 1}\frac{1}{3^{n+1}(3^{n+1}-3)},\qquad \mathcal{T}-\frac{1}{6}=\sum_{n\geq 1}\frac{1}{3^{n+1}(3^{n+1}-1)} $$
are pretty close, namely
$$ \frac{1}{6}-\frac{8}{9}\mathcal{T}=\sum_{n\geq 1}\frac{2}{3^{n+1}(3^{n+1}-1)(3^{n+1}-3)}\geq \frac{1}{216}$$
(leading to $\mathcal{T}\leq\frac{35}{192}$) but
$$ \frac{1}{27}\left(\frac{1}{6}-\frac{8}{9}\mathcal{T}\right)-\left(\frac{1}{6}-\frac{8}{9}\mathcal{T}-\frac{1}{216}\right)=16\sum_{n\geq 1}\frac{1}{3^{n+2}(3^{n+2}-1)(3^{n+2}-3)(3^{n+2}-9)}\geq \frac{1}{18954} $$
leads to $\mathcal{T}\geq \frac{11821}{64896}$, hence $\mathcal{T}=\color{green}{0.182}\ldots$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2502560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
coefficient of $x^{17}$ in $(x^2+x^3+x^4+x^5+x^6+x^7)^3$ coefficient of $x^{17}$ in $(x^2+x^3+x^4+x^5+x^6+x^7)^3$ - I'm reviewing for an exam and don't understand the answer, C(11+3−1,11)−C(3,1)×C(5+3−1,5).
|
Using the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ we obtain
\begin{align*}
\color{blue}{[x^{17}](x^2+x^3+\cdots+x^7)^3}&=[x^{17}]x^6(1+x+\cdots+x^5)^3\tag{1}\\
&=[x^{11}]\left(\frac{1-x^6}{1-x}\right)^3\tag{2}\\
&=[x^{11}]\left(1-\binom{3}{1}x^6\right)\sum_{n=0}^\infty \binom{n+3-1}{n}x^n\tag{3}\\
&=\left([x^{11}]-\binom{3}{1}[x^{5}]\right)\sum_{n=0}^\infty \binom{n+3-1}{n}x^n\\
&\color{blue}{=\binom{11+3-1}{11}-\binom{3}{1}\binom{5+3-1}{5}}\tag{4}\\
&=\binom{13}{2}-3\binom{7}{2}\\
&=15
\end{align*}
Comment:
*
*In (1) we factor out $x^6$.
*In (2) we apply the rule $[x^p]x^qA(x)=[x^{p-q}]A(x)$ and use the finite geometric series formula.
*In (3) we expand $(1-x^6)^3$ up to $x^6$ since other terms do no contribute to $x^{11}$.
*In (4) we select the coefficient of $x^{11}$ and $x^5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2504632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Help solving variable separable ODE: $y' = \frac{1}{2} a y^2 + b y - 1$ with $y(0)=0$ I am studying for an exam about ODEs and I am struggling with one of the past exam questions. The past exam shows one exercise which asks us to solve:
$$y' = \frac{1}{2} a y^2 + b y - 1$$with $y(0)=0$
The solution is given as
$$y(x) = \frac{2 \left( e^{\Gamma x} - 1 \right)}{(b + \Gamma)(e^{\Gamma x} - 1) + 2\Gamma},$$ with $\Gamma = \sqrt{b^2 + 2a}$
I am really getting stuck at this exercise and would love to have someone show me how this solution is derived.
One thing I did find out is that this ODE is variable separable. That is,
$$y' = g(x)h(y) = (1) \cdot (\frac{1}{2} ay^2 + by - 1),$$ and therefore the solution would result from solving
$$\int \frac{1}{\frac{1}{2} ay^2 + by - 1} dy = \int dx + C,$$
where C is clearly zero because $y(0) = 0$.
I am now getting stuck at solving the left integral. Could anyone please show me the steps?
UPDATE
So I came quite far with @LutzL solution, however my answers seems to slightly deviate from the solution given above. These are the steps I performed (continuing from @LutzL's answer):
You complete the square $\frac12ay^2+by-1=\frac12a(y+\frac ba)^2-1-\frac{b^2}{2a}$ and use this to inspire the change of coordinates $u=ay+b$ leading to
$$
\int \frac{dy}{\frac12ay^2+by-1}=\int\frac{2\,du}{u^2-2a-b^2}
$$
and for that your integral tables should give a form using the inverse hyperbolic tangent. Or you perform a partial fraction decomposition for
$$
\frac{2Γ}{u^2-Γ^2}=-\frac{1}{u+Γ}+\frac{1}{u-Γ}
$$
and find the corresponding logarithmic anti-derivatives,
$$
\ln|u-Γ|-\ln|u+Γ|=Γx+c,\\ \frac{u-Γ}{u+Γ}=Ce^{Γx},\ C=\pm e^c
$$
which you now can easily solve for $u$ and then $y$.
Given that $y(0) = 0$ we have $u(y(0)) = u(0) = b$ and therefore the final equation becomes
$$\frac{u(0)-Γ}{u(0)+Γ}= \frac{b-Γ}{b+Γ}=Ce^{Γ\cdot0} = Ce^{Γ \cdot 0} = C$$
Now by first isolating $u$ I get
$$u - \Gamma = u C e^{\Gamma x} + \Gamma C e^{\Gamma x} \Rightarrow \\
u \left( 1 - C e^{ \Gamma x} \right) = \Gamma \left( 1 + C e^{\Gamma x} \right) \Rightarrow \\
u = \frac{\Gamma \left( 1 + C e^{\Gamma x} \right)}{\left( 1 - C e^{ \Gamma x} \right)}$$
Now substituting u and C gives
$$ay + b= \frac{\Gamma \left( 1 + \frac{b-Γ}{b+Γ} e^{\Gamma x} \right)}{\left( 1 - \frac{b-Γ}{b+Γ} e^{ \Gamma x} \right)} \Rightarrow \\
y = \frac{\Gamma \left( 1 + \frac{b-Γ}{b+Γ} e^{\Gamma x} \right) - b \left( 1 - \frac{b-Γ}{b+Γ} e^{ \Gamma x} \right)}{a \left( 1 - \frac{b-Γ}{b+Γ} e^{ \Gamma x} \right)}$$
Now using the fact that $\Gamma = \sqrt{b^2 + 2a} \Rightarrow a = \frac{(\Gamma + b)(\Gamma - b)}{2}$ we get that
$$y = \frac{2 \left( \Gamma \left( 1 + \frac{b-\Gamma }{b+\Gamma } e^{\Gamma x} \right) - b \left( 1 - \frac{b-\Gamma }{b+\Gamma } e^{ \Gamma x} \right) \right)}{(\Gamma + b)(\Gamma - b) \left( 1 - \frac{b-\Gamma }{b+\Gamma } e^{ \Gamma x} \right)} \\
= \frac{2 \left( (\Gamma - b) + (\Gamma + b) \frac{b-\Gamma }{b+ \Gamma } e^{\Gamma x} \right)}{(\Gamma + b)(\Gamma - b) \left( 1 - \frac{b-\Gamma }{b+\Gamma } e^{ \Gamma x} \right)} \\
= \frac{2 \left( (\Gamma - b) - (\Gamma + b) \frac{\Gamma - b }{b+ \Gamma } e^{\Gamma x} \right)}{(\Gamma + b)(\Gamma - b) \left( 1 - \frac{b-\Gamma }{b+\Gamma } e^{ \Gamma x} \right)}$$
Now cancelling the terms $(b + \Gamma)$ and $(b - \Gamma)$ wherever possible and multiplying denominator and nominator by -1 gives
$$y = \frac{2 \left( e^{\Gamma x} - 1 \right)}{(\Gamma + b) \left( \frac{b-\Gamma }{b+\Gamma } e^{ \Gamma x} - 1 \right)} $$
So clearly, I got the nominator right, but I can not seem to get the denominator to equal $(b + \Gamma)(e^{\Gamma x} - 1) + 2\Gamma$. Can someone rescue me and show me what I did wrong?
Maybe it helps if I say that x is always positive?
| write $$\int \frac{\frac{dy}{dx}}{\frac{1}{2}ay(x)^2+by(x)-1}dx=\int1dx$$
$$\frac{2\tan^{-1}\left(\frac{b+ay(x)}{\sqrt{-2a-b^2}}\right)}{\sqrt{-2a-b^2}}=x+C$$
$$y(x)=\frac{-b+\sqrt{-2a-b^2}\tan\left(\frac{1}{2}\sqrt{-2a-b^2}(x+C)\right)}{a}$$
| {
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"url": "https://math.stackexchange.com/questions/2504760",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
If $a + \frac{b}{2} + \frac{c}{3} + \frac{d}{4} = 0$ then relevant cubic has root between $0$ and $1$ If $a + \frac{b}{2} + \frac{c}{3} + \frac{d}{4} = 0$ then prove the function $$P(x)=a+bx+cx^2+dx^3$$ has a root somewhere between $0$ and $1$.
If it has a root between and $0$ and $1$ then I could show that it changes signs between $0$ and $1$. But some cubics have roots where they don't change signs.
$$P(1)=a+b+c+d\\P(0)=a$$
So
$$P(1)=a+b+c+d=\\=a+b+c+d-(a + \frac{b}{2} + \frac{c}{3} + \frac{d}{4})\\=\frac{b}{2} + \frac{2c}{3} + \frac{3d}{4}\\P(0)=a=a-(a + \frac{b}{2} + \frac{c}{3} + \frac{d}{4})\\= -\frac{b}{2} -\frac{c}{3} - \frac{d}{4}$$
I feel like I'm close.
Although I'm sure there is a solution using the formula for cubic roots, I don't think I am supposed to use it, because the next part of the question (which I will hopefully be able to do myself) asks to generalize the result to any polynomial.
| .Consider the polynomial defined by : $Q(x) = ax + \frac{bx^2}{2} + \frac{cx^3}{3} + \frac{dx^4}{4}$. Note that $Q(0) = 0$, since $Q(x)$ has constant term $0$. Furthermore, $Q(1) = a + \frac b2 + \frac c3 + \frac d4 = 0$. Therefore, we have located two roots of $Q$, $0$ and $1$.
Therefore, by Rolle's theorem, the derivative of $Q$ has a root between $0$ and $1$. But you can see that $Q'(x) = P(x)$, so that $P(x)$ has a root between $0$ and $1$.
| {
"language": "en",
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What does $A = L + D + U$ look like? Exercise:
Consider the problem
\begin{split}
10u_1 + u_2 &= 1\\
u_1 + 10u_2 &= 10
\end{split}
with the solution $(u_1,u_2)^T = (0,1)^T$. For a general system of equations
$$Au = b$$
with an $n\times n$ matrix $A$, which consists of a lower triangular submatrix $L$, the diagonal $D$ and the uppertriangular submatrix $U\,(A=L+D+U)$, a Gauss–Seidel iteration is defined by $$u^{i+1} = D^{-1}(b - Lu^{i+1} - Uu^i)$$
and a Jacobi iteration is defined by $$u^{i+1} = D^{-1}(b - Lu^i - Uu^i)$$
Perform:
a) Four Gauss-Seidel iterations.
b) Four Jacobi iterations.
Question:
*
*What does $A = L + D + U$ mean? I suppose that I doesn't mean that $a_{ij} = l_{ij} + d_{ij} + u_{ij}$? If that is what it's supposed to mean, do we get $$L = \begin{bmatrix}0 & 0\\
1 &0 \end{bmatrix}, \,D = \begin{bmatrix}10 & 0\\
0&10\end{bmatrix},\,U = \begin{bmatrix} 0&1\\
0&0\end{bmatrix} $$
Thanks in advance!
| Your $L$, $D$, and $U$ are correct. Note that $a_{ij} = l_{ij} + d_{ij} + u_{ij}$ is true, however, two of the three terms are always zero; the upper right entry would be $a_{ij} = 0 + 0 + 1$.
In general, $\color{red}A = \color{blue}L + \color{orange}D + \color{green}U$ looks like
$$
\color{red}{\begin{pmatrix}
a_{1,1} & a_{1,2} & \ldots & a_{1,n-1} & a_{1,n}
\\ a_{2,1} & a_{2,2} & \ldots & a_{2,n-1} & a_{2,n}
\\ \vdots & \vdots & \ddots & \vdots & \vdots
\\ a_{n-1,1} & a_{n-1,2} & \ldots & a_{n-1,n-1} & a_{n-1,n}
\\ a_{n,1} & a_{n,2} & \ldots & a_{n,n-1} & a_{n,n}
\end{pmatrix}}
=\\
\color{blue}{\begin{pmatrix}
0 & 0 & \ldots & 0 & 0
\\ a_{2,1} & 0 & \ldots & 0 & 0
\\ \vdots & \vdots & \ddots & \vdots & \vdots
\\ a_{n-1,1} & a_{n-1,2} & \ldots & 0 & 0
\\ a_{n,1} & a_{n,2} & \ldots & a_{n,n-1} & 0
\end{pmatrix}}
+\\
\color{orange}{\begin{pmatrix}
a_{1,1} & 0 & \ldots & 0 & 0
\\ 0 & a_{2,2} & \ldots & 0 & 0
\\ \vdots & \vdots & \ddots & \vdots & \vdots
\\ 0 & 0 & \ldots & a_{n-1,n-1} & 0
\\ 0 & 0 & \ldots & 0 & a_{n,n}
\end{pmatrix}}
+\\
\color{green}{\begin{pmatrix}
0 & a_{1,2} & \ldots & a_{1,n-1} & a_{1,n}
\\ 0 & 0 & \ldots & a_{2,n-1} & a_{2,n}
\\ \vdots & \vdots & \ddots & \vdots & \vdots
\\ 0 & 0 & \ldots & 0 & a_{n-1,n}
\\ 0 & 0 & \ldots & 0 & 0
\end{pmatrix}}
$$
Note that in this case $L$ and $U$ are strictly triangular matrices. A triangular matrix may have non-zero entries in its diagonal, e.g.: $\begin{pmatrix}1&1\\0&1\end{pmatrix}$ is an upper triangular matrix (but not a strictly upper triangular matrix).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve for $a,b,c,d \in \Bbb R$, given that $a^2+b^2+c^2+d^2-ab-bc-cd-d+\frac 25 =0$ Today, I came across an equation in practice mock-test of my coaching institute, aiming for engineering entrance examination (The course for the test wasn't topic-specific, it was a test of complete high school mathematics). It was having four variables and only one equation. While analyzing my test paper, this is the only problem I (and my friends too) couldn't figure out even after giving this problem several hours. So I came here for some help.
Question : Solve for $a,b,c,d \in \Bbb R$, given that $$a^2+b^2+c^2+d^2-ab-bc-cd-d+\frac 25 =0$$
Since only one equation is given, there must be involvement of making of perfect squares, such that they all add up to $0$. Thus, resulting in few more equations. But how to?
I tried a lot of things, such as making $(a-b)^2 $ by adding the missing terms and subtracting again, but got no success.
Thanks!
| It's
$$a^2-ab+\frac{b^2}{4}+\frac{3}{4}b^2-bc+\frac{1}{3}c^2+\frac{2}{3}c^2-cd+\frac{3}{8}d^2+\frac{5}{8}d^2-d+\frac{2}{5}=0$$ or
$$\left(a-\frac{b}{2}\right)^2+\frac{3}{4}\left(b-\frac{2c}{3}\right)^2+\frac{2}{3}\left(c-\frac{3d}{4}\right)^2+\frac{5}{8}d^2-d+\frac{2}{5}=0$$ or
$$\left(a-\frac{b}{2}\right)^2+\frac{3}{4}\left(b-\frac{2c}{3}\right)^2+\frac{2}{3}\left(c-\frac{3d}{4}\right)^2+\frac{5}{8}\left(d-\frac{4}{5}\right)^2=0,$$
which gives the answer.
| {
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"source": "stackexchange",
"question_score": "5",
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"answer_id": 3
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A problem :$(edf)^3(a+b+c+d+e+f)^3\geq(abcdef)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}+\frac{1}{f})^3$ Hello during a problem I have this to solve :
Let $a,b,c,d,e,f$ be real positiv number such that $a\geq b\geq c\geq d\geq e\geq f\geq 1$ then :
$$(edf)^3(a+b+c+d+e+f)^3\geq(abcdef)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}+\frac{1}{f})^3$$
I did not find any counter-example (with Pari-Gp and Wolfram alpha) so maybe it's true
Advice :I do not think that Am-Gm will works on it .
Thanks a lot
| The note of Alex says that it remains to prove that
$$(a_1+a_2+a_3+3)^3\geq a_1a_2a_3\left(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+3\right)^3$$ for all $a_i\geq1$.
Indeed, let $a_1=a^3$, $a_2=b^3$ and $a_3=c^3$.
Thus, it's enough to prove that
$$a^3+b^3+c^3+3\geq abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+3\right)$$ or
$$a^2b^2c^2(a^3+b^3+c^3+3)\geq a^3b^3+a^3c^3+b^3c^3+3a^3b^3c^3$$ or
$$a^2b^2c^2\sum_{cyc}(a^3-abc)\geq\sum_{cyc}(a^3b^3-a^2b^2c^2)$$ or
$$a^2b^2c^2(a+b+c)\sum_{cyc}(a-b)^2\geq(ab+ac+bc)\sum_{cyc}c^2(a-b)^2$$ or
$$\sum_{cyc}(a-b)^2c^2(a^2b^2(a+b+c)-ab-ac-bc)\geq0.$$
Now, since our inequality is symmetric, we can assume $a\geq b\geq c$.
Thus, $$\sum_{cyc}(a-b)^2c^2(a^2b^2(a+b+c)-ab-ac-bc)\geq$$
$$\geq(a-c)^2b^2(a^2c^2(a+b+c)-ab-ac-bc)+(b-c)^2a^2(b^2c^2(a+b+c)-ab-ac-bc)\geq$$
$$\geq(b-c)^2a^2(a^2c^2(a+b+c)-ab-ac-bc)+(b-c)^2a^2(b^2c^2(a+b+c)-ab-ac-bc)=$$
$$=(b-c)^2a^2(c^2(a^2+b^2)(a+b+c)-2(ab+ac+bc))\geq$$
$$\geq(b-c)^2a^2((a+b)(a+b+c)-2(ab+ac+bc))=(b-c)^2a^2(a^2+b^2-ac-bc)\geq0$$
and we are done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Fourier cosine transform and K-Bessel function I am looking for a reference on the link between the asymptotic expansion of a functions and its Fourier cosine transform asymptotics.
More precisely I would like to find a reference to have the expansion in zero and infinity of (for $Re(a)<\frac{1}{2}$):
$$F(y)= \int_0^{\infty} \frac{1}{\sqrt{x}} K_a(\frac{1}{x}) \cos(xy) dx$$
By integration by parts it is easy to show that for $n$ positive interger $F(y)= o(\frac{1}{y^n})$ for $y \to \infty$, but any reference to use the fact that $\frac{1}{\sqrt{x}} K_a(\frac{1}{x}) \sim_0 \sqrt{\frac{\pi}{2}}e^{-\frac{1}{x}}$ for $x \to 0$ and show that $F(y)$ decreases like $e^{-c \sqrt{y}}$ for $y\to0$?
And to show that $F(y) \sim k_1 y^{-\frac{1}{2}+a} + k_2 y^{-\frac{1}{2}-a}$ for $y \to 0$ using that $\frac{1}{\sqrt{x}} K_a(\frac{1}{x}) \sim c_1 x^{-\frac{1}{2}+a} + c_2 x^{-\frac{1}{2}-a}$ for $x \to \infty$.
I found only a reference to deduce the asymptotic expansion at infinity of a cosine transform depending of serie expansion in zero of initial function.(Here there is no serie expansion near zero as $\frac{1}{\sqrt{x}} K_a(\frac{1}{x}) \sim_0 \sqrt{\frac{\pi}{2}}e^{-\frac{1}{x}}$ as no serie expansion in zero...)
| To obtain the asymptotic behaviour of this integral near $y=0$, the Mellin transform method can be used. One has
\begin{align}
\mathcal{M} \left[\cos \left(z\right),s\right]&=\Gamma (s)\cos\left( \frac{\pi}{2}s \right)\\
\mathcal{M} \left[x^{-1/2}K_a\left( x^{-1} \right),1-s\right]&=2^{-5/2+s}\Gamma (\frac{s}{2}-\frac{1}{4}+\frac{a}{2})\Gamma (\frac{s}{2}-\frac{1}{4}-\frac{a}{2})
\end{align}
The second expression is obtained by transformation of the well known Mellin transform of the Bessel function (multiplication by $x^{-1/2}$ and change $x\to x^{-1}$). The first identity is valid for $0<s<1$ and the second for $s>a+\tfrac{1}{2}$. It comes
\begin{equation}
F(y)=\frac{1}{2i\pi}\int_{c-i\infty}^{c+i\infty}x^{-s}\Gamma (s)\cos\left( \frac{\pi}{2}s \right)2^{-5/2+s}\Gamma (\frac{s}{2}-\frac{1}{4}+\frac{a}{2})\Gamma (\frac{s}{2}-\frac{1}{4}-\frac{a}{2})\,ds
\end{equation}
where $s+\frac{1}{2}<s<1$.
To evaluate this integral, one may close the contour by a large semi-circle on the left side of the vertical line $s=c$. It gives the asymptotic expansion of the integral, as the poles are situated at $s=\frac{1}{2}+a-2n, \frac{1}{2}-a-2n,-2n$, with $n=0,1,2...$ Taking into account the first three contributions ($n=0$), one obtains
\begin{equation}
F(y)\sim K_1x^{-\tfrac{1}{2}-a}+K_1x^{-\tfrac{1}{2}+a}+K_3
\end{equation}
Expressions for the constants are obtain from the calculus of the residues
\begin{align}
K_1&=2^{-a-\frac{1}{2}}\sqrt{\pi}\Gamma(2a)\left( \cos(\frac{\pi}{2}a)-\sin(\frac{\pi}{2}a)\right)\\
K_2&=2^{a-\frac{1}{2}}\sqrt{\pi}\Gamma(-2a)\left( \cos(\frac{\pi}{2}a)+\sin(\frac{\pi}{2}a)\right)\\
K_3&=2^{3/2}\frac{2\Gamma(\frac{3+2a}{4})\Gamma(\frac{3-2a}{4})}
{1-4a^2}
\end{align}
| {
"language": "en",
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"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $\lim_{n\rightarrow\infty} \frac{(n+1)^{2n^2+2n+1}}{(n+2)^{n^2+2n+1} n^{n^2}}$ Woflram gives $\frac{1}{e}$ as the limit, but I failed to obtain it. Please help.
$$\lim_{n\rightarrow\infty} \frac{(n+1)^{2n^2+2n+1}}{(n+2)^{n^2+2n+1}n^{n^2}}$$
| Consider $$A=\frac{(n+1)^{2n^2+2n+1}}{(n+2)^{n^2+2n+1}n^{n^2}}$$ and take logarithms
$$\log(A)=(2n^2+2n+1)\log(n+1)-(n^2+2n+1)\log(n+2)-n^2\log(n)$$ Now, rewrite
$$\log(n+1)=\log(n)+\log\left(1+\frac 1n \right)=\log(n)+\frac{1}{n}-\frac{1}{2 n^2}+O\left(\frac{1}{n^3}\right)$$
$$\log(n+2)=\log(n)+\log\left(1+\frac 2n \right)=\log(n)+\frac{2}{n}-\frac{2}{n^2}+O\left(\frac{1}{n^3}\right)$$ and replace. You should arrive to
$$\log(A)=-1+\frac{2}{n}+\frac{3}{2 n^2}+O\left(\frac{1}{n^3}\right)$$ Now, using
$$A=e^{\log(A)}\implies A=\frac 1 e\left(1+\frac{2}{n}+\frac{7}{2 n^2}+O\left(\frac{1}{n^3}\right)\right)$$
| {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Show recursion in closed form I've got following sequence formula:
$ a_{n}=2a_{n-1}-a_{n-2}+2^{n}+4$
where $ a_{0}=a_{1}=0$
I know what to do when I deal with sequence in form like this:
$ a_{n}=2a_{n-1}-a_{n-2}$
- when there's no other terms but previous terms of the sequence.
Can You tell me how to deal with this type of problems?
What's the general algorithm behind solving those?
| $$\begin{align}
a_{n+2} - 2a_{n+1} + a_{n} &= 4 \cdot 2^{n} + 4 \\
a_{n+1} - 2a_{n} + a_{n-1} &= 2 \cdot 2^{n} + 4 \\
a_{n} - 2a_{n-1} + a_{n-2} &= 1 \cdot 2^{n} + 4 \\
\end{align}$$
$$
\begin{bmatrix}
1 & -2 & 1 & 0 & 0 \\
0 & 1 & -2 & 1 & 0 \\
0 & 0 & 1 & -2 & 1 \\
\end{bmatrix}
\begin{bmatrix}
a_{n + 2} \\ a_{n + 1} \\ a_{n} \\ a_{n - 1} \\ a_{n - 2}
\end{bmatrix} =
\begin{bmatrix}
4 & 1 \\
2 & 1 \\
1 & 1 \\
\end{bmatrix} \begin{bmatrix} 2^n \\ 4 \end{bmatrix}$$
$$
\begin{bmatrix} 1 & -3 & 2 \end{bmatrix}
\begin{bmatrix}
1 & -2 & 1 & 0 & 0 \\
0 & 1 & -2 & 1 & 0 \\
0 & 0 & 1 & -2 & 1 \\
\end{bmatrix}
\begin{bmatrix}
a_{n + 2} \\ a_{n + 1} \\ a_{n} \\ a_{n - 1} \\ a_{n - 2}
\end{bmatrix} =
\begin{bmatrix} 1 & -3 & 2 \end{bmatrix}
\begin{bmatrix}
4 & 1 \\
2 & 1 \\
1 & 1 \\
\end{bmatrix} \begin{bmatrix} 2^n \\ 4 \end{bmatrix}$$
$$
\begin{bmatrix} 1 & -3 & 2 \end{bmatrix}
\begin{bmatrix}
1 & -2 & 1 & 0 & 0 \\
0 & 1 & -2 & 1 & 0 \\
0 & 0 & 1 & -2 & 1 \\
\end{bmatrix}
\begin{bmatrix}
a_{n + 2} \\ a_{n + 1} \\ a_{n} \\ a_{n - 1} \\ a_{n - 2}
\end{bmatrix} =
\begin{bmatrix} 0 \end{bmatrix}$$
So the characteristic equation of the linear recursion is
$$(x^2 - 3x + 2)(x^2 - 2x + 1)$$
with roots $[2, 1, 1, 1]$, so the general solution is
$$a_n = C_0 2^n + (C_1 + C_2n + C_3n^2)1^n$$
where the coefficients $C$ can be determined form the initial conditions.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Solve $X^2=A$, where X is a 2 by 2 matrix and A is a known matrix A = $
\begin{pmatrix}
1 & 1 \\
-1 & 1 \\
\end{pmatrix}
$
. I wrote X = $
\begin{pmatrix}
a & b \\
c & d \\
\end{pmatrix}
$. So $X^2$=$
\begin{pmatrix}
a^2+bc & ab+bd \\
ac+bc & bc+d^2 \\
\end{pmatrix}
$ = $
\begin{pmatrix}
1 & 1 \\
-1 & 1 \\
\end{pmatrix}
$
Unfortunatelly I don't know how to continue from here and maybe someone can help be find the matrix X
| Hints:
From top-right and bottom-left elements we get:
$$(a+b)d = (a+b)c = -1 \Rightarrow c=-d$$
The top-left and bottom-right elements give us:
$$a^2 + bc = bc + d^2 = 1 \Rightarrow a^2 = d^2 \Rightarrow a = \pm d$$
Can you take it from here?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Taylor inside an integral I know the following integral should be:
$$ \int_{0}^{1} \frac{dx}{\sqrt{1-x^2-\epsilon(1-x^3)}} \approx \pi/2 + \epsilon$$
for $\epsilon$ small. What I do is set $-\epsilon(1-x^3)=y$ and then since $y$ it's always greatly smaller than $(1-x^2)$ I do Taylor expansion around $y=0$:
$$ \frac{dx}{\sqrt{1-x^2+y}} \approx \frac{dx}{\sqrt{1-x^2}}-\frac{y \cdot dx}{2 \cdot (1-x^3)^{3/2}} $$
Then I recover $y=-\epsilon(1-x^3)$ and I get the correct answer. The problem is that I don't know if that's mathematically correct because $y$ depends on $x$.
| You are on safer ground factoring the $1-x$ out of the square root as follows:
$$\begin{align} I(\epsilon) &= \int_0^1 \frac{dx}{\sqrt{1-x^2-\epsilon (1-x^3)}} \\ &= \int_0^1 dx \frac{(1-x)^{-1/2}}{\sqrt{1+x-\epsilon(1+x+x^2)}} \\ &= \int_0^1 dx \, \frac{x^{-1/2}}{\sqrt{2-x-\epsilon (3-3 x+x^2)}} \\ &=2 \int_0^1 dx \left [ 2-x^2 - \epsilon (3-3 x^2+x^4)\right ]^{-1/2} \\ &= 2 \int_0^1 dx (2-x^2)^{-1/2} \left [1-\epsilon \frac{3-3 x^2+x^4}{2-x^2} \right ]^{-1/2} \end{align}$$
I hope you see where this is heading. You may now expand the term in brackets in a Taylor expansion in $\epsilon$ knowing that the term involving $\epsilon$ is small over the entire region of integration. Use trig substitution and the answer is straightforwardly...
$$I(\epsilon) = \frac{\pi}{2} + \epsilon + O(\epsilon^2)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Linear Algebra exercise problem: intersection of two subspaces I’m trying to solve a linear algebra exercise, but my solution does not match the solution given on the exercise book. This is the exercise:
Given $U = \langle(1,0,1,2), (1,1,2,3)\rangle$, a subspace of $\Bbb R^4$, and $S$, solution of the system
$$\begin{cases}
x+z-t=0 \\
y=1
\end{cases}
$$
Find the intersection $U\cap S$.
First, I determined the form of the generic vector $\mathbf u\in U$, using scalars $a$ and $b$, thus obtaining $u=(a+b,b,a+2b,2a+3b)$. Then, I solved the system using a matrix in canonical form,
$$
\begin{bmatrix}
1 & 0 & 1 & -1 & 0 \\
0 & 1 & 0 & 0 & 1 \\
\end{bmatrix}
$$
obtaining $\mathbf s\in S$, $\mathbf s=(t-z,1,z,t)$.
To find the intersection, I equalled the two generic vectors, $\mathbf u=\mathbf s$, getting $(a+b, b, a+2b, 2a+3b) = (t-z, 1, z, t)$. I solved this equation by setting up a system:
$$\begin{cases}
a+b=t-z \\
b=1 \\
a+2b=z \\
2a+3b=t
\end{cases}
$$
With the corresponding matrix:
$$
\begin{bmatrix}
1 & 0 & -1 & 1 & -1 \\
0 & 1 & 0 & 0 & 1 \\
1 & 0 & 0 & -1 & -2 \\
2 & 0 & -1 & 0 & 3
\end{bmatrix}
$$
Solving this matrix gave me a solution in form of a linear variety, $U\cap S=(-2,1,-1,0)+\langle(1,0,2,1)\rangle$, but this answer differs from the one on the textbook, which is $U\cap S=(1,1,2,3)+\langle(1,0,1,2)\rangle$. Can you help me understand where I made a mistake?
Thank you in advance for your help!
P. S. Sorry if I have misspelled some entity, but English is not my native language!
| The first row of the $4\times5$ matrix, the last entry should be $0$ rather than $-1$.
The last two rows of the matrix correspond to equations $x-t=-2$ and $2x-z=3$. How do these equations come from the given equations? There seems to be some confusion.
Here is how the problem should be worked:
$$\begin{pmatrix}
x\\y\\z\\t
\end{pmatrix}=a\begin{pmatrix}
1\\0\\1\\2
\end{pmatrix}+b\begin{pmatrix}
1\\1\\2\\3
\end{pmatrix}=\begin{pmatrix}
a+b\\b\\a+2b\\2a+3b
\end{pmatrix}$$
Since $y=1$ we have
$$\begin{pmatrix}
x\\y\\z\\t
\end{pmatrix}=\begin{pmatrix}
a+b\\b\\a+2b\\2a+3b
\end{pmatrix}
=\begin{pmatrix}
a+1\\1\\a+2\\2a+3
\end{pmatrix}
$$
Then
\begin{equation}
x+z-t=
(a+1)+(a+2)-(2a+3)=0
\end{equation}
So the intersection is the line
$$\begin{pmatrix}
x\\y\\z\\t
\end{pmatrix}=a\begin{pmatrix}
1\\0\\1\\2
\end{pmatrix}+\begin{pmatrix}
1\\1\\2\\3
\end{pmatrix}
$$
Notice, also, that $S$ is not a subspace of $\mathbb{R}^4$ since it does not contain the zero vector.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to evaluate the integral $\int_0^{\pi}\frac{a^n\sin^2x+b^n\cos^2x}{a^{2n}\sin^2x+b^{2n}\cos^2x}dx$?
Evaluate the integral $$\int_0^{\pi}\frac{a^n\sin^2x+b^n\cos^2x}{a^{2n}\sin^2x+b^{2n}\cos^2x}dx.$$
I have no idea.
$$\int_0^{\pi}\dfrac{a^n\sin^2x+b^n\cos^2x}{a^{2n}\sin^2x+b^{2n}\cos^2x}dx=\int_0^{\pi/2}\dfrac{a^n\sin^2x+b^n\cos^2x}{a^{2n}\sin^2x+b^{2n}\cos^2x}dx=\int_0^{\pi/2}\dfrac{a^n\tan^2x+b^n}{a^{2n}\tan^2x+b^{2n}}dx$$
I try the substitution $\tan{x}=t$, but it doesn’t work.
| Your approach does work if you would do it correctly. Assume $a\neq 0$ and $b\neq 0$, why? Otherwise the integral is boring. Also assume that $|a|\neq |b|$ for the same reason. Let:
\begin{align}
I=\int^{\pi}_0 \frac{a^n\sin^2(x)+b^n\cos^2(x)}{a^{2n}\sin^2(x)+b^{2b}\cos^2(x)}dx = 2\int^{\pi/2}_0 \frac{a^n\sin^2(x)+b^n\cos^2(x)}{a^{2n}\sin^2(x)+b^{2b}\cos^2(x)}dx
\end{align}
Set $\tan(x) =t$ and get $\sin(x)=\frac{t^2}{1+t^2}$ and $\cos(x)=\frac{1}{1+t^2}$ and $dx = \frac{dt}{1+t^2}$ so that the original integral $I$ becomes:
\begin{align}
I = 2\int^\infty_0 \frac{a^nt^2+b^n}{(1+t^2)(a^{2n}t^2+b^{2n})}dt = \frac{1}{a^{2n}} \int^{\infty}_{-\infty} \frac{a^nt^2+b^n}{(1+t^2)(t^2+\frac{b^{2n}}{a^{2n}})} dt
\end{align}
Using a semi cirkel contour in the upper half plane and using the Residue Theorem we get:
\begin{align}
I = \frac{1}{a^{2n}} 2\pi i \left( \text{Res}_{z=i} \frac{a^nz^2+b^n}{(1+z^2)(z^2+\frac{b^{2n}}{a^{2n}})} + \text{Res}_{z=i|b|^n/|a|^n} \frac{a^nz^2+b^n}{(1+z^2)(z^2+\frac{b^{2n}}{a^{2n}})}\right)
\end{align}
Calculating these residues:
\begin{align}
\text{Res}_{z=i} \frac{a^nz^2+b^n}{(1+z^2)(z^2+\frac{b^{2n}}{a^{2n}})} = \frac{-a^n+b^n}{2i (-1+\frac{b^{2n}}{a^{2n}})}
\end{align}
And the other one:
\begin{align}
\text{Res}_{z=i|b|^n/|a|^n} \frac{a^nz^2+b^n}{(1+z^2)(z^2+\frac{b^{2n}}{a^{2n}})} = \frac{-a^n\frac{b^{2n}}{a^{2n}}+b^n}{2i\frac{|b|^n}{|a|^n}(1-\frac{b^{2n}}{a^{2n}})}
\end{align}
If we put everything together we get:
\begin{align}
I = \frac{\pi}{a^{2n}}\left( \frac{-a^n+b^n}{ (-1+\frac{b^{2n}}{a^{2n}})} + \frac{-a^n\frac{b^{2n}}{a^{2n}}+b^n}{\frac{|b|^n}{|a|^n}(1-\frac{b^{2n}}{a^{2n}})}\right)
\end{align}
Simplifying a bit gives us:
\begin{align}
I = \frac{\pi}{(b^{2n}-a^{2n})} \left( b^n-a^n+\frac{|a|^n}{|b|^n}\left( \frac{b^{2n}}{a^{n}}-b^n\right)\right)
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Area under random walk The problem - we have random walk set of functions such that:
$$
f(0) = 0, f(x) = f(i) + \alpha_i(x-i), x \in (i,i+1], i=0,\dots,n-1
$$
where $\alpha_i \in \{-1, +1\}$ choosed uniformly with plobability $\dfrac{1}{2}$.
Need to find probability that $P(\int\limits_0^n f(x) dx = 0)$.
Thank you
| As results above justify, consider sum $S = \sum\limits_{i=0}^{n-1}(n-i+\frac{1}{2})\alpha_i = (n+\frac{1}{2})\sum\limits_{i=0}^{n-1}\alpha_i - \sum\limits_{i=0}^{n-1}i \alpha_i$
Denote as $n^{+}$ and $n^{-}$ number of steps up and down respectively. Clear, $n=n^{+}+n^{-}$. Then:
$$
(n+\frac{1}{2})\sum\limits_{i=0}^{n-1}\alpha_i = (n+\frac{1}{2})(n^{+}-n^{-})
$$
Also, we can work around another term:
$$
\sum\limits_{i=0}^{n-1}i\alpha_i = \sum\limits_{i: \alpha_i = 1}i - \sum\limits_{j: \alpha_j = -1}j = \frac{n(n-1)}{2}-2\sum\limits_{j: \alpha_j = -1}j
$$
Therefore:
$$
S = (n+\frac{1}{2})[n^{+}-n^{-}] - \frac{n(n-1)}{2} + 2\sum\limits_{j: \alpha_j = -1}j
$$
$$
P(S=0) = P\left(\frac{n(n-1)}{2} - (n+\frac{1}{2})[n^{+}-n^{-}]= 2\sum\limits_{j: \alpha_j = -1}j \right) = P\left(\frac{1}{4}(n(n-1) - (2n+1)(n^{+}-n^{-})) = \sum\limits_{j: \alpha_j = -1}j\right)
$$
Also, $P(S=0) = \sum\limits_{k=0}^{n-1}P(S=0|n^{-}=k)P(k)$
Fix $n^{-}=k$ first. Then we have:
$P(S=0|k) = P\left(\frac{1}{4}(n(n-1) - (2n+1)(2n-k)) = \sum\limits_{j: \alpha_j = -1}j\right)$
Denote $G(k) = \frac{1}{4}(n(n-1) - (2n+1)(2n-k)$, as $k$ fixed it's fixed number. And $\sum\limits_{j: \alpha_j = -1}j$ is sum of positive $k$ terms. The number of ways to present $G(k)$ in the form of sum positive k terms is $$\binom{G(k)+k-1}{G(k)-k}$$
Number of all possible ways to choose $k$ moments of decreasing from $n$ moments is $$\binom{n}{k}$$
Hence
$$
P(S=0|k) = \frac{\binom{G(k)+k-1}{G(k)-k}}{\binom{n}{k}}
$$
And
$$
P(S=0) = \sum\limits_{k=0}^{n-1} \frac{\binom{G(k)+k-1}{G(k)-k}}{\binom{n}{k}} P(k) = \sum\limits_{k=0}^{n-1} \frac{\binom{G(k)+k-1}{G(k)-k}}{\binom{n}{k}} \binom{n}{k}\frac{1}{2^n} = \frac{1}{2^n}\sum\limits_{k=0}^{n-1}\binom{G(k)+k-1}{G(k)-k}
$$
| {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Integration using reduction formulas issue. Show $(2n+1)I_n=2na^2I_{n-1}$ where $I_n=\int_{0}^{a}(a^2-t^2)^ndt$. I have a question where I am given $I_n=\int_{0}^{a}(a^2-t^2)^ndt$ and I am asked to show
$$(2n+1)I_n=2na^2I_{n-1}$$
I then decided to let $u=(a^2-t^2)^n$ and $\dfrac{dv}{dt}=1$. I proceeded to use integration by parts giving:
$$[_{a}^{0}t(a^2-t^2)^n]-(-2n)\int_{0}^{a}t(a^2-t^2)^{n-1}dt$$
I can spot that I have $I_{n-1}$ in the second half of the equation however I have a $t$ which I have no idea how to get rid of,any ideas?
| If you will allow for the use of the beta function $\text{B}(m,n)$, where the beta function is defined by
$$\text{B}(m,n) = \int^1_0 x^{m - 1} (1 - x)^{n - 1} \, dx,$$
we can also solve your problem.
Starting with your integral
$$I_n = \int^a_0 (a^2 - t^2)^n \, dt,$$
where I will assume $a > 0$ and $n > - 1$, setting $x = t^2/a^2, dt = a/(2 \sqrt{x}) dx$ while for the limits of integration $(0,a) \mapsto (0,1)$ so that the integral becomes
$$I_n = \frac{a^{2n + 1}}{2} \int^1_0 x^{-1/2} (1 - x)^n \, dx = \frac{a^{2n + 1}}{2} \int^1_0 x^{\frac{1}{2} - 1} (1 - x)^{(n + 1) - 1} \, dx.$$
As the integral $I_n$ is now exactly in the form for the beta function, it can be written as
$$I_n = \frac{a^{2n + 1}}{2} \text{B} \left (\frac{1}{2}, n + 1 \right ).$$
Shifting the index $n \mapsto n - 1$ yields
$$I_{n - 1} = \frac{a^{2n - 1}}{2} \text{B} \left (\frac{1}{2}, n \right ).$$
Making use of the following property for the beta function, namely
$$\text{B}(m,n) = \frac{\Gamma (m) \Gamma (n)}{\Gamma (m + n)},$$
where $\Gamma (x)$ is the gamma function, we have
\begin{align*}
\frac{I_n}{I_{n - 1}} &= \frac{\frac{a^{2n + 1}}{2} \text{B} \left (\frac{1}{2}, n + 1 \right )}{\frac{a^{2n - 1}}{2} \text{B} \left (\frac{1}{2}, n \right )}\\
&= a^2 \cdot \frac{\Gamma \left (\frac{1}{2} \right ) \Gamma (n + 1)}{\Gamma \left (n + \frac{3}{2} \right )} \cdot \frac{\Gamma \left (n + \frac{1}{2} \right )}{\Gamma \left (\frac{1}{2} \right ) \Gamma (n)}.
\end{align*}
And since $\Gamma (x + 1) = x \Gamma (x)$, we can write this as
\begin{align*}
\frac{I_n}{I_{n - 1}} &= a^2 \cdot \frac{n \Gamma (n)}{\left (n + \frac{1}{2} \right ) \Gamma \left (n + \frac{1}{2} \right )} \cdot \frac{\Gamma \left (n + \frac{1}{2} \right )}{\Gamma (n)} = \frac{2n a^2}{(2n + 1)},
\end{align*}
or
$$(2n + 1) I_n = 2n a^2 I_{n - 1},$$
as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2532996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Finding the trace of $T^4$ and $T^2$?
Q. Let $V$ denote the (complex) vector space of complex polynomials of degree at most $9$ and consider the linear operator $T:V \to V$ defined by $$T(a_0+a_1 x+a_2 x^2+\cdots+a_9 x^9)=a_0 +(a_2 x +a_1 x^2)+(a_4 x^3+ a_5 x^4 + a_3 x^5)+(a_7 x^6 + a_8 x^7 + a_9 x^8 + a_6 x^9).$$
(a) What is the trace of $T^4$?
(b) What is the trace of $T^2$?
My approach :
First off we form matrix of $T$ under the standard basis $\{1,x,x^2,...,x^9\}.$ We get,
$$
\left(\begin{matrix}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
\end{matrix}\right)
$$
Now $\text {tr}(T)=1$. We calculate $\det T$ by observing number of row exchanges that convert it to an identity matrix. Number of such exchanges is $6$. Thus $\det T=(-1)^6=1.$
We have relation of eigenvalues with trace and determinants as follows,
$\lambda_1+\lambda_2+...+\lambda_{10}=\text {tr} (T)=1$ and $\lambda_1 \lambda_2 \cdots \lambda_{10}=\det T=1.$
Also eigenvalues of $T^4$ and $T^2$ are ${\lambda_1}^4,{\lambda_2}^4,...,{\lambda_{10}}^4$ and ${\lambda_1}^2,{\lambda_2}^2,...,{\lambda_{10}}^2$ respectively.
In the $3 \times 3$ and $4 \times 4$ cases of this problem, I found $1$ as a common eigenvalue and other eigenvalues to be either $-1$ or $\pm i$ or $\frac {-1 \pm \sqrt 3i}2.$ So eigenvalues are coming out to be $n-$th roots of unity but I am not able to find a pattern among them.
My plan is using $\text {tr} (T^2)={\lambda_1}^2 + {\lambda_2}^2 +...+{\lambda_{10}}^2$ and $\text {tr}(T^4)={\lambda_1}^4+{\lambda_2}^4+...+{\lambda_{10}}^4.$
Knowing all these things is not helping me to find a way to tackle the problem. Is there some concept that I am missing to apply?
Moreover are there any shorter elegant ways to approach this problem?
| Hint: The matrix is diagonal by block
$$ T =
\begin{pmatrix}
A & 0 & 0 & 0 \\
0 & B & 0 & 0 \\
0 & 0 & C & 0 \\
0 & 0 & 0 & D \\
\end{pmatrix}
$$
with $$ A = I_{1,1}, B =
\begin{pmatrix}
0 & 1 \\
1 & 0 \\
\end{pmatrix}, C = \begin{pmatrix}
0 & 1 & 0 \\
0 & 0 & 1\\
1 & 0 & 0 \\
\end{pmatrix}, D = \begin{pmatrix}
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
1 & 0 & 0 & 0 \\
\end{pmatrix}$$
since $T$ is diagonal by block :
$$T^n = \begin{pmatrix}
A^n & 0 & 0 & 0 \\
0 & B^n & 0 & 0 \\
0 & 0 & C^n & 0 \\
0 & 0 & 0 & D^n \\
\end{pmatrix}$$
$$Tr(T^n)=Tr(A^n)+Tr(B^n)+Tr(C^n)+Tr(D^n)$$
$$Tr(T^2) = 1+2+0+0 = 3$$
$$Tr(T^4) = 1+2+0+4 = 7$$
You can think of this in term of application and subspaces : there are 4 subspaces of $V$ that are stable to $T$, if you define restriction of $T$ in these subspaces you have much simpler applications (that are only circle permutation moreover)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2533800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Integrate $\sin^4(4x)/\sin^2(x)$ I would like to compute
$$\int_{0}^{2\pi} \frac{\sin^4(4x)}{\sin^2(x)} \mathrm{d}x.$$
Wolfram|Alpha is able to compute an antiderivative explicitly so I do not think use of the residue theorem is needed, but I'm interested in any approach.
| We can compute the integral by Fourier series. Consider
\begin{align}
f(x) = \frac{\sin^2(4x)}{\sin x}
\end{align}
then the sine series expansion of $f$ is given by
\begin{align}
f(x) = \sin x + \sin 3x + \sin 5x + \sin 7 x.
\end{align}
In particular, we see
\begin{align}
\int^{2\pi}_0 f^2(x)\ dx = \int^{2\pi}_0 \sin^2 x+ \sin^2(3x) + \sin^2(5x) + \sin^2(7x)\ dx
\end{align}
since
\begin{align}
\int^{2\pi}_0 \sin(nx) \sin(mx) \ dx = 0 \ \ \text{ if } \ \ n\neq m.
\end{align}
Lastly, we have that
\begin{align}
\int^{2\pi}_0 \sin^2(n x)\ dx = \frac{1}{2}\int^{2\pi}_0 1-\cos(2nx)\ dx = \pi.
\end{align}
Thus, it follows
\begin{align}
\int^{2\pi}_0 \frac{\sin^4(4x)}{\sin^2 x}\ dx = 4\pi.
\end{align}
Remark: The hard work lies in finding the sine series expansion.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the result of the root expression, is my answer correct or not? Suppose $a < 0 < b$. Then what is the result of:
$\sqrt{(a-b)^2} + \sqrt[6]{ b^6 } = ?$
I have a solution but I can't be sure if I did a mistake, because I usually do! My solution:
Call $a = -c$ for some $0 < c $. Then,
$=\sqrt{(-c-b)^2} + \sqrt[6]{ b^6 }$
$=\sqrt{(- (c+b))^2} + \sqrt[6]{ b^6 }$
$=\sqrt{(c+b)^2} + \sqrt[6]{ b^6 }$
$=(c+b) + b$
$= -a +2b$
| That seems good to me. You should be careful that $\sqrt{(c+b)^2} = |c + b|$. But as $c$ and $b$ are both positive, $|c + b| = c+ b$.
You can actually be more direct:
$\sqrt{(a-b)^2} + \sqrt[6]{ b^6 } = |a-b| + |b|$.
As $b > 0$ we know $|b| = b$. As $a < 0$ and $b > 0$ then then $a- b < a < 0$. So $|a-b| = b -a$
So $\sqrt{(a-b)^2} + \sqrt[6]{ b^6 } = |a-b| + |b| = b -a + b = -a + 2b$.
But your reasoning was just fine. But it may be worth stating, you replaced $a$ with $-c$ to make all variables positive.
.....
And just to be thorough:
$|a - b| + |b| = $
1)$ a - b + b=a$; if $b \ge 0$ and $a-b \ge 0$ i.e. if $b \ge 0$ and $a \ge b$, i.e if $0 \le b \le a$.
2) $a - b - b = a - 2b$; if $b \le 0$ and $a-b \ge 0$ i.e. if $b \le 0$ and $a \ge b$. (whether $a$ is bigger or small or equal to zero doesn't matter so long as $a \ge b$.
3)$b-a + b = -a + 2b$ if $b\ge0$ and $a-b \le 0$ i.e. if $a \le b$ and $b \ge 0$ (whether $a$ is bigger or smaller or equal to zero doesn't matter; Note; your problem was a subset of this case.)
4) $b-a - b = -a$ if $b\le 0$ and $a-b \le 0$ i.e. if $a \le b \le 0$.
(Note: Those four cases are not mutually exclusive. If $b=0$ then 1) $|a-b| + |b| = a$ and 2) $|a-b| + |b| = a-2b = a$ are compatible, as are 3) $|a-b| + |b| = -a + 2b = -a$ and 4) $|a -b| + |b| = -a$.)
(If $a-b = 0$ i.e. $a= b$ then 1) $|a-b| + |b| = a$ and 3) $|a-b| +|b| = -a + 2b = -a + 2a = a$ are compatible, as are 2) $|a-b| + |b|=a-2b = a-2a = -a$ and 4) |a-b|+|b| = -a$.
(if $a-b =0$ and $b=0$ then all four are compatible $|a-b| + |b| = 0 =a=-a = a-2b = -a + 2b$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2539919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Find real solutions in $x$,$y$ for the system $\sqrt{x-y}+\sqrt{x+y}=a$ and $\sqrt{x^2+y^2}-\sqrt{x^2-y^2}=a^2.$
Find all real solutions in $x$ and $y$, given $a$, to the system:
$$\left\{
\begin{array}{l}
\sqrt{x-y}+\sqrt{x+y}=a \\
\sqrt{x^2+y^2}-\sqrt{x^2-y^2}=a^2 \\
\end{array}
\right.
$$
From a math olympiad. Solutions presented: $(x,y)=(0.625 a^2,0.612372 a^2)$ and $(x,y)=(0.625 a^2,-0.612372 a^2)$. I tried first to make the substitution $u=x+y$ and $v=x-y$, noticing that $x^2+y^2=0.5((x+y)^2+(x-y)^2)$ but could not go far using that route. Then I moved to squaring both equations, hoping to get a solution, but without success.
Hints and answers are appreciated. Sorry if this is a duplicate.
| Short solution:
WLOG, $a=2$ (as $x,y\propto a^2$).
By squaring the first equation and rearranging, you draw
$$\sqrt{x^2-y^2}=2-x.$$
And from the second,
$$\sqrt{x^2+y^2}=4+\sqrt{x^2-y^2}=6-x.$$
Now by summing the squares of the LHS,
$$2x^2=(2-x)^2+(6-x)^2$$
which gives$$x=\frac52.$$
Then from the first identity
$$y=\pm\sqrt{x^2-(2-x)^2}=\pm\sqrt6.$$
For general $a$,
$$\color{green}{x=\frac52\frac{a^2}4,\\y=\pm\sqrt6\frac{a^2}4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2542647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
If $p$ is prime, $p\ne3$ then $p^2+2$ is composite I'm trying to prove that if $p$ is prime,$p\ne3$ then $p^2+2$ is composite. Here's my attempt:
Every number $p$ can be put in the form $3k+r, 0\le r \lt 3$, with $k$ an integer. When $r=0$, the number is a multiple of 3, so that leaves us with the forms $3k+1$ and $3k+2$. The first one will be even when $k$ is odd, and the second one will be even when $k$ is even. So we will see what happens for each form in the case that $k$ is even (for the first form) and $k$ is odd (for the second form):
*
*$p=3k+1$, $k$ is even
Since $k$ is even, we can write it as $k=2q$ for some $q$. Then
$p^2+2=(3(2q)+1)^2+2=(6q+1)^2+2=6^2q^2+12q+1+2=3(12q^2+4q+1)$
So $p^2+2$ is composite.
*$p=3k+2$, $k$ is odd
Then, $k$ can be written as $k=2q+1$, for some $q$. Then
$p^2+2=(3(2q+1)+2)^2+2=(6q+5)^2+2=6^2q^2+60q+25+2=3(12q^2+10q+9)$
And again, $p^2+2$ is composite.
QED
Is that a correct proof? Is not the same that comes in the answer books.
| if we have $$p\equiv 1 \mod 3$$ then we get $$p^2+2\equiv 0\mod 3$$ and with
$$p\equiv 2\mod 3$$ we get $$p^2+2\equiv 4+2\equiv 0\mod 3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2543345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
$\tan(x) = 3$. Find the value of $\sin(x)$ I’m trying to figure out the value for $\sin(x)$ when $\tan(x) = 3$. The textbook's answer and my answer differ and I keep getting the same answer. Could someone please tell me what I'm doing wrong?
1.) $\tan(x) = 3$, then $\frac{\sin(x)}{\cos(x)} = 3$.
2.) Then $\cos(x) = \frac{1}{3}\sin(x)$
3.) $\sin^2 (x) + \left(\frac{1}{3}\sin(x)\right)^2 = 1$ //Pythagorean identity substitution.
4.) $\left(\frac{4}{3}\sin(x)\right)^2 = 1$ //Combining like terms
5.) $\frac{16}{9}\sin^2(x) = 1$ //Square the fraction so I can move it later.
6.) $\sin^2(x) = \frac{1}{\frac{16}{9}}$ //Divide both sides by $\frac{16}{9}$
7.) $\sin^2(x) = \frac{1}{1} * \frac{9}{16} = \frac{9}{16}$ //divide out the fractions
8.) $\sin(x) = \pm \frac{3}{4}$ //square root both sides.
So $\sin(x) = \pm \frac{3}{4}$ but the book says this is wrong. Any help is much appreciated.
| Another way to think of it is visualizing SOH-CAH-TOA. (I pronounce it "so-cuh-tow-uh".)
Use the relationships on a particular triangle to get the answer.
Tangent is opposite over adjacent (TOA), so choose your vertical side to be $3$ and your horizontal side to be $1$. This right triangle has the correct value for the tangent of the larger acute angle. (One vertex is on the origin, and the tall right triangle is in the first quadrant.)
This means the hypotenuse is $\sqrt{3^2+1^2} = \sqrt{10}$.
Now, for the sine, you take opposite over hypotenuse (SOH): $3/\sqrt{10}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2546705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do you factorize a polynom in $\mathbb{Z}_2$? How can you efficiently factorize a polynom (in $\mathbb{Z}_2$) ?
Example in this answer:
$$x^{16}-x=x (x + 1) (x^2 + x + 1) (x^4 + x + 1) (x^4 + x^3 + 1) (x^4 + x^3 + x^2 + x + 1)$$
How do you do this by hand?
| Partial answer:
We know that $x-1 =x+1$, so we have:
\begin{eqnarray}
x^{16}-x &=& x(x^{15}-1)\\
&=& x(x^5-1)\underbrace{(x^{10}+x^5+1)}_{p(x)}\\
&=& x(x-1)(x^4+x^3+x^2+x+1)p(x)\\
\end{eqnarray}
Now we have to factor somehow $p(x)$...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2547821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that $x^5 + 4x^4 + 2x^3 + x^2 + 4x + 54$ is irreducible over $\mathbb{Z} / 7\mathbb{Z}$ I wish to show that $$f(x) = x^5 + 4x^4 + 2x^3 + x^2 + 4x + 54$$ is irreducible over $\mathbb{Z} / 7\mathbb{Z}$. Simply testing every element, it is easy to see that there are no roots and thus no linear factors, but I'm not sure how to proceed from there.
| You can try case by base: 5 = 1+4 = 2+3 and $F = \mathbb{Z}/7\mathbb{Z}$ is a field so, there are three options: $f = x^5 + 4x^4 + 2x^3 +x^2 + 4x + 54$ is irreducible, it decompose int terms of degree 1 and 4 or terms of degree 2 and 3. As you checked, the second case is impossible because it has no root in $F$. Let's try the other case:
$$
f = (x^2 + b x + c)(x^3 + dx^2 + ex + f) = x^5 + (d+b)x^4 + (e+bd+c)x^3 + (f+be+cd)x^2 + (bf+ce)x + cf
$$
So
$$
b+d = 4\\
e+bd+c = 2\\
f+be+cd = 1\\
bf+ce = 4\\
cf=54
$$
Or
$$
\left[
\begin{array}{l}
1& 0 &0\\
b &1 &0\\
c &b &1\\
0 &c &b\\
0 &0 &c
\end{array}
\right]
\left[
\begin{array}{l}
d\\
e\\
f
\end{array}
\right]
=
\left[
\begin{array}{l}
4-b\\
2-c\\
1\\
4\\
54
\end{array}
\right]
$$
So, we need to verify if this linear system have a solution;
$$
\left[
\begin{array}{l}
1& 0 &0\\
b &1 &0\\
c &b &1\\
0 &c &b\\
0 &0 &c
\end{array}
\right]
\begin{array}{l}
4-b\\
2-c\\
1\\
4\\
54
\end{array}
\\ \simeq
\left[
\begin{array}{l}
1& 0 &0\\
0 &1 &0\\
0 &0 &1\\
0 &0 &0\\
0 &0 &0
\end{array}
\right]
\begin{array}{l}
4-b\\
2-c - b(4-b)\\
1 - c(4-b) - b(2-c - b(4-b))\\
4 - c(2-c - b(4-b)) - b(1 - c(4-b) - b(2-c - b(4-b)))\\
54 - c(1 - c(4-b) - b(2-c - b(4-b)))
\end{array}
$$
We need
$$
4 - c(2-c - b(4-b)) - b(1 - c(4-b) - b(2-c - b(4-b))) = 0\\
54 - c(1 - c(4-b) - b(2-c - b(4-b))) = 0
$$
And you can use a computer to check all 49 cases in (b,c) and see that this system has no solution in $F$.
Then, this case of decomposition is impossible too and $f$ is irreducible.
Note, this is a completely computacional way to do that. Essentially, I'm using that $F$ is a finite field and we can really check all it's elements.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Fair and unfair coins There are 4 fair coins and 1 unfair coin that has only heads. We choose a coin and flip it three times. The result is HHH. What is the probability that the fourth flip is H?
| We would expect the probability to be greater than $\frac{1}{2}$. We must solve for $P(\text{Heads})$. We have
$$P(\text{Heads})=P(\text{Heads}|\text{fair})\cdot P(\text{fair})+P(\text{Heads}|\text{unfair})\cdot P(\text{unfair})$$
However, we cannot just say $P(\text{fair})=\frac{4}{5}$ because we are given that the first $3$ tosses are heads.
$$P(\text{fair|}HHH)=\frac{P(\text{fair} \cap HHH)}{P(HHH)}=\frac{\frac{4}{5}\cdot\frac{1}{2}^3}{\frac{4}{5}\cdot\frac{1}{2}^3+\frac{1}{5}\cdot\left(1\right)^3}=\frac{1}{3}$$
So we have that the probability that the coin was fair when we rolled those $3$ heads was $\frac{1}{3}$ and the probability that it was unfair is $\frac{2}{3}$.
Thus we have
$$\begin{align*}
P(\text{Heads})
&=P(\text{Heads|fair})\cdot P(\text{fair})+P(\text{Heads|unfair})\cdot P(\text{unfair})\\\\
&= \frac{1}{2}\cdot\frac{1}{3}+1\cdot\frac{2}{3}\\\\
&=\frac{5}{6}
\end{align*}$$
Which agrees with Nicola's answer.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Prove that $\sum\limits_{r=0}^\infty\,\dfrac{(-1)^r\,n}{r!\,(r+1)!}\,\prod\limits_{k=1}^r\,\left(n^2-k^2\right)=(-1)^{n+1}$
If n is a positive integer, prove that:
$$\begin{align}n - {\frac{n(n^2-1^2)}{2!}}+{\frac{n(n^2-1^2)(n^2-2^2)}{2!3!}} + \ldots\phantom{aaaaaaaaaa} &&\\
+(-1)^r{\frac{n(n^2-1^2)(n^2-2^2)\cdots(n^2-r^2)}{r!(r+1)!}}+\ldots &=& (-1)^{n+1}\end{align}$$
I tried by simplifying it to $\binom{n}{1}\binom{n}{0}-\binom{n}{2}\binom{n+1}{1}+\binom{n}3\binom{n+2}{2}-\ldots$ , or $\sum(-1)^r\binom{n}{r+1}\binom{n+r}{r}$. But I don't know what to do from here.
| $$(-1)^r{\frac{n(n^2-1^2)(n^2-2^2)...(n^2-r^2)}{r!(r+1)!}}=\dfrac{n(n-1)(n-2)\cdots(n-r)}{(r+1)!}\cdot\dfrac{-(n+1)\cdot-(n+2)\cdots-(n+r)}{r!}$$
Think of the coefficient of $x$ in $$(1+x)^n\left(1+\dfrac1x\right)^{-(n+1)}=\dfrac{x^{n+1}}{1+x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2553579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to calculate this hard integral $\int_0^{\infty} \frac{\arctan(x)\sqrt{\sqrt{1+x^2}-1}}{x\sqrt{1+x^2}}\,dx$? How to prove that $\displaystyle \int_0^{\infty} \dfrac{\arctan(x)\sqrt{\sqrt{1+x^2}-1}}{x\sqrt{1+x^2}}\,dx= \frac{1}{4}\pi^2\sqrt{2}-\sqrt{2}\ln^2\left(1+\sqrt{2}\right)$ ?
It's very difficult and I have no idea.
| This can be reduced to a class of integral related to polylogarithm, and is related to Legendre-chi function $\chi_2$. The formula
$$\int_{0}^{\frac{\pi}{2}} \arctan(r \sin\theta) \, d\theta
= 2 \chi_{2} \left( \frac{\sqrt{1+r^{2}} - 1}{r} \right) $$
gives $$\tag{1}\int_{0}^{\frac{\pi}{2}} \arctan( \sin\theta) \ d\theta
= 2 \chi_{2} \left( \sqrt{2}-1 \right) = \frac{\pi^2}{8} - \frac{1}{2}\log^2(\sqrt{2}+1)$$
where we used a famous special value of $\chi_2$. For a "quick introduction" about $\chi_2$, see Sangchul Lee's here.
Return to your problem, denote your integral by $I$, note that
$$\int \frac{\sqrt{\sqrt{1+x^2}-1}}{x\sqrt{1+x^2}} dx = \sqrt 2 \arctan \left[ {\frac{{\sqrt {\sqrt {1 + {x^2}} - 1} }}{{\sqrt 2 }}} \right]$$
so integration by parts gives
$$I = \frac{\sqrt{2}\pi^2}{4}-\sqrt{2}\int_0^\infty {\arctan \left[ {\frac{{\sqrt {\sqrt {1 + {x^2}} - 1} }}{{\sqrt 2 }}} \right]} \frac{{dx}}{{1 + {x^2}}}$$
making $x\mapsto \tan x$ into the last integral gives
$$I = \frac{\sqrt{2}\pi^2}{4}-\sqrt{2}\underbrace{\int_0^{\pi/2} {\arctan \left( {\frac{{\sin \frac{x}{2}}}{{\sqrt {\cos } x}}} \right)dx}}_J $$
Denote $$F(r)=\int_0^{\pi/2} {\arctan \left( {\frac{{a\sin \frac{x}{2}}}{{\sqrt {\cos } x}}} \right)} dx$$
Then compute $F'(r)$ and do an indefinite integral gives
$$\begin{aligned} F'(r) &= \int_0^{\pi /2} {\frac{{\sqrt {\cos x} \sin (\frac{x}{2})}}{{\cos x + {a^2}\sin {{(\frac{x}{2})}^2}}}} dx \\
& = - \left[ {\frac{{2a\arctan \left[ {\frac{{a\cos (\frac{x}{2})}}{{\sqrt {1 - {a^2}} \sqrt {\cos x} }}} \right]}}{{(2 - {a^2})\sqrt {1 - {a^2}} }} + \frac{{2\sqrt 2 }}{{2 - {a^2}}}\ln \left[ {\sqrt 2 \cos (\frac{x}{2}) + \sqrt {\cos x} } \right]} \right]_0^{\pi /2} \\ &= - \frac{{\pi a}}{{(2 - {a^2})\sqrt {1 - {a^2}} }} + \frac{{2a\arctan \left[ {\frac{a}{{\sqrt {1 - {a^2}} }}} \right]}}{{(2 - {a^2})\sqrt {1 - {a^2}} }} + \frac{{2\sqrt 2 \ln (\sqrt 2 + 1)}}{{2 - {a^2}}}\end{aligned}$$
Then to get $J$, we integrate the whole expression with respect to $a$ from $0$ to $1$, only the second integral deserves attention, applying $a\mapsto \sin x$:
$$\begin{aligned}\int_0^1 {\frac{{\arctan \left[ {\frac{a}{{\sqrt {1 - {a^2}} }}} \right]}}{{(2 - {a^2})\sqrt {1 - {a^2}} }}da} &= \int_0^{\pi /2} {\frac{{x\cos x}}{{2 - {{\sin }^2}x}}dx} \\&=- \int_0^{\frac{\pi }{2}} {xd\left[ {\arctan (\cos x)} \right]} \\&= \int_0^{\frac{\pi }{2}} {\arctan (\cos x)dx} \end{aligned}$$
Then use $(1)$. Combining all above evaluation should give
$$J=\int_0^{\pi/2} {\arctan \left( {\frac{{\sin \frac{x}{2}}}{{\sqrt {\cos } x}}} \right)}dx=\ln^2(\sqrt{2}+1)$$ this proves your result.
Probably there is some smart way to reduce $J$ directly into $(1)$, I hope someone can point on how.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2557405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 4,
"answer_id": 0
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Determine tranformation matrix of linear transformation
A linear transformation between polynomial spaces $f: P_2(\mathbb{R}) \to P_2(\mathbb{R})$ is given by $$f(p(x))=3 \cdot p(1)-x^2 \cdot p(0)+(x-1) \cdot p'(1)$$ Determine the transformation matrix with respect to the monomial basis $(1,x,x^2)$
I tried to approach it this way:
The general form of these polynomials is $ax^2+bx+c$. We have $$P(1)=a+b+c \text{ and } P(0)=c \text{ and } P'(1)=2a+b$$
Therefore $f(p(x))=3(a+b+c)-cx^2+2ax+bx-2a-b$$
I therefore thought that the transformation matrix would be $$\begin{bmatrix} 1 & 2 & 3 \\ 2 & 1 & 0 \\ 0 & 0 & -1 \end{bmatrix}$$
What am I doing wrong?
| Well, if you have a basis $\beta =\{1, x, x^2\}$, and we need to determine the matrix of $f$ wrt $\beta$, then let's first write out $f$ in the way that makes it the most obvious how it's transforming it's coordinates:
$$
\begin{aligned}
f(ax^2+bx+c) = f((a, b, c)) & = 3(a + b + c) - cx^2 + (2a + b)x - (2a + b) \\
& = (-c, 2a+b, 3(a+b+c)-(2a+b)) \\
& = (-c, 2a+b, a+2b+3c)
\end{aligned}
$$
I'm using $(a,b,c)$ to denote $ax^2+bx+c$ to emphasize the fact that these are coordinates in $P_2(\mathbb R)$. So, if our matrix for $f$ is
$$
[f]_\beta = \begin{pmatrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{pmatrix}
$$
then notice that the first coordinate in $[f]_\beta(a,b,c)^\text{T}$ is given by the dot product of the first row with $(a, b, c)$, so $a_{11}a + a_{12}b + a_{13}c = -c$, implying $a_{13}=-1$, and $a_{11}=a_{12}=0$. We can continue this process for each row, since the $i$th coordinate of $[f]_\beta(a,b,c)^\text{T}$ is given by the dot product of the $i$th row with $(a, b, c)$. Hence
$$
a_{21}a + a_{22}b + a_{23}c = 2a + b \\
a_{31}a + a_{32} b + a_{33}c = a + 2b + 3c
$$
from which we can see that $(a_{21}, a_{22}, a_{23}) = (2, 1, 0)$, and $(a_{31}, a_{32}, a_{33}) = (1, 2, 3)$. Hence
$$
[f]_\beta = \begin{pmatrix}0 & 0 & -1 \\ 2 & 1 & 0 \\ 1 & 2 & 3\end{pmatrix}
$$
So you were on the right track, you just had your rows in the reverse order.
| {
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Prove that $\sum_{n=0}^N\binom{2N-n}N2^n(n+1)=(1+2N)\binom{2N}N$ I used WolframAlpha to calculate a sum but it didn't show me the way :( Anybody has a hint or a solution for proving this sum?
$$\sum_{n=0}^N\binom{2N-n}N2^n(n+1)=(1+2N)\binom{2N}N$$
| $$\begin{align}
\sum_{n=0}^N\binom {2N-n}N2^n(n+1)
&=\sum_{n=0}^N\binom {2N-n}N\sum_{j=0}^n \binom nj(n+1)
&&\scriptsize\text{using }\sum_{j=0}^n \binom nj=2^n\\
&=\sum_{n=0}^N\binom {2N-n}N\sum_{j=0}^n \binom {n+1}{j+1}(j+1)\\
&=\sum_{n=0}^N\binom {2N-n}N\sum_{j=1}^{n+1} \binom {n+1}{j}j\\
&=\sum_{j=1}^{N+1}j\sum_{n=0}^{j-1}\binom {2N-n}N\binom {n+1}j
&&\scriptsize (0\le n<j\le N+1)\\
&=\sum_{j=1}^{N+1}j\binom {2N+2}{N+1+j}
&&\scriptsize \text{using}\sum_n\binom {a-n}b\binom {c+n}d=\binom{a+c+1}{b+d+1}\\
&=\frac 12(N+1)\binom {2N+2}{N+1}
&&\scriptsize\text{using (*) }\\
&=\frac 12(N+1)\cdot \frac {2N+2}{N+1}\cdot \binom {2N+1}{N}\\
&=(N+1)\cdot \binom {2N+1}{N+1}\\
&=(N+1)\cdot\frac {2N+1}{N+1}\cdot \binom {2N}N\\
&=\color{red}{(1+2N)\binom {2N}N}\qquad \blacksquare
\end{align}$$
*See derivation below. Putting $n=N+1$ gives the result used above.
$$\small\begin{align}
\sum_{r=1}^n\binom{2n}{n+r}r
&=\sum_{j=n+1}^{2n}\binom {2n}j(j-n)\\
&=\sum_{j=n+1}^{2n}\binom {2n}jj-n\sum_{j=n+1}^{2n}\binom {2n}j\\
&=n2^{2n-1}-n\cdot \frac 12\left(\left(\sum_{j=0}^{2n}\binom {2n}j\right)-\binom {2n}n\right)\\
&=n2^{2n-1}-\frac 12n\left(2^{2n}-\binom {2n}n\right)&&
\hspace{2.5cm}\\
&=\frac 12n\binom {2n}n\end{align}$$
Note that
$$\begin{align}
\frac 12(n+1)\binom {2n}{n+1}
&=\frac 12 (n+1)\frac {(2n)!}{(n+1)!(n-1)!}\\
&=\frac 12 \cdot \frac {(2n)!}{n!(n-1)!}\cdot\color{grey}{ \frac nn}
\qquad\hspace{3cm}\\
&=\frac 12 n\cdot \frac {(2n)!}{n!n!}\\
&=\frac 12 n\binom {2n}n
\end{align}$$
Note also that
$$\small\begin{align}
\sum_{n}\binom {a-n}b\binom {c+n}d
&=\sum_n\binom {a-n}{a-b-n}\binom {c+n}{c+n-d}\\
&=\sum_n(-1)^{a-b-n}\binom {-b-1}{a-b-n}(-1)^{c+n-d}\binom {-d-1}{c+n-d}
&&\text{(upper negation)}\\
&=(-1)^{a-b+c-d}\sum_n\binom {-b-1}{a-b-n}\binom {-d-1}{c-d+n}\\
&=(-1)^{a-b+c-d}\binom {-b-d-2}{a-b+c-d}
&&\text{(Vandermonde)}\\
&=(-1)^{a-b+c-d}(-1)^{a-b+c-d}\binom {a+c+2-1}{a-b+c-d}
&&\text{(upper negation)}\\
&=\binom {a+c+1}{a-b+c-d}\\
&=\binom {a+c+1}{b+d+1}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2558601",
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"source": "stackexchange",
"question_score": "5",
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} |
Maximum and minimum value of a function.
Let us consider a function $f(x,y)=4x^2-xy+4y^2+x^3y+xy^3-4$. Then find the maximum and minimum value of $f$.
My attempt. $f_x=0$ implies $8x-y+3x^2y+y^3=0$ and $f_y=0$ implies $-x+8y+x^3+3xy²=0$ and $f_{xy}=3x^2+3y^2-1$. Now $f_x+f_y=0$ implies $(x+y)((x+y)^2+7)=0$ implies $x=-y$ as $x$ and $y$ are reals. Now putting it in $f_x=0$ get the three values of $x$ as $0$, $(-3+3\sqrt{5})/2$ and $(-3-3\sqrt{5})/2$. And then $f_{xy}(0,0)<0$ implies $f$ has maximum at $(0,0)$. But at other two points $f_{xy}$ gives the positive value. So how can I proceed to solve the problem.
Please help me to solve it.
| $f_x+f_y=\left(x^3+3 x y^2-x+8 y\right)+\left(3 x^2 y+8 x+y^3-y\right)=\\=x^3+3 x^2 y+3 x y^2+7 x+y^3+7 y=(x+y) \left(x^2+2 x y+y^2+7\right)=0$
$y=-x$. Substitute in $f_x=0$ and get
$9 x-4 x^3=0\to x_1=0;\;x_2=\dfrac{3}{2};\;x_3=-\dfrac{3}{2}$
and $y_1=0;\;y_2=-\dfrac{3}{2};\;y_3=\dfrac{3}{2}$
Now the Hessian, for the second partial derivative test
$H(x,y)=\left(
\begin{array}{rr}
6 x y+8 & 3 x^2+3 y^2-1 \\
3 x^2+3 y^2-1 & 6 x y+8 \\
\end{array}
\right)$
and $\det H(x,y)=-9 x^4+18 x^2 y^2+6 x^2+96 x y-9 y^4+6 y^2+63$
so we have
$\det H(x_1,y_1)=63>0$ and $f_{xx}=8>0$ thus $(0,0)$ is a local minimum
$\det H(x_2,y_2)=\det H(x_3,y_3)=-126<0$ so $\left(\frac{3}{2}, -\frac{3}{2}\right)$ and $\left(-\frac{3}{2}, \frac{3}{2}\right)$ are saddle points
Hope this helps
$$...$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2558988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Convergence of $\sum_{n=0}^{+ \infty} \frac{3}{2n² -1}$
Does $$\sum_{n=0}^{+ \infty} \frac{3}{2n² -1}$$
converge?
I did the following:
$$\frac{3}{2n²-1} = \frac{3}{2n²}\left(\frac{1}{1-\frac{1}{2n²}}\right)$$
and because $$\frac{1}{1-\frac{1}{2n²}} \to 1$$
the sequence is bounded, and hence, there exists $M \in \mathbb{R}^{+}$
such that:
$$\left|\frac{1}{1-\frac{1}{2n²}}\right| = \frac{1}{1-\frac{1}{2n²}} < M$$
Hence:
$$\frac{3}{2n²-1} < \frac{3M}{2n²}$$
and because $\sum n^{-2}$ converges, by the comparison test it follows that the given sequence converges.
Is this correct? Is there an easier approach?
Thanks in advance.
| Your series is absolutely convergent by comparison with $\sum_{n\geq 1}\frac{3}{n^2}=\frac{\pi^2}{2}$, for instance.
You may also compute it in a explicit way. If we start with
$$ \frac{\sin(\pi z)}{\pi z}=\prod_{n\geq 1}\left(1-\frac{z^2}{n^2}\right) $$
and apply $\frac{d}{dz}\log(\cdot)$ to both sides, we get
$$-\frac{1}{z}+\pi\cot(\pi z) =\sum_{n\geq 1}\frac{2z}{z^2-n^2} $$
and by evaluating both sides at $z=\frac{1}{\sqrt{2}}$
$$ 1-\frac{\pi}{\sqrt{2}}\,\cot\left(\frac{\pi}{\sqrt{2}}\right)=\sum_{n\geq 1}\frac{1}{n^2-\frac{1}{2}} $$
so:
$$\boxed{ \sum_{n\geq 1}\frac{3}{2n^2-1} = \color{red}{\frac{3}{2}-\frac{3\pi}{2\sqrt{2}}\,\cot\left(\frac{\pi}{\sqrt{2}}\right)}\approx 4.036518.} $$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Mathematical induction proof that $f(n)=\frac{1}{2}+\frac{3}{2}(-1)^n$
The function $f(n)$ for $n=0,1...$ has the recursive definition $$f(n)= \begin{cases} 2 & \text {for n=0} \\ -f(n-1)+1 & \text{for n=1,2...} \end{cases}$$
Prove by induction that the following equation holds: $$f(n)=\frac{1}{2}+\frac{3}{2}(-1)^n$$
So, I begin by checking that the basic step holds
*
*$f(0)=\frac{1}{2}+\frac{3}{2}(-1)^0=2$ OK
*Assume that the equation holds for a given $n$
*Show that n+1 holds: $f(n+1)=\frac{1}{2}+\frac{3}{2}(-1)^{n+1} \Rightarrow f(n+1)=\frac{1}{2}+\frac{3}{2}(-1)^{n} \cdot (-1) = -f(n)-\frac{1}{2}$
I get kind of stuck here. Any advice on how I should approach this?
| Your last step is close, but not quite
$$ f(n+1) = \frac{1}{2} - \frac{3}{2}(-1)^n = 1 - \left(\frac{1}{2} + \frac{3}{2}(-1)^n\right) = -f(n) + 1 $$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Line integral over a non-central ellipse. I have to find the integral $$ \int_C ydx + x^2dy$$ over the curve $C$ given as a intersection of plane $z=0$ and surface $\frac{x^2}{a^2} + \frac{y^2}{b^2}=\frac{x}{a}+\frac{y}{b} $, curve $C$ is positively oriented $(a\geq b>0)$.
This is what i have this far:
given plane is $xy$ plane, it's intersection with this surface (whatever it is in the three-dimensional space) should be some sort of ellipse (non-origin ellipse) obviously, and the limits of integration should be from $0$ to $2\pi$. Now, after little bit of algebraic manipulation of the surface equation i got the following equation: $$\frac{(x-\frac{a}{2})^2}{\frac{a^2(a^2+b^2)}{4}} + \frac{(x-\frac{b}{2})^2}{\frac{b^2(b^2+a^2)}{4}}=1$$
Which is indeed an ellipse eqation, now , i suppose i should introduce polar coordinates here, in order to get the parametric equations for this curve, for this ellipse they should look something like this:
$x=\frac{a}{2}+\frac{a^2(a^2+b^2)}{4}\cos t \\y=\frac{b}{2}+\frac{b^2(a^2+b^2)}{4}\sin t \\ dx=-\frac{a^2(a^2+b^2)}{4}\sin t \\dy=\frac{b^2(a^2+b^2)}{4}\cos t $
Now, all i should do is to insert this into the given integral, but, i am not quite sure is this legitimate approach. Any suggestions or comments if this is incorrect is appreciated or if it is correct, any advice on how to do this more easily is appreciated too.
| By Green's theorem,
$$I:=\int_C ydx + x^2dy=\iint_E (2x-1)dxdy=|E|(2\bar{x}-1)$$
where $E$ is the domain bounded by $C$ in the $xy$-plane, $|E|$ is its area and $\bar{x}$ is the $x$-coordinate of the centroid of $E$.
Since $C$ is the ellipse given by (check your algebraic manipulations!!)
$$\frac{(x-\frac{a}{2})^2}{\frac{a^2}{2}} + \frac{(y-\frac{b}{2})^2}{\frac{b^2}{2}}=1$$
then
$$|E|=\pi\frac{a}{\sqrt{2}}\cdot\frac{b}{\sqrt{2}}=
\frac{\pi ab}{2}\quad\text{and}\quad\bar{x}=\frac{a}{2}.$$
Therefore
$$I=\frac{\pi ab(a-1)}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2566607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the spectral decomposition of $A$ $$
A= \begin{pmatrix} -3 & 4\\ 4 & 3
\end{pmatrix}
$$
So i am assuming that i must find the evalues and evectors of this matrix first, and that is exactly what i did.
The evalues are $5$ and $-5$, and the evectors are $(2,1)^T$ and $(1,-2)^T$
Now the spectral decomposition of $A$ is equal to $(Q^{-1})^\ast$ (diagonal matrix with corresponding eigenvalues) * Q
$Q$ is given by [evector1/||evector1|| , evector2/||evector2||]
and for Q i got the matrix
$$
Q= \begin{pmatrix} 2/\sqrt{5} &1/\sqrt{5} \\ 1/\sqrt{5} & -2/\sqrt{5}
\end{pmatrix}
$$
the inverse of Q is the matrix...
$$
\begin{pmatrix} 2 \sqrt{5}/5 & \sqrt{5}/5 \\ \sqrt{5}/5 & -2 \sqrt{5}/5
\end{pmatrix}
$$
and the diagonal matrix with corresponding evalues is
$$
A= \begin{pmatrix} 5 & 0\\ 0 & -5
\end{pmatrix}
$$
so now i found the spectral decomposition of $A$, but i really need someone to check my work.
Did i take the proper steps to get the right answer, did i make a mistake somewhere?
| I think of the spectral decomposition as writing $A$ as the sum of two matrices, each having rank 1. Let $A$ be given. Then compute the eigenvalues and eigenvectors of $A$. Then
$$
A = \lambda_1P_1 + \lambda_2P_2
$$
where $P_i$ is an orthogonal projection onto the space spanned by the $i-th$ eigenvector $v_i$.
In your case, I get $v_1=[1,2]^T$ and $v_2=[-2, 1]$ from Matlab. Then we have:
$$
\left[ \begin{array}{cc}
-3 & 4 \\
4 & 3\\
\end{array} \right] =
5\left[ \begin{array}{cc}
1/5 & 2/5 \\
2/5 & 4/5\\
\end{array} \right] -
5\left[ \begin{array}{cc}
4/5 & -2/5 \\
-2/5 & 1/5\\
\end{array} \right]
$$
Each $P_i$ is calculated from $v_iv_i^T$.
| {
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In how many ways can $3$ red, $3$ blue, and $3$ green balls be arranged so that no two balls of the same colour are consecutive (up to symmetry)? Make sequence of $9$ balls from 3 red, 3 blue, 3 green, in such a way that no two balls of the same colour are next to each other. In how many different ways can you do this? (symmetric arrangements must be counted only once).
To start with, all possible arrangements are $9! = 362880$.
| We have nine positions to fill with three blue, three green, and three red balls. We can fill three of the nine positions with blue balls in $\binom{9}{3}$ ways, three of the remaining positions with green balls in $\binom{6}{3}$ ways, and the remaining three positions with red balls in $\binom{3}{3}$ ways. Hence, the number of distinguishable arrangements of three blue, three green, and three red balls is
$$\binom{9}{3}\binom{6}{3}\binom{3}{3} = \frac{9!}{3!3!3!}$$
The factors of $3!$ in the denominator represent the number of ways balls of the same color can be permuted among themselves within a given arrangement since permuting balls of the same color among themselves does not produce an arrangement that is distinguishable from the given arrangement.
From these, we must exclude those arrangements in which there is at least one pair of adjacent balls of the same color.
A pair of adjacent balls of the same color: There are three ways to pick the color. We have eight objects to arrange, the block of two adjacent balls, the other ball of that color, and the other six balls. We have eight positions to fill.
Say the block consists of blue balls. Then we can fill three of those eight positions with green balls in $\binom{8}{3}$ ways, three of the remaining positions with red balls in $\binom{5}{3}$ ways, place the block in one of the two remaining positions in $2$ ways, and place the other blue ball in the final open position in one way. Hence, there are
$$\binom{8}{3}\binom{5}{3}2! = \frac{8!}{3!3!}$$
such arrangements. Since there are three ways of selecting the color, there are
$$\binom{3}{1}\frac{8!}{3!3!}$$
arrangements in which a pair of adjacent balls are of the same color.
Two pairs of adjacent balls of the same color: There are two cases.
Both pairs of adjacent balls are of the same color: This means that all three balls of that color are adjacent. Thus, we have seven objects to arrange, the block of three balls of the same color and the other six objects. Since there are three ways to select the color, the number of arrangements in which there are two pairs of adjacent balls of the same color is
$$\binom{3}{1}\frac{7!}{3!3!}$$
Two colors in which there is a pair of adjacent balls of that color: There are $\binom{3}{2}$ ways to select the colors of the pairs. We have seven objects to arrange, the two blocks, the two single balls of those colors, and the three balls of the other color. Thus, there are
$$\binom{3}{2}\frac{7!}{3!}$$
arrangements in which there are two colors in which there is a pair of adjacent balls of that color.
Three pairs of adjacent balls of the same color: There are again two cases.
Two pairs of adjacent balls of the same color and one pair of adjacent balls of a different color: There are three ways to pick the color in which there are two pairs of adjacent balls and two ways to pick the other color in which there is one pair of adjacent balls of that color. We have six objects to arrange, the block of three balls, the pair, the other ball of that color, and the three balls of the remaining color. Hence, there are
$$\binom{3}{1}\binom{2}{1}\frac{6!}{3!}$$
arrangements of this type.
Three colors in which there is a pair of adjacent balls of that color: We have six objects to arrange, the three blocks and the three individual balls. Since the objects are all distinct, there are
$$6!$$
arrangements of this type.
Four pairs of adjacent balls of the same color: We again have two cases.
Two colors in which there are two pairs of adjacent balls of that color: There are $\binom{3}{2}$ ways to select the two colors. We have five objects to arrange, the two blocks of three adjacent balls of the same color and the three balls of the third color. Hence, there are
$$\binom{3}{2}\frac{5!}{3!}$$
arrangements of this type.
One color in which there are two pairs of adjacent balls of that color and two other colors in which there is one pair of adjacent balls of that color: There are three ways to select the color with two pairs of adjacent balls of that color. We have five objects to arrange, the block of three balls, the two blocks of two balls, and the other two balls. Since the five objects are distinct, there are
$$\binom{3}{1}5!$$
arrangements of this type.
Five pairs of adjacent balls of the same color: There must be two colors in which there are two pairs of adjacent balls of that color and there must also be a pair of adjacent balls of the third color. There are $\binom{3}{2}$ ways of selecting the two colors in which there are two pairs of adjacent balls of that color. We have four objects to arrange, the two blocks of three balls, the block of two balls, and the other ball of that color. Since the objects are distinct, there are
$$\binom{3}{2}4!$$
arrangements of this type.
Six pairs of adjacent balls of the same color: There must be two pairs of adjacent balls of the same color in each of the three colors. Hence, we have three objects to arrange, a block of three blue balls, a block of three green balls, and a block of three red balls. Since these objects are distinct, there are
$$3!$$
arrangements of this type.
By the Inclusion-Exclusion Principle, the number of distinguishable arrangements of three blue, three green, and three red balls in which no two balls of the same color are adjacent is
$$\frac{9!}{3!3!3!} - \binom{3}{1}\frac{8!}{3!3!} + \binom{3}{1}\frac{7!}{3!3!} + \binom{3}{2}\frac{7!}{3!} - \binom{3}{1}\binom{2}{1}\frac{6!}{3!} - 6! + \binom{3}{2}\frac{5!}{3!} + \binom{3}{1}5! - \binom{3}{2}4! + 3!$$
This brings us to the question of symmetry. Notice that none of these $174$ arrangements can be a palindrome since for the two colors that do not occupy the middle position, there must be an odd number of balls of that color on one side of the middle ball and an even number of balls of that color on the other side of the middle ball. If we equate two arrangements that can be obtained through reflection, we are left with $87$ distinguishable arrangements of the balls.
| {
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derivative of $\frac 2x \sin(x^3)$ by definition The function: $$\frac 2x \sin(x^3)$$ when $x\neq0$, and $0$ while $x=0$.
I need to find if the function is derivation at $x=1$.
First step was to check if the function is continuous. there is just 1 side of limit to check (since its the same function around $x=1$, so I compared the limit of the function with $f(1)$:
$$\lim_{x\to 1}\space \frac{2}{x}\sin(x^3) = 2\sin(1) = f(1)$$
first question: was this step necessary?
next:
$$f'(1) =\lim_{h\to 0}\space \frac{ \frac2{h+1}\sin((h+1)^3)-2\sin(1)}h$$
$$=\lim_{h\to 0}\space \frac{ \frac{2\sin((h+1)^3)}{h+1}-2\sin(1)}h$$
and I know about $\lim_{x\to 0}\frac {\sin(x)}x = 1$, so:
$$=\lim_{h\to 0}\space \frac{ \frac{2\sin((h+1)^3) \cdot (h+1)^2}{(h+1)^3}-2\sin(1)}h$$
and then:
$$=\lim_{h\to 0}\space \frac{ 2(h+1)^2-2\sin(1)}h$$
and now I don't know how to get rid of the $h$ denominator
| When you have
$$\frac{\frac{2 \sin \left((h+1)^3\right)}{h+1}-2 \sin (1)}{h}$$
you must rewrite for all $h\neq0$ to
$$\frac{\sin \left(\frac{1}{2} \left((h+1)^3-1\right)\right)}{\frac{1}{2} \left((h+1)^3-1\right)}\frac{2 \left(h^2+3 h+3\right) \cos \left(\frac{1}{2} \left((h+1)^3-1\right)+1\right)}{h+1}-\frac{2 \sin (1)}{h+1}$$
and then take the limit by substituting $h =0$ except for the first fraction which has known limit equal to $1$.
When rewriting you simply has to follow the proof for the derivative of a product, a composition, and sine.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2570328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Why product of primes $+1$ is not divisible by any of the product primes? Given $n$ prime numbers, $p_1, p_2, p_3,\ldots,p_n$, then $p_1p_2p_3\cdots p_n+1$ is not divisible by any of the primes $p_i, i=1,2,3,\ldots,n.$ I dont understand why. Can somebody give me a hint or an Explanation ? Thanks.
| First approach:
Let $P= 2\times3\times5\times7\times11\times13.$ Then:
The next number after $P$ that is divisible by $2$ is $P+2.$
The next number after $P$ that is divisible by $3$ is $P+3.$
The next number after $P$ that is divisible by $5$ is $P+5.$
The next number after $P$ that is divisible by $7$ is $P+7.$
The next number after $P$ that is divisible by $11$ is $P+11.$
The next number after $P$ that is divisible by $13$ is $P+13.$
So $P+1$ is not divisible by any of those.
Second approach:
If $(2\times3\times5\times7\times11\times13) + 1$ is divided by $2$ then the quotient is $\underbrace{3\times5\times7\times11\times 13}_{\large\text{excluding 2}}$ and the remainder is $1.$
If $(2\times3\times5\times7\times11\times13) + 1$ is divided by $3$ then the quotient is $\underbrace{2\times5\times7\times11\times 13}_{\large\text{excluding 3}}$ and the remainder is $1.$
If $(2\times3\times5\times7\times11\times13) + 1$ is divided by $5$ then the quotient is $\underbrace{2\times3\times7\times11\times 13}_{\large\text{excluding 5}}$ and the remainder is $1.$
If $(2\times3\times5\times7\times11\times13) + 1$ is divided by $7$ then the quotient is $\underbrace{2\times3\times5\times11\times 13}_{\large\text{excluding 7}}$ and the remainder is $1.$
If $(2\times3\times5\times7\times11\times13) + 1$ is divided by $11$ then the quotient is $\underbrace{2\times3\times5\times7\times 13}_{\large\text{excluding 11}}$ and the remainder is $1.$
If $(2\times3\times5\times7\times11\times13) + 1$ is divided by $13$ then the quotient is $2\underbrace{\times3\times5\times7\times11}_{\large\text{excluding 13}}$ and the remainder is $1.$
(Appendix: $(2\times3\times5\times7\times11\times13) + 1 = 59\times509.$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2573458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 1
} |
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