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Homework: Sum of the cubed roots of polynomial Given $7X^4-14X^3-7X+2 = f\in R[X]$, find the sum of the cubed roots.
Let $x_1, x_2, x_3, x_4\in R$ be the roots. Then the polynomial $X^4-2X^3-X+ 2/7$ would have the same roots. If we write the polynomial as $X^4 + a_1X^3 + a_2X^2 +a_3X + a_4$ then per Viete's theorem:
$a_k = (-1)^k\sigma _k(x_1,x_2,x_3,x_4), k\in \{1,2,3,4\}$, where $\sigma _k$ is the $k$-th elementary symmetrical polynomial. Therefore:
$x_1+x_2+x_3+x_4 = 2$
$x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4 = 0\ (*)$
$x_1x_2x_3 +x_1x_2x_4+x_1x_3x_4+x_2x_3x_4 = 1$
$x_1x_2x_3x_4 = 2/7$
Now how to determine the sum of the cubed roots?
$2^3 = 8= (x_1+x_2+x_3+x_4)(x_1+x_2+x_3+x_4)^2 = (x_1+x_2+x_3+x_4)(x_1^2+x_2^2+x_3^2+x_4^2 + 2(*))$
Here's where things go out of hand:
$(x_1+x_2+x_3+x_4)(x_1^2+x_2^2+x_3^2+x_4^2) = (x_1^3 + x_2^3 + x_3^3+x_4^3) + x_1^2(x_2+x_3+x_4)+x_2^2(x_1+x_3+x_4)+x_3^2(x_1+x_2+x_4)+x_4^2(x_1+x_2+x_3) = 8$
What should I do here?
| Rewrite the equation $7X^4-14X^3-7X+2 = 0$ this way: $7X^3-14X^2-7+\frac{2}{X} = 0$. Replacing $X$ by $X_1$ then $X_2$ etc. and getting the sum will have:
$$
7\sum{X_i^3} -14\sum{X_i^2}-7\sum{X_i}+2\sum{\frac{1}{X_i}}=0 \tag1
$$
From $(1)$ should be easy to get the cube root's sum.
Note that $\sum{\frac{1}{X_i}}$ can be easily calculated using Viete.
| {
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"url": "https://math.stackexchange.com/questions/1533363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Evaluate $\lim_{x \to 0} \left(\frac{ \sin x }{x} \right)^{\frac{1}{x^2}}$ $$\lim_{x \to 0} \bigg(\frac{\sin x}{x} \bigg)^{\frac{1}{x^2}}$$
The task should be solved by using Maclaurin series so I did some kind of asymptotic simplification
$$\lim_{x \to 0} \bigg(\frac{\sin x}{x} \bigg)^{\frac{1}{x^2}} \approx \lim_{x \to 0} \bigg(\frac{x - \frac{x^3}{6}}{x} \bigg)^{\frac{1}{x^2}} \approx \lim_{x \to 0} \bigg(1 - \frac{x^2}{6} \bigg)^{\frac{1}{x^2}}$$ How can we say answer that is $e$ in the power of $-\frac{1}{6}$. I want some proving of that fact.
Thank you.
| If you saw something variable shows on the exponent, first take logarithm usually helps. Write
\begin{align}
& \left(\frac{\sin x}{x}\right)^{1/x^2} \\
= & \exp\left[\frac{1}{x^2}\log\left(\frac{\sin x}{x}\right)\right] \\
= & \exp\left[\frac{1}{x^2}\log\left(\frac{x - \frac{x^3}{6} + o(x^3)}{x}\right)\right] \\
= & \exp\left[\frac{1}{x^2}\log\left(1 - \frac{1}{6}x^2 + o(x^2)\right)\right] \\
= & \exp\left[\frac{1}{x^2}\left(- \frac{1}{6}x^2 + o(x^2)\right)\right] \\
= & \exp \left(-\frac{1}{6} + o(1)\right) \\
\to & e^{-1/6}
\end{align}
as $x \to 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1535827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Limit with trigonometric function
Find $$\lim_{x \to \pi/4}\frac{1-\tan(x)}{\cos(2x)}$$
without L'Hôpital.
$$\lim_{x \to \pi/4}\frac{1-\tan(x)}{\cos(2x)}=\lim_{x \to \pi/4}\frac{\cos^2(x)+\sin^2(x)-\frac{\sin(x)}{\cos(x)}}{\cos^2(x)-\sin^2(x)}$$
How can I continue?
| Note that
$$
1-\tan x = \frac{\cos x - \sin x}{\cos x}=\frac{(\cos x - \sin x)(\cos x+ \sin x)}{\cos x(\cos x+ \sin x)}=\frac{\cos^2 x - \sin^2 x}{\cos x (\cos x+ \sin x)}=
\frac{\cos 2x}{\cos x (\cos x+ \sin x)}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1543143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Possible values of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ given $x+y+z=1$ Given $x,y,z$ are real numbers and $x+y+z=1$ and $x$ is not equal to $z$, if $ {1\over x} + {1\over y} + {1\over z} = m $, which of the following values of $m$ are possible?
(A) 1
(B) 2
(C) 3
(D) All of these.
I don't know where to start. I thought of taking $x=y$ but there doesn't seem to be any pattern to obtain the answer.
| (1) Consider the function $f(x)=x^3-x^2-2x+2=(x-1)(x^2-2)$ we have three real roots and $x+y+z=-{1\over -1}=1,{1\over x}+{1\over y}+{1\over z}=-{2\over-2}=1$.
(2) Consider the function $f(x)=x^3-x^2-40x+20$. Set the derivative equal $0$ we get $x=4$ and $x=-{10\over3}$ are two local extrema. When $x=4, f(x)<0$, when $x=-{10\over 3}, f(x)>0$ so there are three real roots. We have $x+y+z=1$ and ${1\over x}+{1\over y}+{1\over z}=-{-40\over20}=2$.
(3) Consider the function $f(x)=x^3-x^2-40x+{40\over3}$. We still have the same local extrema as (2) and when $x=4, f(x)<0$ and when $x=-{10\over 3}, f(x)>0$ so there are three real roots. We have $x+y+z=1$ and ${1\over x}+{1\over y}+{1\over z}=-{-40\over{40\over3}}=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1544109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Square of a sum equals sum of cubes Consider a sequence $(a_n)$ of positive numbers such that
$$(a_1+\cdots+a_n)^2=a_1^3+\cdots +a_n^3,\quad n\ge 1.$$
Prove that $a_n=n$ for all $n\ge 1$.
| The question should specify that the sequence is increasing since $\{1,2,4,3,5\}$ is a counterexample.
Let's prove this by induction. The base case of $a_1 = 1$ holds since $a_1^2 = a_1^3$. Now assume that $(a_1,\ldots,a_k) = (1,\ldots,k)$ for some $k$. Then we have that $(1+2+\cdots+k+a_{k+1})^2 = \Bigg(\dfrac{(1+k)k}{2}+a_{k+1}\Bigg)^2$. Also, we have that $1^3+2^3+\cdots+a_{k+1}^3 = \dfrac{(1+k)^2k^2}{4}+a_{k+1}^3$. Then, setting them equal we obtain $\Bigg(\dfrac{(1+k)k}{2}+a_{k+1}\Bigg)^2=\dfrac{(1+k)^2k^2}{4}+a_{k+1}^3 \longrightarrow a_{k+1}(k^2+k)+a_{k+1}^2-a_{k+1}^3 = 0 \longrightarrow a_{k+1}(1+k-a_{k+1})(k+a_{k+1}) = 0$.
Since the sequence is positive, out of the solutions $a_{k+1} = 0, -k,$ and $k+1$, only $k+1$ works thus finishing the induction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1545745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Converting Polar Equation to Cartesian Equation problem So I have
1. $$\frac{r}{3\tan \theta} = \sin \theta$$
2. $$r=3\cos \theta$$
What would be the Cartesian equation???
| the first
$$\frac{r}{3\tan \theta} = \sin \theta$$
$$\frac{r\cos \theta}{3\sin \theta} = \sin \theta$$
$$\frac{x}{3\sin \theta} =\sin \theta$$
$$x =3\sin^2 \theta$$
$$r^2x =3r^2\sin^2 \theta$$
$$x(x^2+y^2)=3y^2$$
The second
$$r=3\cos \theta$$
$$r^2=3r\cos \theta$$
$$r^2=3x$$
$$x^2+y^2=3x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1547097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Proving that $\binom{n}{0}+\binom{n-1}{1}+\binom{n-2}{2}+\cdots =F_{n+1}$ where $F_{n+1}$ is the $n+1$ th Fibonacci number I have to proove this this identity which connects Fibonacci sequence and Pascal's triangle:
$$\begin{pmatrix}n\\0\end{pmatrix}+\begin{pmatrix}n-1\\1\end{pmatrix}+\dotsm+\begin{pmatrix}n-\lfloor\frac{n}{2}\rfloor\\\lfloor\frac{n}{2}\rfloor\end{pmatrix} = F_{(n+1)}$$
Example: ${6\choose0} + {5\choose1} + {4\choose2} + {3\choose3} = 13 = F_{7}$
| Using the recursion for Pascal's Triangle, $\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}$, we get
$$
\begin{align}
a_n
&=\binom{n}{0}+\binom{n-1}{1}+\binom{n-2}{2}+\dots\\
&=\binom{n-1}{0}+\binom{n-2}{1}+\binom{n-3}{2}+\dots\\
&+\binom{n-1}{-1}+\binom{n-2}{0}+\binom{n-3}{1}+\dots\\[4pt]
&=a_{n-1}+a_{n-2}
\end{align}
$$
Noting that $a_0=1$ and $a_1=1$, we get $a_n=F_{n+1}$.
If there is concern over the term $\binom{n-1}{-1}$, consider the edge of Pascal's Triangle. Looking at it as $\binom{n}{0}=\binom{n-1}{0}+0$ and $\binom{n}{0}=\binom{n-1}{0}+\binom{n-1}{-1}$, we get $\binom{n-1}{-1}=0$. Furthermore, we have the identity $\binom{n-1}{k-1}=\frac kn\binom{n}{k}$. Set $k=0$ and we get $\binom{n-1}{-1}=0$. However we look at it, we get
$$
\binom{n-1}{-1}=0
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\lim\limits_{n\rightarrow\infty}\left(\frac{n^2 + 1}{n^2 - 2}\right)^{n^2}$ $$\lim\limits_{n\rightarrow\infty}\left(\frac{n^2 + 1}{n^2 - 2}\right)^{n^2}$$
Trying to solve this. At first thought it was $1$, but in wolfram it's $e^3$. Thanks
| Since $$\lim _{ n\rightarrow \infty }{ { \left( 1+\frac { 1 }{ n } \right) }^{ n } } =e$$
so
$$\lim _{ n\rightarrow \infty }{ { \left( \frac { { n }^{ 2 }+1 }{ { n }^{ 2 }-2 } \right) }^{ { n }^{ 2 } } } =\lim _{ n\rightarrow \infty }{ { \left( \frac { { n }^{ 2 }-2+3 }{ { n }^{ 2 }-2 } \right) }^{ { n }^{ 2 } } } =\lim _{ n\rightarrow \infty }{ { \left( 1+\frac { 3 }{ { n }^{ 2 }-2 } \right) }^{ { n }^{ 2 } } } =\\ =\lim _{ n\rightarrow \infty }{ { \left( 1+\frac { 1 }{ \frac { { n }^{ 2 }-2 }{ 3 } } \right) }^{ \frac { { n }^{ 2 }-2 }{ 3 } \frac { { 3n }^{ 2 } }{ { n }^{ 2 }-2 } } } ={ e }^{ \lim _{ n\rightarrow \infty }{ \frac { 3{ n }^{ 2 } }{ { n }^{ 2 }-2 } } }={ e }^{ \lim _{ n\rightarrow \infty }{ \frac { 3{ n }^{ 2 } }{ { n }^{ 2 }\left( 1-\frac { 2 }{ { n }^{ 2 } } \right) } } }={ e }^{ 3 }$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1550349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the value of $s$ for which the matrix is positive definite I am given the following Matrix
$$
A= \begin{pmatrix}
s & -4 & -4 \\
-4 & s & -4 \\
-4 & -4 & s \\
\end{pmatrix}
$$
and asked to find $s$ such that $A$ is positive definite. In finding it, I simply set the determinant to $0$ and solved for $s$, and in doing so, I got $0 = s^2-4s-32$ which solves to $s=-4,8$. I'm not sure whether this is the correct process though, and if I use either value for $s$ I have at least one $0$ for each eigenvalue of $A$. Am I on the right track? Thanks in advance.
| The matrix $A$ will be positive definite if and only if all its eigenvalues are positive. You have
$$ A = \left( \begin{matrix} s & -4 & -4 \\ -4 & s & -4 \\ -4 & -4 & s \end{matrix} \right) = \left( \begin{matrix} -4 & -4 & -4 \\ -4 & -4 & -4 \\ -4 & -4 & -4 \end{matrix} \right) + \left( \begin{matrix} s + 4 & 0 & 0 \\ 0 & s + 4 & 0 \\ 0 & 0 & s + 4 \end{matrix} \right) = B + (s+4)I. $$
The eigenvalues of $B$ are $0$ (with multiplicity $2$) and $-12$ (with multiplicity $1$) and so the eigenvalues of $A$ are $s + 4$ (with multiplicity $2$) and $s - 8$ (with multiplicity $1$). Thus, $A$ will be positive definite if and only if $s + 4 > 0$ and $s - 8 > 0$ - that is, if and only if $s > 8$.
In the answer, we used the observation that if $B$ is a diagonalizable matrix with eigenvalues $\lambda_1, \ldots, \lambda_n$ and $p(X) = a_nX^n + \ldots + a_0$ is a polynomial, then $p(B)$ (which is also diagonalizable) has eigenvalues $p(\lambda_1), \ldots, p(\lambda_n)$. This is easily proved by checking that if $v_i$ is an eigenvector of $B$ with $Bv_i = \lambda_i v_i$ then it is also an eigenvector of $p(B)$ and $p(B)v_i = p(\lambda_i) v_i$. In our case, $p(X) = X + (s + 4)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1550489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\tan\frac{\pi}{16}+\tan\frac{5\pi}{16}+\tan\frac{9\pi}{16}+\tan\frac{13\pi}{16}$ Find the value of the expression $\tan\frac{\pi}{16}+\tan\frac{5\pi}{16}+\tan\frac{9\pi}{16}+\tan\frac{13\pi}{16}$
I identified that $\frac{\pi}{16}+\frac{13\pi}{16}=\frac{5\pi}{16}+\frac{9\pi}{16}=\frac{14\pi}{16}$
$\tan(\frac{\pi}{16}+\frac{13\pi}{16})=\tan(\frac{5\pi}{16}+\frac{9\pi}{16})$
$\frac{\tan\frac{\pi}{16}+\tan\frac{13\pi}{16}}{1-\tan\frac{\pi}{16}\tan\frac{13\pi}{16}}=\frac{\tan\frac{5\pi}{16}+\tan\frac{9\pi}{16}}{1-\tan\frac{5\pi}{16}\tan\frac{9\pi}{16}}$
But i am stuck here.Please help me.Thanks.
| let $\frac{\pi}{16}=A$ then $tan13A=-tan3A$ and $tan9A=-tan7A$
so
$$S=tanA-tan3A+tan5A-tan7A=\frac{-sin2A}{cosAcos3A}+\frac{-sin2A}{cos5Acos7A}$$
$$S=-2sin2A\left(\frac{1}{2cosAcos3A}+\frac{1}{2cos5Acos7A}\right)$$
$$S=-2sin2A\left(\frac{1}{cos4A+cos2A}+\frac{1}{cos12A+cos2A}\right)$$
$$S=-2sin2A\left(\frac{1}{\frac{1}{\sqrt{2}}+cos2A}+\frac{1}{cos2A-\frac{1}{\sqrt{2}}}\right)$$
$$S=\frac{-4sin2Acos2A}{cos^22A-\frac{1}{2}}=\frac{-8sin2Acos2A}{2cos^22A-1}=\frac{-4sin4A}{cos4A}=-4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1552737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Infinite series equality $\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{3x^2}{1+x^3}+\frac{4x^3}{1+x^4}+\cdots$ Prove the following equality ($|x|<1$).
$$\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{3x^2}{1+x^3}+\frac{4x^3}{1+x^4}+\cdots\\
=\frac{1}{1-x}+\frac{3x^2}{1-x^3}+\frac{5x^4}{1-x^5}+\frac{7x^6}{1-x^7}+\cdots\\$$
| Integrate both sides. Let L.H.S be $f(x)$ and R.H.S be $g(x)$.
(1). L.H.S
$$\int \frac{ix^{i-1}}{1+x^{i}} = \ln (1+x^i)$$
This gives $$\int f(x) dx= \sum_{i=1}^\infty \ln (1+x^i) = \ln (\prod_{i=1}^\infty (1+x^i))$$
(2). R.H.S
$$\int \frac{(2i+1)x^{2i}}{1-x^{2i+1}} = -\int \frac{-(2i+1)x^{2i}}{1-x^{2i+1}} = -\ln (1-x^{2i+1})$$
This gives $$\int g(x) dx=\sum_{i=0}^\infty -\ln (1-x^{2i+1}) = -\ln (\prod_{i=0}^\infty (1-x^{2i+1}))$$
(3). Finishing the problem.
Since $f(0)=g(0)=0$, it now suffices to show that $$ \prod_{i=1}^\infty (1+x^i) \cdot \prod_{i=0}^\infty (1-x^{2i+1}) = 1$$
Now here is the part that I'm not sure if it is true or not.
Let $$h(x)=\prod_{i=1}^\infty (1+x^i) \cdot \prod_{i=0}^\infty (1-x^{2i+1})$$
Of course, the domain of $h(x)$ is $|x|<1$.
Now we have $$h(x)=\prod_{i=1}^\infty (1+x^i) \cdot \prod_{i=0}^\infty (1-x^{2i+1})=\prod_{i=1}^\infty (1+x^{2i}) \cdot \prod_{i=0}^\infty (1+x^{2i+1}) \cdot \prod_{i=0}^\infty (1-x^{2i+1}) =\prod_{i=1}^\infty (1+x^{2i}) \cdot \prod_{i=0}^\infty (1-x^{4i+2})=h(x^2)$$
Therefore, for all $|x|<1$, we have $$h(x)=h(x^2)=h(x^4)= \cdots \lim_{n \to \infty} h(x^{2^n}) = h(0) = 1$$ as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1554777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 0
} |
Finding the (complex) solutions $z$ to $z^4 = w$
What are possible solutions to $$z^4 = w$$ where $z \in \mathbb{C} $ and $w \in \mathbb{R}$?
Here is my attempt:
Write both numbers in polar form: $r^4(\cos 4 \theta + i\sin 4 \theta) = w(\cos 0 + i \sin 0) $.
Working with the conventions of $r \geq 0$ and $0 \leq \theta < 2\pi$ I get the following:
$r^4 = w \iff r = w^\frac{1}{4}$
and
$4\theta = n2\pi$ where $n \in \mathbb{Z}$.
$\Rightarrow \theta = \frac{n}{2}\pi$
Thus we get four possible solutions:
$z = w^\frac{1}{4}(\cos 0 + i \sin 0) = w^\frac{1}{4}$
$z = w^\frac{1}{4}(\cos \frac{\pi}{2} + i \sin \frac{\pi}{2})= iw^\frac{1}{4}$
$z = w^\frac{1}{4}(\cos \pi + i \sin \pi) = -w^\frac{1}{4}$
$z = w^\frac{1}{4}(\cos \frac{3\pi}{2} + i \sin \frac{3\pi}{2}) = -iw^\frac{1}{4}$
Is this legit, something missing or very much not rigorous?
| Assume $z\in\mathbb{C}$ and $w\in\mathbb{R}$:
$$z^4=w\Longleftrightarrow$$
$$z^4=|w|e^{\arg(w)i}\Longleftrightarrow$$
$$z^4=we^{0i}\Longleftrightarrow$$
$$z=\left(we^{2\pi ki}\right)^{\frac{1}{4}}\Longleftrightarrow$$
$$z=\sqrt[4]{w}e^{\frac{2\pi ki}{4}}\Longleftrightarrow$$
$$z=\sqrt[4]{w}e^{\frac{\pi ki}{2}}$$
With $k\in\mathbb{Z}$ and $k:0-3$
So the solutions are:
$$z_0=\sqrt[4]{w}e^{\frac{\pi\cdot 0i}{2}}=\sqrt[4]{w}e^{0}=\sqrt[4]{w}$$
$$z_1=\sqrt[4]{w}e^{\frac{\pi\cdot 1i}{2}}=\sqrt[4]{w}e^{\frac{\pi i}{2}}=i\sqrt[4]{w}$$
$$z_2=\sqrt[4]{w}e^{\frac{\pi\cdot 2i}{2}}=\sqrt[4]{w}e^{\pi i}=-\sqrt[4]{w}$$
$$z_3=\sqrt[4]{w}e^{\frac{\pi\cdot 3i}{2}}=\sqrt[4]{w}e^{\frac{3\pi i}{2}}=-i\sqrt[4]{w}$$
So we can conclude that your solutions are:
$$z=\pm\sqrt[4]{w} \space\space\vee\space\space z=\pm i\sqrt[4]{w}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1554941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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What is the integral of $\frac{x-1}{(x+3)(x^2+1)}$? I've worked with partial fractions to get the integral in the form $$\int\frac{A}{x+3} + \frac{Bx + C}{x^2+1}\,dx$$ Is there a quicker way?
| $\int \frac{x-1}{(x+3)(x^2+1)} dx= \int\frac{A}{x+3} + \frac{B(2x) + C}{x^2+1}\,dx $
and
$ x-1= A(x^{2}+1)+ (B(2x)+C)(x+3)$
which is simple.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1555299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Distribution of $Y$ given $X$ Let $X_1$ and $X_2$ be the numbers on two independent fair-die rolls. Let $X$ be the minimum and $Y$ the maximum of $X_1$ and $X_2$.
$(a)$ Find the distribution of $Y\mid X$
Here is my work:
Here is what I know:
$$P(Y = y\mid X=x) = \frac{P(Y=y, X=x)}{P(X=x)}$$
and $P(X=x)$ can be found by the following equation $\sum_y p_{X,Y}(x,y)$
So here is what I did:
\begin{array}{cc|cccc}
&&&&&y \\
&& 1 & 2 & 3 & 4 & 5 & 6 & P(X=x)\\
\hline
&1 & \frac{1}{36} & \frac{2}{36} & \frac{2}{36} & \frac{2}{36} & \frac{2}{36} & \frac{2}{36} & \frac{11}{36} \\
&2 & 0 & \frac{1}{36} & \frac{2}{36} & \frac{2}{36} & \frac{2}{36} & \frac{2}{36} & \frac{9}{36} \\
x&3 & 0 & 0 & \frac{1}{36} & \frac{2}{36} & \frac{2}{36} & \frac{2}{36} & \frac{7}{36} \\
&4 & 0 & 0 & 0 & \frac{1}{36} & \frac{2}{36} & \frac{2}{36} & \frac{5}{36} \\
&5 & 0 & 0 & 0 & 0 & \frac{1}{36} & \frac{2}{36} & \frac{3}{36} \\
&6 & 0 & 0 & 0 & 0 & 0 & \frac{1}{36} & \frac{1}{36}
\end{array}
So I said:
$$
P(Y=y\mid X=x) =
\begin{cases}
\frac{1}{36} \div P(X=x) & \text{for } x = y \\[4pt]
\frac{2}{36} \div P(X=x) & \text{for } x = 1,2,3,4,5,6, \ y > x
\end{cases}
$$
Am I correct?
|
So I said, $$P(Y=y\mid X=x) =
\begin{cases}
\frac{1}{36} \cdot P(X=x) & \text{for } x = y \\[4pt]
\frac{2}{36} \cdot P(X=x) & \text{for } x = 1,2,3,4,5,6, \ y > x
\end{cases}$$
Am I correct?
Almost but, no. You know that the conditional probability is the joint divided by the marginal, so why did you multiply rather than divide?
Also notice that: $\mathsf P(X=x) = \frac{13-2x}{36}$
$$\begin{align}\mathsf P(Y{=}y\mid X{=}x) & = \begin{cases} \frac{1}{36} \div P(X=x) & \text{for } x\in\{1\,\ldots\,6\}{\times}\{x\}\\[1ex] \frac{2}{36} \div P(X=x) & \text{for } (x,y)\in \{1\,\ldots\,5\}{\times}\{x{+}1\,\ldots\,6\} \end{cases} \\[2ex] & = \begin{cases}\frac{1}{13-2x} & \text{for } (x,y)\in\{1\,\ldots\,6\}{\times}\{x\}\\[1ex] \frac{2}{13-2x} & \text{for } (x,y)\in \{1\,\ldots\,5\}{\times}\{x{+}1\,\ldots\,6\} \end{cases}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1555435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Double Angle identity??? The question asks to fully solve for $$\left(\sin{\pi \over 8}+\cos{\pi \over 8}\right)^2$$
My question is, is this a double angle formula? And if so, how would I go about to solve it?
I interpreted it this way; $$\left(\sin{\pi \over 8}+\cos{\pi \over 8}\right)^2$$
$$=2\sin{\pi \over 4}+\left(1-2\sin{\pi \over 4}\right)$$
Have I done this right so far? I feel I have not.
| Yes, you can use double angle identity to simplify as follows $$\left(\sin\frac{\pi}{8}+\cos\frac{\pi}{8}\right)^2=\sin^2\frac{\pi}{8}+\cos^2\frac{\pi}{8}+2\sin\frac{\pi}{8}\cos\frac{\pi}{8}$$
$$=\left(\sin^2\frac{\pi}{8}+\cos^2\frac{\pi}{8}\right)+2\sin\frac{\pi}{8}\cos\frac{\pi}{8}$$
$$=1+2\sin\frac{\pi}{8}\cos\frac{\pi}{8}$$
using double angle identity, $2\sin A\cos A=\sin 2A$
$$=1+\sin2\left(\frac{\pi}{8}\right)$$
$$=1+\sin\frac{\pi}{4}$$
$$=1+\frac{1}{\sqrt 2}=\color{red}{\frac{2+\sqrt 2}{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1559257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 2
} |
Sum of a Series With Denominators of the form $(2^i) (3^j)(5^k)$ Can anyone solve this?
Find the sum of the series $1 + \frac{1}{2} +\frac{1}{3} + \frac{1}{4} + \frac{1}{5}+ \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \cdots,$ where the denominators are of the form $(2^i) (3^j)(5^k)$?
The test came with the next answer choices:
a) $\frac{7}{2}$
b) $1$
c) $3$
d) $\frac{15}{4}$
e) $\frac{3}{2}$
I thought this would tend to infinity. Maybe the test answer choices were wrong..
| Consider $$(1 + \frac{1}{2} + \frac{1}{4} + \ldots + \frac{1}{2^n} + \ldots)(1 + \frac{1}{3} + \frac{1}{9} + \ldots + \frac{1}{3^n} + \ldots)(1 + \frac{1}{5} + \frac{1}{25} + \ldots + \frac{1}{5^n} + \ldots)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1559946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Why is my $\epsilon$-$\delta$ proof incorrect? Prove that $f(x)= \sqrt{4+x^2}$ is continuous at $x_0$ using the $\epsilon -\delta$ definition of continuity.
Textbook proof:
$|\sqrt{4+x^2}-\sqrt{4+x^2}|=\frac{|4+x^2-(4+{x_0}^2)|}{\sqrt{4+x^2}+\sqrt{4+{x_0}^2}}=\frac{|x^2-{x_0}^2|}{\sqrt{4+x^2}+\sqrt{4+{x_0}^2}}=\underbrace{\frac{|x+x_0|}{\sqrt{4+x^2}+\sqrt{4+{x_0}^2}}}_{K(x,x_0)}|x-x_0|$
$K(x,x_0)=\frac{|x+x_0|}{\sqrt{4+x^2}+\sqrt{4+{x_0}^2}}\leq \frac{|x|+|x_0|}{\sqrt{4+x^2}+\sqrt{4+{x_0}^2}}=\frac{|x|}{\sqrt{4+x^2}+\sqrt{4+{x_0}^2}}+\frac{|x_0|}{\sqrt{4+x^2}+\sqrt{4+{x_0}^2}}$
$\leq\frac{|x|}{\sqrt{4+x^2}}+\frac{|x_0|}{\sqrt{4+{x_0}^2}}$
$\frac{|x|}{\sqrt{4+x^2}}\leq 1$ and $\frac{|x_0|}{\sqrt{4+{x_0}^2}}\leq 1\Rightarrow K(x,x_0)\leq 2\Rightarrow |f(x)-f(x_0)|\leq2|x-x_0|<2\delta=\epsilon$
So, for all $\epsilon$ we can choose a $\delta$ such that $\delta=\frac{\epsilon}{2}$.
My proof:
$|\sqrt{4+x^2}-\sqrt{4+x^2}|=\frac{|4+x^2-(4+{x_0}^2)|}{\sqrt{4+x^2}+\sqrt{4+{x_0}^2}}=\frac{|x^2-{x_0}^2|}{\sqrt{4+x^2}+\sqrt{4+{x_0}^2}}\leq |x^2-{x_0}^2|$
$=|x-x_0||x-x_0+2x_0|<\delta^2+2|x_0|\delta$
Case $\delta<1$:
$\delta^2+2|x_0|\delta<\delta(1+2|x_0|)=\epsilon$
$\Rightarrow$ If $1+2|x_0|>\epsilon$ we have $\delta=\frac{\epsilon}{1+2|x_0|}$
Case $\delta=1$:
$\Rightarrow$ If $1+2|x_0|\leq\epsilon$ we have $\delta = 1$
Why do I get a different result for delta? What did I do wrong?
| Regarding your proof, there is nothing wrong with it. Only that dealing with $\delta=1$ is not needed, as you can always request $\delta<1$.
As for why the "results" are different? There is no result. No one says that for a given $f$ and a given $x_0$ and a given $\varepsilon$, there is a single $\delta$. Any $\delta$ less than a given $\delta$ will work, so there is not obvious choice of a "best" $\delta$ or anything like that.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1563236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solving $\int \left ( \frac{5x^2+3x-2}{x^3+2x^2} \right )$ via partial fractions I'm having issues with the Partial Fractions method:
$$\int \left ( \frac{5x^2+3x-2}{x^3+2x^2} \right )dx$$
I am doing this in a way I was taught, which somehow I feel is a bit different from other methods online.
So, first of all, we must factorize the denominator:
$$x^3+2x^2 = (x+2)\cdot x^2$$
Great. So now we write three fractions:
$$\frac{A}{x^2} + \frac{B}{x} + \frac{C}{x+2}$$
Eventually we conclude that
$$A(x+2)+B(x+2)(x)+C(x^2) = 5x^2+3x-2$$
So now we look at what happens when $x = -2$:
$$C = 12$$
When $x = 0$:
$$A = -1$$
And now we are missing $B$, but we can just pick an arbitrary number for $x$ like... $1$:
$$B = -1$$
We replace the values here:
$$\int \frac{-1}{x^2}dx + \int \frac{-1}{x}dx + \int \frac{12}{x+2}dx$$
Which results in
$$\frac{1}{x}-\ln(x)+12\ln(x+2)+K$$
But I fear the answer actually is
$$\frac{1}{x}+2\ln(x)+3\ln(x+2)+K$$
Can you tell me what did I do wrong, and what should I have done?
| You have an algebra error. When you put $x = -2$, the constraint equation collapses to $4C = 12$, so $C = 3$.
Your method for determining $B$ is correct – the constraining identity must hold for any $x$. Putting $x = 1$ is a fine choice as the arithmetic is easier. However the incorrect value for $C$ will affect the value you determine for $B$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1565338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
Finding all possible integer values of $a,b,c$, given $2$ equations. If $a^2=bc+1$ and $b^2=ac+1$, how do I find all possible values of $a,b,c$. Assuming $a,b,c$ all are integers?
| Subtract one equation from the other. Then
$$a^2 - b^2 = bc - ac$$
$$(a + b)(a - b) = -(a - b)c$$
$$(a + b)(a - b) + (a - b)c = 0$$
$$(a - b)(a + b + c) = 0$$
Considering the case where $a = b$: We end up with $a^2 = ac + 1$ so that $a(a - c) = 1$. Since $a, c$ are integers, we only have to consider two cases: $a, a-c = \pm1$.
Considering the case where $a + b + c = 0$: We get $c = -a - b$ so that
$$a^2 = b(-a - b) + 1$$
$$a^2 = -ab - b^2 + 1$$
$$a^2 + b^2 + ab - 1 = 0$$
$$a^2 + ba + b^2 - 1 = 0$$
Viewing this as a quadratic equation in $a$, we know that its discriminant must be nonnegative for real solutions of $a$ to exist. That is, $$b^2 - 4(b^2 -1)\geq0$$
$$-3b^2 + 4 \geq 0$$
$$3b^2 \leq 4$$
and using a similar argument,
$$3a^2 \leq 4$$
Here, we also only have to consider a few cases : namely only $b = 0, \pm 1, a = 0,\pm1$. Note that each possible pair of $a,b$ should be checked against the original equation $a^2 + ab + b^2 -1=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1565812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
If $a_{n+1}=\frac {a_n^2+5} {a_{n-1}}$ then $a_{n+1}=Sa_n+Ta_{n-1}$ for some $S,T\in \Bbb Z$. Question
Let $$a_{n+1}:=\frac {a_n^2+5} {a_{n-1}},\, a_0=2,a_1=3$$
Prove that there exists integers $S,T$ such that $a_{n+1}=Sa_n+Ta_{n-1}$.
Attempt
I calculated the first few values of $a_n$: $a_2=7,a_3=18, a_4=47$ so I'd have the system of diophantine equations:
$$
7=3S+2T\\
18=7S+3T\\
47=18S+7T
$$
Now: it seems that all of the $a_i$ are pairwise coprime, so these equations should always have solutions, but how could I check that the intersection of all the solutions is not $\emptyset$?
| From the initial data and recursion, compute $a_0=2,a_1=3,a_2=7,a_3=18$.
If we have an $S$ and $T$ so that $a_{n+1}=Sa_n+Ta_{n-1}$, then
$$
\begin{bmatrix}3&2\\7&3\end{bmatrix}\begin{bmatrix}S\\T\end{bmatrix}=\begin{bmatrix}7\\18\end{bmatrix}\implies\begin{bmatrix}S\\T\end{bmatrix}=\begin{bmatrix}3&2\\7&3\end{bmatrix}^{-1}\begin{bmatrix}7\\18\end{bmatrix}=\begin{bmatrix}3\\-1\end{bmatrix}
$$
Now, to verify that the linear recursion $a_{n+1}=3a_n-a_{n-1}$ implies
$$
a_{n+1}a_{n-1}-a_n^2=5\tag{1}
$$
Assume the linear recursion, then
$$
\begin{align}
&\left(a_{n+2}a_n-a_{n+1}^2\right)-\left(a_{n+1}a_{n-1}-a_n^2\right)\\
&=(3a_{n+1}-a_n)a_n-a_{n+1}^2-a_{n+1}a_{n-1}+a_n^2\\
&=3a_{n+1}a_n-a_{n+1}^2-a_{n+1}a_{n-1}\\
&=a_{n+1}(3a_n-a_{n-1})-a_{n+1}^2\\
&=a_{n+1}^2-a_{n+1}^2\\
&=0\tag{2}
\end{align}
$$
Thus, the linear recursion and $(1)$ imply that
$$
a_{n+2}a_n-a_{n+1}^2=5\tag{3}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1566162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 4
} |
$\lim\limits_{n\to\infty}$ $(1+ {2\over n})^n$ = $e^2$ If I set N=$2\over n$, the equation becomes $(1+ N)^{2\over N}$, which I can take the natural log of and work down until I reach $e^2$. However, my teacher wants me to use subsequences, starting with $x_n$ = $(1+ {1\over n})^n = e$, to prove this and I am stuck.
| Let $a_n$ =$(1+2/n)^n$ and $a_{n+1}$=$1+$($\frac{2}{n+1})^{n+1} $
Proceed by their binomial expansion.
$a_n$=$1+n$($\frac{2}{n})$+n($\frac{n-1}{2!})$ $(\frac{2}{n})^{2}$+⋯+$\frac{n(n-1)(n-2)…}{n!}$ ($\frac{2}{n})^{n}$
$a_n=$1+2+($\frac{1}{2!})$ $2^2$(1-$\frac{1}{n})$+($\frac{1}{3!})$ $2^3$(1-$\frac{1}{n})$(1-$\frac{2}{n})$+⋯+($\frac{1}{n!})$ $2^n$(1-$\frac{1}{n})$(1-$\frac{2}{n})$...(1-$\frac{n-1}{n})$
$a_{n+1}=$1+2+($\frac{1}{2!})$ $2^2$(1-$\frac{1}{n+1})$+($\frac{1}{3!})$ $2^3$(1-$\frac{1}{n+1})$(1-$\frac{2}{n+1})$+⋯+($\frac{1}{{n+1}!})$ $2^n$(1-$\frac{1}{n+1})$(1-$\frac{2}{n+1})$...(1-$\frac{n}{n+1})$
For n≥2,$a_{n+1}$ >a_n Hence Monotone increasing.
$a_n ≤ 1+2+(2/2!)+(2/3!)+⋯
a_n ≤1+(2+(2/2^2) +(2/2^3) +⋯)
a_n ≤ 1+2(1+1/2^2 +1/2^3 +⋯)
By using sum to infinity, $a_n ≤1+2(2(1-1/2^n ))
$a_n ≤1+2(2-2^(1-n) )≤1+4-2^(2-n)≤5;as n→∞.
Hence convergent and bounded above by 5.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1566602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluating Limits - Finding Multiple Results Only one of which is Correct When I calculate the limit
$$\lim_{x\to\infty}\sqrt{x^2+x} - \sqrt {x^2-x}$$
I get $2$ answers for this question: $1$ and $0$ but $1$ is the right answer. I don't know why this is the case, however. If you multiply by the conjugate divided by the conjugate (1), you take the radical out of top and get it in the bottom and then if you factor out $x$ from both and cancel it with top u get $2/2$ which is $1$. But if you just factor you get
$$\lim_{x\to\infty} x (\sqrt {1+ 1/x} - \sqrt{1 - 1/x}.$$
This simplifies to 0. So how would you know which method to use if you didn't know the right answer?
| $
x\bigl(
\sqrt{1+\tfrac1x}
-
\sqrt{1-\tfrac1x}
\bigr)
$
as $x\to\infty$ is of the form $\infty\cdot0$, which is indeterminate hence your wrong conclusion.
A correct way to get rid of the indeterminate form is as you suggested earlier: indeed,
\begin{align}
\sqrt{x^2+x} - \sqrt{x^2-x}
={}&
\frac{
\left(\sqrt{x^2+x} - \sqrt{x^2-x}\right)
\left(\color{red}{\sqrt{x^2+x} + \sqrt{x^2-x}}\right)
}{
\color{red}{\sqrt{x^2+x} + \sqrt{x^2-x}}
}
\\
={}&
\frac{2x}{
\sqrt{x^2+x} + \sqrt{x^2-x}
}
\\
={}&
\frac{2x}{
x\left(\sqrt{1+\frac1x} + \sqrt{1-\frac1x}\right)
}
\\
={}&
\frac{2}{
\sqrt{1+\frac1x} + \sqrt{1-\frac1x}
}
\xrightarrow{x\to+\infty}
\frac 22
=
1
\end{align}
By doing so you passed from an indeterminate to a well definite form.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1574088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Induction Proof with Fibonacci How do I prove this?
For the Fibonacci numbers defined by $f_1=1$, $f_2=1$, and $f_n = f_{n-1} + f_{n-2}$ for $n ≥ 3$, prove that $f^2_{n+1} - f_{n+1}f_n - f^2_n = (-1)^n$ for all $n≥ 1$.
| the inductive step:
$$
f^2_{n+1} - f_{n+1}f_n - f^2_n = (-1)^n \Leftrightarrow \\
f_{n+1}f_{n-1} - f_n^2 =(-1)^n \Leftrightarrow \\
(f_{n+2}-f_n)(f_{n+1}-f_n) -f_n^2 = (-1)^n \Leftrightarrow \\
f_{n+2}f_{n+1}-f_nf_{n+3} = (-1)^n \Leftrightarrow \\
f_{n+1}^2+f_nf_{n+1}-f_nf_{n+1}-f_nf_{n+2}=(-1)^n \Leftrightarrow \\
f_{n+2}f_n-f_{n+1}^2 = (-1)^{n+1} \Leftrightarrow \\
f_{n+2}^2-f_{n+2}f_{n+1}-f_{n+1}^2 =(-1)^{n+1}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1574290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Solve $x^2\equiv -3\pmod {\!91}$ by CRT lifting roots $\!\bmod 13\ \&\ 7$ Question 1) Solve $$x^2\equiv -3\pmod {13}$$
I see that $x^2+3=13n$. I don't really know what to do? Any hints?
The solution should be $$x\equiv \pm 6 \pmod {13}$$
Question 2) $\ $ [note $\bmod 7\!:\ x^2\equiv -3\equiv 4\iff x\equiv \pm 2.\,$ Here we lift to $\!\!\pmod{\!91}\ $ -Bill]
Given $x\equiv \pm 6 \pmod {13}$ and $x\equiv \pm 2 \pmod {7}$ find solutions $\pmod {91}$. I see that $91=13 \times 7$, does it mean I have to use chinese remainder theorem on 4 equations? If,so $x=6\times 13\times 7 \times 7\times (13\times 7 \times 7)^{-1}...$
| For Question 2), you can use the Chinese Remainder Theorem on the one equation
$$
13x+7y=1\tag{1}
$$
Using the Extended Euclidean Algorithm, an implementation of which is in this answer, compute
$$
\begin{array}{r}
&&1&1&6\\\hline
1&0&1&-1&7\\
0&1&-1&2&-13\\
13&7&6&1&0
\end{array}\tag{2}
$$
that is,
$$
13(-1)+7(2)=1\tag{3}
$$
From $(3)$, we can get
$$
\begin{align}
14&\equiv1\pmod{13}\\
14&\equiv0\pmod{7}
\end{align}\tag{4}
$$
and
$$
\begin{align}
-13&\equiv0\pmod{13}\\
-13&\equiv1\pmod{7}
\end{align}\tag{5}
$$
Now adding $6$ times $(4)$ to $2$ times $(5)$, we get
$$
\begin{align}
58&\equiv6\pmod{13}\\
58&\equiv2\pmod{7}
\end{align}\tag{6}
$$
Adding other multiples of $(4)$ and $(5)$, we can get the other answers you are looking for.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1575576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Finding Lyapunov function for a given system of differential equations I am being introduced to the Lyapunov functions in order to determine the stability of a given system. I know that finding a Lyapunov function is not easy, so I would like to ask for any trick or hint in order to find a Lyapunov function for
$$ \left\{\begin{array}{l}x'=-4y+x^2,\\y'=4x+y^2\end{array}\right. $$
at $(0,0)$. I have tried combinations of $x^{2n}$ and $y^{2m} $ and also products of $x$ and $y$ but got nothing clear. Also, I've searched the phase plot for the system and it is clear that $(0,0)$ is a stable point (not asymptotically stable). Thanks in advance.
| It is not quite clear why you put a bounty on this question, since @Evgeny answered it in the best possible way. However, if you are looking for a Lyapunov function, here it is (up to an additive constant):
$$
L(x,y)=\frac{2^{2/3} \left(1-\frac{x (x+4) \left(x^2-12 x+8 y+48\right)}{\sqrt[3]{x^3 (x+4)^3} \left(x^2-4 y\right)}\right) \left(\frac{x (x+4) \left(x^2-12 x+8 y+48\right)}{2 \sqrt[3]{x^3 (x+4)^3} \left(x^2-4 y\right)}+1\right) \left(\left(1-\frac{x (x+4) \left(x^2-12 x+8 y+48\right)}{\sqrt[3]{x^3 (x+4)^3} \left(x^2-4 y\right)}\right) \log \left(2^{2/3} \left(1-\frac{x (x+4) \left(x^2-12 x+8 y+48\right)}{\sqrt[3]{x^3 (x+4)^3} \left(x^2-4 y\right)}\right)\right)+\left(\frac{x (x+4) \left(x^2-12 x+8 y+48\right)}{\sqrt[3]{x^3 (x+4)^3} \left(x^2-4 y\right)}-1\right) \log \left(2\ 2^{2/3} \left(\frac{x (x+4) \left(x^2-12 x+8 y+48\right)}{2 \sqrt[3]{x^3 (x+4)^3} \left(x^2-4 y\right)}+1\right)\right)-3\right)}{9 \left(-\frac{\left(x^2-12 x+8 y+48\right)^3}{2 \left(x^2-4 y\right)^3}+\frac{3 x (x+4) \left(x^2-12 x+8 y+48\right)}{2 \sqrt[3]{x^3 (x+4)^3} \left(x^2-4 y\right)}-1\right)}-\frac{x (x+4) \left(4 \log \left(x^2-4 x+16\right)+x\right)}{18 \sqrt[3]{2} \sqrt[3]{x^3 (x+4)^3}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1575951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
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Green's Theorem on Line Integral I am asked to find the line integral for the following field:
$$F = (e^{y^2}-2y)i + (2xye^{y^2}+\sin(y^2))j$$
On the line segment with points $(0,0),(1,2)$ and $(3,0)$. I have to do it with Greens theorem. This is the setup I have so far.
$$\int_C F \cdot dr = \iint_D \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\, dA$$
$$\frac{\partial Q}{\partial x} = ye^{y^2} \qquad \frac{\partial P}{\partial y}=2ye^{y^2}-2$$
Line from $(0,0)$ to $(1,2)$ is $y = 2x$. Line from $(1,2)$ to $(3,0)$ is $y = -x + 3$
$$\int_0^1\int_0^{2x}ye^{y^2}-2ye^{y^2}-2 + \int_1^3\int_0^{-x+3}ye^{y^2}-2ye^{y^2}-2 $$
$$\int_0^1\int_0^{2x}-ye^{y^2}-2 \, dy\,dx+ \int_1^3\int_0^{-x+3}-ye^{y^2}-2 \,dy\,dx$$
$$\int_0^1\left[\frac{e^{y^2}}{2} - 2y\right]_0^{2x} + \int_1^3\left[-\frac{e^{y^2}}{2} - 2y\right]_0^{-x+3}$$
$$\int_0^1\frac{e^{4x^2}}{2}-\frac{1}{2}-4x \,dx+ \int_1^3-\frac{e^{(-x+3)^2}}{2}-2(-x+3)+\frac{1}{2}\,dx$$
Let $u = -x + 3$ and $du = -1$.
$$\int_0^1\frac{e^{4x^2}}{2}-\frac{1}{2}-4x \,dx+ \int_1^3\frac{e^{(u)^2}}{2}+2(u)-\frac{1}{2}\,du$$
$$\left[\frac{e^{4x^2}}{8x}-\frac{x}{2}-2x^2 \right]_0^1+ \left[\frac{e^{(u)^2}}{4u}+u^2+\frac{u}{2}\right]_1^3$$
$$\left[\frac{e^{4x^2}}{8x}-\frac{x}{2}-2x^2 \right]_0^1+ \left[\frac{e^{(-x+3)^2}}{4(-x+3)}+(-x+3)^2+\frac{-x+3}{2}\right]_1^3$$
$$\frac{e^{4}}{8}-\frac{1}{2}-2 \,-\frac{e^{4}}{8}-4-1$$
$$\frac{1}{2}-2 \,-5$$
But the answer key says the answer is $-3$. Where have I gone wrong?
| The partial derivative of $Q$ regarding $x$ lacks a factor $2$.
| {
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"url": "https://math.stackexchange.com/questions/1578033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Differentiating $x^2=\frac{x+y}{x-y}$ Differentiate:
$$x^2=\frac{x+y}{x-y}$$
Preferring to avoid the quotient rule, I take away the fraction:
$$x^2=(x+y)(x-y)^{-1}$$
Then:
$$2x=(1+y')(x-y)^{-1}-(1-y')(x+y)(x-y)^{-2}$$
If I were to multiply the entire equation by $(x-y)^2$ then continue, I get the solution. However, if I continue the following, I don't. Likely some place I erred, but I cannot figure out where:
Expansion:
$$2x=(x-y)^{-1}+y'(x-y)^{-1}-(x+y)(x-y)^{-2}+y'(x+y)(x-y)^{-2}$$
Preparing to isolate for $y'$:
$$2x-(x-y)^{-1}+(x+y)(x-y)^{-2}=y'(x-y)^{-1}+y'(x+y)(x-y)^{-2}$$
$$2x-(x-y)^{-1}+(x+y)(x-y)^{-2}=y'[(x-y)^{-1}+(x+y)(x-y)^{-2}]$$
Isolating $y'$:
$$y'=\frac{2x-(x-y)^{-1}+(x+y)(x-y)^{-2}}{(x-y)^{-1}+(x+y)(x-y)^{-2}}$$
Multiple top and bottom by $(x-y)$:
$$y'=\frac{2x(x-y)-1+(x+y)(x-y)^{-1}}{1+(x+y)(x-y)^{-1}}$$
Then, inserting $x^2$ into $(x+y)(x-y)^{-1}$, I get:
$$y'=\frac{2x(x-y)-1+x^2}{1+x^2}$$
While the answer states:
$$y'=\frac{x(x-y)^2+y}{x}$$
Which I do get if I multiplied the entire equation by $(x-y)^2$ before. It does not seem to be another form of the answer, as putting $x=2$, the denominator cannot match each other. Where have I gone wrong?
| You should first find $y$ in terms of $x$ as follows $$x^2=\frac{x+y}{x-y}$$ $$x^3-x^2y=x+y$$$$ (1+x^2)y=x^3-x$$
$$y=\frac{x^3-x}{1+x^2}$$
now, differentiate w.r.t. $x$, $$\frac{dy}{dx}=\frac{(1+x^2)(3x^2-1)-(x^3-x)(2x)}{(1+x^2)^2}$$
$$=\frac{x^4+4x^2-1}{(1+x^2)^2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $x+y+z=6$ and $xyz=2$, then find $\cfrac{1}{xy} +\cfrac{1}{yz}+\cfrac{1}{zx}$
If $x+y+z=6$ and $xyz=2$, then find the value of $$\cfrac{1}{xy}
+\cfrac{1}{yz}+\cfrac{1}{zx}$$
I've started by simply looking for a form which involves the given known quantities ,so:
$$\cfrac{1}{xy} +\cfrac{1}{yz} +\cfrac{1}{zx}=\cfrac{yz\cdot zx +xy \cdot zx +xy \cdot yz}{(xyz)^2}$$
Now this might look nice since I know the value of the denominator but if I continue to work on the numerator I get looped :
$$\cfrac{yz\cdot zx +xy \cdot zx +xy \cdot yz}{(xyz)^2}=\cfrac{4\left(\cfrac{1}{xy}+\cfrac{1}{zy}+\cfrac{1}{zy}\right)}{(xyz)^2}=\cfrac{4\left(\cfrac{(\cdots)}{(xyz)^2}\right)}{(xyz)^2}$$
How do I deal with such continuous fraction ?
| $$x+y+z=6$$
Divide both sides with $xyz$
$$\frac{x+y+z}{xyz}=\frac{6}{xyz}$$
$$\frac{x}{xyz}+\frac{y}{xyz}+\frac{z}{xyz}=\frac{6}{2}$$
$$\frac{1}{yz}+\frac{1}{xz}+\frac{1}{xy}=3$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$\lim_{x\to 0}\frac{2+\cos x}{x^3\sin x}-\frac{3}{x^4}=\frac{1}{60}$ without using L Hospital rule or series expansion Prove that $\lim_{x\to 0}\frac{2+\cos x}{x^3\sin x}-\frac{3}{x^4}=\frac{1}{60}$ without using L Hospital rule or series expansion.
I tried $\lim_{x\to 0}\frac{2+\cos x}{x^3\sin x}-\frac{3}{x^4}=\lim_{x\to 0}\frac{2+\cos x}{x^4\lim_{x\to 0}\frac{\sin x}{x}}-\frac{3}{x^4}$
$=\lim_{x\to 0}\frac{2+\cos x}{x^4}-\frac{3}{x^4}=\lim_{x\to 0}\frac{\cos x-1}{x^4}=\lim_{x\to 0}\frac{-2\sin^2 \frac{x}{2}}{x^4}=\lim_{x\to 0}\frac{-2\sin^2 \frac{x}{2}}{4x^2(\frac{x}{2})^2}$
$=\lim_{x\to 0}\frac{-1}{2x^2}$
Now it has turned into limit does not exist.I dont know where have i made mistake,because as per my knowledge my every step is correct.If i have made mistake please correct me
Please help me.Thanks.
| As per the method suggested by Lab bhattacharjee, I got the answer to my problem.Thanks to him.
Let $L=\lim_{x\to 0}\frac{2+\cos x}{x^3\sin x}-\frac{3}{x^4}......(1)$
So $L=\lim_{x\to 0}\frac{2+\cos 2x}{8x^3\sin 2x}-\frac{3}{16x^4}$
$L=\lim_{x\to 0}\frac{1+2\cos^2x}{16x^3\sin x\cos x}-\frac{3}{16x^4}$
$16L=\lim_{x\to 0}\frac{1+2\cos^2x}{x^3\sin x\cos x}-\frac{3}{x^4}......(2)$
Subtracting $(1)$ from $(2)$,we get
$15 L=\lim_{x\to 0}\frac{1-2\cos x+\cos^2 x}{x^3\sin x}$
$15 L=\lim_{x\to 0}\frac{(1-\cos x)^2}{x^4\lim_{x\to 0}\frac{\sin x}{x}}$
$15L=\lim_{x\to 0}\frac{4\sin^4\frac{x}{2}}{x^4}$
$15L=\lim_{x\to 0}\frac{4\sin^4\frac{x}{2}}{16(\frac{x}{2})^4}$
$L=\frac{1}{60}\lim_{x\to 0}(\frac{\sin\frac{x}{2}}{\frac{x}{2}})^4$
$L=\frac{1}{60}$
| {
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"url": "https://math.stackexchange.com/questions/1581770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Line tangent to circle
A circle with a radius of $2$ units has its center at $(0,0)$. A circle with a radius of $7$ units has its center at $(15,0)$. A line tangent to both circles intersects the $x$-axis at $(x,0)$. What is the value of $x$? Express your answer as a common fraction.
My problem with this question is that there are $4$ such tangent lines, so how do I know which one to pick?
| So there is solution using the $Ax+By+C=0$ notation for lines. Given a circle with center $(x_c,y_c)$ and radius $r$ the all the possible tangents are given by the line:
$$ x (\cos\psi) + y (\sin \psi) -(x_c \cos\psi + y_c \sin \psi + r) = 0 $$
In fact, every scalar multiple of the $(A,B,C)$ coefficients also describe the same line and we are going to use this fact below to match the tangent lines for the two circles.
For the two circles the two general tangent lines are:
$$\begin{align}
x (\cos \psi_1) + y ( \sin \psi_1) - 2 & =0 \\
x (\cos \psi_2) + y ( \sin \psi_2) - (7+15 \cos \psi_2) & = 0
\end{align}$$
If we multiply the first equation with $\lambda = \frac{7}{2} + \frac{15}{2} \cos\psi_2$ and then subtract it from the second equation you will get
$$ x \left(\cos\psi_2 - \lambda \cos\psi_1 \right) + y \left( \sin\psi_2 - \lambda \sin\psi_1 \right) = 0 $$
This must be true, regardless which point $(x,y)$ along the tangent line is used. So the above is alike a set of equations for the two angles
$$\begin{align} \cos \psi_2 & = \lambda \cos \psi_1 \\ \sin \psi_2 & = \lambda \sin \psi_1 \end{align}$$
This has two solutions, $\lambda=1$ and $\lambda=-1$. The first one gives the outside tangents, and the second the inside tangents. So the two tangents are found by $$ \cos \psi_2 = -\frac{1}{3} \\ \cos \psi_2 = - \frac{3}{5} $$
Solving the second tangent line equation for $x$ using $y=0$ and the angles above gives us
$$ x= \frac{7}{\cos \psi_2} + 15 = \begin{cases} -6 & \lambda=1 \\ \frac{10}{3} & \lambda=-1 \end{cases} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1584672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Double sum of products of integers up to $n$ Suppose that $S$ is defined by
$$
S(n) = \sum_{i=0}^{n} \sum_{j=0}^{i} ij.
$$
I'm confused as to how $S(3) = 25$ from this summation. Can anyone expand on it as to how to get the answer?
| \begin{align*}
S(3) &= \sum_{i=0}^3\sum_{j=0}^i ij \\
&= 0\sum_{j=0}^0 j + 1\sum_{j=0}^1 j + 2\sum_{j=0}^2 j + 3\sum_{j=0}^3 j \\
&= 0 + (0+1) + 2(0+1+2) + 3(0+1+2+3) \\
&= 25.
\end{align*}
More generally,
\begin{align*}
S(n) &= \sum_{i=0}^n\sum_{j=0}^i ij \\
&= \sum_{i=0}^n i\cdot\frac{i(i+1)}{2} \\
&= \frac{1}{2}\sum_{i=0}^n i^3+i^2 \\
&= \frac{1}{2}\left(\frac{n^2(n+1)^2}{4} + \frac{n(n+1)(2n+1)}{6}\right) \\
&= \frac{1}{24}(3 n^4+10 n^3+9 n^2+2 n).
\end{align*}
| {
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"url": "https://math.stackexchange.com/questions/1585132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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How find gcd polynomials? How to find gcd of polynomials $gcd(x^3+x^2-x-1,3x^2+2x-1)$ ??
I divide of polynomials. It worked like this $\frac 13 x - \frac19$,$ R\left( x\right) =-\frac89x -\frac89$
| HINT:
If $f(x)$ divides both,
$f(x)$ must divide $$3(x^3+x^2-x-1)-x(3x^2+2x-1)=x^2-2x-3=(x-3)(x+1)$$
Now $3x^2+2x-1=(3x-1)(x+1)$
OR
$x^3+x^2-x-1=(x+1)(x^2-1)=(x+1)^2(x-1)$
and
$3x^2+2x-1=(3x-1)(x+1)$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Conditions under which $a+b+c$ divides $1-abc$ What are the conditions such that $a+b+c$ divides $1-abc$, where $(a, b, c)$ are nonzero integers ?
| The triples $(a,b,c)$ with the property that $a+b+c$ divides $1-abc$ can be characterized as triples $(a,b,d-a-b)$, where $d$ is a divisor of $1+ab(a+b)$. This is because $a+b+c=d$ for such triples, while
$$1-abc=1-ab(d-a-b)=1+ab(a+b)-abd$$
For example, if $a=b=4$, we have $1+ab(a+b)=129$, which has $8$ divisors: $d=\pm1,\pm3,\pm43$ and $\pm129$, giving triples
$$\begin{align}
&(4,4,-7)\\
&(4,4,-9)\\
&(4,4,-5)\\
&(4,4,-11)\\
&(4,4,35)\\
&(4,4,-51)\\
&(4,4,121)\\
&(4,4,-137)
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1586758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Number of ways of forming 4 letter words using the letters of the word RAMANA
Question: Find the number of ways of forming 4 letter words using the letters of the word "RAMANA"
This can be solved easily by taking different cases.
*
*All 3 'A's taken: remaining one letter can be chosen in $^3C_1$ ways. Total possibilities $=^3C_1\cdot\frac{4!}{3!}=12$
*Only 2 'A's taken: remaining two letters out of {R,M,N} can be chosen in $^3C_2$ ways. Total possibilities $= ^3C_2\cdot\frac{4!}{2!}=36$
*Only one A: Number of ways: $4!=24$
Total $=72$.
But my teacher solved it like this. He found the coefficient of $x^4$ in $4!\cdot(1+\frac{x}{1!})^3(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!})$ which also came out to be $72$.
Why does this work? Also, if I avoid the factorials, I get number of combinations. That is, number of combinations $=$ coefficient of $x^4$ in $(1+x)^3(1+x+x^2+x^3)$
| Suppose you have a $4$ letter string composed of, say, $1$ distinct and $3$ identical letters
There would be $\frac{4!}{1!3!}$ permutations, also expressible as a multinomial coefficient, $\binom{4}{1,3}$
Similarly, for $2$ distinct, $2$ identical, and $3$ distinct, $1$ identical,
it would be $\binom{4}{2,2}\;$ and $\binom{4}{3,1}$ respectively.
In the polynomial expression $4!(1+x/1!)^3(1+x+x^2/2!+x^3/3!)$,
the 4! corresponds to the numerator, whatever the combination; the first term in $( )$ corresponds to choosing one or more from $R,M,N$; and the other term corresponds to choosing $1,2,$ or $3 A's$
It will become evident why this approach works if we expand the first term in ( ), and compare serially with your case approach by just using the appropriate coefficients to get terms in $x^4$
$4!(1 + 3x + 3x^2 + x^3)(1 + x + x^2/2! + x^3/3!)$
To find the coefficient of $x^4$, consider the three cases that produce $x^4$
One from $R,M,N, 3A's : 4!\cdot3\cdot\frac{1}{3!} = 12$
Two from $R,M,N, 2A's : 4!\cdot3\cdot\frac1{2!} = 36$
Three from $R,M,N, 1A : 4!\cdot1\cdot 1 = 24$
Coefficient of $x^4 = 12+36+24 = 72$
We can now clearly see why the coefficient of $x^4$ in the expression automatically gives all possible permutations of $4$ letters
| {
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If $x-y = 5y^2 - 4x^2$, prove that $x-y$ is perfect square Firstly, merry christmas!
I've got stuck at a problem.
If x, y are nonzero natural
numbers with $x>y$ such that
$$x-y = 5y^2 - 4x^2,$$
prove that $x - y$ is perfect square.
What I've thought so far:
$$x - y = 4y^2 - 4x^2 + y^2$$
$$x - y = 4(y-x)(y+x) + y^2$$
$$x - y + 4(x-y)(x+y) = y^2$$
$$(x-y)(4x+4y+1) = y^2$$
So $4x+4y+1$ is a divisor of $y^2$.
I also take into consideration that $y^2$ modulo $4$ is $0$ or $1$ (I don't know if this can help.)
So how do I prove that $4x+4y+1$ is a perfect square (this would involve $x-y$ - a perfect square)? While taking examples, I couldn't find any perfect square with a divisor that is $M_4 + 1$ and is not perfect square.
If there are any mistakes or another way, please tell me.
Some help would be apreciated. Thanks!
| Generalization. Let $a,b$ be integers. If there exists consecutive integers $c,d$ such that $a-b=a^2c-b^2d$, then $|a-b|$ is a perfect square.
Proof. If $c=d+1$, we have $a-b=a^2(d+1)-b^2d=(a-b)(a+b)d+a^2$, so $$a^2=(a-b)(1-d(a+b))$$ Now let $g=\text{gcd}(a-b,1-d(a+b))$. We have $g^2|a^2$, so $g|a$. Now we have $g|b$, and we have $g|1$, so $g=1$.
Since $a-b$ and $1-d(a+b)$ are coprime and their multiple is a perfect square, we are done.
The case $c+1=d$ is handled similarly. $\blacksquare$.
| {
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"timestamp": "2023-03-29T00:00:00",
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$3^{n+1}$ divides $2^{3^n}+1$ Describe all positive integers,n such that $3^{n+1}$divides $2^{3^n}+1$.
I am little confused about what the question asks-if it asks me to find all such positive integers, or if it asks me to prove that for every positive integer n,$3^{n+1}$ divides $2^{3^n}+1$. Kindly clarify this doubt and if it's the former part, please verify my solution-n=1.
| This actually looks like an induction proof. Let $P(n)$ be the statement : $$3^{n+1} \ \mid \ 2^{3^n} + 1 $$
Then we can verify that it is indeed true for $n = 1$ since $9$ divides itself. So suppoose $P(n)$ is true. Now,
$$2^{3^{n+1}} + 1 = \left({2^{3^n}}\right)^3 + 1 $$
Notice that $$\left({2^{3^n} + 1}\right)^3 = \left({2^{3^n}}\right)^3 + 3 \cdot \left({2^{3^n}}\right)^2 + 3 \cdot \left({2^{3^n}}\right) + 1 $$
So,
\begin{align}
\left({2^{3^n}}\right)^3 + 1 & = \left({2^{3^n} + 1}\right)^3 - 3 \cdot \left({2^{3^n}}\right)^2 - 3 \cdot \left({2^{3^n}}\right) \\
& = \left({2^{3^n} + 1}\right)^3 - 3 \cdot \left[{\left({2^{3^n } + 1}\right)^2 - 2\left({2^{3^n}}\right) - 1}\right] - 3 \cdot \left({2^{3^n} }\right) \\
& = \left({2^{3^n} + 1}\right)^3 - 3 \cdot \left({2^{3^n } + 1}\right)^2 + 3 \cdot \left({2^{3^n } + 1}\right) \\
& = \left({3^{n+1} \cdot k}\right)^3 - 3 \left({3^{n+1} \cdot k}\right)^2 + 3 \left({3^{n+1} \cdot k}\right) \\
& = 3^{n +2} \left({3^{2n+1} \cdot k^3 - k^2 + k}\right)
\end{align}
which implies $$3^{n+2} \ \mid \ 2^{3^{n+1}} + 1 $$
And so the result is true for every $n$.
So all the integers satisfy the given condition and that is the answer to the relevant question.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "11",
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prove this inequality with $a+b+c=1$ Let $a,b,c>0,a+b+c=1$,show that
$$\left(\sqrt{\dfrac{a+b}{c}}+\sqrt{\dfrac{b+c}{a}}+\sqrt{\dfrac{c+a}{b}}\right)^2\ge \dfrac{16}{3(a+b)(b+c)(c+a)}$$
| This inequality I think some time,Now I solve it.Following is my solution.
Use Holder inequality we have
$$\left(\sqrt{\dfrac{a+b}{c}}+\sqrt{\dfrac{b+c}{a}}+\sqrt{\dfrac{c+a}{b}}\right)^2\cdot\sum_{cyc}c(a+b)^2\ge (a+b+b+c+c+a)^3$$
so
$$\left(\sqrt{\dfrac{a+b}{c}}+\sqrt{\dfrac{b+c}{a}}+\sqrt{\dfrac{c+a}{b}}\right)^2\ge\dfrac{8(a+b+c)^3}{\displaystyle\sum_{cyc}c(a+b)^2}=\dfrac{8}{6abc+\displaystyle\sum_{cyc}c(a^2+b^2)}$$
it suffices to show that
$$\dfrac{8}{6abc+\displaystyle\sum_{cyc}c(a^2+b^2)}\ge\dfrac{16}{3(a+b)(b+c)(c+a)}$$
since
$$(a+b)(b+c)(c+a)=2abc+\sum_{cyc}c(a^2+b^2)$$
it suffices to show
$$\sum_{cyc}c(a^2+b^2)\ge 6abc$$
it is clear AM-GM inequality
By Done
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $x^n+y^n=2015$ Determine the natural numbers $x,y,n$ matching equality
$$x^n+y^n=2015.$$
I noticed for $n = 1$ the equation has solutions $(x, 2015-x), x$ integer.
For $n = 2$, given that $x$ and $y$ are different parities taking $x = 2k$ and $y=2m + 1$ we come to contradiction.
What must be done to $n\geq3$?
| Since $2015=5\cdot 13\cdot 31$, and the multiplicities of the primes congruent $3$ modulo $4$ are odd (only $p=31$ to consider here), it follows that $2015$ cannot be represented as the sum $x^2+y^2$, by Fermat's theorem. This does not only work for $2015$. Also, characterisation of two cubes is known, see here. Of course, $2015$ is still small enough to check all cases. For bigger numbers however, one would like to apply more number theory.
Edit: In case we would like to have more positive integer solutions to $x^n+y^n=m$, then $m=4097$ is the smallest number (besides $2$) which is a sum of two squares and two positive cubes. Indeed, the equation $x^n+y^n=4097$ has integer solutions for $n=1,2,3,4,6,12$:
$$
4097=1+4096=1^2+64^2=1^3+16^3=1^4+8^4=1^6+4^6=1^{12}+2^{12}.
$$
| {
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"url": "https://math.stackexchange.com/questions/1593105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Solve integral $\int{\frac{x^2 + 4}{x^2 + 6x +10}dx}$ Please help me with this integral:
$$\int{\frac{x^2 + 4}{x^2 + 6x +10}}\,dx .$$
I know I must solve it by substitution, but I don't know how exactly.
| $$\int{\frac{x^2 + 4}{x^2 + 6x +10}}dx=x-6\int{\frac{x+1}{x^2 + 6x +10}}dx$$
$$\int{\frac{x+1}{x^2 + 6x +10}}dx=\int{\frac{x+1}{(x+3)^2+1}}dx$$ now we can use $u$ substitution where $u=x+3$ and $du = dx$ so
$$\int{\frac{u-2}{u^2+1}}du=\int{\frac{u}{u^2+1}}du-2\int{\frac{1}{u^2+1}}du=\frac{\ln|u^2+1|}{2}-2\arctan(u)+C$$ now just substitute whatever u is.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
Prove that $\sum_{n=1}^{\infty}\left(\frac{1}{(8n-7)^3}-\frac{1}{(8n-1)^3}\right)=\left(\frac{1}{64}+\frac{3}{128\sqrt{2}}\right)\pi^3$ Prove that $$\sum_{n=1}^{\infty}\left(\frac{1}{(8n-7)^3}-\frac{1}{(8n-1)^3}\right)=\left(\frac{1}{64}+\frac{3}{128\sqrt{2}}\right)\pi^3$$
I don't have an idea about how to start.
| This answer uses the hint using the polygamma function
$$
\psi^{(2)}(z) = -\int_0^1 \frac{t^{z-1}}{1-t}\ln^2t dt.
$$
First, using the expansion of $(1-t)^{-1}$, the polygamma function $\psi^{(2)}(x)$ can be written as
\begin{align}
\psi^{(2)}(z) &= -\sum_{n=0}^\infty \int_0^1 t^{n+z-1}\ln^2 tdt \cr
&= -\sum_{n=0}^\infty \int_0^\infty s^2 e^{-(n+z)s}ds
\qquad \qquad \qquad (t=e^{-s}) \cr
&= -2\sum_{n=0}^\infty \frac{1}{(n+z)^3}.
\end{align}
Therefore,
$$
\sum_{n=1}^{\infty}\left(\frac{1}{(8n-7)^3}-\frac{1}{(8n-1)^3}\right) = \frac{1}{1024}\left(\psi^{(2)}\left(\frac{7}{8}\right)
-\psi^{(2)}\left(\frac{1}{8}\right)\right).
$$
Using the reflection relation
$$
\psi^{(2)}(1-z)-\psi^{(2)}(z) = \pi\frac{d^2}{dz^2}\cot \pi z
$$
with $z=1/8$, the sum can be written as
$$
\left.\frac{\pi}{1024}\frac{d^2}{dz^2}\cot \pi z\right|_{z=1/8}
=\frac{\pi^3}{512}\left(
\cot\frac{\pi}{8}+\cot^3\frac{\pi}{8}\right).
$$
Finally, using trigonometric identities (half angle),we can get $\cot \pi/8 = 1+\sqrt{2}$, and
$$
\sum_{n=1}^{\infty}\left(\frac{1}{(8n-7)^3}-\frac{1}{(8n-1)^3}\right)
=\frac{\pi^3}{512}\Big(
1+\sqrt{2}+(1+\sqrt{2})^3
\Big)=\frac{\pi^3}{256}(4+3\sqrt{2}).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Trigonometric Ratios Of Multiple Angles If $2\tan A=3\tan B$ then prove that $$\tan(A-B)=\frac{\sin2B}{5-\cos 2B}.$$
I found that $\tan A=\frac{3}{2} \tan B$ and after that used the formula of $\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$ but could not reach to the required answer.
| We have
$$\tan\left(A-B\right) = \dfrac{\tan(A)-\tan(B)}{1+\tan(A)\tan(B)} = \dfrac{\tan(B)/2}{1+\dfrac32 \tan^2(B)} = \dfrac{\tan(B)}{2+3\tan^2(B)} = \dfrac{\dfrac{\sin(B)}{\cos(B)}}{2+3 \cdot \dfrac{\sin^2(B)}{\cos^2(B)}}$$
Hence, we have
$$\tan\left(A-B\right) = \dfrac{\sin(B)\cos(B)}{2\cos^2(B)+3\sin^2(B)} = \dfrac{\sin(2B)}{4+2\sin^2(B)} = \dfrac{\sin(2B)}{5-\cos(2B)}$$
where we made use of $\sin(2B) = 2\sin(B) \cos(B)$, $\cos(2B) = 1-2\sin^2(B)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
solution of nested radical $\sqrt{7+2\sqrt{7-2\sqrt{7-2x} } } =x$ This question is from my friend. he think that there is trigonometry involved to this equation.
$\sqrt{7+2\sqrt{7-2\sqrt{7-2x} } } =x$
that is from $\ \ \ x = \sqrt{7+2\sqrt{7-2\sqrt{7-2\sqrt{7+2\sqrt{7-2\sqrt{7-2\sqrt...} } }} } }$
I try to compare between this nested radical and Ramanujan's nested radicals form ,but solution is for all positive terms.
Thank you for all comments , thank you so much
| $$\begin{align}
\sqrt{7+2\sqrt{7-2\sqrt{7-2x} } } &=x\\
7+2\sqrt{7-2\sqrt{7-2x} } &=x^2\\
2\sqrt{7-2\sqrt{7-2x} } &=x^2-7\\
4\left(7-2\sqrt{7-2x} \right) &=x^4-14x^2+49\\
-8\sqrt{7-2x} &=x^4-14x^2+21\\
64(7-2x) &=x^8-28x^6+238x^4-588x^2+441\\
0 &=x^8-28x^6+238x^4-588x^2+128x-7\\
0&=\left(x^2+2x-7\right)\left(x^3-x^2-9x+1\right)^2
\end{align}$$
where a CAS helped with the last factoring. Based on the initial equation, $$\sqrt{7}<x<\sqrt{7+2\sqrt{7}}$$ and you can separately analyze that $x^2+2x-7$ has no zeros in this range. The other factor, $x^3-x^2-9x+1$, has one positive zero in this interval. As the root of a cubic, it is expressible in radicals if you want to use Cardano's method.
If the original equation has a solution at all, it must be this root. It remains to check that this solution actually solves the original equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Find the summation $\frac{1}{1!}+\frac{1+2}{2!}+\frac{1+2+3}{3!}+ \cdots$ What is the value of the following sum?
$$\frac{1}{1!}+\frac{1+2}{2!}+\frac{1+2+3}{3!}+ \cdots$$
The possible answers are:
A. $e$
B. $\frac{e}{2}$
C. $\frac{3e}{2}$
D. $1 + \frac{e}{2}$
I tried to expand the options using the series representation of $e$ and putting in $x=1$, but I couldn't get back the original series. Any ideas?
| Your sum is $$\sum_{n = 1}^{\infty} \sum_{k = 1}^{n} \frac {k} {n!} = \sum_{n = 1}^{\infty} \frac {n + 1} {2 (n - 1)!} = \sum_{n = 0}^{\infty} \frac {n + 2} {2 n!} = \frac {3e} {2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 3
} |
If $n$ and $m$ are odd integers, show that $ \frac{(nm)^2 -1}8$ is an integer I am trying to solve:
If $n$ and $m$ are odd integers, show that $ \frac{(nm)^2 -1}8$ is an integer.
If I write $n=2k+1$ and $m=2l+1$ I get stuck at
$$\frac{1}{8}(16k^2 l^2 +4(k+l)^2 +8kl(k+l)+4kl+2(k+l))$$
| Expanding the whole thing, you get
$$
(2k+1)^2(2\ell+1)^2 -1 = 16 k^2 \ell^2+16 k^2 \ell+4 k^2+16 k \ell^2+16 k \ell+4 k+4 \ell^2+4 \ell
$$
Modulo 8, the terms multiplied by 16 disappear, and this becomes
$$
4 k^2+4 k+4 \ell^2+4 \ell = 4k(k+1) + 4\ell(\ell+1)
$$
and it only remains to see that both $\ell(\ell+1)$ and $k(k+1)$ are even to conclude. (since then $8$ divides both $4\ell(\ell+1)$ and $4k(k+1)$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 4
} |
I have a problem with solving this integral by partial fraction I have an example like this :
$\int \frac{1}{x^{4}}dx$ = $\int x^{-4}dx$ = $\frac{1}{-4 + 1}x^{-4 + 1}+C$ = $\frac{1}{-3}x^{-3}+C$
What if this?
$\int \frac{1}{2x-10}dx$ = ?
| $\displaystyle \int\! x^r \, dx=\frac{x^{r+1}}{r+1}+C$ for $\color{red}{r \ne -1}$.
Let $u = 2x-10$ then $dx = \frac{1}{2}du$ and
$\displaystyle \int \frac{1}{2x-10}dx=\frac{1}{2}\int \frac{1}{u}\,du =\frac{1}{2}\ln|u|+C = \frac{1}{2}\ln|2x-10|+C. $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
For all $n >0$, $ \left(1+\frac{1}{n} \right)^n = 1+ \sum_{k=1}^n\bigl[ \frac{1}{k!} \prod_{r=0}^{k-1}(1-\frac{r}{n}) \bigr]$ I am working through some problems on induction and I have been stuck on this one for a while. If anyone has any hints. I can show it is true for the $n=1$ and $n=2$ case but I am having difficulty on the induction step.
This is what I have so far,
$ (1+\frac{1}{n} )^n = 1+ \sum\limits_{k=1}^n[ \frac{1}{k!} \prod\limits_{r=0}^{k-1}(1-\frac{r}{n}) ]$
$n =1 $ case
$ (1+\frac{1}{1} )^1 = 1+ \sum\limits_{k=1}^1[ \frac{1}{k!} \prod\limits_{r=0}^{k-1}(1-\frac{r}{1}) ]$
$ 2 = 1+ \frac{1}{1!} \prod\limits_{r=0}^{0}(1-\frac{r}{1}) ]= 1+(1-\frac{0}{1})=1+1=2$
So, 2=2. True
$n=2$ case
$ (1+\frac{1}{2} )^2 = 1+ \sum\limits_{k=1}^2[ \frac{1}{k!} \prod\limits_{r=0}^{k-1}(1-\frac{r}{2}) ]$
$ (1+\frac{1}{2} )^2 = 1+ [ \frac{1}{1!} \prod\limits_{r=0}^{0}(1-\frac{r}{2}) ]+\frac{1}{2!} \prod\limits_{r=0}^{1}(1-\frac{r}{2}) ]$
$ (1+\frac{1}{2} )^2 = 1+ [ \frac{1}{1!} (1-\frac{0}{2}) ]+[\frac{1}{2!} (1-\frac{1}{2}) ]$
$ (1+\frac{1}{2} )^2 = 1+ [ \frac{1}{1!} (1) ]+[\frac{1}{2!} (\frac{1}{2}) ]$
$ (1+\frac{1}{2} )^2 = 1+ 1+ \frac{1}{4} ]$
And finally, $\frac{9}{4} = \frac{9}{4}$
So the n=2 step is also true.
Induction Step
Show that the $(n+1)^{th}$ case is true when the $n^{th}$ case is true.
Now, I assume the $n^{th}$ case is true. So,
$ (1+\frac{1}{n} )^n = 1+ \sum\limits_{k=1}^n[ \frac{1}{k!} \prod\limits_{r=0}^{k-1}(1-\frac{r}{n}) ]$
then I multiply both sides by $ (1+\frac{1}{n} )^{n+1} $, but I haven't gotten far from there. Any hints would be great!
| s(n+1)
Hints:
1.(1 + 1/n+1)^n+1 = (1 + 1/n)^n(1+ 1/n)...
2.Split up the sums
3.Use the binomial Theorem
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find Complex Roots
Question:
Find the complex roots of $$ {(z^{12} -1)\over (z^4-1)(z^3-1)} = 0 $$
What I have attempted:
$$ {(z^{12} -1)\over (z^4-1)(z^3-1)} = 0 $$
$$ {(z^{6} -1)(z^{6} +1)\over (z^2-1)(z^2+1)(z^3-1)} = 0 $$
$$ {(z^{3} -1)(z^3 +1)(z^6+1)\over (z+1)(z-1)(z^2+1)(z^3-1)} = 0 $$
$$ {(z^3 +1)(z^6+1)\over (z+1)(z-1)(z^2+1)} = 0 $$
$$ {(z+1)(z^2-z+1)(z^6+1)\over (z+1)(z-1)(z^2+1)} = 0 $$
$$ {(z^2-z+1)(z^6+1)\over (z-1)(z^2+1)} = 0 $$
How should I continue?
| A fraction is zero if and only if its numerator is zero, and well-defined if and only if its denumerator is non-zero.
Therefore
$$
\frac{z^{12} - 1}{(z^4 - 1)(z^3 - 1)} = 0
\iff
z^{12} = 1, z^4 \neq 1, z^3 \neq 1.
$$
The solutions of the first condition are given by the $12$-th root of unity, namely
$$
z = e^{k 2 \pi i / 12}
\qquad
k = 0, 1, \dotsc, 11.
$$
We have to exlude the fourth roots of unity and third roots of unity for the other two conditions.
These are
$$
z
= e^{k 2 \pi i / 4}
= e^{(3k) 2 \pi i /12}
\quad\text{for $k = 0, 1, 2, 3$}
$$
and
$$
z
= e^{k 2 \pi i / 3}
= e^{(4k) 2 \pi i / 12}
\quad\text{for $k = 0, 1, 2$}.
$$
Thus all we are left with is
$$
z
= e^{k 2 \pi i / 12}
\qquad
k \in \{1,2,5,7,10,11\}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integer Square Root Algorithm I'm currently working my way through David M. Bressoud's "Factorization and Primality Testing", and I'm struggling with an exercise (exercise 5.7) that asks the reader to prove that the following algorithm produces the greatest integer less than or equal to the square root of $n$:
\begin{align}
\text{INITALIZE:} \quad &\text{READ} \;n \\
&a \leftarrow n \\
&b \leftarrow \lfloor (n+1)/2 \rfloor \\\\
\text{MYSTERY_LOOP:} \quad &\text{WHILE} \: b < a \; \text{DO} \\
&\quad a \leftarrow b \\
&\quad b \leftarrow \lfloor (a \times a + n) / (2 \times a) \rfloor \\\\
\text{TERMINATE:}\quad &\text{WRITE} \; a
\end{align}
I know from real analysis that the sequence $x_{m+1} = \frac{x_m^2 + n}{2 x_m}$ converges to the square root of $n$ (for any sensible $x_0$), so morally, I can believe that the above algorithm converges to the integer square root of $n$.
How do we formally prove that the above algorithm works (i.e. terminates in a finite number of steps, such that $a^2 \leq n$ and $(a+1)^2 >n$), and moreover, why start the iterations with $b = \lfloor\frac{n+1}{2} \rfloor$? Is $\lfloor\frac{n+1}{2} \rfloor$ just a very crude approximation? What is its significance?
I have tried splitting the floor into a real part minus a part between 0 and 1 (i.e. $\lfloor x \rfloor = x - \{x\}$), but I haven't been able to get very far with this.
In addition, do we have the same quadratic convergence that we get when we compute the (ordinary) square root using Newton's method?
| Using the notation of the $a$ and $b$ is only necessary when programming.
You have correctly identified that the algorithm gives a sequence of values, which I will call (as you did) the $x_m$.
We thus have the relationship $x_{m+1} = \lfloor \frac{x_m^2 + n}{2 x_m} \rfloor$
Starting with $x_0=n$ gives $x_{1} = \lfloor \frac{x_0^2 + n}{2 x_0} \rfloor=\lfloor \frac{n^2 + n}{2 n} \rfloor = \lfloor \frac{n+1}{2} \rfloor$, which answers one of your questions. It also demonstrates that the change from $x_0$ to $x_1$ is always a decrease (unless n=1).
Given two successive terms of the sequence $x_m$ and $x_{m+1}$, there are three possible outcomes:
1) $x_{m+1}=x_m$
2) $x_{m+1}<x_m$
3) $x_{m+1}>x_m$
Let's consider these in turn.
1) $x_{m+1}=x_m$
$\lfloor \frac{x_m^2 + n}{2 x_m} \rfloor=x_m$
The $floor$ function can be dealt with by expressing $\lfloor \frac{x_m^2 + n}{2 x_m} \rfloor=\frac{x_m^2 + n}{2 x_m}-p$ where $0 \le p < 1$
Thus we have $\frac{x_m^2 + n}{2 x_m}-p=x_m$
Rearranging gives $\frac{n-x_m^2}{2 x_m}=p$
Since $0 \le p < 1$, we can say that $0 \le \frac{n-x_m^2}{2 x_m} < 1$
Considering this in two parts, we have first:
$0 \le \frac{n-x_m^2}{2 x_m}$
$x_m$ is positive, so $0 \le {n-x_m^2}$
$x_m^2 \le n$
$x_m \le \sqrt n$
Then:
$\frac{n-x_m^2}{2 x_m} < 1$
${n-x_m^2} < {2 x_m}$
$n<x_m^2 + 2 x_m$
Completing the square,
$n+1<x_m^2 + 2 x_m+1$
$n+1<(x_m + 1)^2$
$\sqrt {n+1}<x_m + 1>$
$\sqrt {n+1}-1<x_m $
Putting the two inequalities back together gives us $\sqrt {n+1}-1<x_m \le \sqrt n$
It can be shown (binomial expansion) that $\sqrt {n+1}-1>\sqrt n -1 $ so we have:
$\sqrt n - 1 <x_m \le \sqrt n$
So if the sequence converges to a value, then this will be the largest integer less than or equal to the square root of $n$.
It could be, however, that the values oscillate without converging. What about those situations?
Moving on:
2) $x_{m+1}<x_m$
Following the same method as before, we have
$\frac{x_m^2 + n}{2 x_m}-p<x_m$
$\frac{n - x_m^2}{2 x_m}<p$
$\frac{n - x_m^2}{2 x_m}<1$
${n - x_m^2}<{2 x_m}$
$n<x_m^2+2 x_m$
$n+1<x_m^2+2 x_m+1$
$n+1<(x_m+1)^2$
$\sqrt {n+1}<x_m+1$
$x_m>\sqrt {n+1}-1$
As before, this gives $x_m>\lfloor \sqrt n \rfloor$
If there are two successive decreases in value, then both $x_m$ and $x_m+1$ will be greater than $x_m>\lfloor \sqrt n \rfloor$, but the values will be moving closer to $x_m>\lfloor \sqrt n \rfloor$, thus having convergence.
3) $x_{m+1}>x_m$
$\frac{x_m^2 + n}{2 x_m}-p>x_m$
$\frac{n - x_m^2}{2 x_m}>p$
We know that $p \ge 0$, so $\frac{n - x_m^2}{2 x_m} \ge 0$
$n - x_m^2 \ge 0$
$x_m^2 \le n$
$x_m \le \sqrt n$
This means that an increase in the values of the sequence only happens if the value in the sequence is less than or equal to the square root. There cannot be an infinite run of increases because eventually the value will become greater than the square root. Therefore there must eventually be a decrease in the values of the sequence.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How can I integrate this? (for calculate value of L-function ) I want to calculate the definite integral:
$$
\int_{0}^{1} \frac{x+x^{3}+x^{7}+x^{9}-x^{11}-x^{13}-x^{17}-x^{19}}{x(1-x^{20})}dx.
$$
Indeed, I already know that $\int_{0}^{1} \frac{x+x^{3}+x^{7}+x^{9}-x^{11}-x^{13}-x^{17}-x^{19}}{x(1-x^{20})}dx=L(\chi,1)=\frac{\pi}{\sqrt{5}}$ where $\chi$ is the Dirichlet character for $\mathbb{Z}[\sqrt{-5}]$.
I have some trouble with this actual calculation.
My calculation is as following:
\begin{eqnarray}
&&\int_{0}^{1} \frac{x+x^{3}+x^{7}+x^{9}-x^{11}-x^{13}-x^{17}-x^{19}}{x(1-x^{20})}dx \cr
&=& \int_{0}^{1} \frac{1+x^{6}}{1-x^{2}+x^{4}-x^{6}+x^{8}} dx\cr
&=& \int_{0}^{1} \frac{x^{2}(1+x^{2})(\frac{1}{x^{2}}-1+x^{2})}{x^{4}(\frac{1}{x^{4}}-\frac{1}{x^{2}}+1-x^{2}+x^{4})}dx
\end{eqnarray}
Substitute $x-\frac{1}{x}=t$. Then, $(1+\frac{1}{x^{2}})dx= dt$.
So The integral is
$$
\int_{-\infty}^{0} \frac{t^{2}+1}{t^{4}+3t^{2}+1}dt
$$
However, it has different value with the value which is calculated by the site
http://www.emathhelp.net/calculators/calculus-2/definite-integral-calculator/?f=%28t%5E%7B2%7D%2B1%29%2F%28t%5E%7B4%7D%2B3t%5E%7B2%7D%2B1%29&var=&a=-inf&b=0&steps=on
So I've just given up to try more.
Is there any fine solution?
| You may notice that, through the substitution $x=\frac{1}{z}$, we have:
$$ I=\int_{0}^{1}\frac{1+x^6}{1-x^2+x^4-x^6+x^8}\,dx = \int_{1}^{+\infty}\frac{1+z^6}{1-z^2+z^4-z^6+z^8}\,dz $$
hence:
$$ I = \frac{1}{2}\int_{0}^{+\infty}\frac{1+z^6}{1-z^2+z^4-z^6+z^8}\,dz = \frac{1}{4}\int_{\mathbb{R}}\frac{1+z^6}{1-z^2+z^4-z^6+z^8}\,dz $$
and the last integral can be easily computed through the residue theorem.
Let $Z$ be the set of the primitive $20$th roots of unity with positive imaginary part, i.e.:
$$Z=\left\{\exp\left(\frac{\pi i }{10}\right),\exp\left(\frac{3\pi i }{10}\right),\exp\left(\frac{7\pi i }{10}\right),\exp\left(\frac{9\pi i }{10}\right)\right\}=\{\zeta_1,\zeta_2,\zeta_3,\zeta_4\}.$$
We have:
$$\begin{eqnarray*} I &=& \frac{2\pi i}{4}\sum_{j=1}^{4}\text{Res}\left(\frac{1+z^6}{1-z^2+z^4-z^6+z^8},z=\zeta_i\right)\\&=&\frac{2\pi i}{4}\left(-\frac{i}{2\sqrt{5}}-\frac{i}{2\sqrt{5}}-\frac{i}{2\sqrt{5}}-\frac{i}{2\sqrt{5}}\right)\\&=&\color{red}{\frac{\pi}{\sqrt{5}}}\end{eqnarray*}$$
as wanted. Ultimately, that $\sqrt{5}$ just depends on a well-known quadratic Gauss sum (have a look at this Wikipedia entry, too), but the identity $L(\chi,1)=\frac{\pi}{\sqrt{5}}$ is also a consequence of Kronecker's formula, since the ring of integers of $\mathbb{Z}[\sqrt{5}]$ has class number one.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Finding the sum of the infinite series whose general term is not easy to visualize: $\frac16+\frac5{6\cdot12}+\frac{5\cdot8}{6\cdot12\cdot18}+\cdots$ I am to find out the sum of infinite series:-
$$\frac{1}{6}+\frac{5}{6\cdot12}+\frac{5\cdot8}{6\cdot12\cdot18}+\frac{5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}+...............$$
I can not figure out the general term of this series. It is looking like a power series as follows:-
$$\frac{1}{6}+\frac{5}{6^2\cdot2!}+\frac{5\cdot8}{6^3\cdot3!}+\frac{5\cdot8\cdot11}{6^4\cdot4!}+.....$$
So how to solve it and is there any easy way to find out the general term of such type of series?
| This is $\frac{1}{2} \left(_2F_1(2/3,1;1;\frac{1}{2}) - 1\right)$ where $_2F_1$ is the hypergeometric function. For these parameters this evaluates to $\frac{1}{2} \left(_2F_1(2/3,1;1;\frac{z}{2}) - 1\right) = \frac{1}{2} \left( (1 - \frac{z}{2})^{-2/3} - 1\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
If $a^2 + b^2 = 1$, show there is $t$ such that $a = \frac{1 - t^2}{1 + t^2}$ and $b = \frac{2t}{1 + t^2}$ My question is how we can prove the following:
If $a^2+b^2=1$, then there is $t$ such that $$a=\frac{1-t^2}{1+t^2} \quad \text{and} \quad b=\frac{2t}{1+t^2}$$
| As $a^2 + b^2 = 1$, there exists a $\theta$, such that $a = \cos\theta$ and $b=\sin\theta$
$$ a= \cos\theta = \frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}$$
$$ b= \sin\theta = \frac{2\tan\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}$$
so exists $t = \tan\frac{\theta}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Pove that the angle between planes in which origin lies is acute if $a_1a_2+b_1b_2+c_1c_2<0$ Suppose we have two planes $$a_1x+b_1y+c_1z+d_1=0$$ and $$a_2x+b_2y+c_2z+d_2=0$$ where $d_1,d_2 >0 \ or \ <0$ then prove that the angle between planes in which origin lies is acute if $$a_1a_2+b_1b_2+c_1c_2<0$$
The angle between the planes is given by $$cos\theta=\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}.\sqrt{a_2^2+b_2^2+c_2^2}} $$
In this the denominator $\sqrt{a_1^2+b_1^2+c_1^2}.\sqrt{a_2^2+b_2^2+c_2^2}$ is positive, hence the sign of $cos\theta $ depends only on $a_1a_2+b_1b_2+c_1c_2$
How do I proceed further ?
|
The angle between the planes is given by $$cos\theta=\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}.\sqrt{a_2^2+b_2^2+c_2^2}} \tag1$$
More precisely, one of the angles between the planes is given by $(1)$.
So, if you mean that $\theta=\text{(the angle between the planes in which origin lies)}$, then we have
$$\cos\theta=\color{red}{\pm}\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}.\sqrt{a_2^2+b_2^2+c_2^2}} $$
The following will use a different approach.
Let $P_i : a_ix+b_iy+c_iz+d_i=0$, and let $Q_i$ be a point on $P_i$ such that $OQ_i$ is perpendicular to $P_i$. Also, let $\theta$ be the angle between the planes in which origin $O$ lies, and let $\alpha=\angle{Q_1OQ_2}$.
$\qquad\qquad\qquad\qquad\qquad$
Then, since
$$OQ_1\ :\ \frac{x}{a_1}=\frac{y}{b_1}=\frac{z}{c_1}$$
we have
$$Q_1\left(\frac{-d_1a_1}{a_1^2+b_1^2+c_1^2},\frac{-d_1b_1}{a_1^2+b_1^2+c_1^2},\frac{-d_1c_1}{a_1^2+b_1^2+c_1^2}\right)$$
Similarly,
$$Q_2\left(\frac{-d_2a_2}{a_2^2+b_2^2+c_2^2},\frac{-d_2b_2}{a_2^2+b_2^2+c_2^2},\frac{-d_2c_2}{a_2^2+b_2^2+c_2^2}\right)$$
Hence,
$$\begin{align}&\text{$\theta$ is acute}\\&\iff \text{$\alpha$ is obtuse}\\&\iff \cos\alpha\lt 0\\&\iff \frac{\vec{OQ_1}\cdot\vec{OQ_2}}{\left|\vec{OQ_1}\right|\cdot\left|\vec{OQ_2}\right|}\lt 0\\&\iff \vec{OQ_1}\cdot\vec{OQ_2}\lt 0\\&\iff \frac{(-d_1a_1)(-d_2a_2)+(-d_1b_1)(-d_2b_2)+(-d_1c_1)(-d_2c_2)}{(a_1^2+b_1^2+c_1^2)(a_2^2+b_2^2+c_2^2)}\lt 0\\&\iff d_1d_2(a_1a_2+b_1b_2+c_1c_2)\lt 0\\&\iff a_1a_2+b_1b_2+c_1c_2\lt 0\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find the Fourier's coefficient $a_n$ of the Fourier's series of $\sin(x)$ on $(0,\pi]$, $0$on $(-\pi,0]$ Considering $g(x)$, periodical with a period of $2\pi$ defined by
\begin{equation*}
g(x)=
\begin{cases}
0 & \text{for $x \in (-\pi;0]$} \\
\sin(x) & \text{for $x \in [0;\pi)$}
\end{cases}
\end{equation*}
I want to find the Fourier's series $S(g)$ of $g$.
\begin{align*}
a_n&=\frac{1}{\pi}\int\limits_0^\pi \sin(x)\cos(x)d(x)\\
&=[sin^2(x)]-\int\limits_0^\pi \cos^2(x)d(x)\\
&=-\frac{1}{2}\int\limits_0^\pi (\cos(2x)+1)d(x)\\
&= \sin(2x)+\frac{x}{2}\\
\end{align*}
I've the feeling, seeing the solution of \begin{equation*}
g(x)=
\begin{cases}
0 & \text{for $x \in (-\pi;0]$} \\
x\sin(x) & \text{for $x \in [0;\pi)$}
\end{cases}
\end{equation*}
That I'm wrong...
| For
\begin{equation*}
g(x)=
\begin{cases}
0 & \text{$x \in [-\pi,0)$} \\
\sin(x) & \text{$x \in [0,\pi)$}
\end{cases}
\end{equation*}
Write
$$g(x) = \frac12 a_0 + \sum_{n=1}^{\infty} \left (a_n \cos{n x} + b_n \sin{n x} \right ) $$
$$a_0 = \frac1{\pi} \int_0^{\pi} dx \, \sin{x} = \frac{2}{\pi}$$
$$a_1 = \frac1{\pi} \int_0^{\pi} dx \, \sin{x} \cos{x} = 0$$
$$b_1 = \frac1{\pi} \int_0^{\pi} dx \, \sin^2{x} = \frac12$$
For $n \gt 1$:
$$\begin{align}a_n &= \frac1{\pi} \int_0^{\pi} dx \, \sin{x} \cos{n x}\\ &= \frac1{2 \pi} \int_0^{\pi} dx \, \left [\sin{(n+1) x} - \sin{(n-1) x} \right ]\\ &= \frac1{2 \pi} \left [\frac{1-(-1)^{n+1}}{n+1} - \frac{1-(-1)^{n-1}}{n-1} \right ] \\ &= -\frac{1+(-1)^n}{\pi (n^2-1)} \end{align}$$
$$\begin{align}b_n &= \frac1{\pi} \int_0^{\pi} dx \, \sin{x} \sin{n x}\\ &= \frac1{2 \pi} \int_0^{\pi} dx \, \left [\cos{(n-1) x} - \cos{(n+1) x} \right ]\\ &= 0 \end{align}$$
Thus,
$$g(x) = \frac1{\pi} + \frac12 \sin{x} - \frac{2}{\pi} \sum_{n=1}^{\infty} \frac{\cos{2 n x}}{4 n^2-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
integration $\int \frac{-4x+5}{\sqrt{12x-4x^2-8}}$
$$\int \frac{-4x+5}{\sqrt{12x-4x^2-8}}$$
My attempt:
$-4x^2+12x-8=-(2x-3)^2+1$
$$\begin{align}\int \frac{-4x+5}{\sqrt{1-(2x-3)^2}}\,dx&=\int \frac{-4x}{\sqrt{1-(2x-3)^2}}\,dx+\int \frac{5}{\sqrt{1-(2x-3)^2}}\,dx\\
&=
-4\int \frac{x}{\sqrt{1-(2x-3)^2}}\,dx+5\int \frac{1}{\sqrt{1-(2x-3)^2}}\,dx\\
&=-4\int \frac{x}{\sqrt{1-(2x-3)^2}}\,dx+5\arcsin(2x-3)
\end{align}$$
$$-4\int \frac{x}{\sqrt{1-(2x-3)^2}}\,dx$$
$u=2x-3 \rightarrow du=2dx$
$x=\frac{u+3}{2}$
$$-4\cdot\frac{1}{2}\int \frac{\frac{u+3}{2}}{\sqrt{1-(u)^2}}du=-\int \frac{u+3}{\sqrt{1-(u)^2}}du$$
$s=u^2$
$ds=2udu$
$$-\frac{1}{2}\int \frac{1}{\sqrt{1-s}}ds+\int \frac{1}{\sqrt{1-s}}ds=-\sqrt{1-(2x-3)^2}+5\arcsin(2x-3)-3\arcsin(2x-3)+c=-\sqrt{1-(2x-3)^2}+2\arcsin(2x-3)+c$$
But it is wrong according to Wolfram
| Noting that $-4x+5=-2(2x-3)-1$, you get:
$$\int \frac {-2(2x-3)}{\sqrt{1-(2x-3)^2}}\,dx -\int\frac{1}{\sqrt{1-(2x-3)^2}}\,dx$$
In the first, let $u=1-(2x-3)^2$, then $du=-4(2x-3)\,dx$ and in the second, let $v=2x-3$ and you get the integrals:
$$\frac{1}{2}\int \frac{du}{\sqrt{u}} - \frac{1}{2}\int \frac{dv}{\sqrt{1-v^2}}$$
This gives you $\sqrt{u} + \frac{1}{2}\arcsin(v)=\sqrt{12x-4x^2-8} - \frac{1}{2}\arcsin(2x-3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
limit as n goes to infinity of $\frac{\sqrt{n^3-3}-\sqrt{n^3+2n^2+3}}{\sqrt{n+2}}.$
How do you go about solving $$\lim_{n\to\infty}\frac{\sqrt{n^3-3}-\sqrt{n^3+2n^2+3}}{\sqrt{n+2}}.$$
I know that I have to fix the top so that it is not $(\infty - \infty$),
but if I multiple it by
$$\frac{\sqrt{n^3-3}+\sqrt{n^3+2n^2+3}}{\sqrt{n^3-3}+\sqrt{n^3+2n^2+3}},$$
the bottom part becomes very ugly and extremely hard to deal with.
| When you are at the stage
$$
\lim_{n\to\infty}\frac{-2n^2-6}{
\sqrt{n+2}\,(\sqrt{n^3-3}+\sqrt{n^3+2n^2+3})
}
$$
pull $n^2$ from the numerator, $n$ from $n+2$ and $n^3$ from the other two radicals, so you have
$$
\lim_{n\to\infty}\frac{n^2\left(-2-\dfrac{6}{n^2}\right)}{
\sqrt{n}\,\sqrt{1+\dfrac{2}{n}}\,
\sqrt{n^3}\left(
\sqrt{1-\dfrac{3}{n^3}}+\sqrt{1+\dfrac{2}{n}+\dfrac{3}{n^3}}
\right)
}
$$
Cancel $n^2$ with $\sqrt{n}\,\sqrt{n^3}$ and you're done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Real numbers $x,y$ satisfies $x^2+y^2=1.$If the minimum and maximum value of the expression $z=\frac{4-y}{7-x}$ are $m$ and $M$ Real numbers $x,y$ satisfies $x^2+y^2=1.$If the minimum and maximum value of the expression $z=\frac{4-y}{7-x}$ are $m$ and $M$ respectively,then find $2M+6m.$
Let $x=\cos\theta$ and $y=\sin\theta$,because $\sin^2\theta+\cos^2\theta=1$.
Then we need to find the minimum and maximum value of the expression $\frac{4-\sin\theta}{7-\cos\theta}$.
I differentiated it and equated it to zero to find the critical points or points of extrema.
They are $\theta_1=\arcsin(\frac{1}{\sqrt{65}})-\arctan(\frac{7}{4})$ and $\theta_2=\arccos(\frac{1}{\sqrt{65}})+\arctan(\frac{4}{7})$
I found $\frac{4-\sin\theta_1}{7-\cos\theta_1}$ and $\frac{4-\sin\theta_2}{7-\cos\theta_2}$.
$\frac{4-\sin\theta_1}{7-\cos\theta_1}=\frac{3}{4}$ and $\frac{4-\sin\theta_2}{7-\cos\theta_2}=\frac{5}{12}$
This method is full of lengthy calculations.I want to know is there an elegant solution possible for this problem which is short and easy.
| HINT:
Let $\dfrac{4-\sin \theta}{7-\cos \theta}=u$
Use Weierstrass Substitution $t=\tan\dfrac\theta2$ to form a Quadratic Equation in $t$ on rearrangement.
As $t$ is real, the discriminant must be non-negative
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1603335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
} |
What is the ratio of empty to filled volume of the glass? The base diameter of a glass is $20$% smaller than the diameter at the rim. The glass is filled to half of the height. Then what is the ratio of empty to filled volume of the glass ?
| Let $r_1, r_2$ and $r_3$ be the top, bottom and middle radius respectively, and h be the hight of the cone.
Volume of glass= $\frac{1}{3}\pi h (r_1+r_2+r_1 r_2)$.
Let $r_1=10$, then $r_2=8$ and $r_3=9$.
So the ratio of volume of empty portion of glass to volume of filled portion of glass,
$=\frac{\frac{1}{3} \pi\frac{h}{2} (r_1+r_3+r_1 r_3)}{\frac{1}{3} \pi \frac{h}{2}(r_2+r_3+r_2 r_3)}$
$=\frac{\frac{1}{3} \pi (10+9+10.9)}{\frac{1}{3} \pi (8+9+9.8)}$
$=\frac{271}{217}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1603507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Find $\int_{1}^{2}\frac{x-1}{x^2\sqrt{x^2+(x-1)^2}}$ Find $\int_{1}^{2}\frac{x-1}{x^2\sqrt{x^2+(x-1)^2}}$
I tried to solve it by using the property $\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$
Let $I=\int_{1}^{2}\frac{(x-1)dx}{x^2\sqrt{x^2+(x-1)^2}}$
$I=\int_{1}^{2}\frac{(2-x)dx}{(3-x)^2\sqrt{(3-x)^2+(2-x)^2}}$
But this does not seem to work here.I am stuck.What should i do?
| First step is put $x = u+\dfrac{1}{2}\Rightarrow \sqrt{x^2+(x-1)^2}=\sqrt{\left(u+\dfrac{1}{2}\right)^2+\left(u-\dfrac{1}{2}\right)^2}=\sqrt{2u^2+\dfrac{1}{2}}=\sqrt{2}\cdot \sqrt{u^2+\dfrac{1}{4}}$. Second step is to put $u =\dfrac{1}{2}\tan \theta$. I think we can go further at this point.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Now am I doing induction correctly? Recursion: $L_n = L_{n-1} + n$ where $L_0 = 1$.
We guess that solution is $L_n = \frac{n(n+1)}{2} + 1$.
Base case: $L_0 = \frac{0(0+1)}{2} + 1 = 1$ is true.
Inductive step: Assume $L_n = \frac{n(n+1)}{2} + 1$ is true for some $n$. We will show that $L_{n+1} = \frac{(n+1)(n+2)}{2} + 1$ given that $L_n = L_{n-1} + n$ is true.
$L_{n+1} = \frac{(n+1)(n+2)}{2} + 1 = L_n + (n+1)$
$L_n = \frac{(n+1)(n+2)}{2} + 1 - (n+1)$
$L_n = \frac{(n+1)(n+2)}{2} + \frac{2}{2} - \frac{2n+2}{2} = \frac{n^2+3n+2 + 2 - 2n - 2}{2}$
$L_n = \frac{n^2+n+2}{2} = \frac{n^2+n}{2} + 1 = \frac{n(n+1)}{2} + 1$
This completes the proof.
Is everything in place for a correct induction proof? Is anything wrong? Backwards? Unclear? Awkward?
|
Base case: $L_0 = \frac{0(0+1)}{2} + 1 = 1$ is true.
Inductive step: Assume $L_n = \frac{n(n+1)}{2} + 1$ is true for some $n$. We will show that $L_{n+1} = \frac{(n+1)(n+2)}{2} + 1$ given that $L_n = L_{n-1} + n$ is true.
Fine.
$L_{n+1} = \frac{(n+1)(n+2)}{2} + 1 = L_n + (n+1)$
Don't start with $L_{n+1}=\frac{(n+1)(n+2)}{2}+1$ which is what you have to prove.
$$\begin{align}L_{n+1}&=L_n+n+1\\&=\frac{n(n+1)}{2}+1+n+1\\&=\frac{n(n+1)}{2}+\frac{2(n+1)}{2}+1\\&=\frac{n+1}{2}(n+2)+1\\&=\frac{(n+1)(n+2)}{2}+1\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a function satisfying $f(x+y^3)=f(x)+f(y)^3$
Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a function satisfying $f(x+y^3)=f(x)+f(y)^3$ for all $x,y\in\mathbb{R}$. If $f'(0)\ge0$, find $f(10)$.
If $x=y=0$, $f(0+0)=f(0)+f(0)$. So $f(0)=0$.
If $x=0$, $f(0+y^3)=f(0)+f(y)^3$. So $f(y^3)=f(y)^3$.
So $f(x)=x^n$ or $f(x)=\mathrm{constant}$.
But I am not getting the given answer if $f(x)=x^n$.
| Since $f'(0)$ exists, we know that $f'(0) = \displaystyle\lim_{y \to 0}\dfrac{f(y)-f(0)}{y-0} = \lim_{y \to 0}\dfrac{f(y)}{y}$ exists.
Then, for any $x \in \mathbb{R}$, we have $\dfrac{f(x+y^3)-f(x)}{y^3} = \dfrac{f(y)^3}{y^3} = \left(\dfrac{f(y)}{y}\right)^3 \to f'(0)^3$ as $y \to 0$.
Therefore, $f'(x)$ exists for all $x \in \mathbb{R}$ and $f'(x) = f'(0)^3$ for all $x \in \mathbb{R}$.
Then, since $f(0) = 0$, we have $f(x) = f'(0)^3x$ for all $x \in \mathbb{R}$.
Setting $x = 0$ gives us $f'(0) = f'(0)^3$, and thus, $f'(0) = -1, 0, 1$.
Since we are given $f'(0) \ge 0$, we must have $f'(0) = 0$ or $1$.
Thus, $f(x) = 0$ or $f(x) = x$, (both satisfy the constraints of the problem). So there are two possible values for $f(10)$, specifically, $f(10) = 0$ and $f(10) = 10$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1605162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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If in a triangle $\alpha$ is pi/3 then $\frac 1c+\frac 1b\ge\frac2a$. If in a triangle $\alpha$ is pi/3 then $\frac 1c+\frac 1b\ge\frac2a$.
Since $\alpha$ is pi/3 that means $ b\le a \le c $.
Then i used AM and HM ,so $\frac 1c+\frac 1b\ge\frac {4}{c+b} $ .
and since $\alpha$ is pi/3 and $ b\le a \le c $ then $ \frac b2 + \frac c2 \le a $ ,then we replace c+b with 2a and get what we wanted.
This is my work i think it's not a correct proof( or it's ok?),any idea how(else?) to prove this.
| From the Law of Sines, we know $\frac{\sin\alpha}{a} = \frac{\sin\beta}{b} = \frac{\sin\gamma}{c}$, so
$$ \frac{1}{b}+\frac{1}{c} = \sin\alpha\left(\frac{1}{\sin\beta}+\frac{1}{\sin\gamma}\right)\frac{1}{a}.$$
It suffices to prove
$$\sin\alpha\left(\frac{1}{\sin\beta}+\frac{1}{\sin\gamma}\right)\ge 2.$$
Since $\alpha = \frac{\pi}{3}$, we have $\beta+\gamma = \frac{2\pi}{3}$, and so $\sin\alpha = \sin\left(\frac{\beta+\gamma}{2}\right)$. As such, it suffices to prove
$$\sin\left(\frac{\beta+\gamma}{2}\right)\left(\frac{1}{\sin\beta}+\frac{1}{\sin\gamma}\right)\ge 2 \iff \frac{1}{2}\left(\frac{1}{\sin\beta}+\frac{1}{\sin\gamma}\right)\ge\frac{1}{\sin\left(\frac{\beta+\gamma}{2}\right)} $$
which follows from Jensen's inequality, since $\frac{1}{\sin}$ is convex on $(0,\pi)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the maximum value of $\sqrt6 xy + 4yz$ given $x^2 + y^2 + z^2 = 1?$ Problem:
Let $x$,$y$ and $z$ be real numbers such that $x^2 + y^2 + z^2 = 1.$ Find the maximum possible value of $\sqrt6 xy + 4yz$
I don't know how to proceed with the question. Applying AM-GM inequality doesn't work because the second equation gives RHS dependent on $y$.
| Let $a,b\in\mathbb{R}$. We shall maximize $axy+byz$ subject to $x^2+y^2+z^2=1$ ($x,y,z\in\mathbb{R}$).
By Cauchy-Schwarz, $axy+byz=(ax+bz)y\leq \left(\sqrt{a^2+b^2}\sqrt{x^2+z^2}\right)|y|$. By AM-GM, $\sqrt{x^2+z^2}\,|y|\leq \frac{\left(x^2+z^2\right)+y^2}{2}=\frac{1}{2}$. That is, $axy+byz\leq \frac{\sqrt{a^2+b^2}}{2}$. The equality holds iff $\left(x,y,z\right)=\pm\left(\frac{a}{\sqrt{2\left(a^2+b^2\right)}},\frac{1}{\sqrt{2}},\frac{b}{\sqrt{2\left(a^2+b^2\right)}}\right)$.
(Similarly, we also have the lower bound $axy+byz\geq -\frac{\sqrt{a^2+b^2}}{2}$. The equality holds iff $\left(x,y,z\right)=\pm\left(\frac{a}{\sqrt{2\left(a^2+b^2\right)}},-\frac{1}{\sqrt{2}},\frac{b}{\sqrt{2\left(a^2+b^2\right)}}\right)$.)
| {
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"source": "stackexchange",
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Limit of: $\lim_{n \to \infty}(\sqrt[3]{n^3+\sqrt{n}}-\sqrt[3]{n^3-1})\cdot \sqrt{(3n^3+1)}$ I want to find the limit of: $$\lim_{n \to \infty}(\sqrt[3]{n^3+\sqrt{n}}-\sqrt[3]{n^3-1})\cdot \sqrt{(3n^3+1)}$$
I tried expanding it by $$ \frac{(n^3+n^{1/2})^{1/3}+(n^3-1)^{1/3}}{(n^3+n^{1/2})^{1/3}+(n^3-1)^{1/3}} $$
but it didn't help much. Wolfram says the answer is $\frac{3^{1/2}}{3}$.
Any help would be greatly appreciated.
| Remember the identity: $a^3-b^3=(a-b)\left(a^2+ab+b^2\right)$.
$$\left(\sqrt[3]{n^3+\sqrt{n}}-\sqrt[3]{n^3-1}\right)\sqrt{3n^3+1}$$
$$=\frac{\left(\left(n^3+\sqrt{n}\right)-\left(n^3-1\right)\right)\sqrt{3n^3+1}}{\sqrt[3]{\left(n^3+\sqrt{n}\right)^2}+\sqrt[3]{\left(n^3+\sqrt{n}\right)\left(n^3-1\right)}+\sqrt[3]{\left(n^3-1\right)^2}}$$
$$=\frac{\left(1+\frac{1}{\sqrt{n}}\right)\sqrt{3+\frac{1}{n^3}}}{\sqrt[3]{\left(1+\frac{1}{n^2\sqrt{n}}\right)^2}+\sqrt[3]{\left(1+\frac{1}{n^2\sqrt{n}}\right)\left(1-\frac{1}{n^3}\right)}+\sqrt[3]{\left(1-\frac{1}{n^3}\right)^2}}$$
$$\stackrel{n\to \infty}\to \frac{(1)\sqrt{3}}{1+1+1}=\frac{1}{\sqrt{3}}$$
| {
"language": "en",
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"source": "stackexchange",
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Prove that if $a,b,$ and $c$ are positive real numbers, then $\frac{a^3}{a^3+2b^3}+\frac{b^3}{b^3+2c^3}+\frac{c^3}{c^3+2a^3} \geq 1.$
Prove that if $a,b,$ and $c$ are positive real numbers, then $$\dfrac{a^3}{a^3+2b^3}+\dfrac{b^3}{b^3+2c^3}+\dfrac{c^3}{c^3+2a^3} \geq 1.$$
This question seems hard since we aren't given any other information about $a,b,c$. I think expanding it might got somewhere, but I don't see Cauchy-Schwarz or AM-GM leading anywhere.
| Let $x=a^3$, $y=b^3$, $z=c^3$. By Cauchy-Schwarz inequality:
$$\left(\sum_{\text{cyc}}\frac{x}{x+2y}\right)\ge \frac{(x+y+z)^2}{\left(\sum_{\text{cyc}} x(x+2y)\right)}=1$$
Equality holds iff $\frac{\frac{x}{x+2y}}{x(x+2y)}=\frac{\frac{y}{y+2z}}{y(y+2z)}=\frac{\frac{z}{z+2x}}{z(z+2x)}$, i.e. iff $x+2y=y+2z=z+2x$,
i.e. iff $y=2z-x=2x-z$, i.e. iff $x=y=z$, i.e. iff $a=b=c$.
It's quite a standard trick for similar inequalities.
First, notice that $a^2+b^2+c^2\ge ab+bc+ca$ for all $a,b,c\in\mathbb R$, because this is equivalent to $\frac{1}{2}\left((a-b)^2+(b-c)^2+(c-a)^2\right)\ge 0$, which is true (or by the Rearrangement Inequality).
E.g., if $a,b,c>0$ and $n\in\mathbb Z^+$, then:
$$\sum_{\text{cyc}}\frac{a}{b+nc}\ge \frac{(a+b+c)^2}{\sum_{\text{cyc}}a(b+nc)}=\frac{\sum_{\text{cyc}}a^2+2\sum_{\text{cyc}}ab}{(n+1)\sum_{\text{cyc}}ab}$$
$$\ge \frac{\sum_{\text{cyc}}ab+2\sum_{\text{cyc}}ab}{(n+1)\sum_{\text{cyc}}ab}=\frac{3}{n+1}$$
$n=1$ gives Nesbitt's Inequality.
| {
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"timestamp": "2023-03-29T00:00:00",
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Volume between hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and line $x = 2a$ around $y$ axis I'm trying to calculate the volume between the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and the line $x = 2a$ around the $y$ axis using two methods but I'm getting different answers:
*
*Using volume of a solid of revolution with circular-ring method (the usual formula is multiplied by $2$ because half the interval is defined for the integral):
$$
\begin{eqnarray}
V &=& 2 \pi \int_0^{b\sqrt{3}} [(2a)^2 - (\frac{a}{b}\sqrt{y^2+b^2})^2] \, \textrm{d}y \\
&=& \frac{2 \pi a^2}{b^2} \left[3 b^2 y - \frac{y^3}{3}\right]_0^{b \sqrt{3}} \\
&=& 4 \sqrt{3} \pi a^2 b \\
\end{eqnarray}
$$
*Using volume of a solid of revolution with cylindrical-shell method:
$$
\begin{eqnarray}
V &=& 2 \pi \int_a^{2a} [x (\frac{2b}{a} \sqrt{x^2 - a^2})] \, \textrm{d}x \\
&=& 4 \pi a b \left[ \frac{(x^2 - a^2)^{3/2}}{3 a^3} \right]_a^{2a} \\
&=& 4 \sqrt{3} \pi a b \\
\end{eqnarray}
$$
The second answer is the one from my book but I'd really like to understand why the first answer is different.
| The second integral is wrong.
\begin{equation*}
\int \left(x\left(\frac{2b}{a}\sqrt{x^2-a^2}\right)\right)\,dx
= \frac{2b}{3a}\int 3x\sqrt{x^2-a^2}\,dx
= \frac{2b}{3a}(x^2-a^2)^{3/2},
\end{equation*}
so the definite integral is
\begin{equation*}
V = 2\pi\int_a^{2a} \left(x\left(\frac{2b}{a}\sqrt{x^2-a^2}\right)\right)\,dx
= \frac{4b\pi}{3a}(x^2-a^2)^{3/2}\big\lvert_a^{2a}
= 4 \sqrt{3} \pi a^2 b .
\end{equation*}
| {
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Prove the inequality $\tan{\frac{\pi\sin{x}}{4\sin{\alpha}}}+\tan{\frac{\pi\cos{x}}{4\cos{\alpha}}} > 1$
Prove the inequality $$\tan{\dfrac{\pi\sin{x}}{4\sin{\alpha}}}+\tan{\dfrac{\pi\cos{x}}{4\cos{\alpha}}} > 1$$
for any $x, \alpha$ with $0 \leq x \leq \dfrac{\pi}{2}$ and $\dfrac{\pi}{6} < \alpha < \dfrac{\pi}{3}$.
The best idea I had was to use the identity $\tan(A+B) = \dfrac{\tan(A)+\tan(B)}{1-\tan(A)\tan(B)}$. Thus we may say that $\tan \left({\dfrac{\pi \sin{x}}{4 \sin{\alpha}}}+\dfrac{\pi\cos{x}}{4\cos{\alpha}} \right) \left(1-\tan{\dfrac{\pi\sin{x}}{4\sin{\alpha}}} \tan{\dfrac{\pi\cos{x}}{4\cos{\alpha}}} \right) = \tan{\dfrac{\pi\sin{x}}{4\sin{\alpha}}}++\tan{\dfrac{\pi\cos{x}}{4\cos{\alpha}}}.$ Then I just have to show its greater than $1$ on these intervals.
| We could also do this without casework using Jensen inequality, and using the fact that the first derivative of $\tan(x)$ is always positive and the second derivative is positive when $0 < tan(x) < \frac{\pi}{2}$.
So, by Jensen we have,
$\tan{\left(\dfrac{\pi\sin{x}}{4\sin{\alpha}}\right)}+\tan{\left( \dfrac{\pi\cos{x}}{4\cos{\alpha}}\right)}
\geq 2 \tan{\frac{\pi}{8} \left(\frac{\sin{x}}{\sin{\alpha}}+\frac{\cos{x}}{\cos{\alpha}}\right) } =2\tan{\frac{\pi}{8} \left(\frac{\sin(x+ \alpha)}{\sin(\alpha)\cos{\alpha}}\right)} \geq 2\tan{\frac{\pi}{8}\left(\frac{1}{cos \alpha}\right)} \geq 2\tan {\frac{\pi}{8}\left(\frac{1}{\frac{1}{2}}\right)} \geq 2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Extreme of $\cos A\cos B\cos C$ in a triangle without calculus.
If $A,B,C$ are angles of a triangle, find the extreme value of $\cos A\cos B\cos C$.
I have tried using $A+B+C=\pi$, and applying all and any trig formulas, also AM-GM, but nothing helps.
On this topic we learned also about Cauchy inequality, but I have no experience with it.
The answer according to Mathematica is when $A=B=C=60$.
Any ideas?
| Assume without loss of generality that $\displaystyle A , B, C $ belongs to the first quadrant.By the identity $\displaystyle sen^2x+cos^2x=1$, we can see that:
\begin{equation*}
tanx+cotx=\frac{1}{senxcosx}
\end{equation*}
So:
\begin{equation}
\frac{sen\frac{\beta}{2}sen\frac{\gamma}{2}}{sen\frac{\alpha}{2}cos\frac{\alpha}{2}}=\left(tan\frac{\alpha}{2}+cot\frac{\alpha}{2}\right)sen\frac{\beta}{2}sen\frac{\gamma}{2}
\end{equation}
\begin{equation}
\frac{cos\frac{\beta}{2}cos\frac{\gamma}{2}}{sen\frac{\alpha}{2}cos\frac{\alpha}{2}}=\left(tan\frac{\alpha}{2}+cot\frac{\alpha}{2}\right)cos\frac{\beta}{2}cos\frac{\gamma}{2}
\end{equation}
Add so:
$\\ \\ \displaystyle \left(tan\frac{\alpha}{2}+cot\frac{\alpha}{2}\right)sen\frac{\beta}{2}sen\frac{\gamma}{2}+\left(tan\frac{\alpha}{2}+cot\frac{\alpha}{2}\right)cos\frac{\beta}{2}cos\frac{\gamma}{2}=\frac{cos\left(\frac{\beta}{2}-\frac{\gamma}{2}\right)}{sen\frac{\alpha}{2}cos\frac{\alpha}{2}}\leq \frac{1}{sen\frac{\alpha}{2}cos\frac{\alpha}{2}} \\ \\ $
We conclude that:
\begin{equation}
\left(tan\frac{\alpha}{2}+cot\frac{\alpha}{2}\right)\left(cos\frac{\beta}{2}cos\frac{\gamma}{2}+sen\frac{\beta}{2}sen\frac{\gamma}{2}\right)\leq \frac{1}{sen\frac{\alpha}{2}cos\frac{\alpha}{2}}
\end{equation}
Apply Ravi transfomation, we get:
\begin{equation*}
\left(\sqrt{\frac{xy}{z(x+y+z)}}+\sqrt{\frac{z(x+y+z)}{xy}}\right)\times
\end{equation*}
\begin{equation*}
\left( \sqrt{\frac{x(x+y+z)}{(x+z)(x+y)}}\sqrt{\frac{y(x+y+z)}{(x+y)(y+z)}}+ \sqrt{\frac{yz}{(x+z)(x+y)}} \sqrt{\frac{xz}{(x+y)(y+z)}}\right)
\end{equation*}
\begin{equation}
\leq \frac{(x+y)(y+z)}{\sqrt{xyz(x+y+z)}}
\end{equation}
Its easy see that:
\begin{equation}
(xy+z(x+y+z))\left((x+y+z)\sqrt{xy}+z\sqrt{xy}\right)\leq (x+y)^2(y+z)\sqrt{(x+z)(y+z)}
\end{equation}
The above inequality is equivalent to:
\begin{equation}
(x+z)(y+z)\sqrt{xy}\left(x+y+2z\right)\leq (x+y)^2(y+z)\sqrt{(x+z)(y+z)}
\end{equation}
Which with due cancellation reduces to the inequality below:
\begin{equation}
(x+z)\sqrt{xy}\left(x+y+2z\right)\leq (x+y)^2\sqrt{(x+z)(y+z)}
\end{equation}
By symmetry we conclude the inequalities:
\begin{equation}
(x+y)\sqrt{yz}\left(2x+y+z\right)\leq (y+z)^2\sqrt{(x+z)(x+y)}
\end{equation}
\begin{equation}
(y+z)\sqrt{xz}\left(x+2y+z\right)\leq (x+z)^2\sqrt{(y+z)(x+y)}
\end{equation}
Multiplying, we get:
\begin{equation}
xyz(2x+y+z)(x+2y+z)(x+y+2z)\leq [(x+z)(x+y)][(y+z)(x+y)][(x+z)(y+z)]
\end{equation}
Supose $\displaystyle xy+xz+yz=1$, note that x,y and z will be cotangents of angles of an acute triangle, hence it follows
$\\ \displaystyle [(x+z)(x+y)][(y+z)(x+y)][(x+z)(y+z)]=$
$\\ \displaystyle [x^2+xy+xz+yz][y^2+xy+xz+yz][z^2+xy+xz+yz]=$
$\\ \displaystyle [x^2+1][y^2+1][z^2+1]=[cot^2\alpha'+1][cot^2\beta'+1][cot^2\gamma'+1]=csc^2\alpha' csc^2\beta' csc^2\gamma' \\ \\$
From where we can see that:
\begin{equation*}
\cot\alpha \cot\beta \cot\beta \left(2\cot\alpha+\cot\beta+\cot\beta\right)\left(\cot\alpha+2\cot\beta+\cot\beta\right)\left(\cot\alpha+\cot\beta+2\cot\beta\right) \leq
\end{equation*}
\begin{equation}
csc^2\alpha' csc^2\beta' csc^2\gamma'
\end{equation}
Extracting the cube root on both sides of the inequality, we have:
\newpage
\begin{equation*}
(\cot\alpha \cot\beta \cot\beta)^{\frac{1}{3}} \left(2\cot\alpha+\cot\beta+\cot\beta\right)^{\frac{1}{3}}\left(\cot\alpha+2\cot\beta+\cot\beta\right)^{\frac{1}{3}}\left(\cot\alpha+\cot\beta+2\cot\beta\right)^{\frac{1}{3}}
\end{equation*}
\begin{equation}
\leq csc^{\frac{2}{3}}\alpha' csc^{\frac{2}{3}}\beta' csc^{\frac{2}{3}}\gamma'
\end{equation}
Look at the expression:
$$\\ \\ \sqrt[3]{(2\times a+b+c)(a+2\times b+c)(a+b+2\times c)}$$
So:
$$ \displaystyle a+a+b+c\geq 4\sqrt[4]{a^2bc}$$
$$ \displaystyle a+b+b+c\geq 4\sqrt[4]{ab^2c}$$
$$ \displaystyle a+b+c+c\geq 4\sqrt[4]{abc^2}$$
And see, taking the product:
$$\\ \\ (2\times a+b+c)(a+2\times b+c)(a+b+2\times c)\geq 64\sqrt[4]{4a^4b^4c^4}=64abc$$
And so:
$$\\ \\ (2\times a+b+c)(a+2\times b+c)(a+b+2\times c)\geq 64abc$$
Extracting the cube root, we have:
$$\\ \\ \sqrt[3]{(2\times a+b+c)(a+2\times b+c)(a+b+2\times c)}\geq4\sqrt[3]{abc} \\ \ \ $$
Make the replacement $\displaystyle a=\cot(\alpha),b=\cot(\beta),c=\cot(\gamma)$, and so
$$\\ \\ \sqrt[3]{( 2\cot(\alpha)+\cot(\beta)+\cot(\gamma) )(\cot(\alpha)+2\cot(\beta)+\cot(\gamma))(\cot(\alpha)+\cot(\beta)+2\cot(\gamma))}\geq$$
$$4\sqrt[3]{\cot(\alpha)\cot(\beta)\cot(\gamma)} \\ \ \ $$
Multiplying by
$\displaystyle (\cot\alpha \cot\beta \cot\beta)^{\frac{1}{3}}$, we get:
\newpage
\begin{equation*}
4(\cot\alpha \cot\beta \cot\beta)^{\frac{2}{3}} \leq
\end{equation*}
\begin{equation}
(\cot\alpha \cot\beta \cot\beta)^{\frac{1}{3}}\left(2\cot\alpha+\cot\beta+\cot\beta\right)^{\frac{1}{3}}\left(\cot\alpha+2\cot\beta+\cot\beta\right)^{\frac{1}{3}}\left(\cot\alpha+\cot\beta+2\cot\beta\right)^{\frac{1}{3}}
\end{equation}
We arrive by transitivity to the inequality below:
\begin{equation}
4(\cot\alpha \cot\beta \cot\beta)^{\frac{2}{3}} \leq \csc^{\frac{2}{3}}\alpha \csc^{\frac{2}{3}}\beta \csc^{\frac{2}{3}}
\end{equation}
We can assume without loss of generality that the angles are in the first quadrant, so we can extract the root and preserve the sign of inequality, because by this hypothesis all terms are positive... our inequality is equivalent to the inequality required in the problem.Done.For more solution enter to the link https://www.overleaf.com/read/qyrxbsjhhjst
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 8,
"answer_id": 2
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Is there an easy way to compute the determinant of matrix with 1's on diagonal and a's on anti-diagonal? \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & a \\ 0 & 1 & 0 & 0 & a & 0 \\0 & 0 & 1 & a & 0 & a \\0 & 0 & a & 1 & 0 & a \\0 & a & 0 & 0 & 1 & 0 \\a & 0 & 0 & 0 & 0 & 1 \\ \end{bmatrix}
Thanks
| Partition the matrix into blocks $\begin{bmatrix}A&B\\C&D\end{bmatrix}$, where
$A = \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$, $B = \begin{bmatrix}0&0&a\\0&a&0\\a&0&a\end{bmatrix}$, $C = \begin{bmatrix}0&0&a\\0&a&0\\a&0&0\end{bmatrix}$, $D = \begin{bmatrix}1&0&a\\0&1&0\\0&0&1\end{bmatrix}$.
The block matrix determinant formula states that
$\det \begin{bmatrix}A&B\\C&D\end{bmatrix} = \det(A)\det(D-CA^{-1}B)$.
Trivially, $A = I_3$ is the $3 \times 3$ identity matrix, so $\det(A) = 1$. Also, $A^{-1} = I_3$.
Thus, $D-CA^{-1}B = D-CB$ $= \begin{bmatrix}1&0&a\\0&1&0\\0&0&1\end{bmatrix} - \begin{bmatrix}0&0&a\\0&a&0\\a&0&0\end{bmatrix}\begin{bmatrix}0&0&a\\0&a&0\\a&0&a\end{bmatrix}$ $= \begin{bmatrix}1&0&a\\0&1&0\\0&0&1\end{bmatrix} - \begin{bmatrix}a^2&0&a^2\\0&a^2&0\\0&0&a^2\end{bmatrix}$ $= \begin{bmatrix}1-a^2&0&a-a^2\\0&1-a^2&0\\0&0&1-a^2\end{bmatrix}$.
Since $D-CA^{-1}B$ is upper-triangular, its determinant is simply the product of the diagonal elements, i.e. $\det(D-CA^{-1}B) = (1-a^2)^3$.
Therefore, $\det \begin{bmatrix}A&B\\C&D\end{bmatrix} = \det(A)\det(D-CA^{-1}B) = 1 \cdot (1-a^2)^3 = (1-a^2)^3$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Probability that $2^a+3^b+5^c$ is divisible by 4
If $a,b,c\in{1,2,3,4,5}$, find the probability that $2^a+3^b+5^c$ is divisible by 4.
For a number to be divisible by $4$, the last two digits have to be divisible by $4$
$5^c= \_~\_25$ if $c>1$
$3^1=3,~3^2=9,~3^3=27,~3^4=81,~ 3^5=243$
$2^1=2,~2^2=4,~2^3=8,~2^4=16,~2^5=32$
Should I add all possibilities? Is there a simpler method?
| $2^{\color\red{a}}+3^\color\green{b}+5^\color\magenta{c}\equiv0\pmod4\iff$
*
*$\Big(\big(\color\red{a}=1\big)\wedge\big((\color\green{b}=2)\vee(\color\green{b}=4)\big)\Big)\vee$
*$\Big(\big(\color\red{a}\neq1\big)\wedge\big((\color\green{b}\neq2)\wedge(\color\green{b}\neq4)\big)\Big)$
Therefore, the probability is $\dfrac{\color\red1\cdot\color\green2\cdot\color\magenta5+(5-\color\red1)\cdot(5-\color\green2)\cdot\color\magenta5}{5\cdot5\cdot5}=\dfrac{14}{25}$
| {
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"source": "stackexchange",
"question_score": "11",
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Prove that $\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right)=\left(\frac{p}{3}\right)$ Why does $\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right)$ equal $\left(\frac{p}{3}\right)$?
| Use quadratic reciprocity :
$$ \left (\frac{p}{3} \right )=\left (\frac{3}{p} \right ) \cdot (-1)^{\frac{3-1}{2} \cdot \frac{p-1}{2}}$$
But we also know that : $$(-1)^{\frac{p-1}{2}}=\left (\frac{-1}{p} \right )$$ and thus the conclusion follows :
$$\left (\frac{p}{3} \right )=\left (\frac{3}{p} \right ) \cdot \left (\frac{-1}{p} \right )$$ by combining the two relations.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Writing sigma notation $\sum^n_{i=1} \frac {i}{2^i}$ in closed form What would be a way to find the closed form of $\frac {1}{2} + \frac {2}{4}+\frac {3}{8}+\cdots+\frac {n}{2^n}=\sum^n_{i=1} \frac {i}{2^i}=s$
I've looked at $\frac {s}{2}=\frac {1}{4} + \frac {2}{8}+\frac {3}{16}+\cdots+\frac {n}{2^{n+1}}$
And then $s-\frac {s}{2}=\frac {s}{2}=\frac {1}{2} + \frac {1}{4}+\frac {1}{8}+\cdots+\frac {1}{2^n}-\frac {n}{2^{n+1}}$
Any idea where to go from here?
Please, if you are sure of where I'm trying to go with this, just ask.
| $$
\sum_{i=1}^n i x^i = \sum_{i=1}^n x \frac d {dx} x^i = x \frac d {dx} \sum_{i=1}^n x^i = x \frac d {dx}\ \frac{x - x^{n+1}}{1-x} = \cdots
$$
(Now evaluate the derivative and then substitute $\dfrac 1 2$ for $x$.)
| {
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Solving the absolute value inequality $\big| \frac{x}{x + 4} \big| < 4$ I was given this question and asked to find $x$:
$$\left| \frac{x}{x+4} \right|<4$$
I broke this into three pieces:
$$
\left| \frac{x}{x+4} \right| = \left\{
\begin{array}{ll}
\frac{x}{x+4} & \quad x > 0 \\
-\frac{x}{x+4} & \quad -4 < x < 0 \\
\frac{x}{x+4} & \quad x < −4
\end{array}
\right.
$$
Solving,
$$4>\frac{x}{x+4}$$
$$4x+16>x$$
$$3x>-16$$
$$x>-\frac{16}{3}=-5.3$$
and
$$4>-\frac{x}{x+4}$$
$$4x+16>-x$$
$$5x>-16$$
$$x>-3.2$$
The answer is $x<-5.3$ and $x>-3.2$. What am I doing wrong?
| Your mistake was not considering the restrictions on $x$ when you solved the individual cases.
Case 1: $x \geq 0$
If $x > 0$, then $x + 4 > 0$, so the direction of the inequality is preserved if we multiply by $x + 4$. Hence,
\begin{align*}
\frac{x}{x + 4} & < 4\\
x & < 4(x + 4)\\
x & < 4x + 16\\
-16 & < 3x\\
-\frac{16}{3} & < x
\end{align*}
as you found. However, the restriction that $x \geq 0$ means $x \geq 0$ and $x > -\dfrac{16}{3} \Longrightarrow x \geq 0$. You did not account for the restriction $x \geq 0$.
Also, note that $-\dfrac{16}{3} \approx 5.3$. The statement you made that $-\dfrac{16}{3} = 5.3$ is false since the decimal approximation you used does not actually equal the fraction.
Case 2: $-4 < x < 0$
Since $-4 < x < 0$, $x + 4 > 0$. Thus, the direction of the inequality is preserved if we multiply both sides of the inequality by $x + 4$. Hence,
\begin{align*}
-\frac{x}{x + 4} & < 4\\
-x & < 4(x + 4)\\
-x & < 4x + 16\\
-16 & < 5x\\
-\frac{16}{5} & < x
\end{align*}
as you found. The restriction that $-4 < x < 0$ means $-4 < x < 0$ and $-\dfrac{16}{5} < x \Longrightarrow -\dfrac{16}{5} < x < 0$. You did not account for the restriction $x < 0$.
Case 3: $x < -4$
If $x < -4$, then $x + 4 < 0$, so the direction of the inequality is reversed if we multiply both sides of the inequality by $x + 4$, which you did not take into account.
\begin{align*}
\frac{x}{x + 4} & < 4\\
x & > 4(x + 4)\\
x & > 4x + 16\\
-16 & > 3x\\
-\frac{16}{3} & > x
\end{align*}
The restriction that $x < -4$ means that $x < -4$ and $x < -\dfrac{16}{3} \Longrightarrow x < -\dfrac{16}{3}$.
Combining the results of the three cases yields
$$x \geq 0~\text{or}-\frac{16}{5} < x < 0~\text{or}~x < -\frac{16}{3} \Longrightarrow x > -\frac{16}{5}~\text{or}~x < -\frac{16}{3}$$
Hence, the solution set is
$$S = \left(-\frac{16}{5}, \infty\right) \cup \left(-\infty, -\frac{16}{3}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1614060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
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How to solve this definite integral $\int_0^\pi \frac{\cos^9(x)}{\sin^3(x)+\cos^3(x)}dx$ I'm having trouble evaluating the following integral:
$$
\int^\pi_0 \frac{\cos^9(x)}{\sin^3(x)+\cos^3(x)}dx
$$
I tried to convert it into an algebraic function by multiplying the numerator and denominator by $\sec^{11}(x)$ and substituting $\tan(x)=t$ as
$$
\int^\pi_0 \frac{\cos^9(x)\cdot \sec^{11}(x)}{(\sin^3(x)+\cos^3(x))\cdot \sec^{11}(x)}dx
$$
$$
\int^\pi_0 \frac{\sec^2(x)}{(\tan^3(x)+1)\cdot (\tan^2(x)+1)^4}dx
$$
Substituting $\tan(x)=t$,
$$
\int^0_0 \frac{dt}{(t^3+1)\cdot (t^2+1)^4}
$$
But now both upper and lower limit become $0$ so apparently this is not the right approach, so how do I go about solving it?
| Assuming the usage of the Cauchy principle value, let's define
$$ I_1 := \int_0^\frac\pi2 \frac{\cos^9(x)}{\cos^3(x)+\sin^3(x)} {\rm d}x
\\ I_2 := \text{p. v.} \int_\frac\pi2^\pi \frac{\cos^9(x)}{\cos^3(x)+\sin^3(x)} {\rm d}x
$$
For $I_1$, by making a substitution and adding it to itself,
$$
I_1 = \int_0^\frac\pi2 \frac{\sin^9(x)}{\cos^3(x)+\sin^3(x)} {\rm d}x
\\ = \frac12 \int_0^\frac\pi2 \frac{\cos^9(x) +\sin^9(x)}{\cos^3(x)+\sin^3(x)} {\rm d}x
\\ = \frac12 \int_0^\frac\pi2 \left( \sin^6(x) + \cos^6(x) - \sin^3(x)\cos^3(x) \right) {\rm d}x
$$
For $I_2$, by making some substitutions,
$$
I_2 = \text{p. v.} \int_0^\frac\pi2 \frac{\sin^9(x)}{\sin^3(x)-\cos^3(x)} {\rm d}x
\\ = \text{p. v.} \int_0^\frac\pi2 \frac{-\cos^9(x)}{\sin^3(x)-\cos^3(x)} {\rm d}x
\\ = \frac12 \text{p. v.} \int_0^\frac\pi2 \frac{\sin^9(x)-\cos^9(x)}{\sin^3(x)-\cos^3(x)}
\\ = \frac12 \int_0^\frac\pi2 \left( \sin^6(x) + \cos^6(x) + \sin^3(x)\cos^3(x) \right) {\rm d}x
$$
Therefore,
$$
I_1 + I_2 = \text{p. v.} \int_0^\pi \frac{\cos^9(x)}{\cos^3(x)+\sin^3(x)} {\rm d}x
\\ = \int_0^\frac\pi2 \sin^6(x) + \cos^6(x) {\rm d}x
\\ = \int_0^\frac\pi2 \frac58 + \frac38 \cos(4x) {\rm d}x
\\ = \frac{5\pi}{16}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1614744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Which Has a Larger Volume a Cylinder or a Truncated Cone? The following link describes a programming problem. However, I am unable to work out the maths for the problem.
We are given $r$ – radius of lower base and $s$ – slant height. The figure can be cylinder or truncated cone. You have to find as largest volume as possible to carry oil respect to given information.
You are given two numbers that less than 100: radius $r$ of lower base and slant height $s$. The slant height is the shortest possible distance between the edges of two bases.
EDIT suggested :
Given $ \sqrt{(R-r)^2 + h^2 }$ and $ r,$ find $ h/r $ ratio.
$~~~~~~~~~~~~~~~~$
| If the constant slant height and base radius are $s, r$ respectively and the variable angle which the slanting face makes with the vertical is $\theta$ then the height and top radius of the frustum are
$$h=s\cos\theta, R=r+s\sin\theta$$
The volume of the frustum is $$V=\frac13 \pi h (R^2+Rr+r^2)$$
Now $$\frac{dh}{d\theta}=-s\sin\theta, \frac{dR}{d\theta}=s\cos\theta$$
Volume is maximum when
$$\frac{dV}{d\theta}=\frac13 \pi [\frac{dh}{d\theta}(R^2+Rr+r^2)+h(2R\frac{dR}{d\theta}+r\frac{dR}{d\theta})=0$$
$$-s\sin\theta (R^2+Rr+r^2)+s\cos\theta(2Rs\cos\theta+rs\cos\theta)=0$$
$$-s\sin\theta (3r^2+3rs\sin\theta+s^2\sin^2\theta)+s^2(1-\sin^2\theta)(3r+2s\sin\theta)=0$$
$$- (3r^2\sin\theta+3rs\sin^2\theta+s^2\sin^3\theta)+(3rs+2s^2\sin\theta-3rs\sin^2\theta-2s^2\sin^3\theta)=0$$
$$3x+(2-3x^2)y-6xy^2-3y^3=0$$
where $x=r/s$ and $y=\sin\theta$.
In general the cubic equation will have to be solved numerically. In the special case $s=r$ so that $x=1$ we have $3-y-6y^2-3y^3=0$. The only real root is $y=\sin\theta\approx 0.56298, \cos\theta\approx 0.82647$. Then the maximum volume is $V\approx 4.33$ as found in the link. Whereas for the cylinder $V=\pi\approx 3.14$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the domain of $\frac{1}{f(x)}$ in this case? Suppose $f(x) = \frac{x-1}{x-2}$ ,
Then $\frac{1}{f(x)} = \frac{1}{\frac{x-1}{x-2}}$
So would the domain be all real numbers excluding 1 and 2 or would the domain include 2? Since
$\frac{1}{\frac{x-1}{x-2}} = \frac{x-2}{x-1}$ ,
I guess I should also ask whether the denominator this case is $\frac{x-1}{x-2}$ or just $x-1$.
| One typically understands "$\frac{1}{f(x)}$" as the composition of the two functions $x \mapsto f(x)$ and $x \mapsto \frac{1}{x}$. So we have to apply $f(x)$ first, hence 2 is not in the domain.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1616241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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To find the sum of the series $\,1+ \frac{1}{3\cdot4}+\frac{1}{5\cdot4^2}+\frac{1}{7\cdot4^3}+\ldots$ The answer given is $\log 3$.
Now looking at the series
\begin{align}
1+ \dfrac{1}{3\cdot4}+\dfrac{1}{5\cdot4^2}+\dfrac{1}{7\cdot4^3}+\ldots &=
\sum\limits_{i=0}^\infty \dfrac{1}{\left(2n-1\right)\cdot4^n}
\\
\log 3 &=\sum\limits_{i=1}^\infty \dfrac{\left(-1\right)^{n+1}\,2^n}{n}
\end{align}
How do I relate these two series?
| hint
Consider the power series
$$
f(x) := \sum\limits_{n=1}^\infty \dfrac{x^n}{2n-1}
$$
Investigate what differential equation it satisfies. Looking at it suggests that this is
maybe easier for
$$
g(x) := \frac{f(x^2)}{x} = \sum\limits_{n=1}^\infty \dfrac{x^{2n-1}}{2n-1}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1616688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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} |
$\epsilon - \delta$ proof of limit of $\arctan \left(\frac {x+z}{y}\right)$ as $(x, y, z) \to (1, 2, -3)$ Find the limit and prove the limit for
$$\lim_{(x,y,z)\rightarrow(1,2,-3)}\arctan \left(\frac{x+z}{y}\right).$$
This is a homework problem. While I am comfortable with general epsilon-delta proof techniques, I am at a loss for inverse trigonometric inequalities. I was trying to use the trigonometric identity that
$$\arctan(\alpha)+\arctan(\beta)= \arctan\left(\frac{\alpha +\beta}{1-\alpha \beta}\right),$$
but trying to select the correct $\alpha$ and $\beta$, the argeument of the arctan became ugly. I could use some suggestions on arctan inequalities and identities.
| We have
$$\begin{align}
\left|\arctan\left(\frac{x+z}{y}\right)-\arctan\left(-1\right) \right|&=\left|\arctan\left(\frac{x+y+z}{x-y+z}\right)\right|\\\\
&=\left|\arctan\left(\frac{(x-1)+(y-2)+(z+3)}{x-y+z}\right)\right|\\\\
&\le\left|\frac{(x-1)+(y-2)+(z+3)}{x-y+z}\right|\\\\
&\le \frac{3\sqrt{(x-1)^2+(y-2)^2+(z+3)^2}}{|x-y+z|} \tag1
\end{align}$$
First we choose $\delta =1$ so that if $\sqrt{(x-1)^2+(y-2)^2+(z+3)^2}<\delta =1$, then $0<x<2$, $-3<-y<-1$, and $-4<z<-2$.
Therefore, with $\delta=1$, we see that $\frac{1}{|x-y+z|}<1$. Then, from $(1)$ we have
$$\begin{align}
\left|\arctan\left(\frac{x+z}{y}\right)-\arctan\left(-1\right) \right|&\le
3\sqrt{(x-1)^2+(y-2)^2+(z+3)^2} \\\\
&<\epsilon
\end{align}$$
whenever $\sqrt{(x-1)^2+(y-2)^2+(z+3)^2}<\min \left(1,\epsilon/3\right)$. And we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1617744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How to prove that all odd powers of two add one are multiples of three
For example
\begin{align}
2^5 + 1 &= 33\\
2^{11} + 1 &= 2049\ \text{(dividing by $3$ gives $683$)}
\end{align}
I know that $2^{61}- 1$ is a prime number, but how do I prove that $2^{61}+1$ is a multiple of three?
| If $k$ is odd, then $2^k$ is $2*2^{k-1}$. If $n$ is not a multiple of 3 then $n^2-1$ is since $n^2-1 = (n-1)(n+1)$ and at least one of them is a multiple of 3. This means that $2*2^{k-1} - 2$ is a multiple of $3$ as is $2*2^{k-1} + 1$.
Hope I helped.
| {
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"url": "https://math.stackexchange.com/questions/1618741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "41",
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"answer_id": 4
} |
Prove the excircle and right triangle inequality
Let $\triangle{ABC}$ be a right triangle with $\angle{A}= 90^{\circ}$. Let $D$ be the intersection of the internal angle bisector of $\angle{A}$ with side $BC$ and $I_a$ be the center of the excircle of the triangle ABC opposite to the vertex $A$. Prove that $$\dfrac{DA}{DI_a} \leq \sqrt{2}-1.$$
I tried using reflections since $\angle{BI_aC} = 45^{\circ}$ we know that $I_a$ lies on a circle with $A$ as the center. I couldn't really find out how to incorporate the ratio. In case anyone wants a diagram to see this I drew one (ignore the scratch work to the side of it).
| Not the best solution but still:
Let E be point where excircle touches triangle $ABC$.
From right triangle $ EDI_a $ we have $ DI_a \geq r_a $
Using $ r_a=\frac{A}{s-a}$ and $A= \frac{1}{2}DA \cdot b \sin 45 + \frac{1}{2}DA \cdot c \sin 45 = \frac{\sqrt2}{4} DA \cdot (b+c) $ where A is area of triangle $ABC$ we get
$ DI_a \geq r_a = \frac{A}{s-a} $
$ \iff DA \leq DI_a \sqrt{2} \cdot \frac{b+c-a}{b+c} $
$ \iff DA \leq DI_a \sqrt2 (1-\frac{a}{b+c}) $
And since $ \frac{a}{b+c} = \frac{1}{\sin \alpha + \cos \alpha } \geq \frac{1}{\sqrt2}$ (Because $\sin \alpha + \cos \alpha \leq \sqrt2$)
$ \iff DA \leq DI_a \sqrt2(1-\frac{a}{b+c}) \leq DI_a (\sqrt2-1)$
Q.E.D.
| {
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"url": "https://math.stackexchange.com/questions/1618916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What does the homogeneous system of equations represent under certain conditions?
Consider the following linear equations
$ax+by+cz=0,bx+cy+az=0,cx+ay+bz=0$
1) $a+b+c \neq o$ and $a^2+b^2+c^2=ab+bc+ca$
2) $a+b+c \neq o$ and $a^2+b^2+c^2 \neq ab+bc+ca$
3) $a+b+c = o$ and $a^2+b^2+c^2 \neq ab+bc+ca$
4) $a+b+c = o$ and $a^2+b^2+c^2 = ab+bc+ca$
Now I need to match these with the following options.
a) The equations represent planes meeting only at a single point.
b) The equations represent the line $x=y=z$
c) The equations represent identical planes
d) The equations represent the whole 3D space.
I found out the determinant using Cramer's rule i.e. $-(a^3+b^3+c^3-3abc)$.So using Cramer's rule whenever the determinant is 0 there should be infinite solutions.After that I'm confused as to how to proceed.
| The determinant is :
$$\begin{align}\begin{vmatrix}
a&b&c\\\ b&c&a \\\ c&a&b
\end{vmatrix}&=3abc-a^3-b^3-c^3
\\
&=-\frac 12(a+b+c)\left((a-b)^2+(b-c)^2+(c-a)^2\right)
\\
&=-(a+b+c)(a^2+b^2+c^2-ab-bc-ca)
\end{align}
$$
1)This one implies $a=b=c\ne0$ so they are identical planes.
2)Here the determinant is non-zero so these are planes meeting at single point.
3) Here the determinant is zero so there are infinite solutions. And if you put any point $(x,x,x)$ in the equation it will always satisfy, so the intersection of these planes is the line $x=y=z$
4)Here $a=b=c=0$ so any point in the space is satisifed.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Show that $a_n=\frac{n+1}{2n}a_{n-1}+1$
Show that $a_n=\frac{n+1}{2n}a_{n-1}+1$ given that:
$a_n=1/{{n}\choose{0}}+1/{{n}\choose{1}}+...+1/{{n}\choose{n}}$
The hint says to consider when $n$ is even and odd. When $n=2k$ I get:
$$a_{n}=1/{{2k}\choose{0}}+1/{{2k}\choose{1}}+...+1/{{2k}\choose{2k}}$$
$$=1+1/{{2k}\choose{1}}+...+1/{{2k}\choose{2k}}$$
$$=1+1/(2k{{2k-1}\choose{0}})+1/(\frac{2k}{2}{{2k-1}\choose{0}})...+1/(\frac{2k}{2k}{{2k-1}\choose{2k-1}})$$
$$=1+\frac{1}{2k}(1/{{2k-1}\choose{0}}+1/{{2k-1}\choose{1}}+1/{{2k-1}\choose{2k-1}})$$
which should be $\frac{2k+1}{4k}a_{n-1}+1$ in the end.
I used:
${{n}\choose{k}}=\frac{n}{k}{{n-1}\choose{k-1}}$ and tried ${{n-1}\choose{k-1}}={{n-1}\choose{n-k}}$
| First, observe that the given recursion can be written as $$2n(a_n - 1) - (n+1)a_{n-1} = 0.$$ Second, observe that $$a_n - 1 = \sum_{k=0}^{n-1} \binom{n}{k}^{\!-1},$$ since the final term of $a_n$ is always $1$. Therefore, the relationship to be proven is equivalent to showing $$0 = S_{n-1} = \sum_{k=0}^{n-1} \frac{2n}{\binom{n}{k}} - \frac{n+1}{\binom{n-1}{k}}.$$ Simplify the summand by factoring out $\binom{n-1}{k}^{\!-1}$: $$\frac{2n}{\binom{n}{k}} - \frac{n+1}{\binom{n-1}{k}} = \frac{n-1-2k}{\binom{n-1}{k}}.$$ Now, with the substitution $m = n-1-k$, we observe that the RHS sum $S_{n-1}$ is now $$S_{n-1} = \sum_{m=n-1}^0 \frac{n-1 - 2(n-1-m)}{\binom{n-1}{n-1-m}} = \sum_{m=0}^{n-1} \frac{-(n-1 - 2m)}{\binom{n-1}{m}},$$ where in the last step we use the reflection identity $$\binom{n}{m} = \binom{n}{n-m}.$$ Therefore, $S_{n-1} = -S_{n-1}$, from which it follows that $S_{n-1} = 0$, proving the required recursion. No need to use odd/even cases.
| {
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"timestamp": "2023-03-29T00:00:00",
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Erasing numbers from the front of the row Numbers $1,2,\ldots,k$ are written in this order in a row. For $i=1,\ldots,k$, in the $i$th step, a random variable $V_i$ is drawn uniformly from the interval $[0,2i]$. If $V_i$ is greater than the first remaining number, that number is erased. What is the expected number of numbers that will be erased?
For example, if $k=1$, then we have the number $1$ and $V_1$ drawn from $[0,2]$, so $1/2$ numbers will be erased in expectation.
If $k=2$, then with probability $1/2$ we have $V_1>1$ and it is 50-50 whether the second number is erased. Otherwise $V_1<1$ and with probability $3/4$ the first number is erased. So the answer is $(1/2)(3/2)+(1/2)(3/4)=9/8$.
For $k=3$, a similar case analysis shows that the answer is $85/48$ (if I calculated correctly.) It could be that no closed form can be found in general. If so, upper/lower bounds would still be interesting.
| Let $E_{b,k}(a)$ be the expected number of erased numbers when you have already erased $a$ numbers and you are at the $b$th shot, out of $k$ shots.
Those can be defined for $1 \le a,b \le k+1$ by induction :
When you don't have any shot left, $E_{k+1,k}(a) = a$.
When you do you have $E_{b,k}(a) = \frac 1 {2b}((a+1) E_{b+1,k}(a) + (2b-a-1) E_{b+1,k}(a+1))$.
It turns out that every $E_{b,k}$ is an affine function :
Suppose $E_{b+1,k}(a) = pa + q$ for some $p,q \in \Bbb R$.
Then $E_{b,k}(a) = \frac 1{2b} ((a+1)(pa + q)+(2b-a-1)(p(a+1)+q))
= (1-\frac 1 {2b})pa + (q+(1-\frac 1 {2b})p)$.
So, focusing on those coefficients, we have $p_{b,k} = (1-\frac 1 {2b}) p_{b+1,k}$ and $q_{b,k} = q_{b+1,k} + p_{b,k}$
Since $p_{k,k}=1$ and $q_{k,k}=0$ we have :
$p_{b,k} = \prod_{j=b}^k (1-\frac 1 {2j})$ and
$q_{b,k} = \sum_{j=b}^k p_{j,k}$.
From there it's easy to find that when we fix $b$, the sequences $(q_{b,l})_{k \ge b-1}$ satisfy the recurrence relation $q_{b,k+1} = (1-\frac 1 {2k+2})(1 + q_{b,k})$.
Since we are interested in $E_{1,k}(0)=q_{1,k}$, we get the sequence (I'll rename it $(a_k)$) that starts with
$a_0 = 0$
$a_1 = \frac 12$
$a_2 = (\frac 12 + 1)\frac 34 = \frac 98$
$a_3 = (\frac 98 + 1)\frac 56 = \frac {85}{48}$ and so on.
Let $f_k(x) = (x+1)(1-\frac 1{2k+2})$.
We can check that $f_k(\frac {2k-1}3) = \frac {2(k+1)-1}3$ and $f_k(\frac 23 k) < \frac 23 (k+1)$.
Since the $f_k$ are increasing, this implies by induction that forall $k$, $\frac {2k-1}3 < a_k \le \frac 23k$.
Letting $b_k$ be the error term $a_k - \frac {2k-1}3$, we have $b_{k+1} = (1-\frac 1 {2k+2})b_k = \frac {2k+1}{2k+2}b_k$. Then for $k \ge 1$, $b_k = \frac {(2k-1)!!}{(2k)!!}b_0 = \frac{(2k-1)!}{2^{2k-1}(k-1)!k!}b_0 = 2^{1-2k}\binom{k-1}{2k-1}b_0$
Since $b_0 = \frac 13$, we get the closed form $a_k = \frac13 ((2k-1)+2^{1-2k}\binom{k}{2k-1})$
As for the asymptotics of $(a_k)$, one can use Stirling's formula to see that it has a development in $1/\sqrt k$ at any order.
$a_k = \frac 13 (2k -1 + \frac 1 {\sqrt{k \pi}} - \frac 1 {8\sqrt{k\pi}k} + \frac 1 {128\sqrt{k\pi}k^2} + \ldots + o(k^{n- \frac 12}))$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Equation with radicals and reciprocals Find all $x\in\mathbb{R}$ satisfying $$x=\sqrt{x-\frac{1}{x}}+\sqrt{1-\frac{1}{x}}.$$
Multiply both sides by $x^{1/2}$ to get $$x^{3/2} = \sqrt{x^2-1} + \sqrt{x-1}.$$Making the substitution $a = \sqrt{x^2 - 1}$, $b=\sqrt{x-1}$, we have $a+b = x^{3/2}$ and $a^2 - b^2 = x^2 - x$, so $a-b = (a^2-b^2)/(a+b) = \frac{x-1}{x^{1/2}}$. Then solving for $b$ gives $b = \frac{x^{3/2} - x^{1/2} + x^{-1/2}}{2}$. Then, we get the equation $$\frac{x^{3/2} - x^{1/2} + x^{-1/2}}{2} = \sqrt{x-1},$$which we can expand/rearrange to get $$x^4 - 2x^3 - x^2 + 2x + 1 = 0,$$which factors as $$\left(x^2 - x - 1\right)^2 = 0 \implies x^2 - x - 1 = 0,$$which has solutions $x = \frac{1\pm\sqrt{5}}{2}$, but note that we must have $x \ge 1$ by the problem, so $\boxed{x = \frac{1+\sqrt{5}}{2}}$.
I would be interested in seeing a slicker solution, though.
| Square both sides first: $x^2 = x-\dfrac{2}{x} + 1+ 2\sqrt{x-1-\dfrac{1}{x}+\dfrac{1}{x^2}}\Rightarrow x\left(x-1-\dfrac{1}{x}+\dfrac{1}{x^2}\right)+\dfrac{1}{x} =2\sqrt{x-1-\dfrac{1}{x}+\dfrac{1}{x^2}}$. but by AM-GM inequality, the $LHS \geq RHS$. Thus equality occurs or the equation holds if $x\left(x-1-\dfrac{1}{x}+\dfrac{1}{x^2}\right) = \dfrac{1}{x}$. From this you have it back that $x^2 - x - 1 = 0 \Rightarrow x = ...$
| {
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Induction Proof for $F_{2n} = F^2_{n+1} - F^2_{n-1}$ As stated in the tag, I'm trying to prove by induction the claim $F_{2n} = F^2_{n+1} - F^2_{n-1}$, where $F_{n}$ is the $n^{th}$ Fibonacci number. I've spent hours on the inductive step without substantial progress, and am hoping someone can provide a path to the desired result.
To facilitate the proof, we're given $F_{2n-1} = F^2_{n} + F^2_{n-1}$
(Also, in various past attempts not reflected here I tried to use the following substitutions whenever it seemed prudent: $F_{n} = F_{n-1} + F_{n-2} = F_{n+1} - F_{n-1}$; and $(x^{2} - y^{2}) = (x - y)(x + y)$).
For the inductive step, let $n\geq 1$ and assume $F_{2n} = F^2_{n+1} - F^2_{n-1}$. For $n+1$ we want to show $F_{2(n+1)} = F^2_{n+2} - F^2_{n}$. So
\begin{align}
F_{2(n+1)} &= F_{2n + 2} \\
&= F_{2n+1} + F_{2n} \\
&= (F_{2n} + F_{2n-1}) + F_{2n} \\
&= 2 F_{2n} + F_{2n - 1} \\
&= 2 (F^2_{n+1} - F^2_{n-1})
+ F^2_{n} + F^2_{n-1}
&&\text{by the inductive hypothesis} \tag{1} \\
&=...F^2_{n+2} - F^2_{n}
\end{align}
I have two challenges. The first is squeezing $-2 F^2_{n}$ out of (1). The only reasonable way I can see to do that is to expand the $-F^2_{n-1}$ term in $2 F^2_{n+1} - F^2_{n-1}$ to $-(F_{n} - F_{n-2})^2$. If in instead I combine it with $F^2_{n-1}$ then it seems impossible to generate the $-2 F^2_{n}$ term needed to offset $F^2_{n}$ and produce $-F^2_{n}$.
The second challenge is to find $(F^2_{n+2}$ in (1), and the only way I can see to do that is to expand the $F^2_{n+1}$ term in $2(F^2_{n+1} - F^2_{n-1})$to $(F_{n+2} - F_{n})^2$. Doing that not only generates $2 F^2_{n}$ which totally screws things up, but it also generates a bunch of intermediate terms I can't get rid of, and it's driving me crazy.
| Method 1. By induction on $n$ we have $F_n=(a^n-b^n)/\sqrt 5$ where $a=(1+\sqrt 5)/2$ and $b=(1-\sqrt 5)/2=-1/a.$ Plug this into your formula.
Method 2. Let $M$ be the $2\times2$ matrix with top row $(1,1)$ and bottom row $(1,0).$ By induction on $n,$ the top row of $M^n$ is $(F_{n+1},F_n)$ and the bottom row of $M^n$ is $(F_n,F_{n-1}).$ Consider that $M^{2 n} =(M^n)^2$.
Method 3. Your method. For brevity let $$F_{n-1}=a, \quad F_n=b,\quad F_{n+1}=c,\quad F_{n+2}=d.$$ From your 3rd line and the inductive hypothesis, and given $F_{2 n-1}=F_n^2+F_{n-1}^2$, we have $$F_{2 n+2}=2 F_{2 n}+F_{2 n-1}=2 (c^2- a^2) +(b^2+a^2)=$$ $$=2 c^2+b^2-a^2 =2 (d-b)^2+b^2-(c-b)^2=$$ $$=2(d-b)^2 +b^2-((d-b)-b)^2=d^2-b^2=F_{n+2}^2-F_n^2.$$
| {
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Find $N$ so that the sequence is the product of three consecutive numbers
Find the smallest natural number $N$ such that $13 \cdot 17 \cdot N$ is the product of three consecutive natural numbers.
$x(x+1)(x+2) = 13 \cdot 17 \cdot N$.
So let $x=N$, then, $N+1 = 13$ and $N+2 = 17$.
I am unsure how to proceed, can I get a hint?
| By finding the gcd of 13 and 17 you can find a solution to $13x-17y=1$. In this case $x=4,y=3$.
Consider $N=3\cdot 4 \cdot 50$
$$13 \cdot 4 =52,17\cdot 3=51$$
$$x=50 \rightarrow x(x+1)(x+2)=50\cdot 51\cdot 52 = 13\cdot 17 \cdot (3\cdot 4\cdot 50)=13\cdot 17 \cdot N$$
Hence smallest value of $N=3\times 4\times 50 = 600$
Solutions for $13x-17y=2$ would be of the form $x=8+13k,y=6+17k$.For $k=-1,x=-5,y=-11$ which is the lowest that we can get for the given case. Hence there doesn't exist any lower solution.
| {
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How to evaluate $\lim _{x\to \infty }\left(x\sqrt{x^2-x}-x^2\cos\left(\frac{1}{\sqrt{x}}\right)\right)$? I have problems to solve this limit
$$\lim _{x\to \infty }\left(x\sqrt{x^2-x}-x^2\cos\left(\frac{1}{\sqrt{x}}\right)\right)$$
I tried with Taylor:
$$\lim _{x\to \infty }\left(x^2\sqrt{1-\frac{1}{x}}-x^2\left(1-\frac{1}{2x}+\frac{1}{24x^2}\right)\right)=\lim _{x\to \infty }\left(x^2-x^2+\frac{x}{2}-\frac{1}{24}\right)=\infty$$
that is the wrong result, in must be $\color{red}{-\frac{1}{6}}$
| Full solution:
$$\lim _{x\to \infty }\left(x\sqrt{x^2-x}-x^2\cos\left(\frac{1}{\sqrt{x}}\right)\right)= \lim _{x\to \infty }x^2\left(\sqrt{1-\frac{1}{x}}-\cos\left(\frac{1}{\sqrt{x}}\right)\right)=$$$$=\lim _{t\to 0}\frac{\sqrt{1-t^2}-\cos{t}}{t^4} =\lim _{t\to 0}\frac{1-t^2-\cos^2{t}}{t^4(\sqrt{1-t^2}+\cos{t})}=\frac{1}{2}\lim _{t\to 0}\frac{\sin^2t-t^2}{t^4}=$$$$=\frac{1}{2}\lim _{t\to 0}\frac{\sin t+t}{t}\cdot\lim _{t\to 0}\frac{\sin t-t}{t^3}=\frac{1}{2}\cdot2\cdot(-\frac{1}{6})=-\frac{1}{6}.$$
For the last limit to apply the rule of L'Hospital.
| {
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Evaluate $\int_{0}^{1}\frac{\sqrt[4]{x (1-x)^{3}}}{(1+x)^{3}}\mathrm{d}x$ How to evaluate the two integrals below:
$$\int_{0}^{1}\frac{\sqrt[4]{x\left ( 1-x \right )^{3}}}{\left ( 1+x \right )^{3}}\mathrm{d}x$$
and
$$\int_{0}^{1}\frac{\sqrt[3]{x\left ( 1-x \right )^{2}}}{\left ( 1+x \right )^{3}}\mathrm{d}x$$
After a long time thinking,I still in trouble.
| Using the definition of Beta function
$$B\left ( p,q \right )=\frac{\Gamma \left ( p \right )\Gamma \left ( q \right )}{\Gamma \left ( p+q \right )}=\int_{0}^{1}x^{p-1}\left ( 1-x \right )^{q-1}\mathrm{d}x~ ~ ~ ~ \left ( \Re q>0,\Re p>0 \right )$$
In general,
$$I\left ( m,n \right )=\int_{0}^{1}\frac{\sqrt[n]{x^{m}\left ( 1-x \right )^{n-m}}}{\left ( 1+x \right )^{3}}\mathrm{d}x$$
So in order to translate your integral into Beta function,you may perform a change of variable like this:
$$t=\frac{2x}{1+x},~ ~ ~ 1-t=\frac{1-x}{1+x},~ ~ ~ \mathrm{d}t=\frac{2\mathrm{d}x}{\left ( 1+x \right )^{2}}$$
Hence we have,
\begin{align*}
\int_{0}^{1}\frac{\sqrt[n]{x^{m}\left ( 1-x \right )^{n-m}}}{\left ( 1+x \right )^{3}}\mathrm{d}x &=\int_{0}^{1}\left ( \frac{x}{1+x} \right )^{\frac{m}{n}}\left ( \frac{1-x}{1+x} \right )^{\frac{n-m}{n}}\frac{\mathrm{d}x}{\left ( 1+x \right )^{2}} \\
&=2^{-\frac{n+m}{n}}\int_{0}^{1}t^{\frac{m}{n}}\left ( 1-t \right )^{\frac{n-m}{n}}\mathrm{d}t \\
&=\frac{2^{-\frac{n+m}{n}}}{\Gamma \left ( 3 \right )}\Gamma \left ( \frac{m+n}{n} \right )\Gamma \left ( \frac{2n-m}{n} \right ) \\
&=2^{-\frac{2n+m}{n}}\cdot \frac{m}{n}\cdot \frac{n-m}{n}\cdot \Gamma \left ( \frac{m}{n} \right )\cdot \Gamma \left ( 1-\frac{m}{n}\right )\\
&=2^{-\frac{2n+m}{n}}\cdot \frac{m\left ( n-m \right )}{n^{2}}\cdot \frac{\pi }{\sin\left ( \frac{m\pi }{n} \right )}
\end{align*}
In the particular case of $m=1,n=4~and~3$,we can easily get
$$\Large\color{blue}{I\left ( 1,4 \right )=\frac{3\sqrt[4]{2}}{64}\pi }$$
$$\Large\color{blue}{I\left ( 1,3\right )=\frac{\pi }{18}\frac{\sqrt[3]{4}}{\sqrt{3}}}$$
| {
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Evaluate the triple integral $\iiint (x^2+y^2+z^2)\,dx\,dy\,dz$ I have to evaluate this integral. It is enough for me to know the correct limits to integration.
$$ \iiint_W (x^2 + y^2 + z^2) \,\mathrm dx\,\mathrm dy\,\mathrm dz$$
Conditions:
$$x\ge 0,\quad y \ge 0 ,\quad z \ge 0,\quad 0 \le x + y + z \le a,\quad (a>0)$$
My ideas:
As I have
$0 \le x + y + z \le a$,
I can say that $$0 \le x + y + z$$
$$x \le - z - y $$
so my limits to integration could be:
$$\int _{ z }^{ a }{ } \int _{ -z-y }^{ y }{ } \int _{ 0 }^{ -z-y }{ (x^2 + y^ 2+ z^ 2) \,\mathrm dx\,\mathrm dy\,\mathrm dz } $$
Can somebody help/correct me?
| *
*Your manipulation of the inequality is wrong: $0\leq x+y+z$ implies $x\geq-y-z$, not $\leq$.
*As @Hagen pointed out, you should NOT have bounds that depend on the variable of integration. Perhaps you meant to order the $d$s differently?
*Again as @Hagen pointed out in his answer, the bounds are much easier.
Here is how to find the bounds:
*
*$z\geq0$, but since all other variables are surely $z\leq a$, so $0\leq z\leq a$ for the outer integral.
*$y\geq0$, and $x+y+z\leq a$ surely implies $x+y\leq a-z$, but $x\geq0$ so you must have $y\leq a-z$.
*$x\geq0$, and $x+y+z\leq a$ equates to $x\leq a-y-z$.
So your integral becomes:
\begin{align*}
\int\limits_0^a\int\limits_0^{a-z}\int\limits_0^{a-z-y}(x^2+y^2+z^2)dxdydz={}&\int\limits_0^a\int\limits_0^{a-z}\left[(y^2+z^2)(a-z-y-0)+\frac{(a-z-y)^3}{3}\right]dydz={} \\
{}={}&\int\limits_0^a\int\limits_0^{a-z}\left[ay^2+az^2-y^2z-z^3-y^3-yz^2+\frac{(a-z-y)^3}{3}\right]dydz={} \\
{}={}&\int\limits_0^a\left[a\frac{(a-z)^3}{3}+a(a-z)z^2-\frac{(a-z)^3}{3}z-(a-z)z^3\right.-{} \\
&{}+\left.\frac{(a-z)^4}{4}-\frac{(a-z)^2}{2}z^2-\frac{(a-z-(a-z))^4}{12}+\frac{(a-z)^4}{12}\right]dydz={} \\
{}={}&-\frac{a}{12}(a-z)^4\Bigg|_0^a+\frac{a^2}{3}z^3\Bigg|_0^a-\frac{a}{4}z^4\Bigg|_0^a+\frac{z}{12}(a-z)^4\Bigg|_0^a-{} \\
&{}+\int\limits_0^a\frac{(a-z)^4}{12}dz-\frac a4z^4\Bigg|_0^a+\frac{z^5}{5}\Bigg|_0^a+\frac{(a-z)^5}{20}\Bigg|_0^a+{} \\
&{}+\frac{(a-z)^3}{6}z^2\Bigg|_0^a-\int\limits_0^a\frac{(a-z)^3}{3}zdz-\frac{(a-z)^5}{60}\Bigg|_0^a={} \\
{}={}&\frac{a^5}{12}+\frac{a^5}{3}-\frac{a^5}{4}+\overline{\frac{(a-z)^5}{60}\Bigg|_0^a}-\frac{a^5}{4}+\frac{a^5}{5}-\frac{a^5}{20}+{} \\
&{}+\frac{(a-z)^4}{12}z\Bigg|_0^a-\int\limits_0^a\frac{(a-z)^4}{12}dz-\overline{\frac{(a-z)^5}{60}\Bigg|_0^a}={} \\
{}={}&a^5\left(\frac{1}{12}+\frac13-\frac14-\frac14+\frac15-\frac{1}{20}\right)+\frac{(a-z)^5}{60}\Bigg|_0^a={} \\
{}={}&a^5\left(\frac{1}{6}-\frac{1}{10}+\frac{1}{60}\right)=\frac{a^5}{12}.
\end{align*}
I hope I didn't get lost in the calculations.
| {
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Prove that the number 14641 is the fourth power of an integer in any base greater than 6? Prove that the number $14641$ is the fourth power of an integer in any base greater than $6$?
I understand how to work it out, because I think you do
$$14641\ (\text{base }a > 6) = a^4+4a^3+6a^2+4a+1= (a+1)^4$$
But I can't understand why they had to specify that the base is greater than 6? Is that because if it's 4, then 4 will be cancelled out in the equation and if it's 6, it will be too?
Please advise.
Sorry for asking such a trivial question.
| As a matter of personal preference, I like $b$ to represent the base rather than $a$.
So then $(b + 1)^2 = b^2 + 2b + 1$. And $(b^2 + 2b + 1)^2 = b^4 + 4b^3 + 6b^2 + 4b + 1$.
These facts are true whether $b$ is an ordinary positive integer greater than 1 or a more "exotic" number, like $\sqrt{-2}$. But whether $(b + 1)^4$ gets represented as 14641 in base $b$, that's a slightly different story.
Consider for example $b = 2$. Indeed $3^4 = 2^4 + 4 \times 2^3 + 6 \times 2^2 + 4 \times 2 + 1$. But the problem is that here it turns out that $6 > b^2$ and $4 > b$. Then $4b^3$ requires more than four binary digits to represent, $6b^2$ requires more than three bits and $4b$ requires more than two bits. So the binary representation of 81 is 1010001 rather than 14641.
Some of these problems persist through $b = 6$, because although $4b^3 < b^4$, we still have $6b^3 > b^3$. Then 2401 is 15041 in base 6.
The smallest integer $b$ satisfying $4b^3 < b^4$, $6b^2 < b^3$ and $4b < b^2$ is $b = 7$. Indeed 4096 in base 7 is 14641.
| {
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Solve $3x(1-x^2)y^2\frac{dy}{dx}+(2x^2-1)y^3=ax^3$ I am solving this linear Differential equation which can be easily solve by using the formulas for the Bernoulli's Equations
I have solved till
$$\frac{dy}{dx}+\frac{(2x^2-1)y^3}{3x(1-x^2)y^2}=\frac{ax^3}{3x(1-x^2)y^2}$$
$$y^2\frac{dy}{dx}+\frac{(2x^2-1)y^3}{3x(1-x^2)}=\frac{ax^3}{3x(1-x^2)}$$
Substituting $y^3=t$
so the equation will be
$$\frac{1}{3}\frac{dt}{dx}+\frac{(2x^2-1)t}{3x(1-x^2)}=\frac{ax^3}{3x(1-x^2)}$$
after this the integrating factor is
$$\frac{1}{x\sqrt{1-x^2}}$$
But I am unable to solve it forward.
| After substitution $t(x) = y^3(x)$
$$x(1-x^2) t'(x) + (2x^2-1) t(x) = ax^3.$$
Let's find a particular solution in form $t_0(x) = bx:$
$$bx - bx^3 + 2bx^3 - bx = ax^3 \Rightarrow b = a$$
So the full solution has form $t(x) = ax + z(x),$
$$x(1-x^2) z'(x) + (2x^2-1) z(x) = 0 \Rightarrow$$
$$\log z = - \int \frac{2x^2 -1}{x(1-x^2)} dx =
\int \frac{1 - x^2}{x(1-x^2)} dx - \int \frac {x^2}{x(1-x^2)} dx =
\log x + \frac 1 2 \log\left(1-x^2\right) +C \Rightarrow
$$
$$t(x) = C x \sqrt{1 -x^2} + ax \Rightarrow y(x) = \left(C x \sqrt{1-x^2} + ax\right)^{\frac 1 3}$$
| {
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Solve $\sqrt[3]{7x+19}+\sqrt[3]{7x-19}=\sqrt[3]{2}$ by algebraic methods I was trying to solve this equation without using calculus.
Is it possible to be solved by elementary algebraic methods?
$$\sqrt[3]{7x+19}+\sqrt[3]{7x-19}=\sqrt[3]{2}$$
| Let $u=\sqrt[3]{q+\sqrt{p^{3}+q^{2}}}$, $v=\sqrt[3]{q-\sqrt{p^{3}+q^{2}}}$,
we have $y=u+v$ satisfying $y^{3}+3py=2q \ldots \ldots (*)$.
Take $y=\sqrt[3]{2}$, $q=7x$ and $p^{3}+q^{2}=19^{2}$ or $p^{3}=19^{2}-(7x)^{2}$.
Substitute into $(*)$,
\begin{align*}
2+3\sqrt[3]{19^{2}-(7x)^{2}} \times \sqrt[3]{2} &= 2(7x) \\
\sqrt[3]{19^{2}-(7x)^{2}} &= \frac{14x-2}{3\sqrt[3]{2}} \\
19^{2}-(7x)^{2} &= \frac{(14x-2)^{3}}{54}
\end{align*}
By factor theorem or Cardano formula,
$\displaystyle x=\frac{7}{4}, \frac{-8\pm 3i\sqrt{15}}{7}$ subject to correct choices of cubic roots in the original equation.
P.S.: May rewrite as $\sqrt[3]{19+7x}-\sqrt[3]{19-7x}=\sqrt[3]{2}$ for easy computer verification.
| {
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Rational Points on Fibonacci-like Sequence of Polynomials Let $\{a_n\}$ be a sequence of polynomials in $\mathbb{Q}[x,y]$ with $a_0=0,a_1=1$, and
$$a_n=xa_{n-1}+ya_{n-2}$$
The first few look like
$$a_3:y+x^2$$
$$a_4:2xy+x^3=x(2y+x^2)$$
$$a_5:y^2+3x^2y+x^4$$
$$a_6:3xy^2+4x^3y+x^5=x(y+x^2)(3y+x^2)$$
Note that $a_6=0$ has many obvious rational points, for example $(3,-9)$.
I conjecture: For $n>6$, $a_n=0$ has no nontrivial rational points. How do/would I prove this?
| First of all, you can do the same sort of trick to get a closed form for this as we use for any other linear recurrence.
The roots of $U^2-xU -y=0$ are $u_1,u_2=\frac{x\pm\sqrt{x^2+4y}}{2}$.
And $a_n = bu_1^n + cu_2^n$ for some $b,c\in R=\mathbb Q(x,y)[\sqrt{x^2+4y}]$.
Solving for $b$ and $c$, we get $b=-c$ and $b(u_1-u_2)=b\sqrt{x^2+4y}=1$.
So $$a_n = \frac{1}{\sqrt{x^2+4y}}\left(\left(\frac{x+\sqrt{x^2+4y}}{2}\right)^n - \left(\frac{x-\sqrt{x^2+4y}}{2}\right)^n\right)$$
Since $u_1u_2=-y$, $u_1^n=u_2^n$ for specific rational $x,y$ means that $u_1^{2n}=(-y)^n$, so $-\frac{u_1^2}{y}$ must be an $n$th root of $1$. Now:
$$-\frac{u_1^2}{y} = \frac{-x^2-2y - x\sqrt{x^2+4y}}{2y}$$ which, if $x,y$ are rational, is quadratic over $\mathbb Q$. The only primitive roots of unity that are at most quadratic over $\mathbb Q$ are the square roots, fourth roots, the cube roots, and the sixth roots.
Note that this also shows that if $(x_0,y_0)$ is a rational point of $a_n$, then it is a rational point of $a_{kn}$ for any $k$. This actually extends to any point, because the argument that $-\frac{u_1^2}{y}$ must be a root of unity holds true whenever $u_1^n=u_2^n$ and $y\neq 0$.
It's worth adding in the separate case when $x^2+4y=0$, which is the case when $u_1=u_2=\frac{x}{2}$ and then the closed formula is of the form:
$$a_n = (bn+c)\left(\frac{x}{2}\right)^n$$
So $c=0$ and $bx/2 = 1$ so $b=\frac{2}{x}$ and the formula in this case is:
$$a_n = n\left(\frac{x}{2}\right)^{n-1}$$
This is only zero when $x=0$ and hence $y=0$.
One way to see that $a_n$ is a factor of $a_{nk}$ is to note that:
$$a_{nk}=\frac{1}{\sqrt{x^2+4y}}(u_1^{nk}-u_2^{nk}) = \frac{1}{\sqrt{x^2+4y}}(u_1^n-u_2^n)\left(...\right)$$
And just show the elided section must be a polynomial in $x,y.$ It's not hard - it is a combination of expressions of the form $u_1^k+u_2^k$, which causes all the square roots to cancel out.
What happens in the cases given? That is, when $x^2=-y,-2y,-3y$?
Then you get:
$$\begin{align}
(x,y)=(1,-1)&\implies -\frac{u_1^2}{y} = \frac{1-\sqrt{3}}{-2}=\frac{-1+\sqrt{3}}{2}\\
(x,y)=(2,-2)&\implies-\frac{u_1^2}{y}=\frac{0-2\sqrt{-4}}{-4}=i\\
(x,y)=(3,-3)&\implies-\frac{u_1^2}{y}=\frac{-3-3\sqrt{-3}}{-6}=\frac{1+\sqrt{3}}{2}
\end{align}$$
Also, for $a_2$, we get any $(0,y)$ is a root, and $-\frac{u_1^2}{y}=-1$.
So you have all the quadratic and linear roots of unity represented, excepting $1$ itself.
| {
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"url": "https://math.stackexchange.com/questions/1638223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
$\sin2(x) - \tan(x) = 0$ , solve for $-180\le x\le 180$ I have been unable to solve the following question,
If $$\sin(2x) - \tan(x) = 0$$
Find $x$ , $-\pi\le x\le \pi$
So far my workings have been
Use following identity:
$$\sin(2x) = 2\sin(x)\cos(x)\\2\sin(x)\cos(x) - \tan(x) = 0\\2\sin(x)\cos(x) - \frac{\sin(x)}{\cos(x)} = 0\\
2\frac{\sin(x)\cos(x)}{1} - \frac{\sin(x)}{\cos(x)} = 0$$
Then cross multiply to give :
$$-\sin x+((2\cos(x)\sin(x))\cos(x))/\cos(x) = 0$$
$$-\sin x+(2\cos^2(x)\sin(x))/ \cos(x) = 0$$
However, I have been unable to get any further.
If someone could help me find a solution to this question it would be very much appreciated.Thank you.
| Notice, $$\sin 2x-\tan x=0$$
$$\frac{2\tan x}{1+\tan^2x}-\tan x=0$$
$$\tan x\left(\frac{1-\tan^2 x}{1+\tan^2x}\right)=0$$
$$\color{blue}{\tan x\cos 2x=0}$$
Now, solving for $x$,
$$\tan x=0\iff x=n\cdot 180^\circ$$
where $n$ is any integer
For given interval $[-180^\circ, 180^\circ]$, setting $n=-1, 0, 1 $, one should get $$\color{blue}{x=-180^\circ, 0, 180^\circ}$$
or $$\cos 2x=0\iff 2x=(2n-1)\cdot 90^\circ\ \ $$$$or \ \ x=(2n-1)\cdot 45^\circ$$
where $n$ is any integer
For given interval $[-180^\circ, 180^\circ]$, setting $n=-1, 0, 1, 2 $, one should get $$\color{blue}{x=-135^\circ, -45^\circ, 45^\circ, 135^\circ}$$
hence, the complete solution is
$$\color{red}{x\in\{-180^\circ, -135^\circ, -45^\circ, 0, 45^\circ, 135^\circ, 180^\circ\}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1639081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Combinatorics: How do you find the coefficient in the given expression? The question asks me to find the coefficient of the term $x^6y^4$ in the expression $(xy^2+x^2+3y)^7$. This was pretty simple. This is how I did it:
$$(xy^2+x^2+3y)^7 = \sum_{a+b+c = n} (xy^2)^a + (x^2)^b + (3y)^c $$
$$ x^{(a+2b)}, y^{(2a+c)}, 3^c = x^6y^4 $$
From this we get, $$ a+2b = 6, 2a+c = 4, a+b+c = 7$$
Solving these will give us $a = 0,b=3, c=4$. After plugging in the values you get 2852 as the final answer.
Now this is where I am stuck, $x^6y^4$ in the expression $(xy^2+x^2+3y+4)^7$. In this question we have an extra term 4 and this is really confusing me. I tried doing it in the same way:
$$(xy^2+x^2+3y+4)^7 = \sum_{a+b+c+d = n} (xy^2)^a + (x^2)^b + (3y)^c + 4^d$$
$$ x^{(a+2b)}, y^{(2a+c)}, 3^c, 4^d = x^6y^4 $$
From this we get, $$ a+2b = 6, 2a+c = 4, a+b+c+d = 7$$
I tried solving this but I couldn't get the right answer. Am I doing it correctly. How do I solve it?
| You were almost there, mate.
Examine: $a+2b = 6,\, 2a+c = 4,\, a+b+c +d = 7$ for $0\leq a,b,c,d\leq 7$
We have $b=3-a/2$, $c=4-2a, d=7-a-b-c$ so we list the possibilities: $$\begin{cases}a=0, b=3, c=4, d=0 \\[1ex] a=2, b=2, c=0, d=3\end{cases}$$
So the term we require is: $x^{(0+6)}\, y^{(0+4)}\, 3^4\, 4^0 + x^{(2+4)}\, y^{(4+0)}\, 3^0\, 4^3$
Hence the coefficient of $x^6y^4$ is: $3^4+4^3$
That is all.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1639708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solve a linear system of equation involving some recursion $$
\begin{align*}
x_{1} &= 1 + x_{2}\\
x_{2} &= 1 + \frac{1}{2} x_{3} + \frac{1}{2} x_{1}\\
&\vdots\\
x_{i} &= 1 + \frac{1}{2} x_{i+1} + \frac{1}{2} x_{1}\\
&\vdots\\
x_{n-2} &= 1 + \frac{1}{2} x_{n-1} + \frac{1}{2} x_{1}\\
x_{n-1} &= 1 + \frac{1}{2} x_{n} + \frac{1}{2} x_1 \\
x_{n} &= 0 \\
\end{align*}
$$
EDIT: I found one way to solve this. It's simply plugging successive $x_i$ into the first equation. When you reach $x_n$, since $x_n = 0$, you end up with an equation of just $x_1$ which you can solve (using geometric sum) to get $x_1 = 3\times 2^{n-2} - 2$. The rest then is easy.
If you have a more elegant solution, please share.
| Disclaimer
I'm gonna write couple of terms and then general equation for both forward and backward substitution, so you'll need to use mathematical induction to actually prove them
First, substitute $x_1$ to the second equation
$$
2 x_2 = 2 + x_3 + x_1 = 2 + x_3 + 1 + x_2 = 3 + x_3 + x_2 \implies x_2 = 3\cdot 2^0 + x_3
$$
Now, substitute both $x_2$ and $x_1$ to the equation for $x_3$
$$
2x_3 = 2 + x_4 + x_1 = 2 + x_4 + 1 + x_2 = 3 + x_4 + 3 + x_3 = 3 \cdot 2^1 + x_4 + x_3 \implies x_3 = 3 \cdot 2^1 + x_4
$$
Now you can prove (using mathematical induction) that
$$
x_{n-1} = 3 \cdot 2^{n-3} + x_n
$$
Now, do backward substitution by using $x_n = 0$
$$
x_{n-1} = 3 \cdot 2^{n-3}
$$
and then
$$
x_{n-2} = 3 \cdot 2^{n-4} + x_{n-1} = 3 \cdot 2^{n-4} + 3 \cdot 2^{n-3} = 3 \cdot 2^{n-4} (2^0 + 2^1)
$$
and one more
$$
x_{n-3} = 3 \cdot 2^{n-5} + x_{n-2} = 3 \cdot 2^{n-5} + 3 \cdot 2^{n-4}(2^0 + 2^1) = 3 \cdot 2^{n-5} (2^0 + 2^1 + 2^2)
$$
Now, you can prove that
$$
x_{n - k} = 3 \cdot 2^{n-k-2} (2^0 + 2^1 + \ldots + 2^{k-1})
$$
In the parenthesis is nothing but simple geometric progression, so
$$
x_{n-k} = 3 \cdot 2^{n-k-2} (2^k - 1)
$$
You can do that all the way until $k = n - 2$ to find $x_2$
$$
x_2 = 3 \cdot 2^0 (2^{n-2} - 1) = 3(2^{n-2}-1)
$$
and finally
$$
x_1 = 1 + x_2 = 3 \cdot 2^{n-2} - 2
$$
Summary
\begin{align}
x_1 &= 3 \cdot 2^{n-1} - 2 \\
x_i &= 3 \cdot 2^{i-2} (2^{n-i} - 1)\quad \text{for}\ 2 \le i < n\\
x_n &= 0
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1639894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Solving $(4y+2x-5)dx+(6y+4x-1)dy=0$ using 2 methods produced 2 different answers! $$(4y+2x-5)dx+(6y+4x-1)dy=0,y(-1)=2$$
First method:
$$\frac{dy}{dx}=-\frac{4y+2x-5}{6y+4x-1}$$
let $Y=y-\frac{9}{2}$, $dY=dy$ and $X=x+\frac{13}{2}$, $dX=dx$;
$$\frac{dY}{dX}=-\frac{4Y+2X}{6Y+4X}$$
let $u=\frac{Y}{X}$ , $Y'=u'X+u$;
$$u'X+u=-\frac{4u+2}{6u+4}$$
$$u'X=\frac{-6u^2-8u-2}{6u+4}$$
$$\int\frac{(6u+4)du}{6u^2+8u+2}=\int-\frac{dX}{X}$$
$$\frac{1}{2}\ln|3u^2+4u+1|=-\ln|X|+Constant$$
$$\frac{1}{2}\ln|3(\frac{y-\frac{9}{2}}{x+\frac{13}{2}})^2+4(\frac{y-\frac{9}{2}}{x+\frac{13}{2}})+1|=-\ln|x+\frac{13}{2}|+Constant$$
Solving for $y(-1)=2$, $Constant=0.89587977346$;
$$\frac{1}{2}\ln|3(\frac{y-\frac{9}{2}}{x+\frac{13}{2}})^2+4(\frac{y-\frac{9}{2}}{x+\frac{13}{2}})+1|=-\ln|x+\frac{13}{2}|+0.89587977346$$
Second method:
let $M=4y+2x-5$ , $N=6y+4x-1$
$$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}=4\rightarrow Exact$$
$$\varnothing=4xy+x^2-5x+3y^2-y=Constant$$
Solving for $y(-1)=2$, $Constant=8$;
$$\varnothing=4xy+x^2-5x+3y^2-y=8$$
I plotted those two answers on MATHEMATICA and found that result:
How could the results be different?
| The indefinite integration formula
$$
\int \frac{dU}{U} = \ln |U| + c
\tag{1}
$$
(correctly) summarizes two formulas that hold in disjoint intervals,
$$
\left.
\begin{aligned}
\int \frac{dU}{U} &= \ln U + c &&(U > 0), \\
\int \frac{dU}{U} &= \ln (-U) + c && (U < 0).
\end{aligned}\right\}
\tag{2}
$$
(I'm deliberately using $U$ as a variable since it does not appear in your situation.)
In any particular definite integral (such as in your case, solving an ODE with an initial condition), only one choice of sign in (2) is correct.
As a result, your first method (using (1)) led to an equation of the form $|U| = C$, i.e., $U = \pm C$, when it should have led (using one equation from (2)) to $U = C$.
As visual confirmation, the dashed curve you expected comprises two branches of the hyperbola $4xy + x^{2} - 5x + 3y^{2} - y = C$; the "extra" portion comprises the two branches of the hyperbola $4xy + x^{2} - 5x + 3y^{2} - y = -C$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1640208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What is the integral value of $\frac{\tan 20^\circ+\tan40^\circ+\tan80^\circ-\tan60^\circ}{\sin40^\circ}$? I have tried possibly all approaches.
I first expressed $80$ as $60+20$ and $40$ as $60-20$ and then used trig identities.I later used conditional identities expressing $\tan 20^\circ+\tan40^\circ+\tan120^\circ$ as $\tan 20^\circ \tan40^\circ \tan120^\circ$. But I really can't get to the end of it .
Please help.
| Using $\tan20^\circ\cdot\tan40^\circ\cdot\tan80^\circ=\tan60^\circ$ (Proof)
$$\tan20^\circ+\tan40^\circ+\tan80^\circ-\tan60^\circ$$
$$=\tan20^\circ+\tan40^\circ+\tan80^\circ-\tan20^\circ\cdot\tan40^\circ\cdot\tan80^\circ$$
$$=\tan20^\circ(1-\tan40^\circ\cdot\tan80^\circ)+\tan40^\circ+\tan80^\circ$$
$$=(1-\tan40^\circ\cdot\tan80^\circ)\left(\tan20^\circ+\dfrac{\tan40^\circ+\tan80^\circ}{1-\tan40^\circ\cdot\tan80^\circ}\right)$$
$$=\dfrac{\cos(40^\circ+80^\circ)}{\cos40^\circ\cos80^\circ}\left(\tan20^\circ+\tan(40^\circ+80^\circ)\right)$$
$$=\dfrac{\cos120^\circ}{\cos40^\circ\cos80^\circ}\cdot\dfrac{\sin(20^\circ+120^\circ)}{\cos20^\circ\cdot\cos120^\circ}$$
$$=\dfrac{\sin40^\circ}{\cos20^\circ\cos40^\circ\cos80^\circ}\text{ Using }\sin(180^\circ-A)=\sin A$$
Now use Upon multiplying $\cos(20^\circ)\cos(40^\circ)\cos(80^\circ)$ by the sine of a certain angle, it gets reduced. What is that angle?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1640343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Show that for each $n \geq 2$, $\left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right) \cdots \left(1 - \frac{1}{n^2}\right) = \frac{n + 1}{2n}$ Need to show that for each $n \in \mathbb{N}$, with $n \geq 2$,
$$\left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right) \cdots \left(1 - \frac{1}{n^2}\right) = \frac{n + 1}{2n}$$
How to start the proof by induction? Is there any way to show this?
| Let $$P = \prod^{n}_{r=2}\left(1-\frac{1}{r^2}\right) =\prod^{n}_{r=2}\left(1-\frac{1}{r}\right)\cdot \left(1+\frac{1}{r}\right) $$
So $$\prod^{n}_{r=2}\left(1-\frac{1}{r^2}\right) = \prod^{n}_{r=2}\left(1-\frac{1}{r}\right)\cdot \prod^{n}_{r=2}\left(1+\frac{1}{r}\right)$$
Now Open These two products.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1640628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
A formula for length of representation of a number in a "base" without zeros If you had 2 items the sequence would go like this:
$$1,1,2,2,2,2,3,3,3,3,3,3,3,3,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,5,5,5,5,5,5,5,5,5,5,5, \ldots$$
This is $\lfloor\log_2(n+2)\rfloor$.
What if I wanted to do for 3 items which goes like this: $$1,1,1,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,4,4,\ldots$$
A visual representation for $3$ items as $(A,B,C)$:
f(0) = 1: A ──┐
f(1) = 1: B ├─ 3^1
f(2) = 1: C ──┘
f(3) = 2: AA ─┐
f(4) = 2: AB │
f(5) = 2: AC │
f(6) = 2: BA │
f(7) = 2: BB ├─ 3^2
f(8) = 2: BC │
f(9) = 2: CA │
f(10) = 2: CB │
f(11) = 2: CC ─┘
f(12) = 3: AAA ┐
f(13) = 3: AAB │
f(14) = 3: AAC │
f(15) = 3: ABA │
f(16) = 3: ABB ├─ 3^3
f(17) = 3: ABC │
f(18) = 3: ACA │
f(19) = 3: ACB │
f(20) = 3: ACC │
...
| Hint The number of elements of length $k$ is $3^k$, so the number of elements of length $\leq k$ is $$3 + 3^2 + \cdots 3^k = \frac{3}{2} (3^k - 1) .$$ Thus, the $n$th element in the sequence $A, B, C, AA, AB, \ldots$ has length $k$ iff $$\tfrac{3}{2} (3^{k - 1} - 1) < n \leq \tfrac{3}{2}(3^k - 1) .$$
Rearranging gives that the $k$ for which the previous inequalities hold is precisely $$\left\lceil \log_3 \left(\tfrac{2}{3} n + 1\right) \right\rceil .$$ Since the indexing in the question begins with $0$, i.e, since $a$ is the $(a + 1)$st number in the list, the function $f$ is $$f(a) := \left\lceil \log_3 \left(\tfrac{2}{3} (a + 1) + 1\right) \right\rceil .$$ It is straightforward to generalize this to arbitrary numbers of symbols $(A, B, C, \ldots)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1642226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\frac{1}{n+1} + \frac{1}{n+3}+\cdots+\frac{1}{3n-1}>\frac{1}{2}$ Without using Mathematical Induction, prove that $$\frac{1}{n+1} + \frac{1}{n+3}+\cdots+\frac{1}{3n-1}>\frac{1}{2}$$
I am unable to solve this problem and don't know where to start. Please help me to solve this problem using the laws of inequality. It is a problem of Inequality.
Edit: $n$ is a positive integer such that $n>1$.
| Let
$$ a_n = \frac{1}{n+1}+\frac{1}{n+3}+\ldots+\frac{1}{3n-1}. $$
We have $a_1=\frac{1}{2}$ and:
$$\begin{eqnarray*} a_{n+2}-a_n &=& \frac{1}{3n+5}+\frac{1}{3n+3}+\frac{1}{3n+1}-\frac{1}{n+1}\\&>&\frac{3}{3n+3}-\frac{1}{n+1} = 0\end{eqnarray*} $$
so the claim is trivial, since $a_2>a_1$ and $a_{n+2}>a_n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1642847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
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