Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Fibonacci sum for $\pi$: $\sum_{n=1}^\infty\frac{F_{2n}}{n^2\binom{2n}{n}}=\frac{4\pi^2}{25\sqrt5}$ Wikipedia's "List of formulae involving $\pi$" entry states that
$$\sum_{n=1}^\infty\frac{F_{2n}}{n^2\binom{2n}{n}}=\frac{4\pi^2}{25\sqrt5}.$$
Why is this true?
If $\varphi=(1+\sqrt5)/2$ and $\psi=(1-\sqrt5)/2$, then we ... | In this answer, it is shown that
$$
\frac{\arcsin(x)}{\sqrt{1-x^2}}
=\sum_{k=0}^\infty\frac1{(2k+1)}\frac{4^k}{\binom{2k}{k}}x^{2k+1}\tag1
$$
Integrating $(1)$ and multiplying by $4$ yields
$$
\begin{align}
2\arcsin^2(x)
&=\sum_{k=0}^\infty\frac1{(2k+1)(2k+2)}\frac{4^{k+1}}{\binom{2k}{k}}x^{2k+2}\\
&=\sum_{k=0}^\infty\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3472167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Contradict the uniqueness of least-squares solution
Find a least-squares solution of $Ax=b$ for
$A=\begin{pmatrix}
1 & 1 & 0 & 0 \\
1 & 1 & 0 & 0 \\
1 & 0 & 1 & 0 \\
1 & 0 & 1 & 0 \\
1 & 0 & 0 & 1 \\
1 & 0 & 0 & 1 \\
\end{pmatrix}
,b=\begin{pmatrix}
-3\\
-1\\
0\\
2\\
5\\
1
\end{pmatrix}$
My approach:
$$A^{T}A=\beg... | Matrix $A$ is not full rank column (since $C_1=C_2+C_3+C_4$) and therefore we have one free parameter for the solution which is not unique.
Indeed recall that least square method corresponds to find solutons for $Ax=p$ where $p$ is the projection of $b$ onto $Col(A)$. Therefore since $A$ is not full rank column the res... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3472740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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What is the two-parametric solution of the biquadratic Diophantine equation? On the following biquadratic Diophantine equation, Leonard Eugene Dickson mentioned a parameter equation in his works (HOTTON Vol-II), but did not give it in detail
$$a^4+6a^2b^2+b^4=c^4+6c^2d^2+d^4$$
A. Gérardin noted that, to find two right... | I am deeply indebted to a French friend for his help.
\begin{align*}
\begin{cases}
x=3 t^7+7 t^6+\,\,\,3 t^5+19 t^4-15 t^3+37 t^2+9 t+1\\
y=\,\,\,t^7-9 t^6+37 t^5+15 t^4+19 t^3-\,\,\,3 t^2+7 t-3\\
\alpha=\,\,\,t^7+9 t^6+37 t^5-15 t^4+19 t^3+\,\,\,3 t^2+7 t+3\\
\beta=3 t^7-7 t^6+\,\,\,3 t^5-19 t^4-15 t^3-37 t^2+9 t-1
\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3473444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Is there a closed form for $3\cdot\frac{3}{\sqrt{6}}\cdot\frac{3}{\sqrt{6+\sqrt{6}}}\cdot\frac{3}{\sqrt{6+\sqrt{6+\sqrt{6}}}}\cdots$? Inspired by Vieta's Formula for $\pi$,
$$\pi=2\cdot\frac{2}{\sqrt{2}}\cdot\frac{2}{\sqrt{2+\sqrt{2}}}\cdot\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2}}}}\cdots$$
I became interested in a more gener... | Just a comment, as the comment section is really getting long, you may really want to know the connection why it is $2$ in the radical that gives this amazing result. It is due to it's connection to trigonometric functions. More specifically, a quick research into this can give that a result, (link)
$$\frac{\sin x}{x}=... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
} |
Solve system of equations $\frac{x}{y}+\frac{y}{x}=\frac{10}{3}$, $x^2-y^2=8$
Solve the system:
$$\begin{array}{l}\dfrac{x}{y}+\dfrac{y}{x}=\dfrac{10}{3} \\
x^2-y^2=8\end{array}$$
First, we have $x \ne 0$ and $y \ne 0$. We can rewrite the first equation as $$\dfrac{x^2+y^2}{xy}=\dfrac{10}{3}$$
What should I do nex... | Denote: $\frac xy=a$. Then from the first equation:
$$a+\frac1a=\frac{10}{3} \Rightarrow 3a^2-10a+1=0 \Rightarrow a=\frac13,3 \Rightarrow \\
x=3y,y=3x$$
Plug these into the second equation:
$$(3y)^2-y^2=8\Rightarrow y=\pm 1 \Rightarrow x=\pm 3\\
x^2-(3x)^2=8 \Rightarrow x^2=-1 \Rightarrow \emptyset.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3476697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Calculate the sum: $C_{2n}^n+2C_{2n-1}^n+4C_{2n-2}^n+...+2^nC_n^n$ Calculate this sum:$$C_{2n}^n+2C_{2n-1}^n+4C_{2n-2}^n+...+2^nC_n^n.$$
What I tried:
$$ C^n_{2n}=\frac{(2n)!}{(n!)^2}$$
$$ 2C^n_{2n-1}=\frac{2(2n-1)!}{n!(n-1)!}=\frac{2n(2n)!}{n!n!(2n)}=\frac{(2n)!}{(n!)^2}$$
$$ 4C^n_{2n-2}=\frac{4(2n-2)!}{n!(n-2)!}=\fra... | Let
\begin{align}
f(n)
&=\sum_{k=0}^n2^k\binom{2n-k}{n}\\
&=\sum_{h=0}^n2^{n-h}\frac{(n+h)!}{n!h!}
\end{align}
Then
\begin{align}
f(n+1)
&=\sum_{h=0}^{n+1}2^{n+1-h}\frac{(n+1+h)!}{(n+1)!h!}\\
&=\sum_{h=0}^{n+1}2\frac{n+1+h}{n+1}2^{n-h}\frac{(n+h)!}{n!h!}\\
&=2\sum_{h=0}^{n+1}2^{n-h}\frac{(n+h)!}{n!h!}+2\sum_{h=0}^{n+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3477228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Calculate $ f [f (x)] $ Where $ f (x) = \frac {1} {1 + x ^ 2} $, $ f [f (x)] $ is equal to:
a
I was factoring and found $f(f(x))= \frac{1}{1+(\frac{1}{1+x^2})^2} = \frac{1}{\frac{x^4+2x^2+2}{x^4+2x^2+1}}= \frac{x^4+2x^2+1}{x^4+2x^2+2}$, but the answer is x. I am wrong?
| Your calculation of
$$f(f(x))=\frac{x^4+2x^2+1}{x^4+2x^2+2}$$
is correct. Both the numerator and denominator are fourth degree polynomials with a coefficient of $1$ before the $x^4$ term. Therefore, there is a horizontal asymptote at $y=1$.
So, the answer cannot be $f(f(x))=x$. As the graph of $f(f(x))$ is bounded abo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3477710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Value of $x$ in floor Inequality: $\lfloor \sin^{-1}(x)\rfloor >\lfloor \cos^{-1}(x)\rfloor$
Find value of $x$ for which $\displaystyle \lfloor \sin^{-1}(x)\rfloor >\lfloor \cos^{-1}(x)\rfloor,$ is
What I try $\displaystyle \sin^{-1}(x)\in \bigg[-\frac{\pi}{2},\frac{\pi}{2}\bigg]$ and $\displaystyle \cos^{-1}(x)\in ... | $$\lfloor\arcsin x\rfloor>\lfloor\arccos x\rfloor\iff\arcsin x\ge\lfloor\arccos x\rfloor+1.$$
As the arc sine cannot exceed $\dfrac\pi2$ and $\arccos x\ge0$, we have the only case
$$\lfloor\arccos x\rfloor=0\to x>\cos1\land \arcsin x\ge 1\implies x\ge\sin 1.$$
Graphically:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3478170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Trapezoid Volume and Surface Area
Given $AB = 6$, $DC = 12$, $CG = 8$. Find the surface area and volume of the object.
I tried solving for h first. By inspection, you can see that the height is equal to the side of the 45°.
Using $h=(6-h)(\tan 60°)$. I got $h=9-3 \sqrt{3}$ then I plug it in to solve the volume.
$$V =... | The volume is completely correct. Since this is a prism, its volume is simply the area of trapezium $ABCD$ (or $EFGH$) multiplied by $CG$. So the volume is $\frac 12(12+6)(9-3\sqrt 3) \times 8 \approx 273.88$
For the area, just sum the areas of the $4$ rectangles and $2$ (identical) trapeziums.
$ABEF$ has area $8 \tim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3480615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Proof by mathematical induction proof verification If $n\in\Bbb N$ then $\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{n}{(n+1)!}=1-\frac{1}{(n+1)!}$.
The base case where $n=1$ is clearly true: $\frac{1}{2!}=\frac{1}{2}$ and $1-\frac{1}{(1+1)!}=\frac{1}{2}$.
Now to show that $S_k\Rightarrow S_{k+1}$, we assume that ... | Your proof is correct. Very well done. However, I would strongly recommend familiarising yourself with using sums as an easier way to express your work.
We are proving:
$$\sum_{r=1}^n\frac{r}{(r+1)!}=1-\frac{1}{(n+1)!} \tag A$$
1) Your base case was done well.
2) Assumption step, assume true for $n=k$
$$\sum_{r=1}^k\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3481040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
value of $n$ in limits
If $\displaystyle \lim_{x\rightarrow 0}\frac{x^n\sin^{n}(x)}{x^{n}-(\sin x)^{n}}$ is a finite non zero number.
Then value of $n$ is equals
What I try:
$$\sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots $$
$$\lim_{x\rightarrow 0}\frac{x^n\bigg(x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots... | The denominator factors as
$$(x-\sin x)\sum_{k=0}^{n-1} x^{n-k-1}\sin^kx\sim ax^3x^{n-1}$$ while the numerator is $\sim bx^{2n}$.
So $2n=n+2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3481642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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Prove that $R$ is reflexive, symmetric, and transitive. Define a relation $R$ on $\Bbb Z$ by declaring that $xRy$ if and only if $x^2\equiv y^2\pmod{4}$. Prove that $R$ is reflexive, symmetric, and transitive.
Suppose $x\in\Bbb Z$. Then $x^2\equiv x^2\pmod {4}$ means that $4\mid (x^2-x^2)$, so $x^2-x^2=4a$ where $a=0\i... | Your solution is ok. I’d just like to point out that your problem is in fact just a specific case of a much more general phenomenon:
Let $A$, $B$ be sets, $f:A\to B$ a function, and $R\subseteq B\times B$ an equivalence relation. Define a relation $S\subseteq A\times A$ such that $$(x,y)\in S\Leftrightarrow (f(x),f(y)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3488525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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Evaluate $\int \frac{\cos^2x}{1+\tan x}dx$ Evaluate $\int \dfrac{\cos^2x}{1+\tan x}dx$
Here are my various unsuccessful attempts:-
Attempt $1$:
$$\tan x=t$$
$$\sec^2 x=\dfrac{dt}{dx}$$
$$\int \dfrac{dt}{(1+t^2)^2(1+t)}$$
$$\ln(1+t)=y$$
$$\dfrac{1}{1+t}=\dfrac{dy}{dt}$$
$$\int \dfrac{dy}{\left(1+(e^y-1)^2\right)^2}$$
$$... | Your partial fraction decomposition fails because the appropriate choice for numerators are not constants, but instead $$\frac{1}{(1+t^2)^2(1+t)} = \frac{A}{1+t} + \frac{Bt + C}{1+t^2} + \frac{Dt + E}{(1+t^2)^2}.$$ When the denominator is an integer power of a quadratic polynomial, the numerator is a linear function o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3489368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Finding the determinant of a matrix given by three parameters.
Show that for $a,b,c\in\mathbb R$ $$\begin{vmatrix}b^2+c^2&ab&ac\\ba&c^2+a^2&bc\\ca&cb&a^2+b^2\end{vmatrix} = \begin{vmatrix}0&c&b\\c&0&a\\b&a&0\end{vmatrix}^2 = 4a^2b^2c^2. $$
There must be some trick, like using elementary row operations, to get the det... | Let $b=ax,c=axy$. Then:
$$\small\begin{vmatrix}b^2+c^2&ab&ac\\ba&c^2+a^2&bc\\ca&cb&a^2+b^2\end{vmatrix} =\begin{vmatrix}a^2x^2(1+y^2)&a^2x&a^2xy\\a^2x&a^2(1+x^2y^2)&a^2x^2y\\a^2xy&a^2x^2y&a^2(1+x^2)\end{vmatrix} =a^6x^2\begin{vmatrix}1+y^2&1&y\\1&1+x^2y^2&x^2y\\y&x^2y&1+x^2\end{vmatrix} =\\
\small\stackrel{R_1-R_2\to R... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3489938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Po-Shen Loh's new way of solving quadratic equation Quadratic equation, $ax^2+bx+c=0$ and its solution is quadratic equation, $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
Now setting $a=1$ then we have $x^2+bx+c=0$
$$x=\frac{-b\pm \sqrt{b^2-4c}}{2}$$ rewrite as
$$x=-\frac{b}{2}\pm \sqrt{\left(\frac{b}{2}\right)^2-c}$$
In thi... | For $b^2-4ac\geq0$ and $a=1$ they are the same:
$$\frac{-b+\sqrt{b^2-4ac}}{2a}=\frac{-\frac{b}{2}+\sqrt{\frac{b^2-4ac}{4}}}{a}=\frac{-\frac{b}{2}+\sqrt{\left(\frac{b}{2}\right)^2-ac}}{a}=-\frac{b}{2}+\sqrt{\left(\frac{b}{2}\right)^2-c}$$ and
$$\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{-\frac{b}{2}-\sqrt{\frac{b^2-4ac}{4}}}{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3490752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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How many ways can we get a number by addition if each part of the addition has to be smaller or equal to a set value? For example, if we need to get 5 with the largest number we can use being 3, we can use:
*
*3 + 2
*3 + 1 + 1
*2 + 2 + 1
*2 + 1 + 1 + 1
*1 + 1 + 1 + 1 + 1
Is there any way to find out the soluti... | We can calculate the number of ways with the help of generating functions. We encode the usage of
*
*zero or more $1$s as $1+x+x^2+x^3+\cdots=\frac{1}{1-x}$
*zero or more $2$s as $1+x^2+x^4+x^6+\cdots=\frac{1}{1-x^2}$
*zero of more $3$s as $1+x^3+x^6+x^9+\cdots=\frac{1}{1-x^3}$
and we look for the coefficient of... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\lim_{n\to\infty}\frac{e^{\frac{1}{n^2}}-\cos \frac{1}{n}}{\frac{1}{n}\log(\frac{n+1}{n})-(\sin\frac{2}{n})^2}$
$$\lim_{n\to\infty}\frac{e^{1/n^2}-\cos \frac{1}{n}}{\frac{1}{n}\log(\frac{n+1}{n})-(\sin\frac{2}{n})^2}, \quad n \in \mathbb{N}$$
I'm stuck trying to evaluate this limit. I have tried applying l'Hôpital's... | The best way to solve limits like these is to split them in to standard limits.
$\lim\limits_{n\to\infty}\frac{e^{\frac{1}{n^2}}-\cos {\frac{1}{n}}}{\frac{1}{n}\log (\frac{n+1}{n})-(\sin (\frac{2}{n}))^2}$
Hence, we can split the term in the numerator to apply standard logarithmic and trigonometric limits.
$\lim\limit... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Factoring $1 - y-x^2-y^2-yx^2+y^3$ I was struggling to factorize this expression, I need it cleaned to bracket form but I was struggling to understand the process to get the answer? Is there a special formula or technique for factorizing these sorts of expressions?
$$1 - y-x^2-y^2-yx^2+y^3$$
Answer: $$ \ ( 1+y)[(1-y)^2... | Note $x^2$ is the only $x$-term,
$$1 - y-x^2-y^2-yx^2+y^3$$
$$=1-y-y^2+y^3 -(1+y)x^2$$
$$=(1-y) -y^2(1-y) -(1+y)x^2$$
$$=(1-y)(1-y^2)-(1+y)x^2 =(1+y)(1-y)^2-(1+y)x^2 $$
$$= ( 1+y)[(1-y)^2-x^2] $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3495171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Find real $a$, $b$, $c$, $d$ satisfying $(1-a)^2+(a-b)^2+(b-c)^2+c^2=\frac{1}{4}$
If real numbers $a$, $b$, $c$, $d$ satisfy
$$(1-a)^2+(a-b)^2+(b-c)^2+c^2=\frac{1}{4}$$
then find $(a,b,c,d)$.
What I try:
$$1+2a^2+2b^2+2c^2-2a-2ab-2bc=\frac{1}{4}$$
$$8a^2+8b^2+8c^2-8a-8ab-8bc+7=0$$
How do I solve it? Help me, pl... | Apply CS inequality: $ LHS \ge \dfrac14\cdot ( 1- a + a - b + b - c + c )^2 = \dfrac14 = RHS $ with equality occurs when $1 - a = a - b = b - c = c$. From this you can find $a, b, c, d$. Specifically, $ b = 2c, a = 2b - c = 4c - c = 3c \implies 1 = 2a - b = 6c - 2c = 4c \implies c = \dfrac14, b = 2c = \dfrac12, a = 3c ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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evaluate the integral $\int x^3(5x^2+3)^5 dx$ How to evaluate for the integral:
$$\int x^3(5x^2+3)^5 dx$$
What I've tried:
For example,
(1) $u = x^2$ whence $du = 2x$ $dx$
(2)$u = ax^2+b$ whence $du = 2ax$ $dx$
Therefore,
(3) $u = 5x^2+3$ whence $du = 10x$ $dx$
$$I = \int \frac{du}{10}u^5 = \frac{1}{10} \int \frac{u^6... | There are a few small mistakes in your reasoning. First you forgot the $dx$. Your initial expression should be
$$I=\int x^3(5x^2+3)^5\ dx$$
Applying the change of variables
$$u=5x^2+3 \qquad du=10xdx$$
Then
$$I=\int x^3(5x^2+3)^5\ dx=\frac1{10}\int x^2(5x^2+3)^5\ 10xdx$$
We will need to get rid of $x^2$ as well.
$$u=5x... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is this solution correct for equation $3x^2-4y^2=13$? Prove equation $3x^2-4y^2=13$ has no integer solution.
Solution: Suppose (y, 3)=1, We have:
$y≡( 1, 2) \mod (3)$
⇒ $4y^2≡( 1, 2) \ mod(3)$
$3x^2≡0 \mod (3)$
The common remainder between $3x^2$ and $4y^2$ is 0 , so we may write:
⇒ $3x^2-4y^2≡ 0 \ mod(3)$
But, $13≡1 \... | If $3 \mid y$, then $3 \mid -4y^2$ and since $3 \mid 3x^2$, you would have $3 \mid 13$, which is not true. Thus, as you stated, $3 \not\mid y$, which means $y^2 \equiv 1 \pmod 3$.
As such, $3x^2 - 4y^2 \equiv -y^2 \equiv 2 \pmod 3$. Note, you made a mistake when you stated "common remainder between $3x^2$ and $4y^2$ is... | {
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"timestamp": "2023-03-29T00:00:00",
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$50\cos^2 x + 5\cos x = 6\sin^2 x$, find $\tan x$
$50\cos^2 x + 5\cos x = 6\sin^2 x$
Find $\tan x$
I used $\cos^2 x + \sin^2 x = 1$ to get the equation $$56\cos^2 x + 5\cos x -6 = 0$$
I then solve this to get $\cos x = \dfrac27, -\dfrac38$
Then I used generic trig ratios to get $\tan x = \pm\dfrac{3\sqrt5}{2}, \pm\df... | It's correct. The signs are necessary because $\cos x=\frac 27$ doesn't tell us which quadrant (either 1st or 4th) the argument is in. You can also use the identity
$$1+\tan^2x=\frac{1}{\cos^2 x}\Rightarrow \tan x=\pm\sqrt{\frac{1}{\cos^2 x}-1} $$
to directly compute $\tan x$ from $\cos x$.
| {
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"url": "https://math.stackexchange.com/questions/3501737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Summation to n terms of series \begin{gather}
In\ a\ problem\ given\ to\ us\ by\ our\ teacher,\ one\ part\ of\ it\ requires\ the\ summation\ of\ the\ \notag\\
following\ sequence\ to\ a\ finite\ number\ of\ terms\ ( n) , \notag\\
t_{r\ } \ =\ \frac{1}{r\times 2^{r}} \notag\\
\notag\\
S\ =\ \frac{1}{1\times 2} +\frac{... | Considering the actual probem
$$t_{r} =\frac{r+2}{r( r+1)} x^r=2\frac{x^r}{r}-\frac{x^r}{r+1}=2\frac{x^r}{r}-\frac 1x\frac{x^{r+1}}{r+1}$$
$$\sum_{r=1}^n t_r=2\sum_{r=1}^n \frac{x^r}{r}-\frac 1x\sum_{r=1}^n\frac{x^{r+1}}{r+1}$$ Differentiate both sides with respect to $x$
$$\left(\sum_{r=1}^n t_r \right)'=2\sum_{r=1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3503883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Intuitive way to prove that $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \cdots = 0$ I’m trying to come up with an intuitive way to prove that $$\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \cdots = 0$$ That is, I want to prove that $$\prod_{n=1}^\infty \frac {(1)^n}{2}=0$$
I thought of this wa... | Let $p_n = \prod_{i=1}^n \frac12 = \left(\frac12\right)^n$. Then $\frac{p_{n+1}}{p_n} = \frac12<1$ so $\{p_n\}$ is decreasing, and the product of positive numbers is a positive number, so $p_n>0$ for all $n$. Since $p_n$ is decreasing and bounded below, it converges to its infimum. Now, given $\varepsilon>0$, we may ch... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3504616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Factoring $3n^3 - 39n^2 + 360n + 20$ I am wondering how to factor
$$f(n) = 3n^3 - 39n^2 + 360n + 20$$ the right way. I think the factors are equal to
$$(n - 39.9762)(n - 12.0791)(n + 0.055248)$$
| Set
$$
P(x)=3x^3-39x^2+360 x+20.
$$
Then set $x=y+\frac{13}{3}$, then
$$
\frac{1}{3}P(x)=P_1(y):=y^3+\frac{191}{3}y+\frac{9826}{27}
$$
If $\rho_0=\sqrt[3]{A}+\sqrt[3]{B}$ is the real root of $P_1(y)=0$, then
$$
\rho_0^3=A+B+3(AB)^{1/3}(\sqrt[3]{A}+\sqrt[3]{B})=s+3p^{1/3}\rho_0.
$$
Hence $s=A+B=-\frac{9826}{27}$, $p=A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3507142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Is there an easy way to see that ${1\over5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12} > 1$? The sum $$\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12}$$ is just a bit larger than $1$. Is there som... | By C-S we obtain:
$$\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12}=$$
$$=\frac{17}{60}+\frac{17}{66}+\frac{17}{70}+\frac{17}{72}\geq\frac{17\cdot4^2}{60+66+70+72}=\frac{68}{67}>1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3514103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 1
} |
JBMO TST macedonia Number Theory
Find all $x,y$ positive integers $$x + y^2 + (\gcd(x, y))^2 = xy \cdot
\gcd(x, y)$$
I tried supposing $\gcd(x,y)=d$ and letting $x=ad$, $y=bd$ but didn't get anything the closest thing I think is usefull that $(b+1)^2+4a^2b$ is a perfect square.
| Your substitutions give the equation
$$ad+b^2d^2+d^2=abd^3$$
Modulo $bd^2-1$ we obtain $$ad+b+d^2\equiv ad.$$ Then $bd^2-1$ is a factor of $b+d^2$ and so $bd^2-1\le b+d^2$. Either $d=1$ or $d\ge2$ and $$b\le \frac {d^2+1}{d^2-1}=1+ \frac {2}{d^2-1}<2.$$
IF $b=1.$
Then $a+2d=ad^2$. Therefore $a|2d$ and $d|a$ and so $a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3518735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Determine all six exact values if $\sqrt[6]{\frac{2i}{1+i}}$. Determine all six exact values if $\sqrt[6]{\frac{2i}{1+i}}$, expressing each in simplified, exact rectangular form.
What I have done:
I set $\sqrt[6]{\frac{2i}{1+i}}=z$ and put the both to the sixth power to get ${\frac{2i}{1+i}}=z^6$. I then multiplied by ... | Use polar coordinates and De Moivre Theorem
Let $w = \frac {2i}{1+i}=\frac {2i(1-i)}{(1+i)(1-i)} =\frac {2+ 2i}{2} = 1 + i$
$w = re^{i(\theta + 2\pi k)} = r(\cos \theta + i\sin \theta) = 1 + i$ where
$r = \sqrt {1^2 + 1^2} =\sqrt 2$ and $\theta = \arctan \frac 11 = \frac \pi 4$.
So $w = \sqrt 2*e^{i(\frac \pi 4+ 2\pi ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3519234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\sqrt{a^2+4}+\sqrt{b^2+4}\leq\frac{\sqrt{2}}{4}(a+b)^2$ with $\frac1a+\frac1b=1$. Let $ a, b> 0 $ and $\frac{1}{a}+\frac{1}{b}=1.$ Prove that$$\sqrt{a^2+4}+\sqrt{b^2+4}\leq\frac{\sqrt{2}}{4}(a+b)^2.$$
Obviously $a+b=ab\ge4$. Other than that, I do not know what trick to use to deal with the constraint. The Lagra... | You can show it just by squaring, using GM-AM and applying the two facts
*
*$(1)$: $a+b = ab$ and
*$(2)$: $ab \geq 4$
LHS:
$$\left(\sqrt{a^2+4}+\sqrt{a^2+4}\right)^2 =a^2+b^2+8 + 2\sqrt{(a^2+4)(b^2+4)}\stackrel{GM-AM}{\leq}2(a^2+b^2+8)$$
RHS:
$$\left(\frac{\sqrt 2}{4}(a+b)^2\right)^2 \stackrel{(1)}{=} \frac 18 (a+b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3521160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
Evaluate a triple integral with spherical coordinates $$\iiint_{Q}z\ dV$$ Where $Q$ is the common region of the spheres $x^{2}+y^{2}+z^{2}\leq 1$ and $x^{2}+y^{2}+(z-1)^{2}\leq 1$
I have tried nothing.
| We have to use spherical coordinates $(x=r\sin\theta\cos\varphi, \quad y=r\sin\theta\sin\varphi, z=r\cos\theta)$, so we will have $$x^2+y^2+z^2 \le 1 \Leftrightarrow r^2 \le 1 \Leftrightarrow r \le 1$$ and $$x^2+y^2+(z-1)^2 \le 1 \Leftrightarrow x^2+y^2+z^2 \le 2z \Leftrightarrow r^2 \le 2r\cos\theta \Leftrightarrow r ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3523733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How do we integrate the expression $\frac{b^{3}}{1 - \theta(1-b)^{2}}$ where $0\leq b\leq 1$ and $\theta\in(0,1)$? MY ATTEMPT
In order to solve this integral, I have tried using the integration by parts method, as suggested by the expression
\begin{align}
\int_{0}^{1}\frac{b^{3}}{1-\theta(1-b)^{2}}\mathrm{d}b = \int_{0... | In a similar way, we can make this:
$b^3=\dfrac{-1}{\theta}(b-\theta b+2\theta b^2-\theta b^3-b+\theta b-2\theta b^2)$
$\dfrac{b^3}{1-\theta (1-b)^2}=\dfrac{-1}{\theta}(b+\dfrac{-b+\theta b-2\theta b^2}{1-\theta (1-b)^2})$
We can write the last term as $2\dfrac{\frac{-b}{2}+\frac{b\theta}{2}-\theta b^2}{1-\theta (1-b)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3525288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Square numbers and divisibility I'm trying to prove the following statement:
If $kn+1$ and $(k+1)n+1$ are both perfect squares, then $2k+1$ divides n, where k and n are positive integers.
Case $k=1$:
Put $n+1=x^2$, $2n+1=y^2$. Quadratic residues can only be $0, 1 \mod 3$.
Hence
$n \equiv 1 \mod 3 \Rightarrow x^2 = n+... | This answer explains how your question relates to the Pell's equation and points you to various online sources which can help you to solve your problem.
The equation you have, where I've switched the $x$ and $y$ around, of
$$(k+1)y^2 - kx^2 = 1 \implies kx^2 - (k+1)y^2 = -1 \tag{1}\label{eq1A}$$
is a form of a generali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3526173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Representing $f(x)=\frac{1}{x^2}$ as power series, with powers of $(x+2)$ As stated in the title, I want $f(x)=\frac{1}{x^2}$ to be expanded as a series with powers of $(x+2)$.
Let $u=x+2$. Then $f(x)=\frac{1}{x^2}=\frac{1}{(u-2)^2}$
Note that $$\int \frac{1}{(u-2)^2}du=\int (u-2)^{-2}du=-\frac{1}{u-2} + C$$
Therefore... | Narrowly on your question: no, it is not correct. But it is almost correct.
The error is that just before the end you forgot to get the range for $n$ correct. When you differentiate a term in $x^0$ you get 0, not a term in $x^{-1}$.
As a quite separate point there are other ways of getting a power series for $\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3527004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Numbers $x,y,z$ are such that $x+y+z=0$ and $x^2+y^2+z^2=1$. Show that at least one of $xy,xz,yz$ is not bigger than $-\frac{1}{3}$. So i had to prove:
Numbers $x,y,z$ are such that $x+y+z=0$ and $x^2+y^2+z^2=1$. Show that at least one of $xy,xz,yz$ is not bigger than $-\frac{1}{3}$.
Here is my proof:
So we have
$(x... | I found another, easier answer, so I thought it might be worth posting it.
Without loss of generality we may assume that $x \leqslant y \leqslant z$. From our assumptions we see that $x<0$ and $0<z<1$. If $0 \,\leqslant y$, then
$$1=x^2+y^2+z^2=(-y-z)^2+y^2+z^2=2y^2+2z^2+2yz \geqslant 2y^2+2y^2+2y^2=6y^2$$
If $y<0%$:
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3527542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Edwards Calculus chapter 4 problem 81 I can see that the terms arise from the differentiation of $\log(1+x^{2^r} )$ and tried to do it but couldn't figure it out in the end.
Show that
$$
\frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \frac{8x^7}{1+x^8} + \ldots = \sum_{n=0}^{\infty} \frac{2^{n} x^{(2^n -1)}... | Let's assume
$$S=\dfrac{1}{1+x}+\dfrac{2x}{1+x^2}+\dfrac{4x^3}{1+x^4}+...$$
The common term of the series, as seen from the pattern, is $\displaystyle \frac{2^{n} x^{(2^n -1)}}{1+x^{(2^n)}}$ for $n=0,1,2,3,...$ and the sum is then $\displaystyle\sum_{n=0}^{\infty} \frac{2^{n} x^{(2^n -1)}}{1+x^{(2^n)}}$.
Now, noting t... | {
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"url": "https://math.stackexchange.com/questions/3527860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Solve $\sin x + \cos x = \sin x \cos x.$ I have to solve the equation:
$$\sin x + \cos x = \sin x \cos x$$
This is what I tried:
$$\hspace{1cm} \sin x + \cos x = \sin x \cos x \hspace{1cm} ()^2$$
$$\sin^2 x + 2\sin x \cos x + \cos^2 x = \sin^2 x \cos^2x$$
$$1 + \sin(2x) = \dfrac{4 \sin^2 x \cos^2x}{4}$$
$$1 + \sin(2x) ... | Using auxiliary angle, we first change the equation
$$
\sin x+\cos x=\sin x \cos x \tag*{(*)}
$$
equivalently to
$$
\sqrt{2} \cos \left(x-\frac{\pi}{4}\right)= \frac{\sin 2 x}{2}
$$
Putting $y=x-\dfrac{\pi}{4} $ gives a quadratic equation in $\cos y $
$$
\sqrt{2} \cos y=\frac{1}{2}\left(2 \cos ^{2} y-1\right)
$$
Solv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3530949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 3
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Combination of problem having identical objects without using generating function approach
A person goes in for an examination in which there are $4$ papers with a maximum of $m$ marks from each paper. The number of ways in which one can get $2m$ marks is
My Attempt
Number of ways of partitioning $n$ identical object... | You should consider $\#\{\text{strictly more than $m$ marks for paper $j$}\}$. So we start with $m+1$ marks in paper $j$ and then distribute the remaining $m-1$ marks among the 4 papers. Thus we get $$\frac{(2m+3)(2m+2)(2m+1)}{3\cdot2}-4\frac{(m+2)(m+1)m}{3\cdot2}$$
$$=\frac{m+1}{3}(4m^2+8m+3-2m^2-4m)$$
$$=\frac{(m+1)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3531319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solve $T(n) = 2^nT(n/2) + n^n$ Let $$T(n) = 2^nT(n/2) + n^n$$ Can't be solved using the master theorem, because the equation doesn't satisfy $$T(n) = aT(n/b) + f(n)$$
How would you approach this to find the time complexity?
| If
$T(n) =
2^nT(n/2) + n^n
$,
then,
putting $2^m$ for $n$,
$T(2^m)
= 2^{2^m}T(2^{m-1}) + (2^m)^{2^m}
= 2^{2^m}T(2^{m-1}) + 2^{m2^m}
$.
Let
$U(m) = T(2^m)$.
then
$U(m)
= 2^{2^m}U(m-1) + 2^{m2^m}
$.
Since
$\sum 2^m = 2^{m+1}$,
divide by
$2^{2^{m+1}}$
to get
$\dfrac{U(m)}{2^{2^{m+1}}}
= \dfrac{2^{2^m}U(m-1)}{2^{2^{m+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3533187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Algebraic and Trigonometric expression is $>0$ for all real $x$
Prove that $$2x^2\sin x+2x\cos x+2x^2+1$$ is always positive for all real $x$.
From completing the square method
Write $1$ as $\sin^2 x+\cos^2 x$
$$x^4+2x^2\sin x+\sin^2 x+x^2+2x\cos x+\cos^2 x+x^2-x^4$$
$$(x^2+\sin x)^2+(x+\cos x)^2+x^2(1-x^2)$$
$\bulle... | For $\sin{x}+1>0$ we obtain: $$2x^2\sin x+2x\cos x+2x^2+1=2(1+\sin{x})x^2+2x\cos{x}+1\geq0$$ because
$$\frac{\Delta}{4}=\cos^2x-2(1+\sin{x})=-(1+\sin{x})^2<0.$$
If $\sin{x}+1=0$ so $$2x^2\sin x+2x\cos x+2x^2+1=1>0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3533800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Integrate $\int \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}}dx$
Evaluate
$$\int \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$$
I have tried substitution of $\sin x$ as well as $\cos x$ but it is not giving an answer.
Do not understand if there is a formula for this or not.
| Recall that $$\begin{align}\sin(x) &= \frac{\tan(x)}{\sec(x)} \\ \cos(x) &= \frac1{\sec(x)} \\ \sec^2(x) &= 1 + \tan^2(x)\end{align}$$
Then,
$$\int\frac{\sqrt{\sin(x)}}{\sqrt{\sin(x)} + \sqrt{\cos(x)}}\,\mathrm dx\equiv\int\sec^2(x)\cdot\frac{\sqrt{\tan(x)}}{\sqrt{\tan(x)} \left(1 + \tan^2(x)\right) + 1 + \tan^2(x)}\,\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3535688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Find $f(x) = \lim_{n\to\infty}\frac{\lfloor x \rfloor + \cdots + \lfloor x^n \rfloor}{x^n}$ I found this cool problem in a textbook. I googled it and used MSE's search tool to check if it has been asked before or not, but it seems that it hasn't been asked before.
Find the function $f(x)$ that the following limit defi... | Solution: We know that $x-1<\lfloor x \rfloor \leq x$, hence
$$x + \cdots + x^n - n < \lfloor x \rfloor + \cdots + \lfloor x^n \rfloor \leq x + \cdots + x^n$$
Dividing both sides by $x^n$
$$\frac{x + \cdots + x^n}{x^n} - \frac{n}{x^n} < \frac{\lfloor x \rfloor + \cdots + \lfloor x^n \rfloor}{x^n} \leq \frac{x + \cdots ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3538941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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About the sequence: $1, 3, 8, 24, 29, 87, 92, ?...$ find $a_n$ This is a classic and curious recurrence sequence used in logic tests. Your rule can be determined as follows:
for $n$ even: $a_n = 3a_{n-1}$
for $n$ odd: $a_n = a_{n-1} + 5$
That is, the ratio alternates with each term.
Will there be a single formula for $... | For an even value
\begin{eqnarray*}
a_{2n}=3a_{2n-1}=5+3 a_{2n-2}.
\end{eqnarray*}
Iterate this
\begin{eqnarray*}
a_{2n}=5+3\times 5 +3^2 \times 5 + \cdots 3^{n-1} \times 5 +3^n.
\end{eqnarray*}
Sum the geometric series
\begin{eqnarray*}
a_{2n}=5\frac{3^{n}-1}{3-1}+3^n = \frac{7 \times 3^n -5}{2}.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3542118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Deriving $\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$ using Fourier Series for $f(x)=|x|$ One can find that $$f(x) = \frac{\pi}{2} + \sum_{n=1}^{\infty}\frac{2(-1+(-1)^n)}{\pi n^2}\cos(nx)$$
Now look at the case for $x=0$. We can find that $\sum_{n \text{ odd}, n \geq 1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{8}$.... | Rearranging gives
\begin{eqnarray*}
\sum_{n=1}^{\infty} \frac{1+(-1)^{n+1}}{n^2} =\frac{\pi^2}{4}.
\end{eqnarray*}
Now let
\begin{eqnarray*}
\sum_{n=1}^{\infty} \frac{1}{n^2} =x.
\end{eqnarray*}
So
\begin{eqnarray*}
\sum_{n=1}^{\infty} \frac{1}{(2n)^2} =\frac{x}{4} \\
\sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} =\frac{3x}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3543368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $a^{b} \equiv 3 \,( \text{mod}\, 4)$ implies $a,b$ odd. So far I've shown that
$a^{b} \equiv 3 \,( \text{mod} \, 4) \implies a^{b} \,\text{odd}\implies a \, \text{odd}$. I also know that since $a^{b} \equiv 3 \, (\text{mod} \, 4)$, there exists prime $p \mid a^{b}$ such that $p \equiv 3 \, (\text{mod} \, 4)... | As you've shown, $a$ must be odd as any even value can't be congruent to $3$ modulo $4$. Thus, either $a \equiv 1 \pmod 4$, which doesn't work as $a^b \equiv 1 \pmod 4$ for any $b$, or $a \equiv 3 \pmod 4$. For the second case, note $a^2 \equiv 3^2 \equiv 1 \pmod 4$. Thus, for any even $b = 2c$ for some integer $c$, yo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3543665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Interval of convergence of a series How to get the interval of convergence for the given function,
$$f(x) = \frac{1}{2+x-x^2}$$
I have computed the Maclaurin series and the generalized power series as follows however I am unable to proceed with the valid interval of convergence. Kindly help.
$$\frac{1}{2+x-x^2} = \sum_... | We can also calculate the partial fraction decomposition
\begin{align*}
\frac{1}{2+x-x^2}=\frac{1}{3(x+1)}-\frac{1}{3(x-2)}\tag{1}
\end{align*}
and obtain this way geometric series. We see the radius of convergence of the geometric series at $x=0$ is $1$ resp. $2$. So, the radius of convergence of the le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3544273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
recursive sequence such as $x_{n+2}-2*x_{n+1}+4*x_{n}=0$ I have the recursive sequence such as $x_{n+2}-2*x_{n+1}+4*x_{n}=0$
We are looking for some solutions in the form $x_n=A* \lambda^n $
The equation becomes $A* \lambda ^{n+2}-2A \lambda ^{n+1}+4* A\lambda ^n=0
\longrightarrow P( \lambda)=\lambda ^2 - 2\lambda ... | The sine term's coefficient needs to be real for $x_n$ to be real. If $x_n$ is allowed to be non-real complex, so are the coefficients.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3545531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$m=\sum_{k=1}^{2017}k^3\binom{n}{k}$. Find m. $$m=\sum_{k=1}^{2017}k^3\binom{n}{k}$$
Find m.
I was given this question for a math class I'm taking and I don't really know how to start or what identities to use. Could someone give me a hint?
| The problem seems to mix up $n$ and $2017$ somehow. We can show
\begin{align*}
\sum_{k=1}^nk^3\binom{n}{k}=2^{n-3}n^2(n+3)\tag{1}
\end{align*}
and a special case is evaluating (1) at $n=2\,017$.
We obtain
\begin{align*}
\color{blue}{\sum_{k=1}^n}&\color{blue}{k^3\binom{n}{k}}=n\sum_{k=1}^nk^2\binom{n-1}{k-1}\tag{2}\... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
minimum value of $f(x)=\frac{(x^2-x+1)^3}{x^6-x^3+1}$
If $f:\mathbb{R}\rightarrow \mathbb{R}.$ Then minimum value of $$\displaystyle f(x)=\frac{(x^2-x+1)^3}{x^6-x^3+1}$$
what i try
If $x\neq 0,$ Then divide numerator and denominator by $x^3$
$$f(x)=\frac{\bigg(x+\frac{1}{x}-1\bigg)^3}{x^3+\frac{1}{x^3}-1}$$
put $\... | Here is a 'non-calculus' solution.
We want to minimize the function $f:(-\infty,-2] \cup [2,\infty) \to \mathbb{R}$
$$f(t) = \frac{(t-1)^3}{t^3-3t-1}$$
Let's pretend we don't know calculus and that we're really clever to spot that for $t=1+\sqrt{3}$, we have $f(t)=2\sqrt{3}-3$. So, we will prove:
$$\frac{(t-1)^3}{t^3-3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3552636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Solving an integral: $\int \frac{x}{x^3-1}\,\mathrm dx$ I'm trying to solve this integral:
$$\int \frac{x}{x^3-1}\,\mathrm dx$$
What I did was:
$$\int \frac{x}{(x-1)(x^2+x+1)}\,\mathrm dx.$$
$$\frac{x}{(x-1)(x^2+x+1)} = \frac{a}{x-1}+ \frac{bx+c}{x^2+x+1}$$
Then I got this in the numerator:
$$ax^2+ax+a+bx^2-bx+cx-c $$
... | You can write that
$$
x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}=\frac{3}{4}\left(1+\left(\frac{2}{\sqrt{3}}\left(x+\frac{1}{2}\right)\right)^2\right)
$$
and you can then put your fraction under the form
$$
\frac{u'}{1+u^2}
$$
that integrated to an arctan.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3553050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Prove that $6 | (a+b+c)$ if and only if $6 | (a^3 + b^3 +c^3).$ I have tried the question but not sure if my solution is correct or not...
My try..
\begin{align}a^3 + b^3 + c^3 = (a+b+c)^3 - 3(a+b)(b+c)(c+a)\end{align}
So , if
\begin{align}6 |(a^3 + b^3 + c^3)\end{align}
Then,
\begin{align}6 | [(a+b+c)^3 - 3(a+b)(b+c)(... | The reason for your incorrect solution has been pointed out by Iris in the comments.
\begin{align}
&(a^3+b^3+c^3) - (a+b+c) = (a^3 - a) + (b^3-b) + (c^3-c) \\ &\quad \quad \ = (a-1)a(a+1) + (b-1)b(b+1) + (c-1)c(c+1)
\end{align}
The set of terms in $(a-1)a(a+1)$ contains at least one multiple of $2$ and one multiple of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3554694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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A simple equation involving trigonometry If $$\cos\theta+ \sin\theta + \tan\theta=\sqrt2+1$$
then what is the maximum value of$$\sin\theta,\sin^2\theta?$$
| Let $a=\sin \theta,\ b=\cos \theta$. Then $a^2+b^2=1$ and
$$a+b+\frac{a}{b}=\sqrt{2}+1$$
Notice that:
$$a+b\leq |a+b| \leq \sqrt{2(a^2+b^2)} = \sqrt{2}$$
and thus we deduce that $\dfrac{a}{b}\geq 1$. So $a,b$ have the same sign. Since we want the maximum of $a$, we are only interested in the case when $a$ and $b$ are ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3555832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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sum of binomial series with alternate terms
Evaluation of series $\displaystyle \sum^{n}_{k=0}(-4)^k\binom{n+k}{2k}$
what I tried:
from Binomial Identity
$$\binom{n+k}{2k}=\binom{n+k-1}{2k}+\binom{n+k-1}{2k-1}$$
series is $$\sum^{n}_{k=0}(-4)^k\binom{n+k-1}{2k}+\sum^{n}_{k=0}(-4)^k\binom{n+k-1}{2k-1}$$
let $\displays... | Let
$$S_n=\displaystyle \sum^{n}_{k=0}(-4)^k\binom{n+k}{2k}\tag{*}$$
This issue can be connected in a natural way to Chebyshev polynomials of the second kind, with explicit expression see Wikipedia article.
$$U_n(x)=\sum_{k=0}^n(-2)^k\binom{n+k+1}{2k+1}(1-x)^k\tag{1}$$
If we take $x=-1$ in (1),
$$U_n(-1)=\sum_{k=0}^n... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Find all the relative extrema of $f(x)=x^4-4x^3$ Find all the relative extrema of $f(x)=x^4-4x^3$
$Solution:$
Step 1: Solve $f'(x)=0$.
$f'(x)=4x^3-12x^2=0$
$\rightarrow$ $4x^2(x-3)=0$
$\rightarrow$ $x=0$ and $x=3$
Step 2: Draw a number line and evaluate the sign of the derivative on each section (I don't know how to d... | Solution without derivative.
By AM-GM $$x^4-4x^3=3\cdot\frac{x^4}{3}+27-4x^3-27\geq4\sqrt[4]{\left(\frac{x^4}{3}\right)^3\cdot27}-4x^3-27=$$
$$=4|x^3|-4x^3-27\geq-27.$$
The equality occurs for $x=3,$ which says that we got a minimal value.
The maximum does not exist because $$\lim\limits_{x\rightarrow+\infty}(x^4-4x^3)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3556380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Taylor's Polynomial Evaluation Hi guys now started off doing some taylor expansion wanted some help with the following question:
$$ f:[0,\infty)\to \mathbb{R}$$
$$ f(x) = 2e^{-x/2} + e^{-x} $$
Determine the nth degree taylor's polynomila $T_n$ of f about c = 0
My attempt to this question can be seen below;
$$P(x) = f(x... | If you start with the usual
$$e^{-t}=\sum_{n=0}^\infty \frac{(-1)^n}{n!} t^n$$ and apply it twice
$$f(x)=2 e^{-x/2}+e^{-x}=2\sum_{n=0}^\infty \frac{(-1)^n}{2^n\,n!}x^n+\sum_{n=0}^\infty \frac{(-1)^n}{n!} x^n=\sum_{n=0}^\infty \frac{(-1)^n}{n!}\left(1+2^{-n+1} \right)x^n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3559213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find minimum of $a^2+b^2+\frac{a^2b^2}{(a+b)^2}-\frac{2(a^2+ab+b^2)}{a+b}$ If $a,b$ are real numbers, find the minimum value of:
$$a^2+b^2+\frac{a^2b^2}{(a+b)^2}-\frac{2(a^2+ab+b^2)}{a+b}$$
This is what I did: I tried some values and I set $a=0$. Then, it becomes a quadratic of $b$:
$$b^2-2b$$
Here, the minimum is $-1$... | If you see this as a map $\mathbb{R}^2 \to \mathbb{R}$, it is differentiable on $\mathbb{R}^2 \backslash \{a=-b\}$. You can calculate the critical points and the hessian to determine if it is a minimum.
This only works to find a local minimum, so you should also look at the behaviour around the line $a=-b$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3562535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Evaluate $\int _0^{2 \pi }\int _0^{2 \pi }\log (3-\cos (x+y)-\cos (x)-\cos (y))dxdy$ How to prove
$$\small\int _0^{2 \pi }\int _0^{2 \pi }\log (3-\cos (x+y)-\cos (x)-\cos (y))dxdy=
-4 \pi ^2 \left(\frac{\pi }{\sqrt{3}}+\log (2)-\frac{\psi ^{(1)}\left(\frac{1}{6}\right)}{2 \sqrt{3} \pi }\right)$$
Where $\psi^{(1)}$ deno... | This is easily tackled by a series of substitution, let $I$ be your integral, then
$$\begin{aligned}I &= \int_0^{2\pi} \int_0^{2\pi} \log(3-2\cos \frac y2 \cos(x+\frac y2)-\cos y) dxdy\\
&= \int_0^{2\pi} \int_0^{2\pi} \log(3-2\cos \frac y2 \cos x-\cos y) dxdy\\
&= 4\pi\int_0^{\pi} \log\left[ \frac{1}{2} \left(-\cos \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3562786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Taylor approximation is different from the actual value The problem is to use the $n$th Taylor polynomial to approximate the value of $F(x)=x^{2}\cos{x}$ with $n=3,x_{0}=-2,x=-1.09$.
I found that the $3$rd Taylor polynomial with remainder term is \begin{align*}
x^{2}\cos{x} =& 4\cos(2)+[-4\cos(2)+4\sin(2)][x+2]+[-\cos... | Expand just the cosine series around $x=-2$ and you will have:
$$x^2cos x = x^2\left(cos(2) + sin(2)[x+2] - \frac{cos(2)[x+2]^2}{2} - \frac{sin(2)[x+2]^3}{6}\right )$$
$$\\$$
(Expanding the $x^2$ in a Taylor polynomial and multiplying everything by brute force also works, but the Taylor series for $x^2$ will just reduc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3563017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Limit of $\lim_{x\to 3} \frac{3-x}{6\sin \frac{\pi x}{18}-x}$ I want to solve a limit without l'Hospital, just with algebraic manipulation:
$$\lim_{x\to 3} \frac{3-\sqrt{6+x}}{6\sin \frac{\pi x}{18}-x}$$
I started with:
$$\lim_{x\to 3} \frac{3-\sqrt{6+x}}{6\sin \frac{\pi x}{18}-x}=\lim_{x\to 3} \frac{(3-\sqrt{6+x})(3+\... | I assume you meant elementary limits are allowed. I'll use:
$$\lim_{x\to 0} \frac{\sin x}{x}=1$$
Instead of the fraction, it's easier to look at its inverse:
$$
\begin{aligned}
\lim_{x\to 3} \frac{6\sin \frac{\pi x}{18}-x}{3-x}&=1+\lim_{x\to 3}\frac{6\sin \frac{\pi x}{18}-3}{3-x}\\
&= 1+ 6\lim_{x\to 3}\frac{\sin \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3563204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Prove or Disprove this statement . I can't find a counter-example to the following statement :
Let $a,b,c>0$ such that $a+b+c=1$ and $a\geq b\geq c$ and $13a^2+5b^2\geq 13b^2+5c^2\geq 13c^2+5a^2$ then $\exists n>1$ such that :$$\frac{a^3}{n}+ \frac{(n-1)(13a^2+5b^2)}{54n}\geq \frac{13a^2+5b^2}{54}$$
$$\Big(\frac{a^3}... | Put $N=54$, $m=n-1>0$, $x=13b^2+5c^2$, $y=13c^2+5a^2$, and $z=13a^2+5b^2$. Then the second inequality transforms to $rm+s\ge 0$, where $r=N(a^3x+b^3z)-2xz$ and $s=N^2a^3b^3-xz$. The third inequality transforms to
$um^2+vm+w\ge 0$, where $u=N(a^3xy+b^3yz+c^3xz)-3xyz$, $v=N^2(a^3b^3y+a^3c^3x+b^3c^3z)-3xyz$, and $w=N^3a^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3563335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding pattern in a sequence of polynomials So I have a set of polynomials, with variables $n_0, n_1, n_2$. I would like to figure out the general formula for these polynomials given the number.
$$f(2) = n_{0} + 2 n_{1} + n_{2}$$
$$f(3) = n_{0}^{3} + 3 n_{0} n_{1}^{2} + 3 n_{1}^{2} n_{2} + n_{2}^{3}$$
$$f(4) = n_{0... | I think it is
$$
f(k)=\sum\limits_{j = 0}^{k} \binom{k}{j}n_0^{\frac{(k - j)(k - j - 1)}{2}} n_1^{(k-j)j} n_2^{\frac{j(j - 1)}{2}}
$$
for $k\geq 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3565816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate the summation $\sum_{n=1}^{10} n \left( \frac{1^2}{1 + n} + \frac{2^2}{2 + n} + ....+\frac{10^2}{10 + n}\right)$ $$P = \sum_{n=1}^{10} n \left( \frac{1^2}{1 + n} + \frac{2^2}{2 + n} + ....+\frac{10^2}{10 + n}\right)$$
My attempt: Get the constant square terms out and evaluate the summations left in $n$
$$P = 1... | $P=S(10)$ where $$S(m)=\sum_{n=1}^m\sum_{k=1}^m\frac{nk^2}{k+n}=\frac{1}{2}\sum_{n=1}^m\sum_{k=1}^m\frac{nk^2+kn^2}{k+n}=\frac{1}{2}\sum_{n=1}^m\sum_{k=1}^m nk=\frac{1}{2}\left(\sum_{n=1}^m n\right)^2\color{LightGray}{=\ldots}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3568642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Prove $\sqrt[3]{\frac{(a^4+b^4)(a^4+c^4)(b^4+c^4)}{abc}}\ge \sqrt{\frac{(a^3+bc)(b^3+ca)(c^3+ab)}{1+abc}}$ If $a,b,c>0$, prove that:
$$\sqrt[3]{\frac{(a^4+b^4)(a^4+c^4)(b^4+c^4)}{abc}}\ge \sqrt{\frac{(a^3+bc)(b^3+ca)(c^3+ab)}{1+abc}}$$
My try: I used $a^4+b^4 \geq ab(a^2+b^2)$ to get:
$$(a^4+b^4)(a^4+c^4)(b^4+c^4) \geq... | Raising to the $6$th power, the inequality is equivalent with:
$$(1+abc)^3(a^4+b^4)^2(b^4+c^4)^2(c^4+a^4)^2 \geq a^4b^4c^4(a^3+bc)^3(b^3+ca)^3(c^3+ab)^3$$
This is wrong for large variables as Michael Rozenberg stated. However, I found that
$$(1+abc)^3(a^4+b^4)^2(b^4+c^4)^2(c^4+a^4)^2 \geq a^2b^2c^2(a^3+bc)^3(b^3+ca)^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3569404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Integrate $\int \frac{1}{1+x+x^4}dx$
Find $$\int \frac{{\rm d}x}{1+x+x^4}.$$
WA gives a result, which introduces complex numbers. But according to the algebraic fundamental theorem, any rational fraction can be integrated over the real number field.
How to do this?
| Factorize
$$x^4 +x+ 1= \left(x^2+ax+\frac{a^3-1}{2a}\right)\left(x^2-ax+\frac{a^3+1}{2a}\right)$$
where $a$ satisfies $a^6-4a^2-1=0$, or
$$a=\frac2{\sqrt[4]3}\sqrt{\cos\left(
\frac13\cos^{-1}\frac{3\sqrt3}{16} \right)}
$$
Then, decompose the integrand
$$\frac1{1+x+x^4}
= \frac a{2a^6+1} \bigg(\frac{2a^2x+2a^3+1}{x^2+ax... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3572565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
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Prove $\sum_{n=1}^{\infty}\frac{n-1}{2^{n+1}}=\frac{1}{2}$, without using any $\sum\frac{n}{2^{n}}$ Problem: Prove $\sum_{n=1}^{\infty}\frac{n-1}{2^{n+1}}=\frac{1}{2}$, without using any $\sum\frac{n}{2^{n}}$.
I manage to make progress, but then end up always getting a $\sum\frac{n}{2^{n}}$. For example, $\sum_{n=1}^{\... | Render
$S=\dfrac{0}{2^1}+\dfrac{1}{2^2}+\dfrac{2}{2^3}+...$
Divide by $2$:
$S/2=\dfrac{0}{2^2}+\dfrac{1}{2^3}+\dfrac{2}{2^4}+...$
Now take the difference between terms with like denominators, e.g. $1/2^2$ from the first equation minus $0/2^2$ from the second. This gives
$S-S/2=\dfrac{0}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3580273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Using the necessary and sufficient conditions compute all minima and maxima of a given function
Using the necessary and sufficient conditions compute all minima and maxima of
$$f(x,y) = x^2 + 8x\cos(y) − 2x + 10\cos(y) − 8\cos^2(y) + 8\cos^3(y)$$
for $(x,y) \in \mathbb{R}^2$.
My attempt:
Take the partial deriva... | If $2x+8\cos(y)−2=0$ then
$$\cos(y)=\frac{1-x}{4}.$$
Moreover $-8x-10+16\cos(y)-24\cos^2(y)=0$ implies
$$4x=-5+8\cos(y)-12\cos^2(y)=-5+2(1-x)-(3/4)(1-x)^2$$
that is
$$x^2+6x+5=0$$
which has two real solutions: $x=-1$ and $x=-5$.
Hence
1) if $x=-1$ then $\cos(y)=1/2$ and we have further critical points:
$$(-1,\pi/3+2... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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For non-negative reals $a$, $b$, $c$, show that $3(1-a+a^2)(1-b+b^2)(1-c+c^2)\ge(1+abc+a^2b^2c^2)$ A 11th grade inequality problem:
Let $a,b,c$ be non-negative real numbers. Prove that
$$3(1-a+a^2)(1-b+b^2)(1-c+c^2)\ge(1+abc+a^2b^2c^2)$$
Do you have any hints to solve this inequality? Any hints would be fine.
I t... | The Macavity's beautiful idea we can release also by C-S:
$$2(1-a+a^2)(1-b+b^2)\geq2\sqrt{\frac{1+a^4}{2}}\cdot\sqrt{\frac{1+b^4}{2}}\geq1+a^2b^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3582641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Integrating odd Legendre polynomials using generating function I must show using generating function of Legendre polynomials, that
\begin{align}
\int_0^1 P_{2n+1}(x)\phantom{1}dx = (-1)^n\frac{(2n)!}{2^{2n+1}n!(n+1)!}
\end{align}
My attempt is to change the generating function $\Phi(u,x)=(1-2ux+u^2)^{-1/2}=\sum_{n=0}^{... | Expand the result of the integral of the generating function with Taylor series, using
$$\sqrt{1+z} = \sum_{n=0}^\infty \binom{\frac{1}{2}}{n}z^n = 1+\sum_{n=0}^\infty \frac{(-1)^n\binom{2n}{n}z^{n+1}}{2^{2n+1}(n+1)}$$
As follows:
\begin{equation}
\begin{split}
I &= \int_0^1 (1-2ux+u^2)^{-1/2}dx\\
&= \left[-\frac{1}{u}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3586831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove $\tan^{-1}m+\tan^{-1}n=\cos^{-1}\frac{1-mn}{\sqrt{1+m^2}\sqrt{1+n^2}}$ (part of sum) I'm looking for a specific clarification for a part of the solution in proving the following identity.
$\tan^{-1}m+\tan^{-1}n=\cos^{-1}\frac{1-mn}{\sqrt{1+m^2}\sqrt{1+n^2}}$
Here I'm taking, $\theta=\tan^{-1}m;$ $-\pi/2<\theta<\p... | I don't have means to put up the image for my plot so bear with me.
Take a right-angled triangle $ABC$ such that AB is the height $=m$ and BC is the base $=1$. $\widehat{ABC}=90^\circ$ and $\widehat{ACB}=\tan^{-1} m$
Let $AC=y$. Draw a line $AE$ perpendicular to $AC$ at $A$ equal to $ny$. $\widehat{EAC}=90^\circ$ and $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3587758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proving inequality by induction produces an unclear answer I need to prove following assertion by induction: $∀n ∈ N , ∀ a,b >0: a^{n-1}b \leq (\frac{(n-1)a+b}{n})^n$
After successfully proving the assertion for n = 1, I establish the following assumption:
$a^{k-1}b \leq (\frac{(k-1)a+b}{k})^k$
Now when I try to p... | The inequality is
$a^{n-1}b
\leq (\frac{(n-1)a+b}{n})^n
= (a+\frac{b-a}{n})^n
= a^n (1+\frac{b/a-1}{n})^n
$
or
$b/a
\le
(1+\frac{b/a-1}{n})^n
$
or,
letting $r = b/a$,
$r
\le (1+(r-1)/n)^n
$
or,
letting $r = 1+ns$,
$1+ns
\le (1+s)^n
$.
This is Bernoulli's inequality
which is easily proved
by induction.
The indu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3588659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Inequality from Israel TST Let $a, b, c, d$ be nonnegative numbers such that $a+b+c+d=18.$
Prove that:
$$\sqrt{\frac{a}{b+6}}+\sqrt{\frac{b}{c+6}}+\sqrt{\frac{c}{d+6}}+\sqrt{\frac{d}{a+6}}\leq5\sqrt{\frac{2}{7}}$$
These are my attempts to solve the Problem:
There is an equality case in $(a, b, c, d) = (8, 1, 8, 1)$ and... | Let $\{a,b,c,d\}=\{x,y,z,t\}$ such that $x\geq y\geq z\geq t$.
Thus, since $(\sqrt{x},\sqrt{y},\sqrt{z},\sqrt{t})$ and $\left(\frac{1}{\sqrt{t+6}},\frac{1}{\sqrt{z+6}},\frac{1}{\sqrt{y+6}},\frac{1}{\sqrt{x+6}}\right)$ are same ordered,
by Rearrangement, AM-GM and by Jensen for $f(x)=\sqrt{x+12}$ and for $g(x)=-\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3591099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Can someone confirm if my answer is right?( complex numbers) I'm fairly new to complex numbers and there wasn't much I could find online but could someone see if my answer is right? we need to find $z = a+bi$ given that $z^2 = -144+12i\sqrt{8}$
for LHS:
$$z^2 = (a+bi)^2 = a^2 + 2abi - b^2$$
$$a^2 +2abi - b^2 = -144+12... | Your calculation are correct, in fact $a,b$ are solutions to the following system:
$$\left\{\begin{matrix}
a^2-b^2=-144
\\ 2ab=12\sqrt{8}
\end{matrix}\right.$$
From here, it's very easy to see that the solutions are:
$$\left\{\begin{matrix}
a=\frac{7}{5}
\\ b=\frac{30\sqrt{8}}{7}
\end{matrix}\right.$$
and:
$$\left\{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3591265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find a set $C(N)$ of $n \times n$ matrices that commute with $\it{N}$ I have a $n \times n$ matrix $N$, where $n \in \mathbb{N}$.
$N = \begin{bmatrix}
0 & 1 & 0 & \dots & 0 & 0 \\
0 & 0 & 1 & \dots & 0 & 0 \\
0 & 0 & 0 & \dots & 0 & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & 0 & \dots & 0 & 1 ... | Consider a generic matrix $A=(a_{i,j})_{i,j}$ and impose $AN=NA$. You have to be very careful, but you obtain these condition:
\begin{equation}
AN=
\left[
\begin{matrix}
0 & a_{1,1} & a_{1,2} & \cdots & \cdots & a_{1,n-1}\\
0 & a_{2,1} & a_{2,2} & \cdots & \cdots & a_{2,n-1}\\
0 & a_{3,1} & a_{3,2} & \cdots & \cdots & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3591427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
A three-parameters identity involving Stirling numbers of both kinds Let $n, m, k $ be three natural numbers, ${n \brack k}$ and ${n \brace k}$ the Stirling numbers of first and second kind respectively.
We have:
$$ \tag{*} {n-1 \choose m}{n-m \brack k}= \sum_i (-1)^{i-m}{k-1+i
\choose k-1}{i \brace m}{n \brack i+k... | We seek to verify that
$$\sum_{q=m}^{n-k} (-1)^{q-m} {k-1+q\choose k-1}
{q\brace m} {n\brack q+k}
= {n-1\choose m} {n-m\brack k}.$$
Using the standard EGFs the LHS becomes
$$\sum_{q=m}^{n-k} (-1)^{q-m} {k-1+q\choose k-1}
q! [z^q] \frac{(\exp(z)-1)^m}{m!}
n! [w^n] \frac{1}{(q+k)!} \left(\log\frac{1}{1-w}\right)^{q+k}
\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3592240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Line Integral with Change of Variables I'm a bit rusty on my computational math, and genuinely can't solve this question which is frustrating me.
QUESTION:
$$\int_Csin(y)dx+xcos(y)dy$$ where C is the ellipse defined as follows: $x^2+xy+y^2=1$
MY ATTEMPT:
Define variable $u= \sqrt{\frac{3}{4}}x+\sqrt{\frac{3}{4}y}$
Defi... | The integral is path independent because
$$\nabla(x\sin y) = (\sin y, x\cos y)$$
So, the integral along the closed contour $C$ vanishes:
$$\int_C (\sin y\,dx + x\cos y\,dy) =0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3594026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solve for $x$: $\log_{\frac {3}{4}}(\log_{8} (x^2+7))+\log_{\frac {1}{2}} (\log_{\frac {1}{4}} (x^2+7)^{-1})=-2$ Solve for $x$: $\log_{\frac {3}{4}}(\log_{8} (x^2+7))+\log_{\frac {1}{2}} (\log_{\frac {1}{4}} (x^2+7)^{-1})=-2$
My Attempt:
$$\log_{\frac {3}{4}}(\frac {1}{3} \log_2 (x^2+7))+\log_{\frac {1}{2}} (\frac {1}... | I want to do some replacement but it's going to be tricky.
$\log_8 (x^2+7)=a$
$8^a=x^2+7$
$2^{3a}=x^2+7$
$2^{-3a}=(x^2+7)^{-1}$
$(2^{-2})^{\frac{3}{2}a}=(x^2+7)^{-1}$
$\left(\dfrac{1}{4} \right)^{\frac{3}{2}a}=(x^2+7)^{-1}$
$\log_{\frac{1}{4}} (x^2+7)^{-1}=\dfrac{3}{2}a$
The whole thing changes to:
$\log_{\frac{3}{4}} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3602742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to find the common factor I am currently stomped at this problem $(2 x^2-(y+z) (y+z-x))/(2 y^2-(z+x) (z+x-y))$ I am supposed to factor it and I can't find a way how to get to the answer.
It says that
$(2 x^2+x y-y^2+x z-2 y z-z^2)/(2 y^2-x^2+x y-2 x z+y z-z^2)$ has a common factor of x+y+z. Can anyone elaborate o... | Consider the numerator $2x^2-(y+z)(y+z-x).$ The first thing to notice here is the sum $y+z.$ You may set $y+z=p$ if it makes the following more convenient to follow. Now expanding gives $$2x^2-(y+z)^2+x(y+z).$$ You almost have a difference of squares -- indeed you do already, but it doesn't help much, so we split into ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3602894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find $\lim_{(x,y)\rightarrow(0,0)} \cos(\frac{x^2-y^2}{\sqrt{x^2+y^2}})$ $\lim_{(x,y)\rightarrow(0,0)} \cos(\frac{x^2-y^2}{\sqrt{x^2+y^2}})$
I tried replacing x and y with several values and kept getting 1 so I tried:
$$0 \le |\cos(\frac{x^2-y^2}{\sqrt{x^2+y^2}})| \le |\cos(|(\frac{x^2-y^2}{\sqrt{x^2+y^2}}|)|\le |\cos(... | HINT
Consider the polar coordinates $x = r\cos(\theta)$ and $y = r\sin(\theta)$. Then we get
\begin{align*}
\lim_{(x,y)\rightarrow(0,0)}\cos\left(\frac{x^{2}-y^{2}}{\sqrt{x^{2}+y^{2}}}\right) = \lim_{r\rightarrow 0}\cos(r\cos(2\theta)) = \ldots
\end{align*}
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3603843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Inequality involving the angle bisectors of a triangle
Let $l_a,l_b,l_c$ denote the lengths of angle bisectors of a triangle with sides $a,b,c$ and semiperimeter $s$. I am looking for the best constant $K>0$ such that
$$l_a^2+l_b^2+l_c^2> K s^2.$$
I found that $K=2/3$ works, but I suspect that best constant is $K=8... | Let $a=b=1$ and $c\rightarrow2^-$.
Thus, $K<\frac{8}{9}.$
We'll prove that $\frac{8}{9}$ it's a best constant.
Indeed, we need to prove that:
$$\sum_{cyc}\left(\frac{2bc\cos\frac{\alpha}{2}}{b+c}\right)^2\geq\frac{8}{9}\cdot\frac{(a+b+c)^2}{4}$$ or
$$\sum_{cyc}\left(\frac{2bc\sqrt{\frac{1+\frac{b^2+c^2-a^2}{2bc}}{2}}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3604661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Find the exact value of integration $ \int_0^1 \frac{1}{\sqrt{1-x}+\sqrt{x+1}+2} dx$ Can you help me find the exact value for integration with the given steps?
$$ \int_0^1 \frac{1}{\sqrt{1-x}+\sqrt{x+1}+2} dx$$
Some of my attempts as indefinite Integral
$$
\int \frac{1}{2+\sqrt{1-x}+\sqrt{1+x}} \, dx\approx \left(\sqrt... | Setting $x=\cos(2t)$, we have:
$1-x=2\sin^2t\;\;$ and $\;\;1+x=2\cos^2t$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3607599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find $\lim _{n\to \infty}\frac{n-1}{n-2}$ I'm taking a university real analysis course and I have been tasked with proving that the sequence $x_n = \frac{n-1}{n-2}$ converges using first principles.
First fix $\epsilon >0$.
Using the algebra of limits we find the limit in question.
$x_n = \frac{n-1}{n-2} = \frac{1-\fra... | Your proof is correct, but I think you should explicitly specify a choice of $N$ as a function of $\epsilon$. Something like this: For any $\epsilon$ and for all $n' \geq N = \max\{\frac{4}{\epsilon}, 3\}$, then
$$
\left\lvert \frac{n'-1}{n'-2} - 1 \right\rvert = \frac{1}{n' - 2} \leq \frac{3}{N} = \frac{3}{4} \epsilon... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3610575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Methods to solve $\int _0^{\infty }\frac{x^{\frac{4}{5}}-x^{\frac{2}{3}}}{\ln \left(x\right)\left(x^2+1\right)}\:dx$
Find $$\int _0^{\infty }\frac{x^{\frac{4}{5}}-x^{\frac{2}{3}}}{\ln \left(x\right)\left(x^2+1\right)}\:dx.$$
I'd like to know in what ways can one approach this integral that can be found here, since th... | Consider
$$J(a)=\int_0^\infty\frac{x^a-1}{(x^2+1)\ln x}dx$$
where $a\in\Bbb C$ with $|\Re a|<1$.
Then
$$J'(a)=\int_0^\infty\frac{x^a}{x^2+1}dx.$$
Using the keyhole contour you can see that
$$J'(a)=\frac{2\pi i}{1-e^{2\pi a}}\left(\frac{e^{\frac{\pi a}{2}}}{2i}+\frac{e^{\frac{3\pi a}{2}}}{-2i}\right)=\frac{\pi}{2}\sec\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3612139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
For what $x$ and $y$ polynomial has maximum value?
For what $x,y\in\mathbb R$ does the polynomial
$$-5x^2-2xy-2y^2+14x+10y-1$$
attain a maximum?
My attempt:
I called $\alpha$ maximum value.
$$-5x^2-2xy-2y^2+14x+10y-1\leqslant\alpha$$
$$-5x^2-2xy-2y^2+14x+10y-1-\alpha\leqslant 0$$
$$5x^2+2xy+2y^2-14x-10y+1+\alpha\... | let $k$ be a maximal value.
Thus, for any $x$ and $y$ we have $$-5x^2-2xy-2y^2+14x+10y-1\leq k$$ or
$$5x^2+2(y-7)x+2y^2-10y+1+k\geq0,$$ which gives
$$(y-7)^2-5(2y^2-10y+1+k)\leq0$$ or
$$9y^2-36y-44+5k\geq0,$$ which gives $$18^2-9(-44+5k)\leq0,$$ which gives $$k\geq16.$$
The equality occurs for $$y=\frac{36}{2\cdot9}=2$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3612880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Row Echelon, Partial Fractions, and Numerator Coefficients I am trying to get the numerator values for the partial fraction decomposition of:
$$\dfrac{x^2+1}{x(x-1)(x+1)(x-4)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}+\frac{D}{x-4}$$
I really started hitting speed bumps on this one, and was forced to take it slow so I de... | Lets use Wolfram Alpha to get a final result of the original expression using partial fractions.
We see that we would get
$$A = \dfrac{1}{4}, B = -\dfrac{1}{3}, C = -\dfrac{1}{5} , D = \dfrac{17}{60}$$
We want to solve the partial fraction expansion of
$$\dfrac{x^2+1}{x(x-1)(x+1)(x-4)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3613022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Sum of Beta functions. Given the summation
$$ S_{n} = \sum_{k=0}^{n-1} \binom{n+k-1}{k} \, ( B(a+k, b+n) + B(a+n, b+k) )$$
is it possible to show that it equals the suspected value of $B(a, b)$?
For the first two $n$ values it can be shown that
\begin{align}
S_{1} &= B(a, b+1) + B(a+1, b) = B(a,b) \\
S_{2} &= B(a, b+2)... | A solution using induction. We have $S_n=\int_0^1 x^{a-1}(1-x)^{b-1}Q_n(x)\,dx$, where \begin{align}
Q_n(x)&:=(1-x)^n P_n(x)+x^n P_n(1-x),\\
P_n(x)&:=\sum_{k=0}^{n-1}\binom{n+k-1}{k}x^k,\\
P_{n+1}(x)&=\sum_{k=0}^{n}\binom{n+k-1}{k}x^k+\sum_{k=1}^{n}\binom{n+k-1}{k-1}x^k
\\&=P_n(x)+\binom{2n-1}{n}x^n+x\left(P_{n+1}(x)-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3614965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Evaluating $\lim_{x\to1^+}\frac{\sqrt{x+1}+\sqrt{x^2 -1}-\sqrt{x^3+1}}{\sqrt{x-1}+\sqrt{x^2 +1}-\sqrt{x^4+1}}.$ I don't know how to solve these kind of limit problems. I was wondering if someone could help me about it to learn them. $$\lim_{x\to 1^+}\frac{\sqrt{x+1}+\sqrt{x^2 -1}-\sqrt{x^3+1}}{\sqrt{x-1}+\sqrt{x^2 +1}-... | Hint
You can always write $f(x)-g(x)$ as $$\frac{f(x)^2-g(x)^2}{f(x)+g(x)}.$$
So, taking $$f(x)=\sqrt{x+1}+\sqrt{x^2+1}\quad \text{and}\quad g(x)=\sqrt{x^3+1},$$ at the numerator and $$f(x)=\sqrt{x-1}+\sqrt{x^2+1}\quad \text{and}\quad g(x)=\sqrt{x^4+1},$$
at the denominator should do the work.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3615791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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How to prove this inequality: $\sum_{i=1}^{n}\sum_{j=1}^{n}\text{lcm}(i,j)\ge\frac{7}{8}n^3$?
Let $n$ be postive integers. Show that
$$\sum_{i=1}^{n}\sum_{j=1}^{n}[i,j]\ge\dfrac{7}{8}n^3\,,$$ where $[a,b]$ denote the least common multiple of $a$ and $b$.
For $n=1,2,3 $, it is clear. How to prove this inequality? ... | Let $S_n:=\sum\limits_{i=1}^n\,\sum\limits_{j=1}^n\,\text{lcm}(i,j)$. We have $S_1=1\geq \dfrac{7}{8}\cdot 1^3$ and $S_2=7\geq \dfrac{7}{8}\cdot 2^3$. Assume that $n>2$ is an integer such that $$S_{n-1}\geq \dfrac{7}{8}\,(n-1)^3\,.$$
Then,
$$S_{n}-S_{n-1}=n+2\,\sum_{k=1}^{n-1}\,\text{lcm}(k,n)\,.$$
Since $n>3$,
$$S_n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3617211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Expected Waiting Time in a Queuing System $(M | M | 2 | 5)$
In the queuing system $(M | M | 2 | 5)$, the input flow rate is $240$ requests per hour, the average service time for one request is $30$ seconds. Find the average waiting time for an application in the queue.
How do I find that?
| Let $\pi_k$ denote the steady-state probability that there are $k$ requests. We have $\lambda=4\,\texttt{min}^{-1}$ and $\mu=2\,\texttt{min}^{-1}$. Write $\rho:=\dfrac{\lambda}{\mu}=2$. We have the balance equations
$$\lambda\,\pi_0=\mu\,\pi_1\,,$$
and
$$\lambda\,\pi_k=2\mu\,\pi_{k+1}$$
for $k=1,2,3,4$. That is,
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3617336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the curve of a family of curves with the following conditions
Let $\mathfrak{J}$ be the family of curves such that their normal line at each point is tangent to the parabola of equation $$y=kx^2$$ that passes through that point. Find the curve $C\in\mathfrak{J}$ that passes through $(0,1)$.
My work: $$\mathfrak... | This line in your attempt is correct:
$$yy'=-\frac x 2$$
Then you have :
$$ 2yy'=-x \implies (y^2)'=-x$$
Integrate:
$$y^2=-\frac {x^2}2+c$$
$$2y^2+ {x^2}=2c$$
$$2y^2+ {x^2}=2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3621252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
A type of Combinatorial equality：$\sum_{k=0}^{n}\binom{n}{k} \cos\frac{k}{2}\pi=2^{\frac{n}{2}}\cos\frac{n}{4}\pi.$ When computing the Taylor series of the function $f(z)=e^z\cos z,$
I use two methods:
On the one hand, using Cauchy product,
\begin{align*}
e^z\cos z
&=\left(\sum_{n=0}^{\infty}\frac{z^n}{n!}\right)
\lef... | There is a fundamental way to prove this.
I used a simple binomial expansion and Euler's identity to solve this...
Here's the solution:
$$
\begin{aligned}
&\sum_{k=0}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right) \cos \frac{k \pi}{2}=?\\
&\text { Let } T_{k}=\left(\begin{array}{l}
n \\
k
\end{array}\right)\left\{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3621412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How to determine direct solution of determinant? How to show that the determinant of the following $(2n+1)×(2n+1)$ matrix $A$?
\begin{equation}
\det A = \begin{array}{|cccccccccc|cc}
1 & -1 & 0 & \dots & 0 & 0 & 0 & \dots & 0 & 0 & & {\color{blue}{\text{row }1}}\\
-1 & 2 & -1 & \dots & 0 ... | For each $X\in\text{Mat}_{m\times m}(\mathbb{R})$, $a,b\in\mathbb{R}$, and $i,j=1,2,\ldots,m$, the notation $$Y:=(R_i\leftleftarrows a\, R_i+b\, R_j)(X)$$
means $Y$ is obtained from $X$ by changing the $i$-th row of $X$ to $a$ times the $i$-th row of $X$ plus $b$ times the $j$-th row of $X$. Let $m:=2n+1$. Consider
... | {
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"url": "https://math.stackexchange.com/questions/3621758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Sequences and series that I cant partial fraction decomposition I have to calculate the partial sum for an equation. How can I calculate the sum for
$$\sum_{n=1}^{\infty}\frac{1}{16n^2-8n-5}$$
And I think that is not simple as $$\sum_{n=1}^{\infty}\frac{1}{16n^2-8n-3}.$$
TIA
| For computing partial fractions, use the one you know how to work as a model for the other. The method is the same.
The zeroes of $16n^2-8n-3$ are $\frac{8\pm\sqrt{256}}{32}$, so $\frac34$ and $-\frac14$, which means you can write the polynomial as $16(n-\frac34)(n+\frac14) = (4n-3)(4n+1)$. Your candidate for the parti... | {
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"url": "https://math.stackexchange.com/questions/3625093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Use mathematical induction to prove that for all integers $n \geq 3,\, 2n + 1 < 2^{n}$ This is what I've got so far.
Let $P(n)$ be the statement that $2n + 1 < 2^n$
Basis:
Let $n = 3$. Show that $P(3)$ is true.
$2(3) + 1 = 7$ and $2^3 = 8$.
Since $7 < 8$, $P(3)$ is true.
Inductive Hypothesis:
Suppose that $P(k)$ is tr... | You're doing fine up until near end. Note that for $k \ge 3$, (actually, for any $k \gt 1$), you have $2 \lt 2^k$, so
$$2^k + 2 \lt 2^k + 2^k = 2^{k + 1} \tag{1}\label{eq1A}$$
with which you can then finish your inductive proof.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving a complicated looking inequality in a simple way This is again a search for alternative proofs:
Let $0 <s \le 1$, and suppose that $0 <a,b $ satisfy
$$ ab=s,a+b=1+\sqrt{s}. \tag{1}$$
I have a proof for the assertion
$$ 2(1-\sqrt s)^3 \le |a-1|^3+|b-1|^3, \, \, \, \text{for every } \, s \ge \frac{1}{9}$$
but it... | Let $\sqrt{s}=t$.
Thus, $\frac{1}{3}\leq t\leq 1.$
Since $a$ and $b$ are roots of the equation $$x^2-(1+t)x+t^2=0,$$ we need to prove that
$$\left|\frac{1+t+\sqrt{(1+t)^2-4t^2}}{2}-1\right|^3+\left|\frac{1+t-\sqrt{(1+t)^2-4t^2}}{2}-1\right|^3\ge2(1-t)^3$$ or
$$\left|\frac{\sqrt{(1-t)(1+3t)}-(1-t)}{2}\right|^3+\left|\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3629718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Find all polynomials $P(x)$ such that $2 P\left(2 x^{2}-1\right)=P(x)^{2}-2$ Question -
(Romania, 1990)Find all polynomials
$P(x)$ such that
$$
2 P\left(2 x^{2}-1\right)=P(x)^{2}-2
$$
for all $x \in \mathbb{R}$.
Solution
Putting $x=1,$ we get a quadratic equation in $P(1)$ and hence $P(1)=1 \pm \sqrt{3} .$ If $P(1)=1+\... | Firstly, when using induction, you need the initial/base case. We conclude that $(x-1) = (x-1)^1$ divides the polynomial $Q(x)$.
Note that I will be using the following notation; $Q(x) = (x-1) \cdot Q^{(1)}(x)$, and in general \begin{equation}Q^{(n)}(x) = (x-1) \cdot Q^{(n+1)}(x)\end{equation} or \begin{equation}Q(x) =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3630051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Simplifying $\frac{ \cos^2 x - \sin^2 x }{\sin{2x}}$ Please consider the following problem and my answer to it:
Problem:
Simplify the following expression:
$$ \frac{ \cos^2 x - \sin^2 x }{\sin{2x}} $$
Answer:
$$ \frac{ \cos^2 x - \sin^2 x }{\sin{2x}} = \frac{ \cos^2 x - \sin^2 x }{ 2 \sin x \cos x} $$
$$ \frac{ \cos^2 ... | $\cos^2x -\sin^2 x = \cos (2x)$. So the result can be written as $\cot(2x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3630952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve the equation $y-3px+ayp^2=0$ where $p=\frac {dy}{dx}$ Solve the equation $y-3px+ayp^2=0$ where $p=\frac {dy}{dx}$
My Attempt:
Given
$$y-3px+ayp^2=0$$
$$3px=y+ayp^2$$
$$x=\frac {1}{3} \cdot \frac {y}{p} + \frac {a}{3} \cdot yp$$
This is solvable for x. Differentiating both sides with respect to $y$
$$\frac {dx}{dy... | From @Rebellos work $$\int \frac{dy}{y}=\int \frac{dp}{p}+\int \frac{dp}{p(2-ap^2)}$$
In the third integral let $p=1/t \implies -dt/t^2$, then $2t^2-a=u$, we get
$$\ln y= \ln p-\frac{1}{4}\ln u \implies y(p)=\frac{Cp\sqrt{p}}{(2-ap^2)^{1/4}}~~~~(1)$$
Inserting this in the given ODE, we get
$$x(p)=\frac{y+ayp^2}{3p}~~~~... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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How to prove that $a + b \neq 2^{n+1} (2c+1) $ with $ab = 4^n - 1$ and $a,b, c, n \in \mathbb{N}$? How to prove that $a + b \neq 2^{n+1} (2c+1) \quad \text{; with } ab = 4^n - 1 \text{ and } a,b, c, n \in \mathbb{N}$ (without zero)?
I already know that:
\begin{align}
a,b &\equiv 1 \pmod{2} \\
a + b &\equiv 0\pmod{4}
\e... | Define $v_2(n)$ to be the exponent of the highest power of $2$ that divides $n$, for example $v_2(24)=v_2(2^3\cdot 3)=3$ and $v_2(7)=0$.
Suppose that $a+b=2^{n+1}(2c+1)$. This gives $$v_2(a+b)=n+1.$$ It is not hard to see that $$n+1=v_2(4^n+a+b)=v_2((a+1)(b+1))=v_2(a+1)+v_2(b+1).$$ As seen above, WLOG $b\equiv_4 1$, so... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve $x+yp=ap^2$ where $p=\frac {dy}{dx}$ Solve $x+yp=ap^2$ where $p=\frac {dy}{dx}$
My Attmept:
$$x+yp=ap^2$$
$$x=ap^2-yp$$
This is solvable for $x$ so differentiating both sides w.r.t $y$
$$\frac {dx}{dy} = 2ap\cdot \frac {dp}{dy} - y\cdot \frac {dp}{dy} -p$$
$$\frac {1}{p} + p = (2ap-y) \cdot \frac {dp}{dy}$$
$$\fr... | Usually you get fastest to the relevant equations if you start with the derivative by $p$, using $\dot y(p)=p\dot x(p)$. Then
$$
\dot x+p\dot y+y=2ap
$$
Multiply with $p$ and eliminate $x$
$$
(1+p^2)\dot y+py=2ap^2\implies \sqrt{1+p^2}y(p)=\int \frac{2ap^2}{\sqrt{1+p^2}}dp
$$
Now substitute $p=\sinh(u)$ in the last int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3634280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Is there a $4$-by-$4$, rank $3$, positive semidefinite matrix with $a_{ii}=3$, $|a_{12}|\neq 1$, and principal minors having minimal eigenvalue $1$? Could anyone help me search for a positive semidefinite matrix $\left(a_{i,j}\right)_{4\times4}$ of rank 3 with $a_{i,i}=3$
and its all 3 by 3 principal minor matrices hav... | Suppose $A$ is a rank-$3$ positive semidefinite $4\times4$ matrix such that the minimum eigenvalues of its principal $3\times3$ submatrices are equal to all $1$. We will prove that
$$
A=D\pmatrix{3&1&1&1\\ 1&3&-1&-1\\ 1&-1&3&-1\\ 1&-1&-1&3}D^\ast
$$
for some unitary diagonal matrix $D$ (and hence all of its off-diagona... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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How to solve exercises with polynomial with 2 parameters having all real roots? $f=2x^{4}+4x^{3}+3x^{2}+bx+c$ has all real roots, find b,c (b,c are from R).
Thanks a lot, I tried with substitution, I don't know, is there something with derivative? please help thanks
| $f(x) = 2x^4+4x^3+3x^2+bx+c$
If $f(x)$ has all $4$ roots to be real and complex, then it's discriminate $$\Delta = (-72)b^2+176b^3-108b^4-864c-2304bc+1536b^2c+2304c^2-3072bc^2+2048c^3$$ must be real, $\Delta > 0$
Also if you divide $f(x)$ by 2, and make a translation $x = y-\frac{1}{2}$
$2x^4+4x^3+3x^2+bx+c = 0$
$x^4+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3635523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.