Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Prove that there are no such positive integers $a,b,c,d$ such that $a^2 + b^2 = 3(c^2 + d^2)$ Prove that there are no positive integers a, b, c, d such that $a^2 + b^2 = 3(c^2 + d^2)$. Hint: What can you say about divisibility of a and b by 3? Look at solution with smallest possible a.
| Well, $A^2+B^2=1,0$ ( mod 4) so it couldn't be $A^2+B^2=4k+3$. But (4k+1)$(4k+3)=16k^2+16k+3=4(4k^2+4k)+3=4r+3$, while (4k+3)$(4k+3)=16k^2+24k+8+1=4(4k^2+6k+2)+1=4s+1$. So if a prime $p|A^2+B^2$ and p=4k+3, $p^2|A^2+B^2!$ But the exponent of number 3 is odd while it should be even. Hence there no positive integers A,B,... | {
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"timestamp": "2023-03-29T00:00:00",
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How do I find the highest and lowest points made by the union of these two functions using Lagrange Multipliers?
Find the highest and lowest points made by the union of these two functions using Lagrange Multipliers.
$x^2+y^2+z^2 = 16$
$(x+1)^2+(y+1)^2+(z+1)^2 = 27$
I got the basics down, I used the first function as... | I think you are missing something. As @copper.hat noted you don't have nothing to optimize. I think that the two equations you uploaded are constrains, but the function you want to maximize/minimize isn't posted.
If that's not the case, add those two and you'll get that both minima and the maxima of the function is $27... | {
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Is there a base for each positive integer where this number can be represented a string of n (base - 1) digit? How are called positive integer numbers that have the following property of being represented as:
$$
N = \sum_{k=0}^n{(B-1)B^k} = (B - 1)\sum_{k=0}^n{B^k}
$$
with $N$ a positive integer number, $B$ the base in... | When it is $1$'s repeated, it is called a repunit. When it is $n$ repeated, it is called a repdigit.
We want $n=(b-1)\sum_{k=0}^{m-1}b^k=b^m-1 $
So whenever $n+1$ is an $m^{th}$ power, $n$ may be expressed as $m\ $ $b-1$'s
| {
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Coordinate Geometry Oblique Coordinates Problem This is a elementary geometry problem which I have tried to solve using coordinate geometry but it is resulting in an impossible and impractical result. Maybe I have some misconceptions with oblique coordinates. So please help me with my misconceptions.
Suppose $\trian... | There are two difficulties here, as the post stands. One is that it appears as though there was an earlier version of the solution posted, which was then revised, leaving it in a form that is now internally inconsistent. Unless, for some reason, two coordinate axes are being used which are not perpendicular to one an... | {
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Putnam Series Question I'm studying for the Putnam Exam and am a bit confused about how to go about solving this problem.
Sum the series
$$
\sum_{m = 1}^{\infty} \sum_{n = 1}^{\infty} \frac{m^2n}{3^m(n3^m + m3^n)}.
$$
I've tried "splitting" the expression to see if a geometric sum pops up but that didn't get me any... | Let
$$
S = \sum_{m=1}^\infty \sum_{n=1}^\infty \frac{m^2n}{3^m(n3^m + m3^n)}= \sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{\frac{3^m}{m}\left( \frac{3^m}{m} + \frac{3^n}{n} \right)}.
$$
Then we see by symmetry, that
$$
2S = \sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{\left( \frac{3^m}{m} + \frac{3^n}{n} \right)}\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/578611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How find this Fibonacci sequence sum $\sum_{k=0}^{\infty}\frac{1}{F_{2^k}}$ let sequence $\{F_{n}\}$ such $$F_{1}=1,F_{2}=1,F_{m+1}=F_{m}+F_{m-1},m\ge 2$$
Find this value
$$I=\sum_{k=0}^{\infty}\dfrac{1}{F_{2^k}}$$
My try: I know this
$$F_{n}=\dfrac{1}{\sqrt{5}}\left(\left(\dfrac{\sqrt{5}+1}{2}\right)^n-\left(\dfr... | In fact, this is summable. This surprises me, because of results like those in this question at MO, which in short discusses how we don't know if the sum of the reciprocals of all the Fibonacci numbers is transcendental, and other related unknown results.
This question boils down to the following:
Claim: $\displaystyl... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How prove this $\frac{a}{(bc-a^2)^2}+\frac{b}{(ca-b^2)^2}+\frac{c}{(ab-c^2)^2}=0$ let $a,b,c\in \mathbb{R}$, if such
$$\dfrac{1}{bc-a^2}+\dfrac{1}{ca-b^2}+\dfrac{1}{ab-c^2}=0$$
show that
$$\dfrac{a}{(bc-a^2)^2}+\dfrac{b}{(ca-b^2)^2}+\dfrac{c}{(ab-c^2)^2}=0$$
Does this problem has nice methods?
My idea:let
$$(ca-b^2)(... | Hint: if we have $$\dfrac{1}{bc-a^2}+\dfrac{1}{ca-b^2}+\dfrac{1}{ab-c^2}=0$$ Then let $$\dfrac{1}{bc-a^2} = x; \dfrac{1}{ca-b^2} =y; \dfrac{1}{ab-c^2} = z$$ So you have something more manageable, then note that if $x+y+z=0$, then $x = -(y+z)$.
Finally, note that $$\dfrac{a}{(bc-a^2)^2} = \dfrac{a}{bc-a^2}\cdot \dfrac{... | {
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Can $a^2+b+2$ and $b^2+4a$ both be perfect squares? Are there any positive integers $a$ and $b$ so that $a^2+b+2$ and $b^2+4a$ are both perfect squares?
| Suppose that $a\le b$. Since $b^2$ is a perfect square, and $b^2+4a\lt (b+2)^2$, $b^2+4a$ cannot be a perfect square unless $b^2+4a=(b+1)^2$, which is impossible.
Now suppose that $a\gt b$. Then $a^2+b+2$ is too small to be a perfect square, since the smallest perfect square greater than $a^2$ is $a^2+2a+1$.
| {
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Integration of trigonmetric function $(a\cos x + b\sin x + c)/(1 - d\cos x)^{2}$
Problem: Prove that the indefinite integral $$\int \frac{a\cos x + b\sin x + c}{(1 - d\cos x)^{2}}\,dx$$ is rational function of $\sin x, \cos x$ if and only if $ad + c = 0$.
My Try: Looking at the problem it is obvious that if $a = c = ... | Let
$$E=\frac{b}{d(d\cos x-1)}-\frac{2\left(cd+a\right)}{d^2-1} \cdot\frac{\tan \left(\frac x2\right)}{(d+1)\tan^2 \left(\frac x2\right)-d+1}
$$
$$F=\frac{ad+c}{(d^2-1)^{3/2}}\;\log \left|\frac{\sqrt{\frac{d+1}{d-1}}\tan\left(\frac x2\right)+1}{\sqrt{\frac{d+1}{d-1}}\tan\left(\frac x2\right)-1}\right|$$
Then, if $d^2>1... | {
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"timestamp": "2023-03-29T00:00:00",
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For how many integers $a$ is $\frac{2^{10} \cdot 3 ^8 \cdot 5^6}{a^4}$ an integer? In Mathleague $11316$ Target #$4$, the question is:
For how many integers $a$ is $$\frac{2^{10} \cdot 3 ^8 \cdot 5^6}{a^4}$$ an integer?
| Hint: Factor $a$ into primes; it's evident that these primes must be $2$, $3$, and $5$ (why?). Then $a = 2^d \cdot 3^b \cdot 5^c$, so that
$$a^4 = 2^{4d} \cdot 3^{4b} \cdot 5^{4c}$$
What conditions on $d, b, c$ are necessary and sufficient to make this a divisor of $2^{10} \cdot 3^8 \cdot 5^6$?
| {
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Can someone help me simplify this trig expression?
$$( \tan x+ \sec x )( \cot x-\cos x ) $$
I got stuck after a few steps of converting and adding and what not.
| A simplification:
$$
\begin{align*}
\left( \tan x + \sec x \right) \left( \cot x - \cos x \right) &= \left(\tan x + \frac{1}{\cos x} \right)\left( \frac{1}{\tan x } - \cos x\right) \\
&=\frac{\tan x}{\tan x}-\cos x \tan x +\frac{1}{\cos x \tan x}-\frac{\cos x}{\cos x} \\
&=1-\cos x \tan x +\frac{1}{\cos x \tan x}-1 \\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solution to a linear Ordinary Differential Equation (ODE) with variable coefficients. I need to solve a series of linear ODEs that are ``close'' to the Euler equation , in a certain sense. One example of my equations is given below:
\begin{equation}
8 \xi^3 \frac{d^3 r_\xi}{d \xi^3} + 36 \xi^2 \frac{d^2 r_\xi}{d \xi^2... | Let us use a ``method of variation of constants''. Since we know that for small values of $\xi$ the solution behaves as: $C_3/\xi$ why do we not make the constant $C_3$ a variable. Thus we conjecture:
\begin{equation}
r_\xi := \frac{f_\xi}{\xi}
\end{equation}
Substituting the above into the original ODE yields:
\begin{... | {
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calculation of $\int\frac{1}{\sin^3 x+\cos^3 x}dx$ and $\int\frac{1}{\sin^5 x+\cos^5x}dx$
Solve the following indefinite integrals:
$$
\begin{align}
&(1)\;\;\int\frac{1}{\sin^3 x+\cos^3 x}dx\\
&(2)\;\;\int\frac{1}{\sin^5 x+\cos^5 x}dx
\end{align}
$$
My Attempt for $(1)$:
$$
\begin{align}
I &= \int\frac{1}{\sin^3 x+\c... | For the second integral, decompose the integrand as follows
\begin{align}
&\frac52\int\frac{1}{\sin^5 x+\cos^5 x}dx\\
=&\int\frac{2}{\sin x+\cos x}-\frac{\sin x + \cos x}{2\sin x\cos x-\sec\frac{\pi}5}
-\frac{\sin x + \cos x}{2\sin x\cos x -\sec\frac{3\pi}5}\ dx\\
=&\sqrt2\tanh^{-1}\frac {\sin x- \cos x}{\sqrt2}
-\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/595038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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"answer_id": 2
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Finding $\int_{0}^{2\pi} \frac {(1+2\cos\theta)^n\cos(n\theta)}{3+2\cos\theta} \operatorname{d}\theta$ If $n$ is a positive integer find
$$
\int_{0}^{2\pi} \frac {(1+2\cos\theta)^n\cos(n\theta)}{3+2\cos\theta} \operatorname{d}\theta
$$
I know that I have to use contour integral with a circle of radius 1 centered at the... | Note that
$$\cos{n \theta} = \frac12 \left ( z^n + z^{-n}\right )$$
so upon your substitution, you get
$$-\frac{i}{2} \oint_{|z|=1} \frac{dz}{z} \frac{(1+z+z^{-1})^n (z^n+z^{-n})}{3+z+z^{-1}}$$
This can be rearranged to get
$$-\frac{i}{2} \oint_{|z|=1} \frac{dz}{z^{2 n}} \frac{(1+z+z^2)^n (1+z^{2 n})}{1+3 z+z^2} $$
Th... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Prove that $\sin^2(A) - \sin^2(B) = \sin(A + B)\sin(A -B)$ This is my attempt:
$$
\begin{align}
& \phantom{={}}\sin^2(A) - \sin^2(B) = \sin(A + B)\sin(A -B) \\[8pt]
& = (\sin(A)\cos(B)+\cos(A)\sin(B))(\sin(A)\cos(B) - \cos(A)\sin(B)) \\[8pt]
& = (\sin(A)+\sin(B))(\cos(B)+\cos(A))(\sin(A)-\sin(B))(\cos(B)-\cos(A)) \\[8p... | $x^2-y^2=(x+y)(x-y)$, we have
$$\sin^2A-\sin^2B\\
=(\sin A+\sin B)(\sin A-\sin B)\\
=2\sin\frac{A+B}2\cos\frac{A-B}2\cdot2\cos\frac{A+B}2\sin\frac{A-B}2\\
=2\sin\frac{A+B}2\cos\frac{A+B}2\cdot2\sin\frac{A-B}2\cos\frac{A-B}2\\
=\sin(A+B)\sin(A-B)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Help with hard complex numbers We had the topic of complex numbers for my senior math team meet this week, and I wasn't able to solve two of the problems.
1.) $z=i^{\displaystyle \left(i^{\displaystyle \left(i^{(2)}\right)}\right)}$ and $a$ is the real part of $z$, find the lowest positive value of $\ln(a)$
[ I... | Okay, the second one took me some time but I solved it, sorry if I do not know how to write math in this program.
I will use 2 pieces of data, the first:
$(e^x)^i = cos(x)+i*sin(x)$ [Euler's identity]
The second [I don know the name, but is easy to prove]:
if you add the n nth roots of 1 you get 0.
$1+(-1) = 0$
$1+e^{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/605110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
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Which is bigger, $1+3\sqrt{2}$ or $3\sqrt{3}$? Which is bigger, $a = 1+3\sqrt2$ or $b = 3\sqrt3$?
To find out result, I am doing: $(3\sqrt3)^2-(1+3\sqrt2)^2=8-2\sqrt{18}$.
But how can I find if the value of $8-2\sqrt{18}$ is positive or negative?
Thank you.
| Hint:
$$\color{blue}8-\color{red}{2\sqrt{18}}=\color{blue}{\sqrt{64}}-\color{red}{\sqrt{4}\sqrt{18}}=\color{blue}{\sqrt{64}}-\color{red}{\sqrt{72}}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Comparison of $S_{n}$ and $T_{n}$, where $S_{n} = \sum_{k=1}^{n}\frac{n}{n^2+nk+k^2}$ and $T_{n} = \sum_{k=1}^{n-1}\frac{n}{n^2+kn+k^2}$ Let $\displaystyle S_{n} = \sum_{k=1}^{n}\frac{n}{n^2+nk+k^2}$ and $\displaystyle T_{n} = \sum_{k=1}^{n-1}\frac{n}{n^2+kn+k^2}$ for $n=1,2,3,\dots$
Then which of the following options... | \begin{align*}
T_n&=\sum_{k=0}^{n-1}\frac{n}{n^2+kn+k^2}-\frac{1}{n}\\
&<\lim_{n\rightarrow \infty}\left(\sum_{k=0}^{n-1}\frac{n}{n^2+kn+k^2} -\frac{1}{n}\right)= \int_{0}^{1}\frac{1}{1+x+x^2}dx+0=\frac{\pi}{3\sqrt{3}}
\end{align*}
OR
\begin{align*}
T_n&=\sum_{k=1}^{n}\frac{n}{n^2+kn+k^2}-\frac{1}{3n}\\
&<\lim_{n\right... | {
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"timestamp": "2023-03-29T00:00:00",
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Limit of $\sum_{i=1}^n \left(\frac{{n \choose i}}{2^{in}}\sum_{j=0}^i {i \choose j}^{n+1}\right)$ I'm trying to calculate the limit for the sum of binomial coefficients:
$$S_{n}=\sum_{i=1}^n \left(\frac{{n \choose i}}{2^{in}}\sum_{j=0}^i {i \choose j}^{n+1} \right).$$
| Simply multiplying the largest term by the number of terms in the sum yields
$$
\begin{align}
\frac{\binom{n}{i}}{2^{in}}\sum_{j=0}^i\binom{i}{j}^{n+1}
&\le(i+1)\binom{i}{i/2}\binom{n}{i}\left(\binom{i}{i/2}2^{-i}\right)^n\tag{1}
\end{align}
$$
using $\binom{i}{i/2}=\binom{i}{(i\pm1)/2}=\frac12\binom{i+1}{(i+1)/2}$ for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/608296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that, for $p > 3$, $(\frac 3p) = 1$ when $p \equiv 1,11 \pmod{12}$ and $(\frac 3p) = -1$ when $p \equiv 5,7 \pmod{12}$ $(\frac 3p)$ is the Legendre symbol here, not a fraction, if that wasn't clear.
This is what I have so far:
We know by Gauss's Lemma that $(\frac qp) = (-1)^v$ where
$$v = \#\left\{1 \le a \le \f... | By Quadratic Reciprocity $$\left(\frac{3}{p}\right)=\left(\frac{p}{3}\right)(-1)^{(p-1)(3-1)/4}=\left(\frac{p}{3}\right)(-1)^{(p-1)/2}=\left(\frac{p}{3}\right)\left(\frac{-1}{p}\right).$$ We will use the fact that $-1$ is a quadratic residue modulo $p$ if $p\equiv 1\pmod{4}$ and is not if $p\equiv -1\pmod{4}$. Since $3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/608481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Divisibility by large powers Show that $2^{111} + 1$ divides $2^{555} + 1$ but does not divide $2^{444} + 1$.
My try:
$(2^{111}+1)^5 = 2^{555}+5*2^{444}+10*2^{333}+10*2^{222}+5*2^{111}+1$
$2^{555} + 1$ = $(2^{111}+1)^5$ - $(5*2^{444}+10*2^{333}+10*2^{222}+5*2^{111})$
= $(2^{111}+1)^5$ - ($5*2^{333}(2^{111}+1)+5*2^{222}... | $a^{odd}+b^{odd}$ is always divisible through $a+b$, and $a^{even}-b^{even}$ is always divisible through $a-b$.
$$a^{2n+1}+b^{2n+1}=(a+b)\cdot\sum_{k=0}^{2n}(-1)^k\cdot a^k\cdot b^{2n-k}$$
E.g., for $n=1$ we have the well-known formula $a^3+b^3=(a+b)(a^2-ab+b^2)$. Likewise, for $n=2$ we have $a^5+b^5=(a+b)(a^4-a^3b+a^2... | {
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"timestamp": "2023-03-29T00:00:00",
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How to prove that $(1+\cos a)/(\sin a)=(\sin a)/(1-\cos a)$? How can I prove this relation $(1+\cos a)/(\sin a)=(\sin a)/(1-\cos a)$ ?
I tried to start from relation $\cos^2a+\sin^2a=1$ but relation went crazy with lot of $\cos$ and $\sin$ and $\sin^2$.
| $$\begin{align}\dfrac{1+\cos a}{\sin a} & = \dfrac{(1+\cos a)\cdot(1-\cos a)}{\sin a(1-\cos a)}\\ \, \\
&=\dfrac{1-\cos^2a}{\sin a(1-\cos a)}\quad\text{since: $(a+b)\cdot(a-b)=a^2-b^2$}\\\,\\
&=\dfrac{\cos^2a+\sin^2a-\cos^2a}{\sin a(1-\cos a)}\quad\text{since: $\cos^2a+\sin^2a=1$}\\\,\\
&=\dfrac{\sin^2a}{\sin a(1-\cos ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/614109",
"timestamp": "2023-03-29T00:00:00",
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Probability that $5$ divides $a^2+b^2$ is $\frac9{25}$
Two positive integers $a$ and $b$ are randomly selected with replacement, then prove that the probability of $(a^2 +b^2)/5$ being positive integer is $9/25$.
I found out the pattern of the last digits of $a^2$ , $b^2$ but then the sample space isn't finite so I ... | The idea of this question is that $a$ has $20\%$ chances of being each of $0$ or $1$ or $2$ or $3$ or $4$ modulo $5$, hence $a^2$ has $20\%$ chances of being $0$, $40\%$ to be $1$ and $40\%$ to be $4$ modulo $5$. The same holds for $b^2$ hence, assuming that $a$ and $b$ are independent, $a^2+b^2$ is $0$ modulo $5$ when... | {
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"timestamp": "2023-03-29T00:00:00",
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Why is $\left(\sum \limits_{n=1}^{\infty}\frac1{n^3}\right)^3 \gt \sum \limits_{n=1}^{\infty}\frac1{n^4} $?
$$\displaystyle \left(\sum \limits_{n=1}^{\infty}\frac1{n^3}\right)^3 \gt \sum_{n=1}^{\infty}\frac1{n^4} $$
Why is this true ?
The LHS can be written either as it is or like this: $\displaystyle \sum \limits_{... | In general a power of a sum is different from the sum of powers, $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$ coincides with $a^3 + b^3$ only if either (at least) one of $a,b$ is zero or $a+b = 0$. For sums with more terms or series, conditions for equality aren't so straightforward. Here we have
$$1 < \sum_{n=1}^\infty \frac{1}{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/616899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How prove this equation $A^2+B^2=C^2+D^2$ define:
plane $W:Ax+By+Cz+D=0$ and the
hyperboloid of one sheet $U:x^2+y^2-z^2=1$
if $$W\bigcap U=l_{1},W\bigcap U=l_{2}$$
where $l_{1},l_{2}$ be two straight lines
show that :$$A^2+B^2=C^2+D^2$$
My try: since
$$\begin{cases}
Ax+By+Cz+D=0\\
x^2+y^2-z^2=1
\end{cases}$$
t... | Here's an answer with a bit of a geometric flavor.
Clearly, the hyperboloid contains no lines that are parallel to any of the coordinate axes; it also contains no points $(x,y,z)$ such that $x^2 + y^2 < 1$. Consequently, an embedded line $\ell$ must pass through a point $P$ on the $xy$-plane, which must in fact lie on ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/618429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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show that $\int_0^1 (\sqrt[3]{1-x^7}-\sqrt[7]{1-x^3}) \, dx=0$ show that
$$I=\int_0^1 (\sqrt[3]{1-x^7}-\sqrt[7]{1-x^3}) \, dx=0$$
I find this is Nice equalition!
My try: let
$$\sqrt[3]{1-x^7}=t\Longrightarrow x=\sqrt[7]{1-t^3}$$
so
$$dx=-\dfrac{3}{7}t^2(1-t^3)^{-\dfrac{6}{7}} \, dt$$
so
$$I=\frac{3}{7}\int_0^1 \frac{t^... | The antiderivative of $(1-x^a)^{1/b}$, at least for integer values of $a$ and $b$, is given by $x{\rm Hypergeometric2F1}[1/a, -(1/b), 1 + 1/a, x^a]$.
The integral of $(1-x^a)^{1/b}$ between 0 and 1 is then given by
$(\Gamma[1 + 1/a] \Gamma[1 + 1/b]) / \Gamma[1 + 1/a + 1/b]$
Then, the two integrals are equal and your id... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/619333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
} |
question on integrals Let $\displaystyle A=\int_0^1 \frac{dx}{1+x^8}$. Then which of the following are true:
1) $A\lt 1$,
2) $A\gt 1$,
3) $A\lt \frac{\pi}{4... | If $0\lt x\lt 1$, we have
$$\frac{1}{1+x^2}\lt \frac{1}{1+x^8}\lt \frac{1}{1+0^2}.$$
Hence, we have
$$\frac{\pi}{4}=\int_0^1\frac{1}{1+x^2}dx\lt\int_0^1\frac{1}{1+x^8}dx\lt\int_0^1\frac{1}{1+0^2}dx=1.$$
That leads that the answer $4)$ and $1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/619494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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If $n\in\mathbb N$ and $4^n+2^n+1$ is prime, prove that there exists an $m\in\mathbb N\cup\{0\}$ such that $n=3^m$. If $n\in\mathbb N$ and $4^n+2^n+1$ is prime, prove that there exists an $m\in\mathbb N\cup\{0\}$ such that $n=3^m$.
I.e. if $4^n+2^n+1$ is prime, prove that $n=3^m$, where $m\in\mathbb N\cup\{0\}$.
I don'... | Hint. For prime $p\not=3$ we have
$$(x^2+x+1)\mid (x^{2p}+x^p+1)$$
Proof.
\begin{align}x^{2p}+x^p+1&=\frac{x^{3p}-1}{x^p-1}\\&=\frac{(x^3-1)\left(\sum\limits_{k=0}^{p-1}x^{3k}\right)}{x^p-1}\\&=\frac{(x^2+x+1)\left(\sum\limits_{k=0}^{p-1}x^{3k}\right)}{(x^p-1)/(x-1)}\end{align} Now it suffices to prove \begin{align}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/620086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
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Solving $\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=c$ Question:
let $a,b,c$ be positive constants. Find $u=u(x,y)$ if is satisfies the partial differential equation
$$\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2}=c$$
and the boundary condition
$$u=0,\dfrac{x^2}{a^2}+\df... | Hint. Try
$$
u(x,y)=\frac{x^2}{a^2}+\frac{y^2}{b^2}-1.
$$
Find a solution, and then show uniqueness.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/621192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How find this $\left(\frac{1}{x^2+a^2}\right)^{(n)}$ Prove that
$$\left(\dfrac{1}{x^2+a^2}\right)^{(n)}=(-1)^{(n)}n!\dfrac{\sin{[(n+1)\cdot \mathrm{arccot}{(x/a)}]}}{a(x^2+a^2)^{(n+1)/2}}$$
my try:
since
$$\dfrac{1}{x^2+a^2}=\dfrac{1}{2ai}\left(\dfrac{1}{x-ai}-\dfrac{1}{x+ai}\right),i=\sqrt{-1}$$
so
$$\left(\dfrac{1}{x... | First Method
Let $$\mathcal{L}\{f(t)\}=\int_0^\infty f(t)\operatorname{e}^{-xt}\operatorname{d}t=F(x)$$ be the Laplace transform of $f(t)$.
For $f(t)=\sin(at)u(t)$ then we have $$F(x)=\int_0^\infty \sin(at)\operatorname{e}^{-xt}\operatorname{d}t=\frac{a}{x^2+a^2}.$$
Recalling that $\mathcal{L}\{t^n f(t)\}=(-1)^n F^{(n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/622323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Sum of an infinite series - obscure sum I am trying to show the following:
I can get to the following result:
I would appreciate any help with solving this problem.
Thanks in advance.
| First thing to do is replace for $$x=\frac{s}{2}$$
$$\implies \frac{-q^2}{8\pi e} \left( \frac{1}{2\frac{s}{2}} + \sum_{n=1}^{\infty} \left(\frac{ns}{(ns)^2-(\frac{s}{2})^2} - \frac{1}{ns} \right) \right)$$
$$\frac{-q^2}{8\pi e} \left( \frac{1}{s} + \sum_{n=1}^{\infty} \left( \left(\frac{s}{s^2}\right)\frac{n}{(n)^2-(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/624713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integration step need some explanation (Calculus) $\displaystyle{\frac15\int_0^1\left[v(1+v^2)^5-v^{11}\right]dv=\frac 15\left[\frac 12\cdot\frac 16(1+v^2)^6-\frac{1}{12}v^{12}\right]_0^1}$
Can someone explain how to get the RHS from LHS. It's the first term after the integral sign that I am confused about.
Thanks.
| Remember that integration is linear so:
$$\int x(1+x^2)^5-x^{11} \, \operatorname{d}\!x = \int x(1+x^2)^5 \, \operatorname{d}\!x - \int x^{11} \, \operatorname{d}\!x$$
Hopefully you already know that
$$\int x^{11} \, \operatorname{d}\!x = \frac{1}{12}x^{12}+C$$
You need to show that
$$\int x(1+x^2)^5 \, \operatorname... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/626445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Why is it that $2^{10} + 2^{9} + 2^{8} + \cdots + 2^{3} + 2^2 + 2^1 = 2^{11} - 2$? Why is it that $$2^{10} + 2^{9} + 2^{8} + \cdots + 2^{3} + 2^2 + 2^1 = 2^{11} - 2?$$
| Try using the fact that $x^n + x^{n - 1} + \cdots + x + 1 = \frac{x^{n + 1} - 1}{x - 1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/628501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 5
} |
Find $\int_0^{\frac{\pi}{4}}e^{\sec^2 x}dx$ How can we find
$$\int_0^{\frac{\pi}{4}}e^{\sec^2 x}dx$$
I tried $t=\frac{\pi}{4}-x$ but this seems not work. Any hints?
| $\int_0^\frac{\pi}{4}e^{\sec^2 x}~dx$
$=\int_0^\frac{\pi}{4}\sum\limits_{n=0}^\infty\dfrac{\sec^{2n}x}{n!}~dx$
$=\int_0^\frac{\pi}{4}dx+\int_0^\frac{\pi}{4}\sum\limits_{n=1}^\infty\dfrac{\sec^{2n}x}{n!}~dx$
$=[x]_0^\frac{\pi}{4}+\int_0^\frac{\pi}{4}\sum\limits_{n=1}^\infty\dfrac{\sec^{2n-2}x}{n!}~d(\tan x)$
$=\dfrac{\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/630472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Inequality $\cos x+\cos y+\cos z=0$ and $\cos{3x}+\cos{3y}+\cos{3z}=0$ prove that $\cos{2x}\cdot \cos{2y}\cdot \cos{2z}\le 0$ Let $x,y,z$ be real numbers such that $\cos x+\cos y+\cos z=0$ and $\cos{3x}+\cos{3y}+\cos{3z}=0$ prove that $\cos{2x}\cdot \cos{2y}\cdot \cos{2z}\le 0$.
| Setting $\cos x=a$ etc.
We have
$\displaystyle a+b+c=0\ \ \ \ (1)$
$\displaystyle\implies a^3+b^3+c^3=(a+b)^3-3ab(a+b)+c^3=(-c)^3+-3ab(-c)+c^3$
$\displaystyle\implies a^3+b^3+c^3=3abc\ \ \ \ (2)$
Again as $\displaystyle\cos3x=4\cos^3x-3\cos x\implies,$
$\sum\cos3x=0\implies 4(a^3+b^3+c^3)=3(a+b+c)=0\ \ \ \ (3)$
Fro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/632565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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Let $n=x^2+y^2$; $n=2^{2k}m$ or $n=2^{2k+1}m$ with $m$ odd. Prove that $2^{k}$ divides both $x$ and $y$. Let $n=x^2+y^2$ where x,y are integers, be one of the forms $n=2^{2k}m$ respectively $n=2^{2k+1}m$ with m odd. Prove that $2^{k}$ divides both x and y.
| We prove something a little stronger. Let $2^a$ be the highest power of $2$ that divides both $x$ and $y$. We show that if $2^{2k}$ or $2^{2k+1}$ is the highest power of $2$ that divides $x^2+y^2$, then $a=k$.
Since $2^a$ is the highest power of $2$ that divides both $x$ and $y$, we have $x=2^a s$ and $y=2^at$, where... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/632930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Prove that $ \lim_{n \rightarrow \infty } \frac{n+6}{n^2-6} = 0 $. My attempt:
We prove that $ \lim\limits_{n \rightarrow \infty } \dfrac{n+6}{n^2-6} =0$.
It is sufficient to show that for any $ \epsilon \in\textbf{R}^+ $, there exists an $ K \in \textbf{R}$ such that for any $ n > K $
$ \displaystyle \left| \frac{... | It is sufficient to use the fact that $\displaystyle \frac{n+6}{n^2-6}\sim_{\infty}\frac{n}{n^2}=\frac{1}{n}$ then you have the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/633047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Solving $2\cos(x) = \sin(x)$ How would you solve equations of the form $ a \sin (x+b) = \sin (x)$?
Eg. $ 2 \cos(x) = \sin(x) $
I realy have no idea how I would solve this kind of equations.
| For your initial question, you can use the angle-sum rule for $\sin(\alpha + \beta):$ $$\sin(\alpha \pm \beta) = \sin \alpha \cos \beta \pm \cos \alpha \sin \beta$$
So $$a \sin (x+b) = \sin (x) \iff a (\sin x \cos b + \cos x \sin b) = \sin x$$
We can divide through by $\cos x$, provided $\cos x \neq 0$, noting that $a,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/633672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that: $\cos(x) -1 < -\frac{x^2}{2} + \frac{x^4}{24}$
Prove that:
$$\cos(x) -1 < -\frac{x^2}{2} + \frac{x^4}{24}$$ for $x \ne 0$
I need to prove this using Cauchy's mean value theorem.
What I did:
$f(x) = \cos(x) -1$
$$g(x) = -\frac{x^2}{2} + \frac{x^4}{24}$$
If I can show that:
$\frac{f(x)}{g(x)}\lt 1$ that wou... | I think it will be easier to show that: $$ \cos x - 1 + \frac{x^2}2 < \frac{x^4}{24} $$
Define $f(x) = \cos x - 1 + \frac{x^2}2$, $g(x) = \frac{x^4}{24}$.
Now using the Cauchy MVT: $$ \frac{f(x)-f(0)}{g(x)-g(0)} = \frac{6(c-\sin c)}{c^3} $$
Now just show that $ h(x) = \frac{6(x-\sin x)}{x^3} $ is bounded above globably... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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What is the integrating factor of $2xy'+x^{2}e^{1-x^2}y=2$ I know how to start solving it, its dividing everything by $2x$, but I can't solve
$$\int \dfrac{x^2e^{1-x^2}}{2x}\,dx$$
| $$\int \frac{x^2 e^{1-x^2}}{2x}dx=\int \frac{xe^{1-x^2}}{2}dx=-\frac{1}{2}\int\frac{-2xe^{1-x^2}}{2}dx$$
Let $u=1-x^2$ then $du=-2xdx$
$$=-\frac{1}{4}\int-2xe^{1-x^2}dx=-\frac{1}{4}\int e^udu =-\frac{1}{4}e^u$$
But $u=1-x^2$
$$\int \frac{x^2 e^{1-x^2}}{2x}dx=-\frac{1}{4}e^{1-x^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/635229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Solving a challenging differential equation How would one go about finding a closed form analytic solution to the following differential equation?
$$\frac{d^2y}{dx^2} +(x^4 +x^2+x+c)y(x) =0 $$
where $c\in\mathbb{R}$
| This is the way I would do it.
Let $p(x)=x^4+x^2+x+c$. Put $z=y'$. Then the equation is equivalent to the system
$$
\begin{bmatrix}y\\ z\end{bmatrix}'=\begin{bmatrix}0&1\\-p(x)&0\end{bmatrix}\,\begin{bmatrix}y\\ z\end{bmatrix}.
$$
Let us write $A(x)$ for the $2\times 2$ matrix above. For a given vector $\begin{bmatri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/635314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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If $a$ and $b$ are cathets a right triangle whose hypotenuse is $1$ determine the highest value of $2a + b$ Can some one help me out on where to go?
If $a$ and $b$ are cathets a right triangle whose hypotenuse is $1$ determine the highest value of $2a + b$ ?
| Let the two sides be $\sin(x)$ and $\cos(x)$.
So we want to maximize $2 \cos(x) + \sin(x)$.
We can write (2,1) in polar form as
$$
2 = R\cos(y),~ 1 = R\sin(y)~~~~\Rightarrow R = \sqrt{5}$$
Hence
$$
2 \cos(x) + \sin(x) = R (\cos(x) \cos(y) + \sin(x) \sin(y)) =\sqrt{5} \cos(x-y)$$
We really do not need to find $x$ or $y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/636361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Positive recurrent, zero recurrent and transient states How do we prove that a state in a Markov chain process is positive recurrent, zero recurrent or transient? For example, if we have a transition matrix
$$P=\left(\begin{array}{ccc}
\frac{1}{3} & \frac{1}{3} & \ 0 & \frac{1}{3} \\
0 & \frac{1}{2} & \frac{1}{2} ... | The path $1\to4\to2\to3\to1$ enumerates the state space and has positive probability hence every state is positive recurrent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/638513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
minimum value of $\cos(A-B)+\cos(B-C) +\cos(C-A)$ is $-3/2$ How to prove that the minimum value of $\cos(A-B)+\cos(B-C) +\cos(C-A)$ is $-3/2$
| As Berci suggested ,$x=A-B,y=B-C \to f=\cos {x} +\cos{y}+cos{(x+y)}$
$f=2\cos{\dfrac{x+y}{2}} \cos{\dfrac{x-y}{2}}+2\left(\cos{\dfrac{x+y}{2}}\right)^2-1 \ge 2\cos{\dfrac{x+y}{2}}+2\left(\cos{\dfrac{x+y}{2}}\right)^2-1 =2\left(\cos{\dfrac{x+y}{2}}+\dfrac{1}{2}\right)^2-\dfrac{3}{2} \ge -\dfrac{3}{2}$
first " $\ge$ " : ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/639890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Find all solutions of $z^5+a^5=0$ The task is as follows:
Find all solutions of $z^5+a^5=0$, where $a$ is a positive real number.
My initial attempt
(which leads nowhere)
My guess is that i'll have to find the 5 5th roots of $-z^5$:
$w_1 = |-z^5|^5(cos(\frac{\theta}{5})+isin(\frac{\theta}{5})) \\
w_1 = |z|(cos(\frac{... | To make our life a bit easier, start by writing $z=r(\cos\theta+i\sin\theta)$ for some $r>0$ and some real $\theta,$ so that $$-a^5=z^5\\-a^5=r^5(\cos\theta+i\sin\theta)^5\\-a^5=r^5(\cos5\theta+i\sin5\theta)$$ Now, taking the modulus of both sides gives us $$a^5=r^5,$$ so we'll need $r=a,$ and so $z=a(\cos\theta+i\sin\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/640965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Diophatine equation $x^2+y^2+z^2=t^2$ Probably duplicate but I don't find: I'd like to solve the diophantine equation
$$x^2+y^2+z^2=a^2$$
which has solutions, by exemple $1^2+2^2+2^2=3^2$ or $2^2+3^2+6^2=7^2$. Every such solution gives a rational point on the unit sphere.
Is there a complete description of the solution... | We'll start with Pythagorean triples to see the pattern. The complete rational solution to $x_1^2+x_2^2 = y_1^2$ has the form,
$$((a^2-b^2)t)^2+(2abt)^2 = ((a^2+b^2)t)^2\tag{1}$$
where $t$ is a scaling factor.
Proof: For any solution where $x_1+y_1 \neq 0$ , one can always find rational {$a,b,t$} using the formulas $a... | {
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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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Solve the system using Gaussian elimination with back-substitution or Gauss-Jordan elimination Here is the system:
$$
\left\{
\begin{aligned}
x_1-3x_3&=-2 \\
3x_1+x_2-2x_3&=5 \\
2x_1+2x_2+x_3&=4
\end{aligned}
\right.
$$
This is my very first problem actually using a matrix so here is my attempt
First I setup the augmen... | From your step: And then $(-1)R_2+R_3->R_3$
If you look at where you have $−11+4$.
| {
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"url": "https://math.stackexchange.com/questions/648351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Induction: show that $\sum\limits_{k=1}^n \frac{1}{\sqrt{k}} < 2 \sqrt{n}$ for all n $\in Z_+$ The question:
show by using induction that $\sum\limits_{k=1}^n \frac{1}{\sqrt{k}} < 2 \sqrt{n}$ for all n $\in Z_+$
My attempt at a solution:
The base case $n = 1$ is true.
First we use the inductive assumption that the stat... | Note that, by induction hypothesis
$$
\sum_{k=1}^{n+1} \frac{1}{\sqrt{k}} < 2\sqrt{n} + \frac{1}{\sqrt{n+1}}
$$
The RHS is $<2\sqrt{n+1}$ iff
$$
\frac{1}{\sqrt{n+1}} < 2(\sqrt{n+1} - \sqrt{n})
$$
$$
\Leftrightarrow \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1}} < 2(n+1-n) = 2
$$
$$
\Leftrightarrow 1 + \sqrt{\frac{n}{n+1}} < ... | {
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"url": "https://math.stackexchange.com/questions/648907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 1
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Identifying a quotient group (NBHM-$2014$)
Let $\mathbb C^*$ denote the multiplicative group of non-zero complex numbers and let $P$ denote the subgroup of positive real numbers. Identify the quotient group.
My thought $$\frac{\mathbb C^*}{P}=\{P,-P,iP,-iP\}.$$ Is it right?
| I am not sure if this is in any way different than B.S's way. i am just omitting maps and trying to see just as a quotient.
We see $x+iy\in \mathbb{C}$ as $(\sqrt{x^2+y^2})(\dfrac{x}{\sqrt{x^2+y^2}}+i\dfrac{y}{\sqrt{x^2+y^2}})$
Now, As $\sqrt{x^2+y^2}\in \mathbb{R}$ when we see $(\sqrt{x^2+y^2})(\dfrac{x}{\sqrt{x^2... | {
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Solve $2000x^6+100x^5+10x^3+x-2=0$ One of the roots of the equation $2000x^6+100x^5+10x^3+x-2=0$ is of the form $\frac{m+\sqrt{n}}r$, where $m$ is a non-zero integer and $n$ and $r$ are relatively prime integers.Then the value of $m+n+r$ is?
Tried to use the fact that another root will be $\frac{m-\sqrt{n}}r$ as coeff... | $$2000x^6+100x^5-200x^4+200x^4+10x^3-20x^2+20x^2+x-2$$
$$(2000x^6+200x^4+20x^2)+(100x^5+10x^3+x)-(200x^4+20x^2+2)$$
$$x^2(2000x^4+200x^2+20)+\frac{x}{20}(2000x^4+200x^3+20x)-\frac{1}{10}(2000x^4+200x^2+20)$$
$$=(x^2+\frac{x}{20}-\frac{1}{10})(2000x^4+200x^2+20)=0$$
| {
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"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 0
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Showing $\frac{x}{1+x}<\log(1+x)0$ using the mean value theorem I want to show that $$\frac{x}{1+x}<\log(1+x)<x$$ for all $x>0$ using the mean value theorem. I tried to prove the two inequalities separately.
$$\frac{x}{1+x}<\log(1+x) \Leftrightarrow \frac{x}{1+x} -\log(1+x) <0$$
Let $$f(x) = \frac{x}{1+x} -\log(1+x).$$... | As a consequence of MVT, there is a $\xi\in(0,x)$, such that
$$
\log(1+x)=\log(1+x)-\log 1=x\cdot \left(\log(1+x)\right)'_{x=\xi}=x\cdot\frac{1}{1+\xi}<x.
$$
Let $y=\frac{x}{1+x}$. Then there is $\xi\in\big(0,y\big)$, such that
\begin{align}
\log(1+x)&=-\log\left(\frac{1}{1+x}\right)=\log 1-\log\left(1-\frac{x}{1+x}\ri... | {
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Maximum and minimum distance from the origin
Find the maximum and minimum distances from the origin to the curve $5x^3+6xy+5y^2-8=0$
My attempt:
We have to maximise and minimise the following function $x^2+y^2$ with the constraint that $5x^3+6xy+5y^2-8=0$.
Let
$$F(x,y)=x^2+y^2+\lambda(5x^2+6xy+5y^2-8)$$
$$\frac{\del... | You have found the points that, under the contraint $5x^2+6xy+5y^2−8=0$, minimise $x^2+y^2$and therefore minimise the euclidean distance between the origin and the point $(x,y)$ that is
$$d_{euclidean}\left( (0,0), (x,y) \right) \doteq \sqrt{(x-0)^2+(y-0)^2} = \sqrt{x^2+y^2}$$
So your minimum distance is the distance... | {
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"source": "stackexchange",
"question_score": "4",
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Solve the difference quotient
So I tried a variety of answers and they all came up incorrect. I guess I don't understand difference quotient because I feel like it's pointless and a slow way to do the problem. That being said... can't get the answer
| You are expected to write
$$\frac{f(x)-f(4)}{x-4}=\frac{\frac{x+6}{x+4}-\frac{10}{8}}{x-4},\tag{1}$$
and then simplify.
We first simplify the numerator. Bringing it to the common denominator $(x+4)(8)$, we get
$$ \frac{x+6}{x+4}-\frac{10}{8}=\frac{(x+6)(8)-(x+4)(10)}{(x+4)(8)}=\frac{-2x+8}{(x+4)(8)}.$$
This simplifies... | {
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"source": "stackexchange",
"question_score": "1",
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Prove trig identity: $\tan(x) + \cot(x) = \sec(x) \csc(x)$ wherever defined I appreciate the help.
My attempt:
$$
\begin{align}
\tan(x) + \cot(x) &= \frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)} \\
&= \frac{\sin^2(x)}{\cos(x) \sin(x)}+\frac{\cos^2(x)}{\cos(x) \sin(x)} \\
&= \frac{\sin^2(x)+\cos^2(x)}{\cos(x) \sin... | An alternative approach writes $t=\tan x/2$ so $$\tan x+\cot x=\frac{2t}{1-t^2}+\frac{1-t^2}{2t}=\frac{1+t^2}{1-t^2}\frac{1+t^2}{2t}=\sec x\csc x.$$
| {
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"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 3
} |
Computing the limit of $n \cdot \arccos \left( \left(\frac{n^2-1}{n^2+1}\right)^{\cos (1/n)} \right)$ I need to solve this earth's wonder:
$$\lim_{n \rightarrow \infty} \left[n \; \arccos
\left( \left(\frac{n^2-1}{n^2+1}\right)^{\cos \frac{1}{n}} \right)\right]$$
I have tried to write down it using $e^{v \ln u}$,and ... | The function given here is really bit complex (at least in typing). We first need to check if we know any fundamental limit associated with $\arccos$ function. Clearly there isn't any, but we do have the standard limit $$\lim_{x \to 0}\frac{1 - \cos x}{x^{2}} = \lim_{x \to 0}\dfrac{2\sin^{2}(x/2)}{x^{2}} = \frac{1}{2}$... | {
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"source": "stackexchange",
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How prove $3(a^4+b^4+c^4)+2(a+b+c)abc\ge 5(a^2b^2+b^2c^2+a^2c^2)$ let $a,b,c>0$ and such $abc=1$,show that
$$3(a^4+b^4+c^4)+2(a+b+c)\ge 5\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)$$
my idea: maybe can use AM-GM inequality,
$$3(a^4+b^4+c^4)a^2b^2c^2+2(a+b+c)(a^2b^2c^2)\ge 5(a^2b^2+b^2c^2+a^2c^2)$$
$$\Long... | Consider Schur's inequality for $r = 2$:
$$a^4 + b^4 + c^4 + (a + b + c)abc \geqslant a^3b + a^3c + b^3a + b^3c + c^3a + c^3b$$
plus three AM-GMs:
$$
\begin{aligned}
a^3b + b^3a &\geqslant 2a^2b^2\\
b^3c + c^3b &\geqslant 2b^2c^2\\
a^3c + c^3a &\geqslant 2a^2c^2
\end{aligned}
$$
This gives you, after multiplying by $2$... | {
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"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Solve $t^4+4 t^3+6 t^2+4 t-32 t^{1/4}+1 = -16 $ I'm trying to solve the following equation: $$(t+1)^4 - 32 t^{\frac{1}{4}}=-16 $$ where t $\geq 0$,
which is equivalent to $$t^4+4 t^3+6 t^2+4 t-32 t^{\frac{1}{4}}+1 = -16 $$
Wolfram Alpha tells that the equation is equivalent to $$t+1=2t^{\frac{1}{4}} $$ if we assume tha... | $u^{12}+2u^9+3u^8+4u^6+4u^5+3u^4+8u^3+4u^2+2u+17=u^8(u^4+2u+3)+(x+1)^2(4x^4-4x^3+7x^2-2x+1)+2(x+1)+14$
$u^4+2u+3 \ge 2u^2+2u+2 \ge \dfrac{3}{2}$
$4x^4-4x^3+7x^2-2x+1=x^2(2x-1)^2+6(x-\dfrac{1}{6})^2+\dfrac{5}{6} > \dfrac{5}{6}$
$(x+1)^2(4x^4-4x^3+7x^2-2x+1)+2(x+1)+14>\dfrac{5}{6}(x+1)^2+2(x+1)+14=\dfrac{5}{6}(x+1+\dfrac... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Prove that $\left(\frac12(x+y)\right)^2 \le \frac12(x^2 + y^2)$ Prove that $$\left(\frac12(x+y)\right)^2 \leq \frac12(x^2 + y^2)$$
I've gotten that
$$\left(\frac12(x+y)\right)^2 \ge 0 $$
but stumped on where to go from here...
| let investigate $$\frac{(x+y)^2}{4}-\frac{x^2+y^2}{2}=\frac{x^2+y^2+2xy-2x^2-2y^2}{4}=\frac{-(x-y)^2}{4}$$ then it is obvious that we have $$\frac{(x+y)^2}{4}-\frac{x^2+y^2}{2}\leq 0.$$ So $$\frac{(x+y)^2}{4}\leq\frac{x^2+y^2}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/665206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 5
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How to use the generating function $F(x) =x/(1-x-x^2).$ The generating function for the Fibonacci sequence is
$$F(x) =x/(1-x-x^2).$$
To work out the 20th value of the sequence I understand you somehow expand this and look at the coefficient of $x^{20}$. How exactly do you do this?
| You can do the following:
*
*Solve the equation $1-x-x^2=0$. The roots are $\frac{-1-\sqrt{5}}{2}$ and $\frac{-1+\sqrt{5}}{2}$.
*Then you have $1-x-x^2=-(x-\frac{-1-\sqrt{5}}{2})(x-\frac{-1+\sqrt{5}}{2})=-(x+\frac{1+\sqrt{5}}{2})(x+\frac{1-\sqrt{5}}{2})$
*$\frac{-x}{(x+\frac{1+\sqrt{5}}{2})(x+\frac{1-\sqrt{5}}{2}... | {
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"source": "stackexchange",
"question_score": "2",
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Show if $a^2+b^2 \le 2$ then $a+b \le 2$ If $a^2+b^2 \le 2$ then show that $a+b \le2$
I tried to transform the first inequality to $(a+b)^2\le 2+2ab$ then $\frac{a+b}{2} \le \sqrt{1+ab}$ and I thought about applying $AM-GM$ here but without result
| Suffices to show if $a^2+b^2 = 2$ then $a+b \leq 2$. From the constraint consider
$$
f(a) = a + \sqrt{2-a^2}
$$
and we need to prove $f(a) \leq 2$ over $[0,\sqrt{2}]$.
$$
f'(a) = 1 - \frac{a}{\sqrt{2-a^2}} \Leftrightarrow a = 1
$$
which is a maximum by 1st derivative test.
Since $f(0), f(1), f(\sqrt{2})$ are $\sqrt{2},... | {
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"question_score": "1",
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How to show that $ 0< a\leq\cos^2(\theta)\leq b<1$ in this problem? The inequality $2\cos^4(\theta/2)-2\cos^2(\theta/2)+1/4\leq 0$ means that $\cos^2(\theta/2)$ lies between the roots of $2x^2-2x+1/4$ i. e., we can conclude that
$$
\frac{2-\sqrt{2}}{4}\leq\cos^2(\theta/2)\leq\frac{2+\sqrt{2}}{4}.
$$
The question is: ... | You can use the "double angle formula"
$$
\cos(\theta) = 2\cos^2(\theta / 2) - 1
$$
so
$$
\cos^2(\theta) = (2\cos^2(\theta / 2) - 1)^2
$$
Then we write
\begin{align*}
\frac{2-\sqrt{2}}{4} &\leq \cos^2(\theta/2)\leq\frac{2+\sqrt{2}}{4} \\
\frac{2-\sqrt{2}}{2} &\leq 2\cos^2(\theta/2)\leq\frac{2+\sqrt{2}}{2} \\
\frac{-\sq... | {
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"question_score": "1",
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"answer_id": 0
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if $(a+\sqrt{a^2+1})$ and $(b+\sqrt{b^2+1})$ are converse then prove that a and b are opposites $(a+\sqrt{a^2+1})\,(b+\sqrt{b^2+1})=1$ is supposed to equal: $b = -a$ but how do i get that? I've been trying to solve for like 2 days now.
| First approach. You can do it quite directly. This is not an elegant path, but it's easy to grasp and it is reliable. Regard your equality as an equation, where $b$ is fixed and $a$ is an unknown. How would you go about solving it?
It seems that the natural thing to do is to get $\sqrt{a^2+1}$ on one side of the equati... | {
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"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Three Surface Integrals Could someone assist with the following three surface integrals?
Q1 The portion of the cone $z=\sqrt{x^2+y^2}$ that lies inside the cylinder $x^2+y^2 =2x$.
Q2 The portion of the paraboloid $z=1-x^2-y^2$ that lies above the $xy$-plane.
Q3 The portion of the paraboloid $2z = x^2+y^2$ that is... | You have $z=f(x,y)$ and the surface area is $$\int\int_A\sqrt{1+f_x^2+f_y^2}dxdy$$ where $A$ is the projection of the surface area on the $xy$-plane.
Use polar coordinates and so $dA=rdrd\theta$
*
*$r=0\to r=2\cos\theta,\theta=-\pi/2\to\pi/2$
*$r=0\to r=1,\theta=0\to2\pi$
*$r=0\to r=\sqrt8,\theta=0\to2\pi$
For t... | {
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Prove for any integer $a$, one of the integers $a$, $a+2$, $a+4$ is divisible by $3$. Prove for any integer $a$, one of the integers $a$, $a+2$, $a+4$ is divisible by $3$.
I know I would use the division algorithm but I am really confused how to go about this. Step by step explanation please? thank you so much!
| For every number $a$: $a \equiv {0, 1, 2} \pmod{3}$
In the first case: $a \equiv 0 \pmod{3}$
For the second one:
$a \equiv 1 \pmod{3}$ =>
$a+2 \equiv 3 \pmod{3} \equiv 0 \pmod{3}$
For the third one:
$a \equiv 2 \pmod{3}$ =>
$a+4 \equiv 6 \pmod{3} \equiv 0 \pmod{3}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Proves of identities in inverse trigonometry Can someone please help me prove the following results from inverse trigonometry?
$$\tan^{-1}x + \tan^{-1}y = \pi + \tan^{-1}\frac{x+y}{1-xy}( x>0, y>0, xy>1)$$
and
$$\tan^{-1}x + \tan^{-1}y = -\pi + \tan^{-1}\frac{x+y}{1-xy} ( x<0, y< 0, xy>1)$$
I know the prove for $\tan... | Hint: Let $\tan^{-1} x = a$ and $\tan^{-1} y = b$. Then $x = \tan a$ and $y = \tan b$.
$$\frac{x+y}{1-xy} = \frac{\tan a + \tan b}{1- \tan a \tan b}$$
$$\frac{x+y}{1-xy} = \tan {(a+b)}$$
| {
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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How to find the determinant of this $3 \times 3$ Hankel matrix? Today, at my linear algebra exam, there was this question that I couldn't solve.
Prove that
$$\det \begin{bmatrix}
n^{2} & (n+1)^{2} &(n+2)^{2} \\
(n+1)^{2} &(n+2)^{2} & (n+3)^{2}\\
(n+2)^{2} & (n+3)^{2} & (n+4)^{2}
\end{bmatrix} = -8$$
Clearly, ca... | Recall that $a^2-b^2=(a+b)(a-b)$. Subtracting $\operatorname{Row}_1$ from $\operatorname{Row}_2$ and from $\operatorname{Row}_3$ gives
$$
\begin{bmatrix}
n^2 & (n+1)^2 & (n+2)^2 \\
2n+1 & 2n+3 & 2n+5 \\
4n+4 & 4n+8 & 4n+12
\end{bmatrix}
$$
Then subtracting $2\cdot\operatorname{Row}_2$ from $\operatorname{Row}_3$ gives... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "37",
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How find this maximum $f(x)=\sqrt{\frac{1+x^2}{2}}+\sqrt{x}-x$ let $x\ge 0$,Find the follow maximum
$$f(x)=\sqrt{\dfrac{1+x^2}{2}}+\sqrt{x}-x$$
I find
$$f'(x)=\dfrac{1}{2}\left(\dfrac{\sqrt{2}x}{\sqrt{x^2+1}}+\dfrac{1}{\sqrt{x}}-2\right)$$
so
$$f'(x)=0\Longrightarrow x=1$$
so I think
$$f(x)_{max}=f(1)$$
But I this me... | Simply consider that $\sqrt{x}$ is a concave function, hence, for any $a,b\geq 0$, $\sqrt{a}+\sqrt{b}\leq \sqrt{2(a+b)}$. By taking $a=x$ and $b=\frac{1+x^2}{2}$ you get $f(x)\leq 1$ with equality only when $x=\frac{1+x^2}{2}$, i.e. $x=1$, as wanted.
| {
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If $x+\frac1x=5$ find $x^5+\frac1{x^5}$. If $x>0$ and $\,x+\dfrac{1}{x}=5,\,$ find $\,x^5+\dfrac{1}{x^5}$.
Is there any other way find it?
$$
\left(x^2+\frac{1}{x^2}\right)\left(x^3+\frac{1}{x^3}\right)=23\cdot 110.
$$
Thanks
| Hint: Expand (binomial formula) $\left(x+\frac{1}{x}\right)^3$ and then $\left(x+\frac{1}{x}\right)^5$. (No need for even powers.)
| {
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"url": "https://math.stackexchange.com/questions/678650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Let $\theta=\frac{2\pi}{5}$. Show $2\cos(2\theta)+2\cos(\theta)+1=0$. I have been at this for a while. Any ideas?
| I think this is a simple way. By using De Moivre’s formula, we obtain the fifth time angle formula:
\begin{equation}
\cos(5\theta)=16\cos^5\theta-20\cos^3\theta+5\cos\theta
\end{equation}
Set $\theta=2\pi/5$, we have:
\begin{equation}
\cos(2\pi)=16\cos^5\theta-20\cos^3\theta+5\cos\theta\\
16\cos^5\theta-20\cos^3\theta+... | {
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Show that the sequence $a_{n+1}=a_{n}+\dfrac{a^2_{n}}{n^2}$ is upper bounded Let $\{a_{n}\}$ be defined with $a_{1}\in(0,1)$, and
$$a_{n+1}=a_{n}+\dfrac{a^2_{n}}{n^2}$$
for all $n\gt 0$. Show that the sequence is upper bounded.
My idea: since
$$a_{n+1}=a_{n}\left(1+\dfrac{a_{n}}{n^2}\right)$$
then
$$\dfrac{1}{a_{n+... | Here is a solution: Since $(a_{n})$ is monotone increasing, either it remains bounded or it diverges to $+\infty$. Assume that $a_{n} \nearrow +\infty$.
Observation 1. Referring to OP's identity
$$ \frac{1}{a_{n}} - \frac{1}{a_{n+1}} = \frac{1}{a_{n} + n^{2}}, $$
we have
$$ \frac{1}{a_{n}} - \frac{1}{a_{m+1}} = \sum_{k... | {
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"timestamp": "2023-03-29T00:00:00",
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Sum of four squares not a prime Let $ a, b, c, d $ be natural numbers such that $ ab=cd $. Prove that $ a^2+b^2+c^2+d^2 $ is not a prime.
I am clueless on this one. I tried contradiction, but didn't get anywhere.
Can you help?
Edit: I understand natural numbers to be strictly positive, excluding $0 $.
| Let $N := a^2 + b^2 + c^2 + d^2$
Note that $bd(a^2 + c^2) = ac(b^2 + d^2)$, while $ac < a^2+c^2$, so that $$lcm(a^2 + c^2, b^2 + d^2) < (a^2 + c^2)(b^2 + d^2)$$ This tells us that $a^2 + c^2$ and $b^2 + d^2$ are not coprime, and so $$ 1 < gcd(a^2 + c^2, b^2 + d^2)< N $$
Finally this $gcd|N$, which completes the proof... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Have I done something wrong in solving the following pair of equations? Question given:
Solve,$3x^2-5y^2-7=0\\3xy-4y^2-2=0$
What I have done so far:
$$
3xy-4y^2-2=0\\\frac{3xy}{y}-\frac{4y2}{y}-\frac{2}{y}=0\\3x-4y-\frac{2}{y}=0\\3x=4y+\frac{2}{y}\\x=\frac{(4y+\frac{2}{y})}{3}
$$By substituting $x=\frac{(4y+\frac{2}{y}... | What you did is perfectly correct. You only missed to continue the simplifcations since $a$ is either $4$ or $1$ and then $y$ is ...
Good job !
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/685821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the value of $a+b+c$ If $$3.\sqrt{5.\sqrt[3]{37}-16}=\sqrt[3]a-\sqrt[3]b-c$$
What is the value of $a+b+c$?
EDIT: I forgot to mention that $a$, $b$ and $c$ are positive integers.
I tried squaring and then cubing but it got very lengthy. Is there some elegant method to do it?
| Let $\theta=\sqrt[3]{37}$. If we put $\alpha=\theta^2-2\theta-2 \approx 2.4$, then
$$
\begin{array}{lcl}
\alpha^2 &=& (\theta^2-2\theta-2)^2 \\
&=& \theta^4 - 4 \theta^3 + 8 \theta + 4 \\
&=& 37\theta -4\times 37+8\theta+4 \\
&=& 45\theta -144 \\
&=& 9(5\theta-16)
\end{array}
$$
It follows that $3\sqrt{5\theta-16}=\a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/686279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Partial Fraction Decomposition of $\frac{x^2}{(x-1)(x+1)^2}$ I am suppose to find the Partial Fraction Decomposition of
$$\dfrac{x^2}{(x-1)(x+1)^2}$$
My work so far:
$$\dfrac{x^2}{(x-1)(x+1)^2} = \dfrac{A}{x-1} + \dfrac{B}{x+1} + \dfrac{C}{(x+1)^2}$$
Thus,
$$ x^2 = A(x+1)^3 + B(x-1)(x+1)^2 + C(x+1)(x-1)$$
With x = 0 we... | After your "thus" it must be
$$x^2=A(x+1)^2+B(x-1)(x+1)+C(x-1)$$
and not what you wrote there. Now substitute $\;x=1\;$ in both sides and get
$$1^2=A\cdot2^2\implies A=\frac14$$
and then substitute $\;x=-1\;$ and get ...etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/687545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Difference of consecutive cubes never divisible by 5. This is homework from my number theory course.
Since $(x+1)^3-x^3=3x^2+3x+1$ and $x^3-(x+1)^3=-3x^2-3x-1$, to say that the difference of two cubes is divisible by 5 is the same as saying that $3x^2+3x+1\equiv 0\mod 5$ or $-3x^2-3x-1\equiv 0\mod 5$. Both of these sta... | $$x^2+x\equiv 3\pmod 5\\
\implies x^2+x+4^{-1}\equiv 3+4^{-1}\pmod 5\\
\implies (x+2^{-1})^2=(x+3)^2\equiv 2\pmod 5$$
The quadratic residues modulo $5$ are $0,1,4$ and this list does not include $2$. QED
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/687975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 6
} |
Find limit of this decreasing sequence $$a_n=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right) \cdots
\left(1-\frac{1}{n^2}\right)
$$
I have proved that this sequence is decreasing. However I am trying to figure out how to find its limit.
| Answer:
$$\left(\frac{3}{2}\frac{4}{3}\frac{5}{4}\cdots\frac{n}{n-1}\frac{n+1}{n}\right)\cdot\left(\frac{1}{2}\frac{2}{3}\frac{3}{4}\cdots\frac{n-2}{n-1}\frac{n-1}{n}\right)$$
After mass cancellations, pull the $$\frac{n+1}{2}\text{ and }\frac{1}{n}$$
$$\frac{n+1}{2}\cdot\frac{1}{n}$$
Limit of this function tending to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/688782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Asymptotic solution of recurrence equation I need help to find asymptotic solution of this recurrence equation
$T(n)=\sqrt{n}T(\sqrt{n})+cn$ where $c$ is constant.
| Here are two thoughts, either one may help.
$$\begin{split}
T(n) &= n^{1/2} T\left(n^{1/2}\right) + cn \\
&= n^{1/2} \left[ n^{1/4} T\left(n^{1/4}\right)+cn^{1/2} \right] + cn\\
&= n^{1/2+1/4} T\left(n^{1/4}\right) + cn^{1/2+1/2} + cn\\
&= n^{1/2+1/4} \left[ n^{1/8} T\left(n^{1/8}\right)+cn^{1/4} \right]... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/689887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Integrate $\frac{x^2-1}{x^2+1}\frac{1}{\sqrt{1+x^4}}dx$ An antiderivative from Spivak
$$\int \frac{x^2-1}{x^2+1}\frac{1}{\sqrt{1+x^4}}dx$$
The idea I had was to write the first factor as $\left(1-\dfrac{2}{x^2+1}\right)$, but I don't see how that's helping!
| $$
\begin{align}
\int\frac{x^2 - 1}{x^2 + 1}\frac{1}{\sqrt{x^4+1}}\,\mathrm{d}x &=-\int\frac{1-x^2}{1+x^2}\frac{\mathrm{d}x}{\sqrt{\dfrac{\left(1+x^2\right)^2 + \left(1-x^2\right)^2}{2}}}\\
&=-\frac{\sqrt{2}}{2}\int\frac{1-x^2}{1+x^2}\frac{1}{\sqrt{1 + \left(\dfrac{1-x^2}{1+x^2}\right)^2}}\frac{2}{1 + x^2}\,\mathrm{d}x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/692118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find square roots upto infinte times Evaluate :
$\sqrt{1+ 2 \sqrt{1+3 \sqrt{1+\dots\infty}}}$
Is it possible to solve in the following way :
Let
$x=\sqrt{1+ 2 \sqrt{1+3 \sqrt{1+\dots\infty}}}$
$x^2= 1+ 2 \sqrt{1+3 \sqrt{1+\dots\infty}}$
next ??????
I don't know
| Note that $(x+1)^2=1+x(x+2)$, now taking the principal square root of both sides and re-substituting the resulting expression for $x+1$ multiple times gives us that:
$$(x+1)=\sqrt{1+x(x+2)}$$
$$(x+1)=\sqrt{1+x\sqrt{1+(x+1)(x+3)})}$$
$$(x+1)=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)(x+4)}}}$$
$$(x+1)=\sqrt{1+x\sqrt{1+(x+1)\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/695873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Fibonacci proof by Strong Induction Prove by strong induction that for a ∈ A we have
$F_a + 2F_{a+1} = F_{a+4} − F_{a+2}.$
$F_a$ is the $a$'th element in the Fibonacci sequence
| Do you consider the sequence starting at 0 or 1? I will assume 1.
If that is the case, $F_{a+1} = F_a + F_{a-1}) $ for all integers where $a \geq 3$.
The original equation states $F_{a+1} = (F_a) + F_{a-1} $.
.
$F_{a+1} = (F_a) + F_{a-1} $
$-(F_a) = -F_{a+1}+ F_{a-1} $
$F_a = F_{a+1}- F_{a-1}$. This equation is impor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/699901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $a,b,c>0$ and $a+b+c=1$ prove inequality: $\frac a{b^2 +c} + \frac b{c^2+a} + \frac c{a^2+b} \ge \frac 94$ If $a,b,c>0$ and $a+b+c=1$ prove inequality: $$\frac a{b^2 +c} + \frac b{c^2+a} + \frac c{a^2+b} \ge \frac 94$$
| Answer:
Multiplying a in the numerator and the denominator, you get
$$LHS = \frac{\sum a^{2}}{\sum (ac+b^{2}a)}$$
Applying the Cauchy-Schwarz inequality in the below steps
\begin{align}
&\geqslant\frac{(\sum a)^{2}}{\sum ac + \sum b^{2}a}\\
&\geqslant \frac{1}{\sum ac + \frac{1}{3}\sum a \sum a^2} = \frac{1}{3\sum ac ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/703518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Proof this curious trigonometric identity Proof that $$\cos^2{10^\circ} + \cos^2{50^\circ} - \sin{40^\circ}\sin{80^\circ} = \frac{3}{4}$$
I notice that $10^\circ + 80^\circ = 90^\circ$, and $50^\circ +40^\circ = 90^\circ$. I tried doing some manipulation but my efforts were futile. Any hints?
| Convert into a polynomial of $\cos(40^\circ)$:
$$
\begin{align}
&\cos^2(10^\circ)+\cos^2(50^\circ)-\sin(40^\circ)\sin(80^\circ)\\
&=\sin^2(80^\circ)+\sin^2(40^\circ)-\sin(40^\circ)\sin(80^\circ)\tag{1}\\
&=4\sin^2(40^\circ)\cos^2(40^\circ)+\sin^2(40^\circ)-2\sin^2(40^\circ)\cos(40^\circ)\tag{2}\\
&=1-2\cos(40^\circ)+3\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/703798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Product of reflections is a rotation, by elementary vector methods Let $\mathbf{u}$ and $\mathbf{v}$ be two 3D unit vectors. The transform that performs reflection in the plane normal to $\mathbf{u}$ is given by
$$
T_{\mathbf{u}}(\mathbf{x}) = \mathbf{x} - 2(\mathbf{x} \cdot \mathbf{u})\mathbf{u}
$$
and similarly, refl... | Write $c := \cos\theta$, $s := \sin\theta$, $\mathbf{w} := \mathbf{u}\times\mathbf{v} = s\mathbf{n}$, and $T := T_\mathbf{u}\left(T_\mathbf{v}\right)$, so that we have ...
$$\begin{align}
T(\mathbf{x}) &=\mathbf{x}-2(\mathbf{x}\cdot\mathbf{u})\mathbf{u}-2(\mathbf{x}\cdot\mathbf{v})\mathbf{v} + 4c(\mathbf{x}\cdot\mathbf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/704094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Integral $\int_0^\infty F(x)\,F\left(x\,\sqrt2\right)\frac{e^{-x^2}}{x^2} \, dx$ involving Dawson's function I need your help evaluating this integral:
$$I=\int_0^\infty F(x)\,F\left(x\,\sqrt2\right)\frac{e^{-x^2}}{x^2} \, dx,\tag1$$
where $F(x)$ represents Dawson's function/integral:
$$F(x)=e^{-x^2}\int_0^x e^{y^2} \,... | Notice that for $a>0$, we have $$F(ax) = e^{-a^{2}x^{2}}\int_{0}^{ax} e^{y^{2}} \mathrm dy = e^{-a^{2}x^{2}} \int_{0}^{a} u e^{x^{2}u^{2}} \, \mathrm du . \tag{1}$$
Then using $(1)$, we get
$$ \begin{align} I &= \int_{0}^{\infty} F(x) F(x \sqrt{2}) \, \frac{e^{-x^{2}}}{x^{2}} \, \mathrm dx \\&= \int_{0}^{\infty} \int_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/704917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 2,
"answer_id": 0
} |
How to show $\sum_{i=1}^{n} \binom{i}{2}=\binom{n+1}{3}$?
Show that $\,\displaystyle\sum_{i=1}^{n} \binom{i}{2}=\binom{n+1}{3}$.
I'm thinking right now (though not getting anywhere with it) that I want to expand out the summation portion to $i!/2!(i-2)!$ and simplify from there? Not sure if that will help, not to men... | $$\sum_{i=2}^n\binom{i}{2}=\binom{2}{2}+\binom{3}{2}+\binom{4}{2}+\binom{5}{2}+\cdots+\binom{n}{2}$$
Since $\binom22=\binom33$
$$\sum_{i=2}^n\binom{i}{2}=\binom{3}{3}+\binom{3}{2}+\binom{4}{2}+\binom{5}{2}+\cdots+\binom{n}{2}$$
Pascal's Triangle Identity states that
$$\binom{n-1}k+\binom{n-1}{k-1}=\binom{n}{k}$$
So,
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/707256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 4
} |
the probability that exactly 4 dice are sixes given that none are fives. My answer key tells me the answer is (10 Choose 4)(1/5)^4 (4/5)^6.
But isn't conditional probability of P(A|B) = P(A and B)/P(B)? Why isn't this answer divided by (5/6)^10? Any help with my intuition? Thanks!
Edit: Sorry, left out some information... | The most straight forward way to interpret that none are fives is simply consider the possible outcomes reduced. That is, you know each roll was one of $\{1,2,3,4,6\}$. So knowing this, the probability that exactly four were $6$ is given as
$$ {10 \choose 4} \left( \frac{1}{5} \right) ^4 \left( \frac{4}{5} \right) ^6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/709949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integral $\int_0^1 \left(\arctan x \right)^2\,dx$ Evaluate
$$\int_0^1 \left(\arctan x \right)^2\,dx$$
The answer should be
$${\pi^2\over16} + \frac{\pi\ln(2)}{4} -C$$
where $C$ is Catalan's constant.
How do I proceed?
I tried doing integration by parts twice and got stuck at $$\int_0^1{\frac{\log\left(\frac1{x}+x\r... | To get rid of the inverse, we let $y=\arctan x$ and transform the integral into
$$
\begin{aligned}
I &=\int_{0}^{\frac{\pi}{4}} y^{2} \sec ^{2} y d y \\
&=\int_{0}^{\frac{\pi}{4}} y^{2} d(\tan y) \\
& \stackrel{IBP}{=} \left[y^{2} \tan y\right]_{0}^{\frac{\pi}{4}}-2 \int_{0}^{\frac{\pi}{4}} y \tan y d y \\
&=\frac{\pi^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/713971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 8,
"answer_id": 6
} |
Solve for $x$ in the given determinant. Solve for $x$.
$$
\begin{vmatrix}
x^2-a^2&x^2-b^2&x^2-c^2\\
(x-a)^3&(x-b)^3&(x-c)^3\\
(x+a)^3&(x+b)^3&(x+c)^3\\
\end{vmatrix}=0.
$$
I could factorise each term, but could not find three factors in a single row or column. Atmost, there were $2$. I cannot think of any m... | I used brute force to solve this.
When we calculate:
$$
\begin{vmatrix}
x^2-a^2&x^2-b^2&x^2-c^2\\
(x-a)^3&(x-b)^3&(x-c)^3\\
(x+a)^3&(x+b)^3&(x+c)^3\\
\end{vmatrix} = 0
$$
We get:
$$-8 a^3 b^2 x^3+8 a^3 c^2 x^3+8 a^2 b^3 x^3+24 a^2 b x^5-8 a^2 c^3 x^3-24 a^2 c x^5-24 a b^2 x^5+24 a c^2 x^5-8 b^3 c^2 x^3+8 b^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/714180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$x$-coordinates of points where tangent line is horizontal or vertical. Using implicit differentiation For the implicit equation $x^3 + y^3 - xy^2 = 8$
Determine the exact x-coordinates of all points where the tangent line is horizontal or vertical
I figured out $\dfrac{dy}{dx} =\dfrac{y^2 - 3x^2}{ 3y^2 - 2xy}$
I kno... | As OP already found, the expression for the slope of a tangent line to a point on the curve in question is
$$ y \ ' \ = \ \frac{y^2 \ - \ 3x^2}{3y^2 \ - \ 2xy} \ = \ \frac{y^2 \ - \ 3x^2}{y \ (3y \ - \ 2x)} \ \ . $$
Something useful to check for immediately is whether this rational function can be "indeterminate", that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/717476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
evaluation of $\int \cos (2x)\cdot \ln \left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)dx$
Compute the indefinite integral
$$
\int \cos (2x)\cdot \ln \left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)\,dx
$$
My Attempt:
First, convert
$$
\frac{\cos x+\sin x}{\cos x-\sin x} = \frac{1+\tan x}{1-\tan x} = \tan \left(\fr... | Let
\begin{equation*}
I=\int \cos 2x\cdot\ln \left|\frac{\cos x+\sin x}{\cos x-\sin x}\right|\,dx.
\end{equation*}
Using the following identity
\begin{equation*}
\cos 2x=2\cos ^{2}x-1
\end{equation*}
and the substitution
\begin{equation*}
u=\cos x,
\end{equation*}
we get
\begin{equation*}
I=\int \frac{1-2u^{2}}{\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/718719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 6,
"answer_id": 5
} |
Inequality $\sum_{cyc}\frac{(a+b)^{3}}{\sqrt[3]{2(a+b)(a^{2}+b^{2})}}\ge{12}$ when $ab+bc+ca=6$ Let $a$, $b$ and $c$ be non-negative reals such that $ab+bc+ca=6$.
Prove that:
$$\frac{(a+b)^{3}}{\sqrt[3]{2(a+b)(a^{2}+b^{2})}}+\frac{(b+c)^{3}}{\sqrt[3]{2(b+c)(b^{2}+c^{2})}}+\frac{(c+a)^{3}}{\sqrt[3]{2(c+a)(c^{2}+a^{2})}}... | A simple and an elegant solution:
first,
$$\sqrt{\frac{a^2+b^2}{2}} \leq \frac{a+b}{2}$$ or $$a^2+b^2 \leq 2 \cdot \frac{(a+b)^2}{4}=\frac{(a+b)^2}{2}$$
So $$2(a+b)(a^2+b^2) \leq 2(a+b)\frac{(a+b)^2}{2}=(a+b)^3$$
now: $$\frac{1}{\sqrt[3]{2(a+b)(a^2+b^2)}} \geq \frac{1}{\sqrt[3]{(a+b)^3}}=\frac{1}{a+b}$$
Our inequality... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/719523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Limit question involving L'Hospitals rule I need help in solving the below question using L'Hospitals rule:
$$ \lim_{x \rightarrow 0} \left( \frac{1}{x^2}-\cot^2 x \right) $$
I'm getting infinity as the final answer. Thanks in advance.
| $$
\lim_{x \rightarrow 0} \left( \frac{1}{x^2}-\cot^2 x \right) \\
= \lim_{x \rightarrow 0} \left( \frac{1}{x^2}-\frac{\cos^2 x}{\sin^2 x} \right) \\
= \lim_{x \rightarrow 0} \left( \frac{\sin^2 x}{x^2 \sin^2 x}-\frac{x^2 \cos^2 x}{x^2 \sin^2 x} \right) \\
= \lim_{x \rightarrow 0} \left( \frac{\sin^2 x - x^2 \cos x}{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/721742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluate the integral. $\int sin^3 (5x) dx$ The bounds are from $(\pi/5)$ to $0$. I know we use a pythagorean identity for this. So $u=\sin^2 (x)$ and $du = \sin (2x) dx$. But I'm helping trouble solving this problem.
| \begin{align*}
\int \sin^3(5x) \, dx &= \int \sin^2(5x) \cdot \sin(5x) \, dx \\
&= \int \left( 1-\cos^2(5x) \right)\sin(5x) \, dx \\
&= \int \left( \sin(5x)-\cos^2(5x)\sin(5x) \right) \, dx \\
&= -\frac{\cos(5x)}{5}-\int \cos^2(5x)\sin(5x) \, dx \\
\end{align*}
For the second integral, let $u=\cos(5x)$.
\begin{align*}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/721797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solving the functional equation $f\left(x^2+f(y)\right)=f(x)^2+y^4+2f(xy)$ Problem: find all functions $f:\mathbb{R}\to\mathbb{R}$ such that
$$f\left(x^2+f(y)\right)=f(x)^2+y^4+2f(xy),\ \ \ \forall x,y\in\mathbb{R}\text.$$
| It can be shown that the only function $ f : \mathbb R \to \mathbb R $ satisfying
$$ f \left( x ^ 2 + f ( y ) \right) = f ( x ) ^ 2 + y ^ 4 + 2 f ( x y ) \tag 0 \label 0 $$
for all $ x , y \in \mathbb R $ is $ f ( x ) = x ^ 2 $. It's straightforward to check that this is indeed a solution. We prove that it's the only o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/722789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
standard Taylor series using substitution Find Taylor series using substitution about $0$ for $f(x)=\frac{125}{(5+4x)^3}$ by writing $\frac{125}{(5+4x)^3}=\frac{1}{(1+\frac{4}{5}x)^3}$?
Determine a range of validity for this series.
| As Claude did,
$$\frac{1}{1+y}=1-y+y^2-y^3+y^4-y^5+...+(-1)^ny^n...$$ and take second derivative of this expension to get,
$$ \frac{2}{(1+y)^3}=(\frac{1}{1+y})^{''}=2-6y+12y^2-20y^3...n(n-1)(-1)^ny^{n-2}...$$
$$\frac{1}{(1+y)^3}=1-3y+6y^2-10y^3...{n\choose2}(-1)^{n+1}y^{n-2}... $$
Then replace $y$ by $4x\over5$ then y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/723199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Matrix Exponential using the Cayley-Hamilton theorem For the matrix $$P=\left( \begin{matrix} 0&1&0 \\ 0&0&1 \\ 0&1&0 \end{matrix} \right)$$ how do you find $e^{Pt}$ using the Cayley-Hamilton theorem?
I have found it by diagonalising $P$, but the question states to use the Cayley-Hamilton theorem.
| If you calculate the characteristic polynomial $\chi_A(x)=x^3-x$, you know from Cayley-Hamilton theorem that $P^3-P=0$, i.e. $P^3=P$.
This implies that $P=P^3=P^5=\dots$ and $P^2=P^4=P^6=\dots$. Hence
$$
\begin{align}
e^{Pt}
&= I+Pt + \frac{P^2t^2}2 + \frac{P^3t^3}{3!} + \dots = \\
&= I + P\left(t+\frac{t^3}{3!}+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/724647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to prove that the angle between two sides of that triangle is less than 60 degree? The product of two sides of triangle is equal to 8*(R*r) where R is circumradius of this triangle, and r is inradius of this triangle.
How to prove that the angle between two sides of that triangle is less than 60 degree?
| Suppose $ab = 8Rr$. We have $abc = 4RA$ where $A$ is the area of the triangle, hence $2rc = A$. The area of the triangle is also given by $rs$ where $2s=a+b+c$. It follows that $2rc = rs$ thus $4c=a+b+c$ or $3c=a+b$. It follows that $$ \cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{9a^2+9b^2-(a+b)^2}{18ab} = \frac{8a^2+8b^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/726512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Volume within the sphere Find the volume of the solid that lies within the sphere ,
above the xy plane, and outside the cone
My problem is finding the integral function and the limits
| First, here is a sketch showing that the cone cuts out washers at each value of $z$. The inner radius is given by the cone and the outer radius by the sphere. In cylindrical coordinates, the cone is given by $z = 7r$ and the sphere by $r^2 + z^2 = 1 \rightarrow r^2 = 1 - z^2$. The $z$ coordinate goes from $z = 0$ to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/728583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How find this value $\frac{S_{\Delta ABC}}{\overrightarrow{OA}\cdot\overrightarrow{OB}}$
In the plane have $4$ point $O,A,B,C$ such
$$\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=\overrightarrow{0},2|\overrightarrow{OA}|=2|\overrightarrow{OB}|+\overrightarrow{OC}$$,and let
$$\theta=<\overrightarrow{... | The answer as given by
orion
is repeated here for convenience.
This answer is certainly incomplete, but let it be assumed that it is correct so far.
$$
c^2+(a-b)^2-2c(a-b)\cos\theta=a^2 \\
c^2+(a-b)^2+2c(a-b)\cos\theta=b^2 \\
(2c)^2=a^2+b^2-2ab\cos\alpha
$$
Quote: "but $a,b,c$ can be scaled without affecting the outcom... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/728740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find$f$ s.t. $f(1)=2$, $f(2)= 4$, $f(3)= 6$ and $f(4)= \pi$. Find a function where $f(1)=2$, $f(2)= 4$, $f(3)= 6$ and $f(4)= \pi$.
I got $\dfrac16(x-3)(x-2)(x-1)\pi$ as a start to get rid of $\pi$.
| Your way of dealing with $\pi$ is very smart, but apparently you only took it because $\pi$ is irrational, and you are expecting to get the "nicer" values of $2,4,6$ in different ways. Here's something you can do, use seperate terms that are zero at $1,2,\pi$ and $6$ at $3$. Try it for yourself. Add all the terms. You ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/731601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
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.