Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Show that $64xy=-1$
Let $a=\frac{\pi}{13}$. If $$\begin{aligned} x &= \cos(2a)\cos(5a)\cos(6a)\\ y &= \cos(a)\cos(3a)\cos(9a)\end{aligned}$$ then show that $$64xy = -1$$
I'm trying to use
$$\cos(A)\cos(B)=\frac{\cos(A+B)+\cos(A-B)}{2}$$
then use summation of $\cos(2n+1)$. I need help, please. Thank you.
| As $\cos(\pi-A)=-\cos A$
$\displaystyle xy=(-1)^2\prod_{r=0}^{(13-1)/2-1}\cos(2r+1)a$
We can prove something more generic $$\prod_{r=0}^{n-1}\cos\dfrac{(2r+1)\pi}{2n+1}=\dfrac{(-1)^p}{2^n}\text{ where }p=\left\lfloor\dfrac n2\right\rfloor$$
Proof: Using this , $$\cos(2n+1)x=2^{2n}\cos^{2n+1}x-\cdots+(-1)^n(2n+1)\cos x$... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given three non-negatve numbers $a,b,c$. Prove that $1+a^{2}+b^{2}+c^{2}+4abc\geqq a+b+c+ab+bc+ca$ .
Given three non-negatve numbers $a, b, c$. Prove that:
$$1+ a^{2}+ b^{2}+ c^{2}+ 4abc\geqq a+ b+ c+ ab+ bc+ ca$$
Let $t= a+ b+ c$, we have to prove
$$\left(\!\frac{1}{t^{3}}- \frac{1}{t^{2}}+ \frac{1}{t}\!\right)\sum ... | Partial answer:
Let $a = \frac{1}{2} + a'$, $b = \frac{1}{2} + b'$ and $c = \frac{1}{2} + c'$. Then, from Empy2's hint we have:
$$8a'b'c' + (a'+b')^2 + (b'+c')^2 + (c' + a')^2 ≥ 0$$
which is true if $a', b', c' ≥ 0$. This leaves only the cases when any and only one or all three of $a, b, c$ satisfy $0 < a,b,c < \frac{1... | {
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"timestamp": "2023-03-29T00:00:00",
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Why do all primes $n>3$ satisfy $\,309\mid 20^n-13^n-7^n$ Solve the following... $309|(20^n-13^n-7^n)$ in $\mathbb{Z}^+$.
I invested lotof time to it and finally went to WolframAlpha for help by typing...
Solve $309k=20^n-13^n-7^n$ over the integers. It returned the following... Note that all the primes are generated... | $20^6 \equiv 13^6 \equiv 7^6 \equiv 229 \bmod 309$, so $20^n \equiv 13^n + 7^n \mod 309$ if and only if $20^{n+6} \equiv 13^{n+6} + 7^{n+6} \bmod 309$. Since it does work for $n=1$ and $n=5$, but not for $0$, $2$, $3$ or $4$, we find that $20^n \equiv 13^n + 7^n \bmod 309$ if and only if $n \equiv 1$ or $5 \bmod 6$. ... | {
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How to compute a Jacobian using polar coordinates? Consider the transformation $F$ of $\mathbb R^2\setminus\{(0,0)\}$ onto itself defined as
$$
F(x, y):=\left( \frac{x}{x^2+y^2}, \frac{y}{x^2+y^2}\right).$$
Its Jacobian matrix is
$$\tag{1}
\begin{bmatrix} \frac{y^2-x^2}{(x^2+y^2)^2} & -\frac{2xy}{(x^2+y^2)^2} \\ -\fra... | I don't think there is any contradiction here.
Consider the volume form
$$ \omega_{\rm Cart} = dx \wedge dy.$$
Your first calculation shows that the pullback $F^\star(\omega_{\rm Cart})$ is given by
$$ F^\star(\omega_{\rm Cart}) = - \frac{1}{(x^2+y^2)^2}\omega_{\rm Cart}.$$
Now consider the volume form
$$ \omega_{\rm P... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "7",
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Number of Collatz steps for Mersenne numbers I noticed that for all $k \in \mathbb{N} \geq 1$ the following is true (I tested up to $2^{5000}$):
$\text{Collatz_Steps}(2^{2k+1} - 1) + 1 = \text{Collatz_Steps}(2^{2k+2} - 1)$
Where $\text{Collatz_Steps}$ is the number of steps required to reach $1$.
Is there a proof for... | Proof sketch:
For a number $2^n-1,$ the first $2n$ steps are the $3x+1$ step and the $x/2$ step in alternating order, until you reach $3^n-1.$
$3^n-1$ is even, so the next step is the $x/2$ step again. We have
$$
\frac{3^n-1}{2} = \sum_{j=0}^{n-1} 3^j
$$
which is odd if $n$ is odd and even if $n$ is even. So for odd $n... | {
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If $f(x)$ Polynomial with real coefficient and $f(0)=1$, $f(2)+f(3)=125$ and $f(x)*f(2x^2)=f(2x^3+x)$ then what is the value of $f(5)$? If $f(x)$ Polynomial with real coefficient and $f(0)=1$, $f(2)+f(3)=125$
and
$f(x)*f(2x^2)=f(2x^3+x)$
then what is the value of $f(5)$?
.
What I tried
$$f(0)=1$$
put x=1 $$f(1)*f(2)=f(... | Here's a somewhat unfulfilling solution.
According to Solve $f(x)f(2x^2) = f(2x^3+x)$, the polynomial $f(x)$ is of the form $f(x)=(x^2+1)^n$ for some $n$.
Using $f(2)+f(3)=125$ we obtain $n=2$, so $f(x)=(x^2+1)^2$. Plugging in $x=5$ we conclude
$$f(5)=(5^2+1)^2=26^2=676.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve: $ z^5 = \bar{z} * (- \frac{1}{2} + \frac{\sqrt3}{2} i)$. P.S. if $z = x+y*i$, $\bar{z} = x - y*i$. Solve: $ z^5 = \bar{z} \left(- \frac{1}{2} + \frac{\sqrt3}{2}i\right)$.
P.S. if $z = x+y i$, $\bar{z} = x - y i$.
Seems trivial but I can not solve it. I tried to write $z, \bar{z}$ using $x$ and $y$ and I got noth... | You can use polar coordinates. So let $r$ and $\theta$ be such that $z=re^{i\theta}$. Then $\bar{z}=re^{-i\theta}$. Furthermore,
$$-\frac{1}{2}+\frac{\sqrt{3}}{2}i=e^{\frac{2\pi}{3}i}.$$
Thus, we get
$$r^5e^{5i\theta}=re^{-i\theta}\cdot e^{\frac{2\pi}{3}i}\Leftrightarrow r^4e^{6i\theta}=e^{\frac{2\pi}{3}},$$
so that $... | {
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Minimum value of $4$-digit number divided by sum of its digits? If x was a positive 4 digit number and you divide it by the sum of its digits to get the smallest value possible, what is the value of x? For example (1234 = 1234/10)
I got 1099 as my answer however I don't know if this is right or how to prove it.
| Write the $4$-digit number as $x = 1000a+100b+10c+d$, where $a,b,c,d$ are integers from $0$ to $9$ (and $a \neq 0$.)
You want to minimize $A = \dfrac{1000a+100b+10c+d}{a+b+c+d}$, which can be written as
$$
\begin{align*}
A &= \dfrac{999a+99b+9c}{a+b+c+d}+1\\
&\geq \dfrac{999a+99b+9c}{a+b+c+9} + 1 & \text{(since the num... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "9",
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Square root of square as 2/2 power From school I know that
$\sqrt{x^2} = |x|$.
But if we rewrite the above equation in another way
$\sqrt{x^2} = (x^2)^{\frac{1}{2}} = x^{\frac{2}{2}} = x^1 = x$
then we get another answer.
How is it possible that rewritting an equation changes the answer? And which answer is right, $x$... | Let x be a positive number.
So the square of x is $x^2$
Also the square of -x is $(-x)^2 = (-1)^2 x^2 = x^2$.
So we have, $x^2 = (\pm x)^2$
Hence, $\sqrt{x^2} = \sqrt{(\pm x)^2 } = |x|$. As x is positive, |x| = x and as -x is negative |-x| = -(-x) = x
This is simply x because we know in prior that, x is positive.
Now ... | {
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Value of $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}+\frac{1}{\sqrt{1+c^2}}+\frac{1}{\sqrt{1+d^2}}$
If $$\frac{a+b+c+d}{\sqrt{(1+a^2)(1+b^2)(1+c^2)(1+d^2)}}= \frac{3\sqrt{3}}{4}$$ for $a,b,c,d>0$
Then
Value of $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}+\frac{1}{\sqrt{1+c^2}}+\frac{1}{\sqrt{1+d^2}}$ is
Try: us... | Another way for the proof of the inequality $(x^2+3)(y^2+3)(z^2+3)(t^2+3)\geq16(x+y+z+t)^2.$
We obtain: $$(x^2+3)(y^2+3)=x^2y^2+3(x^2+y^2)+9=$$
$$=(xy-1)^2+(x-y)^2+2(x+y)^2+8\geq2((x+y)^2+4).$$
By the same way $$(z^2+3)(t^2+3)\geq2((z+t)^2+4).$$
Id est, by C-S $$\prod_{cyc}(x^2+3)\geq4((x+y)^2+4)(4+(y+z)^2)\geq$$
$$\ge... | {
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"timestamp": "2023-03-29T00:00:00",
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Probability distribution of a uniform distribution to the third power I have to find a explicit form of probability distribution of $X^3$, if $X \ \mathtt{\sim} \ U[a, b], \ -\infty < a < b < \infty$.
So far I've succesfully done a simpler version, when it's $\ U[0, a], \ \ 0 < a < \infty$:
$F_{X^3}(x) = \mathbb{P}(X^3... | Tip: Don't use $x$ both in the domain's bounds and as the integration's variable. That inevitably leads to confusion. Additionally, we instinctively associate lower and upper case letters as referring to the same thing, so that's the best one to change.
Now if the support for $X$ is $a\leq X\leq b$, then the suppor... | {
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What is all N that make $2^N + 1$ divisible by 3? When $N = 1$, $2^N + 1 = 3$ which is divisible by $3$.
When $N = 7$, $2^N + 1 = 129$ which is also divisible by $3$.
But when $N = 2$, $2^N + 1 = 5$ which is not divisible by $3$.
A quick lookout can determine that when $N$ is an odd number, $2^N + 1$ is divisible by $3... | If $2^N+1$is divisible by $3$ then in other words $2^N+1$ is a multiple of $3$ .
$2^N+1=3k \implies 2^N=3k-1$ so we want $3k-1$ to be even then we want $3k$ to be odd.
$3k$ is odd only when $k$ is odd.
| {
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"url": "https://math.stackexchange.com/questions/3211860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the limit without using Lhopital this question came out on my analysis exam: Evaluate
$$\lim_{x\to 0}\left(\frac{5^{x^2}+7^{x^2}}{5^x+7^x}\right)^{\frac{1}{x}} $$
I did it using L'hopital rule but is there another way to do this?
| The major part can be handled using squeezing and AM-GM but I need one derivative to calculate the limit
*
*$\color{blue}{(\star)}: \lim_{x\to 0}\left( \frac{5^x + 7^x}{2}\right)^{\frac{1}{x}} = \sqrt{35}$
This is quickly verified when taking the logarithm:
*
*$\frac{\log (5^x + 7^x) - \log 2}{x} \stackrel{x\to ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solve $4+\frac{1}{x}-\frac{1}{x^2}$ using quadratic formula I am to solve for x using the quadratic formula:
$$4+\frac{1}{x}-\frac{1}{x^2}=0$$
The solution provided in the answers section is: $\dfrac{-1\pm\sqrt{17}}{8}$ whereas I arrived at something entirely different: $$\dfrac{\frac{1}{x}\pm\sqrt{\frac{1}{x^2}+\fr... | From
$$4+\left(\frac1x\right)-\left(\frac1x\right)^2=0$$
using the standard formula blindly,
$$\frac1x=\frac{1\pm\sqrt{17}}2$$
and obviously
$$x=\frac2{1\pm\sqrt{17}}.$$
Though this seems to contradict the expected answer, consider
$$\frac2{1\pm\sqrt{17}}=\frac{2(1\mp\sqrt{17})}{1-(\sqrt{17})^2}=\frac{-1\pm\sqrt{17}}8... | {
"language": "en",
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Remove the discontinuity of $f(x) = \frac{x^2-11x+28}{x-7}$ at $f(7)$? I need to remove the discontinuity of $f(x) = \frac{x^2-11x+28}{x-7}$ at $f(7)$, and find out what $f(7)$ equals. I am not sure what I've done wrong, but I'm getting 33, which the website I'm using tells me is wrong.
Please help me figure out what I... | To explain why you get the wrong answer, in your second last line, the last term of the numerator is $-49(x-4)$ and not $-49(x+4).$
Faster approach:
$$\frac{x^2-11x+28}{x-7}=\frac{(x-7)(x-4)}{x-7}=x-4$$
Hence to remove the discontinuity at $7$, the right value to redefine at $f(7)$ should be $7-4$.
Factorization should... | {
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Can the formula $\frac\pi2=(\frac21)^{1/2}(\frac{2^2}{1\cdot3})^{1/4}(\frac{2^3\cdot4}{1\cdot3^3})^{1/8}\cdots$ prove the irrationality of $\pi$? A less known product formula for $\pi$, due to Sondow, is the following:
$$
\frac{\pi}{2}= \left(\frac{2}{1}\right)^{1/2} \left(\frac{2^2}{1\cdot3}\right)^{1/4} \left(\frac{2... | Dirichlet’s approximation theorem can’t be applied here, as this product formula doesn’t give a sequence of rational approximations due to the fractional exponents.
We can show that this product formula is equivalent to the Wallis product, which does give you a sequence of rational approximations:
Let $a_n(x)$ be the p... | {
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Number of possible integer values of $x$ for which $\frac{x^3+2x^2+9}{x^2+4x+5}$ is integer How many integer numbers, $x$, verify that the following
\begin{equation*}
\frac{x^3+2x^2+9}{x^2+4x+5}
\end{equation*}
is an integer?
I managed to do:
\begin{equation*}
\frac{x^3+2x^2+9}{x^2+4x+5} = x-2 + \frac{3x+19}{x^2+4x+5}
... | Let $t=x+2$, then $$t^2+1\mid 3t+13$$ and thus $$t^2+1\mid t(3t+13)-3(t^2+1)= 13t-3$$
so $$t^2+1\mid 13(3t+13)-3(13t-3) = 178$$
So $$t^2+1\in\{1,2,89,178\}\implies t=\pm 1,0 \implies x\in\{-1,-2,-3\}$$
| {
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Evaluate $\lim\limits_{x \to 0}\frac{e^{(1+x)^{\frac{1}{x}}}-(1+x)^{\frac{e}{x}}}{x^2}$ Solution
Expanding $(1+x)^{\frac{1}{x}}$ at $x=0$ by Taylor's Formula，we obtain
\begin{align*}
(1+x)^{\frac{1}{x}}&=\exp\left[\frac{\ln(1+x)}{x}\right]=\exp \left(\frac{x-\frac{x^2}{2}+\frac{x^3}{3}+\cdots }{x}\right)\\
&=\exp \left... | Since both terms, say $A, B$, in numerator tend to same limit $e^e$ we can write $$A-B=B\cdot\frac{\exp (\log A - \log B) - 1}{\log A - \log B} \cdot(\log A - \log B) $$ and thus the numerator can be replaced by $e^e(\log A - \log B) $ or $$e^e\left((1+x)^{1/x}-e\cdot\frac{\log (1+x)}{x}\right)$$ Applying the same tech... | {
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"timestamp": "2023-03-29T00:00:00",
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The odds in rolling two dice together. Two dice are rolled. I want to see the odds of the following:
$1.$ A sum of $5$.
$2.$ A sum of $8$ or $10$.
$3.$ A sum less than $6$.
$4.$ Not a sum of $7$.
Solution: $1.$ A sum of $5$. $5 = 1+4,2+3,3+2,4+1$. So the odds is $4/36 = 1/9.$
$2.$ A sum of $8$ or $10$. We can express ... | Yes, it states all of the ways the event can happen, then all of the possible events, yielding correct probability.
| {
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How many are the non-negative integer solutions of $ + + + = 16$ where $x < y$?
How many are the non-negative integer solutions of $ + + + = 16$ where $x < y$?
Anyone can explain how to think to approach this type of problem?
The answer is 444.
| Consider cases and find a pattern.
When $x=0$:
$$0+y+z+w=16, y\ge 1, z,w\ge 0 \Rightarrow \\
t+1+z+w=16,t,z,w\ge 0 \Rightarrow \\
t+z+w=15, t,z,w\ge 0 \Rightarrow \\
{15+3-1\choose 3-1}$$
Note: It was used Stars and Bars.
When $x=1$:
$$1+y+z+w=16, y\ge 2, z,w\ge 0 \Rightarrow \\
t+2+z+w=15,t,z,w\ge 0 \Rightarrow \\
t... | {
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Different methods give different answers. Let A,B,C be three angles such that $ A=\frac{\pi}{4} $ and $ \tan B \tan C=p $. Let $A,B,C$ be three angles such that $ A=\frac{\pi}{4} $ and $ \tan B \tan C=p $. Find all possible values of $p$ such that $A,B$ and $C$ are angles of a triangle.
case 1- discriminant
We can re... | I think that every approach should give correct solution. Different solutions appear if to do mistakes.
Conditions to $\angle B$ and $\angle C$ are WLOG
\begin{cases}
\angle B+\angle C = \dfrac{3\pi}4\\[4pt]
\tan\angle B\tan\angle C = p\\[4pt]
\angle B\in\left(0,\dfrac{3\pi}8\right]\\
\angle C\in\left[\dfrac{3\pi}8,\... | {
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"timestamp": "2023-03-29T00:00:00",
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The integral solution of $x^{2}-y^{3}=1 (x>1,y>1) $? I know it's a special case of catalan's conjecture.Wiki says its only solution is $(3,2)$.But I cannot work it out.Any help will be appreciated.
| Not finished!
Write $$(x-1)(x+1)=y^3$$
Case 1. $x$ is even then $x+1$ and $x-1$ are relatively prime so $$ x-1 = a^3$$ $$x+1 = b^3$$ where $a<b$ are relatively prime and $ab=y$. Now we have $$2=(b-a)(b^2+ab+a^2)$$
and so:
*
*$b-a=1$ and $b^2+ab+a^2 =2$ so $$ 3a^2+3a+1 =2$$ and this is impossibile.
*$b-a=2$ and $b... | {
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Evaluate $(\sum_{k=1}^{7} \tan^2(\frac{k\pi}{16})) - \tan^2(\frac{4\pi}{16})$ The question is:
Evaluate
$$\displaystyle \left(\sum_{k=1}^{7} \tan^2\left(\frac{k\pi}{16}\right)\right) - \left(\tan^2\frac{4\pi}{16}\right)$$
The given answer:$34$
What I've tried:
$$\displaystyle \left(\sum_{k=1}^{7} \tan^2\left(\f... | Using $\displaystyle \tan^2(x)+\cot^2(x)=\frac{\sin^4(x)+\cos^4(x)}{\sin^2(x)\cdot \cos^2(x)}=\frac{1-2\sin^2 x\cos^2 x}{\sin^2 x\cos^2 x}.$
So we have $\displaystyle =4\csc^2(2x)-2=2+4\cot^2(2x).$
So our expression is
$\displaystyle 2+4\cot^2(\pi/8)+2+4\cot^2(\pi/4)+2+4\cot^2(3\pi/8)$
$\displaystyle 6+4+4\bigg[\cot^2(... | {
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"answer_id": 0
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Prove that $\frac{1}{AM \cdot BN} + \frac{1}{BN \cdot CP} + \frac{1}{CP \cdot AM} \le \frac{4}{3(R - OI)^2}$.
$(O, R)$ is the circumscribed circle of $\triangle ABC$. $I \in \triangle ABC$. $AI$, $BI$ and $CI$ intersects $AB$, $BC$ and $CA$ respectively at $M$, $N$ and $P$. Prove that $$\large \frac{1}{AM \cdot BN} +... | As pointed out in the comments the inequality:
$$\large \frac{1}{AM \cdot BN} + \frac{1}{BN \cdot CP} + \frac{1}{CP \cdot AM} \le \frac{4}{3(R - OI)}$$
Is not homogeneous and therefore cannot be correct. Take any triangle and any point and even if the given inequality is satisfied fot this configuration then after scal... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3227215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Proving that $\sum_{k=0}^{2n} {2k \choose k } {2n \choose k}\left( \frac{-1}{2} \right)^k=4^{-n}~{2n \choose n}.$ I have happened to have proved this sum while attempting to prove another summation. Let
$$S_n=\sum_{k=0}^{2n} {2k \choose k } {2n \choose k}\left( \frac{-1}{2} \right)^k$$
${2k \choose k} $ is the coeff... | We will use
$$
\frac1{1-x}\left(\frac{x}{1-x}\right)^k=\sum_{n=0}^\infty\binom{n}{k}x^n\tag1
$$
and
$$
(1-4x)^{-1/2}=\sum_{k=0}^\infty\binom{2k}{k}x^k\tag2
$$
Extracting the even part of $(1)$
$$
\begin{align}
\sum_{n=0}^\infty\binom{2n}{k}x^{2n}
=\frac12\left[\frac1{1-x}\left(\frac{x}{1-x}\right)^k+\frac1{1+x}\left(-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3229256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
} |
Prove $(1 + \frac{3p-3}{p^2-1}) \prod_{\substack{q=3\\q\ \text{prime}}}^{l(p)}(1 + \frac{q+1}{q-1} \frac{1}{p-1})$ goes to 1 $\lim p\to\infty$ Let function $l(p)$ be defined as the largest prime number less than $p$.
For example: $l(7)=5, l(11)=7, l(17)=13$.
Let the function $f(p)$ be defined as follows:
\begin{eqnarra... | Let $P(n)$ be the set of primes less than $n.$
We have $\lim_{n\to \infty}1+\frac {3n-3}{n^2-1}=1.$ So it suffices to prove that $\lim_{n\to \infty}g(n)=1$ where $$g(n)=\prod_{3\le q\in P(n)}\left(1+\frac {q+1}{q-1}\frac {1}{n-1}\right).$$
Let (as usual) $\pi(n)$ be the number of primes less than $n.$ By elementary mea... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3229974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Prove $\frac{b-a}{6}≤\sqrt{1+b} - \sqrt{1+a}≤\frac{b-a}{4}$ if $3\leq aI don't really know if I should use brute force or some kind of theorem, it comes on a calculus past exam and it says:
suppose: $3≤a<b≤8$
prove that $$\frac{b-a}{6}≤\sqrt{1+b} - \sqrt{1+a}≤\frac{b-a}{4}$$
| As Martin R suggested, multiply all sides by conjugate of the middle term (note: $3≤a<b≤8$):
$$\frac{b-a}{6}≤\sqrt{1+b} - \sqrt{1+a}≤\frac{b-a}{4} \iff \\
\frac{\overbrace{\require{cancel}\cancel{b-a}}^{1}}{6}(\sqrt{1+b} + \sqrt{1+a})≤\overbrace{\cancel{b-a}}^{1}≤\frac{\overbrace{\cancel{b-a}}^{1}}{4}(\sqrt{1+b} + \sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3233155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Using complex numbers to find $\cos 5\theta$ I have seen a few examples of finding $\cos 5\theta$ before on here but I was wondering how to just find it by equating the real parts?
So far I have: $$\cos5\theta=\Re(cos\theta+i\sin\theta)^5$$
But how do I know which is the real part and expand from this further? Any hint... | So here we will use De Moivre's theorem here:
As you stated we will use the fact that $\cos(5\theta)+i\sin(5\theta)=(\cos(\theta)+i\sin(\theta))^5$, so get this just in terms of $\cos(5\theta)$ we can write $\cos(5\theta)=\Re(\cos(\theta)+i\sin(\theta))^5$
So expanding this we have:
$$\cos(5\theta)=\Re(\cos^5\theta+5i\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3236172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Showing that generating function coefficients agree with recurrence relation I have the recurrence relation $a_n = 3a_{n-1} + 4a_{n-2}$ with $a_0 = 1, a_1 = 0$. The solution to this is $a_n = \frac{4^n}{5} + \frac{4(-1)^n}{5}$. The generating function is $g(x) = \frac{1-3x}{1-3x-4x^2} = (1-3x)(\sum_{i=0}^{\infty}(-1)^i... | Take your generating function, write it as partial fractions:
$\begin{equation*}
g(z)
= \frac{4}{5 (1 + z)} + \frac{1}{5 (1 - 4 z)}
\end{equation*}$
Thus (two geometric series) you have directly:
$\begin{equation*}
[z^n] g(z)
= \frac{4}{5} \cdot (-1)^n + \frac{1}{5} \cdot 4^n
\end{equation*}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3239360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Solve for $x, y \in \mathbb R$: $x^2+y^2=2x^2y^2$ and $(x+y)(1+xy)=4x^2y^2$
Solve the following system of equations. $$\large \left\{ \begin{aligned} x^2 + y^2 &= 2x^2y^2\\ (x + y)(1 + xy) &= 4x^2y^2 \end{aligned} \right.$$
From the system of equations, we have that
$$\left\{ \begin{align*} (x + y)^2 \le 2(x^2 + y^2)... | \begin{align*}
(x+y)^2(1+xy)^2&=(4x^2y^2)^2\\
(x^2+y^2+2xy)(1+xy)^2&=16x^4y^4\\
(2x^2y^2+2xy)(1+xy)^2&=16x^4y^4\\
2xy(1+xy)^3&=16x^4y^4
\end{align*}
So, $xy=0$ or $1+xy=2xy$.
$xy=0$ or $1$.
If $xy=0$, $x=y=0$.
If $xy=1$, $x=y=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3239963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
prove $\ln(1+x^2)\arctan x=-2\sum_{n=1}^\infty \frac{(-1)^n H_{2n}}{2n+1}x^{2n+1}$ I was able to prove the above identity using 1) Cauchy Product of Power series and 2) integration but the point of posting it here is to use it as a reference in our solutions.
other approaches would be appreciated.
| actually, you can do the product directly, given well-known series
$$\begin{aligned}
\arctan x & = \sum_{n=0}^{\infty} {\frac{(-1)^n x^{2n+1}}{2n+1}}\\
\ln(1+x^2) & = \sum_{n=1}^{\infty} {\frac{(-1)^{n+1} x^{2n}}{n}}
\end{aligned}$$
obviously their product has no even order items, set
$$\arctan x \ln (1+x^2) = \sum_{m=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3240779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
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$(X^2-1)^2+(X^2-1)-12=0$ has 2 solutions only or 3? I am asked to solve for X using substitution:
$(X^2-1)^2+(X^2-1)-12=0$
Let $U = X^2-1$
$U^2+U-12=0$
$U^2+4U-3U-12=0$
$U(U+4)-3(U+4)=0$
$(U+4)(U-3)$
Then:
$U-3=0$
$U=3$
$X^2-1=3$
$X^2=4$
$X=\pm2$
This is the provided solution by my textbook.
However:
$U+4=0$
$U=-4$
$X^... | Probably your textbook is asking for only real solutions and $\pm2$ are only real solutions while $i\sqrt{3}$ is an imaginary solution
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3240924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Determine the stability of an equilibrium point for a given Lagrangian
I have to determine the equilibria and the stability for a Lagrangian given by
\begin{align}
\frac{\dot{x}^2+\dot{y}^2}{2}-V(x,y)
\end{align}
where $V(x,y)=-(x^2+y^2)^4+(2x^2+y^2)^2-1$.
The origin $(0,0)$ is an equilibria, but since the Hessian o... | We have that
$$(2x^2+y^2)^2=x^4+2x^2(x^2+y^2)+(x^2+y^2)^2$$
And if $0<q<1$, we have that $q>q^2$, so we have that (for $0<x^2+y^2<1$)
$$(x^2+y^2)^2 > (x^2+y^2)^4$$
Which means that
$$x^4+2x^2(x^2+y^2)+(x^2+y^2)^2>x^4+2x^2(x^2+y^2)+(x^2+y^2)^2 +(x^2+y^2)^4$$
So
$$(2x^2+y^2)^2-(x^2+y^2)^4 > x^4+2x^2(x^2+y^2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3242797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Use real integral to calculate $\int_R \frac{x^2 \cos (\pi x)}{(x^2 + 1)(x^2 + 2)}dx$ Problem :
Evaluate $$\int_{-\infty}^{\infty} \frac{x^2 \cos (\pi x)}{(x^2 + 1)(x^2 + 2)}dx$$
Use only real integral.
What I did :
$$\int_{-\infty}^{\infty} \frac{x^2 \cos (\pi x)}{(x^2 + 1)(x^2 + 2)}dx$$
$$=2\int_{0}^{\infty} \frac{... | We can consider a more general function
$$I(t)=2\int_0^\infty\frac{x^2\cos(tx)}{(x^2+1)(x^2+2)}\,\mathrm dx$$
and take its Laplace transform,
$$\begin{align*}
\mathcal{L}\left\{I(t)\right\} &= 2s\int_0^\infty\frac{x^2}{(x^2+s^2)(x^2+1)(x^2+2)}\,\mathrm dx\\
&= 2s\int_0^\infty\left( \frac{1}{(1-s^2)(x^2+1)}+\frac{2}{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3243131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
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Theorem of Pythagoras - Incorrect Derivation I am trying to solve below question from Coursera Intro to Calculus (link)
A right-angled triangle has shorter side lengths exactly $a^2-b^2$ and
$2ab$ units respectively, where $a$ and $b$ are positive real numbers
such that $a$ is greater than $b$. Find an exact expre... | Let $$a' = a^{2} - b^{2}$$ $$ b' = 2ab $$
From the Pythagorean Theorem, $$ c'^{2} = a'^{2} + b'^{2} $$
Substitute both $a'$ and $b'$ into the above equation:
$$ c'^{2} = (a^{2} - b^{2})^{2} + (2ab)^{2} = a^{4} - 2a^{2}b^{2} + b^{4} + 4a^{2}b^{2} = a^{4} + 2a^{2}b^{2} + b^{4}$$
Solve for $c'$ by taking the square root:
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3245496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Find all numbers x which satisfy $|x^2 + 2| = |x^2 − 11|$. This question is taken from book: Exercises and Problems in Calculus, by
John M. Erdman, available online, from chapter 1.1, question $4$.
Request help, as not clear if my approach is correct.
(4) Find all numbers x which satisfy $|x^2 + 2| = |x^2 − 11|$.
Ha... | Perhaps simpler (fewer cases to distinguish):
Note that $|a|=|b|$ (with real numbers $a,b$) means that $a=b$ or $a=-b$.
So here either $x^2+2=x^2-11$, which is absurd, or $x^2+2=-(x^2-11)$, which leads to $2x^2=9$ and so $$x=\pm\frac 32\sqrt 2.$$
Verification: If $x=\pm\frac32\sqrt 2$, then $x^2=\frac 92$ and $|x^2+2|=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3247025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Show that : $\displaystyle\int_0^{\infty}\frac{\ln(1+x^2)\operatorname{arc\,cot} x}{x}=\frac{π^3}{12}$ I need to prove this :
$I=\displaystyle\int_0^{\infty}\frac{\ln(1+x^2)\operatorname{arc\mkern2mucot} x}{x}=\frac{π^3}{12}$
My try :
$I\displaystyle\int_0^{1}\frac{\ln(1+x^2)\operatorname{arc\mkern2mucot} x}{x}$
$+\d... | The key here is the identity:
$$\mathrm{arccot}x + \arctan x = \arctan \frac{1}{x} + \arctan x = \frac{\pi}{2} \quad \quad \text{forall} \quad x>0$$
Thus,
\begin{align*}
\int_{0}^{\infty} \frac{\ln \left ( 1+x^2 \right ) \mathrm{arccot}x }{x} \, \mathrm{d}x &= \int_{0}^{1} \frac{\ln \left ( 1+x^2 \right ) \mathrm{arc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3250077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Elementary Second partial derivative If $V=\frac{xy} {(x^2+y^2)^2}$ and $x=r\cos\theta$, $y=r\sin\theta$ show that $\frac{\partial^2V} {\partial{r^2}}+\frac{1}{r}\frac{\partial{V}} {\partial{r}}+\frac{1} {r^2} \frac{\partial^2{V}}{\partial{\theta^2}}=0$
My attempt :
$\frac{\partial {V}} {\partial{x}} =y\cdot\frac{y^2-3... | We have, by substituting our expressions for $x$ and $y$ $$V=\frac{r^2\cos(\theta)\sin(\theta)}{(r^2\cos^2(\theta)+r^2\sin^2(\theta))^2}=\frac{r^2\cos(\theta)\sin(\theta)}{(r^2(\cos^2(\theta)+\sin^2(\theta)))^2}=\frac{r^2\cos(\theta)\sin(\theta)}{r^4}=\frac{\cos(\theta)\sin(\theta)}{r^2}.$$ We can even utilise a double... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3251673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How can I prove this limit converges to the Euler-Mascheroni constant? I'm trying to prove that
$$\lim_{N\to\infty}\;2\left[ \int_0^N \frac{\text{erf}(x)}{x}\,dx - \ln(2N) \right] = \gamma,$$
where $\gamma$ is the Euler-Mascheroni constant.
This looks similar to the definition of $\gamma$,
$$\gamma\equiv \lim_{N\to\inf... | I figured it out thanks to the hint and some further research. The only "extra" result I need is
$$\frac{\Gamma'(z)}{\Gamma(z)}=-\frac{1}{z}-\gamma+z\sum_{k=1}^\infty \frac{1}{k}\cdot\frac{1}{z+k},$$
which is fine with me because Gamelin proves it in Ch. 14 of Complex Analysis.
Starting from my original expression,
$$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3252696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prob. 5, Sec. 6.2, in Bartle & Sherbert's INTRO TO REAL ANALYSIS, 4th ed: How to show this function is strictly decreasing using derivative Here is Prob. 5, Sec. 6.2, in the book Introduction To Real Analysis by Robert G. Bartle & Donald R. Sherbert, 4th edition:
Let $a > b > 0$ and let $n \in \mathbb{N}$ satisfy $n \... | There is an error; the factorisation $x^{1/n}(x-1) - x(x-1)^{1/n} = x^{1/n} (x-1)^{1/n} ((x-1)-x)$ is incorrect and should be $x^{1/n}(x-1) - x(x-1)^{1/n} = x^{1/n} (x-1)^{1/n} ((x-1)^\frac{n-1}{n}-x^\frac{n-1}{n})$.
To correctly prove this, note that since $\frac{1}{n}-1$ is negative we have $x^{\frac{1}{n}-1}$ is dec... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3255422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Evaluating $\sum_{k=1}^{n} \frac{1}{k(k+1)}$ I have just started learning sums. I need to evaluate the following sum:
$$S_{n} = \sum_{k=1}^{n} \frac{1}{k(k+1)}$$
$$a_{1} = \frac{1}{2}, a_{2} = \frac{1}{6}, a_{3} = \frac{1}{12}, a_{4} = \frac{1}{20}, a_{5} = \frac{1}{30}, a_{6} = \frac{1}{42}, ...$$
I tried splitting it... | "Another way" to write things down, playing with index translation in the discrete sum:
Starting from your last equation:
$$S_{n} = \sum_{k=1}^{n} \frac{1}{k} - \sum_{k=1}^{n} \frac{1}{k+1}$$
you can translate indices $k=k'+1$ in the first sum:
$$S_{n} = \sum_{k'=0}^{n-1} \frac{1}{k'+1} - \sum_{k=1}^{n} \frac{1}{k+1}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3256519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Integral $\int_{-\infty}^{+\infty} \frac { e^{t}( e^{2t} -1)} {t(1+e^{2t})^2 }dt$
How to prove that
$$\int_{-\infty}^{+\infty} \frac { \pi e^{t}( e^{2t} -1)} {4t(1+e^{2t})^2 }d t=C$$
where $C$ is Catalan's constant.
Here is one of my ideas. Let: $$F(a)=\int_{-\infty}^{+\infty} \frac { \pi e^{at}( e^{2t} -1)} ... | $$I=\int_{-\infty}^\infty \frac{e^t(e^{2t}-1)}{t(1+e^{2t})^2}dt\overset{e^t=x}=\int_0^\infty \frac{(x^2-1)}{(x^2+1)^2 \ln x}dx$$
Consider: $$I(a)=\int_0^\infty \frac{(x^a-1)}{(x^2+1)^2 \ln x}dx\Rightarrow I'(a)=\int_0^\infty \frac{x^a}{(x^2+1)^2}dx\overset{x^2=y}=\frac12 \int_0^\infty \frac{y^{a/2}y^{-1/2}}{(y+1)^2}dy$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3264074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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How do I solve quadratic double inequalities? I have two questions involving quadratic double inequalities.
Firstly, what are the steps to get the solution for the following?
$0\le(x+2)^2\le4$
My my thought was to separate the inequality into
$0\le(x+2)^2$ and $(x+2)^2\le4$
Which would then allow me to take the squa... | Option:
1) Set $y=x+2$;
$0 \le y^2 \le 4.$
$f(y)=√y$ is an increasing function:
$0\le \sqrt{y^2} \le 2$;
With $ \sqrt{y^2}=|y| $ we get:
$0\le |y| \le 2$;
$-2 \le y\le 2$, or $-2 \le x+2 \le 2$.
2) Rewrite:
$-2 \le y \le 2$ as $|y| \le 2$.
Note $|y| \ge 0$.
Hence:
$|y| \le 2 $ implies
$ |y||y|\le 2 |y| \le 2\cdot 2$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3266093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
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An identity associated with the centroid of a triangle
Four year ago, I am looking for a proof of my identity as follows:
Let $ABC$ be a triangle, let $G$ be the centroid of $ABC$. Let $P$ be any point on the plane. Let $H, N, O$ on the plane such that: $\overrightarrow{HN}:\overrightarrow{NG}:\overrightarrow{GO}=3:1... | Michael's proof is more elegant, but it is also possible to hack it out with vectors.
Let $G$ be the origin of a cartesian coordinate system, and let boldface lowercase letter denotes the position of the uppercase letter. Then $\mathbf{a}+\mathbf{b}+\mathbf{c}=\vec{0}$, $\mathbf{h}=4\mathbf{n}=-2\mathbf{o}$, so
\begin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3266670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Development of the sine function into the power series I would like to check my solution for the development of the real sine function, $f(x) = \sin x$ into the power series. Here is my solution:
First, we have that $\mathcal{D}_f = \mathbb{R}$. Now, we have to check if the given function is infinitely diferentiable an... | Take this as a supplement, relying on the (omnipotent) power series of the exponential function
$$\exp( z ) \:=\:\sum_{k=0}^\infty \frac{z^k}{k!}\:=\:1 + z + \frac{z^2}{2} + \frac{z^3}{6} + \frac{z^4}{24} + \ldots$$
in the complex plane, which is given instead of developed here.
Papa Rudin's Prologue(*) is dedicated to... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Radical equation solve $\sqrt{3x+7}-\sqrt{x+2}=1$. Cannot arrive at solution $x=-2$ I am to solve $\sqrt{3x+7}-\sqrt{x+2}=1$ and the solution is provided as -2.
Since this is a radical equation with 2 radicals, I followed suggested textbook steps of isolating each radical and squaring:
$\sqrt{3x+7}-\sqrt{x+2}=1$
$(3x+... |
$\sqrt{3x+7}-\sqrt{x+2}=1$
$3x+7=(1 \color{red}{\mathbf{ \,-\, }}\sqrt{x+2})^2$ # square both sides
You want: $3x+7=(1 \color{blue}{\mathbf{ \, + \,}}\sqrt{x+2})^2$
Note: not only $x=-2$ solves this equation, also $x=-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3273876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to determine whether $\Bbb P ( B_{t_1} \in [x - c , x + c ], \ldots , B_{t_n} \in [x -c , x + c ])$ is decreasing in $x$? Let $\phi_t (z) := \frac{1}{\sqrt{2\pi t}} e^{-\frac {z^2} {2t}}$. By intuition $\omega : [0,\infty ) \to [0,1]$
\begin{align}
\omega (x) &:= \Bbb P ( B_{t_1} \in [x - c , x + c ], \ldots , B_{t... | $\newcommand{\PRx}[2]{\Bbb{P}_{#1} (#2)}$
Let $c>0$ and $G\subset (0,\infty)$ a finite set of time points. The function $\omega : \Bbb R \to [0,1]$ given by $\omega (x) := \Bbb P_x ({\vert{B_s}\vert \leq c\ \forall s \in G})$ is radial nonincreasing.
The strategy is induction over $\vert G \vert$. For $\vert G \vert =1... | {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\frac{9}{1!}+\frac{19}{2!}+\frac{35}{3!}+\frac{57}{4!}+\frac{85}{5!}+......$
Prove that $$\frac{9}{1!}+\frac{19}{2!}+\frac{35}{3!}+\frac{57}{4!}+\frac{85}{5!}+......=12e-5$$
$$
e=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+......
$$
I have no clue of where to start and I am not able t... | Your series is apparently
$$\sum_{n=1}^{\infty} \frac{3n^2+n+5}{n!}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3275356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Sum of fractions. Let $a_1 < a_2 <a_3 < a_4$ be positive integers such that $\sum_{i=1}^{4} \frac {1}{ a_i}$=$ 11/6$. Then $a_4-a_2=?$
I've tried to find the $a_i$'s but couldn't.
Is there a general way to solve such equations?
| Note that if $2\leq a_1<a_2<a_3<a_4$ then
$$\frac{77}{60}=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\geq \sum_{i=1}^{4} \frac {1}{ a_i}=\frac{11}{6}$$
which is a contradiction. Hence $a_1=1$.
Now we are left with $2\leq a_2<a_3<a_4$ such that
$$\frac{1}{a_2}+\frac{1}{a_3}+\frac{1}{a_4}=\frac{5}{6}.$$
Similarly i... | {
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Solutions to a system of three equations with Pythagorean triples Is there any solution to this system of equations where $x,y,z,s,w,t\in\mathbb{Z}$, none are $0$.
\begin{align*}
x^2+y^2=z^2\\\
s^2+z^2=w^2\\\
x^2+t^2=w^2
\end{align*}
EDIT: Thank you zwim for the answer. Maybe I should explain where this came from. Well... | Time... began to write, and I will add a few words....
Do there exist four distinct integers such that the sum of any two of them is a perfect square?
This is equivalent to solving the following system of equations:
$$\left\{\begin{aligned}& b+a=x^2 \\&b+c=y^2\\&b+f=z^2\\&a+c=e^2\\&a+f=j^2\\&c+f=p^2\end{aligned}\right.... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A First Course in Mathematical Analysis, David Alexander Brannan, ch 1, problem 11 Self studying Brannan's book. For some problem, I am not seeing how they got the solution.
Theorem 3. Arithmetic Mean-Geometric Mean Inequality:
For any positive real numbers $a_{1},a_{2},\dots,a_{n}$ we have
$$
(a_{1}a_{2}\dots a_{n})^{... | The AM of $1$ and $n$ copies of $1+\frac1n$ is
$$\frac1{n+1}\left(1+\left(1+\frac1n\right)+
\cdots+\left(1+\frac1n\right)\right)=\frac1{n+1}
\left(1+n\left(1+\frac1n\right)\right)=\frac1{n+1}\left(1+n+1\right)
=\frac{n+2}{n+1}.$$
| {
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Evaluate $S=\sum_{n=2}^\infty\frac{_nC_2}{(n+1)!}.$
Prove that $$S=\sum_{n=2}^\infty\frac{_nC_2}{(n+1)!}=\frac{e}{2}-1.$$
$$
S=\sum_{n=2}^\infty\frac{_nC_2}{(n+1)!}=\sum_{n=2}^\infty\frac{n!}{2(n-2)!(n+1)!}=\sum_{n=2}^\infty\frac{1}{2(n+1)(n-2)!}\\
=\frac{1}{2}\bigg[\frac{1}{3.0!}+\frac{1}{4.1!}+\frac{1}{5.2!}+\frac{... | Hint: Take two derivation of both side of
$$\dfrac{e^x-1-x-\dfrac12x^2}{x}=\sum_{n=2}^{\infty}\dfrac{x^n}{(n+1)!}$$
then let $x=1$.
| {
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How to factor a fourth degree polynomial I'm working on a math problem but I am having a hard time figuring out the method used by my textbook to make this factorization:
$$x^4 + 10x^3 + 39x^2 + 70x + 50 = (x^2 + 4x + 5)(x^2 + 6x + 10)$$
I've tried to see if this equation can be factored by grouping or by long division... | For all real $k$ we obtain:
$$x^4+10x^3+39x^2+70x+50=$$
$$=(x^2+5x+k)^2-25x^2-k^2-10kx-2kx^2+39x^2+70x+50=$$
$$=(x^2+5x+k)^2-((2k-14)x^2+(10k-70)x+k^2-50).$$
Now, we'll choose $k$ such that we'll get a difference of squares.
For which we need $$25(k-7)^2-(2k-14)(k^2-50)=0$$ or
$$(k-7)(2k^2-25k+75)=0$$ or
$$(k-7)(k-5)(2... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
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"answer_id": 5
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$bc(b-c) + ca(c-a) + ab(a-b) = -(b-c)(c-a)(a-b)$ My book has used the equality $$bc(b-c) + ca(c-a) + ab(a-b) = -(b-c)(c-a)(a-b)$$ But its proof is not given.
When I open the LHS, I get $b^2c-bc^2+c^2a-ca^2+a^2b-ab^2$. I do not know how to proceed next.
| $bc(b-c)+ca(c-a)+ab(a-b)=c[b(b-c)+a(c-a)]+ab(a-b)=c[b^2-bc+ac-a^2]+ab(a-b)=c[b^2-a^2+ac-bc]+ab(a-b)=c[(b-a)(b+a)-c(b-a)]-ab(b-a)=c(b-a)(b+a-c)-ab(b-a)=(b-a)[c(b+a-c)-ab]=-(a-b)(bc+ac-c^2-ab)=-(a-b)[c(b-c)-a(b-c)]=-(a-b)(c-a)(b-c)$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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minimum value of of $(5+x)(5+y)$
If $x^2+xy+y^2=3$ and $x,y\in \mathbb{R}.$
Then find the minimum value of $(5+x)(5+y)$
What I try
$$(5+x)(5+y)=25+5(x+y)+xy$$
$x^2+xy+y^2=3\Rightarrow (x+y)^2-3=xy$
I am finding $f(x,y)=22+5(x+y)+(x+y)^2$
How do I solve it? Help me please
| Hint: From the equation $$y^2+xy+x^2-3=0$$ we get
$$y_{1,2}=-\frac{x}{2}\pm\sqrt{3-\frac{3}{4}x^2}$$ so you will obtain the two functions
$$f_1(x)=(5+x)\left(5-\frac{x}{2}+\sqrt{3-\frac{3}{4}x^2}\right)$$
or
$$f_2(x)=(5+x)\left(5-\frac{x}{2}-\sqrt{3-\frac{3}{4}x^2}\right)$$
to consider.
Doing this we get $$(5+x)(5+y... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving the following limit without L'Hospital's rule: $\lim_{x\to 0} \frac{\sin(x^2+2)-\sin(x+2)}{x} $
I have been trying to solve the following limit $$\lim_{x\to 0} \frac{\sin(x^2+2)-\sin(x+2)}{x}.$$
I came across the right answer as shown by the steps below, but I would to check if the steps are correct or if so... | The second limit can be evaluated as follows:
$\quad \quad \quad \lim_{x\to0} \frac{sin2(cosx^2 - cosx)}{x}$
$\quad \quad \quad \lim_{x\to0} sin2 \cdot (\frac{cosx^2 - cosx}{x})$
$\quad \quad \quad \lim_{x\to0} sin2 \cdot (\frac{cosx^2}{x} - \frac{cosx}{x})$
$\quad \quad \quad \lim_{x\to0} sin2 \cdot (\frac{1}{x} \cdo... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$ \int_0^\frac{\pi}{2}\ln^n\left(\tan(x)\right)\:dx$ I'm currently working on a definite integral and am hoping to find alternative methods to evaluate. Here I will to address the integral:
\begin{equation}
I_n = \int_0^\frac{\pi}{2}\ln^n\left(\tan(x)\right)\:dx
\end{equation}
Where $n \in \mathbb{N}$. We first observe... | Slightly different way - use well known results
$$\int_0^{\frac{\pi}{2}}\tan^ay\;dy=\frac{\pi}{2}\frac{1}{\sin \frac{\pi}2(a+1) } $$
( this integral is considered in this site probably many times.)
$$\frac{1}{\sin x}=\frac{1}{x}+\sum _{n=1}^\infty (-1)^n\left ( \frac{1}{x-n\pi}+ \frac{1}{x+n\pi}\right )$$
and differen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3294446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Compute $\gcd(a+b, 2a+3b)$ if $\gcd(a,b) = 1$ A question from a problem set is asking to compute the value of $\gcd(a+b, 2a+3b)$ if $\gcd(a+b) = 1$, or if it isn't possible, prove why.
Here's how I ended up doing it:
$\gcd(a,b) = 1$ implies that for some integers $x$, and $y$, that $ax+by = 1$.
Let $d = gcd(a+b, 2a+3b... | It's simple if you use matrices. Suppose a positive integer $d$ divides both $a + b$ and $2a + 3b$. This can be restated as $d$ dividing both entries of the vector
\begin{equation}\begin{bmatrix} 1 & 1 \\ 2 & 3\end{bmatrix}\begin{bmatrix} a \\ b\end{bmatrix}\end{equation}
Multiplying on the left by the inverse matrix, ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find polynomials $p(x) $ such that $(x+10)p(2x)=(8x-32)p(x+6) $ when $p(1)=210$ Hi I have polynomials as below form.
$(x+10)p(2x)=(8x-32)p(x+6)$, when $p(1)=210$
I try to assign some $x$ eg. $0.5$ , $-6$ but can not find the correct relation. Any advice or guidance on this would be greatly appreciated, Thanks.
| Let $S$ be the set of zeroes of $p$.
If $x \in S$, $x \ne -4$ then $$\underbrace{(x+4)}_{\ne 0}p(2x-12) = 8(x-10)p((x-6)+6) = 8(x-10)p(x) = 0$$
so $2x-12 \in S$.
Notice that $8 \in S$ since plugging in $x = 4$ gives $14p(8) = 8(4-4)p(10) = 0$.
Therefore $4 = 2\cdot 8 -12 \in S$ and then $-4 = 2\cdot 4 - 12 \in S$ so $\... | {
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Probability to get a card in a card game Yesterday, while playing a card game, we had an argument about probabilities.
The game settings (for 4 players): there are 2 red cards ($R$) and 3 blue cards ($B$). At the beginning of a round, those 5 cards are shuffled and one is taken out. Then the card are distributed one by... |
What is the probability that two red cards are present?
As you observed, the probability that two red cards are present is equal to the probability that a blue card was removed, which is
$$\Pr(\text{two red cards are present}) = \Pr(\text{a blue card is removed}) = \frac{3}{5}$$
as you found
As for your failed attem... | {
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1025th term of the sequence $ 1,2,2,4,4,4,4,8,8,8,8,8,8,8,8, ... $ Consider the following sequence - $$ 1,2,2,4,4,4,4,8,8,8,8,8,8,8,8, ... $$
In this sequence, what will be the $ 1025^{th}\, term $
So, when we write down the sequence and then write the value of $ n $ (Here, $n$ stands for the number of the below term) ... | The first term is $2^0=1$; the next $2^1=2$ terms are equal to $2^1=2$; the next $2^2=4$ terms are equal to $4$ and so on. Since
$$
2^0+2^1+2^2+\dots+2^n=2^{n+1}-1
$$
the term at place $1023$ is $512$. The next $2^{10}$ terms are equal to $1024$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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"answer_id": 3
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Modular arithmetic. $x \equiv 1 \pmod {m^k}$ implies $x^m \equiv 1 \pmod {m^{k+1}}$ Show that for $k \gt 0$ and $m \ge 1$, $x \equiv 1 \pmod {m^k}$ implies $x^m \equiv 1 \pmod {m^{k+1}}$
This question has already been asked in SE (Show that for $k \gt 0$ and $m \ge 1$, $x \equiv 1 \pmod {m^k}$ implies $x^m \equiv 1 \pm... | We have $x^m - 1 = (x-1) (x^{m-1} + x^{m-2} + \cdots + 1)$. Now, by hypothesis, $m^k \mid x-1$. On the other hand, since $k > 0$, we also have $x \equiv 1 \pmod{m}$, so $x^{m-1} \equiv 1 \pmod{m}$, $x^{m-2} \equiv 1 \pmod{m}$, ..., $1 \equiv 1 \pmod{m}$. Therefore, $x^{m-1} + x^{m-2} + \cdots + 1 \equiv 1 + 1 + \cdo... | {
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$a, b$ are the integer part and decimal fraction of $\sqrt7$ find integer part of $\frac{a}{b}$
$a, b$ are the integer part and decimal fraction of $\sqrt7$
find integer part of $\frac{a}{b}$
using calculator :
$\sqrt 7$ = 2.645
$\frac{a}{b} = \frac{2}{0.6} = \frac{20}{6} = 3.333$
integer part of $\frac{a}{b} = 3$
ho... | Note that $$ 2.5<\sqrt 7<3$$
$$ \frac {a}{b} = \frac {2}{\sqrt 7 -2} = \frac {2(\sqrt 7+2)}{3}$$
$$3= \frac {2( 2.5+2)}{3} < \frac {2(\sqrt 7+2)}{3} <\frac {2( 3+2)}{3}<4$$
Thus the integer part of $\frac {2}{\sqrt 7 -2}$ is $3$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the Range of $y=\frac{x+a}{x^2+bx+c^2}$ Given that $x^2-4cx+b^2 \gt 0$ $\:$ $\forall$ $x \in \mathbb{R}$ and $a^2+c^2-ab \lt 0$
Then find the Range of $$y=\frac{x+a}{x^2+bx+c^2}$$
My try:
Since $$x^2-4cx+b^2 \gt 0$$ we have Discriminant
$$D \lt 0$$ $\implies$
$$b^2-4c^2 \gt 0$$
Also
$$x^2+bx+c^2=(x+a)^2+(b-2a)(x+a... | If a number $p$ is in the range of the $y$ you have given, it means $\frac{x+a}{x^2+bx+c^2} = p$ for some $x$.
Or, in other words, that equation has a real solution.
=> $px^2+bpx+pc^2 = x+a$ For a real value of $x$
=> the discriminant of $px^2+(bp-1)x+pc^2-a$ is non-negative.
After some rearrangement we have any $p$... | {
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"timestamp": "2023-03-29T00:00:00",
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Simplify $\frac{1}{h}\Big( \frac{1}{(x-a-h)^2} - \frac{1}{(x-a)^2}\Big)$
Simplify $\frac{1}{h}\Big( \frac{1}{(x-a-h)^2} - \frac{1}{(x-a)^2}\Big)$
I would like a little help, so that I can finish solving this exercise.
So far, I've got
$$
\frac{\frac{1}{(x-a-h)^2} - \frac{1}{(x-a)^2}}{h}
= \frac{\frac{(x-a)^2 - (x-a-h... | hint
Begin by putting
$$x-a=b$$
then
$$\frac{1}{(b-h)^2}-\frac{1}{b^2}=$$
$$\frac{b^2-(b-h)^2}{b^2(b-h)^2}=$$
$$\frac{h(2b-h)}{b^2(b-h)^2}.$$
And if we divide by h, one find
$$\frac{2(x-a)-h}{(x-a)^2(x-a-h)^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3305064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Minimizing $9 \sec^2{x} + 16 \csc^2{x}$ Find the minimum value of $$9 \sec^2{x} + 16 \csc^2{x}$$
My turn :
Using AM-GM
$$9\sec^2{x} + 16\csc^2{x} \geq 2 \sqrt{144 \sec^2{x} \csc^2{x}}$$
$$9 \sec^2{x} + 16\csc^2{x} \geq 24 \sec{x} \csc{x} $$
But the equality sign holds iff
$$9 \sec^2{x} = 16\csc^2{x}$$
Then $$ \tan{x} =... | @Michael Rozenberg has already given an answer using Cauchy-Schwarz inequality. Here's my answer using the AM-GM inequality.
Write $\sec^2 x = 1 + \tan^2 x$ and $\csc^2 x = 1 + \cot^2 x$. Then
$$9\sec^2 x + 16\csc^2 x = 9(1+\tan^2 x) + 16(1+ \cot^2 x) = 25 + 9\tan^2 x + \frac{16}{\tan^2 x}.$$
Then by AM-GM inequality
$... | {
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How many different matrices $A$ and $B$ are possible such that product $AB$ is defined? Two matrices $A$ and $B$ have in total $6$ different elements (none repeated). How many different matrices $A$ and $B$ are possible such that product $AB$ is defined?
My attempts
$1^{st} Attempt$
Considering that there are $6$ dif... | This answer was revised in response to the posting of the solution. I initially overlooked that the product of a matrix with two elements and a matrix with four elements could be formed in four ways. The method in the posted solution is preferable to the one written below.
Your strategy is correct. However, saying t... | {
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"timestamp": "2023-03-29T00:00:00",
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The Curve $y = \frac{x^2}2$ divides the curve $x^2 + y^2 = 8$ in two partss. Find the area of smaller part. Question: The Curve $y = \frac{x^2}2$ divides the curve $x^2 + y^2 = 8$ in two parts. Find the area of smaller part.
This question is from Basic Mathematics. Please explain how I can solve it according to class 1... | I can substitute: $(x^2/2)^2+x^2=8$. I obtain: $x^4+4x^2-32=0$. Let $t=x^2$: $t^2+4t-32=0$. In other words: $(t+8)(t-4)$. $x^2=-8$ is impossible; $x^2=4$ has two solutions: $x=2$ or $x=-2$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find all positive integers whose square ends in $444$? The question is from the $1995$ British Mathematical Olympiad.
I’ve figured out that $38^2$ ends with a $444$. So $N^2 \equiv 38^2 \pmod{ 1000}$. Consequentially $N^2 - 38^2 \equiv 0 \pmod{ 1000}$. Using difference of two squares we see that $(N-38)(N+38) \e... | $$N^2\equiv444\pmod{1000}\implies N^2\equiv4\pmod{10}$$
$\implies N=10m\pm2$
$N^2=(10m\pm2)^2=4\pm40m+100m^2\equiv4\pm40m\pmod{100}$
We need $4\pm40m\equiv44\pmod{100}$
$4+40m\equiv44\pmod{100}\implies40m\equiv40\pmod{100}\iff100|40(m-1)\iff5|2(m-1)$
$m=5r+1$(say)
$N=10(5r+1)+2=50r+12$
$N^2=(50r+12)^2=2500r^2+1200r+144... | {
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"url": "https://math.stackexchange.com/questions/3309167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $z+\frac{1}{z} = 2\cos(\theta)$ and $z^n+\frac{1}{z^n} = 2\cos (n\theta)$. Hence express $\cos^6 (\theta)$ Show that $z+ \frac{1}{z} = 2\cos (\theta)$ and $z^n+\frac{1}{z^n} = 2\cos (n\theta)$
Hence express $\cos^6(\theta)$
So I have nailed down the proof:
Setting the value of $z=\cos(\theta) + i\sin(\theta)$... | I assume you want to express $\cos^6(x)$ in terms of a linear combination of terms from $\{\cos(kx)\}_{k=0}^6$, as this is a standard question. By the binomial theorem,
$$2^6\cos^6(x)=(z+z^{-1})^6=(z^6+z^{-6})+6(z^4+z^{-4})+\binom{6}{2}(z^2+z^{-2})+\binom{6}{3}$$
Which reduces to
$$2^5\cos^6(x)=\cos(6x)+6\cos(4x)+15\co... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to get the value of $A + B ?$ I have this statement:
If $\frac{x+6}{x^2-x-6} = \frac{A}{x-3} + \frac{B}{x+2}$, what is the value of $A+B$ ?
My attempt was:
$\frac{x+6}{(x-3)(x+2)} = \frac{A(x+2) + B(x-3)}{(x-3)(x+2)}$
$x+6=(x+2)A + B(x-3)$: But from here, I don't know how to get $A + B$, any hint is appreciated.
| Put $x=-2$ and $x=3$ to get $B=-\frac{4}{5}$ and $A=\frac{9}{5}$ respectively.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3310929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$34!=295232799cd96041408476186096435ab000000$, find $a$ ,$b$, $c$, and $d$ There was a number theory question that I have to do for homework.
$34!=295232799cd96041408476186096435ab000000$, find $a$ ,$b$, $c$, and $d$
I know $b=0$ because $10^7\big|34!$ only. But how can I find other variables?
Remark:
No electronic c... | Sum of digits is divisible by $9$, which gives
$$a+c+d\equiv5(\mod9).$$
$$2-9+5-2+3-2+7-9+9-c+d-9+6-0+4-1+4-0+8-4+7-6+1-8+6-0+9-6+4-3+5-a\equiv0(\mod11),$$
which gives
$$a+c-d\equiv-1(\mod11).$$
From here we can get $a+c=2$ and $d=3$.
Now, since $a\in\{2,6\}$, we obtain $a=2$, $c=0$ and we are done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Partial Fractions: Why does this shortcut method work? Suppose I want to resolve $1/{(n(n+1))}$ into a sum of partial fractions. I solve this by letting $1/{(n(n+1))} = {a/n} + {b/(n+1)}$ and then solving for $a$ and $b$, which in this case gives $a=1$ and $b=-1$.
But I learnt about a shortcut method. It says suppose $... | I'll just start with $a$ as an example. Multiply both sides by $n$ to get
$$\frac 1{(n+1)(n+2)} = a + \frac {bn}{n+1} + \frac {cn}{n+2}$$
Since this is true for all $n$, plug in $0$. The $b$ and $c$ terms become $0$, leaving $a$ in the formula described above:
$$\frac 1{(0+1)(0+2)} = a$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Can i factor this expression: $x^3+y^3+z^3$ I have the following numerical expression, which is exactly equal to $1$
Text version:
(-(7 2^(2/3))/38 + 3^(2/3)/19 + (23 5^(2/3))/95 + (5 6^(1/3))/19 + 3/38 (5^(1/3) 6^(2/3)) - 10^(1/3)/19 - 3/190 (3^(1/3) 10^(2/3)) - (4 15^(1/3))/19 - 6/95 (2^(1/3) 15^(2/3)))*(2^1/3+3^1/3... | By Eisenstein's criterion and Gauss' lemma, $x^3+y^3+z^3$ is irreducible in $k[x,y,z]$ for any field $k$ not of characteristic $3$. So, for example, it is irreducible in $\mathbb C[x,y,z]$.
| {
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Modular arithmetic $(2n+1)x \equiv -7 \pmod 9$ Find a solution $(2n+1)x \equiv -7 \pmod 9$
I’m sure this is trivial but I still have doubts about it.
I know the equation has solution for certain $n \in \mathbb {Z}$. Actually I have tried a few and got a similar results (with Diophantine equations ). I wonder if there’s... | $9$ is not prime so it has $0$ divisors and you cant solve $3x \equiv k\pmod 9$ unless $k$ is a multiple of $3$.
Basically if $\gcd(m, n) = 1$ there will always be a solution (and only one solution) to $mx \equiv 1\pmod n$. We can notate that solution as $m^{-1}$. ( So for example $5^{-1} = 2\pmod 9$ because $2*5 \equ... | {
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Calculating value of series by taking the difference Apologies if this is a basic question!
I'm trying to understand a solution to a problem I was solving. The author suggests a trick to calculate expected value by multiplying the expected value series by 0.5 (line 2) and taking the difference (line 3):
$E(X) = 0.5^1 ... | We have $$E = 0.5^1 + 2\cdot0.5^2 + 3\cdot0.5^3 + 4\cdot0.5^4+\cdots$$
$$0.5E = 0.5^2 + 2\cdot0.5^3 + 3\cdot0.5^4+\cdots$$
Combining terms with equal powers of $0.5$,$$E - 0.5E = 0.5^1 + 0.5^2(2-1) + 0.5^3(3-2) + 0.5^4 (4-3) \cdots$$
$$\implies 0.5E = 0.5^1 + 0.5^2+0.5^3\cdots$$
| {
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Prove that $\int_{-1}^{1}\frac{\log(1+x)}{1+x^2}dx = \frac{\pi}{4}\log(2)-\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}$ I have to prove that
$$
I=\int_{-1}^{1}\frac{\log(1+x)}{1+x^2}dx=\frac{\pi}{4}\log(2)-\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}
$$
I know that
$$ I_{+}=\int_{0}^{1}\frac{\log(1+x)}{1+x^2}dx=\frac{\pi}... | Yet another way is to use Feynman's trick of differentiating under the integral sign. Let
$$I(a) = \int_{-1}^1 \frac{\ln (1 + ax)}{1 + x^2} \, dx.$$
Note that $I(0) = 0$ and we require $I(1)$. Now
\begin{align}
I'(a) &= \int_{-1}^1 \frac{x}{(1 + ax)(1 + x^2)} \, dx\\
&= \frac{a}{1 + a^2} \int_{-1}^1 \frac{dx}{1 + x^2} ... | {
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"source": "stackexchange",
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Computation Of Integrals Computer the Integral: $$\int\frac{2x+1}{(x-1)(x-2)}dx$$
Now using partial fraction we can write $$\frac{2x+1}{(x-1)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2}$$, So we get $$\frac{2x+1}{(x-1)(x-2)}=\frac{A(x-2)+B(x-1)}{(x-1)(x-2)}$$ Now for all $x$ not equal to $1, 2$ we can cancel out the denominator ... | You have
$$2x+1=A(x-2)+B(x-1) \tag{1}\label{eq1}$$
As for why substituting values of $x = 1$ and $x = 2$ works, this is because \eqref{eq1} is an identity and, thus, must be true for all values of $x$. With the original equation involving the denominators, due to continuity for all $x \neq 1,2$, the numerators must sti... | {
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Prove $ (x^{3} + x^{4} +y^{3} + y^{4} + z^{3} + z^{4})^{3} \le (x^{9} + y^{9} + z^{9})[ (1+x)^{3/2} + (1+y)^{3/2} + (1+z)^{3/2} ]^{2} $ Prove
$$ (x^{3} + x^{4} +y^{3} + y^{4} + z^{3} + z^{4})^{3} \le (x^{9} + y^{9} + z^{9})[ (1+x)^{3/2} + (1+y)^{3/2} + (1+z)^{3/2} ]^{2} $$
if $x,y,z \ge 0$
I made this problem using H... | Directly citing the Hölder inequality is the shortest solution. If one wants to reduce this to some somewhat more elementary arguments, one can use the inequality of the arithmetic and cubic mean.
For non-negative $a_k$ and positive $b_k$ we get, using the weighted mean value inequalities, that is, Jensen's inequality ... | {
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equation simplification. $(5y-1)/3 + 4 =(-8y+4)/6$ Simplification of this equation gives two answers when approched by two different methods.
Method 1 Using L.C.M( least common multiple)
$(5y-1)/3 + 4 =(-8y+4)/6$
$(5y-1+12)/3 = (-8y+4)/6$
$5y-11 = (-8y+4)/2$
$(5y-11)2= (-8y+4)$
$10y-22 = -8y+4$
$18y=26$
$y = 26/18=13/... | Your first method multiplies $(5y-1+12)/3$ by $3$ to give $5y-11$ when it should give $5y+11$
| {
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Line$ 3x-4y+k$ touches a circle $x^2+y^2-4x-8y-5 = 0$ at $(a,b)$, then find $k+a+b =?$
The line $3x-4y+\lambda=0, (\lambda > 0)$ touches the circle $x^2+y^2-4x-8y-5 = 0$ at $(a,b)$. Find the value of $\lambda+a+b$.
I tried solving in the following manner:
Clearly the circle has center at $(2,4)$ with a radius of $5... | We can take advantage of the fact that the line touches the circle at a single point:
$$y=\frac34x+\frac{\lambda}{4} \Rightarrow \\
x^2+\left(\frac34x+\frac{\lambda}{4}\right)^2-4x-8\left(\frac34x+\frac{\lambda}{4}\right)-5 = 0 \Rightarrow \\
25x^2+2(3\lambda-80)x+\lambda^2-32\lambda-80=0 \quad (1)$$
The quadratic equa... | {
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Find all the integer pairs $(x,y)$ that satisfy the equation $7x^2-40xy+7y^2=(|x-y|+2)^3$ Find all the integer pairs $(x,y)$ that satisfy the equation
$7x^2-40xy+7y^2=(|x-y|+2)^3$
it is clear the equation is symmetric therefore you can assume w.l.o.g that $x\ge y$ which makes the new equation equal to
$(x-y+2)^3=7x^2-... | First notice that $7x^2-40xy+7y^2\le\frac{27}{2}|x-y|^2$ and equality holds only if $x+y=0$. (You can show this directly by expanding)
Now, let $t=|x-y|$ then $(t+2)^3\le\frac{27}{2}t^2$ and $t\ge 0$.
Since $(t+2)^3-\frac{27}{2}t^2=(t-4)^2(t+\frac{1}{2})$, this holds only if $t=4$, and even when $t=4$, you get equality... | {
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find min of $a+b$ given sum of $a\geq b\geq c\geq d$ is 9 and square sum is 21 Suppose $a\geq b\geq c\geq d>0$ and all are real numbers, and $a+b+c+d=9,a^2+b^2+c^2+d^2=21$, how to find the minmum of $a+b$?
What I attempted:
I can show $b\geq 1.5$ and $a\leq 3$.for $r\geq 0$, I consider $\sum(a-r)^2=\sum a^2-2r\sum a+4r... | Your solution is right and very nice!
The equality occurs for $x=4$ and $$(18x-x^2-2y-39)(2y-x^2)=0.$$
For $2y-x^2=0$ we obtain $y=8$, $c=d=2$, $a=3$ and $b=2$.
For $18x-x^2-2y-39=0$ we obtain $y=8.5$, $c=2.5$, $d=1.5$ and $a=b=2.5.$
We got all cases of equality occurring, as you said.
| {
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Eliminating parameter $\beta$ from $x=\cos 3 \beta + \sin 3 \beta$, $y = \cos \beta - \sin \beta$
Based on the given parametric equations:
$$\begin{align}
x &=\cos 3 \beta + \sin 3 \beta \\
y &= \cos \beta \phantom{3}- \sin \beta
\end{align}$$
Eliminate the parameter $\beta$ to prove that $x-3y+2y^3=0$.
What I got so... | Hint: Write your system in the form
$$x=\sqrt{2}\sin\left(\frac{\pi}{4}-\beta\right)(1+2\sin(2\beta))$$
$$y=\sqrt{2}\sin\left(\frac{\pi}{4}-\beta\right)$$
then you will get $$x=y(1+2\sin(2\beta))$$ and you can eliminate $\beta$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = \frac{3x-4}{x^2+5}$ is injective I'm working on proving that $f: \Bbb{R} \rightarrow \Bbb{R}$ defined by $f(x) = \frac{3x-4}{x^2+5}$ is injective. However I'm stuck.
Assuming that $f(x_1) = f(x_2)$:
$$f(x_1) = f(x_2)$$
$$\frac{3x_1-4}{x_1^2+5} = \frac{3x_2-4}{... | $$f(x) = \frac{3x-4}{x^2+5}$$
$$f(0)=-\dfrac 45 = f\left(-\dfrac{15}{4}\right)$$
You would have had an easier time solving
\begin{align}
\dfrac{3x-4}{x^2+5} &= k \\
3x-4 &= kx^2 + 5k \\
kx^2 - 3x +(5k+4) &= 0 \\
x &= \dfrac{3 \pm \sqrt{9-16k-20k^2}}{2k}
\end{align}
So $x$ is double valued for all
$$-\dfra... | {
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"timestamp": "2023-03-29T00:00:00",
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Please explain the formula for the sum of the cubes and the difference: $a^3 - b^3$ and $a^3 + b^3$? I have not yet fully mastered the formula for accelerated multiplication. I was able to understand everything except the last two. I would like to deal with them. Why have : $$ a^3 + b^3 = (a+b)(a^2 -ab + b^2)^* $$ and ... | Let $x=a/b$. We want to factor $b^3(x^3\pm1)$. Note that $1$ is a root of $x^3-1$, so $x-1$ is a factor of $x^3-1$. You could find the quotient by polynomial long division. Also, $-1$ is a root of $x^3+1$, so $x+1$ is a factor of $x^3+1$, and again you could find the quotient by polynomial long division.
(From th... | {
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proving that $a + b \sqrt {2} + c \sqrt{3} + d \sqrt{6} $ is a subfield of $\mathbb{R}$ The question is given below:
My questions are:
1- How can I find the general form of the multiplicative inverse of each element?
2-How can I find the multiplicative identity?
3-Is the only difference between the field and the sub... | Regarding 1: Note that if we multiple your general element by its conjugate with respect to $\sqrt{3}$,
$$ (a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6})(a + b\sqrt{2} - c\sqrt{3} - d\sqrt{6}) \\ = a^2 + 2b^2 - 3c^2 -6d^2 + (2ab-6cd)\sqrt{2} \text{,} $$
which we can simplify further by multiplying with this new expression'... | {
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"timestamp": "2023-03-29T00:00:00",
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Integrating $\int_a^b\left[ \left(1 - \frac{a}{x} \right)\left(\frac{b}{x}-1 \right) \right]^\frac{1}{2}dx$ We are interested in the following integration:
$$ \int_a^b \left[ \left(1 - \frac{a}{x} \right)\left(\frac{b}{x}-1 \right) \right]^{1/2} dx. $$
I try to use substitution with:
$$u = \left(1-\tfrac{a}{x}\right)\l... | Rescale the integration range with the following variable change and the shorthand $q$
$$u = \frac{x-a}{b-a}, \>\>\>\>\>\>q=\frac{b}{a}-1$$
to simplify the original integral
$$ I = \int_a^b \left[ \left(1 - \frac{a}{x} \right)\left(\frac{b}{x}-1 \right) \right]^{1/2} dx=aq^2\int_0^1 \frac{\sqrt{u(1-u)}}{1+qu} du\tag{1}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $a$, $b$, $c$ are sides of a triangle, prove $2(a+b+c)(a^2+b^2+c^2)\geq3(a^3+b^3+c^3+3abc)$
$a$, $b$, $c$ are sides of a triangle, prove:
$$2(a+b+c)(a^2+b^2+c^2)\geq3(a^3+b^3+c^3+3abc)$$
What I have tried:
$$
⇔2\sum (a+b)ab\geq \sum a^3+9abc
$$
so I can't use
$$\sum a^3+3abc\geq \sum ab(a+b)$$
| WLOG $$c=\max\{a,b,c\}\implies a=x+u,b=x+v\text{ and }c=x+u+v (x>0, u,v\ge 0)$$
Then your inequality is equivalent to
$$x(u^2-uv+v^2)+2(u+v)(u-v)^2\ge 0*\text{true}*$$
| {
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Convolution of $f(x)={1 \over 2} \chi_{[-1,1]}*\chi_{[-5,5]}$ Convolution of $f(x)={1 \over 2} \chi_{[-1,1]}*\chi_{[-5,5]}$. This is my first exercise, it's basically an interpolation between the two functions. So here my results:
Calling $u={1 \over 2} \chi_{[-1,1]}$ and $v=\chi_{[-5,5]}$ I can calculate the different... | $f(x)=\frac{1}{2}\int_{-\infty}^{+\infty} X_{[-5,5]}(y)X_{[-1,1]}(x-y)dy$
$$X_{[-1,1]}(x-y)=1 \Longleftrightarrow -1 \leq x-y \leq 1 $$$$\Longleftrightarrow x-1\leq y \leq x+1 \Longleftrightarrow y \in[x-1,x+1]\Longleftrightarrow X_{[x-1,x+1]}(y)=1$$
Thus $X_{[-1,1]}(x-y)=X_{[x-1,x+1]}(y)$
Also $G(y)=X_{[-5,5]}(y)X_{[x... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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Proving a result by making discriminant zero
If the roots of given Quadratic equation
$$a(b-c)x^2 +b(c-a)x + c(a-b)=0$$
are equal, prove the following:
$$\frac{2}{b}=\frac{1}{a}+\frac{1}{c}$$.
MY approach:
Method 1: put Discriminant=0 and get stuck.
Method2:add $acx$ on both sides and get $(x-1)$ as factor... | Following your first approach, we have that the discriminant is
$$\begin{align}
\Delta&=b^2(c−a)^2-4a(b−c)c(a−b)\\
&=b^2(c−a)^2+4ac(b^2-b(a+c)+ac)\\
&=b^2(c+a)^2+(2ac)^2-2b(a+c)(2ac)\\
&=(b(a+c)-2ac)^2
\end{align}.$$
and by letting $\Delta=0$, it is easy to show that $\frac{2}{b}=\frac{1}{a}+\frac{1}{c}$.
As regards yo... | {
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Prove that $\int_0^1 \frac{x^2}{\sqrt{x^4+1}} \, dx=\frac{\sqrt{2}}{2}-\frac{\pi ^{3/2}}{\Gamma \left(\frac{1}{4}\right)^2}$ How to show
$$\int_0^1 \frac{x^2}{\sqrt{x^4+1}} \, dx=\frac{\sqrt{2}}{2}-\frac{\pi ^{3/2}}{\Gamma \left(\frac{1}{4}\right)^2}$$
I tried hypergeometric expansion, yielding $\, _2F_1\left(\frac{1}{... | As I said in my comment I believe by making the substitution:
$$x^2=\tan(t)$$
you get:
$$\int_0^1\frac{x^2}{\sqrt{x^4+1}}dx=\frac{1}{2}\int_0^{\pi/4}\sin(t)^{1/2}\cos(t)^{-3/2}dt=\frac{1}{4}\int_0^{1/\sqrt[4]{2}}u^{-1/4}(1-u)^{-5/4}du=B\left(2^{-1/4};\frac34,-\frac14\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3349364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
The sum of infinite series $(1/2)(1/5)^2 + (2/3)(1/5)^3 + (3/4)(1/5)^4..................$ Initially, I broke this series as $1/2 = 1-1/2, 2/3 = 1-1/3, 3/4 = 1-1/4$, then I got two series as it is
$$= (1/5)^2 + (1/5)^3 + (1/5)^4 +.......-[1/2(1/5)^2 + 1/3(1/5)^3 + 1/4(1/5)^4 +.....]$$
After solving the first part, we re... | It is given:
$$\frac12\cdot \left(\frac15\right)^2+\frac23\cdot \left(\frac15\right)^3+\frac34\cdot \left(\frac15\right)^4+\cdots+\frac{n}{n+1}\cdot \left(\frac15\right)^{n+1}+\cdots$$
Consider the function:
$$f(x)=\frac12\cdot x^2+\frac23\cdot x^3+\frac34\cdot x^4+\cdots+\frac{n}{n+1}\cdot x^{n+1}+\cdots$$
We want to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3350958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Find all $x\in \mathbb{Z}$ with $x \equiv 2 \mod 4$ , $x \equiv 8\mod9$ and $x \equiv 1\mod5$
Find all $x\in \mathbb{Z}$ with $x \equiv 2 \mod 4$ , $x \equiv 8\mod9$ and $x \equiv 1\mod5$
My attempt:
$\gcd\left(9,5,4\right) = 1 = 9r+5s+4\times0$
$9 = 1\times5+4$
$5 = 1\times4+1$
so $1 = 5-(9-5) = 2\times5 - 1\times9$... | I am using the method for solving this system of equations as outlined here.
Here, we have $a_1 = 2, a_2 = 8, a_3 = 1$. Additionally, $n_1 = 4, n_2 = 9, n_3 = 5$. Now, $N = n_1 \cdot n_2 \cdot n_3 = 180$.
Hence, $y_1 = 45, y_2 = 20, y_3 = 36$. Now, we are supposed to find the values of $z_i$ for $i = 1, 2, 3$. I'll ou... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3354136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
Circles and squares and polygon
If i replace the square with circle, i can very easily find the area of circle and get some approximation for the area of the square.
Even the square is doable, but i was thinking that can we find the area of the regular polygon which is made to sit in place of the square ?
Since, i don... | A solution for a regular polygon with one edge a chord of the large circle and the vertices of the opposite edge on the smaller circles
Consider coordinates with $O$ as origin and the line of symmetry as the $x$-axis. Let the edge length of the regular polygon be $2L$ and let the distance between opposite edges be $D$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3354783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Proving $n^2(n^2-1)(n^2+1)=60\lambda$ such that $\lambda\in\mathbb{Z}^{+}$ I'm supposed to prove that the product of three successive natural numbers the middle of which is the square of a natural number is divisible by $60$. Here's my attempt.
My Attempt:
$$\text{P}=(n^2-1)n^2(n^2+1)=n(n-1)(n+1)[n(n^2+1)]$$
It is now ... | Use the Chinese Remainder theorem to show this product is congruent to $0\bmod 3, 4$ and $5$. You'll determine first what the squares are, modulo these numbers.
A sketch for the case of the modulus $4$:
Every number $n\equiv 0,1,2$ or $3\: (\equiv -1)\bmod 4$. So
$$n^2\equiv 0^2=0,\: 1^2=1, 2^2\equiv 0\quad\text{or}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3356512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
$\alpha$, $\beta$, $\gamma$ are the root of $x^3-x^2+px-1=0$. $(\alpha^3+1)(\beta^3+1)(\gamma^3+1)=2019$. $\alpha$, $\beta$, $\gamma$ are the root of $x^3-x^2+px-1=0$. $(\alpha^3+1)(\beta^3+1)(\gamma^3+1)=2019$. What is the product of all possible value of $p$?
Note that $p$ could be a complex number.
I tried some bas... | $$(x^3-1)^3=(x^2-px)^3$$
$$(x^3)^3-1-3x^3(x^3-1)=(x^3)^2-p^3(x^3)-3px^3(x^3-1)$$
Replace $x^3+1$ with $y$
$$(y-1)^3-3(y-1)(y-2)=(y-1)^2-p^3(y-1)-3p(y-1)(y-2)$$
$$y^3+(\cdots)y^2+(\cdots)y-1-6+1-p^3+6p=0$$
Now apply Vieta's formula
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3359868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find all solutions of $a,b,c,d$ if $a+b+c+d=12$ and $abcd = 27 +ab+ac+ad+bc+bd+cd$ Find all solutions of $a,b,c,d>0$ if
$$a+b+c+d=12$$
$$abcd = 27 +ab+ac+ad+bc+bd+cd$$
Attempt:
$a + b + c + d = 12 \implies 3 \ge \sqrt[4]{abcd} \implies 81 \ge abcd$ by AM-GM, eqyality when $a=b=c=d$. Also
$$ abcd - 27 = ab + ac + ad ... | You are fine : essentially, you got $$a+b+c+d \geq 4 \sqrt[4]{abcd} \implies 81 \geq abcd$$
from the first line. Then from the next few, you got :
$$
ab+bc+cd+da+ac+bd \geq 6 \sqrt[6]{a^3b^3c^3d^3} = 6 \sqrt{abcd}
$$
Combining this with $ab+bc+cd+da+ac+bd = abcd - 27$ gives that $abcd - 27 \geq 6 \sqrt{abcd}$, which c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3360569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find $\int \frac{\sin2x+2\tan x}{\cos^6x+6\cos^2x+4}dx$
Find $\displaystyle\int \frac{\sin2x+2\tan x}{\cos^6x+6\cos^2x+4}dx$
My approach is as follow
$\cos^2x=t$;
$\sin2x\ dx=-dt$
Therefore,
\begin{align}
\int \frac{\sin2x+2\tan x}{\cos^6x+6\cos^2x+4}dx&=\int \frac{\sin2x+\frac{2\tan x\cos^2x}{\cos^2x}}{\cos^6x+6\co... | Let $$I=-\int \frac{1+1/t}{t^3+6t+4} dt=\int \frac{1+t}{12t(t^3+6t+4)} dt=\int \left(\frac{1}{4t}-\frac{1}{12} \frac{3t^3+6}{(t^3+6t+4)}\right) dt$$ $$=\frac{\ln t}{4}-\frac{1}{12} \ln
~(t^3+6t+4)+C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3362116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Finding the Taylor Series Expansion using Binomial Series, then obtaining a subsequent Expansion. Hi all doing some prep work for a course and really struggling to wrap my head around some of the revision questions. Some help would be really appreciated. There are two halfs, part one is as follows:
Consider the functi... | $$g(x) =\frac{x^2}{1+x^2}=\frac{x^2+1-1}{1+x^2}=1-\frac{1}{1+x^2}$$ make $x^2=-t$ to face $f(t)$, simplify and, at the end, replace $t$ by $-x^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3362581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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.