Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Finding a limit to negative infinity with square roots: $\lim\limits_{x\to -\infty}(x+\sqrt{x^2+2x})$ Find the limit of the equation
$$\lim_{x\to-\infty} (x+\sqrt{x^2 + 2x})$$
I start by multiplying with the conjugate:
$$\lim_{x\to-\infty} \left[(x+\sqrt{x^2 + 2x})\left({x - \sqrt{x^2 + 2x}\over x - \sqrt{x^2+2x}}\righ... | Notice that since $x$ is tending to $-\infty$ then $x=-\sqrt{x^{2}}$. So when you let $x$ enter the square root in the denominator the minus that was originally there becomes a plus due to cancellation. That is:
$$\frac{-2}{\frac{x-\sqrt{x^{2}+2x}}{x}}=\frac{-2}{1-\frac{1}{x}\sqrt{x^{2}+2x}}=\frac{-2}{1+\sqrt{1+\frac{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/881145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Solve the equation: $\frac{z}{z-5}+\frac{1}{3}=-\frac{5}{5-z}$ Solve the equation: $\frac{z}{z-5}+\frac{1}{3}=-\frac{5}{5-z}$
First $z$ cannot be equal to $5$.
First, I multiplied $z$ with $3$, $1$ with $z-5$ and $-5$ with both. Eliminating the denominators gives me:$3z + z-5 = - 15$. Simplified, I'm left with: $4z-5=-... | This equation has no solution. Change sides to see that
$$\frac{z}{z-5}+(1/3)+\frac{5}{5-z}=0 \\
Or,\frac{z}{z-5}+(1/3)+\frac{-5}{z-5}=0 \\
Or, \frac{z-5}{z-5}+(1/3)=0$$
Which is impossible.
| {
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"timestamp": "2023-03-29T00:00:00",
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Decompose a fraction in a sum of two Let's say that I have this fraction:
$$ \frac{2x}{x^2+4x+3}$$
I would like to decompose in two fraction:
$$ \frac{A}{x+3} + \frac{B}{x+1}$$
Which is the procedure for that? :)
| $$ \frac{2x}{x^2+4x+3}=\frac{2x}{x^2+x+3x+3}=\frac{2x}{x(x+1)+3(x+1)}=$$
$$=\frac{2x}{(x+3)(x+1)}=\frac{A}{x+3}+\frac{B}{x+1}$$
$$2x=Ax+A+Bx+3B=(A+B)x+A+3B$$
You needs to solve the system
$$A+B=2,A+3B=0$$
| {
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"url": "https://math.stackexchange.com/questions/882733",
"timestamp": "2023-03-29T00:00:00",
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Evaluate the limit of $\ln(\cos 2x)/\ln (\cos 3x)$ as $x\to 0$ Evaluate Limits
$$\lim_{x\to 0}\frac{\ln(\cos(2x))}{\ln(\cos(3x))}$$
Method 1 :Using L'Hopital's Rule to Evaluate Limits (indicated by $\stackrel{LHR}{=}$. LHR stands for L'Hôpital Rule)
\begin{align*}
\lim _{x\to \:0}\left(\frac{\ln \left(\cos \left(2x\... | If you know $\lim_{x\to0}\frac{\ln(1+ x)}{x} = 1$, then
\begin{align}
\lim_{x\to 0} \frac{\ln(\cos 2x)}{\ln(\cos 3x)} = \lim_{x\to 0} \frac{\ln(1 + \cos 2x - 1)}{\cos 2x -1} \frac{\cos 3x -1} {\ln(1 + \cos 3x - 1)} \frac{\cos 2x -1}{\cos 3x -1}
\end{align}
We have $\lim_{x\to 0} \frac{\ln(1 + \cos 2x - 1)}{\cos 2x -1} ... | {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate indefinite integral. $\int\frac{dy}{\sqrt{169 + y^2}}$
Evaluate $$\int\frac{dy}{\sqrt{169 + y^2}}$$
I have solved the problem, but don't seem to be getting the right answer. I get $\ln|\sqrt{13+y}+y|+C$ as the answer, and it is not the answer, how can I solve this?
| Use the substitution $x=13\tan{u}$, then $dx=13\sec^2{u} \ du$
\begin{align}
\int\frac{dx}{\sqrt{169+x^2}}
&=\int\frac{13\sec^2{u} \ du}{\sqrt{169+169\tan^2{u}}}\\
&=\int\sec{u} \ du\\
&=\ln|\sec{u}+\tan{u}|+c\\
&=\ln\left|\frac{x}{13}+\sqrt{1+\left(\frac{x}{13}\right)^2}\right|+c
\end{align}
Alternatively, we can use ... | {
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Evaluate $\lim_{x \to -\infty} \left(\frac{\sqrt{1+x^2}-x}{x} \right)$
Evaluate $$\lim_{x \to -\infty} \left(\frac{\sqrt{1+x^2}-x}{x} \right)$$
I tried by taking $x^2$ out of the root by taking it common.
i.e: $$\lim_{x \to -\infty} \left(\frac{x\sqrt{\frac{1}{x^2}+1}-x}{x} \right)$$
and then cancelling the x in nume... | Hint: Multiply numerator and denominator by $\sqrt{1+x^2} + x$.
$$\lim_{x \to -\infty} \left(\frac{\sqrt{1+x^2}-x}{x} \right)\cdot \frac{\sqrt{1+x^2} +x}{\sqrt{1 + x^2}+x} = \lim_{x\to -\infty}\frac{1 + x^2 - x^2}{x(\sqrt{1 + x^2} + x)}$$
$$= \lim_{x\to -\infty}\frac 1{x\sqrt{1 + x^2} + x^2}$$
Can take it from here?
| {
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If $4x^2+y^2=40$ and $xy=-6$, find the value of $2x+y$. If $4x^2+y^2=40$ and $xy=-6$, find the value of $2x+y$.
I tried the following,
As we know, $(a+b)^2=a^2+b^2+2ab$
Therefore, $(4x^2+y^2)^2=(4x^2)^2+(y^2)+2(4x^2*y^2)$
$=16^4+y^4+8x^2y^2$
What should I do know? Please help. I am stuck.
| $$(2x+y)^2=4x^2+4xy+y^2=(4x^2+y^2)+4xy=40+4(-6)=16 \Rightarrow 2x+y=\pm 4$$
| {
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If $\gcd(a,b)=1$, then every odd factor of $a^2 + 3b^2$ has this same form EDIT: Please see EDIT(2) below, thanks very much.
I want to prove by infinite descent that the positive divisors of integers of the form $a^2+3b^2$ have the same form. For example, $1^2+3\cdot 4^2=49=7^2$, and indeed $7,49$ can both be written i... | This is false as stated. $4$ has an expression as $a^2 + 3 b^2,$ with $a=b=1,$ but $2$ does not. $25$ has such an expression, with $a=5,b=0,$ but $5$ itself does not. The same applies to any $q^2,$ where $q \equiv 2 \pmod 3$ is prime.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove the inequality $\frac 1a + \frac 1b +\frac 1c \ge \frac{a^3+b^3+c^3}{3} +\frac 74$ Inequality
Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=3$. Prove the following inequality
$$\frac 1a + \frac 1b +\frac 1c \ge \frac{a^3+b^3+c^3}{3} +\frac 74.$$
I stumbled upon this question some days ago and been... | here is a elementary method:
WLOG, let $a=$Min{$a,b,c$} $\implies a \le1 ,x=bc$
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)=3((a+b+c)^2-3ab-3ac-3bc)=3(9-3a(3-a)-3bc)$
LHS$=\dfrac{a(3-a)+bc}{abc}=\dfrac{a(3-a)+x}{ax},$
RHS$= 9-3a(3-a)-3bc+abc+\dfrac{7}{4}=\dfrac{43}{4}-3a(3-a)+(a-3)x$
LHS-RHS$=\dfrac{(3-a)}{x}\lef... | {
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"timestamp": "2023-03-29T00:00:00",
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How prove $\angle ABC=60^o$ for triangle ABC if $BD = BF = AB-AC$ and $\frac{1}{{AD}}+\frac{1}{{CF}}=\frac{1}{{BD}}$? In acute triangle $ABC$ with $AB > AC$,we consider points $D$, $F$ (internal) of $AB$, $BC$ respectively, such that $BD = BF = AB-AC$.
If: $\frac{1}{{AD}}+\frac{1}{{CF}}=\frac{1}{{BD}}$, to how prove th... | From $BD = BF = AB-AC$ you can deduce that $AD = AC$.
Let $r = BD = BF$ and $R = AC = AD$.
From the other equation we deduce that $CF = \displaystyle\frac{Rr}{R-r}$.
Using the law of cosines:
$$(R+r)^2 + \left(r + \frac{Rr}{R-r}\right)^2 - R^2 = 2(R+r)\left(r + \frac{Rr}{R-r}\right)\cos B$$
Or equivalently:
$$(R+r)^2(R... | {
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"timestamp": "2023-03-29T00:00:00",
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Proving these trigonometric sums $\sum\limits_{k=0}^{n-1}\sin\frac{2k^2\pi}{n}=\frac{\sqrt{n}}{2}\left(\cos\frac{n\pi}{2}-\sin\frac{n\pi}{2}+1\right)$ Can someone help me to prove that:
$$ \sum_{k=0}^{n-1}\sin\frac{2k^2\pi}{n}=\frac{\sqrt{n}}{2}\left(\cos\frac{n\pi}{2}-\sin\frac{n\pi}{2}+1\right)$$
$$\sum_{k=0}^{n-1}\c... | Let $A = \sum_{k=0}^{n-1}\cos\frac{2k^2\pi}{n}$ and $B = \sum_{k=0}^{n-1}\sin\frac{2k^2\pi}{n}$ then we are looking to show:
$$ \sum_{k=0}^{n-1} e^{2\pi i k^2/n} = A + iB = e^{\tfrac{\pi i n}{2}}\sqrt{n}\tfrac{1+i}{2} $$
I believe this is the Gauss sum
I have seen the Gauss sum for primes $p$ but not necessarily for ... | {
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"url": "https://math.stackexchange.com/questions/895419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Trigonometric formula simplifies to $\sin x\cos x[\tan x+\cot x]$ Again, I have a little trouble figuring out how we got from the first step to the next one. It would be really appreciated if someone could help me out.
$$
\begin{split}LHS &= \cos\left(\frac{3\pi}{2}+x\right)\cos(2\pi +x)\left[\cot\left(\frac{3\pi}{2}-... | Replace the terms on the left side of your left side of the identity with these, and you will end up with the right hand side of your identity.
$$\require{cancel}\cos \left( \frac{3\pi}2+x \right)=\cancelto{0}{\cos \frac{3\pi}2}\cos x - \cancelto{1}{\sin \frac{3\pi}2}\sin x=\sin x$$
$$\require{cancel}\cos \left(2\pi+x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/898382",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Proving that a number is non-negative? The numbers $a$,$b$ and $c$ are real.
Prove that at least one of the three numbers
$$(a+b+c)^2 -9bc \hspace{1cm} (a+b+c)^2 -9ca \hspace{1cm} (a+b+c)^2-9ab$$
is non-negative.
Any hints would be appreciated too.
| Hint:
If all three numbers are negative, then:
$$ab > \left(\frac{a+b+c}{3}\right)^2 \hspace{1cm} ac > \left(\frac{a+b+c}{3}\right)^2 \hspace{1cm} bc > \left(\frac{a+b+c}{3}\right)^2 \hspace{1cm}$$
Therefore, if we multiply the three inequalities:
$$a^2b^2c^2 > \left(\frac{a+b+c}{3}\right)^6$$
Or equivalently:
$$\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/900315",
"timestamp": "2023-03-29T00:00:00",
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Prove $ x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1)$ So what I am trying to prove is for any real number x and natural number n, prove $$x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1)$$
I think that to prove this I should use induction, however I am a bit stuck with how to implement my induction hypothesis. My base case is when $n=2$ ... | You may just open hooks in right half of expression:
$$(x - 1)(x^{n - 1} + x^{n - 2} + \dots + x + 1) = x * (x^{n - 1} + x^{n - 2} + \dots + x + 1) - 1 * (x^{n - 1} + x^{n - 2} + \dots + x + 1) = (x^n + x^{n - 1} + \dots + x^2 + x) - (x^{n - 1} + x^{n - 2} + \dots + x + 1) = x^ n - 1 $$
No problems)
| {
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"timestamp": "2023-03-29T00:00:00",
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greatest common divisor and solution in integers The greatest common divisor of 203 and 147; $gcd(203,147)=7$.
Thus how can we find all the solution in integers $x,y$ of the equation $203x + 147y=7$?
| First off we can use the Extended Euclidean Algorithm to find some "first" values of x and y.
$$\begin{array}{rclcrcllr}
1 & \cdot & 203 & + & 0 & \cdot & 147 & = & 203 \\
0 & \cdot & 203 & + & 1 & \cdot & 147 & = & 147 \\
1 & \cdot & 203 & - & 1 & \cdot & 147 & = & 56 \\
-2 & \cdot & 203 & + & 3 & \cdot & 147 & = & 35... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $f$ is differentiable in $(0,0)$ if and only if $\lim_{t\to0+} g(t)$ exists Let $g:[0,\infty)\to\mathbb{R}$ be a mapping and $f(x,y)=xg(\sqrt{x^2+y^2})$ for all $(x,y)\in\mathbb{R^2}$.
Prove that $f$ is differentiable in $(0,0)$ $\iff$ $\lim_{t\to0+} g(t)$ exists.
My attempt:
$\implies:$ We suppose that $f$ ... | For the first implication, if $f$ is differentiable at $(0,0)$, then in particular it admits a partial derivative with respect to $x$ at $0$, so that the following limit exists :
$$
\frac{\partial f}{\partial x}(0,0) = \lim_{x \to 0^+} \frac{f(x,0) - f(0,0)}{\sqrt{x^2+0^2}} = \lim_{x \to 0^+} \frac{x g\left( \sqrt{x^2... | {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating the limits $\lim_{(x,y)\to(\infty,\infty)}\frac{2x-y}{x^2-xy+y^2}$ and $\lim_{(x,y)\to(\infty,8)}(1+\frac{1}{3x})^\frac{x^2}{x+y}$ I got the following problem:
Evaluate the following limits or show that it does not exist:
$$\lim_{(x,y)\to(\infty,\infty)}\frac{2x-y}{x^2-xy+y^2}$$
and
$$\lim_{(x,y)\to(\infty,... | By setting:
$$x:=r\cos\theta \\ y:=r\sin\theta$$
You can turn: $$\lim_{\substack{
x\rightarrow\infty\\
y\rightarrow\infty}}
{f(x,y)}$$ into: $$\lim_{r\rightarrow\infty}{g(r,\theta)}.$$
So that:
$$ \frac{2x-y}{x^2-xy+y^2}=\frac{1}{r}\cdot\frac{2\cos\theta-\sin\theta}{(1-\cos\theta\sin\theta)}\,\overset{r\rightarrow\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/903875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to find the polynomial such that ... Let $P(x)$ be the polynomial of degree 4 and $\sin\dfrac{\pi}{24}$, $\sin\dfrac{7\pi}{24}$, $\sin\dfrac{13\pi}{24}$, $\sin\dfrac{19\pi}{24}$ are roots of $P(x)$ .
How to find $P(x)$?
Thank you very much.
Thank you every one.
But consider this problem.
Find the polynomial with d... | HINT: If $r$ is a root of a polynomial then $(x-r)$ is a factor. You have four roots.
| {
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"timestamp": "2023-03-29T00:00:00",
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Using sum/product of roots of a cubic equation to solve given expression $(a+b)^3+ (b+c)^3 + (c+a)^3$ If $a, b, c$ are the roots of the equation $7x^3- 25x +42 =0$, then the value of the expression $(a+b)^3+ (b+c)^3 + (c+a)^3$ is?
I tried to solve this but wasn't able to simplify the term to be able to apply sum and pr... | Using Vieta's formula $\displaystyle a+b+c=0,(-1)^3abc=42$
$$\implies \sum (a+b)^3=-\sum c^3$$
Method $\#1:$
As $a,b,c$ individually satisfy the given equation, we can write $$7a^3=25a-42\text{ etc.}$$
Again using Newton's
Power Sum formula $$7\sum a^3=25\sum a-3\cdot42=\cdots$$
Method $\#2:$
Like If $a,b,c \in R$ are... | {
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How prove this inequality $\sum\limits_{cyc}\frac{1}{a+3}-\sum\limits_{cyc}\frac{1}{a+b+c+1}\ge 0$ show that:
$$\dfrac{1}{a+3}+\dfrac{1}{b+3}+\dfrac{1}{c+3}+\dfrac{1}{d+3}-\left(\dfrac{1}{a+b+c+1}+\dfrac{1}{b+c+d+1}+\dfrac{1}{c+d+a+1}+\dfrac{1}{d+a+b+1}\right)\ge 0$$
where $abcd=1,a,b,c,d>0$
I have show three variable... | Partial Proof:
For general case of n variables, the inequality converts to:
$$\sum_i^n \frac1{1+a_1+a_2+\cdots+a_n-a_i}\le \sum_i^n\frac1{n-1+a_i}$$
Similiar to the given proof we can convert $\frac {a_1}{a_1+(n-1)}$ like this:
$$\frac{a_1}{a_1+(n-1)}=\frac{a_1}{a_1+(n-1)(a_1a_2\cdots a_n)^{1/n}}$$
Dividing numerator... | {
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To find maximum value If $A>0,B>0$ and $C>0$ and further it is known that $A+B+C=\frac{5\pi}{4}$,then find the maximum value of $\sin A+\sin B+\sin C$
| From a purely algebraic point of view, elimating $C$ from the constraint leads to maximizing $$F=\sin \left(A+B-\frac{\pi }{4}\right)+\sin (A)+\sin (B)$$ Taking derivatives $$F'_A=\cos \left(A+B-\frac{\pi }{4}\right)+\cos (A)=0$$ $$F'_B=\cos \left(A+B-\frac{\pi }{4}\right)+\cos (B)=0$$ This implies $A=B$ and then $$\co... | {
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"timestamp": "2023-03-29T00:00:00",
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Is it true that $\int_0^1 \lfloor x^{-1} \rfloor^{-1} x^n dx = \frac{1}{n+1}(\zeta(2)+\zeta(3) + \dots + \zeta(n+2) ) - 1$? This question is inspired by the formula $$\displaystyle\int_0^1{\left\lfloor{1\over x}\right\rfloor}^{-1}\!\!dx={1\over2^2}+{1\over3^2}+{1\over4^2}+\cdots = \zeta(2)-1,$$
see for instance this qu... | Letting $t=x^{-1}$ so $dx=-t^{-2}dt$ this integral can be rewritten as:
$$\begin{align}\int_{1}^\infty \lfloor t\rfloor^{-1} t^{-(n+2)}dt &= \sum_{k=1}^\infty \frac{1}k\int_{k}^{k+1}t^{-(n+2)}dt \\&= \frac{1}{n+1}\sum_{k=1}^\infty \frac{1}{k}\left(k^{-(n+1)}-(k+1)^{-(n+1)}\right)
\end{align}$$
That's pretty much the ex... | {
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Simplify the derivative of $y=\frac{3x}{\sqrt{x-3}}$ First of I'm still not totally clued up on how to format all this math properly so some of it may look a bit strange, any help in fixing it would be greatly appreciated. Anyhow onto the math...
So I have to differentiate
$$y= \frac{3x}{\sqrt(x-3)}$$
And I used the q... | $$\frac{dy}{dx}=\frac{d}{dx} \left ( \frac{3x}{(x-3)^{\frac{1}{2}}} \right)=\frac{(3x)'(x-3)^{\frac{1}{2}}-3x \cdot ((x-3)^{\frac{1}{2}})'}{x-3}=\frac{3(x-3)^{\frac{1}{2}}-\frac{3}{2}x(x-3)^{\frac{1}{2}-1}}{x-3}=\frac{3(x-3)^{\frac{1}{2}}-\frac{3}{2}x(x-3)^{-\frac{1}{2}}}{x-3}=\frac{2(x-3)^{\frac{1}{2}} \cdot (3(x-3)^{... | {
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"timestamp": "2023-03-29T00:00:00",
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Expressing the area as a function :) Express the area A of an equilateral triangle as a function of the height of the triangle.
Thanks :)
I am not sure where to even start on how to answer this problem.
| $$\text{ Area of an triangle }=\frac{1}{2} \cdot (\text{ base } ) \cdot (\text{ height })$$
The identity of an equilateral triangle is that all sides are equal to $x$.
The height is $AD$. The triangle $ABD$ is a right-angled triangle. Therefore, we can use the Pythagorean Theorem.
$$(AD)^2+(BD)^2=x^2 \Rightarrow (AD)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/908123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Hard Definite integral involving the Zeta function Prove that: $$\displaystyle \int_{0}^{1}\frac{1-x}{1-x^{6}}{\ln^4{x}} \ {dx} = \frac{16{{\pi}^{5}}}{243\sqrt[]{{3}}}+\frac{605\zeta(5)}{54} $$
I was able to simplify it a bit by substituting ${y = -\ln{x}}$ and some further mathematical manipulation but was not able to... | The series
$$ 24\sum_{k=0}^\infty \left(\frac{1}{(6k+1)^5}-\frac{1}{(6k+2)^5}\right)$$
$$ =\frac{24}{6^5}\sum_{k=0}^\infty \left(\frac{1}{(k+1/6)^5}-\frac{1}{(k+2/6)^5}\right)$$
Can be evaluated by the polygamma function $$ =\frac{24}{6^5} \left(\frac{-\psi^4(1/6)}{24} - \frac{-\psi^4(1/3)}{24} \right) $$
$$=\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/909977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Secant line slope question I am asked to:
Show the expression for the slope of the secant line through $y=x^2+3x$ at $x=3$ and $x=3+h$ is $msec=9+h$
\begin{align*}
msec &= \dfrac{f(x+h)-f(x)}{h} \\
&= \dfrac{(x+h)^2+3(x+h)-(x^2+3x)}{h}\\
&=\dfrac{(x+h)(x+h)+3x+3h-x^2-3x}{h}\\
&=\dfrac{2xh+h^2+3h}{h}\\
&=\dfrac{h(2x... | You are missing less that you think. On the line where you conclude with $2x+h+3$, you need to stop, and insert $x=3$ to get $9+h$, and then conclude. There is nothing else left to do. There is no second part.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/910462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If the equation $2x^2-7x+12=0$ has two roots alpha and beta ,then the value of alpha/beta+beta/alpha is If the equation $2x^2-7x+12 =0$ has two roots $\alpha$ and $\beta$ ,
then the value of $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}$ is
note $x=\frac{-7+\sqrt{47}}{4},\frac{-7-\sqrt{47}}{4}$
then
$$\frac{\frac{-7+\sq... | $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}=\frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}$
since $\alpha+\beta=\frac{7}{2}$ and $\alpha\beta=6$, you can compute the value by substituting.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/910717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Evaluating $\ln(\cos x))$ using Taylor expansion
Evaluate $\ln(\cos x)$ at $x_0=0$ and with the order of $n=4$.
Noticing that $\ln(\cos x) = \ln(1+ \cos x - 1)$ we can use $\ln(1+x)$ Taylor series.
Now, I've read I should use:
$$\ln(1+x) = x - \frac{x^2}{2} + R_2(x)$$
$$\cos x -1 = -\frac{x^2}{2} + \frac{x^4}{4!} + R... | So far so good---the key observations remaining are:
*
*In the contribution from the $u^2$ term of $\ln (1 + u)$, we can see that the only way to get a term of order $\leq 4$ in $x$ is $-\tfrac{1}{2}\left(-\frac{x^2}{2}\right)^2 = -\frac{x^4}{8}$.
*The term $R_2\left(-\frac{x^2}{2} + \cdots\right)$ is already order... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is there an injective function such that $f(x^2)-f^2(x)\ge \frac{1}{4}$? The exercise asks me this:
Is there an injective function such that $f(x^2)-f^2(x)\ge \frac{1}{4}$?
ps: $f: \mathbb{R}\to \mathbb{R}$
I really don't know how to start :c, I appreciate hints.
| There is no such injective function. To solve it we first see that if $x^2 = x$ (which is the case for $x=0$ and $x=1$) then $f(x^2) = f(x)$ and $f(x^2) - f^2(x) \geq \frac{1}{4}$ becomes (for $x=0$ or $x=1$)
$$f(x) - f^2(x) - \frac{1}{4} = -\left(f(x)- \frac{1}{2}\right)^2 \geq 0$$
but this is only possible ($-a^2 \ge... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove $\frac{ab}{1+c^2}+\frac{bc}{1+a^2}+\frac{ca}{1+b^2}\le\frac{3}{4}$ if $a^2+b^2+c^2=1$
Ff $a,b,c$ are positive real numbers that $a^2+b^2+c^2=1$ ,Prove: $$\frac{ab}{1+c^2}+\frac{bc}{1+a^2}+\frac{ca}{1+b^2}\le\frac{3}{4}$$
Additional info: I'm looking for solutions and hint that using Cauchy-Schwarz and AM-GM be... | $\displaystyle \frac{ab}{1+c^2}+\frac{bc}{1+a^2}+\frac{ca}{1+b^2}\le\frac{3}{4}$ is equivalent to
EDIT: Do not pursue this method. It is wrong! Left as warning.
$\displaystyle \frac{a^2+b^2}{1+c^2}+\frac{b^2+c^2}{1+a^2}+\frac{c^2+a^2}{1+b^2}\le\frac{3}{2}$, as $ab\leq\frac{1}{2}(a^2+b^2)$.
This is equivalent to:
$\disp... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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How find the minimum $\frac{(5y+2)(2z+5)(x+3y)(3x+z)}{xyz}$,if $x,y,z>0$
let $x,y,z>0$, find the minimum of the value
$$\dfrac{(5y+2)(2z+5)(x+3y)(3x+z)}{xyz}$$
I think we can use AM-GM inequality to find it.
$$5y+2=y+y+y+y+y+1+1\ge 7\sqrt[7]{y^5}$$
$$2x+5=x+x+1+1+1+1+1\ge 7\sqrt[7]{x^2}$$
$$x+3y=x+y+y+y\ge 4\sqrt[4... | Hint:
$5y+2 \geq 2\sqrt{10y}$
$2z+5 \geq 2\sqrt{10z}$
$x+3y \geq 2\sqrt{3xy}$
$3x+z \geq 2\sqrt{3xz}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/913664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Process to show that $\sqrt 2+\sqrt[3] 3$ is irrational How can I prove that the sum $\sqrt 2+\sqrt[3] 3$
is an irrational number ??
| If $x=\sqrt{2}+\sqrt[3]3$, then
$$(x-\sqrt{2})^3=x^3-3\sqrt{2}x^2+6x-2\sqrt{2}=3$$
Thus
$$x^3+6x-3=\sqrt{2}(3x^2+2)$$
And
$$\frac{x^3+6x-3}{3x^2+2}=\sqrt{2}$$
But if $x$ is a rational, then so is the left hand side of the above equality. However we know $\sqrt{2}$ is not rational. Contradiction, so $x$ is irrational.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/913934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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"answer_id": 2
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Solve the following equation I need help to solve the following equation.
$13x=4x^2$
My attempt:
$13x=4x^2$
$4x^2-13x=0$
$\frac{4x^2-13x}{4}=0$
$x^2-\frac{13x}{4}+0=0$
$x=-\frac{-13}{2} \begin{matrix} + \\ - \end{matrix} \sqrt{(\frac{-13}{2})^2}$
this results to $x1 = 8.125$ $x2 = -4.875$
The correct result is $x1 = ... | $13x=x^2$
If $x\neq 0$ we can divide both sides by $x$ to obtain:
$\frac{13x}{x}=\frac{4x^2}{x}\Rightarrow 13=4x$
Then we divide both sides by $4$ to obtain:
$x = \frac{13}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/914188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Index of Summation Shift? Power Series and Differential Equations I have never had to index shift a summation series before, and it seems relatively straightforward, however, I am looking at an example in my textbook that doesn't make sense. I am wondering if someone might be able to outline the steps that appear to b... | Ok, just distribute to see:
\begin{align}
&(1+2x^2)\sum_{n=2}^{\infty}n(n-1)a{_{n}}x^{n-2} + 6x\sum_{n=1}^{\infty}na_{n}x^{n-1}+2\sum_{n=0}^{\infty}a_{n}x^n =\\
&=\sum_{n=2}^{\infty}n(n-1)a{_{n}}x^{n-2} +2x^2\sum_{n=2}^{\infty}n(n-1)a{_{n}}x^{n-2}+
\sum_{n=1}^{\infty}6na_{n}xx^{n-1}+\sum_{n=0}^{\infty}2a_{n}x^n \\
&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/914468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simpler closed form for $\sum_{n=1}^\infty\frac{\Gamma\left(n+\frac{1}{2}\right)}{(2n+1)^4\,4^n\,n!}$ I'm trying to find a closed form of this sum:
$$S=\sum_{n=1}^\infty\frac{\Gamma\left(n+\frac{1}{2}\right)}{(2n+1)^4\,4^n\,n!}.\tag{1}$$
WolframAlpha gives a large expressions containing multiple generalized hypergeomet... | By now, I've found a closed-form by doing some integral evaluation, a lot of hypergeometric, polylogarithm and polygamma manipulation.
$$
S = \sqrt{\pi}\left(\frac{\pi}{12}\zeta(3)+\frac{1}{192\sqrt3}\psi^{(3)}\left(\tfrac13\right)-\frac{\pi^4}{72\sqrt3}-1\right).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/915054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
"answer_count": 2,
"answer_id": 1
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Simplify rational expression How do I simplfy this expression?
$$\dfrac{\frac{x}{2}+\frac{y}{3}}{6x+4y}$$
I tried to use the following rule $\dfrac{\frac{a}{b}}{\frac{c}{d}}=\frac{a}{b}\cdot \frac{d}{c}$
But I did not get the right result.
Thanks!!
| $$\begin{align}
\frac{\frac{x}{2} + \frac{y}{3}}{6x+4y} &= \frac{\frac{x}{2} + \frac{y}{3}}{\frac{6x+4y}{1}}\\
&= \frac{\frac{x}{2} + \frac{y}{3}}{\frac{6x+4y}{1}}\cdot\frac{\frac{1}{6x+4y}}{\frac{1}{6x+4y}}\\
&= \frac{\bigg(\frac{x}{2}+\frac{y}{3}\bigg)\cdot\frac{1}{6x+4y}}{\frac{6x+4y}{6x+4y}}\\
&= \frac{\bigg(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/915154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Divisibility rule of 11
Let $M$ be a natural number with $n+1$ digits; represented by $M=a_{n}a_{n-1}\cdots a_{2}a_{1}a_{0}$
Show $M$ is divisible by $11$ if and only if
$$(a_{0}+a_{2}+a_{4}+\ldots)-(a_{1}+a_{3}+a_{5}+\ldots)$$ is also a divisible by 11
*
*here is Solution By Mr Américo Tavares
Let
$$M=a_{n}a_{... | $$10=11-1=11k-1 \\100=99+1=9(11)+1=11q+1\\1000=1001-1=11m-1\\10000=9999+1=11n+1\\\text{and so}\\M=a_{n}...a_{3}a_{2}a_{1}a_{0}=a_{0} +10a_{1}+100a_{2}+1000a_{3}+...=\\M=a_{0} +(11-1)a_{1}+(99+1)a_{2}+(1001-1)a_{3}+...=\\M=a_{0} +(-1)a_{1}+(1)a_{2}+(-1)a_{3}+...+11(1a_{1}-9a_{2}+99a_{3}....)=\\M=a_{0} +(-1)a_{1}+(1)a_{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/917207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Series representation of function with fractions, logarithms, squares and cosines. I'm looking for a series representation for
$$\dfrac x{x^2+(\log \cos x)^2}$$
Where $x\in(0,\pi/2)$
Note: Both finite and infinite series are accepted.
I have tried taylor series, but it requires the $n$th derivative, which is not trivia... | For a truncated series, I suppose that we could use $$\cos(x) \simeq 1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}$$ then use $$\log(1+y)\simeq y-\frac{y^2}{2}+\frac{y^3}{3}-\frac{y^4}{4}$$ Replace $$y=-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}$$ to get $$\log(\cos(x))\simeq-\frac{x^2}{2}-\frac{x^4}{12}-\frac{x^6}{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Closed form for $1 + 3 + 5 + \cdots +(2n-1)$ What is the closed summation form for $1 + 3 + 5 + \cdots + (2n-1)$ ?
I know that the closed form for $1 + 2 + 3+\cdots + n = n(n+1)/2$ and I tried plugging in $(2n-1)$ for $n$ in that expression, but it didn't produce a correct result:
$(2n-1)((2n-1)+1)/2$
plug in 3
$(2n-... | Let $$S_n=1+2+3+...+(2n-1)$$
Clearly
$S_1=1=1^2$
$S_2=1+3=2^2$
$S_3=1+3+5=9=3^2$
Suppose $S_n=n^2,$ then $S_{n+1}=S_n+(2n+1)=(n+1)^2$
By mathematical induction $$S_n=n^2 ,\forall n\in \mathbb{N}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/922714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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if $x+y+z=2$ and $xy+yz+zx=1$,Prove $x,y,z \in \left [0,\frac{4}{3} \right ]$
if $x+y+z=2$ and $xy+yz+zx=1$,Prove $x,y,z \in \left [0,\frac{4}{3} \right ]$
things i have done: first thing to do is to show that $x,y,z$ are non-negative. $$xy+yz+zx=1 \Rightarrow zx=1-yz-xy \Rightarrow zx=1-y(z+x)\Rightarrow zx=1-y(2-y... | Consider the polynomial:
$$ p(t)=(t-x)(t-y)(t-z) = t^3-2t^2+t+k = t(t-1)^2+k. $$
We know that $p(t)$ has three real roots, hence $-k=xyz$ is bounded between the two values of $t(t-1)^2$ in its stationary points. A stationary point obviously occurs for $t=1$, the other one occurs for $t=1/3$. The three reals roots of $p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/923451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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How many integers between $1$ and $10^n$ contain $14$? How can we derive a formula to calculate the number of integers between $1$ and $10^n$ that contain the number $14$ (as a string)?
For example, there are $20$ integers from $1$ to $1000$ that contain at least one $14$:
"14","114","140","141","142","143","144","145... | There are $\binom{n-k}{k}$ arrangements of $k$ "$14$"s and $10^{n-2k}$ other digits.
Thus, inclusion-exclusion says there are
$$
\sum_{k=1}^{\lfloor n/2\rfloor}(-1)^{k-1}\binom{n-k}{k}10^{n-2k}\tag{1}
$$
numbers from $1$ to $10^n$ which contain "$14$".
Here are some of the first few values.
$$
\begin{array}{r|l}
n&\tex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/924399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
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Find the quadratic equation equation of $x_1, x_2$.
Let $x_1 = 1 + \sqrt 3, x_2 = 1-\sqrt 3$. Find the quadratic equation $ax^2+bx+c = 0$ which $x_1, x_2$ are it's solutions.
By Vieta's theorem:
$$x_1\cdot x_2 = \frac{c}{a} \implies c=-2a$$
$$x_1 + x_2 = -\frac{b}{a} \implies b = -2a$$
Therefore, $b=c$
So we have... | Is $x_1$ and $x_2$ are the solution than you can write
$$(x_1-x)(x_2-x)=0$$
which is
$$x^2-(x_1+x_2)\cdot x+x_1x_2=0$$
Thus $a=1$, $b=-x_1-x_2$ and $c=x_1x_2$
| {
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"url": "https://math.stackexchange.com/questions/925016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Show that the odd prime divisors of $n^2+n+1$ are of form $6k+1$ (exclude 3) Show that the odd prime divisors of $n^2+n+1$ are of form $6k+1$ (exclude 3)
I have started like below:
$n^2+n+1\equiv 0 \pmod {p_i}$
$(n+1)^2\equiv n \pmod {p_i}$
Any hints/help on how to proceed from here ?
| Let $p$ be an odd prime divisor of $n^2+n+1$
Then $n^2+n+1 \equiv 0 \pmod{p}
\implies n^3-1 \equiv 0 \pmod{p}
\implies n^3 \equiv 1 \pmod{p}$
So $ord_pn=3$ (obvious!) and so $3|\phi(p)=p-1$
Note that $p-1$ is even. So $2|p-1$.
Since $gcd(2, 3)=1$, so $6|p-1\implies p-1=6k, k\in\Bbb{Z}\implies p=6k+1$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 2
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Inequality $\frac{\sqrt a+\sqrt b+\sqrt c}{2}\ge\frac{1}{\sqrt a}+\frac{1}{\sqrt b}+\frac{1}{\sqrt c}$ with weird condition I want to prove the following inequality:
$$\frac{\sqrt a+\sqrt b+\sqrt c}{2}\ge\frac{1}{\sqrt a}+\frac{1}{\sqrt b}+\frac{1}{\sqrt c}$$
Where $a,b,c$ are positive reals and with the horrible condi... | Proceeding from your last line: by the power mean inequality (or more simply the AM-HM actually, which is a special case) it is sufficient that $(\sqrt\frac{2w}{u+v}+\sqrt \frac{2u}{v+w}+\sqrt \frac{2v}{w+u})^2 \leq 9$ (left to you! it's not very hard). Motivation: the power mean inequality can give us an inequality be... | {
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"url": "https://math.stackexchange.com/questions/926793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Find $ \int \frac{\sin^2x}{1+\sin^2x}\hspace{1mm}\mathrm{d}x$ Find $$\int \dfrac{\sin^2x}{1+\sin^2x}\hspace{1mm}\mathrm{d}x$$
I cannot figure out how start this problem, can anyone explain
| $$\frac{\sin^2x}{1+\sin^2x}=1-\frac1{1+\sin^2x}=1-\frac{\sec^2x}{2\tan^2x+1}$$
Set $\tan x=u$
Alternatively,
$$\frac{\sin^2x}{1+\sin^2x}=\frac{\tan^2x}{\sec^2x+\tan^2x}$$
$$=\frac{\tan^2x\sec^2x}{(\sec^2x+\tan^2x)\sec^2x}=\frac{\tan^2x\sec^2x}{(\tan^2x+1+\tan^2x)(\tan^2x+1)}$$
Set $\tan x=u$ to find $$\int\frac{\sin^... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Find values of $a$ and $b$ that make the function continuous everywhere. I need some help with this question:
Find the values of $a$ and $b$ that make $f$ continuous everywhere.
$$f(x)=\begin{cases}
x^2 − 4/x-2, &\text{if }x < 2\\
ax^2-bx+1, &\text{if } 2 ≤ x ≤ 3\\
4x - a + b, &\text{if } x ≥ 3\end{cases}$$
I started... | If $f$ is continuous then :
$f(2)=\underset{x\rightarrow2}{\lim}f(x)=2^2-\frac{4}2-2=a(2^2)-b(2)+1$
Therefore $4a-2b+1=4-2-2=0$ hence $4a-2b+1=0$.
Same reasoning around $x=3$ :
$a(3^2)-b(3)+1=4(3)-a+b$ (ie) $9a+3b+1=12-a+b$ therefore $10a-4b=11$
You just need to solve $\begin{cases}4a-2b=-1\\10a-4b=11\end{cases}$ to f... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Comparing the size of square roots How to compare the size of following numbers without using the calculator?
$a=\sqrt{2}+\sqrt{6}+\sqrt{7},$
$b=\sqrt{3}+\sqrt{4}+\sqrt{8},$
$c=\sqrt{5}+\sqrt{5}+\sqrt{5}$
| Looking at the three expressions, one easily notices that the sum of the radicants is $15$ in all three cases. This together with the concavity of the square root ($\sqrt{\lambda a+(1-\lambda)b}\ge\lambda\sqrt{a}+(1-\lambda)\sqrt{b}$ for all $0\le\lambda\le1$) helps to compare the numbers. Note that the concavity of th... | {
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Geometry question with three triangles Any help on this question would be great, I'm not sure how to solve it.
Thanks in advance!
Given $\triangle ABC$, let $A'$ be the point $\frac{1}{3}$ of the way from $B$ to $C$, as shown. Similarly, $B'$ is the point $\frac{1}{3}$ of the way from $C$ to $A$, and $C'$ is the point ... | This is nicely done in barycentric coordinates. We have:
$$ A' = \frac{2B+C}{3},\quad B'=\frac{2C+A}{3},\quad C'=\frac{2A+B}{3} $$
hence:
$$ A''=\frac{2B'+C'}{3} = \frac{4A+B+4C}{9},\quad B''=\frac{4A+4B+C}{9},\quad C''=\frac{A+4B+4C}{9} $$
The last relation implies:
$$ A''-B'' = \frac{1}{3}(C-B),\quad B''-C''=\frac{1}... | {
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Solve the inequality: $|x^2 − 4| < 2$ This is a question on a calculus assignment our class received, I am a little confused on a few parts to the solution, can someone clear a few things up with it?
Since $x^2-4 = 0$ that means $x = 2$ and $x = -2$ are turning points.
When $x < -2$ :
$x^2-4<2$
$x^2<6$
$x < \sqrt{6}$ a... | In this case, we have $ |x^2 - 4| < 2 $, which leads to $ (x^2 - 4)^2 < 4 $, because $ |a| = \sqrt{a^2} $ where the root is the positive result.
Continuing, $ (x^2 - 4)^2 - 4 < 0$ and $ x^4 - 8x^2 + 12 < 0 $ or $ (x^2 - 6)(x^2 - 2) < 0 $.
This is only satisfied when $ x^2 > 2 $ and $ x^2 < 6 $. This is possible when $... | {
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Combinations of $6$-digit natural numbers In each of the following 6-digit natural numbers: $333333,225522,118818,707099$,
every digit in the number appears at least twice. Find the number of such 6-digit natural numbers.
This is how I'm intending to do.
1) Find the total number of 6-digit combinations.
2) Subtract the... | First let us also accept numbers starting with a $0$.
Let $d$ denote the number of distinct digits in the number. For $d>3$ there are $0$ possibilities.
For $d=1$ there are $10$ choices of the digit and every choice leads to $1$ possibility.
For $d=3$ there are $\binom{10}{3}=120$
choices for the digits and each choic... | {
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Square in Interval of Primes Denote by $a_n$ the sum of the first $n$ primes. Prove that there is a perfect square between $a_n$ and $a_{n+1}$, inclusive, for all $n$.
The first few sums of primes are $2$, $5$, $10$, $17$, $28$, $41$, $58$, $75$. It seems there is a perfect square between each pair of successive sums. ... | (waving my hands frantically to keep me aloft:)
$a_n
\approx \sum_{k=1}^n k\ln k
\approx \int_{1}^n k \ln k\ dk
\approx n^2\ln n
$
and
$a_{n+1}-a_n
\approx n\ln n
$
so if
$b_n
= \lfloor \sqrt{a_n}\rfloor
$,
$b_n
\approx n\sqrt{\ln n}
$.
The next square
after $b_n^2$
is
$(b_n+1)^2
=b_n^2+2b_n+1
\approx b_n^2+2n\sqrt{\l... | {
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"source": "stackexchange",
"question_score": "2",
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What is the limit of this specific function? Please evaluate the following limit for me:
$$\lim_{x \to -1} \frac{\sqrt{x^2+8}-3}{x+1} $$
I'd tried my best to solve this but unfortunately, it's too difficult for me. I tried multiplying by its conjugate and factoring the $x^2$ out but I can't get rid of that $x+1$ in the... | Problem:
$$\lim_{x\to-1}\frac{\sqrt{x^2+8}-3}{x+1}$$
Rationalize the numerator by multiplying by its conjugate
$$=\lim_{x\to-1}\frac{\left(\sqrt{x^2+8}-3\right)\left(\sqrt{x^2+8}+3\right)}{\left(x+1\right)\left(\sqrt{x^2+8}+3\right)}$$
Multiply out the numerator
$$=\lim_{x\to-1}\frac{\sqrt{x^2+8}^2+\boxed{3\sqrt{x^2+8}... | {
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"answer_id": 6
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Evaluate the integral $\int_0^{1/4}\frac{x-1}{\sqrt{x}-1}\mathrm dx$ so I have this Integral I have to solve without a calculator.
$$\int_0^{1/4}\dfrac{x-1}{\sqrt{x}-1}\mathrm dx.$$
How would I go about finding the antiderivative of that fraction?
| Problem:
$\int_0^\frac{1}{4}\frac{x-1}{\sqrt{x}-1}dx$
For the integrand $\frac{x-1}{\sqrt{x}-1}$, substitute $u=\sqrt{x}$ and $du=\frac{1}{2\sqrt{x}}dx$. This gives a new lower bound $u=\sqrt{\frac{1}{4}}=\frac{1}{2}$:
$=2\int_0^\frac{1}{2}\frac{u\left(u^2-1\right)}{u-1}du$
For the integrand $\frac{u\left(u^2-1\right... | {
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Taylor expansion square Consider the following expansion $$\sqrt{1+x} = 1 + \dfrac{1}{2}x - \dfrac{1}{8}x^2 + \dfrac{1}{16}x^3 .. $$
Show this equation holds by squaring both sides and comparing terms up to $x^3$.
I wonder, how can I square the right hand side?
| Hint:
\begin{align}
(a + b + c + \ldots )^2 & = (a+b+c+\ldots) \times (a+b+c+\ldots) \\
& = a^2 + ba + ca + \ldots + a b + b^2 + cb + \ldots ac + bc +c ^2 + \ldots
\end{align}
Can you take it from here?
| {
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Integrating $x^3\sqrt{ x^2+4 }$ Trying to integrate $\int x^3 \sqrt{x^2+4 }dx$, I did the following
$u = \sqrt{x^2+4 }$ , $du = \dfrac{x}{\sqrt{x^2+4}} dx$
$dv=x^3$ , $v=\frac{1}{4} x^4$
$\int udv=uv- \int vdu$
$= \frac{1}{4} x^4\sqrt{x^2+4 } - \int \frac{1}{4} x^4\dfrac{x}{\sqrt{x^2+4}} dx$ ---> i'm stuck her... | Use substitution method
First, let $u = x^2$
Then you will have $du = 2x dx$
$$ \int\,x^3\sqrt{x^2+4}\,dx = \dfrac{1}{2}\int\,u\sqrt{u+4}\,du $$
Then let $s = u+4$, which implies $ds=du$
You should be able to get the answer
If you are right, you should get the following:$\frac{ 1}{5} (x^2+4)^\frac {5}{2} -\frac{ 4}{3} ... | {
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Evaluate $\lim_{x\ \to\ \infty}\left(\,\sqrt{\,x^{4} + ax^{2} + 1\,}\, - \,\sqrt{\,x^{4} + bx^{2} +1\,}\,\right)$
I rewrote the function to the form
$$
x^{2}\left(\,
\sqrt{\,1 + \dfrac{a}{x^{2}} + \dfrac{1}{x^{4}}\,}\,-\,
\sqrt{\,1 + \dfrac{b}{x^{2}} + \dfrac{1}{x^{4}}\,}\,\right)
$$
and figured that the answer wo... | You could also use the basic Taylor Series expansion $\left( 1 + \frac{1}{t}\right)^{\lambda} = 1+\frac{\lambda}{t} + O(1/t^2)$ as follows:
$$x^2 \Big( \underbrace{ \sqrt{1 + \dfrac{a}{x^2} +\dfrac{1}{x^4}} }_{1+\frac{a}{2x^2} + O(\frac{1}{x^3})} - \underbrace{ \sqrt{1 + \dfrac{b}{x^2} +\dfrac{1}{x^4}} }_{1+\frac{b}{2x... | {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Sum of squares of two integers divisible by five Supposing $x,y$ are natural numbers,
what is the probability that the sum of their squares are divisible by 5?
I am getting $1/3$ as squares can only end with $0,1,4,5,6,9$. So $36$ pairs are possible. And $12$ pairs { $(0,0),(1,4),(4,1),(4,6),...$} have sums which end i... | Given $x, y$, only the last digits are relevant in determining if $x^2 + y^2$ is divisible by $5$. The last digit can be $0, 1, \ldots, 9$ and all of them are equally likely. Taking the last two digits of $x, y$, we get a sample space of size $10 \times 10 = 100$, where each of them are equally likely.
Now, we count ... | {
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"source": "stackexchange",
"question_score": "2",
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Integrating $\int \frac {dx}{\sqrt{4x^{2}+1}}$ $\int \dfrac {dx}{\sqrt{4x^{2}+1}}$
I've been up to this one for quite a while already, and have tried several ways to integrate it, using substituion, with trigonometric as well as hyperbolic functions. I know(I think) I'm supposed to obtain:
$\dfrac {1}{2}\ln \left| 2\sq... | Let $x = \frac{1}{2} \tan\theta$ so that $dx=\frac{1}{2}\sec^2\theta d\theta$. Then
$\begin{align*}
\int\frac{dx}{\sqrt{4x^2+1}} &= \frac{1}{2} \int \frac{\sec^2 \theta}{\sec\theta} d\theta \\
&= \frac{1}{2} \int\sec\theta d\theta \\
&= \frac{1}{2} \log|\sec\theta + \tan\theta| + C
\end{align*}$
for some real constant ... | {
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"source": "stackexchange",
"question_score": "3",
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Is my outcome wrong? (Evaluating a logarithm) $$log\sqrt [ 4 ]{ x^2+y^2 } $$
$$log\sqrt { x+y } $$
$$logx^{ 1/2 }+log^{ 1/2 }$$
$$\frac { 1 }{ 2 }log (x+y)$$
The answer key saids:
$$\frac { 1 }{ 4 } log(x^2+y^2)$$
| I'm afraid your answer is incorrect.
You should get this:
\begin{align*}
&\log \sqrt[4]{x^2+y^2}\\
&=\log(x^2+y^2)^{\frac{1}{4}}\\
&=\dfrac{1}{4}\log(x^2+y^2)
\end{align*}
You should note that $(a+b)^n\neq a^n+b^n$ in general, so $(a^2+b^2)^{\frac{1}{4}}\neq a^{\frac{1}{2}}+b^{\frac{1}{2}}$
| {
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"source": "stackexchange",
"question_score": "1",
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Least value of an Expression? Find the least value of $\dfrac {3a}{b+c} + \dfrac{4b}{a+c}+ \dfrac{5c}{a+b}$ for positive $a, b, c$.
I tried using the Cauchy-Schwarz inequality, but could not proceed after a bad equation which couldn't be further solved .
| You can minimize it by using Cauchy-Schwarz, but yes, it gets pretty bad. Firstly you can modify the expression as follows:
$$\frac {3a}{b+c} + \frac{4b}{c+a}+ \frac{5c}{a+b}=\frac {3(a+b+c)}{b+c} + \frac{4(a+b+c)}{c+a}+ \frac{5(a+b+c)}{a+b}-12=(a+b+c)\cdot \left(\frac {3}{b+c} + \frac{4}{c+a}+ \frac{5}{a+b}\right)-12=... | {
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"source": "stackexchange",
"question_score": "3",
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Why is $1 \times 3 \times 5 \times \cdots \times (2k-3) = \frac{(2k-2)!}{2^{(k-1)}(k-1)!}$ In order to find out the Catalan numbers from their generating function you have to evaluate the product above.
Here is what I thought:
\begin{align*}
1 \times 3 \times 5 \times...\times (2k-3) &= \frac{1 \times 2 \times 3 \time... | Note that all even numbers in the numerator cancel out with the denominator. But in your calculation, $2k-2$ from denominator doesn't cancel out.
The correct way to arrive at the answer is:
$$\begin{align}1 \times 3 \times 5 \cdots \times (2k-3) &= 1 \times \dfrac{2}{2} \times 3 \times \dfrac{4}{4} \times \cdots \times... | {
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"source": "stackexchange",
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Missing root of equation $\tan(2x)=2\sin(x)$ I tried to solve the equation $\tan(2x)=2\sin x$ and got the roots $x=n2\pi$ and $x=\pm \frac {2\pi}3 +n2\pi$. It seems that $x=n\pi$ is also a root but for some reason I didn't get that one out of my equation. Could you tell me where I went wrong?
$$
\tan(2x)=2\sin x
$$
$$
... | You can't just cancel out $\sin(x)$ in your steps because $\sin(x)$ can also be zero so you have to factor out even $\sin(x)$. $\sin(x)=0$ implies $x=n \pi $ is also the solution.
| {
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Evaluate $\int\frac{x^3}{\sqrt{81x^2-16}}dx$ using Trigonometric Substitution $$\int\frac{x^3}{\sqrt{81x^2-16}}dx$$
I started off doing $$u =9x$$ to get $$\int\frac{\frac{u}{9}^3}{\sqrt{u^2-16}}dx$$
which allows for the trig substitution of $$x = a\sec\theta$$
making the denomonator $$\sqrt{16\sec^2\theta-16}$$
$$\tan^... | Set $9x=4\sec\theta\implies81x^2-16=(4\tan\theta)^2$
$$\int\frac{x^3}{\sqrt{81x^2-16}}dx$$
$$=\int\frac{4^3\sec^3\theta}{9^3\cdot4\tan\theta\cdot(\text{sign of}\tan\theta)}\frac49\sec\theta\tan\theta\ d\theta$$
$$=\frac{4^3}{9^4\cdot(\text{sign of}\tan\theta)}\int(1+\tan^2\theta)\sec^2\theta\ d\theta$$
Hope you can tak... | {
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"source": "stackexchange",
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Integrate by partial fraction decomposition $$\int\frac{5x^2+9x+16}{(x+1)(x^2+2x+5)}dx$$
Here's what I have so far...
$$\frac{5x^2+9x+16}{(x+1)(x^2+2x+5)} = \frac{\mathrm A}{x+1}+\frac{\mathrm Bx+\mathrm C}{x^2+2x+5}\\$$
$$5x^2 + 9x + 16 = \mathrm A(x^2+2x+5) + (\mathrm Bx+\mathrm C)(x+1)=\\$$
$$\mathrm A(x^2+2x+5) + \... | It should be $A=3,B=2,C=1$ and you get:
$$\int \frac{3}{x+1} dx + \int \frac{2x+2}{x^2+2x+5} dx -\int \frac{1}{x^2+2x+5}dx$$
The second is easy by using $u=x^2+2x+5$ and other answerers have covered the last.
The key is writing $\frac{2x+1}{x^2+2x+5} = \frac{2x+2}{x^2+2x+5} - \frac{1}{x^2+2x+5}$.
It's probably easier t... | {
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"source": "stackexchange",
"question_score": "4",
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If $ab+bc+ca=1$, then $\frac{((a+b)^2+1)}{(c^2+2)}+\frac{((b+c)^2+1)}{(a^2+2)}+\frac{((c+a)^2+1)}{(b^2+2)} \geq 3$ Let $\displaystyle a, b, c> 0, ab+bc+ca=1$. Prove that the following inequality holds:
$$\frac{((a+b)^2+1)}{(c^2+2)}+\frac{((b+c)^2+1)}{(a^2+2)}+\frac{((c+a)^2+1)}{(b^2+2)} \geq 3.$$
I tried using the Cauc... | Hint: By CS inequality,$$\sum_{cyc} \frac{(a+b)^2}{(c^2+2)}+\sum_{cyc} \frac1{(c^2+2)}\ge \frac{4(a+b+c)^2+9}{(a^2+b^2+c^2)+6}$$
and then its enough to show
$$4(a+b+c)^2+9 \ge 3(a^2+b^2+c^2)+18 $$
which should be easy using the constraint given.
| {
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"question_score": "2",
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How would you solve $\int \frac{x}{x^2 - 4x + 5} dx$ What is the tip for integrating that integral? I completed the square on the bottom to make it $$\frac{x}{(x-2)^2 + 1}$$
but it doesn't seem helpful. Any tips?
Thanks.
| You can continue by letting $u=x-2$, $x=u+2$, $dx=du$ to get
$\displaystyle\int\frac{u+2}{u^2+1} du=\frac{1}{2}\int\frac{2u}{u^2+1} du+2\int\frac{1}{u^2+1} du=\frac{1}{2}\ln(u^2+1)+2\arctan u + C$
$\;\;=\frac{1}{2}\ln(x^2-4x+5)+2\arctan(x-2)+C$
| {
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Number of solutions of this trigonometric equation.
Q. Find the number of solutions of the equation $\sin(x) + 2\sin(2x) - \sin(3x) = 3$, in the interval $x\in (0,\pi)$.
I tried clubbing the $\sin(x)$ and $\sin(3x)$ terms together but got nothing. I also tried the $\sin(x)$ with $\sin(2x)$ and $\sin(2x)$ with $\sin... | We have
\begin{align}
2\sin 2x+\sin x-\sin 3x &=4\sin x\cos x-2\sin x \cos 2x=\\
&=\sin x(4\cos x -4\cos^2 x+2)=\\
&=\sin x \left[3-(2\cos x-1)^2\right].
\end{align}
Since $\sin x\ge 0$ in the interval $[0,\pi]$ and the second factor is bounded above by $3$, we must have simultaneously $\sin x=1$, $2\cos x=1$, which i... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Expressing a summation using matrix algebra Consider the $r \times n$ matrix
$$\begin{pmatrix}
X_{11} & X_{12} & \cdots & X_{1n} \\
X_{21} & X_{22} & \cdots & X_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
X_{r1} & X_{r2} & \cdots & X_{rn}
\end{pmatrix}\text{.}$$
Define
$$\begin{align*}
&\bar{X} = \dfrac{\sum\limits_... | Here are some expressions, although not necessarily easier to memorize!
Let $\mathbf{1}_{n}$ denote the column vector of all ones and length $n$,
and $\mathbf{I}$ the $n\times n$ identity matrix. Then,
$$
\hat{v}^{S} = \text{Tr}\left( \mathbf{X}^{T}\mathbf{X} \left(\mathbf{I}_{n}- \frac{1}{n}\mathbf{1}_{n}\mathbf{1}_{n... | {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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Prove that $\sum_{k=2}^{\infty} \frac{k^s}{k^{2s}-1}> \frac{3\sqrt{3}}{2}(\zeta(2s)-1),\space s>1$ What ways would you propose for getting the inequality below?
$$\sum_{k=2}^{\infty} \frac{k^s}{k^{2s}-1}> \frac{3\sqrt{3}}{2}(\zeta(2s)-1),\space s>1$$
The left side may be written as
$$\sum_{k=2}^{\infty} \frac{k^s}{k^... | From $n^s\ge3^s>2^s+1$ when $n\ge3$, we have
\begin{align}
&(\zeta(s)-1)-(2^s+1)(\zeta(2s)-1)\\&=\sum_{n=2}^\infty\frac1{n^s}\left(1-\frac{2^s+1}{n^s}\right)\\&=\sum_{k=1}^\infty\frac{1}{2^{ks}}\left(1-\frac{2^s+1}{2^{ks}}\right)+\sum_{\substack{n\ge3\\n\ne2^k}}\frac{1}{n^{s}}\left(1-\frac{2^s+1}{n^s}\right)\\&>\sum_{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/958142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
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Integrating $\cos^3(x)$ My attempt at integrating $\cos^3(x)$:
$$\begin{align}\;\int \cos^3x\mathrm{d}x &= \int \cos^2x \cos x \mathrm{d}x
\\&= \int(1 - \sin^2 x) \cos x \mathrm{d}x
\\&= \int \cos x dx - \int \sin^2x \cos x \mathrm{d}x
\\&= \sin x - \frac {1}{3}\sin^3x + C\end{align}$$
My question is how does integra... | You can use this bit of trickery.
$\cos 3x = 4\cos^3 x - 3\cos x\\
\cos^3x = \frac 14 \cos 3x + \frac 34\cos x$
And that is easy to integrate.
But, to your actual question. $\int \sin^2x\cos x \ dx$
let $u = \sin x, du = \cos x$
$\int u^2 \ du$
| {
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"source": "stackexchange",
"question_score": "3",
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"answer_id": 4
} |
Finding $\sum_{k=1}^{\infty} \left[\frac{1}{2k}-\log \left(1+\frac{1}{2k}\right)\right]$ How do we find $$S=\sum_{k=1}^{\infty} \left[\frac{1}{2k} -\log\left(1+\dfrac{1}{2k}\right)\right]$$
I know that $\displaystyle\sum_{k=1}^{\infty} \left[\frac{1}{k} -\log\left(1+\dfrac{1}{k}\right)\right]=\gamma$, where $\gamma$ is... | Notice that $$S=\sum^{\infty}_{k=1}\Big(\frac{1}{2k}-\frac{1}{2}\log(1+\frac{1}{k})+\frac{1}{2}\log(1+\frac{1}{k})-\frac{2}{2}\log(1+\frac{1}{2k})\Big)$$
Therefore
$$S=\frac{1}{2}\gamma+\frac{1}{2}\sum^{\infty}_{k=1}\log\Big(\frac{1+\frac{1}{k}}{(1+\frac{1}{2k})^2}\Big)$$
In other words
\begin{align}
S=\frac{1}{2}\gam... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/959413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 4
} |
A closed form for $\int_0^1 \frac{\left(\log (1+x)\right)^3}{x}dx$? I would like some help to find a closed form for the following integral:$$\int_0^1 \frac{\left(\log (1+x)\right)^3}{x}dx $$ I was told it could be calculated in a closed form. I've already proved that $$\int_0^1 \frac{\log (1+x)}{x}dx = \frac{\pi^2}{1... | First note that
\begin{align}
\int \frac{\ln^{2}(1+x)}{x} \, dx = - 2 \, Li_{3}(1+x) + 2 \, Li_{2}(1+x) \, \ln(1+x) + \ln(-x) \, \ln^{2}(1+x)
\end{align}
for which
\begin{align}
I_{2} = \int_{0}^{1} \frac{\ln^{2}(1+x)}{x} \, dx = \frac{\zeta(3)}{4}.
\end{align}
Now,
\begin{align}
\int \frac{\ln^{3}(1+x)}{x} \, dx = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/965504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 6,
"answer_id": 1
} |
Evaluating $ \sum\frac{1}{1+n^2+n^4} $ How to evaluate following expression?
$$ \sum_{n=1}^{\infty}\frac{1}{1+n^2+n^4}$$
I doubt it is a telescopic Sum.
| First, we have
$$
\begin{align}
\frac1{z^4+z^2+1}
&=\frac1{12}\left(
\frac{-3-i\sqrt3}{z-e^{\pi i/3}}
+\frac{3+i\sqrt3}{z-e^{4\pi i/3}}
+\frac{3-i\sqrt3}{z-e^{2\pi i/3}}
+\frac{-3+i\sqrt3}{z-e^{5\pi i/3}}
\right)\tag{1}
\end{align}
$$
Let $\gamma$ be the rectangle
$$
[-1-i,1-i]\cup[1-i,1+i]\cup[1+i,-1+i]\cup[-1+i,-1-i]... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/966942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 4
} |
How to calculate the value of the special integral I get
$${\left. \frac{\partial ^2}{\partial n^2} \left( \frac{\partial ^2}{\partial m^2} B(m,n) \right) \right|_{m = \frac{1}{2},n = 0}} = \int_0^1 \frac{\ln^2 x \ln^2 (1 - x)}{\sqrt x (1 - x)} \, dx =\text{ ?}$$
but how to calculate the value of beta function derivat... | For the integral:
$${\left. \frac{\partial ^2}{\partial n^2} \left( \frac{\partial ^2}{\partial m^2} B(m,n) \right) \right|_{m = \frac{1}{2},n = 0}} = \int_0^1 \frac{\ln^2 x \ln^2 (1 - x)}{\sqrt x (1 - x)} \, dx $$
Since
\begin{align}
\partial_{m} B(m,n) = B(m,n) \left[ \psi(m) - \psi(m+n) \right]
\end{align}
then
\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/967412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Finding $ \lim_{(x,y) \to (0,0)} \frac{\sin^2(xy)}{3x^2+2y^2} $ $$ \lim_{(x,y) \to (0,0)} \frac{\sin^2(xy)}{3x^2+2y^2} $$
If I pick $ x = 0$ I get:
$$ \lim_{(x,y) \to (0,0)} \frac{0}{2y^2} = 0$$
So if the limit exists it must be $0$
Now for ${(x,y) \to (0,0)}$ I have $xy \to 0$
So I can use the Taylor series of $sin(t)... | Or $\sin^2(xy)\leq(xy)^2$ and $3x^2\geq 2x^2$ thus $\frac{\sin^2(xy)}{3x^2+2y^2}\leq \frac {(xy)^2}{2(x^2+y^2)}=xy\cdot \frac {xy}{2(x^2+y^2)}\leq xy\cdot \frac {1}{4}\to 0$ because $\frac {xy}{x^2+y^2}\leq \frac {1}{2}\Leftrightarrow 2xy\leq x^2+y^2\Leftrightarrow 0\leq (x-y)^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/967716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
$n^{3} + 2n$ is divisible by $3$. Is this induction proof correct? Question: Prove by means of the principle of induction that for every $n ∈ N$ the
number $n^{3} + 2n$ is divisible by $3$.
Proof
Denote "$n^{3} + 2n$ is divisible by 3" by $P(n)$. Check $P(n)$ for an arbitrary $n$, for instance $n=1$. $1^{3}+2*1=3*1$ an... | I suggest a slight rewording.
Proof
Denote the statement
$n^{3} + 2n$ is divisible by 3
by $P(n)$. We check $P(n)$ for an $n=1$: $1^{3}+2*1=3*1$. Thus $P(1)$ holds.
Induction step: Assume $P(n)$ is is true, let $n ∈ N$. Then $(n+1)^{3}+2(n+1) =(n+1)(n^{2}+2n+1)+2(n+1)=(n+1)(n^{2}+2n+3)=n^{3}+2n^{2}+3n+n^{2}+2n+3=n^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/971599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
No. of integral solutions of $x_1+x_2+x_3+x_4=20.$ I've to solve a no. of questions of this type but don't get how to do it:
Determine the no. of integral solutions of $x_1+x_2+x_3+x_4=20.$ given the constraint that
$$1\leq x_1\leq 6,0\leq x_2\leq7,4\leq x_3\leq 8 ,2\leq x_4\leq 6.$$
I made the follow... | One method that leads to a general solution of these types of problems is to sum each geometric series and then deal with the resulting product. In this case, your generating function would be
$$\frac{x(1-x^6)}{1-x}\cdot
\frac{(1-x^8)}{1-x}\cdot
\frac{x^4(1-x^5)}{1-x}\cdot
\frac{x^2(1-x^5)}{1-x}.$$
Then, the coefficien... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/972894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Computing the limit of an expression I have the following question. Determine the limit of the following expression.
$\lim_{x \rightarrow 0} \frac{cos(x \sqrt{2}) - \frac{1}{1+x^2}}{x^4}$.
My attempt to this question is the following. Let the function $f(x) = \frac{cos(x \sqrt{2}) - \frac{1}{1+x^2}}{x^4} = \frac{\cos(x... | We have :
$$\mathop {\lim }\limits_{x \to {x_0}} \left( {f(x) + g(x)} \right) = \mathop {\lim }\limits_{x \to {x_0}} f(x) + \mathop {\lim }\limits_{x \to {x_0}} g(x) ~~~~ \left(*\right) $$
ONLY IF:
$$\mathop {\lim }\limits_{x \to {x_0}} f(x) ~ and
\mathop {\lim }\limits_{x \to {x_0}} g(x) ~ exist $$
Back to this pro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/975077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
When $a\to \infty$, $\sqrt{a^2+4}$ behaves as $a+\frac{2}{a}$? What does it mean that $\,\,f(a)=\sqrt{a^2+4}\,\,$ behaves as $\,a+\dfrac{2}{a},\,$
as $a\to \infty$?
How can this be justified?
Thanks.
| This is correct:
$$
\sqrt{a^2+4}=a\sqrt{1+\frac{4}{a^2}}=a+a\left(\sqrt{1+\frac{4}{a^2}}-1\right)=a+a\frac{1+\frac{4}{a^2}-1}{\sqrt{1+\frac{4}{a^2}}+1}\\=a+\frac{4}{a}\cdot\frac{1}{\sqrt{1+\frac{4}{a^2}}+1}\approx a+\frac{2}{a}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/975587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Solve $\sqrt{3x}+\sqrt{2x}=17$ This is what I did:
$$\sqrt{3x}+\sqrt{2x}=17$$
$$\implies\sqrt{3x}+\sqrt{2x}=17$$
$$\implies\sqrt{3}\sqrt{x}+\sqrt{2}\sqrt{x}=17$$
$$\implies\sqrt{x}(\sqrt{3}+\sqrt{2})=17$$
$$\implies x(5+2\sqrt{6})=289$$
I don't know how to continue. And when I went to wolfram alpha, I got:
$$x=-289(2\s... | I think your question is about $\sqrt{x}(\sqrt{3}+\sqrt{2})=17$. Then
$$
\begin{array}{rcl}
\sqrt{x}(\sqrt{3}+\sqrt{2}) & = & 17 \\
(\sqrt{x}(\sqrt{3}+\sqrt{2}))^2 & = & 17^2 \\
x(\sqrt{3}^2+2\sqrt{3}\sqrt{2}+\sqrt{2}^2) & = & 289 \\
x(3+2\sqrt{3}\sqrt{2}+2) & = & 289 \\
x(5+2\sqrt{3\times2}) & = & 289\\
x(5+2\sqrt{6})... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/977429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Double Integration Inequality I've been trying to work out the following. Could anyone please show me how to do this?
Let $D$ be the domain bounded by $y=x^2+1$ and $y=2$. Prove the inequality
$$\frac{4}{3}\le \iint_D(x^2+y^2)\,\mathrm{d}A\le\frac{20}{3}.$$
Thank you.
| First notice that in $D$ we have $1 \le x^2 + y^2 \le 5$. This implies $$ |D| \le \iint_Dx^2 + y^2 \le 5|D|$$
Now we only need to find $|D|$ ( = area of $D$), which is of course equal to $$4 - \int_{-1}^1(x^2 + 1) = 4 - 2\int_0^1(x^2 + 1) = 4 - \frac{2}{3} - 2 = \frac{4}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/978644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proof of Nesbitt's Inequality: $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}$? I just thought of this proof but I can't seem to get it to work.
Let $a,b,c>0$, prove that $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}$$
Proof: Since the inequality is homogeneous, WLOG $a+b+c=1$. So $b+c=1-a$, and... | Observe that
$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2} \iff (\frac{a}{b+c}+1)+(\frac{b}{c+a}+1)+(\frac{c}{a+b}+1)= (a+b+c)(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b})\ge \frac{9}{2}.$
By Titu's lemma
$\frac{1^2}{b+c}+\frac{1^2}{c+a}+\frac{1^2}{a+b}\ge \frac{(1+1+1)^2}{(a+b)+(b+c)+(c+a)} = \frac{9}{2(a+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/980751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 5
} |
Evaluating a series of hypergeometric functions I would like to prove (or disprove) the following statement:
$$
\sum_{n=0}^\infty \left[\frac{{}_2{\rm F}_1\left(\frac{1}{2},\frac{1-n}{2};\frac{3}{2};1\right)}{n!}\right] = \frac{\pi}{2} \left[ \sum_{a=0}^\infty \frac{1}{4^a (a!)^2} + \frac{1}{2} \sum_{b=0}^\infty \frac{... | Semiclassical has told me to post an answer to my own question using the technique i have explained to him, now for me to go any further i should state that the original function i am dealing with is:
$
{}_2F_1(\frac{1}{2}, \frac{1-n}{2}; \frac{3}{2};\cos^2 (x))
$
therefore using the power series of a hyper geometric... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/981633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
One Step in an Integration I was doing the integration
$$
\int\frac{1}{(u^2+a^2)^2}du
$$
and I had a look at the lecturer's steps where I got stuck in the following step:
$$
\int\frac{1}{(u^2+a^2)^2}du=\frac{1}{2a^2}\left(\frac{u}{u^2+a^2}+\int\frac{1}{u^2+a^2}du\right).
$$
I guess it is integrating this by parts, but ... | The trick is using the substitution $u = a\tan x $, Then we have
$$ \int \frac{du}{(u^2 +a^2)^2 } = \int \frac{du}{(a^2 \tan^2 x + a^2)^2}= \int \frac{du}{(a \sec x)^2}$$
and $ du = a \sec^2 x dx $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/982938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove $\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}\mathrm{d}\theta = \int_0^{2\pi}\frac{\cos \theta}{(1-a\cos \theta)^3}\,\mathrm{d}\theta$ While doing some mathematical modelling of planetary orbits I have come up with two definite integrals $D_1$ and $D_2$ which appear to produce the same result when I tes... | Hint:
Observe that,
$$\frac{\partial}{\partial a}\left[\frac{1}{\left(1-a\cos{\theta}\right)^2}\right]=\frac{2\cos{\theta}}{\left(1-a\cos{\theta}\right)^3}.$$
Thus, we can simplify the integral we have to compute via the technique of differentiating under the integral sign:
$$\begin{align}
I(a)
&=\int_{0}^{2\pi}\frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/983119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Condition for trigonometric inequality I want to prove the following statement:
Suppose $\frac{1}{4}(\cos(\theta_1)+\cos(\theta_2))^2+\lambda^2(a\sin(\theta_1)+b\sin(\theta_2))^2\leq 1$ holds for all $\theta_1,\theta_2\in[-\pi,\pi]$, then we should have $\lambda^2(a^2+b^2)\leq \frac{1}{2}$.
This problem appears when I ... | You want some bound for the expression $\lambda^2(a^2 + b^2)$.
We have
$$
\frac{1}{4}(\cos(\theta_1)+\cos(\theta_2))^2+\lambda^2(a\sin(\theta_1)+b\sin(\theta_2))^2\leq 1
$$
and with two other angles
$$
\frac{1}{4}(\cos(\theta_3)+\cos(\theta_4))^2+\lambda^2(a\sin(\theta_3)+b\sin(\theta_4))^2\leq 1
$$
Let us choose the a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/984216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
How to show that the integral of bivariate normal density function is 1? How to show the following?
$$\large \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{1}{2 \pi \sqrt{1-\rho^2}} e^{-\frac{x^2+y^2-2 \rho x y}{2(1-\rho^2)}} dx\ dy=1$$
| Complete the square of the exponent $x^2+y^2-2\rho xy = \left(x-\rho y\right)^2+y^2(1-\rho^2)=:u^2+y^2(1-\rho^2)$
$$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{1}{2 \pi \sqrt{1-\rho^2}} e^{-\frac{x^2+y^2-2 \rho x y}{2(1-\rho^2)}} dx dy=\\
=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{1}{2 \pi \sqrt{1-\rh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/985552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How can I find the complex numbers satisfying this condition? For a given complex number $a$ with $|a|\ge1,$ I want to find the all complex numbers on the unit circle such that $$\dfrac{z}{(a-z\bar a)^2}\in\mathbb{R}$$ and satisfying the condition $$\dfrac{z}{(a-z\bar a)^2}\le-\dfrac{1}{4}.$$
I express $z$ as $x+iy,$ ... | Write $a = re^{i\varphi}$ with $r > 0$ and $\varphi \in \mathbb{R}$. An easy computation shows that
$$f(z) = \frac{z}{(a-\overline{a}\cdot z)^2} = \frac{1}{r^2}\frac{e^{-2i\varphi}z}{(1-e^{-2i\varphi}z)^2} = \frac{1}{r^2} K(e^{-2i\varphi}z),$$
where
$$K(w) = \frac{w}{(1-w)^2}$$
is the Koebe function. It is relatively w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/987142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Does $\sum_{n=0}^\infty (-1)^n (e-(1+\frac{1}{n})^n)$ converge absolutely, conditionally, or diverge? I've been given the hint to use the binomial theorem and show that $e-\left(1+\frac{1}{n}\right)^n > \frac{1}{2n}$ for $n \geq 2$. So I've written
\begin{align*}
e-\left(1+\frac{1}{n}\right)^n = e - \sum_{k=0}^n {n \ch... | Converge absolutely? WRONG. Because as you said, $e-\left(1+\frac{1}{n}\right)^n > \frac{1}{2n}$.
Let us prove this inequality.
\begin{align*}e-\left(1+\frac{1}{n}\right)^n>{}&\left(1+\frac{1}{2n}\right)^{2n}-\left(1+\frac{1}{n}\right)^n \\
={}&\left(1+\frac{1}{n}+\frac{1}{4n^2}\right)^{n}-\left(1+\frac{1}{n}\right)^n\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/990668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find $a^2 + b^2+c^2$ Given
$a^2+2b = 7$
$b^2+4c = -7$
$c^2+6a = -14$
Find $a^2 + b^2 + c^2$
The answer was an Integer
I tried to solve it by making $a$ the subject of the equation and substituting in others but equations became too complex(got $a^8$!) and difficult to solve.
| HINT:
Completing the square,
$$(a+3)^2+(b+1)^2+(c+2)^2=7-7-14+9+1+4=0$$
Now if $a,b,c$ are real, what can we derive from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/991525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq \frac{9}{a+b+c}$ for positive $a,b,c$
Show that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq \frac{9}{a+b+c}$, if $a,b,c$ are positive.
Well, I got that $bc(a+b+c)+ac(a+b+c)+ab(a+b+c)\geq9abc$.
| Approach 1: Write the desired inequality as
$$
(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq 9\tag{i}
$$
and use the AM-GM inequality $x+y+z\geq 3(xyz)^{1/3}$ to each sum in the parentheses above.
Edit: just a few more approaches (among potentially many others). Hope you'll find this useful.
Approach 2:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/991885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Can we express the following ordinary generating function? I wish to express the following power series
$$ \sum_{k \ge 0} \binom{n-k}{m} x^k$$
where $n,m$ are positive integer such that $0< m \le n$
| $$
\begin{align}
\sum_{k=0}^\infty\binom{n-k}{m}x^k
&=\sum_{k=0}^n\binom{n-k}{m}x^k+\sum_{k=n+1}^\infty\binom{n-k}{m}x^k\\
&=\sum_{k=0}^n\binom{n-k}{m}x^k+\sum_{k=1}^\infty\binom{-k}{m}x^{k+n}\\
&=\sum_{k=0}^n\binom{n-k}{m}x^k+(-1)^m\sum_{k=1}^\infty\binom{m-1+k}{m}x^{k+n}\\
&=\sum_{k=0}^n\binom{n-k}{m}x^k+(-1)^m\sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/994619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to prove $\frac 1{2+a}+\frac 1{2+b}+\frac 1{2+c}\le 1$? How to prove this inequality ? $$\frac 1{2+a}+\frac 1{2+b}+\frac 1{2+c}\le 1$$
for $a,b,c>0 $ and $a+b+c=\frac 1a+\frac 1b+\frac 1c$.
I do not know where to start. I need some idea and advice on this problem.Thanks
| Let $\sum\limits_{cyc}\frac{1}{2+a}>1$ and $a=ka'$ such that $k>0$ and
$$\frac{1}{2+a'}+\frac{1}{2+b}+\frac{1}{2+c}=1.$$
Hence,
$$\frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c}>1=\frac{1}{2+a'}+\frac{1}{2+b}+\frac{1}{2+c}$$ or
$$\frac{1}{2+ka'}>\frac{1}{2+a'},$$
which gives $k<1$.
In another hand,
$$a'+b+c-\frac{1}{a'}-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1000451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Help with epsilon delta proof $x^3$ is near $27$ when $x$ is near $3$ but $x$ is not equal to $3$.
So I have
$$0<|x-3|<\delta \implies 3-\delta<x<3+\delta$$
$$|x^3-27|<\epsilon \implies|(x-3)(x^2+3x+3^2)|<\epsilon$$
$$=(x^2+3x+9)|(x-3)|<\epsilon \implies28|(x-3)|<\epsilon$$
$$=|(x-3)|<\epsilon/28$$
How do I prove $x^2+... | You can continue by saying that $|(x-3)(x-3+9)|+27\le|x-3|(|x-3|+9)+27<\delta(\delta+9)+27$,
so now you just have to make sure that $\delta(\delta+9)\le1$ by, say, choosing $\delta$ so that it satisfies $\delta\le\frac{1}{18}$
since then $\delta^2\le\frac{1}{2}$ and $9\delta\le\frac{1}{2}$
(along with any other condi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1001686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Integration of $ \int \frac{2\sin x + \cos x}{\sin x + 2\cos x} dx $ How can I integrate this by changing variable?
$$ \int \frac{2\sin x + \cos x}{\sin x + 2\cos x} dx $$
Thanks.
| Let $a$ be such that $(2a)^2+a^2=1$. Let $\theta$ be such that $\cos(\theta)=a$ and $\sin(\theta)=2a$. Then you have
$$
\begin{align}
\int \frac{2\sin x + \cos x}{\sin x + 2\cos x} dx
&=\int \frac{2a\sin x + a\cos x}{a\sin x + 2a\cos x} dx\\
&=\int \frac{\sin(\theta)\sin x + \cos(\theta)\cos x}{\cos(\theta)\sin x + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1004318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
The limit: $\lim_{x\rightarrow0}\frac{1-\cos2x}{\sin3x\cdot\ln(1+\sin4x)}$ $$\lim_{x\rightarrow0}\frac{1-\cos2x}{\sin3x\cdot\ln(1+\sin4x)}$$
My steps
$\lim_{x\rightarrow0}\frac{1-\cos2x}{\sin3x\cdot\ln(1+\sin4x)} = \lim_{x\rightarrow0}\frac{1-\cos2x}{2x}\cdot2x\cdot\frac{3x}{\sin3x}\cdot\frac{1}{3x}\cdot\frac{1}{\ln(1+... | We have $1-\cos(2x)\sim \frac{(2x)^2}{2}=2x^2$ as $x\to 0$; $\sin(3x)\sim 3x$, $\ln(1+\sin(4x))\sim \sin(4x)\sim 4x$ as $x\to 0$. Thus
$$\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos \left( {2x} \right)}}{{\sin \left( {3x} \right).\ln \left( {1 + \sin \left( {4x} \right)} \right)}} = \mathop {\lim }\limits_{x \to 0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1005961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solution for equation with inclusion-exclusion principle. Using the principle of inclusions and exclusions count how many solutions to the equation
$$ x + y + z = 12 $$
$$ 1 \le x \le 5$$ $$ -2 \le y \le 4$$ $$ 0 \le z \le 5 $$
$$ x,y,z \in \mathbb{Z}$$
| Let $a = x - 1, b = y + 2, c = z$. Then our equation becomes $a+b+c=13 \ (\star)$ with constraints: $$\begin{align*}
0&\le a \le 4 \\
0&\le b \le 6 \\
0&\le c \le 5
\end{align*}$$
Let $U$ be the set of all nonnegative integer solutions of $(\star)$. This is $|U| = \binom{15}{2}.$
Let the following sets be subsets of $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1006780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What's wrong with the calculation with polar coordinates here? Suppose $x=r\cos t$ and $y=r\sin t$. I did the following calculation:
$$
\begin{align}
&x^4+y^4=(r\cos t)^4+(r\sin t)^4\\
=&r^4(\sin^4t+\cos^4t)\\
=&r^4[(\sin^2t+\cos^2t)^2-2\sin^2t\cos^2t]\\
=&r^4(1-2\sin^2t\cos^2t)=r^4-2r^4\sin^2t\cos^2t
\end{align}
$$
On... | Notice that $2x^2y^2 = 2(r\sin t)^2(r\cos t)^2 = 2r^4\sin^2t\cos^2t$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1007903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find $a$, $b$ and $c$ in $\frac{e^{ax}}{2+bx}=\frac{1}{2}+\frac{x^2}{4}-cx^3$
Find the values of the positive constants $a$, $b$ and $c$ given that when $x$ is sufficiently small for terms in $x^4$, and higher powers of $x$, to be neglected then:
$$
\frac{e^{ax}}{2+bx}=\frac{1}{2}+\frac{x^2}{4}-cx^3 \space\space\tex... | Hint: Don't divide, Multiply!
$$\frac{e^{ax}}{2+bx}=e^{ax}(2+bx)^{-1}=\left(1+ax+\frac{a^2x^2}{2}+\frac{a^3x^3}{6}+\cdots\right)\left(\frac{1}{2}-\frac{bx}{4}+\frac{b^2x^2}{8}-\frac{b^3x^3}{16}\cdots\right)$$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1013253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Related Rates - Growth of the side of a triangle recently I saw a question on a book but I personally don't agree with the answer. I would appreciate if you could help me achieve the final result. It follows:
Two sides of a triangle measure $12m$ and $15m$.
The angle between the two sides grow at the rate of $2^o/min$... | Let $x=x(t)$ be the third side, and $\theta$ the angle between the first two sides. Using the Law of Cosines,
when $\theta = 60^\circ$
$$x^2 = 15^2+12^2- 2 \cdot 15 \cdot 12 \cdot \cos(60^\circ)$$
$$x = 3 \cdot \sqrt{21}$$
For the Related Rate,
$$[x^2]' = 0-2 \cdot 15 \cdot 12 [ \cos \theta]'$$
$$2x \cdot x' = 2 \cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1014051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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.