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Problems based on triangles and trigonometry. In an acute triangle $x,y,z$ are the given angles where $\cos x=\tan y$, $\cos y = \tan z$ and $\cos z = \tan x$. Find the sum of sines in the triangle.
Could be done by substituting values of $\sin$ function but in vain. Can anyone please help me?
| The question states that it's given that
in an acute triangle
$x,y,z$ are the given angles where
\begin{align}
\cos x&=\tan y
\tag{1}\label{1}
,\\
\cos y&=\tan z
\tag{2}\label{2}
,\\
\cos z&=\tan x
\tag{3}\label{3}
.
\end{align}
There are many ways to prove that
there is no valid triangle with such properties.
For... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3682606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find the radius of the circle given in the picture below. This is the image of the question. I am not able to get how to find the radius. Please help with that.
This is my try. I can't proceed now after it.
Thanks
|
Given: $CD=2,\,BD=3,\,AB=\sqrt{11}$.
Let's define $\angle B=\angle DBE,\, CE=d$.
$2R=\frac{\sqrt{11}}{\cos A}$, $BE=\frac{\sqrt{11}}{\sin A}, \angle DBE=\angle DAE$, thus $CBE\sim CAD$
$$\frac{CB}{CA}=\frac{BE}{AD}=\frac{CE}{CD}$$
$$\frac{5}{d+2R}=\frac{\frac{\sqrt{11}}{\sin A}}{AD}=\frac{d}{2}$$
By applying the cosin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3687003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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Integral solution to a³+3ab²=4c³ Please help if integral solutions to this equation exists or not
A³+3AB²=4C³
Such that A,B,C are disninct
I thought we may be possible to prove solutions exists if and only if A=B=C condition is satisfied.
| Note that
$$(A-B)^3=A^3-3A^2B+3AB^2-B^3 \\
(A+B)^3=A^3+3A^2B+3AB^2+B^3$$
This gives
$$A^3+3AB^2= \frac{1}{2} \left((A+B)^3+(A-B)^3 \right)$$
Therefore, if you have a solution
$$(A-B)^3+(A+B)^3=(2C)^3$$
Use FLT to deduce that all solutions satisfy $A=B$ or $A=-B$ or $A=0$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Technique for simplifying, e.g. $\sqrt{ 8 - 4\sqrt{3}}$ to $\sqrt{6} - \sqrt{2}$ How to find the square root of an irrational expression, to simplify that root. e.g.:
$$
\sqrt{ 8 - 4\sqrt{3} } = \sqrt{6} - \sqrt{2}
$$
Easy to verify:
\begin{align}
(\sqrt{6} - \sqrt{2})^2 = 6 - 2\sqrt{12} +2
= 8 - 4 \sqrt{3}
\end{alig... | If I recall correctly, you make the assumption that your expression takes the form $\sqrt{a}\pm\sqrt{b}$:
$\sqrt{8-4\sqrt{3}} = \sqrt{a}\pm\sqrt{b}$
$8-4\sqrt{3} = 8-\sqrt{48} = a\pm2\sqrt{ab}+b$
We can see that the irrational part must be assigned the negative sign.
Equating rational and irrational parts:
$a+b = 8$
$-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3689064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to evaluate $\lim_{n \to \infty} \sqrt[n]{\frac{n^3}{2^n + 5^n}}$? I need to resolve the convergence of $$\sum_{n = 1}^{\infty} \frac{n^3}{2^n + 5^n}$$ where $n \in \mathbb{N}$.
Mainly, I need advice concerning Cauchy's convergence criterion because I just can't evaluate
$$\lim_{n \to \infty} \sqrt[n]{\frac{n^3}{2^... | Just like started, first we switch to the exponent of $e$:
$$\lim_{n\rightarrow\infty}(\frac{n^3}{2^n+5^n})^{1/n} = \exp{(\lim_{n\rightarrow\infty}\frac{\ln(\frac{n^3}{2^n+5^n})}{n}})$$
Using l'Hospitals rule we need to evaluate the derivative of the numerator:
$$\lim_{n\rightarrow\infty}(\frac{n^3}{2^n+5^n})^{-1}\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3689715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Solve PDE $xzu_x+yzu_y-(x^2+y^2)u_z=0$ using characteristics method I was asked to find the characteristics of the following EDP and then to solve it.
$$xzu_x+yzu_y-(x^2+y^2)u_z=0$$
I've reached the following characteristic system $$x'=-\frac{xz}{x^2+y^2}\qquad y'= -\frac{yz}{x^2+y^2}
$$ but i dont know how to continue... | Multiplying the first equation by $x$ and the second one by $y$ gives
$$
\tfrac12 (x^2)' = -\frac{x^2 z}{x^2 + y^2}, \qquad \tfrac12 (y^2)' = -\frac{y^2 z}{x^2 + y^2} .
$$
The sum leads to $(r^2)' = -2z$ where $r^2 = x^2 + y^2$. Multiplying the first equation by $y$ and the second one by $x$ gives
$$
yx' = -\frac{xy z}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3691069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Need to find maximum in order to prove that the function converges uniformly. So, here is the function: $f_n = \sqrt{n}\left(\sqrt{x+\frac{1}{n}}-\sqrt{x}\right)$
I found that the limit for it is $0$. Now I want to show whether it converges uniformly. So first I tried to find a maximum.
$$f'_n(x) = \frac{1}{2}\sqrt{n}\... | Your sequence of functions converges pointwise to $0$ on $\mathbb{R}_+^*$. Thus, it converges uniformly to zero if $||f_n||_{\infty}= \sup_{x}|f_n(x)| \to 0$. But $f_n(x) \to 1$ as $x \to 0$, so $||f_n||_{\infty}\geqslant 1$ for all $n$ and $(f_n)$ does not converge uniformly
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why does Stolz- Cesaro fail to evaluate the limit of $\dfrac{n + n^2 + n^3 + n^4 + \ldots + n^n}{1^n + 2^n + 3^n + 4^n + \ldots +n^n}$, I need to find the limit of the sequence
$\dfrac{n + n^2 + n^3 + n^4 + \ldots + n^n}{1^n + 2^n + 3^n + 4^n + \ldots +n^n}$,
My strategy is to use Stolz's Cesaro theorem for this seque... | Divide nunberator and denominator by $n^n$. So your question consists of two limits, numerator and denominator, we'll deal with them separately.
For the numerator the limit would become $lim_{n \to \infty} 1+\frac{1}{n}+\ldots+\frac{1}{n^n} = 1*\frac{(1/n)^{n+1}-1}{(1/n)-1} = \lim_{h \to 0} \frac{h^{1+1/h} -1}{h-1} = -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3695175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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How to factor a polynomial with complex roots of the form $a+bi$, where $a\neq 0$? (Not just find the root) For example, the quadratic formula reveals that the roots of $x^2 - 4x + 5$ are $x = 2\pm 2i $
But how do we use these roots to actually factor $x^2 - 4x + 5$?
My best guess was that, since $x = 2\pm 2i $, we wo... | You got the wrong roots in the first place. The roots are $2 \pm i$, please check your work (application of the quadratic formula).
Once you correct that, your factorisation will be correct. But you made another mistake in expanding the product of the factors.
$(x-2-i)(x-2+i) = (x-2)^2 - i(x-2)+i(x-2) +(i)(-i) = (x-2)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3698321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Using the Maclaurin series for $\frac{1}{1-x}$ to find $\frac{x}{1+x^2}$ Suppose I know the Maclaurin series for $$\frac{1}{1-x}=1+x+x^2+x^3+...= \sum_{n=0}^{\infty}x^n \tag{1}$$
then I can find the Maclaurin series for $\frac{1}{(1-x)^2}$ by the substitution $x\to x(2-x)$, which is obtained by solving the following eq... | Because
$$u^2+1=ux(1-x)$$
$$u^2-ux(1-x)+1=0$$
has no real solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3703913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Limit of $\frac{\cos{x}-1-\frac{x^2}{2}}{x^4+y^4}$ as it goes to the origin How can I solve a limit like this one:
$$\lim_{(x,y) \rightarrow (0,0)}\frac{\cos{x}-1-\frac{x^2}{2}}{x^4+y^4}$$
If I do $x=0$ so it aproaches through $y$ or $y=0$ so it approaches through $x$ it doesn't really help because of the division over... | If $x = 0$ there is no division by zero ! Since when $x = 0$
$$f(x ,y) = \frac{\cos(x) - 1 -x^2/2}{x^4 + y^4} = \frac{1-1 - 0}{y^4} = 0$$
Therefore
$$ \lim_{y \to 0} f(0,y) = \lim_{y \to 0} 0 = 0.$$
If $y = 0$ then
$$ f(x,0) = \frac{\cos(x)-1-x^2/2}{x^4}.$$
Since the numerator and the denominator both converge to $0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3704021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
For any $n ≥ 5,$ the value of $1+ \frac{1}2 + \frac{1}3+···+\frac{1}{2^n −1}$ lies between
QUESTION: For any $n ≥ 5,$ the value of $$1+ \frac{1}2 + \frac{1}3+···+\frac{1}{2^n −1}$$ lies between
$(A)$ $0$ and $\frac{n}2$
$(B)$ $\frac{n}2$ and $n$
$(C)$ $n$ and $2n$
$(D)$ none of the above.
MY APPROACH: This is what I... | Divide the sum into blocks with denominators running from $2^m$ through $2^{m+1}-1$, where $0\le m<n$:
$$\begin{align*}
\sum_{k=1}^{2^n-1}\frac1k&=\underbrace{1}+\underbrace{\frac12+\frac13}+\underbrace{\frac14+\frac15+\frac16+\frac17}+\ldots+\underbrace{\frac1{2^{n-1}}+\ldots+\frac1{2^n-1}}\\
&=\sum_{m=0}^{n-1}\sum_{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3704520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $(a^2+b^2+c^2)^3 \geqq 9(a^3+b^3+c^3)$ For $a,b,c>0; abc=1.$ Prove$:$ $$(a^2+b^2+c^2)^3 \geqq 9(a^3+b^3+c^3)$$
My proof by SOS is ugly and hard if without computer$:$
$$\left( {a}^{2}+{b}^{2}+{c}^{2} \right) ^{3}-9\,abc \left( {a}^{3}+{b} ^{3}+{c}^{3} \right)$$
$$=\frac{1}{8}\, \left( b-c \right) ^{6}+{\frac {117... | Yes, SOS helps:
$$(a^2+b^2+c^2)^3-9(a^3+b^3+c^3)=(a^2+b^2+c^2)^3-9abc(a^3+b^3+c^3)=$$
$$=\frac{1}{2}\sum_{cyc}(2a^6+6a^4b^2+6a^4c^2-18a^4bc+4a^2b^2c^2)=$$
$$=\frac{1}{2}\sum_{cyc}(2a^6-a^4b^2-a^4c^2+7a^4b^2+7a^4c^2-14c^4ab-4a^4bc+4a^2b^2c^2)=$$
$$=\frac{1}{2}\sum_{cyc}(a-b)^2((a+b)^2(a^2+b^2)+7c^4-2abc(a+b+c))=$$
$$=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3704808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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If $\frac{t^2-x^2}{(t^2+x^2)^2}+\frac{(1-x)^2-t^2}{((1-x)^2+t^2)^2}=0$ for real $t$ and $0
$$\frac{t^2-x^2}{(t^2+x^2)^2}+\frac{(1-x)^2-t^2}{((1-x)^2+t^2)^2}=0$$
where $0<x<1$ and $t\in\mathbb{R}$. Prove that $x=1/2$.
It is evident that $x=1/2$ satisfies the above equation. Please help.
| Seems the fastest way to show this is indeed considering the polynomial left after clearing the denominator and dividing by $2(x-\frac{1}{2})$
$$f(x)=-3 t^4 - 2 t^2 x^2 + 2 t^2 x - t^2 + x^4 - 2 x^3 + x^2$$
Considering $$f'(x)= -2 (2 x - 1) (t^2 - x^2 + x)$$ we see that as $(t^2+x(1-x))>0$ for $0<x<1$ thus the function... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What does it mean for an ODE to be conservative? What does it mean for an ODE to be conservative?
For example, I already read somewhere that the equation
$$w\cdot y''-y+y^{2k+1}=0,$$
with $w>0$ and $k\in \mathbb{N}$ constants fixeds, is conservative. In practice, what does this mean?
| Given the differential equation
$wy'' - y + y^{2k + 1} = 0, \tag 1$
we may multiply it through by $y'$:
$wy'y'' - yy' + y^{2k + 1}y' = 0, \tag 2$
and observe that
$\left ( \left ( \dfrac{w}{2} y' \right )^2 \right )' = wy''y', \tag 3$
and
$\left ( -\dfrac{y^2}{2} + \dfrac{y^{2k + 2}}{2k + 2} \right )' = - yy' + y^{2k +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3708674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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$p^2+1=q^2+r^2$. Strange phenomenon of primes Problem:
Find prime solutions to the equation
$p^2+1=q^2+r^2$
I welcome you to post your own solutions as well
I have found a strange solution which I can't understand why it works(or what's the math behind it.) Here it is through examples
Put $r=17$(prime)
Now $17^2-1=16\t... | This seems to hold much more broadly than just for primes.
Let$$r^2-1=ab$$where $b>a$, and let $p=\frac{b+a}{2}$ and $q=\frac{b-a}{2}$.
Then the equation $p^2+1=q^2+r^2$ becomes$$\left(\frac{b+a}{2}\right)^2+1=\left(\frac{b-a}{2}\right)^2+ab+1$$and we have$$\frac{b^2+2ab+a^2}{4}+1=\frac{b^2-2ab+a^2+4ab}{4}+1=\frac{b^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3710018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to Prove : $ \gamma +\ln\left(\frac{\pi}{4}\right) = \sum_{n=2}^{\infty} \frac{(-1)^{n} \zeta{(n)}}{2^{n-1}n} $ How to Prove :
$$ \gamma +\ln\left(\frac{\pi}{4}\right) = \sum_{n=2}^{\infty} \frac{(-1)^{n} \zeta{(n)}}{2^{n-1}n} $$
I have tried looking at Series definitions of the Polygamma function from which we ca... |
Lemma:
Let $f(z)=\sum_{n=2}^{\infty} a_nz^n$ be convergent with radius $>1.$ Then:
$$\sum_{n=2}^{\infty} a_n\zeta(n)=\sum_{k=1}^{\infty} f\left(\frac1k\right)$$
Proof:
$$\begin{align}\sum_{n=2}^{\infty} a_n\zeta(n)&=\sum_{n=2}^{\infty} a_n\sum_{k=1}^{\infty} \frac{1}{k^n} \\
&=\sum_{k=1}^{\infty}\sum_{n=2}^{\infty}a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3712992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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The summation $\sum_{n\geqslant1} \frac1n\sum_{d\mid n}\frac{d}{n^2+d}.$
I wish to evaluate $\sum\limits_{n\geqslant1}\frac1n\sum\limits_{d\mid n}\frac{d}{n^2+d}.$
Some observations:
Let $f(n)=\sum\limits_{d\mid n}\frac{d}{n^2+d}$.
Then $f(p)=\frac{p^2+p+2}{(p^2+1)(p+1)},\ f(2)=\frac8{15},\ f(4)=\frac{283}{765}.\ \... | By switching the order of summation and by applying the inverse Laplace transform we have
$$S=\sum_{n\geq 1}\sum_{d\mid n}\frac{d}{n(n^2+d)} = \sum_{d\geq 1}\sum_{k\geq 1}\frac{1}{k(k^2 d^2+d)}=\int_{0}^{+\infty}\frac{ds}{e^s-1}\sum_{k\geq 1}\frac{1-e^{-s/k^2}}{k}\,ds=\sum_{k\geq 1}\frac{H_{1/k^2}}{k} $$
and now we may... | {
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"timestamp": "2023-03-29T00:00:00",
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What kind of inequality is this? What kind of inequality is this?
$|z^2 + 1|$ is greater than or equal to $|z|^2 - 1$
why $-1$? , and why not $-100\,000$?
does $z$ have to be a complex variable?
correction : sorry it was not $|z^2|$, it was $|z|^2$
| In the reals,
$$z^2+1\ge z^2$$ holds (and the modulus is not even necessary).
But in the complex, a counter-example is
$$|i^2+1|=|i^2|-1=0.$$
Let $z^2=a+ib$ and let us look for the minimum of
$$\sqrt{(a+1)^2+b^2}-\sqrt{a^2+b^2}.$$
We cancel the gradient,
$$\begin{cases}\dfrac{a+1}{\sqrt{(a+1)^2+b^2}}=\dfrac{a}{\sqrt{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3717821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is there a formula for adding say: $A\sin^{-1}C-B\sin^{-1}D$ where A, B, C and D are constants? I'm looking for a formula or a method that can deal with the case when $A\ne B$.
$$A\sin^{-1}C-B\sin^{-1}D$$
There is a formula for when $A=B$:
$$\sin^{-1}(C)-\sin^{-1}(D)=\sin^{-1}(C\sqrt{1-D^2}-D\sqrt{1-C^2})$$
The specifi... | I don’t know about the general case, but for your specific case you can do the following:
First notice that $$ \sin(4x) = 2\sin 2x\cos 2x \\= 4\sin x \cos x (1-2\sin^2 x) \\= 4\sin x(1-2\sin^2x)\sqrt{1-\sin^2 x} $$
So if $4\sin^{-1} x = y$, then
$$\sin(4\sin^{-1} x) = \sin y = 4x(1-2x^2)\sqrt{1-x^2} \\ \implies y=\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3720988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why $\int_{ \mathbb{R}^2 } \frac{dx\,dy }{(1+x^4+y^4)} $ converges? Why $\int_{ \mathbb{R}^2 } \frac{dx\,dy }{(1+x^4+y^4)} $ converges?
Apparently this integral is quite similar to the integral
$\iint_{\mathbb R^2} \frac{dx \, dy}{1+x^{10}y^{10}}$ diverges or converges?
and it converges.
So this is quite remarkable tha... | Answer. Yes.
Note that
$$
1+x^4+y^4\ge 1+\frac{1}{2}(x^2+y^2)^2
$$
and hence, using polar coordinates ($x=r\cos\theta, \,y=r\sin\theta$), we have
$$
\int_{\mathbb R^2}\frac{dx\,dy}{1+x^4+y^4}\le \int_{\mathbb R^2}\frac{dx\,dy}{1+\frac{1}{2}(x^2+y^2)^2}=
\int_0^{2\pi}\int_0^\infty\frac{r\,dr\,d\theta}{1+\frac{1}{2}r^4}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3727789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to Prove $\int_{0}^{\infty}\frac {1}{x^8+x^4+1}dx=\frac{π}{2\sqrt{3}}$ Question:- Prove that
$\int_{0}^{\infty}\frac {1}{x^8+x^4+1}dx=\frac{π}{2\sqrt{3}}$
On factoring the denominator we get,
$\int_{0}^{\infty}\frac {1}{(x^4+x^2+1)(x^4-x^2+1)}dx$
Partial fraction of the integrand contains big terms with their long... | Taking $x\mapsto \frac{1}{x}$ transforms the integral into $I= \int_0^{\infty} \frac{x^6}{1+x^4+x^8} d x$ and then taking the average of them gives $$
\begin{aligned}
I & =\frac{1}{2} \int_0^{\infty} \frac{x^6+1}{x^8+x^4+1} d x \\
& =\frac{1}{2} \int_0^{\infty} \frac{\left(x^2+1\right)\left(x^4-x+1\right)}{\left(x^4+x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3728939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Integrate $\int \frac{x^2-1}{x^3 \sqrt{2x^4-2x^2+1}} \mathop{dx}$ Evaluate $$\int \frac{x^2-1}{x^3 \sqrt{2x^4-2x^2+1}} \mathop{dx}$$
I tried $u=\sqrt{2x^4-2x^2+1}$, $u=\dfrac{1}{x}$ and $u=\sqrt{x}$ but none of these worked.
A friend gave this and said it's from IIT JEE and answer is $$\dfrac{\sqrt{2x^4-2x^2+1}}{2x^2}+... | Let $I=\int \frac{x^2-1}{x^3 \sqrt{2x^4-2x^2+1}} \mathop{dx}$. Divide Numerator and denominator of integrand by $x^2$ to get: $I=\int \frac{x^{-3}-x^{-5}}{\sqrt{2-2x^{-2}+x^{-4}}} \mathop{dx}$. Substitute $y=2-2x^{-2}+x^{-4}$ so that $dy=4(x^{-3}-x^{-5})dx$ and hence $I=\int \frac{1}{4 \sqrt{y}} \mathop{dy}=\frac{y^{1/... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Integrate $\int\frac{2\cos x-\sin x}{3\sin x+5\cos x }\,dx$ How can I evaluate this integral $$\int\dfrac{2\cos x-\sin x}{3\sin x+5\cos x } \mathop{dx}$$
I tried using the half angle formula
$$\sin x=\dfrac{2\tan\dfrac x2}{1+\tan^2\dfrac x2}, \cos x=\dfrac{1-\tan^2\dfrac x2}{1+\tan^2\dfrac x2}$$
substituted and simplif... | The trick that I use for solving integrals with $\sin{x}$ and $\cos{x}$ with varying coefficients in the numerator and denominator as follows.
For your integral, consider the two easy integrals:
\begin{align*}
\int \frac{3\sin{x}+5\cos{x}}{3\sin{x}+5\cos{x}} \; \mathrm{d}x&=x+\mathrm{C}\\
\int \frac{ 3\cos{x}-5\sin{x}}... | {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Solve for $y$ in $\frac{dy}{dx}-\frac{3y}{2x+1}=3x^2$ I saw a challenge problem on social media by a friend, solve for $y$ in $$\frac{dy}{dx}-\frac{3y}{2x+1}=3x^2$$
I think this is an integration factor ODE
$$\frac{1}{{(2x+1)}^{\frac{3}{2}}} \cdot \frac{dy}{dx}-\frac{3y}{{(2x+1)}^{\frac{5}{2}}}=\frac{3x^2}{{(2x+1)}^{\f... | Since the ODE is linear $$(2x+1)y'-3y=3x^2(2x+1)$$
We can proceed without integration factor solving homogeneous equation first:
$\dfrac{y'}{y}=\dfrac{3}{2x+1}\ $ gives $\ y=C\,(2x+1)^{3/2}$
And then find a polynomial of degree $3$ satisfying RHS:
$(2x+1)(3ax^2+2bx+c)-3(ax^3+bx^2+cx+d)=3x^2(2x+1)\iff\begin{cases}3a-6=0... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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Asymptotic analysis of $\sum_{n=-\infty}^\infty \tan^{-1} \left(\frac{D}{2n+1}\right) \log\left(\frac{D}{|2n+1|}\right) \frac{1}{n+3/4}$ For large positive constant $D$, I want an asymptotic evaluation of the sum
$$\sum_{n=-\infty}^\infty \tan^{-1} \left(\frac{D}{2n+1}\right) \log\left(\frac{D}{|2n+1|}\right) \frac{1... | We follow the analogous development in this answer. Let $S(D)$ be given by
$$\begin{align}
S(D)&=\sum_{n=-\infty}^\infty \frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{|2n+1|}\right)}{n+3/4}\\\\&=\sum_{n=0}^\infty \frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{2n+1}\right)}{n+3/4}+\sum_{n=-... | {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Find the sum of all positive integers $n$ such that when $1^3+2^3+3^3 +\dots+ n^3$ is divided by $n+5$ the remainder is $17.$
Find the sum of all positive integers $n$ such that when $1^3+2^3+3^3 +\dots+ n^3$ is divided by $n+5$ the remainder is $17.$
Letting $k= n+5$ we get that $1^3+2^3+3^3 +\dots+ (k-5)^3 \equiv 1... | If $S_m=\sum_{r=1}^mr^3$
$$2S_{n+4}=\sum_{r=1}^{n+4}r^3+(n+5-r)^3\equiv0\pmod{n+5}$$ as $r^3+(n+5-r)^3\equiv0\pmod{n+5}, 1\le r\le n+4$
$$\implies2S_n\equiv-2[(n+4)^3+(n+3)^3+(n+2)^3+(n+1)^3]\pmod{n+5}$$
$$\equiv-2[(n+5-1)^3+(n+5-2)^3+(n+5-3)^3+(n+5-4)^3]$$
$$\equiv2(1^3+2^3+3^3+4^3)$$
$$\implies2S\equiv200\pmod{n+5}\t... | {
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"url": "https://math.stackexchange.com/questions/3737447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
prove $\sum\cos^3{A}+64\prod\cos^3{A}\ge\frac{1}{2}$ In every acute-angled triangle $ABC$,show that
$$(\cos{A})^3+(\cos{B})^3+(\cos{C})^3+64(\cos{A})^3(\cos{B})^3(\cos{C})^3\ge\dfrac{1}{2}$$
I want use Schur inequality
$$x^3+y^3+z^3+3xyz\ge xy(y+z)+yz(y+z)+zx(z+x)$$
then we have
$$x^3+y^3+z^3+6xyz\ge (x+y+z)(xy+yz+zx)$... | HINT
Firstly, let
$$\tan \dfrac A2 = x,\quad \tan \dfrac B2 = y,\quad s=(x+y)^2,\quad p=xy,\quad x,y \in(0,1),\tag1$$
then
\begin{align}
&\tan\dfrac C2 = \cot\left(\dfrac A2+\dfrac B2\right) = \dfrac{1-xy}{x+y} = \dfrac{1-p}{\sqrt s}\in(0,1),\\[4pt]
&\cos A = \dfrac{1-x^2}{1+x^2},\quad \cos B = \dfrac{1-y^2}{1+y^2},\qu... | {
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Doubt about how to compute $\sum_{n=1}^{2019}{\frac{1+2+3+4...+n}{1^3+2^3+3^3...+2019^3}}$ In the book I am reading, I have encountered the following sum.
$$S = \sum_{n=1}^{2019}{\frac{1+2+3+4...+n}{1^3+2^3+3^3...+2019^3}}$$
From here, I factored the denominator since it does not seem to be dependent on $n,$ and I rewr... | Like you mentioned, the denominator is independent of $n,$ hence we can consider it separately. We have that $D = 1^3 + 2^3 + \cdots + 2019^3 = \sum_{i = 1}^{2019} i^3.$ Observe that $S$ can be written as $$\begin{align*} S = \frac 1 D \sum_{n = 1}^{2019} (1 + 2 + \cdots + n) &= \frac 1 D \sum_{n = 1}^{2019} \sum_{k = ... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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prove formula by induction I need to prove by induction that $\sum_{n=2}^{m} \frac{1}{n^2 - 1} = \frac{1}{2}(1+\frac{1}{2}-\frac{1}{m}-\frac{1}{m+1})$.
I've seen some similar questions about the convergence of the infinite series, however none about the finite case, I've tried in several ways to factorize or add things... | Note: If
$\sum_{n=2}^{m} \frac{1}{n^2 - 1} = \frac{1}{2}(1+\frac{1}{2}-\frac{1}{m}-\frac{1}{m+1})$
Then $\sum_{n=2}^{m+ 1} \frac{1}{n^2 - 1} = (\sum_{n=2}^{m} \frac{1}{n^2 - 1}) + \frac {1}{(m+1)^2 - 1} =$
$(\frac{1}{2}(1+\frac{1}{2}-\frac{1}{m}-\frac{1}{m+1})) + \frac 1{(m^2 + 2m +1)-1}=$
$(\frac{1}{2}(1+\frac{1}{2}... | {
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"source": "stackexchange",
"question_score": "1",
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how to integrate $\int\frac{1}{x^2-12x+35}dx$? How to integrate following
$$\int\frac{1}{x^2-12x+35}dx?$$
What I did is here:
$$\int\frac{dx}{x^2-12x+35}=\int\frac{dx}{(x-6)^2-1}$$
substitute $x-6=t$, $dx=dt$
$$=\int\frac{dt}{t^2-1}$$
partial fraction decomposition,
$$=\int{1\over 2}\left(\frac{1}{t-1}-\frac{1}{t+1}\ri... | You can do that without substitution. Use partial fractions by factorizing denominator:$x^2-12x+35=(x-5)(x-7)$
$$\int \frac{dx}{x^2-12x+35}=\int \frac{dx}{(x-7)(x-5)}$$
$$=\int\frac12\left( \frac{1}{x-7}-\frac{1}{x-5}\right)dx$$
$$=\frac12(\ln\left| x-7\right|-\ln\left| x-5\right|)$$
$$=\frac12\ln\left| \frac{x-7}{x-5}... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Improper integral $\int_{-\infty}^\infty \frac{1}{x^n + 1}$ for $n$ integer, especially $n$ odd The exercise is to analyze the integral $\int_{-\infty}^\infty \frac{1}{x^n + 1}$ for $n$ integer.
For $n$ even, the integrand is well defined, and I discovered that the integral converges in this case.
My problem is analysi... | Let
\begin{equation}
I=\int\limits_{-\infty}^{+\infty} \frac{1}{x^{n}+1}\,dx
\end{equation}
for some positive integer $n$. For any odd positive integer, the integral diverges, so we will only have to compute the case for even positive integers. Given that $n$ is even, we can write the integral as follows:
\begin{equat... | {
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Let f(x) be a polynomial satisfying $\lim_{x \to\infty} \frac{x^4f(x)}{x^8+1}=3,$ $f(2)=5 $ and $f(3)=10$, $f(-1)=2$ and $f(-6)=37$ Let f(x) be a polynomial satisfying $\lim_{x \to\infty} \frac{x^4f(x)}{x^8+1}=3,$ $f(2)=5 $ and $f(3)=10$, $f(-1)=2$ and $f(-6)=37$ .
And then there are some question related to this.
In ... | Another possibility is to use Lagrange polynomials, and express $f(x)$ as a function that has the given four values in the given points plus a term that produces the given leading term coefficient, but vanishes in all four points:
\begin{align}
f(x)=\
&y_1\frac{(x-x_2)(x-x_3)(x-x_4)}{(x_1-x_2)(x_1-x_3)(x_1-x_4)}+\\
&y... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Is the radius of convergence related to the ratio limit or half of the interval of convergence? I have a series $S$ with general terms $a_n=\frac{(-1)^n(x-1)^n}{(2n-1)2^n}$, $n\ge 1$:
$$S = \sum_{n=1}^\infty \frac{(-1)^n(x-1)^n}{(2n-1)2^n}$$
Finding the ratio $\left|\frac{a_{n+1}}{a_n}\right|$ and then finding the limi... | Note that a power series takes the form
$$\sum_{n=0}^\infty a_n(x-x_0)^n$$
In your case you have
$$a_n=\begin{cases}\frac{(-1)^n}{(2n-1)2^n}&n\ne0\\0&n=0\end{cases}\qquad x_0=1$$
If you calculate the limit you call $N$ we get
$$N=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\frac12$$
So the radius of convergence i... | {
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"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Find $2f(x)\cdot f(x-8) - 3f(x+12) - 2 = 0$
Function $f$ $\in \mathbb{R}$ is odd and has a period of $4$. On a $[0,2]$ segment function $f$ is defined as $f(x)= 4x - 2x^2$. Find the set of solutions for the equation: $$2f(x)\cdot f(x-8) - 3f(x+12) - 2 = 0$$
So, here's my attempt: function has a period of $4$ means t... | In your solution, you make the following error
$$f(x) \space \text{is odd} \implies f(x-8) = -f(x+8)$$
This is not true, $f$ being odd means that $f(-x) = -f(-x)$, hence $f(x-8) = -f(-x+8)$
Now, since the period is 4, what we can say is that $f(x-8) = f(x-4) = f(x)$
Hence, replacing $f(x) = z$
$$2z^2 - 3z-2 = 0$$
$$\im... | {
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If the coefficients of a quadratic equation are odd numbers, show that it cannot have rational roots If the coefficients of a quadratic equation $$ax^2+bx+c=0$$ are all odd numbers, show that the equation will not have rational solutions.
I am also not sure if I should consider $c$ as a coefficient of $x^0$, suppose if... | Take $a=1,b=3,c=2$ to get the rational solutions $-2,-1$. So the statement is false unless $c$ is also required to be odd.
Now consider squares modulo $8$. Any odd number has the form $8n+1$, $8n+3$, $8n+5$, or $8n+7$ (these are abbreviated as $\equiv1,3,5,7\bmod8$). So an odd number squared is
$$1^2=1$$
$$3^2=9=8\cdot... | {
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"answer_id": 4
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Another way to solve $\int \frac{\sin^4(x)}{1+\cos^2(x)}\ dx$ without the substitution $y=\tan\left(\frac{x}{2}\right)$? Is there another way to solve an integral $$\int \frac{\sin^4(x)}{1+\cos^2(x)}\ dx$$ without the substitution $y=\tan\left(\frac{x}{2}\right)$?
$\large \int \frac{\sin^3(x)}{1+\cos^2(x)}\ dx$ is easi... | Denote
$${I_n = \int\frac{\sin^{2n}(x)}{1 + \cos^2(x)}dx}$$
Then
$${I_{n}=\int\sin^2(x)\frac{\sin^{2n-2}(x)}{1+\cos^2(x)}dx=\int(1-\cos^2(x))\frac{\sin^{2n-2}(x)}{1+\cos^2(x)}dx}$$
If you expand this, you get
$${=I_{n-1} - \int \cos^2(x)\frac{\sin^{2n-2}(x)}{1+\cos^2(x)}dx=I_{n-1}-\int \sin^{2n-2}(x) - \frac{\sin^{2n-2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
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Converting parametric equation to Cartesian How can I convert these parametric equations into a Cartesian equation (I think that is what its called). Wolfram alpha doesn't seem like it can handle this one.
$$
x=\cos{t}+\cos{-6t}\\
y=\sin{t}+\sin{-6t}
$$
Is there a name for this operation?
| I think you're looking for an implicit equation for this parametric plane curve. The process of turning a system of parametric equations into a system of implicit equations is called implicitization. See $\S3.3$ of Cox, Little, and O'Shea's Ideals, Varieties, and Algorithms for more on this.
Let $u = \cos(t)$ and $v = ... | {
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"source": "stackexchange",
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Given $\cos(a) +\cos(b) = 1$, prove that $1 - s^2 - t^2 - 3s^2t^2 = 0$, where $s = \tan(a/2)$ and $t = \tan(b/2)$ Given $\cos(a) + \cos(b) = 1$, prove that $1 - s^2 - t^2 - 3s^2t^2 = 0$, where $s = \tan(a/2)$ and $t = \tan(b/2)$.
I have tried using the identity $\cos(a) = \frac{1-t^2}{1+t^2}$. but manipulating this s... | Hint:
$$\dfrac{1-s^2}{1+s^2}+\dfrac{1-t^2}{1+t^2}=1$$
$$\iff\dfrac{1-s^2}{1+s^2}=1-\dfrac{1-t^2}{1+t^2}$$
$$\iff(1-s^2)(1+t^2)=2t^2(1+s^2)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find range of $f(x)=\frac{5}{\sin^2x-6\sin x\cos x+3\cos^2x}$ Find range of $f(x)=\frac{5}{\sin^2x-6\sin x\cos x+3\cos^2x}$
My attempt :
\begin{align*}
f(x)&=\dfrac{5}{9\cos^2x-6\sin x\cos x+\sin^2x-6\cos^2x}\\
&= \dfrac{5}{(3\cos x+\sin x)^2-6\cos^2x}
\end{align*}
The problem is if I'm going to use
$$-1\leqslant\sin x... | Another way:
$$y(\sin^2x -6\sin x\cos x+3\cos^2x)=5$$
Divide both sides by $\cos^2x$
$$y\tan^2x-6y\tan x+3y=5(1+\tan^2x)$$
Rearrange to form a quadratic equation in $\tan x$ which is real
So, the discriminant must be $\ge0$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Find x intercepts of a higher degree polynomial $2x^4+6x^2-8$ I am to factor and then find the x intercepts (roots?) of $2x^4+6x^2-8$
The solutions are provided as 1 and -1 and I am struggling to get to this.
My working:
$2x^4+6x^2-8$ =
$2(x^4+3x^2-4)$
Focus on just the right term $(x^4+3x^2-4)$:
Let $u$ = $x^2$, then:... | you are almost done
$$(x^2+4)(x^2-1)=0$$
$$x^2=-4, \ x^2=1$$
$$x=\pm2 i, \ x=\pm 1$$
considering the real values, the x-intercepts are $x=-1, y=0$ and $x=1, y=0$
x-intercepts in point form: (-1,0) and (1,0)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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On the diophantine equation $x^{m-1}(x+1)=y^{n-1}(y+1)$ with $x>y$, over integers greater or equal than two I don't know if the following diophantine equation (problem) is in the literature. We consider the diophantine equation $$x^{m-1}(x+1)=y^{n-1}(y+1)\tag{1}$$ over integers $x\geq 2$ and $y\geq 2$ with $x>y$, and o... | My approach is to restrict the exponents to $m = 2, n = 3$, because both of your solutions have these values, and due to $m - 1 = 1$, this could makes the question different, as the LHS becomes simply a product of two consecutive numbers.
This approach of mine might not be very useful, as I thought I can come up with s... | {
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"source": "stackexchange",
"question_score": "12",
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Prove that $\tan^{-1}\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}=\frac{\pi}{4}+\frac 12 \cos^{-1}x^2$ Let the above expression be equal to $\phi$
$$\frac{\tan \phi +1}{\tan \phi-1}=\sqrt{\frac{1+x^2}{1-x^2}}$$
$$\frac{1+\tan^2\phi +2\tan \phi}{1+\tan^2 \phi-2\tan \phi}=\frac{1+x^2}{1-x^2}$$
$$\frac{1+\t... | Domain of $\tan^{-1}\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}$ is $x \in (-1,1]$, nothing wrong with that.
But range of the above term(argument of $\arctan$) is $(1,\infty)$ so this means, when you assume it be $\phi$, it is restricted to the interval $\left[ \frac{\pi}{4},\frac{\pi}{2}\right]$
This ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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"answer_id": 2
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How to evaluate $\int _0^{\infty }\frac{\ln \left(x^3+1\right)}{\left(x^2+1\right)^2}\:dx$ without complex analysis This particular integral evaluates to,
$$\int _0^{\infty }\frac{\ln \left(x^3+1\right)}{\left(x^2+1\right)^2}\:dx=\frac{\pi }{8}\ln \left(2\right)-\frac{3\pi }{8}+\frac{\pi }{3}\ln \left(2+\sqrt{3}\right)... | I am going to prove it by integration by parts and partial fractions.
$$
\begin{aligned}
\int_{0}^{\infty} \frac{\ln \left(1+x^{3}\right)}{\left(1+x^{2}\right)^{2}} d x
&=-\int_{0}^{\infty} \frac{\ln \left(1+x^{3}\right)}{2 x} d\left(\frac{1}{1+x^{2}}\right)\\
&=-\left[\frac{\ln \left(1+x^{3}\right)}{2 x\left(1+x^{2}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3761814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
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How to evaluate $\int x^5 (1+x^2)^{\frac{2}{3}}\ dx$? I am trying to evaluate $\int x^5 (1+x^2)^{\frac{2}{3}} dx$
This is apparently a binomial integral of the form $\int x^m (a+bx^k)^ndx$. Therefore, we can use Euler's substitutions in order to evaluate it. Since $\dfrac{m+1}{k} = \dfrac{5+1}{2} = 3 \in \mathbb{Z}$ w... | HINT:
Let $1+x^2=t^3\implies 2xdx=3t^2dt$ or $xdx=\frac{3}{2}t^2dt$
$$\int x^5(1+x^2)^{2/3}dx=\int (x^2)^2(1+x^2)^{2/3}xdx$$
$$=\int (t^3-1)^2(t^3)^{2/3}\ \frac{3t^2}{2}dt$$
$$=\frac32\int(t^3-1)^2t^4 dt $$
$$=\frac{3}{2}\int (t^{10}-2t^7+t^4)dt$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3762071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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How to evaluate $\int \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}} dx$? I am trying to evaluate
$$\int \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}} dx \quad (1)$$
The typical way to confront this kind of integrals are the conjugates i.e:
$$\int \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}} dx = $$... | $$I=\int \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}} \,dx $$
Rationalize:
$$\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}=\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\times\frac{-\sqrt{1+x}-\sqrt{1-x}}{-\sqrt{1+x}-\sqrt{1-x}}$$
$$=\frac{\sqrt{(1+x)(1-x)}+1}{x}$$
$$=\frac{\sqrt{1-x^2}}{x}+\frac 1x$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3763369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is it possible to show that the fifth roots of 1 add up to 0 simply by using trigonometric identities? You can't use geometric sums, minimal polynomials, pentagon, and exact values with radicals.
All the five, fifth-roots of unity are :$1,\left(\cos \left(\frac{2 \pi}{5}\right)+i \sin \left( \frac{2\pi}{5}\right)\righ... | $$z^5=1$$
$$z^5-1=0$$
$$(z-1)(z^4+z^3+z^2+z+1)=0$$
Note that the parenthesis with 5 terms are the roots but as we know the complex is a filed no zero divisors thus the sum of the roots is zero!
Plug in $e^{\frac{2\pi i }{5}}=z$
$$(e^{\frac{2\pi i }{5}}-1)(e^{\frac{8\pi i }{5}}+ e^{\frac{6\pi i }{5}}+ e^{\frac{4\pi i }... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3764372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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Find the constant for $\int_{0}^{1} {\frac{\mathrm{d}x}{\sqrt{(1-x^2)(1-(kx)^4)}}} \sim C\ln(1-k)$ I encounter a problem for Elliptic integral, to find the exact $C$ for
$$\int_{0}^{1} {\frac{\mathrm{d}x}{\sqrt{(1-x^2)(1-(kx)^4)}}} \sim C\ln(1-k)$$
as $k\uparrow1\;(0<k<1)$.
to establish such asymptotic behavior around ... | I will propose a creative approach based on Fourier-Legendre expansions. In $L^2(0,1)$ we have$^{(*)}$
$$ K(x)=\sum_{n\geq 0}\frac{2}{2n+1}P_n(2x-1),\qquad -\log(1-x)=1+\sum_{n\geq 1}\left(\frac{1}{n}+\frac{1}{n+1}\right)P_n(2x-1) $$
so
$$ K(x)+\frac{1}{2}\log(1-x)=\frac{3}{2}-\sum_{n\geq 1}\frac{P_n(2x-1)}{2n(n+1)(2n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3765050",
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"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Find the area between $f(x) = x^2+3x+7 $ and $g(x) = xe^{x^3+4}$ for $x \in [3,5]$.
Calculate the area between the two functions, $f(x)$, $g(x)$, for $x \in [3,5]$.
$$f(x)=x^2+3x+7$$
$$g(x)=xe^{x^3+4}$$
To determine the area between the functions I used the formula $A= \int_a^b|f(x)-g(x)|dx$. Therefore, I have:
\begi... | You can do this:
$$The \ area \ between \ the \ two \ functions = \int_{3}^{5}{xe^{x^3+4}}dx \ - \ \int_3^5{x^2+3x+7}dx$$ Because $\forall x \in [3, 5] \ \ \ xe^{x^3+4} > x^2+3x+7$.
You can look here: https://www.desmos.com/calculator/csh4alwmeu
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that if $a$ and $b$ have the same sign then $|a + b| = |a| + |b| $, and if $a$ and $b$ have opposite signs then $|a+b| < |a| + |b|$ I'm considering the different cases for $a$ and $b$
Case 1) $ a\geq 0$ and $ b\geq 0$
Given both terms are positive, $ a + b \geq 0 $
$$
|a+b| = a + b = |a| + |b|\\
$$
Case 2) $ a< 0$... | Something similar to yours:
The first part was, indeed, easy.
For the second part:
Without loss of generality, take $|a|>|b|~~~~~~~~~(*)$
*
*$a>0, b<0 \overset{(*)}\implies a+b>0$. Then, indeed, $\underbrace{|a+b|}_{=a+b} < \underbrace{|a|}_{=a}+\underbrace{|b|}_{=-b}$ since $b<-b$.
*$a<0, b>0 \overset{(*)}\impli... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given $c =\sqrt {a^2-b^2}$, $c= ea$, conclude that $\frac {2\pi ab}{(a+c)T} =\frac {2\pi a}T\sqrt {\frac {1-e}{1+e}}$ Given $c =\sqrt {a^2-b^2}$, $c= ea$, conclude that $$\frac {2\pi ab}{(a+c)T} =\frac {2\pi a}T\sqrt {\frac
{1-e}{1+e}}$$
Sorry for the awful formatting of this question. I know this is a pretty simple p... | Since $\;c =\sqrt {a^2-b^2},\;$ we get $\;b=\sqrt {a^2-c^2}$.
Moreover,
$$\frac {2\pi ab}{(a+c)T} =\frac{2\pi a\sqrt{a^2-c^2}}{T(a+c)}=\frac{2\pi a}T\sqrt{\frac
{(a+c)(a-c)}{(a+c)^2}}=\frac{2\pi a}T\sqrt{\frac
{a-c}{a+c}}=\frac{2\pi a}T\sqrt{\frac
{a-ea}{a+ea}}=\frac{2\pi a}T\sqrt{\frac
{a(1-e)}{a(1+e)}}=\frac{2\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3770406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving the system $\sqrt{x} + y = 7$, $x + \sqrt{y} = 11$ I want to solve the following nonlinear system of algebraic equations. Indeed, I am curious about a step by step solution for pedagogical purposes. I am wondering if you can come up with anything. I tried but to no avail.
\begin{align*}
\sqrt{x} + y &= 7 \\
x +... | We have
$$\begin{align*}
\sqrt{x} + y &= 7 \\
x + \sqrt{y} &= 11
\end{align*}$$
Under the constraints
$$0\le x \le 11,\quad 0\le y \le 7$$
A contour plot shows a single point of intersection
We can isolate the square root of $x$ in first equation, square both sides (this causes us to have extraneous roots that we hav... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3772635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Given ellipse of axes $a$ and $b$, find axes of tangential and concentric ellipse at angle $t$ Let’s say I have an ellipse with horizontal axis $a$ and vertical axis $b$, centered at $(0,0)$.
I want to compute $a’$ and $b’$ of a smaller ellipse centered at $(0,0)$, with the axes rotated by some angle $t$, tangent to th... | Let the oblique ellipse be
$$
\frac{x^2}{a^2}+
\frac{y^2}{b^2}-1=k
\left(
\frac{x\sin \theta}{a}-
\frac{y\cos \theta}{b}
\right)^2$$
which touches the standard ellipse at $(a\cos \theta,b\sin \theta)$.
This can be easily verified by plugging the point $(x',y')=(a\cos \theta,b\sin \theta)$ in the both ellipses and a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3773593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Question on Cardano's Method of Solving Cubic Polynomial Equations I'm having trouble with part of a question on Cardano's method for solving cubic polynomial equations. This is a multi-part question, and I have been able to answer most of it. But I am having trouble with the last part. I think I'll just post here the... | Let $w(\alpha) = \cos \alpha + i\sin \alpha$. Then
$$w(\alpha) w(\beta) = (\cos\alpha + i \sin \alpha)(\cos \beta + i\sin \beta) \\ =\cos\alpha \cos \beta - \sin \alpha \sin \beta +i(\cos\alpha \sin \beta + \sin \alpha \cos \beta) = \cos(\alpha + \beta) + i \sin(\alpha + \beta) \\= w(\alpha + \beta) .$$
An easier way t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3773856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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EGMO 2014/P3 : Prove that there exist infinitely many positive integers $n$ such that $\omega(n) = k$ and $d(n)$ does not divide $d(a^2+b^2)$ We denote the number of positive divisors of a positive integer $m$ by $d(m)$ and the number of distinct prime divisors of $m$ by $\omega(m)$. Let $k$ be a positive integer. Prov... | $\boxed{\text{Complete solution}}$
(The merit of the following solution is that it gives an explicit construction for $n$ with given $k$ satisfying the conditions.)
Let $p_m$ denote the $m^{th}$ prime with $p_1=2,p_2=3,\ldots$ and so on. Take, for $k>1$, $$n=2^{p-1}p_2p_3\cdots p_k$$ for some suitable prime $p$ and wor... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $a_n=100a_{n-1}+134$, find least value of n for which $a_n$ is divisible by $99$
Let $a_{1}=24$ and form the sequence $a_{n}, n \geq 2$ by $a_{n}=100 a_{n-1}+134 .$ The first few terms are
$$
24,2534,253534,25353534, \ldots
$$
What is the least value of $n$ for which $a_{n}$ is divisible by $99 ?$
We have to find.... | Your method can work quite well. For all $1 \lt i \le n$, note $a_i - a_{i-1}\equiv 35 \pmod{99}$ means each $a_i$ is congruent to $35$ more than the previous one of $a_{i-1}$. Thus, starting from $a_1$ and repeating this $n - 1$ times, you get
$$a_n \equiv a_1 + (n - 1)35 \equiv 35n - 11 \equiv 0 \pmod{99} \tag{1}\lab... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\frac{1}{2} (x-1) x + y$ is a bijection. (on p.45 Munkres Topology 2nd Edition) I am reading "Topology 2nd Edition" by James R. Munkres.
On p.45, Munkres leaves it to the readers to show that $g$ is bijection:
Show that $g(x, y) = \frac{1}{2} (x-1) x + y$ is a bijection from $\{(x, y) \in \mathbb{Z}_{+} \t... | Your proofs are fine.
Here is a quicker proof of injectivity. Suppose $T_{x-1}+y=T_{x'-1}+y'$ for $y\le x$ and $y'\le x'$ where $T_x=x(x+1)/2$ is the $x$th triangular number. Without loss of generality let $x<x'$ and thus $$T_{x-1}+y\ge T_x+1\implies y\ge(T_x-T_{x-1})+1=x+1$$ which is a contradiction. Hence $x=x'$ from... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3775670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\ddot{x}+(2x^2+\dot{x}^2-1)\dot{x}+x=0$ has a periodic solution. Prove that $\ddot{x}+(2x^2+\dot{x}^2-1)\dot{x}+x=0$ has a periodic solution.
Here's my partial solution: Let $\dot{x}=y$. Then $\ddot{x}=\dot{y}=-(2x^2+y^2-1)y-x=-2x^2y-y^3+y-x$.
So we have $\dot{x}=y$ and $\dot{y}=-2x^2y-y^3+y-x$.
Convert to ... | Hint.
$$
\cases{
\dot x_1 = x_2\\
\dot x_2 = -x_1-(2x_1^2+x_2^2-1)x_2
}
$$
or
$$
\cases{
\dot x_1 x_1= x_1x_2\\
\dot x_2 x_2= -x_1x_2-(2x_1^2+x_2^2-1)x_2^2
}
$$
adding the equations
$$
\frac 12(x_1^2+x_2^2)' = -(2x_1^2+x_2^2-1)x_2^2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3776913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $a^2 + b^2 + c^2 = 1$, what is the the minimum value of $\frac {ab}{c} + \frac {bc}{a} + \frac {ca}{b}$?
Suppose that $a^2 + b^2 + c^2 = 1$ for real positive numbers $a$, $b$, $c$. Find the minimum possible value of $\frac {ab}{c} + \frac {bc}{a} + \frac {ca}{b}$.
So far I've got a minimum of $\sqrt {3}$. Can anyo... | For $a=b=c=\frac{1}{\sqrt3}$ we obtain a value $\sqrt3$.
We'll prove that it's a minimal value.
Indeed, we need to prove that:
$$\sum_{cyc}\frac{ab}{c}\geq\sqrt{3(a^2+b^2+c^2)}$$ or
$$\sum_{cyc}a^2b^2\geq\sqrt{3a^2b^2c^2(a^2+b^2+c^2)}$$ or
$$\sum_{cyc}(a^4b^4-a^4b^2c^2)\geq0$$ or
$$\sum_{cyc}c^4(a^2-b^2)^2\geq0$$ and w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3777113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Calculation of $\int\frac{e^{-\sin^2(x)}\tan^3(x)}{\cos(x)}dx=$
$$\int\frac{e^{-\sin^2(x)}\tan^3(x)}{\cos(x)}\,dx=\text{?}$$
My work :
$$\int\frac{e^{-\sin^2(x)} \tan^3(x)}{\cos(x)} \, dx = \left\{\left(\int e^{-\sin^2(x)}\tan^3(x)\,dx\right)\frac{1}{\cos(x)}\right\}-\int\frac{\sin(x)}{\cos^2(x)}\left(\int e^{-\sin^2... | You may try this.
Let $t=\sin^2(x)$, then $$ \int\frac{e^{-\sin^2(x)}\tan^3(x)}{\cos(x)}\,dx=\int2 \dfrac{e^{-t}t^2 }{(1-t)^2}dt,$$
Let $1-t=-u$, then $$ \int\frac{e^{-\sin^2(x)}\tan^3(x)}{\cos(x)}= \int \dfrac{2e^{1-u}(1+u)^2}{u^2}du.$$
NB: Think about $ \Gamma$ -function.
| {
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"timestamp": "2023-03-29T00:00:00",
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To prove $a+b+c \ge ab+bc+ca$ when $abc=1$ The actual question is not in the heading, it's this-
Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that
$$ {a \over \sqrt{7+b+c}} + {b \over \sqrt{7+c+a}} + {c \over \sqrt{7+a+b}} \geq 1$$
To complete a proof of this inequality after a deft application o... | We need to prove
$$\tag{1} (a+b+c)^3\geq7(a+b+c)+2(ab+bc+ca).$$
But from know inequality $3(ab+bc+ca) \leqslant (a+b+c)^2.$ Therefore the proof is completed if
$$\tag{2} (a+b+c)^3\geq7(a+b+c)+\frac23(a+b+c)^2.$$
Let $x = a+b+c \geqslant 3\sqrt[3]{abc} = 3,$ ienquality $(2)$ become
$$x^3 \geqslant 7x+\frac{2x^2}{3},$$
o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3778382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Square equal to sum of three squares For which integers $n$ there exists integers $0\le a,b,c < n$ such that $n^2=a^2+b^2+c^2$?
I made the following observations:
*
*For $n=1$ and $n=0$ those integers doesn't exist.
*If $n$ is a power of 2 those integers doesn't exist. Let $n=2^m$ with $m>0$ the smallest power of 2... | Some Pythagorean triples:
$3^2+4^2=5^2$
$5^2+12^2=13^2$
So: $3^2+4^2+12^2=13^2$
Generalize that:
$(3t)^2+(4t)^2+(12t)^2=(13t)^2$
$n=13t$ , $t> 0 $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3780950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
How to find $a$, $b$, $c$ such that $P(x)=ax^3+bx^2+cx$ and $P\left(x\right)-P\left(x-1\right)=x^2$ I'm trying to find $a$, $b$ and $c$ such that $P(x)=ax^3+bx^2+cx$ and $P\left(x\right)-P\left(x-1\right)=x^2$.
After expanding the binomial in $P(x-1)$, I end up getting
$3ax^2-3ax+2bx+a-b=x^2$. What next? Using $3a = 1$... |
For the sake of a different method:
$$
\begin{cases}
P(x)-P(x-1)=x^2 \\
P'(x)-P'(x-1)=2x \\
P''(x)-P''(x-1)=2
\end{cases}
$$
or writing derivatives explicitly:
$$
\begin{cases}
a\big(x^3-(x-1)^3\big)+b\big(x^2-(x-1)^2\big)+c\big(x-(x-1)\big)=x^2~~~~~~~(1) \\
3a\big(x^2-(x-1)^2\big)-2b\big(x-(x-1)\big)=2x~~~~~~~~~... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3781197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove: $\int_0^2 \frac{dx}{\sqrt{1+x^3}}=\frac{\Gamma\left(\frac{1}{6}\right)\Gamma\left(\frac{1}{3}\right)}{6\Gamma\left(\frac{1}{2}\right)}$ Prove:
$$
\int_{0}^{2}\frac{\mathrm{d}x}{\,\sqrt{\,{1 + x^{3}}\,}\,} =
\frac{\Gamma\left(\,{1/6}\,\right)
\Gamma\left(\,{1/3}\,\right)}{6\,\Gamma\left(\,{1/2}\,\right)}
$$
First... | An elementary solution: Consider the substitution
$$t = \frac{{64 + 48{x^3} - 96{x^6} + {x^9}}}{{9{x^2}{{(4 + {x^3})}^2}}}$$
$t$ is monotonic decreasing on $0<x<2$, and $$\tag{1}\frac{{dx}}{{\sqrt {1 + {x^3}} }} = -\frac{{dt}}{{3\sqrt {1 + {t^3}} }}$$
this can be verified by explictly computing $(dt/dx)^2$ and compare ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3781324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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Finding the determinant of a $5\times 5$ matrix Let
$$A = \left[\begin{array}{rrrrr}6 & 2 & 2 & 2 & 2 \\ 2 & 6 & 2 & 2 & 2 \\ 2 & 2 & 6 & 2 & 2 \\2 & 2 & 2 & 6 & 2 \\ 2 & 2 & 2 & 2 & 6\end{array}\right] \in {M}_{5}(\mathbb{R})$$
Which of following options is $\det(A)$ ?
*
*$4^4 \times 14$
*$4^3 \times 14$
*$4^2 \... | As, dimension of nullspace of $(A-4I)$ is $4$. So, Geometric multiplicity of eigenvalue $4$ is $4$, as matrix $A$ is symmetric, so, $A$ must be diagonalizable, and hence, Algebraic multiplicity and Geometric multiplicity of eigenvalue $4$ is same, so, Algebraic multiplicity of eigenvalue $4$ is $4$.
And, as each row su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3783140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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On proving $a^3+b^3+c^3-3abc \geq 2\left({b+c\over 2}-a\right)^3$. This problem was a "warm-up" problem by the author. Note: $a, b, c$ are non-negative numbers.
$$a^3+b^3+c^3-3abc \geq 2\left({b+c\over 2}-a\right)^3$$
I tried to remove the $2$ from ${b+c\over 2}$ and got this-
$$ 4(a^3+b^3+c^3-3abc) \geq (b+c-2a)^3 $$
... | If $b+c<2a$ then $$LHS \geq 0 \ge RHS.$$
If $b+c \geq 2a.$ We write inequality as
$$4(a^3+b^3+c^3-3abc) \geq (b+c-2a)^3,$$
or
$$4(a+b+c)(a^2+b^2+c^2-ab-bc-ca) \geq (b+c-2a)^3.$$
Because $a+b+c = 3a+(b+c-2a) \geq b+c-2a,$ so we will show that
$$4(a^2+b^2+c^2-ab-bc-ca)\geq (b+c-2a)^2,$$
equivalent to
$$3(b-c)^2 \geq 0.$... | {
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"timestamp": "2023-03-29T00:00:00",
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} |
Let $A, B$ be skew-symmetric matrices such that $AB = -BA$. Show that $AB = 0$
Let $A, B$ be skew-symmetric matrices such that $AB = -BA$. Show that $AB = 0$.
I know that $AB$ is skew-symmetric,because $$(AB)^t=B^tA^t=BA=-AB$$
but I don't know how show that $AB=0$.
| Let
$$ A = \begin{pmatrix}
0 & 0 & 0 & -1 \\
0 & 0 & -1 & 0 \\
0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0
\end{pmatrix}
\quad\text{and}\quad
B = \begin{pmatrix}
0 & -1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 0 & -1 \\
0 & 0 & 1 & 0
\end{pmatrix}.$$
Then
$$ AB = -BA = \begin{pmatrix}
0 & 0 & -1 & 0 \\
0 & 0 & 0 & 1 \\
1 & 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3786905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the number of points of non differentiability in $(x^2-3x+2)(|x^3-6x^2+11x-6|)$ The expression is factorized as $$(x-2)(x-1)(|(x-1)(x-2)(x-3)|)$$
I expected the points to be $1,2,3$, put looking at the graph, it’s only $x=3$
I kinda figured out the reason for this, because the expression would end up simplifying t... | $$F(x)=(x-1)(x-2)|(x-1)(x-2)(x-3)|.$$ Only $x=3$ is the point of non-differentiability. At $x=1,2$ is it differentiable. At these points, we have $(x-a)|x-a|$ which is differentiable at $x=a$.
Note that $g(x)=(x-a)|x-a|= (x-a)(a-x), ~if~ x<a$ and $g(x)=(x-a)^2, ~if ~x\ge a$, hence both the left and right derivatives ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3787202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
How to prove $\frac{a^{n+1}+b^{n+1}+c^{n+1}}{a^n+b^n+c^n} \ge \sqrt[3]{abc}$? Give $a,b,c>0$. Prove that: $$\dfrac{a^{n+1}+b^{n+1}+c^{n+1}}{a^n+b^n+c^n} \ge \sqrt[3]{abc}.$$
My direction: (we have the equation if and only if $a=b=c$)
$a^{n+1}+a^nb+a^nc \ge 3a^n\sqrt[3]{abc}$
$b^{n+1}+b^na+b^nc \ge 3b^n\sqrt[3]{abc}$
$c... | WLOG let $a \geq b \geq c$ , then $\sqrt[3]a \geq \sqrt[3]b \geq \sqrt[3]c$. Let $x = \sqrt[3]a, y = \sqrt[3]b , z =\sqrt[3]c$ , then $x \geq y\geq z > 0$ and the inequality is equivalent to :
$$
x^{3n+3} + y^{3n+3} + z^{3n+3} \geq x^{3n+1}yz + y^{3n+1}xz + z^{3n+1}xy
$$
by Muirhead's inequality, since the sequence ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3787573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
find the complex integral: $\int_0^\infty \frac{z^6}{(z^4+1)^2}dz$. Problem with integral formula.... Question I am trying to find the complex integral: $\displaystyle\int_0^\infty \frac{z^6}{(z^4+1)^2}dz$.
My Attempt (and eventual question): $\int_0^\infty \frac{z^6}{(z^4+1)^2}dz=\frac{1}{2}\int_\infty^\infty\frac{z^6... | What do you mean by "complex integral"? This is a real integral. Okay, you can evaluate it using residues.
Consider $C$, the closed curve consisting of the line segment from $-R$ to $R$ union with semicircle of radius $R$ in the upper half plane
$$\oint_C \frac{z^6}{(z^4+1)^2}\, dz =
\int_0^\pi \frac{R^6 e^{6i\theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3787771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
For $a>1$, show that $\frac{1}{1+x}-\frac{1}{1+ax} \leq \frac{\sqrt{a}-1}{\sqrt{a}+1}$, $x \geq 1$ I'm self-learning the analysis I "by Herbert Amann" and got stuck in this problem. It's in Chapter IV Taylor's theorem.
For $a>1$ and $x\geq 1$ show that $$\frac{1}{1+x}-\frac{1}{1+ax} \leq \frac{\sqrt{a}-1}{\sqrt{a}+1}$... | Let
$$f : x \mapsto \frac{1}{1+x} - \frac{1}{1+ax}$$
$f$ is differentiable on $[1, +\infty)$ and
$$f'(x)= -\frac{1}{(1+x)^2}+\frac{a}{(1+ax)^2}$$
so $$f'(x)\geq 0 \Leftrightarrow \frac{a}{(1+ax)^2} \geq \frac{1}{(1+x)^2}$$
$$\Leftrightarrow \sqrt{a}(1+x) \geq 1+ax \Leftrightarrow \frac{1-\sqrt{a}}{\sqrt{a}-a} \geq x$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3789967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
If $x$, $y$ and $z$ satisfy $xy=1$, $yz=2$ and $xz=3$, what is the value of $x^2+y^2+z^2$? Express your answer as a common fraction.
If $x$, $y$ and $z$ satisfy $xy=1$, $yz=2$ and $xz=3$, what is the value of $x^2+y^2+z^2$? Express your answer as a common fraction.
I tried going along the path of computing $(x+y+z)... | HINT. You can take the equations pairwise. For example, you have $xy=1$ and $yz=2$. So for example,
$$
\dfrac{1}{2}= \dfrac{xy}{yz}=\dfrac{x}{z}
$$
But then $z=2x$. But you know $xz=3$. Can you substitute and find $x$ or $z$? Can you repeat this process for $y$? Knowing their values it should be routine to find $x^2+y^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3790065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Is there a closed form for $\sum_{n=1}^\infty\frac{2^{2n}H_n}{n^3{2n\choose n}}?$ I found
$$\sum_{n=1}^\infty\frac{2^{2n}H_n}{n^3{2n\choose n}}=-8\int_0^{\pi/2}x^2\cot x\ln(\cos x)\ dx=I\tag1.$$
Mathematica failed to find $I$, so I am not sure if there is closed form for it. I am just giving it a try here.
First idea c... | $$S=-8 \text{Li}_4\left(\frac{1}{2}\right)+\frac{\pi ^4}{90}-\frac{1}{3} \log ^4(2)+\frac{4}{3} \pi ^2 \log ^2(2)$$
Proof $1$. This. Proof $2$. This. Proof $3$. This. Bonus: $$\small \int_0^{\frac{\pi }{2}} x^3 \cot (x) \log (\cos (x)) \, dx=\frac{3}{2} \pi \text{Li}_4\left(\frac{1}{2}\right)+\frac{9}{16} \pi \zeta (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3792235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Finding Equilibrium Points of Dynamical System I am wondering how to find the equilibrium points from these $2$ equations
\begin{align}
\frac{\mathrm{d}x}{\mathrm{d}t} & =
0.4x\left(1 - \frac{x}{1000} + \frac{y}{1000}\right)
\\[2mm]
\frac{\mathrm{d}y}{\mathrm{d}t} & =
0.1y\left(1 + \frac{x}{1000} - \frac{y}{500}\right)... | We want to simultaneously find the zeros of
$$0.4x\left(1-\frac{x}{1000}+\frac{y}{1000}\right) = 0\\ 0.1y\left(1+\frac{x}{1000}-\frac{y}{500}\right) = 0$$
We can see that each equation is of the form $a b = 0$, so each of those being zero, will satisfy the equation and we just need for them to do that simultaneously fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3792705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding $\lim_{x \to \infty} (x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}{2}}}{3})$ $\lim_{x \to \infty} (x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}{2}}}{3})$ this limit according to wolframalpha is equal to $0$.
So this is my work thus far
$\lim_{x \to \infty} (x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}... | By binomial approximation
$$(x^2+1)^{\frac{3}{2}}=(x^2)^{\frac{3}{2}}\left(1+\frac1{x^2}\right)^{\frac{3}{2}} = x^3+\frac32 x +O\left(\frac1{x}\right)\implies \frac{2(x^2+1)^{\frac{3}{2}}}{3} = \frac{2x^{3}}{3}+x+O\left(\frac1{x}\right)$$
therefore
$$x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}{2}}}{3}=O\left(\frac1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3794325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 2
} |
If $ \bigtriangleup ABC$: $\angle CAB = \frac{\pi}{2}$, with height $AD$ and median $AK$. Prove $\angle BAD = \angle BCA = \angle KAC.$
If $\triangle ABC$ is a triangle and $\angle CAB = \frac{\pi}{2}$, with height $AD$ and median $AK$; suppose that $D$ is between $B$ and $K$.
*
*Prove that $\angle BAD = \angle BCA ... |
Consider the circumcircle of $\triangle ABC$. Since $\angle A=\frac{\pi}{2}$, it subtends the diameter, thus $K$ is the circumcenter and $$KA=KB=KC\tag{1}$$
*
*Since $\triangle KCA$ is isosceles, $\angle KCA=\angle KAC$.
In $\triangle ABD$ $\ \ \angle D=\frac{\pi}{2}$, thus $\angle BAD=\frac{\pi}{2}-\angle ABD$, but... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3794783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find all 3 number solutions for $x[(x-2)^2+1]=6$
Find all 3 number solutions for $x[(x-2)^2+1]=6$
I used trial and error method to find integer solutions for $x$, and found that 1 possible solution is $x=3$.
However, there are 2 other non-integer solutions and I do not know how to find them.
I appreciate any help, th... | Looking for integer solutions, the equation $x[(x-2)^2+1]=6$ is equivalent to
$$\begin{cases}x=2,\\(x-2)^2+1=3, \end{cases}\qquad\text{or}\qquad\begin{cases}x=3,\\(x-2)^2+1=2. \end{cases}$$
The second equation in the first system implies that $(x-2)^2\equiv -1\mod 3$. Unfortunately, the only squares mos. $3$ are $0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3794900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Where I made a mistake during factoring $x^6+x^5+x^4+x^3+x^2+x+1$? In order to factor the expression, due to symmetry of coefficients if $r_1,r_2,r_3$ are zeros of $x^6+x^5+x^4+x^3+x^2+x+1$ then $\frac{1}{r_1}, \frac{1}{r_2} , \frac{1}{r_3}$ are also zeros. So we can rewrite:
$$x^6+x^5+x^4+x^3+x^2+x+1=(x^2-(r_1+\frac{... | To factor over $\mathbb{R}$, note that:
$$
\begin{align}
x^6+\cdots+1
&=\frac{1}{x-1}\left(x^7-1\right)
\end{align}
$$
where $x^7-1$ has its roots being the complex seventh roots of unity. One such root is of course $1$, while the other six pair up with their complex conjugate:
$$
\begin{align}
&x^6+\cdots+1\\
&=\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3798699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Simple number theory in one unknown Given $n$ is an integer and $0 \leq n \leq 2000$, how many $n$ are there such that $\lfloor \sqrt {n} \rfloor$ divides $n$ and $\lfloor \sqrt {n+1} \rfloor$ divides $n+1$?
I'm new in number theory and I'm getting trouble assuming $\lfloor \sqrt {n} \rfloor$ as $k$ such as $k$ is also... | For a positive integer $ n \geq 1 $, let $ d_{n} = gcd(\left \lfloor{\sqrt{n}}\right \rfloor, \left \lfloor{\sqrt{n+1}}\right \rfloor) $. Then since from the hypothesis we know that $ \left \lfloor{\sqrt{n}}\right \rfloor \mid n $ and $ \left \lfloor{\sqrt{n+1}}\right \rfloor \mid n+1 $ we get the following:
$$ d_{n} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3799585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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Recurrence and modular arithmetic incorrect solution
Let $a_{10} = 10$, and for each integer $n >10$ let $a_n = 100a_{n - 1} + n$. Find the least $n > 10$ such that $a_n$ is a multiple of $99$. (Source: 2017 AIME I)
This is my solution:
We wish to find the least $n$ such that $a_n\equiv 0\pmod{99},$ with the recurren... | You did well through the conclusion $n(n+1)\equiv90\bmod99$, but your solution of that was lacking.
I would solve that congruence using the Chinese remainder theorem;
$99$ is the product of $11$ and $9$, which are relatively prime.
$n(n+1)\equiv0\bmod9$ and $n(n+1)\equiv2\bmod11$.
$n\equiv0 $ or $8\bmod9$ and $n\equiv1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3800174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solving $\sqrt[3]{x+1} - \sqrt[3]{x-1} = \sqrt[3]{x^2-1}$ for real $x$
Solve the equation in the Real number system:
$$\sqrt[3]{x+1} - \sqrt[3]{x-1} = \sqrt[3]{x^2-1}$$
I have attempted using $(A-B)^3 = A^3 - B^3 - 3.A.B.(A-B)$ with $A = \sqrt[3]{x+1}$ , $B = \sqrt[3]{x-1}$ and $(A-B) = \sqrt[3]{x^2-1}$, however I en... | We can use $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc).$$
$a^2+b^2+c^2-ab-ac-bc=0$ for $a=b=c$ only.
Since $$\sqrt[3]{x+1}=- \sqrt[3]{x-1} =-\sqrt[3]{x^2-1}$$ gives $x=0$ and $0$ is not a root of the equation, we need to remove the number $0$ and we obtain:
$$x+1-(x-1)-(x^2-1)-3\sqrt[3]{x+1}\cdot\sqrt[3]{x-1}\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3800380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Find the argument of $z = {\left( {2 + i} \right)^{3i}}$ $z = {\left( {2 + i} \right)^{3i}}$
My approach is as follow
$z = {\left( {2 + i} \right)^{3i}} = {\left( {{{\left( {2 + i} \right)}^3}} \right)^i} = {\left( {8 + {i^3} + 12i - 6} \right)^i} = {\left( {2 + 11i} \right)^i}$
$\ln z = i\ln \left( {2 + 11i} \right)$
... | Another possible approach:
Taking principal logarithm both side of $z=(2+i)^{3i}$,
$Log z=3i[\ln \sqrt{2^2+1^2}+i ~tan^{-1}(1/2)]=3i[\ln \sqrt 5+i ~tan^{-1}(1/2)]$.
Thus $Log z=-3 tan^{-}(1/2)+i~3 \ln \sqrt 5$ and so $z=e^{-3tan^{-1}(1/2)} \cdot e^{i3 \ln \sqrt 5}$.
This gives us $z=e^{-3tan^{-1}(1/2)} (cos(3 \ln \sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3800640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Prove that $1<\frac{1}{1001}+\frac{1}{1002}+\cdots+\frac{1}{3001}<\frac{4}{3}$ Prove that $1<\frac{1}{1001}+\frac{1}{1002}+\cdots+\frac{1}{3001}<\frac{4}{3}$
Using AM- HM inequality,
$\left(\sum_{k=1001}^{3001} k\right)\left(\sum_{k=1001}^{3001} \frac{1}{k} \right) \geq(2001)^{2}$
But $\sum_{k=1001}^{3001} k=(2001)^{2... | Since $$\frac{1}{(1002+i)(3000-i)}<\frac{1}{1002\cdot3000}$$ for any $i\in[1,998]$ and $$\frac{1}{1001\cdot3001}+\frac{1}{2001^2}<\frac{2}{1002\cdot3000},$$we obtain:
$$\sum_{k=1001}^{3001}\frac{1}{k}=\sum_{k=1}^{1001}\left(\frac{1}{k+1000}+\frac{1}{3000-k+2}\right)=$$
$$=4002\sum_{k=1}^{1001}\frac{1}{(k+1000)(3000-k+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3803855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
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If $a$, $b$, $c$ are the roots of $x^3-6x^2+3x+1=0$, find all possible values of $a^2b+b^2c+c^2a$
Let $a$, $b$, $c$ be the roots of
$$x^3 - 6x^2 + 3x + 1 = 0$$ Find all possible values of $a^2 b + b^2 c + c^2 a$. Express all the possible values, in commas.
I've already tried to bash out all the roots, Vieta's Formula... | Hint : Let $A=a^2 b + b^2 c + c^2 a$ and $B=a^2 c + b^2 a + c^2 b$. Now calculate $A+B$ and $AB$.
Now consider the quadratic whose roots are $A$ and $B$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3805574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Experienced mathematicians simplifying messy algebra $$\frac{pa}{n}\left(p\frac{a-1}{N-1}+q\frac{b+1}{N+1}\right)+p\left(1-\frac{a}{N}\right)\left(\frac{pa}{N-1}+\frac{qb}{N+1}\right)+\frac{qb}{N}\left(p\frac{a+1}{N+1}+q\frac{b-1}{N-1}\right)+q\left(1-\frac{b}{N}\right)\left(\frac{pa}{N+1}+\frac{qb}{N-1}\right)=\frac{(... | $$\frac{pa}{N}\left(p\frac{a-1}{N-1}+q\frac{b+1}{N+1}\right)+p\left(1-\frac{a}{N}\right)\left(\frac{pa}{N-1}+\frac{qb}{N+1}\right)+\frac{qb}{N}\left(p\frac{a+1}{N+1}+q\frac{b-1}{N-1}\right)+q\left(1-\frac{b}{N}\right)\left(\frac{pa}{N+1}+\frac{qb}{N-1}\right)= \\
\frac{pa}{N}\left(\frac{pa}{N-1} -\frac{p}{N-1}+\frac{qb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3806463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Determine dlog in quotient rings of polynomial rings Question:
Determine $\operatorname{dlog}_x (x^2 + 1)$ in $\Bbb Z_5[x]/\langle\,x^3 + x + 1\,\rangle$
So I know the elements of $F = \Bbb Z_5[x]/\langle\,x^3 + x + 1\,\rangle $ are of the form $ax^2 + bx + c \bmod x^3 + x + 1$ ($a, b, c \in \Bbb Z_5$). I know how to... | *
*$x^1 = x$.
*$x^2= x^2$.
*$x^3 \equiv -x-1 \equiv 4x+4$. Here we use that $x^3 + x +1 \equiv 0$ in this field (assuming it is a field, haven't checked), and coefficients are mod 5.
*$x^4 \equiv 4x^2 + 4x$.
*$x^5 \equiv 4x^3 + 4x^2 \equiv 4(-x-1)+ 4x^2 \equiv 4x^2 + x +1$.
*$x^6 \equiv 4x^3 + x^2 + x \equiv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3807092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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For $\alpha\in(0^\circ;90^\circ)$ simplify $\sin^2\alpha+\tan^2\alpha+\sin^2\alpha.\cos^2\alpha+\cos^4\alpha$ For $\alpha\in(0^\circ;90^\circ)$ simplify $\sin^2\alpha+\tan^2\alpha+\sin^2\alpha\cdot\cos^2\alpha+\cos^4\alpha.$
My try: $\sin^2\alpha+\tan^2\alpha+\sin^2\alpha\cdot\cos^2\alpha+\cos^4\alpha=\sin^2\alpha+\dfr... | You're almost there!
$$\dfrac{\sin^2\alpha\cdot\cos^2\alpha+\sin^2\alpha+\overbrace{\sin^2\alpha\cdot\cos^4\alpha+\cos^6\alpha}}{\cos^2\alpha}$$
$$=\dfrac{\sin^2\alpha\cdot\cos^2\alpha+\sin^2\alpha+\cos^4(\alpha)(\cos^2(\alpha)+\sin^2(\alpha))}{\cos^2\alpha}$$
$$=\dfrac{\sin^2\alpha+ \overbrace{\sin^2\alpha\cdot\cos^2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3807240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
For $\alpha\in(0^\circ;90^\circ)$ simplify $E=\frac{\sin^2\alpha}{\sin\alpha-\cos\alpha}-\frac{\sin\alpha+\cos\alpha}{\tan^2\alpha-1}$ For $\alpha\in(0^\circ;90^\circ)$ simplify $E=\dfrac{\sin^2\alpha}{\sin\alpha-\cos\alpha}-\dfrac{\sin\alpha+\cos\alpha}{\tan^2\alpha-1}.$
My try: $E=\dfrac{\sin^2\alpha}{\sin\alpha-\cos... | $$\dfrac{\sin^2\alpha}{\sin\alpha-\cos\alpha}-\dfrac{\sin\alpha+\cos\alpha}{\tan^2\alpha-1}=\dfrac{\sin^2\alpha}{\sin\alpha-\cos\alpha}-\dfrac{\cos^2\alpha(\sin\alpha+\cos\alpha)}{\sin^2\alpha-\cos^2\alpha}=$$
$$=\dfrac{\sin^2\alpha}{\sin\alpha-\cos\alpha}-\dfrac{\cos^2\alpha}{\sin\alpha-\cos\alpha}=\sin\alpha+\cos\alp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3807369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Finding an expression and proof for $\sqrt{1-4x}$ This is similar to $$\frac{1}{\sqrt{1-4x}}=\sum_{n\geq0}{2n\choose n}x^n$$
However I want to find an expression the same way for $$\sqrt{1-4x}$$ rather than $$\frac{1}{\sqrt{1-4x}}$$
Here's my thoughts so far:
$$(1-4x)^\frac{1}{2}=\sum_{n\geq0}{\frac{1}{2}\choose n}(-4)... | Notice that $$\frac{d}{dx}\sqrt{1-4x}=\frac{-2}{\sqrt{1-4x}}$$
So now, $$\frac{-1}{2}\frac{-2}{\sqrt{1-4x}}=\sum_{n\geq 0}\binom{2n}{n}x^n$$ and integrating with respect to $x$ yields $$\sum_{n\geq 0}\binom{2n}{n}\frac{x^{n+1}}{n+1}=-\frac{1}{2}\sqrt{1-4x} +C $$ so we get that $$\sqrt{1-4x}+C=-2\sum_{n\geq 0}\binom{2n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3807819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the smallest 3 digit number which satisfies the following conditions
The smallest 3 digit number n such that if the 3 digits are $a,b$ and $c$, then
$n = a+b+c+ab+bc+ac+abc$.
I tried $n = 100a + 10b + c$ (say)
Plugged that into the given equation but couldn't simplify it further. I have no idea how should I pro... | To find a lower bound for the answer we calculate necessary conditions for $a = 1$ first. Then we want to solve $$100 + 10b + c = 1 + b + c + b + bc + c+ bc = 1+ 2b + 2c+ 2bc,$$
i.e. $99 + 8b =c + 2bc = (1+2b)c $ and thus $$c = \frac{99+8b}{1+2b}.$$
From this we obtain $1+2b \mid 99 + 8b$, so $1+2b \mid 99+8b - 4\cdot ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3811021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Implicit derivative of $x^2+y^2=(2x^2+2y^2-x)^2$ Hi there.
I have the following function and want to calculate $y'$:
$x^2+y^2=(2x^2+2y^2-x)^2$
I've used implicit differentiation to solve it and my answer is:
$y'=\frac{(-4x^3+3x^2-4xy^2+y^2)}{y(8x^2-4x+8y^2-1)}$
However, my calculation gets huge ( I use the chainrule) a... | $$x^2+y^2=(2x^2+2y^2-x)^2$$
Differentiate with respect to the variable $x$:
$$2x+2yy'=2(2x^2+2y^2-x)(4x+4yy'-1)$$
$$x+yy'=(2x^2+2y^2-x)(4x+4yy'-1)$$
$$(x+yy')(1-4(2x^2+2y^2-x))=-(2x^2+2y^2-x)$$
$$(x+yy')=-\dfrac {(2x^2+2y^2-x)}{(1-4(2x^2+2y^2-x))}$$
Finally we get;
$$y'=\dfrac 2y\dfrac {(4x^3+4y^2x-3x^2-y^2)}{(1-4(2x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3812115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving $6(x^3+y^3+z^3)^2 \leq (x^2+y^2+z^2)^3$, where $x+y+z=0$ Question :
Let $x,y,z$ be reals satisfying $x+y+z=0$. Prove the following inequality:$$6(x^3+y^3+z^3)^2 \leq (x^2+y^2+z^2)^3$$
My Attempts :
It’s obvious that $x,y,z$ are either one negative and two positive numbers or one positive and two negative num... | $$z=-(x+y)$$
$$\implies 6 [x^3 + y^3 - (x+y)^3]^2 - [x^2 + y^2 +(x+y)^2]^3$$
$$=-8 x^6 - 24 x^5 y + 6 x^4 y^2 + 52 x^3 y^3 + 6 x^2 y^4 - 24 x y^5 - 8 y^6$$
$$=-2\,(x-y)^2 \, (2 x^2 + 5 x y + 2 y^2)^2\leq 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3817541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
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If $a, b, c, d>0$ and $abcd=1$ prove that an inequality holds true If $a, b, c, d>0$ and $abcd=1$ prove that:
$$\frac{a+b+c+d}{4}\ge\frac{1}{a^3+b+c+d}+\frac{1}{a+b^3+c+d}+\frac{1}{a+b+c^3+d}+\frac{1}{a+b+c+d^3}$$
I attempted to solve it in the following way:
$$\begin{equation}\frac{1}{a^3+b+c+d}+\frac{1}{a+b^3+c+d}+\f... | We can not finish it because the inequality, which you'll get after your step is wrong.
Indeed, the degree of the left side is $1$ and the degree of your expression is $\frac{3}{2},$
which says that after homogenization we'll get a wrong inequality for $d\rightarrow0^+$.
Indeed, it's enough to prove that:
$$\frac{a+b+c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3819122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\lim\limits_{x^2 + y^2 \to +\infty} x^2 -2xy + 2y^2 = +\infty$
Prove that $$\lim_{x^2 + y^2 \to +\infty} x^2 -2xy + 2y^2 = +\infty$$
My attempt:
$$x^2 + 2y^2 = x^2+y^2 + y^2 \implies \lim_{x^2 + y^2 \to +\infty}x^2 +2y^2 = +\infty$$
Then, from Cauchy-Schwarz:
$$x^2 + 2y^2 \geq 2\sqrt2xy \geq 2xy $$
Thus,
$$x... | Because $$x^2-2xy+2y^2\geq\frac{1}{3}(x^2+y^2)\rightarrow+\infty.$$
$$x^2-2xy+2y^2\geq\frac{1}{3}(x^2+y^2)$$ it's
$$3x^2-6xy+6y^2\geq x^2+y^2$$ or
$$2x^2-6xy+5y^2\geq0,$$ which is true because $$3^2-2\cdot5<0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3822564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Show that if $n$ divides $a^n-b^n$ then $n$ divides $\frac{a^n-b^n}{a-b}$
Let $a,b,n \in \mathbb Z^+$. Show that if $n$ divides $a^n-b^n$ then $n$ divides $\frac{a^n-b^n}{a-b}$.
This is from Apostol’s Introduction to Analytic Number Theory, Chapter $5$, exercise $13$.
It is trivial when $\gcd(n,a-b) = 1.$
It is also ... | Let $d=\gcd(n,a-b)$. The explicit expression
$$\frac{a^n-b^n}{a-b}=a^{n-1}+a^{n-2}b+a^{n-3}b^2+\ldots+a^2b^{n-3}+ab^{n-2}+b^{n-1},$$
shows that, because $a\equiv b\pmod{d}$, we also have
$$a^{n-1}+a^{n-2}b+a^{n-3}b^2+\ldots+a^2b^{n-3}+ab^{n-2}+b^{n-1}\equiv na^{n-1}\equiv0\pmod{d}.$$
That is to say $d$ divides $\tfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3825775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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On the integer solutions of some exponential equations I am trying to find the non-negative integer solutions of the following exponential equations:
$2^a=3^b+1\quad\color{blue}{(1)}$
$3^a=2^b+1\quad\color{blue}{(2)}$
I have found a way to obtain the non-negative integer solutions of these two exponential equations, bu... | Here's a condensed proof:
If $a$ and $b$ are integers such that $2^a=3^b+1$ then reducing mod $3$ shows that $a$ is even if $b>0$, say $a=2c$. Then
$$3^b=2^a-1=2^{2c}-1=(2^c-1)(2^c+1),$$
and so the two factors are both powers of $3$. They differ by $2$ so they are $1$ and $3$, so $b=c=1$, and $a=2$. If $b=0$ then clear... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3827429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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The polynomial $ x^7 + x^2 +1$ is divisible by (A) $ x^5 - x^4 + x^2 -x +1 \quad$ (B) $ x^5 + x^4 +1 \quad$ (C) $ x^5 + x^4 + x^2 +x +1\quad$
(D) $ x^5 - x^4 + x^2 +x +1$
My effort: Looking at the polynomial, I know that it will have only one real root, which is negative. All other 6 roots should be imaginary. And that... | Polynomial $x^7+x^2+1$ has higher term coefficient $1$ and also constant term coefficient $1$.
Thus if we divide it by a polynomial of degree $5$ of the form $(x^5+\cdots+1)$ as $A,B,C,D$ proposition are, it will be $x^2+ax+1$, only the central coefficient is unknown, the other two are forced to $1$.
If you multiply th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3828174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
$ \lim_{x \to 0}x \tan (xa+ \arctan \frac{b}{x})$ I have to evaluate the following limit
$$ \lim_{x \to 0}x \tan (xa+ \arctan \frac{b}{x})$$
I tried to divide tan in $\frac{sin}{cos}$ or with Hopital but I can't understand where I'm making mistakes.
The final result is:
$\frac{b}{1-ab}$ if $ab \ne 1$
$- \infty$ if $ab... | If you perform the trigonometric expansion, you should end with
$$y=x \tan \left(a x+\tan ^{-1}\left(\frac{b}{x}\right)\right)=x\frac{b \cos (a x)+x \sin (a x)}{x \cos (a x)-b \sin (a x)}$$ and now, there is a problem because of the denominator.
Using Taylor expansions, we have
$$y=x \frac {b+ \left(a-\frac{a^2 b}{2}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3828453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Solving $z^4=(2+3i)^4$ To solve the equation, I calculated right side:
$z^4=(2+3i)^4=(-5+12i)^2=-119-120 i$
And then I get the correct answer:
$z_k=\underbrace{\sqrt[8]{119^2+120^2}}_{\sqrt{13}} \times Cis(\cfrac{\pi+\tan^{-1}(\frac{120}{119})}{4}+\cfrac{k \pi}{2}), k=0,1,2,3$
But, I am looking for a way to solve the e... | Hint:
Use the fact
$$x^2-a^2=(x-a)(x+a)$$
and
$$x^2+a^2=(x-ai)(x+ai)$$
so
$$z^4-(2+3i)^4=0$$
$$\left ( x^2-(2+3i)^2 \right )\left ( x^2+(2+3i)^2 \right )=0$$
$$\left ( x-(2+3i) \right )\left ( x+(2+3i) \right )\left ( x-(2+3i)i \right )\left ( x+(2+3i)i \right )=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3831147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 4
} |
Finding range of a equation with given condition. We are given three real number $$a,b,c$$. With the condition $$(a^2+1)(b^2+1)(c^2+1)=9$$
We have to find the number of integers in the range of $$(ab+bc+ca)$$
Ans given is 7.
| $(a^2+1)(b^2+1)(c^2+1)=((a+b)^2+(ab-1)^2)(c^2+1)=(ab+bc+ca-1)^2+(a+b+c-abc)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3832844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.