Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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What is the value of $\alpha^{8}+\beta^{8}+\gamma^{8}$ if $\alpha$, $\beta$ and $\gamma$ are roots of the equation $x^3+x-1$? What is the value of $\alpha^{8}+\beta^{8}+\gamma^{8}$ if $\alpha$, $\beta$ and $\gamma$ are roots of the equation $x^3+x-1$? Is there a shorter way of finding the answer apart from finding the ... | $a,b,c$ be the roots of $x^3+x-1=0$, then $a^3=1-a \implies a^8=\frac{(1-a)^3}{a}=\frac{1-a^3-3a+3a^2}{a}=\frac{1-(1-a)-3a+3a^2}{a}$
$$\implies a^8=3a-2,\implies a^8+b^8+c^8=3(a+b+c)-6=-6$$
as the sum of the roots is zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3853577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Hint to prove $\sin^4(x) + \cos^4(x) = \frac{3 + \cos(4x)}{4}$ Could go from LHS to RHS by adding zero but I need to know how to do this WITHOUT knowing the half-angle formula. So from RHS to LHS, you an expand $\cos4x$ twice. I get as close as
$$\frac{ \cos^4x + \sin^4x + 3(1 - 2\sin^2x\cos^2x)}{4}$$
| Answer:
After Euler formula We know that:
$Cos(x) ^4 = \frac{cos(4x)}{8}+\frac{cos(2x)}{2}+\frac{3}{8}$
And
$sin(x) ^4 =\frac{cos(4x)}{8}-\frac{cos(2x)}{2}+\frac{3}{8}$
So,
$ Cos(x) ^4 +sin(x) ^4=\frac{cos(4x)}{4}+\frac{3}{4} =\frac{cos(4x)+3}{4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3854825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplify $\frac{x^3+1}{x+\sqrt{x-1}}$
Simplify $$A=\dfrac{x^3+1}{x+\sqrt{x-1}}.$$
Firstly, $x-1\ge0$ and $x+\sqrt{x-1}\ne0:\begin{cases}x-1\ge0\\x+\sqrt{x-1}\ne0\end{cases}.$ The first inequality is equivalent to $x\ge1$. Can we use that in the second inequality? I mean can we say that $x+\sqrt{x-1}>0$ because $x>0$ ... | We have:
$$x^2-x+1 = x^2-(x-1) = (x-\sqrt{x-1}) (x+\sqrt{x-1})$$
Can you do anything with that?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3857023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve this integral $I = \int\dfrac{\cos^3x}{\sin x + \cos x}dx$? $\displaystyle\int\dfrac{\cos^3x}{\sin x + \cos x}dx$
I added $J =\displaystyle \int\dfrac{\sin^3x}{\sin x + \cos x}dx$
then $I + J = \displaystyle\int\dfrac{\cos^3x + \sin^3x}{\sin x + \cos x}dx = x + \dfrac{1}{2}\cos2x + C$
but I can't find how ... | Reducing the degree may be also a good idea...
$$
\begin{aligned}
I &=
\int\frac{\cos^3x}{(\cos x + \sin x)}\;dx =
\int\frac{\cos^3x(\cos x - \sin x)}{(\cos x + \sin x)(\cos x - \sin x)}\;dx
\\
&=
\int\frac{\cos^4x - \cos^3x\sin x}{\cos^2 x - \sin^2 x}\;dx
=
\int\frac{(\cos^2x)^2}{\cos 2x }\;dx
-
\int\frac{\cos^2x\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3857749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Rationalize the denominator of $2\over{2 - \sqrt[4]{2}}$?
Rationalize the denominator of $2\over{2 - \sqrt[4]{2}}$.
Here's my progress. Let $x = \sqrt[4]{2}$. Then our expression can be written as $x^4/(x^4 - x)$, which simplifies to $x^3/(x^3 - 1)$. Multiply top and bottom by $(x^3 + 1)$ to get $x^3(x^3 + 1)/(x^6 - ... | I would not use wolfram alpha. Rather rely on the back of the book for steps if possible. Or better yet, do it by hand first. Wolfram alpha can you give another result that's equivalent but looks different.
Typically when you see a radical in a denominator of a fraction we prefer to rationalize denominator. So in this ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Let $P(z) = az^3+bz^2+cz+d$ , where $a, b, c, d $ are complex numbers with $|a| = |b| = |c| = |d| = 1.$
Let $P(z) = az^3+bz^2+cz+d$
, where $a, b, c, d $ are complex numbers with $|a| = |b|
= |c| = |d| = 1.$ Show that $|P(z)| ≥ \sqrt{6}$ for at least one complex number z satisfying
$|z| = 1.$
Attempt
By triangle ineq... | The following is inspired by Bound on a complex polynomial on AoPS.
For $|z| = 1$ we have $\overline z = 1/z$, so that expanding $|P(z)|^2 = P(z)\overline{P(z)}$ gives
$$
|P(z)|^2 = 4 + 2 \operatorname{Re} \left( a \overline b z + a \overline c z^2
+ a \overline d z^3 + b \overline c z + b \overline d z^2 + c \overl... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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$\lim_{x \rightarrow 1} \frac{ x^2-1 + \sqrt{x^3+1} - \sqrt{x^4+1} }{ x-1 + \sqrt{x+1} - \sqrt{x^2+1} } $ Calculate below limit
$$\lim_{x \rightarrow 1} \frac{ x^2-1 + \sqrt{x^3+1} - \sqrt{x^4+1} }{ x-1 + \sqrt{x+1} - \sqrt{x^2+1} } $$
Using L'Hôpital's rule might be too tedious. I wonder if there is a trick given the ... | $$ \frac{ x^2-1 + \sqrt{x^3+1} - \sqrt{x^4+1} }{ x-1 + \sqrt{x+1} - \sqrt{x^2+1} }=\frac{x+1-\frac{x^3}{ \sqrt{x^3+1} +\sqrt{x^4+1}}}{1-\frac{x}{ \sqrt{x+1}+\sqrt{x^2+1}}}\rightarrow\frac{2-\frac{1}{2\sqrt2}}{1-\frac{1}{2\sqrt2}}=$$
$$=\frac{4\sqrt2-1}{2\sqrt2-1}=\frac{15+2\sqrt2}{7}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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We have sides $a,b,c$ of triangle $ABC$ which are equal to $\sqrt{6}, \sqrt{3}, 1$, Prove that angles $A=90+B$ We have sides $a,b,c$ of triangle $ABC$ which are equal to $\sqrt{6}, \sqrt{3}, 1$, Prove that angles $A=90+B$
I just did the question above in the following way:
$t=\frac{\sqrt{6}+\sqrt{3}+1}{2}$
From Heron w... | Cosine law for $A$ and $B$: $$6=3+1-2\sqrt3\cos A$$ $$3=6+1-2\sqrt6\cos B$$ $$\therefore\quad\cos A=-\frac{1}{\sqrt3},\qquad \cos B=\sqrt{\frac{2}{3}}$$
$$\therefore\quad\cos(A-B)=-\sqrt\frac{1}{3}\sqrt{\frac{2}{3}}+\sqrt{\frac{2}{3}}\sqrt\frac{1}{3}=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3865515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Where am I missing the factor $\frac12$ in my fourier series expansion?
Question: A periodic function $f(t)$, with period $2\pi$ is defined as,
$$
f(t) = \begin{cases}
0 & \text{ if } -\pi<t<0, \\
\pi & \text{ if } 0<t<\pi.
\end{cases}
$$
Find the Fourier series expansion of $f$.
Below is my working:
Since $f$ i... | To see that the coefficients are wrong,
The Fourier series for $f(x)$ may be written
\begin{aligned}
f(x) &= a_0 + \sum_{n=1} a_n \cos n x + \sum_{n=1} b_n \sin n x
\end{aligned}
The coefficients may be calculated
\begin{aligned}
a_0 &= \frac{\int_{-\pi}^{\pi} f(x) dx}{\int_{-\pi}^{\pi} 1 dx} \\
a_n &= \frac{\int_{-\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3866784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Claculate limit $\lim_{x\to 0}\frac{1-(\cos(1-\sqrt{\frac{\sin(x)}{x}}))}{x^4}$ I have a problem to calculte this limit:
$$\lim_{x\to 0}\frac{1-(\cos(1-\sqrt{\frac{\sin(x)}{x}}))}{x^4}$$
I used Taylor expansion for $\sin(x), \cos(x)$ and considered also $1-\cos(\alpha)=2\sin^2(\frac{\alpha}{2})$ and $\alpha=2-2\sqrt{\f... | \begin{align}
\lim_{x\to 0}\frac{1-\left\{\cos\left(1-\sqrt{\dfrac{\sin(x)}{x}}\right)\right\}}{x^4}
&=\lim_{x\to0}\underbrace{\dfrac{1-\left\{\cos\left(1-\sqrt{\dfrac{\sin(x)}{x}}\right)\right\}}{\left(1-\sqrt{\dfrac{\sin x}{x}}\right)^2}}_{=\frac12}\times\left(\dfrac{1-\sqrt{\dfrac{\sin x}{x}}}{x^2}\right)^2\\
&=\dfr... | {
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Use mathematical induction to prove that (n+2)(n+3)(n+7) is divisible by 6. Use mathematical induction to prove that $q(n)=(n+2)(n+3)(n+7)$ is divisible by $6$.
I have already proved the base case at n=1. I need help on the second part to prove $n=k+1$.
What I did: $(n+2)(n+3)(n+7)=6P$
\begin{align*}
((k+2)+1)&((k+1)+3... | This is fine so far. You have divisibility by $3$ and only need to check that $k^2+9k+18$ is even to get divisibility by $6$. This is true because $k^2$ and $k$ are either both even or both odd.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3871936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
solve ;$\sqrt{\frac{1-4\cos^2 4x}{8\cos (2x-2\pi/3)}}=\cos (2x-\pi/6)$ This is one more of my unsolved trigonometry questions:
solve ;$$\sqrt{\frac{1-4\cos^2 4x}{8\cos (2x-2\pi/3)}}=\cos (2x-\pi/6)$$
My Try
provided that $\cos (2x-2\pi/3)\neq 0$ and squaring both sides $$1-4\cos^2 4x=8\cos (2x-2\pi/3)\cos^2(2x-\pi/... | It's been a tour de force, mate. Even with the help of Wolfram Mathematica.
After expanding with standard formulae and subbed $2x=u$ I got this mess
$$-2 \sin ^4 u-\sqrt{3} \sin u-2 \cos ^4 u+2 \cos ^3 u+\cos u+12 \sin ^2 u \cos ^2 u-6 \sin ^2 u \cos u-1=0$$
I substituted $$\sin u=\frac{2t}{1+t^2};\;\cos u=\frac{1-t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3872447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Converting $A8B34_{16}$ to octal. According to online calculators $A8B34_{16}=2505464_{\ 8}$, yet I keep getting $2005464_{\ 8}$. I just want to know where I'm going wrong.
$A8B34_{16}=10( 16^4)+8( 16^3) + 11( 16^2) + 3( 16) + 4$
$=10(2^48^4)+8(2^38^3)+11(2^28^2)+6(8)+4$
$=160(8^4)+8^5+44(8^2)+6(8)+4$
$=(8^2+7[8])(8^4... | We are lucky here that $8$ and $16$ are powers of two. It suffices to break down the numbers into bits and regroup:
$$A8B34_{16}=1010,1000,1011,0011,0100_2$$
$$=10,101,000,101,100,110,100_2=2505464_8$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3872912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For what value of c is the estimator consistent? Suppose $X_1,\ldots,X_n$ are i.i.d. normally distributed with unknown mean $\mu$ and unknown variance $\sigma^2$. Let $\bar{X}_n$ be the sample mean. Consider estimating $\sigma$. For any given constant $c$, define the estimator\footnote{followed this video for both pa... | Notice that $\operatorname E(X_i-\overline X) = 0$ and
\begin{align}
\operatorname{var}(X_i-\overline X) & = \operatorname{var}(X_i) + \operatorname{var}(\overline X) - 2\operatorname{cov}(X_i,\overline X) \\[8pt]
& = \sigma^2 + \frac{\sigma^2} n - \frac{2\sigma^2} n \\[8pt]
&= \frac{n-1} n \sigma^2.
\end{align}
So
\be... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Recurrence relation $a_n = 4a_{n-1} - 3a_{n-2} + 2^n + n + 3$ with $a_{0} = 1$ and $a_{1} = 4$ This is a nonhomogeneous recurrence relation, so there is a homogeneous and a particular solution.
Homogenous:
$a_n - 4a_{n-1} + 3a_{n-2} = 0$
$r^2 - 4r + 3 = 0$
$(r - 3)(r - 1)$
$a_n^h = \alpha(3^n) + \beta(1^n)$
This is whe... | Here's an alternative approach. Let $A(z)=\sum_{n\ge 0} a_n z^n$ be the ordinary generating function for $a_n$. Then the recurrence relation implies that
\begin{align}
A(z) - a_0 - a_1 z
&= \sum_{n\ge 2}\left(4a_{n-1} - 3a_{n-2} + 2^n + n + 3\right)z^n \\
&= 4z \sum_{n\ge 2} a_{n-1} z^{n-1} - 3z^2 \sum_{n\ge 2} a_{n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3873779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $ f(f(x)) \geq 0$ for all real x Let $f(x)= a x^2 + x +1 , x \in \mathbb{R} $. Find all values of parameter $a \in \mathbb{R} $ such that $f(f(x)) \geq 0 $ holds for all real $x$.
$f(x)> 0 $ iff $a> 0 $ and $ 1- 4a \leq 0$ which gives $a \geq \frac{1}{4} $ . But we have:
$f(f(x))= a ( a x^2 + x +1)^2 + ... | Firstly, $1-4a\leq0$ is valid and since $a=0$ is not valid, it's enough to check $0<a<\frac{1}{4}$, which gives
$$ax^2+x+1=\frac{-1+\sqrt{1-4a}}{2a}$$ and
$$ax^2+x+1=\frac{-1-\sqrt{1-4a}}{2a}$$ have no real roots.
It's enough to work with the first equation, which gives:
$$1-4a\left(1-\frac{-1+\sqrt{1-4a}}{2a}\right)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3874988",
"timestamp": "2023-03-29T00:00:00",
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Summation of $n$th partial products of the square of even numbers diverges, but for odd numbers they converge in this series I'm looking at. Why? So I have the two following series:
$$\sum_{n=1}^\infty \frac{\prod_{k=1}^n(2k)^2}{(2n+2)!}$$
$$\sum_{n=0}^\infty \frac{\prod_{k=0}^n(2k+1)^2}{(2n+3)!}$$
I figured out the $n... | Elaborating after @Erick Wong's comments.
You properly found that
$$a_n=\frac{4^n(n!)^2}{(2n+2)!}$$ Take logarithms
$$\log(a_n)=n \log(4)+2\log(n!)-\log((2n+2)!)$$ Use Stirling approximation twice and continue with Taylor series to find
$$\log(a_n)=\left(\frac{3}{2} \log \left(\frac{1}{n}\right)+\log \left(\frac{\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3875640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the roots of the polynomial $x^3-2$. Find the roots of the polynomial $x^3-2$.
If $\alpha$ be the root of this polynomial i.e., $\alpha^3=2$,
then $(\zeta \alpha)^3=2$, where $\zeta$ is $3^{rd}$ root of unity.
Hence the solutions of this equations are
$$\zeta{2}^\frac{1}{3}$$, $\zeta$ a $p^{th}$ root of unity
And... | Let's say $a > 0$ is a real number.
The roots of $x^3 = a$ are these
$x_1 = a^\frac{1}{3} ( \cos{2\pi/3} + i \cdot \sin{2\pi/3} )$
$x_2 = a^\frac{1}{3} ( \cos{4\pi/3} + i \cdot \sin{4\pi/3} )$
$x_3 = a^\frac{1}{3} ( \cos{6\pi/3} + i \cdot \sin{6\pi/3} )$
We can simplify these formulas further...
The root $x_3$ happens ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$x+1$ is always invertible in $\Bbb Z_{x^3}$.
Claim: For any $x > 1$ and $x \in \Bbb N$, $x+1$ is always invertible in $\Bbb Z_{x^3}$.
Proof: We know that $x^3 +1 = (x+1)(x^2-x+1)$. Since $x > 1$ and $x \in \Bbb N$, we have $x^2-x+1 >0$ and $(x+1)(x^2-x+1) = x^3 +1 \equiv 1 \mod x^3$. Hence $x+1$ is always invertible... | You proved the Bezout equation $\,1 = (\color{#c00}{x+1})f(x) - \color{#c00}{x^3}\,$ for $\,f(x) \in \Bbb Z[x]\,$
This implies that $\,\color{#c00}{x+1}\,$ and $\,\color{#c00}{x^3}\,$ are coprime in every ring, since any common divisor divides the RHS so also the LHS $= 1$, i.e. every common divisor is a unit (invertib... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3879147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Proving $\frac{7 + 2b}{1 + a} + \frac{7 + 2c}{1 + b} + \frac{7 + 2a}{1 + c} \geqslant \frac{69}{4}$. Here's the inequality
For positive variables, if $a+b+c=1$, prove that
$$
\frac{7 + 2b}{1 + a} +
\frac{7 + 2c}{1 + b} +
\frac{7 + 2a}{1 + c} \geqslant
\frac{69}{4}
$$
Here equality occurs for $a=b=c=\frac{1}{3}$ wh... | Another way:flipping the inequality it suffices to prove $$\sum_{cyc}\frac{7a-2b}{1+a}\le 15/4...(3)$$.
Indeed by jensen on $f(x)=\frac{x}{1+x}$ $$\sum_{cyc} \frac{a}{1+a}\le 3\sum_{cyc}\frac{\frac{(a+b+c)}{3}}{1+\frac{(a+b+c)}{3}}=3/4...(1)$$
This can also be done like this :
as $$\frac{x}{1+x}\le \frac{9x+1}{16}$$ w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3881347",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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find all $(x,y,z)$ such that $27^{3x^2 + 2y}+27^{3y^2 + 2z}+27^{3z^2 + 2x}=1$
Find all ($x,y,z$) such that
$$ 27^{3x^2 + 2y}+27^{3y^2 + 2z}+27^{3z^2 + 2x}=1$$
I am a high school student and would appreciate it if anyone could solve it using high school mathematics.
here is what I have tried till now:
$$ 3x^2 +2y < 0 ... | Hint:By am-gm $$1\ge3\sqrt[3]{{27}^{3(x^2+y^2+z^2)+2(x+y+z)}} = 3^{3(x^2+y^2+z^2)+2(x+y+z)+1}..(1)$$
But $$3(x^2+y^2+z^2)\ge {(x+y+z)}^2$$ when x=y=z
substituting in (1)
$$1\ge 3^{{(x+y+z+1)}^2}$$
which is possible only when $x+y+z=-1$ and $x=y=z$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3881531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Difference between anti-derivative and indefinite integral My teacher gave me the following integral to evaluate:
$$\int \frac{x^2}{(x\sin(x)+\cos(x))^2}dx$$
After half an hour of uselessly fumbling around with trig identities I gave up and plugged it into an integral calculator: https://www.integral-calculator.com/. H... | Computing the Indefinite Integral
$$
\begin{align}
&\int\frac{x^2}{(x\sin(x)+\cos(x))^2}\,\mathrm{d}x\\
&=\int\frac{x^2}{(x\sin(x)+\cos(x))^2}\frac{\mathrm{d}(x\sin(x)+\cos(x))}{x\cos(x)}\tag1\\
&=-\int\frac{x}{\cos(x)}\,\mathrm{d}\frac1{x\sin(x)+\cos(x)}\tag2\\
&=-\frac{x}{\cos(x)}\frac1{x\sin(x)+\cos(x)}+\int\frac1{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3882899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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(AIME 1994) $ \lfloor \log_2 1 \rfloor + \lfloor \log_2 2 \rfloor + \ldots + \lfloor \log_2 n \rfloor = 1994 $
$($AIME $1994)$ Find the positive integer $n$ for which
$$ \lfloor \log_2 1 \rfloor + \lfloor \log_2 2 \rfloor + \lfloor \log_2 3 \rfloor + \ldots + \lfloor \log_2 n \rfloor = 1994 $$ where $\lfloor x \rflo... | As noted, the sequence goes like
$$ 0,\underset{2}{\underbrace{1,1}},\underset{4}{\underbrace{2,2,2,2}},\underset{8}{\underbrace{3,3,3,3,3,3,3,3}},4,4,\ldots$$
i.e, every natural number $k$ occurs $2^k$ times.
So desired is $$ \sum k\cdot 2^k =1994$$
It's quick enough to attack directly:
$$ 1\cdot2 + 2\cdot4 + 3\cdot8 ... | {
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"source": "stackexchange",
"question_score": "4",
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How to find the dot product using the law of cosines I'm working with the following problem:
We have a triangle with sides $AB=3$ and $BC=2$, the angle $ABC$ is 60 degrees. Find the dot product $AC \cdotp AB$
Since we don't actually know the side $AC$ my first step is to calculate this side via the law of cosines.
$$AC... | $$\vec{AB}=\vec{AC}+\vec{CB}$$
and
$$\vec{AC}\cdot\vec{AB}=\vec{AB}^2-\vec{CB}\cdot\vec{AB}=9-3\cdot2\cdot\cos60°=6.$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Why is $\sum_{k=1}^{\infty}\left(\frac{1}{3}\right)^{k}=\frac{1/3}{2/3}$ $\sum_{k=1}^{\infty}\left(\frac{1}{3}\right)^{k}=\frac{1/3}{2/3}$ what theorem or algebra leads to this equality?
EDIT: The sum should have been infinite.
| Let $S = \sum_{k=1}^\infty \left(\frac{1}{3}\right)^k$. Then,
$$ S = \frac{1}{3} + \left(\frac{1}{3}\right)^2 + \left(\frac{1}{3}\right)^3 + \ldots = \frac{1}{3} \left[ 1 + \underbrace{\frac{1}{3} + \left(\frac{1}{3}\right)^2 + \ldots}_{=S} \right] =
\frac{1}{3} \left[ 1 + S \right]$$
So, we have $S = \dfrac{1}{3} + ... | {
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"answer_count": 2,
"answer_id": 1
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$\lim_{n\to\infty}\sqrt{n^3}(\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n})$ I need to find $\lim_{n\to\infty}{\sqrt{n^3}(\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n})}$ without using L'Hopital's rule, derivatives or integrals.
Empirically, I know such limit exists (I used a function Grapher and checked in wolfram) and it's equal to $-\frac{1}{4... | Multiplying $\sqrt{n+1} - \sqrt n$ by $\dfrac{\sqrt{n+1}+\sqrt n}{\sqrt{n+1}+\sqrt n}$ yields $\dfrac 1 {\sqrt{n+1}+\sqrt n}.$
Similarly $\sqrt n - \sqrt{n-1} = \dfrac 1 {\sqrt n + \sqrt{n-1}}.$
So then we have
\begin{align}
& \big(\sqrt{n+1} - \sqrt n\big) - \big(\sqrt n - \sqrt{n-1} \big) \\[8pt]
= {} & \frac 1 {\sqr... | {
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"answer_id": 2
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How to simplify $\frac {\sin 3A - \cos 3A}{\sin A + \cos A} + 1$? So I started by using $\sin 3A$ and $\cos 3A$ identities and then I added the lone $1$ to the trigonometric term. (Done in the picture below)
But after this I don't have any clue on how to proceed.
$$=\frac{3 \sin \theta-4 \sin ^{3} \theta-\left(4 \cos ^... | Recall the cube sum identity: $x^3+y^3 = (x+y)(x^2-xy+y^2)$
Continuing from where you left off:
\begin{align}&\quad4\frac {\sin\theta+\cos\theta-\sin^3\theta-\cos^3\theta}{\sin\theta+\cos\theta}\\
&=4-4\frac {\sin^3\theta+\cos^3\theta}{\sin\theta+\cos\theta}\\
&=4-4(\sin^2\theta - \sin\theta\cos\theta+\cos^2\theta)\\
&... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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Growth rate of $f(n) = \frac{1-\cos(\frac{n\pi}{n+1})}{1-\cos(\frac{\pi}{n+1})}$ I encountered this expression when studying the growth of condition numbers on a linear system. I was trying to verify that the order of the growth is $O(n^2)$. However I'm stuck upon proving this. Any hints would be appreciated. Thanks.
| Hint
$$1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right)$$
Then $$\begin{aligned}
f(n) &= \frac{1-\cos(\frac{n\pi}{n+1})}{1-\cos(\frac{\pi}{n+1})}\\
&= \frac{\sin^2\left(\frac{n \pi }{2(n+1)}\right)}{\sin^2\left(\frac{\pi}{2(n+1)}\right)}\\
& \le \frac{1}{\sin^2\left(\frac{\pi}{2(n+1)}\right)}\\
&\simeq \frac{\pi^2}{4}n^2
... | {
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determine $\iiint zdv$ using spherical coordinates Let $D$ be the region enclosed by the surface $z = \sqrt{4-x^2-y^2}$ and $z = -\sqrt{x^2+y^2}$
Determine:
$$\iiint zdv$$
using Spherical coordinates.
I'm trying to solve the above but I'm unsure how to go about it.
My attempt has been:
$$\int_0^{2\pi}\int_0^{2}\int_{-\... | Combine $z = \sqrt{4-x^2-y^2}$ and $z = -\sqrt{x^2+y^2}$ to get $z=-\sqrt2$, which corresponds to the boundary $\cos\theta = \frac zr =-\frac{\sqrt2}2$, or $\theta =\frac{3\pi}4$, in spherical coordinates.
Then, the integral is
$$\int_V zdv = 2\pi \int_{\frac{3\pi}4}^\pi\int_0^{2}(r\cos\theta )\>r^2\sin\theta drd\theta... | {
"language": "en",
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Show that $1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+3)}{3}$ using the identity $1^2+2^2+\dots + n^2=\frac{n(n+1)(2n+1)}{6}$ without induction? I've been trying to solve the following problem:
Show that $1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+1)}{3}$ using the identity $1^2+2^2+\dots + n^2=\frac{n(n+1)(2n+1... | You have a typo. Call the known sum $P_n$. The desired sum is$$\begin{align}P_{2n+1}-4P_n&=\frac{(2n+1)(2n+2)(4n+3)-4n(n+1)(2n+1)}{6}\\&=\frac{(n+1)(2n+1)}{3}(4n+3-2n)\\&=\frac{(n+1)(2n+1)(2n+3)}{3}.\end{align}$$
| {
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"source": "stackexchange",
"question_score": "1",
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Systems of Congruences \begin{cases} \overline{xyz138} \equiv 0 \mod7 \\ \overline{x1y3z8} \equiv 5 \mod11 \\ \overline{138xyz} \equiv 6 \mod13 \end{cases}
I worked my way up to this:
\begin{cases} 2000x+200y+20z \equiv 2 \mod7 \\ 100000x+1000y+10z \equiv 4 \mod11 \\ 2000x+200y+20z \equiv 7 \mod 13 \end{cases}
I tried ... | If your workings are correct, you can rewrite your equations as:
\begin{cases}
20(\overline{xyz}) &\equiv 2 \pmod 7\\
10(x+y+z) &\equiv 4 \pmod {11}\\
20(\overline{xyz}) &\equiv 7 \pmod {13}
\end{cases}
which reduces to
\begin{cases}
\overline{xyz} &\equiv 5 \pmod 7\\
x+y+z &\equiv 7 \pmod {11}\\
\overline{xyz} &\equiv... | {
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"source": "stackexchange",
"question_score": "3",
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How do you prove that $\int \frac{1}{\sqrt{x^2+r^2}} \,dx = \ln{\left\lvert x + \sqrt{x^2+r^2} \right\rvert + C}$? I would like to prove that
$$\int \frac{1}{\sqrt{x^2+r^2}} \,dx = \ln{\left\lvert x + \sqrt{x^2+r^2} \right\rvert} + C$$
I tried with applying derivative with respect to $x$ to $\ln{\left\lvert x + \sqrt{x... | HINT
What about the substitution $x = r\sinh(u)$? In such case, one has that
\begin{align*}
\int\frac{1}{\sqrt{x^{2}+r^{2}}}\mathrm{d}x & = \int\frac{r\cosh(u)}{\sqrt{r^{2}\sinh^{2}(u) + r^{2}}}\mathrm{d}u = \int1\mathrm{d}u = u + c
\end{align*}
Can you take it from here?
EDIT
Since the $\sinh$ function is bijective, t... | {
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"question_score": "1",
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Let $x,y \in R$ such that $|x+y| + |x-y| = 2$
Let $x,y \in R$ such that $|x+y| + |x-y| = 2$, then find maximum values of $x^2 - 6x + y^2$ and $x^2 + y^2 + 10y$.
How do I go about solving this question? Is it possible to find all real values of $x$ and $y$ from the first equation? Please help.
| Guide:
Notice that $|x+y|+|x-y|=2$ describes the boundary of a square.
$x^2-6x+y^2=(x-3)^2+y^2-9$
The largest value of $y^2$ is $1$. Given that $-1 \le x \le 1$. Try to solve the $-1 \le x \le 1$, how would you optimize $(x-3)^2$.
| {
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"source": "stackexchange",
"question_score": "2",
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Finding the key matrix of a 2x2 Hill Cipher I'm trying to find the Hill Cipher key from the following given info:
(1,3)^T is encrypted as (-9, -2)^T, and (7, 2)^T is encrypted as (-2, 9)^T.
However I don't seem to end up with the correct answer as the encryption is invalid when I try to use the key. Below is my solutio... | I fully agree with
$$\begin{bmatrix} a & b\\ c & d\end{bmatrix} = \begin{bmatrix} -9 & -2\\ -2 & 9\end{bmatrix} {\begin{bmatrix} 1 & 7\\ 3 & 2\end{bmatrix}}^{-1}$$
and indeed the determinant of the right hand side matrix equals $$2- 21 \pmod{26} = -19 = 7 \pmod{26}$$ which is relatively prime to $26$ so has an inverse.... | {
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"source": "stackexchange",
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How to solve $\int \sqrt{1+\sin x}\, dx$? It's easy to get this:
$$\int \sqrt{1+\sin x}\, dx \\= \int \sqrt{ \sin^2{\frac{x}{2}} + \cos^2{\frac{x}{2}} + 2\sin{\frac{x}{2}}\cos{\frac{x}{2}}}\,\, dx \\ = \int \left | \sin{\frac{x}{2}} + \cos{\frac{x}{2}} \right |\, dx \\= \sqrt{2} \int \left | \sin{\left ( \frac{x}{2} +... | Hint: $|\sin x|=(-1)^n \sin x$ for $\pi n \leqslant x < \pi (n+1)$, where $n=0, \pm 1,\pm2, \cdots$.
To keep antiderivative continuous in $x=\pi (n+1)$ you need to solve
$$\left((-1)^{n+1}\cos x +C_n\right) \Big|_{x=\pi (n+1)}=\left((-1)^{n+2}\cos x +C_{n+1}\right)\Big|_{x=\pi (n+1)}$$
taking some $C=C_0$.
| {
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"source": "stackexchange",
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"answer_id": 1
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IMO 2019 Problem N5 Solution 2 IMO 2019 Problem N5 Solution 2.
https://www.imo-official.org/problems/IMO2019SL.pdf
Let $a$ be a positive integer. We say that a positive integer $b$ is $a$-good if $\binom{an}{b}-1$ is divisible by $an+1$ for all positive integers $n$ with $an\geqslant b$. Suppose $b$ is a positive inte... | I think that if $n$ satisfies $an\ge b+2,\binom{an}{b}\equiv 1\pmod{an+1}$ and $\binom{an}{b+2}\not\equiv 1\pmod{an+1}$, then $an+1$ is a composite number.
To prove this, it is sufficient to prove that if $n$ is such that $an+1$ is prime satisfying $an\ge b+2$ and $\binom{an}{b}\equiv 1\pmod{an+1}$, then $\binom{an}{b+... | {
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"source": "stackexchange",
"question_score": "2",
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Circle $O$ is tangential to triangle $ABC$ at $B$ and passes $C$. Express $AO$ as $\frac{a√b}{c√d}$ Let $ABC$ be a triangle with side lengths $AB = 7, BC = 8, AC = 9$. Draw a circle tangent to $AB$ at $B$ and passing through $C$. Let the center of the circle be $O$. The length of $AO$ can be expressed as $\frac{a√b}{c√... |
According to the cosine rule
$$\cos B = \frac{7^2+8^2-9^2}{2\cdot 7\cdot8}=\frac27,\>\>\>\>\>\sin B= \frac{3\sqrt5}7
$$
Then, the radius $OB$ is
$$OB = \frac{BC}{2\cos\angle OBC}=\frac4{\sin B}=\frac{28}{3\sqrt5}
$$
Per the Pythagoras' Theorem
$$AO=\sqrt{AB^2+BO^2}=\sqrt{7^2+ \frac{28^2}{45}}=\frac{7\sqrt{61}}{3\sqrt5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3918900",
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Determine conic section $x^2-4xy+4y^2-6x-8y+5=0$ and its center So I got a task to determine the conic section of the following:
$$x^2-4xy+4y^2-6x-8y+5=0$$
I started using matrices and got to the equation :
$$5x`^2 + \frac {10x`}{\sqrt5}-\frac {20y`}{\sqrt5}+5=0$$
I completed the square and got to the equation :
$$\fra... | The given equation of the conic is
$ x^2 - 4 x y + 4 y^2 -6x - 8y + 5 = 0 $
If we define $r = [x, y]^T $ then the given conic equation can be written in matrix-vector form as
$ r^T A x + b^T x + c = 0 $
where
$ A = \begin{bmatrix} 1 && - 2 \\ -2 && 4 \end{bmatrix} $
$ b = [-6, -8] $
$ c = 5
First we need to check if $A... | {
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"source": "stackexchange",
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Small angle approximation on cosine The problem is
Using the small angle approximation of cosine, show that $3-2\cos(x)+4\cos^2(x)\approx 5-kx^2$ where k is a positive constant
I did solve it by using $\cos^2(x)=1-\sin^2(x)$ on the $\cos^2(x)$, by plugging $\sin^2(x)\overset{x\to 0}{\approx}x^2$ and $\cos(x)\overset{... | Directly subbing $\cos^2x=(1-x^2/2+\cdots)^2$ should work out, provided you expand properly:
$$\cos^2x=1-2(x^2/2)+\dots=1-x^2+\cdots$$
$$1-\sin^2x=1-(x-\cdots)^2=1-x^2+\cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3927646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Matrix Difference Equations Using Undetermined Coefficients?
$\begin{cases}x[n + 1] = -x[n] + 3 \\ y[n + 1] = -y[n] - e^{-n}\end{cases}$
Although I realize this system happens to be decoupled, I want to solve it using a general technique which doesn't depend on this contingency. I first solve the homogeneous system,... | $\def\A{{\bf A}}
\def\B{{\bf B}}
\def\C{{\bf C}}
\def\M{{\bf M}}
\def\I{{\bf I}}
\def\O{{\bf 0}}
\def\x{{\bf x}}
\def\y{{\bf y}}$There is a minor sign error in your work. Note that $b_2 = -e/(1+e)$. Taking this into account, you will arrive at the correct solution.
Below we consider the technique in a fairly general ... | {
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"url": "https://math.stackexchange.com/questions/3928116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Bound on the coordinates of torsion points on the elliptic curve: $y^2 = x^3 - k^2x + k^3$ Consider the elliptic curve $y^2 = x^3 - k^2x + k^3$ where $k$ is a non-zero integer.
I've come across a question that asks to show that if $(x,y)$ is a rational torsion point on the elliptic curve, then:
$$\vert{y}\vert \le 5\ve... | For $y$, $y^2|-23k^6$, then $|y|^2\leq |23k^6|\leq|25k^6|$, so $|y|\leq 5|k^3|$.
For $x$, we know $|x^3 - k^2x + k^3|\leq 25|k|^6$, thus $|x^3 - k^2x| \leq|k^3|+25|k|^6$, which is also $|x||x ^2- k^2|\leq|k^3|+25|k|^6$.
If $|x|>3k^2$, then $|x||x ^2- k^2|>3k^2(9k^4-k^2)$, so $3k^2(9k^4-k^2)<|k^3|+25|k|^6$, we now get $... | {
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To find the position of the ant after 2020 moves is $(p, q)$, An ant is moving on the coordinate plane. Initially it was at (6, 0). Each move
of the ant consists of a counter-clockwise rotation of 60◦ about the origin fol-
lowed by a translation of 7 units in the positive x-direction. If the position of
the ant after 2... | In the complex plane, a rotation of $60^\circ$ is achieved by multiplying by $e^{i\frac{\pi}{3}} = \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} = \frac{1}{2} + i \frac{\sqrt{3}}{2}.$ For ease of notation, let $\alpha = e^{i\frac{\pi}{3}}$.
Let $z_0 = 6$ be the initial position of the ant and $z_k$ be the location of the a... | {
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"source": "stackexchange",
"question_score": "1",
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Interesting solutions for cyclic infinite nested square roots of 2 I have derived cosine values for following cyclic infinite nested square roots of 2 ( Hereafter simply referred as $cin\sqrt2$)
$cin\sqrt2[1-]$ represents $\sqrt{2-\sqrt{2-...}}$
$cin\sqrt2[1-1+]$ represents $\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2+...}}}}$
$ci... | We can multiply the fraction in the even-$n$ case by $2^n-1$ on top and bottom, yielding
$$\operatorname{cin}\sqrt2[n-n+]=2\cos\frac{2^n\pi(2^n-(-1)^n)}{3(2^{2n}-(-1)^n)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3945950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\lim_{x\rightarrow 0} \frac{\left( \cosh x \right) ^{\sin x}-1}{\sinh x\cos \left( \sin \left( x \right) -1 \right)}$ Evaluate the limit
$$
\lim_{x\rightarrow 0} \frac{\left( \cosh x \right) ^{\sin x}-1}{\sinh x(\cos \left( \sin \left( x \right) \right)-1)}
$$
My Attempt: I tried to use L'Hôpital's rule to ... |
And the method of Taylor's Series is too complicated here.
Not really, actually. Using the Taylor series approximations of $\cos$, $\sin$, $\ln$, $\exp$, and $\cosh$ (all standard) and composing them as we go along:
$$\begin{align*}
\frac{\left( \cosh x \right) ^{\sin x}-1}{\sinh x(\cos \left( \sin \left( x \right) ... | {
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"source": "stackexchange",
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What are the equations of the three straight lines represented by $x^3+bx^2y+cxy^2+y^3=0$ when $b+c=-2$? I am given that $$x^3 + bx^2y + cxy^2 + y^3 = 0$$ represents three straight lines if $b + c = -2$.
Is there a way to find the equations of the three lines separately?
I tried factorizing the equation but wasn't able... | Observe that all three straight lines pass through the origin $(0, 0)$
Let the three straight lines are
$y = m_1x$, $y = m_2x$ and $y = m_3x$
Then
$$ (y -m_1x)(y -m_2x)(y -m_3x) \equiv x^3 + bx^2y + cxy^2 + y^3$$
Comparing the coefficients, we have
$m_1 m_2 m_3 = -1$
$m_1 m_2 + m_1 m_3 + m_2 m_3 = b$
$m_1 + m_2 + m_3 =... | {
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"source": "stackexchange",
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Find the derivative using the definition of derivative (limit).
Given $f(x)=\dfrac{5x+1}{2\sqrt{x}}$. Find $\dfrac{df(x)}{dx}=f'(x)$ using
the definition of derivative.
I have tried as below.
\begin{align*}
f'(x)&=\lim\limits_{h\to 0} \dfrac{f(x+h)-f(x)}{h}\\
&= \lim\limits_{h\to 0}
\dfrac{\dfrac{5(x+h)+1}{2\sqrt... | First, note that we don't need to worry about the derivative of $f$ when $x=0$, because your function isn't defined when $x=0$.
Using the Binomial expansion, as $h \to 0,$ which garuntees that $|\frac{h}{x}| < 1,$
$ \sqrt{x+h} = \sqrt{x} \sqrt{1 + \frac{h}{x}} = \sqrt{x} \left(1 + \frac{h}{x}\right)^\frac12 = \sqrt{x}\... | {
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"source": "stackexchange",
"question_score": "4",
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Finding the indefinite integral of $\int_0^\pi\sqrt{\sin^3x - \sin^5x}dx$ It's not hard to find that $\int\sqrt{\sin^3x - \sin^5x}dx = \frac{2\sin^{5/2}x}{5} + C$,
But the indefinite integral $\int_0^\pi\sqrt{\sin^3x - \sin^5x}dx = \frac{4}{5}$ does not conform to $\cfrac{2\sin^{5/2} \pi}{5} - \cfrac{2\sin^{5/2} 0}{5} ... | I'm going to elaborate on Kavi's answer:
The problem here is that the antiderivative $\frac{2}{5}\sin^{\frac{5}{2}}(x)+C$ is only valid when $x\in\left[0,\frac{\pi}{2}\right]$.
Looking at the Desmos plot I provided, it can be seen that the derivative of $\frac{2}{5}\sin^{\frac{5}{2}}(x)$ does not agree with $\sqrt{\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3949460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Evaluating: $\lim_{x \to \infty} \left(\frac {3x+2}{4x+3}\right)^x$ Limit i want to solve: $\lim_{x \to \infty} \left(\frac {3x+2}{4x+3}\right)^x$
This is how i started solving this limit:
*
*$\lim_{x \to \infty} \left(\frac {3x+2}{4x+3}\right)^x$
*$\left(\frac {3x+x-x+2+1-1}{4x+3}\right)^x$
*$\left(\frac {4x+3}{... | We have:
$$\lim_{x\to\infty}\bigg(\frac{3x + 2}{4x + 3}\bigg)^{x} = \lim_{x\to\infty}\bigg(\frac{3 + \frac{2}{x}}{4 + \frac{3}{x}}\bigg)^{x} = \bigg(\frac{3}{4}\bigg)^{\infty} = \boxed{0}$$
Your mistake was in assuming that $\frac{4x+3}{x + 1}\to\infty$ as $x\to\infty$, which meant you could apply the definition of $e$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3952733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
How can I prove this reduction formula for $\int^1_0{x(1-2x^4)^n}dx$ The exercise in my textbook states
You are given that $$I_n=\int^1_0{x(1-2x^4)^n}dx$$
Show that $$I_n=\frac{(-1)^n}{4n +2} + \frac{2n}{2n+1}I_{n-1}$$
I have started out by splitting the power
$$\int^1_0{x(1-2x^4)^n}dx=\int^1_0{x(1-2x^4)(1-2x^4)^{n... | $$I_n=\int_0^1x(1-2x^4)^ndx$$
$u=x^2\Rightarrow dx=\frac{du}{2x}$ and so:
$$I_n=\frac12\int_0^1(1-2u^2)^ndu$$
$$2I_n=\int_0^1(1+\sqrt{2}u)^n(1-\sqrt{2}u)^ndu$$
If we do IBP:
$$a'=(1+\sqrt{2}u)^n\Rightarrow a=\frac{1}{\sqrt{2}(n+1)}(1+\sqrt{2}u)^{n+1}$$
$$b=(1-\sqrt{2}u)^n\Rightarrow b'=-\sqrt{2}n(1-\sqrt{2}u)^{n-1}$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3953054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to integrate $\int \frac{\cos(x) + \sqrt 3}{1+ 4 \sin(x+\frac{\pi}{3}) + 4\sin^2(x+\frac{\pi}{3})}{\rm d}x$ The question given is to calculate
$$\int \frac{\cos(x) + \sqrt 3}{1+ 4 \sin(x+\frac{\pi}{3}) + 4\sin^2(x+\frac{\pi}{3})}{\rm d}x$$
My attempt
I managed to figure out that the denominator is given out as a pe... | Hint
Let $y=\dfrac{a\cos x+b\sin x +c}{1+\sin x+\sqrt3\cos x}$
Find $\dfrac{dy}{dx}$
Compare
$-(a\cos x+b\sin x+c)(\cos x-\sqrt3\sin x)+(1+\sin x+\sqrt3\cos x)(-a\sin x+b\cos x)$
with $\cos x+\sqrt3$
to find $a,b,c$
For example, by comparing the coefficient of $\cos x\sin x$ we get $b=\sqrt3a$
By $\sin x,a=-\sqrt3c$
So... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3954413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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Find the minimum value of the ratio $\frac{S_1}{S_2}$ using given data
3 points $O(0, 0) , P(a, a2 ) , Q(b, b2 )$ are on the parabola $y=x^2$. Let S1 be the area bounded by the line PQ and the parabola and let S2 be the area of the triangle OPQ, then find min of $\frac{S1}{S2}$
$$S_1 = \frac 12 (a+b)(a^2+b^2) -\int_{... | $$
\dfrac{S_1}{S_2}=\frac{\frac{1}{2} \left(a^2 b+a b^2\right)+\frac{1}{6} \left(a^3+b^3\right)}{\frac{1}{2} \left(a^2 b+a b^2\right)}=\frac{(a+b)^2}{3 a b}\geq \dfrac{4ab}{3ab}=\dfrac{4}{3}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3954911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\displaystyle\left|\int_L \frac{z^3}{z^2+1} dz\right| \le \frac{9\pi}{8}$ where $L=\big(z: |z|=3, Re(z) \ge 0\big)$ I tried solving this by using the fact that the length of L is half the circumference of a circle with radius 3, which would mean that
if
$$\left|\frac{z^3}{z^2+1}\right| \le \frac{3}{8}$$ thi... | I believe the inequality you stated is not true. Firstly, if $|z|=3$, by the reverse triangle inequality we get that $|z^2+1|=|z^2-(-1)| \geq | |z^2|-1| = |3^2-1| = 8$. Taking reciprocals gives us that $\dfrac{1}{|z^2+1|} \leq \dfrac{1}{8}$.
If we multiply both sides of the inequality by $|z^3|$, we obtain for $|z|=3$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3958533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Series convergence or divergence $\sum_{n=1}^{\infty}\frac{(-1)^{n}\cos(3n)}{\sqrt{n^2 + 2}}$ Can you give me some hint for the series
\begin{align*}
\sum_{n=1}^{\infty}\frac{(-1)^{n}\cos(3n)}{\sqrt{n^2 + 2}}
\end{align*} I have tried Leibniz test. It didn't help me.
| Hint: use Dirichlet's test. See more at Wikipedia
It's clear that $\left\{\dfrac 1 {\sqrt{n^2 + 2}}\right\}$ is monotonic and converges to 0
Now the task is to prove $-\cos 3 + \cos 6 - \cos 9 + \cos 12 + ...$ is bounded
Recognize that $\cos 6 - \cos 3 = -2\sin \frac 9 2 \sin \frac 3 2$
$\cos 12 - \cos 9 = -2\sin \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3958971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $\lim\limits_{x\to0} xf(x) = 0$? I am confused about the following proof on a textbook I'm reading.
Suppose $f:(0,b] \to \Bbb R$ is continuous, positive, and integrable on $(0,b]$. Suppose further that as $x \to 0$ from the right, $f(x)$ increases monotonically to $+\infty$. Prove that $\lim\limits_{x\to0} xf(x) ... | As written, this is not true. However, I am not sure if meant to include that $f(x)$ is a decreasing function as you mention later in your proof that it is a monic function. If this condition is not included, then the theorem is false. Consider the function
$$f(x)=\begin{cases}
g_n(x) & \frac{1}{n^3}\leq x \leq ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3960879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is $x=2,y=13$ the unique solution? Problem:
Find all positive integers $x$ and $y$ satisfying:
$$12x^4-6x^2+1=y^2.$$
If $x=1, 12x^4-6x^2+1=12-6+1=7,$ which is not a perfect square.
If $x=2, 12x^4-6x^2+1=192-24+1=169=13^2$, which is a perfect square. Thus, $x=2,y=13$ is a solution to the given Diophantine equation.
Howe... | This is an addition to my previous answer, which by pointed out by Mike doesn't sufficiently solve the problem.
We use the result, $$3(2x-1)^2(2x+1)^2 = (2y-1)(2y+1).$$
We have that $\gcd(2a-1,2a+1)=1$ for positive integer $a$ (by Euclidean Algorithm if you need convincing, however this is trivial enough to state).
It ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3963607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 0
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Fermat factorisation in one step? Apparently if $N$ has a factor within $\sqrt[4] N$ of $\sqrt N$ then Fermat factorisation works in one step.
Specifically this would mean that for $r = \lfloor \sqrt N \rfloor +1$ we have
$$r^2 - N = s^2$$
for some integer $s$.
I've tried bounding $r^2 -N$ in-between $s^2\pm1$ for some... | A small correction: The algorithm starts with $r = \bigl\lceil \sqrt{N}\,\bigr\rceil$.
Usually, that's the same as $\lfloor \sqrt{N}\rfloor + 1$, but when $N$ is a square it's a difference that matters. A square $N$ of course has a factor within $\sqrt[4]{N}$ of $\sqrt{N}$, but $(\lfloor \sqrt{N}\rfloor + 1)^2 - N = 2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3964455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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show this $\lceil \frac{n}{1-a_{n}}\rceil =n+1$ let $a_{n}$ be squence such $a_{1}=2-\dfrac{\pi}{2}$, and
$$a_{n+1}=2a_{n}+\dfrac{a_{n}-1}{n},n\in N^{+}$$
show that
$$f(n)=\lceil \dfrac{n}{1-a_{n}}\rceil =n+1$$
I try:since
$$f(1)=\lceil \dfrac{1}{\dfrac{\pi}{2}-1}\rceil=2$$
and $$a_{2}=2a_{1}+a_{1}-1=3a_{1}-1=5-\dfrac{... | We write the recurrence as
$$
a_{n+1}=\frac{2n+1}n a_n - \frac1n.
$$
Then we divide both sides by $\prod_{j=1}^n \frac{2j+1}j$. We obtain
$$
\frac{a_{n+1}}{\prod_{j=1}^n \frac{2j+1}j} = \frac{a_n}{\prod_{j=1}^{n-1}\frac{2j+1}j}-\frac1{(2n+1)\prod_{j=1}^{n-1} \frac{2j+1}j}.
$$
Repeatedly applying this, we have
$$
\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3965314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
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On Lyapunov stability of a solution I have recently encountered the following exercise: prove or disprove that the solution of the ODE $\dot x = -\frac{2x}{t} + t^2$ that satisfies $x(1)=1$ is Lyapunov stable. I have proved that the solution is stable, however, the correct answer (as per my teacher) is the opposite.
My... | You obtained the following equation
$$\frac{d}{dt}(t^2 x) = t^4 \quad (1) $$
then you found the following general solution:
$$t^2 x(t) = \frac{t^5}{5} + C $$
which is also true, but you can not treat $C$ as an initial condition of the original differential equation. To use the defintion of stability, you sould integra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3967432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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$\frac{a^2+3b^2}{a+3b}+\frac{b^2+3c^2}{b+3c}+\frac{c^2+3a^2}{c+3a}\geqslant 3$ Let $a,b,c>0$ and $a^2+b^2+c^2=3$, prove
$$\frac{a^2+3b^2}{a+3b}+\frac{b^2+3c^2}{b+3c}+\frac{c^2+3a^2}{c+3a}\geqslant 3$$
This inequality looks simple but I do not know how to solve it. The straightforward method is to bash the inequality wi... | By Cauchy-Schwarz Inequality we have
$\begin{align}
~~~~~~\sum_\text{cyc}\left(a+3b\right)\sum_\text{cyc}\dfrac{a^{2}+3b^{2}}{a+3b}\ge\left(\sum_\text{cyc}\sqrt{a^{2}+3b^{2}}\right)^{2}\cdot
\end{align}$
So it suffices to show that
$\begin{align}
~~~~~~&\left(\sum_\text{cyc}\sqrt{a^{2}+3b^{2}}\right)^{2}\ge3\sum_\text{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3970317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
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How to find the closed form of $\sum_{k=0}^n\frac{1}{k+1} {\binom n k}^2$ How to sum the following identity: $$\sum_{k=0}^n\frac{1}{k+1}{\binom n k}^2$$ I know the answer is $\frac{(2n+1)!}{{(n+1)!}^2}$ but I have tried a lot to find the closed form through any algebraic way or combinatorial but not able to find it ou... | consider this polynomial (or OGF).
$$
\sum_{k=0}^n \binom{n}{k}^2 \frac{1}{k+1}=[x^{n+1}]\sum_{k=0}^n \binom{n}{n-k}x^{n-k}\times \binom{n}{k}\frac{x^{k+1}}{k+1}
$$
It's the convolution(product) of $(1+x)^n$ and $F(x)$ where
$$
\begin{aligned}
F(x)&=\sum_{k=0}^n \binom{n}{k}\frac{x^{k+1}}{k+1}\\
F'(x)&=\sum_{k=0}^n \bi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3970922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$3^{1234}$ can be written as $abcdef...qr$. What is the value of $q+r$? It was possible to find $3^{15} ≡ 7\pmod{100}$. Knowing that $7^{4k}≡1\pmod{100}, (3^{15})^{80}≡3^{1200}≡(7)^{80}≡1\pmod{100}$.
$(3^{15})(3^{15})3^{1200}≡1\cdot 7\cdot 7\pmod{100}, 3^{1230}\cdot 3^{4}≡49\cdot 81\pmod{100}≡69\pmod{100}$ to get $6+9=... | You can directly find the last $2$ digits of $3^{1234}$ by writing it as $3^{1232}\cdot 9=(81)^{308}\cdot 9$.
Last two digits of $(81)^{308}=(80+1)^{308}\equiv 41 \pmod{100} $. Hence $41\cdot 9 \equiv 69 \pmod{100}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3971549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Given that $p$ is a prime, and the sum of all positive divisors of $p^4$ is a perfect square, find the possible number of primes $p$ .
Given that $p$ is a prime, and the sum of all positive divisors of $p^4$ is a perfect square, find the possible number of primes $p$ .
What I Tried: I know that the only divisors of $... | Just some thoughts. Still editing...
Note that the sum $1+p+p^2+p^3+p^4$ is odd, so $k^2\equiv1\bmod 8$, and hence, $p(1+p+p^2+p^3)$ is a multiple of $8$.
If $p=2$, this isn't the case, so $p$ must be odd, and hence, $1+p+p^2+p^3$ must be a multiple of $8$.
mod $8$, $1+p+p^2+p^3=1+p+1+p$, so $2+2p$ is a multiple of $8$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3975198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Eliminate $\theta$ and prove $x^2+y^2=1$ We have:$${ \begin{cases}{2x=y\tan\theta+\sin\theta} \\ {2y=x\cot\theta+\cos\theta}\end{cases} }$$
And want to prove $x^2+y^2=1$
My works:
I multiplied first equation by $\cos\theta$ and second one by $\sin\theta$ and get:
$${ \begin{cases}{2x\cos\theta=y\sin\theta+\sin\theta\co... | You are almost there. You already have $$2x\cos\theta=y\sin\theta+\sin\theta\cos\theta \tag1$$
and $$x\cos \theta = y \sin \theta \tag 2$$
Then $$x\cos \theta = y \sin \theta = 2x\cos \theta - y \sin \theta = \sin \theta \cos \theta$$
and you know $\sin \theta \ne 0, \cos \theta \ne 0$. Can you end it now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3975949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
If $p$ and $q$ are solutions of the equation $x \tan x = 1$, show the integral of $\cos^2 px$ entirely in terms of $p$ I am working through a pure maths text book out of interest. I have finished the chapter on integration and differentiation of trigonometric functions and am doing the end of chapter questions. This i... | $$I=\int_0^1 \cos ^2(p x) \, dx=\frac{\sin (2 p)}{4 p}+\frac{1}{2}$$
it is given that $\tan p=\frac{1}{p}$
From the formula
$$\sin \alpha=\frac{2t}{1+t^2};\;t=\tan\frac{\alpha}{2}$$
we get
$$\sin 2p=\frac{2\tan p}{1+\tan^2 p}=\frac{2\cdot\frac{1}{p}}{1+\left(\frac{1}{p}\right)^2}=\frac{2 p}{p^2+1}$$
and finally
$$I=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3976109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Proving $n\leq3^{n/3}$ for $n\geq0$ via the Well-Ordering Principle [2] I know this question was already asked in here, but it was never marked as answered and all the solutions base themselves on the fact that $3(m-1)^3 < m$, what comes from assuming $3^m < m$ and it's not clear to me.
I tried multiple ways to underst... | I already accepted @fleablood's answer. I'd just like to share my own proof after some work on it as further reference for other people working on this problem.
Theorem: $n \leq 3^{n/3}\text{ for }n \geq 0$
Proof:
Let C be the set of counterexamples to the theorem, namely
$$C ::= \{n \in \mathbb{N} | n > 3^{\frac{n}{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3979260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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What am I doing wrong on this question about "varies directly" and "varies inversely"? I don't get this problem:
$R$ varies directly as $S$ and inversely as $T$. When $R = \frac{4}{3}, T = \frac{9}{14}$, $S = \frac{3}{7}$. Find $S$ when $R = \sqrt{48}$ and $T = \sqrt{75}$.
My attempt:
We know
$$ R = c_1 S$$
and
$$ R = ... | You can’t treat the relationship between $R$ and $S$ and the relationship between $R$ and $T$ as if they were independent: the hypothesis actually means that
$$R=c\cdot\frac{S}T$$
for some constant $c$.
Thus, the relationship $R=c_1T$ holds only when $T$ is held constant, and the relationship $R=\frac{c_2}T$ holds only... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $x, y, z \in (0, 1)$ and $x+y+z = 2$, prove that $8(1-x)(1-y)(1-z) \leq xyz$. If $x, y, z \in (0, 1)$ and $x+y+z = 2$, prove that $8(1-x)(1-y)(1-z) \leq xyz$.
I found this in a Facebook group.
I start by doing the math in the LHS:
$8(1-x-y+xy-z+xz+yz-xyz) = 8(1-2+xy+xz+yz-xyz) = 8(xy+xz+yz-1-xyz)$.
Then we set $z=2-... | as $1-x=\frac{1}{2}(y+z-x)$ and similarly for others we have to prove $$(x+y-z)(y+z-x)(x+z-y)\le xyz$$ which is schur's inequality
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3982410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Analytic Proof of Wilson's Theorem I'm trying to find a reference for the following (analytic) proof of Wilson's Theorem. I would appreciate it immensely if some of you knew and told me the person who first came up with it. Here is the proof:
We start by considering the Maclaurin series of the function $f(x)=\ln \left(... | Here is a related argument with power series.
For $|x| < 1$ there is an identity
$$
e^x = \prod_{n \geq 1} (1 - x^n)^{-\mu(n)/n}.
$$
This can also be viewed as an identity of formal power series, and checked by verifying that the logarithmic derivatives ($f \leadsto f'/f$) of both sides are the same and that they start... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3984517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the value of $k$ in $\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}} \ge \sqrt{a}+\sqrt{b}+\sqrt{c}+k$ It is also given that $abc = 1$.
I used AM-GM inequality to reach till
$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}} \ge \sqrt{\frac{bc}{a}} + \sqrt{\frac{ac}{b}} + \sqrt{\frac{ab... | We will prove the homogeneous inequality (where we used $abc=1$) $$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}} \ge \sqrt{a}+\sqrt{b}+\sqrt{c}+3\sqrt[6]{abc}$$ Since equality is attainable for $a=b=c$, this would show that the desired $k$ is $3$. Observe that, in virtue of AM-GM $$\frac{\frac{a}{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3984909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Equivalence class for the following relation A relation of $\mathbb{R}$ is defined as $a\sim b : a^4-b^2=b^4-a^2$
Show that $\sim$ is equivalence relation (I have done this part)
Determine the equivalence class $[-1]_\sim$
Prove or disprove: Every equivalence class in $\mathbb{R}/\sim$ contains exactly 2 real numbers.
... | Follow definitions.
$[-1] = \{b\in \mathbb R|(-1)^2 - b^2=b^4 -(-1)^2\}=$
$\{b\in \mathbb R| 1-b^2 = b^4 -1\}=$
$\{b\in \mathbb R| b^4+b^2-2 = 0\}=$
$\{b\in \mathbb R| (b^2+2)(b-1)(b+1)=0\}=$
$\{b\in \mathbb R| b^2 = -2\lor b=1\lor b=-1\}=$
$\{1,-1\}$
As for the second:
$[a] = \{b\in \mathbb R| a^4-b^2 = b^4 -a^2\}=$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3985631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
how to compute a series whose terms are a rational function times an exponential function? How can I compute the following series?
\begin{equation}
\sum_{n=1}^\infty\frac{n+10}{2n^2+5n-3}\left(-\frac{1}{3}\right)^n
\end{equation}
I manipulated the term and got
\begin{equation}
\frac{n+10}{2n^2+5n-3}\left(-\frac{1}{3}\r... | $$S=\sum_{n=1}^\infty\frac{n+10}{2n^2+5n-3}\left(-\frac{1}{3}\right)^n$$
$$S=\sum_{n=1}^{\infty} \left(\frac{3}{2n-1}-\frac{1}{n+3}\right)(-3^{-1})^{n}=S_1-S_2$$
$$S_2=-27\sum_{n=1}^{\infty} \frac{(-3^{-1})^{n+3}}{n+3}$$
Use $\ln(1+z)=\sum_{k=1}^{\infty} (-1)^{k-1} \frac{z^k}{k}$
$$=-27\sum_{m=4}^{\infty}\frac{(-1)^m(3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3987217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Calculate limit using Maclaurin series $$\lim_{x \to 0}\frac{1-(\cos{x})^{\sin{x}}}{x^3}$$
So what I have tried is defining three functions, $f(x)=\cos{x}, g(x) = \sin{x}, h(x) = x^3$
What I did next was do Maclaurin expansion for three functions up until the third order:
$f(x) = 1 - \frac{x^2}{2} + o(x^3)$
$g(x) = x -... | We can derive the limit using series expansion as follows:
\begin{align*}
\color{blue}{\lim_{x\to 0}}&\color{blue}{\frac{1-\left(\cos x\right)^{\sin x}}{x^3}}\\
&=\lim_{x\to 0}\frac{1-\exp\left((\sin x)\ln\left(\cos x\right)\right)}{x^3}\\
&=\lim_{x\to 0}\frac{1-\exp\left(\left(x-\frac{x^3}{6}+\mathcal{O}\left(x^4\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3987408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solving an equation that contain floor function Let suppose I have three positive integers $a, b,c$ and one unknown $x$ ($x$ is also a positive integer). Solve for the smallest $x$ that satisfies this equation
$$\left \lfloor\frac{x}{a} \right\rfloor + \left \lfloor \frac{x}{b} \right\rfloor \geq c$$
where $\lfloor x\r... | Assume that $a,b,c$ are positive
.
Let $y=\frac{abc}{a+b}$. Then $$\frac{y}{a}+\frac{y}{b}=c.$$
Therefore $$\left \lfloor\frac{y}{a} \right\rfloor + \left \lfloor \frac{y}{b} \right\rfloor$$ will be either $c$ or $c-1$.
If it is $c$
Then $y$ is a solution. We will still have a solution when $y$ is increased, until $y$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3988157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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I found this $\binom{2p^2}{k}\equiv 2\binom{p^2}{k}\pmod {p^2},1\le ktoday,when I deal this problem,I found this result:
let $p>3$ be prime number, show that
$$\binom{2p^2}{k}\equiv 2\binom{p^2}{k}\pmod {p^2},1\le k<p^2$$
I found this when $k=1,2,3$ it is easy to prove it,because
$k=1$
$$\binom{2p^2}{k}=\binom{2p^2}{1}... | Let $n$ be a positive integer and $1\le k<n$. By a straightforward counting argument we have the identity
$$\binom{2n}{k}=\sum_{a,b\ge 0, a+b=k}\binom{n}a\binom{n}b.$$
Setting $n=p^2$ and rewriting the equation gives
$$\binom{2p^2}k=2\binom{p^2}{k}+\sum_{a,b\ge 1, a+b=k}\binom{p^2}a\binom{p^2}b.$$
It therefore suffices... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3990027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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On the integral $\int_{0}^{1} \frac{\mathrm{d}x}{\sqrt{3x^4+2x^2+3}}$ I came across this integral
$$\mathcal{J} = \int_{0}^{1} \frac{\mathrm{d}x}{\sqrt{3x^4+2x^2+3}}$$
According to W|A it equals $\frac{1}{2}$. However, I cannot find a way to crack it. It smells like a Beta integral , but I do not see any obvious subs. ... | A trigonometric substitution does indeed help here: Consider $f \colon [0,1] \to \mathbb{R},$
\begin{align}
f (a) &= \int \limits_0^1 \frac{\mathrm{d} x}{\sqrt{1 + 2 a x^2 + x^4}} = \int \limits_0^1 \frac{\mathrm{d} x}{\sqrt{(1 + x^2)^2 - 2 (1-a) x^2}} \\
&\!\!\!\!\!\!\!\!\stackrel{x = \tan\left(\frac{t}{2}\right)}{=} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3994657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
$\lim\limits_{n\to\infty}n\big(\sum_{k=1}^n\frac{k^2}{n^3+kn}-\frac{1}{3}\big)$? calculate $$\lim\limits_{n\to\infty}n\left(\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}-\dfrac{1}{3}\right).$$
I got it $$\lim\limits_{n\to\infty}\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}=\lim\limits_{n\to\infty}\dfrac{1}{n}\sum\limits_{k=1}^n\dfra... | There is another using generalized haromonic numbers since
$$\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}=n^3 \left(H_{n^2+n}-H_{n^2}\right)+\frac{1}{2} \left(-2 n^2+n+1\right)$$
$$S_n=-\dfrac{1}{3}+\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}=\frac{1}{6} \left(-6 n^2+3n+1\right)+n^3 \left(H_{n^2+n}-H_{n^2}\right)$$ Using asymptot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3996425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Integration of $\int_0^{\infty}\frac{e^{(\lambda +is)t}-e^{-(\lambda +is)t}}{e^{\pi t}-e^{-\pi t}}dt$ Integration of $$\int_0^{\infty}\frac{e^{(\lambda +is)t}-e^{-(\lambda +is)t}}{e^{\pi t}-e^{-\pi t}}dt$$
I am thinking about this integral from last half and hour but still can't figure it out how to solve this. what i ... | Considering $\displaystyle \mathcal{I}(\alpha, \beta) = \int_{0}^{\infty} \frac{\sinh\alpha z}{\sinh \beta z}\,\mathrm{d}z$, we show that $\displaystyle \mathcal{I}(\alpha, \beta) = \frac{\pi}{2\beta} \tan\left(\frac{\alpha \pi}{2\beta}\right)$.
Then your integral is $\displaystyle \mathcal I(\lambda+is, \pi)$. Startin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4002019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Proving $\sum_{cyc}\frac{\sqrt{s(s-a)}}{a}\geq\frac{3\sqrt{3}}2$ for $a$, $b$, $c$, $s$ the sides and semi-perimeter of a triangle
If $a, b, c$ are the lengths of the sides of a triangle and s is the semiperimeter, prove that:
$$\sum_{cyc}\frac{\sqrt{s(s-a)}}{a}\geq \frac{3\sqrt{3}}{2}$$
My attempt: $$\sum_{cyc} cos ... | Let $a = y+z,\,b = z+x,\,c = x+y$ for $x,\,y,\,z>0,$ we have
$$\sum \frac{\sqrt{s(s-a)}}{a} = \sum \frac{\sqrt{\frac{a+b+c}{2}\left(\frac{a+b+c}{2}-a\right)}}{a} = \sum \frac{\sqrt{x(x+y+z)}}{y+z}.$$
We will show that
$$\sum \frac{\sqrt{x(x+y+z)}}{y+z} \geqslant \frac{3\sqrt 3}{2}.$$
Indeed, suppose $x+y+z=3$ then the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4005064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Inequality $a^2b(a-b) + b^2c(b-c) + c^2a(c-a) \geq 0$ Let $a,b,c$ be the lengths of the sides of a triangle. Prove that:
$$a^2b(a-b) + b^2c(b-c) + c^2a(c-a) \geq 0.$$
Now, I am supposed to solve this inequality by applying only the Rearrangement Inequality.
My Attempt:
W.L.O.G. let $a \geq b \geq c.$
Then, L.H.S. = $a ... | Method 1:As Chrystomath used Ravi's substituition we are left to prove $$\frac{y^2}{z}+\frac{x^2}{y}+\frac{z^2}{x}\ge x+y+z$$ Now
*
*$x\ge y\ge z\implies x^2\ge y^2\ge z^2,\frac{1}{x}\le \frac{1}{y} \le \frac{1}{z}$ $$x^2\left(\frac{1}{x}\right)+y^2\left(\frac{1}{y}\right)+z^2\left(\frac{1}{z}\right)\le \frac{y^2}{z}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4008310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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a random variable has the density $f\left(x\right)=a+bx^{2}$ . Determine and b so that its mean will be 2/3 A random variable has the density $f\left(x\right)=a+bx^{2}$ with $0<x<1$. Determine and b so that its mean will be 2/3.
I'm a little confued trying to get a and b. This is what I already tried:
$$P\left(0<X<1\... | You are mixing up the $P(X\in(0,1))=1$ and $E(X)=\frac 2 3$. The other post presumably showed you the expected value correctly, so $P(X\in(0,1))=\int_{-1}^1 f(x) dx=\int_0^1(a+bx^2)dx=\left[ax+\frac{bx^3}{3}\right]_{x=0}^1=a+\frac b 3=1$
Solving, you obtain $a=\frac 1 3, b=2$.
To check:
$E(X)=\int_{-\infty}^\infty xf(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4008556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Prove that $2<\int^2_1ln(x^2+1)+x<3$ To show that $2<\int^2_1ln(x^2+1)+x$ , I took the derivative of the integrand and got that it's positive always in $[1,2]$, so its strictly increasing and the minimum in $[1,2]$ is at $x=1$, so $ln(2)+1<ln(x^2+1)+x$, and $ln(2)<ln(e)=1$ so: $ln(2)+1<2< ln(x^2+1)+x$ in $(1,2]$.
EDIT... | I began to solve this before Bacha's solution appeared, so I did not delete.
Hope I'm correct.
$$\int\ln(x^2+1)dx=(*)\\let\:f=\ln(x^2+1),g=x\\then\:df=\frac{2x}{x^2+1}dx,dg=dx\\\text{using integrating by parts where }\int fdg=dg-\int gdf\text{ we get}\\(*)=x\ln(x^2+1)-\int\frac{2x^2}{x^2+1}dx=x\ln(x^2+1)-2\int \left(1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4008677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int_{0}^\infty\frac{2}{({M-1})!}\sqrt{\frac{a}{x}}K_1\left(2\sqrt{\frac{a}{x}}\right)x^{M-1}\frac{e^{-\frac{x}{b}}}{b^M}\,dx$ How to evaluate this integral or at least rewrite it as special (non-elementary) function step by step ?
$$\int_{0}^\infty\frac{2}{({M-1})!}\sqrt{\frac{a}{x}}K_1\left(2\sqrt{\frac{a}{... | It seems useful to rearrange the integral:
\begin{align}
I&=\int_{0}^\infty\frac{2}{({M-1})!}\sqrt{\frac{a}{x}}K_1\left(2\sqrt{\frac{a}{x}}\right)x^{M-1}\frac{e^{-\frac{x}{b}}}{b^M}\,dx\\
&=\frac{2b^{-M}\sqrt{a}}{({M-1})!}\int_{0}^\infty K_1\left(2\sqrt{\frac{a}{x}}\right)x^{M-3/2}e^{-\frac{x}{b}}\,dx\\
&=\frac{2}{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4010424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Famous or common mathematical identities that yield $1$ To me the most common identity that comes to mind that results in $1$ is the trigonometric sum of squared cosine and sine of an angle:
$$
\cos^2{\theta} + \sin^2{}\theta = 1 \tag{1}
$$
and maybe
$$
-e^{i\pi} = 1 \tag{2}
$$
Are there other famous (as in commonly us... | Here are two well-known infinite series whose sum is $1$:
$$\sum_{n=1}^{\infty} \frac{1}{2^n} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \cdots = 1$$
$$\sum_{n=1}^{\infty} \frac{1}{n(n+1)} = \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \frac{1}{4 \cdot 5} + \frac{1}{5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4011521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Question on finding maximum of AM GM inequality How can i solve for minimum value of function $f(x)=\frac{x^2+x+1}{x^2-x+1} $ using AM GM inequality.
I am aware about the AM GM inequality, but am not able to split the terms into the correct form in order to apply the inequality
| we prove the answer is $\frac{1}{3}$ which happens at $x=-1$ for proving it we have :
$$\frac{x^2+x+1}{x^2-x+1} \ge \frac{1}{3}$$
if and only if
$$3x^2+3x+3 \ge x^2-x+1$$
if and only if
$$2x^2+4x+2=2(x+1)^2 \ge 0$$
[edit] if you think that it is not AM_GM here is a solution which is the same but the use of AM_GM is mor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4013300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$x^3 + y^3 +3x^2 y^2 =x^3y^3$ Find all possible values to $\frac{x+y}{xy}$ $x,y \in \mathbb{R}\setminus\{0\}$
Indeed the original question said :Find all possible values to: $$\frac{1}{x} + \frac{1}{y}$$
But it’s the same thing.
My Attempt:
$$x^3 + y^3 +3x^2 y^2 =x^3y^3 \iff (x+y)(x^2-xy+y^2)=x^3y^3 -3x^2y^2$$
$$\frac{... | The following is equivalent to @NN2's solution, only written a bit differently.
First substitute $x=1/a, y=1/b$, so that the equation becomes
$$ \tag{*}
a^3+b^3+3ab - 1 = 0 \, .
$$
We are looking for the possible values of $a+b$, this suggests to introduce $S=a+b$ and $P = ab$. Then
$$
0 = a^3+b^3+3ab - 1 = S^3+3PS +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4017341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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Prove that $\angle AEF =90^\circ$ given a square $ABCD$ Let $ABCD$ be a square and $E\in CD$ such that $DE=EC$. Let $F\in BC$ such that $BC=4FC$. Prove that $\angle AEF =90^\circ$.
My attempt:
Proving that $\angle AEF =90^\circ$ is the same as proving that $\triangle AEF$ is a right triangle. In other words, we wish t... | Ignoring the similar triangles for variety...
Let the side length of the square be $s$. We have $|AF|^2 = |AB|^2 +|BF|^2 = s^2 + \left(\frac 34 s\right)^2 = \frac{25}{16}s^2$
Then $|AE|^2 = |AD|^2 +|DE|^2 = s^2 + \left(\frac 12 s\right)^2 = \frac{5}{4}s^2 = \frac{20}{16}s^2$
and $|EF|^2 = |EB|^2 +|BF|^2 = \left(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4018025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 4
} |
Finding the value of $ax^4+by^4$
If $\quad a+b=23 , \quad ax+by=79,\quad ax^2+by^2=217,\quad
ax^3+by^3=691\quad$ find the value of $ax^4+by^4$.
Here is my attempt:
$$(a+b)(x+y)=ax+by+ay+bx\rightarrow 23(x+y)=79+(ay+bx)$$
$$(ax+by)(x+y)=ax^2+by^2+axy+bxy\rightarrow79(x+y)=217+23 xy$$
In each equation I have two unkno... | Answer to OP's comment:
For solving the following system:
$$79(x+y)=23xy+217$$$$ 217(x+y)=79xy+691$$
We first eliminate $(x+y) $ from both equations:
$$ (x+y) = \frac{1}{79} (23 xy + 217)\tag{1}$$
And,
$$ x+y = \frac{1}{217} (79 xy + 691) \tag{2}$$
We get:
$$ \frac{1}{217} (79 xy + 691)= \frac{1}{79} (23 xy + 217)$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4018900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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Establishing $\cos^2 4\alpha=\cos^2 2\alpha-\sin 6\alpha \sin 2\alpha$ without the use of Werner or prostaferesis formulas I have this identity:
$$\bbox[5px,border:2px solid #138D75]{\cos^2 4\alpha=\cos^2 2\alpha-\sin 6\alpha \sin 2\alpha} \tag 1$$
If I write this like as:
$$(\cos 4\alpha+\cos 2\alpha)\cdot (\cos 4\alp... | OK, I think I have something a little more straightforward. If I am using some of the forbidden fruit, I apologize. First of all, as hamam already noted, I'm going to let $2\alpha = x$ to save my sanity. So I want to show that
$$\cos^2 2x = \cos^2 x - \sin 3x\sin x.$$
Aside from the double-angle formula for $\cos 2x$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4019530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Prove $\int_0^\infty\frac{\ln x}{x^3-1}\mathrm{d}x=\frac{4\pi^2}{27}$ Proof of the integral
$$\int_0^\infty\frac{\ln x}{x^3-1}\mathrm{d}x=\frac{4\pi^2}{27}$$
I try to substitute $u = \ln x$. Then $x = e^u,\>\mathrm{d}x = e^u\mathrm{d}u$ and the limits $(0,\infty)\to (-\infty,\infty)$.
The integral becomes $$\int_{-\inf... | Split the integral to simplify as follows
\begin{align}\int_0^\infty\frac{\ln x}{x^3-1}{d}x
= &\int_0^1\frac{\ln x}{x^3-1}{d}x + \int_1^\infty\overset{x\to \frac1x}{\frac{\ln x}{x^3-1}}{d}x
=\int_0^1\frac{(1+x)\ln x}{x^3-1}{d}x \\
= &\int_0^1\frac{(x^2+x+1)\ln x}{x^3-1}{d}x
- \int_0^1\overset{x^3\to x}{\frac{x^2\ln x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4023908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 7,
"answer_id": 4
} |
Solving recurrence with substitution I am trying to solve this recurrence $T(n) = 4T(n − 2) + 2^{2n}$ by using substitution and knowing $T(1)=1,T(2)=2$ and here is my attempt:Expanding $T(n)=4T(n-2)+2^{2n}$ using $T(n-2)=4T(n-4)+2^{2(n-2)}$, we get $T(n)=4(4T(n-4)+2^{2(n-2)})+2^{2n}=4^2T(n-4)+2^{2n-2}+2^{2n}=4^2(4T(n-6... | Hint
$$T(n)=4T(n-2)+4^{n}$$
$$T(n)=4^2T(n-4)+4^{n}+4^{n-1}$$
by induction, find that
$$T(n)=4^kT(n-2k)+4^{n}+4^{n-1}+...+4^{n-(k-1)}.$$
Now, take $n=2k$ and $n=2k+1$ and find your general formula.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4027800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Prove that $\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\leq 3+\sqrt{3}$ Problem. (Nguyễn Quốc Hưng) Let $0\le a,b,c\le 3;ab+bc+ca=3.$ Prove that $$\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\leq 3+\sqrt{3}$$
I have one solution but ugly, so I 'd like to find another. I will post my solution in the answer.
| Some thoughts:
WLOG, assume $a\ge b \ge c$.
By Cauchy-Bunyakovsky-Schwarz inequality, we have
$$\sqrt{a+b} + \sqrt{b + c} + \sqrt{c + a}
\le \sqrt{ \left( \frac{a + b}{2} + (b+c) + \frac{c+a}{\sqrt{3}}\right)(2 + 1 + \sqrt{3})}.$$
It suffices to prove that
$$\frac{a + b}{2} + (b+c) + \frac{c+a}{\sqrt{3}} \le 3 + \sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4028695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Coefficients of $p\circ p$ for polynomial $p(x)=\sum_{k=0}^d a_kx^k$ Let $p(x)=\sum_{k=0}^d a_kx^k$ be a polynomial of degree $d$. Then
\begin{align*}
(p\circ p)(x) &= \sum_{k=0}^da_k\left(\sum_{j=0}^da_j x^j\right)^k\\[3pt]
&=\sum_{k=0}^d \sum_{\substack{k_0+\cdots+k_d=k\\[3pt]k_0,\dots,k_d\geqslant0}}a_k\frac{k!}{k_0... | Not quite the answer you're looking for, but for fun taking a cubic $p(z) = a_0 + a_1 z + a_2 z^2 + a_3 z^3$, $p(p(z)) = c_0 + \cdots + c_9 z^9$ where:
\begin{align}
c_0 &= a_{0}^{3} a_{3} + a_{0}^{2} a_{2} + a_{0} a_{1} + a_{0} \\
c_1 &= 3 \, a_{0}^{2} a_{1} a_{3} + 2 \, a_{0} a_{1} a_{2} + a_{1}^{2} \\
c_2 &= 3 \,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4029832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Finding $A$, $B$, $C$ such that $(2x-A)(2x+4)-Bx+25=(x-C)^2-Cx^2$ for all $x$
Find $A$, $B$, $C$ such that
$$(2x-A)(2x+4)-Bx+25=(x-C)^2-Cx^2$$
for all $x$.
This is what I've got so far.
$$\begin{align}
\text{L.H.S.}
&= (2x-A)(2x+4)-Bx+25 \tag1\\
&= 4x^2-2Ax+8x-4A-Bx+25 \tag2\\
&= 4x^2-(2A+8)x-4A-Bx+25 \tag3 \\ \\
\t... | Since it is true for all $x$, this is an identity and hence all you need to do is compare coefficients as mentioned in comments. So, we have $$\begin{aligned} 4 &=1-C \\ 2A+B-8&=2C \\ -4A+25&=C^2\end{aligned}$$
Which gives you $C=-3,A=4,B=-6$.
Note that you have done some mistakes in sign.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4031469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove $\sqrt{x} - \sqrt{x-1}>\sqrt{x+1} - \sqrt{x}$ for $x\geq 1$? Intuitively when $x$ gets bigger, $\sqrt{x+1}$ will get closer to $\sqrt{x}$, so their difference will get smaller.
However, I just cannot get a proper proof.
| $$\sqrt{x} - \sqrt{x-1}>\sqrt{x+1} - \sqrt{x}\quad| +\sqrt{x}$$
$$2\sqrt{x} > \sqrt{x-1}+\sqrt{x+1} \quad|^2$$
$$4x>2x+ 2\sqrt{(x-1)(x+1)} \quad| -2x $$
$$2x>2\sqrt{(x-1)(x+1)} \quad| /2$$
$$x=\sqrt{x^2-1} \quad|^2$$
$$x^2>x^2-1 \quad|-x^2$$
$$0>-1$$
The transfoemations are valid in both directions, top down and bottom... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4032599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Is there a known closed form solution to $\int_0^1\frac{\ln(1+x^{2n})}{1+x^2} \,dx$? $\require{begingroup} \begingroup$
$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}\def\Catalan{\mathsf{Catalan}}$
Related question
Is there a known closed form solution to
\begin{ali... | Just to put in more closed form what @Varu Vejalla has evaluated
$I=\int_0^1\frac{\ln(1+x^{2n})}{1+x^2} dx$
$\int_0^\infty\frac{\ln(1+x^{2n})}{1+x^2} dx=\frac{1}{2}\int_{-\infty}^\infty\frac{\ln(1+x^{2n})}{1+x^2} dx=\int_0^1\frac{\ln(1+x^{2n})}{1+x^2} dx+\int_1^\infty\frac{\ln(1+x^{2n})}{1+x^2} dx=I+\int_1^\infty\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4036975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 2
} |
Solve for integer values of $x,y,z$: $\frac{xy}{z} + \frac{xz}{y} + \frac{yz}{x} = 3$ Solve for integer values of $x,y,z$;
$$\frac{xy}{z} + \frac{xz}{y} + \frac{yz}{x} = 3.$$
My attempt:
*
*Note that all $x, y, z$ are non-zero, otherwise a denominator would be zero.
*Mutliplying by $xyz$ gives:
$$
x^2y^2+x^2z^2+y^2... | If $(x,y,z)$ is a solution then so is $(|x|,|y|,|z|)$, so without loss of generality $x,y,z>0$, and after rearranging we may assume without loss of generality that $x\geq y\geq z$. Then $\tfrac xy,\tfrac xz,\tfrac yz\geq1$, and hence
$$3=\frac{xy}{z}+\frac{xz}{y}+\frac{yz}{x}> x\frac yz\geq x,$$
and similarly $y,z<3$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4041487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find the minimum of $\sqrt{4y^2-12y+10}+\sqrt{18x^2-18x+5}+\sqrt{18x^2+4y^2-12xy+6x-4y+1}$ Find the minimum of $$f(x,y)=\sqrt{4y^2-12y+10}+\sqrt{18x^2-18x+5}+\sqrt{18x^2+4y^2-12xy+6x-4y+1}$$ It seems that $f_x=f_y=0$ is very hard to compute. Is there any easier idea?
| Complete the square in the first two terms and let $Y=2y-3$ and $X=3x-3/2$ to get $$f(X,Y)=\sqrt{Y^2+1}+\sqrt{2X^2+\frac12}+\sqrt{2\left(X+\frac12\right)^2+\left(Y+\frac12\right)^2-2XY+\frac74}$$ so that \begin{align}f_X&=\frac{2X}{\sqrt{2X^2+1/2}}+\frac{2X-Y+1}{\sqrt{2(X+1/2)^2+(Y+1/2)^2-2XY+7/4}}=0\\f_Y&=\frac Y{\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4043163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 0
} |
limit $\lim_{x \to \infty} \frac{\sqrt{x+2} - \sqrt{x+1}}{\sqrt{x+1} - \sqrt{x}}$ How can I find the limit to infinity of this function? As this is a $0/0$ equation, I tried using the L'Hôpital's rule in this but ended up making it more complex. I've also tried rationalising the denominator but it didn't lead to anywhe... | \begin{gather*}
\lim _{x\rightarrow \infty }\frac{\sqrt{x+2} -\sqrt{x+1}}{\sqrt{x+1} -\sqrt{x}}\\
=\lim _{x\rightarrow \infty }\frac{\sqrt{1+\frac{2}{x}} -\sqrt{1+\frac{1}{x}}}{\sqrt{1+\frac{1}{x}} -1}\\
\end{gather*}
(Taking $\sqrt{x}$ common from both numerator and denominator)
Now, from the binomial series expansion... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4043952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to find the nth pair from a generated list of pairs? I have a mathematical question applied on informatic (python). I would like to find the fastest way to get the pair from a given index.
As example, we have all pair combinations of values from 0 to 10:
impor itertools
combinations = list(itertools.combinations((i... | Let $n$ play the role of the number $10$: the pairs are: $(0,1),(0,2),\ldots,(0,n-1),(1,2),(1,3),\ldots,(1,n-1),\ldots,(n-2, n-1)$ (all $n\choose 2$ of them).
Let's find a (zero-based) $k$th pair: say it is $(x,y)$. We have $n-1$ pairs starting with $0$, $n-2$ pairs starting with $1$, etc., ending with one pair startin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4046051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solving system of equation using Macaulay2 As an algebraic curve, the Klein quartic can be viewed as a projective algebraic curve over the complex numbers $\mathbb{C}$, defined by the following quartic equation in homogeneous coordinates $[x:y:z]$ on $\mathbb{P}^2_{\mathbb{C}}$:
$$x^3 y + y^3 z + z^3 x = 0.$$
Now we wa... | I divided by $z^4$ all equations and set $$\frac{x}{z}=X;\;\frac{y}{z}=Y$$
so I got
$$
\begin{cases}
-X^4+3 X^2 Y^2-3 X Y^2+Y=0\\
-3 X^2 Y+3 X^2+X Y^3-1=0\\
X^3-3 X Y-Y^4+3 Y^2=0\\
\end{cases}
$$
solved by Mathematica I got these solutions
$$
\begin{array}{rr}
X & Y\\
\hline
1 & 1 \\
0.307979 & 1.55496 \\
0.643104 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4046318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"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.