Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Evaluating $\int_{0}^{\pi/2}\frac{x\sin x\cos x\;dx}{(a^{2}\cos^{2}x+b^{2}\sin^{2}x)^{2}}$ How to evaluate the following integral
$$\int_{0}^{\pi/2}\frac{x\sin x\cos x}{(a^{2}\cos^{2}x+b^{2}\sin^{2}x)^{2}}dx$$
For integrating I took $\cos^{2}x$ outside and applied integration by parts.
Given answer is $\dfrac{\pi}{4ab^... | You can use this property : $$\int_a^b f(x)\hspace{1mm}dx = \int_a^b f(a+b-x)\hspace{1mm}dx$$
To prove this property : Substitute $a+b-x = u$
Let us have $$I = \int_0^{\pi/2}\dfrac{x\sin x\cos x}{(a^2\cos^2 x+b^2\sin^2 x)^2}\hspace{1mm}dx\rightarrow (1)$$
After applying the property, you will get
$$I = \int_0^{\pi/2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1068649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 4
} |
Closed form of $\int_0^\infty \ln \left( \frac{x^2+2kx\cos b+k^2}{x^2+2kx\cos a+k^2}\right) \;\frac{\mathrm dx}{x}$ Today I discussed the following integral in the chat room
$$\int_0^\infty \ln \left( \frac{x^2+2kx\cos b+k^2}{x^2+2kx\cos a+k^2}\right) \;\frac{\mathrm dx}{x}$$
where $0\leq a, b\leq \pi$ and $k>0$.
... | Here is a complex-analytic method: Notice that
$$ \int_{0}^{\infty} \log\left( \frac{x^{2} + 2x\cos b + 1}{x^{2} + 2x\cos a + 1} \right) \frac{dx}{x}
= 2 \int_{0}^{\infty} \Re \frac{\log(1 + e^{ib}x) - \log(1 + e^{ia}x)}{x} \, dx. \tag{1} $$
Let $R$ be a positive large number. Then the function $z \mapsto \log(1+z)/z$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1069376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
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Putnam definite integral evaluation $\int_0^{\pi/2}\frac{x\sin x\cos x}{\sin^4 x+\cos^4 x}dx$
Evaluate $$\int_0^{\pi/2}\frac{x\sin x\cos x}{\sin^4 x+\cos^4 x}dx$$ Source : Putnam
By the property $\displaystyle \int_0^af(x)\,dx=\int_0^af(a-x)\,dx$:
$$=\int_0^{\pi/2}\frac{(\pi/2-x)\sin x\cos x}{\sin^4 x+\cos^4 x}dx=\fr... | \begin{aligned}
I&= \int_{0}^{\frac{\pi}{2}} \frac{x \sin x \cos x}{1-\frac{\sin ^{2} 2 x}{2}} d x\\&=\int_{0}^{\frac{\pi}{2}} \frac{x \sin 2 x}{1+\cos ^{2} 2 x} d x
\\&= \int_{0}^{\frac{\pi}{2}} \frac{\left(\frac{\pi}{2}-x\right) \sin 2 x}{1+\cos ^{2} 2 x} d x \quad \text{ (By }x \mapsto \frac{\pi}{2}-x.)\\ &=\frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1072316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 4,
"answer_id": 3
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Proving inequality $(x^2+y^2)(y-1)+yx-y^2<0$ I have an inequality which came out of Lyapunov function for system of ODE's: $$(x^2+y^2)(y-1)+yx-y^2<0.$$ To prove stability of my solution, I have to prove that the inequalty is true in area $0<x^2 +y^2<1$. I know it must be simple, but I am still in trouble. Any help is... | If $x = y$, then $$(x^2 + y^2)(y - 1) + yx - y^2 = 2x^2(x - 1) < 0$$ since $x < 1$. Now suppose $x \neq y$. If $x$ and $y$ are positive, then $0 < x^2 < x$ and $y^3 < y^2$, thus
\begin{align}(x^2 + y^2)(y - 1) + yx - y^2 &= x^2y + y^3 - x^2 - y^2 + yx - y^2\\
& < x^2y - x^2 - y^2 + yx \\
& < 2xy - (x^2 + y^2)\\
& < 0.\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1072790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determinant of a 4x4 matrix with trigonometric functions I am stuck with my homework from math. I should calcutate the determinant of a matrix:
$$\begin{bmatrix}
\sin(x) & \sin(2x) & \cos(x) & \cos(2x)\\
\cos(x) & 2\cos(2x) & -\sin(x)& -2\sin(2x)\\
-\sin(x)& -4\sin(2x)& -\cos(x)& -4\cos(2x)\\
-\cos(x)& -8\cos(2x)& \si... | Call the matrix $M$. Then by using @r9m's suggestion, we interchange the two middle columns (this switches the sign of the determinant) and apply two row replacements (this doesn't change the determinant) in order to obtain a zero lower left submatrix:
\begin{align*}
|M|
&= -\begin{vmatrix}
\sin x & \cos x & \sin 2x &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1073852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Exact value of $\frac{\arccos(1-2\tan^2\alpha)}{2\arcsin(\tan\alpha)}$ Let $\alpha\in\left(0,\dfrac\pi2\right)$. What is the exact value of
$$\dfrac{\arccos(1-2\tan^2\alpha)}{2\arcsin(\tan\alpha)}$$
Firstly, I tried to simplify $1-2\tan^2\alpha$ and got
$$\dfrac{3\cos^2\alpha-2}{\cos^2\alpha}$$
What is next step? Is th... | Let $\theta = \arcsin (\tan \alpha) \to \sin \theta = \tan \alpha \to 1 - 2\tan^2\alpha = 1 - 2\sin^2\theta = \cos 2\theta \to \arccos \left(1-2\tan^2\alpha\right) = \arccos (\cos 2\theta) = 2\theta \to L = \dfrac{2\theta}{2\theta} = 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1074982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Evaluation of $\int \frac{\sqrt{1+x^4}}{1-x^4}dx$ Evaluation of $\displaystyle \int \frac{\sqrt{1+x^4}}{1-x^4}dx$
$\bf{My\; Try::}$ Given $\displaystyle \int\frac{\sqrt{1+x^4}}{1-x^4}dx\;,$ Then We can write the above Integral as $$\displaystyle \int\frac{\left(1+x^4\right)}{\left(1-x^4\right)\sqrt{1+x^4}}dx = \frac{1... | Thanks Claude Leibovici and Lucian, Using Lucian Hint.
$\bf{Another \; Try::}$ Let $\displaystyle \mathcal{I} = \int \frac{\sqrt{1+x^4}}{1-x^4}dx\;,$ Now Let $x^2=\tan \phi\;,$
Then $\displaystyle 2xdx = \sec^2 \phi d\phi \Rightarrow dx = \frac{\sec^2 \phi}{2\sqrt{\tan \phi}}d\phi$.
So $$\displaystyle \mathcal{I} = \i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1076348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Explanation solution partial-fraction of $\frac{x^2 + 2}{x^2 - 1}$ The partial fraction of $\dfrac{x^2+2}{x^2-1}$ is $1 + \dfrac{3}{2}\cdot(\dfrac{1}{x-1}-\dfrac{1}{x+1})$.
I understand how you get $\dfrac{3}{2}\cdot(\dfrac{1}{x-1}-\dfrac{1}{x+1})$ but from where does the $1 +$ come?
| Hint. you may write
$$
\frac{x^2+2}{x^2-1}=\frac{(x^2-1)+3}{x^2-1}=\frac{x^2-1}{x^2-1}+\frac{3}{x^2-1}=\color{red}{1+}\frac{3}{x^2-1}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1076437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $\int \ln(1 + e^x)\ \mathrm dx$ Evaluate the following indefinite integral.
$$\int\ln(1 + e^x) \mathrm dx$$
My attempt ::
Using integration by-parts,
\begin{align}
\int\ln(1 + e^x)\cdot 1\ \mathrm dx &= x\ln(1 + e^x) - \int \frac{x\cdot e^x\ \mathrm dx}{1 + e^x}\\
&= x\ln(1+e^x) - \frac{x^2\cdot e^x}{2(1+e^x)}... | Let $$\mathcal{ I}=\int \ln(1 + e^x)\ \mathrm dx$$
By substituting $e^x=-t\iff e^x\,\mathrm dx=\dfrac{\mathrm dt}{t}$
$$\begin{align}
\mathcal{ I}
&=\int \frac{\ln(1 -t)}{t}\ \mathrm dt
=-\int \frac{1}{t}\sum_{n=1}^{\infty}\frac{t^n}{n}\ \mathrm dt
=-\int \sum_{n=1}^{\infty}\frac{t^{n-1}}{n}\ \mathrm dt\\
&=-\sum_{n=1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1076805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find Sum of $\sum_{n=2}^\infty \ln\left(1-\frac{1}{n^2}\right)$. Prove that it converges. Question : For $$\sum_{n=2}^\infty \ln\left(1-\frac{1}{n^2}\right)$$
a. Prove it converges
b. Find the sum
My Try
$
= \sum_{n=2}^\infty \ln\left(1-\frac{1}{n^2}\right)\\
= \ln(1 - \frac{1}{4} ) + \ln(1 - \frac{1}{9} ) + \ln ( 1 -... | Hint:
Rewrite it as:
$$\sum_{n=2}^{\infty} ln(\frac{n^{2} - 1}{n^{2}}) = \sum_{n=2}^{\infty} ln( \frac{(n-1)(n+1)}{n^{2}}) = \sum_{n=2}^{\infty} [ ln(n-1) + ln(n+1) - 2ln(n)]$$
So we have: $ln(1) + ln(3) - 2ln(2) + ln(2) + ln(4) - 2ln(3) + ...$
Think about this as a telescoping series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1079174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Calculate $\sin{55}-\sin{19}+\sin{53}-\sin{17}$ without calculator So here is a trigonometric series.
$$\sin{55^\mathrm{o}}-\sin{19^\mathrm{o}}+\sin{53^\mathrm{o}}-\sin{17^\mathrm{o}}$$
Strange isn't it, and I have to calculate the total result of the series (without calculator). I don't think Maclaurin series will ... | You can use the following product-sum trig identity:
$$\sin (a+b) - \sin (a-b) = 2 \cos a \sin b$$
$$\sin (a+b) + \sin (a-b) = 2 \sin a \cos b$$
So, $$(\sin 55^\circ - \sin 17^\circ) + (\sin 53^\circ - \sin 19^\circ)$$
$$(2 \cos 36^\circ \sin 19^\circ) + (2 \cos 36^\circ \sin 17^\circ)$$
$$2 \cos 36^\circ (\sin 19^\cir... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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What is the value of $\sum_{n=0}^{\infty}(-\frac{1}{8})^n\binom{2n}{n}$ What is the value of $$\sum_{n=0}^{\infty}\left(-\frac{1}{8}\right)^n\binom{2n}{n}\;?$$
EDIT
I bumped into this series when inserting $\overrightarrow{r_1}=\left(\begin{array} {c}0\\0\\1\end{array}\right)$ and $\overrightarrow{r}=\left(\begin{array... | As mentioned in Dr. Graubner's comment, the sum is $\sqrt{\frac23}$.
In fact, this is a special case of a sort of famous Taylor series expansion:;
$$\frac{1}{\sqrt{1-4z}} = \sum_{n=0}^\infty \binom{2n}{n} z^n$$
which appeared as limiting example of several theorems in complex analysis.
To compute the series ourselves, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integration by differentiating under the integral sign $I = \int_0^1 \frac{\arctan x}{x+1} dx$ $$I = \int_0^1 \frac{\arctan x}{x+1} dx$$
I spend a lot of my time trying to solve this integral by differentiating under the integral sign, but I couldn't get something useful. I already tried:
$$I(t) = \int_0^1 e^{-tx}\frac... | $$I(a) = \int \frac{\arctan (ax)}{x+1}dx \Rightarrow I'(a) = \int \frac{x}{(x + 1)(a^2x^2+1)} dx$$
Now $$\frac{x}{(x + 1)(a^2x^2+1)} =\frac{A}{(x + 1)} + \frac{Bx + C}{(a^2x^2+1)}$$
then $A = \frac{-1}{a^2 + 1}$, $B = \frac{a^2}{a^2 + 1}$ and $C = \frac{1}{a^2+1}$. From this we have
$$\begin{align}I'(a) &= \int \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1082491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 7,
"answer_id": 3
} |
Find the value of : $\lim_{ x \to \infty} \left(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)$ I need to calculate the limit of the function below:
$$\lim_{ x \to \infty} \left(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)$$
I tried multiplying by the conjugate, substituting $x=\frac{1}{t^4}$, and both led to nothing.
| A useful result for a brute force solution to these sorts of computations is $\sqrt{1+ \theta} = 1+ {1 \over 2} \theta + r (\theta)$, where $\lim_{\theta \to 0} { r(\theta)\over \theta} = 0$
Note that $\sqrt{x+\sqrt{x+ \sqrt{x}}} = \sqrt{x} \sqrt{1+ \sqrt{{1 \over x}+\sqrt{1 \over x^3}
}}$, and so
$\sqrt{x+\sqrt{x+ \sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1082649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 3
} |
Why is $\cos(x/2)+2\sin(x/2)=\sqrt5 \sin(x/2+\tan^{-1}(1/2))$ true? According to Wolfram Alpha the following equality holds:
$$\cos\left(\frac{x}{2}\right)+2\sin\left(\frac{x}{2}\right)=\sqrt5 \sin\left(\frac{x}{2}+\tan^{-1}\left(\frac{1}{2}\right)\right)$$
I also checked it numerically. Why is it true?
| Let us consider the general case $$A=a \sin(x)+b\cos(x)$$ Without changing anything, we can rewrite $$A=\sqrt{a^2+b^2}\Big(\frac{a}{\sqrt{a^2+b^2}} \sin(x)+\frac{b}{\sqrt{a^2+b^2}} \cos(x)\Big)$$ Now, let us define$$\frac{a}{\sqrt{a^2+b^2}}=\cos(\phi)$$ so $$\sin^2(\phi)=1-\cos^2(\phi)=1-\frac{a^2}{a^2+b^2}=\frac{b^2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1082979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Show that $(1+a_1x+\ldots+a_rx^r)^k=1+x+x^{r+1}q(x)$
Fixed $k\ge 1$. Show that for each $r$, you can find $a_1,\cdot\cdot\cdot,a_r\in \mathbb{F}$ such that :$$(1+a_1x+\cdot\cdot\cdot+a_rx^r)^k=1+x+x^{r+1}q(x)$$
where $q(x)$ is a polynomial.
Any ideas?
I tried using induction as follows:
for the base case, $r=1:$
... | If the field $\mathbb{F}$ has characteristic $p \neq 0$, and $p \mid k$, then the requested $a_1,\dots, a_r$ do not exist: writing $k=pk'$, we have
\begin{align*}
(1+a_1x+\cdots+a_rx^r)^k &= ((1+a_1x+\cdots+a_rx^r)^p)^{k'}\\
&= (1+a_1^px^p+\cdots+a_r^px^{rp})^{k'}
\end{align*}
which has a zero coefficient of $x$ and h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Minimizing Sum of Reciprocals Find the minimum value, in terms of $k$ of $\frac{1}{x_1}+…+\frac{1}{x_n}$ if $x_1^2+x_2^2+…+x_n^2=n$ and $x_1+x_2+…+x_n=k$, where $\sqrt{n} < k \leq n$.
I tried the am-hm, but how to relate with the sum of squares?
| Checking the case $n = 2$
First condition:
$$
g(x,y) = x + y = k \iff y = - x + k
$$
Second condition:
$$
h(x,y) = x^2 + y^2 = 2 \iff y = \pm \sqrt{2 - x^2}
$$
This gives
\begin{align}
2 &= x^2 + (-x + k)^2=2 x^2 -2kx + k^2 \iff \\
1 &= x^2 - kx + k^2/2 = (x - k/2)^2 + k^2/4 \iff \\
x &= \frac{k\pm\sqrt{4-k^2}}{2} \qua... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
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Integration by substitution. Integrating $$\int (x^2+2)(x-1)^7 dx$$
Let $ u = x^2 + 2$ then $ \frac{du}{dx} = 2x$
I can't see where to go after this?
I may have chosen the wrong substitution. Would it help if I let let $ u = \sqrt{u-2} $?
| First note that it would be possible to expand this polynomial and integrate term by term, though this would be fairly annoying. However, if we make the substitution $u=x-1$, we can get rid of some of the work:
$\begin{align} \int(x^2+2)(x-1)^7\,dx&=\int((u+1)^2+2)u^7\,du
\\&=\int(u^2+2u+3)u^7\,du
\\&=\int u^9+2u^8+3u^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1087537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Limit involving square roots, more than two "rooted" terms The limit is
$$\lim_{x\to\infty} \left(\sqrt{x^2+5x-2}-\sqrt{4x^2-3x+7}+\sqrt{x^2+7x+5}\right)$$
which has a value of $\dfrac{27}{4}$.
Normally, I would know how to approach a limit of the form
$$\lim_{x\to\infty}\left( \sqrt{a_1x^2+b_1x+c_1}\pm\sqrt{a_2x^2+b_2... | Hint: $\sqrt{ax^2+bx+c} = \dfrac{\sqrt{a+\dfrac{b}{x}+\dfrac{c}{x^2}}}{\dfrac{1}{x}} = \dfrac{\sqrt{a+by+cy^2}}{y}$,and use L'hospitale rule. Note $y \to 0$. Repeat this process for each term of the sum and simplify to a common fraction and proceed to L'hospitale rule.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1090307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Divisibility of polynomial
Prove that:
$(x^2+x+1) \mid (x^{6n+2}+x^{3n+1}+1) $ and
$(x^2+x+1) \mid (x^{6n+4}+x^{3n+2}+1) $.
I saw proof in book with third roots of unity but i didn't understand it, so i want to see other solution but i dont have idea for that.
| $x^{6n+2}+x^{3n+1}+1 = (x^2+x+1) + x^2(x^{6n}-1) + x(x^{3n}-1)$
We know, $x^3-1\mid x^{3n}-1$ and $x^3-1 \mid x^{6n} - 1$ and $x^2+x+1 \mid x^3-1$ which proves the claim.
As for the second part note that $x^4+x^2+1 = (x^2+x+1)(x^2-x+1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1091108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Find $\sin^3 a + \cos^3 a$, if $\sin a + \cos a$ is known
Given that $\sin \phi +\cos \phi =1.2$, find $\sin^3\phi + \cos^3\phi$.
My work so far:
(I am replacing $\phi$ with the variable a for this)
$\sin^3 a + 3\sin^2 a *\cos a + 3\sin a *\cos^2 a + \cos^3 a = 1.728$. (This comes from cubing the already given state... | i see that you already have several hints pointing you towards an answer. i will try a different method, perhaps a lengthier one. we will use the addition formula $$\sin (a+b) = \sin a \cos b + \sin b \cos a, \cos (a+b) = \cos a \cos b - \sin a \sin b$$ for $\sin()$ and $\cos().$
from $$\cos(t -45^\circ) = \cos 45^\ci... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1092396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 4
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When is $(x-1)(y-1)(z-1)$ a factor of $xyz-1$? Let $x$, $y$, $z$ be three natural numbers such that $1< x< y< z$. For how many sets of values of $(x,y,z)$, is $(x-1)(y-1)(z-1)$ a factor of $xyz-1$?
I noticed that $(x-1)(y-1)(z-1)=(xyz-1)-z(x+y-1)-xy+x+y$.
But i don't know how to proceed from here. Any clues?
| Well, at least there are solutions: $x = 2, y = 4, z = 8$ and $x = 3, y = 5, z = 15$
$(x-1)(y-1)(z-1) = x y z - 1 - p(x, y, z)$ with $p(x, y, z) = x y + y z + x z - x - y - z\,.$
If $(x-1)(y-1)(z-1)$ divides $x y z -1$, it also must divide $p(x, y, z)$.
We can prove easily that $(x-1)(y-1)(z-1) > \lvert p(x, y, z) \rv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1092874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to prove inequality $\;\frac{x+y+1}{\sqrt{x^2+x}+\sqrt{y^2+y}}\le2\;,\;\;\text{for}\;\;x,y\ge 1\;$ I have this great problem: to prove
$$\frac{x+y+1}{\sqrt{x^2+x}+\sqrt{y^2+y}}\le 2\;,\;\;\forall\;x,y\ge1\;?$$
Multiplicate by conjugated I get left side as
$$\frac{\sqrt{x^2+x}-\sqrt{y^2+y}}{x-y}\le 2$$
but I can't g... | Since the denominator of the LHS is positive you are allowed to cross multiply. So it can be simplified as follows:-$$\begin{align}&(x+y+1)^2\le 4\left(\sqrt{x^2+x}+\sqrt{y^2+y}\right)^2\\&=>x^2+y^2+2xy+2x+2y+1\le 4x^2+4y^2+4x+4y+8\sqrt{(x^2+x)(y^2+y)}\\&=>2x^2+2y^2+(x-y)^2+2x+2y+8\sqrt{(x^2+x)(y^2+y)}-1\ge 0\end{align... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1093041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Optimization—Finding the Area of the Largest Isoceles Triangle
I managed to solve $(a)$. Since the area of a triangle is determined by $\frac{1}{2}$ base $\times$ height, and we already know the height, we just have to solve for the base. Using Pythagorean theorem, we can deduce base as $\sqrt{36-h^2}$. Therefore, the... | For solving $(b)$, we can find the total height $(h_T)$ of the isosceles triangle by adding the value for $h$ found in $(a)$ to the other height $6$:
$$h_T=6+3=9.$$
So, we can say that the total height of the triangle is $9$ units. Then we can find the base:
$$b=\sqrt{36-3^2}=\sqrt{36-9}=\sqrt{27}=3\sqrt{3}.$$
We ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1094908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proving the remainder when a polynomial is divided by an integer. How should I go around proving that $\forall x \in \mathbb{Z}$, the remainder when $x^2+2x$ is divided by $3$ is $0$ or $2$?
Do I use the division algorithm for this one?
| You have $x^2 + 2x = (x + 1)^2 - 1$. Now one of the following happens for any $x \in \mathbb{Z}$.
$$x \equiv 0 \mod 3 \Rightarrow x+ 1 \equiv 1 \mod 3 \Rightarrow (x+1)^2 - 1 \equiv 0 \mod 3$$
$$x \equiv 1 \mod 3 \equiv (x+1)^2 \equiv 4 \equiv 1 \mod 3 \Rightarrow (x+1)^2 - 1 \equiv 0 \mod 3$$
$$x \equiv 2 \mod 3 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1095058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
A fair die is rolled $N > 6$ times. What is the probability the last 6 rolls were exactly $1,2,3,4,5,6$? You rolls a fair die $N > 6$ times and you want to rolls the sequence $1,2,3,4,5,6$ in this order.
What is the probability that the last 6 rolls were (in consecutive order) $1,2,3,4,5,6$, (So, on the Nth roll you ge... | For seven rolls the last six being $1,2,3,4,5,6$ rules out having $1,2,3,4,5,6$ before that, so the chance you get the first $1,2,3,4,5,6$ starting on roll $2$ is $1/6^6$. For any number of rolls below $12$ the odds of ending with the first $1,2,3,4,5,6$ are $1/6^6$. It is only with $12$ or more rolls that you can ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1095568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
System of recursive equation. Let's consider:
$$u_o = -1, v_0 = 3$$
$$\begin{cases} u_{n+1} = u_n + v_n \\ v_{n+1} = -u_n + 3v_n \end{cases}$$
I tried:
$$x^n = u_n , y^n = v_n$$
$$\begin{cases} x^{n+1} = x^n + y^n \\ y^{n+1} = -x^n + 3y^n \end{cases}$$
$$\begin{cases} x = 1 + \frac{y^n}{x^n} \\ y = -\frac{x^n}{y^n} +... | I don’t necessarily recommend it, but in this case the original problem is easily solved by rather elementary ad hoc methods. Adding the two recurrences, we see that $u_{n+1}+v_{n+1}=4v_n$. From the first recurrence we know that $u_{n+1}+v_{n+1}=u_{n+2}$, so $u_{n+2}=4v_n$, and $u_n=4v_{n-2}$ for $n\ge 2$. Substituting... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1097238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
The sum of square and cube roots is an algebraic function How can we prove that the complex valued function $\sqrt z + \sqrt[3] z$ is an algebraic function?
Context
According to my understanding, an algebraic function can be defined as a root of a polynomial equation, I just cannot find a polynomial equation which has ... | In general, the sum of two algebraic function is algebraic. To find an explicit algebraic equation with root $\sqrt z + \sqrt[3] z$ expand the product
$$\prod_{k \in \{0,1\}, l \in \{0,1,2\}}( u - ( (-1)^k \sqrt{z} + \omega^l \sqrt[3]{z} ))$$
$\omega = -\frac{1}{2} + i \frac{\sqrt{3}}{2}$ is a third root of $1$. All t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1098493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Finding big O of a function How do I find Big O of function which are polynomial fractions
$$f(x) = \frac {x^4 + x^2 + 1}{x^3 + 1}$$
The same question is posted here (Finding Big-O with Fractions) but i dont understand the explanation on how from the following we concluded that it is order x
$$f(x) = \frac {x^4 + x^2 +... | For every $x\geqslant1$, $x\leqslant f(x)\leqslant x+1$. Can you conclude?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1100315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve $x+y-2=0$ and $y= x^3$ for $x$ and $y$ This is not a full question but needs to be solved to evaluate area using double integral. I need to solve
$$x+y=2$$
$$y=x^3$$
If I put $y= x^3$ in first equation I get messy $x+ x^3 -2=0$ to solve to find what $x$ is? How to do it using viable methods from calculus, or an... | Clearly, $x=1$ satisfies $x^3+x-2=0$
Alternatively, $x^3+x-2=x^3-1+x-1=(x-1)(x^2+x+1)+(x-1)=\cdots$
the rest $\dfrac{x^3+x-2}{x-1}=0$ is a Quadratic equation, right?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1101575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find a basis for the subspace $\left\{\begin{bmatrix}x & y \\ z & t\end{bmatrix}, x-y-z = 0\right\}$ The exercise gives me the subspace
$$\left\{\begin{bmatrix}x & y \\ z & t\end{bmatrix}, x-y-z = 0\right\}$$
and ask me to show that these two sets are basis for this subspace:
$$B = \left\{\begin{bmatrix}1 & 1 \\ 0 & 0\... | This is similar to what you did with B:
$$
\begin{bmatrix}x & y \\ z & t\end{bmatrix} = x\begin{bmatrix}1 & 0 \\ 1 & 0\end{bmatrix} + (-y) \begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix} + t\begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1101927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Limit $\lim \limits_{x \to 2}{\frac{\sqrt{x^3 - 3x^2 + 4}-x +2}{x^2 - 4}}$ I try to calculate $\lim \limits_{x \to 2}{\frac{\sqrt{x^3 - 3x^2 + 4}-x +2}{x^2 - 4}}$. So, $\frac{\sqrt{x^3 - 3x^2 + 4}-x +2}{x^2 - 4} = \frac{(x-2)x}{(x+2)(x+\sqrt{(x-2)^2(x+1)}-2)}$ but I don't know what to do next.
| $$
\begin{aligned}
\lim _{x\to 2}\left(\frac{\sqrt{x^3\:-\:3x^2\:+\:4}-x\:+2}{x^2\:-\:4}\right)
& = \lim _{t\to 0}\left(\frac{\sqrt{\left(t+2\right)^3\:-\:3\left(t+2\right)^2\:+\:4}-\left(t+2\right)\:+2}{\left(t+2\right)^2\:-\:4}\right)
\\& = \lim _{t\to 0}\left(\frac{\left|t\right|\sqrt{\left(t+3\right)}-t}{t\left(t+4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1102589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Evaluate to find the sum of an infinite series $∑_{n=1}^\infty$ $n\over2^{n-1}$
or
1 + $2\over2$ + $3\over4$ + $4\over8$ + $5\over16$ + $\ldots$
How to go about evaluating the above, showing that it sums to 4?
| $\sum_{n\geq 1} \frac{n}{2^{n-1}} = f'(1)$ where $f(x) = 2 \sum_{n\geq 0} \frac{x^n}{2^n} = 2 \frac{1}{1 - \frac{x}{2}}$ so that $f'(x) = 2 \frac{\frac{1}{2}}{\left(1-\frac{x}{2}\right)^2}$ and $f'(1) = 4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1103011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Proving $\sin^2A \equiv \cos^2B + \cos^2C + 2 \cos A\cos B\cos C$ As the title,
By considering $\bigtriangleup$ABC, Prove
$$\sin^2A \equiv \cos^2B + \cos^2C + 2 \cos A\cos B\cos C$$
Thanks
| this is just the $\cos$ rule and the $\sin$ rule combined
$a^2 = b^2 + c^2 - 2bc \cos A$ and $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ where $R$ is the radius of the circumcircle of $ABC.$
the cosine rule becomes
$\begin{align}
\sin^2 A &= \sin^2 B + \sin^2 C-2\sin B \sin C \cos A \\
&=\cos^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1105391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Easiest way to calculate determinant 5x5 witx x I would like to calculate this determinant:
\begin{vmatrix}x&1&0&0&0\\4&x&2&0&0\\0&3&x&3&0\\0&0&2&x&4\\0&0&0&1&x\end{vmatrix}
| Hint By adding all the rows to first you get
$$\begin{vmatrix}x&1&0&0&0\\4&x&2&0&0\\0&3&x&3&0\\0&0&2&x&4\\0&0&0&1&x\end{vmatrix} = \begin{vmatrix}x+4&x+4&x+4&x+4&x+4\\4&x&2&0&0\\0&3&x&3&0\\0&0&2&x&4\\0&0&0&1&x\end{vmatrix} \\
=(x+4)\begin{vmatrix}1&1&1&1&1\\4&x&2&0&0\\0&3&x&3&0\\0&0&2&x&4\\0&0&0&1&x\end{vmatrix}=(x+4)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1109560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$\int_0^\infty\int_0^\pi\frac{k^2(e^{-it\sqrt{k^2+m^2}}-e^{it\sqrt{k^2+m^2}})\sin(\theta)}{e^{-ikx\cos{\theta}}\sqrt{k^2+m^2}}d\theta dk$ $$\int_0^\infty\int_0^\pi\frac{k^2\left(e^{-it\sqrt{k^2+m^2}}-e^{it\sqrt{k^2+m^2}}\right)\sin(\theta)}{e^{-ikx\cos{\theta}}\sqrt{k^2+m^2}}d\theta dk$$
I saw this Integral at Quora, a... | You can write:
$$\int_0^\infty\int_0^\pi\frac{k^2\left(e^{-it\sqrt{k^2+m^2}}-e^{it\sqrt{k^2+m^2}}\right)\sin(\theta)}{e^{-ikx\cos{\theta}}\sqrt{k^2+m^2}}d\theta dk$$
As:
$$\int_0^\infty\int_0^\pi\frac{k^2\left(-2 i \sin{(t \sqrt{k^2+m^2})}\right)\sin(\theta)}{e^{-ikx\cos{\theta}}\sqrt{k^2+m^2}}d\theta dk = \int_0^\inft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1111590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Evaluation of a general trigonometric integral How can I evaluate the integral
$$\int\sin^k(x)\ dx$$
in which I don't know if $k$ is an even or an odd number?
| Let $f(x) = \sin^k x$.
We have $\sin x = \frac{e^{ix} - e^{-ix}}{2i}$. So
$$
\begin{align*}
f(x) &= \frac{1}{(2i)^k}(e^{ix} - e^{-ix})^k \\
&=\frac{1}{(2i)^k} \sum_{j = 0}^k \binom{k}{j} (e^{ix})^{k-j} (-e^{-ix})^j \\
&=\frac{1}{(2i)^k} \sum_{j = 0}^k (-1)^j \binom{k}{j} e^{(k-2j)ix} \\
&=\frac{1}{(2i)^k} \left(\sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1113610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
factor the following expression $25x^2 +5xy -6y^2$ How to factor
$$25x^2 +5xy -6y^2$$
I tried with $5x(5x+y)-6y^2$. I'm stuck here.
I can't continue.
| let me expand on my comment of using quadratic formula to factor an expression like this. of course, you need to have the quadratic formula that
$x = \dfrac{-b + \pm \sqrt{b^2 - 4ac}}{2a}$ are the solutions of
$ax^2 + bx + c = 0.$
we will look at first the equation $25x^2 +5xy - 6y^2 = 0$ later on we will see how we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1115172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 5
} |
For $c<1$, let $f_c(x)=x^2+c$, determine the period-2 points? Ok so I know I need to set $$f_c^2(x)=x$$ so:
$$(x_0^2+c)^2+c = x \iff x_0^4+2cx_0^2-x_0+c^2+c=0$$
But how do I then solve this? Ok I have solved for the four set of roots, now you can see there are two roots for $c<\frac{1}{4}$ and four for $c<-\frac{3}{4}$... | Notice there is no $x^3$ term.
Try something of the form $(x^2+Ax+B)(x^2-Ax+D)$
$$(x^2+Ax+B)(x^2-Ax+D)=x^4+(D+B-A^2)x^2+(AD-AB)x+BD$$
So we know
$$D+B-A^2=2c$$
$$AD-AB=-1$$
$$BD=c^2+c$$
Which has solution $A=1$, $B=c+1$, and $D=c$.
So $x^4+2cx^2-x+c^2+c=(x^2+x+c+1)(x^2-x+c)$
I assume you can take it from there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1115771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Is this equality true? Why? Why not? Let $$ \lim_{a\to 0} \frac{1}{2} \left( \left( \sum_{n=-\infty}^\infty \frac{1}{(n+a)^2} - \frac{1}{a^2} \right) \right) = \sum_{n=1}^\infty \frac{1}{n^2}$$
I already know that $\sum_{n=-\infty}^\infty \frac{1}{(n+a)^2} = \frac{\pi^2}{\sin^2(\pi a)}$. With that in mind, we know that... | Since
$$\sum_{n\in\mathbb{Z}}\frac{1}{(n+a)^2}=\frac{\pi^2}{\sin^2(\pi a)}$$
as a function of $a$, is an even meromorphic function with a double pole in zero, as the square of a meromorphic function with a simple pole in zero:
$$ \frac{\pi}{\sin(\pi z)}=\frac{1}{z}+\zeta(2) z+\ldots $$
it happens that:
$$ \lim_{z\to 0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1116670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How prove $\sin \left( \alpha+\frac{\pi }{n} \right) \cdots \sin \left( \alpha+\frac{n\pi }{n} \right) =-\frac{\sin n\alpha}{2^{n-1}}$? How prove
$$\prod_{k=1}^{n}\sin \left( \alpha+\frac{\pi k }{n}\right) =-\frac{\sin n\alpha}{2^{n-1}}$$
for $n \in N$?
| Using De Moivre's formula for odd $n=2m+1$,
and writing $\cos x=c,\sin x=s$
$$i\sin(2m+1)x=(i s)^{2m+1}+\binom{2m+1}2(i s)^{2m-1}c^2+\binom{2m+1}4(i s)^{2m-3}c^4+\cdots+\binom{2m+1}{2m}(is)c^{2m}$$
$$=i^{2m+1}[s^{2m+1}-\binom{2m+1}2s^{2m-1}(1-s^2)+\binom{2m+1}4s^{2m-3}(1-s^2)^2+\cdots+\binom{2m+1}{2m}(-1)^ms(1-s^2)^m]$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1118062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
How do I simplify this expression about factorization? I am trying to simplify this
$$\frac{9x^2 - x^4} {x^2 - 6x +9}$$
The solution is
$$\frac{-x^2(x +3)}{x-3} = \frac{-x^3 - 3x^2}{x-3} $$
I have done $$\frac{x^2(9-x^2)}{(x-3)(x-3)} = \frac{x^2(3-x)(3+x)}{(x-3)(x-3)} $$
but I do not find a way to simplify
How can I... | Observe that
$$
9x^2-x^4=-x^2(x^2-9)=-x^2(x-3)(x+3).
$$
Thus, we have that
$$
\frac{9x^2-x^4}{x^2-6x+9}=\frac{-x^2(x-3)(x+3)}{(x-3)^2}=-x^2\left(\frac{x+3}{x-3}\right).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1122635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Prove that if $a$,$b$,$c$ are non-negative real numbers such that $a+b+c =3$, then $abc(a^2 + b^2 + c^2)\leq 3$ Prove that if $ a,b,c $ are non-negative real numbers such that $a+b+c = 3$, then $$ abc(a^2 + b^2 + c^2) \le 3 $$
My attempt :
I tried AM-GM inequality, tried to convert it to $a+b+c$, but I think I cannot ... | Note that $(ab+bc+ac)^2 \ge 3abc(a+b+c) = 9abc$
Since, $(ab+bc+ac)^2 = \sum\limits_{cyc} a^2b^2 + 2\sum\limits_{cyc}ab^2c \ge 3abc(a+b+c)$
Hence, $$\begin{align}abc(a^2 + b^2 + c^2) &\le \frac{1}{9}(ab+bc+ac)^2(a^2 + b^2 + c^2) \tag{1} \\&\le \frac{1}{9}\left(\frac{a^2 + b^2 + c^2+2ab+2bc+2ac}{3}\right)^3\tag{2} \\ &= ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1124019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find the area of the triangle under certain preconditions
With vertices $(0, 0)$, $(b, a)$, $(x, y)$, prove the area of this triangle is $\frac{|by - ax|}{2}$.
We know area of a triangle = $\frac{rh}{2}$. ($r$ is the base.) Well, we have $r =$ the distance from the origin to $(x, y)$ and the height is the line perpen... | If we denote the point $(x,y)$ by $(x_1,y_1)$ to avoid confusion, we can take the base of the triangle to be $\;\;\;B=\sqrt{x_1^2+y_1^2}$.
The line through $(0,0)$ and $(x_1,y_1)$ has equation $y=\frac{y_1}{x_1}x$ (assuming $x_1\ne0$), so the perpendicular line through $(b,a)$ has equation $y-a=-\frac{x_1}{y_1}(x-b)$ o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1124221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Evaluating $\int_0^1 \frac {x^3}{\sqrt {4+x^2}}\,dx$ How do I evaluate the definite integral $$\int_0^1 \frac {x^3}{\sqrt {4+x^2}}\,dx ?$$ I used trig substitution, and then a u substitution for $\sec\theta$.
I tried doing it and got an answer of: $-\sqrt{125}+12\sqrt{5}-16$, but apparently its wrong.
Can someone hel... | Hint:
$$\int_0^1\frac{x^3}{\sqrt{4+x^2}}dx=\frac{1}{2}\int_0^1\frac{x^22x}{\sqrt{4+x^2}}dx=\int_0^1\frac{(x^2+4-4)(4+x^2)'}{\sqrt{4+x^2}}dx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1124546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
How does $n(n-1)(n-2)\cdots(n-m+1) \cdot \frac{(n-m)(n-m-1)\cdots1}{(n-m)(n-m-1)\cdots1} = \frac{n(n-1)(n-2)\cdots1}{(n-m)(n-m-1)\cdots1} $ I'm having issues understanding how the previous line goes to the net line.
$$
\text{Assume } m \le n \\
n(n-1)(n-2)...(n-m+1) \cdot \frac{(n-m)(n-m-1)\cdots1}{(n-m)(n-m-1)\cdots1}... | Look at a concrete example, say $n=6$ and $m=3$. Then $n-m+1=6-3+1=4$, so we have
$$\begin{align*}
6\cdot5\cdot4&=6\cdot5\cdot4\cdot\color{brown}1\\\\
&=6\cdot5\cdot4\cdot\color{brown}{\frac{3\cdot2\cdot1}{3\cdot2\cdot1}}\\\\
&=\frac{6\cdot5\cdot4\cdot\color{brown}{3\cdot2\cdot1}}{\color{brown}{3\cdot2\cdot1}}\\\\
&=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1125778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int^\infty_0 t^{a+b-1}(t+1)^{-b-1} U(a+2,a-b+2,ct)dt$
Evaluate
$$
\int^\infty_0 t^{a+b-1}\left(t+1\right)^{-b-1} U\left(a+2,a-b+2,ct\right)dt
$$
under the condition $a>0$, $b>0$ and $c>0$, where $U(\cdot,\cdot,\cdot)$ denotes the confluent hypergeometric function of the second kind. All of $a$, $b$ and ... | (Too big for comment)
It's probably a matter of taste whether this counts as simpler (and it certainly isn't shorter), but the second confluent hypergeometric function can be reduced to a linear combination of confluent hypergeometric functions of smaller rank by working through their respective integral representation... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1125967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Let $f(x)=x^2+12x+30$. Solve $f(f(f(f(f(x)))))=0$ Here is my solve, is it correct?
I figured out that we can restate $f(f(x))$ as
$((x+r)(x+s)+r)((x+r)(x+s)+s)$
thus
$f(f(f(f(f(x)))))=0$
is
$(x+r)^2(x+s)^2(4s+3r)(4r+3s)$
from vieta's
$0=(x+r)(x+s)(36^2-r^2)$ and we get $x=-6$
Did I make a mistake somewhere?
I'd al... | We have $f(x) = (x+6)^2 - 6$, therefore $f^2 (x) = (x+6)^4 - 6$, $f^3 (x) = (x+6)^8 - 6$, and so on.
So $f^5 (x) = (x+6)^{32} - 6 = 0$, giving the (real) solutions $x=-6 \pm \sqrt[32]{6}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1127043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
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no. of real roots of the equation $ 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+............+\frac{x^7}{7} = 0$
The no. of real roots of the equation $\displaystyle 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+............+\frac{x^7}{7} = 0 $
$\bf{My\; Try::}$ First we will find nature of graph of function $\displaystyle ... | The simplest way of seeing that it can have only one real root is IMO to look at the derivative. To that end we use the formula for a geometric sum
$$
f'(x)=1+x+x^2+\cdots+x^6=\frac{x^7-1}{x-1}
$$
showing that $f'(x)>0$ whenever $x\neq1$, because the numerator and denominator both change signs only at $x=1$. Of course,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1128932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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$a^2 + b^2 + c^2 = 1 ,$ then $ab + bc + ca$ gives =? In a recent examination this question has been asked, which says:
$a^2+b^2+c^2 = 1$ , then $ab + bc + ca$ gives = ?
What should be the answer? I have tried the formula for $(a+b+c)^2$, but gets varying answer like $0$ or $0.25$, on assigning different values to vari... | It depends on $a+b+c$:
$$
(a+b+c)^2-2(ab+bc+ca)=1\\
ab+bc+ca=\dfrac{(a+b+c)^2-1}{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1129070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
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Proving $ab(a+b)+ac(a+c)+bc(b+c)$ is even
Prove that $\forall a,b,c\in \mathbb N: ab(a+b)+ac(a+c)+bc(b+c)$ is even
I tried to simplify the expression to something that would always yield an even number: $ (a+b+c)(ab+ac+bc)-3abc$ but that's just a sum of numbers that are divisible by $3$...
Is there a way to do this w... | We will use that for any $A$ the numbers $A$, $3A$ and $A^3$ all have the same parity. Then writing $S$ for our expression, we see
$$\begin{align*}
S &= ab(a+b)+bc(b+c)+ca(c+a) \\ &\equiv 3ab(a+b) + 3bc(b+c)+3ca(c+a) \\
&= (a+b)^3 - a^3 - b^3 + (b+c)^3 - b^3 - c^3 + (c+a)^3 - c^3 - a^3 \\
&\equiv (a+b)^3 + (b+c)^3 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1129149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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How to compute $\int_{0}^{\infty}dx\:\frac{\exp(-ax^2+bx)}{x+1}\:\text{ for }\: a>0, b\in \mathbb{C}$? As the title says I am trying to compute the integral $I=\displaystyle\int_{0}^{\infty}dx\:\frac{\exp(-ax^2+bx)}{x+1}$ where $a>0$ and $b$ is a complex number. For the special case of $b=-2a$, we have $I=-\displaysty... | Hint:
$I(b)=\int_0^\infty\dfrac{e^{-ax^2+bx}}{x+1}dx$
$\dfrac{dI(b)}{db}=\int_0^\infty\dfrac{xe^{-ax^2+bx}}{x+1}dx$
$\therefore\dfrac{dI(b)}{db}+I(b)=\int_0^\infty e^{-ax^2+bx}~dx$
$=\int_0^\infty e^{-a\left(x^2+\frac{bx}{a}\right)}~dx$
$=\int_0^\infty e^{-a\left(x^2+\frac{bx}{a}+\frac{b^2}{4a^2}-\frac{b^2}{4a^2}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1129527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Taylor expansion of $\frac{(-1)^n}{\ln n(1+\frac{1}{n\ln n}+o(\frac{1}{n\ln n})}$ I can't get the right terms:
$$\frac{(-1)^n}{\ln n + \frac{(-1)^n}n + o(\frac1n)}=\frac{(-1)^n}{\ln n} - \frac1{n\ln^2 n}+o\left(\frac1{n\ln^2 n}\right)$$
My thoughts
$$\frac{(-1)^n}{\ln n(1+\frac{1}{n\ln n}+o(\frac{1}{n\ln n})}$$
note th... | We have $$\frac{(-1)^n}{\ln n + \frac{(-1)^n}{n} + o(\frac{1}{n})} = \frac{(-1)^n}{\ln n} \cdot \frac{1}{1 + \frac{(-1)^n}{n\ln n} + o(\frac{1}{n\ln n})} = \frac{(-1)^n}{\ln n}\left(1 - \frac{(-1)^n}{n\ln n} + o\left(\frac{1}{n\ln n}\right)\right),$$
which multiplies out to $$\frac{(-1)^n}{\ln n} - \frac{1}{n\ln^2 n} ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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I need the proving of $x\log(x)=(\frac{x-1}{x})+\frac{3}{2!}(\frac{x-1}{x})^2+\frac{11}{3!}(\frac{x-1}{x})^3+...\frac{S_{n}}{n!}(\frac{x-1}{x})^n$ I could find the role of Striling Numbers in the natural logarithm function as follows
$$x\log(x)=(\frac{x-1}{x})+\frac{3}{2!}(\frac{x-1}{x})^2+\frac{11}{3!}(\frac{x-1}{x}... | Suppose we seek to show that
$$x\log x = \sum_{n=1}^\infty \frac{1}{n!} \left[n+1\atop 2\right]
\left(\frac{x-1}{x}\right)^n.$$
Recall the species equation for decompositions into disjoint cycles:
$$\mathfrak{P}(\mathcal{U}\mathfrak{C}(\mathcal{Z}))$$
which gives the generating function
$$G(z, u) = \exp\left(u\log\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1130883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
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Does $\int_0^\infty \frac{\cos x}{1+x}$ absolutly converge?
Does the following indefinite integral converge?
$\int_0^\infty \frac{\cos x}{1+x}$
converges, absolutely converges?
i can say that by the Dirichlet test it does converge.
i am trying to prove it diverges absolutely (seems close to $\frac{1}{x}$)
unsuccess... | Consider intervals $[2k\pi-\frac{\pi}{4},2k\pi+\frac{\pi}{4}]$ for $k \in \mathbb{N}$, then:
$$\int_{0}^{\infty}\frac{|\cos(x)|}{1+x}dx>\int_{\bigcup_{k \in \mathbb{N}}[2k\pi-\frac{\pi}{4},2k\pi+\frac{\pi}{4}]}\frac{|\cos(x)|}{1+x}dx=\sum_{k=1}^{\infty}\int_{2k\pi-\frac{\pi}{4}}^{2k\pi+\frac{\pi}{4}}\frac{|\cos(x)|}{1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1132226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Find coefficient of x in a generating function The problem is as follows:
$\text{Determine the coef. of } x^{10} \text{ in } (x^3 + x^5 + x^6)(x^4 + x^5 + x^7)(1+x^5+x^{10}+x^{15}+...)$
I factored out some $x$'s, to get $x^3(1+x^2+x^3)x^4(1+x+x^3)(1+x^5+x^{10}+x^{15}+...)$and then combined the factored terms to get $... | Hint :
coefficient of $x^3$ in $(1+x^2+x^3)(1+x+x^4)(1+x^5+x^{10}+x^{15}+...)$ is $2$, because
$x^3=x^3\times 1 \times 1, x^3= x^2\times x \times 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1132304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For positive operators $A$ and $B$ with $A^6=B^6$ show that $A=B$ Since $A$ and $B$ are positive, I managed to show that $A^6$ and $B^6$ are positive.
Now, I can use the fact that there exists a unique square root of both of those and since they're equal, their roots must be equal, so $A^3=B^3$. But what now?
I'm guess... | Hint: Subtracting $A^2B$ on both sides leads to $A^2(A-B) = (B^2-A^2)B$. When subtracting $B^2A$ on both sides one obtains $(A^2-B^2)A = B^2(B-A)$. Since $A,B$ are arbitrary, $B-A$ vanishes if and only if $A^2-B^2 = 0$. Then apply the roots and you are done.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Having trouble solving part two of an equation: Part one consisted of proving that
$$\frac{x-1}{x-3} = 1+ \frac{2}{x-3}$$
I completed this and here is my working:
$$let \frac{x-1}{x-3}= LHS$$
$$RHS=\frac{x-3}{x-3} + \frac{2}{x-3}$$
$$\frac{2+(x-3)}{(x-3)}$$
$$\frac {x-3+2}{x-3}$$
$$\frac {x-1}{x-3} = LHS$$
Part two th... | The hence suggests:
Let $x=x-2$ the first equation becomes
$$
\frac{(x-2)-1}{(x-2)-3} = 1+\frac{2}{(x-2)-3} \\
\frac{x-3}{x-5}=1+\frac{2}{x-5}
$$
This process can be done with the other fractions in part 2. Notice that this will be of benefit because the $1$ will cancel because of the subtraction.
$$
1-\frac{2}{x-3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1137364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Having problems finding $x$ in terms of $a$ and $b$. I have attempted this question but have not found a solution. I am currently stuck. Hints on how I may go further would be helpful. Thank You in advance.
The Question:
$$\frac{a^2b}{x^2} + \left(1+\frac{b}{x}\right)a = 2b+ \frac{a^2}{x}$$
What I have done so far that... | Bring $a\left(1+\frac{b}{x}\right)+\frac{a^2b}{x^2}$ together using the common denominator $x^2$. Bring $2b+\frac{a^2}{x}$ together using the common denominator $x$:
$$
\frac{a^2b+abx+ax^2}{x^2}=\frac{a^2+2bx}{x}.
$$
Now cross multiply:
$$
x(a^2b+abx+ax^2)=x^2(a^2+2bx).
$$
Then expand out the terms of the left- and rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1137445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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How to integrate $\frac{1}{x\sqrt{1+x^2}}$ using substitution? How you integrate
$$\frac{1}{x\sqrt{1+x^2}}$$
using following substitution? $u=\sqrt{1+x^2} \implies du=\dfrac{x}{\sqrt{1+x^2}}\, dx$
And now I don't know how to proceed using substitution rule.
| Maybe another substitution is interesting:
For $x>0, v=\frac{1}{x}$
$$
\int \frac 1 {x\sqrt{1+x^2}} \,dx = \int \frac {1} {x^2\sqrt{\frac{1}{x^2}+1}} \,dx = -\int\frac{dv}{\sqrt{v^2+1}} = - \ln(v+\sqrt{v^2+1})=$$
$$ =- \ln\left(\frac{1}{x}+\sqrt{\frac{1}{x^2}+1}\right) + C.
$$
For $x<0, v=\frac{1}{x}$
$$
\int \frac 1 {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1137842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 4
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finding the max of $f(x)=\sqrt{(x^2-4)^2+(x-5)^2}-\sqrt{(x^2-2)^2+(x-1)^2}$ I need to find the max of $$f(x)=\sqrt{(x^2-4)^2+(x-5)^2}-\sqrt{(x^2-2)^2+(x-1)^2}$$
When $x$ is a real number.
What i did is to simplify: $$f(x)=\sqrt{x^4-7x^2-10x+41}-\sqrt{x^4-3x^2-2x+5}$$.
Then i compute: $$f'(x)=\frac{-5-7x+2x^3}{\sqrt{41-... | You have $\frac{-5-7x+2x^3}{\sqrt{41-10x-7x^3+x^4}}+\frac{1+3x-2x^3}{\sqrt{5-2x-3x^2+x^4}}=0$ so
$\frac{-5-7x+2x^3}{\sqrt{41-10x-7x^3+x^4}}=-\frac{1+3x-2x^3}{\sqrt{5-2x-3x^2+x^4}}$ then
$\frac{(-5-7x+2x^3)^2}{41-10x-7x^3+x^4}=\frac{(1+3x-2x^3)^2}{5-2x-3x^2+x^4}$ and finally $(5-2x-3x^2+x^4)(-5-7x+2x^3)^2=(41-10x-7x^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1141449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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When is $2^x+3^y$ a perfect square?
If $x$ and $y$ are positive integers, then when is $2^x+3^y$ a perfect square?
I tried this question a lot but failed. I tried dividing cases into when $x,y$ are even/odd, but still have no idea what to do when they are both odd. I tried looking into when is $2^x+3^y$ is of the fo... | If $x=1$, $2+3^y$ cannot be a square since it is $\equiv 2\pmod{3}$.
This gives $x\geq 2$, so $2^x+3^y\equiv (-1)^y\pmod{4}$ and $y$ must be even. Since in this case $3^y\equiv 1\pmod{8}$, we must have $x\geq 3$ and
$$ 2^x + 3^{2z} = A^2 $$
with $A$ odd, or:
$$ 2^x = (A-3^z)(A+3^z) \tag{1} $$
so both $A-3^x$ and $A+3^x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1142313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Proof by induction and combinations I think I am stuck on this, I am not sure if I'm going down the correct path or not. I am trying to algebraically manipulate $p(k+1)$ so I can use $p(k)$ but I am unable to do so, so I am not sure if my math is bad or I am going about it the incorrect way.
15b.) Prove by induction t... | I don't see how you can use induction here. Here are two direct proofs.
The first proof is just a calculation:
$$
\begin{align*}
\frac{n-k+1}{k} \binom{n}{k-1} &= \frac{n-k+1}{k} \frac{n!}{(k-1)!(n-k+1)!} \\ &=
\frac{n-k+1}{(n-k+1)!} \frac{1}{k(k-1)!} n! \\ &=
\frac{1}{(n-k)!} \frac{1}{k!} n! \\ &=
\frac{n!}{(n-k)!k!} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1144331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof by induction; inequality $1\cdot3+2\cdot4+3\cdot5+\dots+n(n+2) \ge \frac{n^3+5n}3$ Ok so I'm kind of struggling with this:
The question is:
"Use mathematical induction to prove that
1*3 + 2*4 + 3*5 + ··· + n(n + 2) ≥ (1/3)(n^3 + 5n) for n≥1"
Okay, so P(1) is true as 1(1+2)=3 and (1/3)(1^3 + 5)=2
Assuming P(k) is ... | Assume that statement is true for number unto k. We need to show that it is true for k+1.
$$
1*3+...+k(k+2)+(k+1)(k+3)\geq \frac{k^3+5k}{3}+(k+1)(k+3)=\frac{k^3+3k^2+17k+9}{3}\geq\frac{k^3+3k^2+8k+6}{3}=\frac{(k+1)^3+5(k+1)}{3}
$$
for last inequality expand both sides and see that
$$
\frac{(k+1)^3+5(k+1)}{3}=\frac{k^3+... | {
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"timestamp": "2023-03-29T00:00:00",
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Proof that the $\lim\limits_{x \to 2}\dfrac{1}{x} = \dfrac{1}{2}$ using the $\epsilon$-$\delta$ definition of limits (verification). Prove that the $\lim\limits_{x \to 2}\dfrac{1}{x} = \dfrac{1}{2}$ using the $\epsilon-\delta$ definition of limits.
$$
\\ \begin{align}
\\ &\textrm{Let } \forall \epsilon > 0
\\ &\textrm{... | Here you have to show that for each $ \epsilon >0$, there exists $ \delta >0 $ such that for each $ x\in Domn(\frac{1}{x}) $ if $ 0<|x-2|<\delta $ then $ \left | \frac{1}{x}-\frac{1}{2}\right |<\epsilon $.
So begin with arbitrary $ \epsilon >0 $.
Notice that if $ 0<|x-2|<1 $ then $ 1<|x|<3 $ and hence $\frac{1}{3}<\fr... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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inequalites of an acute triangle angles $ 180^{180}*a^b*b^c*c^a \le (a^2+b^2+c^2)^{180} $ If $a,b,c$ are an acute angle of triangle the prove that
$ 180^{180}*a^b*b^c*c^a \le (a^2+b^2+c^2)^{180} $
No idea
| To solve this you should know (this fact)[https://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means]
Hint:
$180 = a + b + c$
Then we can use the fact that:
$\frac{a\cdot b + b\cdot c + c\cdot a}{a+b+c} \ge (a+b+c)(a^b\cdot b^c \cdot c^a)^{\frac{1}{a+b+c}} $
After you understood how to use all this fact... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Two inequalities in a triangle I'm trying to prove that in a triangle with side lengths $a,b,c$, median lengths $m_a, m_b, m_c$ and circumdiameter $D$ the following inequality holds:
$$
\frac{a^2+b^2}{m_c}+\frac{b^2+c^2}{m_a}+\frac{c^2+a^2}{m_b}\le 6D
$$
In order to prove it, I tried to use $m_a\ge h_a$ etc. which was ... | For the second one :
$ \frac{a^2+b^2}{h_c} = \frac{a^2+b^2}{\frac{2S}{c} } = \frac{c(a^2+b^2)}{2S} \geq \frac{2abc}{2S}= \frac{abc}{S} =4R$ because $a^2+b^2 \geq 2ab$ and $4R=\frac{abc}{S} $
analogously you prove for other 2 terms so we get :
$\sum_{cyc} \frac{a^2+b^2}{h_c} \geq 3 \cdot 4R = 12R = 6D $
For the other... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1149731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Strong induction with Fibonacci numbers I have two equations that I have been trying to prove. The first of which is:F(n + 3) = 2F(n + 1) + F(n) for n ≥ 1.For this equation the answer is in the back of my book and the proof is as follows:1) n = 1: F(4) = 2F(2) + F(1) or 3 = 2(1) + 1, true.2) n = 2: F(5) = 2F(3) + F(2)... | They assume that $f(k+3) = 2f(k+1) + f(k)$, then consider $f((k+1)+3) = f(k+4)$.
Setting aside our hypothesis for a moment, we know by definition that
$$f((k+1) + 3) = f(k+4) = f(k+3) + f(k+2) = f((k+1)+2) + f((k+1)+1)$$
Then, substituting what we know about $f(k+3)$:
$$f((k+1) + 3) = f((k+1) + 1) + 2f(k+1) + f(k)$$
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1149963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that f is unique, and express f(x) in closed form.
Let $f : [−1, 1] \to R$ be a continuous function such that
(i) $f(x) = \frac {2−x^2}{2}f(\frac {x^2}{2-x^2}) $ for every $x$ in [−1, 1],
(ii) $f(0) = 1$, and
(iii) $\lim_{x\to 1^{-}}\frac {f(x)}{\sqrt{1−x}}$ exists and is finite.
Prove that f is unique, and expr... | The third condition is actually a hint.
It implies that $\sqrt{1-x}$ is a factor of $f(x)$,
and it is not difficult to figure out that
in order to fit the rest of conditions
the other factor should be $\sqrt{1+x}$.
Indeed, the closed form expression is:
\begin{align}
f(x)&=\sqrt{1-x^2}.
\end{align}
Check on condition... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to calculate sin(65) without a calculator. I know about the sum and difference formula but I can't think of two values which will be able to use for sin(65). Therefore, I come to the question: How to calculate sin(65) without a calculator.
| Let $i = \sqrt{-1}$. Then from this link, with the methodology explained here:
$$ \sin 65^\circ = -\left(-\frac12 + \frac i2 \sqrt{3}\right) \left( -\frac{1}{32} \sqrt{6} \left(1- \frac{\sqrt 3}{3}\right) + \frac{i}{32} \sqrt{-6 \left( 1 - \frac{\sqrt{3}}{3} \right)^2 + 16 }\right)^{1/3} -\left(-\frac12 - \frac i2 \sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1151521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Closer form for $\int_0^\infty\frac{(\arctan{x})^2\log^2({1+x^2})}{x^2}dx$ I Would like to know the value of this integral. $$\int_0^\infty\frac{(\arctan{x})^2\log^2({1+x^2})}{x^2}dx$$
I think
$$I=\frac{a}{b}(\pi^3\ln2)+\frac{c}{d}(\pi\ln^32)+\frac{e}{f}(\pi\ln^22)+\frac{g}{h}(\pi\ln2)+\frac{i}{j}(\pi^3)+\frac{k}{m}\z... | Let $\displaystyle\small\gamma=\lim_{R\to\infty}[-R,R]\cup Re^{i[0,\pi]}$. Observe that
$$\small\oint_\gamma\frac{\ln^4(1-iz)}{z^2}dz=\frac{1}{8}\int^\infty_0\frac{\ln^4(1+x^2)}{x^2}dx-3\int^\infty_0\frac{\ln^2(1+x^2)\arctan^2{x}}{x^2}dx+2\int^\infty_0\frac{\arctan^4{x}}{x^2}dx=0$$
since
*
*The integral over the arc... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 2,
"answer_id": 0
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How should you prove product rules by induction? For example:
$$\prod_{i=2}^n\left(1-\frac{1}{i^2}\right)=\frac{n+1}{2n}$$
For every $n$ greater than or equal to $2$
my approach for this was that I need to prove that:
$$ \left(1-\frac{1}{n^2}\right)\left(1-\frac{1}{(n+1)^2}\right)=\frac{n+1+1}{2(n+1)}$$
is this the ri... | Hint $\ $ No ingenuity is required: by telescopy the proof reduces to this one-line calculation
$\qquad\ \ $ if $\rm\, \ f(k) = \dfrac{k\!+\!1}{2k\ }\ $ then $\rm\,\ \dfrac{f(k)}{f(k\!-\!1)} =\, \dfrac{k\!+\!1}{2k\ }\dfrac{2(k\!-\!1)}{k\, }\,=\,\dfrac{k^2\!-\!1}{k^2}\, =\, 1-\dfrac{1}{k^2}\,\ $ thus
Multiplicative Tel... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1154218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Un-Simplifying a fraction, i.e. computing partial fraction decomposition $\frac{3x^2+17x}{x^3+3x^2+-6x-8}$
I need to find the value of C in the form of
$\frac{A}{x+1} + \frac{B}{x-2} + \frac{C}{x+4}$
which is based on the fraction give at the top.
I can get so far to do the following:
$A(x^2+2x-8) + B(x^2+5x+4) + C(x^... | $$
\begin{align*}
\left(A+B+C\right) x^2 &= 3x^2 \\
\left(2A + 5B -C\right)x &= 17x \\
-8A + 4B - 2C &= 0
\end{align*}
$$
Dividing the first equation by $x^2$ and the second by $x$ will get you a linear system of equations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1156046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
How does one show that for $k \in \mathbb{Z_+},3\mid2^{2^k} +5$ and $7\mid2^{2^k} + 3, \forall \space k$ odd.
For $k \in \mathbb{Z_+},3\mid2^{2^k} +5$ and $7\mid2^{2^k} + 3, \forall \space k$ odd.
Firstly,
$k \geq 1$
I can see induction is the best idea:
Show for $k=1$:
$2^{2^1} + 5 = 9 , 2^{2^1} + 3 = 7$
Assume for... | We have $2^2\equiv 1\pmod 3$ now one notes that $$2^{2^k}=(2^2)^{2^{k-1}}\equiv 1\pmod 3$$ so $\forall k$ we have $$2^{2^k}+5\equiv 6\equiv 0\pmod 3$$
We also have $2^{2^2}\equiv 2\pmod 7$ this means $2^{2^k}\equiv 2^{2^{k-2}}\pmod 7$ So when $k$ is odd $2^{2^k}\equiv 2^2\pmod 7$ and therefore $2^{2^k}+3\equiv 0\pmod 7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1156440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Expressing in terms of symmetric polynomials. How to express $$a^7+b^7+c^7$$ in terms of symmetric polynomials ${\sigma}_{1}=a+b+c$, ${\sigma}_{2}=ab+bc+ca$ and ${\sigma}_{3}=abc$ ?
| We can take a slightly different approach. We have that $a,b,c$ are the roots of the polynomial:
$$ p(x)=x^3-\sigma_1 x^2+\sigma_2 x-\sigma_3 $$
hence the eigenvalues of the companion matrix:
$$ M = \left(\begin{array}{ccc} 0 & 0 & \sigma_3 \\ 1 & 0 & -\sigma_2 \\ 0 & 1 & \sigma_1\end{array}\right) $$
so:
$$\begin{eqna... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1158579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Three terms from a sequence Find the number of ways of choosing three numbers from the set ${1,2,3,…,20}$, so that the sum of the three numbers is divisible by $3$. I wrote $a+b+c=3k$, where $k=1,2,...20$, till $k=6$, there were no problems, but then restrictions started($a<21$). Even if I come to know of the correct m... | We only need concern ourselves with congruence $\bmod 3$.
There are $7$ numbers that are $1\bmod 3$
There are $7$ numbers that are $2\bmod 3$
There are $6$ numbers that are $0\bmod 3$
We now count the ways to add to $0\bmod 3$:
Three zeros:
$0+0+0$ (There are $\binom{6}{3}$ of these)
Two zeros:
One zero:
$0+1+2$ (Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1161183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find the solutions of $x^2\equiv -1\pmod{5}$ and $x^2\equiv -1\pmod{13}$ Find the solutions of $x^2\equiv -1\pmod{5}$ and $x^2\equiv -1\pmod{13}$
I know that they are both soluble since $5\equiv 1\pmod{4}$ and $13\equiv 1\pmod{4}$
What is the method to solving this simultaneous equation.
Looking for a standard method t... | by Wilson's theorem, for prime $p$:
$$
(p-1)! \equiv_p -1
$$
if $p \equiv_4 1$ then $p-1 = 4n$ and
$$
(p-1)! = \prod_{k=1}^{2n} k \prod_{k=2n+1}^{4n} k =\prod_{k=1}^{2n} k\prod_{k=1}^{2n}(p- k) \equiv_p (-1)^{2n} \left( \prod_{k=1}^{2n} k \right)^2 = ((2n)!)^2
$$
i.e.
$$
\left( \left(\frac{p-1}2 \right)! \right)^2 \equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1161523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
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Solving for $z^2 = x^2 -xy + y^2$ Recently, I came across the following solution to finding integer solutions for $z^2 = x^2 - xy + y^2$:
*
*$x = k(-n^2 -2mn)$
*$y = k(m^2 - n^2)$
*$z = k(mn + m^2 + n^2)$
I've been scratching my head trying to figure out how to derive this solution.
I initially thought that thi... | Here is a way to find a parametrization.
First, notice that an integer solution means a rational solution $1=\left(\frac{x}{z}\right)^2-\frac{x}{z}\cdot\frac{y}{z}+\left(\frac{y}{z}\right)^2$ (when $z=0$ the only possible solution is $x=y=z=0$). So, we can reduce this problem to looking for rational points on the elli... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1162231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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To find the close form for the last value of a matrix product Suppose I have two matrices
$$A = \begin{bmatrix} 0 & 1 \\ a_2 & a_1\end{bmatrix}$$
$$B = \begin{bmatrix} 0 & 1 \end{bmatrix}$$
Then $AB^T = \begin{bmatrix} 1 \\ a_1\end{bmatrix}$, the last term is $a_1$
Now I expand matrix $A$ and $B$ a little bit to get
$... | It's more helpful to write $\begin{bmatrix} 1 \\ a_1 \\ a_2 + a_1^2 \\ a_3 + 2 a_1a_2 + a_1^3 \end{bmatrix}$.
The next one is $\begin{bmatrix} 1 \\ a_1 \\ a_2 + a_1^2 \\ a_3 + 2 a_1a_2 + a_1^3\\a_4+2a_1a_3+3 a_1^2a_2 +a_2^2+a_1^4 \end{bmatrix}$.
The next one is $\begin{bmatrix} 1 \\ a_1 \\ a_2 + a_1^2 \\ a_3 + 2 a_1a_2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1165948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Closed-form solution for 3D rotation angles given pre- and post-image I'm working on some math involving a pinhole camera model, and I've run in to the following problem: given only $x$, $y$, $z$, $A$, $B$, and $C$, I need to solve for the angles $\theta$ and $\phi$ in the following system of equations:
$$
\begin{bmatr... | There are two possible solutions for each angle, only one of which will work best
$$\begin{aligned}
\theta & = 2 \arctan\left( -\frac{\sqrt{x^2+y^2-B^2}+x}{y+B}\right) \\
\theta & = 2 \arctan\left( \frac{\sqrt{x^2+y^2-B^2}-x}{y+B}\right) \\
\end{aligned}$$
$$\begin{aligned}
\varphi & = 2 \arctan\left( -\frac{\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1166882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Find point on x-axis that is a minimum distance from 2 points Find the point on the x-axis the sum of whose distances from (2, 0) and
(0, 3) is a minimum.
Let the point be denoted (x, 0) and the sum of the distances be D.
By Pythagoras' Theorem,
$$D = \sqrt{(x-2)^2+(0-0)^2} + \sqrt{(x-0)^2+(0-3)^2}$$
$$D = \sqrt{x^2-4... | This is a great algebra problem.
Even though Jason's remark is correct, it does NOT address the full issue here.
Why not use the TI for a change.
Put in for $y1$ the first D function with the square roots.
You can look at the graph and clearly see that the function makes an abrupt turn at its minimum $x=2$. In other wo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1167971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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solving cubic diophantine equation Can someone show me how to find all solutions in positive integers to the diophantine equation: $$x^3 + y^3 = 35$$ I know how to do it algebraically, but I want to know how you solve it in number theory.
My Algebraic Approach:
$x^3 + y^3 = (x + y)\cdot (x^2 - xy + y^2) = 35$.
The only... | COMMENT.-You have solved easily the problem algebraically and you want to know how solve it in number theory. Well your curve is an elliptic curve of rank $1$ whose generator is the point $(3,2)$ which is the only point of integer coordinates and it is determined using your algebraic way. For the rest it has an infinit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1168613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Show that f is linear Let $f : \mathbb R \to \mathbb R$ be a solution of the additive Cauchy functional equation satisfying the condition
$$f(x) = x^2 f(1/x)\quad \forall x \in \mathbb R\setminus \{0\}.$$
Then show that $f(x) = cx,$ where $c$ is an arbitrary constant.
| Let $F(x)=f(x)-xf(1)$
For some $x\neq 0$, $F(\frac{1}{x})=f(\frac{1}{x})-\frac{1}{x}f(1)$.
Hence for $x\neq 0$ $$\begin{align}
x^2F(\frac{1}{x})&=x^2f(\frac{1}{x})-xf(1)\\&=f(x)-xf(1)\\&=F(x)\end{align}$$ and of course $F(1)=0$ and $F$ is additive.
Let us prove that $\forall x\in \mathbb R, F(x)=-F(-x)$
Indeed, $0=F(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1169058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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find all integer $x$ such that $7x\equiv 2x$ (mod 8) I am trying to find all integer $x$ such that $7x\equiv 2x$ (mod 8)
First, I have $$ 7x-2x=8k \hspace{0.1in} (\text{where} \hspace{0.1in}k\in\mathbb{z}) $$
$$5x=8k$$ $$x=\frac{8k}{5}$$
Does $x=\frac{8k}{5}$ right? if not, can someone give me a hit or a suggestion to... | I think you went one step too far, you already had the answer. When you got to $5x = 8k$, you should have realized that what you're looking for are the multiples of $8$. The only $x$ that will work are $x$ that are multiples of $8$.
And since $8$ is such a small number, it's no problem to check the eight possibilities ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1174703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding $\sin 2x, \cos 2x, \tan 2x$, & finding all solutions of $x$? I am a senior in high school at the moment taking a pre-calculus course. I was out a few days this week and last due to illness and I have a big test tomorrow. Please help me figure out how to do these two problems on my review sheet, and please show ... | The double angle formulas for sine, cosine, and tangent are
\begin{align*}
\sin(2x) & = 2\sin x\cos x\\
\cos(2x) & = \cos^2x - \sin^2x\\
& = 2\cos^2x - 1\\
& = 1 - 2\cos^2x\\
\tan(2x) & = \frac{2\tan x}{1 - \tan^2x}
\end{align*}
Of course, if you know $\sin(2x)$ and $\cos(2x)$, you can compute $\tan(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1176976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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$n^2(n^2-1)(n^2-4)$ is always divisible by 360 $(n>2,n\in \mathbb{N})$ How does one prove that $n^2(n^2-1)(n^2-4)$ is always divisible by 360? $(n>2,n\in \mathbb{N})$
I explain my own way:
You can factorize it and get $n^2(n-1)(n+1)(n-2)(n+2)$.
Then change the condition $(n>2,n\in \mathbb{N})$ into $(n>0,n\in \mathbb{N... | $n(n+1)(n+2)$, and $(n+2)(n+3)(n+4)$ are $2$ products of $3$ consecutive natural numbers hence each divisible by $3! = 6$, thus the product divisible by $6\cdot 6 = 36$, hence it is divisible by $9$, and it is divisible by $5! = 120$ since the product contains $5$ consecutive natural numbers $n(n+1)(n+2)(n+3)(n+4)$, t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1178495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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Counting the number of integers less than $x$ that are relatively prime to a primorial $p\#$ Let $p \ge 5$ be a prime. Let $x \ge 20$ be an integer. Let $p\#$ be the primorial for $p$.
Let $|\{ i \le x \, \wedge \gcd(i,p\#)=1\}|$ be the count of integers less than or equal to $x$ that are relatively prime to $p\#$.
... | Yes, it is true.
For $x = 6k \equiv 0 \bmod 6$, there are exactly $2k=\frac{x}{3}$ numbers $i\le x$ that are coprime to $6 = 3\#$ - every $i\equiv 1 \text{ or }5 \bmod 6$. Clearly for $x=3k$ it is also true that there are exactly $\frac{x}{3}$ numbers $i\le x$ that are coprime to $6$.
If we assume $x\ge 5$ and we also ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1179294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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How is $RN =\frac{2}{3}BC$ in this rectangle?
I tried this question first by letting $AB = 6p$ and $BC = 6q$. Then I let $R$ the midpoint of $GB$ and $N$ the midpoint of $FE$. I got stuck so I looked at the memo for a hint and it turns out I followed the memo exactly but the next step said that $RN = \frac{2}{3}BC$. H... | We are free to assume that $ABCD$ is the unit square.
Then assuming $BE=x$ we have $GF=\frac{2}{3}x$ by Thales' theorem and we must have:
$$\frac{GF+BE}{MF+CE}=\frac{\frac{2}{3}x+x}{1-\frac{2}{3}x+1-x}=2,$$
from which it follows that $x=\frac{4}{5}$. That implies $IH=\frac{4}{15}$. Since $AI=\frac{1}{3}$,
$$\frac{[AIH... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1179442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Prove $7$ divides $13^n- 6^n$ for any positive integer I need to prove $7|13^n-6^n$ for $n$ being any positive integer.
Using induction I have the following:
Base case:
$n=0$: $13^0-6^0 = 1-1 = 0, 7|0$
so, generally you could say:
$7|13^k-6^k , n = k \ge 1$
so, prove the $(k+1)$ situation:
$13^{(k+1)}-6^{(k+1)}$
$... | We write
$n=1$ and get
$13 - 6 =7$( a multiple of $7$)
Moving to the induction hypothesis,
We assume that $n=k$, therefore
$13^k - 6^k = 7b$($b$ shows that $7$ is a multiple of $13^k -6^k$ by $b$ times)
Taking $k=1$ from above ($n=k=1$) then we get :
$13^1 - 6^1=13 - 6=7$.
Take one side of the equation and use t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1181297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 6
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Stirling numbers: Combinatorial proof of an identity How to prove the following combinatorially ?
\begin{equation}
{n+1 \brace k+1}=\sum_{m=k}^{n}(k+1)^{n-m}{m \brace k}.
\end{equation}
My question is how are only ( n - m ) elements being considered to be placed on the (k + 1) boxes ? And that too without any combinati... | By way of enrichment here is a proof using generating functions.
Suppose we seek to evaluate
$$\sum_{m=k}^n (k+1)^{n-m} {m\brace k}
= (k+1)^n \sum_{m=k}^n (k+1)^{-m} {m\brace k}.$$
This is
$$(k+1)^n \sum_{m=0}^{n-k} (k+1)^{-m-k} {m+k\brace k}$$
or
$$(k+1)^{n-k} \sum_{m=0}^{n-k} (k+1)^{-m} {m+k\brace k}.$$
What we have ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1181911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Lower bound on $|a+b \sqrt{2} + c \sqrt{3}|$ I have stumbled across this question:
Let $a,b,c$ be integers, not all $0$ such that $\max(|a|,|b|,|c|)<10^6$. Prove that $|a+b \sqrt{2} + c \sqrt{3}| > 10^{-21}$.
Could anybody help by solving this? Elementary solution is preferred.
| Let $$\begin{cases}f_{1}=a+b\sqrt{2}+c\sqrt{3}\\
f_{2}=a-b\sqrt{2}+c\sqrt{3}\\
f_{3}=a-b\sqrt{2}-c\sqrt{3}\\
f_{4}=a+b\sqrt{2}-c\sqrt{3}\end{cases}$$
It is clear $$f_{1}f_{2}f_{3}f_{4}\in Z,a,b,c\in Z$$
since $a,b,c$ are integer,and not all 0 ,so $f_{k}\neq 0,k=1,2,3,4$.and Note
$$\max\{|a|,|b|,|c|\}<10^6\Longrightarro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1182523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
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Evaluate $∫_γ \frac{z^2+1}{z(z^2+4)} dz$ Where $γ(t)=re^{it}$ with $0≤t≤2π$ for all possible value of $r$, $0Evaluate
$∫_γ \frac{z^2+1}{z(z^2+4)} dz$
Where $γ(t)=re^{it}$ with $0≤t≤2π$ for all possible value of $r$, $0<r<2$ and $2<r<∞$
Theorem: Let $f: G \to \mathbb C$ be analytic, suppose $B(a,r) \subset G(r>0)$. If... | From you theorem, you can write
\begin{align}
g(z) &= \int_{\gamma}\frac{z^2+1}{z(z^2+4)}dz\\
& = 2i\pi\biggl[\int_{\gamma}\frac{(z^2+1)/(z^2+4)}{z}dz +\int_{\gamma}\frac{(z^2+1)/(z^2+2zi)}{z-2i}dz+\int_{\gamma}\frac{(z^2+1)/(z^2-2zi)}{z+2i}\biggr]\\
&= 2i\pi [f_1(0)+f_2(2i)+f_3(-2i)]\tag{1}
\end{align}
where $f_1(z) =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1185642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculating arc length $y=x^2$ I picked this example for practice and got stuck with it. Someone moderate me if I am in the right path.
I need to calculate the length of arc s, on the section of the curve $y=x^2$ with $0≤x≤1$
My Workings:
The formula is $s=\int^b_a\sqrt{1+[f'(x)]^2}dx$
I work out everything under the $... | Use the substitution $2x = \sinh u$. This gives you $\frac{dx}{du} = \frac{1}{2}\cosh u$.
So you get: $$ \int_0^1 (1+4x^2)^\frac{1}{2} dx= \int_0^{\sinh^{-1}2} (1+\sinh^{2} u)^\frac{1}{2} \frac{1}{2} \cosh udu = \frac{1}{2} \int_0^{\sinh^{-1}2}\cosh^{2} udu = \frac{1}{2} \int_0^{\sinh^{-1}2} \frac{1}{2} (\cosh2u +1)du ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1187358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find centroid of region of two curves Find the coordinates (to three decimal places) of the centroid of $y=2^x$, and $y=x^2$.
EDIT: $(0\le x\le2)$
I understand this with a triangle, not with curves.
| The computation of the centroid in $R^2$, of a region bounded by two continuous functions, goes, by definition, as follows. (Note that, over $[0,2]$, $x^2 \le 2^x$.)
First, one has to calculate the area, $\mathscr a$, of the region $$A=\{x,y\ ;\ 0\le x\le 2, \ x^2\le y\le 2^x\}.$$
$ \color{white}{bbbbbbbbbbb}$
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1187887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integrate $\int_{\frac{\sqrt{2}}3}^{\frac 23} \frac{dx}{x^5\sqrt{9x^2-1}}$
$$\int_{\frac{\sqrt{2}}3}^{\frac 23} \frac{dx}{x^5\sqrt{9x^2-1}}$$
What I did was trig substitution:
$$x=\frac 13 \sec \theta$$ $$dx=\frac 13 \sec \theta \tan \theta \, d\theta$$
Then my integral becomes $$\int_{\frac 13 \sec\frac{\sqrt{2}}{3}... | Use:
$$\frac{1}{\sec^4(x)} = \cos^4(x) = (\cos^2(x))^2 = (\frac{1 + \cos(2x)}{2})^2 = \frac14(1 + 2\cos(2x) + \cos^2(2x)) = \frac14(1 + 2\cos(2x) + \frac{1 + \cos(4x)}{2}) = etc. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1189974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to simplify $(a^2+ab+b^2)/(a+\sqrt{ab}+b)$ How can I simplify as much as possible:
$$\frac{a^2+ab+b^2}{a+\sqrt{ab}+b}$$
Also, first post here, looking forward to sticking around!
| $$\frac{a^2 + ab + b^2}{a + \sqrt{ab} + b}= \frac{(a^3-b^3)/(a-b)}{(a^{3/2} -b^{3/2})/(a^{1/2} - b^{1/2}) } =\frac{a^{3/2} + b^{3/2}}{a^{1/2} + b^{1/2}}=a-\sqrt{ab} + b$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1190659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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A sum expressed by a Kampe de Feriet function. Let $a_1$,$a_2$, $a_3$ and $b_1$,$b_2$, $b_3$ be real numbers subject to $1+b_1+b_2 - b_3 > 0 $. By generalizing the result from A sum involving a ratio of two binomial factors. we have shown that the following identity holds:
\begin{eqnarray}
&&S:= \sum\limits_{i=0}^{m-1... | We will compute the sum in question in a slightly different way and only then analyze the large-$m$ behavior of the result. The idea is to take the second binomial factor in the numerator and rewrite it as a linear combination of factors that can be absorbed into the first binomial factor. We start with the following ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1192441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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I'm having trouble with induction. Prove $1 + 2^3 + 3^3 + ... + n^3 = \frac{((n^2)(n+1)^2)}4$ I started a new course and I'm expected to know this stuff, and I'm having trouble learning some on my own.
I'm stuck with this problem:
Prove $1 + 2^3 + 3^3 + ... + n^3 = \frac{((n^2)(n+1)^2)}4$ using induction (the $1+2^3..... | This question often causes a lot of confusion simply because of the cumbersome factoring.
Let $P(n)$ be the proposition that $$\sum_{i=1}^n i^3 = \frac{(n^2)(n+1)^2}{4}.$$
Base Case, $P(1)$: $\sum_{i=1}^1 i^3 = 1$ and $\frac{(1^2)(1+1)^2}{4} = \frac{4}{4} = 1$ so $P(1)$ is true.
Inductive Hypothesis: Assume that $P(k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1196090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Inverse of this function I have given
$$ f(x) = \frac{1}{x}\left(\left(\frac{1}{x}\right)^{a} + 1\right)^{-\frac{1}{a}}$$
And am trying to invert it, but despairing.
Perhaps it helps (but I wouldn't know) that $f$ comes from
$$
g(m,n) = mn (m^a + n^a)^{-1/a}\\
f(x) = g(1/x, 1)\\
$$
| Hint:
$f(x) = \frac{1}{x} \bigg( \bigg( \frac{1}{x} \bigg)^a +1 \bigg)^{-\frac{1}{a}}$
$= \bigg( \bigg( \frac{1}{x} \bigg)^a \bigg)^{\frac{1}{a}} \bigg( \bigg( \frac{1}{x} \bigg)^a +1 \bigg)^{-\frac{1}{a}} $
$ = \bigg( \frac{ \big( \frac{1}{x} \big)^a}{\big( \frac{1}{x} \big)^a+1} \bigg)^{\frac{1}{a}}$
$=\bigg( 1- \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1197645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Continued Fraction for Root 5 How can I find the continued fraction expansion for the square root of 5. Do this without the use of a calculator and show all the steps.
| It is the same idea as $\sqrt{7}$ and $\sqrt{3}$, or any $a+b\sqrt{c}$.
First determine the integer part of $\sqrt{5}$. We know that $2<\sqrt{5}<3$.
*
*Then it is only about extracting the integer part of improper fractions,
*taking reciprocals of proper fractions
*and multiplying numerator and denominator by th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1198833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"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.