Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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How to prove the identity involving Sinc-series? Here the Sinc function is defined as: ${\rm sinc} (x):= \sin(x)/x$.
I found the following identity by numerical experiments, but how to prove it?
Let:$$
f(x;x_0):={\rm sinc}\ x_0+\sum_{n=1}^\infty {\rm sinc}\ (nx+x_0)+\sum_{n=1}^\infty {\rm sinc}\ (nx-x_0), 0<x<2\pi, x_... | I believe instead that $f$ depends on the value of $x_0$. Here is what I thought: let $x = \pi \in (0,2\pi)$ and let $x_0 \in (0, \pi)$.
For $N \geq 1$, we have:
\begin{align}
\sum_{k = 1}^N \frac{\sin (k \pi + x_0)}{k \pi + x_0} + \frac{\sin (k \pi - x_0)}{k \pi - x_0}
&= \cos x_0 \cdot \left( \sum_{k = 1}^{N} ({-1})^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2586397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Number of real solutions of $f(f(f(x)))=1$, where $f(x) = x-x^{-1}$ given $f(x) = x-x^{-1}$ ,then number of real solution of $f(f(f(x)))=1$
$f(x)=\frac{x^2-1}{x}$, $f(f(x))=\frac{(f(x))^2-1}{f(x)} = \frac{(x^2-1)^2-x^2}{x(x^2-1)}=\frac{x^4-3x^2+1}{x^3-x}$
$f(f(f(x))) = \frac{(f(x))^4-3(f(x))^2+1}{(f(x))^3-f(x)} = \fra... | Notice that $f$ is an odd function $f(-x)=-f(x)$.
We also have the relation $f(\frac 1x)=-f(x)=f(-x)$
From these two we can deduce $$f(f(f(-\frac 1x)))=f(f(f(x)))$$
Whenever $x$ is solution of $f^{\circ[3]}(x)=1$ then $-\frac 1x$ is solution too, so we can restrict our search to only positive roots.
The equation $f(x)=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2586733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solving $\cos x + \cos 2x - \cos 3x = 1$ with the substitution $z = \cos x + i \sin x$ I need to solve
$$\cos x+\cos 2x-\cos 3x=1$$
using the substitution$$z= \cos x + i \sin x $$
I fiddled around with the first equation using the double angle formula and addition formula to get
$$\cos^2 x+4 \sin^2x\cos x-\sin^2 x=1$... | Hint:
$$\cos x+\cos2x=2\cos\dfrac{3x}2\cos\dfrac x2$$
$$1+\cos3x=2\cos^2\dfrac{3x}2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2586848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
Find minimum value of $\sum \frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}$ If $a,b,c$ are sides of triangle Find Minimum value of
$$S=\sum \frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}$$
My Try:
Let $$P=\sqrt{a}+\sqrt{b}+\sqrt{c}$$
we have $$S=\sum \frac{1}{\frac{\sqrt{b}}{\sqrt{a}}+\frac{\sqrt{c}}{\sqrt{a}}-1}$$
$$S=\sum ... | $$\sqrt{b}+\sqrt{c}=\sqrt{b+c+2\sqrt{bc}}>\sqrt{b+c}>\sqrt{a},$$
which says that all denominators are positives.
Now, by C-S $$\sum_{cyc}\frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}=\sum_{cyc}\frac{a}{\sqrt{ab}+\sqrt{ac}-a}\geq$$
$$\geq\frac{(\sqrt{a}+\sqrt{b}+\sqrt{c})^2}{\sum\limits_{cyc}(\sqrt{ab}+\sqrt{ac}-a)}=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2587231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluate $\int (2x+3) \sqrt {3x+1} dx$ Evaluate $\int (2x+3) \sqrt {3x+1} dx$
My Attempt:
Let $u=\sqrt {3x+1}$
$$\dfrac {du}{dx}= \dfrac {d(3x+1)^\dfrac {1}{2}}{dx}$$
$$\dfrac {du}{dx}=\dfrac {3}{2\sqrt {3x+1}}$$
$$du=\dfrac {3}{2\sqrt {3x+1}} dx$$
| $$\int \left( 2x+3 \right) \sqrt { 3x+1 } dx=\\ 3x+1={ t }^{ 2 }\\ x=\frac { { t }^{ 2 }-1 }{ 3 } \\ dx=\frac { 2t }{ 3 } dt\\ \int { \left( \frac { 2{ t }^{ 2 }-2 }{ 3 } +3 \right) } { t }\frac { 2t }{ 3 } dt=\frac { 2 }{ 9 } \int { \left( 2{ t }^{ 2 }+7 \right) { t }^{ 2 }dt } =\frac { 4 }{ 9 } \int { { t }^{ 4 }dt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2588847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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Find $\sum_{k=1}^\infty\frac{1} {k(k+1)(k+2)(k+3)}$ I have to solve this series transforming it into a telescopic sequence
$$\sum_{k=1}^\infty\frac{1}
{k(k+1)(k+2)(k+3)}$$
But I'm lost in the calculation!
| There is a general solution for this type of problem: $$\sum_{k=1}^{\infty} \frac{1}{k(k+1)\cdot\dotso\cdot (k+N)}.$$
Observe that
$$\sum_{k\geq 1} \frac{1}{k(k+1)\cdot\dotso\cdot (k+N)}=\sum_{k\geq 1} \frac{(k-1)!}{(k+N)!}=\frac{1}{N!}\sum_{k\geq 1}\frac{\Gamma(k)\Gamma(N+1)}{\Gamma(k+N+1)}=\frac{1}{N!}\sum_{k\geq 1} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2589176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
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Taylor series of $\ln(1+x)$ bounded for 2nd order
Prove that for all $x > 0 \Rightarrow $ $x - \frac{x^2}{2} < \ln(1+x) < x- \frac{x^2}{2} + \frac{x^3}{3}$
My Attempt -
I want to show the boundaries of the Maclaurin Series of $\ln(1+x)$ up to $x^2$. As such -
$P_n(x) = f(x) + \dfrac{f'(x)}{1!}(x)+\dfrac{f''(x)}{2!... | Let $f (x)=\ln (1+x) $.
$$f'(x)=\frac {1}{1+x} $$
$$f''(x)=\frac {-1}{(1+x)^2} $$
$$f^{(3)}(c)=\frac {2}{(1+c)^3} $$
we know that $0 <c <x $, so
$$0<f^{(3)}(c)<2$$
and $$0<R_n (x)<\frac {2x^3}{6} $$
Done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2592253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Integer solutions of $a^6 + 4 b^3 = 1$ This question is simplified case of
Integer solutions of $a^6 + 4 b^3 = c^6$.
Is there integer solutions for
$$a^6 + 4 b^3 = 1$$
The way I am trying to prove is based on Fermat's Theorem, but I can not get the final result.
| From $a^6+4b^3=1$, we see that $a$ must be odd. Let's rewrite the equation as
$$(a^3-1)(a^3+1)=-4b^3$$
Since $a^3$ is odd, we have $\gcd(a^3-1,a^3+1)=2$, so we have
$$\begin{align}
a^3-1&=2c^3\\
a^3+1&=2d^3
\end{align}$$
where $cd=-b$ with $\gcd(c,d)=1$. But this implies
$$a^3=c^3+d^3$$
which, as Fermat observed, has... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2592893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Prove: For $\triangle ABC$, if $\sin^2A + \sin^2B = 5\sin^2C$, then $\sin C \leq \frac{3}{5}$. We have a triangle $ABC$. It is given that $\sin^2A + \sin^2B = 5\sin^2C$. Prove that $\sin C \leq \frac{3}{5}$.
Let's say that $BC = a$, $AC=b$, $AB=c$.
According to the sine law,
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \fra... | By the law of sines and AM-GM we obtain $$5c^2=a^2+b^2\geq2ab,$$ which gives
$$\frac{c^2}{ab}\geq\frac{2}{5}.$$
In another hand, by the law of cosines we obtain:
$$\cos{C}=\frac{a^2+b^2-c^2}{2ab}=\frac{2c^2}{ab}\geq\frac{4}{5}.$$
Id est, $$\sin{C}=\sqrt{1-\cos^2C}\leq\sqrt{1-\left(\frac{4}{5}\right)^2}=\frac{3}{5}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2593739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Value of Indefinite Integralsl involving Trigonometric function
Finding $\displaystyle \int\frac{\ln(\cot x)}{\bigg(\sin^{2009}x+\cos^{2009} x\bigg)^2}\cdot (\sin^{2008}(2x))dx$
Try: $$\int \frac{\ln(\cot x)}{\bigg(1+\tan^{2009}(x)\bigg)^2}\cdot \tan^{2008}(x)\cdot \sec^{2010}(x)dx$$
Now substuting $\tan^{2009} x=t$ ... | Using $\sin(2x)=2\sin x\cos x$, we have
$$\begin{align}&\int\frac{\ln(\cot x)}{(\sin^{n+1}x+\cos^{n+1} x)^2}\cdot (\sin^{n}(2x))\ \mathrm dx\\\\&=\int\frac{\ln(\cot x)}{(\sin^{n+1}x+\cos^{n+1} x)^2}\cdot 2^n\sin^nx\cos^nx\ \mathrm dx\\\\&=-2^n\int\frac{\ln(\cot x)}{(\sin^{n+1}x+\cos^{n+1} x)^2}\cdot \sin^{2n+2}x\cdot\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2610763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
} |
convergence of the series $\sum_{n=1}^\infty \frac {\sqrt{n^2+1}-n}{\sqrt{n}}$
I'm studying the convergence of the series
$$\sum_{n=1}^\infty \frac {\sqrt{n^2+1}-n}{\sqrt{n}}$$
*
*$\frac {\sqrt{n^2+1}-n}{\sqrt{n}}>0, \forall n \ge 1$
*$\lim_{n \to +\infty}\frac {\sqrt{n^2+1}-n}{\sqrt{n}}=\lim_{n \to +\infty}\fr... | Now, $$\frac{1}{\sqrt{n}(\sqrt{n^2+1}+n)}<\frac{1}{2\sqrt{n^3}}$$ and since $\frac{3}{2}>1$ it converges.
Because
$$\sum_{n=1}^{+\infty}\frac{\sqrt{n^2+1}-n}{\sqrt{n}}<\sum_{n=1}^{+\infty}\frac{1}{2\sqrt{n^3}}$$ and since $\sum\limits_{n=1}^{+\infty}\frac{1}{2\sqrt{n^3}}$ converges, we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2610963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Find a power series for $\frac{z^2}{(4-z)^2}$. Find a power series for $$\frac{z^2}{(4-z)^2}$$
What I did is:
Since
$$\sum^{\infty}_{k=0}z^k=\frac{1}{1-z}$$
Take the derivative, so
$$\sum^{\infty}_{k=1}kz^{k-1}=\frac{1}{(1-z)^2}$$
So
$$\frac{z^2}{(4-z)^2}=\frac{z^2}{4^2(1-\frac{z}{4})^2}=\frac{z^2}{4^2}\sum^{\infty}_{k... | For $|z|<4$ we have:$$\frac{1}{4-z}=\sum_{n=0}^{\infty}\frac{z^n}{4^{n+1}}\to \frac{1}{(4-z)^2}=\sum_{n=0}^{\infty}\frac{(n+1)z^n}{4^{n+2}}\to\frac{z^2}{(4-z)^2}=\sum_{n=0}^{\infty}\frac{(n+1)z^{n+2}}{4^{n+2}}$$
and for $|z|>4$ similarly: $$\frac{z^2}{(4-z)^2}=\sum_{n=0}^{\infty}\frac{(n+1)4^n}{z^{n}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2611571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Changing modulus in modular arithmetic Is it true that
$$a\equiv b\pmod{m}\implies\frac{a}{n}\equiv\frac{b}{n} \pmod{\frac{m}{n}},$$ where $a, b, m, n, \frac{a}{n}, \frac{b}{n}, \frac{m}{n}\in\mathbb{N}$? If so, how do I prove it?
| From the definition of modulo:
$$a \equiv b \pmod m \implies a=km+b \tag{1}$$
$$\frac{a}{n} \equiv {\frac{b}{n}} \pmod {\frac{m}{n}} \implies \frac{a}{n}=k \frac{m}{n} + \frac{b}{n} \tag{2}$$
Observe $(2)$.
\begin{align}
\frac{a}{n}&=k \frac{m}{n} + \frac{b}{n} \tag {2} \\
\frac{a}{n} \cdot n&=k \frac{m}{n} \cdot n + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2611739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
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Solve $ \int_\frac{-π}{3}^{\frac{π}{3}}\frac{\cos^2(x)-\sin^2(x)}{\cos^2(x)}dx$ I came across this question in my textbook and have been trying to solve it for a while but I seem to have made a mistake somewhere.
$$ \int_\frac{-π}{3}^{\frac{π}{3}}\frac{\cos^2(x)-\sin^2(x)}{\cos^2(x)}dx$$ and here is what I did. First I... | To simplify, try this approach of $\tan^2 x = \sec^2 x - 1$ to get
$$\int_{-\pi/3}^{\pi/3}(2-\sec^2 x)dx = 2 \cdot \int^{\pi/3}_{0}(2-\sec^2 x)dx$$
To get $2 \cdot[2x-\tan x]_{0}^{\pi/3}$ = $\frac{4\pi}{3} - 2\sqrt{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2612141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Young Tableaux generating function The number of young tableaux of $n$ cells is known to satisfy the recurrence $a_{n+1} = a_{n} + na_{n-1}$. I am trying to find the generating function but I keep getting something dependent on $n$. Here's what I did so far:
Denote by $f(x) = \sum_{n\geq 1}a_nx^n$. We have $\sum_{n \g... | Using $a_{n+1} = a_{n} + n \, a_{n-1}$ with $a_{0} = 1$ then the following exponential generating function can be obtained.
Let
$$B(t) = \sum_{n=0}^{\infty} \frac{a_{n}}{n!} \, t^{n}$$
then
\begin{align}
\sum_{n=0}^{\infty} \frac{a_{n+1}}{n!} \, t^{n} &= \sum_{n=0}^{\infty} \frac{(n+1) \, a_{n+1}}{(n+1)!} \, t^{n} = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2613328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Arc length of curve of intersection between cylinder and sphere
Given the sphere $x^2+y^2+z^2 = \frac{1}{8}$ and the cylinder $8x^2+10z^2=1$, find the arc length of the curve of intersection between the two.
I tried parametrizing the cylinder (the task specifies this as a hint). My attempt:
$$x(t) = \frac{1}{\sqrt{8}... | Apparently I was right.
If we define $r(t) = (x, y, z)$ where
$$x = \frac{1}{\sqrt{8}}sin(t)$$
$$z = \frac{1}{\sqrt{10}}cos(t)$$
$$ y = \pm \sqrt{\frac{cos(2t)+1}{4\sqrt{5}}}$$
we find
$$|r(t)| = \sqrt{(\frac{1}{\sqrt{8}}sin(t))^2 + (\frac{1}{\sqrt{10}}cos(t))^2+(\frac{\sqrt{cos(2t)+1}}{4*\sqrt{5}})^2} = \frac{1}{2\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2613408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
I have a inequality, I don't know where to start Show that for $x,y,z > 0$ the inequality is true:
$\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}+x+y+z \geq \frac{(x+y)^2}{y+z}+\frac{(y+z)^2}{z+x}+\frac{(z+x)^2}{x+y}$
I have tried Holder, but i had no luck. Please give me a hint of how to start.
| The hint.
It's $$\sum_{cyc}\left(\frac{x^2}{y}-2x+y\right) \geq \sum_{cyc}\left(\frac{(y+z)^2}{z+x}-2(y+z)+(z+x)\right)$$ or
$$\sum_{cyc}\frac{(x-y)^2}{y}\geq\sum_{cyc}\frac{(x-y)^2}{z+x}$$ or
$$\sum_{cyc}(x-y)^2S_z\geq0,$$
where $S_z=\frac{1}{y}-\frac{1}{z+x}.$
Now, by C-S $$\sum_{cyc}S_x=\sum_{cyc}\left(\frac{1}{x}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2614390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Calculating the integral $\int \sqrt{1+\sin x}\, dx$. I want to calculate the integral $\int \sqrt{1+\sin x}\, dx$.
I have done the following:
\begin{equation*}\int \sqrt{1+\sin x}\, dx=\int \sqrt{\frac{(1+\sin x)(1-\sin x)}{1-\sin x}}\, dx=\int \sqrt{\frac{1-\sin^2 x}{1-\sin x}}\, dx=\int \sqrt{\frac{\cos^2x}{1-\sin ... | Rewrite the given integral using trigonometric/hyperbolic substitutions :
$$\int \sqrt{1+\sin x} dx ={\displaystyle\int}\sqrt{2}\cos\left(\dfrac{2x-{\pi}}{4}\right)\,\mathrm{d}x$$
Apply the substitution :
$$u=\dfrac{2x-{\pi}}{4} \to dx = 2du$$
which means the integral becomes equal to :
$$=\class{steps-node}{\cssId{s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2614504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Convergence of the sequence $ \sqrt {n-2\sqrt n} - \sqrt n $ Here's my attempt at proving it:
Given the sequence $$ a_n =\left( \sqrt {n-2\sqrt n} - \sqrt n\right)_{n\geq1} $$
To get rid of the square root in the numerator:
\begin{align}
\frac {\sqrt {n-2\sqrt n} - \sqrt n} 1 \cdot \frac {\sqrt {n-2\sqrt n} + \sqrt n}{... | It is true, but formally I think it would be more elegant to say
$\frac{\sqrt{n-2\sqrt{n}}}{\sqrt{n}}=\sqrt{\frac{n-2\sqrt{n}}{n}}=\sqrt{1-\frac{2}{\sqrt{n}}}\longrightarrow1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2614626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
The general problem of finding $\int \frac{dx}{(ax^2 + bx + c)^2}$ I know that $\displaystyle\int \dfrac1{ax^2+bx+c}\,\mathrm dx$ is easily solvable using completing the square, but my question is how would would find
$$\displaystyle\int\frac{1}{\left(ax^2+bx+c\right)^2}\,\mathrm dx$$
I have tried using the same approa... | If the quadratic factor ($a \neq 0$) that appears in the denominator can be factored into linear factors over the reals, then a partial fraction decomposition can be used. I will assume this is doable for you so will only concentrate on the case where the quadratic factor in the denominator is irreducible over the real... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2618657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove via induction $\sum_{k=2}^{n}{\frac{k-1}{k!}} = \frac{n!-1}{n!}, \forall n \in \mathbb{N}, n \ge 2$ I have to prove by induction, that
$$\sum_{k=2}^{n}{\frac{k-1}{k!}} = \frac{n!-1}{n!}, \forall n \in \mathbb{N}, n \ge 2$$
$\begin{align}
\sum_{k=2}^{n+1}{\frac{k-1}{k!}} &= \sum_{k=2}^{n}{\frac{k-1}{k!} + \frac{(... | From your third line:
\begin{align}
\frac{n!-1}{n!} + \frac{n}{(n+1)!} &= \frac{(n!-1)(n+1)}{n!(n+1)}+\frac{n}{(n+1)!} \\
&= \frac{(n+1)!-(n+1)+n}{(n+1)!} \\
&=\frac{(n+1)!-1}{(n+1)!}
\end{align}
Don't forget to check base case.
To go on from where you stop, divide numerator and denominator by $n!$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $ax^2+bx+c=0$ has $2$ different solutions in $(0,1)$ then prove $a\geq 5$. Say $a,b,c$ are integers, $a>0$. Suppose $ax^2+bx+c=0$ has $2$ different solutions in $(0,1)$ then prove $a\geq 5$. Find an example for $a=5$.
I am struggling with this for some time with no success.
I try Vieta's formula $0<x_1+x_2 = -{b\... | Let $z=-b$. We know that the minimum is going to be at $(\frac{z}{2a},c-\frac{z^2}{4a})$. (For proof plug $\frac{z}{2a}$, the extremum, into $ax^2-zx+c$). We must find values of $a,b,c$ such that $a>c>0,z>0,a+c>z,c-\frac{z^2}{4a}<0$. Rewriting the last one, we get $z^2>4ac$. However, because everything is in integers w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2619761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving that the sequence $\{\frac{3n+5}{2n+6}\}$ is Cauchy. I'm not quite sure how to tackle these kinds of questions in general, but I tried something that I thought could be right. Hoping to be steered in the right direction here!
Let $\{\frac{3n+5}{2n+6}\}$ be a sequence of real numbers. Prove that this sequence i... | If you do have to prove it with the ε-$n_0$ method, it's much simpler that that:
First rewrite $a_n$ as
$$a_n=\frac{3n+5}{2n+6}=\frac32-\frac2{n+3}.$$
Then, if $m,n>n_0$,
$$\bigl|a_m-a_n\bigr|=2\,\biggl|\frac1{m+3}-\frac1{n+3}\biggr|=\frac{
2\,| m-n|}{(m+3)(n+3)}\le 2\biggl|\frac1m-\frac1n\biggr|<2\biggl(\frac1m+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2620493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
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$\arg(\frac{z_1}{z_2})$ of complex equation If $z_1,z_2$ are the roots of the equation $az^2 + bz + c = 0$, with $a, b, c > 0$; $2b^2 > 4ac > b^2$; $z_1\in$ third quadrant; $z_2 \in$ second quadrant in the argand's plane then, show that $$\arg\left(\frac{z_1}{z_2}\right) = 2\cos^{-1}\left(\frac{b^2}{4ac}\right)^{1/2}$$... | Since $a,b,c$ are real then the roots are $\dfrac{-b\pm i\sqrt{4ac-b^2}}{2a}$.
Given the quadrants locations, $z_1$ is the root with $-$ sign.
Thus $\dfrac{z_1}{z_2}=\dfrac{-b-i\sqrt{4ac-b^2}}{-b+i\sqrt{4ac-b^2}}=\dfrac{(b+i\sqrt{4ac-b^2})^2}{4ac}=\dfrac{(b^2-2ac)+ib\sqrt{4ac-b^2}}{2ac}$
It is quite easy to see that th... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving the area of the circle using sticks. I was just trying to prove the area of the circle but couldn't reach any conclusion.so here i went----- I know
For 2 identical sticks making a regular polygon we have the regular polygon as square i.e with 4 sides. For 3 identical sticks which are able to make aregular poly... | I suppose the diameter is the length of the stick, let it be $d$, then we need to prove that the area of the circle with infinite sticks is $\pi\, \frac{d^2}{4}$.
The angle of the internal triangle formed with $n$ sticks is $\frac{n-1}{n}\times \frac{\pi}{2}$
Hence,
\begin{align*}
\text{base} &= d \, \cos{\left(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2625144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Difference between minus one and plus one induction? I recently started a Combinatorics class, in which my teacher (grad student) has instructed us to Prove by induction that $$1^2+2^2+\ldots+n^2 = \frac{n(n+1)(2n+1)}{6} = \frac{2n^3+3n^2+n}{6}$$ this is trivial in the fact that it has been solved many times before, ho... | I think the only reason s/he did it this way is s/he wanted to have the final simplification end in the form $\frac {2n^3 + 3n^2 + n}6$
Had he done $P(n) \implies P(n+1)$ it would have involve a lot of factoring to get in the form $\frac {2(n+1)^3 + 3(n+1)^2 + (n+1)}6$.
Try it:
$1 + 2 + .... + n^2 = \frac {2n^3 + 3n^2 ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\lim\limits_{x\to\infty}\frac{\sqrt{4x^2+x^4}+3x^2}{x^2-5x}$ Can anyone help me solve this?
I know the answer is 4, but I don't really know how do I find the biggest power of $x$ when there's a square root.
$$\lim_{x\to\infty}\frac{\sqrt{4x^2+x^4}+3x^2}{x^2-5x}$$
| As an alternative, note that
$$\frac{\sqrt{x^4}+3x^2}{x^2-5x}=\frac{4x^2}{x^2-5x}\le\frac{\sqrt{4x^2+x^4}+3x^2}{x^2-5x}\le\frac{\sqrt{(x^2+2)^2}+3x^2}{x^2-5x}=\frac{4x^2+2}{x^2-5x}$$
thus for squeeze theorem
$$\frac{\sqrt{4x^2+x^4}+3x^2}{x^2-5x}\to4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2625933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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If $a,b,c$ be in Arithmetic Progression, If $a,b,c$ be in Arithmetic Progression, $b,c,a$ in Harmonic Progression, prove that $c,a,b$ are in Geometric Progression.
My Attempt:
$a,b,c$ are in AP so
$$b=\dfrac {a+c}{2}$$
$b,c,a$ are in HP so
$$c=\dfrac {2ab}{a+b}$$
Multiplying these relations:
$$bc=\dfrac {a+c}{2} \dfrac... | Since $a,b,c$ are in arithmetic progression, we get
\begin{align*}
&c-b=b-a\\[4pt]
\implies\;&a = 2b - c\\[4pt]
\end{align*}
Since $b,c,a$ are in harmonic progression, we get
\begin{align*}
&\frac{1}{a}-\frac{1}{c} = \frac{1}{c}-\frac{1}{b}\\[4pt]
\implies\;&a = \frac{bc}{2b-c}\\[4pt]
\implies\;&a = \frac{bc}{a}\\[4pt]... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Equation of Normal I'm struggling to get the same answer as the book on this one. I wonder if someone could please steer me in the right direction?
Q. Find the equation of the normal to $y=x^2 + c$ at the point where $x=\sqrt{c}$
At $x = \sqrt{c}$ then $y = 2c$, so the point given is $(\sqrt{c},2c)$
The gradient of the... | HINT:
The gradient is $2x$ but you have that $x=\sqrt{c}$ so the gradient is equal to $2\sqrt c$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Probability that $a^2+b^2+c^2$ divisible by $7$
Three numbers $a,b,c\in\mathbb{N}$ are choosen randomly from the set of natural numbers. The probability that $a^2+b^2+c^2$ is divisible by $7$ is
Try:any natural number when divided by $7$ gives femainder $0,1,2,3,4,5,6$
So it is in the form of $7k,7k+1,7k+2\cdots ,7k... | What remainders do the squares of numbers give when divided by 7 these are $$(7k)^2 \rightarrow 0 \\ (7k+1)^2 \rightarrow 1 \\ (7k+2)^2 \rightarrow 4 \\ (7k+3)^2 \rightarrow 2 \\ (7k+4)^2 \rightarrow 2 \\ (7k+5)^2 \rightarrow 4 \\(7k+6)^2 \rightarrow 1 \\$$
Now the sum of remainders must add upto a multiple of 7, this ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Largest constant for an inequality with powers of three variables Let real $(a,b,c)$, with $a+b+c=0$, and let real positive $p$ with $p\ne1$ and $p\ne2$.
With some constant $C = C(p)$, the following inequality must hold:
$$
(a^2)^p + (b^2)^p + (c^2)^p \ge C \cdot (a^{2} + b^{2} + c^{2} )^p
$$
It is understood that $... | By homogeneity, let $a^2 + b^2 + c^2 = 1$. Then we need to establish
$$
(a^2)^p + (b^2)^p + (c^2)^p \ge C(p)
$$
The two conditions $a^2 + b^2 + c^2 = 1$ and $a+b+c=0$ form a unit circle in $(a,b,c)$-space. Let's parametrize this circle as follows:
$$
a = \frac{-2}{\sqrt{6}}\cos(\phi)\\
b = \frac{2}{\sqrt{6}}\cos(\phi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2631115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve $x^2 + x = y^4 + y^3 + y^2 + y$ over integers. I am trying to solve solve $x^2 + x = y^4 + y^3 + y^2 + y$ over the integers. So far I have decomposed it into $x(x + 1) = y(y + 1)(y^2 + 1)$ and noticed that both sides of the equation are nonnegative. Furthermore $GCD(x, x + 1) = GCD(y, y + 1) = GCD(y, y^2 + 1) = G... | Hint: If you multiply both sides by $4$ and add $1$, you get
$$(2x+1)^2=4y^4+4y^3+4y^2+4y+1=(2y^2+y+1)^2-(y^2-2y).$$
This results in two squares that are very close together. Can you prove that they are too close together when $y$ is sufficiently large?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2631349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Problem with $\sin 90^\circ =1$ We know
$$\sin\theta = \frac{\text{perpendicular}}{\text{hypotenuse}}$$
then $\sin 90^\circ = 1$ refers in this case perpendicular = hypotenuse. But, then, the base becomes $0$ according to the Pythagorean triplet law. How is this possible?
| The Pythagorean Theorem asserts that $a^2 + b^2 = c^2$ such that $c$ represents the hypotenuse, namely the side of a triangle that is directly opposite the $90^\circ$ angle. There has to be a $90^\circ$ angle because this theorem only applies to right triangles. If $\sin 90^\circ = 1$, since $$\sin(\cdot) = \frac{\text... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2633648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Polynomial such that $f''(x) \rightarrow2$ as $x\rightarrow\infty$ given some values what is $f(1)$? Let $f$ be a polynomial such that $f''(x) \rightarrow2$ as $x\rightarrow\infty$, the minimum of f is attained at $3$, and $f(0)=3$, Then $f(1)$ equals.
$(A) \ 1$
$(B) \ 2$
$(C) -1$
$(D) -2$
I am not sure how to deal wi... | Dave showed that the degree of the polynomial is $2$. So we have $f(x)=a_2x^2+a_1x+a_0$.
We know that $f''(x)=2a_2=2\implies a_2=1\implies f(x)=x^2+a_1x+a_0$
We also know that there is a minima at $x=3\implies f'(3)=0\implies f'(3)=2\overbrace x^3+a_1=6+a_1=0\implies a_1=-6$ hence we have $f(x)=x^2-6x+a_0$.
We also kno... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2636558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is it possible to know whether the following series converges without its formula? Given this series:
$$\dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{10} + \dfrac{1}{14} + \dfrac{1}{18} + \ldots$$
Is it possible to know if the series converges without its formula? I tried using the comparison test with both $1/n$ and $1/n^2... | $$\dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{10} + \dfrac{1}{14} + \dfrac{1}{18} + \ldots > \\
\dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{12} + \dfrac{1}{16} + \dfrac{1}{20} + \ldots = \\
\dfrac{1}{4}\left(\dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5} + \ldots \right)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Finding the maximum value without using derivatives
Find the maximum value of $$f(x)=2\sqrt{x}-\sqrt{x+1}-\sqrt{x-1}$$ without using derivatives.
The domain of $f(x)$ is $x \in [1,\infty)$. Then, using derivatives, I can prove that the function decreases for all $x$ from $D(f)$ and the maximum value is $f(1)= 2 - \sq... | Some other manipulation perhaps?
$$\begin{align}
f(x)&=2\sqrt{x}-\sqrt{x+1}-\sqrt{x-1} \\
&=\sqrt{x}\left[2-\frac{\sqrt{x+1}}{\sqrt{x}}-\frac{\sqrt{x-1}}{\sqrt{x}}\right] \\
&=\sqrt{x}\left[2-\left(\sqrt{1+\frac{1}{x}}+\sqrt{1-\frac{1}{x}}\right)\right]
\end{align}$$
With $\left(\sqrt{1+\frac{1}{x}}+\sqrt{1-\frac{1}{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2637601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
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How many ways can the digits $2,3,4,5,6$ be arranged to get a number divisible by 11 How many ways can the digits $2,3,4,5,6$ be arranged to get a number divisible by $11$
I know that the sum of the permutations of the digits should be divisible by 11. Also, the total number of ways the digits can be arranged is $5! = ... | Hint. By the divisibility rule by $11$ we have to count the arrangements $d_1,d_2,d_3,d_4,d_5$ of the digits $2,3,4,5,6$ such that $d_1+d_3+d_5-(d_2+d_4)$ is divisible by $11$. Notice that
$$-2=2+3+4-(5+6)\leq d_1+d_3+d_5-(d_2+d_4)\leq 4+5+6-(2+3)=10$$
therefore we should have $d_1+d_3+d_5=d_2+d_4=\frac{2+3+4+5+6}{2}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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All possible Jordan normal forms with a given minimal polynomial
Suppose $A \in M_{n\times n}(\Bbb R)$ has minimal polynomial $$(\lambda-1)^2(\lambda+1)^2$$ and $n\leq 6$. Find all possible Jordan normal forms.
If we denote the $n \times n$ Jordan block
$$J_\alpha (n) = \begin{bmatrix}\alpha&1&\cdots&\cdots\\0&\alph... | Any of the Jordan blocks cannot be of size $3$ (or more) because it would then correspond to a factor $(\lambda\pm 1)^{3\text{ or more}}$ in the minimal polynomial. However, to get to the minimal polynomial with factors $(\lambda\pm 1)^2$, there must be at least one Jordan block of size 2. Therefore, the solutions for ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Understanding quadratic reciprocity I'm reading this proof and I don't think I understand the notion of quadratic reciprocity properly.
$q$ is a prime. $p = 2q + 1$.
I'm referring to section (6) which claims:
$p\equiv 2\pmod{5}$ so by quadratic reciprocity $5$ is a quadratic non-residue modulo $p$.
I know that $$\lef... | As $p\equiv 2\mod 5$, $\biggl(\dfrac p5\biggr)=\biggl(\dfrac 25\biggr)=-1$, so by the quadratic reciprocity law $\biggl(\dfrac 5p\biggr)=-1$.
Second point: if $q\equiv \pm2\mod 5$ and $p=2q+1$, then $p\equiv 2\cdot2+1\equiv 0$ or $p\equiv 2(-2)+1\equiv1+1=2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do I evaluate $\int_{1}^{2} \int_{0}^{z} \int_{0}^{y+z} \frac{1}{(x+y+z)^3}\text{d}x\text{d}y\text{d}z$?
$$\int_{1}^{2} \int_{0}^{z} \int_{0}^{y+z} \frac{1}{(x+y+z)^3}\text{d}x\text{d}y\text{d}z$$
This is the answer WolframAlpha gives:
substituting it with polar coordinates makes it worse..
| You just need to evaluate the integrals from inside. When you are integrating with respect to $x$, the other variables ($y$ and $z$) are constants:
$$\int_{1}^{2} \int_{0}^{z} \int_{0}^{y+z} \frac{1}{(x+y+z)^3}\text{d}x\text{d}y\text{d}z=$$$$\int_{1}^{2} \int_{0}^{z}\left.-\frac{1}{2(x+y+z)^2}\right|_{x=0}^{x=y+z}\text... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2641829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Real Fundamental- System/Matrix of a Differential equation We consider:
$$y''' - 2y'' + 2y' - y = 0$$
The real solution to this equation is:
$$y(x) = c_3e^{x} + c_2e^{x/2}sin\left(\frac{\sqrt{3}x}{2}\right) + c_1e^{x/2}cos\left(\frac{\sqrt{3}x}{2}\right)$$
How do we now represent it as a fundamental- system/matrix ? ... | Proceeding from
$$X' = Ax = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & -2 & 2 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}$$
we find the eigenvalues $\lambda_1 = 1, \lambda_2 = \frac{1}{2}\left(1 + i\sqrt{3}\right), \lambda_3 = \frac{1}{2}\left(1 - i\sqrt{3}\right)$, which correspond to the followin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2644520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Find the Maclaurin Series for $e^{\arctan x}$ up to $\mathcal{O}(n^5)$ The question is to find the value of the Maclaurin series expansion of $e^{\arctan x}$ up to (but not including) $\mathcal{O}(n^5)$.
I tried using the Maclaurin series for $e^{u}$ then subbing $\arctan x$ into it, and I got $$\sum_{n=0}^{\infty } ... | https://en.wikipedia.org/wiki/Exponential_formula
\begin{align}
\arctan x & = x - \frac{x^3} 3 + \frac{x^5} 5 - \frac{x^7} 7 + \cdots \\[10pt]
& = a_1 x + a_2 \frac{x^2} 2 + a_3 \frac{x^3} 6 + a_4 \frac{x^4}{24} + \cdots \\[10pt]
a_1 & = 1 \\
a_2 & = 0 \\
a_3 & = -1/3 \\
a_4 & = 0 \\
a_5 & = +1/5 \\
& \,\,\,\vdots \\[1... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find all the solutions $\in \Bbb R^+$ that satisfy $x(6-y)=9$, $y(6-z)=9$, $z(6-x)=9$ Find all the solutions $\in \Bbb R^+$ that satisfy $x(6-y)=9$, $y(6-z)=9$, $z(6-x)=9$
My try
I found that $0 \lt x,y,z \lt 6$
Multiplying the equations we got $x(6-y)y(6-z)z(6-x)=9^3$
$x(6-x)y(6-y)z(6-x)=9^3$
And here is the problem,... | You got till $x(6-x)y(6-y)z(6-x)=9^3$.
Now it just remains to note by AM-GM, $x(6-x)y(6-y)z(6-x) \leqslant \left(\frac16(x+6-x+y+6-y+z+6-z) \right)^6=3^6=9^3$ with equality possible iff $x=6-x=y=6-y=z=6-z$. So $x=y=z=3$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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Solving general cubic without complicated substitutions I am trying to solve a general cubic without complicated substitutions:
$$
ax^3+bx^2+cx +d = 0
$$
Assuming nonzero $a$, after dividing both sides by $a$, moving the $\frac{d}{a}$ term to the RHS and performing further simplifications I get:
$$
(x + \frac{b}{3a}) [... | Consider an equation of degree $3$ over $\mathbb{C}$: $ R(x)=x^3+ux^2+vx+w=0$ where $u,v,w\in\mathbb{C}$.
Putting $ x=y-u/3$, we obtain an equation in the form $ y^3+px+q=0$.
If $ p=0$, then it's easy, else:
Putting $ y=z+b/z$, we obtain an equation in the form
(1) $ z^6+a_4z^4+a_3z^3+a_2z^2+a_0=0$.
The system $ a_4=a... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Probability of $3+3$ cards, out of $6$ cards drawn from a solitaire
A solitaire consists of $52$ cards. We take out $6$ out of them (wihout repetition). Find the probability there are $3+3$ cards of the same type (for example, $3$ "1" and $3$ "5").
Attempt.
First approach. There are $\binom{13}{2}$ ways to choose $2... | There are $13$ possible numbers/people. We need to pick $3$, so the probability is $\binom{4}{3} \cdot \binom{13}{2} \cdot \binom{4}{3}$. The total probability of choosing 6 cards is $\binom{13}{2}$. The total probability is $\frac{4 \cdot 78 \cdot 4}{20358520}$
, which equals $\frac{1248}{20358520} \Rightarrow \appr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2648160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
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How to find the indefinite integral? $$\int\frac{x^2}{\sqrt{2x-x^2}}dx$$
This is the farthest I've got:
$$=\int\frac{x^2}{\sqrt{1-(x-1)^2}}dx$$
| Hint:
As $\dfrac{d(2x-x^2)}{dx}=2-2x$
$$\dfrac{x^2}{\sqrt{2x-x^2}}=\dfrac{x^2-2x+2x-2+2}{\sqrt{2x-x^2}}$$
$$=-\sqrt{1-(x-1)^2}-\dfrac{2-2x}{\sqrt{2x-x^2}}+\dfrac2{\sqrt{1-(x-1)^2}}$$
Now use $\#1,\#8$ of this
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Simplifying tensor equation I was given a following tensor equation to simplify and solve:
$$A^s \cdot (A^s \otimes A^a \otimes I) \cdot A^s$$ where:
$A^s$- symmetric part of the representation; $A^s=\frac{1}{2}(A-A^T)$
$A^a$- anti-symmetric part of the representation; $A^a=\frac{1}{2}(A+A^T)$
$I$ - Kronecker delta, he... | I went to ask about this and got a solution. As all of the tensors here are $T^2$, using tensor properties, the whole equation can be simplified to:
$$(A^s \cdot A^s) \cdot A^a \cdot (I \cdot A^s) = \text{||using associative property||} = (A^s \cdot A^s)(I \cdot A^s) \cdot A^a$$
Because, as I said above, the tensors ar... | {
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Using integration to solve a formula for the area of a ellipse Problem:
Set up a definite integral to find the area of an ellipse with axis lengths $a$ and $b$. Use a trigonometric substitution to find a formula for the area. What happens if $a=b$? Does this agree with a Geometry formula for a circle? Explain.
$$\frac... | Hint...You just need to change the limits to $0$ and $\frac {\pi}{2}$ and use the identity $\cos^2\theta=\frac 12(1+\cos2\theta)$ and you will be finished
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2649479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Problem on row Echelon form Consider a $3\times 3$ matrix $$A =\begin{bmatrix}1 & 2 & -1\\2&1&0\\ 3&0&1\\\end{bmatrix}.$$ I have to find nonsingular matrix $P$ such that $PA$ is in row reduced Echelon form.
I am not able to get any idea to solve this problem. I understand Echelon form of a matrix. But what exactly sho... | We can obtain P in this way by left multiplication
$$P(I|A)=(P|PA)$$
thus consider $(I|A)$
$$\left[\begin{array}{rrr|rrr} 1 & 0 & 0 & 1 &2 & -1\\ 0 & 1 & 0 & 2 &1 &0 \\ 0 & 0 & 1 & 3 &0 &1\end{array}\right]$$
and by row operations
$$\left[\begin{array}{rrr|rrr} 1 & 0 & 0 & 1 &2 & -1\\ -2 & 1 & 0 & 0 &-3 &2 \\ -3 & 0 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2649661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Inequality Proof $\frac{a}{1-a^2}+\frac{b}{1-b^2}+\frac{c}{1-c^2}\geq \frac{3\sqrt{3}}{2}$ Let $a,b,c\in \mathbb{R}^+$, and $a^2+b^2+c^2=1$, show that:
$$ \frac{a}{1-a^2}+\frac{b}{1-b^2}+\frac{c}{1-c^2}\geq \frac{3\sqrt{3}}{2}$$
| Here is a hint: it is enough to show for $x\in (0, 1)$,
$$f(x) = \left(\frac{x}{1-x^2}-\frac{\sqrt3}2\right)-\frac{3\sqrt3}2\left(x^2-\frac13 \right) \geqslant 0$$
$$\iff \frac{x(\sqrt3x+2)(3x-\sqrt3)^2}{6(1-x^2)} \geqslant 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2650458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding value of Quadratic If the quadratic equations $3x^2+ax+1=0$ and $2x^2+bx+1=0$ have a common root, then the value of $5ab-2a^2-3b^2$ has to be find.
I tried by eliminating the terms but ended with $(2a-b)x=1$. Can you please suggest how to proceed further?
| The common root is where two curves of functions $y_1=3x^2+ax+1$ and $y_2=2x^2+bx+1$ intersect; so we may write:
$3x^2+ax+1=2x^2+bx+1$ ⇒ $x^2 +(a-b)x=0$ ⇒ $ x[x+(a-b)]=0$
$x=0$ is not acceptable
$x= b-a$ can be a solution; plugging this in one of equations we get:
$3(b-a)^2 +a(b-a) +1=0$
Reducing this relation we ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to solve the inequality analytically?
Solve the inequality analytically.
$$ \log_x{(\sqrt{x^2 + 2x - 3} + 2)} \cdot \log_5{(x^2 + 2x - 2)} \ge \log_x{4} $$
My solution
$$
\left.
\left\{
\begin{array}{l}
x > 0 \\
x \ne 1 \\
x^2 + 2x - 3 \ge 0 \\
x^2 + 2x - 2 > 0
\end{array... | You have correctly eliminated $x \leqslant 1$. For $x> 1$, we may take powers of $x$ on both sides and have the equivalent:
$$(x^2+2x-3)^{\log_5(x^2+2x-2)} \geqslant 4$$
Now if $b=x^2+2x-2 \in (1, 5)$, then in the LHS, the exponent is $\in (0, 1)$, while the base is $\in (0, 4)$. Clearly the LHS will be less than RHS ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Bayes's rule and unfair coin | Solution Explanation
There are three coins in a bag. Two of them are fair. One has heads on
both sides. A coin selected at random shows heads in two successive
tosses.
What is the conditional probability of obtaining another head in the
third trial given the fact that the first tw... | Bayes' Rule says
$$P(A|B) = \frac{P(B|A)\cdot P(A)}{P(B)}$$
so in your question, it is
$$P(HHH|HH) = \frac{P(HH|HHH)\cdot P(HHH)}{P(HH)}$$
Now, notice that
$$P(HH|HHH) = 1$$
because it simply says "if we know that three successive tosses are resulted in $HHH$, what is the probability that first two of them are result... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2655238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Probability Problem based on the a set of numbers. Determine if the number is divisible by three Three distinct numbers are selected at random from the set $\{1,2,3,4,5,6\}$. What is the probability that their product is divisible by $3$?
I think that since because $3$ and $6$ are the only numbers that divide into thre... | We need to find the probability that at least one of the numbers is divisible by three. This is the same as
$$\begin{align*}
1-P(\text{none of the numbers are divisible by three})
&=1-\left(\frac{4}{6}\cdot\frac{3}{5}\cdot\frac{2}{4}\right)\\\\
&=1-0.2\\\\
&=0.8
\end{align*}$$
In order to get three numbers that are not... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove determinant is negative Prove that the determinant $\Delta$ is negative
$$
\Delta=\begin{vmatrix}
a & b & c \\
b & c & a \\
c & a & b
\end{vmatrix}<0
$$
where $a,b,c$ are positive and $a\neq b\neq c$.
My Attempt:
Applying Sarrus' rule,
$$
\begin{matrix}
a&b&c&a&b\\
b&c&a&b&c\\
c&a&b&c&a
\end{matrix}
$$
$$
\Delt... | By the AM-GM inequality:
$$
\frac{a^3+b^3+c^3}{3} \ge \sqrt[3]{a^3b^3c^3} = abc
$$
The strict inequality holds unless $\,a=b=c\,$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $76$ raised to any integer power and then divided by $100$ (ignoring the remainder) is always divisible by $57$ Related (but different) question: Is it true that $76^n=76\pmod{100}$ for all $n>0$?
If you raise $76$ to an integer power ($n\ge2$) and ignore the last two digits, it appears that you always have ... | Another way to look at it:
$\frac {76^n - 76}{100} = \frac {76(76^{n-1} - 1)}{100} = \frac {19(76^{n-1}-1)}{25} = 19* \frac {(76 -1)(76^{n-2} + 76^{n-3} + ...... + 76 + 1)}{25} =$
$19*\frac {75(76^{n-2} + 76^{n-3} + ...... + 76 + 1)}{25} = 3*19*(76^{n-1} + 76^{n-2} + ...... + 76 + 1)$
So that's that.
But there so some ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How many positive integer solutions satisfy the condition $y_1+y_2+y_3+y_4 < 100$ In preparation for an upcoming test I have come across the following problem and I am looking for some help with it just in case a question of its kind comes up on a evaluation. Thanks!
How many positive integer solutions satisfy the con... | Lets say the question was:
$$y_1+y_2+y_3+y_4 = 4$$
There is only $1$ solution:
$$1+1+1+1 = 4$$
Now lets assume it was:
$$y_1+y_2+y_3+y_4 = 5$$
There are $4$ solutions:
$$1+1+1+2=5$$
$$1+1+2+1 = 5$$
$$1+2+1+1 = 5$$
$$2+1+1+1 = 5$$
Next assume it was:
$$y_1+y_2+y_3+y_4 = 5 $$
There are $10$ solutions:
$$1+1+1+3 = 6$$
$$1... | {
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"timestamp": "2023-03-29T00:00:00",
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Divisibility of a Polynomial Prove that for all $n$, $121\mid n^2+3n+5$.
I thought proving $n^2 +3n+5\pmod{121}\equiv 0$ had no solutions would be a good start, so I completed the square of $n^2+3n+5($, and arrived at $(n+62)^2 \pmod{121} \equiv 3718$. Replacing $n+62$ by $x$, I lastly arrived at $x^2 \equiv 88\pmod{1... | I think you want to prove that $121\nmid(n^2+3n+5)$, that is, $121$ does not divide $n^2+3n+5$.
First of all ask yourself whether $n^2+3n+5$ is a multiple of $11$. The equation $n^2+3n+5\equiv0\pmod{11}$ can be rewritten
$$
n^2-8n+16=(n-4)^2\equiv0\pmod{11}
$$
which only holds for $n\equiv4\pmod{11}$.
Now we know that ... | {
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"timestamp": "2023-03-29T00:00:00",
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How to find the sum of the following series from $n=0$ to $n=99$ Find the summation of the series from $n=0$ to $n=99$. The question was given in the format of
$$(1\cdot 2)+(3\cdot 4)+(5\cdot 6)+\dots +(99\cdot 100).$$
I was able to generalise it but could not solve it. Help!! the general formula is summation (2n+2)!/... | We have that
$$n(n-1)=\frac{n^3-(n-1)^3}{3}-\frac{1}{3}.$$
Therefore
$$\begin{align}
(1\cdot 2)+(3\cdot 4)+(5\cdot 6)+\dots +(99\cdot 100)&=\sum_{n=1}^{50}(2n-1)(2n)=4\sum_{n=1}^{50}n(n-1)+2\sum_{n=1}^{50}n\\
&=4\sum_{n=1}^{50}\frac{n^3-(n-1)^3}{3}-\frac{4\cdot 50}{3}+50\cdot 51\\
&=\frac{4}{3}\left(50^3-0^3\right)-\fr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Formulating LPP from given optimal table in Simplex method I have a problem set in my assignment sheet that asks to formulate a Linear Programming Problem from a given optimal solution and we know that method used is Simplex method.
What is the correct procedure for going in reverse order? I know our basic thinking to ... | To explain why the basis matrix can be read from the given simplex tableau, I introduce these additional symbols:
*
*$A \in M_{m\times n}(\Bbb R)$ ($m\le n$) has rank $n$ and current basis matrix $B$.
*$x_B$ denotes the current basic solution.
*$c_B$ denotes the reduced objective function so that $c^T x=c_B^T x_B$... | {
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"timestamp": "2023-03-29T00:00:00",
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>Finding range of $f(x)=\frac{\sin^2 x+4\sin x+5}{2\sin^2 x+8\sin x+8}$
Finding range of $$f(x)=\frac{\sin^2 x+4\sin x+5}{2\sin^2 x+8\sin x+8}$$
Try: put $\sin x=t$ and $-1\leq t\leq 1$
So $$y=\frac{t^2+4t+5}{2t^2+8t+8}$$
$$2yt^2+8yt+8y=t^2+4t+5$$
$$(2y-1)t^2+4(2y-1)t+(8y-5)=0$$
For real roots $D\geq 0$
So $$16(2y-1)... | You can simplify the expression as
$$\frac12+\frac1{2(\sin x+2)^2}$$ and the extreme values are
$$\frac12+\frac1{2\cdot3^2},\\\frac12+\frac1{2\cdot1^2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Alternative way to calculate the sequence Given $a>0$ that satisfies $4a^2+\sqrt 2 a-\sqrt 2=0$.
Calculate $S=\frac{a+1}{\sqrt{a^4+a+1}-a^2}$.
Attempt:
There is only one number $a>0$ that satisfies $4a^2+\sqrt 2 a-\sqrt 2=0$, that is
$a=\frac{-\sqrt{2}+\sqrt{\Delta }}{2\times 4}=\frac{-\sqrt{2}+\sqrt{2+16\sqrt{2}}}{8}$... | Hint: Use
$$4a^2+\sqrt 2 a-\sqrt 2=0 \implies a^2=\dfrac{\sqrt{2}}{4}-\dfrac{\sqrt{2}}{4}a$$
and
$$a^4 = \dfrac{2}{16}-2\dfrac{\sqrt{2}}{4}\dfrac{\sqrt{2}}{4}a+\dfrac{2}{16}a^2$$
$$= 1/8-a+1/8\left[\dfrac{\sqrt{2}}{4}-\dfrac{\sqrt{2}}{4}a\right].$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2669096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Evaluate $ \int_{0}^{\pi/2} \frac{\sin(nx)}{\sin(x)}\,dx $ For every $odd$ $n \geq 1$ the answer should be $\pi/2$
For every $even$ $n \geq 2$ the possible answers are :
$A )$ $ 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots + (-1)^{n/2+1}\frac{1}{n-1} $
$B )$ $ 3(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots + (-1)^{n/... | For $n$ even, say $2m$, we can write
$$\begin{align}
\int_0^{\pi/2}\frac{\sin(2mx)}{\sin(x)}\,dx&=\text{Im}\int_0^{\pi/2}\frac{e^{i2mx}}{\sin(x)}\,dx\\\\
&=2\text{Re}\int_0^{\pi/2}\frac{e^{i2mx}}{e^{ix}-e^{-ix}}\,dx\tag1
\end{align}$$
Then, letting $z=e^{ix}$ and using long division reveals that
$$\begin{align}
\text{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Odds of winning a superlottery The problem:
There's a lottery (or a "superlottery", I'm not really sure how that's different). In order to play, I select 8 numbers from the first 90 positive integers (so, 1-90 inclusive). Also, a computer selects 12 numbers from the first 90 positive integers. If all of my 8 numbers ar... | Your solution is correct and your friend made a common mistake.
Suppose the winning numbers are $\{1,2,3,4,5,6,7,8,9,10,11,12\}$.
The first number selected must be in this set with probability $\frac{12}{90}$.
The second number selected must be one of the remaining numbers with probability $\frac{11}{89}$.
and so fort... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2676897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solutions to the equation $x^4+3y^4=z^2$ It was proven that the equation $x^4+2y^4=z^2$ has no non-trivial solutions in integers. What about the equation
$x^4+3y^4=z^2$? It has a solution $x=1,y=1, z=1$. Can we find all solutions?
| Yes, the equation $x^4+3y^4 = z^2$ is birationally equivalent to an elliptic curve, hence we can find all its solutions.
However, if an easy proof there are infinitely many integer solutions with $\gcd(x,y)=1$ will suffice, then given an initial solution to,
$$x^4+dy^4=z^2$$
then, in general, further ones can be genera... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2680710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Area of largest inscribed rectangle in an ellipse. Can I square the area before taking the derivative? So say I have an ellipse defined like this:
$$\frac{x^2}{9} + \frac{y^2}{4} = 1$$
I have to find the largest possible area of an inscribed rectangle.
So the area ($A$) of a rectangle is $2x2y=4xy$. Also we can redefin... | You are over-complicating it. Affine maps preserve the ratios of areas and, in a circle, the inscribed squares are pretty obviously the largest inscribed quadrilaterals. Their area is $\frac{2}{\pi}$ times the area of the circle. By applying $\varphi:(x,y)\mapsto(ax,by)$ the ellipse with equation $\frac{x^2}{a^2}+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2684375",
"timestamp": "2023-03-29T00:00:00",
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Finding equation and centre of circle through 3 points using matrices Based on answer given here: Get the equation of a circle when given 3 points. We can find equation of circle through points $(1,1), (2,4), (5,3)$ by taking:
$\left|\begin{array}{cccc}
x^2+y^2&x&y&1\\
1^2+1^2&1&1&1\\
2^2+4^2&2&4&1\\... | This comes from the general equation of a circle
$$x^2+y^2+Ax+By+E=0$$ which gives since $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ are points in the circle
$$x^2+y^2+Ax+By+E=0\\x_1^2+y_1^2+Ax_1+By_1+E=0\\x_2^2+y_2^2+Ax_2+By_2+E=0\\x_3^2+y_3^2+Ax_3+By_3+E=0$$ This is equivalent to
$$\left (\begin{matrix}x^2+y^2&x&y&1\\x_1^2+y_1^... | {
"language": "en",
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Solving the congruence $9x \equiv 3 \pmod{47}$ For this question $9x \equiv 3 \pmod{47}$.
I used euler algorithm and found that the inverse is $21$ as $21b-4a=1$
when $a=47$ and $b=9$
I subbed back into the given equation:
\begin{align*}
(9)(21) & \equiv 3 \pmod{47}\\
189 & \equiv 3 \pmod{47}\\
63 & \equiv 1 \pmod{47}
... | you can write $$x\equiv \frac{3}{9}=\frac{1}{3}\equiv \frac{48}{3}\equiv 16\mod 47$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to deal with $(-1)^{k-1}$ It's a problem on mathematical induction.
$$1^2-2^2+3^2-.....+(-1)^{n-1}n^2=(-1)^{n-1}\frac{n.(n+1)}{2}$$
I have proved it for values of $n=1,2$.
Now I assume for $n=k$
$$P(k):1^2-2^2+3^2-.....+(-1)^{k-1}k^2=(-1)^{k-1}\frac{k.(k+1)}{2}$$.
$$P(k+1):1^2-2^2+3^2-.....+(-1)^{k-1}k^... | You forgot a minus sign.
$\displaystyle P(k+1):1^2-2^2+3^2-.....+(-1)^{k-1}k^2+(-1)^k(k+1)^2\\=(-1)^{k-1}\dfrac{k.(k+1)}{2}+(-1)^k(k+1)^2\\=\dfrac{(k+1)}{2} [(-1)^{k-1}.k +2(-1)^k (k+1)]\\=\dfrac{(k+1)}{2} [-(-1)^{k}.k +2(-1)^k (k+1)]\\=(-1)^k\dfrac{(k+1)}{2}[-k+2k+2]\\=(-1)^k\dfrac{(k+1)}{2}[k+2]\\=(-1)^k\dfrac{(k+1)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2688026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Number of divisors of the number $2079000$ which are even and divisible by $15$
Find the number of divisors of $2079000$ which are even and divisible by $15$?
My Attempt: Since they are divisible by $15$ and are even, $2$ and $5$ have to included from the numbers prime factors.
$2079000 = 2^3 \cdot 3^3 \cdot 5^3 \cdo... | Using your method, which probably isn't the best:
$2079000 =2^3*3^3*5^3*7*11$ so all factors are of form $2^a3^b*5^c*7^d*11^e$ where $a,b,c$ are between $0$ and $3$ and $d,e$ are between $0$ and $1$ exclusively.
So there $4*4*4*2*2$ factors.
But to be even the $a$ must be at least $1$ and divisible by $15$ they must b... | {
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Fraction and simplification
solve: $\frac{1}{x(x-1)} + \frac{1}{x} = \frac{1}{x-1}$
What are the possible answers ?
(A) -1 (B) Infinitely Many Solutions (C) No solution (D) 0
The answer from where i've referred this is (B), but when i simplify it I get (D)
My solution:
$$\frac{1}{x(x-1)} + \frac{1}{x} = \frac{1... | You have got an identity, L.H.S. = R.H.S. which will hold for all values of X in domain of equation.
This implies Infinite solution as domain of the equation is infinite.
In this case $X ={0,1}$ are not in domain of equation.
| {
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} |
general equation to find cubic polynomial from two minimums? I tried researching and found that I can use a system of linear equations and solve by an inverse matrix to find the cubic equation given 4 points which satisfy the function f(x) of the general form $f(x)=ax^3+bx^2+cx+d$
I can also find a cubic of the form $a... | First find a quadratic that has zeros at the $x$-values of your points. Then integrate. Then adjust the constant to get the correct $y$-values. Using your example, we need a quadratic that passes through the points $(-1,0)$ and $(2,0)$.
So it must have the form $g(x+1)(x-2) = gx^2-gx -2g$.
Integrate this to get $f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2694934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find all positives integers $n$ such that $n^3+1$ is a perfect square A solution as follows:
$n^3+1=x^2$
$n^3=x^2-1$
$n^3=(x-1)(x+1)$
$x-1=(x+1)^2~~or~~x+1=(x-1)^2$
$x^2+x+2=0~~or~~x^2-3x=0$
$x(x-3)=0$
$x=0~~or~~x=3~~\Longrightarrow~~n=2$
Does it cover all possible solutions? How to prove that 2 is the only which solve... | Hint: use that $n^3+1=(n+1)(n^2-n+1)$ so $n+1=n^2-n+1$.
As @Arthur pointed out, it's possible that both factors are not equal. Let's say $n+1=km^2$ and $n^2-n+1=k$. Then we have $n=km^2-1$, $n^2-n+1=k^2m^4-2km^2+1-km^2+1+1=k^2m^4-3km^2+3=k$ or $km^2(km^2-3)=k-3$. The last equation does not have a solution for positive ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2698849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Evaluating $\sum\limits_{n=1}^\infty \frac{(-1)^{n+1}}{n}$ I have this series:
$$\sum_{n=1}^\infty s_n=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{(-1)^{n+1}}{n}+...$$
The sum of the first N terms is:
$$S_N=\sum_{n=1}^N s_n=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{(-1)^{n+1}}{n}$$
It converges for Leibniz's ... | HINT:
Note that
$$S_{2N}=\sum_{n=1}^{2N}\frac{(-1)^{n+1}}{n}=\sum_{n=1}^N \left(\frac{1}{2n-1}-\frac{1}{2n}\right)$$
and
$$S_{2N+1}=\sum_{n=1}^{2N+1}\frac{(-1)^{n+1}}{n}=\sum_{n=1}^N \left(\frac{1}{2n-1}-\frac{1}{2n}\right)+\frac1{2N+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2699080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $23x \equiv 1 \mod 120$ using Euclidean Algorithm I want to do inverse modulo to solve the equation
$$23x \equiv 1 \mod 120$$
And to do that I used the extended Euclidean Algorithm.
$$120=23\times5+5$$
$$23=5 \times4+3$$
$$5=3\times1+2$$
$$3=2\times1+1$$
Which makes $1 \equiv3-2\times1=3-2$. I then substituted t... | We wish to solve the congruence $23x \equiv 1 \pmod{120}$.
You correctly obtained
\begin{align*}
120 & = 5 \cdot 23 + 5\\
23 & = 4 \cdot 5 + 3\\
5 & = 1 \cdot 3 + 2\\
3 & = 1 \cdot 2 + 1\\
2 & = 2 \cdot 1
\end{align*}
Now, if we work backwards, we can obtain $1$ as a linear combination of $23$ and $120$.
\begin{align*... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2703270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove $4^n-1$ is divisible by $3$, for all $n\in\Bbb N$? Prove $4^n-1$ is divisible by $3$, for all $n\in\Bbb N$?
I started by assuming there exists some $k\in\Bbb N$ s.t. $4^n-1=3k\iff \dfrac{4^n}3-\dfrac 13=k$, so for $k$ to be a natural number, $4^n\equiv 1\mod 3$ must be true, but this tells us no new information, ... | The simples way is using modular arithmetic, under which
$$4^n-1\equiv 1^n-1\equiv 1-1\equiv 0\mod 3$$
Another way is to write
$$4^n = (3+1)^n$$
and use the binomial formula to get
$$(3+1)^n = 3^n +{n\choose 1} 3^{n-1} + \cdots + {n\choose n-1} 3^1 + 1\\=3\left(3^{n-1} +{n\choose 1} 3^{n-2} + \cdots + {n\choose n-1} 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2704691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
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Expressing $\tan 20°$ in terms of $\tan 35°$ If $\tan 35^\circ = a$, we are required to express $\left(\frac{\tan 145^\circ - \tan 125^\circ}{1 + \tan 145^\circ\tan 125^\circ}\right)$ in terms of $a$.
Here's one way to solve this: $$\frac{\tan 145^\circ - \tan 125^\circ}{1 + \tan 145^\circ\tan 125^\circ} = \tan (145^\... | If $\dfrac{1-a^2}{2a}=\dfrac{a-b}{1+ab}$
$$\iff\dfrac{a^3-3a}{1-3a^2}=\dfrac1b$$
If $a=\tan x,b=2-\sqrt3$
$$-\tan3x=\tan(-3x)=2+\sqrt3=\csc30^\circ+\cot30^\circ=\cdots=\cot15^\circ=\tan75^\circ$$
$\implies-3x=180^\circ n+75^\circ$ where $n$ is any integer
$\implies-x\equiv25^\circ,(60+25)^\circ,(120+25)^\circ\pmod{180^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2705899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Spivak's Calculus: Chapter 1, Problem 18b (Quadratic determinant less than zero) The problem in question is as follows:
18b Suppose that $b^2 -4c \lt 0$. Show that there are no numbers $x$ that satisfy $x^2 + bx + c = 0$; in fact, $x^2 + bx + c \gt 0$ for all $x$. Hint: complete the square.
Trying to apply the hint, I ... | A completely informal approach:
consider $x^2+bx+c$, now differentiate with respect to $x$ twice. This gives a positive value,which indicates that the parabola is upturned. Now consider the fact that $b^2<4c$, clearly gives imaginary solutions. Now if the parabola doesn't touch the $X$ axis and is upturned it must lie ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2706487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 5
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Compute the intersection of perpendicular lines Lines Ax + By = C and Bx – Ay = 0 are perpendicular.
I am trying to understand how their intersection is computed below:
$$ x = \frac{AC}{A^2 + B^2}$$
$$ y= \frac{BC}{A^2 + B^2}$$
I tried solving for y first but I am not sure how to continue:
$$ Ax+By-C= Bx-Ay $$
$$By+Ay... | Assuming that $(A,B)\neq(0,0)$:\begin{align}\left\{\begin{array}{l}Ax+By=C\\Bx-Ay=0\end{array}\right.&\implies A(Ax+BY)+B(Bx-Ay)=AC\\&\iff(A^2+B^2)x=AC\\&\iff x=\frac{AC}{A^2+B^2}.\end{align}From this and from $Bx=Ay$, you get that$$y=\frac{BC}{A^2+B^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2707315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What's the algebraic trick to evaluate $\lim_{x\rightarrow \infty} \frac{x \sqrt{x}+\sqrt[3]{x+1}}{\sqrt{x^{3}-1}+x}$? $$\lim_{x \rightarrow \infty} \frac{x \sqrt{x}+\sqrt[3]{x+1}}{\sqrt{x^{3}-1}+x}$$
I got the first half:
$$\frac{x\sqrt{x}}{\sqrt{x^{3}-1}+x}=\frac{x\sqrt{x}}{\sqrt{x^{3}(1-\frac{1}{x^3})}+x}=\frac{1}{\... | In the numerator, the terms are of order $x^{3/2}$ and $x^{1/3}$, so that the first dominates (the terms are added, there is no cancellation). In the denominator, $x^{3/2}+x^1$.
So the expression is virtually $$\frac{x^{3/2}}{x^{3/2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2709342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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$x_{0}= \cos \frac{2\pi }{21}+ \cos \frac{8\pi }{21}+ \cos\frac{10\pi }{21}$ Prove that $x_{0}= \cos \frac{2\pi }{21}+ \cos \frac{8\pi }{21}+ \cos\frac{10\pi }{21}$ is a solution of the equation
$$4x^{3}+ 2x^{2}- 7x- 5= 0$$
My try:
If $x_{0}$, $x_{1}$, $x_{2}$ be the solutions of the equation then
$$\left\{\begin{matri... | As above $ 4x^{3}+ 2x^{2}- 7x- 5$ has a rational root of -1 so the problem reduces after polynomial division to proving $\cos(\frac{2 \pi}{21}) + \cos(\frac{8 \pi}{21}) + \cos(\frac{10 \pi}{21})$ is a root of $4x^{2} - 2x - 5$
First some observations since these are all 21st roots of unity
Let $\omega = \frac{2 \pi}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2711150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Proving $\frac{a}{a+ b}+ \frac{b}{b+ c}+ \frac{c}{2c+ 5a}\geq \frac{38}{39}$ $$a, b, c \in \left [ 1, 4 \right ] \text{ }
\frac{a}{a+ b}+ \frac{b}{b+ c}+ \frac{c}{2c+ 5a}\geq \frac{38}{39}$$
This is an [old problem of mine in AoPS]
(https://artofproblemsolving.com/community/u372289h1606772p10020940).
First solution
$$a... | We can use also the following way.
Full expanding gives:
$$(5a+41c)b^2+(200a^2-71ac-37c^2)b+5a^2c+41ac^2\geq0,$$ which is obvious for
$200a^2-71ac-37c^2\geq0.$
But for $200a^2-71ac-37c^2<0$ we obtain $a<c$ and it's enough to prove that
$$(200a^2-71ac-37c^2)^2-4(5a+41c)(5a^2c+41ac^2)\leq0$$ or
$$(c-a)(4a-c)(10000a^2+537... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2712732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Evaluating $\lim_{x\to \infty } (x +\sqrt[3]{1-{x^3}} ) $ $$\lim_{x\to \infty } (x +\sqrt[3]{1-{x^3}} ) $$
What method should I use to evaluate it. I can't use the ${a^3}$-${b^3}$ formula because it is positive. I also tried to separate limits and tried multiplying with $\frac {\sqrt[3]{(1-x^3)^2}}{\sqrt[3]{(1-x^3... | Use the identity
$$
a^3+b^3=(a+b)(a^2-ab+b^2)
$$
with $a=x$ and $b=\sqrt[3]{1-x^3}$, to get that
$$
1=(x+\sqrt[3]{1-x^3})(x^2-x\sqrt[3]{1-x^3}+(1-x^3)^{2/3}).
$$
Thus
$$
\lim_{x\to\infty} (x+\sqrt[3]{1-x^3})=\lim_{x\to\infty}\frac{1}{(x^2-x\sqrt[3]{1-x^3}+(1-x^3)^{2/3})}
=\lim_{x\to\infty}\frac{1}{x^2(1-(x^{-3}-1)^{1/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2716247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 3
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Explain how to solve this trigonometric limit without L'Hôpital's rule? In my previous class our professor let us the following limit:
$$
\lim_{x\to0} \frac{\tan(x)-\sin(x)}{x-\sin(x)}
$$
He solved it by applying L'Hôpital's rule as follow:
$
\lim_{x\to0} \frac{\sec^2(x) - \cos(x)}{1-\cos(x)} = \lim_{x\to0} \frac{\cos^... | HINT
By Taylor's expansion
$$ \frac{\tan(x)-\sin(x)}{x-\sin(x)}= \frac{x+\frac{x^3}3-x+\frac{x^3}6+o(x^3)}{x-x+\frac{x^3}6+o(x^3)}=\frac{\frac{x^3}2+o(x^3)}{\frac{x^3}6+o(x^3)}=\frac{\frac{1}2+o(1)}{\frac{1}6+o(1)}\to 3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2717908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $\left ( a+ b \right )^{2}\left ( b+ c \right )^{2}\left ( c+ a \right )^{2}\geq 64abc$ Give $a, b, c$ be positive real numbers such that $a+ b+ c= 3$. Prove that:
$$\left ( a+ b \right )^{2}\left ( b+ c \right )^{2}\left ( c+ a \right )^{2}\geq 64abc$$
My try
We have:
$$\left ( a+ b \right )^{2}\geq 4ab$$
$$\l... | Consider the polynomial
$$p(x) = (x-a)(x-b)(x-c) = x^3 - Ax^2 + Bx - C
\quad\text{ where }\quad
\begin{cases}
A &= a + b + c\\
B &= ab + bc + ca\\
C &= abc
\end{cases}
$$
We are given $A = 3$. By AM $\ge$ GM, we have
$$1 \ge C \implies \sqrt{C} \ge C$$
Since all the roots of $p(x)$ is real. By Newton's inequalities, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2721251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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$\int_{0}^{1} \sqrt[]{1+t^4}dt$ I tried by using $t^2$ = $tan\theta$ and then by inserting $t^4$ by $tan^2\theta$ in $\int_{0}^{1} \sqrt[]{1+t^4}dt$, I get $dt$ = $\frac{sec^2\theta\times d\theta}{2\times \sqrt[]{tan\theta}}$ & $\sqrt[]{1+t^4}$ = $sec\theta$. Thus the integration becomes $\int_{0}^{\frac{\pi}{4}} \fra... | \begin{align}
\int_{0}^{1}\sqrt{1+t^4}\,\mathrm{d}t
&=\int_{0}^{1}\frac{1+t^4}{\sqrt{1+t^4}}\,\mathrm{d}t\\
&=\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{1+t^4}}+\int_{0}^{1}\frac{t^4}{\sqrt{1+t^4}}\,\mathrm{d}t\\
&=\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{1+t^4}}+\int_{0}^{1}t\cdot\frac{t^3}{\sqrt{1+t^4}}\,\mathrm{d}t\\
&=\int_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2721939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Eigenvalues of matrix with ijth entry as $i+j$
Let $A$ be a matrix with entries as $(a_{ij})=i+j$. I would like to find out eigenvalues of $A$.
I noticed that after reducing the matrix to Echelon form only two non zero rows will be left and thus rank(A)=2, that means $n-2$ eigenvalues are zero. Let $\lambda_1, \lambd... | The conjecture of Carl Schildkraut concercing $\lambda_1 \lambda_2$ is correct up to a minus sign. I assume $1$-indexed rows and columns, however. Note that $\lambda_1^2$ and $\lambda_2^2$ are the two non-zero eigenvalues of $A^2$, and since $A$ is symmetric, $AA = A^T A$. Using the formula $\mathrm{Tr}(X^TY) = \sum_{i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2723456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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I need the steps to solving this limit without using l´Hopital rule I've tried many ways of solving this limit without using l'Hopital and I just can't figure it out. I know the answer is $3/2 \sin (2a).$
$$\lim_{x\,\to\,0} \frac{\sin(a+x)\sin(a+2x) - \sin^2(a)} x$$
Thank you!
| You could also have used Talor series around $x=0$ and even got more than the limit itself
$$\sin(a+k x)=\sin (a)+k \cos (a)\,x-\frac{1}{2} k^2 \sin (a)\,x^2+O\left(x^3\right)$$ making
$$\sin(a+ x)\sin(a+2 x)=\sin ^2(a)+3 \sin (a) \cos (a)\,x+ \left(2 \cos ^2(a)-\frac{5 }{2}\sin
^2(a)\right)\,x^2+O\left(x^3\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2724226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Finding eigenvalues and eigenvectors and then determining their geometric and algebraic multiplcities I have the following matrix:
$A = \begin{bmatrix}
1 && 7 && -2 \\
0 && 3 && -1 \\
0 && 0 && 2 \end{bmatrix}$
and I am trying to find the eigenvalues and eigenvectors followed by their respective geometric... | $$\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}\iff \begin{bmatrix}y\\z\\0\end{bmatrix} =\begin{bmatrix}0\\0\\0\end{bmatrix}$$
meaning that $y=z=0$, while leaving $x$ to be whatever you want, results in a kernel vector.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2724790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Solve pair of equations : $ x(y-\sin{y}) = -\pi$ and $ x(1-\cos{y}) = 2$ I was trying to solve the following pair of equations :
$$ x(y-\sin{y}) = -\pi$$
$$ x(1-\cos{y}) = 2$$
My attempt :
We get, $x \sin{y} = xy + \pi$ and $x \cos y = x - 2$ , then by squaring and eliminating $x^2$ from both sides we get, $x^2 y^2 + 2... | I find it easier to use $\sin(x)$ instead of $\sin(y)$ so lets change $y\leftrightarrow x$ to get the system:
$$y=\frac{-\pi}{x-\sin(x)} \tag{1}$$
$$y = \frac{2}{1-\cos(x)} \tag{2}$$
cross multiply noting that $x\neq 2n\pi$, let
$$f(x)=2(x-\sin x)+\pi(1-\cos x ) $$
Where we need to solve $f(x) = 0$. We know for $x>0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2726078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Calculating $X^2$ of a Maxwell-distributed RV I'm trying to calculate $X^2$, where $X$ is a Maxwell-distributed RV; which means it has the density function $\phi$ with
$$\phi(x) = \frac{2}{d^3 \sqrt{2\pi}} x^2 \exp\left(- \frac{x^2}{2d^2}\right) \text{ for } x \ge 0.$$
In the lecture we proved that the density functio... | When the distribution of $X$ has support $X \ge 0$, the transformation $Y = X^2$ is already one-to-one, thus we can simply write $$f_Y(y) = \frac{f_X(\sqrt{y})}{2\sqrt{y}}, \quad y > 0.$$ That is why you have an extra factor of $2$: you have presumed that $X$ has negative support when it does not.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2726734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to find the value of $x$ that satisfies $3x=4$ in $\mathbb Z/5\mathbb Z$? Let $\mathbb Z_5 = \mathbb Z/5\mathbb Z$.
The value of $x$ which satisfies the equation $3x = 4\bmod 5$ is...?
The answer is $3$. I understand why the answer is $3$, but not how it was derived. Is there an equation or process I can use that w... | You need to find the inverse of $3$ modulo $5$. By Bézout’s identity, there are integers $x$ and $y$ such that $3x+5y=1$, which you can find with the (reverse) Euclidean algorithm:
\begin{align}
\color{red}{5}&=\color{red}{3}\cdot1+\color{red}{2}\\
\color{red}{3}&=\color{red}{2}\cdot1+\color{red}{1}\\
\color{red}{2}&=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2727545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Compute the surface area of the unit sphere $x^2+y^2+z^2=1$ Compute the surface area of the unit sphere $x^2+y^2+z^2=1$
The following is a solution the book suggests:
The upper hemisphere is the graph of the function $\varphi(x,y)=\sqrt{1-x^2-y^2}.$
A little calculation yields $$\sqrt{1+(\partial_x\varphi)^2+(\par... | Note that
$$(\partial_x\varphi)^2=\frac{x^2}{\varphi^2}$$
$$(\partial_y\varphi)^2=\frac{y^2}{\varphi^2}$$
and then
$$\sqrt{1+(\partial_x\varphi)^2+(\partial_y\varphi)^2}=\frac1{\sqrt{1-x^2-y^2}}$$
and thus
$$\int_0^1\int_0^{2\pi}\frac {r}{\sqrt{1-r^2}}d\theta dr=2\pi $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2727897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Value of $k$ in ratio of two definite integration
If $\displaystyle I_{1}=\int^{1}_{0}\frac{x^{\frac{5}{2}}(1-x)^{\frac{7}{2}}}{12}dx$ and $\displaystyle I_{2}=\int^{1}_{0}\frac{x^{\frac{5}{2}}(1-x)^{\frac{7}{2}}}{(3+x)^8}dx$ and $I_{1}=k(108\sqrt{3})I_{2}$. Then $k$ is
Try: put $x=\sin^2\theta$. Then $dx=2\sin\thet... | Just try to convert $I_1$ into $I_2$ form
Substituting $x = \cfrac{4y}{3+y}$
$$I_1 =\int_0^1 \cfrac{4^{\cfrac{5}2}3^{\cfrac72} x^{\cfrac52}(1-x)^{\cfrac72}}{(3+x)^6} \cfrac{4(3+x)-4x}{(3+x)^2}dx\\ \hspace{-1cm}= 32*27*\sqrt 3 *12*I_1 \\ \hspace{-1cm} =10368\sqrt 3*I_1 \\ \hspace{0cm}= 96*108*\sqrt 3 I_1$$
$\impl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2729376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Using the $5$-th order Maclaurin polynomial of $f(x) = e^x$ to approximate $f(-1)$
Find the 5th-order Maclaurin polynomial $P_5(x)$ for $f(x) = e^x$.
I got
$$P_5(x) = 1 + x +\frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \frac{x^5}{120} + O(x^6) $$
From this answer, I'm supposed to approximate $f(-1)$, correct to ... | Yes, it looks like your method for evaluating $f(-1)$ there is correct. I'd avoid the notation $O((-1)^6)$, and instead just write $$f(-1)\approx1 + (-1) + {(-1)^2\over2} + {(-1)^3\over6} + {(-1)^4\over24} + {(-1)^5\over120}$$
and so on.
However it's worth pointing out that this isn't an approximation within 5 decimals... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2729617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Showing the sequence converges $a_{1}=\frac{1}{2}$, $a_{n+1}=\frac{1}{2+a_{n}}$ Showing the sequence converges $a_{1}=\frac{1}{2}$, $a_{n+1}=\frac{1}{2+a_{n}}$.
I already know that if $(a_{n})$ converges then it does to $\sqrt{2}-1$.But i dont't know how to prove that this sequence cenverges.
EDIT
I think that the su... | Let $b_n:=a_{2n-1}$ and $c_n:=a_{2n}$ for $n\in\mathbb{N}$. Since $a_n$ is bounded because $a_n\leqslant 1/2$ and $a_n>0$ for all $n$, so are the sequences $b_n$ and $c_n$. Now we show that $b_n$ is monotonic decreasing while $c_n$ is monotonic increasing.
First note that $b_n>\sqrt{2}-1$ for all $n$. For $n=1$ it is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2729836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 4
} |
Finding probability -picking at least one red, one blue and one green ball from an urn when six balls are selected Six balls are to be randomly chosen from an urn containing $8$ red, $10$ green, and 12 blue balls.
What is the probability at least one red ball, one blue and one green ball is chosen?
Sample space = $\bin... | Your approach is not correct.
In 'Only red and green" you don't exclude situations "Only red" and "only green".
Also each of the situations "Only x" is counted twice - in situation "Only x and y" and "only x and z"
Thus the number of situations where there is a color missing should be:
$$\left(\binom{8}{6} + \binom{10}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2730747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Stuck on a recursively defined converging sequence problem. I am given a simple quadratic equation
$$x^2-x-c=0, x>0, c>0$$
and then we define a sequence $\{x_n\}$ with $x_1>0$ fixed and then, if $n$ is an index for which $x_n$ has been defined, we define
$$x_{n+1}=\sqrt{c+x_n}$$.
With that I am asked to prove that $\{x... | $\begin{array}\\
x_{n+1}-x_n
&=\sqrt{x_n+c}-x_n\\
&=(\sqrt{x_n+c}-x_n)\dfrac{\sqrt{x_n+c}+x_n}{\sqrt{x_n+c}+x_n}\\
&=\dfrac{x_n+c-x_n^2}{\sqrt{x_n+c}+x_n}\\
\end{array}
$
If
$f(x) = x^2-x-c$,
$f'(x) = 2x-1$,
so $f'(x) > 0$
for $x > \frac12$.
The roots of $f(x)$
are
$x
=\dfrac{1\pm\sqrt{1+4c}}{2}
$,
so the positive roo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2733016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Using complex exponential to show the indefinite integration of sin(x)sinh(x) dx Use the complex exponential to evaluate the indefinite integral of
$\sin x \sinh x$.
Express your answer in terms of trigonometric and/or hyperbolic functions
The attached photo is what I have tried so far
| Method I
Using integration by parts twice,
\begin{align}
I &=\int \sin ax \sinh bx \, dx \\
&= \frac{\sin ax \cosh bx}{b}-\int \frac{\cosh bx}{b} d(\sin ax) \\
&= \frac{\sin ax \cosh bx}{b}-\frac{a}{b} \int \cos ax \cosh bx \, dx \\
&= \frac{\sin ax \cosh bx}{b}-\frac{a}{b}
\left[
\frac{\cos ax \sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2733525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Computing Minima for degree 8 polynomial Is there any better method than derivative or trial and error to calculate minima of $$f(x)=x^8-8x^6+19x^4-12x^3+14x^2-8x+9$$
The minima occurs at $x=2$ and $f(2)=1$
We were given option too $-1,9,6,1$ if that helps
| By splitting conveniently the terms (by completing squares from left to right) we obtain
$$\begin{align}f(x)&=x^8-8x^6+(16+3)x^4-12x^3+(12+2)x^2-8x+(8+1)
\\&=(x^8-8x^6+16x^4)+(3x^4-12x^3+12x^2)+(2x^2-8x+8)+1\\
&=x^4(x^2-4)^2+3x^2(x-2)^2+2(x-2)^2+1\geq 1=f(2).\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2733785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.