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Calculate a limit of exponential function Calculate this limit:
$$
\lim_{x \to \infty } = \left(\frac{1}{5} + \frac{1}{5x}\right)^{\frac{x}{5}}
$$
I did this:
$$
\left(\frac{1}{5}\right)^{\frac{x}{5}}\left[\left(1+\frac{1}{x}\right)^{x}\right]^\frac{1}{5}
$$
$$
\left(\frac{1}{5}\right)^{\frac{x}{5}}\left(\frac{5}{5}\right)^\frac{1}{5}
$$
$$
\left(\frac{1}{5}\right)^{\frac{x}{5}}\left(\frac{1}{5}\right)^\frac{5}{5}
$$
$$
\lim_{x \to \infty } = \left(\frac{1}{5}\right)^\frac{x+5}{5}
$$
$$
\lim_{x \to \infty } = \left(\frac{1}{5}\right)^\infty = 0
$$
Now I checked on Wolfram Alpha and the limit is $1$
What did I do wrong? is this the right approach? is there an easier way?:)
Edit:
Can someone please show me the correct way for solving this? thanks.
Thanks
| Write
$$
\left(\frac{1+x}{5x}\right)^{x/5} = e^{\frac{x}{5}\ln\left(\frac{1+x}{5x}\right)} = e^{-\frac{x}{5}\ln\left(\frac{5x}{1+x}\right)} = e^{-\frac{x}{5}\ln\left(5-\frac{5}{1+x}\right)}
$$
Observing that $\ln\left(5-\frac{5}{1+x}\right)\xrightarrow[x\to\infty]{}\ln 5$, can you conclude by composing limits?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1075652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
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 $$\displaystyle \int\frac{\left(1+x^4\right)}{\left(1-x^4\right)\sqrt{1+x^4}}dx = \frac{1}{2}\int\left\{\frac{1}{1-x^2}+\frac{1}{1+x^2}\right\}\cdot \frac{(1+x^4)}{\sqrt{1+x^4}}dx$$
So Integral is $$\displaystyle = \frac{1}{2}\int\frac{1+x^4}{(1-x^2)\sqrt{1+x^4}}dx+\frac{1}{2}\int\frac{1+x^4}{(1+x^2)\sqrt{1+x^4}}dx$$
Now Let $$\displaystyle I = \int\frac{1+x^4}{(1-x^2)\sqrt{1+x^4}}dx$$ and $$\displaystyle J = \int\frac{1+x^4}{(1+x^2)\sqrt{1+x^4}}dx$$
Now How can I evaluate $I$ and $J\;,$ plz help me
Thanks
| We can write the integral as
$$
I=\int\frac{\sqrt{1+x^4}}{1-x^4}\textrm{d}x=\int\frac{\left(1+x^4\right)}{\left(1-x^4\right)\sqrt{1+x^4}}\textrm{d}x.
$$
Let $t=\frac{1+x^4}{1-x^4}$ so that $x^4=\frac{t+1}{t-1}$ and $\textrm{d}x=\pm\left(\frac{t+1}{t-1}\right)^{\frac{1}{4}}\cdot\frac{-1}{2(t-1)(t+1)}\textrm{d}t$.
The integral above becomes
$$
\pm \int t\left(\frac{t-1}{2t}\right)^{\frac{1}{2}}\left(\frac{t+1}{t-1}\right)^{\frac{1}{4}}\frac{-1}{2(t-1)(t+1)}\textrm{d}t=\mp\frac{1}{2\sqrt 2}\int t (t^2-1)^{-3/4}\textrm{d}t
$$
where the last integral is
$$
\int t (t^2-1)^{-3/4}\textrm{d}t
=2\sqrt[4]{t^2-1}+\text{constant}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1076348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
Transforming linear combination of the cosine and sine function In the proof of Transforming $a\cos\left(\, x\,\right)+b\sin\left(\, x\right)$ to $r\cos\left(\,\phi - x\,\right)$
\begin{align}
a\cos\left(\, x\,\right) + b\sin\left(\, x\,\right)
&=\,\sqrt{\,a^{2} + b^{2}\,}\,
\left[\,\frac{a}{\,\sqrt{\, a^{2} + b^{2}\,}\,}\,\cos\left(\, x\,\right)
+\frac{b}{\,\sqrt{\, a^{2} + b^{2}\,}\,}\,\sin\left(\, x\,\right)\,\right]\,
\\[2mm]&=\,\sqrt{\,a^{2} + b^{2}\,}\,\left[\,
\cos\left(\,\phi\,\right)\cos\left(\, x\,\right) + \sin\left(\,\phi\,\right)\sin\left(\,x\,\right)
\,\right]
\\[2mm]&=\,\sqrt{\, a^{2} + b^{2}\,}\,\cos\left(\,\phi - x\,\right)
\end{align}
why did we factor out $\,\sqrt{\, a^{2} + b^{2}\,}\,$ where did this idea came from ?. It seems out of the blue.
| The idea behind the proof comes from noting that
$$\left(\dfrac{a}{\sqrt{a^2+b^2}}\right)^2+\left(\dfrac{b}{\sqrt{a^2+b^2}}\right)^2=1.\tag1$$
This is analogous to the Pythagorean identity
$$\cos^2\phi+\sin^2\phi=1.\tag2$$
So for all $a$ and $b$ we can get an expression that satisfies $(2)$ by just factoring out $\sqrt{a^2+b^2}$, but in what way will this be useful? From $(1)$ and $(2)$ we can say that there exists an angle $\phi$ such that
$$\cos\phi=\dfrac{a}{\sqrt{a^2+b^2}}\qquad\sin\phi=\dfrac{b}{\sqrt{a^2+b^2}},$$ or $$\cos\phi=\dfrac{b}{\sqrt{a^2+b^2}}\qquad\sin\phi=\dfrac{a}{\sqrt{a^2+b^2}}.$$
All of this will permit us to transform $a\cos x+b\sin x$ into an expression of the form
$$\text{constant}\cdot(\sin\phi\cos x\pm\cos\phi\sin x),$$
or $$\text{constant}\cdot(\cos\phi\cos x\pm\sin\phi\sin x).$$
Those two are the result of the angle-addition formulas, which will enable us to write it in the form $r\cos(\phi\pm x)$ or $r\sin(\phi'\pm x)$.
In conclusion, the proof boils down to determining how to transform the coefficients of $\cos$ and $\sin$ such that they are exactly equal to the $\cos$ and $\sin$ of an angle. The rest follows from the angle-addition formulas.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1076500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Sum of roots: Vieta's Formula The roots of the equation $x^4-5x^2+2x-1=0$ are $\alpha, \beta, \gamma, \delta$. Let $S_n=\alpha^n +\beta^n+\gamma^n+\delta^n$
Show that $S_{n+4}-5S_{n+2}+2S_{n+1}-S_{n}=0$
I have no idea how to approach this. Could someone point me in the right direction?
| Hint. $$ \begin {align*} S_{n+4} - 5S_{n+2} + 2S_{n+1} - S_n &= \alpha^{n+4} + \beta^{n+4} + \gamma^{n+4} + \delta^{n+4} - 5 \cdot \left( \alpha^{n+2} + \beta^{n+2} + \gamma^{n+2} + \delta^{n+2} \right) + 2 \cdot \left( \alpha^{n+1} + \beta^{n+1} + \gamma^{n+1} + \delta^{n+1} \right) - \left( \alpha^n + \beta^n + \gamma^n + \delta^n \right). \end {align*} $$Now, see what you can factor out. The coefficients should have given you this suspicion.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1077648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Power series for the rational function $(1+x)^3/(1-x)^3$
Show that $$\dfrac{(1+x)^3}{(1-x)^3} =1 + \displaystyle\sum_{n=1}^{\infty} (4n^2+2)x^n$$
I tried with the partial frationaising the expression that gives me
$\dfrac{-6}{(x-1)} - \dfrac{12}{(x-1)^2} - \dfrac{8}{(x-1)^3} -1$
how to proceed further on this having doubt with square and third power term in denominator.
| As shown in this answer, $\binom{-n\vphantom{1}}{k}=(-1)^k\binom{n+k-1}{k}$, therefore,
$$
\begin{align}
\frac{(1+x)^3}{(1-x)^3}
&=(1+x)^3\sum_{k=0}^\infty\binom{-3}{k}(-x)^k\\
&=(1+x)^3\sum_{k=0}^\infty\binom{k+2}{k}x^k\\
&=\sum_{k=0}^\infty\sum_{j=0}^3\binom{3}{j}\binom{k-j+2}{k-j}x^k\\
&=1+\sum_{k=1}^\infty(4k^2+2)x^k
\end{align}
$$
since, for $k\ge1$,
$$
\begin{align}
&\overbrace{\binom{k+2}{k}}^{j=0}+\overbrace{3\binom{k+1}{k-1}}^{j=1}+\overbrace{3\binom{k}{k-2}}^{j=2}+\overbrace{\color{#C00000}{\binom{k-1}{k-3}}}^{j=3}\\[3pt]
&=\binom{k+2}{2}+3\binom{k+1}{2}+3\binom{k}{2}+\color{#C00000}{\binom{k-1}{2}}\\[3pt]
&=\frac{k^2+3k+2}2+\frac{3k^2+3k}2+\frac{3k^2-3k}2+\frac{k^2-3k+2}2\\[9pt]
&=4k^2+2
\end{align}
$$
Note that the red coefficients do not match for $k=0$; $\binom{-1}{-3}=0$ while $\binom{-1}{2}=1$. The rest match for $k=0$ and all match for $k\ge1$ because $\binom{n\vphantom{1}}{k}=\binom{n\vphantom{1}}{n-k}$ for $n\ge0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1078092",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
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 - \frac{1}{16})\\
= \ln(\frac{3}{4}) + \ln (\frac{8}{9}) + \ln ( \frac{15}{16})\\
= -.287 + -.117 + -.064 = -.50
$
it converges to $-\frac{1}{2}$.
| Hint
$$\ln\left(1-\frac1{n^2}\right)=\ln\left(\frac{(n-1)(n+1)}{n^2}\right)=\ln\left(\frac{n-1}{n}\right)-\ln\left(\frac{n}{n+1}\right)=u_n-u_{n+1}$$
and then telescope.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1079174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
prove that $a^2 b^2 (a^2 + b^2 - 2) \ge (a + b)(ab - 1)$ Good morning
help me to show the following inequality
for all $a$, $b$ two positive real numbers
$$a^2 b^2 (a^2 + b^2 - 2) \ge (a + b)(ab - 1)$$
thanks you
| Let $a=\frac{x^2}{z},\, b=\frac{y^2}{z}$ then $x,y \geqq 0$ and $z>0.$ Then we have$:$
$$z^6 \cdot (\text{LHS}-\text{RHS})={x}^{3}{y}^{3} \left( x-y \right) ^{2} \left( {x}^{3}y+{x}^{2}{y}^{2}+ x{y}^{3}+{z}^{2} \right) + \left( xy+z \right) \left( {x}^{2}{y}^{2}+ xyz+{z}^{2} \right) \left( xy-z \right) ^{2} \left( {x}^{2}+{y}^{2} \right) \geqq 0$$
Another proof$,$ you can see here.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
$b_{n}$ is increasing I think there is misunderstanding in my last post because its contain three questions so i will post question by question step by step
An inequality for the product $\prod_{k=2}^{n}\cos\frac{\pi }{2^{k}}$
Let $n\geq 2\quad a_{n}=\prod\limits_{k=2}^{n}\cos\left(\dfrac{\pi }{2^{k}}\right)$ and $b_{n}=a_{n}\cos\left(\dfrac{\pi }{2^{n}}\right)$
Show that $b_{n}$ is increasing
my attempts :
*
*$b_{n+1}-b_{n}=a_{n+1}\cos\left(\dfrac{\pi }{2^{n+1}}\right)-a_{n}\cos\left(\dfrac{\pi }{2^{n}}\right)=a_{n}\cos\left(\dfrac{\pi }{2^{n}}\right)\cos\left(\dfrac{\pi }{2^{n+1}}\right)-a_{n}\cos\left(\dfrac{\pi }{2^{n}}\right)$
*$\dfrac{b_{n+1}}{b_n}=\cos(\pi/2^{n+1})$
im stuck here in these two ways
any help would be appreciated
| Notice that $\dfrac{b_{n+1}}{b_n}=\dfrac{\cos^2\left(\dfrac{\pi}{2^{n+1}}\right)}{\cos\left(\dfrac{\pi}{2^n}\right)}=\dfrac{\dfrac{1+\cos\left(\dfrac{\pi}{2^n}\right)}{2}}{\cos\left(\frac{\pi}{2^n}\right)}=\dfrac{1+\cos\left(\dfrac{\pi}{2^n}\right)}{2\cos\left(\dfrac{\pi}{2^n}\right)}>\dfrac{2\cos\left(\dfrac{\pi}{2^n}\right)}{2\cos\left(\dfrac{\pi}{2^n}\right)}=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to solve IVP using Laplace transform(of matrix)? $$x' = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & -2 \\ 3&2 & 1\end{bmatrix} x, ~~ x{(0)} = \begin{bmatrix} 2 \\ -1 \\ 1\end{bmatrix}$$
I am having very hard time solving this question using Laplace transform. I don't know how to do it for $3\times 3$ matrix. Would anyone be able to show me step by step how to do it for one of the cases? I have no clue how to do the inverse.
| We are given:
$$X'(t) = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & -2\\ 3 & 2 & 1\end{bmatrix} \begin{bmatrix} x(t) \\ y(t)\\ z(t)\end{bmatrix}, ~~ X(0) = \begin{bmatrix} 2 \\ -1\\ 1\end{bmatrix}$$
We can write this as:
$$\tag 1 \begin{align} x' &= x \\ y' &= 2x+y-2z \\ z' &= 3x +2y + z \end{align}$$
Taking the Laplace transform of $(1)$ yields:
$$\begin{align} s x(s) - x(0) &= x(s) \\ s y(s) - y(0) &= 2x(s) + y(s) - 2z(s) \\ s z(s) - z(0) &= 3 x(s) + 2y(s) + z(s)\end{align}$$
This reduces to the system:
$$\begin{bmatrix} s-1 & 0 & 0 \\ -2 & s-1 & 2\\ -3 & -2 & s-1\end{bmatrix} \begin{bmatrix} x(s) \\ y(s)\\ z(s)\end{bmatrix} = \begin{bmatrix} 2 \\ -1\\ 1 \end{bmatrix}$$
All that is needed is to solve for $x(s),y(s),z(s)$ and then find the inverse Laplace Transform.
You could have also used many methods to solve this system, including eigenvalues/eigenvectors, matrix exponential, etc.
Update
*
*The inverse of the matrix is:
$$\begin{bmatrix} \frac{1}{s-1} & 0 & 0 \\
\frac{2 (s-4)}{s^3-3 s^2+7 s-5} & \frac{s-1}{s^2-2 s+5} & -\frac{2}{s^2-2 s+5} \\
\frac{3 s+1}{s^3-3 s^2+7 s-5} & \frac{2}{s^2-2 s+5} & \frac{s-1}{s^2-2 s+5} \end{bmatrix}$$
The inverse matrix times the column vector yields:
$$\begin{bmatrix} x(s) \\y(s)\\z(s) \end{bmatrix} =
\begin{bmatrix} \frac{2}{s-1} \\ \frac{2 s}{s^2-2s+5}-\frac{3}{s-1} \\
\frac{s (s+2)+5}{(s-1) (s^2-2s+5)}\end{bmatrix}$$
Next, just find the inverse Laplace Transform of $x(s), y(s), z(s)$ and you are done.
I will assume you can take it from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Proving that a number is an integer. Prove that the following number is an integer:
$$\left( \dfrac{76}{\dfrac{1}{\sqrt[\large{3}]{77}-\sqrt[\large{3}]{75}}-\sqrt[\large{3}]{5775}}+\dfrac{1}{\dfrac{76}{\sqrt[\large{3}]{77}+\sqrt[\large{3}]{75}}+\sqrt[\large{3}]{5775}}\right)^{\large{3}}$$
How can I prove it?
| we have $$ \left( 76\, \left( \left( \sqrt [3]{77}-\sqrt [3]{75} \right) ^{-1}-
\sqrt [3]{5775} \right) ^{-1}+ \left( 76\, \left( \sqrt [3]{77}+\sqrt
[3]{75} \right) ^{-1}+\sqrt [3]{5775} \right) ^{-1} \right) ^{3}
$$
after expanding we obtain
$$438976\, \left( \left( \sqrt [3]{77}-\sqrt [3]{75} \right) ^{-1}-
\sqrt [3]{5775} \right) ^{-3}+17328\,{\frac {1}{ \left( \left( \sqrt
[3]{77}-\sqrt [3]{75} \right) ^{-1}-\sqrt [3]{5775} \right) ^{2}
\left( 76\, \left( \sqrt [3]{77}+\sqrt [3]{75} \right) ^{-1}+\sqrt [3
]{5775} \right) }}+228\,{\frac {1}{ \left( \left( \sqrt [3]{77}-
\sqrt [3]{75} \right) ^{-1}-\sqrt [3]{5775} \right) \left( 76\,
\left( \sqrt [3]{77}+\sqrt [3]{75} \right) ^{-1}+\sqrt [3]{5775}
\right) ^{2}}}+ \left( 76\, \left( \sqrt [3]{77}+\sqrt [3]{75}
\right) ^{-1}+\sqrt [3]{5775} \right) ^{-3}
$$
computing this we obtain
$$616$$
wow
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1085491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 1
} |
Finding the area of a triangle, given the distance between center of incircle and circumscribed circle Consider the following depiction:
$ABC$ is an isosceles triangle ($AB=AC$), where the two angles opposite the equal sides are equal $\beta$ ($\beta>60$), and $AD$ perpendicular to $BC$.
$O$ is the center of the circumscribed circle of $ABC$, and $M$ is the center of the incircle of $ABC$.
I want to express the area of the triangle $BOM$ and the radius of the incircle of $ABC$, in terms of $k$ and $\beta$.
This is what I tried: I denoted the radius of the circumscribed circle as $R$, and the radius of the incircle as $r$.
The area of $BOM$ can be obtained by: $\frac{k\cdot BD}{2}$, and $BD=\frac{BC}{2}$, I get that the area is actually: $\frac{k\cdot BC}{4}$.
Now from Law of sines I get $BC=\sin(180-2\beta)\cdot 2 R$.
So this reduces to problem to expressing $R$ (which is $OA$) in terms of $\beta$ and $k$. This is where I'm stuck. I couldn't find a way to do that.
I know that $OA=OB=R$, and that $AOB$ is also isosceles triangle, and I have a feeling it is important, but I can't figure out what am I missing...
Hints and guidelines would be very appreciated.
| Assuming $BD=1$, we have $AB=AC=\frac{1}{\cos\beta}$, $[ABC]=\tan\beta$,
$$ R = \frac{AB}{2\sin\beta} = \frac{1}{\sin(2\beta)}$$
and:
$$ r = \frac{2[ABC]}{AB+AC+BC} = \frac{\tan\beta}{1+\frac{1}{\cos\beta}}=\frac{\sin\beta}{1+\cos\beta}=\tan\frac{\beta}{2}. $$
Moreover, by Euler's theorem we have:
$$ k^2 = OM^2 = R^2-2Rr = \frac{1}{4\sin^2\beta\cos^2\beta}-\frac{1}{\cos\beta(1+\cos\beta)} $$
so:
$$ k^2 = \frac{(1-2\cos\beta)^2}{4\cos^2\beta(1-\cos^2\beta)} $$
and:
$$ k = \frac{1-2\cos\beta}{\sin(2\beta)}. $$
This gives, in general,
$$ BD = \frac{\sin(2\beta)}{1-2\cos\beta} k, $$
so:
$$ [BOM]=\frac{k}{2}BD = \frac{\sin\beta\cos\beta}{1-2\cos\beta} k^2,\tag{1} $$
and:
$$ r = \frac{\sin\beta}{1+\cos\beta} BD = \frac{2(1-\cos\beta)\cos\beta}{1-2\cos\beta} k.\tag{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1086757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How do I integrate: $\int\sqrt{\frac{x-3}{2-x}} dx$? I need to solve:
$$\int\sqrt{\frac{x-3}{2-x}}~{\rm d}x$$
What I did is:
Substitute: $x=2\cos^2 \theta + 3\sin^2 \theta$. Now:
$$\begin{align}
x &= 2 - 2\sin^2 \theta + 3 \sin^2 \theta \\
x &= 2+ \sin^2 \theta \\
\sin \theta &= \sqrt{x-2} \\
\theta &=\sin^{-1}\sqrt{x-2}
\end{align}$$
and, $ cos \theta = \sqrt{(3-x)} $
$ \theta=\cos^{-1}\sqrt{(3-x)}$
The integral becomes:
$$\begin{align}
&= \int{\sqrt[]{\frac{2\cos^2 \theta + 3\sin^2 \theta-3}{2-2\cos^2 \theta - 3\sin^2 \theta}} ~~(2 \cos \theta\sin\theta)}~{\rm d}{\theta}\\
%
&= \int{\sqrt[]{\frac{2\cos^2 \theta + 3(\sin^2 \theta-1)}{2(1-\cos^2 \theta) - 3\sin^2 \theta}}~~(2 \cos \theta\sin\theta)}~{\rm d}{\theta} \\
%
&= \int\sqrt[]{\frac{2\cos^2 \theta - 3\cos^2 \theta}{2\sin^2 \theta - 3\sin^2 \theta}}~~(2 \cos \theta\sin\theta) ~{\rm d}\theta \\
%
&= \int\sqrt[]{\frac{-\cos^2 \theta }{- \sin^2 \theta}}~~(2 \cos \theta\sin\theta) ~{\rm d}\theta \\
%
&= \int \frac{\cos \theta}{\sin\theta}~~(2 \cos \theta\sin\theta)~{\rm d}\theta \\
%
&= \int 2\cos^2 \theta~{\rm d}\theta \\
%
&= \int (1- \sin 2\theta)~{\rm d}\theta \\
%
&= \theta - \frac {\cos 2\theta}{2} + c \\
%
&= \sin^{-1}\sqrt{x-2} - \frac {\cos 2(\sin^{-1}\sqrt{x-2})}{2} + c
\end{align}$$
But, The right answer is :
$$\sqrt{\frac{3-x}{x-2}} - \sin^{-1}\sqrt{3-x} + c $$
Where am I doing it wrong?
How do I get it to the correct answer??
UPDATE:
I am so sorry I wrote:
= $\int 2\cos^2 \theta .d\theta$
= $\int (1- \sin 2\theta) .d\theta$
It should be:
= $\int 2\cos^2 \theta .d\theta$
= $\int (1+ \cos2\theta) .d\theta$
= $ \theta + \frac{\sin 2\theta}{2} +c$
What do I do next??
UPDATE 2:
= $ \theta + \sin \theta \cos\theta +c$
= $ \theta + \sin \sin^{-1}\sqrt{(x-2)}. \cos\cos^{-1}\sqrt{(3-x)}+c$
= $ \sin^{-1}\sqrt{(x-2)}+ \sqrt{(x-2)}.\sqrt{(3-x)}+c$
Is this the right answer or I have done something wrong?
| to shorten the working, since the function is only well-defined for $x \in (2,3)$ we may substitute
$$
3-x \to s
$$
giving
$$
I=\int\sqrt{\frac{x-3}{2-x}} .dx = - \int\sqrt{\frac{s}{1-s}} .ds
$$
now substitute
$$
s \to \sin^2 \theta \\
ds \to 2\sin\theta \cos\theta\cdot d\theta
$$
so
$$
I = -\int2\sin^2 \theta \cdot d\theta = \int (\cos 2\theta-1)\cdot d\theta=\frac12\sin2\theta-\theta+c \\
= \sqrt{s(1-s)} - \sin^{-1}s \\
=\sqrt{(3-x)(x-2)} - \sin^{-1} \sqrt{3-x} + c
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1089410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
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
$$\lim_{x\to\infty}\left( \sqrt{a_1x^2+b_1x+c_1}\pm\sqrt{a_2x^2+b_2x+c_2}\right)$$
(provided it exists) by using the expression's conjugate, but this problem has me stumped.
I've considered using the conjugate
$$\sqrt{x^2+5x-2}-\sqrt{4x^2-3x+7}-\sqrt{x^2+7x+5}$$
and a term like this one,
$$\sqrt{x^2+5x-2}+\sqrt{4x^2-3x+7}-\sqrt{x^2+7x+5}$$
but that didn't seem to help simplify anything.
Edit: I stumbled across something at the last second that lets me use the conjugate approach. The expression can be rewritten as
$$\sqrt{x^2+5x-2}-\sqrt{4x^2-3x+7}+\sqrt{x^2+7x+5}\\
\sqrt{x^2+5x-2}-2\sqrt{x^2-\frac{3}{4}x+\frac{7}{4}}+\sqrt{x^2+7x+5}\\
\left(\sqrt{x^2+5x-2}-\sqrt{x^2-\frac{3}{4}x+\frac{7}{4}}\right)+\left(\sqrt{x^2+7x+5}-\sqrt{x^2-\frac{3}{4}x+\frac{7}{4}}\right)$$
which approaches
$$\frac{23}{8}+\frac{31}{8}=\frac{27}{4}$$
| You can also use Taylor series. Start writing $$A=\sqrt{x^2+5x-2}-\sqrt{4x^2-3x+7}+\sqrt{x^2+7x+5}$$ $$\frac Ax=\sqrt{1+\frac{5x-2}{x^2}}-2\sqrt{1-\frac{3x-7}{4x^2}}+\sqrt{1+\frac{7x+5}{x^2}}$$ Now, remember that, when $y$ is small compared to $1$, $\sqrt{1+y}=1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)$.
So replace $y$ by $\frac{5x-2}{x^2}$ for the first radical, by $-\frac{3x-7}{4x^2}$ for the second radical and by $\frac{7x+5}{x^2}$ for the third radical. So, in the order, each radical write $$\sqrt{1+\frac{5x-2}{x^2}}=1+\frac{5}{2 x}-\frac{33}{8 x^2}+O\left(\left(\frac{1}{x}\right)^3\right)$$ $$\sqrt{1-\frac{3x-7}{4x^2}}=1-\frac{3}{8 x}+\frac{103}{128 x^2}+O\left(\left(\frac{1}{x}\right)^3\right)$$ $$\sqrt{1+\frac{7x+5}{x^2}}=1+\frac{7}{2 x}-\frac{29}{8 x^2}+O\left(\left(\frac{1}{x}\right)^3\right)$$ All of the above makes $$\frac Ax=\frac{27}{4 x}-\frac{599}{64 x^2}+O\left(\left(\frac{1}{x}\right)^3\right)$$ that is to say $$A=\frac{27}{4}-\frac{599}{64 x}+O\left(\left(\frac{1}{x}\right)^2\right)$$ which gives the limit but also tells how it is approched.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1090307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Consecutive cubes equal to a square $\frac{1}{8}ab(a^2+b^2-1) = y^2$, and Pythagorean triples If we wish that the sum of $b$ consecutive cubes with initial cube $c_k=\tfrac{1}{2}(1+a-b)$ is equal to a square, then we have the rather simple equation,
$$F_k=\tfrac{1}{8}ab(a^2+b^2-1) = y^2$$
It seems only three integral families are known that solve it:
$$\begin{align}
F_1(m):\quad a,\,b &= m+1,\;\; m\\
F_2(n):\quad a,\,b &= 16n^3+24n^2+8n,\;\; 2n+1\\
F_3(n):\quad a,\,b &= 4n^4-1,\;\; 2n^2\\
\end{align}$$
Examples.
*
*$F_1(22)$ and $F_1(27)$
$$1^3+2^3+\dots+22^3 = 253^2$$
$$1^3+2^3+\dots+27^3 = 378^2$$
*$F_2(1)$
$$23^3+24^3+25^3 = 204^2$$
*$F_3(2)$:
$$28^3+29^3+\dots+35^3 = 504^2$$
This implies that $F_1(c_2-1)+F_2(n)$ or $F_1(c_3-1)+F_3(n)$ are in a Pythagorean triple,
$$F_1(22)+F_2(1) = 253^2+204^2 =325^2$$
$$F_1(27)+F_3(2) = 378^2+504^2 =630^2$$
Question: Can anyone give a fourth integral family $F_4$ where initial cube $c\neq0,1$?
| As pointed out by Barto, appropriate substitutions can transform the problem into solving a Pell-like, or even a Pell equation. For example, let,
$$a =2bn^2,\quad y = bnu/2$$
and $F_k=\frac{1}{8}ab(a^2+b^2-1) = y^2\,$ transforms into the Pell,
$$u^2-(4n^4+1)b^2 = -1\tag1$$
An initial solution is $u_0 = 2n^2,\; b_0 = 1$ from which we can get an infinite more. For example, the third generation is,
$$u = 2n^2(3+16n^4),\quad b = 1+16n^4$$
which implies the initial cube,
$$c = n^2(1-4n^2)^2$$
Example:
Let $n = 2$, so $b=257$, with initial cube $c=30^2=900$, and end cube $900+256=1156$,
$$900^3+901^2+902^3+\dots+1156^3 = 532504^2$$
And so on, for an infinite family of polynomials generated from $u_0,\,b_0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1091437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Trace of a matrix $A$ with $A^2=I$ Let $A$ be a complex-value square matrix with $A^2=I$ identity.
Then is the trace of $A$ a real value?
| $$A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\implies A^2=\begin{pmatrix}a^2+bc&b(a+d)\\c(a+d)&d^2+bc\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}\implies$$
$$\begin{align}a^2+bc=1=d^2+bc\implies a=\pm d\end{align}$$
But we also have that
$$b(a+d)=0=c(a+d)\implies \begin{cases}a=-d\\or\\b=c=0\end{cases}$$
In the first case
$$a=-d\implies \text{Tr.}\,A=0$$
In the second case:
$$b=c=0\implies a=\pm d=\pm 1\implies \;\text{Tr.}\,A=0\;\;or\;\;\text{Tr.}\,A=\pm2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1091929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
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?
| There are no other solutions besides $(2,4,8)$ and $(3,5,15)$, already
discovered in aRaRa's answer.
Let $f(x,y,z)=\frac{xyz-1}{(x-1)(y-1)(z-1)}$ for integers $z>y>x>1$. We want
to know when $k=f(x,y,z)$ is an integer. Note first that $k$ cannot equal $1$,
because this would imply $z=\frac{x+y-xy}{x+y-1}$ whence $xy\leq 1$ which is impossible.
Now
$$
f(x,y,y+1)-f(x,y,z)=\frac{(xy-1)(z-(y+1))}{y(x-1)(y-1)(z-1)}\geq 0 \tag{1}
$$
and
$$
f(x,x+1,x+2)-f(x,y,y+1)=\frac{t\bigg((2x^2+2x-1)t+(2x^3+4x^2-1)\bigg)}{(x-1)x(x+1)(y-1)y}\geq 0 \tag{2}
$$
where $t=y-(x+1)$, so that $f(x,y,z)\leq f(x,x+1,x+2)=\phi(x)=\frac{x^3 + 3*x^2 + 2*x - 1}{x^3 - x}$.
If $x\geq 4$, we have $f(x,x+1,x+2)=\frac{119}{60}-\frac{(x-4)(100+(x-1)(59x+15))}{60x(x^2-1)}$,
so $k\leq \frac{119}{60} <2$ and hence $k=1$ which is impossible as already shown above.
So $x$ can only be $2$ or $3$. As $\phi(2)=\frac{23}{6}<4$ and $\phi(3)=\frac{59}{24}<3$,
we have $k<4$ when $x=2$ and $k<3$ when $x=3$.
Note that $f(x,y,z)=k$ can be rewritten as
$$
z=\frac{k(x+y)-kxy-k+1}{k(x+y)-(k-1)xy-k} \tag{3}
$$
When $x=2$ and $k=2$, (3) yields $z=\frac{3}{2}-y$ which is impossible.
When $x=2$ and $k=3$, (3) yields $z=\frac{3y-4}{y-3}$ which is possible iff $y=4$.
When $x=3$ and $k=2$, (3) yields $z=\frac{4y-5}{y-4}$ which is possible iff $y=5$.
This concludes the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1092874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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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 get any more, and I can't use derivatives or multiple limits since we didn't study these yet.
Thanks
| Given $x,y\ge 1$.
$\therefore x<\sqrt{x^2+x}$ and $y<\sqrt{y^2+y}$ $\implies$ $x+y<\sqrt{x^2+x}+\sqrt{y^2+y}$
$$\frac{x+y+1}{x+y}>\frac{x+y+1}{\sqrt{x^2+x}+\sqrt{y^2+y}} $$ and
$\frac{x+y+1}{x+y}=1+\frac{1}{x+y}\le \frac{3}{2}$, because $x+y\ge2$
$$\therefore \frac{x+y+1}{\sqrt{x^2+x}+\sqrt{y^2+y}} <2 $$
| {
"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": 0
} |
a way to integrate: $\int (\sqrt{x} +3)/(2+ x^ \frac{1}{3}) dx$ Im looking for a way to integrate:
$$
\int \frac{ \sqrt{x} +3}{2+ x^ \frac{1}{3}} dx
$$
that would make it efficient and not too difficult...
Any suggestions?
| Let $x=u^6$, $dx=6u^5du$.
\begin{align*}
&\int\frac{\sqrt x+3}{2+\sqrt[3]x}\,dx\\
=&\int\frac{u^3+3}{u^2+2}6u^5\,du\\
=&6\int\frac{u^8+3u^5}{u^2+2}\,du\\
=&6\int\left(u^6-2u^4+3u^3+4u^2-6u-8+\frac{12u+16}{u^2+2}\right)\,du\\
=&6\int\left(u^6-2u^4+3u^3+4u^2-6u-8+6\frac{2u}{u^2+2}+8\frac{1}{\left(\frac{\sqrt2}{2}u\right)^2+1}\right)\,du\\
=&\frac67u^7-\frac{12}{5}u^5+\frac92u^4+8u^3-18u^2-48u+36\ln\left(u^2+2\right)+48\sqrt2\arctan\left(\frac{\sqrt2}{2}u\right)+C\\
=&\frac67x^{7/6}-\frac{12}{5}x^{5/6}+\frac92x^{2/3}+8x^{1/2}-18x^{1/3}-48x^{1/6}+\\
&36\ln\left(x^{1/3}+2\right)+48\sqrt2\arctan\left(\frac{\sqrt2}{2}x^{1/6}\right)+C
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1095751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
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:
$$x^n = u_n , y^n = v_n$$
$$\begin{cases} x^{n+1} = x^n + y^n \\ y^{n+1} = -x^n + 3y^n \end{cases}$$
$$\begin{cases} x = 1 + \frac{y^n}{x^n} \\ y = -\frac{x^n}{y^n} + 3 \end{cases}$$
And I don't know how to continue. And it is the basic problem.
The second problem:
(*)The question which occured while attemption solving it, and basically it is not connected with main problem.
We have $y = 4-x$ and now $f =\frac{y^n}{x^n}= (\frac{4-x}{x})^n $
Lets observe that $n \to \infty$ so the $\lim_{n \to \infty} f = 0$
So we can ignore it in our system of equation? It is my doubt. I'm nearly convinced it is not correct but I can't convince of myself why.
Help me, please!
| $$\begin{cases}
u_o = -1 \\
v_0 = 3 \\
u_{n+1} = u_n + v_n \\
v_{n+1} = -u_n + 3v_n
\end{cases}$$
So:
$$\begin{bmatrix}
u_{n+1} \\
v_{n+1} \\
\end{bmatrix} =
\begin{bmatrix}
1 & 1 \\
-1 & 3
\end{bmatrix}
\begin{bmatrix}
u_{n} \\
v_{n} \\
\end{bmatrix}$$
In order to make it nonrecursive, convert the sequence of matrix multiplications to a $\text{matrix}^\text{integer}$ exponent:
$$\begin{bmatrix} u_{n} \\ v_{n} \\ \end{bmatrix} = \begin{bmatrix} 1 & 1 \\
-1 & 3 \end{bmatrix} ^ n \begin{bmatrix} u_{0} \\ v_{0} \\ \end{bmatrix}$$ And if you want to write it without matrices, a Jordan decomposition works:
$$
\begin{align}\begin{bmatrix} u_{n} \\ v_{n} \\ \end{bmatrix} &= \left(\begin{bmatrix} -1 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} -1 & 1 \\ -1 & 0 \end{bmatrix}^{-1}\right)^n\begin{bmatrix} u_{0} \\ v_{0} \\ \end{bmatrix} \\
&= \begin{bmatrix} -1 & 1 \\ -1 & 0 \end{bmatrix}\begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix}^n\begin{bmatrix} 1 & 1 \\ -1 & 3 \end{bmatrix}\begin{bmatrix} u_{0} \\ v_{0} \\ \end{bmatrix} \\
&= \begin{bmatrix} -1 & 1 \\ -1 & 0 \end{bmatrix}\begin{bmatrix} 2^n & n2^{n-1} \\ 0 & 2^n \end{bmatrix}\begin{bmatrix} 1 & 1 \\ -1 & 3 \end{bmatrix}\begin{bmatrix} u_{0} \\ v_{0} \\ \end{bmatrix} \\
&= \begin{bmatrix} u_0\left(2^n + n2^{n-1}\right) - v_0 n2^{n-1} \\ v_0\left(2^n + n2^{n-1}\right) - u_0 n2^{n-1} \end{bmatrix}
\end{align}$$
| {
"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": 0
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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
| Since $A+B+C=\pi$, we can write
$$
\sin^2 A=\sin^2(\pi-(B+C))=\sin^2(B+C)
$$
Expand with the sum formula:
$$
\sin^2B\cos^2C+2\sin B\sin C\cos B\cos C+\cos^2B\sin^2C
$$
Transform $\sin^2$ into $\cos^2$:
$$
\cos^2C-\cos^2B\cos^2C+2\sin B\sin C\cos B\cos C+\cos^2B-\cos^2B\cos^2C
$$
Reduce and collect:
$$
\cos^2B+\cos^2C-2\cos B\cos C(\cos B\cos C-\sin B\sin C)
$$
Finish up.
| {
"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": 1
} |
calculate expected value of the product of two non independent random variables I've got two independent bernoulli distributed random variables $X$ and $Y$ with parameter $\frac{1}{2}$. Based on those I define two new random variables
$X' = X + Y , E(X') = 1$
$Y' = |X - Y|, E(Y') = \sum_{x=0}^1\sum_{y=0}^1|x-y|*P(X=x)*P(Y=y) = \frac{1}{2}$
How can I calculate E(X'Y')? As X' and Y' and not independent (e.g. it is impossible for Y' to assume 1 if X' is 0) I must not use the sum of all possible outcomes multiplied with their likelihood as I did for $E(Y')$ but I cannot find another formula to calculate the expected value.
| Someone else can answer more authoritatively for the general case, but for a small experiment such as this one can we build up all possible values of $X' \cdot Y'$ from the four possible outcomes of $(X,Y)$?
$$
\begin{array}{l|l|l|l|l}
(X,Y) & X' & Y' & X' \cdot Y' & P(\ \ ) \\
\hline
(0,0) & 0 & 0 & 0 & \frac{1}{4} \\
(0,1) & 1 & 1 & 1 & \frac{1}{4} \\
(1,0) & 1 & 1 & 1 & \frac{1}{4} \\
(1,1) & 2 & 0 & 0 & \frac{1}{4}
\end{array}
$$
So $P(X'Y'=0) = P(X'Y'=1) = \frac{1}{2}$ and $E(X'Y') = 0 \cdot \frac{1}{2} + 1 \cdot \frac{1}{2} = \frac{1}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1105971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How many ordered triples $(a, b, c)$ exist? How many ordered triples $(a, b, c)$ of positive integers exist with the property
that $abc = 500$?
Breaking it up, $500 = 2^2\cdot5^3$
$abc = 2^2 \cdot 5^3 = 2\cdot 2 \cdot 5 \cdot 5 \cdot 5$
But how can I use this?
| We solve a slightly more general problem, the number of triples $(a,b,c)$ of positive integers such that $abc=n=2^s5^t$. The solution generalizes nicely to $k$-tuples, and any positive $n$, given the prime power factorization of $n$.
The answer is the number of ways to distribute the $s$ $2$'s between $a$, $b$, and $c$, times the number of ways to distribute the $t$ $5$'s between $a$, $b$, and $c$.
By Stars and Bars (please see Wikipedia) there are $\binom{s+3-1}{3-1}$ ways to distribute the $2$'s, and $\binom{t+3-1}{3-1}$ ways to distribute the $5$'s.
In our particular case we get $\binom{4}{2}\binom{5}{2}$.
Remark: The solution is overkill. For $n=500$, we can count explicitly the ways to distribute the $2$'s, and the ways to distribute the $5$'s. Let's do the $5$'s. They all could go to one person ($3$ ways). Or else they could be distributed evenly ($1$ way). Or $2$ could go to one person, and $1$ to another ($(3)(2)$ ways). So there are $10$ ways to distribute the $5$'a. A similar but simpler analysis shows there are $6$ ways to distribute the $2$'s.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1115295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to go about solving this inequality question? $\cos(3x-\pi/3) \leq (1/2).$
Here is what I have done so far...
Let $3x-\pi/3 = X$. So I need to solve $\cos(X) \leq 1/2$. Which is all $X$ from $\pi/3$ to $5\pi/3$, so--
$\pi/3 \leq X \leq 5\pi/3 \quad\longrightarrow\quad \pi/3 \leq 3x-\pi/3 \leq 5\pi/3.$
So should I now solve for $x$? or...
| It's more correct to say that
$$ \frac{\pi}{3} + 2n\pi \le 3x - \frac{\pi}{3} \le \frac{5\pi}{3} + 2n\pi, \quad n \in \mathbb{Z} $$
Solving for $x$:
$$ \frac{2\pi}{3} + 2n\pi \le 3x \le 2\pi + 2n\pi $$
$$ \frac{2}{9} + \frac{2n\pi}{3} \le x \le \frac{2\pi}{3} + \frac{2n\pi}{3} $$
or
$$\frac{2 + 6n}{9}\,\pi \le x \le \frac{2 + 2n}{3}\,\pi$$
What this means is $x$ lies in not one interval but a family of intervals, equally spaced by $\frac{2\pi}{3}$. For example $\frac{2\pi}{9} \le x \le \frac{2\pi}{3}$ is a solution but $\frac{8\pi}{9} \le x \le \frac{4\pi}{3}$ and $\frac{14\pi}{9} \le x \le 2\pi$ are also solutions.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is this logically valid? $$1+\frac{1}{2}+\frac{1}{3}.....+\frac{1}{n-1} > ln(n)$$
and so, necessarily, $$1+\frac{1}{2}+\frac{1}{3}.....+\frac{1}{n-1}+\frac{1}{n} = 1+\frac{1}{2}+\frac{1}{3}.....+\frac{1}{n} > ln(n)$$
assuming that $n \in \mathbb {N}$.
| Imagine an inequality $$\frac{1}{n}>0.$$ We can add the same quantity on both sides $$\left(1+\frac{1}{2}+...+\frac{1}{n-1}\right)+\frac{1}{n}>0+\left(1+\frac{1}{2}+...+\frac{1}{n-1}\right)$$
Now we use your first inequality and the transitivity of $>$, which says that if $a>b$ and $b>c$, then $a>c$, where $a:=\left(1+\frac{1}{2}+...+\frac{1}{n-1}\right)+\frac{1}{n}$, $b:=\left(1+\frac{1}{2}+...+\frac{1}{n-1}\right)$ and $c:=\ln(n)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1116971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $2^{3^n} + 1$ is divisible by 9, for $n\ge1$
Prove that $2^{3^n} + 1$ can be divided by $9$ for $n\ge 1$.
Work of OP: The thing is I have no idea, everything I tried ended up on nothing.
Third party commentary: Standard ideas to attack such problems include induction and congruence arithmetic. (The answers will illustrate, among others, that in this case both approaches work nicely.)
| Now, since we are trying to prove something for all n, we should look immediately to proof by induction. For $n = 1$, the result is trivial as $2^{3^1}+1 = 9$. Next, we assume that $2^{3^{(n-1)}} + 1$ is divisible by $9$, say $2^{3^{(n-1)}} + 1=9\cdot k$. So, $2^{3^{(n-1)}}=9\cdot k -1$. Now, we examine $2^{3^n}+ 1$ which can be written as $2^{3^{n-1}\cdot3}+1$ which can be written as ${(2^{3^{n-1}}})^3+1$. Now, since $2^{3^{(n-1)}} = 9 \cdot k -1$ by our induction hypothesis, we have that ${(2^{3^{n-1}}})^3 + 1={(9 \cdot k - 1})^3 +1 = (729\cdot {k^3}- 243\cdot k^2 +27 \cdot k -1)+1 = 729\cdot {k^3}- 243\cdot k^2 +27 \cdot k = 9\cdot (81 k^3 -27 k^2 +3k)$ and so it is divisible by $9$.
| {
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"timestamp": "2023-03-29T00:00:00",
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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 help check my error?
| Following your way:
$$\begin{align}\int_0^1 \frac {x^3}{\sqrt {4+x^2}}\,dx &=8\int_0^{\arctan(1/2)}\tan^3\theta\sec\theta{\rm d}\theta\tag{$x=2\tan\theta$}\\&=8\int_1^{\sqrt{5}/2}(t^2-1){\rm d}t\tag{$t=\sec\theta$}\\&=8\left(\frac{t^3}3-t\right)\Bigg|_1^{\sqrt5/2}\\&=\frac83\left(\frac{5\sqrt5}8-1\right)-8\left(\frac{\sqrt5}2-1\right)\\&=\large \frac{16-7\sqrt5}3
\end{align}$$
| {
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How to use U substitution for the integral $\int\frac{8}{49+x^2}\,dx$? So, the following is my problem.
$$\int\frac{8}{49+x^2}\,dx$$
I understand this. I should first take out the constant which is 8 so it'll be
$$8\int\frac{1}{49+x^2}\,dx$$
Then I should factor out the 49 no? So it'll be
$$\frac{8}{49}\int\frac{1}{1+\frac{x^2}{49}}\,dx$$
Let $u=\dfrac{x}{7}$ and thus $du=\dfrac{1}{7}\,dx$. I'm stuck at this point.
| Here are the steps
$$\int\frac{8}{x^2+49}dx= 8\int\frac{1}{x^2+7^2}dx
$$
$$=\frac{8}{7^2} \int\frac{1}{\frac{x^2}{7^2}+1}dx =\frac{8}{7^2} \int\frac{1}{\left(\frac{x}{7}\right)^2+1}dx $$
Let $u=\frac{x}{7}$, then $du=\frac{1}{7}dx$. So now
$$ \frac{8}{7} \int\frac{1}{u^2+1}du= \frac{8}{7} \arctan u+C $$
$$ = \frac{8}{7} \arctan\left(\frac{x}{7}\right)+C $$
| {
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Find all integers $x$ such that $x^2+3x+24$ is a perfect square. Find all integers $x$ such that $x^2+3x+24$ is a perfect square.
My attempt:
$x^2+3x+24=k^2$
$3(x+8)=(k+x)(k-x)$
Now, do I find solution treating cases? But that doesn't seem very easy. Please help.
| Complete the square to get $(x+3/2)^2 + 87/4$. We want this to be a square itself, so
$$(x+3/2)^2 + 87/4 = k^2.$$
This is the same as
$$(2x+3)^2 + 87 = 4k^2.$$
Now consider the difference of squares $(2k - 2x - 3)(2k + 2x + 3) = 87 = 3\cdot 29$. Now there are only a few cases to check.
*
*$2k + 2x + 3 = 3$ and $2k - 2x - 3 = 29$ means $4k = 32$, $k = 8$.
*$2k + 2x + 3 = 87$ and $2k - 2x - 3 = 1$ means $4k = 88$, $k = 22$.
Do these give you any integer solutions for $x$?
| {
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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 variables.
How to approach such question?
| Just knowing that $a^2+b^2+c^2=1$ is not enough to determine the value of $ab+bc+ca$. For example, if $a=b=0$ and $c=1$, then $ab+bc+ca = 0$. On the other hand, if $a=b=c=\frac{1}{\sqrt3}$, then $ab+bc+ca = 1$. In fact, using $$(a+b+c)^2=a^2+b^2+c^2 + 2(ab+bc+ca),$$ you get that $$ab+bc+ca=\frac{(a+b+c)^2 - 1}{2}.$$
| {
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How to derive this infinite product formula? Show:
$$\prod_{n=0}^{\infty}\left(1 + x^{2^n}\right) = \frac{1}{1-x}$$
I tried numerous things, multiplying by $x$, dividing, but none of that worked. Also, I realized that:
$$\prod_{n=0}^{\infty} \left(1 + x^{2^n}\right) = \sum_{n=0}^{\infty} x^n$$
But I cannot prove the relation. I get:
$$(1 + x)(1+x^2)(1+x^4)(1+x^8)...$$
But for a general $n$ it is more difficult.
| The partial products are
$$1+x,$$
$$(1+x)+x^2(1+x)=1+x+x^2+x^3,$$
$$(1+x+x^2+x^3)+x^4(1+x+x^2+x^3)=1+x+x^2+x^3+x^4+x^4+x^5+x^6+x^7$$
$$\cdots$$
Every time the number of terms doubles.
Doesn't that ring a bell ?
| {
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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
$$x\log(x)=(\frac{x-1}{x})+\frac{3}{2!}(\frac{x-1}{x})^2+\frac{11}{3!}(\frac{x-1}{x})^3+...\frac{S_{n}}{n!}(\frac{x-1}{x})^n$$
Where $S_n$= absolute Striling Numbers of first kind (0,1,3,11,50,274....)
this series numerically checked without any problem, but I need the proving. Any help?
| We have
$$x\log(x) = x[-\log(1/x)] = x\left[-\log\left(1 - \frac{x-1}{x}\right)\right] = x\sum_{m = 0}^\infty \frac{1}{m+1}\left(\frac{x-1}{x}\right)^{m+1},$$
and $$x = \dfrac{1}{\frac{1}{x}} = \dfrac{1}{1 - \frac{x-1}{x}} = \sum_{m = 0}^\infty \left(\frac{x - 1}{x}\right)^m.$$ Thus
$$x\log(x) = \sum_{m = 0}^\infty \left(\frac{x-1}{x}\right)^m \sum_{m = 0}^\infty \frac{1}{m}\left(\frac{x-1}{x}\right)^m = \sum_{m = 0}^\infty \sum_{k = 0}^m \frac{1}{k+1}\left(\frac{x-1}{x}\right)^{m+1},$$
which is the same as $$\sum_{m = 0}^\infty \frac{S_m}{m!}\left(\frac{x-1}{x}\right)^{m+1}$$
| {
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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 $x^7(1+x^2+x^4)(1+x+x^3)(1+x^5+x^{10}+x^{15}+...)$
Now I don't know what to do; usually it ends up factoring to $(1+x+x^2+...)$, but that doesn't appear to be the case here.
| From the sum $\frac{1}{1-x^5}$ you only need to consider the first two terms - other will give you a higher power. $10 = x_1 +x_2+ x_3$ (with terms all three sums). Can you idenify $x_k$?
| {
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Matrix $\mathbf C$ such that $\mathbf A = \mathbf B \mathbf C=\mathbf C\mathbf B$ Let
$$\mathbf A =\begin{pmatrix}1&1&1\\1&2&2\\1&2&3\end{pmatrix} \text{and }\mathbf B=\begin{pmatrix}1&0&0\\1&1&0\\1&1&1\end{pmatrix}$$
Then which of following is true
*
*There exists a matrix $\mathbf C$ such that $\mathbf A=\mathbf B\mathbf C=\mathbf C\mathbf B$
*There exists no matrix $\mathbf C$ such that $\mathbf A=\mathbf B\mathbf C$
*There exists a matrix $\mathbf C$ such that $\mathbf A=\mathbf B\mathbf C$ but $\mathbf A \ne \mathbf C\mathbf B$
*There exists no matrix $\mathbf C$ such that $\mathbf A=\mathbf C\mathbf B$.
My attempt: since $\mathbf B$ is invertible I see option 1 to be correct. But it can't be that easy. Also, $\mathbf A$ is symmetric and $\mathbf B$ is triangular so think there some trick here. So I need suggestions. Thanks.
| I consider $$C=\left(\begin{array}{ccc}a&b&c\\d&e&f\\g&h&i \end{array}\right),$$ then
$$BC=\left(\begin{array}{ccc}a&b&c\\a+d&b+e&c+f\\a+d+g&b+e+h&c+f+i \end{array}\right),$$
and
$$CB=\left(\begin{array}{ccc}a+b+c&b+c&c\\d+e+f&e+f&f\\a+d+g&h+i&i \end{array}\right).$$
If, we want $A=BC$, we get
$$C=\left(\begin{array}{ccc}1&1&1\\0&1&1\\0&0&1 \end{array}\right),$$
in the case $A=CB$ we have
$$C=\left(\begin{array}{ccc}0&0&1\\-1&0&2\\-1&-1&3 \end{array}\right).$$
Therefore:
1) False
2) False
3) True
4) False
Regards!!
| {
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Let $x,y,z>0,xyz=1$. Prove that $\frac{x^3}{(1+y)(1+z)}+\frac{y^3}{(1+x)(1+z)}+\frac{z^3}{(1+x)(1+y)}\ge \frac34$ Let $x,y,z>0$ and $xyz=1$. Prove that $\dfrac{x^3}{(1+y)(1+z)}+\dfrac{y^3}{(1+x)(1+z)}+\dfrac{z^3}{(1+x)(1+y)}\ge \dfrac34$
My attempt:
Since it is given that $xyz=1$, I tried substituting $x=\dfrac{a}{b},y=\dfrac{b}{c},z=\dfrac{c}{a}$. But the expansion looked really ugly and I didn't think I could make out anything out of it.
So, I made another attempt, if each element was greater than $\dfrac{1}{4}$, we could have a solution, so, treating that way, I get $4x^3\ge 1+x+y+xy, 4y^3\ge 1+x+z+xz, 4z^3\ge 1+x+y+xy$. Using AM-GM I get an equality.
So, please help. Thank you.
| I think you can do this using Lagrange multipliers. Define the function:
$$f(x,y,z)= \frac{x^3}{(1+y)(1+z)} +\frac{y^3}{(1+x)(1+z)} + \frac{z^3}{(1+x)(1+y)} - \frac{3}{4}.$$
We want to show that $f(x,y,z)\geq 0$ for all $x,y,z\in \mathbb R^3$ such that $xyz=1$. Define the Lagrangian:
$$L(x,y,z,\lambda) = f(x,y,z)- \lambda g(x,y,z)$$
where $g(x,y,z)=0$ is the function defining the set where you look at: that is $xyz=1$ and hence $g(x,y,z)=xyz-1$. As a result,
$$L(x,y,z,\lambda) = f(x,y,z)- \lambda (xyz-1).$$
Differentiating $L$ w.r.t. all variables we get
\begin{align*}
\partial_x L(x,y,z,\lambda) =& \frac{3x^2}{(1+y)(1+z)} -\frac{y^3}{(1+x)^2(1+z)} - \frac{z^3}{(1+x)^2(1+y)} -\lambda y z = 0\\
\partial_y L(x,y,z,\lambda) =& -\frac{x^3}{(1+y)^2(1+z)} +\frac{3y^2}{(1+x)(1+z)} -\frac{z^3}{(1+x)(1+y)^2} - \lambda xz=0\\
\partial_z L(x,y,z,\lambda) =& -\frac{x^3}{(1+y)(1+z)^2} -\frac{y^3}{(1+x)(1+z)^2} + \frac{3z^2}{(1+x)(1+y)} - \lambda x y = 0\\
\partial_{\lambda} L(x,y,z,\lambda) =& 1-xyz=0
\end{align*}
Now try to exploit the symmetries or similarities in order to solve this system. The critical points can be maxima, minima or saddle points. If you happen to find only minima and such points satisfy $f(x,y,z)\geq 0$ then you will have proven the inequality. Since you are proving that the function $f$ has minima in the set $\{(x,y,z)\in \mathbb{R}^3 \, xyz=1\}$ and all of them are positive.
Be careful that the function is continuous and the set you are looking at is closed.
Another possibility is to subsitute the vaue for $z=1/(xy)$ (it is ok since $x,y,z\neq 0$) in the inequality and define a two-variable function $f(x,y)$ and do exactly the same without using Lagrange Method. Maybe that is easier.
| {
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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 then asks me to hence solve:
$$\frac{x-1}{x-3}-\frac{x-3}{x-5} = \frac{x-5}{x-7}−\frac{x-7}{x-9}$$
I have attempted to sub in part one into part two but my answers always turn out to be incorrect.
Any tips in the matter would be much appreciated.
I'd like to thank anyone who comments and answers in advance, your help is much appreciated.
| Start by realizing you can get a common denominator (and simplify once you have):
$$
-\frac{4}{(x-5)(x-3)}=-\frac{4}{(x-9)(x-7)}.
$$
Now take the reciprocal of both sides:
$$
-\frac{1}{4}(x-5)(x-3)=-\frac{1}{4}(x-9)(x-7).
$$
Now expand out the terms on the left- and right-hand sides:
$$
-\frac{x^2}{4}+2x-\frac{15}{4}=-\frac{x^2}{4}+4x-\frac{63}{4}.
$$
To make it easy to solve for $x$ (i.e., try to solve for when one of the sides equals $0$), subtract $-\frac{x^2}{4}+4x-\frac{63}{4}$ from both sides, obtaining
$$
12-2x=0 \Longleftrightarrow 2(6-x)=0 \Longleftrightarrow x=6.
$$
Thus, your equality is true when $x=6$.
| {
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How do I solve $\sum_{n=5}^\infty \frac{12}{16 \, n^{2} + 40 \, n + 21}$ How do I solve
$\displaystyle\sum_{n=5}^\infty \displaystyle\frac{12}{16 \, n^{2} + 40 \, n + 21}$ ?
A tip was to turn this into a telescoping series but I don't see how I could do this. Are there any guidelines in how I could transform this series into a telescoping series?
| $\begin{align} \displaystyle\frac{12}{16 \, n^{2} + 40 \, n + 21}&=\displaystyle\frac{12}{(4n+3)(4n+7)}\\&=3\frac{4}{(4n+3)(4n+7)}\\&=\displaystyle3{\huge[}\frac{1}{(4n+3)}-\frac{1}{(4n+7)}{\huge]}\end{align}$
$\begin{align}\therefore \displaystyle\sum_{n=5}^\infty \displaystyle\frac{12}{16 \, n^{2} + 40 \, n + 21}&=3\displaystyle\sum_{n=5}^\infty \displaystyle\frac{1}{(4n+3)}-\frac{1}{(4n+7)}\\&=3{\huge[}\frac{1}{23}-\frac{1}{27}+\frac{1}{27}-\frac{1}{31}+\cdots+\\&\frac{1}{(4n+3)}-\frac{1}{(4n+7)}+\frac{1}{(4n+7)}-\frac{1}{(4n+11)}+\cdots{\huge]}\\&=\frac{3}{23}\end{align}$
| {
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Find $ \sum_{n=2}^{+\infty}$ $\frac{1}{n^{3}-n}$ Find:
$$ \sum_{n=2}^{+\infty}\frac{1}{n^{3}-n}$$
I tried to resolve into partial fractions to see if there are some cancellations ,but that did not helped me .How do i do this ?Thanks
| We know that the series at hand is convergent absolutely, so we can safely play with the sum.
$$\sum_{i=2}^\infty\frac{1}{n^3 - n} = -\sum_{i=2}^\infty\frac{1}{n} + 0.5\sum_{i=2}^\infty\frac{1}{n+1} + 0.5\sum_{i=2}^\infty\frac{1}{n-1} = -\sum_{i=2}^\infty\frac{1}{n} + 0.5\sum_{i=3}^\infty\frac{1}{n} + 0.5\sum_{i=1}^\infty\frac{1}{n} = -\sum_{i=1}^\infty\frac{1}{n} + 1 + 0.5\sum_{i=1}^\infty\frac{1}{n} - 0.5 - 0.25 + 0.5\sum_{i=1}^\infty\frac{1}{n} = 1 - 0.75 = 0.25 = \frac{1}{4}$$.
As $n \rightarrow \infty$, the limit is $\frac14$.
| {
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Prove $n^2+4n+3$ is not prime for $n \in \mathbb{Z}^{+}$. I am trying to write a proof for this theorem:
For every positive integer $n$, $n^2+4n+3$ is not a prime.
Proof: Let $n \in \mathbb{Z}^{+}$. Note that $$n^2+4n+3=(n+1)(n+3)>1\text{,}$$
and $n+1 >1$ and $n+3 >1$.
Let $a = n+1$ and $b = n+3$. Then we have $$\dfrac{(n+1)(n+3)}{a}>\dfrac{1}{a}$$ and $$\dfrac{(n+1)(n+3)}{b}>\dfrac{1}{b}\text{.}$$
Therefore, $n^2+4n+3$ is not prime. $\square$
I don't think my proof is right and miss many things. Can anyone give me a hit or show me how to write a better proof for this question?
| One easy way of going about this is simply to consider the parity of $n$:
$n$ is even: We have that
$$
(2\ell)^2+4(2\ell)+3=4\ell^2+8\ell+3=\underbrace{(2\ell+1)(2\ell+3)}_{\text{composite}}.
$$
Thus, $n^2+4n+3$ is not a prime when $n$ is even.
$n$ is odd: We have that
$$
(2\ell+1)^2+4(2\ell+1)+3=\underbrace{2(2\ell^2+6\ell+4)}_{\text{composite}}.
$$
Thus, $n^2+4n+3$ is not a prime when $n$ is odd.
Consequently, when $n$ is even or odd, we have that $n^2+4n+3$ is not a prime. $\Box$
| {
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Simplify the sum Consider the sum of two polynomials
$$
\sum_{k=0}^{n-1} {{{n-1} \choose {k}}^2 z^{2k}}+\sum_{k=0}^{n-2} {{n-2} \choose {k}} {{n} \choose {k+1}} z^{2k+1}=\sum_{i=0}^{2n-1}a_i z^i.
$$
I want to find the exact expression for $a_i$. Of course I may divide it into even and odd parts but I hope there is a formula for $a_i$ .
|
A sum $A(z)=\sum_{k=0}^{2n}a_kz^k$ can be split into even and odd parts via
\begin{align*}
\frac{1}{2}\left(A(z)+A(-z)\right)
&=\frac{1}{2}\left(\sum_{k=0}^{2n}a_kz^k+\sum_{k=0}^{2n}a_k(-z)^k\right)\\
&=\sum_{k=0}^{2n}\frac{1+(-1)^k}{2}a_kz^k\\
&=\sum_{k=0}^na_{2k}z^{2k}
\end{align*}
and
\begin{align*}
\frac{1}{2}\left(A(z)-A(-z)\right)=\sum_{k=0}^{2n}\frac{1-(-1)^k}{2}a_kz^k
=\sum_{k=0}^{n-1}a_{2k+1}z^{2k+1}
\end{align*}
Therefore the LHS of your expression can be written as
\begin{align*}
\sum_{k=0}^{n-1}\binom{n-1}{k}^2&z^{2k}+\sum_{k=0}^{n-2}\binom{n-2}{k}\binom{n}{k+1}z^{2k+1}\\
&=\sum_{k=0}^{2n-2}\left(\binom{n-1}{k}^2\frac{1+(-1)^k}{2}
+\binom{n-2}{k}\binom{n}{k+1}\frac{1-(-1)^k}{2}\right)z^k
\end{align*}
and could now be transformed in order to find a convenient representation.
Note: On your RHS you could replace the upper bound of the sum with $2n-2$.
| {
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Calculating the limit: $\lim \limits_{x \to 0}$ $\frac{\ln(\frac{\sin x}{x})}{x^2}. $ How do I calculate $$\lim \limits_{x \to 0} \dfrac{\ln\left(\dfrac{\sin x}{x}\right)}{x^2}\text{?}$$
I thought about using L'Hôpital's rule, applying on "$\frac00$," but then I thought about $\frac{\sin x}{x}$ which is inside the $\ln$: it's not constant as $x$ goes to $0$, then I thought that maybe this what caused that my calculating of the limit wasn't true.
Can someone clarify how we calculate the limit here?
Note: I'd like to see a solution using L'Hôpital's rule.
| Using taylor series you have
$$\color{#05f}{\frac{\sin x}{x}} = \sum_{n=0}^{\infty}(-1)^{n} \frac{x^{2n}}{(2n+1)!} = 1 - \frac{x^2}{6} + \frac{x^4}{120} + O(x^6)$$
and $$\ln \Bigg(\color{#05f}{\frac{\sin x}{x}}\Bigg) = - \frac{x^2}{6} - O(x^4)$$
$$\lim _{x\to 0 } \frac{- \frac{x^2}{6} - O(x^4)}{x^2}= \lim_{x\to 0} -\frac{1}{6} - O(x^2) = \color{#f05}{-\frac{1}{6}}$$
Edit:
By L'hospital you have
$$\begin{align}\require{cancel}\frac{d}{dx} \ln \Bigg(\frac{\sin x}{x}\Bigg) &= \frac{\color{#f05}{\cancel x}}{\sin x}\frac{x\cos x - \sin x}{x^{\color{#f05}{\cancel{2}}}} = \cot x - \frac{1}{x} \\&\implies \frac{d^2}{dx^2} \ln \Bigg(\frac{\sin x}{x}\Bigg) = \frac{1}{x^2} - \csc^2 x \color{#085}{\to -\frac{1}{3}} \end{align}$$
as $x \to 0$.
Then
$$\lim _{x\to 0 } \frac{\frac{1}{x^2} - \csc^2 x}{2}= {-\frac{1}{2}\frac{1}{3}} = \color{#f05}{-\frac{1}{6}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1147074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 3
} |
Simplify a series of $\cos x$ or $\sin x$ Suppose we have this series
$$\sum\limits_{k=0}^n a^k \cos(kx) = 1 + a\cos(x) + a^2 \cos(2x) + a^3 \cos(3x) + \cdots + a^n\cos(nx)$$
What should we do to simplify this?
| let $a$ be a real number and $z = \cos t + i \sin t$.
this looks like the real part of $$1 + az + (az)^2 + \dots + (az)^n = \frac{1- (az)^{n+1}}{1-az} $$ thanking the real part, we get
$$1 +a\cos t + a^2 \cos 2t + \dots+a^n \cos nt = \text{real part} \frac{1- (az)^{n+1}}{1-az}$$
finding the real part of $$\frac{1- (az)^{n+1}}{1-az}=
\frac{1-a^{n+1}\cos (n+1)t - ia^{n+1}\sin(n+1)t }{1 - a\cos t -ia \sin t}
=\frac{[1-a^{n+1}\cos(n+1)t][1-a\cos t]-a^{n+2}\sin (n+1)t \sin t + i\dots}{(1-a\cos t)^2+a^2\sin^2t} = \frac{1-a^{n+1}\cos(n+1)t-a\cos t+a^{n+2}\cos(n+2)t + i\dots}{(1-2a\cos t+a^2)}$$
this is what i come up with
$$ 1 +a\cos t + a^2 \cos 2t + \dots+a^n \cos nt = \frac{1-a^{n+1}\cos(n+1)t-a\cos t+a^{n+2}\cos(n+2)t}{(1-2a\cos t+a^2)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1147717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
If $f(x) = 3x^2-x + 2$, find $f(a)$ and $[f(a)]^2$ If $f(x) = 3x^2-x^2$, find $f(a)$ and $[f(a)]^2$
Also, $2f(a) = 3x2(a)^2-2(a)+2 = 6a^2-2a+2$
The book says the answer is $6a^2-2a+4$. Why is that? Is the book wrong?
| what you are computing in your work sheet is $f(f(a).$ that is usually is different from $[f(a)]^2.$ you were asked to compute $[f(a)]^2 = (3a^2 - a + 2)^2 = 9a^4 + \dots +4.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1147821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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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
| Taking $180^{\text{th}}$ roots and using $a+b+c=180^\circ$, the inequality is equivalent to
$$(a^b b^c c^a)^{\frac1{a+b+c}} \le \frac{a^2+b^2+c^2}{a+b+c}$$
Use weighted AM-GM in the form
$$(a^b b^c c^a)^{\frac1{a+b+c}} \le \frac{b\cdot a + c\cdot b+a\cdot c}{b+c+a}$$
and $ab+bc+ca \le a^2+b^2+c^2$ to conclude.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1149128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Maximisation: Half circle area inscribed within isoceles triangle? Given an isosceles triangle with the line of symmetry along $x=0$, and the odd side along $y=0$, How can I optimize the maximum parabolic area inscribed within the triangle? The "half -circular" area has points of tangency to the triangle, one of each side of the line of symmetry. What I am having trouble is, I simply don't know how to start, since I don't know the dimensions of of the triangle which the problems usually give. I have tried looking only at its "first quadrant", trying to find a negative parabola tangent to the line (triangle side) with a negative slope, but I still can't figure it out. Thanks in advance!
| Let the triangle have side with equation $y_1=c-mx$. Start here.
Let the parabola have equation $y_2=b-ax^2$.
Gradients are $\dfrac{dy_1}{dx}=-m$ and $\dfrac{dy_2}{dx}=-2ax$
Gradients equal when $m=2ax$, this is ${x}=\dfrac{m}{2a}$.
Must also have same $y$ value, so
$$c-m\frac{m}{2a}=b-a\frac{m}{2a}\frac{m}{2a}$$
$$c-\frac{m^2}{2a}=b-\frac{m^2}{4a}$$
$$4ac-2m^2=4ab-m^2$$
$$4a(c-b)=m^2.$$
$$a=\frac{m^2}{4(c-b)}$$
Area under curve is $\int_{-\sqrt{\frac{b}{a}}}^{\sqrt{\frac{b}{a}}} {b-ax^2}dx$
$$A_{curve}=\frac{4}{3} \sqrt{\frac{b^3}{a}}$$
$$A_{triangle}=\frac{\frac{2c}{m} c}{2}=\frac{c^2}{m}$$
$$\frac{A_{curve}}{A_{triangle}}=\frac{4m}{3c^2} \sqrt{\frac{b^3}{a}}$$
$$=\frac{4m}{3c^2} \sqrt{\frac{4b^3(c-b)}{m^2}}$$
$$=\frac{8}{3c^2} \sqrt{b^3(c-b)}$$
Differentiate this wrt b
$$\frac{d{Area}_{proportion}}{db}=\frac 1 2 {(cb^3-b^4)}^{\frac{-1}{2}} (3cb^2-4b^3)$$
Minimums means $ (3cb^2-4b^3)=0$
$$3c=4b$$
$$b=\frac {3c}{4}$$
Substituting gives $\frac{A_{curve}}{A_{triangle}}=\frac{\sqrt{3}}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1151283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determine the probability that each of the 8 members serves on at least one of the three committees. A certain group has 8 members. In January, 3 members are selected at random to serve on a committee. In February, 4 members are selected at random and independently of the first selection to serve on another committee. In March, 5 members are selected at random and independently of the previous 2 selections to serve on a third committee. Determine the probability that each of the 8 members serves on at least one of the three committees.
So I am really unsure about this problem, I put the committees choice in the denominator, to represent the total choices. My reasoning for the numerator were you can choose any 3 people from the 8, then from the remaining 5 who haven't had positions you have to choose 4 and then there is 1 left who hasn't had a position so he has to be one of the 5 selected in March, and the remaining 4 can be any from the
$\frac{{8 \choose 3}{5 \choose 4}{1 \choose 1}{7 \choose 4}}{{8 \choose 3}{8 \choose 4}{8 \choose 5}}$
So I need to use the inclusion exclusion principle?
| WLOG, assume that the members $1$ to $5$ belong to the third committee.
The probability that $4$ of those members belong to the second committee, is
$\frac{\binom{5}{4}}{\binom{8}{4}}=\frac{1}{14}$. In this case, all the remaining
members must belong to the first committee, so $\frac{1}{14}\times\frac{1}{\binom{8}{3}}=\frac{1}{784}$ is the probability in this case.
The probability that $3$ of those members belong to the second committee, is
$\frac{\binom{5}{3} \times \binom{3}{1}} {\binom{8}{4}}=\frac{3}{7}$. In
this case, the $2$ remaining members must belong to the first committee, so
$\frac{3}{7}\times \frac{6}{\binom{8}{3}}=\frac{9}{196}$ is the
probability in this case.
The probability that $2$ of those members belong to the second committee, is
$\frac{\binom{5}{2} \times \binom{3}{2}} {\binom{8}{4}}=\frac{3}{7}$. In this
case, the remaining member must belong to the first committee, so $\frac{3}{7}\times\frac{3}{8}=\frac{9}{56}$ is the probability in this case.
Finally, the probability that $1$ of those members belongs to the second committee
is $\frac{1}{14}$. In this case, all members already belong to at least one
committee.
In total, the probability is $\frac{219}{784}=0.2793$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1151363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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 right approach? Because when i try and work out the algebra i keep on hitting a wall.
\begin{align}
\left(1-\frac{1}{n^2}\right)\left(1-\frac{1}{(n+1)^2}\right)&=1-\frac 1{(1-n)^2}-\frac 1{n^2}-\frac 1{n^2(n+1)} \\
&=\frac{n^2}{(n+1)^2}-\frac 1{(n+1)^2} \\
&=\frac{n^2-1}{(n+1)^2}-\frac 1{n^2}+\frac 1{n^2(n+1)^2} \\
&=\frac{n^2(n^2-1)}{n^2(n+1)^2}-\frac{(n+1)^2}{n^2(n+1)^2} \\
&=\frac{n^2(n^2-1)-(n+1)^2}{n^2(n+1)^2}+\frac 1{n^2(n+1)^2} \\
&=\frac{n^2(n^2-1)-(n+1)^2+1}{n^2(n+1)^2}
\end{align}
| Check that the equation holds for $n=2$. Assuming that it holds for some $n$:
$$\begin{align}\prod_{i=2}^n\left( 1-\frac1{i^2}\right)&=\frac{n+1}{2n}\\
\end{align}$$
And so
$$\begin{align}
\prod_{i=2}^{n+1}\left( 1-\frac1{i^2}\right)&=\frac{n+1}{2n}\left( 1-\frac1{(n+1)^2}\right)\\
&=\frac{n+1}{2n}-\frac{n+1}{2n(n+1)^2}\\
&=\frac{(n+1)^2}{2n(n+1)}-\frac{1}{2n(n+1)}\\
&=\frac{(n+1)^2-1}{2n(n+1)}\\
&=\frac{n^2+2n}{2n(n+1)}\\
&=\frac{n+2}{2(n+1)}\\
\end{align}$$
If the equation holds for some $n$, it also holds for $n+1$. Since it holds for $2$, it also holds for $3$, and since it holds for $3$, it also holds for $4$, and so on. The statement is true for all natural numbers greater than or equal to $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1154218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
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:
$A(x^2+2x-8) + B(x^2+5x+4) + C(x^2-x-2) = 3x^2+17x$
No clue on my next step or even if this is the right step.
| $$A(x^2+2x-8) + B(x^2+5x+4) + C(x^2-x-2) = 3x^2+17x.$$
Setting $x=-4$,
$$0A+0B+18C=-20.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1156046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can I determine the position of the apex of an irregular tetrahedron I have an irregular tetrahedron the base of which is an equilateral triangle. Knowing the lengths of all sides I need to then determine the position of the apex.
Is there anyone that can offer a bit of guidance on this.
| Doing this in $\mathbb{R}^3$, let the three points of the equilateral triangular base of the tetrahedron be $(0,0,0)$, $(0,2\ell,0)$, and $(\sqrt{3}\ell,a,0)$. Suppose the lengths of the edges of the tetrahedron between each of these point and our apex are $a$, $b$, and $c$ respectively. The apex (if it exists) will be the intersection of the three spheres given by the following equations:
$$\begin{cases}
x^2 + y^2 + z^2 = a^2 \\
x^2 + (y-2\ell)^2 + z^2 = b^2 \\
(x-\sqrt{3}\ell)^2 + (y-\ell)^2 + z^2 = c^2 \\
\end{cases}$$
After some rough scratch work, I ended up getting our apex as the point
$$\left(
\frac{a^2+b^2-2c^2+4\ell^2}{4\sqrt{3}\ell},\quad
\frac{a^2-b^2+4\ell^2}{4\ell}, \quad
\sqrt{a^2-x^2-y^2}
\right)$$
The $z$-coordinate is kinda gross, so I left it here in terms of $x$ and $y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1159508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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 to use with this type of problem.
| Hint: $x^2 \equiv -1$ mod $5$ $\iff$ $x^2 \equiv 4$ mod $5$ $\iff$ $(x-2)(x+2) \equiv 0$ mod $5$, whence $x \equiv 2$ or $x \equiv -2$ mod $5$.
$x^2 \equiv -1$ mod $13$ $\iff$ $x^2 \equiv 25$ mod $13$ $\iff$ $(x-5)(x+5) \equiv 0$ mod $13$, whence $x \equiv 5$ or $x \equiv -5$ mod $13$.
Now cross-solve these $4$ possibilities with the Chinese Remainder Theorem.
Example: if $x \equiv 2$ mod $5$ and $x \equiv 5$ mod $13$, this should give you $x \equiv 57$ mod $65$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1161523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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if $x^3-x\in\mathbb{Z}$ and $x^4-x\in\mathbb{Z}$ for some $x\in\mathbb{R}$, then $x\in\mathbb{Z}$. Assume that $x^3-x\in\mathbb{Z}$ and $x^4-x\in\mathbb{Z}$ for some $x\in\mathbb{R}$.
Prove that $x\in\mathbb{Z}$.
my attempt:
Let $a=x^3-x$ and consider polynomial $X^3-X-a$, then $x$ is a root of it and if $x\in\mathbb{Q}$, then obviously $x\in\mathbb{Z}$.
But it doesn't give anything if $x\notin\mathbb{Q}$, so it's surely not the right way.
| Let $x^3-x=a$ and $x^4-x=b$.
Let's do a few steps of the Euclidean algorithm for polynomial GCD:
$x^4-x-b=x(x^3-x-a)+(x^2+(a-1)x-b)\\ \implies x^2+(a-1)x-b=0$
$x^3-x-a=x(x^2+(a-1)x-b)+((1-a)x^2+(b-1)x+a)\\ \implies (1-a)x^2+(b-1)x+a=0$
$0=(1-a)(x^2+(a-1)x-b)-((1-a)x^2+(b-1)x+a)$
$\implies x ((1-a) (a-1)-b+1)-(1-a) b-a=0$
and so $x$ is rational, unless $((1-a) (a-1)-b+1)=0$ and $(1-a) b-a=0$.
The last condition happens for integers $a$ and $b$ iff $a=b=0$, in which case $x=0$ or $x=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1163847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find : $ dy/dx, y=\sqrt{4x^2 - 7x - 2}$ The problem says Find $dy/dx, y=\sqrt{4x^2 - 7x - 2}$
So far I changed it to $(4x^2 - 7x - 2)^{1/2}$
I don't know where to go from there.
| If $y = \sqrt{4x^2 - 7x - 2}$
then
$$\frac{dy}{dx} = \frac{d}{dx}(4x^2 - 7x - 2)^{\frac{1}{2}}
\\
$$
$$
=\frac{1}{2}\frac{d}{dx}(4x^2 - 7x - 2) \times (4x^2 - 7x - 2)^{\frac{-1}{2}}
$$
$$
= \frac{8x-7}{2\sqrt{4x^2 - 7x - 2}}
$$
using the chain rule.
That is, if we have a function $y = u^{\frac{1}{2}}$ then $$\frac{dy}{dx} = \frac{dy}{du}\times\frac{du}{dx}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1165243",
"timestamp": "2023-03-29T00:00:00",
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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 integer factors of $35$ are $(1, 35)$ or $(5, 7)$. There are no integers $x$ and $y$ that add to $1$, so $x + y = 5$ or $7$.
Using $x + y = 5$ we get that $y = 5 - x$, so $y^3 = 125 - 75x + 15x^2 - x^3$ so
$x^3 + y^3 = 15x^2 - 75x + 125 = 35$
or
$$15x^2 - 75x + 90 = 0$$
$$x^2 - 5x + 6 = 0$$
$$(x - 3)(x - 2) = 0$$
So $x = 2$ or $x = 3$
Thus $y = 3$ or $y = 2$ respectively.
Using $x + y = 7$ we get that $y = 7 - x$, so $y^3 = 343 - 147x + 21x^2 - x^3$ so
$$x^3 + y^3 = 21x^2 - 147x + 343 = 35$$
$$21x^2 - 147x + 308 = 0$$
$$3x^2 - 21x + 44 = 0$$
or
$$x = \frac{21 \pm \sqrt{-87}}{6}$$
Since x is complex, this can't be a solution.
So $(x, y) = (2, 3)$ or $(x, y) = (3, 2)$
| To bound all solutions
in integers
(not just to positive integers)
to
$x^3+y^3 = n$:
First,
using the factorization
$x^3+y^3
=(x+y)(x^2-xy+y^2)
$,
we get
possible values
for $x+y$
since $(x+y) | n$.
Then,
since
$x^2-xy+y^2
=(x+y)^2-3xy
$,
we get possible values
for $xy$
($xy
=\dfrac{(x+y)^2-n/(x+y)}{3}
$)
and
this gives $x$ and $y$.
Note (added after a comment):
If $a=x+y$ and $b=xy$ are known,
then,
since
$(r-x)(r-y)
=r^2-r(x+y)+xy
=r^2-ar+b
$,
$x$ and $y$
are the roots of
$r^2-ar+b$.
So use the quadratic formula:
$r
=\dfrac{a\pm \sqrt{a^2-4b}}{2}
$.
(Additional stuff to make the solution explicit)
We have
$a = x+y$
(the divisor of $n$),
$b = xy = ((x+y)^2−n/(x+y))/3=(a^2−n/a)/3$.
Then the discriminant
$d
=a^2-4b
=a^2-4(a^2−n/a)/3
=(4n/a-a^2)/3
=(4n-a^3)/(3a)
$
and the roots are
$\dfrac{a\pm \sqrt{(4n-a^3)/(3a)}}{2}$.
If we adopt the convention
that $x < y$,
then
$x = \dfrac{a- \sqrt{(4n-a^3)/(3a)}}{2}$
and
$y = \dfrac{a + \sqrt{(4n-a^3)/(3a)}}{2}$
.
For this to work,
$(4n-a^3)/(3a)$
must be a perfect square.
Some elementary bounds on $x$ and $y$:
If
$0 < x \le y$,
$\sqrt[3]{n/2} < y \lt \sqrt[3]{n}
$.
If $x < 0 < y$,
since
$x^2-xy+y^2
=x^2-xy+(y/2)^2+3y^2/4
=(x-y/2)^2+3y^2/4
> 3y^2/4
$,
$x+y > 0$
or
$y > -x$.
This can be improved to
$x^2-xy+y^2
=x(x-y)+y^2
=(-x)(y-x)+y^2
> y^2
$
and
$x^2-xy+y^2
<3y^2
$.
For every divisor $d$ of $n$,
$d y^2 < n < 3dy^2$,
so
$\sqrt{n/(3d)} < y < \sqrt{n/d}$.
| {
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"url": "https://math.stackexchange.com/questions/1168613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Sum $\sum_{n=1}^{\infty}\frac{1}{(9n-1)(9n+1)}$ Find the value of
$$\sum_{n=1}^{\infty}\frac{1}{(9n-1)(9n+1)}$$
I have tried to substitute the value of $n$ from $1$ but it is not telescoping series.
| Hint. Recall the following series representation for the digamma function
$$\begin{equation}
\psi(x+1) = -\gamma - \sum_{n=1}^{\infty} \left( \frac{1}{n+x} -\frac{1}{n}
\right), \quad \Re x >-1, \tag1
\end{equation}
$$ where $\gamma$ is the Euler-Mascheroni constant.
Then by partial fraction decomposition we have
$$
\begin{align}
\frac{1}{(9n-1)(9n+1)} &= \frac{1}{2}\left(\frac{1}{9n-1}-\frac{1}{9n+1}\right)\\\\
&=\frac{1}{18}\left(\frac{1}{n-1/9}-\frac{1}{n+1/9}\right)\\\\
&=\frac{1}{18}\left[\left(\frac{1}{n-1/9}-\frac1n\right)-\left(\frac{1}{n+1/9}-\frac1n\right)\right]
\end{align}
$$ then summing from $n=1$ to $+\infty$, using $(1)$, we get
$$
\sum_{n=1}^{\infty}\frac{1}{(9n-1)(9n+1)} =\frac{1}{18}\left(\psi\left(\frac{10}{9}\right)-\psi\left(\frac{8}{9}\right)\right)
$$ equivalently
$$
\sum_{n=1}^{\infty}\frac{1}{(9n-1)(9n+1)} =\frac{1}{2}-\frac{\pi}{18}\cot\left(\frac{\pi}{9}\right)
$$
where we have used Gauss' theorem for the digamma function. A numerical approximation for this series is given by $$\sum_{n=1}^{\infty}\frac{1}{(9n-1)(9n+1)} \approx \color{blue}{0.02047472906319532...}$$
| {
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"url": "https://math.stackexchange.com/questions/1168836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluating $\lim\limits_{x \to \infty } \frac{5 + \sqrt{x^2+5}}{x-6}$ I don't know how to evaluate
$$\lim_{x \to \infty } \frac{5 + \sqrt{x^2+5}}{x-6}$$
*
*It is $\frac{\infty}{\infty}$
*I have tried to multiply by $5 - \sqrt{x^2+5}$, but I reach a wrong result.
*I have tried to take the derivative of $5 + \sqrt{x^2+5}$ and $x-6$, but this also doesn't help.
Probably, I should use another method. Which method do I use?
| Derivative of the numerator: $(5+ \sqrt{x^2+5})'= (\sqrt{x^2+5})'= \frac{(x^2+5)'}{2}\sqrt{x^2+5}^{-1} = \frac{2x}{2 \sqrt{x^2+5}} = \frac{1}{\sqrt{1+\frac{5}{x^2}}}$.
Derivative of the denominator: $(x-6)'=1$.
L'Hospital's rule is possible.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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critical points of the function Find the critical points of the function
$$f(x,y)=(4x-x^2)\cos y$$
first let's determinate Partial derivatives:
$$\dfrac{\partial f}{\partial x}(x,y)=(4-2x)\cos y$$
$$\dfrac{\partial f}{\partial y}(x,y)=-\sin y(4x-x^2) $$
To find the critical points, we solve:
\begin{cases}\dfrac{\partial f}{\partial x}(x,y)=0 & \\
\\
\dfrac{\partial f}{\partial y}(x,y)=0 & \end{cases}
\begin{cases}
(4-2x)\cos y=0 & \\
\\
-\sin y(4x-x^2)=0 & \end{cases}
\begin{cases}
(4-2x)\cos y=0 & \\
\\
(x^2-4x)\sin y=0 & \end{cases}
which gives
$ \begin{cases}
4-2x=0 & \\
\\
x^2-4x=0 & \end{cases} \textrm{or}
\begin{cases}
4-2x=0 & \\
\\
\sin y=0 & \end{cases}
\textrm{or}
\begin{cases}
\cos y=0 & \\
\\
x^2-4x=0 & \end{cases}
\textrm{or}
\begin{cases}
\cos y=0 & \\
\\
\sin y=0 & \end{cases} $
let's treat each case individually
*
*$ \begin{cases}
4-2x=0 & \\
\\
x^2-4x=0 & \end{cases} \implies \begin{cases}
x=2 & \\
\\
x=0 \textrm{ or } x=4 & \end{cases} $ thus The system does not have solution
*$ \begin{cases}
4-2x=0 & \\
\\
\sin y=0 & \end{cases} \implies \begin{cases}
x=2 & \\
\\
\ y=k\pi,\ k \in \mathbb{Z} & \end{cases} \implies (2,k\pi ) $
*$\begin{cases}
\cos y=0 & \\
\\
x^2-4x=0 & \end{cases}\implies \begin{cases}
y=\dfrac{\pi}{2}+k\pi, \ k \in \mathbb{Z} & \\
\\
x=0 \textrm{ or } x=4 & \end{cases}\implies (0,\dfrac{\pi}{2}+k\pi) \textrm{ or } (4,\dfrac{\pi}{2}+k\pi)$
*$\begin{cases}
\cos y=0 & \\
\\
\sin y=0 & \end{cases} \implies$ thus The system does not have solution
The function has three critical points $$(2,k\pi ), (0,\dfrac{\pi}{2}+k\pi) \textrm{ and }(4,\dfrac{\pi}{2}+k\pi)$$
*
*Am i right ?
| Hint: The second equation is $x(x-4)\sin y=0$ which gives $x=0$ or $x=4$ or $y=k\pi$, $k \in \Bbb{Z}$.
Consider all three cases separately by plugging them in to the first equation.
| {
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Evaluate a rational function of $x,y,z$ given two polynomial equations in $x,y,z$ Let $x, y, z$ be real numbers. Given that $$2x(y^2−1)+2y(x^2−1)=(1+x^2)(1+y^2)$$ and $$4z(1−y^2)+4y(1−z^2)=(1+z^2)(1+y^2)$$ Find the value of the following expression:
$$\Bigg(\frac{2x}{1+x^2}−\frac{2z}{1+z^2}\Bigg)^2+\Bigg(\frac{1−z^2}{1+z^2}−\frac{1−x^2}{1+x^2}\Bigg)^2$$ How do I use $x=\tan\alpha$ and $z=\tan\beta$ in this question. I know it will be useful as the expression to be found out is screaming trigonometric substitution. These were the two equations I finally got after substitution-$$(y^2-1)\sin2\alpha-2y\cos2\alpha=1+y^2$$ and $$2(1-y^2)\sin2\beta+4y\cos2\beta=1+y^2$$ and the expression to be found is $-2(\sin2\alpha\sin2\beta-2\cos2\alpha\cos2\beta)$. Now how do I manipulate my equations? Thanks.
| I don't know if a better solution is possible, but here is mine.Consider the following points:
$A(\frac{2x}{1+x^2},\frac{1−x^2}{1+x^2}),B(\frac{1−y^2}{1+y^2},\frac{2y}{1+y^2}),C(\frac{2z}{1+z^2},\frac{1−z^2}{1+z^2})$.
Notice that these points on a unit circle, as for any real number $a$,
$$(\frac{2a}{1+a^2})^2+(\frac{1−a^2}{1+a^2})^2=1$$
According to the assumptions of the problem, we have that
$$AB^2=(\frac{2x}{1+x^2}−\frac{1−y^2}{1+y^2})^2+(\frac{1−x^2}{1+x^2}−\frac{2y}{1+y^2})^2$$$$=2−2⋅\frac{2x(1−y^2)+2y(1−x^2)}{(1+x^2)(1+y^2)}=4$$
Therefore, $AB=2$. On the other hand, we have that
$$BC^2=(\frac{2y}{1+y^2}−\frac{1−z^2}{1+z^2})^2+(\frac{1−y^2}{1+y^2}−\frac{2z}{1+z^2})^2$$$$=2−2⋅\frac{2z(1−y^2)+2y(1−z^2)}{(1+z^2)(1+y^2)}=1$$
Hence, we obtain that $BC=1$. From the condition $AB=2$, it follows that $AB$ is a diameter, thus $∠C=90°$ and $AC^2=3$. Hence,
$$(\frac{2x}{1+x^2}−\frac{2z}{1+z^2})^2+(\frac{1−z^2}{1+z^2}−\frac{1−x^2}{1+x^2})^2=AC^2=3. $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Positive integer solutions of $\frac{1}{a_1}+\frac{2}{a_2}+\frac{3}{a_3}+\cdots+\frac{n}{a_n}=\frac{a_1+a_2+a_3+\cdots+a_n}{2}$
Find all ordered tuples of positive integers $(a_1,a_2,a_3,\ldots,a_n)$ such that
$\frac{1}{a_1}+\frac{2}{a_2}+\frac{3}{a_3}+\cdots+\frac{n}{a_n}=\frac{a_1+a_2+a_3+\cdots+a_n}{2}$
The only thing I have been able to think about is using inequalities. I have tried to apply AM-GM, Titu's lemma etc.. Cauchy-Schwarz gives the following thing:
$$(\frac{1}{a_1}+\frac{2}{a_2}+\frac{3}{a_3}+\cdots+\frac{n}{a_n})(a_1+\cdots a_n) \ge (\sqrt{1}+\sqrt{2}+\cdots \sqrt{n})^2$$
$$(a_1+\cdots a_n)^2\ge 2(\sqrt{1}+\cdots \sqrt {n})^2$$
which doesn't really help us at all.
I have also tried considering smaller cases. For $n=2$,
$$a_1a_2(a_1+a_2)=4a_1+2a_2$$ which tells us that $2a_2=ka_1$ and $8a_1=pa_2=ka_1p\implies kp=8$. This should now give us all the solutions by checking all the cases.
So how can we even begin to attack this problem?
| Partial solution: If $n+1$ is a square, then set $a_1=a_2=\cdots=a_n=\sqrt{n+1}$.
Then the LHS is $$(1+2+\cdots+n)\frac{1}{\sqrt{n+1}}=\frac{n(n+1)}{2\sqrt{n+1}}=\frac{1}{2}n\sqrt{n+1}$$
which agrees with the RHS.
More of a partial solution, following @Stefan4024's induction idea. Start with a solution of the preceding form, then add two more terms to the LHS, namely $\frac{n+1}{b}+\frac{n+2}{b}$. The equation will still balance if $$\frac{1}{b}(n+1+n+2)=\frac{b+b}{2}$$
or $b^2=2n+3$. Hence, we seek $n$ satisfying (1) $n+1=a^2$; and (2) $2n+3=b^2$. We can eliminate the $n$ to get Pell's equation:
$$b^2-2a^2=1$$
This has infinitely many solutions. The smallest is $b=3, a=2$ ($n=3$, 5 terms). The next smallest is $b=17, a=12$ ($n=143$, 145 terms). The next one is $b=99, a=70$ ($n=4899$, 4901 terms).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Floor function right hand limit Show
$$\lim_{x\to 0^+}\frac{x}a\cdot\left\lfloor\frac{b}x\right\rfloor=\frac{b}a\;.$$
I think I should use boundedness of $\left\lfloor\dfrac{b}x\right\rfloor$, $$\frac{b}x-1\le\left\lfloor\frac{b}x\right\rfloor\le\frac{b}x\;.$$
But I am not sure how to proceed with this.
| As you mention, $|\lfloor b/x \rfloor - b/x| \leq 1$, and so
$$
\frac{x}{a} \left|\left\lfloor \frac{b}{x} \right\rfloor - \frac{b}{x}\right| \leq \frac{x}{a}.
$$
We can simplify this to
$$
\left|\frac{x}{a}\left\lfloor \frac{b}{x} \right\rfloor - \frac{b}{a}\right| \leq \frac{x}{a}.
$$
As $x\to0^+$, the right-hand side tends to zero, and so $(x/a)\lfloor b/x \rfloor \to b/a$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find lim sup lim inf How do I find $\lim \sup\text{ or } \lim \inf$ of $ \sin (\frac{n\pi}{5})$ ?
I know the $\sin$ function normally oscillates between $-1$ and $1$ but that obviously is not the answer for $\lim \inf$ and $\lim \sup$.
| Observe that $$\begin{align*}&\left\{ \sin \left(\frac{n\pi}{5}\right)|n\in \mathbb{N}\right\}\\&=\left\{\sin \left(\frac{\pi}{5}\right),\sin \left(\frac{2\pi}{5}\right),\sin \left(\frac{3\pi}{5}\right),\sin \left(\frac{4\pi}{5}\right),\sin \left(\frac{5\pi}{5}\right),\sin \left(\frac{6\pi}{5}\right),\sin \left(\frac{7\pi}{5}\right),\sin \left(\frac{8\pi}{5}\right),\sin \left(\frac{9\pi}{5}\right),\sin \left(\frac{10\pi}{5}\right)\right\} \\&=\left\{0,\sin \left(\frac{\pi}{5}\right),\sin \left(\frac{2\pi}{5}\right),\sin \left(\frac{3\pi}{5}\right),\sin \left(\frac{4\pi}{5}\right),-\sin \left(\frac{\pi}{5}\right),-\sin \left(\frac{2\pi}{5}\right),-\sin \left(\frac{3\pi}{5}\right),-\sin \left(\frac{4\pi}{5}\right)\right\}.\end{align*}$$
Therefore
$$\begin{align*}&\limsup_{n\to\infty} \sin \left(\frac{n\pi}{5}\right)\\&=\max \left\{0,\sin \left(\frac{\pi}{5}\right),\sin \left(\frac{2\pi}{5}\right),\sin \left(\frac{3\pi}{5}\right),\sin \left(\frac{4\pi}{5}\right),-\sin \left(\frac{\pi}{5}\right),-\sin \left(\frac{2\pi}{5}\right),-\sin \left(\frac{3\pi}{5}\right),-\sin \left(\frac{4\pi}{5}\right)\right\}\end{align*} $$ and
$$\begin{align*}&\liminf_{n\to\infty} \sin \left(\frac{n\pi}{5}\right)\\&=\min \left\{0,\sin \left(\frac{\pi}{5}\right),\sin \left(\frac{2\pi}{5}\right),\sin \left(\frac{3\pi}{5}\right),\sin \left(\frac{4\pi}{5}\right),-\sin \left(\frac{\pi}{5}\right),-\sin \left(\frac{2\pi}{5}\right),-\sin \left(\frac{3\pi}{5}\right),-\sin \left(\frac{4\pi}{5}\right)\right\}.\end{align*} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1176432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to show $n^5 + 29 n$ is divisible by $30$ Show that $n^5 + 29 n$ is divisible by $30$.
Attempt:
$n^4 ≡ 1 \pmod 5$ By Fermat Little Theorem
| $$n^5+29n-(n-2)(n-1)n(n+1)(n+2)=5n^3+25n=5(n-1)n(n+1)+30n$$
$$\implies n^5+29n=\underbrace{(n-2)(n-1)n(n+1)(n+2)}_{\text{The product of }5\text{ consecutive integers}}+5\underbrace{((n-1)n(n+1))}_{\text{The product of }3\text{ consecutive integers}}+30n$$
See The product of n consecutive integers is divisible by n factorial
| {
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"timestamp": "2023-03-29T00:00:00",
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If $ f(x) = \left\{\frac{3x}{2}\right\}\;,$ Where $\{x\}=x-\lfloor x \rfloor$. Then real solution of $f(f(x))=0$
If $\displaystyle f(x) = \left\{\frac{3x}{2}\right\}\;,$ Where $\{x\}=x-\lfloor x \rfloor$. Then no. of solution of the equation
$f(f(x))=0$ and $f(f(f(x)))=0$ and $f(f(f(f(x))))=0$.
$\bf{My\; Try::}$ We can write $\displaystyle f(x)= \left\{\frac{3x}{2}\right\}=\frac{3x}{2}-\lfloor \frac{3x}{2} \rfloor = \left\{\begin{matrix}
\displaystyle \frac{3x}{2} \;,& 0\leq x<\displaystyle \frac{2}{3} \\\\
\displaystyle \frac{3x}{2}-1 \;,& \displaystyle \frac{2}{3}\leq x<\displaystyle 1
&
\end{matrix}\right.$
Now for Solution of $f(f(x)) = 0.$ We have used Two cases.
$\bullet$ If $0\leq x<\displaystyle \frac{2}{3} .$ Then we use $\displaystyle f(x) = \frac{3x}{2}.$ So $\displaystyle f(f(x)) = 0\Rightarrow \frac{9x}{4}=0\Rightarrow x=0$..
$\bullet$ If $\displaystyle \frac{2}{3}\leq x<\displaystyle 1 .$ Then we use $\displaystyle f(x) = \frac{3x}{2}-1.$ So $\displaystyle f(f(x)) = 0\Rightarrow \frac{9x}{4}-\frac{5}{2}=0\Rightarrow x=\frac{10}{9}$.
So we get only one solution $x=0$ for $f(f(x))=0$
Is my solution is Right, If not , Then plz explain me how can i solve it.
| Note that $f(f((\frac23)^2))=0$, so $0$ is not the only solution.
$\bullet$ If $0\leq x<\displaystyle \frac{2}{3} .$ Then we use $\displaystyle f(x) = \frac{3x}{2}.$ So $\displaystyle f(f(x)) = 0\Rightarrow \frac{9x}{4}=0\Rightarrow x=0$..
The mistake you are making is that even when $0\le x\le \frac23$, $f(x)$ may not be in the same interval. Such is the case with $(\frac23)^2$, it is strictly between $0$ and $\frac23$, but $f((\frac23)^2)$ is $\frac23$, which is not.
Can you see why this does not yield every solution?
Here is how I found all the solutions to $f(f(x))=0$.
$f(f(x))=0$ if and only if $\frac32f(x)$ is an integer. This is the case if and only if $f(x)$ is some integer multiple of $\frac23$. $f$ can only return values in $[0,1)$, however, so $f(x)$ must be either $0$ or $\frac23$.
$f(x)=0$ if and only if $\frac32x$ is an integer:
$$\frac32x=n\implies x=\frac23n$$
And so, $x=\frac23n$ is a solution to $f(f(x))=0$ for every integer $n$.
$f(x)=\frac23$ if and only if $\frac32x$ is $n+\frac23$ for some integer $n$:
$$\frac32x=n+\frac23\implies x=\frac23n+\left(\frac23\right)^2$$
And so, $x=\frac23n+\left(\frac23\right)^2$ is a solution to $f(f(x))=0$ for every integer $n.$
In conclusion, the solutions to $f(f(x))=0$ are all numbers of the forms $\frac23n+\left(\frac23\right)^2$ and $\frac23n$ for integer $n$.
| {
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$\left(\frac{a}{n}\right)=1$ does not necessarily imply that, $a$ is a quadratic residue $\mod n$
$\left(\frac{a}{n}\right)=1$ does not necessarily imply that, $a$ is a quadratic residue $\mod n$, where $\left(\frac{.}{.}\right)$ is the Jacobi symbol.
For example I know that $\left(\frac{2}{15}\right)=\left(\frac{2}{3}\right)\left(\frac{2}{5}\right)=(-1)^{\frac{3^2-1}{8}}(-1)^{\frac{5^2-1}{8}}=(-1)^1(-1)^3=1$
but $x^2\equiv2\mod 15$ has no integer solutions http://www.wolframalpha.com/input/?i=quadratic+residues+mod+15
Is there a generalization ?
$\textbf{EDIT}$: How can I find all quadratic residues $\mod 15 $ without wolframalpha ?
| For the second part of your question which remains unanswered, one can find all the quadratic residues $\pmod n$ by considering all numbers up to $n$ modulo $n$, squaring all of them and considering the residues $\pmod n$. Notice that any other numbers, for example $n+2$, would have the same quadratic residue as the number reduced $\pmod n$, in this example $2$, would, hence this list would cover all of the residues (In the taken example, $(n+2)^2 \equiv 2^2 \equiv 4 \pmod n$, provided $4<n$ or it would be reduced further). Notice that we can write the last $\lfloor\frac{n}{2}\rfloor$ numbers in our set as negative counterparts of previous numbers which would have the same quadratic residue $\pmod n$. This means that we only need to check upto $\lceil\frac{n}{2}\rceil$ and would cover all residues.
To make this clearer I will do this procedure for $7$.
$\begin{array}{| c | c | c |}\hline \text{n} & n^2 \pmod 7 \\ \hline 1 & 1\\ \hline 2 & 4 \\ \hline 3 & 2 \\ \hline 4 & 2 \\ \hline
\end{array}$
The above table implies that the only quadratic residues$\pmod 7$ are $1,2,4$. One can repeat this process for $15$ to get that the quadratic residues are $1,4,6,9,10$. Notice that this does not include $2$, hence $2$ is a quadratic nonresidue.
| {
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"timestamp": "2023-03-29T00:00:00",
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Does this qualify as a Laurent series? Evaluate the Laurent series around the singularity at $ z_0 = 3$.
$$ \frac{1}{z^2(z-3)} $$
I can apply the geometric series as follows:
$$\frac{1}{z^2}\cdot\frac{1}{z-3}=-\frac{1}{3z^2}\cdot\frac{1}{1+\frac{z}{3}}=\frac{1}{3z^2}\sum_{n=0}^{\infty}(-1)^n\left(\frac{z}{3}\right)^n$$
though I'm not sure that if the final answer, since the Laurent series involves negative powers of $ n $.
What do you think?
| For a Laurent series about the point $z_0=3$, you want the series to be in powers of $z-3$.
$$
\begin{align}
\frac1{z^2(z-3)}
&=\frac1{((z-3)+3)^2(z-3)}\\
&=\frac1{9\left(1+\frac{z-3}3\right)^2}\cdot\frac1{z-3}\\
&=\frac19\frac1{z-3}\left(1-2\frac{z-3}3+3\frac{(z-3)^2}9-4\frac{(z-3)^3}{27}+\dots\right)\\
&=\frac1{9(z-3)}-\frac2{27}+\frac{3(z-3)}{81}-\frac{4(z-3)^2}{243}+\dots\\
&=\sum_{k=0}^\infty\frac{k+1}{(-3)^{k+2}}(z-3)^{k-1}
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove this inequality with $xyz\le 1$
If $x,y,z\ge 0$ and $\color{red}{xyz\le 1}$, show that
$$\color{blue}{\dfrac{x^2-x+1}{x^2+y^2+1}+\dfrac{y^2-y+1}{y^2+z^2+1}
+\dfrac{z^2-z+1}{z^2+x^2+1}\ge 1}.$$
| Multiplying and dividing each of the expressions numerator by $(1 +
x)$, $(1 + y)$ and $(1 + z)$ respectively:
$$
\frac{z^3+1}{(z+1) \left(x^2+z^2+1\right)}+\frac{x^3+1}{(x+1) \left(x^2+y^2+1\right)}+\frac{y^3+1}{(y+1) \left(y^2+z^2+1\right)}
$$
Further simplifying:
$$
1+ \frac{x^2-x+1}{x^2+y^2+1}-\frac{x^2+z}{x^2+z^2+1}+\frac{y^2-y+1}{y^2+z^2+1}
$$
Giving similar treatment to second and fourth terms:
$$
1+\frac{x^3+1}{(x+1) \left(x^2+y^2+1\right)}+\frac{y^3+1}{(y+1) \left(y^2+z^2+1\right)}-\frac{x^2+z}{x^2+z^2+1}
$$
Note that the $1$ in the above expression suggests that this expression could be $\geq1$ as long as the sum of second, third and fourth term is $\geq0$. Simplifying further:
$$
1+\frac{y^2-y+1}{y^2+z^2+1}-\frac{x^4+\left(y^2+z\right) x^2+x+y^2 z+z-1}{\left(x^2+y^2+1\right) \left(x^2+z^2+1\right)}
$$
$$
\Rightarrow 1+\frac{y^3+1}{(y+1) \left(y^2+z^2+1\right)}+\frac{-x^4-x^2 y^2-x^2 z-x-y^2 z-z+1}{\left(x^2+y^2+1\right) \left(x^2+z^2+1\right)}
$$
Note that since the second term is always positive, this means first and second terms will always be greater than 1. Now let's concentrate on third term to see if that is negative and if so, whether it is big enough to make the whole term negative.
We are given that $xyz\le1$, let us consider the upper bound of this inequality and assume $xyz=1$, this implies, $z=\frac{1}{xz}$, which upon substitute in above expression, we get:
$$
\frac{x^2+y^2+1+x y \left(-x^2+\left(2 x^3 y-x^2 y\right)+x y^3+x y-y^2-1\right)}{\left(x^2+y^2+1\right) \left(x^2 \left(x^2+1\right) y^2+1\right)}
$$
How do we find out if the above term is positive or not? Let's try proof by contradiction. The denominator is always positive, so let's concentrate on the numerator and assume that the term is negative.
The approach I took was to consider only the nominator and try to solve the following equations simulataneously:
$$
2 x^4 y^2-x^3 y^2-x^3 y+x^2 y^4+x^2 y^2+x^2-x y^3-x y+y^2+1 < 0
$$
$$x\gt0$$
$$y\gt0$$
And since there does not exist any solution so that means that numerator will be non-negative.
Hence we can conclude that the term will be $\ge1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1192440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "39",
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How to prove that $\prod_{n=0}^\infty \frac{(4n+2)^2}{(4n+1)(4n+3)}=\sqrt{2}$ How to prove that $$\displaystyle\prod_{n=0}^\infty \frac{(4n+2)^2}{(4n+1)(4n+3)}=\sqrt{2}$$
Thanks in advances.
| Rewrite a partial product as
$$\prod_{n=0}^N \frac{(4 n+2)^3 (4 n+4)}{(4 n+1)(4 n+2)(4 n+3) (4 n+4)} = \frac{2^{3 N+3} (2 N+1)!!^3 4^{N+1} (N+1)!}{(4 N+4)!} $$
Use the fact that
$$(2 N+1)!! = \frac{(2 N+1)!}{2^N N!} $$
and
$$M! \approx M^M e^{-M} \sqrt{2 \pi M} \quad (M \to \infty)$$
The rest is careful bookkeeping, and making sure you use the fact that
$$\lim_{M \to \infty} \left ( 1+\frac1{M} \right )^M = e$$
the sought-after result follows.
| {
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"url": "https://math.stackexchange.com/questions/1193869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integral solutions to inverse trigonometric equation. The original question is this : I have to find number of ordered pairs of integral solutions to this :
$$\tan^{-1}{x} + \cos^{-1}{\frac{y}{\sqrt{1+y^2}}} = \sin^{-1}{\frac{3}{\sqrt{10}}} $$
I rearranged this as $\tan^{-1}(\frac{1}{y}) = tan^{-1}{3} - \tan^{-1}x $
And then $\frac{1}{y}=\frac{3-x}{1+3x}$.
Finally reduced it to $$(1+3x).(1-3y)=-10xy$$
The problem is $-10xy = -1.2.5.x.y$
So, do I really have to check $32$ cases? I know some of them can be discarded easily, but still, checking all will take a lot of time.
My question is, is there a different method to solve this? Or is there an easier way to check the cases?
Note : The time constraint of the question is ideally 3-4 minutes. This is a question for IIT-JEE which is an entrance exam for engineering in India.
| $\bf{My\; Solution::}$ Given $$\displaystyle \arctan x + \arccos\left(\frac{y}{\sqrt{1+y^2}}\right) = \arcsin \left(\frac{3}{\sqrt{10}}\right)$$
Now We can Write $$\displaystyle \arccos\left(\frac{y}{\sqrt{1+y^2}}\right) = \arctan\left(\frac{1}{y}\right)$$ and $$\displaystyle \arcsin\left(\frac{3}{\sqrt{10}}\right)=\arctan (3)$$
So Our equation is convert into $$\displaystyle \arctan (x)+\arctan\left(\frac{1}{y}\right)=\arctan(3)$$
So we can write it as $$\displaystyle \arctan \left(\frac{x+\frac{1}{y}}{1-\frac{x}{y}}\right)=\arctan(3)\Rightarrow \frac{xy+1}{y-x}=3$$
So our expression is $$\displaystyle y=\frac{3x+1}{3-x} = 3\left(\frac{x+\frac{1}{3}}{3-x}\right).$$
Now If Given $$\bf{x,y>0}$$ and $$\bf{x,y\in \mathbb{Z}^{+}}\;,$$ Then $$y=\left(\frac{x+\frac{1}{3}}{3-x}\right)>0$$
So We Get $\displaystyle -\frac{1}{3}<x<3$ and $x>0\;,$ Then we Get $0<x<3$
So Integer values of $x$ in This Interval are $x=1\;,2$
Now If $x=1\;,$ Then $\displaystyle y = \frac{3x+1}{3-x} = 2$ and If $x=2\;,$ Then $\displaystyle y=\frac{3x+1}{3-x} = 7$
So We get Only Two Positive Integer ordered pairs of $(x,y) = \left\{(1,2)\;\;,(2,7)\right\}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving $x^2 (x-y)^2 = x^2 - y^2$ using implicit differentiation I have a test tomorrow and I am doing homework to review and study. The problem is to differentiate
$$
x^2 (x-y)^2 = x^2 - y^2.
$$
I tried multiple times; however, every time I try I get the incorrect answer. It would be greatly appreciated if somebody could show me the steps on how to get the answer. (No full answer please; I still want to learn).
The answer according to the book is
$$
\frac{-2x^3 + 3x^2y - xy^2 + x}{x^y - x^3 + y}.
$$
| $\frac{d}{dx}[x^2(x-y)^2 = x^2 - y^2]$
$\Rightarrow 2x(x-y)^2 + x^2\cdot 2(x-y) \cdot (1-\frac{dy}{dx}) = 2x - 2y\cdot\frac{dy}{dx}$
$\Rightarrow 2x^3-4x^2y + 2xy^2 +2x -2x\frac{dy}{dx}-2y + 2y\frac{dy}{dx}= 2x -2y\frac{dy}{dx}$
$\Rightarrow 2x^3 - 4x^2y + 2xy^2 - 2y = (2x-4y)\frac{dy}{dx}$
$\frac{dy}{dx}=\frac{x^3 - 2x^2y+xy^2 -y}{x-2y}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Problem in trigonometry integral $$\int \frac{1}{(9-x^2)^{\frac{3}{2}}}dx$$
Let $x=3\sin (u)$
$dx=3\cos (u)$
So, $(9-x^2)^{\frac{3}{2}}=(9-9\sin ^2(u))^{\frac{3}{2}}$
$=27\cos ^3(u)$
$u=\sin ^-1(\frac{x}{3})$
My problem is how to substitute u into the integral to become $3\int \frac{\sec ^2(u)}{27}$du?
| Using your substitutions:
$$
\int \frac{dx}{(9 - x^2)^{\frac{3}{2}}} \;\; =\;\; \int \frac{3 \cos u du}{27(1- \sin^2u)^{\frac{3}{2}}} \;\; =\;\; \frac{1}{9} \int\frac{\cos u}{\cos^3u} du \;\; =\;\; \frac{1}{9} \int \frac{du}{\cos^2u}.
$$
Now use the fact that $\sec u = \frac{1}{\cos u}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Series of inverse function $A(s) = \sum_{k>0}a_ks^k$ and $A(s)+A(s)^3=s$.
I want calculate $a_5$. What ways to do it most efficiently?
| We can elaborate on the Lagrange inversion concept.
Suppose we have $$A(s) = \sum_{n\ge 0} a_n s^n$$
and $A(s)+A(s)^3=s$ and we seek $a_n.$
Using the Cauchy Residue Theorem to prepare for Lagrange inversion we have that
$$a_n =
\frac{1}{2\pi i}
\int_{|s|=\epsilon}
\frac{1}{s^{n+1}} A(s) \; ds.$$
Now put $A(s)=w$ so that $w+w^3 = s$ and
$$ds = 1 + 3w^2 \;dw.$$
This yields
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{(w+w^3)^{n+1}} w \times (1+3w^2) \; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{n}}
\frac{1}{(1+w^2)^{n+1}} \times (3w^2+3-2) \; dw.$$
The first component here is
$$3 \times\frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{n}}
\frac{1}{(1+w^2)^{n}} \; dw
= 3\times[w^{n-1}] \frac{1}{(1+w^2)^{n}}.$$
This is zero when $n$ is even
and when $n$ is odd it yields
$$3\times (-1)^{(n-1)/2}\times {(n-1)/2+n-1\choose n-1}
= 3 (-1)^{(n-1)/2} {3/2n-3/2\choose n-1}.$$
The second component is
$$-2\times \frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{n}}
\frac{1}{(1+w^2)^{n+1}} \; dw
= -2\times [w^{n-1}] \frac{1}{(1+w^2)^{n+1}}.$$
This is again zero when $n$ is even
and when $n$ is odd it yields
$$-2\times (-1)^{(n-1)/2} {(n-1)/2+n\choose n}
= -2 (-1)^{(n-1)/2} {3/2n-1/2\choose n}.$$
Combining these two yields
$$(-1)^{(n-1)/2}
\left(3 {3/2n-3/2\choose n-1}
-2 {3/2n-1/2\choose n}\right)$$
when $n$ is odd and zero otherwise.
If desired this can be simplified to
$$(-1)^{(n-1)/2}
\left(3-2 \frac{3/2n-1/2}{n}\right)
{3/2n-3/2\choose n-1}
= (-1)^{(n-1)/2}
\frac{1}{n}{3/2n-3/2\choose n-1}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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If Matrix $A^3 = 0$ then what will be $I+A+A^2$ Matrix $A$ is a square matrix such that $$A^3=0$$ then what would be value of $$I+A+A^2=?$$ such that $I$ is unit matrix of same order as $A$
I firstly supposed sum to be $X$ that is
$X=I+A+A^2$ then
$$AX=A+A^2+A^3$$
$$AX=A+A^2$$
and multiplying both sides with $A^{-1}$ we get
$$K=I+A$$ but my answer was incorrect , what is correct solution to this problem?
| If $A^3=0$, then $A^4=0$, $A^5=0$, and so on, so $I+A+A^2=I+A+A^2+A^3+A^4+\cdots$. The infinite sum $I+A+A^2+A^3+A^4+\cdots$ looks a lot like an infinite geometric series. There’s a formula for the sum of an infinite geometric series (of real numbers, if the ratio has absolute value less than $1$), $\frac{a}{1-r}=a+ar+ar^2+\cdots$ so maybe it would be interesting to play with that formula in this case:
$$I+A+A^2+A^3+A^4+\cdots \overset{?}{=}\frac{I}{I-A}.$$
Given that $I+A+A^2+A^3+A^4+\cdots = I+A+A^2$, this becomes
$$I+A+A^2 \overset{?}{=}\frac{I}{I-A}.$$
Even if you don’t believe this makes any sense, it might lead you to look at whether
$$\left(I+A+A^2\right)\left({I-A}\right) =I.$$
It does, so you’ve discovered that when $A^3=0$, $I+A+A^2=\left(I-A\right)^{-1}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solving $3\sin x - \cos x = 2$ for $x \in [0, 2\pi)$ Problem:
Solve $$3\sin x - \cos x = 2, \ \ \ x \in [0, 2\pi)$$
My attempt:
I am able to solve it using Weierstrass substitutions and a good bit of patience, but the problem was given at an exam at a level where such substitutions are not part of the curriculum.
I've tried to find ways to rewrite the equation by using $\sin x / \cos x = \tan x$ which I expect is the "desired" method, but for some reason, my algebra is failing me, and I can't seem to eliminate the $\sin$ and $\cos$ terms.
I've tried squaring both sides, but given the coefficient on $\sin x$, I can't find a way to use that identity either.
So far I end up with $$9\sin^2x - 6\sin x\cos x + \cos^2x = 4$$
Any help appreciated!
| Another approach is to write: $$A \sin(x) + B \cos(x)=C$$ and then divide by $$\sqrt{A^2 + B^2}$$ to get:
$$\frac{A}{\sqrt{A^2+B^2}}\sin(x) + \frac{B}{\sqrt{A^2+B^2}}\cos(x) = \frac{C}{\sqrt{A^2+B^2}}$$
Since $$\left(\frac{A}{\sqrt{A^2+B^2}}\right)^2 + \left(\frac{B}{\sqrt{A^2+B^2}}\right)^2 = 1$$ there is a $\psi$ such that $$\cos(\psi) = \frac{A}{\sqrt{A^2+B^2}} \text{ and } \sin(\psi) = \frac{B}{\sqrt{A^2+B^2}}.$$
Thus we have $$\cos(\psi)\sin(x) + \sin(\psi)\cos(x) = \frac{C}{\sqrt{A^2+B^2}}$$ and using the sum formula we find:
$$\sin(x+\psi) = \frac{C}{\sqrt{A^2+B^2}}.$$
In this case $A=3$ and $B=-1$ so $\sqrt{A^2+B^2} = \sqrt{10}$. Since $A > 0$ and $B < -1$, this means $\psi$ is in the third quadrant. Thus $\arctan(-1/3) = \psi$.
Finally this means $$\sin(x + \arctan(-1/3)) = 2/\sqrt{10}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $y'$ for $\ln(x+y)=\arctan(xy)$
Find $y'$ for $\ln(x+y)=\arctan(xy)$
Here is my attempt at a solution. Is this correct? Any hints or advice would be appreciated.
| Even if the result is correct, I guess that a systematic use of total differentiation makes things slightly easier.
Consider $$F=\log(x+y)-\tan^{-1}(xy)=0$$ So $F'_x~dx+F'_y~dy=0$. I this case, $$F'_x=\frac{1}{x+y}-\frac{y}{x^2 y^2+1}$$ $$F'_y=\frac{1}{x+y}-\frac{x}{x^2 y^2+1}$$ $$\frac{dy}{dx}=-\frac{F'_x}{F'_y}=\frac{\frac{y}{x^2 y^2+1}-\frac{1}{x+y}}{\frac{1}{x+y}-\frac{x}{x^2 y^2+1}}=\frac{y \left(x+y-x^2y\right)-1}{x^2 \left(y^2-1\right)-x y+1}=- \frac{(x+y)y-(1+x^2y^2)}{(x+y)x-(1+x^2y^2)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1205246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Problem in Differential Equation (How to proceed?) $$y\frac{dy}{dx}=\sqrt{1-y^2}$$, y=0 when x=0
$$\frac{\frac{dy}{dx}y}{\sqrt{1-y^2}}=1$$
$$\int\frac{\frac{dy}{dx}y}{\sqrt{1-y^2}}dx=\int 1dx$$
$$-\sqrt{1-y^2}=x+c_1$$
$$1-y^2=(x+c_1)^2$$
$$y^2=1-(x+c_1)^2$$
$$y=\pm \sqrt{1-(x+c_1)^2}$$
When x=0, y=0
$$0=\pm \sqrt{1-c^2_1}$$
| Square both sides:
$$1-y^2=(x+c_1)^2$$
rearrange for $y^2$:
$$y^2=1-(x+c_1)^2$$
take $\pm$ square root:
$$y=\pm\sqrt{1-(x+c_1)^2}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find $p,q$ s.t. $2q^2-p^2=\Box$ and $2p^2-q^2=\Box$ Problem. Find all integers $p,q$ such that $2q^2-p^2$ and $2p^2-q^2$ are perfect squares.
I think this is only true when $p=\pm q$ but I have not been able to prove it.
One approach I tried is letting (wlog) $q=p+t$ with $t>0$ to get $p^2+4pt+2t^2$ and $p^2-2pt-t^2$ are squares. Then completing the square we have $(p+2t)^2-2t^2$ and $(p-t)^2-2t^2$, but I didn't get much from here.
Another thing I tried is setting $2q^2-p^2=x^2$ to get $x^2+p^2=2q^2$, and similarly $y^2+q^2=2p^2$. Then write these as $(x-p)^2+(x+p)^2=(2q)^2$ and $(y-q)^2+(y+q)^2=(2p)^2$, then use the Pythagorean triples formula. But it got a bit messy after that.
| We may assume WLOG $p,q\in \mathbb{N}^+$ and $\gcd(p,q)=1$.
Lemma 1. If $\gcd(a,b)=1$ and $a^2+b^2=2c^2$, then:
$$ \{a,b\} = \{u^2-2uv-v^2,u^2+2uv-v^2\},\quad c=u^2+v^2,\quad \gcd(u,v)=1. $$
Proof. We have that $P=\left(\frac{a}{c},\frac{b}{c}\right)$ is a rational point on the circle $\Gamma:x^2+y^2=2$, hence the line joining $P$ with $(1,1)\in\Gamma$ has a rational slope. The vice-versa also holds: by Vieta's formulas, a line with rational slope through $(1,1)$ intersects $\Gamma$ in a rational point.
Lemma 1. gives that both $p$ and $q$ are the sum of two squares and are represented by the quadratic form $A^2-2B^2$. Moreover, both $3p^2$ and $3q^2$ are represented by the quadratic form $A^2+2B^2$.
Lemma 2. If $\gcd(a,b)=1$ and $a^2+2b^2 = 3c^2$, then:
$$ a=2u^2+4uv-v^2,\quad b=2u^2-2uv-v^2,\quad c=2u^2+v^2,\quad \gcd(u,v)=1.$$
Proof. We have that $P=\left(\frac{a}{c},\frac{b}{c}\right)$ is a rational point on the ellipse $\Gamma:x^2+2y^2=3$, hence the line joining $P$ with $(1,1)\in\Gamma$ has a rational slope. The vice-versa also holds: by Vieta's formulas, a line with rational slope through $(1,1)$ intersects $\Gamma$ in a rational point.
Lemma 2. gives that $p$ and $q$ are both represented by the quadratic form $A^2+2B^2$.
By putting all together, we have that both $p$ and $q$ are represented by the quadratic forms
$$ A^2+B^2,\quad A^2-2B^2,\quad A^2+2B^2.$$
Now it is possible to set a descent argument leading to $p=q=1$. Continues.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Mathematical Induction Proof Question dealing with integers How would you use mathematical induction to prove that
$1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + \dots + n \cdot (n + 1) \cdot (n + 2) = \frac{n(n + 1)(n + 2)(n + 3)}{4}$
I tried proving the base case of $n = 1$ but the left half is much larger than the right half after computing $n = 1$. Can someone please solve this problem and explain how and why the solution is correct? I've been stuck on it for hours.
| The base case for $n=1$ is actually quite easy, since:
$\frac{1\cdot 2\cdot 3 \cdot 4}{4} = 1\cdot 2 \cdot 3$
Assuming that it holds for $n$, we check $n+1$:
$1\cdot 2\cdot 3 + \cdots + n(n+1)(n+2) + (n+1)(n+2)(n+3) = \frac{n(n+1)(n+2)(n+3)}{4} + (n+1)(n+2)(n+3) = \frac{n(n+1)(n+2)(n+3) + 4(n+1)(n+2)(n+3)}{4} = \frac{(n+1)(n+2)(n+3)(n+4)}{4}$
which is what we set out to prove.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sqrt{2011} + \sqrt{2013} + \sqrt{2015} + \sqrt{2017} < 4 \sqrt{2014}$ How to prove that $\sqrt{2011} + \sqrt{2013} + \sqrt{2015} + \sqrt{2017} < 4\sqrt{2014}$ without using calculator?
| The function $f(x)=\sqrt{x}$ is strictly concave for $x>0$ so by Jensen's inequality, we have
$$
\frac{1}{4}f(2011)+\frac{1}{4}f(2013)+\frac{1}{4}f(2015)+\frac{1}{4}f(2017)<f\left(\frac{1}{4}(2011+2013+2015+2017)\right)
$$
which is $\sqrt{2014}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding a limit with two independent variables: $\lim_{(x,y)\to (0,0)}\frac{x^2y^2}{x^2+y^2}$ I must find the following limit: $$\lim_{(x,y)\to (0,0)}\frac{x^2y^2}{x^2+y^2}$$ Substituting $y=mx$ and $y=x^2$, I have found the limit to be $0$ both times, as $x \to 0$. I have thus assumed that the above limit is $0$, and will attempt to prove it. Let $\varepsilon>0$. We have that: $$\left\lvert\frac{x^2y^2}{x^2+y^2}\right\rvert=\frac{x^2y^2}{x^2+y^2}\leq\frac{(x^2+y^2)(x^2+y^2)}{x^2+y^2}=x^2+y^2$$ However, I must find $\delta>0$ such that $0<\sqrt{x^2+y^2}<\delta$, and I cannot see a way to obtain $\sqrt{x^2+y^2}$ in the above inequality to complete the proof. Am I mistaken in my process? Thank you.
| We have $$\frac{x^2y^2}{x^2+y^2}=\frac{1}{\dfrac{1}{x^2}+\dfrac{1}{y^2}}.$$ Note that each term in the denominator on the RHS is positive and independently goes to $\infty$ as $x$ and $y$ approach $0$. So the required limit is $0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Steady state solution for wave equation with gravity For the following wave equation and initial condition where G is a constant due to gravity, how would I go about finding the steady state solution:
$\frac{∂^2u}{∂t^2}= c \frac{∂^2u}{∂x^2}+ G, \quad 0\le x \le L,\: t\gt 0 $
$u(0, t) = 0 ,\quad t > 0,$
$u(L, t) = H ,\quad t > 0, $
$u(x, 0) = 0 ,\quad 0 < x < L$
$\frac{∂u}{∂t}(x, 0) = 0 ,\quad 0 < x < L$
I'm assuming that as t tends to infinity, $\frac{∂^2u}{∂t^2} = 0$ but I've never dealt with wave equations with gravity before so I'm not sure what to do with the G term.
| For the equation $u_{tt} = c^{2} u_{xx} + a$ with the conditions
\begin{align}
u(0, t) &= 0 \hspace{10mm} u(L, t) = H \\
u(x,0) &= 0 \hspace{10mm} u_{t}(x,0) = 0
\end{align}
consider a solution of the form $u(x,t) = F(x) G(t) + f(x)$, where $f(x)$ is the steady-state solution, for which
\begin{align}
u_{xx} &= F'' G + f'' \\
u_{tt} &= F G''
\end{align}
and the equation becomes
\begin{align}
F G'' = c^{2} F'' G + f'' + a.
\end{align}
Now, let $0 = c^{2} f'' + a$ in order to obtain the conditions $f(0) = 0$ and $f(L) = H$. The solution to this equation is
\begin{align}
f(x) = - \frac{a x^{2}}{2 c^{2}} + c_{1} x + c_{2}.
\end{align}
The condition $f(0) = 0$ yields $c_{2} = 0$. The condition $f(L) = H$ yields
$c_{1} = H/L + aL/2c^{2}$ for the general form
\begin{align}
f(x) = \frac{H \, x}{L} - \frac{a \, x(x - L)}{2 \, c^{2}}.
\end{align}
The remaining equations are derived from
\begin{align}
\frac{G''}{G} = - \mu^{2} = c^{2} \frac{F''}{F}
\end{align}
and are
\begin{align}
G'' + \mu^{2} G &= 0 \\
c^{2} F'' + \mu^{2} F &= 0
\end{align}
which have solutions
\begin{align}
G(t) &= A_{1} \cos(\mu t) + B_{1} \sin(\mu t) \\
F(x) &= A_{2} \cos\left( \frac{\mu}{c} \, x \right) + B_{2} \sin\left( \frac{\mu}{c} \, x \right)
\end{align}
The boundary conditions for $F$ are $F(0) = 0$, $F(L) = 0$ which yields
\begin{align}
A_{2} &= 0 \\
0 &= B_{2} \sin\left( \frac{\mu}{c} \, L \right)
\end{align}
The value for $\mu$ is obtained from $0 = \sin(\mu L/c)$ and is
\begin{align}
\mu = \frac{n \pi c}{L}
\end{align}
for $n$ and integer.
The conditions for $G$ are $G(0) = 0$, $G_{t}(0) = 0$, as stated in the problem, lead to the results
\begin{align}
A_{1} &= 0 \\
\mu B_{1} &= 0.
\end{align}
This yields the result for $G(t) = 0$ which says the only solution to this equation for these conditions is that of $f(x)$ which is independent of time and not an actual wave form solutions.
Bypassing the conditions provided the general solution of the form
\begin{align}
u(x,t) = \frac{H \, x}{L} - \frac{a \, x(x - L)}{2 \, c^{2}} + \sum_{n=1}^{\infty} \left( A_{n} \cos\left(\frac{n \pi c t}{L} \right) + B_{n} \sin\left( \frac{n \pi c t}{L} \right) \right) \, \sin\left( \frac{n \pi \, x}{L} \right)
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Denesting a square root: $\sqrt{7 + \sqrt{14}}$ Write:
$$\sqrt{7 + \sqrt{14}} = a + b\sqrt{c}$$
Form. $$7 + \sqrt{14} = a^2 + 2ab\sqrt{c} + b^2c$$
$a^2 + b^2c = 7$ and $2ab = 1$, and $c = 14$
But that doesn't seem right as $a, b,$ wont be integers?
| A simple way to see if a double radical $\sqrt{a\pm \sqrt{b}}$ can be denested is to check if $a^2-b$ is a perfect square. In this case we have:
$$
\sqrt{a\pm \sqrt{b}}=\sqrt{\dfrac{a+ \sqrt{a^2-b}}{2}}\pm\sqrt{\dfrac{a- \sqrt{a^2-b}}{2}}
$$
(you can easely verify this identity).
In this case $a^2-b=35$ is not a perfect square.
Note that if $\sqrt{a+\sqrt{b}}$ can be denested than $a^2-b$ must be positive since, by:
$$
\sqrt{a+\sqrt{b}})= \sqrt{p}+\sqrt{q}
$$
we have (squaring)
$$
a+\sqrt{b}=p+q+2\sqrt{pq}
$$
and for $a,b,q,p \in \mathbb{Q}$ this implies:
$$
p+q=a \qquad \land \qquad \sqrt{b}=2\sqrt{pq} \iff pq=b/4
$$
this means that $p$ and $q$ are solutions of the equation $ x^2-ax+b/4=0$ that has rational solutions only if $\Delta=a^2-b>0$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate:$\int_{0}^{\pi}\frac{x\sin2x\sin(\frac{\pi}{2}\cos x)}{2x-\pi}dx$ Here i have a problem
$$\int_{0}^{\pi}\frac{x\sin (2x)\sin(\frac{\pi}{2}\cos x)}{2x-\pi}dx$$
I have started applying the property of definite integral but stucked after going few steps. Please help.
| Let $A$ denote the value of the integral. First note that since $\sin(2x)/(2x - \pi) = -\sin(2x - \pi)/(2x - \pi) \to -1$ as $x \to \pi/2$, the integral is proper. By a change of variable $u = 2x - \pi$, we have
$$A = \int_{-\pi}^\pi \frac{u + \pi}{2u}\cdot \left[-\sin u \sin\left(\frac{\pi}{2}\cos\frac{\pi + u}{2}\right)\right]\, \frac{du}{2},$$
which is
$$-\frac{1}{4}\int_{-\pi}^\pi \sin u\sin\left(\frac{\pi}{2}\cos \frac{\pi + u}{2}\right)\, du - \frac{\pi}{4}\int_{-\pi}^\pi \sin u \sin\left(\frac{\pi}{2}\cos \frac{\pi + u}{2}\right)\, \frac{du}{u}.$$
Since $\cos (\pi + u)/2 = -\sin u/2$, this simplifies further to
$$\frac{1}{4}\int_{-\pi}^\pi \sin u \sin\left(\frac{\pi}{2}\sin \frac{u}{2}\right)\, du + \frac{\pi}{4}\int_{-\pi}^\pi \sin u \sin\left(\frac{\pi}{2}\sin \frac{u}{2}\right)\, \frac{du}{u}.$$
By symmetry the second integral evaluates to $0$, so we are left with
$$\frac{1}{4}\int_{-\pi}^\pi \sin u \sin\left(\frac{\pi}{2}\sin \frac{u}{2}\right)\, du = \frac{1}{2}\int_0^\pi \sin u \sin\left(\frac{\pi}{2}\sin \frac{u}{2}\right)\, du.$$
Letting $t = \sin \frac{u}{2}$ we have
$$\int_0^\pi \sin u \sin\left(\frac{\pi}{2}\sin \frac{u}{2}\right)\, du = \int_0^1 2t\sqrt{1 - t^2} \sin\left(\frac{\pi}{2}t\right)\, \frac{2\, dt}{\sqrt{1 - t^2}} = 4 \int_0^1 t\sin\left(\frac{\pi}{2} t\right)\, dt.$$
By the change of variable $r = \pi t/2$, we obtain
$$4\int_0^1 t\sin\left(\frac{\pi}{2}t\right)\, dt = \frac{16}{\pi^2}\int_0^{\pi/2} r \sin r\, dr = \frac{16}{\pi^2}(-r\cos r + \sin r)\bigg|_{r = 0}^{\pi/2} = \frac{16}{\pi^2}.$$
Thus
$$A = \frac{1}{2}\cdot \frac{16}{\pi^2} = \frac{8}{\pi^2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove, that every $p$ prime has a multiple, which is smaller than $\frac{p^4}{4}$, it can be written down as the sum of five integer's fourth power. I have a really interesting (and hard) number theory task:
Prove, that every $p$ prime has a multiple(not $0$), which is smaller than $\frac{p^4}{4}$, and it can be written down as the sum of five integer's fourth power.
I have tried it for smaller $p$ numbers and I couldn't even find an example for it. Is this statement true or not, if so, how should I prove it? :)
Thanks!
| As a first step, it is easy to see that the multiple $6p^2$ can be written as the sum of $12$ fourth powers. The reason is, that every integer of the form $6x^2$ can be written this way, because $x=a^2+b^2+c^2+d^2$ is the sum of four squares, and
\begin{align*}
6x^2 & =6(a^2+b^2+c^2+d^2)^2 \\
& =(a+b)^4+(a-b)^4+(a+c)^4+(a-c)^4+(a+d)^2+(a-d)^4 \\
& +(b+c)^4+(b-c)^4+(b+d)^4+(b-d)^4+(c+d)^4+(c-d)^4.
\end{align*}
To show that $rp=a^4+b^4+c^4+d^4+e^4$ for some $r<p^3/4$ one could perhaps use Minkowski's lattice theorem (which is used to prove that every integer is a sum of four squares etc.).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the area of trapezium $ABCD$ is a trapezium in which $AB||CD$. If $P$ is the point of intersection of diagonals $AC$ and $BD$ such that area of triangle $DPC=50cm^2$ and area of triangle $APB=32cm^2$.Then find area of trapezium $ABCD$.
I found that triangle APB and DPC are similar.The ratio of their sides is 4:5.The area of triangles DPC and CPB are equal.Please help me proceed.
| Consider the diagram below:
We are given that $\overline{AB} \parallel \overline{CD}$. If two parallel lines are cut by a transversal, then alternate interior angles are congruent. Thus, $\angle ABP \cong \angle CDP$ and $\angle CAB \cong \angle DCP$. Therefore, $\triangle ABP \sim \triangle CDP$.
We are also given that the area of $\triangle ABP$ is $32~\text{cm}^2$, while the area of $\triangle CDP$ is $50~\text{cm}$. Hence, the ratio of the areas of the similar triangles is
$$\frac{A(ABP)}{A(CDP)} = \frac{32~\text{cm}^2}{50~\text{cm}^2} = \frac{16}{25}$$
The ratio of the areas of similar triangles is the square of the ratio of corresponding sides. Thus,
$$\frac{|AB|}{|CD|} = \frac{|AP|}{|CP|} = \frac{|BP|}{|DP|} = \sqrt{\frac{16}{25}} = \frac{4}{5}$$
Let $E$ be the foot of the altitude from point $P$ to $\overline{AB}$; let $F$ be the foot of the altitude from point $P$ to $\overline{CD}$. Since all right angles are congruent, $\angle BEP \cong \angle DFP$. Since we have also established that $\angle EBP \cong \angle FDP$, $\triangle BEP \sim \triangle DFP$. Thus,
$$\frac{|EP|}{|FP|} = \frac{|BP|}{|DP|} = \frac{4}{5}$$
The area of trapezoid $ABCD$ is
\begin{align*}
A(ABCD) & = \frac{1}{2}(|AB| + |CD|)(|EP| + |FP|)\\
& = \frac{1}{2}\left(|AB| + \frac{5}{4}|AB|\right)\left(|EP| + \frac{5}{4}|EP|\right)\\
& = \frac{1}{2}\left(\frac{9}{4}|AB|\right)\left(\frac{9}{4}|EP|\right)\\
& = \frac{81}{16} \cdot \frac{1}{2} \cdot |AB| \cdot |EP|\\
& = \frac{81}{16} \cdot A(\triangle ABP)\\
& = \frac{81}{16}(32~\text{cm}^2)\\
& = 162~\text{cm}^2
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
polar coordinates and complex numbers Prove that $\dfrac{-1+i\sqrt{3}}{2}$ is a cube root of $1$.
I believe I must use polar coordinates to solve this. Perhaps $z=r\cos(\theta)+i\sin (\theta)$. Any help would be great!
| Hint:
$x$ is a cubic root of $y$ if and only if $x^3 = y$. So see what $x^3$ is. No need to use what you call polar coordinates.
EDIT: $$ \left({\dfrac{-1+i\sqrt{3}}{2}}\right)^3 = \dfrac{(-1)^3 + 3(-1)^2( i\sqrt{3} ) + 3(-1)( i\sqrt{3})^2 + (i\sqrt{3})^3}{8} = \dfrac{-1 + 3 \sqrt 3 \cdot i + 9 - 3 \sqrt 3 \cdot i }{8} = 1 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1220711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculation of real root values of $x$ in $\sqrt{x+1}-\sqrt{x-1}=\sqrt{4x-1}.$
Calculation of x real root values from $ y(x)=\sqrt{x+1}-\sqrt{x-1}-\sqrt{4x-1} $
$\bf{My\; Solution::}$ Here domain of equation is $\displaystyle x\geq 1$. So squaring both sides we get
$\displaystyle (x+1)+(x-1)-2\sqrt{x^2-1}=(4x-1)$.
$\displaystyle (1-2x)^2=4(x^2-1)\Rightarrow 1+4x^2-4x=4x^2-4\Rightarrow x=\frac{5}{4}.$
But when we put $\displaystyle x = \frac{5}{4}\;,$ We get $\displaystyle \frac{3}{2}-\frac{1}{2}=2\Rightarrow 1=2.$(False.)
So we get no solution.
My Question is : Can we solve above question by using comparision of expressions?
Something like $\sqrt{x+1}<\sqrt{x-1}+\sqrt{4x-1}\; \forall x\geq 1?$
If that way possible, please help me solve it. Thanks.
| Alternatively, isolating $ \sqrt{4x-1}$ and then multiplying both sides by $\sqrt{x+1}+\sqrt{x-1}$ makes it easier to conclude the LHS is smaller than the RHS: $$\require\cancel \cancel{x}+1-\cancel{x}+1=2<\sqrt{4x-1}\left(\sqrt{x+1}+\sqrt{x-1}\right).\tag{$\star$}$$
Once we check $(\star)$ holds for $x=1$ we are done, since clearly its RHS is increasing.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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integrate $dx/(a^2 \cos^2x+b^2 \sin^2x)^2$ Integrate $\dfrac{dx}{(a^2 \cos^2x+b^2 \sin^2x)^2}$.
I can go up to the residue formula like in this example here
but then I just can't end up with the result for when $n=2$. I keep messing up my math. Can someone please show me how to get to the answer when $n=2$? I think I've spent 3+ hours trying to get to it and no luck unfortunately. Thank you!
| By setting $x=\arctan t$ we have:
$$I=\int\frac{dx}{(a^2\sin^2 x+b^2\cos^2 x)^2}=\int\frac{1+t^2}{(a^2 t^2+b^2)^2}\,dt$$
and by setting $t=u\frac{b}{a}$ we get:
$$ I = \frac{1}{ab^3}\int\frac{1+\frac{b}{a} u^2}{(1+u^2)^2}\,du $$
so it is enough to check that:
$$\int \frac{du}{(1+u^2)^2} = \frac{1}{2}\left(\frac{u}{1+u^2}+\arctan u\right),\quad \int \frac{u^2\, du}{(1+u^2)^2} = \frac{1}{2}\left(\arctan u-\frac{u}{1+u^2}\right).$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to factorize $x^4+2x^2-x+2$? look at this:
$$x^4+2x^2-x+2$$
How to factorize it?
It should be changed to be in the form of standard factorization formulas.
| You can write
$$x^4+2x^2-x+2=(x^2+ax+b)(x^2+cx+d)$$
Then $a+c=0$, $bd=2$, $ad+bc=-1$ and $b+d+ac=2$. If you want integer values for $a,b,c,d$, you must have $b=\pm2$ or $\pm1$, and afters some checks,
$$x^4+2x^2-x+2=(x^2-x+1)(x^2+x+2)$$
Here is a more systematic approach. From the preceding equations, you have
$$c=-a$$
$$a(b-d)=1$$
$$b+d=2-ac=2+a^2$$
From $b-d=1/a$ and $b+d=2+a^2$, you get
$$b=\frac12(2+a^2+1/a)$$
$$d=\frac12(2+a^2-1/a)$$
Then, the equation $bd=2$ leads to
$$(2+a^2)^2-1/a^2=8$$
With $t=a^2$,
$$t(2+t)^2-8t-1=0$$
Or
$$t^3+4t^2+4t-8t-1=t^3+4t^2-4t-1=0$$
This one is a reciprocal equation, and it has the obvious solution $t=1$. Then you can choose $a=1$ and $c=-1$, then $b=2$ and $d=1$.
As DanielV remarks below, this is a general method to solve the quartic equation
$$x^4+Ax^3+Bx^2+Cx+D=0$$
First, let $x=z-A/4$, by expanding you can check that the coefficient of $z^3$ vanishes, and you have to solve
$$z^4+Uz^2+Vz+W=0$$
Then, as above, you write
$$z^4+Uz^2+Vz+W=(z^2+az+b)(z^2+cz+d)$$
And after expanding the RHS and equating coefficients with same power of $z$, you get $a+c=0$, $b+d+ac=U$, $ad+bc=V$, $bd=W$.
Then
$$c=-a$$
$$b+d=U+a^2$$
$$b-d=-V/a$$
Hence
$$b=\frac12(U+a^2-V/a)$$
$$d=\frac12(U+a^2+V/a)$$
Thus
$$4bd=4W=(U+a^2)^2-V^2/a^2$$
And with $t=a^2$, you have to solve
$$t(U+t)^2-4Wt-V^2=0$$
$$t^3+2Ut^2+(U^2-4W)t-V^2=0$$
This is a cubic equation, so you have proved that to solve a quartic, you just have to solve first a cubic, which is supposed to be easier (and there are indeed methods to solve the cubic).
Then, there remains two trinomials to solve, and you are done.
This method is also described here, along with others. Look here for solutions of the cubic equation.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all the complex numbers $z$ satisfying
Find all the complex numbers $z$ satisfying
$$
\bigg|\frac{1+z}{1-z}\bigg|=1
$$
So far I´ve done this:
$$
z=a+bi \\
\bigg|\frac{(1+a)+bi}{(1-a)-bi}\bigg|=1 \\
\mathrm{expression*conjugated} \\
\bigg|\frac{1+2bi-(a^2)-(b^2)}{1+2a+a^2+b^2}\bigg|=1
$$
| $\bf{My\; Solution}::$ Given $$\displaystyle \left|\frac{1+z}{1-z}\right| = 1\Rightarrow \left|1+z\right| = |1-z| = |z-1|.$$
Now Put $z=x+iy\;,$ Then $$|x+iy+1|=|x+iy-1|$$
So we get $$\sqrt{(x+1)^2+y^2} = \sqrt{(x-1)^2+y^2}\Rightarrow (x+1)^2=(x-1)^2$$.
So we get $x=0$. So all Complex no. are in the form of $z=iy\;,$ where $y\in \mathbb{R}.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof of sum results I was going through some of my notes when I found both these sums with their results
$$
x^0+x^1+x^2+x^3+... = \frac{1}{1-x}, |x|<1
$$
$$
0+1+2x+3x^2+4x^3+... = \frac{1}{(1-x)^2}
$$
I tried but I was unable to prove or confirm that these results are actually correct, could anyone please help me confirm whether these work or not?
| There are many excellent proofs given above.
Another approach would be to note that the summation (LHS) is actually a binomial expansion of the result (RHS). (NB - This would only be applicable if you have to prove that the result is true, instead of having to find the result.)
Hence
$$\begin{align}
\color{blue}{\frac 1{1-x}}&=(1-x)^{-1}\\
&=\sum_{r=0}^{\infty}{-1\choose r}(-x)^r\\
&=\color{lightgray}{1+(-1)(-x)+\frac{-1\cdot-2}{1\cdot 2}(-x)^2+\frac{-1\cdot -2\cdot -3}{1\cdot 2\cdot 3}(-x)^3\cdots}\\
&=\sum_{r=0}^{\infty}(-1)^r{r\choose r}(-x)^r
\qquad\text{using upper negation}\\
&=\color{green}{\sum_{r=0}^{\infty}{r\choose 0}x^r} =\sum_{r=0}^{\infty}x^r \\
&=1+x+x^2+x^3+\cdots+x^r+\cdots \\
&\phantom{=}\text{(Geometric Series)}
\end{align}$$
and
$$\begin{align}
\color{blue}{\frac 1{(1-x)^2}}&=(1-x)^{-2}\\
&=\sum_{r=0}^{\infty}{-2\choose r}(-x)^r\\
&=\color{lightgray}{1+(-2)(-x)+\frac{-2\cdot-3}{1\cdot 2}(-x)^2+
\frac{-2\cdot-3\cdot -4}{1\cdot 2\cdot 3}(-x)^3\cdots}\\
&=\sum_{r=0}^{\infty}(-1)^r{r+1\choose r}(-x)^r
\qquad\text{using upper negation}\\
&=\color{green}{\sum_{r=0}^{\infty}{r+1\choose 1}x^r}=\sum_{r=0}^{\infty}(r+1)x^r \\
&=1+2x+3x^2+4x^3+\cdots+(r+1)x^r+\cdots \\
&\phantom{=}\text{(Arithmetico-Geometric Series)}\\
\end{align}$$
Similarly,
$$\begin{align}
\color{blue}{\frac 1{(1-x)^3}}&=(1-x)^{-3}\\
&=\sum_{r=0}^{\infty}{-3\choose r}(-x)^r\\
&=\color{lightgray}{1+(-3)(-x)+\frac{-3\cdot-4}{1\cdot 2}(-x)^2+
\frac{-3\cdot-4\cdot -5}{1\cdot 2\cdot 3}(-x)^3\cdots}\\
&=\sum_{r=0}^{\infty}(-1)^r{r+2\choose r}(-x)^r
\qquad\text{using upper negation}\\
&=\color{green}{\sum_{r=0}^{\infty}{r+2\choose 2}x^r}=\sum_{r=0}^{\infty}\frac{(r+2)(r+1)}2x^r \\
&=1+3x+6x^2+10x^3+\cdots+\frac{(r+2)(r+1)}2 x^r+\cdots \\
\end{align}$$
This can also be derived by differentiating the result of the Arithmetico-Geometric Series.
By extension,
$$\color{blue}{\frac 1{(1-x)^{n+1}}}=\color{green}{\sum_{r=0}^{\infty}{r+n\choose n}x^r}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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No definite integrals of trigonometry I have big problems with the following integrals:
$$\int\frac{dx}{\sin^6 x+\cos^6x}$$
$$\int\frac{\sin^2x}{\sin x+2\cos x}dx$$
It isn't nice of me but I almost have no idea, yet I tried the trigonometric substitution $\;t=\tan\frac x2\;$ , but I obtained terrible things and can't do the rational function integral.
Perhaps there is exist some trigonometry equalities? I tried also
$$\frac1{\sin^6x+\cos^6x}=\frac{\sec^6x}{1+\tan^6x}=\frac13\frac{3\sec^2x\tan^2x}{1+\left(\tan^3\right)^2}\cdot\overbrace{\frac1{\sin^2x\cos^2x}}^{=\frac14\sin^22x}$$
and then doing parts with
$$u=\frac14\sin^22x\;\;:\;\;u'=\sin2x\cos2x=\frac12\sin4x\\{}\\v'=\frac13\frac{3\sec^2x\tan^2x}{1+\left(\tan^3\right)^2}\;\;:\;\;v=\arctan\tan^3x$$
But it is impossible to me doing the integral of $\;u'v\;$ .
Any help is greatly appreciated
| Let $$I = \int\frac{1}{\sin^6 x+\cos^6 x}dx = \int\frac{1}{(\sin^2 x+\cos^2 x)(\sin^4 x-\sin^2 x\cos^2 x+\cos^4 x)}dx$$
So $$\int\frac{1}{\sin^4 x-\sin^2 x\cos^2 x+\cos^4 x}dx = \int\frac{1}{\sin^2 x\cos^2 x(\tan^2 x-1+\cot^2 x)}dx$$
So $$\int\frac{\sin^2 x+\cos^2 x}{\sin^2 x\cos^2 x(\tan^2 x+\cot^2 x-1)}dx$$
So $$I = \int\frac{\sec^2 x+\csc^2 x}{(\tan x-\cot x)^2+1^2}dx$$
Now Put $(\tan x-\cot x) = t\;, $ Then $(\sec^2 x+\csc^2 x)dx = dt$
So $$I = \int\frac{1}{1+t^2}dt = \tan^{-1}(t)+\mathcal{C} = \tan^{-1}(\tan x-\cot x)+\mathcal{x}$$
| {
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"answer_id": 4
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Is there another way to solve $\int \frac{x}{\sqrt{2x-1}}dx$? $$\int \frac{x}{\sqrt{2x-1}}dx$$
Let $u=2x-1$
$du=2dx$
$$=\frac{1}{2}\int \frac{u+1}{2\sqrt{u}}du$$
$$=\frac{1}{2}\int (\frac{\sqrt{u}}{2}+\frac{1}{2\sqrt{u}})du$$
$$=\frac{1}{4}\int \sqrt{u}du+\frac{1}{4}\int\frac{1}{\sqrt{u}}du$$
$$=\frac{u^{\frac{3}{2}}}{6}+\frac{\sqrt{u}}{2}+c$$
$$=\frac{1}{6}(2x-1)^{\frac{3}{2}}+\frac{1}{2}\sqrt{2x-1}+c$$
$$=\frac{1}{3}(x+1)\sqrt{2x-1}+c$$
Is there another way except this?
| Here's a way with trig (though it is harder than what you wrote).
Draw a right triangle with hypotenuse $\sqrt{2x}$. Put a side of length $1$ adjacent to the acute angle $\theta$. Then the opposite side is $\sqrt{2x-1}$. So $x=\sec(\theta)^2/2$,$dx=\sec(\theta)^2\tan(\theta) d \theta$, and $\sqrt{2x-1}=\tan(\theta)$. So you have
$$\frac{1}{2} \int \sec(\theta)^4 d \theta = \frac{1}{2} \int \left ( \sec(\theta)^2 \tan(\theta)^2 + \sec(\theta)^2 \right ) d \theta = \frac{1}{2} \int (u^2+1) du$$
where $u=\tan(\theta)$. Of course, this boils down to $u=\sqrt{2x-1}$, but the steps in the middle were suggested by the trigonometry.
| {
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"url": "https://math.stackexchange.com/questions/1231507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Is $\mathbb{Q}(\sqrt{2},\sqrt[3]{5})=\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})$? I understand that $\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})\subseteq\mathbb{Q}(\sqrt{2},\sqrt[3]{5})$
But I am struggling to algebraically show that $\sqrt{2}$,$\sqrt[3]{5}\in\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})$ to conclude that $\mathbb{Q}(\sqrt{2},\sqrt[3]{5})\subseteq\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})$
| Let $u = \sqrt{2} + \sqrt[3]{5}$, we have
$$\begin{align}
(u - \sqrt{2})^3 = 5
\iff & u^3 - 3\sqrt{2}u^2 + 6u - 2\sqrt{2} = 5\\
\implies &
\begin{cases}
\sqrt{2} &= \frac{u^3 + 6u - 5}{3u^2 + 2} \in \mathbb{Q}(u)\\
\sqrt[3]{5} &= u - \sqrt{2} \in \mathbb{Q}(u)
\end{cases}
\end{align}
$$
As a result, $\mathbb{Q}(\sqrt{2},\sqrt[3]{5}) \subset \mathbb{Q}(u)$.
The other direction is trivial because
$$\sqrt{2}, \sqrt[3]{5} \in \mathbb{Q}(\sqrt{2},\sqrt[3]{5})
\implies u = \sqrt{2} + \sqrt[3]{5} \in \mathbb{Q}(\sqrt{2},\sqrt[3]{5})
\implies \mathbb{Q}(u) \subset \mathbb{Q}(\sqrt{2},\sqrt[3]{5})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1234930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 0
} |
Does the following series of transformations of inequalities holds? I am to calculate limit of the function $f(x,y)$ i am trying to apply squeeze theorem. Is the following series of transformations of this inequality correct? If not how to do this correctly? i.e. are those inequalities true or false? They should hold for all positive numbers.
$$f(x,y)=\frac{\ln \left(1+x^3+y^3 \right)}{\sqrt{x^2+y^2}} = \frac{\sqrt{x^2+y^2} \ln \left(1+x^3+y^3 \right)}{x^2+y^2} \le \frac{\sqrt{x^2+y^2} \left(x+y
\right)}{x^2+y^2} \le \frac{\sqrt{x^2+y^2} }{1}$$
| Note that your last inequality is equivalent to the inequality
\begin{equation}
x + y \leq x^2 + y^2 \quad \forall x, y > 0,
\end{equation}
which is false (e.g. take $x, y = \frac{1}{2}$).
Perhaps use the following:
*
*For any $\alpha > 0$, $\ln(1 + \alpha) \leq \alpha$.
*$x^3 + y^3 \leq (x^2 + y^2)(x + y) \leq (x^2 + y^2)^{3/2}$, provided $x, y > 0$.
You should find $f(x,y) \leq x^2 + y^2$, $\forall x, y > 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1236360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Linearly independent eigenvectors The matrix A=$\begin{pmatrix}0 & 1 &1 \\1& 0&1 \\ 1 & 1 & 0\end{pmatrix}$
has eigenvalues $\lambda=2$ with algebriac multiplicity $1$ and $\lambda=-1$ with multiplicity $2$
For $\lambda=-1$
$A+I=0 \implies$$\begin{pmatrix}1 & 1 &1 \\1& 1&1 \\ 1 & 1 & 1\end{pmatrix}\begin{pmatrix}x_1 \\x_2 \\ x_3 \end{pmatrix}=\begin{pmatrix}0 \\0 \\ 0 \end{pmatrix}$
so $x_1+x_2+x_3=0$
$3$ possible cases of eigenvectors are:
$\begin{pmatrix}-2 \\1 \\ 1 \end{pmatrix}$, $\begin{pmatrix}1 \\-2 \\ 1 \end{pmatrix}$, $\begin{pmatrix}1 \\-1 \\ 0 \end{pmatrix}$
which are all linearly independent.
So we can have more than $2$ distinct (non-linearly dependent) eigenvectors, even if the algebraic multiplicity is only $2$?
| Compute
$$\begin{pmatrix}-2 \\1 \\ 1 \end{pmatrix} - \begin{pmatrix}1 \\-2 \\ 1 \end{pmatrix} + 3\begin{pmatrix}1 \\-1 \\ 0 \end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1239112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Why does the Gaussian-Jordan elimination work when finding the inverse matrix? In order to find the inverse matrix $A^{-1}$, one can apply Gaussian-Jordan elimination to the augmented matrix
$$(A \mid I)$$
to obtain
$$(I \mid C),$$
where $C$ is indeed $A^{-1}$. However, I fail to see why this actually works, and reading this answer didn't really clear things up for me.
| Let me make it concrete in the following example(@pauly-b's answer is much better).
Suppose that 4 apples(any two are the same price) and 3 bananas(any two are the same price) in city A would cost 10 euros, and 3 apples and 2 bananas 7 euros. Let's calculate how much it would cost for 1 apple and 1 banana in A? We don't need to calculate the price for each fruit, the answer is just 10 minus 7.
$\pmatrix{4&3\\ 3&2} \cdot \pmatrix{a_A\\ b_A} = \pmatrix{10\\ 7}$
We apply Gaussian elimination by $R_1 = R_1 − R_2$
$\pmatrix{1&1\\ 3&2} \cdot \pmatrix{a_A\\ b_A} = \pmatrix{3\\ 7}$
Obviously, the above two equations are equivalent. By the same token we can perform more such operations to make the matrix on the LHS an identity one.
$\pmatrix{1&0\\ 0&1} \cdot \pmatrix{a_A\\ b_A} = \pmatrix{1\\ 2}$
And we get $a_A$ and $b_A$: 1 and 2. We denote the above by
$\left(\begin{array}{cc|c}
4 & 3 & 10 \\
3 & 2 & 7 \\
\end{array}\right)
\rightarrow
\left(\begin{array}{cc|c}
1 & 0 & 1 \\
0 & 1 & 2 \\
\end{array}\right)
$
Let's assume that we are in a different city B and the prices change, and we get
$\pmatrix{4&3\\ 3&2} \cdot \pmatrix{a_B\\ b_B} = \pmatrix{11\\ 8}$
Following the same process, we can get $a_B$ and $b_B$: 2 and 1.
We can combine the two and get
$\pmatrix{4&3\\ 3&2} \cdot \pmatrix{a_A&a_B\\ b_A&b_B} = \pmatrix{10&11\\ 7&8}$
Performing Gaussian elimination we get
$\pmatrix{a_A&a_B\\ b_A&b_B} = \pmatrix{1&2\\ 2&1}$
What if we change $\pmatrix{10&11\\ 7&8}$ in $\pmatrix{4&3\\ 3&2} \cdot \pmatrix{a_A&a_B\\ b_A&b_B} = \pmatrix{10&11\\ 7&8}$ to $\pmatrix{1&0\\ 0&1}$?
We get the following:
$\pmatrix{4&3\\ 3&2} \cdot \pmatrix{a_A&a_B\\ b_A&b_B} = \pmatrix{1&0\\ 0&1}$
After Gaussian elimination we get the inverse of $\pmatrix{4&3\\ 3&2}$: $\pmatrix{-2&3\\ 3&-4}$
And we can denote the above by
$\left(\begin{array}{cc|cc}
4 & 3 & 1&0 \\
3 & 2 & 0&1 \\
\end{array}\right)
\rightarrow
\left(\begin{array}{cc|cc}
1 & 0 & -2 & 3 \\
0 & 1 & 3 & -4 \\
\end{array}\right)
$
References:
*
*Gaussian Elimination
*Going from Gaussian elimination to finding the inverse matrix
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1240055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
"answer_id": 2
} |
Solve $x + y + z = xyz$ such that $x , y , z \neq0$ I came across the equation $x+y+z=xyz$ such that $x , y , z \neq 0$.
I set $x=1, y=2, z=3$ but how can i reach formal mathematical solution without " guessing " the answer ? Thank you
| Let $x = 1$, then $yz = y+z$.
Let $y = z$, then $y\cdot y = y^2 = 2y$.
$\frac{y^2}y = y = 2$.
Using substitution of $x=1$, $y=2$:
$1 + 2 + z = 2z$
$1 + 2 = 2z-z$
$z=3$
$1\times2\times3 = 1 + 2 + 3$
If $x = y = z$, $x$ to $z$ would equal $3^{\frac12}$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1240229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
Deriving the barycentric coordinates of a triangle's orthocenter, using the areal definition of such coordinates Wikipedia's "Altitude (triangle)" entry describes the barycentric coordinates of $\triangle ABC$'s orthocenter as $$(\tan A : \tan B : \tan C)$$
How would you prove this using solely the areal definition of barycentric coordinates?
Thank you.
|
Barycentric Coordinates
In the triangle above,
$$
\begin{align}
CE&=\tan(A)\,AE\tag{1a}\\[6pt]
DE&=\cot(B)\,AE\tag{1b}\\
\frac{DE}{CE}&=\cot(A)\cot(B)\tag{1c}
\end{align}
$$
Explanation:
$\text{(1a)}$: $\triangle AEC$ has right angle at $E$
$\text{(1b)}$: $\triangle DEA\cong\triangle BGA$ and $\triangle DEA$ has a right angle at $E$
$\text{(1c)}$: divide $\text{(1b)}$ by $\text{(1a)}$
The ratio of the altitudes of $\triangle ADB$ and $\triangle ACB$ is the ratio of their areas. Therefore, the area definition of the barycentric coordinates says that the $C$-barycentric coordinate for $D$ is $\cot(A)\cot(B)$. Thus, by similarity,
$$
\begin{align}
D
&=A\,\overbrace{\cot(B)\cot(C)}^\text{$A$-coordinate}+B\,\overbrace{\cot(C)\cot(A)}^\text{$B$-coordinate}+C\,\overbrace{\cot(A)\cot(B)}^\text{$C$-coordinate}\tag{2a}\\[6pt]
&=\frac{A\,\tan(A)+B\,\tan(B)+C\,\tan(C)}{\tan(A)\tan(B)\tan(C)}\tag{2b}\\
&=\frac{A\,\tan(A)+B\,\tan(B)+C\,\tan(C)}{\tan(A)+\tan(B)+\tan(C)}\tag{2c}
\end{align}
$$
Explanation:
$\text{(2a)}$: extend the coordinates by similarity
$\text{(2b)}$: multiply by $\frac{\tan(A)\tan(B)\tan(C)}{\tan(A)\tan(B)\tan(C)}$
$\text{(2c)}$: if $A+B+C=\pi$, then $\tan(A)+\tan(B)+\tan(C)=\tan(A)\tan(B)\tan(C)$
We could stop here because $\text{(2c)}$ gives the proportions $\tan(A):\tan(B):\tan(C)$, but we can compute those tangents given the sides.
Computing the Tangents
In the triangle above, the Pythagorean Theorem says
$$
\begin{align}
\overbrace{a^2-BE^2}^{CE^2}&=\overbrace{b^2-AE^2}^{CE^2}\tag{3a}\\
AE^2-BE^2&=b^2-a^2\tag{3b}
\end{align}
$$
Explanation:
$\text{(3a)}$: Pythagorean Theorem
$\text{(3b)}$: add $AE^2-a^2$
Furthermore,
$$
\begin{align}
AE
&=\frac12((AE-BE)+(AE+BE))\tag{4a}\\
&=\frac12\left(\frac{AE^2-BE^2}{AE+BE}+(AE+BE)\right)\tag{4b}\\
&=\frac12\left(\frac{b^2-a^2}c+c\right)\tag{4c}\\
&=\frac{b^2-a^2+c^2}{2c}\tag{4d}
\end{align}
$$
Explanation:
$\text{(4a)}$: algebra
$\text{(4b)}$: algebra
$\text{(4c)}$: apply $\text{(3b)}$ and $c=AE+BE$
$\text{(4d)}$: algebra
$$
\begin{align}
CE
&=\sqrt{b^2-AE^2}\tag{5a}\\
&=\frac{\sqrt{4b^2c^2-\left(b^2-a^2+c^2\right)^2}}{2c}\tag{5b}\\
&=\frac{\sqrt{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}}{2c}\tag{5c}
\end{align}
$$
Explanation:
$\text{(5a)}$: Pythagorean Theorem
$\text{(5b)}$: apply $\text{(4d)}$
$\text{(5c)}$: factor
Combining $\text{(4d)}$, $\text{(5c)}$, and $\text{(1a)}$, we get $\tan(A)$, and by similarity, $\tan(B)$ and $\tan(C)$:
$$
\begin{align}
\tan(A)&=\frac{\sqrt{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}}{b^2+c^2-a^2}\tag{6a}\\
\tan(B)&=\frac{\sqrt{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}}{c^2+a^2-b^2}\tag{6b}\\
\tan(C)&=\frac{\sqrt{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}}{a^2+b^2-c^2}\tag{6c}
\end{align}
$$
Computing $\boldsymbol{D}$
Applying $\text{(2c)}$ and $(6)$, we get
$$
D=\frac{\frac{A}{b^2+c^2-a^2}+\frac{B}{c^2+a^2-b^2}+\frac{C}{a^2+b^2-c^2}}{\frac1{b^2+c^2-a^2}+\frac1{c^2+a^2-b^2}+\frac1{a^2+b^2-c^2}}\tag7
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1242168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Find the closed-form for $\sum_{i=0}^n(-1)^i(\frac{1}{2})^i$ I start with simplifying:
$$\sum_{i=0}^n(-1)^i(\frac{1}{2})^i=\sum_{i=0}^n(-\frac{1}{2})^i$$
then:
$$S = 1 + (-\frac{1}{2}) + (-\frac{1}{2})^2 + ... +(-\frac{1}{2})^n$$
$$(-\frac{1}{2})S = (-\frac{1}{2}) + (-\frac{1}{2})^2 + ... +(-\frac{1}{2})^n+(-\frac{1}{2})^{n+1}$$
$$(-\frac{1}{2}-1)S = (-\frac{1}{2})^{n+1} - 1$$
am I on the right track?
| Here is the complete answer:
$$\sum_{i=0}^n(-1)^i(\frac{1}{2})^i=\sum_{i=0}^n(-\frac{1}{2})^i$$
then:
$$S = 1 + (-\frac{1}{2}) + (-\frac{1}{2})^2 + ... +(-\frac{1}{2})^n$$
$$(-\frac{1}{2})S = (-\frac{1}{2}) + (-\frac{1}{2})^2 + ... +(-\frac{1}{2})^n+(-\frac{1}{2})^{n+1}$$
$$(-\frac{1}{2}-1)S = (-\frac{1}{2})^{n+1} - 1$$
divide both sides by $(-\frac{1}{2}-1)$, we have:
$$S = \frac{(-\frac{1}{2})^{n+1} - 1}{(-\frac{1}{2}-1)}=\frac{(-\frac{1}{2})^{n+1} - 1}{(-1.5)}=\frac{1-(-\frac{1}{2})^{n+1}}{1.5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1242383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Algebraic Manipulation What is the best method to get the LHS equal to RHS?
$\frac{n(n+1)(n+2)}{3} + (n+1)(n+2) = \frac{(n+1)(n+2)(n+3)}{3}$
Thank you.
| Since $1=\frac 33$, one has $$\begin{align}\frac{n(n+1)(n+2)}{3}+\color{blue}{1}\cdot (n+1)(n+2)&=\frac{n(n+1)(n+2)}{3}+\frac{\color{blue}{3}(n+1)(n+2)}{\color{blue}3}\\&=\frac{n\color{red}{(n+1)(n+2)}}{\color{red}{3}}+\frac{3\color{red}{(n+1)(n+2)}}{\color{red}{3}}\\&=\color{red}{\frac{(n+1)(n+2)}{3}}(n+3)\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1243683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Is upper Hessenberg form preserved through similarity transformation Suppose $X$ is non-singular and $M$ is upper Hessenberg.
Is $X^{-1}MX$ also upper Hessenberg.
| No.
Let
$$M
= \begin{pmatrix}
1 & 4 & 2 & 3 \\
3 & 4 & 1 & 7 \\
0 & 2 & 3 & 4 \\
0 & 0 & 1 & 3 \\
\end{pmatrix}
~~~~~~
X =
\begin{pmatrix}
1 & 0 & 0 & 1 \\
1 & 2 & 3 & -1 \\
1 & 2 & -1 & 0 \\
1 & 1 & 1 & 1
\end{pmatrix}
$$
then
$$X^{-1} =
\begin{pmatrix}
\frac{11}{7} & \frac{3}{7} & \frac{1}{7} & -\frac{8}{7} \\
-\frac{6}{7} & -\frac{1}{7} & \frac{2}{7} & \frac{5}{7} \\
-\frac{1}{7} & \frac{1}{7} & -\frac{2}{7} & \frac{2}{7} \\
-\frac{4}{7} & -\frac{3}{7} & -\frac{1}{7} & \frac{8}{7}
\end{pmatrix}
$$
and we have
$$X^{-1}MX =
\begin{pmatrix}
\frac{132}{7} & \frac{190}{7} & \frac{188}{7} & -\frac{4}{7} \\
-\frac{37}{7} & -\frac{54}{7} & -\frac{72}{7} & \frac{13}{7} \\
-\frac{5}{7} & -\frac{16}{7} & -\frac{5}{7} & \frac{8}{7} \\
-\frac{62}{7} & -\frac{85}{7} & -\frac{97}{7} & \frac{4}{7}
\end{pmatrix}$$
which is not upper Hessenberg.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1245358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Radius of a circle Let $f:\mathbb{R} \rightarrow \mathbb{R}$ defined as $f(x)=x^2+1$. Find a real positive number $r$ such that the graph of $f$ intersects the circle about the origin and radius $r$ into two pieces , of which the one contains the origin and has an area equal to the value of the integral $ \displaystyle \mathcal{I}=\int_1^0 \frac{\ln x}{\sqrt{x}}\,{\rm d}x $.
Path to a solution: It is pretty much obvious that the radius $r$ will be greater than $1$, because if the radius is $1$ then the two graphs attach to each and if $r<1$ then they do not have an intersection.
That integral evaluates to $4$ with a direct calculation. And here comes the most difficult part of all:
The points at which the two graphs intersect are of the form:
$${\rm A}\left ( -\frac{1}{2}\sqrt{-6+2\sqrt{5+4r^2}}, \frac{1}{2}\left ( -1+\sqrt{5+4r^2} \right ) \right ), \;\; {\rm B}\left ( \frac{1}{2}\sqrt{-6+2\sqrt{5+4r^2}}, \frac{1}{2}\left ( -1+\sqrt{5+4r^2} \right ) \right )$$
Thus, the wanted area is:
$${\rm E}\left ( r \right )=\pi r^2 -\int_{-1/2 \cdot \sqrt{-6+2\sqrt{5+4r^2}}}^{1/2 \cdot \sqrt{-6+2\sqrt{5+4r^2}}} \left [ \sqrt{r^2-x^2}-x^2-1 \right ]\,{\rm d}x$$
All we have to do is to solve ${\rm E}(r)=4$. Basically I cannot solve it (at least not by hand).
Do you see an alternative?
P.S: Integrating with respect to $y$ does not have have much difference.
| For first, by setting $x=t^2$:
$$ \int_{1}^{0}\frac{\log x}{\sqrt{x}} = 4\int_{0}^{1}(-\log t)\,dt = 4.\tag{1} $$
Next, let we assume that the circle and the parabola intersect in two points having ordinate $y>1$ and abscissa $x=\pm\sqrt{y-1}$. The area of the parabolic segment defined by these two points is:
$$ \frac{2}{3}(y-1)\cdot 2\sqrt{y-1}\tag{2} $$
hence, since $r^2=y^2+y-1$, the area of the region inside the circle and below the parabola is given by:
$$A_y= \left(\pi+2\arctan\frac{y}{\sqrt{y-1}}\right)\frac{y^2+y-1}{2}+y\sqrt{y-1}-\frac{4}{3}(y-1)\sqrt{y-1}\tag{3}$$
but I do not see many ways to solve $A_y=4$ without the help of a CAS.
Numerically, I got that the right radius should be $r\approx 1.13622\ldots$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1247345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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