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How many squares in base $9$ consist of only ones How many perfect squares exist of the following form: $(1111111....11)_9$?
Let the required number equal be $(111111...1\space (n \space ones))_9$. This can written as:
$9^{n-1} + 9^{n-2} + \dots + 9^2 + 9 + 1 = \frac{9^{n} - 1}{8} = k^2$ for some k.
Since $\frac14$ is a perfect square, $\frac{9^n - 1}{2}$ must be a perfect square as well.
So, the problem boils down to finding all $n$ such that $\frac{9^n - 1}{2}$ is a perfect square.
$\frac{(3^n - 1)(3^n + 1)}{2}$
One such solution seems to be $n = 1$ which yields $4$. This was the only small solution I was able to find by hand. There are, however, others, which I found out when I ran a small computer program of $n$s upto $100$: $33,
34,
35,
36,
37,
38,
39,
40,
41,
42,
43,
44,
45,
46,
47,
48,
49,
50,
51,
52,
53,
54$ amongst others.
Is there any way to identify these by hand?
EDIT: According to the comments, the numerical findings are incorrect. Indeed, upon checking the program, I found that this was the case.
Now, it only remains to be proven that $n=1$ is the only solution. How do I do that? I would prefer an elementary proof.
| The following is to show there are not solutions with $n>1$ of the equation.
The equation $(9^n-1)/8=k^2$ may, after doubling the sides, be factored as
$$\frac{3^n-1}{2}\cdot \frac{3^n+1}{2}=2k^2.$$
The two factors on the left here are adjacent integers and so are coprime. This implies, from the right side of the equation, that there are $u,v$ such that one factor is $2u^2$ and the other is $v^2.$ One gets something by considering the factor which is of the form $2u^2.$
Consider the possibility that $(3^n-1)/2=2u^2.$ This implies that $3^n-1=2\cdot 2u^2=(2u)^2.$ If $n>1$ this is impossible, since by Catalan's "conjecture", (now a theorem; see below) the only nontrivial perfect powers differing by $1$ are $8$ and $9$.
On the other hand it may be that $(3^n+1)/2=2u^2$, which implies $3^n+1=(2u)^2$, and again we get if $n>1$ two perfect powers other than $8,9$ which differ by $1$ against Catalan's conjecture.
Link for the Catalan conjecture/theorem:
http://en.wikipedia.org/wiki/Catalan%27s_conjecture
ADDED: The OP has asked for a simpler proof, not using advanced theorems. I believe the following fits the bill, only based on simple congruence ideas.
For this we need to use the "parity fact" that $(3^n-1)/2$ is odd or even when $n$ is respectively odd or even. The latter can be shown by noting that from $3^{2k}-1=8r$ follows $(3^{2k}-1)/2=4r,$ and multiplying by $3$ and rearranging, $3^{2k+1}-1=3\cdot8r+2,$ so that $(3^{2k+1}-1)/2=3 \cdot 4r+1.$
Recall from the above that we have the two adjacent integers $(3^n-1)/2,\ (3^n+1)/2$ and know that one of them is of the form $2k^2$. If $n$ is odd this is $(3^n+1)/2$ from the parity fact, while if $n$ is even it is the factor $(3^n-1)/2$ which is of form $2k^2$.
Consider the case with $n$ odd and $3^n+1=(2k)^2.$ From this, we have
$$3^n=(2k-1)(2k+1),$$
where the factors on the right are coprime since they are odd and differ by $2$. By unique factorization, it follows that the smaller factor must be $1$ (otherwise we factor $3^n$ into nontrivial coprime factors). This leads to $k=1$ and then to $n=1$.
Consider on the other hand the case with $n$ even and $3^n-1=(2k)^2.$ Since $n \ge 1$ and $n$ is even, we have $n=2m$ with $m\ge 1$, and also
$$(3^m)^2-1=(2k)^2,$$
which is two squares differing by $1$, but the only such squares are $0,1$, whereas here one of the squares is at least $3^2$ since $m \ge 1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/522412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
integrate $ \frac {(x^3 + 36)} {(x^2 + 36)}$ I know I have to use long division first, but I don't really know how to do it in this case
$$\int \frac{x^3 + 36}{x^2 + 36}dx$$
| Using http://en.wikipedia.org/wiki/Partial_fraction_decomposition#Procedure
$$ \frac{x^3+36}{x^2+36} = \frac{x(x^2+36)- 36x +36 }{x^2+ 36} $$
$$ \frac{x^3+36}{x^2+36} = \frac{x(x^2+36)}{x^2+36} - \frac {36x}{x^2+36} + \frac{36}{x^2+36} $$
$$ \int \frac{x^3+36}{x^2+36}dx = \int \frac{x(x^2+36)}{x^2+36} dx -\int \frac {36x}{x^2+36}dx + \int\frac{36}{x^2+36}dx $$
$$ \int \frac{x^3+36}{x^2+36}dx = \int x*dx - 18\int\frac{2x}{x^2+36}dx + \int\frac{36}{x^2+36}dx $$
I hope now you can proceed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/523209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding minimum $\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}$ I would appreciate if somebody could help me with the following problem
Q. Finding maximum minimum
$$\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}(\text{where} ~~x,y,z>0)$$
| $$\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}(\text{where} ~~x,y,z>0)$$
Consider the following :
$a , \frac{1}{a}$ we know that $A.M. \geq G.M.$
$\therefore \frac{a+ \frac{1}{a}}{2} \geq \sqrt{a . \frac{1}{a}}$
$\Rightarrow \frac{a^2+1}{2a} \geq 1 $
$\Rightarrow a^2 + 1 \geq 2a $
$\Rightarrow (a-1)^2 \geq 0$
$\Rightarrow a \geq 1$
$\therefore $ the expression has only minimum value which is 1 and no maximum value.
The expression $$\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}(\text{where} ~~x,y,z>0)$$
has minimum value of 1 + 1 +1 +1 +1 +1+1 = 6
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/524086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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} |
Derivative of $\lceil 1/x \rceil$ I'm looking for the derivative $\frac{d}{dx}\lceil 1/x \rceil$.
I would like to find a real number $1<x \le y$ satisfying the minimum of $\left\lceil \frac{y}{x} \right\rceil x$, when $y$ is a fixed value $>0$.
Is the ceiling function a problem ?
| $\large x > 0:$
\begin{align}
\left\lceil x\right\rceil
&=
\Theta\left(x\right)\Theta\left(1 - x\right)
+
2\Theta\left(x - 1\right)\Theta\left(2 - x\right)
+
3\Theta\left(x - 2\right)\Theta\left(3 - x\right) + \cdots
\\[3mm]&=
\sum_{n = 0}^{\infty}\left(n + 1\right)\Theta\left(x - n\right)
\Theta\left(n + 1 - x\right)
\end{align}
\begin{align}
{{\rm d}\left\lceil x\right\rceil \over {\rm d}x}
&=
\sum_{n = 0}^{\infty}\left(n + 1\right)\left[%
\delta\left(x - n\right)
\Theta\left(n + 1 - x\right)
-
\Theta\left(x - n\right)
\delta\left(n + 1 - x\right)
\right]
\\[3mm]&=
\sum_{n = 0}^{\infty}\left(n + 1\right)\left[%
\delta\left(x - n\right)
\Theta\left(1\right)
-
\Theta\left(1\right)
\delta\left(n + 1 - x\right)
\right]
\\[3mm]&=
\sum_{n = 0}^{\infty}\left(n + 1\right)
\delta\left(x - n\right)
-
\sum_{n = 1}^{\infty}
n\,\delta\left(n - x\right)
=
\sum_{n = 0}^{\infty}\delta\left(x - n\right)
\end{align}
\begin{align}
{{\rm d}\left\lceil x\right\rceil \over {\rm d}x}
&=
\sum_{n = 0}^{\infty}\delta\left(x - n\right)\,,
\qquad
\left\lceil x \right\rceil = 1\ \mbox{when}\ 0 < x < 1
\end{align}
\begin{align}
{{\rm d}\left\lceil 1/x\right\rceil \over {\rm d}x}
&=
\sum_{n = 0}^{\infty}\delta\left({1 \over x} - n\right)\,
\left(-\,{1 \over x^{2}}\right)
=
\sum_{n = 1}^{\infty}
{\delta\left(x - 1/n\right) \over \left\vert -1/x^{2}\right\vert}\,
\left(-\,{1 \over x^{2}}\right)
\end{align}
\begin{align}
\color{#ff0000}{\large{{\rm d}\left\lceil 1/x\right\rceil \over {\rm d}x}}
&\color{#000000}{\large\ =\ }
\color{#ff0000}{\large -\sum_{n = 1}^{\infty}\delta\left(x - {1 \over n}\right)}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/526412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Find the limit -> Infinity with radicals First guess to multiply by $x^{-1.4}$ so the radical in numerator with $x^7$ becomes 1 and other stuff becomes 0. But then denominator becomes $-\infty$. What is the right approach?
$$ \lim_{x \to \infty} \frac{\sqrt[5]{x^7+3} - \sqrt[4]{2x^3 - 1}}{\sqrt[8]{x^7 + x^2 + 1} - x} $$
| For starters: Note, that the denominator has as highest power of $x$ the $x^1$, so we multiply by $1 = \frac {1/x}{1/x}$, giving
$$\frac{\sqrt[5]{x^7+3} - \sqrt[4]{2x^3 - 1}}{\sqrt[8]{x^7 + x^2 + 1} - x} =
\frac{\sqrt[5]{x^2 + 3x^{-5}} - \sqrt[4]{2x^{-1} - x^{-4}}}{\sqrt[8]{x^{-1} + x^{-6} + x^{-8}} - 1}
$$
Now the denominator converges to $-1$ for $x \to \infty$, and the numerator diverges to $\infty$, hence the fraction diverges to $-\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/526976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Is my calculation of $\frac{\partial}{\partial x}\frac{x+y}{\sqrt{y^2-x^2}}$ correct? Is my calculation of the partial derivative (with respect to $x$) of the function $$f(x,y)=\frac{x+y}{\sqrt{y^2-x^2}}$$ correct?
$$f'_x=\frac{\sqrt{y^2-x^2}-(x+y)(\frac{-x}{\sqrt{y^2-x^2}})}{|y^2-x^2|}\\=\frac{\sqrt{y^2-x^2}+\frac{x^2+xy}{\sqrt{y^2-x^2}}}{|y^2-x^2|}\\=\frac{\frac{|y^2-x^2|+x^2+xy}{\sqrt{y^2-x^2}}}{|y^2-x^2|}\\=|y^2-x^2|\frac{|y^2-x^2|+x^2+xy}{\sqrt{y^2-x^2}}$$
| All correct except the last line
$$f'_x=\frac{\sqrt{y^2-x^2}-(x+y)(\frac{-x}{\sqrt{y^2-x^2}})}{|y^2-x^2|}\\=\frac{\sqrt{y^2-x^2}+\frac{x^2+xy}{\sqrt{y^2-x^2}}}{|y^2-x^2|}\\=\frac{\frac{|y^2-x^2|+x^2+xy}{\sqrt{y^2-x^2}}}{|y^2-x^2|}\\=\frac{|y^2-x^2|+x^2+xy}{(y^2-x^2)^{3/2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/527571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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Prove $_4F_3(1/8,3/8,5/8,7/8;1/4,1/2,3/4;1/2)=\frac{\sqrt{2-\sqrt2+\sqrt{2-\sqrt2}}+\sqrt{2+\sqrt2+\sqrt{2+\sqrt2}}}{2\,\sqrt2}$ How do I prove
$$_4F_3\left(\frac18,\frac38,\frac58,\frac78;\ \frac14,\frac12,\frac34;\ \frac12\right)\stackrel?=\frac{\sqrt{2-\sqrt2+\sqrt{2-\sqrt2\phantom{|}}}+\sqrt{2+\sqrt2+\sqrt{2+\sqrt2\phantom{|}}}}{2\,\sqrt2}$$
I encountered this conjecture when working with Fourier transforms of Kelvin functions.
| Some related values, based on Kirill's answer, given by radicals:
$$\begin{align}
{_4F_3}\left(\begin{array}c\tfrac18,\tfrac38,\tfrac58,\tfrac78\\\tfrac14,\tfrac12,\tfrac34\end{array}\middle|\,\frac59\right) &= \frac{1}{4}\sqrt{9+\sqrt{30}+2\sqrt{15+3\sqrt{30}}}\\
{_4F_3}\left(\begin{array}c\tfrac18,\tfrac38,\tfrac58,\tfrac78\\\tfrac14,\tfrac12,\tfrac34\end{array}\middle|\,\frac34\right) &= \frac{1}{2}\sqrt{4+\sqrt{3}+\sqrt{6+4\sqrt{3}}}\\
{_4F_3}\left(\begin{array}c\tfrac18,\tfrac38,\tfrac58,\tfrac78\\\tfrac14,\tfrac12,\tfrac34\end{array}\middle|\,\frac89\right) &= \frac{1}{2}\sqrt{9+\sqrt{6}+\sqrt{12+6\sqrt{6}}}\\
{_4F_3}\left(\begin{array}c\tfrac18,\tfrac38,\tfrac58,\tfrac78\\\tfrac14,\tfrac12,\tfrac34\end{array}\middle|\,9\right) &= \frac{1}{4}\sqrt{\frac{1}{2}-i\sqrt{2}+\sqrt{-6-6i\sqrt{2}}}
\end{align}
$$
It seems that in general for any $z$ rational values the solution is in the form of $\sqrt{x}$ where $x$ is a root of an $8$th-degree polynomial with integer coefficients.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/532260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Checking the equality of an equation(Trigonometery) The equation
$\displaystyle\cos^2\theta=\frac{(x+y)^2}{4xy}$ is only possible when
What i need to compare is x and y and derive a relation between them
i.)x
ii.)x=-y
iii.)x>y
iv.)x=y
Pls Answer
| $$\cos^2\theta=\frac{(x+y)^2}{4xy}\iff \sec^2\theta=\frac{4xy}{(x+y)^2}$$
$$\implies \tan^2\theta=\frac{4xy}{(x+y)^2}-1=-\frac{(x+y)^2-4xy}{(x+y)^2}=-\left(\frac{x-y}{x+y}\right)^2$$
$$\implies \tan^2\theta+\left(\frac{x-y}{x+y}\right)^2=0$$
As $\theta$ is real, $\tan^2\theta\ge0$ and $x,y$ are real $\displaystyle\left(\frac{x-y}{x+y}\right)^2\ge0$
Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
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probability regarding three people throwing a die There are 3 players, A, B, C, taking turns to roll a die in the order ABCABC....
What's the probability of A is the first to throw a 6, B is the second, and C is the third?
The answer said it's 216/1001, but I always got 125/1001
The way I did it was:
Let $X_1$= number of rolls till A throw a 6
$X_2$= number of rolls till B throw a 6
$X_3$= number of rolls till C throw a 6
$P(X_1=i)=\frac{1}{6} (\frac{5}{6})^{i-1}$
$P(X_2=j)=\frac{1}{6} (\frac{5}{6})^{j-1}$
$P(X_3=k)=\frac{1}{6} (\frac{5}{6})^{k-1}$
therefore, $P$(A is the first to throw a 6, B is the second, and C is the third)
$=P(X_1 < X_2 < X_3) $
= $\sum_{i<j<k}^{ } (\frac{1}{6})^3 (\frac{5}{6} )^{(i-1)+(j-1)+(k-1)} $
$= (\frac{1}{6})^3 \sum_{i<j}^{ } (\frac{5}{6} )^{(i-1)+(j-1)} \sum_{k=j+1}^{\infty } (\frac{5}{6} )^{k-1} $
$ = (\frac{1}{6})^3 \sum_{i<j}^{ } (\frac{5}{6} )^{(i-1)+(j-1)} \frac{(\frac{5}{6})^{j}} {1-\frac{5}{6}} $
$ =(\frac{1}{6})^2 \sum_{i<j}^{ } (\frac{5}{6} )^{(i-1)+(j-1)+j}$
$ =(\frac{1}{6})^2 \sum_{i}^{ }(\frac{5}{6} )^{(i-1)}\sum_{j=i+1}^{\infty} (\frac{5}{6} )^{2j-1}$
$ =(\frac{1}{6})^2 \sum_{i}^{ }(\frac{5}{6} )^{(i-1)} \frac{(\frac{5}{6})^{2i+1}} {1-\frac{5}{6}} $
$ =(\frac{1}{6}) \sum_{i}^{ }(\frac{5}{6} )^{(i-1)+2i+1}$
$ =(\frac{1}{6}) \sum_{i=1}^{\infty }(\frac{5}{6} )^{3i}$
$=\frac{125}{546}$
Could anyone help me check where went wrong? Thanks so much!
| Since they go in the order A,B,C, you want $\sum_{i\le j\le k}$.
Detail: We want $$\sum_{0\le i\le j\le k}(5/6)^i(5/6)^j(5/6)^k(1/6)^3.$$
Summing from $k=j$ to $\infty$, we get
$$\sum_{0\le i\le j}(1/6)^2(5/6)^i (5/6)^{2j}.$$
The summation with respect to $j$ yields
$$\sum_{0\le i} (5/6)^{3i},$$
which is $\frac{6^3}{(11)(91)}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
gaussian and mean curvatures I am trying to review, and learn about how to compute and gaussian and mean curvature. Given $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$, how can I compute the gaussian and mean curvatures?
This is what I have so far,
$$K(u, v) = \frac{a^2 b^2 c^2}{[c^2 \sin^2(v) (a^2 \sin^2(u)+b^2 \cos^2(u))+a^2 b^2 \cos^2(v)]^2}$$
Please help me out.
| The parametric equations of an ellipsoid can be written as
\begin{align}
x &=a \cos u \sin v\\
y &=b \sin u \sin v\\
z &=c \cos v
\end{align} for $u \in [0,2\pi)$ and $v \in [0,\pi]$.
In this parametrization, the coefficients of the first fundamental form are
\begin{align}
E &= (b^2\cos^2u+a^2\sin^2u)\sin^2v\\
F &= (b^2-a^2)\cos u\sin u\cos v\sin v\\
G &= (a^2\cos^2u+b^2\sin^2u)\cos^2v+c^2\sin^2v
\end{align}
and of the second fundamental form are
\begin{align}
e &= \frac{abc\sin^2v}{\sqrt{a^2b^2\cos^2v+c^2(b^2\cos^2u+a^2\sin^2u)\sin^2v}}\\
f &= 0\\
g &= \frac{abc}{\sqrt{a^2b^2\cos^2v+c^2(b^2\cos^2u+a^2\sin^2u)\sin^2v}}.
\end{align}
In this parametrization, the Gaussian curvature is
$$
K(u,v)=\frac{eg-f^2}{EG-F^2}=\frac{a^2b^2c^2}{[a^2b^2\cos^2v+c^2(b^2\cos^2u+a^2\sin^2u)\sin^2v]^2 }
$$
and the mean curvature is
\begin{align}
H(u,v) &=\frac{eG-2fF+gE}{2(EG-F^2)}\\
&=\frac{abc[3(a^2+b^2)+2c^2+(a^2+b^2-2c^2)\cos(2v)-2(a^2-b^2)\cos(2u)\sin^2v]}{8[a^2b^2\cos^2v+c^2(b^2\cos^2u+a^2\sin^2u)\sin^2v]^{3/2}}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/540710",
"timestamp": "2023-03-29T00:00:00",
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How prove this $I=\int_{0}^{\infty}\frac{1}{x}\ln{\left(\frac{1+x}{1-x}\right)^2}dx=\pi^2$ Prove this
$$I=\int_{0}^{\infty}\dfrac{1}{x}\ln{\left(\dfrac{1+x}{1-x}\right)^2}dx=\pi^2$$
My try: let
$$I=\int_{0}^{\infty}\dfrac{2\ln{(1+x)}}{x}-\dfrac{2\ln{|(1-x)|}}{x}dx$$
| Allow me to present an approach that uses dilogarithms. Split the integral up into 2 and substitute $x \mapsto \dfrac{1}{x}$ for the second integral. This yields
\begin{align}
\int^\infty_0\frac{1}{x}\ln\left(\frac{1+x}{1-x}\right)^2dx
&=\int^1_0\frac{1}{x}\ln\left(\frac{1+x}{1-x}\right)^2dx+\int^\infty_1\frac{1}{x}\ln\left(\frac{1+x}{1-x}\right)^2dx\\
&=4\int^1_0\frac{\ln(1+x)-\ln(1-x)}{x}dx\\
&=4\left(\operatorname{Li}_2(1)-\operatorname{Li}_2(-1)\right)\\
&=4\left(\frac{\pi^2}{6}+\frac{\pi^2}{12}\right)\\
&=\pi^2
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/541751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 3
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Are these ideals the same? I have already proved that $(X^3-Y^3,X^2Y-X)\subseteq(X^2-Y,X-Y^2)$ since the elements $X^3-Y^3$ and $X^2Y-X $ can be written as a linear combination of $(X^2-Y,X-Y^2)$.
However, I can't write $X^2-Y,X-Y^2$ as a combination of $(X^3-Y^3,X^2Y-X)$. How can I prove that $(X^2-Y,X-Y^2) \subseteq(X^3-Y^3,X^2Y-X)$ or they are not the same ideals?
Thank you very much.
| If $(X^2-Y,X-Y^2) \subseteq(X^3-Y^3,X^2Y-X)$, then $X^2-Y\in (X^3-Y^3,X^2Y-X)$. There exist two polynomials $f,g$ such that $X^2-Y=(X^3-Y^3)f(X,Y)+(X^2Y-X)g(X,Y)$. Send $X$ to $0$ in the equation and find $Y=Y^3f(0,Y)$ $\Leftrightarrow$ $1=Y^2f(0,Y)$. Now send $Y$ to $0$ and obtain $1=0$, a contradiction.
Conclusion: $(X^3-Y^3,X^2Y-X)\subsetneq(X^2-Y,X-Y^2)$.
| {
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Factorize the polynomial $x^3+y^3+z^3-3xyz$ I want to factorize the polynomial $x^3+y^3+z^3-3xyz$. Using Mathematica I find that it equals $(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$. But how can I factorize it by hand?
| Use Newton's identities:
$p_3=e_1 p_2 - e_2 p_1 + 3e_3$ and so $p_3-3e_3 =e_1 p_2 - e_2 p_1 = p_1(p_2-e_2)$ as required.
Here
$p_1= x+y+z = e_1$
$p_2= x^2+y^2+z^2$
$p_3= x^3+y^3+z^3$
$e_2 = xy + xz + yz$
$e_3 = xyz$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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True or false: $a^2+b^2+c^2 +2abc+1\geq 2(ab+bc+ca)$ Is this inequality true?
$a^2+b^2+c^2 +2abc+1\ge2(ab+bc+ca)$, where $a,b,c\gt0$.
Can you find a counterexample for this or not?
| Let $a=x^3$, $b=y^3$ and $c=z^3$.
Hence, by AM-GM and Schur we obtain:
$$a^2+b^2+c^2+2abc+1\geq a^2+b^2+c^2+3\sqrt{a^2b^2c^2}=\sum_{cyc}(x^6+x^2y^2z^2)\geq$$
$$\geq\sum_{cyc}(x^4y^2+y^4x^2)\geq2\sum_{cyc}x^3y^3=2(ab+ac+bc)$$
and we are done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Integral $\int_0^\infty\frac{1}{\sqrt[3]{x}}\left(1+\log\frac{1+e^{x-1}}{1+e^x}\right)dx$ Is it possible to evaluate this integral in a closed form?
$$\int_0^\infty\frac{1}{\sqrt[3]{x}}\left(1+\log\frac{1+e^{x-1}}{1+e^x}\right)dx$$
| We first remark the following identity:
\begin{align*}
\int_{0}^{\infty} \frac{x^{s-1}}{ze^{x} - 1} \, dx
&= \int_{0}^{\infty} \frac{z^{-1}x^{s-1}e^{-x}}{1 - z^{-1}e^{-x}} \, dx \\
&= \sum_{n=1}^{\infty} z^{-n} \int_{0}^{\infty} x^{s-1} e^{-nx} \, dx \\
&= \Gamma(s) \sum_{n=1}^{\infty} \frac{z^{-n}}{n^{s}}
= \Gamma(s)\mathrm{Li}_{s}(z^{-1})
\end{align*}
which initially holds for $|z| > 1$, and then extends holomorphically for $z^{-1} \notin (1, \infty]$ since both sides define holomorphic functions on this region. Now, by integrating by parts,
\begin{align*}
\int_{0}^{\infty} \frac{1}{\sqrt[3]{x}} \left(1+\log\frac{1+e^{x-1}}{1+e^{x}}\right) \, dx
&= \frac{3}{2} \int_{0}^{\infty} x^{2/3} \left( \frac{1}{(-1)e^{x} - 1} - \frac{1}{(-e^{-1})e^{x} - 1} \right) \, dx \\
&= \frac{3}{2} \Gamma\left(\frac{5}{3} \right) \left\{ \mathrm{Li}_{5/3}(-1) - \mathrm{Li}_{5/3}(-e) \right\} \\
&= -\frac{3}{2} \Gamma\left(\frac{5}{3} \right) \left\{ (1 - 2^{-2/3})\zeta(5/3) + \mathrm{Li}_{5/3}(-e) \right\}.
\end{align*}
I can hardly believe that $\mathrm{Li}_{5/3}(-e)$ can be simplified further.
| {
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evaluating $\sum_{n=1}^\infty \frac{n^2}{3^n}$ I am trying to compute the sum $$\sum_{n=1}^\infty \frac{n^2}{3^n}.$$
I would prefer a nice method without differentiation, but if differentiation makes it easier, then that's fine.
Can anyone help me?
Thanks.
| Here is a non-calculus method which works in these cases:
Put
$$S = \frac{1}{3} + \frac{4}{9} + \frac{9}{27} + \frac{16}{81} + \dots$$
then $$\frac{1}{3}S = \frac{1}{9} + \frac{4}{27} + \frac{9}{81} + \dots$$
subtract:
$$\frac{2}{3}S = \frac{1}{3} + \frac{3}{9} + \frac{5}{27} + \dots$$
This has simplified the problem (since the numerators are now linear instead of quadratic), and if you repeat the process, you get the answer, using the sum of a geometric progression.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving that $\frac{1}{2} \le \frac{1}{2^n+1} + \frac{1}{2^n+2} + ... + \frac{1}{2^n + 2^n}$ I'm having trouble proving that:
$$\frac{1}{2} \le \frac{1}{2^n+1} + \frac{1}{2^n+2} + ... + \frac{1}{2^n + 2^n}$$
Edit: The next step is actually a mistake. I've put up a comment to the accepted answer explaining why it's a mistake.
I've tried reducing every single fraction on the right side to $\frac{1}{2^n + 2^n}$, but then I get result:
$$\frac{1}{2} \le \frac{1-\frac{1}{2^n}}{2}$$
which is not true.
I also tried to put this harmonic sequence in Wolfram Alpha and it tells me it equals $\ln(2)$, which is greater than $\frac{1}{2}$.
| For any $\;1\le k\le 2^n\;$ we have that
$$\frac1{2^n+k}\ge\frac1{2^n+2^n}=\frac1{2^{n+1}}\;,\;\;\text{so:}$$
$$\underbrace{\frac1{2^n+1}+\ldots+\frac1{2^n+2^n}}_{2^n\;\text{summands}}\ge\frac{2^n}{2^{n+1}}=\frac12$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Sum of series of ratios I need to calculate
$$\lim_{n \to \infty}\frac{1}{n}\left(\frac{1}{1+n} + \frac{1}{2+n} + \frac{1}{3+n}+\cdots+\frac{3n}{4n}\right).$$
It can be written as
$$\lim_{n \to \infty}\frac{1}{n^2}\left(\frac{1}{\frac{1}{n}+1} + \frac{1}{\frac{2}{n}+1} + \frac{1}{\frac{3}{n}+1}+\cdots+\frac{1}{\frac{4n}{3n}}\right).$$
Now I need to write it as $\frac{1}{n}\sum_{?}^{?}{?}$ and ultimately in definite integral form so that I can evalute the sum of this limit.
| Hint:
$$
\frac{1}{n}\sum_{k=1}^{mn} \frac{1}{\frac{k}{n} + 1} \approx \int_0^m \frac{1}{x+1}dx
$$
| {
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$ 7^{50} \cdot 4^{102} ≡ x \pmod {110} $ The way I would solve this would be: $$ (7^3)^{15} \cdot 7^5 \cdot (4^4)^{25} \cdot 4^2 $$ and take it from there, but I know that this is most likely in an inefficient way. Does anyone have more efficient methods?
| Do prime factorization on $110$. It's $110 = 5 \cdot 2 \cdot 11$.
Now work with the prime factors as moduli.
$$7^{50} \cdot 4^{102} \equiv 0 \pmod 2 \text{ trivial, as 2 \mid 4}$$
Then apply Fermat's Little Theorem so we have:
$$7^4 \equiv 1 \pmod 5 \implies 7^{50} \equiv 49 \equiv 4 \pmod 5$$
$$4^4 \equiv 1 \pmod 5 \implies 4^{102} \equiv 4^2 \equiv 1 \pmod 5$$
Now multiply them and we have:
$$7^{50} \cdot 4^{102} \equiv 4 \cdot 1 \equiv 4 \pmod 5$$
Now repeat the method with the last factor:
$$7^{10} \equiv 1 \pmod {11} \implies 7^{50} \equiv 1 \pmod {11}$$
$$4^{10} \equiv 1 \pmod {11} \implies 4^{102} \equiv 4^2 \equiv 16 \equiv 5 \pmod {11}$$
Multiply them and we have:
$$7^{50} \cdot 4^{102} \equiv 1 \cdot 5 \equiv 5 \pmod {11}$$
Now just apply CRT to the three congruence relation to get the final answer.
| {
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"timestamp": "2023-03-29T00:00:00",
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Combinatorics error correcting code (56) * (36)^4 * (-55) + (35)(67)(-14)^2 mod 17. Find the least non-negative residue of the expression module the given n.
First I just want to make sure I understand what the question wants. To do that, I give a simple case 15 mod 7. 1 is what we looking for right?
Then get back to the question, what is the method for effecticiently finding the answer other than just use calculator to multiply everything and then find the multiple of 7 to minus that member. By the way no use calculator
| The calculation can be done without a calculator. Since $56=3\cdot 17+5$, we have $56\equiv 5\pmod{17}$. We have $36\equiv 2\pmod{17}$, so $36^4\equiv 2^4\equiv -1\pmod{17}$. But $(-1)(-55)=55\equiv 4\pmod{17}$. Thus the first part of our expression is $\equiv 20\equiv 3\pmod{17}$.
A similar calculation shows that the second part is $\equiv 8\pmod{17}$.
For $35\equiv 1\pmod{17}$, and $67\equiv 16\pmod{17}$. We have $-14\equiv 3\pmod{17}$, so $(-14)^2\equiv 9\pmod{17}$. It remains to calculate $(16)(9)$ modulo $17$. One can work directly, multiplying to get $144$, and finding the remainder. But here is a useful trick: $16\equiv -1\pmod{17}$, so $(16)(9)\equiv (-1)(9)=-9\equiv 8\pmod{17}$.
Now add. The sum is $\equiv 11\pmod{17}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $\lim\limits_{x\to0}\left(\frac{1}{\log(x+\sqrt{1+x^2})}-\frac{1}{\log(1+x)}\right) =-\frac12$ How can I prove that
$$\lim_{x\to0}\left(\frac{1}{\log(x+\sqrt{1+x^2})}-\frac{1}{\log(1+x)}\right)=-\frac{1}{2}$$
| The answer can be achieved using simple limit theorems as follows
$\displaystyle \begin{aligned}L &= \lim_{x \to 0}\left(\frac{1}{\log(x + \sqrt{1 + x^{2}})} - \frac{1}{\log(1 + x)}\right)\\
&= \lim_{x \to 0}\frac{\log(1 + x) - \log(x + \sqrt{1 + x^{2}})}{\log(x + \sqrt{1 + x^{2}})\log(1 + x)}\\
&= \lim_{x \to 0}\frac{\log(1 + x) - \log(x + \sqrt{1 + x^{2}})}{\log(1 - 1 + x + \sqrt{1 + x^{2}})\log(1 + x)}\\
&= \lim_{x \to 0}\dfrac{\log(1 + x) - \log(x + \sqrt{1 + x^{2}})}{{\displaystyle \left(x + \sqrt{1 + x^{2}} - 1\right)\dfrac{\log\left\{1 + \left(x + \sqrt{1 + x^{2}} - 1\right)\right\}}{x + \sqrt{1 + x^{2}} - 1}\cdot x\cdot\dfrac{\log(1 + x)}{x}}}\\
&= \lim_{x \to 0}\dfrac{\log(1 + x) - \log(x + \sqrt{1 + x^{2}})}{{\displaystyle \left(x + \sqrt{1 + x^{2}} - 1\right)1\cdot x\cdot 1}}\\
&= \lim_{x \to 0}\dfrac{\log(1 + x) - \log(x + \sqrt{1 + x^{2}})}{x \left(x + \sqrt{1 + x^{2}} - 1\right)}\\
&= \lim_{x \to 0}\dfrac{\log(1 + x) - \log\left(\dfrac{\sqrt{1 + x^{2}} - x}{\sqrt{1 + x^{2}} - x}\cdot\left(x + \sqrt{1 + x^{2}}\right)\right)}{x \left(x + \sqrt{1 + x^{2}} - 1\right)}\\
&= \lim_{x \to 0}\dfrac{\log(1 + x) + \log\left(\sqrt{1 + x^{2}} - x\right)}{x \left(x + \sqrt{1 + x^{2}} - 1\right)}\\
&= \lim_{x \to 0}\dfrac{\log\left\{(1 + x)\left(\sqrt{1 + x^{2}} - x\right)\right\}}{x \left(x + \sqrt{1 + x^{2}} - 1\right)}\\
&= \lim_{x \to 0}\dfrac{\log\left\{\sqrt{1 + x^{2}} - x + x\sqrt{1 + x^{2}} - x^{2}\right\}}{x \left(x + \sqrt{1 + x^{2}} - 1\right)}\\
&= \lim_{x \to 0}\dfrac{\log\left\{1 - 1 + \sqrt{1 + x^{2}} - x + x\sqrt{1 + x^{2}} - x^{2}\right\}}{x \left(x + \sqrt{1 + x^{2}} - 1\right)}\\
&= \lim_{x \to 0}\dfrac{\log\left\{1 + \left(\sqrt{1 + x^{2}} - x + x\sqrt{1 + x^{2}} - x^{2} - 1\right)\right\}\left(\sqrt{1 + x^{2}} - x + x\sqrt{1 + x^{2}} - x^{2} - 1\right)}{\left(\sqrt{1 + x^{2}} - x + x\sqrt{1 + x^{2}} - x^{2} - 1\right)x \left(x + \sqrt{1 + x^{2}} - 1\right)}\\
&= \lim_{x \to 0}\dfrac{\left(\sqrt{1 + x^{2}} - x + x\sqrt{1 + x^{2}} - x^{2} - 1\right)}{x \left(x + \sqrt{1 + x^{2}} - 1\right)}\\
&= \lim_{x \to 0}\dfrac{\left(x^{2} + x\sqrt{1 + x^{2}} - x + \sqrt{1 + x^{2}} - 2x^{2} - 1\right)}{x \left(x + \sqrt{1 + x^{2}} - 1\right)}\\
&= \lim_{x \to 0} 1 + \dfrac{\sqrt{1 + x^{2}} - 2x^{2} - 1}{x \left(x + \sqrt{1 + x^{2}} - 1\right)}\\
&= 1 + \lim_{x \to 0}\dfrac{1 + x^{2} - (2x^{2} + 1)^{2}}{x \left(1 + x^{2} - (1 - x)^{2}\right)}\frac{\sqrt{1 + x^{2}} + 1 - x}{\sqrt{1 + x^{2}} + 2x^{2} + 1}\\
&= 1 + \lim_{x \to 0}\dfrac{-3x^{2} - 4x^{4}}{2x^{2}}\cdot\frac{2}{2} = 1 - \frac{3}{2} = -\frac{1}{2}\end{aligned}$
The above solution looks long because of detailed steps involving algebraic simplification but involves nothing more than logarithmic limit $\lim\limits_{y \to 0}\dfrac{\log(1 + y)}{y} = 1$ and rationalization to handle $\sqrt{1 + x^{2}}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How do you factor $(10x+24)^2-x^4$? I tried expanding then decomposition but couldn't find a common factor between two terms
| Whoa there, Ladies and Gentlemen! We ain't done yet! As pointed out by yanbo and Hugus in their answers, a good first step is to observe that the identitiy
$a^2 - b^2 = (a + b)(a - b) \tag{1}$
may be directly applied to
$(10x+24)^2-x^4 \tag{2}$
and we obtain
$(10x+24)^2-x^4 = (10x + 24 + x^2)(10x + 24 - x^2); \tag{3}$
but $10x + 24 + x^2$ and $10x + 24 - x^2$ can be factored as well! We have
$10x + 24 + x^2 = (x + 4)(x + 6), \tag{4}$
and
$10x + 24 - x^2 = -(x + 2)(x - 12) = (x + 2)(12 - x). \tag{5}$
Apparently $(10x+24)^2-x^4$ may be completely reduced over $\Bbb Z[x]$ to linear factors:
$(10x+24)^2-x^4 = -(x + 4)(x + 6)(x + 2)(x - 12), \tag{6}$
and its roots are $-6, -4, -2 \, \text{and} \, 12$. Cute, the way $10x + 24$ is always a perfect square for any of these integers; but, then again, that's factoring in $\Bbb Z[x]$ for you.
Hope this helps! Cheerio,
and as always,
Fiat Lux!!!
| {
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How prove this inequality $\frac{x^2y}{z}+\frac{y^2z}{x}+\frac{z^2x}{y}\ge x^2+y^2+z^2$ let $x\ge y\ge z\ge 0$,show that
$$\dfrac{x^2y}{z}+\dfrac{y^2z}{x}+\dfrac{z^2x}{y}\ge x^2+y^2+z^2$$
my try:
$$\Longleftrightarrow x^3y^2+y^3z^2+z^3x^2\ge xyz(x^2+y^2+z^2)$$
| let $$E(a,b,c)=a^3b^2+b^3c^2+c^3a^2-abc(a^2+b^2+c^2)$$
then
$$2E(a,b,c)=\sum a^3(b-c)^2-\sum a^2(b^3-c^3)=\sum a^2(b-c)^2(a+c-b)\ge 0$$
because
$$\sum a^2(b^3-c^3)=\sum a^2(b-c)^3$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Integration of $\int\frac{x^2-1}{\sqrt{x^4+1}} \, dx$ Integration of $\displaystyle \int\frac{x^2-1}{\sqrt{x^4+1}} \,dx$
$\bf{My\; Try}$:: Let $x^2=\tan \theta$ and $\displaystyle 2xdx = \sec^2 \theta \, d\theta\Rightarrow dx = \frac{\sec^2 \theta}{2\sqrt{\tan \theta}} \, d\theta$
$$
\begin{align}
& = \int\frac{\tan \theta - 1}{\sec \theta}\cdot \frac{\sec^2 \theta}{2\sqrt{\tan \theta}} \, d\theta = \frac{1}{2}\int \frac{\left(\tan \theta - 1\right)\cdot \sec \theta}{\sqrt{\tan \theta}} \, d\theta \\
& = \frac{1}{2}\int \left(\sqrt{\tan \theta}-\sqrt{\cot \theta}\right)\cdot \sec \theta \, d\theta
\end{align}
$$
Now i did not understand how can i solve it
Help me
Thanks
| For any real number of $x$ ,
When $|x|\leq1$ ,
$\int\dfrac{x^2-1}{\sqrt{x^4+1}}dx$
$=\int(x^2-1)\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{4n}}{4^n(n!)^2}dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{4n+2}}{4^n(n!)^2}dx-\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{4n}}{4^n(n!)^2}dx$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{4n+3}}{4^n(n!)^2(4n+3)}-\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{4n+1}}{4^n(n!)^2(4n+1)}+C$
When $|x|\geq1$ ,
$\int\dfrac{x^2-1}{\sqrt{x^4+1}}dx$
$=\int\dfrac{x^2-1}{x^2\sqrt{1+\dfrac{1}{x^4}}}dx$
$=\int\left(1-\dfrac{1}{x^2}\right)\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{-4n}}{4^n(n!)^2}dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{-4n}}{4^n(n!)^2}dx-\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{-4n-2}}{4^n(n!)^2}dx$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{1-4n}}{4^n(n!)^2(1-4n)}-\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{-4n-1}}{4^n(n!)^2(-4n-1)}+C$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{4^n(n!)^2(4n+1)x^{4n+1}}-\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{4^n(n!)^2(4n-1)x^{4n-1}}+C$
| {
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How to find $x^4+y^4+z^4$ from equation? Please help me.
There are equations: $x+y+z=3, x^2+y^2+z^2=5$ and $x^3+y^3+z^3=7$. The question:
what is the result of $x^4+y^4+z^4$?
Ive tried to merge the equation and result in desperado. :(
Please explain with simple math as I'm only a junior high school student. Thx a lot
| use this
since
$$x^2+y^2+z^2=(x+y+z)^2-2xy-2yz-2xz\Longrightarrow xy+yz+xz=2$$
$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)=(x+y+z)^3-3(xy+yz+xz)(x+y+z)$$
so
$$7-3xyz=27-18\Longrightarrow xyz=-\dfrac{2}{3}$$
use
$$x^4+y^4+z^4=(x+y+z)(x^3+y^3+z^3)-(xy+yz+xz)(x^2+y^2+z^2)+xyz(x+y+z)$$
| {
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Ordinary generating function for $\binom{3n}{n}$ The ordinary generating function for the central binomial coefficients, that is, $$\displaystyle \sum_{n=0}^{\infty} \binom{2n}{n} x^{n} = \frac{1}{\sqrt{1-4x}} \, , \quad |x| < \frac{1}{4},$$
can be derived by using the duplication formula for the gamma function and the generalized binomial theorem.
But what about the ordinary generating function for $ \displaystyle \binom{3n}{n}$?
According to Wolfram Alpha, $$ \sum_{n=0}^{\infty} \binom{3n}{n} x^{n} = \frac{2\cos \left(\frac{1}{3} \arcsin \left(\frac{3 \sqrt{3x}}{2} \right)\right)}{\sqrt{4-27x}} \, , \quad |x| < \frac{4}{27}. $$
Any suggestions on how to prove this?
EDIT:
Approaching this problem using the fact that $$ \text{Res} \Big[ \frac{(1+z)^{3n}}{z^{n+1}},0 \Big] = \binom{3n}{n},$$
I get $$ \sum_{n=0}^{\infty} \binom{3n}{n} x^{n} = -\frac{1}{2 \pi i x} \int_{C} \frac{dz}{z^{3}+3z^{2}+3z - \frac{z}{x}+1},$$
where $C$ is a circle centered at $z=0$ such that every point on the circle satisfies $ \displaystyle\Big|\frac{x(1+z)^{3}}{z} \Big| < 1$.
Evaluating that contour integral would appear to be quite difficult.
|
Proposition : $$ f(z,a) = \int_{0}^{\infty} \dfrac{x^z}{x^2 +2ax +1} \mathrm{d}x = \dfrac{\pi}{\sin \pi z} \dfrac{\sin((1-z) \cos^{-1}(a))}{\sin(\cos^{-1}(a))}$$
Proof : Note that,
$$ \sum_{n=0}^{\infty}{{U}_{n}(a) {(-x)}^{n}} = \dfrac{1}{x^2 +2ax+1} $$
where $ U_{n} (x) $ is the Chebyshev Polynomial of the second kind.
$$ \implies \sum_{n=0}^{\infty}{{U}_{n}(a) \Gamma(n+1) \frac{{(-x)}^{n}}{n!}} = \frac{1}{x^2 +2ax +1} $$
Using Ramanujan Master Theorem, we have,
$$ f(z,a) = \dfrac{\pi}{\sin \pi z} U_{-s} (a) $$
$$ = \dfrac{\pi}{\sin \pi z} \dfrac{\sin((1-z) \cos^{-1}(a))}{\sin(\cos^{-1}(a))} \quad \square $$
Now, using Gamma Triplication Formula,
$$ (3n)! \; = \; \Gamma(3n+1) \; = \; \dfrac{1}{2\pi} 3^{2n+\frac12} \Gamma \left(n+\dfrac13\right)\Gamma \left(n+\dfrac23\right)\Gamma(n+1) $$
and hence
$$ {3n \choose n} \; = \; \frac{3^{3n+\frac12}}{2\pi} \times \frac{\Gamma \left(n+\dfrac13\right)\Gamma \left(n+\dfrac23 \right)}{\Gamma(2n+1)} \; = \;\frac{3^{2n+\frac12}}{2\pi} \operatorname{B} \left(n+\dfrac23,n+\dfrac13\right) $$
Thus,
$$ \begin{array}{rcl}\displaystyle \text{S} \; =\; \sum_{n=0}^\infty {3n \choose n}x^n & = & \displaystyle\frac{\sqrt{3}}{2\pi}\sum_{n=0}^\infty 3^{2n} x^n \int_0^1 u^{n-\frac13}(1-u)^{n-\frac23}\,du \\ & = & \displaystyle \frac{\sqrt{3}}{2\pi} \int_0^1 \frac{u^{-\frac13}(1-u)^{-\frac23}}{1 - 27xu(1-u)}\,du \; =\; \frac{\sqrt{3}}{2\pi}\int_0^1 \frac{u^{-\frac13}(1-u)^{-\frac23}}{1 - 4a^2\, u(1-u)}\,du \end{array} $$
where $a = \dfrac{3}{2} \sqrt{3x} $. Substituting $u = \sin^2\theta$ and then $\tan\theta = t$, we have,
$$\text{S} = \dfrac{\sqrt{3}}{2 \pi} \int_0^\infty \frac{t^{\frac{1}{3}}(1 + t^2)}{(t^2 + 2at + 1)(t^2 - 2at + 1)} \mathrm{d}t $$
Using Partial Fraction and the Proposition, we have,
$$\text{S} = \dfrac{\sqrt{3}}{2\pi} \left[ \dfrac{1}{2} f \left( \dfrac{1}{3} , -a \right) + \dfrac{1}{2} f\left( \dfrac{7}{3} , -a \right) - \dfrac{1}{4a} f\left( \dfrac{4}{3} , -a \right) - \dfrac{1}{4a} f\left( \dfrac{10}{3} , -a \right) + \dfrac{1}{2} f \left( \dfrac{1}{3} , a \right) + \dfrac{1}{2} f\left( \dfrac{7}{3} , a \right) + \dfrac{1}{4a} f\left( \dfrac{4}{3} , a \right) + \dfrac{1}{4a} f\left( \dfrac{10}{3} , a \right)\right] $$
After simplification, we have,
$$ \text{S} = \dfrac{2\cos \left(\frac{1}{3} \sin^{-1} \left(\dfrac{3\sqrt{3x}}{2} \right)\right)}{\sqrt{4-27x}} \quad \square $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/557230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 4,
"answer_id": 3
} |
Imaginary numbers and polynomials question I have a task which I do not understand:
Consider $w = \frac{z}{z^2+1}$ where $z = x + iy$, $y \not= 0$ and $z^2 + 1 \not= 0$.
Given that Im $w = 0$, show that $| z | = 1$.
Partial solution (thanks to @ABC and @aranya):
If I substitute $z$ with $x + iy$ then we have $w = \frac{x + iy}{x^2+2xyi-y^2+1}$ or written slightly different $w= \frac{x + iy}{x^2-y^2+1+2xyi}$ or $w= \frac{x + iy}{(x^2-y^2+1)+(2xyi)}$.
In this format we can multiply with complex conjugate $w= \frac{x + iy}{(x^2-y^2+1)+(2xyi)}\cdot\frac{(x^2-y^2+1)-(2xyi)}{(x^2-y^2+1)-(2xyi)}$.
Then we get $w=\frac{x^3+xy^2-x^2yi-y^3i+x+iy}{(x^2+^2+1)^2-(2xyi)^2} = \frac{(x^3+xy^2+x)+(-x^2yi-y^3i+iy)}{(x^2+^2+1)^2+(2xy)^2} = \frac{(x^3+xy^2+x)+i(-x^2y-y^3+y)}{(x^2+^2+1)^2+(2xy)^2}$.
And as stated above, imaginary part of $w = 0$ meaning $\frac{-x^2y-y^3+y}{(x^2+^2+1)^2+(2xy)^2} = 0$.
As denominator can't be zero, means nominator is zero $-x^2y-y^3+y=0$. Or $y\cdot(-x^2-y^2+1)=0$. As $y\not=0$ implies $-x^2-y^2+1=0$ or $x^2+y^2=1$.
What now?
| You're basically done, since $x^2+y^2=\left|z\right|^2$.
| {
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What is the summation of the following expression? What's the summation of the following expression;
$$\sum_{k=1}^{n+3}\left(\frac{1}{2}\right)^{k}\left(\frac{1}{4}\right)^{n-k}$$
The solution is said to $$2\left(\frac{1}{4} \right)^{n}\left(2^{n+3}-1\right)$$
But I'm getting $$\left(\frac{1}{4} \right)^{n}\left(2^{n+3}-1\right).$$
How is this possible?
$$\sum_{k=1}^{n+3}\left(2 \times\frac{1}{4}\right)^{k}\left(\frac{1}{4}\right)^n\left(\frac{1}{4}\right)^{-k} \rightarrow \left(\frac{1}{4}\right)^n \sum_{k=1}^{n+3} 2^k\left(\frac{1}{4}\right)^k \left(\frac{1}{4}\right)^{-k}\rightarrow \left(\frac{1}{4}\right)^n\left(2^{n+3}-1\right)$$
| Let $n$=0. The sum is then $\sum_{k=1}^{3}\left(\frac{1}{2}\right)^{k}\left(\frac{1}{4}\right)^{-k}= \frac{1}{2}\cdot4+\frac{1}{4}\cdot16+\frac{1}{8}\cdot64=2+4+8=14$. This equals $2\left(\frac{1}{4} \right)^{0}\left(2^{0+3}-1\right)=2\cdot 7$, and it does not equal $\left(\frac{1}{4} \right)^{n}\left(2^{n+3}-1\right)$, which equals 7.
How are you getting that the sum equals $\left(\frac{1}{4} \right)^{n}\left(2^{n+3}-1\right)$?
| {
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Equation $\sqrt{x}+\sqrt{y}=\sqrt{2013}$ in rationals Can we find all rational numbers $x,y$ such that $\sqrt{x}+\sqrt{y}=\sqrt{2013}$?
Certainly possible answers are $(2013,0)$ and $(0,2013)$.
If we square the equation, we get $x+y+2\sqrt{xy}=2013$, so $\sqrt{xy}$ must be rational.
| if $x\neq 0$ then $x,\sqrt{xy},y$ make a geometric progression, so we can write $y=xq^2$ and $q$ is rational.
$x+y+2\sqrt{xy} = 2013$
$x + xq^2+2xq = x(q+1)^2=2013$
$x= \frac{2013}{(q+1)^2}, y=xq^2, q\in \mathbb Q^+\cup\{0\}$
if $x=0$ then $y = 2013$.
| {
"language": "en",
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Integral $\int_{-1}^{1} \frac{1}{x}\sqrt{\frac{1+x}{1-x}} \log \left( \frac{(r-1)x^{2} + sx + 1}{(r-1)x^{2} - sx + 1} \right) \, \mathrm dx$ Regarding this problem, I conjectured that
$$ I(r, s) = \int_{-1}^{1} \frac{1}{x}\sqrt{\frac{1+x}{1-x}} \log \left( \frac{(r-1)x^{2} + sx + 1}{(r-1)x^{2} - sx + 1} \right) \, \mathrm dx = 4 \pi \operatorname{arccot} \sqrt{ \frac{2r + 2\sqrt{r^{2} - s^{2}}}{s^{2}} - 1}. $$
Though we may try the same technique as in the previous problem, now I'm curious if this generality leads us to a different (and possible a more elegant) proof.
Indeed, I observed that $I(r, 0) = 0$ and
$$\frac{\partial I}{\partial s}(r, s) = \int_{0}^{\infty} \left\{ \frac{2\sqrt{y}}{(r-s)y^{2} + 2(2-r)y + (r+s)}+\frac{2\sqrt{y}}{(r+s)y^{2}+ 2(2-r)y + (r-s)} \right\} \,\mathrm dy, $$
which can be evaluated using standard contour integration technique. But simplifying the residue and integrating them seems still daunting.
EDIT. By applying a series of change of variables, I noticed that the problem is equivalent to prove that
$$ \tilde{I}(\alpha, s) := \int_{-1}^{1} \frac{1}{x}\sqrt{\frac{1+x}{1-x}} \log \left( \frac{ 1 + 2sx \sin\alpha + (s^{2} - \cos^{2}\alpha) x^{2}}{ 1 - 2sx \sin\alpha + (s^{2} - \cos^{2}\alpha) x^{2}} \right) \, \mathrm dx = 4\pi \alpha $$
for $-\frac{\pi}{2} < \alpha < \frac{\pi}{2}$ and $s > 1$. (This is equivalent to the condition that the expression inside the logarithm is positive for all $x \in \Bbb{R}$.)
Another simple observation. once you prove that $\tilde{I}(\alpha, s)$ does not depend on the variable $s$ for $s > 1$, then by suitable limiting process it follows that
$$ \tilde{I}(\alpha, s) = \int_{-\infty}^{\infty} \log \left( \frac{ 1 + 2x \sin\alpha + x^{2}}{ 1 - 2x \sin\alpha + x^{2}} \right) \, \frac{\mathrm dx}{x}, $$
which (I guess) can be calculated by hand. The following graph may also help us understand the behavior of this integral.
| Just for references, I remark that the following proposition was proved in my answer:
Proposition. If $0 < r < 1$ and $r < s$, then
$$ I(r, s) := \int_{-1}^{1} \frac{1}{x} \sqrt{\frac{1+x}{1-x}} \log \left( \frac{1 + 2rsx + (r^{2} + s^{2} - 1)x^{2}}{1 - 2rsx + (r^{2} + s^{2} - 1)x^{2}} \right) \, dx = 4\pi \arcsin r. \tag{*} $$
Recently, I found an alternate proof which is much simpler and does not use contour integration technique.
Lemma 1. For any $k = 0, 1, 2, \cdots$ we have
$$ \int_{0}^{1} \frac{x^{2k}}{\sqrt{1-x^{2}}} \, dx
= (-1)^{k} \frac{\pi}{2} \binom{-1/2}{k}. $$
Since this is so famous, we skip the proof.
Lemma 2. For any $z \in \Bbb{C}$ with $|z| \leq 1$, we have
$$ f(z)
:= - \int_{-1}^{1} \frac{1}{x} \sqrt{\frac{1+x}{1-x}} \log(1 - zx) \, dz= \pi \sin^{-1} z - \pi \log \left( \tfrac{1}{2}+\tfrac{1}{2}\sqrt{1-z^{2}} \right) . \tag{1} $$
Proof of Lemma. Expand $-\log(1-zx)$ using the MacLaurin series. Then we have
$$ f(z) = \sum_{n=1}^{\infty} \frac{z^{n}}{n} \int_{-1}^{1} x^{n-1} \sqrt{\frac{1+x}{1-x}} \, dx. \tag{2} $$
To identify the coefficient, we observe that
\begin{align*}
\int_{-1}^{1} x^{n-1} \sqrt{\frac{1+x}{1-x}} \, dx
&= \int_{0}^{1} x^{n-1} \sqrt{\frac{1+x}{1-x}} \, dx + \int_{-1}^{0} x^{n-1} \sqrt{\frac{1+x}{1-x}} \, dx \\
&= \int_{0}^{1} x^{n-1} \frac{(1+x) + (-1)^{n-1}(1-x)}{\sqrt{1-x^{2}}} \, dx
\end{align*}
Dividing the cases based on the parity of $n$, it follows that
$$ \int_{-1}^{1} x^{n-1} \sqrt{\frac{1+x}{1-x}} \, dx = \begin{cases}
\displaystyle 2\int_{0}^{1} \frac{x^{n}}{\sqrt{1-x^{2}}} \, dx, & n \text{ even} \\
\displaystyle 2\int_{0}^{1} \frac{x^{n-1}}{\sqrt{1-x^{2}}} \, dx, & n \text{ odd}.
\end{cases}. $$
Thus by Lemma 1 we know an exact formula for the coefficients of $f(z)$ in $\text{(2)}$, and we obtain
\begin{align*}
f(z)
&= \pi \sum_{k=0}^{\infty} \binom{-1/2}{k} \frac{(-1)^{k} z^{2k+1}}{2k+1} + \pi \sum_{k=1}^{\infty} \binom{-1/2}{k} \frac{(-1)^{k} z^{2k}}{2k} \\
&= \pi \int_{0}^{z} \frac{dw}{\sqrt{1- w^{2}}} + \pi \int_{0}^{z} \left( \frac{1}{\sqrt{1- w^{2}}} - 1 \right) \, \frac{dw}{w}.
\end{align*}
Therefore evaluating the last integral yields $\text{(1)}$ as desired. ////
Proof of Proposition. Now let us return to the proof of our proposition. Let $r = \cos\alpha$ and $s = \cos\beta$ for any $\alpha, \beta \in \Bbb{R}$. Then by a simple application of trigonometry, we find that
$$ 1 \pm 2rsx + (r^{2} + s^{2} - 1)x^{2} = (1 \pm x \cos(\alpha+\beta))(1 \pm x \cos(\alpha-\beta)). $$
So it follows that
\begin{align*}
I(r, s)
&= f(\cos(\alpha+\beta)) + f(\cos(\alpha-\beta)) - f(-\cos(\alpha+\beta)) - f(\cos(\alpha-\beta)) \\
&= 2\pi \sin^{-1}\cos(\alpha+\beta) + 2\pi \sin^{-1}\cos(\alpha-\beta).
\end{align*}
If we restrict our attention to the case $0 < \alpha < \beta < \pi/2$, then it follows that we have
\begin{align*}
I(r, s)
&= 2\pi \sin^{-1}\cos(\alpha+\beta) + 2\pi \sin^{-1}\cos(\alpha-\beta) \\
&= 4\pi ( \tfrac{\pi}{2} - \alpha ) \\
&= 4\pi \arcsin r.
\end{align*}
This completes the proof.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Strange mistakes when calculate limits I have difficulties with calculating the following limits. W|A gives the correct answers for both of them:
$$
\lim_{x \to +\infty} \sqrt{x} \cdot \left(\sqrt{x+\sqrt x} + \sqrt{x - \sqrt x} - 2\sqrt x\right) = \lim_{x \to +\infty} \sqrt{x^2+x\sqrt x} + \sqrt{x^2-x\sqrt x} - 2x = \\
= \lim_{x \to +\infty} x\left(\sqrt{1+ \frac {\sqrt x}{x}} + \sqrt{1 - \frac{\sqrt x}{x}} - 2\right) = \left[t = \frac 1x\right] = \lim_{t \to 0+} \frac{\sqrt{1+\sqrt t} - 1 +\sqrt{1-\sqrt t} - 1}{t} =
\left[c = \sqrt t\right] = \lim_{c \to 0+} \frac{\sqrt{1+c} - 1 +\sqrt{1-c} - 1}{c^2} = \frac 12c + \left(-\frac 12 c\right) = 0
$$
It's because:
$$
\lim_{c \to 0+} \frac{\sqrt{1+c}-1}{c} = \left|k = \sqrt{1+c}\right| = \lim_{k \to 1+} \frac{k-1}{k^2-1} = \lim_{k \to 1+} \frac{1}{k+1} = \frac 12
$$
The second part is similar to it. But the answer is -$\frac 14$.
And the second limit is:
$$
\lim_{x \to 0} \left(\frac{e^x - \sin x}{\sqrt{1 - 2x} + \log(x+1)}\right)^\cfrac{1}{x^2} = \exp\left(\lim_{x \to 0} \frac{1}{x^2}\left(\frac{e^x-\sin x}{\sqrt{1-2x} + \log(x+1)}-1\right)\right) =
\exp\left(\lim_{x \to 0}\frac{e^x - \sin x - \sqrt{1-2x} - \log(x+1)}{x^2} \cdot \frac{1}{\sqrt{1-2x}+\log(x+1)}\right) = \exp\left(\lim_{x \to 0} \left(\frac{e^x-1}{x \cdot x} - \frac{\sin x}{x \cdot x} - \frac{\sqrt{1-2x}-1}{x \cdot x} - \frac{\log(x+1)}{x \cdot x}\right)\right) = \exp\left(\lim_{x \to 0} \left(\frac 1x - \frac 1x + \frac 1x - \frac 1x\right)\right) = e^0 = 1
$$
The answer is $e^\frac 32$. Where am I wrong?
| How do you get from $$\lim\limits_{c \rightarrow 0^+} \frac{\sqrt{1+c} - 1 + \sqrt{1-c} - 1}{c^2}$$ to $$\frac{1}{2}c - \frac{1}{2}c?$$
You cannot just evaluate part of the limit and leave another part. Compare your mistake with "evaluating" $\lim_{x \rightarrow 0} \frac{x^2}{x^2}$ by noting that $\lim_{x \rightarrow 0} \frac{x^2}{x}$ is zero and then concluding that $$\lim_{x \rightarrow 0} \frac{x^2}{x^2} = \lim_{x \rightarrow 0} \frac{0}{x} = 0$$
In terms of Taylor series, $\frac{\sqrt{1+x} - 1}{x}$ is approximated by $\frac{1}{2} - \frac{x}{8} + O(x^2)$. So the limit as $x \rightarrow 0$ is equal to $\frac{1}{2}$, but $\frac{\sqrt{1+x} - 1}{x^2}$ is approximated by the Taylor series $\frac{1}{2x} - \frac{1}{8} + O(x)$. Notice that this is not equal to $\frac{1}{2x}$, which you substitute it with! In a similar way, $\frac{\sqrt{1-x} - 1}{x}$ is approximated by the Taylor series $-\frac{1}{2x} - \frac{1}{8} + O(x)$, and adding these up gives $-\frac{1}{4} + O(x)$, and thus the limit of $\frac{\sqrt{1+x} - 1 + \sqrt{1 - x} - 1}{x^2}$ as $x \rightarrow 0$ is equal to -$\frac{1}{4}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the radius of convergence and the interval of convergence of the series $\sum_{n=1}^\infty \frac{n(x-4)^n}{n^3+1}$ Series is:
$$\sum_{n=1}^\infty \frac{n(x-4)^n}{n^3+1}$$
So, I understand that I use the ratio test to find r, but I can't simplify the equation to the point where I can do this. Here's where I am so far:
$$
\frac{(n+1)(x-4)^{n+1}}{(n+1)^3+1}\bullet\frac{n^3+1}{n(x-4)^n}
$$
which gets me to
$$
\frac{(n+1)(x-4)(n^3+1)}{n[(n+1)^3+1]}
$$
Which is where I'm stuck. How can I simplify this further so I can find R? Do I just need to multiply this out and look at the coefficients of the highest powers of n?
| $$
\lim_{n \to \infty} \left|\frac{(n+1)(x-4)(n^3+1)}{n(n^3+3n^2+3n+2)}\right| < 1 \\
\left|x-4\right|\lim_{n \to \infty} \left|\frac{(n+1)(n^3+1)}{n(n^3+3n^2+3n+2)}\right| < 1 \\
\left|x-4\right| < 1 \\
R = 1 \ \text{and} \ x \in (3, 5)
$$
| {
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"url": "https://math.stackexchange.com/questions/571073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate the limit of the function Please help me to calculate the limit of the function. I do not know where to start.
Thank you.
| To get started, Assuming limit L exists then
$$
\begin{align}
\\ \lim_{x\to0} f(x)^{g(x)} &={\rm L}
\\ \lim_{x\to0} {g(x)} \ln f(x) &=\ln {\rm L}
\end{align}
$$
Where
$$
\begin{align}
f(x)&= \left[\left(\frac{\sin(x+x^2)}{1-x^2}\right)^2-\frac{\arctan(x^2)}{1-x^2}-2x^3+1\right]
\\ g(x) &=-\frac{1}{\left(1-\operatorname{e}^{2x^2}\right)\sinh(3x^2)}.
\end{align}
$$
so $$
\begin{align}
\\ \lim_{x\to0} {g(x)} \ln f(x) &=\ln {\rm L}
\\ \lim_{x\to0} -\frac{\ln\left[\left(\frac{\sin(x+x^2)}{1-x^2}\right)^2-\frac{\arctan(x^2)}{1-x^2}-2x^3+1\right]}{{\left(1-\operatorname{e}^{2x^2}\right)\sinh(3x^2)}}&= \ln {\rm L}
\\ \lim_{x\to0} \frac{\ln\left[\left(\frac{\sin(x+x^2)}{1-x^2}\right)^2-\frac{\arctan(x^2)}{1-x^2}-2x^3+1\right]}{{\left(1-\operatorname{e}^{2x^2}\right)\sinh(3x^2)}}&= \ln {\rm \frac{1}{L}}
\end{align}
$$
The last limit is of the form $\frac{0}{0}$, so appeal to sprite of L'Hopital
$$\begin{align}\\ \lim_{x\to0} \frac{\frac{\rm d}{{\rm d}x}\left(\ln\left[\left(\frac{\sin(x+x^2)}{1-x^2}\right)^2-\frac{\arctan(x^2)}{1-x^2}-2x^3+1\right]\right)}{{\frac{\rm d}{{\rm d}x}\left(\left(1-\operatorname{e}^{2x^2}\right)\sinh(3x^2)\right)}}&= \ln {\rm \frac{1}{L}}
\end{align}
$$
Now either the above can be evaluated, if not repeat L'Hopital for desired result.
| {
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"source": "stackexchange",
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How find this equation solution $2\sqrt[3]{2y-1}=y^3+1$ find this equation roots:
$$2\sqrt[3]{2y-1}=y^3+1$$
My try: since
$$8(2y-1)=(y^3+1)^3=y^9+1+3y^3(y^3+1)$$
then
$$y^9+3y^6+3y^3-16y+9=0$$
Then I can't.Thank you someone can take hand find the equation roots.
| Let $f(y) = \frac12 (y^3+1)$, we have
$$2\sqrt[3]{2y-1} = y^3+1\quad\iff\quad f^{-1}(y) = f(y) \quad\implies\quad y = f(f(y))$$
Since $f(y)$ is a strictly increasing function in $y$, we have
*
*If $f(x) > x$, then $f(f(x)) > x$.
*If $f(x) < x$, then $f(f(x)) < x$.
This means
$$\begin{align}
y = f(f(y)) \implies & y = f(y)\\
\iff & y^3+1 - 2y = 0\\
\iff & (y^2 + y -1 )(y-1) = 0\\
\implies & y = 1\text{ or } \frac{-1\pm \sqrt{5}}{2}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/575903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A closed form for $\int_0^1{_2F_1}\left(-\frac{1}{4},\frac{5}{4};\,1;\,\frac{x}{2}\right)^2dx$ Is it possible to evaluate in a closed form integrals containing a squared hypergeometric function, like in this example?
$$\begin{align}S&=\int_0^1{_2F_1}\left(-\frac{1}{4},\frac{5}{4};\,1;\,\frac{x}{2}\right)^2dx\\\vphantom{=}\\&=\frac{1}{4\pi}\int_0^1\left(\sum_{n=0}^\infty\frac{4n+1}{8^n}\cdot\frac{\Gamma\left(2n-\frac{1}{2}\right)}{\Gamma(n+1)^2}\cdot x^n\right)^2dx\end{align}$$
It is approximately
$$S\approx0.8263551866500213413164525287...$$
| Incomplete solution:
In fact, we have $${}_2F_1(-\frac14,\frac54;1;\frac{x}{2})=\frac{8\sqrt{2}}{\pi\sqrt{2+\sqrt{2x}}}((2+\sqrt{2x})E(\frac{2\sqrt{x}}{\sqrt{2}+\sqrt{x}})-K(\frac{2\sqrt{x}}{\sqrt{2}+\sqrt{x}})).$$
Here $E(m)=\int^1_0\sqrt{\frac{1-mt^2}{1-t^2}}~dt$ and $K(m)=\int^1_0\frac{dt}{\sqrt{(1-t^2)(1-mt^2)}}$ are the complete elliptic integrals
Therefore, by substituting $y=\frac{2\sqrt{x}}{\sqrt{2}+\sqrt{x}}$ (so that $x=\frac{y^2}{(2-y)^2}$), we have $$S=\frac{16}{\pi^2}\int^{2\sqrt{2}-2}_{0}\frac{y}{(2-y)^4}\left(4E(y)-(2-y)K(y)\right)^2~dy.$$
$$%E'(y)=\frac{E(y)-K(y)}{2y}\\K'(y)=\frac{E(y)-(1-y)K(y)}{2y(1-y)}\\\int E(y)dy=\frac{2(1+y)E(y)-2(1-y)K(y)}{3}\\\int K(y)dy=2E(y)-(1-y)K(y)$$
This may be doable with some clever integration by parts.
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
Generalizing a series of numbers I know that I studied this long ago but I can't seem to bring the information to mind. I am looking to construct a general formula for the following sequence of numbers ->
when $x=4$ : $n=(3+x)+2(x-1)$;
when $x=5$ : $n=(3+x)+3(x-1)+2(x-2)+1(x-3)$;
when $x=6$ : $n=(3+x)+4(x-1)+3(x-2)+2(x-3)+1(x-4)$;
etc. I worked out the pattern but can't generalize this into a formula. Can anyone help?
| Note that we can rewrite your statements:
when $x=4:n=(3+x)+2(x-1)$. Obviously, if $x=4$, then we have $n=(3+4)+2(4-1) = 13$.
$\begin{align} x=5:n&=(3+x)+3(x-1)+2(x-2)+1(x-3)\\ &= 3+4+3(4-1)+2(4-2)+1(4-3)\\ &= 7+9+4+1 \\ &= 21\end{align}$
$\begin{align} x=6:n&=(3+x)+4(x-1)+3(x-2)+2(x-3)+1(x-4) \\ &= 3+6+4(6-1)+3(6-2)+2(6-3)+1(6-4) \\ &=9+20+12+6+2 \\ &= 49 \end{align}$
I searched for a sequence containing $13,21,49$ in the Online Encylopedia of Integer Sequences, but I could not find anything that seems particularly helpful. So I'm not sure if there is a particular sequence here.
You could rewrite all integers in terms of $x$:
$$\begin{align}x=4:n &= 1(3+x)+(x-2)(x-1) \\ &= 3+x+x^2-3x-2 \\ &= x^2-2x+1 \\ &= (x-1)(x-1)\end{align}$$
$$\begin{align} x=5: n &= 1(3+x)+(x-2)(x-1)+(x-3)(x-2) + (x-4)(x-3)\\ &= 3+x+x^2-3x-2+x^2-5x+6+x^2-7x+12 \\ &= 3x^2-14x+1 \end{align}$$
You could use $\ldots$ notation to make the process easier: e.g. $(x-5)(x-4) + \ldots +(x-1)$ will help write the longer expressions more compactly.
I know this is not a complete answer, but I hope it helps or gives you some useful ideas.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that $a^2 + b^2 + c^2 \ge a + b + c$.
Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that
$a^2 + b^2 + c^2 \ge a + b + c$.
I'm supposed to prove this use AM-GM, but can't figure it out. Any hints?
| By QM-AM for the first inequality and AM-GM for the second,
$$\frac{a^2+b^2+c^2}{3}\geq\left( \frac{a+b+c}{3} \right)^2\geq\frac{a+b+c}{3}\left( \sqrt[3]{abc} \right)$$
so since $abc=1, a^2+b^2+c^2\geq a+b+c$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
Prove $13|19^n-6^n$ by congruences I am trying to prove $13|19^n-6^n$. With induction its not so bad but by congruences its quite difficult to know how to get started.
Any hints?
| Modular Arithmetic is an extremely useful way of thinking about numbers. It can be used to calculate the last three digits of $2011^{2011}$ by hand. The day of the week, the time of the day, and the flipping of a light switch are all based upon modular arithmetic. The day of the week is in $\mod 7$ (cycles in a period of $7$), the time of the day is in $\mod 24$, and a light switch is either on or off; $\mod 2$.
Two integers are congruent modulo $n$ if their difference is a multiple of $n$. We write this as $$a \equiv b \pmod{n}$$
For example, $36 \equiv 71 \pmod{7}$, because $71 - 36 = 35$, and $\dfrac{35}{7} = 5$.
We can alternatively write this as $a-b = qn$, for an integer $q$, or $a = b + qn$ (by adding $b$ to both sides). Notice that for any integers $x$ and $y$, if $x \equiv y \pmod{n}$, $x$ and $y$ leave the same remainder when divided by $n$. This is sometimes used as another definition for modular arithmetic. Since modular arithmetic is only defined for integers, let all variables be integers unless otherwise specified.
$\textbf{Addition}$
$\textbf{Theorem 1:}$ If $a \equiv b \pmod{n}$, then $a+c \equiv b+c \pmod{n}$.
If $a \equiv b \pmod{n}$, we know that $a-b = qn$ for some integer $q$. This means $(a+c) - (b+c) = a-b = qn$, so $a+c \equiv b+c \pmod{n}$. $\blacksquare$
If we know that $123$ leaves a remainder of $2$ when divided by $11$, and want the remainder when $127$ is divided by $11$, we can simply use Theorem 1: $123 \equiv 2 \pmod{11} \implies 123 + 4 \equiv 2 + 4 \pmod{11}$, so $127 \equiv 6 \pmod{11}$. A similar proof can be used to prove a stronger theorem:
$\textbf{Theorem 2:}$ If $a \equiv b \pmod{n}$, and $c \equiv d \pmod{n}$, then $a+c \equiv b+d \pmod{n}$.
We know that $a-b = qn$ for some integer $q$ and $c-d = rn$ for some integer $r$. Adding the equations gives $(a+c) - (b+d) = qn + rn = (q+r)n$. This means that the difference of $a+c$ and $b+d$ is a multiple of $n$, so $a+c \equiv b+d \pmod{n}$. $\blacksquare$
Theorem 2 can be generalized even further:
$\textbf{Theorem 3:}$ If $a_1 \equiv b_1 \pmod{n}$, $a_2 \equiv b_2 \pmod{n}$, $\cdots , \ a_k \equiv b_k \pmod{n}$ then $a_1 + a_2 + \cdots + a_k \equiv b_1 + b_2 + \cdots + b_k \pmod{n}$.
By the definition of modulos, we get $a_1 - b_1 = q_1 n$, $a_2 - b_2 = q_2 n$, $\cdots , \ a_k - b_k = q_k n$. Adding these, we get $(a_1 + a_2 + \cdots + a_k) - (b_1 + b_2 + \cdots + b_k) = (q_1 + q_2 + \cdots + q_k)n$, so $a_1 + a_2 + \cdots + a_k \equiv b_1 + b_2 + \cdots + b_k \pmod{n}$. $\blacksquare$
Similar theorems hold for muliplication as well, but the proofs are different.
$\textbf{Theorem 4:}$ If $a \equiv b \pmod{n}$, then $ac \equiv bc \pmod{n}$.
If $a \equiv b \pmod{n}$, we know that $a-b = qn$ for some integer $q$. This means $ac-bc = qcn$, so $ac \equiv bc \pmod{n}$. $\blacksquare$
There is also another proof of this:
We know that $$\underbrace{a \equiv b \pmod{n}, \ a \equiv b \pmod{n}, \cdots , \ a \equiv b \pmod{n}}_{\text{c statements}}$$
From Theorem 3, we get $$\underbrace{a+a+a+a+\cdots + a}_{\text{c "a"s}} \equiv \underbrace{b+b+b+b+\cdots + b}_{\text{c "b"s}}$$
or $ac \equiv bc \pmod{n}$. $\blacksquare$
$ \textbf{Theorem 5:}$ If $a \equiv b \pmod{n}$, and $c \equiv d \pmod{n}$, then $ac \equiv bd \pmod{n}$.
We repeatedly use Theorem 4: Since $a \equiv b$, $ac \equiv bc$, and since $c \equiv d$, $bc \equiv bd$. Therefore, $ac \equiv bc \equiv bd$, or $ac \equiv bd \pmod{n}$. $\blacksquare$
$\textbf{Theorem 6:}$ If $a_1 \equiv b_1 \pmod{n}$, $a_2 \equiv b_2 \pmod{n}$, $\cdots , \ a_k \equiv b_k \pmod{n}$ then $a_1 a_2 \cdots a_k \equiv b_1 b_2 \cdots b_k \pmod{n}$.
We repeatedly use Theorem 5: Because $a_1 \equiv b_1 \pmod{n}$, $a_2 \equiv b_2 \pmod{n}$, we get $a_1a_2 \equiv b_1b_2 \pmod{n}$. We also know $a_3 \equiv b_3 \pmod{n}$, so $a_1a_2a_3 \equiv b_1b_2b_3 \pmod{n}$. Also, $a_4 \equiv b_4 \pmod{n}$, so $a_1a_2a_3a_4 \equiv b_1b_2b_3b_4 \pmod{n}$. Continuing to do this will give $a_1 a_2 \cdots a_k \equiv b_1 b_2 \cdots b_k \pmod{n}$. $\blacksquare$
Note that a corollary of Theorem 6, when $a_1 = a_2 = \cdots = a_k$ and $a_1 = a_2 = \cdots = a_k$, is that
$\textbf{Theorem 7:} $ If $a \equiv b \pmod{n}$, then $a^k \equiv b^k \pmod{n}$.
$\textbf{Main Proof:}$
Using this corollary, we have $19^k \equiv 6^k \pmod{13}$, because $19 \equiv 6 \pmod{13}$. By definition, this means that $$13 | (19^k - 6^k)$$ as desired. $\blacksquare$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/582670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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Intuitive explanation for formula of maximum length of a pipe moving around a corner? For one of my homework problems, we had to try and find the maximum possible length $L$ of a pipe (indicated in red) such that it can be moved around a corner with corridor lengths $A$ and $B$ (assuming everything is 2d, not 3d):
My professor walked us through how to derive a formula for the maximum possible length of the pipe, ultimately arriving at the equation $L = (A^{2/3} + B^{2/3})^{3/2}$.
The issue I have is understanding intuitively why this formula works, and exactly what it's doing. I understand the steps taken to get to this point, but there's an odd symmetry to the end result -- for example, is the fact that $\frac{2}{3}$ and its inverse are the only constants used just a coincidence, or indicative of some deeper relationship?
I also don't quite understand how the formula relates, geometrically, to the diagram. If I hadn't traced the steps myself, I would have never guessed that the formula was in any way related to the original problem.
If possible, can somebody give an intuitive explanation as to why this formula works, and how to interpret it geometrically?
Here's how he found the formula, if it's useful:
The formula is found by finding the maximum possible length of the pipe by expressing the length in terms of the angle $\theta$ formed between the pipe and the wall, and by taking the derivative to find when $\frac{dL}{d\theta} = 0$, which is the minimum of $\frac{dL}{d\theta}$ and is therefore when $L$ is the smallest:
$$
L = \min_{0 \leq \theta \leq \frac{\pi}{2}} \frac{A}{\cos{\theta}} + \frac{B}{\sin{\theta}} \\
0 = \frac{dL}{d\theta} = \frac{A\sin{\theta}}{\cos^2{\theta}} - \frac{B\cos{\theta}}{\sin^2{\theta}} \\
0 = \frac{A\sin^3{\theta} - B\cos^3{\theta}}{\sin^2{\theta}\cos^2{\theta}} \\
0 = A\sin^3{\theta} - B\cos^3{\theta} \\
\frac{B}{A} = \tan^3{\theta} \\
\theta = \arctan{\left( \frac{B}{A} \right)^{\frac{1}{3}}} \\
$$
At this point, we can substitute $\theta$ back into the original equation for $L$ by interpreting $A^{1/3}$ and $B^{1/3}$ as sides of a triangle with angle $\theta$ and hypotenuse $\sqrt{A^{2/3} + B^{2/3} }$:
$$
\cos{\theta} = \frac{A^{1/3}}{ \sqrt{A^{2/3} + B^{2/3} }} \\
\sin{\theta} = \frac{B^{1/3}}{ \sqrt{A^{2/3} + B^{2/3} }} \\
\therefore L = A^{2/3} \sqrt{A^{2/3} + B^{2/3} } + B^{2/3} \sqrt{A^{2/3} + B^{2/3} } \\
L = (A^{2/3} + B^{2/3}) \sqrt{A^{2/3} + B^{2/3} } \\
L = (A^{2/3} + B^{2/3})^{3/2} \\
$$
The equation for the formula for the maximum length of the pipe is therefore $L = (A^{2/3} + B^{2/3})^{3/2}$.
| Here's a derivation without trigonometry.
There is also a short discussion of the geometric significance
of the exponents in the answer after the derivation.
Let $L$ be the length of a line touching the inner corner of the
two corridors and extending to the outer wall of each corridor.
Where this line meets the outer wall of each corridor, draw a line
perpendicular to the corridor.
Let $x$ be the distance from the inner corner to the perpendicular line
across corridor $B$
and let $y$ be the distance from the inner corner to the perpendicular line
across corridor $A$.
The result is the figure shown above.
Applying the Pythagorean Theorem to the obvious
right triangle with hypotenuse $L$,
$$ L^2 = (A + x)^2 + (B + y)^2.$$
The longest pipe that can fit around the corner is the smallest
value of $L$ for any value $x > 0$, so let's minimize $L$ as a function of $x$.
But $L$ is minimized when $L^2$ is minimized, so we would like to set
$$ \frac{d}{dx} L^2 = 0. $$
That is,
$$\begin{eqnarray}
0 = \frac{d}{dx} L^2 &=& \frac{d}{dx}\left((A + x)^2 + (B + y)^2\right) \\
&=& 2(A + x) + 2(B + y)\frac{dy}{dx}.
\end{eqnarray}$$
Now, by similar triangles,
$ \dfrac Bx = \dfrac yA .$
That is, $xy = AB.$ Differentiating both sides of this by $x$,
$$ x\frac{dy}{dx} + y = 0,$$
$$ \frac{dy}{dx} = -\frac yx.$$
Substituting for $\frac{dy}{dx}$ in our earlier equation for $\frac{d}{dx}L^2,$
$$ 0 = 2(A + x) + 2(B + y)\left(-\frac yx\right),$$
from which it follows that
$$ (A + x)x = (B + y)y,$$
But (again by similar triangles)
$$ \frac Bx = \frac{B+y}{A+x},$$
$$ B+y = \frac Bx (A+x),$$
and combining this with the fact that $y = \dfrac{AB}{x},$
$$ (B + y)y = \left(\frac Bx (A+x)\right) \frac{AB}{x}
= \frac{AB^2}{x^2}(A + x).$$
That is,
$$ (A + x)x = \frac{AB^2}{x^2}(A + x),$$
$$ x^3 = AB^2,$$
$$ x = A^{1/3}B^{2/3},$$
$$ y = \frac{AB}{x} = A^{2/3}B^{1/3},$$
$$A + x = A + A^{1/3}B^{2/3} = \left(A^{2/3} + B^{2/3} \right)A^{1/3},$$
and $$B + y = B + A^{2/3}B^{1/3} = \left(A^{2/3} + B^{2/3} \right)B^{1/3}.$$
Therefore at the value of $x$ that minimizes $L$,
$$\begin{eqnarray}
L^2 &=& (A + x)^2 + (B + y)^2 \\
&=& \left(A^{2/3} + B^{2/3} \right)^2 A^{2/3} +
\left(A^{2/3} + B^{2/3} \right)^2 B^{2/3} \\
&=& \left(A^{2/3} + B^{2/3} \right)^2 \left(A^{2/3} + B^{2/3}\right) \\
&=& \left(A^{2/3} + B^{2/3} \right)^3
\end{eqnarray}$$
and
$$ L = \left(A^{2/3} + B^{2/3} \right)^{3/2}.$$
Regarding the symmetry of the solution, clearly the length of the pipe
does not change if we swap the labels $A$ and $B$ on the two corridors.
We should suspect from this that there might be some expression for
$L$ in which $A$ and $B$ appear in completely symmetric roles.
Regarding the geometric interpretation of the exponents, the key facts
that lead to the $\frac23$ exponents are the findings that $x^3 = AB^2$
and (by symmetry) $y^3 = BA^2.$
This is what leads to the conclusion that $L$ is the hypotenuse of a
right triangle with legs $A + A^{1/3}B^{2/3}$ and $B + B^{1/3}A^{2/3}$;
it is from the squares of those two lengths that we get terms
in $A^{2/3}$ and $B^{2/3}$.
Finally, note that if we double both $A$ and $B$ simultaneously,
we should expect the entire figure showing the "tightest" angle
of the pipe around the corner to scale up by a factor of $2$,
hence $L$ also doubles.
But we see that $A^{2/3}$ and $B^{2/3}$ scale up only by a factor
of $2^{2/3}$ when we double $A$ and $B$; it is the exponent of $\frac32$
outside the parentheses that gives us a scaling factor
of $\left(2^{2/3}\right)^{3/2} = 2.$
So it is not a mere coincidence that the exponents $\frac23$ and $\frac32$
are multiplicative inverses.
Another way to approach this is to apply the fact that
$B+y = \dfrac Bx (A+x)$
before differentiating $L^2$ rather than afterwards.
That is, we simply substitute for $B+y$ in the formula for $L^2$,
obtaining
$$\begin{eqnarray}
L^2 &=& (A + x)^2 + \frac{B^2}{x^2}(A + x)^2 \\
&=& \left(1 + \frac{B^2}{x^2}\right) (A + x)^2.
\end{eqnarray}$$
We can differentiate this with respect to $x$ to obtain an analytic solution,
or apply numeric methods using $x$ as an independent variable.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 3,
"answer_id": 2
} |
inequality $\frac{1}{1+3a}+\frac{1}{1+3b}+\frac{1}{1+3c}+\frac{1}{1+3d} \geq 1$ Given real numbers $a, b, c, d$ with this condition: $$abcd=1$$
Let's prove this inequality: $$\frac{1}{1+3a}+\frac{1}{1+3b}+\frac{1}{1+3c}+\frac{1}{1+3d} \geq 1$$
Thank everybody!
| The function $f(x ) = \frac{1}{1+3 e^x}$ is convex on $[\ln(\frac{1}{3}),\infty)$ look: convexity proof.
Whenever $y\leq 1/3$ then we have that $\dfrac{1}{1+3y} \geq \dfrac{1}{2}$, therefore whenever more than two of the $a,b,c,d$ are less than $1/3$ then the inequality holds. So, it remains to deal with the case that all $a,b,c,d$ are more than $\frac{1}{3}$ or only one of them is less than $1/3$.
We now deal with the first case, and convexity gives
$$ f(\ln a) + f(\ln b)+ f(\ln c)+f(\ln d) \geq 4 f\left( \frac{1}{4}\ln (abcd) \right) =4f(0)= 1$$
Now, for the second case assume that $d\leq 1/3$ and $a,b,c >1/3$. We use Jensen's again
$$ f(\ln a) + f(\ln b)+ f(\ln c) \geq 3 f\left( \frac{1}{3}\ln (abc) \right) = 3 f\left( \frac{1}{3}\ln (1/d) \right) = \frac{3}{1+3d^{-\frac{1}{3}}}$$
Hence, we need to show
$$ \frac{1}{1+3d}+ \frac{3}{1+3d^{-\frac{1}{3}}}\geq 1 $$
This case is dealt in M. Roseberg's post.
| {
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"timestamp": "2023-03-29T00:00:00",
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Particular solution of a non-homogenous recurrence relation I need some help with the following non-homogenous recurrence relation.
$$a_n-2a_{n-1}+a_{n-2}=n+1$$
$$a_0=0, a_1=1$$
When I solve the associated homogenous equation I use the auxiliary equation $x^2-2x+1=0$ and obtain the root $x=1$. Hence, I obtain the equation $a_n=(A+nB)1^n$. Using the initial conditions I get $a_n=n$.
When it comes to the particular solution I get stuck, however. I thought the solution should be on the form $cn+d$ due to $n+1$. The following calculations show how I then get stuck:
$$a_n-2a_{n-1}+a_{n-2}=n+1$$
$$cn+d-2(c(n-1)+d)+c(n-2)+d=n+1$$
$$cn+d-2(cn-c+d)+cn-2c+d=n+1$$
$$cn+d-2cn+2c-2d+cn-2c+d=n+1$$
$$0=n+1$$
So obviously I am wrong about the form of the particular solution.
| The form of the equation allows us to solve it by using only first-order difference equations (see Remark 1).
Arrange the equation as
$$\begin{cases}[a_{n}-a_{n-1}]-[a_{n-1}-a_{n-2}]=n+1,{\quad}n=3,4,\cdots\\ a_{1}=1,\ a_{0}=0,\end{cases}$$
which is a first-order difference equation in $b_{n}:=[a_{n-1}-a_{n-2}]$ for $n=2,3,\cdots$.
So, we have
$$\begin{cases}b_{n+1}-b_{n}=n+1,{\quad}n=2,3,\cdots\\ b_{2}=1,\end{cases}$$
which implies
$$\begin{aligned}b_{n}=&1\prod_{k=2}^{n-1}1+\sum_{k=2}^{n-1}(k+1)\prod_{\ell=k+1}^{n-1}1\\ =&1+\sum_{k=2}^{n-1}(k+1)\\ =&\sum_{{\color{blue}k=1}}^{n-1}k+(n-2)\\ =&\frac{1}{2}(n^{2}+n-4),\end{aligned}$$
where we have applied Remark 2(i).
Next, we set $c_{n}:=a_{n-2}$ for $n=2,3,\cdots$, then $c_{n+1}-c_{n}=a_{n-1}-a_{n-2}=b_{n}$, i.e.,
$$\begin{cases}c_{n+1}-c_{n}=\frac{1}{2}(n^{2}+n-4),{\quad}n=2,3,\cdots\\ c_{2}=0\end{cases}$$
whose solution is
$$\begin{aligned}c_{n}=&0\prod_{k=2}^{n-1}1+\frac{1}{2}\sum_{k=2}^{n-1}(k^{2}+k-4)\prod_{\ell=k+1}^{n-1}1\\ =&\frac{1}{2}\sum_{k=2}^{n-1}(k^{2}+k-4)\\ =&\frac{1}{2}\sum_{{\color{blue}k=1}}^{n-1}k^{2}+\frac{1}{2}\sum_{{\color{blue}k=1}}^{n-1}k-2(n-2)-{\color{blue}1}\\
=&\frac{1}{2}\frac{1}{6}(n-1)n(2n-1)+\frac{1}{2}\frac{1}{2}(n-1)n-2(n-2)-1\\ =&\frac{1}{6}(n^{3}-13n+18),\end{aligned}$$
where we have applied Remark 2(ii).
As $a_{n}=c_{n+2}$, we get
$$a_{n}=\frac{1}{6}(n^{3}+6n^{2}-n),{\quad}n=2,3,\cdots.\tag*{$\blacksquare$}$$
Remark 1. The first-order difference equation
$x_{n+1}-p_{n}x_{n}=q_{n}$
has the solution
$\displaystyle x_{n}=x_{m}\prod_{k=m}^{n-1}q_{k}+\sum_{k=m}^{n-1}q_{k}\prod_{\ell=k+1}^{n-1}p_{\ell}$
with the convention that the empty product is unity.
Remark 2. The following identities hold.
(i) $\displaystyle\sum_{k=1}^{m}k=\frac{1}{2}m(m+1)$.
(ii) $\displaystyle\sum_{k=1}^{m}k^{2}=\frac{1}{6}m(m+1)(2m+1)$.
| {
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"source": "stackexchange",
"question_score": "2",
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$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\cdots}}}}}$ approximation Is there any trick to evaluate this or this is an approximation, I mean I am not allowed to use calculator.
$$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\cdots}}}}}$$
| Alternatively, let $a_1,\,a_2,\,a_3,\,\cdots,\,a_n$ be the following sequence
$$\sqrt{7},\,\sqrt{7\sqrt{7}},\,\sqrt{7\sqrt{7\sqrt{7}}},\,\cdots,\underbrace{\sqrt{7\sqrt{7\sqrt{7\sqrt{\cdots\sqrt{7}}}}}}_{\large n\,\text{times}}$$
respectively.
Notice that
$$\large a_n=7^{\Large 1-2^{-n}}$$
Hence
$$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{\cdots}}}}}=\large\lim_{n\to\infty}\, a_n=\lim_{n\to\infty}\,7^{1-\Large2^{-n}}=\bbox[3pt,border:3px #FF69B4 solid]{\color{red}{7}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/589288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "69",
"answer_count": 7,
"answer_id": 6
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How find this integral $\int\frac{\sqrt{x-1}\arctan{(x\sqrt{x-1})}}{x}dx$ Find this integral
$$\int\dfrac{\sqrt{x-1}\arctan{(x\sqrt{x-1})}}{x}dx$$
My try: let
$$\arctan{(x\sqrt{x-1})}=t$$
and that's very ugly,Thank you
| $\because$ according to http://integrals.wolfram.com/index.jsp?expr=%28x-1%29%5E%281%2F2%29%2Fx&random=false, $\int\dfrac{\sqrt{x-1}}{x}dx=2\sqrt{x-1}-2\tan^{-1}\sqrt{x-1}+C$
$\therefore\int\dfrac{\sqrt{x-1}\tan^{-1}(x\sqrt{x-1})}{x}dx$
$=\int\tan^{-1}(x\sqrt{x-1})~d(2\sqrt{x-1}-2\tan^{-1}\sqrt{x-1})$
$=(2\sqrt{x-1}-2\tan^{-1}\sqrt{x-1})\tan^{-1}(x\sqrt{x-1})-\int(2\sqrt{x-1}-2\tan^{-1}\sqrt{x-1})~d(\tan^{-1}(x\sqrt{x-1}))$
$\because$ according to http://www.wolframalpha.com/input/?i=d%2Fdx%28arctan%28x%28x-1%29%5E%281%2F2%29%29%29, $\dfrac{d}{dx}(\tan^{-1}(x\sqrt{x-1}))=\dfrac{3x-2}{2\sqrt{x-1}(x^3-x^2+1)}$
$\therefore(2\sqrt{x-1}-2\tan^{-1}\sqrt{x-1})\tan^{-1}(x\sqrt{x-1})-\int(2\sqrt{x-1}-2\tan^{-1}\sqrt{x-1})~d(\tan^{-1}(x\sqrt{x-1}))$
$=2\sqrt{x-1}\tan^{-1}(x\sqrt{x-1})-2\tan^{-1}\sqrt{x-1}\tan^{-1}(x\sqrt{x-1})-\int(2\sqrt{x-1}-2\tan^{-1}\sqrt{x-1})\dfrac{3x-2}{2\sqrt{x-1}(x^3-x^2+1)}dx$
$=2\sqrt{x-1}\tan^{-1}(x\sqrt{x-1})-2\tan^{-1}\sqrt{x-1}\tan^{-1}(x\sqrt{x-1})-\int\dfrac{3x-2}{x^3-x^2+1}dx+\int\dfrac{(3x-2)\tan^{-1}\sqrt{x-1}}{\sqrt{x-1}(x^3-x^2+1)}dx$
$\int\dfrac{3x-2}{x^3-x^2+1}dx$ is just an integral of rational functions and should have close form.
For $\int\dfrac{(3x-2)\tan^{-1}\sqrt{x-1}}{\sqrt{x-1}(x^3-x^2+1)}dx$ ,
Let $u=\sqrt{x-1}$ ,
Then $x=u^2+1$
$dx=2u~du$
$\therefore\int\dfrac{(3x-2)\tan^{-1}\sqrt{x-1}}{\sqrt{x-1}(x^3-x^2+1)}dx$
$=\int\dfrac{2(3u^2+1)\tan^{-1}u}{(u^2+1)^3-(u^2+1)^2+1}du$
Then you can separate it to the terms of $\int\dfrac{\tan^{-1}u}{au+b}du$ or $\int\dfrac{\tan^{-1}u}{pu^2+qu+r}du$ by partial fraction. According to http://pi.physik.uni-bonn.de/~dieckman/IntegralsIndefinite/IndefInt.html, they relate to the polylogarithm function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/589407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
why is $ \sqrt x - \sqrt2 = \sqrt{(x-2)} $ when x tends to 2 $ \sqrt x - \sqrt2 = \sqrt{(x-2)} $ when x tends to $ 2^+ $
i have this problem
lim when x tends to $ 2^+ $
$ \frac{\sqrt x -\sqrt2 +\sqrt{x-2}}{\sqrt{x^2-4}} $
i know i must group $ \sqrt x - \sqrt 2 $ into $ \sqrt{x-2} $ only because that is true when x tends to 2 but then i separate the sums and simplify but the calculator gives me another result here's what i did
$ \frac{\sqrt{x-2} +\sqrt{x-2}}{\sqrt{x^2 -4}} $
$\frac{\sqrt{x-2}}{\sqrt{x^2 -4}} + \frac{\sqrt{x-2}}{\sqrt{x^2 -4}} $
$ \frac{\sqrt{x-2}}{\sqrt{x-2}\sqrt{x+2}} + \frac{\sqrt{x-2}}{\sqrt{x-2}\sqrt{x+2}} $
simplifying
$ \frac{1}{\sqrt{x+2}} + \frac{1}{\sqrt{x+2}} $
this is when x tends to $ 2^+ $
$ \frac{2}{\sqrt{4}} = 1 $
but the calculator gets me $ \frac{\sqrt{2}}{2} $
where did i go wrong?
| You need to evaluate $$\lim_{x \to 2^{+}}\frac{\sqrt{x} - \sqrt{2} + \sqrt{x - 2}}{\sqrt{x^{2} - 4}}$$ and you are not supposed to replace $\sqrt{x} - \sqrt{2}$ by $\sqrt{x - 2}$. This is not allowed through any rule of algebra/calculus. What we do here is not difficult, but a bit of careful and simple algebraical manipulation:
$\displaystyle \begin{aligned}\lim_{x \to 2^{+}}\frac{\sqrt{x} - \sqrt{2} + \sqrt{x - 2}}{\sqrt{x^{2} - 4}} &= \lim_{x \to 2^{+}}\frac{\sqrt{x} - \sqrt{2}}{\sqrt{x^{2} - 4}} + \frac{\sqrt{x - 2}}{\sqrt{x^{2} - 4}}\\
&= \lim_{x \to 2^{+}}\frac{(\sqrt{x} - \sqrt{2})(\sqrt{x} + \sqrt{2})}{(\sqrt{x} + \sqrt{2})\sqrt{x^{2} - 4}} + \frac{\sqrt{x - 2}}{\sqrt{(x - 2)(x + 2)}}\\
&= \lim_{x \to 2^{+}}\frac{x - 2}{(\sqrt{x} + \sqrt{2})\sqrt{(x - 2)(x + 2)}} + \frac{1}{\sqrt{x + 2}}\\
&= \lim_{x \to 2^{+}}\frac{\sqrt{x - 2}}{(\sqrt{x} + \sqrt{2})\sqrt{x + 2}} + \frac{1}{\sqrt{x + 2}} = 0 + \frac{1}{\sqrt{2 + 2}} = \frac{1}{2}\end{aligned}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/589470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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} |
Solve the equation $\lfloor x^2\rfloor-3\lfloor x \rfloor +2=0$
Solve the equation
$$
\lfloor x^2\rfloor-3\lfloor x \rfloor +2=0
$$
where $\lfloor x\rfloor $ denotes floor function.
My Attempt:
Let $x = n+f$, where $n= \lfloor x \rfloor \in \mathbb{Z} $, $f=x-\lfloor x \rfloor = \{x\} $, and $0\leq f<1$. Then using $\lfloor n+f \rfloor = n$ gives us
$$
\begin{align}
\lfloor (n+f)^2 \rfloor -3 \lfloor n+f \rfloor +2 &= 0\\
n^2+\lfloor f^2+2nf \rfloor -3n+2 &= 0
\end{align}
$$
How can I solve the equation from here?
| Note that if $x\lt0$, then $\lfloor x^2\rfloor-3\lfloor x\rfloor+2\ge5$, so any solution $x=n+f$ with $0\le f\lt1$ must have $n\ge0$. Picking up at the OP's equation $n^2+\lfloor f^2+2nf\rfloor-3n+2=0$ and letting $r$ denote the non-negative integer $\lfloor f^2+2nf\rfloor$, we find
$$(2n-3)^2=1-4r$$
Thus $r$ must be $0$, which implies $(2n-3)^2=1$, so that $n=1$ or $2$.
If $n=1$, then $r=0$ means $f^2+2f\lt1$, or $(f+1)^2\lt2$, and thus $0\le f\lt\sqrt2-1$, so that $1\le x\lt\sqrt2$.
If $n=2$, then $r=0$ means $f^2+4f\lt1$, or $(f+2)^2\lt5$, and thus $0\le f\lt\sqrt5-2$, so that $2\le x\lt\sqrt5$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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} |
Simplify $2^{(n-1)} + 2^{(n-2)} + .... + 2 + 1$ Simplify $2^{(n-1)} + 2^{(n-2)} + .... + 2 + 1$
I know the answer is $2^n - 1$, but how to simplify it?
| $$\begin{eqnarray}
\\ \text{Let } 2^{(n-1)} + 2^{(n-2)} + \cdots + 2 + 1 &=& S
\\ 2^{n} + 2^{(n-1)} + .... + 2^2 + 2 &=& 2S \tag {multiply both sides by 2}
\\ 2^{n} + 2^{(n-1)} + .... + 2^2 + 2 +1 &=& 2S +1\tag {add 1 to both sides }
\\ 2^{n} + S &=& 2S +1\tag {replace the term on LHS by S }
\\ 2^{n} &=& S +1\tag {subtract S from both sides }
\\ 2^{n} -1 &=& S
\\ S &=& 2^{n} -1 \quad \square
\end{eqnarray}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/590733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
} |
Let a function $f(x)= \sqrt{(x^2-2x+2)(-x^2+3x+10)}-\sqrt{(-x^{2}+x-1)(x^{2}+x-6)}$ Let a function $f(x)= \sqrt{(x^2-2x+2)(-x^2+3x+10)}-\sqrt{(-x^{2}+x-1)(x^{2}+x-6)}$ and its domain is $D=[a,b+c]$. What's the value of $a^3+b^3+c^3 \over abc$ ?
Any ideas ?
The answer is $\boxed{3}$ .
| in the first term $x^2-2x+2$ is always greater than $0$. hence $-x^2+3x+10$ should be greater than or equal to $0$.
similarly for the second term $-x^2+x-1$ is always less than $0$. hence $x^2+x-6$ should be less than or equal to $0$.
now find the common interval in which both inequality satisfies . the domain will be $[-2,2]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/591154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Fast way to prove that $(a^2+b^2+c^2-ab-ac-bc)^2=(a-b)^2\times (a-c)^2 + (b-c)^2\times(b-a)^2 + (c-b)^2\times(c-a)^2$ What is the most simplest way to prove that $$(a^2+b^2+c^2-ab-ac-bc)^2=(a-b)^2\times (a-c)^2 + (b-c)^2\times(b-a)^2 + (c-b)^2\times(c-a)^2$$
Please!! Thanxx
| For brevity's sake, let $x = a - b$, $y = b - c$, and $z = c - a$. We compute a few products:
$$-xz = (a - b)(a - c) = a^2 - ab - ac + bc$$
$$-xy = (b - a)(b - c) = b^2 - ab - bc + ac$$
$$-yz = (c - a)(c - b) = c^2 - ac - bc + ab$$
Adding these all together gives:
$$ - (xy + yz + zx) = a^2 + b^2 + c^2 - ab - ac - bc $$
which we want to square.
$$
\begin{align*}
(a^2 + b^2 + c^2 - ab - ac - bc)^2 &= (xy + yz + xz)^2 \\
&= x^2y^2 + y^2z^2 + x^2z^2 + 2 xyz^2 + 2 xy^2z + 2 x^2yz \\
&= x^2y^2 + y^2z^2 + x^2z^2 + 2 xyz (x + y + z)
\end{align*}
$$
But $x + y + z$ is zero, so we're left with the desired conclusion:
$$
\begin{align*}
(a^2 + b^2 + c^2 - ab - ac - bc)^2 &= x^2y^2 + y^2z^2 + x^2z^2 \\
&= (a - b)^2(b - c)^2 + (b - c)^2(c - a)^2 + (a - b)^2(c - a)^2
\end{align*}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/593121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to prove $\sum_{n=0}^{\infty} \frac{n^2}{2^n} = 6$? I'd like to find out why
\begin{align}
\sum_{n=0}^{\infty} \frac{n^2}{2^n} = 6
\end{align}
I tried to rewrite it into a geometric series
\begin{align}
\sum_{n=0}^{\infty} \frac{n^2}{2^n} = \sum_{n=0}^{\infty} \Big(\frac{1}{2}\Big)^nn^2
\end{align}
But I don't know what to do with the $n^2$.
| First we compute the sum $S_1=\sum_{n=1}^\infty \frac{n}{2^n}$ using the identity $n=\sum_{k=1}^n1$ $$S_1=\sum_{n=1}^\infty \frac{n}{2^n}=\sum_{n=1}^\infty \sum_{k=1}^n \frac{1}{2^n}=\sum_{ k \leq n }\frac{1}{2^n}=\sum_{k=1}^\infty \sum_{n=k}^\infty \frac{1}{2^n}=\sum_{k=1}^\infty \frac{1/2^k}{1-1/2}=2 \sum_{k=1}^\infty \frac{1}{2^k}=2. $$
Next we compute the sum $S_2=\sum_{n=1}^\infty \frac{n+n^2}{2^{n+1}}$ using the identity $\frac{n+n^2}{2}=\sum_{k=1}^n k$
$$S_2=\sum_{n=1}^\infty \sum_{k=1}^n \frac{k}{2^n}=\sum_{k \leq n} \frac{k}{2^n}=\sum_{k=1}^\infty \sum_{n=k}^\infty \frac{k}{2^n}=\sum_{k=1}^\infty k \frac{1/2^k}{1-1/2}=2 \sum_{k=1}^\infty \frac{k}{2^k}=2S_1=4. $$
Finally the desired sum is $S=2S_2-S_1=6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/593996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "44",
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Solid Revolution question I'm trying to do the following problem:
Find the volume of the solid obtained by rotating the region bounded by the given curves about the $y=3$ axis specified by means of the circular arrow drawn :
$$ y = 3 \sin x, y = 3 \cos x, 0 \le x \le π/4 $$
First off, am I drawing the picture correctly?
Second, am I setting up the integral correctly?
$$
\pi\int_0^{\pi/4} [(3-3\sin x)^2-(3-3\cos x)^2]dx
$$
Did I correctly identify $3-3 \sin x$ as the outer radius? Because it makes sense for it to be that way, but I saw this same question worked out on like yahoo answers and it instead says that $y=3-3 \cos x $ should be the outer radius?
| Yahoo!Answers is notoriously unreliable on math and technical matters; my informal count is that the answer is alright only about half the time. (I suspect the poster overlooked the rotation axis being $ \ y = 3 \ , $ rather than the $ \ x-$ axis.) Your integrand looks fine and reduces to
$$ \ (9 - 18 \sin x + 9 \sin^2 x) \ - \ (9 - 18 \cos x + 9 \cos^2 x) $$
$$ = \ 18 \ (\cos x - \sin x) \ + \ 9 \ (\sin^2 x - \cos^2 x) \ = \ 18 \ (\cos x - \sin x) \ - \ 9 \ \cos 2x \ . $$
The evaluation of the volume is then
$$ \pi \ \left[ \ 18 \ (\sin x + \cos x) \ - \ \frac{9}{2} \sin 2x \ \right]_0^{\pi/4} $$
$$ = \ \pi \ \left( \ [ \ 18 \ (\ \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}) \ - \ \frac{9}{2} \cdot 1 \ ] \ - \ [ \ 18 \ (0 + 1) \ - \ \frac{9}{2} \cdot 0 \ ] \ \right) \ $$
$$ = \ \pi \ ( \ 18 \sqrt{2} \ - \ \frac{9}{2} \ - \ 18 \ ) \ = \ \pi \ ( \ 18 \sqrt{2} \ - \ \frac{45}{2} \ ) \ \ \text{or} \ \ \frac{9 \pi}{2} \ ( \ 4 \sqrt{2} \ - \ 5 \ ) \ \ , $$
confirming JohnD's result in the comment above.
| {
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"url": "https://math.stackexchange.com/questions/596500",
"timestamp": "2023-03-29T00:00:00",
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$a, b \in\Bbb N$, $A=\sqrt{a^2+2b+1}+\sqrt{b^2+2a+1}\in \Bbb N $, to show that $a = b$. If $a$ and $b$ are positive integers such that $A=\sqrt{a^2+2b+1}+\sqrt{b^2+2a+1}\in \Bbb N $, to show that $a = b$.
By contradiction assuming that $a <b$ then it follows that $2 (a +1) <A <2 (b +1)$.
This means that $A = 2a + r,$ $r$ taking values $3, 4, ..., 2b-2a +1$.
From the equality
(1) $\sqrt{a^2+2b+1}+\sqrt{b^2+2a+1}= 2a+r$ following
$(a-b)(b+a-2)=(2a+r)(\sqrt{a^2+2b+1}-\sqrt{b^2+2a+1})$ and then
(2) $\sqrt{a^2+2b+1}-\sqrt{b^2+2a+1}=\frac{(a-b)(b+a-2)}{2a+r} \in \Bbb Q. $
From (1) and (2) obtain $\sqrt{a^2+2b+1}\in \Bbb Q$ and $ \sqrt{b^2+2a+1}\in \Bbb Q$. Finally
$a^2+2b+1=p^2$ and $b^2+2a+1= q^2$, $p, q \in \Bbb N.$
Hence the attempts made not lead to anything conclusive. Does anyone have any suggestions?
| First, if $n\in \Bbb N$ and $\sqrt{n} \in \Bbb Q$, then $\sqrt{n} \in \Bbb N$.
To see why, notice first that if $n$ is a perfect square, there is nothing to prove.
So, suppose $n\in \Bbb N$ and $\sqrt{n} \in \Bbb Q$, and $n$ is not a perfect square. Then there is a prime number $\alpha$ such that $\alpha$ factors in $n$ to an odd power, that is: $\alpha^{2k-1}|n$ but $\alpha^{2k}\not|n$, for some integer $k$.
You have $\sqrt{n}=\frac{p}{q}$, with coprime $p$ and $q$. Then $p^2=nq^2$. Since $\alpha^{2k-1}|n$, you have also $\alpha^{2k-1}|p^2$, thus $\alpha^{k}|p$, but then $\alpha^{2k}|nq^2$, thus $\alpha^{2k}|n$ or $\alpha|q$, but both are impossible.
Hence there is a contradiction, and $n$ must be a perfect square, so $\sqrt{n} \in \Bbb N$.
Edit: There was a mistake above, $\alpha^2$ may divide $n$, the important thing it that there is a prime $\alpha$ with an odd power in the prime factorization of $n$, when $n$ is not a perfect square. For example, $27$ is not a square, but obviously $3^2|27$.
Now, there is no square integer between $b^2$ and $b^2+2b+1=(b+1)^2$. Since $a<b$, $b^2+2a+1$ can't be a perfect square, so its square root can't be a rational, thus you have the contradiction you want.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $\frac{\log(2x+1)-\log 4}{1-\log(3x+2)}=1$ My attempt:$$\frac{\log(2x+1)-\log4}{1-\log(3x+2)}=1$$
$$\frac{\log(2x+1)-\log4}{\log10-\log(3x+2)}=\log10$$
$$\frac{\log\frac{(2x+1)}{4}}{\log\frac{10}{(3x+2)}}=\log10$$
$$\frac{\frac{(2x+1)}{4}}{\frac{10}{(3x+2)}}=10$$
$$\frac{(2x+1)}{4}=10\cdot\frac{10}{(3x+2)}$$(is this even correct ??)
$$\frac{(2x+1)}{4}=\frac{100}{(3x+2)}$$
$$(2x+1)100=4(3x+2)$$
$$200x+100=12x+8$$
$$200x+100=12x+8$$
$$188x=-92$$
$$x=-23/47$$
The solution is 2.
| $$\frac{\log(2x+1)-\log4}{1-\log(3x+2)}=1$$
$$\frac{\log(2x+1)/4}{\log 10/(3x+2)}=1$$
$$\log(2x+1)/4=\log 10/(3x+2)$$
$$(2x+1)/4=10/(3x+2)$$
$$(2x+1)(3x+2)=40$$
$$6x^2+7x-38=0$$
$$x_{1,2}=\frac{-7\pm31}{12}:x_1=2,x_2=-\frac{19}{6}$$
in real field only $x_1=2$ is a solution
| {
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"timestamp": "2023-03-29T00:00:00",
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Show $(1+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{9}+\frac{1}{11}-\cdots)^2 = 1+\frac{1}{9}+\frac{1}{25}+\frac{1}{49} + \cdots$ Last month I was calculating $\displaystyle \int_0^\infty \frac{1}{1+x^4}\, dx$ when I stumbled on the surprising identity:
$$\sum_{n=0}^\infty (-1)^n\left(\frac{1}{4n+1} +\frac{1}{4n+3}\right) = \frac{\pi}{\sqrt8}$$
and I knew
$$\sum_{n=0}^\infty \frac{1}{(2n+1)^2} = \frac{\pi^2}{8}$$
So if I could find a proof that $$\left(\sum_{n=0}^\infty (-1)^n\left(\frac{1}{4n+1} +\frac{1}{4n+3}\right)\right)^2 = \sum_{n=0}^\infty \frac{1}{(2n+1)^2}$$ then this could be a new proof that $\zeta(2)=\frac{\pi^2}{6}$. I've thought over this for almost a month and I'm no closer on showing this identity.
Note: Article on the multiplication of conditionally convergent series: http://www.jstor.org/stable/2369519
| Let $a_k = (-1)^k \left(\frac{1}{4k+1} + \frac{1}{4k+3}\right)$ and $b_k = \frac{1}{(4k+1)^2} + \frac{1}{(4k+3)^2}$. The goal is to show that: $$ \left(\sum_{i=0}^\infty a_i\right)^2 = \sum_{i=0}^\infty b_i $$
The key observation that I missed on my previous attempt is that: $$ \sum_{i=0}^n a_i = \sum_{i=-n-1}^n \frac{(-1)^i}{4i+1} $$ This transformation allows me to then mimic the proof that was suggested in the comments by @user17762.
\begin{align*} \left(\sum_{i=0}^n a_i\right)^2 - \sum_{i=0}^n b_i
&= \left(\sum_{i=-n-1}^n \frac{(-1)^i}{4i+1}\right)^2 - \sum_{i=-n-1}^n \frac{1}{(4i+1)^2} \\
&= \sum_{\substack{i,j=-n-1 \\ i \neq j}}^n \frac{(-1)^i}{4i+1}\frac{(-1)^j}{4j+1} \\
&= \sum_{\substack{i,j=-n-1 \\ i \neq j}}^n \frac{(-1)^{i+j}}{4j-4i}\left(\frac{1}{4i+1}-\frac{1}{4j+1} \right) \\
&= \sum_{\substack{i,j=-n-1 \\ i \neq j}}^n \frac{(-1)^{i+j}}{2j-2i} \cdot \frac{1}{4i+1} \\
&= \frac{1}{2}\sum_{i=-n-1}^n \frac{(-1)^i}{4i+1} \sum_{\substack{j=-n-1 \\ i \neq j}}^n \frac{(-1)^j}{j-i} \\
&= \frac{1}{2}\sum_{i=-n-1}^n \frac{(-1)^i }{4i+1}c_{i,n} \\
&= \frac{1}{2}\sum_{i=0}^n a_i \,c_{i,n}
\end{align*}
Where the last equality follows from $c_{i,n} = c_{-i-1, n}$. Since $c_{i,n}$ is a partial alternating harmonic sum, it is bounded by its largest entry in the sum: $\left| c_{i,n} \right| \le \frac{1}{n-i+1}$. We also know that $\left|a_i\right| \le \frac{2}{4i+1}$. Apply these two inequalities to get:
\begin{align*}\left| \left(\sum_{i=0}^n a_i\right)^2 - \sum_{i=0}^n b_i \right| &\le \frac{1}{2} \sum_{i=0}^n \frac{2}{4i+1} \cdot \frac{1}{n-i+1} \\ &\le \sum_{i=0}^n \frac{1}{4n+5}\left( \frac{4}{4i+1} + \frac{1}{n-i+1} \right) \\ &\le \frac{1}{4n+5}\left( 5 + \ln(4n+1) +\ln(n+1)\right) \\
& \to 0 ~\text{ as }~ n \to \infty
\end{align*}
This concludes the proof. In fact, with the same idea, you can prove this general family of identities: Fix an integer $m \ge 3$, then:
\begin{align*} & \left( 1 + \frac{1}{m-1} - \frac{1}{m+1} - \frac{1}{2m-1} + \frac{1}{2m+1} + \frac{1}{3m-1} - \cdots \right)^2 \\
=& ~ \left(\sum_{i=-\infty}^\infty \frac{(-1)^i}{im+1}\right)^2 \\
=& ~ \sum_{i=-\infty}^\infty \frac{1}{(im+1)^2} \\
=& ~ 1 + \frac{1}{(m-1)^2} + \frac{1}{(m+1)^2} + \frac{1}{(2m-1)^2} + \frac{1}{(2m+1)^2} + \frac{1}{(3m-1)^2} + \cdots \\
=& ~ \left(\frac{\frac{\pi}{m}}{\sin\frac{\pi}{m}}\right)^2 \end{align*}
The last equality follows from the comment by @Lucian.
| {
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} |
Closed form for $\int_0^\alpha \int_0^{\alpha-x} \frac{1}{[(x+y)(\beta-x)(1-\beta-y)]^{3/2}} \,dy \,dx$ Let $0\leq \alpha \leq \frac{1}{2},$ and let $\beta$ be such that $\alpha \leq \beta \leq 1-\alpha$.
Define $\varphi(\alpha,\beta)$ to be the integral
$$\varphi(\alpha,\beta) = \int_0^\alpha \int_0^{\alpha-x} \frac 1 {[(x+y)(\beta-x)(1-\beta-y)]^{3/2}} \,dy\, dx.$$
It'd be nice to have some closed expression for this integral, but so far I've been unable to come up with one.
Mathematica starts to spit out some really large expressions involving lots of complex numbers, and I've been unable to convince it that the answer should be real.
Does anyone know of a substitution or other method that would yield a nicer expression?
| Let $z = x+y$, we have
$$\varphi(\alpha,\beta) = \int_0^\alpha \frac{dz}{z^{3/2}}\left\{\int_0^z \frac{dy}{((\beta-z+y)(1-\beta-y))^{3/2}}\right\}$$
For fixed $z$, introduce variables $u, v$ such that $u = \beta -z + y$, $v = 1 - \beta-y$. It is easy to check
$$d\left(\frac{u-v}{\sqrt{uv}}\right)
= \left(\frac{2}{\sqrt{uv}} - \frac{(u-v)(v-u)}{2\sqrt{uv}^3}\right) dy
= \frac{(u+v)^2}{2\sqrt{uv}^3} dy = \frac{(1-z)^2}{2\sqrt{uv}^3} dy
$$
and hence
$$\begin{align}
\varphi(\alpha,\beta)
=& 2 \int_0^\alpha \frac{dz}{z^{3/2}(1-z)^2} \left\{
\left[\frac{u-v}{\sqrt{uv}}\right]_{y=0}^{y=z}
\right\}\\
=& 2 \int_0^\alpha \frac{dz}{z^{3/2}(1-z)^2}\left\{
\frac{2\beta-1 +z}{\sqrt{\beta(1-\beta-z)}} - \frac{2\beta-1-z}{\sqrt{(1-\beta)(\beta-z)}}
\right\}\\
=& -4\frac{\partial}{\partial\beta} \int_0^\alpha \frac{dz}{z^{3/2}(1-z)^2}\left\{
\sqrt{\beta(1-\beta-z)} - \sqrt{(1-\beta)(\beta-z)}
\right\}\\
=& -4\frac{\partial}{\partial\beta}\left[
\sqrt{\beta(1-\beta)}
\left( F\left(\frac{1}{1-\beta},\alpha\right) - F\left(\frac{1}{\beta},\alpha\right) \right)
\right]
\end{align}
$$
where
$$F(\mu,z) = \int^z \frac{\sqrt{1-\mu t}}{t^{3/2}(1-t)^2} dt + \text{ constant }.$$
To evaluate $F(\mu,z)$, let us introduce new variables
$$s = \frac{t}{1-\mu t} \;\;\leftrightarrow\;\; t = \frac{s}{1+\mu s}
\quad\implies\quad ds = \frac{dt}{(1-\mu t)^2}
$$
and let $\nu = \mu - 1$. Notice $0 < \beta < 1$ implies both
$\frac{1}{1-\beta}$ and $\frac{1}{\beta}$ are bigger than $1$. This means in this problem,
$\nu$ is always positive. In terms of these variables, we can transform $F(\mu,z)$ as
$$
F(\mu,z)
= \int^z \frac{dt}{(1-\mu t)^2} \sqrt{\frac{1-\mu t}{t}}\left(\frac{1-\mu t}{t}\right)^2 \frac{t}{(1-t)^2}
= \int \frac{ds}{\sqrt{s}} \frac{1+\mu s}{s(1+\nu s)^2}
$$
Notice
$$\begin{align} \frac{1+\mu s}{s(1+\nu s)^2}
=& \frac{(1+\nu s) + s}{s(1+\nu s)^2}
= \frac{1}{s(1+\nu s)} + \frac{1}{(1+\nu s)^2}
= \frac{1}{s} - \frac{\nu}{1+\nu s} + \frac{1}{(1+\nu s)^2}\\
=& \frac{1}{s} + \left(\frac{\partial}{\partial \nu} - 1 \right) \frac{\nu}{1+\nu s}
= \frac{1}{s} + \left(\nu \frac{\partial}{\partial \nu} + 1 - \nu\right)\frac{1}{1 + \nu s}
\end{align}$$
We get
$$\begin{align}
F(\mu, z)
= & -\frac{2}{\sqrt{s}} + \left(\nu \frac{\partial}{\partial \nu} + 1 - \nu\right)
\int \frac{ds}{\sqrt{s}}\frac{1}{1+\nu s}\\
= & -\frac{2}{\sqrt{s}} + \left(\nu \frac{\partial}{\partial \nu} + 1 - \nu\right)
\left[\frac{2}{\sqrt{\nu}}\tan^{-1}\sqrt{\nu s}\right]\\
= & \left(\frac{s}{1+\nu s} - 2\right)\frac{1}{\sqrt{s}} + \frac{1-2\nu}{\sqrt{\nu}}\tan^{-1}\sqrt{\nu s}\\
= & \frac{3z-2}{\sqrt{z}(1-z)}\sqrt{1-\mu z} + \frac{3-2\mu}{\sqrt{\mu - 1}}
\tan^{-1}\sqrt{\frac{(\mu - 1)z}{1-\mu z}}
\end{align}
$$
This leads to
$$\begin{align}
& \sqrt{\beta(1-\beta)}\left(
F(\frac{1}{1-\beta},\alpha) - F(\frac{1}{\beta},\alpha)\right)\\
= & \frac{3\alpha-2}{\sqrt{\alpha}(1-\alpha)}
\left(\sqrt{\beta(1-\beta-\alpha)} - \sqrt{(1-\beta)(\beta-\alpha)}\right)\\
&+(1-3\beta)\tan^{-1}\sqrt{\frac{\beta \alpha}{1-\beta-\alpha}}
-(3\beta-2)\tan^{-1}\sqrt{\frac{(1-\beta)\alpha}{\beta-\alpha}}
\end{align}$$
and as a result,
$$\begin{align}
\varphi(\alpha,\beta)
= &
2\left\{
\frac{3\alpha-2}{\sqrt{\alpha}(1-\alpha)}\left(
\frac{2\beta-1+\alpha}{\sqrt{\beta(1-\beta-\alpha)}}
-\frac{2\beta-1-\alpha}{\sqrt{(1-\beta)(\beta-\alpha)}}
\right)
+6\left(
\tan^{-1}\sqrt{\frac{\beta \alpha}{1-\beta -\alpha}}
+\tan^{-1}\sqrt{\frac{(1-\beta)\alpha}{\beta-\alpha}}
\right)
-\sqrt{\alpha}
\left(
\frac{1-3\beta}{(1-\beta)\sqrt{\beta(1-\beta-\alpha)}}
+ \frac{3\beta-2}{\beta\sqrt{(1-\beta)(\beta-\alpha)}}
\right)
\right\}\\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/599454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Prove that for every $k\geq 1$, $1\cdot 3\cdot(1!)^2+....+k\cdot(k+2)\cdot(k!)^2=((k+1)!)^2-1$ Prove that for every $k\ge1,$
$\displaystyle1\cdot3\cdot(1!)^2+2\cdot4\cdot(2!)^2+....+k\cdot(k+2)\cdot(k!)^2=((k+1)!)^2-1$
I start off with considering induction, but I can't quite get the left side to match the right side. Maybe there is something I'm not seeing or a calculation I have wrong. Any kind of help would be great!
$\displaystyle((n+1)!)^2-1+(n+1)(n+3)((n+1)!)^2=((n+2)!)^2-1$
LHS:
$\displaystyle((n+1)!)^2[(n+1)(n+3)-1]$
$\displaystyle((n+1)!)^2[(n+2)^2-2]$
| Let
$\displaystyle f(k): 1\cdot3\cdot(1!)^2+2\cdot4\cdot(2!)^2+....+k\cdot(k+2)\cdot(k!)^2=((k+1)!)^2-1$ holds true for $k=m$
$\displaystyle\implies 1\cdot3\cdot(1!)^2+2\cdot4\cdot(2!)^2+....+m\cdot(m+2)\cdot(m!)^2=\{(m+1)!\}^2-1 $
For $k=m+1,$
$\displaystyle\implies 1\cdot3\cdot(1!)^2+2\cdot4\cdot(2!)^2+....+m\cdot(m+2)\cdot(m!)^2+\underbrace{(m+1)\{(m+1)+2\}\{(m+1)!\}^2}$
$\displaystyle=\{(m+1)!\}^2-1+\underbrace{(m+1)\{(m+1)+2\}\{(m+1)!\}^2} $
$\displaystyle=\{(m+1)!\}^2\{1+(m+1)(m+3)\}-1$
$\displaystyle=\{(m+1)!\}^2(m+2)^2-1$ as $1+(m+1)(m+3)=(m+2)^2$
$\displaystyle=\{(m+2)\cdot (m+1)!\}^2-1$
$\displaystyle=\{(m+2)!\}^2-1$
So, $f(k)$ will hold true for $k=m+1,$ if it holds true for $k=m$
Now establish the base case i.e., $k=1$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove the following trigonometric identity. $$\frac{\sin x - \cos x +1}{\sin x + \cos x -1}=\frac{\sin x +1}{\cos x}$$
I tried substituting $\sin^2x+\cos^2x = 1$ but I cannot solve it.
| Observe that the Right Hand side $\displaystyle\frac{\sin x+1}{\cos x}=\tan x+\sec x$
So, I want to utilize $\displaystyle\sec^2x-\tan^2x=1$
Dividing the numerator & the denominator by $\cos x,$
$$\frac{\sin x - \cos x +1}{\sin x + \cos x -1}=\frac{\tan x-1+\sec x}{\tan x+1-\sec x}$$
$$=\frac{\tan x+\sec x-(\sec^2x-\tan^2x)}{\tan x+1-\sec x}(\text{ Replacing }1\text{ with } \sec^2x-\tan^2x)$$
$$=\frac{(\sec x+\tan x)-(\sec x+\tan x)(\sec x-\tan x)}{\tan x+1-\sec x}$$
$$=\frac{(\sec x+\tan x)(1-\sec x+\tan x)}{\tan x+1-\sec x}$$
$$=\sec x+\tan x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/604169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
What is $ \lim_{(x,y)\to(2,2)}\frac{x^4-y^4}{x^2 - y^2} $? I have limit:
$$
\lim_{(x,y)\to(2,2)}\frac{x^4-y^4}{x^2 - y^2}
$$
Why is the result $8$ ?
| $$\lim_{x,y\to{2,2}}\frac{x^4-y^4}{x^2-y^2}$$ $$= \lim_{x,y\to{2,2}}\frac{(x^2+y^2)(x^2-y^2)}{x^2-y^2}$$
$$\lim_{x,y\to{2,2}} x^2+y^2 =...$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/604913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Inverse Trig & Trig Sub Can someone explain to me how to solve this using inverse trig and trig sub?
$$\int\frac{x^3}{\sqrt{1+x^2}}\, dx$$
Thank you.
| You can also use integration by part: let $u=x^2$ and $dv=\frac{x}{\sqrt{1+x^2}}$ then you will have:
\begin{align*}
\int udv&=uv-\int vdu\\
&=x^2\sqrt{1+x^2}-\int2x\sqrt{1+x^2}\,dx\\
&=x^2\sqrt{1+x^2}-\frac{2}{3}(1+x^2)^{\frac{3}{2}}+C
\end{align*}
where the last integral was solved by substitution $\ u=1+x^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/606431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Inequality: if $\cos^2a+\cos^2b+\cos^2c=1$, prove $\tan a+\tan b+\tan c\geq 2(\cot a+\cot b+\cot c)$ Let $a,b,c$ be in $\left(0;\dfrac{\pi}2\right)$ such that $\cos^2a+\cos^2b+\cos^2c=1$. I am trying to prove the following inequality: $\tan a+\tan b+\tan c\geq 2\left(\cot a+\cot b+\cot c\right)$, but I do not know how. Does any know help me to show this?
| Lemma:
let $x,y,z>0$,then
$$\dfrac{3}{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}}\le\dfrac{x+y+z}{3}\le\sqrt{\dfrac{x^2+y^2+z^2}{3}}$$
let $$\cos{a}=x,\cos{b}=y,\cos{c}=z,x,y,z>0$$
Note
$$\tan{a}=\dfrac{\sin{a}}{\cos{a}}=\dfrac{\sqrt{1-\cos^2{a}}}{x}=\dfrac{\sqrt{y^2+z^2}}{x}\ge\dfrac{1}{\sqrt{2}}\dfrac{y+z}{x}$$
so
$$\tan{a}+\tan{b}+\tan{c}\ge\dfrac{1}{\sqrt{2}}\left(\dfrac{y}{x}+\dfrac{x}{y}+\dfrac{x}{z}+\dfrac{z}{x}+\dfrac{y}{z}+\dfrac{z}{y}\right)$$
on the other hand,
$$\cot{a}=\dfrac{\cos{a}}{\sin{a}}=\dfrac{x}{\sqrt{y^2+z^2}}\le\dfrac{x\left(\dfrac{1}{y}+\dfrac{1}{c}\right)}{2\sqrt{2}}$$
so
$$\cot{a}+\cot{b}+\cot{c}\le\dfrac{1}{2\sqrt{2}}\left(\dfrac{y}{x}+\dfrac{x}{y}+\dfrac{x}{z}+\dfrac{z}{x}+\dfrac{y}{z}+\dfrac{z}{y}\right)$$
so
$$\tan{a}+\tan{b}+\tan{c}\ge 2(\cot{a}+\cot{b}+\cot{c})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/609197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Triple Euler sum result $\sum_{k\geq 1}\frac{H_k^{(2)}H_k }{k^2}=\zeta(2)\zeta(3)+\zeta(5)$ In the following thread
I arrived at the following result
$$\sum_{k\geq 1}\frac{H_k^{(2)}H_k }{k^2}=\zeta(2)\zeta(3)+\zeta(5)$$
Defining
$$H_k^{(p)}=\sum_{n=1}^k \frac{1}{n^p},\,\,\, H_k^{(1)}\equiv H_k $$
But, it was after long evaluations and considering many variations of product of polylogarithm integrals.
I think there is an easier approach to get the solution, any ideas ?
| Consider the integral $$I= - \int_0^1 \frac{\ln(1-x^2)}{\sqrt{1-x^2}} (\sin^{-1} x)^4 \,dx.$$
Since $$(\sin^{-1} x)^4 = \frac32 \sum_{n=1}^{\infty} \cfrac{2^{2n} H_{n-1}^{(2)}}{n^2 \binom{2n}{n}} \,x^{2 n} \tag{1}$$
and
$$-\int_0^1 \frac{\ln(1-x^2)}{\sqrt{1-x^2}} x^{2n}\,dx= \frac{\pi}{2} \binom{2n}{n} \frac{(H_n + 2\ln2)}{2^{2n}}, \tag{2}$$
we have
$$\begin{align*}
&I= - \frac32 \sum_{n=1}^{\infty} \cfrac{2^{2n} H_{n-1}^{(2)}}{n^2 \binom{2n}{n}} \int_0^1 \frac{\ln(1-x^2)}{\sqrt{1-x^2}} x^{2n}\,dx
\\&= \frac{3 \pi}{4} \sum_{n=1}^{\infty} \frac{H_{n-1}^{(2)}}{n^2} ( H_n +2 \ln2 )
\\& = \frac{\pi^5}{80} \ln2 + \frac{3 \pi}{4} \sum_{n=1}^{\infty} \frac{H_{n-1}^{(2)}\,H_n}{n^2}.
\end{align*}$$
However, substituting $x\mapsto \sin x$ and employing the fourier expansion of $\ln \cos x$:
$$\begin{align*}
& I= -2 \int_0^{\pi/2} x^4 \, \ln\cos x\, dx
\\&= 2 \int_0^{\pi/2} x^4 \left(\ln2 + \sum_{n=1}^{\infty} \frac{(-1)^n \cos(2 x n)}{n} \right)dx
\\&= \frac{\pi^5}{80}\ln2 + 2 \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \int_0^{\pi/2} x^4 \cos(2 x n) dx
\\&= \frac{\pi^5}{80}\ln2 + 2 \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \frac{(-1)^n}{n^2}\left(\frac{\pi^3}{8}-\frac{3 \pi}{4 n^2}\right)
\\&= \frac{\pi^5}{80}\ln2 + \frac{\pi^3}{4}\zeta(3) - \frac{3 \pi}{2} \zeta(5). \end{align*}$$
Therefore,
$$\sum_{n=1}^{\infty} \frac{H_{n-1}^{(2)}\,H_n}{n^2} = 2\zeta(2)\,\zeta(3)-2\zeta(5).$$
Finish off using Euler's formula for $\sum H_n/n^q $.
Notes.
You may find a proof of $(1)$ here, and $(2)$ is just the derivative of a beta function. The swap of the sum and the integral should be justified.
I found this proof while exploring series involving $H_n^{(2)}$. Using the same method, I also obtained the following
related results:
$$\sum_{n=1}^{\infty} \frac{H_{n-1}^{(2)}\,H_{2n}}{n^2} =\frac{11}{4}\zeta(2)\,\zeta(3)-\frac{47}{16}\zeta(5) \tag{3}$$
$$\sum_{n=1}^{\infty} \frac{H_{n-1}^{(2)}\,H_{n}^2}{n^2} = 4 \zeta(3)^2 - \frac{5}{8} \zeta(6) \tag{4}$$
$$\sum_{n=1}^{\infty} \frac{H_n \left(H_{n-1}^{(2)2}-H_{n-1}^{(4)}\right)}{n^2} = 3\,\zeta(3)\,\zeta(4)-4\,\zeta(2)\,\zeta(5)+4\,\zeta(7) \tag{5}$$
and others.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/612181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "44",
"answer_count": 6,
"answer_id": 1
} |
Elevator stop, other approach There is a $10$-floor building, $10$ people get in the elevator in the ground floor and each gets off at one of the floors randomly and independently. What is the probability that the elevator stops at floor $5$?
$$1 - (9/10)^{10}$$
I understand the approach of using the probability that no one gets out on the $3$rd floor, which is $(9/10)^{10}$.
But, my question is how do I approach this problem using the probability that people gets out on the $3$rd floor?
I thought if one person chooses the $5$th floor the probability is $1/10$, and if $2$ people choose the $5$th floor then the probability is $1/9$, and so on.
Thank you.
| Yes, there is an alternative way to solve this, although it's more difficult than the method you mentioned.
Start by considering the case where there are 2 floors and 2 people. The probability of at least one person exiting on our chosen floor, say floor 2, is
$$P(\geq\text{1 person on floor; 2 people and 2 floors}) = \frac{1}{2} + \frac{1}{2}\left(\frac{1}{2}\right) = \frac{3}{4}$$
To see where this came from, we first consider where person 1 gets off. There is a $1/2$ chance the person will exit on our chosen floor, so we add that to our probability. In the event that she doesn't get off on floor 2 (an event with probability $1/2$), we consider the other person. Person 2 has a $1/2$ probability of exiting on floor 2, so we multiply that by the probability that the second person even matters (ie. the probability of the first person not getting off on floor 2).
For 3 people and 3 floors, we'd have
$$P(\geq\text{1 person on floor; 3 people and 3 floors}) = \frac{1}{3} + \frac{2}{3}\left(\frac{1}{3}+\frac{2}{3}\left(\frac{1}{3}\right)\right) = \frac{19}{27}$$
since there's a $1/3$ chance of the first person getting off on our chosen floor, and a $2/3$ probability that they don't and that we have to look at the second person. The same logic is applied to the second person and so on.
For 4 people and 4 floors, we'd have
\begin{align}P(\geq\text{1 person on floor; 4 people and 4 floors}) &= \frac{1}{4} + \frac{3}{4}\left(\frac{1}{4}+\frac{3}{4}\left(\frac{1}{4}+\frac{3}{4}\left(\frac{1}{4}\right)\right)\right) \\
&= \frac{175}{256}\end{align}
For $N$ people and $N$ floors, we'd have
\begin{align}&P(\geq\text{1 person on floor; N people and N floors}) \\ =& \frac{1}{N} + \frac{N-1}{N}\left(\frac{1}{N}+\frac{N-1}{N}\left(\;\;\;\;\left.\ldots+\frac{N-1}{N}\left(\frac{1}{N}\right)\right)\right)\ldots\right)
\end{align}
where the pattern continues until there are $N$ of the fractions $1/N$ present in the expression.
By multiplying this out, we have
\begin{align} P =& \frac{1}{N}+\frac{\left(N-1\right)}{N^2}+\frac{\left(N-1\right)^2}{N^3}+\ldots+\frac{\left(N-1\right)^{N-1}}{N^N} \\
=& \sum_{k=0}^{N-1} \frac{1}{N} \left(\frac{N-1}{N}\right)^k
\end{align}
and using the formula for the sum of a finite geometric series, ie
$$\sum_{k=0}^{n-1} a r^k = \frac{a\left(1-r^n\right)}{1-r}$$
we get,
\begin{align}P =& \frac{\left(\frac{1}{N}\right) \left(1-\left(\frac{N-1}{N}\right)^N\right)}{1-\left(\frac{N-1}{N}\right)} \\
=& 1 - \left(\frac{N-1}{N}\right)^N
\end{align}
which is the same expression as the one given by the more simple method.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/613686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the ratio of areas produced by perpendiculars from the $3$ sides of an equilateral triangle.
A point O is inside an equilateral triangle $PQR$ and the perpendiculars $OL,OM,\text{and } ON$ are drawn to the sides $PQ,QR,\text{and } RP$ respectively. The ratios of lengths of the perpendiculars $OL:OM:ON \text{ is } 1:2:3$.
If $\ \dfrac{\text{area of }LONP}{\text{area of }\Delta PQR}=\dfrac{a}{b}, \quad$ where $a$
and $b$ are integers with no common factors,
what is the value of $a+b$ ?
All that I was able to do is:
Area $LONP=\frac{1}{2} |OL||PL|+\frac{1}{2} |NP||ON|=\frac{1}{2} |OL||PL|+\frac{1}{2} |NP|\ 3|OL|=\frac{1}{2} |OL|\ \left[\ |PL|+3|NP|\ \right]$
Area $PQR=\frac{1}{2} |PR||PQ|\sin 60^o=\frac{\sqrt 3}{4} |PR||PQ|=\frac{\sqrt 3}{4} |NP+RN||PL+LQ|=\frac{\sqrt 3}{4} \left[\ |NP|+|PL|+|RN|+|LQ| \ \right]$
Area $\Delta LON=\frac{1}{2} |ON||OL|\sin 120=\frac{\sqrt 3}{4} |OL|\ 3|OL|=\frac{3\sqrt 3}{4} |OL|^2$
$\mathbf{EDIT : }$Following Suraj M.S 's answer :
$$\begin{align}
\text{Area } \Delta PQR &=\dfrac{3x\ RN}{2}+\dfrac{3x\ PN}{2}+\dfrac{x\ PL}{2}+\dfrac{x\ QL}{2}+ {x\ QM}+{x\ MR} \\
\\
&=\dfrac{3x\ (PN+RN)}{2}+\dfrac{x\ (PL+QL)}{2}+{x\ (QM+MR)}\\
\\
&=\dfrac{3x\ (PR)}{2}+\dfrac{x\ (PQ)}{2}+{x\ (QR)}\\
\\
&=kx(\dfrac{3}{2}+\dfrac{1}{2}+{1})=3kx\\
\end{align}$$
Area $\Delta PQR=\dfrac{1}{2} k^2 \sin 60^o=\dfrac{k^2 \sqrt 3}{4} \implies x=\dfrac{k}{4\sqrt 3}$
$\mathbf{Question: }$How do I now find the area of $LONP$ in terms of $x'$s and/or $k'$s only ?
| Here's a different approach.
Well-known Fact. The sum of the distances from an interior point to the sides of an equilateral triangle is equal to the height of that triangle.
Proof (in case you've never seen it). Using our current triangle, writing $s$ for its side-length, and $h$ for its height:
$$\frac{1}{2} s h = |\triangle PQR| = |\triangle OPQ| + |\triangle OQR| + \triangle ORP| = \frac{1}{2} s(|\overline{OL}|+|\overline{OM}|+|\overline{ON}|)$$
From the Fact, and the given proportionality condition, we have
$$|\overline{OL}| : |\overline{OM}| : |\overline{ON}| : h \;=\; 1 : 2 : 3 : (1+2+3) \;=\; 1:2:3:6 \quad (\star)$$
Now, through $O$, draw lines parallel to the sides of the triangle to create three new equilateral triangles of which $\overline{OL}$, $\overline{OM}$, $\overline{ON}$ are altitudes; call these $\triangle L$, $\triangle M$, $\triangle N$, with side-lengths $\ell$, $m$, $n$.
Observe that $(\star)$ implies
$$\begin{align}
\ell : m : n : s \;&=\; 1 : 2 : 3 : 6\\[4pt]
|\triangle L| : |\triangle M| : |\triangle N| : |\triangle PQR| \;&=\; 1 : 4 : 9 : 36
\end{align}$$
The lines also create a parallelogram with diagonal $\overline{OP}$ (and others with diagonals $\overline{OQ}$ and $\overline{OR}$); call it $\square P$.
Then
$$\begin{align}
|LONP| &= |\square P| + \frac{1}{2}|\triangle L| + \frac{1}{2}|\triangle N| \\[4pt]
&= n |\overline{OL}| + \frac{1}{2} \cdot \frac{1}{36}|\triangle PQR| + \frac{1}{2}\cdot\frac{9}{36}|\triangle PQR| \\[4pt]
&= \frac{3}{6}s \cdot \frac{1}{6} h + \frac{5}{36}|\triangle PQR| \\[4pt]
&= \frac{1}{6} |\triangle PQR| + \frac{5}{36}|\triangle PQR| \\[4pt]
&= \frac{11}{36} |\triangle PQR|
\end{align}$$
So, $\frac{a}{b}=\frac{11}{36}$, whereupon $a+b=47$.
Here's a slicker route to the area ratio. Let $\overline{Q^\prime R^\prime}$ be the added segment through $O$ parallel to $\overline{QR}$, and let $P^\prime$ be the foot of the perpendicular from $P$ to that segment.
Then
$$\begin{align}
|\overline{PP^\prime}| + |\overline{OM}| = h \quad &\implies \quad |\overline{PP^\prime}|:h = (6-2):6 = 2:3 \quad \\[4pt]
&\implies \quad |\triangle PQ^\prime R^\prime| :|\triangle PQR| = 4:9
\end{align}$$
so that
$$\begin{align}
|LONP| &= |\triangle PQ^\prime R^\prime| - \frac{1}{2}|\triangle L| - \frac{1}{2}|\triangle N| \\[4pt]
&= \frac{4}{9}|\triangle PQR| - \frac{5}{36}|\triangle PQR| \\[4pt]
&= \frac{11}{36}|\triangle PQR|
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/615018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Help finding the $\lim\limits_{x \to \infty} \frac{\sqrt[3]{x} - \sqrt[5]{x}}{\sqrt[3]{x} + \sqrt[5]{x}}$ I need help finding the $$\lim_{x \to \infty} \frac{\sqrt[3]{x} - \sqrt[5]{x}}{\sqrt[3]{x} + \sqrt[5]{x}}$$
I did the following:
$$\begin{align*}
\lim_{x \to \infty} \frac{\sqrt[3]{x} - \sqrt[5]{x}}{\sqrt[3]{x} + \sqrt[5]{x}}
=& \lim_{x \to \infty} \frac{(\sqrt[3]{x} - \sqrt[5]{x})(\sqrt[3]{x} + \sqrt[5]{x})}{(\sqrt[3]{x} + \sqrt[5]{x})(\sqrt[3]{x} + \sqrt[5]{x})}\\
\\
=& \lim_{x \to \infty} \frac{(\sqrt[3]{x})^2 - (\sqrt[5]{x})^2}{(\sqrt[3]{x})^2+2\sqrt[3]{x}\sqrt[5]{x}+(\sqrt[3]{x})^2}\\
\\
=& \lim_{x \to \infty} \frac{x^{2/3}-x^{2/5}}{x^{2/3}+2x^{1/15}+x^{2/5}}\\
\\
=& \lim_{x \to \infty} \frac{x^{4/15}}{2x^{17/15}}
\end{align*}$$
Somehow I get stuck. I am sure I did something wrong somewhere.. Can someone please help me out?
| Your last expression is completely wrong. This question is not solved in your way.
HINT:
$$\frac{x^{1/3}-x^{1/5}}{x^{1/3}+x^{1/5}}=\frac{1-x^{(1/5)-(1/3)}}{1+x^{(1/5)-(1/3)}}$$
| {
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} |
Limits of square root $$\lim_{x\to\infty}\left(\sqrt{x+\sqrt{x+\sqrt{x + \sqrt x} }}-\sqrt x\right) $$
(original screenshot)
Compute the limit
Can you please help me out with this limit problem
| Here's a comparatively clean way to do it: $$ \sqrt{x+\sqrt x}-\sqrt x\le\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x}}}}-\sqrt{x}\le\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}-\sqrt{x} $$
Now, let $u=\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}$. Then $$u^2=x+u\implies u=\frac{1+\sqrt{1+4x}}{2}=\frac12+\sqrt{\frac14+x}$$
(Note that $u$ is strictly positive). Now, $$\begin{align}\sqrt{x+\sqrt{x}}-\sqrt{x}&=\frac{x+\sqrt{x}-x}{\sqrt{x+\sqrt{x}}+\sqrt{x}}\\
&=\frac{\sqrt{x}}{\sqrt{x+\sqrt{x}}+\sqrt{x}}\\
&=\frac{1}{\sqrt{1+\frac1{\sqrt x}}+1}\end{align}$$
Thus we have $$\begin{align} \lim_{x\to\infty}\frac{1}{\sqrt{1+\frac1{\sqrt x}}+1}\le\lim_{x\to\infty}\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x}}}}-\sqrt{x}&\le\lim_{x\to\infty}\frac12+\sqrt{\frac14+x}-\sqrt x\\
\frac12\le\lim_{x\to\infty}\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x}}}}-\sqrt{x}&\le\frac12\\
\lim_{x\to\infty}\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x}}}}-\sqrt{x}&=\frac12 \end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 6,
"answer_id": 1
} |
Finding the maximum value of a function on an ellipse Let $x$ and $y$ be real numbers such that $x^2 + 9 y^2-4 x+6 y+4=0$. Find the maximum value of $\displaystyle \frac{4x-9y}{2}$.
My solution: the given function represents an ellipse. Rewriting it, we get $\displaystyle (x-2)^2 + 9(y+\frac{1}{3})^2=1$. To find the maximum of $\displaystyle \frac{4x-9y}{2}$, $x$ should be at its maximum and $y$ at its minimum. Solving the equations, I get that $x = 3$ and $\displaystyle y = -\frac{2}{3}$, but the answer i get is wrong. What am I doing wrong?
| Using parametric form, $\displaystyle x-2=\cos\phi,3\left(y+\frac13\right)=\sin\phi$
$\displaystyle\implies x=2+\cos\phi, 3y=\sin\phi-1 $
$\displaystyle\implies\frac{4x-9y}2=\frac{4(2+\cos\phi)-3(\sin\phi-1)}2=\frac{11+4\cos\phi-3\sin\phi}2$
Can you prove $$-\sqrt{a^2+b^2}\le a\cos \theta+b\sin\theta\le \sqrt{a^2+b^2}?$$
| {
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"answer_id": 2
} |
So close yet so far Finding $\int \frac {\sec x \tan x}{3x+5} dx$ Cruising the old questions I came across juantheron asking for $\int \frac {\sec x\tan x}{3x+5}\,dx$ He tried using $(3x+5)^{-1}$ for $U$ and $\sec x \tan x$ for $dv$while integrating by parts. below is his work.
How can I calculate
$$
\int {\sec\left(x\right)\tan\left(x\right) \over 3x + 5}\,{\rm d}x
$$
My Try:: $\displaystyle \int \frac{1}{3x+5}\left(\sec x\tan x \right)\,\mathrm dx$
Now Using Integration by Parts::
We get
$$= \frac{1}{3x+5}\sec x +\int \frac{3}{(3x+5)^2}\sec x\,\mathrm dx$$
Here he hit his road block.
I tried the opposite tactic
Taking the other approach by parts.
let $$U= \sec x \tan x$$ then$$ du= \tan^2 x \sec x +\sec^3 x$$ and $$dv=(3x+5)^{-1}$$ then $$v=\frac 1 3 \ln(3x+5)$$ Thus $$\int \frac {\sec x \tan x}{3x+5}\,dx= \frac {\ln(3x+5)\sec x \tan x}{3} - \int \frac {\ln(3x+5) [\tan^2 x \sec x +\sec^3 x]}{3} \,dx$$
As you can see I got no further than he did.
So how many times do you have to complete integration by parts to get the integral of the original $\frac {\sec x \tan x}{3x+5} \, dx$ or is there a better way?
| Finally figured this one out.
${let: 3x+5= \sec \theta, x=\frac {\sec\theta+5}{3}, dx=\frac{1}{3}\sec\theta\tan\theta d\theta}$
then we have $${\frac {1}{3}\int\frac{\sec\frac{\sec\theta+5}{3}\tan\frac{\sec\theta+5}{3}}{\sec\theta}\sec\theta\tan\theta d\theta=}$$ $${\frac{1}{3}\int\cos\theta {\sec\frac{\sec\theta+5}{3}\tan\frac{\sec\theta+5}{3}}\sec\theta\tan\theta d\theta}$$ Then
${let: u=\frac{\sec\theta+5}{3}=, du=\frac{1}{3}\sec\theta\tan\theta d\theta}$ $${\frac {1}{9}\cos\theta\int\sec u \tan u du=}$$ $${\frac{1}{9}\cos\theta\sec u+c=}$$ $${\frac{1}{9}\frac{\sec x}{3x+5}+c}$$
${Edit}$
Nope back to the drawing board.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
evaluation of $\lim_{x\rightarrow \infty}\frac{\ln x^n-\lfloor x \rfloor }{\lfloor x \rfloor} = $ (1) $\displaystyle \lim_{x\rightarrow \infty}\frac{\ln x^n-\lfloor x \rfloor }{\lfloor x \rfloor} = $, where $n\in \mathbb{N}$ and $\lfloor x \rfloor = $ floor function of $x$
(2)$\displaystyle \lim_{x\rightarrow \infty}\left({\sqrt{\lfloor x^2+x \rfloor }-x}\right) = , $where $\lfloor x \rfloor = $ floor function of $x$
$\bf{My\; Try}::$ for (1) one :: We can write as $\displaystyle \lim_{x\rightarrow \infty}\frac{n\cdot \ln x-\lfloor x \rfloor }{\lfloor x \rfloor}$
and we can say that when $x\rightarrow \infty$, Then $\lfloor x\rfloor \rightarrow x$
So $\displaystyle \lim_{x\rightarrow \infty}\frac{n\cdot \ln(x)-x}{x} = n\lim_{x\rightarrow \infty}\frac{\ln (x)}{x}-1$
Now Let $\displaystyle L = \lim_{x\rightarrow \infty}\frac{\ln(x)}{x}{\Rightarrow}_{L.H.R} =\lim_{x\rightarrow \infty}\frac{1}{x} = 0$
So $\displaystyle \lim_{x\rightarrow \infty}\frac{n\cdot \ln x-\lfloor x \rfloor }{\lfloor x \rfloor} = n\cdot 0-1 =-1$
$\bf{My\; Try}::$ for (2)nd one::we can say that when $x\rightarrow \infty$, Then $\lfloor x^2+x\rfloor\rightarrow (x^2+x)$
So $\displaystyle \lim_{x\rightarrow \infty}\left({\sqrt{x^2+x}-x}\right) = \lim_{x\rightarrow \infty}\frac{\left({\sqrt{x^2+x}-x}\right)\cdot \left({\sqrt{x^2+x}+x}\right)}{\left({\sqrt{x^2+x}+x}\right)}$
$\displaystyle \lim_{x\rightarrow \infty}\frac{x}{\left(\sqrt{x^2+x}+x\right)} = \frac{1}{2}$
Now my doubt is can we write when $x\rightarrow \infty$, Then $\lfloor x\rfloor \rightarrow x$
and when $x\rightarrow \infty$, Then $\lfloor x^2+x\rfloor\rightarrow (x^2+x)$
please clear me
Thanks
| It is not true that
$\lfloor x \rfloor
\to x
$.
What is true is that
$\frac{\lfloor x \rfloor}{x}
\to 1
$.
All that you need for
(1) is that
$\frac{\ln x}{x}
\to 0
$.
For (2),
note that
$x^2+x
\le
\lfloor x^2+x \rfloor
<
x^2+x+1
$.
Therefore
$\lfloor x^2+x \rfloor
= x^2+x+c
$
where
$0 \le c < 1
$.
You can then write
$\begin{align}
\sqrt{\lfloor x^2+x \rfloor}-x
&=\sqrt{x^2+x+c}-x\\
&=(\sqrt{x^2+x+c}-x)\frac{\sqrt{x^2+x+c}+x}{\sqrt{x^2+x+c}+x}\\
&=\frac{x^2+x+c-x^2}{\sqrt{x^2+x+c}+x}\\
&=\frac{x+c}{\sqrt{x^2+x+c}+x}\\
\end{align}
$
and you can show that this $\to \frac12$
as you did in your answer.
To show that
$\frac{x+c}{\sqrt{x^2+x+c}+x}
\to \frac12$,
note that
$x^2
< x^2+x+c
<x^2+x+1
<(x+1)^2
$,
so
$2x
< \sqrt{x^2+x+c}+x
< 2x+1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
$(1+1/x)(1+1/y)(1+1/z) = 3$ Find all possible integer values of $x$, $y$, $z$ given all of them are positive integers. Find all possible integer values of $x$, $y$, $z$ given all of them are positive integers
and
$$(1+1/x)(1+1/y)(1+1/z) = 3.$$
I know
$(x+1)(y+1)(z+1) = 3xyz$ which is no big deal. I can't move forward now.
morever it is given $x$ is less than or equal to $y$ and $y$ is less than or equal to $z$
| Let $x$ be the smallest of the three.
Case 1: $x=1$.
Then
$$ y=\frac{2\,z+2}{z-2} = 2 + \frac{6}{z-2}$$
So $z-2$ should divide $6$
So the possibilities are $z=3,4,5,8$ If we want $x \le y \le z$, we can eliminate two of them.
case 2: $x=2$
Then
$$ y = \frac{z+1}{z-1} = 1 + \frac{2}{z-1}$$
So $z-1$ should divide 2, so z=2, or 3.
case 3:$x > 2$.
In this case show that $y < x$ and we can stop.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Find $\sum_{n=0}^\infty\frac{2^n}{3^{2^{n-1}}+1}$ Find $\sum_{n=0}^\infty\frac{2^n}{3^{2^{n-1}}+1}$.
I already proved that it converges but I can't find the sum, Although I think it should be $1+\frac{1}{\sqrt3 +1}$, from calculations.
| Start with following infinite product expansion which is valid for $|q| < 1$,
$$\prod_{n=0}^\infty \left(1 + q^{2^n}\right) = \sum_{n=0}^\infty q^n = \frac{1}{1-q}$$
Taking logarithm and apply $q\frac{\partial}{\partial q}$ on both sides, one get
$$\sum_{n=0}^\infty \frac{2^n q^{2^n}}{1 + q^{2^n}} = \frac{q}{1-q}$$
Substitute $q$ by $\frac{1}{\sqrt{3}}$, one obtain
$$\sum_{n=0}^\infty \frac{2^n}{3^{2^{n-1}}+1} = \frac{1}{\sqrt{3}-1} = \frac{\sqrt{3}+1}{2} = 1 + \frac{\sqrt{3}-1}{2} = 1 + \frac{1}{\sqrt{3}+1}$$
the expression you expected.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
How to find coefficient of $x^8$ in $\frac{1}{(x+3)(x-2)^2}$ Using which way can we find coefficient of $x^8$ in $\frac{1}{(x+3)(x-2)^2}$? I have used binomial theorem but failed to find an answer for it.
| You can decompose
$$
\frac{1}{(x+3)(x-2)^2}=\frac{A}{1+x/3}+\frac{B}{1-x/2}+\frac{C}{(1-x/2)^2}
$$
and it's just a matter of finding $A$, $B$ and $C$. For $x=0$, $x=1$ and $x=-1$, we get
$$
\begin{cases}
\frac{1}{12}=A+B+C\\
\frac{1}{4}=\frac{3}{4}A+2B+4C\\
\frac{1}{18}=\frac{3}{2}A+\frac{2}{3}B+\frac{4}{9}C
\end{cases}
$$
or
$$
\begin{cases}
A+B+C=1/12\\
3A+8B+16C=1\\
27A+12B+8C=1
\end{cases}
$$
This gives
$$
A=\frac{1}{75},\quad B=\frac{1}{50},\quad C=\frac{1}{20}
$$
(if I computed right).
Now you can recall that
$$
\frac{1}{1+x/3}=\sum_{k\ge0}(-1)^k\frac{x^k}{3^k},
\qquad
\frac{1}{1-x/2}=\sum_{k\ge0}\frac{x^k}{2^k},
$$
and that $1/(1-x/2)^2$ is proportional to the derivative of $1/(1-x/2)$, precisely
$$
\left(\frac{1}{1-x/2}\right)'=\frac{1}{2}\frac{1}{(1-x/2)^2}.
$$
Therefore
$$
\frac{1}{(1-x/2)^2}=2\sum_{k\ge0}(k+1)\frac{x^k}{2^k}
$$
and the coefficient you're looking for is
$$
A\frac{1}{3^8}+B\frac{1}{2^8}+2C\frac{9}{2^8}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/622228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
remainder when $(x+1)^n$ is divided by $(x-1)^3$, where $n\in \mathbb{N}$ Calculation of remainder when $(x+1)^n$ is divided by $(x-1)^3$, where $n\in \mathbb{N}$
$\bf{My\; Try}::$ Using Division Algorithm:: $p(x) = q(x)\cdot g(x)+r(x)$
Now Let $r(x) = ax^2+bx+c$
So $(x+1)^n=q(x)\cdot (x-1)^3+ax^2+bx+c........................(1)$
Now put $x=1$, we get $\boxed{a+b+c=2^n}$
Now differentitae $(1)$, we get $n(x+1)^{n-1} = q(x)\cdot 3(x-1)^2+(x-1)^3\cdot q^{'}(x)+2ax+b$
again put $x=1$, we get $\boxed{2a+b=n(2)^{n-1}}$
Now again differentitae $(1)$ and then put $x=1$, we get
$\displaystyle \boxed{2a=n(n-1)2^{n-2}\Rightarrow \displaystyle a=\frac{n(n-1)}{2}\cdot 2^{n-2}}$
Similarly we get $\displaystyle b = n(2)^{n-1}-n(n-1)\cdot 2^{n-2}$
Similarly we get $\displaystyle c= 2^{n}+n(n-1)\cdot 2^{n-2}-\frac{n(n-1)}{2}\cdot 2^{n-2}$
So Remainder
$\displaystyle r(x) = \frac{n(n-1)}{2}2^{n-2}x^2+\left\{n(2)^{n-1}-n(n-1) 2^{n-2}\right\}x+2^{n}+n(n-1)\cdot 2^{n-2}-\frac{n(n-1)}{2}2^{n-2}$
is my solution is right , if not then how can i calculate it
or If there is any better method , then plz explain here
Thanks
| Your method is correct. You are essentially computing the first few terms of the Taylor power series expansion of $\,(x+1)^n$ at $\,x=1$ (or, equivalenly, of $\,(z+2)^n$ at $\,z =0,\,$ where $\,z = x-1).\,$ Alternatively, one may compute the expansion by using the Binomial Theorem.
| {
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"question_score": "3",
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"answer_id": 1
} |
How prove this $f(x,y)=(3x+y-7)(1+x^2+xy)+9\ge 0$ Let $x,y>0$,and prove or disprove
$$f(x,y)=(3x+y-7)(1+x^2+xy)+9\ge 0$$
I know
$$f(1,1)=(4-7)(1+1+1)+9=0$$http://www.wolframalpha.com/input/?i=%28%283x%2By-7%29%281%2Bx%5E2%2Bxy%29%2B9%29
since I have use nice methods solve follow inequality
let $a,b>0$,show that
$$9a^2(b+1)-(8a-b-1)(a+b+ab)\ge 0$$
proof:
\begin{align*}&9a^2(b+1)-(8a-b-1)(a+b+ab)\\
&=9(a^2b+a^2)-(8a^2+8ab+8a^2b-ab-b^2-ab^2-a-b-ab)\\
&=a^2b+a^2+ab^2-6ab+b^2+a+b\\
&=(a^2b-2ab+b)+(ab^2-2ab+a)+(a^2-2ab+b^2)\\
&=b(a-1)^2+a(b-1)^2+(a-b)^2\ge 0
\end{align*}
if and only if $a=b=1$
so I think my inequality maybe can use this methods,we can
$$(3x+y-7)(1+x^2+xy)+9=()(x-y)^2+()(x-1)^2+()(y-1)^2$$
Thank you very much!
| Here is something strange, for which I don't really have an entirely rigorous justification, but which works and for which the calcluations seem simpler.
Firstly, (as Christian Blatter's answer does) note that for $x\to\infty$ or $y \to \infty$, we have $f(x,y)\to\infty$ as well. So $f(x, y)$ attains its minimum either on the boundary:
*
*For $x \to 0^+$, we have $f(x, y) \to (y - 7)(1) + 9 \ge (0-7)(1) + 9 = 2$.
*For $y \to 0^+$, we have $f(x, y) \to (3x - 7)(1 + x^2) + 9 \ge (-7)(1) + 9 = 2$.
or in the interior. So far, it's fine. Now here's the strange part. The minimum of $(3x-y)(1+x^2+xy)$ is the solution to
$$\begin{align}
\min \quad &1 + x^2 + xy \\
\text{s.t.} \quad & 3x + y - 7 = c\\
\end{align}
$$
for some $c$. These problems (for any $c$) can be solved by the method of Lagrage multipliers: extrema (in our case, minima) occur at points where the partial derivatives are parallel:
$$ \left(\frac{\partial}{\partial x}(1 + x^2 + xy), \frac{\partial}{\partial y}(1 + x^2 + xy)\right) = (2x + y, x)$$
and similarly
$$ \left(\frac{\partial}{\partial x}(3x + y - 7), \frac{\partial}{\partial y}(3x + y - 7)\right) = (3, 1)$$
Setting them parallel, so that $(2x + y, x) = \lambda(3, 1)$, gives $x = y$.
Among these, $f(x, x) = (1 + 2x^2)(4x - 7) + 9 = 2(x-1)^2(4x+1)$ is minimized (for $x > 0$) at $x = 1$.
The stranger thing is that if we set the partial derivatives equal (instead of parallel) we get $x = y = 1$ directly. Is this coincidence or is there something to this?
| {
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"question_score": "2",
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} |
An equivalent of : $f(x)=\int_0^{+\infty}\frac{e^{-xt}}{(1+t^3)^{1/3}} dt$ $\forall\ x\ \in\ \left]0,+\infty\right[\ $ we put:
$$
{\rm f}\left(x\right)
=
\int_{0}^{\infty}{{\rm e}^{-xt} \over \left(1 + t^{3}\right)^{1/3}}\,{\rm d}t
$$
The question is the question is to find an equivalent of $\,\,{\rm f}\left(x\right)$ when $x \to 0^{+}$.
That means find a simple function $g$ such that when $x \to 0^+$ we have : $f(x)\sim g(x)$ that means : $$\lim_{x \to 0^+} \frac{f(x)}{g(x)} =1$$
| Mathematica gives
$$
\frac{G_{1,4}^{4,1}\left(\frac{s^3}{27}|
\begin{array}{c}
\frac{2}{3} \\
0,0,\frac{1}{3},\frac{2}{3} \\
\end{array}
\right)}{2 \sqrt{3} \pi \Gamma \left(\frac{1}{3}\right)}$$ for the integral, and
expanding this in a power series at $0,$ gets
$$
\frac{-3 \sqrt{3} \Gamma \left(\frac{1}{3}\right) \Gamma
\left(\frac{2}{3}\right) \log (s)-2 \sqrt{3} \gamma \Gamma
\left(\frac{1}{3}\right) \Gamma \left(\frac{2}{3}\right)+\sqrt{3} \log (27)
\Gamma \left(\frac{1}{3}\right) \Gamma \left(\frac{2}{3}\right)+\sqrt{3}
\Gamma \left(\frac{1}{3}\right) \Gamma \left(\frac{2}{3}\right) \psi
^{(0)}\left(\frac{2}{3}\right)}{6 \pi }+\frac{4 \pi ^2 s}{27 \Gamma
\left(\frac{1}{3}\right) \Gamma \left(\frac{4}{3}\right)^2}+O\left(s^2\right),
$$
so the dominant term is
$$
\frac{-3 \sqrt{3} \Gamma \left(\frac{1}{3}\right) \Gamma
\left(\frac{2}{3}\right)}{6\pi} \log (s)
$$
Running FullSimplify on the power series, gets one
$$
\left(-\log (s)+\frac{\pi }{6 \sqrt{3}}-\gamma +\frac{\log
(3)}{2}\right)+\frac{4 \pi ^2 s}{27 \Gamma \left(\frac{1}{3}\right) \Gamma
\left(\frac{4}{3}\right)^2}+O\left(s^2\right),
$$
which agrees with Robert Israel's comment.
| {
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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What is $\lim_{n\to\infty} \frac{1}{n+1}\sum_{k=1}^n \frac{k}{1 + \sqrt{2} +\sqrt{3} +\cdots+\sqrt{k+1}}$? I am stuck on this one; Its a sum and don't know how to calculate the denominator.
$$\lim_{n\to\infty} \frac{1}{n+1}\sum_{k=1}^n \frac{k}{1 + \sqrt{2} +\sqrt{3} +\cdots+\sqrt{k+1}}$$
| Stolz-Cesaro lemma is applied twice successively:
$$L=\lim_{n\to\infty} \frac{1}{n+1}\sum_{k=1}^n \frac{k}{1 + \sqrt{2} +\sqrt{3} +\cdots+\sqrt{k+1}} =\lim_{n\to\infty} \frac{1}{n+2-(n+1)} \frac{n+1}{1 + \sqrt{2} +\sqrt{3} +\cdots+\sqrt{n+2}}=\lim_{n\to\infty}\frac{n+1}{1 + \sqrt{2} +\sqrt{3} +\cdots+\sqrt{n+2}}=\frac{n+2-(n+1)}{1 + \sqrt{2} +\sqrt{3} +\cdots+\sqrt{n+3}-(1 + \sqrt{2} +\sqrt{3} +\cdots+\sqrt{n+2})}=\lim_{n\to\infty}\frac{1}{\sqrt{n+3}}=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/629116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determinant of a special $n\times n$ matrix Compute the determinant of the nun matrix:
$$
\begin{pmatrix}
2 & 1 & \ldots & 1 \\
1 & 2 & \ldots & 1\\
\vdots & \vdots & \ddots & \vdots\\
1 & 1 &\ldots & 2
\end{pmatrix}
$$
For $n=2$, I have$$
\begin{pmatrix}
2 & 1 \\
1 & 2
\end{pmatrix}
$$
Then $det = 3$.
For $n=3$, we have
$$
\begin{pmatrix}
2 & 1 & 1\\
1 & 2 & 1\\
1 & 1 & 2 \\
\end{pmatrix}
$$
Then $det = 4$.
For $n=4$ again we have
$$
\begin{pmatrix}
2 & 1 & 1 & 1 \\
1 & 2 & 1 & 1\\
1 & 1 & 2 & 1\\
1 & 1 & 1 & 2
\end{pmatrix}
$$
Then $det = 5$
How can I prove that the determinant of nun matrix is $n+1$.
| A standard result (http://en.wikipedia.org/wiki/Matrix_determinant_lemma) is $\det(I+AB) = \det(I+BA)$.
Since the matrix above can be written as $I+ e e^T$, where $e$ is a vector of ones, we have $\det(I+ e e^T) = \det(1+ e^T e) = 1+e^Te = n+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/629892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Prove or disprove the implication: Prove or disprove the implication:
$a^2\cdot \tan(B-C)+ b^2\cdot \tan(C-A)+ c^2\cdot \tan(A-B)=0 \implies$
$ ABC$ is an isosceles triangle.
I tried to break down the left hand side in factors, but all efforts were in vain.
Does anyone have a suggestion?
Thank you very much!
| I will follow the method suggested by Blue in his comment to show that the implication does not hold.
The sides $a$, $b$, $c$ are proportional to $\sin A$, $\sin B$ and $\sin C$, respectively, by the sine law, so the equation is equivalent to $S = 0$, where
$$ S = \sin^2 A \tan(B-C) + \sin^2 B \tan(C-A) + \sin^2 C \tan(A - B).$$
Letting $x = \tan A$, $y = \tan B$, and $z = \tan C$, we can rewrite $S$ in terms of $x$, $y$, $z$; for example, we have $\sin^2 A = x^2/(1 + x^2)$ and $\tan(B - C) = (y-z)/(1 + yz)$. (We will ignore for now the special case where $ABC$ is a right triangle, and one of $x$, $y$, $z$, is correspondingly infinite.) Thus
$$ S = \frac{x^2}{1 + x^2} \frac{y-z}{1 + yz} + \frac{y^2}{1 + y^2} \frac{z-x}{1 + zx} + \frac{z^2}{1 + z^2} \frac{x-y}{1 + xy.}$$
Now we note that $S = 0$ when any two of $x, y, z$ are equal, so we expect the numerator of $S$ to be divisible by $(y-z)(z-x)(x-y)$. This is indeed the case (I used a computer for this part, although it is possible to do long division first by $y-z$, then by $z-x$ and finally by $x-y$):
$$S = \frac{(y-z)(z-x)(x-y)(2x^2 y^2 z^2 + x^2 y^2 + y^2 z^2 + z^2 x^2 - 1)}{(1 + yz)(1 + zx)(1 + xy)(1 + x^2)(1 + y^2)(1 + z^2)}.$$
Now that we have recovered the factors $y-z$, $z-x$ and $x - y$, we can split off factors equal to the tangents that originally appeared in $S$, so that
$$S = \frac{y-z}{1 + yz} \frac{z-x}{1 + zx} \frac{x-y}{1 + xy} T = \tan(B-C)\tan(C-A)\tan(A-B) T,$$
where
$$ T = \frac{2x^2 y^2 z^2 + x^2 y^2 + y^2 z^2 + z^2 x^2 - 1}{(1 + x^2)(1 + y^2)(1 + z^2)}.$$
To simplify $T$, we write $X = 1 + x^2$, $Y = 1 + y^2$, $Z = 1 + z^2$, so that $X = 1 + \tan^2 A = \sec^2 A$, and similarly, $Y = \sec^2 B$, $Z = \sec^2 C$. Then by replacing $x^2$, $y^2$ and $z^2$ with $X-1$, $Y-1$, $Z-1$, respectively, where they appear in $T$, we obtain
$$
\begin{align}
T &= \frac{2(X-1)(Y-1)(Z-1) + (X-1)(Y-1) + (Y-1)(Z-1) + (Z-1)(X-1) - 1}{XYZ} \\
&= \frac{2XYZ - XY - YZ - ZX}{XYZ} \\
&= 2 - \frac{1}{X} - \frac{1}{Y} - \frac{1}{Z} \\
&= 2 - \cos^2 A - \cos^2 B - \cos^2 C.
\end{align}
$$
We therefore have the following identity:
$$S = \tan(B-C)\tan(C-A)\tan(A-B) (2 - \cos^2 A - \cos^2 B - \cos^2 C),$$
which corresponds more or less to Blue's factorization. Interestingly, we never used the fact that $A$, $B$ and $C$ were the angles of a triangle. The identity extends to the case where one of the angles is a right angle by continuity.
Now the only question becomes whether it is possible to find a scalene triangle for which
$$\cos^2 A + \cos^2 B + \cos^2 C = 2.$$
If we assume that $A < B < C$, then this amounts to showing that the function
$$f(A,B) = \cos^2 A + \cos^2 B + \cos^2 (A + B)$$
takes the value $2$ somewhere in the interior of the triangle $\Delta$ bounded by the lines $A = 0$, $B = A$ and $A + 2B = \pi$. At the vertex $(0,0)$ of $\Delta$, we find $f(0,0) = 3$, so by continuity, at points in the interior of $\Delta$ near $(0,0)$, we must have $f(A,B) \approx 3$. Likewise, at the vertex corresponding to an equilateral triangle we have $f(\pi/3, \pi/3) = 3/4$, so at some points in the interior of $\Delta$, we must have $f(A, B) \approx 3/4$. Since the interior of $\Delta$ is a connected set, by the intermediate value theorem, we must have $f(A, B) = 2$ for some choice of $(A, B)$ in the interior of $\Delta$, which therefore corresponds to a scalene triangle.
Question: Was it predictable somehow, before the factoring step, that the numerator of $T$ would be expressible in terms of $x^2$, $y^2$ and $z^2$? That is what allowed the method to work as smoothly as it did.
| {
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"timestamp": "2023-03-29T00:00:00",
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Convergence of $ 1+1+\frac{1}{3} + \frac{1}{2} + \frac{1}{3^2} + \frac{1}{2^2} + ...$ How does one use the comparison test to prove that $ 1+1+\frac{1}{3} + \frac{1}{2} + \frac{1}{3^2} + \frac{1}{2^2} + \frac{1}{3^3} + \frac{1}{2^3} + \; ...$ converges?
Is the following argument valid?
$\quad 1+1+\frac{1}{3} + \frac{1}{2} + \frac{1}{3^2} + \frac{1}{2^2} + \frac{1}{3^3} + \frac{1}{2^3} + \; ...$
$< 1 + 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^3} + \;... $
$= 2 + 1 + \frac{1}{2} + \frac{1}{2^2} + \;... $
$= \frac{2}{1-1/2} = 4 $
So the series is bounded above and since every term is strictly positive, the sequence of partial sums is monotone increasing and hence the series converges.
| Using the root test, we can prove easily,
note that,
$$\limsup_{n\to \infty} \sqrt[n]a_n=\lim_{n\to \infty}\sqrt[2n]{\frac 1 {2^n}}=\frac 1 {\sqrt 2}<1 $$
Hence the convergence.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating a trigonometric integral by means of contour $\int_0^{\pi} \frac{\cos(4\theta)}{1+\cos^2(\theta)} d\theta$ I am studying for a qualifying exam, and this contour integral is getting pretty messy:
$\displaystyle I = \int_0^{\pi} \dfrac{\cos(4\theta)}{1+\cos^2(\theta)} d\theta $
I first notice that the integrand is an even function hence
$\displaystyle I = \dfrac{1}{2} \int_{-\pi}^{\pi} \dfrac{\cos(4\theta)}{1+\cos^2(\theta)} d\theta $
Then make the substitutions $\cos(n\theta) = \dfrac{e^{in\theta}+e^{-in\theta}}{2}$, and $z=e^{i\theta}$ to obtain:
$\displaystyle I = \dfrac{1}{2} \int_{||z||=1} \dfrac{\dfrac{z^4+z^{-4}}{2}}{1+\left(\dfrac{z+z^{-1}}{2}\right)^2} \dfrac{-i}{z}dz = -i \int_{||z||=1} \dfrac{z^8+1}{z^3(z^4+6z^2+1)} dz$
Now, assuming this is right so far, this seems "straight-forward" in the sense that I know what to do. However, it gets ugly. I was hoping there might be a better method to evaluate this integral.
Thanks
| Note that $\sin 4\theta$ is an odd function, so we can simplify to
$$\begin{align}
I &= \frac12 \int_{-\pi}^\pi \frac{e^{4i\theta}}{1 + \cos^2\theta}\,d\theta\\
&= \frac{1}{2i}\int_{\lvert z\rvert = 1} \frac{z^4}{1 + \left(\frac{z+z^{-1}}{2}\right)^2}\,\frac{dz}{z}\\
&= \frac{1}{2i}\int_{\lvert z\rvert = 1} \frac{z^5}{z^2+\frac14(z^2+1)^2}\,dz\\
&= -2i \int_{\lvert z\rvert = 1} \frac{z^5}{z^4 + 6z^2+1}\,dz.
\end{align}$$
That looks a little simpler to my untrained eye.
Then we need the zeros of the denominator, which are $\pm \sqrt{-3\pm\sqrt{8}}$, where the inner square root shall be the positive, and the outer can be either square root. The zeros inside the unit disk are $\zeta_\pm = \pm i\sqrt{3-\sqrt{8}}$, both are simple, so
$$\operatorname{Res}\left(\frac{z^5}{z^4+6z^2+1};\,\zeta_\pm\right) = \frac{\zeta_\pm^5}{4\zeta_\pm^3 + 12\zeta_\pm} = \frac{\zeta_\pm^4}{4\zeta_\pm^2 + 12} = \frac{(\sqrt{8}-3)^2}{4(\sqrt{8}-3)+12}=\frac{17-12\sqrt{2}}{8\sqrt{2}},$$
both residues are the same, and
$$I = 8\pi \frac{17-12\sqrt{2}}{8\sqrt{2}} = \pi\frac{17-12\sqrt{2}}{\sqrt{2}} = \frac{\pi}{24+17\sqrt{2}},$$
if I haven't miscalculated.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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pre algebraic factoring with polynomials I really need help solving this particular problem.
$$\frac14x^2y(x-1)^3-\frac54xy(x-1)^2$$
I need help factoring this. It seems like I need to get rid of the fraction but I really just need a little boost.
| Maybe this will help you to see things clearer, but it is the same concept as @NasuSama..
$$\underbrace{\frac14x^2y(x-1)^3}_\text{one expression}-\underbrace{\frac54xy(x-1)^2}_\text{another expression}$$
Let $z=(x-1)$, your expression now becomes:
$$\frac{1}{4}x^2yz^3 - \frac{5}{4}xyz^2$$
You can easily see now that $\frac{1}{4}, x, y, \text{and}, z^2$, are common in both expressions..and so you can "factor out" $\frac{1}{4}xyz^2$ from both expressions. What you are doing when you are "factoring out" is trying to find what expression when multiplied by the term you are "factoring out" $(\frac{1}{4}xyz^2)$, will give you the original expression $(\frac{1}{4}x^2yz^3 - \frac{5}{4}xyz^2)$. This is equivalent to dividing $\frac{1}{4}x^2yz^3 - \frac{5}{4}xyz^2$ by $\frac{1}{4}xyz^2$. So dividing:
$$\frac{{\frac14x^2yz^3}-{\frac54xyz^2}}{\frac{1}{4}xyz^2}=\frac{\frac14x^2yz^3}{\frac{1}{4}xyz^2}-\frac{\frac54xyz^2}{\frac{1}{4}xyz^2}=xz-5$$
Hence
$$\underbrace{\frac{1}{4}xyz^2}_{a} \cdot {(xz-5)} =\underbrace{\frac{1}{4}x^2yz^3 - \frac{5}{4}xyz^2}_{b}$$
Once you have figured out what is common in both expressions, $a$, you simply need to figure out what you can multiply $a$ by to get $b$. Once you get more accustomed to factoring this will come more naturally.
Now re substitute $z=(x-1)$ and you get,
$$\frac{1}{4}xy(x-1)^2 \cdot (x(x-1)-5)=\frac{1}{4}xy(x-1)^2(x^2-x-5)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Supremum and infimum of a set of numbers $\sum_{k=1}^n{\frac{a_k}{a_k+a_{k+1}+a_{k+2}}}$ Let $n \geq 3$ be an arbitrarily fixed integer. Take all the possible finite sequences $(a_1,...,a_n)$ of positive numbers. Find the supremum and infimum of the set of numbers $$\sum_{k=1}^n{\frac{a_k}{a_k+a_{k+1}+a_{k+2}}}$$ where $a_{n+1}=a_1$ and $a_{n+2}=a_2$.
I try to formulate an inequality for the sum, that is $?\leq \sum_{k=1}^n{\frac{a_k}{a_k+a_{k+1}+a_{k+2}}} \leq ?$. But I have no idea. Can anyone give some hints?
| (Write $S(a)=\sum \dots$ for any such sum)
Plug in $q,q^2,\dots,q^n$ for a $q > 0$ and get
$$
S(a) = (n-2)\frac{1}{1+q+q^2}+\frac{q^{n-1}}{q^{n-1}+q^n+q^1}+\frac{q^{n}}{q^{n}+q^1+q^2}
$$
let $q \rightarrow 0,\infty$ yield inf $\leq 1$ and $\sup\geq n-2$
Not let $a_1,\dots,a_n$ be any such sequence, let $A:=\sum a_i$
1) make the denominators bigger:
$$
S(a) \geqslant \frac{a_1}{A}+\dots+\frac{a_n}{A} = \frac{A}{A} = 1
$$
2) observe that for any $x>0$ and quotient $\frac{a}{b}$ where $0 < a < b$
$$
\frac{a}{b} \leqslant \frac{a+x}{b+x}
$$
because $a(b+x)\leqslant b(a+x)$ is equivalent to $ax\leqslant bx$
3) apply 2) picking $x$ such that the sum in the denominator becomes $A$.
$$
S(a)=\sum \frac{a_i}{a_i+a_{i+1}+a_{i+2}}
\leq
\sum \frac{a_i+\left[A-(a_i+a_{i+1}+a_{i+2})\right]}{a_i+a_{i+1}+a_{i+2}+\left[A-(a_i+a_{i+1}+a_{i+2})\right]}$$
and this is
$$
=\sum \frac{A-(a_{i+1}+a_{i+2})}{A} =
\frac{nA-a_2-a_3-a_3-a_4-a_4-a_5-\dots-a_{n-1}-a_n-a_n-a_1-a_1-a_2}{A}
$$
Here every $a_i$ appears exactly twice and therefore $S(a) \leqslant n-2$
| {
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"timestamp": "2023-03-29T00:00:00",
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In how many ways can you arrange a circle of partners so that no partners are touching? There are a lot of similar questions to this but none that is quite the same so I figured I would ask a new question. The problem is you have a group of people that came in pairs, in how many ways can the $N$ people be arranged in a circle such that no person in the circle is holding hands with the partner they came with.
I have worked out by hand that for two couples(four people) the answer is $2$ ways
For three couples ($6$ people) the answer is $32$ ways
Also it should be noted that a rotated circle is not counted as a different arrangement.
If the answer posted below is correct the OEIS sequence is http://oeis.org/A129348 Can any one else confirm that this is correct?
| Here is the calculation for $8:$
Seat one person of the first couple. This fixes the rotation of the circle. Seat the second person of the first couple. The remaining seats can be in groups of $1+5 (2$ ways), $2+4 (2$ ways), or $3+3, (1$ way).
From $1+5$ we can go to $4$ in $4$ ways, $1+3$ in $6$ ways, $2+2$ in $2$ ways, $1+1+2$ in $4$ ways, or $1+1+1+1$ in $2$ ways.
Similarly we can go from $2+4$ to $1+3$ in $8$ ways or to $1+1+2$ in $12$ ways.
We can go from $3+3$ to $1+3$ in $4$ ways, $2+2$ in $8$ ways, $1+1+2$ in $8$ ways, or $1+1+1+1$ in $2$ ways.
Multiplying out, after two couples we can have $4$ in $8$ ways, $1+3$ in $32$ ways, $2+2$ in $12$ ways, $1+1+2$ in $40$ ways or $1+1+1+1$ in $6$ ways.
Finally, from $4$ we can place the last two couples in $8$ ways, $1+3$ in $8$ ways, $2+2$ in $16$ ways, $1+1+2$ in $16$ ways, and $1+1+1+1$ in $24$ ways.
The total comes out $1296$
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate a multiple sum of inverse integers. The question is to calculate a following sum:
\begin{equation}
{\mathcal S}_p(n) :=\sum\limits_{1\le j_1 < j_2 < \dots <j_p \le n-1} \prod\limits_{q=1}^p \frac{1}{n-j_q}
\end{equation}
for $p=1,2,..$ and $n\ge 1$.
From purely combinatorial reasoning we have:
\begin{eqnarray}
{\mathcal S}_1(n) &=& H_{n-1} \\
{\mathcal S}_2(n) &=& \frac{1}{2!} \left(H_{n-1}^2 - H^{(2)}_{n-1} \right) \\
{\mathcal S}_3(n) &=& \frac{1}{3!} \left(H_{n-1}^3 - 3 H_{n-1} H_{n-1}^{(2)} + 2 H_{n-1}^{(3)}\right) \\
{\mathcal S}_4(n) &=& \frac{1}{4!} \left(H^4_{n-1} - 6 H_{n-1}^2 H_{n-1}^{(2)} + 8 H_{n-1} H_{n-1}^{(3)} + 3 H_{n-1}^{(2)} H_{n-1}^{(2)} - 6 H_{n-1}^{(4)}\right) \\
{\mathcal S}_5(n) &=& \frac{1}{5!} \left(H_{n-1}^5 - 10 H_{n-1}^3 H_{n-1}^{(2)} + 20 H_{n-1}^2 H_{n-1}^{(3)} + 15 H_{n-1} ((H_{n-1}^{(2)})^2 - 2 H_{n-1}^{(4)}) - 20 H_{n-1}^{(2)} H_{n-1}^{(3)} + 24 H_{n-1}^{(5)}\right) \\
{\mathcal S}_6(n) &=& \frac{1}{6!} \left(H_{n-1}^6 - 15 H_{n-1}^4 H_{n-1}^{(2)} + 40 H_{n-1}^3 H_{n-1}^{(3)} + 45 H_{n-1}^2 ((H_{n-1}^{(2)})^2 - 2 H_{n-1}^{(4)}) -
24 H_{n-1} (5 H_{n-1}^{(2)} H_{n-1}^{(3)} - 6 H_{n-1}^{(5)}) + 5 (-3 (H_{n-1}^{(2)})^3 + 18 H_{n-1}^{(2)} H_{n-1}^{(4)} + 8 ((H_{n-1}^{(3)})^2 - 3 H_{n-1}^{(6)}) \right)
\end{eqnarray}
where $H_{n-1}^{(r)} := \sum\limits_{j=1}^{n-1} 1/j^r$ is the generalised Harmonic number.
Is it possible to find the result for generic $p\ge 1$?
| Expanding the right hand side of the identity given by achille hui we get ``a compact'' expression for the sum:
\begin{eqnarray}
{\mathcal S}_p(n) &=& \sum\limits_{m=1}^p \frac{(-1)^{-m+p} }{m!} \sum\limits_{p_1+p_2+\dots+p_m=p} \prod\limits_{q=1}^m \frac{H_{n-1}^{(p_q)}}{p_q} \\
&=& \frac{1}{1!} \frac{(-1)^{p-1}}{p} H^{(p)}_{n-1} + \frac{(-1)^{p-2}}{2!} \sum\limits_{p_1=1}^{p-1} \frac{1}{p_1 (p-p_1)} H_{n-1}^{(p_1)} H_{n-1}^{(p-p_1)} + \dots + \frac{1}{p!} H_{n-1}^p
\end{eqnarray}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to get the theta in this projectile equation? The equation for the Parabolic Trajectory is
$$y = x\tan{\theta}- \frac {gx^2}{2V_0^2\cos^2{\theta}}$$
It will be easy to get any variables if there's an angle, but how I can solve if the missing one is the angle? What will be the formula to get the angle? In quadratic way and identity.
Like for example:
$$12.5 = 6.7\tan{\theta} - \frac{(9.81)(6.7)^2}{(2)(24)^2\cos^2{\theta}}$$
$\theta = ?$
Any Help is appreciated!
| Note that due to the Pythagorean Trigonometric Identity : $1 + \tan^2{\theta} = \sec^2{\theta}$,
$$\begin{align}12.5 &= 6.7\tan{\theta} - \frac{(9.81)(6.7)^2}{(2)(24)^2\cos^2\theta}\\
&= 6.7\tan{\theta} - \frac{(9.81)(6.7)^2}{(2)(24)^2}\sec^2\theta \\
&= 6.7\tan{\theta} - \frac{(9.81)(6.7)^2}{(2)(24)^2}(1 + \tan^2\theta)
\end{align}$$
This reduces (after rearranging the terms) into merely a quadratic equation in $\tan{\theta}$. Solve for $\tan{\theta}$, then for $\theta$.
| {
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Factoring $(a+b)(a+c)(b+c)=(a+b+c)(ab+bc+ca)-abc$ How to prove the following equality?
$$(a+b)(a+c)(b+c)=(a+b+c)(ab+bc+ca)-abc$$
I did it
$$\begin{aligned}
a^2b + a^2c + ab^2 + cb^2 + bc^2 + ac^2 + 2abc &=a^2(b + c) + bc(b + c) + a(b^2 + c^2 +2bc)\\&=a^2(b + c) + bc(b+ c) +a(b + c)^2\\&=(b + c)(a^2 + bc + ab + ac)\\&=(b + c)\left[a(a + b) + c(a + b)\right]\\&=(a + b)(b + c)(c + a)
\end{aligned}$$
I just wanted to see some different way to work out with it.
| Set $\displaystyle a+b+c=x$
So, we have $\displaystyle (x-a)(x-b)(x-c)=x^3-x^2(a+b+c)+(ab+bc+ca)x-abc$
Now, $\displaystyle x^3-x^2(a+b+c)=x^2(x-\overline{a+b+c})=x^2\cdot0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/645063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Why does $2^{-n} = 5^n \times 10^{-n}$? If we look at the decimal equivilents of $2^{-n}$, we see they resemble $5^n$ with a decimal point in front of them:
$\begin{align}
2^{-1} &= 0.5 \\
2^{-2} &= 0.25 \\
2^{-3} &= 0.125 \\
2^{-4} &= 0.0625 \\
2^{-5} &= 0.03125 \\
...
\end{align}$
It looks like it's as simple as saying $2^{-n} = 5^n \times 10^{-n}$, and when we calculate that out, it's correct:
$\begin{align}
5^1 \times 10^{-1} &= 5 \times 0.1 = 0.5 \\
5^2 \times 10^{-2} &= 25 \times 0.01 = 0.25 \\
5^3 \times 10^{-3} &= 125 \times 0.001 = 0.125 \\
5^4 \times 10^{-4} &= 625 \times 0.0001 = 0.0625 \\
5^5 \times 10^{-5} &= 3125 \times 0.00001 = 0.03125 \\
...
\end{align}$
I calculated this out for $n = [0, 10]$ and it works out, but I have no idea how to prove it fully.
| $$10^x = (2\cdot 5)^x = 2^x\cdot 5^x$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 4,
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} |
The value of (x-y) of quadrilateral inscribed in circle A quadrilateral that has consecutive sides of length $70,90,110,130$ is inscribed in a circle and also has a circle inscribed in it . The point of tangency of the inscribed circle to the side of length $130$ divides that side into segments of $x$ and $y$ . If $y\geq x$ then what is the value of $(y-x)$
| Call the quadrilateral $ABCD$, with sides $AB=70, BC=90, CD=130, DA=110$, i.e. in the order
$$\begin{array}{ccc}
A&\leftarrow70\rightarrow&B\\
\begin{array}{c}\uparrow\\110\\\downarrow\end{array}&O&\begin{array}{c}\uparrow\\90\\\downarrow\end{array}\\
D&E&C\\
&\leftarrow130\rightarrow&\\
\end{array}$$
For $ABCD$ to be cyclic, $\angle B + \angle D = 180^\circ$, so $\cos\angle B = -\cos\angle D$. Consider square of length of $AC$:
$$\begin{align*}
AC^2 = 130^2+110^2 -2\times130\times110\cos\angle D =& 70^2+90^2-2\times70\times90\cos\angle B\\
\cos\angle D =& \frac{16000}{41200}\\
=& \frac{40}{103}
\end{align*}$$
Similarly,
$$\begin{align*}
BD^2 = 110^2+70^2 -2\times110\times70\cos\angle A =& 130^2+90^2-2\times130\times90\cos\angle C\\
\cos\angle C =& \frac{8000}{38800}\\
=& \frac{20}{97}
\end{align*}$$
Let the centre of the inscribed circle be $O$, and the tangent point of the inscribed circle on $CD$ be $E$. The line $OC$ bisects $\angle BCD$, and the line $OD$ bisects $\angle CDA$. $OE$ is a line perpendicular with $CD$. Consider $\triangle OCD$,
$$OD\sin\angle ODE = OC\sin\angle OCE\\
OD\cos\angle ODE + OC\cos\angle OCE = 130$$
With half-angle formulae,
$$\sin\angle ODE = \sqrt{\frac{1-\cos\angle CDA}{2}}\\
\cos\angle ODE = \sqrt{\frac{1+\cos\angle CDA}{2}}\\
\sin\angle OCE = \sqrt{\frac{1-\cos\angle BCD}{2}}\\
\cos\angle OCE = \sqrt{\frac{1+\cos\angle BCD}{2}}$$
These should be enough to find the ratio $OD:OC$ and hence find out both of $DE$ and $CE$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A moving point has its distance from (1,3) always one-third of its distance from (8,2). Find the equation of its Locus. A moving point has its distance from (1,3) always one-third of its distance from (8,2).
Find the equation of its Locus.
My equation displays a circle formed by the loci, I don't know if it's right. Please help me find the equation, Thank you in advance to those that can answer.
| $$\begin{align*}3^2\left[(x-1)^2+(y-3)^2\right] =& (x-8)^2 + (y-2)^2\\
9\left(x^2-2x+y^2-6y+10\right) =& x^2-16x+y^2-4x+68\\
8x^2-2x+8y^2-50y+22 =& 0\\
x^2-\frac14x+y^2-\frac{25}4y+\frac{11}4=&0\\
\left(x-\frac{1}{8}\right)^2+\left(y-\frac{25}8\right)^2=&\frac{225}{32}\\
\left(x-\frac{1}{8}\right)^2+\left(y-\frac{25}8\right)^2=&\left(\frac{15\sqrt2}{8}\right)^2
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/648792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Explanation for limits equality. $$\mathop {\lim }\limits_{x \to 0} {\left( {\frac{{{a^x} + {b^x}}}{2}} \right)^{\frac{1}{x}}} = \exp \left( {\mathop {\lim }\limits_{x \to 0} \frac{{\frac{{{a^x} + {b^x}}}{2} - 1}}{x}} \right)$$
I am familiar with the "taking the exponent trick" which bring us to:
$$\\exp \left( {\mathop {\lim }\limits_{x \to 0} \left( {\frac{{{a^x} + {b^x}}}{2}} \right) \cdot \frac{1}{x}} \right)
$$
I truly don't get it. I'll be glad for an explanation.
Cheers!
Clarification: $exp(A) = e^A$.
| \begin{align*}
\left(\frac{a^x+b^x}{2}\right)^{\frac 1 x}=\exp\left(\frac 1 x{\color{\red}\ln}\left(\frac{a^x+b^x}{2}\right)\right).
\end{align*}
Now, as $x\to0$, $(a^x+b^x)/2\to (a^0+b^0)/2=(1+1)/2=1$ (assuming that $a,b\neq0$). Also, $\ln(1)=0=1-1$, so that $\lim_{y\to1}\ln y=\lim_{y\to1}(y-1)$, since both functions are continuous at 1. Let $y\equiv(a^x+b^x)/2$. As have seen, $y\to1$ as $x\to0$, so that we have
$$\lim_{x\to0}\ln\left(\frac{a^x+b^x}{2}\right)=\lim_{y\to1}\ln y=0=\lim_{y\to1}(y-1)=\lim_{x\to0}\left(\frac{a^x+b^x}{2}-1\right).$$
Edit: My previous solution contained an error, as pointed out by @AndréNicolas. I apologize. This is a revised version.
From here, we have that (pulling the limit inside is possible, since $x\mapsto\exp(x)$ is continuous)
\begin{align*}
=&\,\lim_{x\to0}\left(\frac{a^x+b^x}{2}\right)^{\frac 1 x}=\lim_{x\to0}\exp\left(\frac 1 x{\color{\red}\ln}\left(\frac{a^x+b^x}{2}\right)\right)=\exp\left(\lim_{x\to0}\frac1x\ln\left(\frac{a^x+b^x}{2}\right)\right)\\
=&\,\exp\left(\lim_{x\to0}\frac1 x\left(\frac{a^x+b^x}{2}-1\right)\right),
\end{align*}
provided that the two limits in the parentheses in the last two expressions are equal. But this is not too difficult to show. We can use L'Hôpital's rule, since both $\ln((a^x+b^x)/2)$ and $(a^x+b^2)/2-1$ converge to $0$ as $x\to 0$, as shown above, and the denominator $x$ naturally converges to zero in a sufficiently smooth way in each case, so that L'Hôpital's rule is applicable. I leave it as an exercise to show that the limit is $(\ln a+\ln b)/2$ in both cases, provided that $a,b>0$. (Hint: Use the fact that $(a^x)'=a^x\cdot\ln a$ for any $a>0$.)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why does this outcome change depending on infinity Why does the outcome of the limit as x approaches infinity of $$\sqrt{x^2+2x}- \sqrt{x^2-2x}$$
which simplifies to
$$\dfrac {4x}{x \left(\sqrt{1 + \frac 2x} + \sqrt {1 - \frac 2x}\right) }= \dfrac {4}{\sqrt{1 + \frac 2x} + \sqrt {1 - \frac 2x}}$$
change depending on whether we take the limit from minus infinity or positive infinity?
If positive, the answer is 2. If negative infinity, the answer is -2??
It seems to me that we get $$ \frac {4}{\sqrt{1} + \sqrt{1}}$$ no matter which side we approach.
| $$\sqrt{x^2} = |x|$$
So when $x$ tends to $- \infty$, $\sqrt{x^2} = -x$.
Hence then answer would be:
$$\frac{4}{-2} = -2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/659725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Cubic spline that interpolates $f$ Given the following cubic spline that interpolates $f$:
$$\left\{\begin{matrix}
(x+3)^3-9(x+3)^2+22(x+3)-10 & ,-3 \leq x < -1\\
(x+1)^3-3(x+1)^2-2(x+1)+6 & , -1\leq x <0\\
ax^3+bx^2+cx+d & ,0\leq x <2\\
(x-2)^3+6(x-2)^2+7(x-2) & ,2 \leq x \leq 3
\end{matrix}\right.$$
It's also given that: $f(-3)=-10, f(-1)=?, f(0)=?, f(2)=0, f(3)=14$.
Which is the value of $c$?
| HINT: Use the fact that $f(x)$ and $f'(x)$ must be continuous at $x=0$ and $x=2$
| {
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"url": "https://math.stackexchange.com/questions/665822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determine for what value of $x$ the series converges $∑_{n=1}^∞ \frac{(3^n+(-2)^n )}{n} (x+1)^n $ Determine for what value of $x$ the series converges
$∑_{n=1}^∞ \frac{(3^n+(-2)^n )}{n} (x+1)^n $
Here is what I got
Using the ratiotest, I got
$D_n =\frac{\frac{(3^{n+1}+(-2)^{n+1} )}{n+1}(x+1)^{n+1}}{\frac {(3^n+(-2)^n )}{n} (x+1)^n}$
$=\frac{n(3^{n+1}+(-2)^{n+1} )}{(n+1)(3^n+(-2)^n)}(x+1) ->3(x+1) <1$ if $x<-2/3$
Is this correct?
I use maple to calculate $lim_{n->\infty} \frac{n(3^{n+1}+(-2)^{n+1} )}{(n+1)(3^n+(-2)^n)}=3$ but I don't know why they got this either.
| $lim_{n->\infty} \frac{n(3^{n+1}+(-2)^{n+1} )}{(n+1)(3^n+(-2)^n)}$
$=lim_{n->\infty} \frac{3^{n+1}n(1+(-2/3)^{n+1} )}{3^nn(1+1/n)(1+(-2/3)^n)}$
$=lim_{n->\infty} \frac{3(1+(-2/3)^{n+1} )}{(1+1/n)(1+(-2/3)^n)}=\frac{3(1+0)}{1+0)(1+0)}=3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/668404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
High degree polynom divisibility The problem is:
Show that $x^{44}+x^{33}+x^{22}+x^{11}+1$ is divisible by $x^4+x^3+x^2+x+1$
I am not sure how to approach this problem so any help would be very appreciated. Thank you in advance.
| It is not hard to see that $x^5-1$ divides $x^{55}-1$.
Thus $(x-1)(x^4+x^3+x^2+x+1)$ divides $(x^{11}-1)(x^{44}+x^{33}+x^{22}+x^{11}+1)$ and therefore
$x^4+x^3+x^2+x+1$ divides $(x^{10}+x^9+\cdots +x+1)( x^{44}+x^{33}+x^{22}+x^{11}+1)$.
But $x^4+x^3+x^2+x+1$ and $x^{10}+x^9+\cdots +x+1$ are relatively prime (use the Euclidean Algorithm).
The result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/670555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Showing that $\lim\limits_{n \to\infty} z_n = A$ implies $\lim\limits_{n \to\infty} \frac{1}{n} (z_1 + z_2 + \ldots + z_n) = A$ In what follows let all values be in $\mathbb{C}$. I'm trying to show that if
$$\lim z_n = A,$$
that then
$$
\lim_{n \to \infty} \frac{1}{n} (z_1 + z_2 + \ldots + z_n) = A.
$$
For ease of notation, let $s_n = \frac{1}{n} (z_1 + z_2 + \ldots + z_n)$.
Attempt:
*
*Let $n \in \mathbb{N}$ be arbitrary and consider that
$$
\left| A - s_n \right| = \left|A - \frac{1}{n} (z_1 + \ldots + z_n) \right| = \left|A - \frac{z_1}{n} - \ldots - \frac{z_n}{n} \right|
$$
so that through repeated applications of the triangle inequality we have that
$$
\left| A - s_n \right| \le \left| A - \frac{z_n}{n} \right| + \left| - \frac{z_{n-1}}{n} - \ldots - \frac{z_{1}}{n} \right| \le \left| A - \frac{z_n}{n} \right| + \left| -\frac{z_{n-1}}{n} \right| + \ldots + \left|- \frac{z_{1}}{n} \right|
$$
*Now as $n \rightarrow \infty$, we have that the numerator of the term $\left| -\frac{z_{n-1}}{n} \right|$ approaches $-A$ while the denominators of all of the terms in the sum $\left| -\frac{z_{n-1}}{n} \right| + \ldots + \left|- \frac{z_{1}}{n} \right|$ approach infinity. \uline{[Gap]}. Then as $n \rightarrow \infty$, we have that $\left| -\frac{z_{n-1}}{n} \right| + \ldots + \left|- \frac{z_{1}}{n} \right|$ approaches $0$.
*On the other hand, we have also that $z_n \rightarrow A$ (by hypothesis) so that the term $\left| A - \frac{z_n}{n} \right|$ can get as close to $\left| A - \frac{A}{n} \right|$ as we'd like. Yet since $\frac{A}{n} \rightarrow 0$, we have that $\left| A - \frac{z_n}{n} \right| \rightarrow \left| A - 0 \right| = \left| A \right|$.
*Then since
$$
\left( \left| -\frac{z_{n-1}}{n} \right| + \ldots + \left|- \frac{z_{1}}{n} \right| \right) \rightarrow 0
$$
and
$$
\left| A - \frac{z_n}{n} \right| \rightarrow \left| A \right|
$$
we have that
$$
|A - s_n| \le \left| A - \frac{z_n}{n} \right| + \left( \left| -\frac{z_{n-1}}{n} \right| + \ldots + \left|- \frac{z_{1}}{n} \right| \right) \rightarrow |A| + 0 = |A|.
$$
Question: My argument doesn't quite work since I have shown only that $|A - s_n| \rightarrow |A|$ and yet we want $|A - s_n| \rightarrow |0|$. Is there a way to keep most of my argument in place and yet actually to prove the desired statement?
| In step 1., "through repeated applications of the triangle inequality we have", more interestingly, that
$$
\left| A - s_n \right|=\left|\frac1n\sum_{k=1}^n(A - z_k) \right| \leqslant\frac1n\sum_{k=1}^n\left| A - z_k \right| .
$$
Then one can proceed.
Later on, in step 4., you fall prey to the common fallacy that since $\displaystyle\left|\frac{z_k}n\right|\to0$ when $n\to\infty$, for each fixed $k$ (which is quite true), one would be sure that $\displaystyle\sum_{k=1}^{n-1}\left|\frac{z_k}n\right|\to0$. Not so, since thenumber of terms is $n-1$, which is unbounded.
| {
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"source": "stackexchange",
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L.C.M. and H.C.F. problem Let $a$ and $b$ be natural numbers with $ab>2$. Suppose that the sum of their highest common factor and least common multiple is divisible by $a+b$. Prove that the quotient is at most $\frac{a+b}4$. When is this quotient exactly equal to $\frac{a+b}4$.
| Let $g=\gcd(a,b)$ and $a=gA$, $b=gB$. Then, $\mathrm{lcm}(a,b)=gAB$ and we have
$$\frac{\mathrm{lcm}(a,b)+\gcd(a,b)}{a+b} = \frac{AB+1}{A+B}$$
Without loss of generality, we can assume $A\geq B$. Let's start with two simple cases:
*
*$A=B$; which implies $A=B=1$ (and $a=b=g$) due to $\gcd(A,B)=1$. The quotient is equal to $1$ and $\frac{a+b}{4}=\frac{g}{2}$. Since $ab\geq 2$, the desired inequality is proven in this case and equality happens only for $g=2$, corresponding to $(a,b)=(2,2)$.
*$A=B+1$; so the quotient reduces to $$\frac{B^2 + B + 1}{2B+1}$$
But this can be integer only if $\frac{B+2}{2B+1}$ is an integer; which happens only for $B=1$. Thus, $(A,B)=(2,1)$ and $(a,b)=(2g,g)$ and the quotient is equal to $1$. The inequality $ab>2$ implies $g\geq 2$, so the inequality $\frac{a+b}{4}\geq 1$ is always strict.
*$A\geq B+2$. Then, we have
$$\begin{eqnarray}
2 & \leq & A-B \\
4 & \leq & (A-B)^2 \\
4(AB+1) & \leq & (A+B)^2 \\
\frac{AB+1}{A+B} & \leq & \frac{A+B}{4} \leq g\frac{A+B}{4} = \frac{a+b}{4}\\
\end{eqnarray}$$
Clearly, the equality can happen only for $A=B+2$ and $g=1$. This corresponds to pairs of consecutive odd numbers, $(a,b)=(2n-1,2n+1)$ for any positive integer $n$.
| {
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$a+b+c=3, a,b,c>0$, Prove that $a^2b^2c^2\ge (3-2a)(3-2b)(3-2c)$ $a+b+c=3, a,b,c>0$, Prove that $$a^2b^2c^2\ge (3-2a)(3-2b)(3-2c)$$
My work:
From the given inequality, we can have,
$a^2b^2c^2\ge 27-18(a+b+c)+12(ab+bc+ca)-8abc$
We can also have,$abc\le \bigg(\dfrac{a+b+c}{3}\bigg)^3=1$
So, $0\ge -36+12(ab+bc+ca)$
Again, we can have, $ab+bc+ca\le b(3-b)+\dfrac{1}{b}$
Now, I have to show that, $b(3-b)+\dfrac{1}{b}\le 3$
How can I prove this now? Please help.
| $$\Longleftrightarrow (a+b+c)^3(-a+b+c)(a-b+c)(a+b-c) \le 27a^2b^2c^2$$
Let $x = -a + b + c, y = a-b+c, z = a+b-c$ and note that at most one of $x,y,z$ can be negative (since the sum of any two is positive). If $x < 0$ then $$(a+b+c)^3xyz \leq 0 < 27a^2b^2c^2$$.
If $x,y,z > 0$, note $x+y+z = a+b+c, x+y = 2c$ etc so our inequality becomes
$$64xyz(x+y+z)^3 \leq 27(x+y)^2(y+z)^2(z+x)^2$$
Note that $$9(x+y)(y+z)(z+x) \geq 8(x+y+z)(xy+yz+zx)$$ and $$(xy+yz+zx)^2 \geq 3xyz(x+y+z)$$. Combining these completes our proof!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/676392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to prove the convergence of a series of prime numbers I have a bit of a problem proving that the series:
$$
\sum_{p\leq x} \frac{p\ln\left(p\right)}{x^2}
$$
where the sum is extended over all prime numbers, converges to 0.5.
Any ideas?
Thanks in advance,
Kijn
| A standard method is to write the sum as a sum over all positive integers $\leqslant x$, multiplying the term with $\pi(n) - \pi(n-1)$, where $\pi$ is the prime counting function, to annihilate the terms for composite $n$, and then rearrange:
$$\begin{align}
\sum_{p\leqslant x} p\log p &= \sum_{n\leqslant x} \bigl(\pi(n)-\pi(n-1)\bigr)n\log n\\
&= \sum_{n \leqslant x} \pi(n)n\log n - \sum_{n\leqslant x-1} \pi(n)(n+1)\log (n+1)\\
&= \pi(x)x\log x - \sum_{n\leqslant x} \pi(n)\left[\log (n+1) + \underbrace{n \log \left(1+\frac{1}{n}\right)}_{1 + O(1/n)}\right] + O(x)\tag{1}\\
&= x^2 - \sum_{n\leqslant x} \underbrace{\pi(n)\log n}_{n + O(n/\log n)} - \underbrace{\sum_{n\leqslant x} \pi(n)}_{O(x^2/\log x)} + O(x^2/\log x)\tag{2}\\
&= x^2 - \sum_{n\leqslant x} n + O(x^2/\log x)\\
&= \frac{x^2}{2} + O\left(\frac{x^2}{\log x}\right).
\end{align}$$
Dividing by $x^2$ yields the proposition.
In $(1)$, we use
$$\pi(x)x\log x - \pi\left(\lfloor x\rfloor\right)\lfloor x+1\rfloor \log \lfloor x+1\rfloor \in O(x),$$
and in $(2)$ we use
$$\pi(x) - \frac{x}{\log x} \in O\left(\frac{x}{(\log x)^2}\right)$$
as well as
$$\sum_{2 \leqslant n \leqslant x} \frac{n}{\log n} \in O\left(\frac{x^2}{\log x}\right).$$
At the end,
$$\sum_{n\leqslant x} n = \frac{1}{2} \lfloor x\rfloor\cdot \lfloor x+1\rfloor = \frac{x^2}{2} + O(x).$$
| {
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How to solve this simple trignometric problem? So this is the question that was given in a textbook and i attempted to win from the book which was saying i was wrong?
If $$\frac{\sin\theta + \cos\theta}{\sin\theta - \cos\theta} = \frac{5}{4}$$ then what is te value of$$\frac{\tan^2\theta + 1}{\tan^2\theta - 1}$$
so i attempted to solve it this way$$\implies\frac{\tan^2\theta + 1}{\tan^2\theta - 1} = \frac{\frac{\sin^2\theta}{\cos^2\theta} + 1}{\frac{\sin^2\theta}{\cos^2\theta} - 1}$$
$$=\frac{\sin^2\theta + \cos^2\theta}{\sin^2\theta - \cos^2\theta}$$$$=\frac{1}{\sin^2\theta - \cos^2\theta}$$$$=\frac{1}{(\sin\theta + \cos\theta)(\sin\theta - \cos\theta)}$$substituting the values from$$=\frac{\sin\theta + \cos\theta}{\sin\theta - \cos\theta} = \frac{5}{4}$$we get$$=\frac{1}{20}$$ but this is wrong answer so where am i wrong and what is the mistake
Note = I am not looking for the solution, i want to know what is wrong with this approach and why
Thanks
Akash
| Observe that $\dfrac a b=\dfrac c d$ doesn't imply $a=c$ and $b=d$.
Now you may proceed as follows,
\begin{align}
&\dfrac{\sin \theta +\cos\theta}{\sin \theta -\cos\theta}=\dfrac54
\\\implies&4\sin \theta +4\cos\theta=5\sin \theta-5\cos\theta
\\\implies&\sin\theta=9\cos\theta.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/676973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to prove this inequality $\frac{a}{\sqrt{b^2+b+1}}+\frac{b}{\sqrt{c^2+c+1}}+\frac{c}{\sqrt{a^2+a+1}}\ge\sqrt{3}\;?$
Let $a,b,c\ge 0$, and assume $a+b+c=3$.
I'd like to show that
$$\frac{a}{\sqrt{b^2+b+1}}+\frac{b}{\sqrt{c^2+c+1}}+\frac{c}{\sqrt{a^2+a+1}}\ge\sqrt{3}$$
My try: Use Hölder inequality
$$\left(\sum_{cyc}\dfrac{a}{\sqrt{b^2+b+1}}\right)\left(a\sqrt{b^2+b+1}\right)^2\ge (a+b+c)^3\,,$$
then we only need to prove this
$$\sum_{cyc}a(b^2+b+1)=\sum_{cyc}(ab^2+ab+a)\le 9$$
but
$$ab^2+bc^2+ca^2+ab+bc+ac\le 6$$
is not true.
So I can't prove it.
| I'm going to use two facts. The first one can be obtained as following:
$$\begin{align}
ab + bc + ca &\leqslant a^2 + b^2 + c^2, \\
3(ab + bc + ca) &\leqslant (a + b + c)^2 = 9, \\
ab + bc + ca &\leqslant 3.
\end{align}$$
And the second one:
$$\frac{1}{\sqrt{x^2 + x + 1}} \geqslant \frac{\sqrt{3}}{2} - \frac{x}{2\sqrt{3}}$$
(here's Wolfram Alpha visualization, a bit later I'm going to prove it more rigorously).
So, putting it all together:
$$
\begin{align}
\frac{a}{\sqrt{b^2 + b + 1}} + \frac{b}{\sqrt{c^2 + c + 1}} + \frac{c}{\sqrt{a^2 + a + 1}} &\geqslant \frac{\sqrt{3}}{2}(a + b + c) - \frac{ab + bc + ca}{2\sqrt{3}} \\
&\geqslant \frac{3\sqrt{3}}{2} - \frac{\sqrt{3}}{2} = \sqrt{3}
\end{align}$$
Lemma. $\displaystyle \frac{1}{\sqrt{x^2 + x + 1}} \geqslant \frac{\sqrt{3}}{2} - \frac{x}{2\sqrt{3}},\ \forall x \geqslant 0$.
Note that $\displaystyle \frac{\sqrt{3}}{2} - \frac{x}{2\sqrt{3}} = \frac{1}{2\sqrt{3}}(3 - x) < 0$ for $x > 3$. So, for these $x$ we get obvious inequality that square root of something is greater than $0$.
For $x \in [0, 3]$:
$$
\begin{align}
\frac{1}{\sqrt{x^2 + x + 1}} \geqslant \frac{\sqrt{3}}{2} - \frac{x}{2\sqrt{3}} &\Longleftrightarrow 12 \geqslant (x^2 + x + 1)(3 - x)^2 \\
&\Longleftrightarrow x^4 - 5x^3 + 4x^2 +3x - 3 \leqslant 0 \\
&\Longleftrightarrow (x - 1)(x^3 - 4x^2 + 3) \leqslant 0 \\
&\Longleftrightarrow (x - 1)^2(x^2 - 3x - 3) \leqslant 0
\end{align}$$
which is true because roots of $x^2 - 3x - 3$ are:
$x_1 = \frac{1}{2}(3 - \sqrt{21}) < 0$,
$x_2 = \frac{1}{2}(3 + \sqrt{21}) > 3$.
So for $x \in [0, 3]$ holds $x^2 - 3x - 3 < 0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/678619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Roots of a certain family of polynomials Let $\xi \in (0,1)$ and $p$ be a positive non-zero integer.Show that in the limit $p \rightarrow \infty$ the following algebraic equation:
\begin{equation}
\frac{x^{p+2} }{\xi} + x^{p+1} + x^p = (-\xi)^p
\end{equation}
has roots:
\begin{equation}
\left\{ x_\xi, \bar{x}_\xi, \left\{ \xi \exp(\imath \frac{2 \pi j}{p}) \right\}_{j=1}^{p}\right\}
\end{equation}
where
\begin{equation}
(1) x_\xi := - \frac{1}{2} \xi +\imath \left(\sqrt{\xi} - \frac{1}{2^3} \xi^{3/2} - \frac{1}{2^7} \xi^{5/2} - \frac{1}{2^{10}} \xi^{7/2} - \frac{5 }{2^{15}} \xi^{9/2} - \frac{7}{2^{18}} \xi^{11/2}- \frac{21}{2^{22}} \xi^{13/2} - \frac{33}{2^{25}} \xi^{15/2} - \frac{429}{2^{31}} \xi^{17/2} - \frac{715}{2^{34}} \xi^{19/2} - \frac{2431}{2^{38}} \xi^{21/2} - \frac{4199}{2^{41}} \xi^{23/2} - \frac{29393}{2^{46}} \xi^{25/2}- O\left(\xi^{27/2}\right)\right)
\end{equation}
How do we find all terms in the expansion of the root $x_\xi$?
On the left, roots of the algebraic equation for $p=50$. On the right, the two complex conjugate ``non-trivial'' roots $x_\xi$ and $\bar{x}_\xi$ along with the series expansion (1).
| I do not see why a polynomial of degree $p+2$ would have $p+3$ roots...
This is a perturbation problem, essentially you have
$q(x)x^p-(-ξ)^p=0$ with the quadratic polynomial $q(x)=\frac1ξx^2+x+1$.
The quadratic polynomial $0=4ξq(x)=(2x+ξ)^2+4ξ-ξ^2$ has roots
$$x=\frac12\left(-ξ\pm\sqrt{ξ^2-4ξ}\right)=-\frac12ξ\pm i\sqrtξ\,\sqrt{1-\tfrac14ξ}$$
and the binomial series for the square root gives the expression for $x_ξ$ and $\bar x_ξ$.
Since $q(-ω^kξ)=1-ξω^k(1-ω^k)$ is bounded away from zero, $-ω^kξ$ is an approximation for the other roots, even if not the best. $ω$ being the primitive $p$-th root of unity.
Keyword to look up: Newton polytope.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/680785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Triangle with sides a,b,c and area There is a triangle with sides $a,b,c$, then his area is:
a) $A$ not greater than $\frac{1}{2}(a^2-ab+b^2)$
b) $B$ greater than $\frac{1}{2}(a^2-ab+b^2)$
c) $C$ greater than $\frac{1}{2}(a^2-\frac{1}{2}ab+b^2)$
d) $D$ less than $\frac{1}{2}(a^2+ab+b^2)$
I am thinking about applying the Heron formula here, because we know all its sides.
| HINT:
The area is $\displaystyle\triangle= \frac12ab\sin\gamma$
As $\displaystyle 0<\sin\gamma\le1, ab\sin\gamma\le ab\iff -ab\sin\gamma\ge -ab$
$\displaystyle\implies A-\triangle=\frac{a^2-ab+b^2-ab\sin\gamma}2\ge\frac{a^2-ab+b^2-ab}2$ which is $\displaystyle\frac{(a-b)^2}2\ge0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/680999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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