Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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How do you integrate $(x+2)\ln(x-3)$? I got $$\left(\frac {x^2}{2} +2x\right)\ln(x-3)-\left(\frac {x^2}{4}-\frac {7x}{2} -\frac {21}{2}\right)\ln(2x-6)$$ as my answer... Not sure If I got it right. Please correct me, thank you!
| $\displaystyle \int (x+2)\ln(x-3)dx=\int x\ln(x-3)dx+\int 2\ln(x-3)dx$
Note that we make use of the formulas $\displaystyle\int \ln u du=u\ln u-u+C$ and $\displaystyle\int u\ln udu=\frac{1}{4}u^2(2\ln u-1)+C$, which can be shown via integration by parts.
For the first integral, let $u=x-3$ and $du=dx$
This yields $\displaystyle\int (u+3)\ln udu=\int (u\ln u+3\ln u )du=\frac{1}{4}u^2(2\ln u-1)+3u\ln u-u+C_1=\frac{1}{4}(x-3)^2(2\ln (x-3)-1)+3(x-3)\ln (x-3)-(x-3)+C_1$
For the second integral, let $u=x-3$ and $du=dx$
This yields $\displaystyle2\int \ln u du=2u\ln u-2u+C_2=2(x-3)\ln (x-3)-2(x-3)+C_2.$
Therefore $$\displaystyle \int (x+2)\ln(x-3)dx=\frac{1}{4}(x-3)^2(2\ln (x-3)-1)+3(x-3)\ln (x-3)-(x-3)+2(x-3)\ln (x-3)-2(x-3)+C$$
You can further simplify the antiderivative, but this would be the general strategy for integration.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1682977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 9,
"answer_id": 1
} |
Solving linear recursive equation $a_n = a_{n-1} + 2 a_{n-2} + 2^n$. I wish to solve the linear recursive equation:
$a_n = a_{n-1} + 2a_{n-2} + 2^n$, where $a_0 = 2$, $a_1 = 1$.
I have tried using the Ansatz method and the generating function method in the following way:
Ansatz method
First, for the homogenous part, $a_n = a_{n-1} + 2a_{n-2}$, I guess $a_n = \lambda^n$ as the solution, and substituting and solving for the quadratic, I get $\lambda = -1, 2$. So, $a_n = \alpha (-1)^n + \beta 2^n$. Then, for the inhomogenous part, I guess $a_n = \gamma 2^n$, to get $\gamma 2^n = \gamma 2^{n-1} + 2\gamma 2^{n-2} + 2^n$, whence $2^n=0$, which means, I suppose, that this guess is not valid. These are the kind of guesses that usually work, so I don't know why it fails in this particular case, and what to do otherwise, so I tried the generating function method.
Generating function method
Let
$$
A(z) = \sum_{i=0}^{\infty} a_k z^k
$$
be the generating function for the sequence $\{ a_n \}_{n \in \mathbb{N} \cup {0}}$. Then, I try to write down the recursive relation in terms of $A(z)$:
$$
A(z) = zA(z) + 2z^2 A(z) + \frac{1}{1-2z} + (1 - 2z),
$$
where the last term in the brackets arises because of the given initial conditions. Then, solving for $A(z)$,
$$
\begin{align}
A(z) &= \frac{1}{(1+z)(1-2z)^2} + \frac{1}{1+z}\\
&= \frac{2}{9}\frac{1}{1-2z} + \frac{2}{3}\frac{1}{(1-2z)^2} + \frac{10}{9}\frac{1}{1+z}\\
&=\frac{2}{9} \sum_{k=0}^{\infty} 2^k z^k + \frac{2}{3} \sum_{k=0}^{\infty} (k+1)2^k z^k + \frac{10}{9} \sum_{k=0}^{\infty} (-1)^k z^k\\
&= \sum_{k=0}^\infty \frac{(3k+4)2^{k+1} + (-1)^k 10}{9} z^k.
\end{align}
$$
So,
$$
a_k = \frac{(3k+4)2^{k+1} + (-1)^k 10}{9}.
$$
But then, $a_1 = 2$, whereas we started out with $a_1 = 1$.
At first, I thought that maybe the generating function method did not work because some of the series on the right hand side were not converging, but they all look like they're converging for $|z| < 1/2$. I rechecked my calculations several times, so I don't think there is any simple mistake like that. It would be great if someone could explain to me what exactly is going wrong here.
| Suppose
$$
a_n=a_{n-1}+2a_{n-2}+2^n\tag{1}
$$
Let $a_n=b_n+\frac23n\,2^n$, then
$$
b_n+\frac23n\,2^n
=b_{n-1}+\frac23(n-1)\,2^{n-1}+2b_{n-2}+2\cdot\frac23(n-2)\,2^{n-2}+2^n\tag{2}
$$
and cancelling, we get
$$
b_n=b_{n-1}+2b_{n-2}\tag{3}
$$
The standard solution to $(3)$ is $b_n=c_1(-1)^n+c_22^n$. Therefore,
$$
a_n=c_1(-1)^n+\left(c_2+\frac23n\right)2^n\tag{4}
$$
Solving for $a_0=2$ and $a_1=1$, we get
$$
a_n=\frac{13}9(-1)^n+\left(\frac59+\frac23n\right)2^n\tag{5}
$$
Comment on the Ansatz Method
Let $Sa_n=a_{n+1}$. If we apply $S-2$ to $(1)$, we get
$$
\begin{align}
(S-2)\left(a_n-a_{n-1}-2a_{n-2}\right)
&=(S-2)2^n\\
&=0\tag{6}
\end{align}
$$
This means
$$
(S+1)(S-2)^2a_n=0\tag{7}
$$
The problem with the Ansatz method, is the exponent of $2$ on $(S-2)$.
The standard solution for $(7)$ is $a_n=c_1(-1)^n+(c_2+c_3n)2^n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1683908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Find the minimum value of $\frac{4}{4-x^2} + \frac{9}{9-y^2} $ Let $x, y ∈ (−2, 2)$ and $xy = −1$. Find the minimum value of $\frac{4}{4-x^2} + \frac{9}{9-y^2} $ ?
My Attempt
let $t=\frac{4}{4-x^2} + \frac{9}{9-y^2} $ , replacing $y$ by $- \frac{1}{x}$ we get $t=\frac{1}{1-(\frac{x}{2})^2} + \frac{1}{1-(\frac{1}{3x})^2} $ . Using AM-HM inequality we get $t(1-(\frac{x}{2})^2 + 1-(\frac{1}{3x})^2) \geq 2^{2}.$ let $ m =(1-(\frac{x}{2})^2 + 1-(\frac{1}{3x})^2)$ and using AM-GM inequality we get $m \leq 5/3$.
But from this point my inequality signs are getting mixed up. Am I on right track?
| here is another elementary method which can also find min.
$f=\dfrac{4}{4-u}+\dfrac{9u}{9u-1},u=x^2=\dfrac{1}{y^2}< 4, y^2 < 4 \implies u> \dfrac{1}{4}$
$9(f-1)u^2+(72-37f)u+4(f-1)=0, \Delta=(72-37f)^2-4*4*9(f-1)^2 \ge 0 $
$\iff (12-5f)(12-7f)\ge 0 \iff f\ge \dfrac{12}{5}$ or $f \le \dfrac{12}{7}$
when $f \le \dfrac{12}{7}, 72-37f>0 \implies u<0$
when $f=\dfrac{12}{5}, u= \dfrac{37f-72}{2*9(f-1)}=\dfrac{2}{3} > \dfrac{1}{4} \implies f_{min}=\dfrac{12}{5}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1686727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
How to evaluate $\int_{0}^{\infty }\frac{x^{4}e^{-2x^{2}}}{\left ( 1+x^{2} \right )^{4}}\mathrm{d}x$ How to evaluate the following integral
$$\int_{0}^{\infty }\frac{x^{4}e^{-2x^{2}}}{\left ( 1+x^{2} \right )^{4}}\mathrm{d}x$$
Can we solve it without using complex analysis method?
| We will illustrate a methodology for evaluating the integral of interest that is based on "Feynmann's Method" for differentiating under the integral.
To begin, we augment the integral of interest by introducing the parameter $a$ in the exponential argument of the integrand. let $I(a)$ denote the integral
$$I(a)=\int_0^\infty \frac{x^4e^{-ax^2}}{(1+x^2)^4} \,dx\tag 1$$
Then, the integral of interest is simply given by
$$I(2)=\int_0^\infty \frac{x^4e^{-ax^2}}{(1+x^2)^4} \,dx$$
Using partial fraction expansion on the term $\frac{x^4e^{-ax^2}}{(1+x^2)^4}$ reveals
$$\frac{x^4}{(1+x^2)^4}=\frac{1}{(1+x^2)^2}-\frac{2}{(1+x^2)^3}+\frac{1}{(1+x^2)^4} \tag 2$$
whereby we can write $(1)$ as
$$\begin{align}
I(a)&=\int_0^\infty \frac{e^{-ax^2}}{(1+x^2)^2}\,dx-2\int_0^\infty \frac{e^{-ax^2}}{(1+x^2)^3}\,dx+\int_0^\infty \frac{e^{-ax^2}}{(1+x^2)^4}\,dx \\\\
&=e^a\int_0^\infty \frac{e^{-a(1+x^2)}}{(1+x^2)^2}\,dx-2e^a\int_0^\infty \frac{e^{-a(1+x^2)}}{(1+x^2)^3}\,dx+e^a\int_0^\infty \frac{e^{-a(1+x^2)}}{(1+x^2)^4}\,dx \tag 3
\end{align}$$
We can evaluate each of the $3$ integrals on the right-hand side of $(3)$ by differentiating under the integral. In particular, we will evaluate the first integral on the right-hand side of $(3)$ and leave the evaluation of the second and third integrals as an exercise.
Let $F_n(a)$ denote the integral
$$F_n(a)=\int_0^\infty \frac{e^{-a(1+x^2)}}{(1+x^2)^n}\,dx$$
The integral $F_0(a)$ is Gaussian and easily evaluated; we find that $F_0(a)$ is
$$\begin{align}
F_0(a)&=\int_0^\infty e^{-a(1+x^2)}\,dx\\\\
&=\frac{\sqrt \pi}{2}\frac{e^{-a}}{\sqrt a}\end{align}$$
INTEGRATING $F_0(a)$
Noting $F_1'(a)=-F_0(a)$, with $F_1(0)=\pi/2$, yields
$$\begin{align}
F_1(a)&=\int_0^\infty \frac{e^{-a(1+x^2)}}{1+x^2}\,dx\\\\
&=\frac\pi 2-\int_0^a F_0(y)\,dy\\\\
&=\frac{\pi }{2}-\frac{\sqrt \pi}{2}\int_0^a \frac{e^{-y}}{\sqrt y}\,dy\\\\
&=\frac{\pi}{ 2}-\frac{\pi}{2}\frac{2}{\sqrt \pi}\int_0^a e^{-t^2}\,dt\\\\
&=\frac{\pi}{2}\text{erfc}(\sqrt a)
\end{align}$$
INTEGRATING $F_1(a)$
Similarly, since $F_2'(a)=-F_1(a)$, with $F_2(0)=\pi/4$, we have
$$\begin{align}
F_2(a)&=\int_0^\infty \frac{e^{-a(1+x^2)}}{(1+x^2)^2}\,dx\\\\
&=\frac\pi 4-\int_0^a F_1(x)\,dx\\\\
&=\frac\pi 4-\frac{\pi}{2}\int_0^a \text{erfc}(\sqrt x)\,dx\\\\
&=\frac\pi 4-\pi \int_0^{\sqrt a} x\,\text{erfc}(x)\,dx\\\\
&=\frac\pi 4-\pi\left(\frac a2\text{erfc}(\sqrt a)-\frac12 \int_0^{\sqrt{a}} x^2\,\text{erfc}'(x)\,dx\right)\\\\
&=\frac\pi 4-\frac\pi 2 a\,\text{erfc}(\sqrt a)-\frac\pi 2 \frac{2}{\sqrt \pi}\int_0^\sqrt a x^2e^{-x^2}\,dx\\\\
&=\frac\pi 4-\frac\pi 2 a\,\text{erfc}(\sqrt a)-\frac\pi 2 \frac{2}{\sqrt \pi}\left(\frac{\sqrt \pi}{4}\text{erf}(\sqrt a)-\frac {\sqrt a}2 e^{-a}\right)\\\\
&=\frac{\pi}{4}\left(1-2a\right)\text{efrc}(\sqrt a)+\frac{\sqrt \pi}{2}\sqrt a e^{-a}
\end{align}$$
Thus, the first term on the right-hand side of $(3)$ is $e^2F_2(2)$.
The evaluation of $F_3(2)$ and $F_4(2)$ follow along the same methodology and is left as an exercise.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1687019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Count triangles in a pentagon with all of its diagonals drawn How many triangles in this picture?
I am able to count one by one, but it takes long time. I wonder whether there is an easy way or systematic way to count. Thank you.
| Ordering the $10$ vertices from top to bottom, and in rows left-to-right, the adjacency matrix of this graph is
$$\left(
\begin{array}{cccccccccc}
0 & 1 & 1 & 1 & 1 & 1 & 1 & 0 & 1 & 1 \\
1 & 0 & 1 & 1 & 1 & 1 & 0 & 1 & 1 & 1 \\
1 & 1 & 0 & 1 & 1 & 1 & 0 & 0 & 1 & 0 \\
1 & 1 & 1 & 0 & 1 & 0 & 1 & 0 & 0 & 1 \\
1 & 1 & 1 & 1 & 0 & 0 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 0 & 0 & 0 & 0 & 1 & 1 & 1 \\
1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 1 \\
0 & 1 & 0 & 0 & 1 & 1 & 1 & 0 & 1 & 1 \\
1 & 1 & 1 & 0 & 1 & 1 & 1 & 1 & 0 & 1 \\
1 & 1 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 0 \\
\end{array}
\right)$$
The trace of the third power of this matrix is $330$, and this counts the triangles in the graph, each six times, giving $55$ triangles But this also counts triangles whose vertices are collinear. This happens when all three vertices are on the same one of the five sides of the “star.” Each star side contains four vertices, so there are ${4\choose3}=4$ such triangles on each of the five star sides. Subtracting these $4\times5=20$ degenerate triangles from the $55$ triangles in the graph leaves $35$ triangles.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1691693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Does there exist a linear transformation from $\mathbb{R}^3 \to\mathbb{R}^3$ such that $T((1,1,1)) = (1,2,3), T((1,2,1) = (1,1,1)$? I tried making the column vectors $(1,1,1)$ and $(1,2,1)$ into a matrix, but because it's $3\times2$, it's noninvertible. Does this mean that I cannot recover the linear transformation without a third vector? Do the vectors that define the transformation $(1,1,1)$ and $(1,2,1)$ have to form a basis for the subspace they're being transformed from for the transformation to be defined? Sorry if I'm mixing things up or wording things in a confusing manner.
| You're looking for a matrix $ \begin{pmatrix}
a&b&c\\
e&f&g\\
h&i&j
\end{pmatrix}$ such that $$\begin{pmatrix}
a&b&c\\
e&f&g\\
h&i&j
\end{pmatrix}
\begin{pmatrix}
1\\
1\\
1
\end{pmatrix}
= \begin{pmatrix}
1\\
2\\
3
\end{pmatrix}
$$
and
$$\begin{pmatrix}
a&b&c\\
e&f&g\\
h&i&j
\end{pmatrix}
\begin{pmatrix}
1\\
2\\
1
\end{pmatrix}
= \begin{pmatrix}
1\\
1\\
1
\end{pmatrix}
.$$ If you do the multiplications you get six equations
$$\begin{cases}
a+b+c=1 \\
e+f+g=2 \\
h+i+j=3 \\
a+2b+c=1 \\
e+2f+g= 1 \\
h+2i+j=1
\end{cases}.$$ You can easily find (by subtraction [e.g. $a+b+c=1$ and $a+2b+c=1$]) that $b=0$, $f=-1$, $i=-2$. Then for example choose $a=1, c=0, e=3,g=0, h=5,j=0$ and you can conclude that $$ \begin{pmatrix}
1&0&0\\
3&-1&0\\
5&-2&0
\end{pmatrix}$$ is a solution to your problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1692162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Series expansion of $\ln(1+(1-x)^{1/2})$ I am practicing series expansions by coming up with some expression, trying to do it by myself and then checking myself with wolfram alpha. However, I have some issues with the following example
I need to expand $\ln(1+(1-x)^{1/2})$. My approach was to first expand
$$(1-x)^{1/2} = 1 -\frac{x}{2} - \frac{x^2}{8} - \frac{x^3}{16}+O(x^3)$$
then I used the expansion of
$$
\ln(1+v) = v-\frac{v^2}{2}+\frac{v^3}{3}+O(v^3)
$$
and instead of $v$, I substituted the series expansion of $(1-x)^{1/2}$. The final result I got was $$
\frac{5x}{6}-\frac{x^2}{2}-\frac{5x^3}{36} +O(x^3).$$
However, in this case Wolfram Alpha alpha gives me the following answer:
$$
\ln 2-\frac{x}{4}-\frac{3 x^2}{32}-\frac{5 x^3}{96}-\frac{35 x^4}{1024}-\frac{63 x^5}{2560}+O(x^6)
$$
Why is my answer different, since it seems to me my approach is right?
Thanks for the help!
| Your first two expansions are correct, for $x$ and $v$ respectively going to $0$. Substituting, you now obtain
$$
\ln( 1 + \sqrt{1-x} ) =
\ln( 1 + 1-\frac{x}{2}-\frac{x^2}{8}-\frac{x^3}{16} + O(x^3) ) =
\ln( 2-\frac{x}{2}-\frac{x^2}{8}-\frac{x^3}{16} + O(x^3) )
$$
and here is your issue: the development of $\ln(1+u)$ is not longer around $0$, but around $1$ (since you have $\ln(2+u) = \ln(1+(1+u))$). To deal with this, you can rewrite:
$$
\ln( 2-\frac{x}{2}-\frac{x^2}{8}-\frac{x^3}{16} + O(x^3) )
= \ln 2 + \ln( 1-\frac{x}{4}-\frac{x^2}{16}-\frac{x^3}{32} + O(x^3) )
$$
using $\ln(ab) = \ln a + \ln b$, and proceed.
Upshot: when composing Taylor expansions around $0$, make sure that each step remains a Taylor expansion around $0$. The expansion of the inner function may shift this.
Another similar example: do a Taylor expansion of $e^{\cos x}$ around 0.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1692333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Alternate method to prove this series in a better way Prove that $\frac{1.2 + 2.3 + 3.4 + .....+ n(n + 1)}{n(n + 3)} \ge \frac{n + 1}{4}$ for $n\ge1$
My attempt :
Breaking the series into two different series
$$ S_1 = \sum_{i = 0}^n i^2 = \frac{n(n + 1) (2n + 1)}{6}$$
$$ S_2 = \sum_{i = 0}^n i = \frac{n(n + 1)}{2}$$
Now $$S_1 + S_2 = \frac{n(n + 1)(n + 2)}{3}$$
Writing down LHS and RHS we get :
$$\frac{(n + 1)(n + 2)}{3(n + 3)} \ge \frac{(n + 1)}{4}$$
Multiplying both sides by $$\frac{12}{n + 1}$$ we get :
$$4(n + 1) \ge 3$$
This will always be true $$. ^ . _ . n\ge1$$
Hence proved.
Is there a mistake in my method? Is there a better method to prove this?
| Since $\left(\frac{n+2}{n+3}\right)$ increases, $\left(\frac{n+2}{n+3}\right)$ has the minimum $\frac{3}{4}$ when $n=1$. Thus we get
\begin{align}
\frac{n(n+1)(n+2)}{3n(n+3)} &= \frac{(n+1)(n+2)}{3(n+3)}\\
&= \frac{n+1}{3}\cdot \frac{n+2}{n+3}\\
&\ge \frac{n+1}{3}\cdot\frac{3}{4}\\
&=\frac{n+1}{4}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1692929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
SVD of (2,1,-2) not ok I'm trying to find the SVD of
$$
\begin{pmatrix}
2&1&-2\\
\end{pmatrix}
$$
I found $$\Sigma , u$$
But on the V matrix I got
$$
\begin{pmatrix}
-\frac{2}{3}&&\frac{1}{\sqrt{2}}&&\frac{-1}{\sqrt{5}} \\
-\frac{1}{3}&&0&&\frac{2}{\sqrt{5}} \\
\frac{2}{3}&&\frac{1}{\sqrt{2}}&&0 \\
\end{pmatrix}
$$
While Wolfram Alpha gives :
result
I checked the eigenvectors on wolfram alpha as well and they are correct:
$$
\begin{pmatrix}
-2&&1&&-1\\
-1&&0&&2\\
2&&1&&0\\
\end{pmatrix}
$$
I realized that the 3rd collumn of V (the one from wolfram alpha) is obtained by doing the cross product of the first two eigenvectors , why is that ?
| A singular value decomposition of a $1\times 3$ matrix can be written almost "by inspection":
$$ \begin{pmatrix} 2 & 1 & -2 \end{pmatrix} = U \Sigma V^* $$
where $U,V$ are unitary (in this case, orthogonal) matrices and $\Sigma$ is a rectangular diagonal matrix with the required singular values on the diagonal:
$$ \Sigma = \begin{pmatrix} \lambda & 0 & 0 \end{pmatrix} $$
Since $U$ is a $1\times 1$ unitary matrix, actually $U = (1)$ is just the $1\times 1$ identity. $\Sigma$ is a $1\times 3$ matrix of equal Euclidean norm to $\begin{pmatrix} 2 & 1 & -2 \end{pmatrix}$, so $\Sigma = \begin{pmatrix} 3 & 0 & 0 \end{pmatrix}$.
The requirement that $V$ be unitary (or orthogonal) does not uniquely specify its entries. So long as:
$$ \begin{pmatrix} 3 & 0 & 0 \end{pmatrix} V^* = \begin{pmatrix} 2 & 1 & -2 \end{pmatrix} $$
it will meet the definition of a singular value decomposition, and from this we see that only the first column of $V$ (top row of $V^*$) is determined:
$$ V^* = \begin{pmatrix} \frac{2}{3} & \frac{1}{3} & -\frac{2}{3} \\
{-} & {-} & {-} \\ {-} & {-} & {-} \end{pmatrix} $$
The bottom two rows of $V^*$ can be filled in with any pair of unit vectors that make up an orthonormal basis for $\mathbb{R}^3$ together with the top row shown. Therefore it is not surprising that the third row might be the cross-product of the first two rows (since the third row will be determined, up to sign, by the first two rows).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1693041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Are $B_1(0)$ and $B_1(0) \setminus X$ homeomorphic? Work in the vector space $\mathbb{R}^2$. I'll write $B_r(y)$ for the open ball of radius $r$ centered at $y$, for all vectors $y$ and all real numbers $r \geq 0$.
Now let $x$ denote a fixed but arbitrary non-zero vector of length $1/2.$ Define $X = \mathbb{R}_{\geq 1} x.$ That is: $$X = \{ax \mid a \in \mathbb{R}_{\geq 1}\}$$
Observe that since $X$ is closed, hence $B_1(0) \setminus X$ is open.
Now equip both $B_1(0)$ and $B_1(0) \setminus X$ with the subspace topology.
Question. Are $B_1(0)$ and $B_1(0) \setminus X$ homeomorphic?
| Yes, $B_1(0)$ and $B_1(0)\setminus\left(\left[\frac 1 2, 1\right)\times\left\{0\right\}\right)$ are homemorphic. Since I'm a masochist, here's a homeomorphism: map $\begin{pmatrix}x\\y\end{pmatrix}\in B_1(0)$ to
$$
\begin{cases}
\Big(1-\frac{x}{\sqrt{1-y^2}}\Big) \begin{pmatrix}y\\\frac 12\sqrt{1-(2y-1)^2}\end{pmatrix}+\frac{x}{\sqrt{1-y^2}}\begin{pmatrix}\frac 12+\frac 12 y\\0\end{pmatrix} & \text{if $x\ge 0$, $y\ge 0$}, \\
\Big(1-\frac{x}{\sqrt{1-y^2}}\Big) \begin{pmatrix}-y\\-\frac 12\sqrt{1-(-2y-1)^2}\end{pmatrix}+\frac{x}{\sqrt{1-y^2}}\begin{pmatrix}\frac 12-\frac 12 y\\0\end{pmatrix} & \text{if $x\ge 0$, $y\le 0$}, \\
\Big(1-\frac{-x}{\sqrt{1-y^2}}\Big) \begin{pmatrix}y\\\frac 12\sqrt{1-(2y-1)^2}\end{pmatrix}+\frac{-x}{\sqrt{1-y^2}}\begin{pmatrix}2y-1\\\sqrt{1-(2y-1)}\end{pmatrix} & \text{if $x\le 0$, $y\ge 0$}, \\
\Big(1-\frac{-x}{\sqrt{1-y^2}}\Big) \begin{pmatrix}-y\\-\frac 12\sqrt{1-(-2y-1)^2}\end{pmatrix}+\frac{-x}{\sqrt{1-y^2}}\begin{pmatrix}-2y-1\\-\sqrt{1-(-2y-1)}\end{pmatrix} & \text{if $x\le 0$, $y\le 0$}.
\end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1694273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Simple method to solve a geometry question for junior high school student Rencently, my sister asked me a geometry question that came from her mock examination, please see the following graph.
Here,
*
*$\angle DOE=45°$
*the length of $DE$ is constant, and $DE=1$. Namely, $OD,OE$ are changeable.
*$\triangle DEF$ is equilateral triangle.
Q: What is the maximum length of $OF$?
My solution
Denote $OD,OE,\angle ODE$ as $x,y,\theta$, respectively.
Via sine theorem
$$
\begin{cases}
ED^{2} = OE^{2} + OD^{2} - 2OE \times OD\cos \angle EOD \\[6pt]
\cos \theta = \dfrac{EO^{2} + ED^{2} - OD^{2}}{2 EO \times ED}
\end{cases}
$$
$$
\begin{align}
1^{2} &= x^{2} + y^{2} - 2xy\cos 45^{\circ} \\
&= x^{2} + y^{2} - \sqrt{2} xy
\end{align}
$$
$$
\implies
\begin{cases}
\color{red}{xy} = \dfrac{x^{2} + y^{2} - 1}{\sqrt{2}} \color{red}{\leq}
\dfrac{x^{2} + y^{2}}{2} \implies x^{2} + y^{2} \color{red}{\leq}
2 + \sqrt{2} \\[6pt]
\cos \theta = \dfrac{x^{2} + 1 - y^{2}}{2x}
\end{cases}
$$
Via cosine theorem
$$
\frac{y}{\sin \theta} = \frac{DE}{\sin \angle EOD} = \frac{1}{\sin
45^{\circ}} \implies \sin \theta = \frac{y}{\sqrt{2}}
$$
$$\begin{align}
OF^{2} &= EO^{2} + EF^{2} - 2EO \times EF\cos \angle OEF \\
&= x^{2} + 1^{2} - 2x\cos(\theta + 60^{\circ}) \\
&= x^{2} + 1 - 2x(\cos \theta \cos 60^{\circ} - \sin \theta \sin
60^{\circ}) \\
&= x^{2} + 1 - 2x\left(\frac{x^{2} + 1 - y^{2}}{2x} \frac{1}{2} -
\frac{y}{\sqrt{2}} \frac{\sqrt{3}}{2}\right) \\
&= \frac{x^{2} + y^{2} + 1}{2} + \frac{\sqrt{3} xy}{\sqrt{2}} \\
&= \frac{x^{2} + y^{2} + 1}{2} + \frac{\sqrt{3}}{\sqrt{2}}
\frac{x^{2} + y^{2} - 1}{\sqrt{2}} \\
&= \frac{(\sqrt{3} + 1)(x^{2} + y^{2})}{2} + \frac{1 - \sqrt{3}}{2}
\\
&\color{red}{\leq} \frac{(\sqrt{3} + 1)(2 + \sqrt{2})}{2} + \frac{1 -
\sqrt{3}}{2} = \frac{1}{2}(3 + \sqrt{3} + \sqrt{2} + \sqrt{6})
\end{align}
$$
However, for junior high school student, she doesn't learn the following formulae:
*
*sine theorem
*cosine therem
*$\cos(x+y)=\cos x \cos y-\sin x \sin y$
*fundamental inequality $x y\leq \frac{x^2+y^2}{2}$
Question
*
*Is there other simple/elegant method to solve this geometry question?
Update
Thanks for MXYMXY's hint
Here, the line $O'F$ pass the center of the circle. Namely, $O'D=OF$
In Rt $\triangle O'OF$, the inequality $O'F>OF$ holds.
| I am assuming that there is a condition that $\angle DOE=45°$ from your graph.
HINT
Note that the set of $O$ such that $\angle DOE=45°$ forms a circle.
And a point on a circle that is farthest away from a fixed point $Q$ holds when the point, the center of the circle, and $Q$ are collinear.
Using this, note the maximum of $\overline {OF}$ holds when $O'D=O'F$.
Computing gives us (credit goes to @AmeetSharma) that the maximum is $$\frac{\overline{DE}}{2}+\frac{\sqrt{2}\overline{DE}}{2}+\frac{\sqrt{3}\overline{DE}}{2}= \dfrac{1+\sqrt{2}+\sqrt{3}}{2}$$
To generalize, if $\angle DOE=\theta$ then the maximum is $$\frac{\cot \theta}{2}+\frac{\sec \theta}{2}+\frac{\sqrt{3}}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1695132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 2
} |
If $\phi : S^n - \{e_{n+1}\} \to \mathbb{R}^n$ is the stereographic projection, how to compute $\phi^{-1}$? If $\phi : S^n - \{e_{n+1}\} \to \mathbb{R}^n$ is the stereographic projection, how to compute $\phi^{-1}$?
If $$\phi(x) = \frac{1}{1 - x_{n+1}}(x^1,\ldots,x^n),$$
how to compute $\phi^{-1}(y)?$
| Let $y = \phi(x) = (y_1, \ldots, y_n)$. As $x \in S^n$, we have
$$ |y|^2 = \frac 1{(1-x_{n+1})^2}\sum_{i=1}^n x_i^2 = \frac{1 - x_{n+1}^2}{(1-x_{n+1})^2} = \frac{1 + x_{n+1}}{1-x_{n+1}} \iff x_{n+1} = \frac{|y|^2-1}{|y|^2 + 1} $$
Hence,
\begin{align*}
\frac 1{1- x_{n+1}} &= \frac {|y|^2 + 1}{|y|^2 + 1 - (|y|^2 - 1)}\\
&= \frac 12(|y|^2 + 1)
\end{align*}
Hence, for $1 \le i \le n$:
$$ x_i = (1 - x_{n+1})y_i = \frac{2y_i}{|y|^2 + 1} $$
That is
$$ x = \phi^{-1}(y) = \left(\frac{2y}{|y|^2 + 1}, \frac{|y|^2 - 1}{|y|^2 + 1}\right). $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1696988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to solve $\cos(x)\cos(2x)\cos(4x)=1/8$ I have to solve $\cos(x)\cos(2x)\cos(4x)=1/8$.
I can express it for $x$ only with $\cos(2x)=\cos^2(x)-\sin^2(x)$ and $\cos(4x)=\cos(2x+2x)$, but it only seems to become a really big expression and I have no clue how to proceed after... Any suggestions?
| Let $z=e^{ix}$. Then your relation says $$\left(z+z^{-1}\right)\left(z^2+z^{-2}\right)\left(z^4+z^{-4}\right)=1$$ That is:
$$
\begin{align}
z^7+z^5+z^3+z+z^{-1}+z^{-3}+z^{-5}+z^{-7}&=1\\
z^{-7}\frac{z^{16}-1}{z^2-1}&=1&\text{(for $z^2\neq1$)}\\
z^{16}-1&=z^9-z^7\\
z^{16}-z^9+z^7-1&=0\\
\left(z^7-1\right)\left(z^9+1\right)&=0
\end{align}$$
So $z$ is either a $7$th root of $1$ or a $9$th root of $-1$. Which means your $x$ is among $$\left\{\frac{2\pi}{7}, \frac{4\pi}{7},\frac{6\pi}{7},\frac{8\pi}{7},\frac{10\pi}{7},\frac{12\pi}{7},\frac{\pi}{9},\frac{3\pi}{9},\frac{5\pi}{9},\frac{7\pi}{9},\frac{11\pi}{9},\frac{13\pi}{9},\frac{15\pi}{9},\frac{17\pi}{9}\right\}$$ or translates by $2\pi k$. We've left out $0\pi/7$ and $9\pi/9$ since they cause $z^2-1$ to equal $0$, invalidating an earlier step here. You can directly check that these angles do not satisfy the original equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1697505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Proving that $2\sqrt 3+3\sqrt[3] 2-1$ is irrational
Prove that $2\sqrt 3+3\sqrt[3] 2-1$ is irrational
My attempt:
$$k=2\sqrt 3+3\sqrt[3] 2-1$$
Suppose $k\in \mathbb Q$, then $k-1\in \mathbb Q$.
$$2\sqrt 3+3\sqrt[3] 2=p/q$$
I'm stuck here and don't know how to procced. I tried to do this:
$$\sqrt 3=\frac{p/q-3\sqrt[3] 2}{2}$$
contradiction, but I'm not at all sure about that. How should I proceed?
| \begin{align*}
2\sqrt 3+3\sqrt[3] 2=\frac{p}{q} &\Rightarrow 3\sqrt[3] 2=\frac{p}{q}-2\sqrt 3\\
&\Rightarrow 54=\left(\frac{p}{q}-2\sqrt 3\right)^3\\
&\Rightarrow 54=\frac{p^3}{q^3}-6\frac{p^2}{q^2}\sqrt{3}+36\frac{p}{q}-24\sqrt{3}\\
&\Rightarrow \sqrt{3}=\frac{p^3+36pq^2-54q^3}{6q(p^2+4q^2)} \in \mathbb{Q}\\
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1698392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
Prime numbers are related by $q=2p+1$ Let primes $p$ and $q$ be related by $q=2p+1$. Prove that there is a positive multiple of $q$ for which the sum of its digits does not exceed $3$.
My work so far:
$p,q -$ primes and $q=2p+1 \Rightarrow \exists n,k \in \mathbb N: n=qk$ and $S(n) \le 3$
$p=2 \Rightarrow q=5 \Rightarrow 5|10=n; S(10)=1 \le3$.
$p=3 \Rightarrow q=7 \Rightarrow 7|21=n; S(21)=3 \le3$.
$p=5 \Rightarrow q=11 \Rightarrow 1|11=n; S(11)=2 \le3$.
$p=7 \Rightarrow q=15=5 \cdot3 $.
$p=11 \Rightarrow q=23 \Rightarrow 23|n; S(n) \le3$.
| (ALMOST an answer) In order for $S(kq)\leq3$, $kq$ must be in one of the following forms:
(1): $$kq=2*10^n+1$$
(2): $$kq=10^n+1$$
(3): $$kq=10^n+10^m+1$$, where $n\neq m$
We will look into the cases one by one and show that there must exist some $n$ or $n,m$ that satisfy at least one of the above cases, following @Peter 's direction.
Case (1):
$2*10^n+1\equiv2p+1$ (mod $q$)
$10^n\equiv p$
Now let $g$ be a primitive root for mod $q$, such that $g^a\equiv10$ and $g^b\equiv p$ (mod $q$), where $1\leq a,b\leq 2p$
Then the question reduces to whether $g^{an}=g^b$ (mod q) is solvable or not. And since the primitive root has order $2p$,
$an\equiv b$ (mod $2p$)
Is only solvable iff
$b\equiv 0$ (mod $GCD(a,2p)$)
So if $a,2p$ are coprime, there must be an integer $0\geq n\geq 2p$ that satisfies $kq=2*10^n+1$
Case (2):
If the above case fails, that means $b\not\equiv 0$ (mod $GCD(a,2p)$)
Then $a$ and $2p$ must not be coprime, since $p$ is a prime, $a=p$ or $a=2$.
$2p\equiv -1\equiv 10^n$ (mod $q$)
Let $g^c\equiv 2$ (mod $q$), $1\leq c\leq 2p$
We know,
$g^p\equiv g^{b+c}\equiv -1$
Because $(g^p)^2\equiv 1$ (mod $q$) but $p\not\equiv 0$ (mod $2p$)
If $g^p\equiv g^{an}$ is solvable
Then,
$p\equiv an$ (mod $2p$) has a solution for $n$, and it has solution iff
$p\equiv 0$ (mod $GCD(a,2p)$)
Sub-case: $a=p$,
$p\equiv 0$ (mod $a=p$) is trivial
So there must be solution for $n$ if $a=p$
Sub-case: $a=2$, there must be solution for $n$ if $b,c$ have the same parity.
In fact, $1 \leq b,c \leq 2p$ implies that $2\leq b+c\leq 4p$,
so $b+c=p$ or $b+c=3p$,
because if $b+c=2p$ or $4p$, it would contradict that $g^{b+c}\equiv -1$ (mod $q$)
So $b+c\equiv 1$ (mod $a=2$), for the case that $a=2$, there will be no solution in the form of (1) or (2).
Case (3):
$kq=10^n+10^m+1$
$2p\equiv 10^n+10^m$ (mod $q$)
$g^{b+c}\equiv -1\equiv g^{2n}+g^{2m}$
I cant figure how to show that this type of equation has solution $n,m$. If you can prove that there exist solutions, then we have covered all possible cases.
(Side note: it can be shown that if $a=p$, then there must be no solution in the form $10^n+10^m+1$ since
$g^{pm}+g^{pn}+1\not\equiv 0$ (mod $q$) since $g^{pm},g^{pn}\equiv \pm 1$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1698878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Finding function based on slope and point I have to find the equation of the function given the following information:
$$f(0) = 3 \ \ \ ; \ \ \ \frac{dy}{dx} = \frac{xy}{1+x^2}$$
So:
$$\frac{dy}{y} = \frac{xdx}{1+x^2}$$
Integrate:
$$\ln |y| + K = \ln|\sqrt{1+x^2}| + K$$
$$y = \sqrt{1+x^2} + K$$
Plug in values to find K
$$3 = \sqrt{1+0^2} + K \ \ \ ; \ \ \ K = 2$$
$$y = \sqrt{1+x^2} + 2$$
Answer in the book is
$$y = 3\sqrt{1+x^2}$$
Where did I go wrong?
| $$y'(x)=\frac{xy(x)}{1+x^2}\Longleftrightarrow$$
$$\frac{y'(x)}{y(x)}=\frac{x}{1+x^2}\Longleftrightarrow$$
$$\int\frac{y'(x)}{y(x)}\space\text{d}x=\int\frac{x}{1+x^2}\space\text{d}x\Longleftrightarrow$$
$$\ln\left|y(x)\right|=\frac{\ln\left|x^2+1\right|}{2}+\text{C}\Longleftrightarrow$$
$$\left|y(x)\right|=\exp\left[\frac{\ln\left|x^2+1\right|}{2}+\text{C}\right]\Longleftrightarrow$$
$$\left|y(x)\right|=\exp\left[\text{C}\right]\exp\left[\frac{\ln\left|x^2+1\right|}{2}\right]\Longleftrightarrow$$
$$\left|y(x)\right|=\text{C}\exp\left[\frac{\ln\left|x^2+1\right|}{2}\right]\Longleftrightarrow$$
$$\left|y(x)\right|=\text{C}\sqrt{\left|1+x^2\right|}$$
Now, use $y(0)=3$:
$$\left|3\right|=\text{C}\sqrt{\left|1+0^2\right|}\Longleftrightarrow$$
$$3=\text{C}\sqrt{1}\Longleftrightarrow$$
$$3=\text{C}\cdot1\Longleftrightarrow$$
$$3=\text{C}\Longleftrightarrow$$
$$\text{C}=3$$
So:
$$\left|y(x)\right|=3\sqrt{\left|1+x^2\right|}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1700704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to compute $I_n= \int_{0}^{\pi/4} x\tan^{2n}x\, \mathrm{d}x $ Consider the following integral
$$I_n= \int_{0}^{\pi/4} x\tan^{2n}x \, \mathrm{d}x $$
Then compute $I_n + I_{n+1}$ as a function of $n$ $\forall n\geq0$.
| First, notice that
$$\begin{align}
I_n + I_{n+1} &= \int_{0}^{\pi/4} x \left( \tan^{2n}x +\tan^{2n+2}x \right) \, \mathrm{d}x \\
&= \int_{0}^{\pi/4} x\tan^{2n}x \left(1+ \tan^2 x\right) \, \mathrm{d}x \\
&= \int_{0}^{\pi/4} x\tan^{2n}x \, \sec^2 x \, \mathrm{d}x \\
&= \dfrac{1}{2n+1} x \tan^{2n+1} x |_{0}^{\pi/4} - \dfrac{1}{2n+1} \int_{0}^{\pi/4} \tan^{2n+1}x \, \mathrm{d}x \\
&= \dfrac{1}{2n+1} \left( \dfrac{\pi}{4} - \int_{0}^{\pi/4} \tan^{2n+1}x \, \mathrm{d}x \right) \\
&\equiv \dfrac{1}{2n+1} \left( \dfrac{\pi}{4} - J_n \right)
\end{align}$$
and for computing $J_n$ I refer you to this post. For the sake of completeness, I just mention the final result
$$\begin{align}
J_n &=\int_{0}^{\pi/4} \tan x ^{2n+1} \mathrm dx \\
&= \left[ \sum_{k=0}^{n-1} \frac{(-1)^k}{2(n-k)}\tan x ^{2(n-k)}+(-1)^{n+1}\ln \cos x \right]_{0}^{\pi/4} \\
&= \sum_{k=0}^{n-1} \frac{(-1)^k}{2(n-k)}+\frac{(-1)^n}{2}\ln2
\end{align}$$
and hence
$$\boxed{
I_n + I_{n+1} = \dfrac{1}{2n+1} \left( \dfrac{\pi}{4} - \sum_{k=0}^{n-1} \frac{(-1)^k}{2(n-k)}-\frac{(-1)^n}{2}\ln2 \right)
}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1701566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Linear Functions of Independent Random Variables: What's the value of b? Suppose that $X_1,X_2,\ldots,X_{25}$ are independent random variables from N$(1, 4)$.
$Y = aX + b$, where $a$ ($a > 0$) and $b$ are constants. $X = \frac{1}{25} \sum_{i=1}^{25} X_i$. We know that $Y$ is a standard normal random variable.
What's the value of $b$?
| Recall that $X_i \sim N(1,4)$, gives $\sum_i X_i \sim N(25, 100)$, hence $X = \frac 1{25}\sum_i X_i \sim N(1,\frac{4}{25})$. This implies that $aX \sim N(a,\frac{4}{25}a^2)$ and $aX + b \sim N(a+b, \frac{4}{25}a^2)$. If $aX+ b \sim N(0,1)$, this implies
$$ \frac{4}{25}a^2 = 1, \qquad a +b = 0$$
as $a > 0$, $\frac{4}{25}a^2 = 1$ gives $a = \frac 52$, hence $b = -a = -\frac 52$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1702246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Sum of all Products on Catalan numbers how can I simplify this?
let:
$$
C_n = {{2n \choose n}\over n+1}
$$
find:
$$
\sum_{P_1 + P_2 + ... + P_k = r} \left(\prod_{j = 1}^k C_{P_j}\right)
$$
thanks!
| Following @RobertIsrael we have the generating function of the Catalan
numbers
$$f(z) = \frac{1-\sqrt{1-4z}}{2z}$$
and the quantity in question is
$$[z^r] f(z)^k.$$
This has the integral representation
$$[z^r] f(z)^k
= \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{r+1}} \left(\frac{1-\sqrt{1-4z}}{2z}\right)^k
\; dz.$$
Now put $1-4z = w^2$ so that $1/4-w^2/4 = z$ and
$dz = -w/2 \; dw$ to get
$$-\frac{1}{2^{k+1}} \frac{1}{2\pi i}
\int_{|w-1|=\gamma}
\frac{4^{r+k+1}}{(1-w^2)^{r+k+1}}
(1-w)^k w
\; dw
\\ = - 2^{2r+k+1} \frac{1}{2\pi i}
\int_{|w-1|=\gamma}
\frac{1}{(1+w)^{r+k+1}} \frac{1}{(1-w)^{r+1}}
w \; dw
\\ = (-1)^r 2^{2r+k+1} \frac{1}{2\pi i}
\int_{|w-1|=\gamma}
\frac{1}{(1+w)^{r+k+1}} \frac{1}{(w-1)^{r+1}}
w \; dw
\\ = (-1)^r 2^{2r+k+1} \frac{1}{2\pi i}
\int_{|w-1|=\gamma}
\frac{1}{(1+w)^{r+k+1}} \frac{1}{(w-1)^{r}}
\; dw
\\ + (-1)^r 2^{2r+k+1} \frac{1}{2\pi i}
\int_{|w-1|=\gamma}
\frac{1}{(1+w)^{r+k+1}} \frac{1}{(w-1)^{r+1}}
\; dw.$$
Now we require the derivative
$$\left(\frac{1}{(1+w)^{r+k+1}}\right)^{(q)}
= (-1)^q \frac{(r+k+q)!}{(r+k)!}
\frac{1}{(1+w)^{r+k+q+1}}.$$
This yields for the two pieces
$$(-1)^r 2^{2r+k+1}
(-1)^{r-1} \frac{(2r+k-1)!}{(r-1)!(r+k)!} \frac{1}{2^{2r+k}}
+ (-1)^r 2^{2r+k+1}
(-1)^{r} \frac{(2r+k)!}{r!(r+k)!} \frac{1}{2^{2r+k+1}}
\\ = -2 {2r+k-1\choose r-1}
+ {2r+k\choose r}
\\ = {2r+k\choose r}
\left(1-2\frac{r}{2r+k}\right)
\\ = \frac{k}{2r+k} {2r+k\choose r}.$$
Here we have used the fact that $\sqrt{1-4z} = 1 - 2z - 2z^2 - \cdots$
so that with $z$ rotating once around the origin the variable $w$
rotates once around the value one, with a contour that may be deformed
to a circle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1702416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find k , if k = $\lim \limits_{n \to \infty} n[(n^3+3n^2+2n+1)^{1/3}+(n^2-2n+3)^{1/2}-2n]$ Find k , if k = $\lim \limits_{n \to \infty} n[(n^3+3n^2+2n+1)^{1/3}+(n^2-2n+3)^{1/2}-2n]$
I converted it to $\lim \limits_{t \to 0}$ and tried using L'Hospital's rule and I got it after differentiating twice. It looks pretty bad. Please give an elegant solution. Thank you.
| Observe that $(n+1)^3=n^3+3n^2+3n+1$, $(n-1)^2=n^2-2n+1$, $2n=(n+1)+(n-1)$
$$\lim \limits_{n \to \infty}n((n+1)^3-n)^{1/3}-(n+1))\\= \lim \limits_{n \to \infty}\frac{-n^2}{((n+1)^3-n)^{2/3}+(n+1)((n+1)^3-n)^{1/3}+(n+1)^2}\\=\frac{-1}{3}$$
$$\lim \limits_{n \to \infty}n((n-1)^2+2)^{1/2}-(n-1))=\lim \limits_{n \to \infty}\frac{2n}{((n-1)^2+2)^{1/2}+(n-1)}=1$$
so $k=1-\frac{1}{3}=\frac{2}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1704236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If the roots of $ax^2+bx+c=0$ are of the form $\frac{m}{m-1}$ and $\frac{m+1}{m}$ then find.. Problem :
If the roots of $ax^2+bx+c=0$ are of the form $\frac{m}{m-1}$ and $\frac{m+1}{m}$ then find the value of $(a+b+c)^2$
My approach :
Let $\alpha, \beta$ are the two roots of the given equation.
then
$\frac{1}{\alpha}+\beta =2$
Also we know that sum of the roots $= \frac{-b}{a}$ and product of the roots $\frac{c}{a}$
but how to find $(a+b+c)^2$ not getting any idea further please help thanks.
| Say the roots are $\frac{m}{m-1}$ and $\frac{m+1}{m}$.
Then we have that,
$$\frac{m}{m-1}+\frac{m+1}{m}=-\frac{b}{a}$$
$$\frac{m}{m-1}\cdot \frac{m+1}{m}=\frac{c}{a}$$
So we can conclude that
$$\frac{2m^2-1}{m(m-1)}=-\frac{b}{a}$$
$$\frac{m+1}{m-1}=\frac{c}{a}$$
So we can write that
$$(a+b+c)^2=a^2(1+\frac{b}{a}+\frac{c}{a})^2$$
$$=a^2\left[1-\frac{2m^2-1}{m(m-1)}+\frac{m+1}{m-1}\right]^2$$
$$=a^2\left[\frac{m^2-m-2m^2+1+m^2+m}{m(m-1)}\right]^2$$
$$=\left[\frac{a}{m(m-1)}\right]^2$$
Now, again we can write that $$(x-\frac{m}{m-1})(x-\frac{m+1}{m})=0$$
$$x^2-(\frac{m}{m-1}+\frac{m+1}{m})x-\frac{m+1}{m-1}=0$$
$$m(m-1)x^2-(2m^2-1)x-(m^2+m)=0$$
By comparison with $ax^2+bx+c=0$, we can say that $a=rm(m-1)$.
So we have that $$(a+b+c)^2=r^2$$
See if you can progress any far with this result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1705798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Determinant of $2 \times 2$ matrix such that $A = A^{-1}$
Let $A$ be a $2 \times 2$ matrix such that $A = A^{-1}$. The value of $\operatorname{det} (A)$ can be:
*
*$\operatorname{det} (A)=-2$
*$\operatorname{det} (A)=-1$
*$\operatorname{det} (A)=0$
*$\operatorname{det} (A)=2$
My attempt:
$$\begin{bmatrix}
a &b \\
c& d
\end{bmatrix} = \frac{1}{ad-bc} \times \begin{bmatrix}
d &-b \\
-c& a
\end{bmatrix}$$
Obviously the determinant cannot be $0$, since an inverse exists, but I'm not sure how to proceed. Any help would be appreciated.
| Your approach is workable, but as other have answered it's not the easiest approach.
If you want to continue you only have to calculate the determinants of the matrices:
$$\det\begin{pmatrix}a & b \\ c& d\end{pmatrix} = ad - cb$$
$$\det{1 \over ad-bc}\begin{pmatrix}d & -b \\ -c & a\end{pmatrix} =\det\begin{pmatrix}{d \over ad-bc}& {-b \over ad-bc} \\ {-c \over ad-bc} & {a \over ad-bc}\end{pmatrix} = \left({1\over ad-bc}\right)^2\left(da-\left(-c\right)\left(-b\right)\right) = {1\over ad-bc}$$
Now you see that $\det (A^{-1}) = 1/\det A$, then you have that if $A=A^{-1}$ that $\det(A) = 1/\det(A)$ which is a second degree equation with solutions $\det(A) = \pm 1$
Another overly complicated approach might be to use that since $A=A^{-1}$ we have that $A^2 = AA = AA^{-1} = I$ so calculating $A^2$ we get
$$A^2 = \begin{pmatrix}a & b \\ c& d\end{pmatrix}\begin{pmatrix}a & b \\ c& d\end{pmatrix}=\begin{pmatrix}a^2 + bc & (a+d)b \\ (a+d)c & d^+bc\end{pmatrix}$$
Here we see that $a^2+bc = d^2+bc=1$ which implies that $a^2=d^2$ which means $a=\pm d$. If $a=d$ we have that $b=c=0$ which means that the diagonal elements becomes $a^2=d^2=1$ which means $a=\pm1$ and $d=\pm1$ which means the determinant is $\pm 1$. If $a=-d$ the diagonal elements becomes $a^2+bc = d^2 + bc = -ad+bc = 1$, which means that $\det A = ad-bc = -1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Cyclic Inequality in n (at least 4) variables Proof of a cyclic inequality.
Let $a_i$ , $i=1..n$, $n \geq 3$ be real numbers with $a_i > 0 $ and $\prod_{i=1}^n a_i= 1$. Denote $S = \sum_{i=1}^n a_i$ (unrestricted).
Prove (or disprove) the following cyclic (i.e. $a_{n+1} = a_{1}$) inequality at least for $n=4$, preferably for general $n \geq 6$:
$$\sum_{i=1; cyc}^n \frac{a_i \; \{S - (a_i + a_{i+1})\}}{a_i + a_{i+1}} \geq \frac{(n-2)\, n }{2}$$
Remarks:
*
*Obviously, if all $a_i$ are equal, $a_i = a$, we have $a = 1$ due to the product condition and hence one gets equality since
$$LHS = \sum_{i=1; cyc}^n \frac{n -2}{2} = \frac{(n -2)n}{2}$$
*The case for $n=3$ is given by
$$ \frac{a\; c}{a +b} + \frac{b\; a}{b +c} + \frac{c\; b}{c +a} \geq \frac{3}{2}$$
or equivalently, due to the product condition,
$$ \frac{1}{b(a +b)} + \frac{1}{c(b +c)} + \frac{1}{a(c +a)} \geq \frac{3}{2}$$
which is known (2008 International Zhautykov Olympiad), for some proofs see here: http://artofproblemsolving.com/community/c6h183916p1010959
*In the meantime (29. April 2016), also the case $n=5$ was proven (take into account the comment below by Macavity Mar 25 at 6:36) by the author, see here: Inequality with five variables
| Here is a proof for general $n$.
As pointed out in the comment by Macavity on Mar 25 at 6:36, the claim is equivalent to:
$$\sum_{cyc}\frac{a_i}{a_i+a_{i+1}}\geq 1+\frac{n(n-2)}{2S}$$
We can remove the requirement $\prod_{i=1}^n a_i=1$ by homogenizing this to (simultaneously defining $L$ and $R$):
$$L = \sum_{cyc}\frac{a_i}{a_i+a_{i+1}}\geq 1+\frac{n(n-2)\sqrt[n]{a_1a_2...a_n}}{2S} = R$$
The discussion following here is a generalization of the discussion for $n=5$ at Inequality with five variables
We write $2 L \geq 2 R$ or $L \geq 2 R- L$ and add on both sides a term $$\sum_{cyc}\frac{a_{i+1}}{a_i+a_{i+1}}$$ which leaves us to show
$$n = \sum_{cyc}\frac{a_i + a_{i+1}}{a_i+a_{i+1}}\geq 2+\frac{n(n-2)\sqrt[n]{a_1a_2...a_n}}{S} + \sum_{cyc}\frac{-a_i + a_{i+1}}{a_i+a_{i+1}}$$
or, in reformulation of our claim,
$$ \sum_{cyc}\frac{-a_i + a_{i+1}}{a_i+a_{i+1}} \leq (n - 2) (1- \frac{n \sqrt[n]{a_1a_2...a_n}}{S} )$$
We will need the following Lemma (required below), which is the above L-R-inequality for 3 variables:
$$ \frac{b-a}{b+a} + \frac{c-b}{c+b} + \frac{a-c}{a+c} \leq (1- \frac{3 \sqrt[3]{a\, b \, c}}{a + b+ c} )$$
This Lemma is, from the above discussion, just a re-formulation of the claim in $L$ and $R$ above, for 3 variables, i.e.
$$ \frac{a}{b+a} + \frac{b}{c+b} + \frac{c}{a+c} \geq 1+\frac{3\sqrt[3]{a \, b \ c}}{2(a+b+c)}$$
By homogeneity, we can demand $abc=1$ and under that restriction, we need to show $$ \frac{a}{b+a} + \frac{b}{c+b} + \frac{c}{a+c} \geq 1+\frac{3}{2(a+b+c)}$$
This reformulates into
$$ \frac{a\; c}{a +b} + \frac{b\; a}{b +c} + \frac{c\; b}{c +a} \geq \frac{3}{2}$$
which is known (see remark 2 in the problem description).
Hence the Lemma holds.
For general $n$, we rewrite the LHS of our above reformulation by adding and subtracting terms $\frac{a_{k}-a_1}{a_{k}+a_1}$, for $k=3 \ldots (n-1)$. Here, the first terms are given by $a,b,c,d,e$ for better readability.
$$ \sum_{cyc}\frac{-a_i + a_{i+1}}{a_i+a_{i+1}} = \\
(\frac{b-a}{b+a} + \frac{c-b}{c+b} + \frac{a-c}{a+c}) + (\frac{c-a}{c+a}+\frac{d-c}{d+c} + \frac{a-d}{a+d}) + (\frac{d-a}{d+a}+ \frac{e-d}{e+d} + \frac{a-e}{a+e})
\\
+ \cdots + (\frac{a_{n-1}-a_1}{a_{n-1}+a_1}+ \frac{a_{n}-a_{n-1}}{a_{n}+a_{n-1}} + \frac{a_1-a_n}{a_1+a_n})
$$
There are in total $n-2$ of these terms with three summands each.
This also holds for any of the $n$ cyclic shifts in $(a_1 \cdots a_n)$, denoted by $cyc(n)$, so we can write
$$ n \sum_{cyc}\frac{-a_i + a_{i+1}}{a_i+a_{i+1}} = \\
\sum_{cyc (n)} (\frac{b-a}{b+a} + \frac{c-b}{c+b} + \frac{a-c}{a+c}) + \sum_{cyc (n)}(\frac{c-a}{c+a}+\frac{d-c}{d+c} + \frac{a-d}{a+d}) + \sum_{cyc (n)} (\frac{d-a}{d+a}+ \frac{e-d}{e+d} + \frac{a-e}{a+e}) +
\cdots +
\sum_{cyc (n)} (\frac{a_{n-1}-a_1}{a_{n-1}+a_1}+ \frac{a_{n}-a_{n-1}}{a_{n}+a_{n-1}} + \frac{a_1-a_n}{a_1+a_n})
$$
Since all triples of terms are in 3 variables only, we can use our Lemma. Then it suffices to show
$$
\sum_{cyc (n)} (1- \frac{3 \sqrt[3]{a\, b \, c}}{a + b+ c} )
+ \sum_{cyc (n)}(1- \frac{3 \sqrt[3]{a\, c \, d}}{a + c+ d} )
+ \sum_{cyc (n)}(1- \frac{3 \sqrt[3]{a\, d \, e}}{a + d+ e} )
+ \cdots
+ \sum_{cyc (n)}(1- \frac{3 \sqrt[3]{a_1\, a_{n-1} \, a_n}}{a_1 + a_{n-1} + a_n} )\\
\leq n (n-2) (1- \frac{n \sqrt[n]{a_1a_2...a_n}}{S} )
$$
which is
$$
\sum_{cyc (n)} (\frac{3 \sqrt[3]{a\, b \, c}}{a + b+ c} )
+ \sum_{cyc (n)}(\frac{3 \sqrt[3]{a\, c \, d}}{a + c+ d} )
+ \sum_{cyc (n)}(\frac{3 \sqrt[3]{a\, d \, e}}{a + d+ e} )
+ \cdots
+ \sum_{cyc (n)}(\frac{3 \sqrt[3]{a_1\, a_{n-1} \, a_n}}{a_1 + a_{n-1} + a_n} )\\
\geq n^2 \, (n-2) (\frac{\sqrt[n]{a_1a_2...a_n}}{S} )
$$
Using Cauchy-Schwarz leaves us with showing
$$
\frac {(\sum_{cyc (n)} \sqrt[6]{a\, b \, c})^2}{\sum_{cyc (n)}(a + b+ c)}
+
\frac {(\sum_{cyc (n)} \sqrt[6]{a\, c \, d})^2}{\sum_{cyc (n)}(a + c+ d)}
+
\frac {(\sum_{cyc (n)} \sqrt[6]{a\, d \, e})^2}{\sum_{cyc (n)}(a + d+ e)}
+ \cdots +
\frac {(\sum_{cyc (n)} \sqrt[6]{a_1\, a_{n-1} \, a_n})^2}{\sum_{cyc (n)}(a_1 + a_{n-1} + a_n)}
\geq \frac{1}{3} n^2 \,(n-2) \frac{\sqrt[n]{a_1a_2...a_n}}{S}
$$
The denominators all equal $3S$, so this becomes
$$ {(\sum_{cyc (n)} \sqrt[6]{a\, b \, c})^2}
+{(\sum_{cyc (n)} \sqrt[6]{a\, c \, d})^2}
+{(\sum_{cyc (n)} \sqrt[6]{a\, d \, e})^2}
+ \cdots + {(\sum_{cyc (n)} \sqrt[6]{a_1\, a_{n-1} \, a_n})^2}\\
\geq n^2 \,(n-2) \sqrt[n]{a_1a_2...a_n}
$$
Using AM-GM gives for the first term
$$
(\sum_{cyc (n)} \sqrt[6]{a\, b \, c})^2 \geq ( n (\prod_{cyc (n)} \sqrt[6]{a\, b \, c} )^{1/n})^2
= n^2 (\prod_{cyc (n)} ({a\, b \, c} ) )^{1/(3 n)} = n^2 \sqrt[n]{a_1a_2...a_n}
$$
By the same procedure, the other term on the LHS are likewise greater or equal than $n^2 \sqrt[n]{a_1a_2...a_n}$. There are $n-2$ of these terms, hence the sum on the LHS after AM-GM equals the RHS, which concludes the proof.
| {
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Prove $\cos(2\theta) + \cos\left(2 \left(\frac{\pi}{3} + \theta\right)\right) +\cos\left(2 \left(\frac{2\pi}{3} + \theta\right)\right) = 0$ It's been a while since I've done trig proofs. I know that
$$\cos(2\theta) + \cos\left(2 \left(\frac{\pi}{3} + \theta\right)\right) + \cos\left(2 \left(\frac{2\pi}{3} + \theta\right)\right) = 0$$
is true. This can be easily seen by plugging in values, for example where $\theta=0$ we get
$$1 -\frac{1}{2} -\frac{1}{2} = 0$$
and the same can be seen for any $\theta$. Is there a straight forward way to prove this statement is true.
| We know that $$\cos 6\theta = 4\cos^{3}2\theta - 3\cos 2\theta$$ and hence $x = \cos 2\theta$ is a root of the cubic equation $$4x^{3} - 3x -\cos 6\theta = 0$$ and clearly the sum of the roots of this equation is $0$. One of the roots is $x= \cos 2\theta$ and other two roots are $\cos((6\theta + 2\pi)/3)$ and $\cos((6\theta + 4\pi)/3)$ and the sum of these $3$ roots is $0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Does |x| = |y| requires checking conditions while solving? I am trying to solve this equation $\lvert 2x \rvert = \lvert x - 2 + y \rvert$ (specifically, find set of all points $(x, y)$ satisfying equation).
$\lvert 2x \rvert = \lvert x - 2 + y \rvert$ is equivalent to $2x = x - 2 + y \lor 2x = -(x - 2 + y)$.
If $2x = x - 2 + y$, then I am assuming that $2x \ge 0$ and $x - 2 + y \ge 0$, but do I have to check if solutions meet those conditions?
In general, while solving any equation with absolute values you have to check conditions, but sometimes you don't have to.
For example,
$$\lvert\lvert x-1\rvert +5\rvert = 7$$
I can write it as $\lvert x - 1\rvert + 5 = 7$ or $\lvert x - 1\rvert + 5 = -7$. In this case, I don't have to check conditons.
|
It is obvious that $2x = x - 2 + y$ or $2x = -(x - 2 + y)$.
True, but is it obvious for you that $\lvert 2x \rvert = \lvert x - 2 + y \rvert$ is equivalent to $2x =\pm( x - 2 + y)$ ?
We have
$$\begin{align}|X|=|Y|&\iff |X|^2=|Y|^2\\&\iff X^2-Y^2=0\\&\iff (X-Y)(X+Y)=0\\&\iff X=\pm Y\end{align}$$
So, $\lvert 2x \rvert = \lvert x - 2 + y \rvert$ is equivalent to $2x = \pm (x - 2 + y)$.
Hence, we can have
$$\begin{align}\lvert 2x \rvert = \lvert x - 2 + y \rvert &\iff 2x = x - 2 + y\quad \text{or}\quad 2x = -(x - 2 + y)\\&\iff y=x+2\quad\text{or}\quad y=-3x+2\end{align}$$
You don't have to check if solutions meet the conditions.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving that $\lim_{x\to 4}\frac{2x-5}{(6-3x)(x-6)} = \frac{1}{4} $
Proving that $$\lim_{x\to 4}\frac{2x-5}{(6-3x)(x-6)} = \frac{1}{4} $$
I need to prove this using delta-epsilon.
$\forall \varepsilon \gt 0,\;\exists\delta \gt 0\;\forall x \in D $ (where $D$ is the domain of the function)
$$ 0 \lt |x-4|\lt \delta \Rightarrow |\frac{2x-5}{(6-3x)(x-6)} - (\frac{1}{4})| < \epsilon $$
$$ 0 \lt |x-4|\lt \delta \Rightarrow |-\frac{(3x-4)(x-4)}{12(x-6)(x-2)}| < \epsilon $$
Great so we have (x-4) in our term, that's usable since we know that |x-4|>0.
How do I go about proving the rest of this? Do I have to leave the minus in the term or can I leave it out considering it's the absolute value?
| When you have rational functions depending on $x$ where the denominator is not approaching to zero, a good idea is try to estimate the value. For instance, note that if $|x-4|<1$, then $3<x<5$ and we have
$$\left|\dfrac{(3x-4)(x-4)}{-12(x-6)(x-2)}\right|=\left|\dfrac{(3x-4)(x-4)}{12(6-x)(x-2)}\right|\leq \dfrac{1}{12(6-5)(3-2)}\left|(3x-4)(x-4)\right|\leq \dfrac{1}{12}(3(5)-4)\left|x-4\right|=\dfrac{11}{12}|x-4|$$
If we want the last expression to be less than $\epsilon$, it will suffice to take $\delta=\frac{12}{11}\epsilon$.
Thus, you can take $\delta=\min\{1,\frac{12}{11}\epsilon\}$, and the proof will follow from the previous analysis.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrating $\frac {1}{\sqrt{6x+x^2}}$ using a specified U-substitution? I'm doing a two part homework question based on finding the integral of $\frac {1}{\sqrt{6x+x^2}}$
The first part was pretty simple, just completing the square, and recognizing it as being arcsine.
The second part is a bit tougher, since it wants the integral to be found using u-substitution, specifically having $u = \sqrt{x}$.
I've tried doing some algebra to enable such substitution, but I can't really see a way to split up the radical and make it happen.
I'm starting to think the question can't be done, and is simply a typo or something in the worksheet, since I found another more obvious error a few problems earlier.
Am I simply missing something, or am I right in assuming the worksheet is wrong?
| As mentioned in the comments by Archis, the integral is $sinh^{-1} \left( \frac{u}{\sqrt{6}} \right) + C$.
\begin{align*}
\int \frac{dx}{\sqrt{x^2 + 6x}} = \int \frac{dx}{\sqrt{(x + 3)^2 - 9}}
\end{align*}
Let $\quad$ $ u = x^{\frac{1}{2}}, \\ du = \frac{1}{2}x^{-\frac{1}{2}}dx \\ dx = 2x^{\frac{1}{2}}du \rightarrow 2udu. $
$\\$
This yields
\begin{align*}
\int \frac{2udu}{\sqrt{(u^2 + 3)^2 - 9}} = \int \frac{2udu}{\sqrt{u^4 + 6u^2}} = \int \frac{2du}{\sqrt{u^2 + 6}}.
\end{align*}
Recall that
\begin{align}
\int \frac{dx}{\sqrt{x^2 + a^2}} = sinh^{-1}\left(\frac{x}{a} \right) + C.
\end{align}
Thus,
\begin{align}
\int \frac{2du}{\sqrt{u^2 + 6}} = 2sinh^{-1}\left(\frac{u}{\sqrt{6}} \right) + C &&\text{where $u = \sqrt{x}$}
\end{align}
| {
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Can anyone give a combinatorial proof of the identity ${n \choose m} + 2{n-1 \choose m}+3{n-2 \choose m}+...+(n-m+1){m \choose m}={n+2 \choose m+2}$ Can anyone give a combinatorial proof of the identity
$${n \choose m} + 2{n-1 \choose m}+3{n-2 \choose m}+\ldots+(n+1-m){m \choose m}={n+2 \choose m+2}$$
I am finding difficult as $n$ is varying instead of $m$
| One way is by induction. If $n = m$, then we have
$$
\binom{n}{m} = \binom{n+2}{m+2}
$$
which is true, as both sides equal $1$. Next, suppose that
$$
\binom{m+k}{m} + 2\binom{m+k-1}{m} + 3\binom{m+k-2}{m} + \cdots
+ (k+1)\binom{m}{m} = \binom{m+k+2}{m+2}
$$
for some $k \geq 0$. Then, for $k+1$, your left-hand side equals
\begin{align}
L & = \binom{m+k+1}{m} + 2\binom{m+k}{m} + 3\binom{m+k-1}{m} + \cdots
+ (k+2)\binom{m}{m} \\
& = \binom{m+k+1}{m} + \binom{m+k}{m} + \binom{m+k-1}{m} + \cdots
+ \binom{m}{m} + \binom{m+k+2}{m+2} \\
& = \binom{m+k+2}{m+1} + \binom{m+k+2}{m+2}
\qquad \leftarrow \text{hockey-stick identity} \\
& = \binom{m+k+3}{m+2}
\end{align}
which establishes the proposition.
| {
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Find the integral $\int \frac{2\sqrt[5]{2x-3}-1}{(2x-3)\sqrt[5]{2x-3}+\sqrt[5]{2x-3}}\mathrm dx$ $$\frac{2\sqrt[5]{2x-3}-1}{(2x-3)\sqrt[5]{2x-3}+\sqrt[5]{2x-3}}=\frac{2\sqrt[5]{2x-3}-1}{2\sqrt[5]{2x-3}(x-1)}=\frac{1}{x-1}-\frac{1}{2\sqrt[5]{2x-3}(x-1)}$$
$$\int \frac{1}{x-1}\mathrm dx=\ln|x-1|+c$$
$$\int \frac{1}{2\sqrt[5]{2x-3}(x-1)}\mathrm dx=\frac{1}{2}\int \frac{1}{\sqrt[5]{2x-3}(x-1)}\mathrm dx$$
Substitution $u=2x-3,du=2dx$ gives
$$\frac{1}{2}\int \frac{1}{\sqrt[5]{2x-3}(x-1)}\mathrm dx=\int \frac{1}{u^{6/5}+u^{1/5}}\mathrm du$$
How to evaluate $\int \frac{1}{u^{6/5}+u^{1/5}}\mathrm du$?
| As @Archis Welankar noted, this integral is familiar. If we let $u=(2x-3)^{\frac15}$, then $2x-3=u^5$, so $dx=\frac52du$. Then
$$\int\frac{2\sqrt[5]{2x-3}-1}{(2x-3)\sqrt[5]{2x-3}+\sqrt[5]{2x-3}}dx=\int\frac{2u-1}{u^6+u}\cdot\frac52u^4du=\frac52\int\frac{2u^4-u^3}{u^5+1}du$$
I calculated this integral a couple of weeks ago. Today I checked that integral by comparing with numerical solution, and it passed verification.
| {
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Find the integral $\int \frac{2x^3+3x^2}{(2x^2+x-3)\sqrt{x^2+2x-3}}\mathrm dx$ $$\int \frac{2x^3+3x^2}{(2x^2+x-3)\sqrt{x^2+2x-3}}\mathrm dx$$$$=2\int \frac{x^3}{(2x^2+x-3)\sqrt{x^2+2x-3}}\mathrm dx+3\int \frac{x^2}{(2x^2+x-3)\sqrt{x^2+2x-3}}\mathrm dx$$
For these two integrals, I have tried Euler substitutions and various types of substitutions of irrational functions, but it seems that those substitutions don't simplify the integral.
What substitution is useful for this type of integrals?
| Hint:
The numerator of the original integral can be written
$$2x^3 + 3x^2 = (2x+3)(x^2)$$
The denominator of the original integral can be written
$$\left(2x^2 + x - 3\right)\sqrt{x^2 + 2x - 3} = (2x+3)(x-1)\sqrt{(x+3)(x-1)}$$
The $(2x+3)$ terms cancel, so the integral becomes
$$\displaystyle\int \frac{x^2}{(x-1)^{3/2}(x+3)^{1/2}}\,\mathrm{d}x$$
| {
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"url": "https://math.stackexchange.com/questions/1719129",
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"question_score": "2",
"answer_count": 4,
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} |
Why is $x^3-x^2+x$ injective? The function $$ f(x) = x^3-x^2+x$$ is injective (as seen on the graph), but by doing f(a) = f(b) I can't get to the point where a = b.
Can this be indicated in an analytical way (without graph) ?
| Let $f(a) = f(b)$, then $f(a)-f(b) = 0 \iff a^3-a^2+a -b^3+b^2-b = 0 \iff a-b = a^2(1-a) -b^2(1+b)$
But this means
$2(a-b) = a^2(1-a) -b^2(1+b) + a - b = (f(a)-f(b)) = 0$, from which $a = b$ follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1719409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
partial fractions
$$\frac{1}{1-x^2}$$
$$\frac{1}{1-x^2}=\frac{a}{1-x}+\frac{b}{1+x}$$
$$1=a+ax+b-bx$$
$$1=a+b+x(a-b)$$
$a+b=1$ and $x(a-b)=0\Rightarrow a-b=0\Rightarrow a=b$
$$2a=1\Rightarrow a=\frac{1}{2}$$
$b=\frac{1}{2}$
$$\frac{1}{1-x^2}=\frac{1}{2(1-x)}+\frac{1}{2(1+x)}$$
Where I went wrong?
| But you are not wrong.
$$\frac{1}{2(1-x)}+\frac{1}{2(1+x)} = \frac{1+x+1-x}{2(1+x)(1-x)} = \frac{2}{2(1+x)(1-x)} = \frac{1}{1-x^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1723331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Solving ode form of $\frac{dy}{dx} = \frac{cx+dy}{ax+by}$. \begin{align}
\frac{dy}{dx} = \frac{cx+dy}{ax+by}
\end{align}
i want to solve this differential equation, in two ways.
First with condition of $ad-bc\neq0$ and without this condition.
I wonder this condition is really necessary.
Following mea43, I have
\begin{align}
&x \frac{dv}{dx} = \frac{c+(d-a)v - bv^2}{a+bv} \\
&\int \frac{1}{x} dx = \int \frac{a+bv}{c+(d-a)v - bv^2} dv =
-\frac{(a+d) \text{ArcTan}\left[\frac{a-d+2 b v}{\sqrt{-a^2-4 b c+2 a d-d^2}}\right]}{\sqrt{-a^2-4 b c+2 a d-d^2}}-\frac{1}{2} \text{Log}[-c+v (a-d+b v)]
\end{align}
where i borrow the results of mathematica. Is there some ways to integrate this?
| If $ad=bc$ then $$\frac{a}{c}=\frac{b}{d}=k$$(let us say that ratio is k )
Then $$\frac{dy}{dx}=\frac{cx+dy}{ax+by}=\frac{1}{k}\frac{ax+by}{ax+by}=\frac{1}{k}$$
Hence $y=\frac{x}{k}+c$
But if $ad-bc\ne 0$
Let $y=vx$ then $\frac{dy}{dx}=v+x\frac{dv}{dx}$.
Hence $$ \frac{dy}{dx}=\frac{cx+dy}{ax+by}$$
$$\Rightarrow v+x\frac{dv}{dx}=\frac{c+dv}{a+bv}$$ we get an entirely different solution
| {
"language": "en",
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"source": "stackexchange",
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Surprising Summation (4): $\frac 12 \sum_{i=1}^n (n+1-i)(n+i)=\sum_{i=1}^n i^2$
Show that
$$\frac 12 \sum_{i=1}^n (n+1-i)(n+i)=\sum_{i=1}^n i^2$$
without expanding the summation to its closed form, i.e. $\dfrac 16n(n+1)(2n+1)$ or equivalent.
e.g.for $n=5$,
$$\frac12\bigg[5(6)+4(7)+3(8)+2(9)+1(10)\bigg]=1^2+2^2+3^2+4^2+5^2 $$
(Note: Whilst looking through similar questions posted earlier, another form of the summation cropped up and I thought I would post it as a question.)
| This answer should not be taken too seriously, since usually we would calculate the series much more efficiently. But since OPs is asking for alternatives ... :-)
Here we show the validity of the identity by deriving a generating function valid for LHS and RHS.
The following is valid
\begin{align*}
\sum_{n=0}^{\infty}\left(\sum_{r=1}^nr^2\right)x^n
&=\sum_{n=0}^\infty\left(\frac{1}{2}\sum_{r=1}^{n}(n+1-r)(n+r)\right)x^n\\
&=\frac{x(1+x)}{(1-x)^4}
\end{align*}
We start with the simpler series:
We obtain
\begin{align*}
\sum_{n=0}^{\infty}\sum_{r=1}^nr^2x^n&=\frac{1}{1-x}\sum_{n=0}^{\infty}n^2x^n\tag{1}\\
&=\frac{1}{1-x}\left(xD_x\right)^2\sum_{n=0}^\infty x^n\tag{2}\\
&=\frac{1}{1-x}\left(xD_x\right)^2\frac{1}{1-x}\tag{3}\\
&=\frac{x(1+x)}{(1-x)^4}
\end{align*}
Comment:
*
*In (1) we use $\frac{1}{1-x}$ the relationship
\begin{align*}
\frac{1}{1-x}\sum_{n=0}^\infty a_n x^n=\sum_{n=0}^\infty\sum_{k=0}^na_k x^n
\end{align*}
*In (2) we use the differential Operator $D_x$ and the relationship
\begin{align*}
\sum_{n=0}^{\infty}na_nx^n=\left(xD_x\right)\sum_{n=0}^{\infty}a_nx^n
\end{align*}
*In (3) we use the geometric series expansion
*In (4) we apply the $(xD_x)$ operator successively. We use here in the following
\begin{align*}
\left(xD_x\right)\frac{1}{1-x}&=\frac{x}{(1-x)^2}\\
\left(xD_x\right)^2\frac{1}{1-x}&=\frac{x(1+x)}{(1-x)^3}\\
\left(xD_x\right)^3\frac{1}{1-x}&=\frac{x(1+4x+x^2)}{(1-x)^4}\\
\end{align*}
Now the generating series for the other series:
We obtain
\begin{align*}
\sum_{n=0}^{\infty}&\frac{1}{2}\sum_{r=1}^n(n+1-r)(n+r)x^n\\
&=\frac{1}{2}\sum_{n=0}^{\infty}\sum_{r=1}^nn(n+1)x^n+\frac{1}{2}\sum_{n=0}^{\infty}\sum_{r=1}^nr(1-r)x^n\\
&=\frac{1}{2}\sum_{n=0}^{\infty}(n^3+n^2)x^n+\frac{1}{2}\cdot\frac{1}{1-x}\sum_{n=0}^{\infty}n(n-1)x^n\\
&=\frac{1}{2}\left(\left(xD_x\right)^3\frac{1}{1-x}+\left(xD_x\right)^2\frac{1}{1-x}\right)\\
&\qquad\qquad+\frac{1}{2}\left(\left(xD_x\right)\frac{1}{1-x}-\left(xD_x\right)^2\frac{1}{1-x}\right)\tag{5}\\
&=\frac{1}{2}\left(\frac{x(1+4x+x^2)}{(1-x)^4}+\frac{x(1+x)}{(1-x)^3}+\frac{x}{(1-x)^3}-\frac{x(1+x)}{(1-x)^4}\right)\\
&=\frac{1}{2}\left(\frac{3x^2+x^3}{(1-x)^4}+\frac{x^2+2x}{(1-x)^3}\right)\tag{6}\\
&=\frac{1}{2}\cdot\frac{2x^2+2x}{(1-x)^4}\\
&=\frac{x(1+x)}{(1-x)^4}
\end{align*}
and the claim follows.
Comment:
*
*In (5) we use the identities stated in the comment section above in (4).
*In (6) we collect according to the same denominator.
| {
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"url": "https://math.stackexchange.com/questions/1726195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Correct way to solve a limit. I have the following limit,
$$\lim_{n\rightarrow \infty}\left \{ \frac{\left ( n+1 \right )\left ( n+2 \right )...3n}{n^{2n}} \right \}^{\frac{1}{n}}$$
My procedure of solving (which is wrong).
Step 1: I break up the expression in the following manner,
$$\lim_{n\rightarrow \infty }\left \{ \left ( 1+\frac{1}{n} \right )\left ( 1+\frac{2}{n} \right )...\left ( \frac{3n}{n} \right ) \right \}^{\frac{1}{n}}$$
Step 2: I apply the limits,
$$\left \{ \left ( 1+\frac{1}{\infty } \right ) \left ( 1+\frac{2}{\infty } \right )...\left ( 2+\frac{1}{\infty } \right )...3\right \}^{\frac{1}{\infty }}$$
which makes it, $$\left \{ 1^{\infty } *2^{\infty }*3\right \}^{0}$$
I am stuck here and could not proceed. Can somebody please help me out?
| A simpler approach is to use the following theorem on sequences:
Theorem: If $a_{n}$ is a sequence of positive terms such that $\lim\limits_{n \to \infty}\dfrac{a_{n + 1}}{a_{n}} = L$ then $\lim\limits_{n \to \infty}\sqrt[n]{a_{n}} = L$.
Let $$a_{n} = \frac{(n + 1)(n + 2)\cdots(n + 2n)}{n^{2n}}$$ and the sequence in question is $b_{n} = \sqrt[n]{a_{n}}$. We have
\begin{align}
\frac{a_{n + 1}}{a_{n}} &= \frac{(n + 2)(n + 3)\cdots(3n + 3)}{(n + 1)^{2n + 2}}\cdot\frac{n^{2n}}{(n + 1)(n + 2)\cdots(n + 2n)}\notag\\
&= \frac{3(3n + 1)(3n + 2)}{(n + 1)^{2}}\left(\frac{n}{n + 1}\right)^{2n}\notag\\
&= 27\cdot\dfrac{\left(1 + \dfrac{1}{3n}\right)\left(1 + \dfrac{2}{3n}\right)}{\left(1 + \dfrac{1}{n}\right)^{2}}\cdot\left\{\left(1 + \dfrac{1}{n}\right)^{n}\right\}^{-2}\notag\\
&\to \frac{27}{e^{2}}\text{ as }n \to \infty\notag
\end{align}
and therefore the sequence in question $b_{n} = a_{n}^{1/n} \to 27/e^{2}$ as $n \to \infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1728025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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General formula for the numerators? Suppose that $a$ is a natural number. The numerator of $\dfrac {1}{a}$ is $1$. The numerator of $\dfrac {1}{a} + \dfrac {1}{a+1}$ is $2a+1$ [Note: Here for our purpose we don't cancel common factors of the numerator and denominator]. The numerator of $\dfrac {1}{a} + \dfrac {1}{a+1} + \dfrac {1}{a+2}$ is $3a^2+6a+2$. And so on. I tried a lot to come up with a general formula for the numerator of the sum of $n$ consecutive terms but I couldn't. I think there should exist some formula for it same as we have a formula for the coefficients of a binomial expansion. Any idea?
Saying in a more mathematical language:
The numerator of $\dfrac {1}{a} + \dfrac {1}{a+1} + \dots + \dfrac {1}{a+n}$ is $b_1+b_2a+ \dots + b_na^n$. What are the $b_i$'s?
| The expression is:
$\frac{1}{a} + \frac{1}{a+1} + \frac{1}{a+2} ... $
The numerator is going to be sum of the product of all the denominators with one term missing.
$(a+1) * (a+2) *.... (a+n)$ [a missing]
$+ a*(a+2)... (a+n)... $ [a+1 missing]
.
.
$+ a*(a+1)... (a+n-1)$ [a+n missing]
Each term can be expressed in terms of factorials.
$= \sum_{i = a}^{a+n} \frac{(a+n)!}{(a-1)!i}$
$ = \frac{(a+n)!}{(a-1)!} \sum_{i = a}^{a+n} \frac{1}{i}$
I don't think there is a closed form for the numerator though.
EDIT: Thanks to Rick Decker's useful comment, I remembered the forgotten Harmonic numbers, so the sum of 1/i becomes H(a+n) - H(a - 1)
Where H(n) is the $n^{th}$ Harmonic number = $\sum_{i=1}^{n} \frac{1}{i}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1732737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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If $11^m\cdot 5^n-3^p\cdot 2^q=1$ where $m,n,p,q$ are non-negative integers,Find all $m,n,p,q$ If $11^m\cdot 5^n-3^p\cdot 2^q=1$ where $m,n,p,q$ are non-negative integers, Find all $m,n,p,q$.
It seems $(1,1,3,1),(0,1,0,2),(0,2,1,3)$ are the only solutions.
Now, the idea is to plug $m=1+x,n=1+y,p=3+z,q=1+r$ with $x,y,z,r \ge 1$
It gives me $55(11^x\cdot 5^y)-54(3^z\cdot 2^r)=1$..How to proceed from here?
| A PARTIAL ANSWER.-Seeking positive solutions $mnpq\ne 0$ (or its impossibility) we prove that if $n>1$ then there are no solutions. Besides we give restrictions on the exponents $m,p,q$
►Consider the equation $$11x-3y=1\qquad(*)$$
A solution of $(*)$ is $(x,y)=(5,18)$ so the general solutions is given by
$$(x,y)=(3t+5,11t+18)$$ It follows the equations
$$\begin{cases}3t+5=11^{m-1}5^n\\11t+18=3^{p-1}2^q\end{cases}\qquad(**)$$
It is seen in $(**)$ that $t$ must be multiple of $5$ (first equation)and of $2$ (second equation) hence $t=10s$ so one has
$$30s+5=11^{m-1}5^n$$
If $n>1$ then $1\equiv 5\pmod {10}$ thus since we are looking for $n$ positive, $\color{red}{n=1}$.
Furthermore, if $p>1$ we get $t=30s$ so one has
$$\begin{cases}900s+5=11^{m-1}5^n\\330s+18=3^{p-1}2^q\end{cases}\qquad(***)$$ and the second equation in $(***)$ gives $8\equiv3^{p-1}2^q\pmod {10}$ so
$$(p-1,q)\in \{(1,4),(2,1),(3,2),(4,3)\}\text{modulo}\space 4$$
The proposed equation becomes in the following
four possibilities to study
$$\begin{cases}5\cdot11^m-3^{2+4r}2^{4s}=1\\5\cdot11^m-3^{3+4r}2^{1+4s}=1\\5\cdot11^m-3^{4r}2^{2+4s}=1\\5\cdot11^m-3^{1+4r}2^{3+4s}=1\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1733117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Least Squares Solution for a Matrix System with Diagonal Matrix Constraint
Let's say I have the following equation
$$AX=B$$
where $A$ is a $8\times 3$ matrix (known), $X$ is a $3\times3$ "diagonal" matrix which represents the coefficients (unknown) and $B$ is a $8\times 3$ matrix (known).
Whenever I solve for $X$ using least squares $X= (A^TA)^{-1} A^T B$, I get a square matrix which is not diagonal.
Any idea how can I force least square solution matrix to be diagonal?
| Decouple your equations to force diagonalization. A simpler problem demonstrates the basics.
Instead of solving
$$
\left(
\begin{array}{cc}
a_{11} & a_{12} \\
a_{21} & a_{22} \\
a_{31} & a_{32}
\end{array}
\right)
\left(
\begin{array}{cc}
x_{11} & x_{12} \\
x_{21} & x_{22}
\end{array}
\right)
%
=
\left(
\begin{array}{cc}
b_{11} & b_{12} \\
b_{21} & b_{22} \\
b_{31} & b_{32}
\end{array}
\right)
$$
solve instead
$$
\left(
\begin{array}{cc}
a_{11} & a_{12} \\
a_{21} & a_{22} \\
a_{31} & a_{32}
\end{array}
\right)
\left(
\begin{array}{cc}
x_{1} & 0 \\
0 & x_{2} \\
\end{array}
\right)
%
=
\left(
\begin{array}{cc}
b_{11} & b_{12} \\
b_{21} & b_{22} \\
b_{31} & b_{32}
\end{array}
\right).
$$
This is two separate problems for $k=1,2$:
$$
\left(
\begin{array}{c}
a_{1k} \\
a_{2k} \\
a_{3k}
\end{array}
\right)
x_{k}
=
\left(
\begin{array}{c}
b_{1k} \\
b_{2k} \\
b_{3k}
\end{array}
\right)
$$
The pseudoinverse of a vector $v$ is $\frac{v^{\mathrm{*}}}{\lVert v \rVert}$. For example
$$
\left(
\begin{array}{c}
a_{1k} \\
a_{2k} \\
a_{3k}
\end{array}
\right)^{\dagger}
=
\frac{
\left(
\begin{array}{ccc}
a_{1k} & a_{2k} & a_{3k}
\end{array}
\right)
}{\lVert a_{k} \rVert}.
$$
The least squares point solution is
$$
x_{k} = a_{k}^{\dagger} b_{k} = \frac{ \sum_{j=1}^{3}a_{jk}b_{jk} } {\sqrt{ \sum_{j=1}^{3}a_{jk}a_{jk} }}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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How do I show that $\frac a{1 - a^2} + \frac b{1 - b^2} + \frac c{1 - c^2} \ge \frac {3 \sqrt 3}2$ For $0 \lt a, b, c \lt 1$, if $ab + bc + ca = 1$, show that $$\frac a{1 - a^2} + \frac b{1 - b^2} + \frac c{1 - c^2} \ge \frac {3 \sqrt 3}2.$$
I want to use trigonometric substitution:
For the angles $A, B, C$ of any acute triangle, $$\tan A + \tan B + \tan C = \tan A \tan B \tan C,$$ $$\frac 1{\tan A \tan B} + \frac 1{\tan B \tan C} + \frac 1{\tan C \tan A} = 1.$$
Also, $\tan A, \tan B, \tan C \gt 0$. So I substitute $a, b, c$ for $\frac 1{\tan A}, \frac 1{\tan B}, \frac 1{\tan C}$ respectively. Then the inequality in question becomes $$\frac {\tan A}{1 - \tan^2 A} + \frac {\tan B}{1 - \tan^2 B} + \frac {\tan C}{1 - \tan^2 C} \le -\frac {3 \sqrt 3}2.$$
Here $A, B, C \not = \frac {\pi}4$ since $a, b, c \not = 1$.
By the trigonometric identity $\tan 2A = \frac {2 \tan A}{1 - \tan^2 A}$, we have
$$\tan 2A + \tan 2B + \tan 2C \le -3 \sqrt 3,$$
where $0 \lt A, B, C \lt \frac {\pi}2$, $A, B, C \not = \frac {\pi}4$, and $A + B + C = \pi$.
How do I proceed?
Edit: The restriction $a, b, c \lt 1$ was added after the question had received some answers, thanks to Michael Rozenberg, who pointed out this mistake.
| hint: Lets use $a = \tan\left(\frac{A}{2}\right)$. You can define $b, c$ similarly, then your inequality becomes: $\tan(A) + \tan(B) + \tan(C) \geq 3\sqrt{3}$, with $0 < A,B,C < \dfrac{\pi}{2}$ and $A+B+C = \pi$. And this inequality is standard result of convexity of $\tan(x)$ over $\left(0,\frac{\pi}{2}\right)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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What is the indefinite integral of $\sqrt[4] {\tan \left( x\right) }$ $$\int\sqrt[4] {\tan \left( x\right) } dx$$
I'm really stuck right now with this integral, so any kind of advice would be appreciated.
Perhaps a pretty nifty substitution that can save me from partial fraction decomposition.
| I usually factor a cyclotomic denominator like this into linear factors,
expand into partial fractions, integrate to logarithms, and then convert
those complex logarithms into logarithms and arctangents, but this time I
thought I would try combining complex conjugate partial fractions before
integrating and see how it went. First the substitution $z^4=\tan x$:
$$\int\sqrt[4]{\tan x}dx=4\int\frac{z^4}{z^8+1}dz$$
And now
$$\frac{4z^4}{z^8+1}=\frac{4z^4}{\prod_{k=0}^7(z-\omega_k)}=\sum_{k=0}^7\frac{A_k}{z-\omega_k}$$
Where $\omega_k=e^{i\theta_k}=\cos\theta_k+i\sin\theta_k$, and $\theta_k=\frac{(2k+1)\pi}8$
And then as @Count Iblis remarked, it's easy to get the numerators by L'Hopital's rule
$$\lim_{z\rightarrow\omega_k}\frac{4z^4}{z^8+1}=\lim_{z\rightarrow\omega_k}\frac{4z^4}{8z^7}=\frac1{2\omega_k^3}=-\frac12\omega_k^5$$
Combining complex conjugate fractions,
$$\begin{align}\frac{4z^4}{z^8+1} & =\sum_{k=0}^7\frac{-\frac12\omega_k^5}{z-\omega_k}=\sum_{k=0}^3\frac{-\frac12\omega_k^5(z-\omega_{7-k})-\frac12\omega_{7-k}^5(z-\omega_k)}{(z-\omega_k)(z-\omega_{7-k})}\\
& =\sum_{k=0}^3\frac{-z\cos5\theta_k}{z^2-2z\cos\theta_k+1}=\sum_{k=0}^3\frac{(z-\cos\theta_k)\cos3\theta_k+\cos\theta_k\cos3\theta_k}{z^2-2z\cos\theta_k+1}\end{align}$$
So
$$\begin{align}\int\sqrt[4]{\tan x}dx & =\sum_{k=0}^3\int\frac{(z-\cos\theta_k)\cos3\theta_k+\cos\theta_k\cos3\theta_k}{z^2-2z\cos\theta_k+1}d\theta\\
& =\sum_{k=0}^3\left[\frac12\cos3\theta_k\ln\left(z^2-2z\cos\theta_k+1\right)+\frac{\cos3\theta_k\cos\theta_k}{\sin\theta_k}\tan^{-1}\left(\frac{z-\cos\theta_k}{\sin\theta_k}\right)\right]+C\\
& =\sum_{k=0}^3\left[\frac12\cos3\theta_k\ln\left(\sqrt{\tan x}-2\sqrt[4]{\tan x}\cos\theta_k+1\right)+\frac{\cos3\theta_k\cos\theta_k}{\sin\theta_k}\tan^{-1}\left(\frac{\sqrt[4]{\tan x}-\cos\theta_k}{\sin\theta_k}\right)\right]+C\end{align}$$
Evaluating the trig functions,
$$\begin{align}\int\sqrt[4]{\tan x}dx & =\frac14\sqrt{2-\sqrt2}\ln\left(\sqrt{\tan x}-\sqrt[4]{\tan x}\sqrt{2+\sqrt2}+1\right)+\\
& \frac12\sqrt{2+\sqrt2}\tan^{-1}\left(\sqrt[4]{\tan x}\sqrt{4+2\sqrt2}-\frac{2+\sqrt2}2\right)-\\
& \frac14\sqrt{2+\sqrt2}\ln\left(\sqrt{\tan x}-\sqrt[4]{\tan x}\sqrt{2-\sqrt2}+1\right)-\\
& \frac12\sqrt{2-\sqrt2}\tan^{-1}\left(\sqrt[4]{\tan x}\sqrt{4-2\sqrt2}-\frac{2-\sqrt2}2\right)+\\
& \frac14\sqrt{2+\sqrt2}\ln\left(\sqrt{\tan x}+\sqrt[4]{\tan x}\sqrt{2-\sqrt2}+1\right)-\\
& \frac12\sqrt{2-\sqrt2}\tan^{-1}\left(\sqrt[4]{\tan x}\sqrt{4-2\sqrt2}+\frac{2-\sqrt2}2\right)-\\
& \frac14\sqrt{2-\sqrt2}\ln\left(\sqrt{\tan x}+\sqrt[4]{\tan x}\sqrt{2+\sqrt2}+1\right)+\\
& \frac12\sqrt{2+\sqrt2}\tan^{-1}\left(\sqrt[4]{\tan x}\sqrt{4+2\sqrt2}+\frac{2+\sqrt2}2\right)+C\end{align}$$
All in all, combining complex conjugates before integrating seemed to work well in this case.
It took much more time to typeset the solution than to solve the integral by hand.
It agrees with the Mathematica solution already posted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1736816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
None of $3,5,7$ can divide $r^4+1$ Let $n=r^4+1$ for some $r$. Show that none of $3,5,$ and $7$ can divide $n$.
I am thinking to use a corollary that "each prime divisor p of an integer of the form $(2m)^4+1$ has the form $8k+1$", but I failed. Anyone can give some hint?
| I'm going to try
to make
the most elementary possible proof.
$\bmod 3$,
$(0, \pm 1)^4
\equiv (0, 1)
$
so
$(0, \pm 1)^4+1
\equiv (1, 2)
$
so
$3 \not \mid r^4+1$.
$\bmod 5$,
$(0, \pm 1, \pm 2)^4
\equiv (0, 1, 1)
$
so
$(0, \pm 1, \pm 2)^4+1
\equiv (1, 2, 2)
$
so
$5 \not \mid r^4+1$.
$\bmod 7$,
$(0, \pm 1, \pm 2, \pm 3)^4
\equiv (0, 1, 2, 4)
$
so
$(0, \pm 1, \pm 2, \pm 3)^4+1
\equiv (1, 2, 3, 5)
$
so
$7 \not \mid r^4+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1737021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 4
} |
Use Laplace Transformations to solve $y''+2y'+5y=3e^{-x}sin(x)$, with $y(0)=0$, $y'(0)=3$ I've gotten this far and I cannot proceed:
$L[y]=\frac{L[3e^{-x}sin(x)]+3}{p^2+2p+5}= \frac{3}{((p+1)^2+1)(p^2+2p+5)}+\frac{3}{p^2+2p+5}$
I'm finding it impossible to find the inverse to solve for $y$. I would guess partial fraction decomposition is next, but I cannot do it for these quantities.
Help?
| It's easier than it looks. Just write out
$$\frac3{(p^p+2p+5)(p^2+2p+2)}=\frac{Ap+B}{p^2+2p+5}+\frac{Cp+D}{p^2+2p+2}$$
Multiply out that denominator
$$3=(Ap+B)(p^2+2p+2)+(Cp+D)(p^2+2p+5))$$
Then compare coefficients of like powers of $p$
$$\begin{align}0 & =A+C&(1)\\
0 &=2A+B+2C+D&(2)\\
0 &=2A+2B+5C+2D&(3)\\
3 &=2B+5D&(4)\end{align}$$
Now subtract twice eq $(1)$ from eqs $(2)$ and $(5)$
$$\begin{align}0 &=B+D&(5)\\
0&=2B+3C+2D&(6)\end{align}$$
Subtract twice eq $(5)$ from eqs $(6)$ and $(4)$
$$\begin{align}0&=3C&(7)\\
3&=3D&(8)\end{align}$$
And back into eq $(1)$ we have
$$\begin{align}0&=A&(9)\end{align}$$
At this point from eq (8), we can see that $D=1$ and then from eq $(5)$, $B=-1$. So
$$\frac3{(p^p+2p+5)(p^2+2p+2)}=\frac{-1}{p^2+2p+5}+\frac{1}{p^2+2p+2}$$
So I hope you can continue from this point.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is the infinite sum $$S = {\frac{2}{5} \\+ \frac{2}{5}\cdot\frac{3}{7} \\+ \frac{2}{5}\cdot\frac{3}{7}\cdot\frac{5}{11} \\+ \frac{2}{5}\cdot\frac{3}{7}\cdot\frac{5}{11}\cdot\frac{9}{13} \\+ \frac{2}{5}\cdot\frac{3}{7}\cdot\frac{5}{11}\cdot\frac{9}{13}\cdot\frac{11}{17} \\+ \frac{2}{5}\cdot\frac{3}{7}\cdot\frac{5}{11}\cdot\frac{9}{13}\cdot\frac{11}{17}\cdot\frac{15}{19} \\+ \frac{2}{5}\cdot\frac{3}{7}\cdot\frac{5}{11}\cdot\frac{9}{13}\cdot\frac{11}{17}\cdot\frac{15}{19}\cdot\frac{17}{23} \\+ \frac{2}{5}\cdot\frac{3}{7}\cdot\frac{5}{11}\cdot\frac{9}{13}\cdot\frac{11}{17}\cdot\frac{15}{19}\cdot\frac{17}{23}\cdot\frac{21}{25} \\+ ...}$$ ?
I tried to figure out the general form of the terms, but this is unfamiliar to me. How should I proceed?
| This is more a hint than a full answer, but here is the full series you are looking for with general terms
$$S = \frac{2}{5} + \sum\limits_{n=2}^\infty \left(\frac25 \prod\limits_{k=1}^{n-1} \frac{6k-(-1)^k-1}{6k+(-1)^k+9}\right).$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
How do I show that $\frac {a^2}b + \frac {b^2}c + \frac {c^2}d + \frac {d^2}a \ge 4$ for $a^2 + b^2 + c^2 + d^2 = 4$? Let $a, b, c, d$ be positive real numbers such that $a^2 + b^2 + c^2 + d^2 = 4$, show that $$\frac {a^2}b + \frac {b^2}c + \frac {c^2}d + \frac {d^2}a \ge 4.$$
My try:
$$\frac {a^2}b + \frac {b^2}c + \frac {c^2}d + \frac {d^2}a \ge a + b + c + d,$$
yet $$a + b + c + d \le \sqrt{4(a^2 + b^2 + c^2 + d^2)} = 4.$$
Thus, direct application of Cauchy-Schwarz inequality is too weak. I tried other methods but with no significant progress:
$$(\frac {a^2}b + \frac {b^2}c + \frac {c^2}d + \frac {d^2}a)^2 \ge \frac {(a^{4/3} + b^{4/3} + c^{4/3})^3}{a^2 + b^2 + c^2 + d^2} = \frac {(a^{4/3} + b^{4/3} + c^{4/3})^3}4.$$
I also observed that
$$(\frac {a^2}b + \frac {b^2}c + \frac {c^2}d + \frac {d^2}a) + (\frac {a^2}c + \frac {b^2}d + \frac {c^2}a + \frac {d^2}b) + (\frac {a^2}d + \frac {b^2}a + \frac {c^2}b + \frac {d^2}c) + (\frac {a^2}a + \frac {b^2}b + \frac {c^2}c + \frac {d^2}d) = 4 (\frac 1a + \frac 1b + \frac 1c + \frac 1d) \ge 16,$$
since $$\frac 1a + \frac 1b + \frac 1c + \frac 1d \ge \sqrt{\frac {(1 + 1 + 1 + 1)^3}{a^2 + b^2 + c^2 + d^2}} = 4.$$
Now my work might seem stupid or off-topic here, but I provide it here because I wish any of these attempts will lead to a solution. Any hints will be appreciated.
| (This is actually from the deleted answer to a different question,
posted here with permission.)
From Cauchy-Schwarz:
$$
\left(\frac {a^2}b + \frac {b^2}c + \frac {c^2}d + \frac {d^2}a \right)
\left( a^2 b + b^2 c + c^2 d + c^2 a \right) \ge (a^2+b^2+c^2+d^2)^2 =16
$$
therefore it suffices to show that
$$ \tag{*}
\left( a^2 b + b^2 c + c^2 d + c^2 a \right) \le 4
$$
Using Cauchy-Schwarz again:
$$
\left( a^2 b + b^2 c + c^2 d + c^2 a \right)^2 \le
(a^ 2+ b^2 + c^2 + d^2)(a^2b^2 + b^2c^2 + c^ 2d^2 + d^2 a^2) \\
= 4 (a^2+c^2)(b^2+d^2) \\
\le 4 \left( \frac {a^2+b^2+c^2+d^2}{2} \right) ^2 = 16
$$
with AM-GM in the last step.
From this $(*)$ follows.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 0
} |
Simplifying radicals inside radicals: $\sqrt{24+8\sqrt{5}}$ Simplify: $\sqrt{24+8\sqrt{5}}$
I removed the common factor 4 out of the square root to obtain $2\sqrt{6+2\sqrt{5}}$, but the answer key says it is $2+2\sqrt{5}$.
Am I missing out on some general rule here?
| We can generalize this into a formula, but you can ask me later if you want. For now, we denest $\sqrt{24+8\sqrt{5}}$.
To denest, you have to assume that the radical can be rewritten as the sum of two other radicals (surds). So we have $$\sqrt{24+8\sqrt{5}}=\sqrt{x}+\sqrt{y}$$
Squaring both sides gives us $$24+8\sqrt{5}=x+y+2\sqrt{xy}$$
So we have $x+y=24$ and $2\sqrt{xy}=8\sqrt{5}$. So $x\cdot y=80$. This can be easily solved by finding two numbers whose sum is $24$ and their product is $80$.
When $x=20$ and $y=4$, the conditions are met. So $\sqrt{24+8\sqrt{5}}=\sqrt{20}+\sqrt{4}=2+2\sqrt{5}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1746207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
} |
Exponential generating function - what happens when there is a leftover term? Find the coefficient on $x^2/2!$ in the following generating function:
$$xe^{3x}-x^2$$
I got this far:
$$x\sum_{n=0}^{\infty}\frac{3^nx^n}{n!}-x^2=\sum_{n=0}^{\infty}\frac{3^nx^{n+1}}{n!}-x^2$$
So, there are two terms here. To find the coefficient of $x^2/2!$, I can see that for the first sigma term, $n = 1$. So we have $\frac{3^1x^2}{1!}=\frac{6x^2}{2!}$. Thus the coefficient for the first term is 6. (is that right?) But what about the coefficient for the term $-x^2$? Is it just $-1$? I guess that $-x^2$ can be rewritten as $-\frac{2x^2}{2!}$, so maybe the coefficient should be $-2$? I would appreciate any clarification on this. Thanks.
| It’s probably simplest first to find the coefficient of $x^2$. The $x^2$ term in the summation
$$\sum_{n\ge 0}\frac{3^nx^{n+1}}{n!}$$
is the $n=1$ term, $\dfrac{3^1x^{1+1}}{1!}=3x^2\;.$ The $x^2$ term in $xe^{3x}-x^2$ is therefore $3x^2-x^2=2x^2$.
To finish the problem, you need to express $2x^2$ as a multiple of $\dfrac{x^2}{2!}$:
$$2x^2=4\cdot\frac{x^2}{2!}\;,$$
so the desired coefficient is $4$. Your approach will also work: $x^2=\dfrac{2x^2}{2!}$, so the $x^2$ term is
$$\frac{6x^2}{2!}-\frac{2x^2}{2!}=4\cdot\frac{x^2}{2!}\;.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1748941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Parallel surface For a regular surface $\mathbf{x} = \mathbf{x}(u,v)$
Define $\mathbf{y}(u,v) = \mathbf{x}(u,v) + t \mathbf{N} (u,v)$ where $\mathbf{N}$ is the unit normal of $\mathbf{x}$
How could I show the following?
$\mathbf{y}_u \times \mathbf{y}_v = (1-2Ht+Kt^2)\mathbf{x}_u \times \mathbf{x}_v$
where $H$ is the mean curvature of $\mathbf{x}$ and $K$ is its Gaussian curvature
|
Fundamental Forms
$$
\begin{pmatrix} E & F \\ F & G \end{pmatrix}=
\begin{pmatrix} \mathbf{x}_{u} \\ \mathbf{x}_{v} \end{pmatrix}
\begin{pmatrix} \mathbf{x}_{u} & \mathbf{x}_{v} \end{pmatrix} $$
$$
\begin{pmatrix} e & f \\ f & g \end{pmatrix}=
-\begin{pmatrix} \mathbf{N}_{u} \\ \mathbf{N}_{v} \end{pmatrix}
\begin{pmatrix} \mathbf{x}_{u} & \mathbf{x}_{v} \end{pmatrix} $$
Let $\begin{pmatrix} \mathbf{N}_{u} \\ \mathbf{N}_{v} \end{pmatrix}=
\mathbb{A} \begin{pmatrix} \mathbf{x}_{u} \\ \mathbf{x}_{v} \end{pmatrix}$
where $\mathbb{A}=
\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}$.
Now
\begin{align*}
\begin{pmatrix} \mathbf{N}_{u} \\ \mathbf{N}_{v} \end{pmatrix}
\begin{pmatrix} \mathbf{x}_{u} & \mathbf{x}_{v} \end{pmatrix} &=
\mathbb{A} \begin{pmatrix} \mathbf{x}_{u} \\ \mathbf{x}_{v} \end{pmatrix}
\begin{pmatrix} \mathbf{x}_{u} & \mathbf{x}_{v} \end{pmatrix} \\
-\begin{pmatrix} e & f \\ f & g \end{pmatrix} &=
\mathbb{A} \begin{pmatrix} E & F \\ F & G \end{pmatrix} \\
\mathbb{A} &=
-\begin{pmatrix} e & f \\ f & g \end{pmatrix}
\begin{pmatrix} E & F \\ F & G \end{pmatrix}^{-1}
\end{align*}
The principal curvatures $k_{1}, k_{2}$ are the eigenvalues of $-\mathbb{A}$. That is
$$H=\frac{k_{1}+k_{2}}{2}=-\frac{1}{2} \operatorname{tr} \mathbb{A}$$
$$K=k_{1} k_{2}=(-1)^{2} \det \mathbb{A}$$
\begin{align*}
\mathbf{y}_{u} &= \mathbf{x}_{u}+t\mathbf{N}_{u} \\
&= \mathbf{x}_{u}+t(a_{11} \mathbf{x}_{u}+a_{12} \mathbf{x}_{v}) \\
\mathbf{y}_{v} &= \mathbf{x}_{v}+t\mathbf{N}_{v} \\
&= \mathbf{x}_{v}+t(a_{21} \mathbf{x}_{u}+a_{22} \mathbf{x}_{v}) \\
\mathbf{y}_{u} \times \mathbf{y}_{v} &=
[1+t(a_{11}+a_{22})+t^{2}(a_{11}a_{22}-a_{12}a_{21})]
\mathbf{x}_{u} \times \mathbf{x}_{v} \\
&= (1-2Ht+Kt^{2}) \mathbf{x}_{u} \times \mathbf{x}_{v}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1749658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integer solutions to $xyz = w^2(x+y+z)$ I'm looking for a way to enumerate all positive integer solutions of the equation
$xyz = w^2(x+y+z)$
where $w \le W$ and $1 \le x \le y \le z$.
Could anyone provide a hint at how to approach this?
| Generalizing individ's answer,
in
$xyz=w^2(x+y+z)
$,
if
$x+y = z$,
then
$xy(x+y)=2w^2(x+y)
$
or
$xy = 2w^2$.
If
$r(x+y) = z$
where $r$ is rational,
then
$xyr(x+y)=(1+r)w^2(x+y)
$
or
$rxy = (1+r)w^2$
or
$xy = (1+\frac1{r})w^2
$.
Therefore,
for each $w^2$,
for any $r$
such that
$w^2/r$
is an integer,
look at all the factorizations of
$w^2(1+1/r)=xy$.
In particular,
if $r = \frac1{n}$
for integer $n$,
let $xy$ go over the
factors of
$(n+1)w^2$,
choose those where
$n$ divides $x+y$,
and let
$z = (x+y)/n$.
To verify,
$xyz
=(n+1)w^2(1/n)(x+y)
=(n+1)w^2(x+y)/n
$
and
$w^2(x+y+z)
=w^2(x+y+(x+y)/n)
=w^2(x+y)(n+1)/n
$.
If $r=1$,
this gives the first solution.
If $r=2$,
we look at the
factors of $xy=3w^2$
with the same parity
(this implies that
$w$ is even),
and let
$z = (x+y)/2$.
For general $n$,
we want
$n | x+(n+1)w^2/x$.
It's getting late
and I'm tired,
so I'll stop here.
| {
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"url": "https://math.stackexchange.com/questions/1750542",
"timestamp": "2023-03-29T00:00:00",
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Catalan numbers formula derivation I'm trying to follow a proof of the Catalan numbers being equal to $\frac{1}{n+1} {2n \choose n}$ from the recurrence relation $C_n = C_0C_{n-1}+C_1C_{n-2}+...+C_{n-2}C_{1}+C_{n-1}C_0$
Now it's seen that the generating function satisfies $xf^2-f+1=0$ so $f=\frac{1-\sqrt{1-4x}}{2x}$ since the other root has a pole at 0 but I'm struggling to see how $f=\frac{1-\sqrt{1-4x}}{2x}$ can be expanded to obtain the necessary power series
I've seen Wikipedia's proof but I don't see how ${ \frac{1}{2} \choose n} = \frac{(-1)^{n+1}}{4^n(2n-1)} {2n \choose n}$ nor how this gets lost into ${2n \choose n}$ via plugging y=-4x and putting it into the expression $f=\frac{1-\sqrt{1-4x}}{2x}$
| Here as supplement to the answer of @MarcusM we show the validity of the binomial identity
\begin{align*}
\binom{\frac{1}{2}}{n}=\frac{(-1)^{n+1}}{4^n(2n-1)} \binom{2n}{n}
\end{align*}
In the following we use as definition of the binomial coefficient
\begin{align*}
\binom{\alpha}{n}=\frac{\alpha(\alpha-1)(\alpha-2)\cdots(\alpha-n+1)}{n!}
\end{align*}
We obtain
\begin{align*}
\binom{\frac{1}{2}}{n}
&=\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)\left(\frac{1}{2}-2\right)\cdots\left(\frac{1}{2}-n+1\right)}{n!}\\
&=\frac{1(-1)(-3)\cdots(3-2n)}{2^nn!}\tag{1}\\
&=\frac{(-1)^{n-1}}{2^nn!}(2n-3)!!\tag{2}\\
&=\frac{(-1)^{n-1}}{2^nn!(2n-1)}(2n-1)!!\tag{3}\\
&=\frac{(-1)^{n-1}}{2^nn!(2n-1)}\cdot\frac{(2n)!}{(2n)!!}\tag{4}\\
&=\frac{(-1)^{n-1}}{2^nn!(2n-1)}\cdot\frac{(2n)!}{2^nn!}\tag{5}\\
&=\frac{(-1)^{n+1}}{4^n(2n-1)}\cdot\frac{(2n)!}{n!n!}\tag{6}\\
&=\frac{(-1)^{n+1}}{4^n(2n-1)}\binom{2n}{n}
\end{align*}
and the claim follows.
Comment:
*
*In (1) we factor out $2^n$
*In (2) we factor out $(-1)^n$ use as convenient notation double factorial
\begin{align*}
(2n)!!&=(2n)(2n-2)(2n-4)\cdots 4\cdot 2\\
(2n-1)!!&=(2n-1)(2n-3)(2n-5)\cdots 3\cdot 1\\
\end{align*}
*In (3) we multiply the expression with $\frac{2n-1}{2n-1}$ and use $(2n-3)!!(2n-1)=(2n-1)!!$
*In (4) we use the identity
\begin{align*}
(2n)!=(2n)!!(2n-1)!!
\end{align*}
*In (5) we use the identity
\begin{align*}
(2n)!!=(2n)(2n-2)\cdots4\cdot2=2^n n!
\end{align*}
*In (6) we do a small rearrangement
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Right triangle inscribed in a square. Find the square area? I hope it's valid to ask for "a more neat solution" of a problem on this network, despite the fact that I don't have a strict definition of the word "neat".
Here is the square and the right triangle inscribed in it.
I did the following:
$$AC = Ah + hB$$
$$4\sin\theta = 4\cos\theta + 3\sin\theta$$
So
$$\tan\theta = 4$$
But
$$\sin\theta = \frac{\tan\theta}{\sqrt{1+\tan^2\theta}}$$
Therefore
$$AC = 4\sin\theta = \frac{16}{\sqrt {17}}$$
$$\text{Area} = \left(\frac{16}{\sqrt {17}}\right)^2$$
| I think your way is good, but we don't need to find $\tan\theta$.
From $4\sin\theta=4\cos\theta+3\sin\theta$, we have
$$\sin\theta=4\cos\theta$$
Squaring the both sides gives
$$\sin^2\theta=16(1-\sin^2\theta)$$
from which we can have
$$\sin^2\theta=\frac{16}{17}\quad\Rightarrow\quad \text{(area)}=16\sin^2\theta=\frac{16^2}{17}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 8,
"answer_id": 4
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order $a$ = 5, $a^3b = ba^3$. show that that $ab = ba$. Let $a, b$ be elements of a group $G$. Suppose that a has order $5$ and that $a^3b = ba^3$. I want to show that that $ab = ba$.
Here is what I think:
We know that we have $a^1, a^2, a^3, a^4, a^5 = 1$. So, $a^4 = a^{-1}$ and $a^3 = a^{-2}$ and $a^{2} = a^{-4}$. Now,
$a^3b = ba^3$ then $a^{-2}b = ba^{-2}$ then $ba^2 = a^2b$ then $ba^{-4} = a^{-4}b$ then $a^{-1}b = ba^{-1}$ then multiply on the left and right by $a$ to get $ba=ab$. Is this correct or is there a simpler approach?
| $b=a^2a^3b=a^2ba^3=a^{-1}a^3ba^3=a^{-1}ba^3a^3=a^{-1}ba$. So you have you result.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove inequality: $\frac x{2+xy+yz}+\frac y{2+yz+zx}+\frac z{2+zx+xy}\le \frac{x+y+z}{x+y+z+xyz}$
Numbers $x,y,z$ satisfy $x\in(0,1], y\in(0,1], z\in(0,1]$. Prove inequality:
$$\frac x{2+xy+yz}+\frac y{2+yz+zx}+\frac z{2+zx+xy}\le \frac{x+y+z}{x+y+z+xyz}$$
My work so far:
$\frac x{2+xy+yz}\le \frac x{x^2+xz+xy+yz}=\frac x{(x+z)(x+y)}$
Then
$$\frac x{2+xy+yz}+\frac y{2+yz+zx}+\frac z{2+zx+xy}\le$$
$$\le \frac x{(x+z)(x+y)}+\frac y{(y+z)(x+y)}+\frac z{(x+z)(z+y)}=$$
$$=\frac{2(xy+yz+zx)}{(x+y)(y+z)(z+x)}$$
| Hint:
$$(1-x)(1-y)(1-z)+(1-zx)\ge 0 \implies 2+ xy+yz \ge x+y+z+xyz$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1755258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Prove if $A$ is positive definite, then $A_{11}$,$A_{22}$ are positive definite. Here ${\mathbf{A}} = \left( {\begin{array}{*{20}{c}}
{{{\mathbf{A}}_{11}}}&{{{\mathbf{A}}_{12}}} \\
{{{\mathbf{A}}_{21}}}&{{{\mathbf{A}}_{22}}}
\end{array}} \right)$.
I can prove $\mathbf{A}_{11}$ is positive definite but need help with $\mathbf{A}_{22}$. By Sylvester’s criterion, $|\mathbf{A}_{11} |>0$ since $|\mathbf{A}_{11} |$ is a principle minor. Thus $\mathbf{A}_{11}$ is invertible and we have
${\mathbf{A}} = \left( {\begin{array}{*{20}{c}}
{{{\mathbf{A}}_{11}}}&{{{\mathbf{A}}_{12}}} \\
{{{\mathbf{A}}_{21}}}&{{{\mathbf{A}}_{22}}}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{\mathbf{I}}&{\mathbf{O}} \\
{{{\mathbf{A}}_{21}}{\mathbf{A}}_{11}^{ - 1}}&{\mathbf{I}}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{{{\mathbf{A}}_{11}}}&{\mathbf{O}} \\
{\mathbf{O}}&{\mathbf{S}}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{\mathbf{I}}&{{\mathbf{A}}_{11}^{ - 1}{{\mathbf{A}}_{12}}} \\
{\mathbf{O}}&{\mathbf{I}}
\end{array}} \right)$
where ${\mathbf{S}} = {{\mathbf{A}}_{22}} - {{\mathbf{A}}_{21}}{\mathbf{A}}_{11}^{ - 1}{{\mathbf{A}}_{12}}$ is the Schur complement. Since
${\left( {\begin{array}{*{20}{c}}
{\mathbf{I}}&{\mathbf{O}} \\
{{{\mathbf{A}}_{21}}{\mathbf{A}}_{11}^{ - 1}}&{\mathbf{I}}
\end{array}} \right)^{\text{T}}} = \left( {\begin{array}{*{20}{c}}
{\mathbf{I}}&{{{\left( {{\mathbf{A}}_{11}^{\text{T}}} \right)}^{ - 1}}{\mathbf{A}}_{21}^{\text{T}}} \\
{\mathbf{O}}&{\mathbf{I}}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{\mathbf{I}}&{{\mathbf{A}}_{11}^{ - 1}{{\mathbf{A}}_{12}}} \\
{\mathbf{O}}&{\mathbf{I}}
\end{array}} \right)$
we have $\mathbf{A}$ is congruent with $\left( {\begin{array}{*{20}{c}}
{{{\mathbf{A}}_{11}}}&{\mathbf{O}} \\
{\mathbf{O}}&{\mathbf{S}}
\end{array}} \right)$. If $$ is positive definite, then $\left( {\begin{array}{*{20}{c}}
{{{\mathbf{A}}_{11}}}&{\mathbf{O}} \\
{\mathbf{O}}&{\mathbf{S}}
\end{array}} \right)$ is positive definite, and $_{11}$,$\mathbf{S}$ are both positive definite, since a block diagonal matrix is positive definite iff each diagonal block is positive definite.
Anyone can help continue to prove $_{22}$ is also positive definite? Thank you!
| To elaborate on Stella's answer, let's say $A_{11}$ is $5\times5$. Then if you take $$x=\begin{bmatrix}a \\ b \\ c \\ d \\e \\0 \\ \cdots \\ 0 \end{bmatrix}$$ then $$x^T A x = \begin{bmatrix}a & b & c & d & e \end{bmatrix}A_{11}\begin{bmatrix}a \\ b \\ c \\ d \\e \end{bmatrix}$$ and so if $x^T A x$ is positive for all nontrivial x then the product with $A_{11}$ must be positive for all nontrivial $a,b,c,d,e$ which shows that $A_{11}$ is positive definite.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1756129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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How do I prove that $\det A_{n+2} = a \det A_{n+1} + b \det A_n$ for matrix $A$?
I have calculated:
$\det A_1=2$, $\det A_2=3$, $\det A_3=4$, so I was putting some numbers in $\det A_{n+2} = a \det A_{n+1} + b\det A_n$ like $n=1$, $n=2$ ($\det$ $n\times n$ matrix) and get that $a=2$, $b=-1$,
but how that can help me to prove that $\det A_{n+2} = a\det A_{n+1} + b\det A_n$?
And the end how do I prove that $\det A_n = n+1$?
| Use induction on $n$. For $n=1$, the result is trivial. Now,
suppose that there is an $N\in\mathbb{N}$ such that $\det A_n=n+1$ for all
$n\le N$. Then for the case $N+1$, we have
\begin{align}
\det A_{N+1}
&=2\det A_N-\det
\begin{pmatrix}
2&1&0&0&&&&&\\
1&2&1&0&&&&&\\
0&1&2&1&&&&&\\
&&&&\ddots\\
&&&&&1&2&1&0\\
&&&&&0&1&2&0\\
&&&&&0&0&1&1
\end{pmatrix}_{N\times N}\\
&=2(N+1)-\det
\begin{pmatrix}
2&1&0&0&&&\\
1&2&1&0&&&\\
0&1&2&1&&&\\
&&&&\ddots\\
&&&&&1&2&1\\
&&&&&0&1&2
\end{pmatrix}_{(N-1)\times(N-1)}\\
&\quad+\det
\begin{pmatrix}
2&1&0&0&&&\\
1&2&1&0&&&\\
0&1&2&1&&&\\
&&&&\ddots\\
&&&&&1&2&0\\
&&&&&0&1&0
\end{pmatrix}_{(N-1)\times(N-1)}\\
&=2N+2-[(N-1)+1]+0\\
&=N+2.
\end{align}
This completes the proof for the second question.
| {
"language": "en",
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"source": "stackexchange",
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} |
Solve the system of equations: $a+b+c=2$, $a^2+b^2+c^2=6$, $a^3+b^3+c^3=8$ If we have \begin{cases} a+b+c=2 \\ a^2+b^2+c^2=6 \\ a^3+b^3+c^3=8\end{cases} then what is the value of $a,b,c$?
| One possibility to determine this---which is by no means perfect, and will not work all of the time!---is to try a few small integers. The best one to focus on is the second equation:
$$
a^2 + b^2 + c^2 = 6
$$
since all of the terms must be positive. Note that if $a = \pm3$, say, then $a^2 = 9$, which means that all terms have to be from the set $\{-2, -1, 0, 1, 2\}$. They can't all be $\pm1$, since that would be to small, so at least one has to be $\pm2$.
Can you see where to go from here? You can also simplify by noting that the solution is symmetric under swapping $a \to b \to c$ (or any other permutation), so you can assume for simplicity that $a$ is the largest in absolute value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1759141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
What does $\Big(\frac{(x+1)^2}{2}\Big)^n-\Big(\frac{(x-1)^2}{2}\Big)^n$ equal to?
Determine the highest degree term of the polynomial $$\Bigg(\frac{(x+1)^2}{2}\Bigg)^n-\Bigg(\frac{(x-1)^2}{2}\Bigg)^n, \quad n\in\mathbb{N}$$
The answer suggests that the highest degree term is equal to $\dfrac{2nx^{2n-1}}{2^n} + \dfrac{2nx^{2n-1}}{2^n} = \dfrac{4n}{2^n}x^{2n-1}$ But I don't know how to get there. I think it is:
$\begin{split}\Bigg(\dfrac{(x+1)^2}{2}\Bigg)^n-\Bigg(\dfrac{(x-1)^2}{2}\Bigg)^n &= \Bigg(\dfrac{(x+1)^2}{2}-\dfrac{(x-1)^2}{2}\Bigg)\Bigg(\Big(\frac{(x+1)^2}{2}\Big)^{n-1}+\ldots +\Big(\frac{(x-1)^2}{2}\Big)^{n-1}\Bigg) \\ &=2x\Bigg(\Big(\frac{(x+1)^2}{2}\Big)^{n-1}+\ldots +\Big(\frac{(x-1)^2}{2}\Big)^{n-1}\Bigg)\end{split}$
| You can find the highest degree term in $(x+1)^{2n}-(x-1)^{2n}$, then divide it by $2^n$. Now apply the binomial theorem:
\begin{align}
(x+1)^{2n}&=x^{2n}+2nx^{2n-1}+\text{lower degree terms}\\[4px]
(x-1)^{2n}&=x^{2n}-2nx^{2n-1}+\text{lower degree terms}
\end{align}
Subtracting we get
$$
(x+1)^{2n}-(x-1)^{2n}=4nx^{2n-1}+\text{lower degree terms}
$$
So the required term is
$$
\frac{4n}{2^n}x^{2n-1}
$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
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Is the sum of reciprocals of all products from $2$ to $n-1$ always $0.5n-1$? I was looking up riddles for my math classes to work on for the end of the year and found the following riddle. http://mathriddles.williams.edu/?p=129
I followed the advice and started working with examples of small numbers and stumbled upon a pattern that I wanted to generalize.
$$\frac{1}{2}(1)=0.5$$
$$\frac{1}{3}\left(1+\frac{1}{2}\right)=0.5$$
$$\frac{1}{4}\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{2\cdot 3}\right)=0.5$$
$$\frac{1}{5}\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{2\cdot 3}+\frac{1}{2\cdot 4}+\frac{1}{3\cdot 4}+\frac{1}{2\cdot 3\cdot 4}\right)=0.5$$
$$\frac{1}{6}\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{2\cdot 3}+\frac{1}{2\cdot 4}+\frac{1}{2\cdot 5}+\frac{1}{3\cdot 4}+\frac{1}{3\cdot 5}+\frac{1}{4\cdot 5}+\frac{1}{2\cdot 3\cdot 4}+\frac{1}{2\cdot 3\cdot 5}+\frac{1}{2\cdot 4\cdot 5}+\frac{1}{3\cdot 4\cdot 5}+\frac{1}{2\cdot 3\cdot 4\cdot 5}\right)=0.5$$
If my pattern doesn't make sense, I'm taking $\frac{1}{n}$ and multiplying it by the sum of the reciprocals of all unique products for $2$ to $n-1$ and it comes out to 0.5 each time up to $n=7$ (I have not tested any higher values). Equivalently, if you multiply both sides by $n$ then subtract $1$, you see that all the reciprocals sum to $0.5n-1$.
I don't know where to start with generalizing this pattern as I have never seen explicit formulas for such sums, so I wanted to see if anyone knew if this was the case for all $n$ and how one could prove or disprove it.
| With more systematic notation, your observation is that
$$ f(n) = \sum_{A\subseteq\{2,3,\ldots,n-1\}}\;\prod_{k\in A} \frac1k = \frac12n $$
(I don't see where you get "$0.5n -1$" out of it; subtracting $1$ doesn't seem to match anything in your examples.)
This does hold for all $n\ge 2$, and we can prove it by mathematical induction on $n$:
When we go from $f(n)$ to $f(n+1)$, the difference is that we now have more $A$s to sum over -- namely, we have all the ones we have before, plus all of the subsets that contain $n$. But each of the new subsets arises as one of the old subsets with $n$ appended, so we can write it as
$$ \begin{align} f(n+1) &= \sum_{A\subseteq\{2,3,\ldots,n-1\}}\; \prod_{k\in A} \frac1k + \sum_{A\subseteq\{2,3,\ldots,n-1\}}\;\frac1n \prod_{k\in A} \frac1k \\
&= f(n) + \frac1n f(n)
\\& = \frac{n+1}{n} f(n)
\\& = \frac{n+1}{n} \frac12 n
\\& = \dfrac12 (n+1) \end{align} $$
which is what is needed for the induction step.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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} |
The sum of the 2nd and 3rd term is 12, and the sum of the 3rd term and 4th term is 60. This is the 9th question in my assignment. It's another confusing question.
Here's the question:
In a geometric progression, the sum of the 2nd and 3rd terms is 12, and the sum of the 3rd and 4th term is 60. Find the common ratio of the first term.
Here's what I did:
I stopped there.
| So you have that
$
\begin{align*}
ar + ar^2 = 12 \\
ar^2 + ar^3 = 60
\end{align*}
$
Then
\begin{align*}
a = \frac{12}{r + r^2} \\
ar^2 + ar^3 = 60
\end{align*}
$\implies$
\begin{align*}
a = \frac{12}{r + r^2} \\
\frac{12}{r + r^2}r^2 + \frac{12}{r + r^2}r^3 = 60
\end{align*}
$\implies$
\begin{align*}
\frac{12}{r + r^2}r^2 + \frac{12}{r + r^2}r^3 = 60\frac{12}{r + r^2}\frac{r + r^2}{12}
\end{align*}
$\implies$
\begin{align*}
12r^2 + 12r^3 = 60(r + r^2)
\end{align*}
$\implies$
\begin{align*}
12r^3 - 48r^2 - 60r = 0
\end{align*}
$\implies$
\begin{align*}
12r^2 - 48r - 60 = 0
\end{align*}
$
Now solve this equation and you will find 2 solutions. Then you should verify which one satisfies the first equation.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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How many $5$-digit numbers (including leading $0$'s) are there with no digit appearing exactly $2$ times? How many $5$-digit numbers (including leading $0$'s) are there with no digit appearing exactly $2$ times? The solution is supposed to be derived using Inclusion-Exclusion.
Here is my attempt at a solution:
Let $A_0$= sequences where there are two $0$'s that appear in the sequence.
...
$A_{9}$=sequences where there are two $9$'s that appear in the sequence.
I want the intersection of $A_0^{'}A_1^{'}...A_9^{'}$= $N-S_1+S_2$ because you can only have at most two digits who are used exactly two times each in a $5$ digit sequence.
$N=10^5$, $S_1=10\cdot \binom{5}{2}\cdot[9+9\cdot 8 \cdot 7]$, and $S_2=10 \cdot 9 \cdot \binom{5}{4} \cdot8$.
The $S_1$ term comes from selecting which of the ten digits to use twice, selecting which two places those two digits take, and then either having the same digit used three times for the other three places, or having different digits used for the other three digits.
The $S_2$ term comes from selecting which two digits are used twice, selecting where those four digits go, and then having eight choices for the remaining spot.
So my answer becomes $10^5 -10 \cdot \binom{5}{2} \cdot [9+9 \cdot 8 \cdot 7]+10 \cdot 9 \cdot \binom{5}{4} \cdot 8$.
Am I doing this correctly?
| The partitions of $5$ showing the admissible multiplicities of occurring digits are $(1,1,1,1,1)$, $(3,1,1)$, $(4,1)$, and $(5)$.
There are $10\cdot9\cdot 8\cdot 7\cdot 6=30\,240$ numbers with five different digits.
There are $10\cdot{9\choose 2}=360$ ways to choose three digits whereby the first chosen digit will be used three times. You can arrange these digits in ${5!\over3!}=20$ ways, so that we get $7200$ numbers with multiplicities $(3,1,1)$.
There are $10\cdot9=90$ ways to choose two digits whereby the first chosen digit will be used four times. The second chosen digit can be placed in five different slots, so that we get $450$ numbers with multiplicities $(4,1)$.
There are $10$ ways to choose a single digit to be used five times.
In all there are $37\,900$ admissible numbers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1765530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $2\int_0^\infty \frac{e^x-x-1}{x(e^{2x}-1)} \, \mathrm{d}x =\ln(\pi)-\gamma $ Let $\gamma$ be the Euler-Mascheroni constant.
I'm trying to prove that
$$2\int_0^\infty \frac{e^x-x-1}{x(e^{2x}-1)} \, \mathrm{d}x =\ln(\pi)-\gamma $$
I tried introducing a parameter to the exponent in the numerator and then differentiating under the integral sign. But doing so seems to result in an integral that doesn't converge.
| $$ \begin{align} 2\int_{0}^{\infty} \frac{e^x-x-1}{x(e^{2x}-1)} \, dx &= 2 \int_{0}^{\infty} \frac{1}{x(e^{2x}-1)} \sum_{n=2}^{\infty} \frac{x^{n}}{n!}\\ & = 2 \sum_{n=2}^{\infty}\frac{1}{n!} \int_{0}^{\infty} \frac{x^{n-1}}{e^{2x}-1} \, dx \\ &= 2 \sum_{n=2}^{\infty}\frac{1}{n!2^{n}} \int_{0}^{\infty} \frac{u^{n-1}}{e^{u}-1} \, du \\ &= 2 \sum_{n=2}^{\infty} \frac{\zeta(n)}{n 2^{n}} \tag{1} \\ &= 2 \sum_{n=2}^{\infty} \frac{1}{n 2^{n}} \sum_{k=1}^{\infty} \frac{1}{k^{n}} \\ &= 2 \sum_{k=1}^{\infty} \sum_{n=2}^{\infty} \frac{1}{n(2k)^{n}} \\ & =2 \sum_{k=1}^{\infty} \left[ \log \left(\frac{2k}{2k-1} \right) - \frac{1}{2k}\right] \\ & =2 \lim_{N \to \infty} \left(\log \left(\frac{(2N)!!}{(2N-1)!!} \right) - \frac{H_{N}}{2} \right) \\ &= 2 \lim_{N \to \infty} \left(\log \left(\frac{2^{2N}(N!)^2}{(2N)!} \right) - \frac{H_{N}}{2} \right) \tag{2} \\ &= 2 \lim_{N \to \infty} \left(\log \left(\frac{2^{2N}(2\pi N) \left(\frac{N}{e} \right)^{2N}}{\sqrt{2 \pi(2N)} \left(\frac{2N}{e} \right)^{2N}} \right) - \frac{H_{N}}{2} \right) \tag{3} \\ &= \lim_{N \to \infty} \left(\log(\pi) + \log(N) - H_{N} \right) \\ &= \log(\pi) - \gamma \end{align}$$
$(1)$ https://en.wikipedia.org/wiki/Riemann_zeta_function#Definition
$(2)$ http://mathworld.wolfram.com/DoubleFactorial.html
$(3)$ https://en.wikipedia.org/wiki/Stirling's_approximation
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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Prove that for $k>1$ $a_n$ is a perfect square I'm having problems with this exercise.
Let $k > 1$ be an integer. We define $(a_n)_{n \in \Bbb N_0}$ as:
$$a_0 = 1$$ $$a_1 = 1$$ $$a_{n+2} = (k^2-2)a_{n+1}-a_n-2(k-2)$$
Prove that $\forall n \in \Bbb N_0, a_n$ is a perfect square.
I'm not sure how to tackle this problem. I want to prove that $a_n=(a+b)^2$ in some way. I was going to attempt with induction, $P(0)$ and $P(1)$ are true, $P(2)=(k-1)^2$... but how should I pick $P(n)$ and $P(n+1)$. Or should I try other method?
I've also tried finding the roots so I could have the closed formula for the sequence but I've ended up with a rather disappointing looking expression: $\frac{(k^2-2) \pm \sqrt{k^4-4k^2-8k+16}}{2}$
Any ideas or suggestions? Thanks!
| just one of those things, given $b_0=1, b_1 = 1,$ and
$$ b_{n+2} = k \, b_{n+1} - b_n, $$ then
$$ a_n = b_n^2. $$
The key step is
$$ b_{n+2} b_n - b_{n+1}^2 = k-2 $$
Let us show how that fits in: we say $a_n = b_n^2.$ We know $b_{n+2} + b_n = k b_{n+1.}$ Square both sides, this gives
$$ b_{n+2}^2 + a_n + 2 b_{n+2}b_n = k^2 a_{n+1}, $$
$$ b_{n+2}^2 + a_n = k^2 a_{n+1} - 2 b_{n+2}b_n, $$
$$ b_{n+2}^2 + a_n = k^2 a_{n+1} - 2 b_{n+2}b_n + 2 b_{n+2}b_n - 2 b_{n+1}^2 - 2 k + 4, $$
$$ b_{n+2}^2 + a_n = k^2 a_{n+1} - 2 b_{n+1}^2 - 2 k + 4, $$
$$ b_{n+2}^2 + a_n = k^2 a_{n+1} - 2 a_{n+1} - 2 k + 4, $$
$$ b_{n+2}^2 + a_n = (k^2 -2) a_{n+1} - 2 k + 4, $$
$$ b_{n+2}^2 = a_{n+2}. $$
In turn, $ b_{n+2} b_n - b_{n+1}^2 = k-2 $ comes from a standard quadratic forms construction. We consider the quadratic form $f(x,y) = x^2 - kxy + y^2.$ Its Hessian matrix is
$$
H =
\left(
\begin{array}{rr}
2 & -k \\
-k & 2
\end{array}
\right)
$$
We define a matrix $A$ (for "automorphism") as
$$
A =
\left(
\begin{array}{rr}
k & -1 \\
1 & 0
\end{array}
\right).
$$
The word automorphism means
$$ A^T H A = H. $$
In turn, from the observation that
$$
\left(
\begin{array}{c}
b_{n+2} \\
b_{n+1}
\end{array}
\right)
\left(
\begin{array}{rr}
k & -1 \\
1 & 0
\end{array}
\right) =
\left(
\begin{array}{c}
b_{n+1} \\
b_n
\end{array}
\right)
$$
we find
$$ b_{n+2}^2 - k b_{n+2} b_{n+1} + b_{n+1}^2 = b_{n+1}^2 - k b_{n+1} b_{n} + b_{n}^2, $$ so that
$$ b_{n+1}^2 - k b_{n+1} b_{n} + b_{n}^2 $$ is independent of $n$ and constant.
From $b_1$ and $b_0$ we find
$$ b_{n+1}^2 - k b_{n+1} b_{n} + b_{n}^2 = 2-k. $$
Finally
$$b_{n+2} b_n - b_{n+1}^2 = (k b_{n+1} - b_n) b_n - b_{n+1}^2 = - b_{n+1}^2 + k b_{n+1} b_n - b_n^2 = -(2-k) = k-2 $$
I asked the OP for the source of the problem, it turns out the part about defining $b_n$ was the hint given; exercise 24. The exercise (review?) set pdf is at PDF
| {
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If $a,b,c>0$ and $abc=1\;,$ Then minimum value of Expression.
If $a,b,c>0$ and $abc=1\;,$ Then minimum value of $$\frac{a^2}{a^2+2}+\frac{b^2}{b^2+2}+\frac{c^2}{c^2+2}$$
$\bf{My\; Try::}$ Using $\bf{Cauchy\; Schwarz}$ Inequality
$$\frac{a^2}{a^2+2}+\frac{b^2}{b^2+2}+\frac{c^2}{c^2+2}\geq \frac{(a+b+c)^2}{a^2+b^2+c^2+6}$$
Now How can I solve after that, Help Required
Thanks
| After your manipulation, the only reasonable claim is that:
$$(a+b+c)^2 \geq a^2+b^2+c^2+6 $$
that is equivalent to:
$$ ab+ac+bc \geq 3 $$
that follows from the AM-GM inequality:
$$ ab+ac+bc = 3\cdot AM(ab,ac,bc) \geq 3\cdot GM(ab,ac,bc) = 3.$$
Equality occurs at $a=b=c=1$, as expected.
| {
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Limit of the nth power of certain partial sums Evaluate
$$\lim_{n\to \infty}\left (\frac{6}{\pi^2}\sum_{k=1}^{n} \frac{1}{k^2} \right )^n.$$
The context is: I just thought it up and thought some members of MSE would like to try it.
| Consider $$A_n=\left (\frac{6}{\pi^2}\sum_{k=1}^{n} \frac{1}{k^2} \right )^n$$ Now, using harmonic numbers $$\sum_{k=1}^{n} \frac{1}{k^2} =H_n^{(2)}$$ Taking logarithms $$\log(A_n)=n\log\Big(\frac{6}{\pi^2}H_n^{(2)} \Big)$$ Now, using asymptotics $$H_n^{(2)}=\frac{\pi ^2}{6}-\frac{1}{n}+\frac{1}{2
n^2}+O\left(\frac{1}{n^3}\right)$$ $$\frac{6}{\pi^2}H_n^{(2)}=1-\frac{6}{\pi ^2 n}+\frac{3}{\pi ^2 n^2}+O\left(\frac{1}{n^3}\right)$$ Using Taylor $$\log\Big(\frac{6}{\pi^2}H_n^{(2)} \Big)=-\frac{6}{\pi ^2 n}+\frac{3 \left(\pi ^2-6\right)}{\pi ^4
n^2}+O\left(\frac{1}{n^3}\right)$$ $$\log(A_n)=n\log\Big(\frac{6}{\pi^2}H_n^{(2)} \Big)=-\frac{6}{\pi ^2}+\frac{3 \left(\pi ^2-6\right)}{\pi ^4
n}+O\left(\frac{1}{n^2}\right)$$ Now, using $A_n=e^{\log(A_n)}$ and Taylor again $$A_n=e^{-\frac{6}{\pi ^2}}+\frac{3 e^{-\frac{6}{\pi ^2}} \left(\pi ^2-6\right)}{\pi ^4
n}+O\left(\frac{1}{n^2}\right)$$ which shows the limit and how it is approached.
Just for illustration purposes, using $n=10$ would lead to $A_{10}\approx 0.551038$ while the above approximation gives $\approx 0.550967$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1771642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Determining the order of convergence of $ X_{n+1} = \frac{(X^3_n + 3aX_n)}{(3X^2_n + \alpha)} $ I need to find the order of convergence for:
$$
X_{n+1} = \frac{(X^3_n + 3aX_n)}{(3X^2_n + \alpha)}
$$
In a previous part we are told $\alpha$ = 2 and $x_0$=1.
I know the first step is to take the limit as $\lim_{n\to \infty}$. From here on I am confused as to the logic of a limit and the convergence as I have very little experience with either. I suspect that the powers of X play a role. When I take the limit, I get:
$$
X = \frac{(X^3 + 6X)}{(3X^2 + 2)}
$$
I called this g(x), I then took the derivative (quotient rule) and tried to solve but am stuck.
Why am I taking the derivative, and is this how I would solve this problem?
| You are studying a special case of the fixpoint iteration $x_{n+1}=g(x_n)$. If your case $g : \mathbb{R} \rightarrow \mathbb{R}$ is given by
\begin{equation}
g(x) = \frac{x^3+6x}{3x^2+2}.
\end{equation}
You are computing the derivative of $g$ in an attempt to determine if the restriction of $g$ to some closed interval $I$ is a contraction, which maps $I$ into itself. This is frequently an excellent strategy, but when the derivative(s) are complicated, alternatives should be explored.
From your equation it is clear that if the iteration converges to $X$, then $X \in \{-\sqrt{2},0,\sqrt{2}\}$. Since $g(x) > 0$ for all $x>0$, and $x_0 = 1 >0$, we see that $X = \sqrt{2}$ is the only viable option. We now find that
\begin{equation}
x_{n+1} - \sqrt{2} = \frac{x_n^3+6x_n-\sqrt{2}(3x_n^2+2)}{3x_n^2 +2}=\frac{x_n^3+3(\sqrt{2})^2x_n^2-3\sqrt{2}x_n^2-(\sqrt{2})^3}{3x_n^2+2}=\frac{(x_n-\sqrt{2})^3}{3x_n^2+2}.
\end{equation}
Now, if $e_n = \left|x_n - \sqrt{2} \right|$ denotes the absolute value of the error at the $n$th stage, then
\begin{equation}
e_{n+1} = \frac{e_n^3}{3x_n^2 +2} \leq \frac{1}{2}e_n^3.
\end{equation}
Since $e_0 = \left|\sqrt{2}-x_0\right|=\sqrt{2}-1 < 1$, it is clear that $e_n \rightarrow 0$ or equivalently that $x_n \rightarrow \sqrt{2}$. As for the exact order of the method, we observe that
\begin{equation}
\frac{e_{n+1} }{e_n^3} = \frac{1}{3x_n^2 + 2} \rightarrow \frac{1}{8} \not = 0, \quad n \rightarrow \infty, \quad n \in \mathbb{N}.
\end{equation}
By definition, this shows that the order is $p=3$.
| {
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"url": "https://math.stackexchange.com/questions/1773641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate a double integral.
To find:
$$I =\int\int_Rx(1+y^2)^{\frac{-1}{2}}dA$$
R is the region in the first quadrant enclosed by $y=x^2$, $y=4$, and $x=0$
$$y=x^2, y=4,x=0, (x= y^\frac{1}{2})$$
$$R=((x,y), 0 \le y \le 4, x^2 \le y \le 4)$$
i.e.
$$R=((x,y), 0 \le x \le 2, x^2 \le y \le 4)$$
$$I=\int_0^2\int_{x^2}^4x(1+y^2)^{\frac{-1}{2}}\,dy\,dx$$
Where to from here?
| As you defined well,
$$
R=\left\{ (x,y) \in \mathbb{R}^2: 0\leq x \leq 2, x^2 \leq y \leq 4 \right\}.
$$
So
$$
I =
\int_{R} \frac{x}{\sqrt{1+y^2}} \mathrm{d}A =
\int_0^2 \left( \int_{x^2}^4 \frac{x}{\sqrt{1+y^2}} \mathrm{d}y \right) \mathrm{d}x
$$
Here, either you do a change of variables or you check in any table that:
$$
\frac{\mathrm{d}}{\mathrm{d}z}\sinh^{-1}(z)=\frac{1}{\sqrt{1+y^2}}
$$
So
\begin{align}
I &= \int_0^2 \left[ x\sinh^{-1}(y) \right] ^{y=4}_{y=x^2} \mathrm{d}x \\
&= \int_0^2 \left( x\sinh^{-1}(4) -x\sinh^{-1}(x^2)\right) \mathrm{d}x = \\
&= \left[\frac{x^2}{2}\sinh^{-1}(4)\right]^{x=0}_{x=2} - \int_0^2x\sinh^{-1}(x^2) \mathrm{d}x \\
&= 2\sinh^{-1}(4) - \frac{1}{2}\left[x^2 \sinh^{-1} \left(x^2\right) -\sqrt{x^4+1}\right]^{x=0}_{x=2}\\
&= 2\sinh^{-1}(4) - \frac{1}{2} \left(4 \sinh ^{-1}(4)+1-\sqrt{17}\right) \\
&= \frac{\sqrt{17}-1}{2}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1773792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
What's the best way to compute $\frac{a^4 + b^4 + c^4}{a^2 + b^2 + c^2}$ So, my teacher gave us this to compute yesterday, and I'm completly confused on how should I proceed :
$$\frac{1^4 + 2012^4 +2013^4}{1^2 + 2012^2 + 2013^2}$$
I've tried several ways, but most of them are very long, for example I've simplified both numbers :
$2012^2 = (2 * 10^3)^2 + 12^2 + 24 * 10^3$
$2013^2 = (2 * 10^3)^2 + 13^2 + 26 * 10^3$
I can't see how this could help me solve this problem . So how how should I go with this kind of problems in general ?
| Hint:
$$\frac{1^4+x^4+(x+1)^4}{1^2+x^2+(x+1)^2}=x^2+x+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1774030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Simplifying the result formula for depressed Cubic After understanding the Cardano's formula for solving the depressed cubic (of the form $x^3+mx=n$, of course), I tried to find the solution of the equation $$x^3+6x=20.$$
After plugging into the formula
$$x=(n/2+\sqrt{ \frac{n^2}{4}+ \frac{m^3}{27} })^{1/3}+(-n/2+\sqrt{ \frac{n^2}{4}+ \frac{m^3}{27} })^{1/3}$$
where $m=6$ and $n=20$, we get
$$x=(10+ \sqrt{108})^{1/3}-(-10+ \sqrt{108})^{1/3}.$$
However, we notice that, without using Cardano's formula, that $x=2$ is the solution for the equation $x^3+6x=20.$
My question is: how does the equation $$x=(10+ \sqrt{108})^{1/3}-(-10+ \sqrt{108})^{1/3}$$ get simplified to $x=2$?
P.S. I understand that it was Niccolo Fontana who first figured out how to solve depressed cubic, to give one the proper credit.
| Let $a^3 = 10+\sqrt{108}=10+6\sqrt{3}, b^3 = -10 + \sqrt{108}=-10+6\sqrt{3}\implies a^3-b^3 = 20 = (a-b)(a^2+ab+b^2)=(a-b)((a-b)^2 + 3ab)$. Observe that $(ab)^3 = 8 \implies ab = 2\implies (a-b)((a-b)^2 + 6) = 20\implies x^3+6x-20 = 0, x = a-b$. Using synthetic division we have: $x^3+6x-20 = 0\implies (x-2)((x+1)^2 + 9) = 0 \implies x = 2$ as claimed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1776260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluating the rational integral $\int \frac{x^2+3}{x^6(x^2+1)}dx $
Evaluate $$\int \frac{x^2+3}{x^6(x^2+1)}dx .$$
I am unable to break into partial fractions so I don't think it is the way to go. Neither is $x=\tan \theta$ substitution. Please give some hints. Thanks.
| I learned something weird in attempting this integral. When you see $x^2+a^2$, the four substitutions that come to mind are $x=a\sec\theta$, $x=a\csc\theta$, $x=a\sinh\theta$, and $x=a\,\text{csch}\,\theta$. The first $3$ leave a big power of a trigonometric or hyperbolic function in the denominator, so I tried $x=\text{csch}\theta$.
$$\begin{align}\int\frac{x^2+3}{x^6(x^2+1)}dx&=-\int\frac{(\text{csch}^2\theta+3)\text{csch}\theta\coth\theta}{\text{csch}^2\theta\coth^2\theta}d\theta\\
&=-\int\frac{(1+3\sinh^2\theta)\sinh^4\theta}{\cosh\theta}d\theta\\
&=\int\left(-3\cosh\theta\sinh^4\theta+2\cosh\theta\sinh^2\theta-2\cosh\theta+2\text{sech}\theta\right)d\theta\end{align}$$
Now, the first $3$ terms above are no problem but the last one has the issue that the other $3$ secants have a common form for their indefinite integrals,
$$\int\sec\theta d\theta=\ln|\sec\theta+\tan\theta|+C$$
$$\int\csc\theta d\theta=-\ln|\csc\theta+\cot\theta|+C$$
$$\int\text{csch}\,\theta d\theta=-\ln|\text{csch}\,\theta+\coth\theta|+C$$
But $\text{sech}\,\theta$ doesn't fit this pattern so I can never remember its integral, so I always have to substitute $z=\tanh\left(\frac{\theta}2\right)$ and end up with
$$\int2\text{sech}\,\theta\,d\theta=4\tan^{-1}\left(\tanh\frac{\theta}2\right)+C=4\tan^{-1}\left(\frac1{x+\sqrt{x^2+1}}\right)+C$$
Now, not only does this look horrible, but it exposes the integrator for having used the unimaginative 'Weierstrass' substitution. This can be fixed up because
$$\begin{align}4\tan^{-1}\left(\frac1{x+\sqrt{x^2+1}}\right)&=2\tan^{-1}\left(\frac{\frac2{x+\sqrt{x^2+1}}}{1-\frac1{(x+\sqrt{x^2+1})^2}}\right)\\
&=2\tan^{-1}\left(\frac{2(x+\sqrt{x^2+1})}{x^2+2x\sqrt{x^2+1}+x^2}\right)\\
&=2\tan^{-1}\frac1x=\pi-2\tan^{-1}x\end{align}$$
So that means that
$$\int2\text{sech}\theta\,d\theta=-2\tan^{-1}x+C$$
But that in turn implies that
$$\int\text{sech}\,\theta\,d\theta=\tan^{-1}\sinh\theta+C_1=-\tan^{-1}\text{csch}\,\theta+C_2$$
OK, so I will have to remember that. Differentiation confirms this identity.
Maybe it would be easier to remember the forms
$$\int\text{csch}\,\theta\,d\theta=-\coth^{-1}\cosh\theta+C$$
$$\int\sec\theta\,d\theta=\tanh^{-1}\sin\theta+C$$
$$\int\csc\theta\,d\theta=-\tanh^{-1}\cos\theta+C$$
Nahh... Getting back to the task at hand,
$$\begin{align}\int\frac{x^2+3}{x^6(x^2+1)}dx&=-\frac35\sinh^5\theta+\frac23\sinh^3\theta-2\sinh\theta-2\tan^{-1}\text{csch}\theta+C\\
&=-\frac3{5x^5}+\frac2{3x^3}-\frac2x-2\tan^{-1}x+C\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1776507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 3
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About the Fourier-Legendre series of $f(x)=e^{-x}$ So for the function $f(x) = \exp(-x)$ I have the formula for the coefficients of
$$f(x) = \sum_{n=0}^{\infty}a_n P_n(x)$$
which is(by using Rodrigues formula)
$$a_n = \frac{2n+1}{2} \int_{-1}^{1}\exp(-x)\frac{1}{2^nn!}\frac{d^n}{dx^n}(x^2-1)^ndx$$
I think my next step is the integrate by parts,
$$a_n = \frac{2n+1}{2} -[\exp(-x)\frac{1}{2^nn!}\frac{d^{n-1}}{dx^{n-1}}(x^2-1)^n]^{1}_{-1} + \frac{2n+1}{2} \int_{-1}^{1}\exp(-x)\frac{1}{2^nn!}\frac{d^{n-1}}{dx^{n-1}}(x^2-1)^ndx = \frac{2n+1}{2} \int_{-1}^{1}\exp(-x)\frac{1}{2^nn!}\frac{d^{n-1}}{dx^{n-1}}(x^2-1)^ndx $$
Since the term in the brackets is zero.
It seems to me that if we do this $n$ times (further $n-1$ times). We obtain.
$$a_n = \frac{2n+1}{2}\int_{-1}^{1}\exp(-x)\frac{1}{2^nn!}(x^2-1)^ndx$$
I think I have an idea on how to proceed.
EDIT:
Let
$$\alpha_n = \int_{-1}^{1}\exp(-x)(x^2-1)^ndx$$
Then $$\alpha_n = 2n\int_{-1}^{1}\exp(-x)x(x^2-1)^{n-1}dx = $$ $$-2n(n-1)\alpha_{n-1} - 2n(n-1)\alpha_{n-2} - 2n(n-1)\int_{-1}^{1}\exp(-x)x(x^2-1)^{n-2}dx$$
I can not seem to rid the $x$ in the last integral.
I have integrated $xe^{-x}$ before integrating the second time then added and subtracted $\alpha_{n-2}$
| You have noted that
\begin{align}
a_n & = \frac{2n+1}{2}\int_{-1}^{1}e^{-x}\frac{1}{2^n n!}\frac{d^{n}}{dx^{n}}(x^2-1)^{n}dx \\
& = \frac{2n+1}{2^{n+1}n!}\int_{-1}^{1}e^{-x}\frac{d^{n}}{dx^{n}}(x^2-1)^ndx
\end{align}
And,
\begin{align}
\int fg^{(n)}dx&=fg^{(n-1)}-f^{(1)}f^{(n-2)}+f^{(2)}{g^{(n-3)}}-\cdots+(-1)^{n}\int f^{(n)}gdx
\end{align}
Because $(x^2-1)^n=(x-1)^n(x+1)^n$, then all derivatives of $(x^2-1)$ of order $k < n$ vanish at the endpoints of $[-1,1]$. So, using a standard binomial expansion of $(x^2-1)^n$,
\begin{align}
a_n & = \frac{2n+1}{2^{n+1}n!}\int_{-1}^{1}e^{-x}(x^2-1)^ndx \\
& = \frac{2n+1}{2^{n+1}n!}\int_{-1}^{1}e^{-x}\sum_{k=0}^{n}{{n}\choose{k}}(-1)^{k}x^{2k}dx \\
& = \frac{2n+1}{2^{n+1}n!}\sum_{k=0}^{n}{{n}\choose{k}}\int_{-1}^{1}e^{-x}x^{2k}dx
\end{align}
Then
\begin{align}
\int e^{-x}x^{2k}dx & = -e^{-x}x^{2k}+\int e^{-x}2kx^{2k-1}dx \\
& = -e^{-x}x^{2k}-e^{-x}(2k)x^{2k-1}+\int e^{-x}(2k)(2k-1)x^{2k-2}dx \\
& = -e^{-x}\{ x^{2k}+(2k)x^{2k-1}+(2k)(2k-1)x^{2k-2}+\cdots+(2k)! \}
\end{align}
It's ugly, but you do get the coefficients once you evaluate the above from $-1$ to $1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Find the range of the function $y = \sqrt{x^2 + 1}− x$? I have a function $y = \sqrt{x^2 + 1} − x$, where the Domain is $(−\infty,+\infty)$.
Explanation for the domain
I need to make sure the domain of the function does not include values of $x$
that will make the square root negative.
This means that I need:
$x^2+1\ge0$
the discriminant is negative and $x \in\mathbb R$. In fact, $x^2\ge 0$, $\forall x\in\mathbb R$ and $1 > 0$.
I wish, if it's possible, to explain the value of the range with an algebraic demonstration. I am using an example to solve the function for $x$ using $y$ as parameter:
$\sqrt{x^2 + 1} − x = y$
$\sqrt{x^2 + 1} = y + x$
irrational equation, therefore:
$$
\begin{cases}
& y + x >= 0\\
& x^2 + 1 =(y+x)^2
\end{cases}
$$
$$
\begin{cases}
& x >= -y\\
& x^2 + 1 =y^2 + x^2 + 2yx
\end{cases}
$$
$$
\begin{cases}
& x >= -y\\
& 1 =y^2 + 2yx
\end{cases}
$$
$$
\begin{cases}
& x >= -y\\
& x = \frac{1 - y^2}{2y}
\end{cases}
$$
now I need to find for which values is $x >= -y$, therefore:
$ \frac{1 - y^2}{2y} < 0$
solving the numerator and denominator
numerator
$1 - y^2 > 0$
$y^2 -1 < 0$
solve for $-1 < y < +1$
denominator
$2y > 0$
solve for $y > 0$
combine the tow solutions the inequality has occurred when
$-1 < y < 0$ or $y> 1$
| To determine the range you must study the variations of $y$ and apply the Intermediate value theorem (it's a continuous function).
Now $\;y'=\dfrac x{\sqrt{x^2+1}}-1=\dfrac {x-\sqrt{x^2+1}}{\sqrt{x^2+1}}<0,\;$ since $\;\lvert x\rvert<\sqrt{x^2+1} $. Hence the function is decreasing.
On the other hand,
*
*$\lim_{x\to-\infty}\sqrt{x^2+1}-x=+\infty+\infty=+\infty$,
*$\lim_{x\to+\infty}\sqrt{x^2+1}-x=\lim_{x\to+\infty}\dfrac{(x^2+1)-x^2}{\sqrt{x^2+1}+x}=0$.
Hence the range of $y$ is $(0,+\infty)$.
A purely computational solution:
As noted in another answer, $y=\dfrac1{x+\sqrt{x^2+1}}$. Hence the range is contained in $(0,+\infty)$. Lets show for any $m>0$, the equation $\dfrac1{x+\sqrt{x^2+1}}=m$ has a solution. Indeed, it is equivalent to
$$\sqrt{x^2+1}=\frac1m-x\iff x^2+1=\Bigl(\frac1m-x\Bigr)^2\enspace\textbf{and}\enspace x \le\frac 1m.$$
Now $$x^2+1=\Bigl(\frac1m-x\Bigr)^2=\frac1{m^2}-\frac 2mx+x^2\iff x=\frac{1-m^2}{2m}$$
and the condition $\;x\le\dfrac1m$ is satisfied iff $\;1-m^2\le 2$, i.e. $m^2\ge -1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1778769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Find the number of solutions of $\sin x+2 \sin 2x- \sin 3x=3$ In $(0 \:\:\pi)$Find the number of solutions of $$\sin x+2 \sin 2x- \sin 3x=3$$
The equation can be written as
$$\sin x+4 \sin x \cos x=3+\sin 3x$$ i.e.
$$\sin x(1+4\cos x)=3+\sin 3x$$ i.e.,
$$\sin x(1+4\cos x)=3+\sin x(3-4\sin^2 x)=3+\sin x(4\cos ^2x-1)$$ so
$$\sin x(4\cos^2 x-4\cos x-2)=-3$$ Any hint from here?
| One easily verifies that
$$f(x):=\sin x+2\sin(2x)-\sin(3x)=2\sin x\bigl(2\cos x-\cos(2x)\bigr)\ .$$
Since the function $h(x):=2\cos x-\cos(2x)$ assumes its maximum ${3\over2}$ at $x={\pi\over3}$ (check this!) one obtains
$$f(x)\leq2\sin x \cdot {3\over2}=3\qquad(0<x<\pi)\ ,$$
whereby equality cannot take place, since $h\bigl({\pi\over2}\bigr)<{3\over2}$.
The conclusion is that the given equation has no solutions in $\ ]0,\pi[\ $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1780727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Evaluation of product of $n$ terms of series , If $n\rightarrow \infty$
If $a_{1},a_{2},a_{3},.....,a_{n}$ are n terms of series such that $$\frac{n+1}{a_{n+1}}-\frac{n-1}{a_{n}} = \frac{2(n+2)}{n}\;,n\geq 1, n\in \mathbb{N}$$
Then $\displaystyle n^4\lim_{n\rightarrow \infty}\prod^{n}_{r=1}a_{r} = $
$\bf{My\; Try::}$ Given $$\frac{n+1}{a_{n+1}}-\frac{n-1}{a_{n}} = \frac{2(n+2)}{n}=2+\frac{4}{n}$$
We can Write it as $$\frac{n+1}{a_{n+1}}-\frac{n}{a_{n}} = \frac{1}{a_{n}}+2+\frac{4}{n}$$
Now Put $n=1,2,3,4,.......n$ and adding, We get
$$\frac{n+1}{a_{n+1}}-\frac{1}{a_{n}} = \left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+......+\frac{1}{a_{n}}\right)+2n+4\left(\frac{1}{1}+\frac{1}{2}+.....+\frac{1}{n}\right)$$
Now How can I solve after that, Help me
Thanks
| HINT :
We can write
$$\frac{n+1}{a_{n+1}}-\frac{n-1}{a_{n}} = \frac{2(n+2)}{n}$$
as
$$b_{n+1}-b_n=n+2$$
where
$$b_n=\frac{\binom n2}{a_n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1783377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{7} -\frac{1}{8}-\frac{1}{9}-\frac{1}{10} -\cdots $ converge $$1-\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}
-\dfrac{1}{8}-\dfrac{1}{9}-\dfrac{1}{10} ... $$
I added parentheses for each sub-sequence with the same sing.
so i got :
$$1-(\dfrac{1}{2}+\dfrac{1}{3})+(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6})-(\dfrac{1}{7}
+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}) ... $$
I want to show that the new sequence is a leibniz sequence and by that conclude that is converge.
I managed to show that each pair of parentheses is greater than:
$$\dfrac{2}{n+1} $$
Cant find a way to proceed.
Thanks for helping.
| 1.in case of non absolute convergence, grouping is not a good idea.
2.Sum $S, \quad \frac {1}{2}<S<1$, using Bolzano-Weisrass theorem convergence follows
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Linear approximation to find $\frac{1}{4.002}$ I am asked to find $\frac{1}{4.002}$ using linear approximation.
The way I proceeded was:
$$
f(x) = \frac{1}{x} \quad a = 4 \quad f(a) = f(4) = 0.25\\
f'(x) = - \frac{1}{x^2} \quad f'(a) = f'(4) = - \frac{1}{16}
$$
So
$\frac{1}{4.002} \approx \frac{1}{4} - \frac{1}{16} (4.002 - 4) = 0.249875$
Is this correct?
If I want to find out how good was my approximation to the real value, what should I do?
Thank you.
| If $x>0$ is close to zero
$$\frac{1}{4+x}=\frac{1}{4}\cdot\frac{1}{1+\frac{x}{4}}=\frac{1}{4}\sum_{n\geq 0}\frac{(-1)^n x^n}{4^n}$$
holds, and since the terms of the last series are decreasing in absolute value,
$$ \frac{1}{4}\left(1-\frac{x}{4}\right) \leq \frac{1}{4+x}\leq \frac{1}{4}\left(1-\frac{x}{4}+\frac{x^2}{16}\right) $$
holds, too. Just plug in $x=\frac{1}{500}$ to get an upper bound for the approximation error by the RHS inequality.
| {
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max of $e$ with $a+b+c+d+e=8$ and $a^2+b^2+c^2+d^2+e^2=16$ Given that a,b,c,d,e are real number such that:
$\begin{cases}
a+b+c+d+e=8\\
a^2+b^2+c^2+d^2+e^2=16
\end{cases}$
determine the maximun value of $e$.
I started like that :
$\max(e)=\max(8-a-b-c-d)=\min(a+b+c+d)$
but i don't kow how to carry on
can you help please ?
| Physics (Classical Mechanics) Solution:
Consider a $1$-dimensional elastic collision of a particle $X$ of mass $4$ moving at velocity $2$ into a particle $E$ of mass $1$, initially at rest. Due to this collision, $X$ breaks into $4$ smaller particles $A$, $B$, $C$, and $D$ (of course, we are ignoring their binding energy, which would lead to an inelastic collision) with identical mass $1$, with velocities $a$, $b$, $c$, and $d$, and $E$ attains a velocity $e$. Let $T_e$ denote the total energy of the particles $A$, $B$, $C$, and $D$ as a function of $e$. Since $T_e$ is at least the kinetic energy of the center-of-mass frame of the particles $A$, $B$, $C$, and $D$, $$T_e\geq \frac{1}{2}\cdot (1+1+1+1)\cdot \left(\frac{a+b+c+d}{4}\right)^2=\frac{(a+b+c+d)^2}{8}\,.$$
The equality case of the inequality above is when $a=b=c=d$ (i.e., when the particles $A$, $B$, $C$, and $D$ are at rest in their center-of-mass frame). By the Conservation Law of Momentum, $$a+b+c+d=4\cdot 2-1\cdot e=8-e\,.$$ By the Conservation Law of Energy, $$T_e=\frac{1}{2}\cdot4\cdot 2^2-\frac{1}{2}\cdot 1\cdot e^2=\frac{16-e^2}{2}\,.$$
Hence,
$$\frac{16-e^2}{2}=T_e\geq \frac{(a+b+c+d)^2}{8}=\frac{(8-e)^2}{8}\,,$$
whence $0\leq e\leq \frac{16}{5}$. The maximum $e=\frac{16}{5}$ is attained iff $a=b=c=d=\frac{6}{5}$. The minimum $e=0$ is attained iff $a=b=c=d=2$.
| {
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Simplify $\frac{1}{\sqrt{4+2\sqrt{3}} - \sqrt{4-2\sqrt{3}}}$ Simplify $$\frac{1}{\sqrt{4+2\sqrt{3}} - \sqrt{4-2\sqrt{3}}}$$
I know there is another easier method except the one I answered. I cannot find it. Can you please help? Thanks in advance.
| My attempt
$$\frac{1}{\sqrt{4+2\sqrt{3}} - \sqrt{4-2\sqrt{3}}}$$
$$ =\frac{\sqrt{4+2\sqrt{3}} + \sqrt{4-2\sqrt{3}}}{4+2\sqrt{3} - (4-2\sqrt{3})} $$
$$ =\frac{\sqrt{4+2\sqrt{3}} + \sqrt{4-2\sqrt{3}}}{4\sqrt{3}} $$
$$ = \sqrt{\left(\frac{\sqrt{4+2\sqrt{3}} + \sqrt{4-2\sqrt{3}}}{4\sqrt{3}} \right)^2}$$
$$ = \sqrt{\frac{4 + 2\sqrt{3} - (4 - 2\sqrt{3}) - 2\sqrt{4^2 - 4\times3}}{48}}$$
$$ = \sqrt{\frac{4 + 2\sqrt{3} + (4 - 2\sqrt{3}) + 2\sqrt{4^2 - 4\times3}}{48}}$$
$$ = \sqrt{\frac{12}{48}} = \sqrt{\frac{1}{4}}$$
$$\mathbf{= \frac{1}{2}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$8^a=3$ and $3^b=5$ and $10^c=5$ then find $c$ in terms of $a$ and $b$. if $8^a=3$ and $3^b=5$ and $10^c=5$ then find $c$ using $a$ and $b$.
My Attempt:
if $8^a=3$ and $3^b=5$ then we can say that $8^{ab}=5$ and then we have $2^{3ab}=10^c$ but i cant solve this equation.
| $$8^a=3\implies3a\log2=\log3$$
Similarly, $$b\log3=\log5$$ and $$\log5=c(\log5+\log2)\iff(c-1)\log5=c\log2$$
$$\implies3a\log2\cdot b\log3\cdot(c-1)\log5=\log3\cdot \log5\cdot c\log2$$
$$\iff3a\cdot b(c-1)=c$$
See: Laws of Logarithms
| {
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Showing that $\sum_{n=0}^{\infty}\frac{2^{n+2}}{{2n\choose n}}\cdot\frac{n-1}{n+1}=(\pi-2)(\pi-4)$ Showing that
(1)
$$\sum_{n=0}^{\infty}\frac{2^{n+2}}{{2n\choose n}}\cdot\frac{n-1}{n+1}=(\pi-2)(\pi-4)$$
see here
(2)
$$(\arcsin(x))^2=\frac{1}{2}\sum_{n=1}^{\infty}\frac{(2x)^{2n}}{n^2{2n\choose n}}$$
Look very much similar to (1). How can I make the use of (2) to solve (1)?
Let try and differentiate (1)
$$\frac{d}{dx}\arcsin(x))^2=\frac{d}{dx}\frac{1}{2}\sum_{n=1}^{\infty}\frac{(2x)^{2n}}{n^2{2n\choose n}}$$
$$\arcsin(x)\cdot\frac{1}{\sqrt{1-x^2}}=2\sum_{n=1}^{\infty}\frac{(2x)^{2n-1}}{n{2n\choose n}}$$
Getting close but not near yet
I am stuck, any help please?
| It is worth considering that:
$$ \frac{1}{\binom{2n}{n}}=\frac{n!^2}{(2n)!}=(2n+1)\cdot B(n+1,n+1)=(2n+1)\int_{0}^{1}x^n(1-x)^n\,dx \tag{1}$$
so:
$$\sum_{n\geq 0}\frac{2^{n+2}}{\binom{2n}{n}}=\int_{0}^{1}\frac{4+8x(1-x)}{(1-2x(1-x))^2}\,dx = 4\int_{-1}^{1}\frac{3-y^2}{(1+y^2)^2}\,dy=8+2\pi\tag{2}$$
as well as:
$$\begin{eqnarray*} \sum_{n\geq 0}\frac{2^{n+2}}{(n+1)\binom{2n}{n}}&=&4\int_{0}^{1}\int_{0}^{1}\frac{1+2xy-2x^2y}{(1-2xy+2x^2 y)^2}\,dx\,dy \\&=&8\int_{0}^{1}\frac{dx}{1-2x(1-x)}+2\int_{0}^{1}\frac{\log(1-2x(1-x))}{x(1-x)}\tag{3}\end{eqnarray*}$$
that is easy to compute through the substitution $y=2x-1$ as in $(2)$.
| {
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Simplifying $3^{-4}$ to $1/81$ How would you solve for $3^{-4}$? I know the answer is $1/81$ but I can't work out how you get there with this one.
| Let's consider the expression
$$\frac{3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3}{3\cdot 3\cdot 3\cdot 3}$$
There are three more $3$s multiplied together on the top than there are on the bottom, and we write it more concisely as $3^3$.
Suppose now that the number of $3$s on top are equal to the number of threes on the bottom. In this case we would write $3^0$ (which simplifies to just $1$).
$$\frac{3\cdot 3\cdot 3\cdot 3}{3\cdot 3\cdot 3\cdot 3}$$
Following this model, we would expect that $3$ raised to a negative power would represent a fraction where the number of $3$s on the bottom exceed the number of $3$s on the top, as follows.
$$\frac{3\cdot 3}{3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3}$$which we would represent as $3^{-4}$.
Continuing we have$$\frac{3\cdot 3}{3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3}=\frac{3\cdot 3}{3\cdot 3\cdot ((3\cdot 3)\cdot (3\cdot 3))}=\frac{3\cdot 3}{3\cdot 3\cdot (9\cdot 9)}=\frac{1}{81}$$
| {
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Find the maximal value of $a+b-c+d$ Let $a, b, c, d$ be real numbers satisfying inequality $$f(x)=a\cos x+b\cos 2x+c\cos 3x+d\cos 4x\le 1$$ holds for $x\in\Bbb{R}$. Find the maximal value of $a+b-c+d$ and determine the values of $a,b,c,d$ when that maximum is attained.
Try $f(0)=a+b+c+d\le 1$, $f(\pi)=-a+b-c+d\le 1$ and
$$f\left(\dfrac{\pi}{2}\right)=-b+d\le 1$$
| We claim that the maximum is 3 given by $a=1,b=\frac{1}{2},c=-1,d=\frac{1}{2}$. Using the usual formulae, we have $\cos x+\frac{1}{2}\cos2x-\cos3x+\frac{1}{2}\cos4x=f(c)$, where $f(c)=c+\frac{1}{2}(2c^2-1)-(4c^3-3c)+\frac{1}{2}(8c^4-8c^2+1)=4c^4-4c^3-3c^2+4c$, where $c=\cos x$. We have $f'(c)=16c^3-12c^2-6c+4=2(2c-1)(4c^2-c-2)$, which as zeros at $c=\frac{1\pm\sqrt{33}}{8},\frac{1}{2}$, so $f'(c)$ is negative for $-1<c-0.593$, positive for $-0.593<c<0.5$, negative for $0.5<c<0.843$ and positive for $c>0.843$. So in the range $[-1,1]$ it has two local minima and a local maximum at $\frac{1}{2}$. We find $f(-1)=f(1)=1$ and $f(\frac{1}{2})=1$, so $f(c)\le1$ for all $c\in[-1,1]$.
It remains to show that we cannot do better than that. The inequality must hold for $x=0$, so we have $a+b+c+d\le1$ (1). It holds for $x=\frac{\pi}{3}$ so $a-b-2c-d\le2$ (2), and it holds for $x=\pi$ so $-a+b-c+d\le1$ (3). Taking (1)+$\frac{2}{3}$(2)+$\frac{2}{3}$(3) we have $a+b-c+d\le3$.
| {
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If $n$ is a power of $13$, does $n \mid 5^n + 8^n$? Motivated by this question:
Is it true that if $n$ is a power of $13$, then $n \mid 5^n + 8^n$ ?
The limited data in oeis/A045597 seems to suggest it is true.
The converse does not hold. The first counterexample is $n=114413= 13^2 \cdot 677$.
The data also suggests that
If $n$ is a power of $13$, then $13n$ is the largest power of $13$ that divides $5^n + 8^n$.
Anyone knows proofs or counterexamples for these two observations?
| An easy elementary proof without valuation $v_{13}$.
$$(a+b)^{13}=a^{13}+b^{13}+13ab(a^{11}+b^{11})+13(6)(ab)^2(a^9+b^9)+13(22)(ab)^3(a^7+b^7)+13(55)(ab)^4(a^5+b^5)+13(99)(ab)^5(a^3+b^3)+13(132)(ab)^6(a+b)\qquad (1)$$ We use induction.
For $n=1$ one has $5^{13}+8^{13}=13(13)(3260216077)$ so the property is true for $13^1$. Suppose it is true for $13^n$ so $5^{13^n}+8^{13^n}=13^nA$.
It follows from $(1)$
$$(5^{13^n}+8^{13^n})^{13}=5^{13^{n+1}}+8^{13^{n+1}}+13(40)((5^{13^n})^{11}+(8^{13^n})^{11})+….+13(132)(40)^6(5^{13^n}+8^{13^n})$$
Since for $k$ odd we have $(5^{13^n})^k+(8^{13^n})^k=(5^{13^n}+8^{13^n})Q_k(5^{13^n},8^{13^n})$ where the form of the second factor is well known it follows
$$(13^nA)^{13}=5^{13^{n+1}}+8^{13^{n+1}}+13(13^nA)\left[40(Q_{11}(5^{13n},8^{13n})+….+132(40)^6Q_1(5^{13^n},8^{13^n})\right]$$ Thus $$13^{n+1}|5^{13^{n+1}}+8^{13^{n+1}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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how many answers do this equation have$3^{2x}-34(15^{x-1})+5^{2x}=0$ How many answers do this equation have?
$3^{2x}-34(15^{x-1})+5^{2x}=0$
My Attempt:$3^{2x}+5^{2x}=34(15^{x-1})$.Now what to do?
| Letting $A = 3^x$ and $B = 5^x$, we get:
$$A^2 - \dfrac{34}{15}AB + B^2 = 0.$$
Since this equation is symmetric in $A$ and $B$, we can consider it as a quadratic in the variable $A$.
Using the quadratic formula:
$$A = \dfrac{\dfrac{34}{15}B \pm \sqrt{\left(\dfrac{34}{15}B\right)^2 - 4B^2}}{2} = \dfrac{\dfrac{34}{15}B \pm \sqrt{\left(\dfrac{16}{15}B\right)^2}}{2} = \dfrac{\dfrac{34}{15}B \pm \dfrac{16}{15}B}{2},$$
from which we obtain
$$A_1 = \dfrac{\dfrac{34}{15}B - \dfrac{16}{15}B}{2} = \dfrac{9}{15}B = \dfrac{3}{5}B,$$
and
$$A_2 = \dfrac{\dfrac{34}{15}B + \dfrac{16}{15}B}{2} = \dfrac{25}{15}B = \dfrac{5}{3}B.$$
Substituting back $A = 3^x$ and $B = 5^x$, we obtain:
$$3^x = \left(\dfrac{3}{5}\right){5^x},$$
or
$$3^x = \left(\dfrac{5}{3}\right){5^x}.$$
From the first equation, we have:
$$3^{x-1} = 5^{x-1}$$
while from the second equation, we get:
$$3^{x+1} = 5^{x+1}.$$
Since $\gcd(3,5) = 1$, then we have either
$$3^{x-1} = 1$$
or
$$3^{x+1} = 1$$
(i.e., either $x = 1$ or $x = -1$).
Conclusion: Since $x = 1$ or $x = -1$ satisfy the original equation, then we have obtained TWO SOLUTIONS.
| {
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Testing convergence of series $\sum_{n=1}^\infty\sin(\pi\sqrt{n^2+k^2})$ Considering
$$\sum_{n=1}^\infty\sin(\pi\sqrt{n^2+k^2})$$
depending on $k$, which can be real. I have absolutely no clue how to proceed. Tried to taylor it, but with no result.
| Hint. Your series is convergent.
As $n$ tends to $+\infty$, we may write
$$
\begin{align}
u_n &:=\sin \left( \pi \sqrt{n^2+k^2 }\right)\\
&=\sin \left( \pi n \:\sqrt{1+\frac{k^2}{n^2}}\right)\\
&=\sin \left( \pi n \:\left(1+\frac{k^2}{2n^2}+\mathcal{O}\left(\frac{1}{n^4}\right)\right)\right)\\
&=\sin \left( \pi n +\frac{\pi k^2}{2n}+\mathcal{O}\left(\frac{1}{n^3}\right)\right)\\
&=(-1)^n\sin \left(\frac{\pi k^2}{2n}+\mathcal{O}\left(\frac{1}{n^3}\right)\right)\\
&=\frac{\pi k^2}2\frac{(-1)^n}{n}+\mathcal{O}\left(\frac{1}{n^3}\right)
\end{align}
$$
giving the convergence of the initial series $\displaystyle \sum u_n$.
| {
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Tough Recurrence Relation I'm trying to find a recurrence equation solution to
$$f(n)=a(n)f(n-1)+f(n-2)$$
with the initial conditions that $f(-1)=0, f(-2)=1$ and $$a(n)=\frac{c}{2}(1+(-1)^n)-\frac{d}{n+1}(1-(-1)^n)$$ with some constants $c$ and $d$.
Alternatively we could write $a(2n)=c$ and $a(2n-1)=-\frac{d}{n}$ and try to solve the cases seperately.
| Let $b = cd$ and we will assume $b \ne 0$.
Let $f_n = f(n)$, the recurrence relation can be rewritten as
$$f_n - f_{n-2} = f_{n-1}\times
\begin{cases}
c, & n \text{ even }\\
-\frac{2d}{n+1},& n\text{ odd }
\end{cases}
\quad\iff\quad
\begin{cases}
f_{2m} - f_{2m-2} &= c f_{2m-1}\\
f_{2m+1} - f_{2m-1} &= -\frac{d f_{2m}}{m+1}
\end{cases}
$$
Let $g_m = f_{2m}$, it is clear once we figure out what $g_m$ are, we can compute $f_{2m-1}$ from $g_m$ and $g_{m-1}$.
It is clear $g_{-1} = g_0 = 1$. Furthermore, for any $m \ge 0$, we have
$$g_{m+1} - 2 g_m + g_{m-1} = (f_{2m+2}-f_{2m}) - (f_{2m} - f_{2m-2}) = c(f_{2m+1} - f_{2m-1}) = -\frac{b g_m}{m+1}$$
Let $g(z) = \sum\limits_{m=0}^\infty g_m z^m$ be the OGF. Multiply both sides of above equation by $z^{m+1}$ and start to sum from $m = 0$, we get
$$(g(z)-1) - 2zg(z) + (z + z^2g(z)) =
(1-z)^2 g(z) + (z-1) = -b\int_0^z g(t)dt
$$
Differentiate $z$ on both sides, we get
$$\begin{align}
& (1-z)^2 g'(z) - 2(1-z) g(z) + 1 = - b g(z)\\
\iff & g'(z) + \left[ -\frac{2}{1-z} + \frac{b}{(1-z)^2}\right] g = - \frac{1}{(1-z)^2}\\
\iff & \frac{d}{dz}\left[ (1-z)^2 e^{bz/(1-z)} g(z) \right] = - e^{bz/(1-z)}\\
\implies & g(z) = \frac{1}{(1-z)^2}e^{-bz/(1-z)}\left[ 1 - \int_0^z e^{bt/(1-t)} dt \right]\tag{*1}
\end{align}
$$
With some algebra, we can express $g(z)$ in terms of
exponential integrals.
$$\frac{1}{1-z} - \frac{b}{(1-z)^2} e^{-b/(1-z)}\left[{\mathrm Ei}\left(\frac{b}{1-z}\right) - {\mathrm Ei}(b)\right]$$
Unluckily, I cannot figure out how to use this to extract a nice expression
of $g_n$. It turns out there is another ugly way.
Recall the definition of Laguerre polynomials $L_n(x)$ in terms of its OGF:
$$\sum_{m=0}^\infty t^m L_m(x) = \frac{1}{1-t} e^{-xt/(1-t)}$$
We find
$$1 - \int_0^z e^{bt/(1-t)} dt
= 1 - \int_0^1 \sum_{m=0}^\infty t^m(1-t) L_m(-b) dt
= 1 - z - \sum_{m=0}^\infty \frac{L_m(-b) - L_{m-1}(b)}{m+1} z^{m+2}
$$
As a result,
$$g(z) = \frac{1}{1-z}\left(\sum_{n=0}^\infty L_n(b) z^n\right)\left((1-z) - z^2 \sum_{m=0}^\infty \frac{L_m(-b) - L_{m-1}(b)}{m+1} z^m \right)$$
Compare coefficients of both sides, we find $g_n$ is a polynomial in $b$ of degree $n$.
$$f_{2n} = g_n = L_n(b) - \underbrace{\sum_{q=0}^{n-2}\frac{L_{q+1}(-b)-L_{q}(-b)}{q+2}\left(\sum_{p=0}^{n-2-q} L_p(b)\right)}_{= 0 \text{ when } n < 2}$$
Using this formula, the first few $g_{n}$ are given below
$$
\begin{array}{rcl}
g_0 &=& 1\\
g_1 &=& 1-b\\
2!g_2 &=& 2-5b+b^2\\
3!g_3 &=& 6-26b+11b^2-b^3\\
4!g_4 &=& 24-154b+102b^2-19b^3+b^4\\
5!g_5 &=& 120-1044b+954b^2-272b^3+29b^4-b^5
\end{array}
$$
They match what one get if we solve the recurrence relations by brute force.
| {
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Write the general term of the periodic sequence $1$, $-1$, $-1$, $1$, $-1$, $-1$, $1$, ..., as $(-1)^{g(n)}$ or other closed form How to put mathematically sequence that changes sign like:
$n = 0\quad f = 1$
$n = 1 \quad f = -1$
$n = 2 \quad f = -1$
$n = 3 \quad f = 1$
$n = 4 \quad f = -1$
$n = 5 \quad f = -1$
$n = 6 \quad f = 1$
.....
In the form of (-1)^(something) or similar analytical expression.
| If you allow yourself other roots of unity, you can use the discrete fourier transform.
I presume you want the sequence to have period $3$. So, we use the cube root of unity $\zeta = \frac{-1 + i \sqrt{3}}{2}$. The DFT $\hat{f}$ is given by
$$ \hat{f}(n) = \sum_{k=0}^2 f(k) \zeta^{kn} $$
which we can tabulate as
*
*$\hat{f}(0) = -1$
*$\hat{f}(1) = 2$
*$\hat{f}(2) = 2$
Then, we can recover $f$ by
$$ f(n) = \frac{1}{3} \sum_{k=0}^2 \hat{f}(k) \zeta^{-kn} $$
which works out to
$$ f(n) = \frac{1}{3} \left(-1 + 2 \zeta^n + 2 \zeta^{-n} \right) $$
which can be expressed in a variety of forms; e.g. since $\zeta = \exp(2 \pi i / 3)$, we have
$$ \cos\left( \frac{2 \pi n}{3} \right) = \frac{\zeta^n + \zeta^{-n}}{2} $$
and thus
$$ f(n) = \frac{1}{3} \left( -1 + 4 \cos\left( \frac{2 \pi n}{3} \right) \right) $$
More generally, you could solve this by picking any three linearly independent functions with period $3$, then solving a system of equations to recover what coefficients you need on them to construct $f$; the DFT is just a direct way to obtain such a solution when the three functions are $1$, $\zeta$, and $\zeta^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1817522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 11,
"answer_id": 9
} |
Minimum number of possible value of x Let $f: R\rightarrow R$ is a function stratifying $f(2-x)=f(2+x)$ and $f(20-x)=f(x)$ for all $x$ belonging to $R$ . If $f(0) =5$ , them minimum number of possible value of $x$ satisfying $f(x) =5$ for $x$ belonging $[0,170]$
|
Define $g(x)=\frac{5}{2}(x-2)$ for $0\le x\le10$ and $\frac{5}{2}(18-x)$ for $10\le x\le18$, and extend it to all $x$ by making it periodic with period 16.
It is easy to check that this satisfies the conditions $g(2-x)=g(2+x)$ and $g(x)=g(20-x)$.
It is clear that in any one period $g(x)$ has two values for which $g(x)=5$. For the interval $[2,18]$ they are $x=4$ and $x=16$. So each of the 10 complete periods in the range $[0,170]$, namely $[2,18],\dots,[146,162]$ contains two values of $x$. The values 0 and 164 are also in the range, so a total of 22.
Now suppose that $f(x)$ is any function satisfying $f(2-x)=f(2+x)$ and $f(x)=f(20-x)$. Then for any $x$ we have $f(16+x)=f(20-(16+x))=f(4-x)=f(2-(x-2))=f(2+x-2)=f(x)$, so $f(x)$ must be periodic with period 16.
Also we have $f(4)=f(2+2)=f(2-2)=f(0)=5$. So the 11 values $0,16,32,\dots,160$ and the 11 values $4,20,36,\dots,164$ are in the range $[0,170]$. Hence there are at least 22 values in the range where $f(x)=5$.
Hence the minimum is 22.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1818071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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limit involving composition of Trigonometric function
Evaluation of limit $\displaystyle \lim_{x\rightarrow 0}\frac{\cos (\tan x)-\cos x}{x^4}\;,$
without using L, Hospital Rule and series expansion.
I have solved it using $\displaystyle \cos (\tan x)-\cos x = -2\sin\left(\frac{\tan x+x}{2}\right)\cdot \sin \left(\frac{\tan x-x}{2}\right)$
Now Using $\displaystyle \lim_{x\rightarrow 0}\frac{\tan x-x}{x^3} = \frac{1}{3}$
But i did not understand how can i solve it without using $\displaystyle $ L hospital rule and series exp.
Help required, Thanks
| $$=\dfrac{\sin\dfrac{\tan x+x}2}{\dfrac{\tan x+x}2}\cdot\dfrac{\dfrac{\tan x+x}2}{\dfrac x2} \cdot\dfrac{\sin\dfrac{\tan x-x}2}{\dfrac{\tan x-x}2}\cdot\dfrac{\tan x-x}{x^3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1819439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Triple Integrals in Spherical Coordinates where (z-2)^2 $$\iiint \frac{1}{\sqrt{x^2+y^2+(z-2)^2}}$$
for $x^2+y^2+z^2 = 1$
I've used spherical coordinates, like this:
$x=\rho sin\phi cos\theta$; $y=\rho sin\phi sin\theta$; $z=\rho cos\phi$ and $J=\rho^2 sin\phi$
$$\rho^2 sin^2\phi cos^2\theta + \rho^2 sin^2\phi sin^2\theta +\rho^2 cos^2\phi = 1$$
$$\rho^2 sin^2\phi + \rho^2 cos^2\phi = 1$$
$$\rho^2 = 1$$
$$\rho = 1$$
and when i put coordinates in my initial integral and simplify it, i get this:
$$\int_0^{2\pi}\int_0^\pi\int_0^1\frac{1}{\sqrt{\rho^2-4\rho cos\phi +4}}\rho^2sin\phi d\theta d\phi d\rho$$
and i can not move from there.
Please, help.
| I was going to suggest the substitution.
$\rho = 2\sin \phi\tan u + 2 \cos \phi$
And that would clean up the radical. But, after thinking about it, I decided that spherical is not the way to go.
Suppose instead you did cylindrical.
$x = r \cos \theta\\
y = r \sin \theta\\
z = z\\
dx\,dy\,dz = r\,dr\,dz\,d\theta$
$\int\int\int\dfrac {r}{\sqrt{r^2 + (z-2)^2}} dr\,dz\,d\theta$
We would like to integrate by r first. Otherwise, we have an inverse tangent to deal with. So, find limits of r in terms of z.
$\int_0^{2\pi}\int_{-1}^1\int_0^{\sqrt{1-z^2}} \dfrac {r}{\sqrt{r^2 + (z-2)^2}} dr\,dz\,d\theta\\
\int_0^{2\pi}\int_{-1}^1 (r^2 + (z-2)^2)^{\frac12}|_0^{\sqrt{1-z^2}} dz\,d\theta\\
\int_0^{2\pi}\int_{-1}^1 [(1-z^2 + (z-2)^2)^{\frac12} - ((z-2)^2)^{\frac12}]dz\,d\theta\\
\int_0^{2\pi}\int_{-1}^1 (-4z+5)^{\frac12} - |z-2|dz\,d\theta$
And you can get home from here.
Suppose we stuck with spherical, and flipped the order of integration!
$\int_0^1\int_0^\pi\int_0^{2\pi}\frac{\rho^2\sin\phi}{\sqrt{\rho^2-4\rho \cos\phi +4}} d\theta d\phi d\rho$
$2\pi\int_0^1\int_0^\pi\frac{\rho^2sin\phi}{\sqrt{\rho^2-4\rho \cos\phi +4}} d\phi d\rho$
$u = \rho^2-4\rho \cos\phi +4\\
du = 4\rho \sin\phi d\phi$
$2\pi\int_0^1\int_{\rho^2 - 4\rho + 4}^{\rho^2+4\rho + 4}\frac{\rho}{4\sqrt{u}} du d\rho\\
2\pi\int_0^1\frac 12 \sqrt{u}\rho|_{\rho^2-4\rho + 4}^{\rho^2+4\rho + 4} d\rho\\
2\pi\int_0^1\frac 12 (|\rho+2|\rho - |\rho-2|\rho) d\rho\\
2\pi\int_0^1 \rho^2\, d\rho\\
\frac{2\pi}{3} \rho^3|_0^1 \\
\frac{2\pi}{3}$
| {
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"url": "https://math.stackexchange.com/questions/1821448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\int_{0}^{\infty}{1\over (1+{\phi^{-2}}x^2)(1+{\phi^{-4}}x^2)}dx={\pi\over 2}$ Prove
$\phi$; golden ratio
$$I=\int_{0}^{\infty}{1\over (1+{\phi^{-2}}x^2)(1+{\phi^{-4}}x^2)}dx={\pi\over 2}\tag1$$
Let $q=\phi^{-2}$ and $p=\phi^{-4}$
$${1\over (1+qx^2)(1+px^2)}={Ax+B\over 1+qx^2}+{Cx+D\over 1+px^2}\tag2$$
Result
$A=0$, $B=\phi$, $C=0$ and $D=-{1\over \phi}$
Sub the result into $(2)$ and we got
$${1\over (1+\phi^{-2}x^2)(1+\phi^{-4}x^2)}={\phi\over 1+\phi^{-2}x^2}-{\phi^{-1}\over 1+\phi^{-4}x^2}$$
Hence
$$I=\int_{0}^{\infty}\left({\phi\over 1+\phi^{-2}x^2}-{\phi^{-1}\over 1+\phi^{-4}x^2}\right)dx\tag3$$
Simplified $(3)\rightarrow (4)$
$$I=\phi^3\int_{0}^{\infty}{1\over \phi^2+x^2}dx-\phi^3\int_{0}^{\infty}{1\over \phi^4+x^2}dx\tag4$$
Recall
$$\int_{0}^{\infty}{1\over a^2+x^2}dx={\pi \over 2a}\tag5$$
Utilise $(5)$ to brings $(4)$ to $(6)$
$$I=\phi^3\cdot{\pi\over 2\phi}-\phi^3{\pi\over 2\phi^2}\tag6$$
$$I={\pi\over 2}(\phi^2-\phi)={\pi\over 2}\tag7$$
This method it is a bit tedious, can someone tackle integral (1) with a quicker method?
| Through the residue theorem, by setting $f(x)=\frac{1}{\left(1+\frac{x^2}{\varphi^2}\right)\left(1+\frac{x^2}{\varphi^4}\right)}$ we have:
$$\begin{eqnarray*} I = \pi i \cdot\!\!\!\!\!\! \sum_{z\in\left\{i\varphi,i\varphi^2\right\}}\!\!\!\text{Res}(f(x),x=z)&=&\pi i\left(\frac{i\varphi^5}{2(\varphi^2-\varphi^4)}+\frac{i\varphi^4}{2(\varphi^4-\varphi^2)}\right)\\&=&\frac{\pi}{2}\cdot(\varphi^2-\varphi)\cdot\frac{\varphi^3}{\varphi^4-\varphi^2}\\&=&\frac{\pi}{2}\cdot 1\cdot 1 = \color{red}{\frac{\pi}{2}}.\end{eqnarray*} $$
An alternative is to prove a more general statement:
$$ \forall A,B>0,\qquad I(A,B)=\int_{0}^{+\infty}\frac{dx}{\left(1+\frac{x^2}{A^2}\right)\left(1+\frac{x^2}{B^2}\right)}=\frac{\pi}{2}\cdot\frac{AB}{A+B}$$
through Lagrange's identity and Glasser's master theorem, or the fact that the Fourier transform of a Cauchy distribution is a Laplace distribution. For instance, Lagrange and Glasser give:
$$\begin{eqnarray*}I(A,B)=\int_{0}^{+\infty}\frac{dx}{\left(\frac{A+B}{AB}x\right)^2+\left(1-\frac{x^2}{AB}\right)^2}&=&\frac{AB}{A+B}\int_{0}^{+\infty}\frac{dx}{x^2+\left(1-\frac{AB x^2}{(A+B)^2}\right)^2}\end{eqnarray*}$$
where $\int_{0}^{+\infty}\frac{dx}{x^2+(1-k^2 x^2)^2}$ constantly equals $\frac{\pi}{2}$ for any $k\in\mathbb{R}\setminus\{0\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1823137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Evaluate $\int_0^9 \frac{\ln(x)}{\sqrt x}$ I need to solve this defined integral:
$$\int_0^9 \frac{\ln(x)}{\sqrt{x}}dx$$
I tried to solve by parts, so:
$$f(x)g(x) = \int f'(x)g(x)dx$$
where:
$$f(x) = \ln(x)$$
$$f'(x) = \frac{1}{x}$$
$$g'(x) = \frac{1}{\sqrt{x}}$$
$$g(x) = 2\sqrt{x}$$
And I get:
$$\ln(x)2\sqrt x - \int\frac{1}{x} \cdot \frac{1}{\sqrt x}$$
How should I proceed?
| Consider
\begin{equation}
I(a) = \int\limits_{0}^{b} x^{a} \mathrm{d} x = \frac{x^{a+1}}{a+1}\bigg|_{0}^{b} = \frac{b^{a+1}}{a+1}
\end{equation}
Then
\begin{equation}
\int\limits_{0}^{b} x^{a} \mathrm{ln}(x) \mathrm{d} x = \frac{\partial I(a)}{\partial a} =
\frac{b^{a+1}}{(a+1)^{2}}[(a+1)\mathrm{ln}(b) - 1]
\end{equation}
and
\begin{align}
\int\limits_{0}^{b} \frac{\mathrm{ln}(x)}{\sqrt{x}} \mathrm{d} x & =
\lim_{a \to -\frac{-1}{2}} \int\limits_{0}^{b} x^{a} \mathrm{ln}(x) \mathrm{d} x \\
& = \lim_{a \to -\frac{-1}{2}} \frac{b^{a+1}}{(a+1)^{2}}[(a+1)\mathrm{ln}(b) - 1] \\
& = 4\sqrt{b}[\mathrm{ln}(\sqrt{b})-1]
\end{align}
For $b=9$ we have
\begin{equation}
\int\limits_{0}^{9} \frac{\mathrm{ln}(x)}{\sqrt{x}} \mathrm{d} x
= 4\sqrt{9}[\mathrm{ln}(\sqrt{9})-1]
= 12(\mathrm{ln}3-1)
\end{equation}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Triangular sides
In a triangle the least angle is $45º$ and the tangents of the angle are in arithmetic progression. If its area is $27\text{cm}^2$, find the length of the sides.
I tried to solve the problem in this way
Let the smallest angle in the triangle $ABC$ (say) be $A=45º$, where $\tan A$, $\tan B $ and $\tan C$ are in $AP$, then according to question
$$\begin{array}
2\tan B=\tan A+\tan C\\
\implies 2\tan B=\frac{\sin A\cos C+\cos A\sin C}{\cos A\cos C}\\
\implies 2\tan B=\frac{\sin B}{\cos A\cos C}\\
\implies \cos B=2\cos A\cos C
\end{array}$$
As we know that $\cos A=\frac{1}{\sqrt2}$ hence
$$\cos B=\sqrt2\cos C$$
squaring both sides would give
$$\cos^2B=2\cos^2C$$
Changing the cosines to sines we get
$$1-\sin^2B=2-2\sin^2C$$
$$=2\sin^2C-\sin^2B=1$$
Now as we already know that $\sin C=\frac{2\Delta}{ab}$ and similarly $\sin B=\frac{2\Delta}{ac}$ where $\Delta$ is the area of the triangle and $a,b,c$ the sides of the respective triangle.
But now when I put the sine formula in the equation and take $\Delta=27\text{cm}^2$ I get stuck with the product of $abc$ what should I do to get the sides.
| HINT:
$\tan A=1,$
If $\tan B=p,\tan C=\tan(135^\circ -B)=\dfrac{p+1}{p-1}$
We have $$\tan A+\tan C=2\tan B\implies1+\dfrac{p+1}{p-1}=2p$$
Can you take it from here using
$$27=\dfrac12bc\sin A\iff bc=54\sqrt2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1825559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Floor sum of reciprocal of square root of first $50$ numbers
Find Sum of $$\bigg\lfloor 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+\cdots+\frac{1}{\sqrt{50}}\bigg\rfloor$$
$\bf{My\; Try::}$ Let $\displaystyle y=f(x) = \frac{1}{\sqrt{x}}\;,$ Then draw that graph in coordinate axis, We get
$$\displaystyle \int_{1}^{51}\frac{1}{\sqrt{x}}dx<\sum^{50}_{k=1}\frac{1}{\sqrt{k}}<1+\int_{1}^{50}\frac{1}{\sqrt{x}}dx$$
So we get $$\displaystyle 2\left(\sqrt{51}-1\right)<\sum^{50}_{k=1}\frac{1}{\sqrt{k}}<1+2(\sqrt{50}-1)$$
So we get $$12.28<\sum^{50}_{k=1}\frac{1}{\sqrt{k}}<13.14$$
But answer given is $12,$ How cai i solve avobe question, Help Required, Thanks
| $$
\sum_{k=1}^{50} \frac 1 {\sqrt{k}} = \int_1^{51} \frac 1 {\sqrt{x}} \, dx + \text{a little bit more} = 2(\sqrt{51} - 1) + \text{a little bit more}.
$$
We just need to show that that "little bit" is small enough. A carefully drawn graph show show you why
$$
\text{that little bit} < \left( 1 - \frac 1 {\sqrt 2} \right) + \left( \frac 1 {\sqrt 2} - \frac 1 {\sqrt 3} \right) + \left( \frac 1 {\sqrt 3} - \frac 1 {\sqrt 4} \right) + \cdots + \left( \frac 1 {\sqrt{50}} - \frac 1 {\sqrt{51}} \right)
$$
and all of the terms cancel out except the first and the last, so
$$
\text{that little bit} < 1 - \frac 1 {\sqrt{51}}.
$$
That doesn't quite do it, so we refine the technique:
$$
\frac 1 {\sqrt 1} = \int_1^2 \frac{dx}{\sqrt x} + \text{a little bit} = 2(\sqrt 2 - 1) + \text{a little bit}.
$$
That little bit can be found numerically and it is less than $60\%$ of $1 - \dfrac 1 {\sqrt 2}$.
In all the latter terms, the $60\%$ would be replaced by something even smaller than $60\%$ (but always bigger than $50\%$ for reasons that should be obvious from looking at the graph).
Thus
$$
\text{little bit} < 0.6\times \left( 1 - \frac 1 {\sqrt 2} \right) + \left( \frac 1 {\sqrt 2} - \frac 1 {\sqrt 3} \right) + \left( \frac 1 {\sqrt 3} - \frac 1 {\sqrt 4} \right) + \cdots + \left( \frac 1 {\sqrt{50}} - \frac 1 {\sqrt{51}} \right)
$$
and that does it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1826197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Prove that for all odd integer $n, n^{4}=1\pmod {16}$ My answer: $n=2k+1$
$n^{4}=(2k+1)^{4}$=$16k^{4}+32k^{3}+8k^{2}+24k+1$.
I do not know how to conclude; really needed help here.
| Let $m=n-1$
$n^4 = m^4+4m^3+6m^2+4m^1+1$
Since $m$ is even, $m=2x$ for some $x\in\mathbb{N}$
$\implies n^4=2^4x^4+4\times2^3x^3+6\times2^2x^2+4\times2x+1$
$16x^4, 32x^3$ are both $0($mod $16)$, leaving $6\times2^2x^2+4\times2x+1$.
If $x$ is odd, $3x+1$ must be even, therefore $8x(3x+1)$ is $0($ mod $16)$.
If $x$ is even, $8x$ must be $0($ mod $16)$.
$\implies n^4=0($ mod $16)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1827165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Complex numbers inside determinant Let $ \begin{vmatrix}6\iota & -3\iota & 1\\ 4 & 3\iota & -1\\ 20 & 3 & \iota \\ \end{vmatrix}= x +\iota y$, then what are the values of $x$ and $y$?
| If you did not notice the linear dependency in the last two columns, then the "standard way" should have given you:
$$\begin{align}\begin{vmatrix}6\iota & -3\iota & 1\\ 4 & 3\iota & -1\\ 20 & 3 & \iota \\ \end{vmatrix}
=&~ 6\iota\begin{vmatrix}3\iota&-1\\3&\iota\end{vmatrix}
-3\iota\begin{vmatrix}-1&4\\\iota&20\end{vmatrix}
+1\begin{vmatrix}4&3\iota\\20 & 3\end{vmatrix}
\\[1ex] =&~ 6\iota(3\iota^2+3)-3\iota(-20-4\iota)+(12-60\iota)
\\[1ex] =&~ 18\iota^3+12\iota^2+18\iota+12\\[1ex] =&~ 0\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1829313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+...}}}}}$ $\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+...}}}}}$
My Attempt: I tried to use the regular way.
$A=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+...}}}}}$
$A^2=1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+...}}}}$
But then I saw that nothing happened twice and I stopped. Any hints?
It's better to don't use limits but if you used no problem. If there is an answer that doesn't use limits will accept.
update1: The question that look likes this wants to find the limit but I want a much easier way to solve it. But if there isn't any answer easier answer the limition one will accept.
update2: You solve the problem when you know the answer is 3 think that you don't know that the answer is 3 then solve.
| First you can guess that this seems to converge to $3$ by using a calculator (or by estimating the radicals). Assuming the limit is three, then one can try to wrtie $3$ as the following down:
$3=\sqrt{1+8}=\sqrt{1+2\cdot4}=\sqrt{1+2\sqrt{16}}=\sqrt{1+2\sqrt{1+15}}=\sqrt{1+2\sqrt{1+3*5}}=\sqrt{1+2\sqrt{1+3\sqrt{25}}}=\sqrt{1+2\sqrt{1+3\sqrt{1+4\cdot 6}}}=\ldots$
This iteration is the same as the formula you stated above, because it holds
$\sqrt{n^2}=\sqrt{1+(n-1)(n+1)}=\sqrt{1+(n-1)\sqrt{(n+1)^2}} \quad \forall n\in \mathbb{N}, n>1$
Thus, the limit is $3$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove inequality: $\frac{P+2004a}{P-2a}\cdot\frac{P+2004b}{P-2b}\cdot\frac{P+2004c}{P-2c}\ge2007^3.$
Let $a,b,c$ the sides of a triangle, $P$ its perimeter. Prove inequality:
$$\frac{P+2004a}{P-2a}\cdot\frac{P+2004b}{P-2b}\cdot\frac{P+2004c}{P-2c}\ge2007^3.$$
My attempt:
1) $P=a+b+c$. Then $\frac{P+2004a}{P-2a}=\frac{2005a+b+c}{b+c-a}$. Here $b+c-a>0 -$ triangle inequality.
2) $\sqrt[3]{xyz}\ge\frac{3}{\frac1x+\frac1y+\frac1z} \Rightarrow xyz\ge \left(\frac{3}{\frac1x+\frac1y+\frac1z}\right)^3$
| Let
$$b+c-a=s,\quad c+a-b=t,\quad a+b-c=u$$
Then, we can have
$$a=\frac{t+u}{2},\quad b=\frac{s+u}{2},\quad c=\frac{s+t}{2}$$
So, using AM-GM inequality and letting $d=1003$,
$$\begin{align}&\frac{P+2004a}{P-2a}\cdot\frac{P+2004b}{P-2b}\cdot\frac{P+2004c}{P-2c}\\\\&=\left(1+d\frac ts+d\frac us\right)\left(1+d\frac st+d\frac ut\right)\left(1+d\frac su+d\frac tu\right)\\\\&=2d^3+3d^2+1+(d^3+d^2+d)\left(\frac su+\frac st+\frac tu+\frac ts+\frac us+\frac ut\right)+d^2\left(\frac{s^2}{tu}+\frac{t^2}{su}+\frac{u^2}{st}\right)\\\\&\ge 2d^3+3d^2+1+6(d^3+d^2+d)\sqrt[6]{1}+3d^2\sqrt[3]{1}\\\\&=(2d+1)^3\\\\&=2007^3\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve without solving by inspection? $\frac{x+5}{x+k}=\frac{-kx+5}{x-1}$ Background: This is from a test review on functions. The original problem was
Find the value of $k$ so that the function $f(x) = \frac{x+5}{x+k}$ will be its own inverse.
I found the answer by inspection, and then tried to solve it through more rigorous means. Continuing from the question title, I cross multiplied and used the quadratic formula on $k.$
$x^2+4x-5=5x+5k-kx^2-k^2x$
$k^2x+k(x^2-5)+(x^2-x-5)=0$
$k = \frac{-x+5\pm\sqrt{x^4-10x^2+25-4x(x^2-x-5)}}{2x}$
$k = \frac{-x+5\pm\sqrt{x^4-4x^3-6x^2+20x+25}}{2x}$
This is where I got stuck. Any ideas?
| You are close but you went in the wrong direction. Rearrange equation 2 as follows
$x^2(k+1) + x(k^2-1) -5k - 5 = 0$
Now factor out an $k+1$ and you get $(k+1)(x^2+x(k-1) - 5) = 0$.
Now this equation must be zero for any $x$ you pick so no value of $k$ will do that except $k=-1$
| {
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"answer_id": 0
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Value of $\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\cdot \left(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\right)$
If $a+b+c=0,$ Then value of $\displaystyle \left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\cdot \left(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\right)$
$\bf{My\; Try::}$ Given $$\displaystyle \left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\cdot \left(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\right) = 1+1+1+\frac{b-c}{a}\left(\frac{b}{c-a}+\frac{c}{a-b}\right)+\frac{c-a}{b}\left(\frac{a}{b-c}+\frac{c}{a-b}\right)+\frac{a-b}{c}\left(\frac{a}{b-c}+\frac{b}{c-a}\right)$$
Now How can I Solve after that, Is there is any less complex method plz explain here , Thanks
| We have from GAVD's computation that
$$
\begin{align}
A &:= \left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\left(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\right)
\\
&= \left(\frac{a}{b-c} + \frac{b}{c-a} + \frac{c}{a-b}\right)\left(\frac{bc(b-c) + ca(c-a) + ab(a-b)}{abc}\right)
\\&=\left(\frac{a}{b-c} + \frac{b}{c-a} + \frac{c}{a-b}\right)\left(-\frac{(a-b)(b-c)(c-a)}{abc}\right)
\\
&= \frac{a(a-b)(a-c)+b(b-c)(c-a)+c(c-a)(c-b)}{abc}
\\
&=\frac{a^3+b^3+c^3-a^2(b+c)-b^2(c+a)-c^2(a+b)+3abc}{abc}\,.
\end{align}$$
Since $a+b+c=0$, we have
$$A=\frac{2\left(a^3+b^3+c^3\right)+3abc}{abc}=\frac{2\left(a^3+b^3+c^3-3abc\right)}{abc}+9\,.$$
As $a^3+b^3+c^3-3abc=(a+b+c)\left(a^2+b^2+c^2-bc-ca-ab\right)=0$, we obtain $A=9$ (given that $a,b,c\neq 0$ are pairwise distinct).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1837865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Find the value of $h$ if $x^2 + y^2 = h$
Consider equation $x^2 + y^2 = h$ that touches the line $y=3x+2$ at some point $P$. Find the value of $h$
I know that $x^2 + y^2 = h$ is a circle with radius $\sqrt{h}$.
Also, since $y = 3x + 2 $ is a tangent, we know that the slope of the radius perpendicular to the tangent is $M_{OP}= -\frac{1}{3}$. I'm not sure how I can determine the value of $h$ though?
| The slope of the perpendicular radius is $-\dfrac 13$ and it passes through the center of the circle, $(0,0)$. So its equation is
$y-0 = -\dfrac 13(x - 0)$, which simplifies to $y = -\dfrac 13x$. Avoiding fractions, we write this as $x = -3y$.
We compute the intersection
\begin{align}
x &= -3y \\
y &= 3x + 2 \\
\hline
y &= -9y + 2\\
y &= \dfrac 15 \\
x &= -\dfrac 35
\end{align}
So $h = x^2 + y^2 = \dfrac{10}{25} = \dfrac 25$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1837931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 10,
"answer_id": 9
} |
If $x,y,z\gt 0$ and $xyz=1$ Then minimum value of $\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}$
If $x,y,z\gt 0$ and $xyz=1$ Then find the minimum value of $\displaystyle \frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}$
$\bf{My\; Try::}$Using Titu's Lemma $$\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\ge \frac{(x+y+z)^2}{2(x+y+z)} = \frac{x+y+z}{2}\ge 3\frac{\sqrt[3]{xyz}}{2} = \frac{3}{2}$$
and equality holds when $$x=y=z=1$$
My question is how can we solve it without the above lemma, like using Jensen's Inequality or other inequality.
Please explain me.
Thanks
| Use $$\frac{a^2}{b}\ge 2a-b$$
Then
$$\frac{x^2}{y+z}=\frac14\cdot\frac{(2x)^2}{y+z}\ge\frac14(4x-y-z)$$
Similarly
$$\frac{y^2}{x+z}\ge\frac14(4y-x-z)$$
$$\frac{z^2}{x+y}\ge\frac14(4z-x-y)$$
Hence
$$\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\geq \frac{1}{4}2(x+y+z) = \frac{x+y+z}{2}\geq 3\frac{\sqrt[3]{xyz}}{2} = \frac{3}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1841084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.