Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
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.
| Using the characteristic equation method, we have the homogeneous part of the given equation,
$$g_n = g_{n-1} + 2g_{n-2}$$
As you have done, the roots of the characteristic equation are $2$ and $-1$, so the solution to the homogeneous part is $c_12^n + c_2(-1)^n$ for some constants $c_1$ and $c_2$. For the nonhomogeneous part, according to the comment by @AndreNicolas, we assume the solution is of the form $c_3n2^n$ and we can write:
$$c_3n2^n = c_3(n-1)2^{n-1} + 2c_3(n-2)2^{n-2} + 2^n \\
\implies c_3 = \frac{2}{3}$$
Note: We guess $c_3n2^n$ for the nonhomogeneous part, and not $c_32^n$, because $2$ is already a root of the characteristic equation of the homogeneous part. In the same way, when we have repeated roots for the homogeneous part (say the root $a$ appears thrice), we use $c_1a^n + c_2na^n + c_3n^2a^n$, etc.
Now,
$$\begin{align}
a_n &= h_n + n\frac{2 \cdot 2^n}{3} \\
&= c_12^n + c_2(-1)^n + \frac{2^{n+1}n}{3}
\end{align}$$
and by substituting for $a_0$ and $a_1$, we get $c_1 = \frac{5}{9}$, $c_2 = \frac{13}{9}$, and
$$a_n = \frac{5\cdot 2^n}{9} + \frac{13 \cdot (-1)^n}{9} + \frac{2^{n+1}n}{3}$$
| {
"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": 2
} |
Solve for $x$ if $4^{\frac{x}{y} + \frac{y}{x}}$ $=$ $32$ and $\log_3(x+y)+\log_3(x-y)=1$
Question:
Solve for $x$ if $4^{\frac{x}{y} + \frac{y}{x}}$ $= 32$ and $\log_3(x+y)+\log_3(x-y)=1$
My attempt: With the first equation
$$4^{\frac{x}{y} + \frac{y}{x}} = 32$$
$$2^{2(\frac{x}{y} + \frac{y}{x})} = 2^5$$
$$ 2(\frac{x}{y} + \frac{y}{x}) = 5 $$
$$ \frac{x}{y} + \frac{y}{x} = \frac{5}{2} $$
$$ \frac{x^2 + y^2}{xy} = \frac{5}{2} $$
Now with the second equation
$$\log_3(x+y)+\log_3(x-y)=1$$
$$\log_3((x+y)(x-y)) = 1 $$
$$ \log_3(x^2-y^2) = 1 $$
$$ x^2-y^2 = 3$$
Now I have 2 equations:
$$ \frac{x^2 + y^2}{xy} = \frac{5}{2} $$
$$ x^2-y^2 = 3$$
Now I am stuck..
|
Sorry- I am on my phone in a cafe and this napkin was the best I could do.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1684743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
How can we prove that a quadratic equation has at most 2 roots? A quad equation can be factored into two factors containing $x $, but how can we prove that there no other sets of different factors yielding OTHER VALUES OF $X $?
| Any quadratic equation has at most 2 roots because any polynomial equation $f(x)$ ($f(x)$ defines the polynomial) will have $\deg f(x)$ roots to it. Because of that, the degree of quadratics is 2, so there are a maximum of 2 roots in $\mathbb{C}$. To prove this, let $\alpha$, $\beta$, and $\gamma$ represent three roots of any quadratic equation in the form $ax^2 + bx + c = 0$. So, each $\alpha$, $\beta$, and $\gamma$ will satisfy said quadratic equation:
$$\begin{align}
a\alpha^2 + b\alpha + c & = 0 \tag 1 \label{1}\\
a\beta^2 + b\beta + c & = 0 \tag 2 \label{2} \\
a\gamma^2 + b\gamma + c & = 0 \tag 3 \label{3}
\end{align}
$$ If you subtract equation $\ref{2}$ from $\ref{1}$, you get
$$\begin{align}
a\alpha^2 + b\alpha + c - (a\beta^2 + b\beta + c) & = 0 \\
\implies a(\alpha^2 - \beta^2) + b(\alpha - \beta) & = 0 \\
a(\alpha - \beta)(\alpha + \beta) + b(\alpha - \beta) & = 0 \\
(\alpha - \beta)(a(\alpha + \beta) + b) & = 0 \\
a(\alpha + \beta) + b & = 0 && (\alpha - \beta \neq 0) \tag 4 \label{4}
\end{align}
$$ If you subtract equation $\ref{3}$ from $\ref{2}$, you get
$$\begin{align}
a\beta^2 + b\beta + c - (a\gamma^2 + b\gamma + c) & = 0 \\
\implies a(\beta^2 - \gamma^2) + b(\beta - \gamma) & = 0 \\
(\beta - \gamma)(a(\beta + \gamma) + b) & = 0 \\
a(\beta + \gamma) + b & = 0 && (\beta - \gamma \neq 0) \tag 5 \label{5}
\end{align}
$$ Now, when you subtract equation $\ref{5}$ from $\ref{4}$, you get
$$\begin{align}
a(\alpha - \gamma) & = 0 \\
\implies \alpha = \gamma
\end{align}
$$ This is an impossible situation because $\alpha$ and $\gamma$ are both distinct roots of the quadratic equation and $a \neq 0$. Therefore, there is a maximum of 2 solutions. Quod erat demonstrandum.
References:
*
*My question related to this: Why does the equation $\frac{-x^2 + 2x}{5x - 4} = 6$ have 2 solutions?
*Show that Quadratic Equations cannot have more than 2 Roots- AskMath
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1688685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Prove using the formal definition of a limit that How would I go about proving this limit?
$\lim_{x\to\infty}\frac{1}{x^4+x^2+5}=0$
so far I have:
$|f(x) -L|< ϵ$ wherever $x > N$
$|\frac{1}{x^4+x^2+5} - 0| < ϵ $ wherever $x > N$
$|\frac{1}{x^4+x^2+5}| < ϵ ,\; x > ∞$, assuming, $x > 0$
taking the absolute value $\frac{1}{x^4+x^2+5} < ϵ$
${x^4+x^2+5}$ > $\frac{1}{ϵ}$
i am not sure whether this is correct, but what would be the next step?
| HINT:
$$\frac{1}{x^4+x^2+5}\le \frac{1}{x^4}<\epsilon$$
whenever $x>B=\epsilon^{-1/4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1690758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
I am stuck on proving $\frac1{2!}+\frac2{3!}+\dots+\frac{n}{(n+1)!}=1-\frac1{(n+1)!}$ by induction, could anyone check my work? I will skip the Base Case step.
This is the questions.
Use mathematical induction to prove that$$\frac{1}{2!}+\frac{2}{3!}+\cdots+\frac{n}{(n+1)!}=1-\frac{1}{(n+1)!}$$for all integers $n\ge 1$.
This is my proof:
$$\sum_{i=1}^n \frac{i}{(i+1)!} = 1-\frac{1}{(n+1)!}$$
$$\sum_{i=1}^k \frac{i}{(i+1)!} = 1-\frac{1}{(k+1)!}$$
$$\sum_{i=1}^{k+1} \frac{i}{(i+1)!} = 1-\frac{1}{(k+2)!}$$
$$\sum_{i=1}^{k+1} \frac{i}{(i+1)!} = \sum_{i=1}^k \frac{i}{(i+1)!} + \frac{k+1}{(k+2)!}$$
$$\sum_{i=1}^{k+1} \frac{i}{(i+1)!} = 1-\frac{1}{(k+1)!} + \frac{k+1}{(k+2)!}$$
And this is where I am stuck, I don't know how to prove that: $$1-\frac{1}{(k+1)!} + \frac{k+1}{(k+2)!} = 1-\frac{1}{(k+2)!}$$
| Note that $(k + 2)! =(k + 2) \cdot (k + 1)! $. You then get for your last equation $$ 1 - \frac{1}{(k + 1)!} + \frac{k + 1}{(k + 2)!} = 1 - \frac{k + 2}{(k + 2)!} + \frac{k + 1}{(k + 2)!} = 1 + \frac{1 - 2 + k - k}{(k + 2)!} = 1 - \frac{1}{(k + 2)!}$$
which proves the statement.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1692644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
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?
| We have
$$
\sum_{k=1}^n k (k+1)
= 2 \sum_{k=1}^{n} \binom{k+1}{2}
= 2 \sum_{k=2}^{n+1} \binom{k}{2}
= 2 \binom{n+2}{3}
= \dfrac13 n (n+1) (n+2)
$$
Therefore
$$
\frac{1\cdot 2 + 2\cdot 3 + \cdots + n(n + 1)}{n(n + 3)}
= \frac{ n (n+1) (n+2)}{3n(n + 3)}
= (n+1)\frac{n+2}{3(n + 3)}
\ge (n+1) \frac{1}{4}
$$
because
$$
\frac{n+2}{3(n + 3)} \ge \dfrac14
$$
for all $n\ge 1$.
| {
"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": 1
} |
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 ?
| Problem
Detailed instructions for computing the SVD abound on Math Stack Exchange. For example: SVD and the columns, SVD -obligation of normalization, how SVD is calculated in reality, Pseudo-inverse of a matrix that is neither fat nor tall?.
This problem is straightforward. Let the target matrix be the covector
$$
\mathbf{A} = \left[ \begin{array}{ccr}
2 & 1 & -2
\end{array} \right]
$$
Find the singular value decomposition
$$
\mathbf{A} = \mathbf{U} \, \Sigma \, \mathbf{V}^{*}
$$
Methods
The first link presents a table showing two paths for finding the SVD for a matrix $\mathbf{A} \in \mathbb{C}^{m \times n}_{\rho}$:
$$
\begin{array}{lll}
%
\text{Operation} &
\text{Row space first} & \text{Column space first} \\\hline
%
\text{1. Construct product matrix} &
\mathbf{W} = \mathbf{A}^{*} \mathbf{A} &
\mathbf{W} = \mathbf{A} \, \mathbf{A}^{*} \\
%
\text{2. Solve for eigenvalues} &
\sigma = \tilde{\lambda} \left( \mathbf{W} \right) &
\sigma = \tilde{\lambda} \left( \mathbf{W} \right) \\
%
\color{blue}{\text{3. Solve for eigenvectors }} w_{k},\ k=1, \rho &
\left( \mathbf{W} - \lambda_{k} \mathbf{I}_{n} \right) w_{k} = \mathbf{0} &
\left( \mathbf{W} - \lambda_{k} \mathbf{I}_{m} \right) w_{k} = \mathbf{0} \\
%
\text{4. Assemble domain matrix} &
\mathbf{V}_{k} = \frac{w_{k}}{\lVert w_{k} \rVert_{2}} &
\mathbf{U}_{k} = \frac{w_{k}}{\lVert w_{k} \rVert_{2}} &
\\
%
\text{5. Compute complementary domain matrix} &
\mathbf{U}_{k} = \sigma_{k}^{-1} \mathbf{A} \mathbf{V}_{k} &
\mathbf{V}_{k} = \sigma_{k}^{-1} \mathbf{A}^{*} \mathbf{U}_{k} &
\\
%
\end{array}
$$
The two product matrices to wrok with are
$$
\mathbf{A} \mathbf{A}^{*} =
\left[
\begin{array}{c}
9 \\
\end{array}
\right]
\qquad
\mathbf{A}^{*} \mathbf{A} =
\left[
\begin{array}{rrr}
4 & 2 & -4 \\
2 & 1 & -2 \\
-4 & -2 & 4 \\
\end{array}
\right].
$$
Solution
The product matrix $\mathbf{A} \mathbf{A}^{*}$ is much easier to work with!
Singular values
The eigenvalue is
$$
\lambda \left( \mathbf{A}\mathbf{A}^{*} \right) = 9
$$
which implies the singular value is
$$
\sigma = \sqrt{\lambda \left( \mathbf{A}\mathbf{A}^{*} \right)} = 3
$$
The matrix of singular values is
$$
\mathbf{S} =
\left[
\begin{array}{c}
3
\end{array}
\right]
$$
There is no need for $0$ padding, so
$$
\Sigma = \mathbf{S}
$$
Matrix $\mathbf{U}$
The eigenvector is already normalized and represents the first column
$$
\mathbf{U} =
\color{blue}{\left[
\begin{array}{c}
1
\end{array}
\right]}
$$
Coloring distinguishes $\color{blue}{range}$ space vectors from $\color{red}{null}$ space vectors.
Column space
$$
\begin{align}
\color{blue}{\mathcal{R}\left(\mathbf{A}\right)}
&=
\text{span} \left\{ \,
\color{blue}{\left[
\begin{array}{c}
1
\end{array}
\right]}
\, \right\} \\
%
\color{red}{\mathcal{N}\left(\mathbf{A}^{*}\right)} &=
\left\{ 0 \right\}
\end{align}
$$
Matrix $\mathbf{V}$
$$
\mathbf{V}_{1}
= \frac{1}{\sigma} \mathbf{A}^{*} \mathbf{U}_{k}
= \frac{1}{3}
\color{blue}{\left[
\begin{array}{r}
2 \\ 1 \\ -2
\end{array}
\right]}
$$
For the $\color{red}{null}$ space vectors, we have options. We could call upon the process of Gram and Schmidt. Or we could eyeball the problem to find any $\color{red}{null}$ space vector, and use the cross product to find the missing vector.
Choosing the latter, we select
$$
\color{red}{\mathbf{V}_{2}} = \frac{1}{\sqrt{2}}
\color{red}{\left[
\begin{array}{r}
1 \\ 0 \\ 1
\end{array}
\right]}
$$
The missing vector is then
$$
\color{blue}{\left[
\begin{array}{r}
2 \\ 1 \\ -2
\end{array}
\right]}
\times
\color{red}{\left[
\begin{array}{r}
1 \\ 0 \\ 1
\end{array}
\right]}
=
\color{red}{\left[
\begin{array}{r}
1 \\ -4 \\ -1
\end{array}
\right]}
$$
Final assembly
$$
\begin{align}
\mathbf{A} &= \mathbf{U} \, \Sigma \mathbf{V}^{*}
% U
\\ &=
\color{blue}{\left[
\begin{array}{c}
1
\end{array}
\right]}
% Sigma
\left[
\begin{array}{c}
3
\end{array}
\right]
% V*
\left[
\begin{array}{ccc}
% c1
\frac{1}{3}
\color{blue}{\left[
\begin{array}{r}
2 \\ 1 \\ -2
\end{array}
\right]}
% c2
\frac{1}{\sqrt{2}}
\color{red}{\left[
\begin{array}{r}
1 \\ 0 \\ 1
\end{array}
\right]}
% c3
\frac{1}{3\sqrt{2}}
\color{red}{\left[
\begin{array}{r}
1 \\ -4 \\ -1
\end{array}
\right]}
\end{array}
\right]
%
\\ &=
\left[ \begin{array}{ccr}
2 & 1 & -2
\end{array} \right]
%
\end{align}
$$
| {
"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": 1
} |
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.
| Cool problem. Looks like some kind of challenge or tournament problem for junior high students.
MXYMXY gave the hint. $OF$ passes through the center of the circle for maximum length.
Let $C$ be the center of the circle. $\angle DCE = 2\times\angle DOE = 90^\circ$ (inscribed angle is half central angle)
Let $X$ be the intersection of $DE$ and $OF$.
$\angle DXO$ is a right angle.
$\triangle DCE$ is an isosceles right triangle. $DC = \dfrac{\sqrt{2}}{2}$ (this is the radius of the circle)
$\angle DCX = 45^\circ$
$CX = DC \cos(45^\circ) = \dfrac{\sqrt{2}}{2}\times\dfrac{\sqrt{2}}{2}=\dfrac{1}{2}$
$XF = 1\times \sin(60) = \dfrac{\sqrt{3}}{2}$
$OF = OC+CX+XF =\dfrac{\sqrt{2}}{2}+\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}$
$\Rightarrow OF = \dfrac{1+\sqrt{2}+\sqrt{3}}{2}$
| {
"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": 3
} |
Prove that $n^{n+1} \leq (n+1)^{n} \sqrt[n]{n!}$ Let $n$ be a positive integer. I conjectured that the following inequality is true
\begin{equation}
n^{n+1} \leq (n+1)^{n} \sqrt[n]{n!} .
\end{equation}
Anyhow I could neither prove nor disprove it. I could only check, by using Stirling's Formula, that the ratio of the right and left members tends to 1 as $n \rightarrow \infty$. Any help is welcome.
| From $$
e^{-1} = \left(e^{-\frac{1}{n+1}}\right)^{n+1} > \left(1-\frac{1}{n+1}\right)^{n+1}(\because e^x \ge x+1)
$$
and
$$
\left(1-\frac{1}{n+1}\right)^{n+1} \cdot \left(1 + \frac{1}{n}\right)^{n+1} = 1
$$
Note
$$e \le (1+\frac{1}{n})^{n+1}=\frac{(1+n)^{n+1}}{n^{n+1}} \Leftrightarrow n^{n+1} \le \frac{(n+1)^{n+1}}{e} $$
Also note
$$n!e^n=\sum_{k=0}^{\infty}{n^k\frac{n!}{k!}} \ge \sum_{k=0}^{n}{n^k\frac{n!}{k!}} \ge \sum_{k=0}^{n}{n^k\binom{n}{k}}= ({n+1})^n \Leftrightarrow \sqrt[n]{n!} \ge \frac{n+1}{e} $$
This gives us that $$(n+1)^{n} \sqrt[n]{n!} \ge \frac{(n+1)^{n+1}}{e} \ge n^{n+1}$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1695669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 4,
"answer_id": 0
} |
How to solve similar with Transformation of a quadratic form into diagonal form? Define $\color{red}{f=f(x),f'=f'(x)}$,where the derivative with respect to $x$ of a function $f(x)$ is denoted $ f'(x)$.
Now give six postive numbers $k_{1},k_{2},k_{3},k_{4},k_{5},k_{6}$, and a quadratic form $$\color{red}{k_{1}f^2_{1}+k_{2}f^2_{2}+k_{3}f_{1}f_{2}+k_{4}f'^2_{1}+k_{5}f'^2_{2}+k_{6}f'_{1}f'_{2}}\tag{1}$$ How to Transformation of a quadratic form into diagonal form: $$A_{1}f^2_{1}+A_{2}f^2_{2}+A_{3}f'^2_{1}+A_{4}f'^2_{2}$$
For example :such
$$f^2_{1}+\dfrac{1}{2}f^2_{2}+\dfrac{1}{2}f'^2_{1}+\dfrac{3}{8}f'^2_{2}+\dfrac{1}{2}f'_{1}f'_{2}\tag{2}$$
because we have
$$f^2_{1}+\dfrac{1}{2}f^2_{2}+\dfrac{1}{2}f'^2_{1}+\dfrac{3}{8}f'^2_{2}+\dfrac{1}{2}f'_{1}f'_{2}=\dfrac{1}{3}(f_{1}+f_{2})^2+\dfrac{1}{3}(f'_{1}+f'_{2})^2+\dfrac{1}{6}(2f_{1}-f_{2})^2+\dfrac{1}{24}(2f'_{1}-f'_{2})^2\tag{3}$$
we Transformation
$$ f_{1}+f_{2}=g_{1}, 2f_{1}-f_{2}=g_{2}$$
then $(3)$ diagonal form
$$\dfrac{1}{3}g^2_{1}+\dfrac{1}{3}g'^2_{1}+\dfrac{1}{6}g^2_{2}+\dfrac{1}{24}g'^2_{2}$$
Unfortunately I don't know any idea expression for this $ (1)$. which could help. I'll be grateful for all useful suggestions.
| You don't use a similarity transformation, the relationship for quadratic forms is called congruence. In matrices, you are solving $P^T M P = D$ diagonal, with $\det P = 1.$ Although it would be possible to force $P$ orthogonal in this instance, there is little advantage.
Very hard to read
$$\color{red}{k_{1}f^2_{1}+k_{2}f^2_{2}+k_{3}f_{1}f_{2}+k_{4}f'^2_{1}+k_{5}f'^2_{2}+k_{6}f'_{1}f'_{2}}\tag{1}$$
$$\color{blue}{k_{1}u^2+k_{2}v^2+k_{3}uv+k_{4}x^2+k_{5}y^2+k_{6}xy }$$
All you need to do is separately complete the squares for the two binary forms.
$$ 4 k_1 (k_{1}u^2+k_{2}v^2+k_{3}uv) = 4 k_1^2 u^2 + 4 k_1 k_3 uv + 4 k_1 k_2 v^2. $$
Compare
$$ (2k_1u + k_3v)^2 = 4 k_1^2 u^2 + 4 k_1 k_3 uv + k_3^2 v^2. $$
$$ 4 k_1 (k_{1}u^2+k_{2}v^2+k_{3}uv) = (2k_1u + k_3v)^2 + (4 k_1 k_2 - k_3^2)v^2 $$
$$ k_{1}u^2+k_{2}v^2+k_{3}uv = \left(\frac{1}{ 4 k_1} \right) (2k_1u + k_3v)^2 + \left(\frac{4 k_1 k_2 - k_3^2}{ 4 k_1} \right) v^2 $$
$$ k_{4}x^2+k_{5}y^2+k_{6}xy = \left(\frac{1}{ 4 k_4} \right) (2k_4x + k_6y)^2 + \left(\frac{4 k_4 k_2 - k_6^2}{ 4 k_4} \right) y^2 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1697227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
prove this inequality $a^n>b^n+c^n$ We know a,b and c are positive and
$a^2=b^2+c^2$
How we can conclude this inequality:
$a^n>b^n+c^n$ , $n>2$
I tried Binomial Theorem but I can't prove this.
Thanks
| If $n=2m$ is even, then this is straightforward:
$$ a^n=(a^2)^m=(b^2+c^2)^m> b^{2m}+c^{2m}=b^n+c^n$$
If $n>2$ is odd, then since $n-1$ is even it follows that
$$ a^{n}=a\cdot a^{n-1}\geq a(b^{n-1}+c^{n-1})$$
using either equality if $n-1=2$, or what we have shown above if $n-1>2$.
Finally, $a^2=b^2+c^2$ with $a,b,c>0$ implies that $a>b$ and $a>c$, hence
$$ a(b^{n-1}+c^{n-1})>b\cdot b^{n-1}+c\cdot c^{n-1}=b^n+c^n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1697689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
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?
| You have some confusion in the use of the constant of integration.
1) Wan you integrate $$\frac{dy}{y} = \frac{xdx}{1+x^2}$$ you cannot write the constant $K$ on both sides ($A+K= B+K$ is the same as $A=B$ without $K$), so this step is:
$$\ln |y| = \ln|\sqrt{1+x^2}| + K$$
2) exponentiating this equation you have:
$$\ y = \exp(\ln|\sqrt{1+x^2}| + K)=\exp(\ln|\sqrt{1+x^2}|)\times \exp(K)=$$
$$
=C\exp(\ln|\sqrt{1+x^2}|)
$$
where $C=\exp(K)$ is a new constant.
3) from the initial condition you can find
$$
3=C\exp(\ln|\sqrt{1+0}|)=C
$$
| {
"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": 1
} |
$(x+1)^2 + (y+1)^2 + xy(x+y+3)=2$ I've came across this problem some hours ago and, although it looks (and possibly is) just some algebra calculus, I can't get on the right track.
Find $x$, $y$ integers such that
$$ (x+1)^2 + (y+1)^2 + xy(x+y+3)=2 $$
Some basic algebra will lead to:
$$
(x+y)^2 + x(xy+2) + y(xy+2) + xy = 0$$
$$(x+y)(xy+x+y+2)=-xy$$
Now, some divisibility properties should end the problem.
Well, maybe my approach isn't the right one, but I'm just stuck at this point.
A piece of advice or a hint would be apreciated.
| Replace $x+1$ by $a$ and $y+1$ by $b$ and you get
$$2=a^2+b^2 +(a-1)(b-1)(a+b+1)+1=a^2b+b^2a-ab =ab(a+b-1)+1.$$
Since $x$ and $y$ are integers, $a,b$ are integers too. Therefore $a,b,a+b-1 \in \{-1,1\}$ and so $a,b \in \{-1,1\}$ and $a+b \in \{0,2\}$. If $a=b=-1$ we get a contradiction. Since the starting equation was symmetric in $a$ and $b$, lets assume $a=1$. Then $b=1$ and $b=-1$ are still valid.
So the only solutions for $(x,y)$ are $(0,0), (2,0), (0,2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1702123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
computing the value of $\sin^2(\frac{\pi}{10}$) The question asks me to prove the following formula: $$\cos5\theta=\cos\theta(16\sin^4\theta-12\sin^2\theta+1)$$ which is pretty straightforward to do using De Moivre's theorem. They further ask me to show that the exact value for $\sin^2\frac{\pi}{10}=\frac{3-\sqrt{5}}{8}$.
I proceed by simply solving the quadratic equation $$16\sin^4\frac{\pi}{10}-12\sin^2\frac{\pi}{10}+1=0$$ However, this yields two answers, namely $\frac{3\pm \sqrt{5}}{8}$. I don't know how to eliminate the answer where it's a plus and the justification from the book is
Since $\sin^2(\frac{\pi}{10})\lt \sin^2(\frac{3\pi}{10})$
I don't understand how that relates to the problem.
| The given identity states $$16 \sin^4 \theta - 12 \sin^2 \theta + 1 = \frac{\cos 5\theta}{\cos \theta}.$$ By substituting $\theta = \pi/10$, we get the first quadratic (i.e., $\sin^2 \theta$ is a root of the quadratic $16z^2 - 12z + 1$), since the RHS is zero as a result of $5\theta = \pi/2$. But we can also note that $\theta = 3\pi/10$ implies $$\cos 5\theta = \cos \frac{3\pi}{2} = 0,$$ so the same quadratic has $\sin^2 \frac{3\pi}{10}$ as a root. Now it becomes a simple matter to decide which root is which.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1707571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How does the determinant link to the cross product For a $2\times 2$ matrix $$\begin{pmatrix}a&b\\c&d \end{pmatrix}
$$
The determinant is given by $ad-bc$. And the cross product of $$\begin{pmatrix} a\\b\\0\end{pmatrix}\times \begin{pmatrix} c\\d\\0\end{pmatrix} =\begin{pmatrix} 0\\0\\ad-bc\end{pmatrix}$$
We can also note that $|a\times b|=|a|b|\sin\theta$ where $\theta$ is the angle between $a$ and $b$. Hence is there any way to relate the determinant to the equation with sine? I recently saw that $$\text{Re}(a)\text{Im}(b)-\text{Im}(a)\text{Re}(b)=|ab|\sin{\text{arg}(a/b)}$$
How would one verify/prove this?
| You can actually define the cross product of two vectors $\mathbf{a}, \mathbf{b} \in \mathbb{R}^3$ to the be unique vector $\mathbf{a} \times \mathbf{b} \in \mathbb{R}^3$ such that
$$
\forall \mathbf{c} \in \mathbb{R}^3, \quad (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c} = \det(\mathbf{a}\,\vert\,\mathbf{b}\,\vert\,\mathbf{c}),
$$
where $(\mathbf{a}\,\vert\,\mathbf{b}\,\vert\,\mathbf{c})$ denotes the $3 \times 3$ matrix whose columns are $\mathbf{a},\mathbf{b},\mathbf{c}$ in that order. In particular, you can recover $\mathbf{a} \times \mathbf{b}$ as
$$
\mathbf{a} \times \mathbf{b} = \det(\mathbf{a}\,\vert\,\mathbf{b}\,\vert\,\mathbf{i})\mathbf{i} + \det(\mathbf{a}\,\vert\,\mathbf{b}\,\vert\,\mathbf{j})\mathbf{j} + \det(\mathbf{a}\,\vert\,\mathbf{b}\,\vert\,\mathbf{k})\mathbf{k},
$$
which can be massaged using determinant identities to give you the usual ghastly explicit formula; in the special case that $\mathbf{a}$ and $\mathbf{b}$ lie in the $xy$-plane, you immediately recover your observation above. Moreover, it immediately follows that $\mathbf{a} \times \mathbf{b}$ is perpendicular to $\mathbf{a}$, $\mathbf{b}$, and any linear combination of $\mathbf{a}$ and $\mathbf{b}$, since by basic determinant identities, including the fact that a square matrix with repeated columns has a vanishing determinant,
$$
(\mathbf{a} \times \mathbf{b}) \cdot (s\mathbf{a}+t\mathbf{b}) = \det(\mathbf{a}\,\vert\,\mathbf{b}\,\vert\,s\mathbf{a}+t\mathbf{b}) = s\det(\mathbf{a}\,\vert\,\mathbf{b}\,\vert\,\mathbf{a}) + t\det(\mathbf{a}\,\vert\,\mathbf{b}\,\vert\,\mathbf{b}) = 0.
$$
Anyhow, the point of all this you're comfortable with basic linear algebra, especially with how the determinant behaves under elementary row and column operations, then you can derive the identity
$$
\|\mathbf{a} \times \mathbf{b}\| = \|\mathbf{a}\|\,\|\mathbf{b}\|\sin\theta_{\mathbf{a},\mathbf{b}}
$$
from the identity
$$
\forall \mathbf{c} \in \mathbb{R}^3, \quad (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c} = \det(\mathbf{a}\,\vert\,\mathbf{b}\,\vert\,\mathbf{c}),
$$
without too much trouble.
For simplicity, let's assume that $\mathbf{a} \neq \mathbf{0}$ and $\mathbf{b} \neq \mathbf{0}$; otherwise the claim is trivial. Actually, let's show that for any $\mathbf{c} \in \operatorname{Span}\{\mathbf{a},\mathbf{b}\}^\perp$, i.e., for any $\mathbf{c}$ perpendicular to both $\mathbf{a}$ and $\mathbf{b}$, that
$$
\left\lvert(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}\right\rvert = \left(\|\mathbf{a}\|\,\|\mathbf{b}\|\sin(\theta_{\mathbf{a},\mathbf{b}})\right)\|\mathbf{c}\|.
$$
If $\mathbf{a} \times \mathbf{b} \neq \mathbf{0}$, then we can plug in $\mathbf{c} = \mathbf{a} \times \mathbf{b}$ to get
$$
\|\mathbf{a} \times \mathbf{b}\|^2 = \left\lvert(\mathbf{a} \times \mathbf{b}) \cdot (\mathbf{a} \times \mathbf{b})\right\rvert = \left(\|\mathbf{a}\|\,\|\mathbf{b}\|\sin(\theta_{\mathbf{a},\mathbf{b}})\right)\|\mathbf{a}\times\mathbf{b}\|,
$$
and hence
$$
\|\mathbf{a} \times \mathbf{b}\| = \|\mathbf{a}\|\,\|\mathbf{b}\|\sin\theta_{\mathbf{a},\mathbf{b}}
$$
If $\mathbf{a} \times \mathbf{b} = \mathbf{0}$, then since $\operatorname{Span}\{\mathbf{a},\mathbf{b}\}^\perp$ is at least $1$-dimensional, take any non-zero vector $\mathbf{c} \in \operatorname{Span}\{\mathbf{a},\mathbf{b}\}^\perp$ to get
$$
0 = \left\lvert(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}\right\rvert = \left(\|\mathbf{a}\|\,\|\mathbf{b}\|\sin(\theta_{\mathbf{a},\mathbf{b}})\right)\|\mathbf{c}\|,
$$
which yields $\sin(\theta_{\mathbf{a},\mathbf{b}}) = 0$ and hence
$$
\|\mathbf{a} \times \mathbf{b}\| = 0 = \|\mathbf{a}\|\,\|\mathbf{b}\|\sin\theta_{\mathbf{a},\mathbf{b}}.
$$
First, by the defining identity for cross products,
$$
(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c} = \det(\mathbf{a}\,\vert\,\mathbf{b}\,\vert\,\mathbf{c}).
$$
Next, since determinants are preserved under column additions (e.g., $\det(\mathbf{a}\,\vert\,\mathbf{b}\,\vert\,\mathbf{c}) = \det(\mathbf{a}\,\vert\,\mathbf{b}+s\mathbf{a}\,\vert\,\mathbf{c})$), we have that
$$
\det(\mathbf{a}\,\vert\,\mathbf{b}\,\vert\,\mathbf{c}) = \det(\mathbf{a}\,\vert\,\mathbf{b}^\prime\,\vert\,\mathbf{c}),
$$
where
$$
\mathbf{b}^\prime := \mathbf{b} - \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\|^2}\mathbf{a}
$$
is the orthogonal projection of $\mathbf{b}$ onto $\operatorname{Span}\{\mathbf{a}\}^\perp$, i.e., onto the plane through the origin with normal vector $\mathbf{a}$; geometrically, if you believe that $\det(\mathbf{a}\,\vert\,\mathbf{b}\,\vert\,\mathbf{c})$ is the signed volume of the parallelepiped spanned by $\mathbf{a},\mathbf{b},\mathbf{c}$, then we're essentially saying that the parallelpiped spanned by $\mathbf{a},\mathbf{b},\mathbf{c}$ has the same volume as the paralleliped spanned by spanned by $\mathbf{a},\mathbf{b}^\prime,\mathbf{c}$ by Cavalieri's principle. Observe, in particular, that $\mathbf{a}$, $\mathbf{b}^\prime$, and $\mathbf{c}$ are pairwise orthogonal by construction.
Next, since $\mathbf{a}$, $\mathbf{b}^\prime$, and $\mathbf{a} \times \mathbf{b}$ are pairwise orthogonal,
$$
\lvert\det(\mathbf{a}\,\vert\,\mathbf{b}^\prime\,\vert\,\mathbf{c})\rvert = \sqrt{\det(\mathbf{a}\,\vert\,\mathbf{b}^\prime\,\vert\,\mathbf{c})^2}\\ = \sqrt{\det\left((\mathbf{a}\,\vert\,\mathbf{b}^\prime\,\vert\,\mathbf{c})^T\right) \det(\mathbf{a}\,\vert\,\mathbf{b}^\prime\,\vert\,\mathbf{c})}\\
= \sqrt{\det\left((\mathbf{a}\,\vert\,\mathbf{b}^\prime\,\vert\,\mathbf{c})^T(\mathbf{a}\,\vert\,\mathbf{b}^\prime\,\vert\,\mathbf{c}) \right)}\\
= \begin{vmatrix}\|\mathbf{a}\|^2&0&0\\0&\|\mathbf{b}^\prime\|^2&0\\0&0&\|\mathbf{c}\|^2\end{vmatrix}^{1/2}\\
= \|\mathbf{a}\|\,\|\mathbf{b}^\prime\|\,\|\mathbf{c}\|.
$$
At last, since the angle $\theta_{\mathbf{a},\mathbf{b}} \in [0,\pi]$ between the non-zero vectors $\mathbf{a},\mathbf{b}$ is given by the formula
$$
\cos \theta_{\mathbf{a},\mathbf{b}} = \frac{\mathbf{a}\cdot\mathbf{b}}{\|\mathbf{a}\|\,\|\mathbf{b}\|}
$$
it follows that
$$
\|\mathbf{b}^\prime\|^2 = \left(\mathbf{b} - \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\|^2}\mathbf{a}\right) \cdot \left(\mathbf{b} - \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\|^2}\mathbf{a}\right)\\
=\|\mathbf{b}\|^2 - 2 \mathbf{b} \cdot \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\|^2}\mathbf{a} + \left\|\frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\|^2}\mathbf{a}\right\|^2\\
=\|\mathbf{b}\|^2 - \frac{(\mathbf{a} \cdot \mathbf{b})^2}{\|\mathbf{a}\|^2}\\
=\|\mathbf{b}\|^2\left(1 - \left(\frac{\mathbf{a}\cdot\mathbf{b}}{\|\mathbf{a}\|\,\|\mathbf{b}\|}\right)^2\right)\\
=\|\mathbf{b}\|^2(1-\cos^2\theta_{\mathbf{a},\mathbf{b}})\\
=\|\mathbf{b}\|^2\sin^2\theta_{\mathbf{a},\mathbf{b}},
$$
and hence that
$$
\left\lvert(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}\right\rvert = \|\mathbf{a}\|\,\|\mathbf{b}^\prime\|\,\|\mathbf{c}\| = \|\mathbf{a}\|\,\|\mathbf{b}\|\sin\left(\theta_{\mathbf{a},\mathbf{b}}\right)\|\mathbf{c}\|,
$$
as was claimed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1707747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Bijection from $[-1,1]\times[-1,1] \rightarrow \{(x,y) \in \mathbb{R}: \sqrt{x^2+y^2} \leq 1\}$ I am trying to find and prove a bijection from the square $[-1,1]\times[-1,1]$ to the unit circle. Given a point $(x,y)$ in the square, my function maps it to the point $(r, \theta)$ in the circle where
$r = \frac{\sqrt{x^2+y^2}}{\sqrt{1+(y/x)^2}}$
and
$\theta = \begin{cases} \arctan(\frac{y}{x}) &\mbox{if } x > 0 \\
\arctan(\frac{y}{x})+\pi & \mbox{if } x < 0 \\
\pi/2 & \mbox{if } x = 0\ \mbox{and}\ y > 0\\
-\pi/2 & \mbox{if } x = 0\ \mbox{and}\ y < 0 \end{cases}$
In other words, the angle of the point on the circle is the same as the 'angle' of the point on the square, and the radius is the 'radius' of the point on the square, scaled down to fit inside the circle.
I'm trying to prove the injection, which means I need to prove that
$f(a, b) = f(c, d) \implies a = c\ \mbox{and}\ b = d$
or
$\frac{\sqrt{a^2+b^2}}{\sqrt{1+(b/a)^2}} = \frac{\sqrt{c^2+d^2}}{\sqrt{1+(d/c)^2}}\ \mbox{and}\ \arctan(b/a) = \arctan(d/c) \implies a = c\ \mbox{and}\ b = d$
How can I do this?
Edit: I realize that my formula for $r$ is slightly incorrect, because when $|y| > |x|$ then the denominator is $\sqrt{1+(x/y)^2}$.
| A different approach could be to stay in Cartesian coordinates by defining
$$
f(x,y)=\left(\frac x{\sqrt{1+(y/x)^2}},\frac y{\sqrt{1+(y/x)^2}}\right)
$$
when we are in the triangular region $0\leq y < x \leq 1$ which bijectively maps $1/8$ of the square to $1/8$ of the circle, defining $f(0,0)=(0,0)$, and then extending to the rest of the shapes by re-using the same idea as for this triangular region. Then we only need to prove bijectivity for this region.
So suppose $f(a,b)=f(c,d)\neq(0,0)$. Then we have the system of equations:
$$
\begin{align}
\frac a{\sqrt{1+(b/a)^2}}&=\frac c{\sqrt{1+(d/c)^2}}\\
&\text{and}\\
\frac b{\sqrt{1+(b/a)^2}}&=\frac d{\sqrt{1+(d/c)^2}}
\end{align}
$$
Squaring all of these expressions and simplifying we see that
$$
\begin{align}
\frac {a^4}{a^2+b^2}&=\frac {c^4}{c^2+d^2}\\
&\text{and}\\
\frac {a^2b^2}{a^2+b^2}&=\frac {c^2d^2}{c^2+d^2}\\
\end{align}
$$
From the second equation we have $b=0\iff d=0$ since we have assumed $a,c>0$. In that case the first equation implies $a=c$.
If $b,d>0$ we see that
$$
\frac{a^4}{c^4}=\frac{a^2b^2}{c^2d^2}\implies\frac ac=\frac bd \iff\frac ab=\frac cd
$$
So we must have $(c,d)=(ka,kb)$ for some $k>0$. Thus
$$
\begin{align}
\frac a{\sqrt{1+(b/a)^2}}&=\frac c{\sqrt{1+(d/c)^2}}\\
&=\frac{ka}{\sqrt{1+(kb/(ka))^2}}\\
&=\frac{ka}{k\sqrt{1/k^2+(b/a)^2}}\\
&=\frac{a}{\sqrt{1/k^2+(b/a)^2}}
\end{align}
$$
implying $k^2=1$ so $k=1$ and therefore $(c,d)=(a,b)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1710907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
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.
| Recall that $\cos(a + b) = \cos a \cos b - \sin a \sin b$
\begin{align}
\cos(\frac{\pi}{3} + \theta) &= \cos \frac{\pi}{3}\cos\theta - \sin \frac{\pi}{3}\sin\theta\\
&= \frac{\sqrt{3}\sin \theta - \cos\theta}{2}\\
\cos(\frac{2\pi}{3} + \theta) &= \cos \frac{2\pi}{3}\cos\theta - \sin \frac{2\pi}{3}\sin\theta\\
&= \frac{\sqrt{3}\sin \theta + \cos\theta}{2}
\end{align}
Remember that $\cos 2\theta = 2\cos^2 \theta- 1$. Now we use the previous results
\begin{align}
\text{L.S} &= \cos 2\theta + \cos(2(\frac{\pi}{3} + \theta)) + \cos(2(\frac{2\pi}{3} + \theta))\\
&= 2\cos^2\theta-1 +2\cos^2(\frac{\pi}{3} + \theta) - 1 + 2\cos^2(\frac{2\pi}{3} + \theta) - 1\\
&= 2\cos^2\theta + 2\bigg(\frac{(\sqrt{3}\sin \theta - \cos\theta)^2}{4} + \frac{(\sqrt{3}\sin \theta + \cos\theta)^2}{4}\bigg) - 3
\end{align}
Recall that $(a + b)^2 + (a - b)^2 = 2a^2 + 2b^2$
\begin{align}
\text{L.S} &= 2\cos^2\theta + 2(\frac{6\sin^2 \theta + 2\cos^2 \theta}{4}) - 3\\
&= 2\cos^2\theta + 3\sin^2\theta + \cos^2\theta - 3\\
&= 3(\sin^2\theta + \cos^2\theta) - 3\\
&= 3 - 3 = 0 = \text{R.S}
\end{align}
Thus concludes the proof. The perfect square pattern saves a lot of work. This method also does not require the sum to product formulas.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1712479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Can you prove that $\frac{1}{n}+\frac{1}{n^2}+\frac{1}{n^3}.... = \frac{1}{n-1}$? I'm a student and, while playing with my calculator, find out that:
$$\frac{1}{n}+\frac{1}{n^2}+\frac{1}{n^3}.... = \frac{1}{n-1}$$
is it a well defined equivalence and what is its name, is there a proof for that?
if we put it this way:
$$1+\frac{1}{n}+\frac{1}{n^2}+\frac{1}{n^3}.... = \frac{n}{n-1}$$
what do you call the last term (the sum), the complementary inverse or reciprocal of 1/n?
| assume $n>1$. Then
$$\sum_{k=0}^{K} \left(\frac{1}{n}\right)^k=1+\frac{1}{n}+...+\frac{1}{n^{K}}=L$$
Notice however, that $\frac{1}{n}\cdot L=\frac{1}{n}+..+\frac{1}{n^{K+1}}$
And then, that when you subtract them, all but the first and last term cancel: $$L-\frac{1}{n}\cdot L=1-\frac{1}{n^{K+1}}\implies L(1-\frac{1}{n})=1-\frac{1}{n^{K+1}} \implies L=\frac{1-1/(n^{K+1})}{(n-1)/(n)}=\frac{n(1-1/(n^{K+1}))}{(n-1)}$$
Now notice that as $K \to \infty$, $\frac{1}{n^{K+1}} \to 0$, so the term cancels. Thus: $$\sum_{k=0}^{\infty} \left(\frac{1}{n}\right)^k=\frac{n}{(n-1)}$$
This is a familiar trick to derive the more general geometric series equation:
geometric series
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1712599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find all such triangles ABC such that $AB+AC =2$cm and $AD+BC = \sqrt{5}$ cm where AD is the altitude through A. Find all such triangles ABC such that $AB+AC =2$cm and $AD+BC = \sqrt{5}$ cm
where AD is the altitude through A.
I got 3 equations but there are 4 variables. So, its not working. Maybe sine rule can work
| Let $BC=a,CA=b,AB=c$. Also, let $S$ be the area of $\triangle{ABC}$. Then, having that
$$S=\frac 12\times a\times AD\quad\Rightarrow\quad AD=\frac{2S}{a}=\frac{bc\sin\theta}{a}$$ where $\theta=\angle{BAC}$,
we have
$$c+b=2$$
$$\frac{bc\sin\theta}{a}+a=\sqrt 5$$
Then,
$$\sin\theta=\frac{a(\sqrt 5-a)}{b(2-b)}$$
By the law of cosines,
$$a^2=b^2+(2-b)^2-2b(2-b)\cos\theta\Rightarrow \cos\theta=\frac{b^2+(2-b)^2-a^2}{2b(2-b)}$$
Now, from $\cos^2\theta+\sin^2\theta=1$, we have
$$\left(\frac{b^2+(2-b)^2-a^2}{2b(2-b)}\right)^2+\left(\frac{a(\sqrt 5-a)}{b(2-b)}\right)^2=1,$$
i.e.
$$4(4-a^2)b^2+8(a^2-4)b+5a^4-8\sqrt 5\ a^3+12a^2+16=0$$
and so we have to have
$$(8(a^2-4))^2-4\cdot 4(4-a^2)(5a^4-8\sqrt 5\ a^3+12a^2+16)\ge 0$$$$\iff (a-2)(a+2) (\sqrt 5\ a-4)^2\ge 0\iff a=\frac{4}{\sqrt 5}\quad\text{or}\quad a\ge 2$$
Also, we have to have
$$a\lt b+c\iff a\lt 2.$$
Thus, $BC=a$ has to be $\frac{4}{\sqrt 5}$, from which we have $AC=AB=1$. This is sufficient.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1712878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
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.
| The absolute value $\lvert x \rvert$ is defined as
$$
\lvert x \rvert =
\begin{cases}
x & \text{if } x \ge 0 \\
-x & \text{if } x < 0
\end{cases}
$$
So one has to treat two cases. Your example involves two such expressions, so up to $2\cdot 2 = 4$ cases can occur.
*
*$-,-$: $x < 0$, $x-2+y < 0 \iff y < 2 - x$. Equality condition is $-2x = -(x-2+y) \iff y = 2 + x$. Combining them gives the condition $2 + x < 2 - x \iff x < 0$. So this case gives the line segment $y = 2 + x$ for $x < 0$.
*$0+,0+$: $x\ge 0$, $x-2+y \ge 0 \iff y \ge 2 - x$. Equality condition is $y = 2 + x$ again. Combination: $2+x \ge 2-x \iff x \ge 0$. So the whole line $y = 2 + x$ is part of the solution.
*$-,0+$: $x<0$, $y \ge 2-x$. Equality: $-2x = x-2+y \iff y = 2-3x$. Combination: $2-3x \ge 2-x \iff 0 \ge 2x \iff x \le 0$. This case gives the line segment $y=2-3x$ for $x < 0$.
*$0+,-$: $x\ge 0$, $y<2-x$. Equality: $y=2-3x$. Combination: $2-3x<2-x\iff 0<2x \iff x > 0$. This gives the line segment $y=2-3x$ for $x > 0$.
Thus the solution are both lines $y = 2 + x$ and $y = 2-3x$.
Do you have to consider all cases like above? The alternatives shown here are to employ properties of $\lvert . \rvert$ to operate on a slightly more abstract level (mathlove's answer) or to manipulate the logical conditions in a different order (Graham's answer). Of the above I prefer that I know I have handled all possible cases.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1714080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to calculate $\left( 1+\tan 5^\circ\right) \left( 1+\tan 10^\circ\right)\left( 1+\tan 15^\circ\right)\cdots\left( 1+\tan 40^\circ\right)$ I curious practical solution.(Step by step)
$\left( 1+\tan 5^\circ\right) \left( 1+\tan 10^\circ\right)\left( 1+\tan 15^\circ\right)\cdots\left( 1+\tan 40^\circ\right)$
Answer is $16$.
| For example, observe that
$$1=\tan 45^\circ =\tan(5^\circ+40^\circ)=\frac{\tan5^\circ+\tan40^\circ}{1-\tan5^\circ\tan40^\circ}.$$
Then we have $\tan5^\circ+\tan40^\circ=1-\tan5^\circ\tan40^\circ$, and thus
\begin{align}
(1+\tan5^\circ)(1+\tan40^\circ)&=1+\tan5^\circ\tan40^\circ+\tan5^\circ+\tan40^\circ\\
&=1+\tan5^\circ\tan40^\circ+1-\tan5^\circ\tan40^\circ\\
&=2.
\end{align}
Hence we conclude that
$$(1+\tan5^\circ)(1+\tan10^\circ)\cdots(1+\tan40^\circ)=2^4=16.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1716859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to tackle this polynomial given as a determinant? Let $$p (x) = \begin{vmatrix}
1 & x & x & \dots & x & x \\
x & 1 & x & \dots & x & x \\
x & x & 1 & \dots & x & x \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
x & x & x & \dots & 1 & x \\
x & x & x & \dots & x & 1 \\
\end{vmatrix} .$$
How to find the (multiple) zeroes, the degree of the polynomial and the initial coefficient, all depending on the natural number $b$?
| Add all columns (from second one) to the first one and then substract first row from all the rest:
$$\begin{vmatrix}1+(n-1)x&x&x&\ldots&x\\0&1-x&0&\ldots&0\\0&0&1-x&\ldots&0\\..&..&..&..&..\\0&0&0&\ldots&1-x\end{vmatrix}=\left(1+(n-1)x\right)(1-x)^{n-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1716970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How do I demonstrate that a polynomial of degree $2$ divides one of degree $n$? Let $f$ and $g$ the polynomials
$$f(x) = (x+1)^{2n-1}+(-1)^n(x+2)^{n+1}\qquad\text{and}\qquad g(x) = x^2 + 3x + 3$$
How do I demonstrate that $g$ divides $f$? I tried finding the roots of $g$ then testing them for $f$, it didn't work ...
I also wrote g as $(x+1)^2 + x + 2$ but I couldn't get anywhere.
| Let $r$ be any of the two roots of $g$; since $g(r) = (r+1) (r+2) + 1$, obviously $r+1 \ne 0$ and we get $r+2 = -\frac 1 {r+1}$. To have $g \mid f$ we should have $f(r) = 0$, so let us compute $f(r)$.
Since
$$f(r) = (r+1)^{2n-1} + (-1)^n (r+2)^{n+1} = (r+1)^{2n-1} + (-1)^n \left(- \frac 1 {r+1} \right)^{n+1} = \\
(r+1)^{2n-1} - \frac 1 {(r+1)^{n+1}} = \frac {(r+1)^{3n} - 1} {(r+1)^{n+1}} ,$$
let us finally show that $(r+1)^{3n} - 1 = 0$. To do this, notice that
$$(r+1)^3 = r^3 + 3r^2 + 3r + 1 = r g(r) + 1 = 1 ,$$
whence it follows that $(r+1)^{3n} - 1= \left((r+1)^3 \right)^n - 1 = 1^n - 1 = 0$, as desired.
This shows that indeed $f(r) = 0$ for each of the two roots $r$ of $g$, so indeed $g \mid f$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1717114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
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:
Let
$$I=\int \frac{2x^3+3x^2}{(2x^2+x-3)\sqrt{x^2+2x-3}}\mathrm dx$$
Then, we have
\begin{align*}
I&=\int \frac{2x^3+x^2-3x+2x^2+x-3+2x+3}{(2x^2+x-3)\sqrt{x^2+2x-3}}\mathrm dx\\
&=\int\frac{x+1}{\sqrt{x^2+2x-3}}\mathrm d x+\int\frac{2x+3}{(2x^2+x-3)\sqrt{x^2+2x-3}}\mathrm dx\\
&=\int\frac{x+1}{\sqrt{x^2+2x-3}}\mathrm d x+\int\frac{2x+3}{(2x+3)(x-1)\sqrt{x^2+2x-3}}\mathrm dx\\
&=\int\frac{x+1}{\sqrt{x^2+2x-3}}\mathrm d x+\int\frac{1}{(x-1)\sqrt{x^2+2x-3}}\mathrm dx\\
\end{align*}
Now, we can make $x+1=2\sec t$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1719129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Find $\lim_{x\to -\infty}\frac{3^{\sin x}+2x+1}{\sin x-\sqrt{x^2+1}}$ Find the value of $\lim_{x\to -\infty}\frac{3^{\sin x}+2x+1}{\sin x-\sqrt{x^2+1}}$
$\lim_{x\to -\infty}\frac{3^{\sin x}+2x+1}{\sin x-\sqrt{x^2+1}}$
Since this is in $\frac{\infty}{\infty}$ form,so i applied L Hospital rule,
$\lim_{x\to -\infty}\frac{3^{\sin x}\cos x\log 3+2}{\cos x-\frac{x}{\sqrt{x^2+1}}}$
But i am stuck here.
|
Note : In the denominator, when we take $x$ out of the square root, we will get $|x|\sqrt{1+\frac{1}{x^2}}$ Which will become = $-x\sqrt{1+\frac{1}{x^2}}$ (As $x$ is negative)
No need for LH here,
Dividing the numerator and denominator by $x$, the given expression becomes:
$$\frac{{3^{\sin x} \over x} + 2 + \frac{1}{x}}{{\sin x \over x} + \sqrt{1+\frac{1}{x^2}}}$$
Now, applying the limit,
$$\lim_{x \to -\infty}\frac{{3^{\sin x} \over x} + 2 + \frac{1}{x}}{{\sin x \over x} + \sqrt{1+\frac{1}{x^2}}}=2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1723115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
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?
| Whats the problem take $1/2$ common so you get $1/2(\frac{1+x+1-x}{1-x^2})=\frac{2}{2(1-x^2)}=\frac{1}{1-x^2}$
| {
"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": 3
} |
Exact solution at t = 2 and x = 1 (PDE)
I need help in this question. I did the following steps:
(a) I considered the characteristic $dx/dt = \frac{3t}{3x^2 +5}$ and solve it using the separation method and obtained $x^3 + 5x = 3/2t^2 + c$ . Then I considered (xo, to) to obtain that $x^3 + 5x = 3/2t^2 + xo^3 + 5xo - 3/2to^2$
(b) Then along the characteristic $du/dt = \frac{-tU}{x^3 +5x +1}$. Plugging in $x^3 + 5x = 3/2t^2 + xo^3 + 5xo - 3/2to^2$ and using the separation method I end up with the following
$U(x,t) = K(3/2t^2 + x0 + 5x0 - 3/2to^2 +1)^-1/3$
(c) When I plug x = 1 and t = 2, I have that the final solution for $U(x,T)$ at that value is $U(1,2) = \frac{1}{5^1/3} = .5848$
But I have the feeling it is not correct.
| Begin by noting that we have PDE:
$$ u_t +\frac{3t}{3x^2+5} u_x = - \frac{t}{x^3 + 5x + 1}u$$
Divide through by $u$ to yield
$$ \frac{u_t}{u} +\frac{3t}{3x^2+5} \frac{u_x}{u} = - \frac{t}{x^3 + 5x + 1}$$
Observe that $\frac{u_t}{u} = \frac{\partial}{\partial t} \ln (u)$ and $\frac{u_x}{u} = \frac{\partial}{\partial x} \ln (u)$. So let $g = \ln (u)$ then the problem becomes:
$$ g_t +\frac{3t}{3x^2+5} g_x = - \frac{t}{x^3 + 5x + 1}$$
And this reduces to
$$ (3x^2 + 5)g_t +3t g_x = - \frac{3tx^2 + 5t}{x^3 + 5x + 1}$$
This is in standard form for attack! since $A(x) g_t + B(t) g_x=0$ has solution $$ F\left( \int B(t) dt - \int A(x) dx \right)$$ and the general solution to $A(x) g_t + B(t) g_x = E(x,y)$ is a single solution to the latter diff EQ PLUS ANY solution to the homogenous case (ask in comments if this isn't clear).
So we wish to find a single solution $\mu$ to:
$$ (3x^2 + 5)g_t +3t g_x = - \frac{3tx^2 + 5t}{x^3 + 5x + 1}$$
And the know that $\mu + F \left( \frac{3}{2}t^2 - x^3 - 5x \right)$ is the general solution for any $F$. We now go backwards (you'll see why in a minute!) to
$$ \frac{1}{3t} g_t + \frac{1}{3x^2 + 5}g_x = - \frac{1}{x^3 + 5x+1}$$
Let $g_t = 0$ (since the right side is entirely dependent on x we can do this) then we have that $g_x = - \frac{3x^2 + 5}{x^3 + 5x + 1}$ which can be integrated with our old friend partial fractions (or more slickly notice that $3x^2 + 5 = \frac{d}{dx}(x^3 + 5x + 1)$) giving us a solution:
$$ \mu = \ln(x^3 + 5x + 1)$$
So the general solution then will be:
$$ g = \mu + F\left(\frac{3}{2}t^2 - x^3 - 5x \right) \rightarrow $$
$$ g = \ln(x^3 + 5x + 1) + F\left(\frac{3}{2}t^2 - x^3 - 5x \right)$$
But $g = \ln u$ so we have that $u = e^g$, thus
$$ u = e^{\ln(x^3 + 5x + 1) + F\left(\frac{3}{2}t^2 - x^3 - 5x \right)}= (x^3 + 5x + 1)F\left(\frac{3}{2}t^2 - x^3 - 5x \right)$$
(the $F$ absorbs the exponent, and the natural logarithm disappears).
Now things get ugly... It's easy to satisfy $u(0,t) = cos(4t)$ as
$$ u = (x^3 + 5x + 1) \cos \left(\sqrt{16t^2 - \frac{32}{3}x^3 - \frac{160}{3}x} \right) $$
But the other condition doesn't have a straightforward fix.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1728469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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 $b_i$ are simple multiples of unsigned Stirling numbers of the first kind. In a common notation, the coefficient of $x^k$ in the combined sum of $n$ terms is $(k+1)\left[{n \atop k+1}\right]$
Take the triangle of values in Wikipedia
which starts
n \ k 0 1 2 3 4
0 1
1 0 1
2 0 1 1
3 0 2 3 1
4 0 6 11 6 1
multiply the values in the triangle by the index at the top and drop the initial column and the indices to get a revised triangle
1
1 2
2 6 3
6 22 18 4
This new triangle corresponds to the coefficients in
*
*$\dfrac{1}{a} = \dfrac{1}{a}$
*$\dfrac {1}{a} + \dfrac {1}{a+1} =\dfrac {1+2a}{a(a+1)}$
*$\dfrac {1}{a} + \dfrac {1}{a+1} + \dfrac {1}{a+2}=\dfrac {2+6a+3a^2}{a(a+1)(a+2)}$
*$\dfrac {1}{a} + \dfrac {1}{a+1} + \dfrac {1}{a+2}+ \dfrac {1}{a+3}=\dfrac {6+22a+18a^2+4a^3}{a(a+1)(a+2)(a+3)}$
and so on
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1732737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Definite integral of $\int_0^\pi \frac{d\theta}{a+\sin^2 \theta} \quad (a>0)$ Definite integral of $$\int_0^\pi \frac{d\theta}{a+\sin^2 \theta} \quad (a>0)$$
I was attempting to do a trig substitution but that did not work out for me I believe because of the $a$. Looking at this to begin with I thought it was going to go smoothly but I keep getting stuck.
| If you don't mind contour integration, let $z=e^{i\theta}$:
$$
\begin{align}
\int_0^\pi\frac{\mathrm{d}\theta}{a+\sin^2(\theta)}
&=\frac12\int_0^{2\pi}\frac{\mathrm{d}\theta}{a+\sin^2(\theta)}\tag{1}\\
&=\frac12\oint_\gamma\frac1{a-\frac14\left(z-\frac1z\right)^2}\frac{\mathrm{d}z}{iz}\tag{2}\\
&=i\oint_\gamma\frac{2z\,\mathrm{d}z}{z^4-(2+4a)z^2+1}\tag{3}\\
&=i\oint_{\tilde\gamma}\frac{\mathrm{d}z}{z^2-(2+4a)z+1}\tag{4}\\
&=2i\oint_{\gamma}\frac{\mathrm{d}z}{(z-w_-)(z-w_+)}\tag{5}\\
&=-\frac{4\pi}{w_--w_+}\tag{6}\\[3pt]
&=\frac\pi{\sqrt{a+a^2}}\tag{7}
\end{align}
$$
Explanation:
$(1)$: integrand is $\pi$ periodic
$(2)$: $z=e^{i\theta}$ and $\gamma$ is a counter-clockwise unit circle
$(3)$: algebra
$(4)$: substitute $z^2\mapsto z$ but now $\tilde\gamma$ circles the origin twice
$(5)$: $w_+=1+2a+2\sqrt{a+a^2}$ and $w_-=1+2a-2\sqrt{a+a^2}$
$\phantom{(5)\text{:}}$ are the roots of $z^2-(2+4a)z+1$
$(6)$: $w_+$ is outside the unit circle
$\phantom{(6)\text{:}}$ the residue of $\frac1{(z-w_-)(z-w_+)}$ at $z=w_-$ is $\lim\limits_{z\to w_-}\frac{z-w_-}{(z-w_-)(z-w_+)}=\frac1{w_--w_+}$
$(7)$: evaluation
Alternatively, we can use the Weierstrass substitution.
$$
\begin{align}
\int_0^\pi\frac{\mathrm{d}\theta}{a+\sin^2(\theta)}
&=\int_0^\infty\frac{\frac{2\,\mathrm{d}t}{1+t^2}}{a+\left(\frac{2t}{1+t^2}\right)^2}\tag{8}\\
&=\int_0^\infty\frac{2\left(1+t^2\right)\mathrm{d}t}{a\left(1+t^2\right)^2+4t^2}\tag{9}\\
&=\int_0^\infty\frac{2\left(1+t^2\right)\mathrm{d}t}{at^4+(4+2a)t^2+a}\tag{10}\\
&=\frac2a\int_0^\infty\frac{\left(1+t^2\right)\mathrm{d}t}{(t^2+s_+^2)(t^2+s_{\vphantom{+}-}^2)}\tag{11}\\
&=\frac1{\sqrt{a+a^2}}\int_0^\infty\left(\frac{s_+}{t^2+s_+^2}+\frac{s_-}{t^2+s_{\vphantom{+}-}^2}\right)\mathrm{d}t\tag{12}\\
&=\frac1{\sqrt{a+a^2}}\left(\frac\pi2+\frac\pi2\right)\tag{13}\\
&=\frac\pi{\sqrt{a+a^2}}\tag{14}
\end{align}
$$
Explanation:
$\phantom{1}(8)$: Weierstrass substitution
$\phantom{1}(9)$: algebra
$(10)$: more algebra
$(11)$: factor the denominator with $s_+=\sqrt{1+\frac1a}+\sqrt{\frac1a}$ and $s_-=\sqrt{1+\frac1a}-\sqrt{\frac1a}$
$(12)$: partial fractions
$(13)$: arctangent integral
$(14)$: simplify
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1733250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
How to prove or disprove $\forall x\in\Bbb{R}, \forall n\in\Bbb{N},n\gt 0\implies \lfloor\frac{\lfloor x\rfloor}{n}\rfloor=\lfloor\frac{x}{n}\rfloor$. How to prove or disprove $\forall x\in\Bbb{R}, \forall n\in\Bbb{N},n\gt 0\implies \left\lfloor\frac{\lfloor x\rfloor}{n}\right\rfloor=\left\lfloor\frac{x}{n}\right\rfloor$.
So we want to prove $\left\lfloor\frac{\lfloor x\rfloor}{n}\right\rfloor\ge\left\lfloor\frac{x}{n}\right\rfloor$ and $\left\lfloor\frac{\lfloor x\rfloor}{n}\right\rfloor\le\left\lfloor\frac{x}{n}\right\rfloor$
Since $\lfloor x\rfloor\le x$, we can just start from here and prove
$\left\lfloor\frac{\lfloor x\rfloor}{n}\right\rfloor\le\left\lfloor\frac{x}{n}\right\rfloor$
But for $\left\lfloor\frac{\lfloor x\rfloor}{n}\right\rfloor\ge\left\lfloor\frac{x}{n}\right\rfloor$, I have no idea how to start.
| We know that integers can be pulled in and out of the floor function. Then WLOG we may assume $0\le x<n$ (otherwise subtract a suitable multiple of $n$). Obviously $0\le\lfloor x\rfloor\le x$.
Then dividing by $n$ and taking the floor,
$$0\le\left\lfloor \frac{\lfloor x\rfloor}n\right\rfloor\le\left\lfloor \frac xn\right\rfloor<1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1733848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
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.
| It's obviously wrong! Try $a=1.01$ and $b=c=\sqrt{2.0201}-1.01$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1735597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
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.
| [Too long for a comment]
Mathematica gives this answer.
$\displaystyle\int \sqrt[4]{\tan (x)} \, dx=\\\cos \left(\frac{\pi }{8}\right) \tan ^{-1}\left(\csc \left(\frac{\pi }{8}\right) \left(\sqrt[4]{\tan (x)}-\cos \left(\frac{\pi }{8}\right)\right)\right)+\\\cos \left(\frac{\pi }{8}\right) \tan ^{-1}\left(\csc \left(\frac{\pi }{8}\right) \left(\sqrt[4]{\tan (x)}+\cos \left(\frac{\pi }{8}\right)\right)\right)-\\\sin \left(\frac{\pi }{8}\right) \tan ^{-1}\left(\sec \left(\frac{\pi }{8}\right) \left(\sqrt[4]{\tan (x)}-\sin \left(\frac{\pi }{8}\right)\right)\right)-\\\sin \left(\frac{\pi }{8}\right) \tan ^{-1}\left(\sec \left(\frac{\pi }{8}\right) \left(\sqrt[4]{\tan (x)}+\sin \left(\frac{\pi }{8}\right)\right)\right)+\\\frac{1}{2} \cos \left(\frac{\pi }{8}\right) \log \left(\sqrt{\tan (x)}+2 \sin \left(\frac{\pi }{8}\right) \sqrt[4]{\tan (x)}+1\right)-\\\frac{1}{2} \cos \left(\frac{\pi }{8}\right) \log \left(\sqrt{\tan (x)}-2 \sin \left(\frac{\pi }{8}\right) \sqrt[4]{\tan (x)}+1\right)-\\\frac{1}{2} \sin \left(\frac{\pi }{8}\right) \log \left(\sqrt{\tan (x)}+2 \cos \left(\frac{\pi }{8}\right) \sqrt[4]{\tan (x)}+1\right)+\\\frac{1}{2} \sin \left(\frac{\pi }{8}\right) \log \left(\sqrt{\tan (x)}-2 \cos \left(\frac{\pi }{8}\right) \sqrt[4]{\tan (x)}+1\right)+C$
I wouldn’t want to have to have worked it out by hand, but it doesn’t look impossible.
| {
"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": 2
} |
Bound on the infinite sum of logarithms Is it possible to show that
$X=\frac 12 \log 3 + \frac 14 \log 4 + \frac 18 \log 5 + \frac{1}{16} \log 6 + \dots < \log 4$?
I think we can do
$\frac 12 X= \frac 14 \log 3 + \frac 18 \log 4 + \frac {1}{16} \log 5 + \frac{1}{32} \log 6 + \dots $
and subtract the two to get
$\frac 12 X=\frac 12 \log 3 + \frac 14 \log \frac 43 + \frac 18 \log \frac 54 + \frac{1}{16} \log \frac 65 + \dots $
So $X = \log 3 + \frac 12 \log \frac 43 + \frac 14 \log \frac 54 + \frac{1}{8} \log \frac 65 + \dots < \log 3 + \frac 12 \log \frac 43 + \frac 12 \log \frac 54 + \frac{1}{2} \log \frac 65 + \dots$.
But the right most expression telescopes to $\frac 12 \log 3$ which is absurd! Where is my mistake and how to prove the original statement?
| A rigorous approach: let $\displaystyle S_n=\sum_{k=0}^n\frac{1}{2^k}=2-\frac{1}{2^n}$
Performing an Abel transformation yields:
$$\begin{align}\sum_{k=3}^n \frac{\ln k}{2^k} &= \ln n S_n - \ln 3 S_2 + \sum_{k=3}^{n-1}\ln\left( \frac{k}{k+1}\right)S_k \\ &=2\ln n -\frac{\ln n}{2^n} + \sum_{k=3}^{n-1}\frac{1}{2^k}\ln\left(\frac{k+1}{k}\right) -2\ln n +2\ln3 - \frac 74 \ln 3 \\ &= (2-\frac 74) \ln 3+ \sum_{k=3}^{n-1}\frac{1}{2^k}\ln\left(1+\frac 1k\right)-\frac{\ln n}{2^n} \\ &\le (2-\frac 74) \ln 3+ \sum_{k=3}^{\infty}\frac{1}{k2^k} \\ &\leq \frac{\ln 3}{4} + \ln 2 - \frac{5}{8}
\end{align}$$
Hence $$\sum_{k=3}^n \frac{\ln k}{2^{k-2}} \leq 4\left( \frac{\ln 3}{4} + \ln 2 - \frac{5}{8} \right) = \ln 48 - \frac 52 <\ln 4$$
I hope you bear no grudge if I don't prove $\displaystyle \ln 48 - \frac 52 < \ln 4$.
Checking with a pocket calculator, $\ln 48 - \frac 52 \sim 1.371$ while $\ln 4\sim 1.386$
You can refine the bound ad libitum if you use sharper inequalities for $\ln(1+x)$.
Using $\ln(1+x)\leq x-\frac{x^2}2+\frac{x^3}3$ and computing $\sum _{k=3}^{\infty } \frac{1}{2^k}\left(\frac{1}{3 k^3}-\frac{1}{2 k^2}+\frac{1}{k}\right)$. with Mathematica yields the much tighter bound:
$$\sum_{k=3}^n \frac{\ln k}{2^{k-2}}\leq \frac{1}{144} \left(42 \zeta (3)-6 \pi ^2-75+8 \log ^3(2)+36 \log ^2(2)-4 \pi ^2 \log (2)+144 \log (2)\right)+\log (3)$$
with $\frac{1}{144} \left(42 \zeta (3)-6 \pi ^2-75+8 \log ^3(2)+36 \log ^2(2)-4 \pi ^2 \log (2)+144 \log (2)\right)+\log (3)\sim \color{red}{1.3397}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1738382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Are these metrics? I want to find if the below functions are metrics. I have worked through each of the three conditions, but am stuck on the positivity of $f(a, b)$ (first condition-see below) and the triangle inequality of $g(a, b)$ (third condition-see below).
Edit: Have now solves these two parts, see the answers for positivity of $f(a, b)$ and the triangle inequality of $g(a, b)$
Please could you help me on these two points?
$f(a, b)=a^2+ab+b^2$
*
*Initially I thought this would not satisfy positivity. But for $f$
to be negative, $ab$ must be negative, and greater in modulus value
to $a^2+b^2$. So for this to work, $a$ and $b$ must be opposite sign
and less than $1$. I have tried small opposite signed values of $a$
and $b$ but have not found $f(a, b)$ to be positivity. Not sure.
*Symmetry. $f(b, a)=b^2+ba+a^2=a^2+ab+b^2$ so this is satisfied
*Triangle inequality. Is $f(a, b)+f(b, c) \geq f(a, c)$ satisfied? Let $X=f(a, b)+f(b, c)-f(a, c)=a^2+b^2+ab+c^2+b^2+cb-a^2-c^2-ac=b^2+ab+cb-ac$. Assume $X \geq 0 $ for all $a, b, c \in \mathbb{R}\implies b^2 +ab+cb \geq ac$. My initial thought is that if we take $b$ to be very small and $a$ and $c$ to be comparatively larger, then we may arrive at a contradiction. Take $b=0.01, a=100, c=100 \implies 0.0001+2 \geq 10000$. Contradiction. Triangle inequality does not hold.
$g(a, b)= \frac{|a^3-b^3|}{2}$
*
*Positivity satisfied since $|x| \geq 0 \forall x \in \mathbb{R}$ and $2 \geq 0$
*$g(b, a)=\frac{|b^3-a^3|}{2}=\frac{|-(-b^3+a^3)|}{2}=\frac{|-1||a^3-b^3|}{2}=\frac{|a^3-b^3|}{2}=g(a, b)$. So symmetry is satisfied.
*Must consider the sign of: $X=g(a, b)+g(b, c)-g(a, c)=\frac{|a^3-b^3|}{2}+\frac{|b^3-c^3|}{2}-\frac{|a^3-c^3|}{2}$ For the triangle inequality to be satisfied $X$ must be positive. I tried to find some counter-examples, such as take $a=100, c=0 \implies \frac{|100^3-b^3|}{2}+\frac{|b^3|}{2}-\frac{|100^3|}{2}$. But if $b$ is positive then the whole thing is positive. If it is negative , then the first two terms are positive and larger than the third. From this reasoning, I believe $X$ is always positive and the triangle inequality holds. Not sure.
| So $$|x|+|y|\ge |x+y|,$$ then
$$\frac{|a^3-b^3|}{2}+\frac{|b^3-c^3|}{2}=\frac{|a^3-b^3|+|b^3-c^3|}{2}\ge$$
$$\ge \frac{|a^3-b^3+b^3-c^3|}{2}=\frac{|a^3-c^3|}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1739007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
solve $x^2 -4x +13 \equiv 0 \pmod{81}$? How do I solve $x^2 -4x +13 \equiv o \pmod{81}$ ?
I know that this is the same as $x^2 -4x +13 \equiv x^2 + 2x + 1 \equiv (x +1)^2\equiv 0\pmod{3^4}$
but why is $x \equiv -1\pmod{3}$ the only solution of the congruence $f(x) \equiv 0\pmod{3}$ ? (why is $x \equiv 2\pmod{3}$ not a solution?)
How do I use hensel's lemma to find/ prove that there are no solutions to this problem?
| We have $\left(x-2\right)^2\equiv -9\left(\text{mod}\,81\right)$. The left-hand side is a multiple of 9, so its square root is a multiple of 3, say $x-2=3y$. Then $y^2\equiv -1\left(\text{mod}\,9\right)$. But integers cannot even solve $y^2\equiv -1\left(\text{mod}\,3\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1741007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Quick, self-contained way to see why $\left({{-1}\over p}\right) = 1$? Let $p$ be a prime number congruent to $1$ modulo $4$. What is a quick and self-contained way to see why$$\left({{-1}\over p}\right) = 1?$$
| If $p$ is any odd prime, then by Wilson's Theorem: $$(p-1)!\equiv 1\cdot 2\cdots\left(\frac{p-1}{2}\right)\left(-\frac{p-1}{2}\right)\cdots (-2)(-1)\pmod{p}$$
$$\equiv (-1)^{\frac{p-1}{2}}\left(\left(\frac{p-1}{2}\right)!\right)^2\equiv -1\pmod{p}$$
If $p\equiv 1\pmod{4}$, then all the solutions of $x^2\equiv -1\pmod{p}$ are $x\equiv \pm\left(\frac{p-1}{2}\right)!\pmod{p}$.
Edit: another proof: Let $p$ be prime, $p\equiv 1\pmod{4}$ and let $g$ be a primitive root mod $p$. Then $\text{ord}_p(g)=p-1$, so $g^{\frac{p-1}{2}}\equiv -1\pmod{p}$. Therefore, all the solutions of $x^2\equiv -1\pmod{p}$ are $x\equiv \pm g^{\frac{p-1}{4}}\pmod{p}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1743469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
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.
| Just another way is to use Holder :
$$\left(\sum_{cyc} \frac{a^2}b \right)^2 \left( \sum_{cyc} a^2b^2\right) \geqslant \left( \sum_{cyc} a^2\right)^3=4^3$$
So it remains to show
$$a^2b^2+b^2c^2+c^2d^2+d^2a^2 = (a^2+c^2)(b^2+d^2) \leqslant \frac14(a^2+b^2+c^2+d^2)^2 = 4 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1743834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 1
} |
Proving a trigonometric identity:
Prove that $\sin \frac{{2\pi }}{7} + \sin \frac{{4\pi }}{7} + \sin
\frac{{8\pi }}{7} = \frac{{\sqrt 7 }}{2}$.
I have tried to square both side and got ${\sin ^2}\frac{{2\pi }}{7} + {\sin ^2}\frac{{4\pi }}{7} + {\sin ^2}\frac{{8\pi }}{7} = \frac{7}{4}$. But I cannot proceed further. Any help would be appreciated.
| Let $\mu=\cos\frac{2\pi}{7}+i\sin\frac{2\pi}{7}$ and so $\mu^7=1$. What we want is $\text{Im}(\mu+\mu^2+\mu^4)$. Consider
\begin{align*}
(\mu+\mu^2+\mu^4)(\mu^3+\mu^5+\mu^6)&=3+\mu+\mu^2+\mu^3+\mu^4+\mu^5+\mu^6\\
&=3-1\\
&=2
\end{align*}
Let $x=\mu+\mu^2+\mu^4$. Then $\mu^3+\mu^5+\mu^6=-1-x$. We then have
\begin{eqnarray}x(-1-x)=2\end{eqnarray}
Thus $\displaystyle x=\frac{-1\pm\sqrt{7}i}{2}$. Note that $\text{Im}(x)=\text{Im}(\mu+\mu^2+\mu^4)$ is positive, therefore the answer must be $\displaystyle\frac{\sqrt{7}}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1745060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"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?
| You can start from the identity
\begin{equation}
24+8 \sqrt{5} =24+8 \sqrt{5}
\end{equation}
And write
\begin{equation}
24+8 \sqrt{5} =20+4+8 \sqrt{5} = (2+2 \sqrt{5})^{2}
\end{equation}
Then:
\begin{equation}
\sqrt{24+8 \sqrt{5}}=\sqrt{(2+2 \sqrt{5})^{2}}=2+2 \sqrt{5}
\end{equation}
| {
"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": 4
} |
How can I get f(x) from its Maclaurin series? I know how to get a Maclaurin series when $f(x)$ is given. I have to find $\sum_{n=0}^{\infty}\frac{f^{(k)}(0)}{k!}x^k$.
But how can I get $f(x)$ from its Taylor series?
The problem is $$f(x) = \sum_{n=0}^{\infty} C_n x^n,$$ where $C_n$ is a Catalan number defined by $C_n = \frac{1}{n+1}\binom{2n}{n}$.
How can I get $f(x)$?
| *
*The Catalan numbers $$C_n=\frac{1}{n+1}\binom{2n}{n}$$ are strongly related with the Central binomial coefficients $\binom{2n}{n}$.
*The binomial identity
\begin{align*}
\frac{1}{2n-1}\binom{2n}{n}=(-1)^{n+1}\binom{\frac{1}{2}}{n}4^n\tag{1}
\end{align*}
provides a relationship with the binomial series expansion
\begin{align*}
\sqrt{1+x}=\sum_{n= 0}^\infty\binom{\frac{1}{2}}{n}x^n\tag{2}
\end{align*}
From (1) we obtain a relationship with the Catalan numbers.
\begin{align*}
\frac{1}{2}\cdot\frac{1}{2n-1}\binom{2n}{n}&=\frac{1}{2}\cdot\frac{1}{2n-1}\cdot\frac{2n}{n}\binom{2n-1}{n-1}\\
&=\frac{1}{2n-1}\binom{2n-1}{n-1}\\
&=\frac{1}{2n-1}\binom{2n-1}{n}\\
&=\frac{1}{n}\binom{2n-2}{n-1}\\
&=C_{n-1}\tag{3}
\end{align*}
We conclude
\begin{align*}
\sum_{n=0}^\infty C_nx^n&=\sum_{n=1}^\infty C_{n-1}x^{n-1}\tag{4}\\
&=\frac{1}{2}\sum_{n=1}^\infty\frac{1}{2n-1}\binom{2n}{n}x^{n-1}\tag{5}\\
&=\frac{1}{2}\sum_{n=1}^\infty(-1)^{n+1}\binom{\frac{1}{2}}{n}4^nx^{n-1}\tag{6}\\
&=-\frac{1}{2x}\sum_{n=1}^\infty\binom{\frac{1}{2}}{n}(-4x)^{n}\tag{7}\\
&=-\frac{1}{2x}\left(\sqrt{1-4x}-1\right)\tag{8}\\
&=\frac{1}{2x}\left(1-\sqrt{1-4x}\right)
\end{align*}
Comment:
*
*In (3) we use twice the identity $\binom{n}{k}=\frac{n}{k}\binom{n-1}{k-1}$ and the symmetry $\binom{2n-1}{n-1}=\binom{2n-1}{n}$
*In (4) we shift the index by one
*In (5) we use the relationship (3)
*In (6) we use the identity (1)
*In (7) we do a small rearrangement
*In (8) we use the identity (2). Since the series starts with $n=1$ we have to subtract $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1746795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
$\int_0^\infty {x^a\over (x^2+1)^2} dx$ where $0$\int_0^\infty {x^a\over (x^2+1)^2} dx$ where $0<a<1$.
I know I can use partial fraction decomposition to obtain two different integrals, but I'm not sure how to integrate them. Any solutions or hints are greatly appreciated.
| If you know about the $\text{B}$ and $\Gamma$ functions, you can first let $x^2=u$ and then $t=\frac1{1+u}$ to get
$$\begin{align}\int_9^{\infty}\frac{x^a}{(x^2+1)^2}dx&=
\frac12\int_0^{\infty}\frac{u^{\frac{a-1}2}}{(1+u)^2}du
=\frac12\int_0^1(1-t)^{\frac{a-1}2}t^{\frac{1-a}2}dt\\
&=\frac12\text{B}\left(\frac{a+1}2,\frac{3-a}2\right)
=\frac12\frac{\Gamma\left(\frac{a+1}2\right)\Gamma\left(\frac{3-a}2\right)}{\Gamma\left(2\right)}\\
&=\frac12\Gamma\left(\frac{a+1}2\right)\left(\frac{1-a}2\right)\Gamma\left(\frac{1-a}2\right)\\
&=\frac{1-a}4\cdot\frac{\pi}{\sin\left(\frac{\pi(1-a)}2\right)}
=\frac{\pi(1-a)}{4\cos\left(\frac{\pi a}2\right)}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1746926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Evaluate: $\csc^2\left(\frac{\pi}{9}\right)+\csc^2\left(\frac{2\pi}{9}\right)+\csc^2\left(\frac{4\pi}{9}\right)$
$$\csc^2\left(\frac{\pi}{9}\right)+\csc^2\left(\frac{2\pi}{9}\right)+\csc^2\left(\frac{4\pi}{9}\right) \;=\; \text{???}$$
$\bf{My\; Try::}$ Let $\displaystyle \frac{\pi}{9} = \theta\;,$ Then $9\theta = \pi\Rightarrow 6\theta = \pi-3\theta$
So $\sin (6\theta) = -\sin (3\theta)\Rightarrow 2\sin 3 \theta \cdot \cos 3 \theta+\sin 3 \theta =0$
So we get $\displaystyle \sin 3 \theta \cdot \left[2\cos 3 \theta +1\right] =0$
So we get $\sin 3\theta = 0$ or $2\cos 3 \theta+1=0$
Now i did not understand how can I convert into $\sin^2$ form, Help me
Thanks
| Like my answer here in Sum of tangent functions where arguments are in specific arithmetic series
$$\tan9x=\dfrac{\binom91t-\binom93t^3+\binom95t^5-\binom97t^7+t^9}{\cdots}$$ where $t=\tan x$
If $\tan9x=0,9x=n\pi$ where $n$ is any integer
So, the roots of $$\binom91t-\binom93t^3+\binom95t^5-\binom97t^7+t^9=0$$ are $\tan\dfrac{r\pi}9$ where $n\equiv0,\pm1,\pm2,\pm3,\pm4\pmod9$
As $\tan0=0,$ the roots of $$\binom91-\binom93t^2+\binom95t^4-\binom97t^6+t^8=0$$ are $\tan\dfrac{r\pi}9$ where $r\equiv\pm1,\pm2,\pm3,\pm4\pmod9$
Putting $c=1/t,$ the roots of $$\binom91c^8-\binom93c^6+\binom95c^4-\binom97c^2+1=0$$ are $\cot\dfrac{r\pi}9$ where $r\equiv\pm1,\pm2,\pm3,\pm4\pmod9$
Putting $c^2=d,$ the roots of $$\binom91d^4-\binom93d^3+\binom95d^2-\binom97d+1=0$$ are $\cot^2\dfrac{r\pi}9$ where $r\equiv1,2,3,4\pmod9$
$$\sum_{r=1}^4\cot^2\dfrac{r\pi}9=\dfrac{\binom93}{\binom91}=?$$
But $r=3\implies\cot^2\dfrac{3\pi}9=3$
Use $\csc^2u=1+\cot^2u$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1749306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Limit of the function series I have to find the value
$$\lim_{k \to \infty}\sum_{n=1}^{\infty}\frac{k^3+n^3}{k^32^n+1}$$
Hint I've got:
Compare $$\frac{k^3+n^3}{k^32^n+1}$$ with $$\frac{1+n^3}{2^n}$$ and use Weierstrass theorem.
I do not know how to apply Weierstrass theorem with the limit of k.
Help guys :)
| You have $\frac{k^3 + n^3}{k^3 2^n +1} -\frac{1}{2^n} = \frac{n^3 - 2^{-n}}{k^3 2^n +1}$ (and $\sum \limits_{n\geqslant 1} \frac{n^3 - 2^{-n}}{k^3 2^n +1}$ converges because the general term is equivalent to $\frac{n^3}{k^3 2^n}$).
So $\sum \limits_{n=1}^{\infty} \frac{k^3 + n^3}{k^3 2^n +1} = \sum \limits_{n=1}^{\infty} \frac{1} {2^n} + \frac{1}{k^3} \cdot \sum \limits_{n=1}^{\infty} \frac{n^3 - 2^{-n}}{2^n + \frac{1}{k^3}}$, and all the sums converge.
So finally the limit seeked is $\sum \limits_{n=1}^{\infty} \frac{1}{2^n} = 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1750145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
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
$$AC = 4\sin\theta = \frac{16}{\sqrt {17}}$$
$$\text{Area} = \left(\frac{16}{\sqrt {17}}\right)^2$$
| Denote: $AB=a, BF=x, BH=y$. Then:
$$\begin{cases}a^2+(a-x)^2=25 \ (1)\\
(a-y)^2+a^2=16 \ (2)\\
x^2+y^2=9 \ (3)\end{cases}$$
$(1)-(2)$:
$$(a-x)^2-(a-y)^2=x^2+y^2 \Rightarrow 2a(y-x)=2y^2 \Rightarrow a=\frac{y^2}{y-x} \ \ \ \ \ \ \ \ \ \ \ (4)$$
$(4)\to (2)$:
$$\begin{align}\left(\frac{y^2}{y-x}-y\right)^2+\left(\frac{y^2}{y-x}\right)^2&=16 \Rightarrow \\
\frac{y^2(x^2+y^2)}{(y-x)^2}&=16 \stackrel{(3)}\Rightarrow \\
9y^2&=16(9-2xy) \stackrel{(3)}\Rightarrow \\
144-9y^2&=32y\sqrt{9-y^2} \Rightarrow \\
1104y^4-11808y^2+20736&=0 \Rightarrow \\
y_1&=\frac{12}{\sqrt{17}}; y_2=\frac{12}{\sqrt{65}} \Rightarrow \\
x_1&=\frac3{\sqrt{17}}; x_2=\frac{21}{\sqrt{65}}.\end{align}$$
Hence, from $(4)$:
$$a=\frac{y^2}{y-x}=\frac{\frac{144}{17}}{\frac9{\sqrt{17}}}=\frac{16}{\sqrt{17}}\\
y-x<0 \Rightarrow a\in \emptyset$$
Reference: WA answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1754133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 8,
"answer_id": 5
} |
Is there a formula for finding the nth number in a sequence with a changing pattern If a sequence has a pattern where +2 is the pattern at the start, but 1 is added each time, like the sequence below, is there a formula to find the 125th number in this sequence? It would also need to work with patterns similar to this. For example if the pattern started as +4, and 5 was added each time.
2, 4, 7, 11, 16, 22 ...
| Let $a_1 = 2$. From the way you defined the sequence you can see that $a_n - a_{n-1} = n$. We can use this to find
\begin{align}
a_n &= a_{n-1} + n\\
&= a_{n-2} + (n-1) + n\\
&= a_{n-3} + (n-2) + (n-1) + n\\
&\vdots \\
&= a_1 + 2 + \cdots + (n - 2) + (n-1) + n
\end{align}
which is just the sum of the natural numbers except 1($1 + 2 + \cdots + n = \frac{n(n+1)}{2}$). So
\begin{equation}
a_n = a_1 + \frac{n(n+1)}{2} - 1 = 2 - 1 + \frac{n(n+1)}{2} = \frac{n^2 + n + 2}{2}
\end{equation}
where $a_1$ is the starting number (in this case 2). This sequence is a quadratic sequence as it exhibits second differences(the difference of the differences is constant).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1754931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to write the following polynomial in $(1-\frac{x}{a}) (1-\frac{x}{b}) (1-\frac{x}{c}) (1-\frac{x}{d})$? I was given the following problem:
Write the polynomial $f(x) = \frac{1}{24} \displaystyle \prod_{i \mathop = 1}^4 (x-i)$ in the form $(1-\frac{x}{a}) (1-\frac{x}{b}) (1-\frac{x}{c}) (1-\frac{x}{d})$
So far I have done this:
$f(x) = \frac{1}{24} \displaystyle \prod_{i \mathop = 1}^4 (x-i) = \frac{1}{24}(x-1)(x-2)(x-3)(x-4)
= \frac{1}{24}(1-x)(2-x)(3-x)(4-x)
= (\frac{1-x}{1}) (\frac{2-x}{2}) (\frac{3-x}{3}) (\frac{4-x}{4})
= ({1-x}) (1-\frac{x}{2}) (1-\frac{x}{3}) (1-\frac{x}{4})$
EDIT: Seems like I got it, thanks for all the help!
| $$\frac{1}{24} \prod_{i = 1}^4 (x-i)=\frac{1}{24} \prod_{i = 1}^4 \left((-i)(1-\frac{x}{i})\right)=\frac{1}{24} \prod_{i = 1}^4 (-i)\prod_{i = 1}^4(1-\frac{x}{i})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1757075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Least-squares when some coefficient is $0$? I'm trying to find least squares approximation $p(x)=c_1x+c_2x^2$ of $f(x)=xe^{x/2}$ in $[0,2]$.
Using the algorithm here, p.7.:
http://www.math.niu.edu/~dattab/MATH435.2013/APPROXIMATION.pdf
I'm able to come up with a $3 \times 3$ matrix $S$ and $3 \times 1$ vector $b$. However this gives me as a solution three coefficients, even though I only have two. What do I need to do to get only two (i.e. no $c_0$)?
| The approximation is
$$
g(x) \approx c_{1} f_{1}(x) + c_{2} f_{2}(x)
$$
over the domain $a < x < b$. Choose the method of least squares. Minimize the sums of the squares of residuals:
$$
r^{2}(c) = \int_{a}^{b} \left(f(x) - c_{1} f_{1}(x) - c_{2} f_{2}(x) \right)^{2} dx.
$$
This creates the linear system
$$
\begin{align}
%
\mathbf{A} c &= G \\
%
\left[
\begin{array}{cc}
\int_{a}^{b} f_{1}(x) \times f_{1}(x) dx & \int_{a}^{b} f_{1}(x) \times f_{2}(x) dx \\
\int_{a}^{b} f_{2}(x) \times f_{1}(x) dx & \int_{a}^{b} f_{2}(x) \times f_{2}(x) dx
\end{array}
\right]
%
\left[
\begin{array}{c}
c_{1} \\ c_{2}
\end{array}
\right]
%
&=
%
\left[
\begin{array}{c}
\int_{a}^{b} f_{1}(x) \times g(x) dx \\
\int_{a}^{b} f_{2}(x) \times g(x) dx
\end{array}
\right]
\end{align}
$$
In this problem $g(x) = xe^{\frac{x}{2}}$, $f_{1}(x) = x$, and $f_{2}(x) = x^{2}.$
For the domain $a= 0$ and $b=2$, the linear system is
$$
\left[
\begin{array}{cc}
\frac{8}{3} & 4 \\
4 & \frac{32}{5} \\
\end{array}
\right]
%
\left[
\begin{array}{c}
c_{1} \\ c_{2}
\end{array}
\right]
%
=
%
\left[
\begin{array}{r}
8 (e-2) \\ -32 (e-3)
\end{array}
\right],
$$
which has the solution
$$
\left[
\begin{array}{c}
c_{1} \\ c_{2}
\end{array}
\right]
%
=
%
\left[
\begin{array}{c}
24 (-19 + 7 e) \\ 300 - 110 e
\end{array}
\right].
%
$$
The following plot shows the residual error function $g(x) - c_{1}x - c_{2}x^{2}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1758075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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 $a,b,c$?
| You may use Newton's formulas for recovering the values of $ab+ac+bc$ and $abc$, in order to compute $a,b,c$ from the factorization of $(x-a)(x-b)(x-c)$. As an alternative, we may understand which conditions on $P_1,P_2,P_3$ grant that if $a+b=P_1$ and $a^2+b^2=P_2$, then $a^3+b^3=P_3$. We have:
$$ P_1 P_2 = P_3 + ab(a+b) = P_3+P_1\cdot\frac{P_1^2-P_2}{2} $$
or:
$$ 2 P_3 = 3 P_1 P_2 - P_1^3.$$
Assuming $c=1$, we have $(P_1,P_2,P_3)=(1,5,7)$ that fulfills the previous condition, hence we may assume $a+b=1$ and $a^2+b^2=5$. The solutions are so given by the permutations of $\color{red}{(-1,1,2)}$.
| {
"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": 5
} |
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:
$\begin{split}\Bigg(\dfrac{(x+1)^2}{2}\Bigg)^n-\Bigg(\dfrac{(x-1)^2}{2}\Bigg)^n &= \Bigg(\dfrac{(x+1)^2}{2}-\dfrac{(x-1)^2}{2}\Bigg)\Bigg(\Big(\frac{(x+1)^2}{2}\Big)^{n-1}+\ldots +\Big(\frac{(x-1)^2}{2}\Big)^{n-1}\Bigg) \\ &=2x\Bigg(\Big(\frac{(x+1)^2}{2}\Big)^{n-1}+\ldots +\Big(\frac{(x-1)^2}{2}\Big)^{n-1}\Bigg)\end{split}$
| Just apply the binomial theorem.
$$\begin{eqnarray*}F_n(x)=\left(\frac{(x+1)^2}{2}\right)^n-\left(\frac{(x-1)^2}{2}\right)^n &=& \frac{1}{2^n}\left[(x+1)^{2n}-(x-1)^{2n}\right]\\&=&\frac{1}{2^n}\sum_{k=0}^{2n}\binom{2n}{k}x^{2n-k}(1^k-(-1)^k) \end{eqnarray*}$$
hence the leading term is $\frac{2\binom{2n}{1}}{2^n}\cdot x^{2n-1}$. You may also notice that:
$$ F_{n+2}(x) = (1+x^2)F_{n+1}(x)-\frac{1}{4}(x^2-1)^2 F_n(x),$$
hence if you prove that the degree of $F_n(x)$ is $2n-1$, the leading coefficient $L_n$ fulfills the recursion:
$$ L_{n+2} = L_{n+1} - \frac{1}{4}L_n $$
hence $L_n = \frac{A+Bn}{2^n}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1759869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Show that the curve $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ form an ellipse If the definition of an ellipse is the set of points $(x,y)$ such that given two focus points $F_1, F_2$ the sum of the distances from $(x,y)$ to each focus point is constant, how can one show that the curve $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1, \quad 0 < b \leq a \quad$ forms an ellipse?
The methods that I know of are to either derive the formula by considering the foci $(-c,0), (c,0)$ and the constant distance $2k$, or to set $(x,y) = (a\cos{v}, b\sin{v})$. Is there some other, relatively easy way to show that the points satisfying the equation form an ellipse?
I am asking because in a book I am reading the author states that a "direct calculation" shows that the curve indeed forms an ellipse, but I do not understand what kind of calculation this might be.
| If $c=\sqrt{a^2-b^2}$ and $F_1=(-c,0)$, $F_2=(c,0)$ are the two foci, then the distance of a generic point $P=(x,y)$ of the ellipse from $F_1$ is
$$
PF_1=\sqrt{(x+c)^2+y^2}=\sqrt{x^2+2cx+c^2+y^2}.
$$
Substituting here $y^2=b^2-{b^2\over a^2}x^2$ one gets
$$
PF_1=\sqrt{x^2+2cx+c^2+b^2-{b^2\over a^2}x^2}=
\sqrt{{c^2\over a^2}x^2+2cx+a^2}
=a+{c\over a}x.
$$
An analogous calculation gives $PF_2=a-{c\over a}x$, so that $PF_1+PF_2=2a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1763292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Geometric inequality involving sides of the triangle I was triying to learn geometric inequalities and I got into this problem:
Let $a,b,c$ be the sides of the $\Delta ABC$. Show that:
$$ \left (\frac S R \right)^2 \le \frac 3 8 \left (\frac {ab \sqrt {ab}} {a+b} + \frac {bc \sqrt {bc}} {b+c} + \frac {ca \sqrt {ca}} {c+a} \right) \le \left (\frac {S} {2r} \right)^2,$$ where S is the area of the triangle, R the circumradius and r the inradius.
| A left inequality.
Let $a=y+z$, $b=x+z$ and $c=x+y$. Hence, $x>0$, $y>0$ and $z>0$.
By AM-GM $\sum\limits_{cyc}\frac{\sqrt{a^3b^3}}{a+b}\geq\frac{3abc}{\sqrt[3]{\prod\limits_{cyc}(a+b)}}=\frac{9abc}{3\sqrt[3]{\prod\limits_{cyc}(a+b)}}\geq\frac{9abc}{2(a+b+c)}$.
$\frac{S}{R}=\frac{4S^2}{abc}$.
Thus, it remains to prove that $\frac{9abc}{2(a+b+c)}\geq\frac{1}{6}\left(\frac{16S^2}{abc}\right)^2$ or
$27(x+y)^3(x+z)^3(y+z)^3\geq512x^2y^2z^2(x+y+z)^3$.
Since $(x+y)(x+z)(y+z)\geq\frac{8}{9}(x+y+z)(xy+xz+yz)\Leftrightarrow\sum\limits_{cyc}z(x-y)^2\geq0$,
it remains to prove that $(xy+xz+yz)^3\geq27x^2y^2z^2$, which is true by AM-GM. Done!
For the proof of the right inequality we can use $\frac{\sqrt{ab}}{a+b}\leq\frac{1}{2}$.
Hence, it remains to prove that $(a+b+c)^2\geq3(ab+ac+bc)$,
which is $\sum\limits_{cyc}(a-b)^2\geq0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1764557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
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.
| Let $s_n$ be the sum of $1$ and the fractions for the $n$ case, e.g.
$$s_4 = 1+\frac{1}{2} + \frac{1}{3} + \frac{1}{2\cdot3}$$
Assume $s_k = k/2$ is true for some $k$.
For the $n=k+1$ case,
$$s_{k+1} = s_k\cdot \left(1+\frac1k\right) = s_n\cdot\frac{k+1}{k} = \frac{k+1}{2}$$
For the $n=2$ case,
$$s_2 = 1 = \frac{2}{2}$$
By induction, $s_n = n/2$ is true for natural numbers $n\ge 2$. i.e.
$$\frac{1}{n}s_n = \frac12$$
Some example of the recursion $s_{k+1} = s_k\cdot\left(1+\frac1k\right) $:
$$\begin{align*}
s_4 &= 1+\frac{1}{2} + \frac{1}{3} + \frac{1}{2\cdot3}\\
&= \left(1 + \frac{1}{2}\right) + \frac{1}{3}\left(1 + \frac{1}{2}\right)\\
s_5 &= 1 + \frac12+\frac13+\frac1{2\cdot3}+\frac14+\frac1{2\cdot4} +\frac1{3\cdot4} + \frac{1}{2\cdot3\cdot4}\\
&= \left(1 + \frac12+\frac13+\frac1{2\cdot3}\right) + \frac14\left(1 + \frac12+\frac13+\frac1{2\cdot3}\right)
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1764677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Time and Work problems $4$ boys and $5$ girls can do $\frac{1}{2}$ work in $6$ days. after this $1$ boy and $2$ girls are added and $\frac{1}{3}$ work is done in $3$ days. how many boys must be added to complete the remaining work in $1$ day?
My Attempt
$4$ boys and $5$ girls can do $\frac{1}{2}$ work in $6$ days.
$5$ boys and $7$ girls can do $\frac{1}{3}$ work in $3$ days.
$5+x$ boys and $7$ girls can do $\frac{1}{6}$ work in $1$ day.
I could just reach upto here. can anyone help me further?
| Let $b$ be the time (in days) a boy needs alone to complete the entire job and $g$ the time (in days) a girl needs alone to do the same. Then:
$$
4\frac{1}{b}+5\frac{1}{g}=\frac{\frac{1}{2}}{6}
$$
and also
$$
5\frac{1}{b}+7\frac{1}{g}=\frac{\frac{1}{3}}{3}
$$
which give $b=g=108$. Thus:
$$
(5+x)\frac{1}{108}+7\frac{1}{108}=\frac{1-\frac{1}{2}-\frac{1}{3}}{1}=\frac{\frac{1}{6}}{1}\Leftrightarrow 6(x+12)=108\Leftrightarrow x=6
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1764774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Expectation of $(X+Y+Z)^3$ Given that X,Y and Z are independent random variables and we are also given the moment generating functions of X,Y and Z, is there an easy way to find the expectation of $(X+Y+Z)^3$?
Any help would be appreciated.
| Given that $X,Y$ and $Z$ are independent random variables, so:
$E(X+Y+Z)^3=E(X^3+Y^3+Z^3+3(X+Y)(Y+Z)(X+Z))$
$= E(X^3+Y^3+Z^3+3X^2Y + 3X^2Z + 3XY^2 + 3Y^2Z + 3XZ^2 + 3YZ^2 + 6XYZ)$
$= E(X^3)+E(Y^3)+E(Z^3)+E(3X^2Y) + E(3X^2Z) + E(3XY^2) + E(3Y^2Z) + E(3XZ^2) + E(3YZ^2) + E(6XYZ)$
$= E(X^3)+E(Y^3)+E(Z^3)+3E(X^2)E(Y) + 3E(X^2)E(Z) + 3E(X)E(Y^2) + 3E(Y^2)E(Z) + 3E(X)E(Z^2) + 3E(Y)E(Z^2) + 6E(X)E(Y)E(Z)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1766394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Does anyone know a reference to best-fitting lines with integral coefficients? I'm writing up a manual on how to generate "nice" Linear Algebra problems; that is, where the solutions tend to be integral. I "discovered" the following fact about the best-fitting line:
Theorem. Let $x_1,x_2,\ldots,x_n$ be integers such that (a) $\displaystyle\sum_{i=1}^n x_i=0$, and (b) $x_i\not=x_j$ for some $i,j$. Then the best-fitting line $y=ax+b$
for the data points $(x_1,y_1), (x_2,y_2), \ldots, (x_n,y_n)$
has integral coefficients iff
$$\sum_{k=1}^n x_i y_i \equiv 0 \pmod {\sum_{i=1}^n {x_i}^2}
\quad\quad\quad\quad{\rm and}\quad\quad\quad\quad
\sum_{k=1}^n y_i \equiv 0 \pmod {n}.$$
This is the sort of result which is nice, was probably discovered before, and which I cannot find any reference to. Has anyone seen this in a paper anywhere?
| The trial function is
$$
y(x) = a_{0} + a_{1} x.
$$
The data will consist of a sequence of measurements $\left\{ x_{k}, y_{k} \right\}_{k=1}^{m}.$
The goal is to create restraints to insure that $a\in\mathbb{Z}^{2}$.
Derivation
The linear system is
$$
\begin{align}
\mathbf{A} a &= y \\
%
\left[ \begin{array}{cc}
\mathbf{1} & x
\end{array} \right]
%
\left[ \begin{array}{c}
a_{0} \\ a_{1}
\end{array} \right]
%
&=
\left[ \begin{array}{c}
y
\end{array} \right]
%
\end{align}
$$
The normal equations are
$$
\begin{align}
\mathbf{A}^{*} \mathbf{A} a &= \mathbf{A}^{*} y \\
%
\left[ \begin{array}{cc}
\mathbf{1} \cdot \mathbf{1} & \mathbf{1} \cdot x \\
x \cdot \mathbf{1} & x \cdot x
\end{array} \right]
%
\left[ \begin{array}{c}
a_{0} \\ a_{1}
\end{array} \right]
%
&=
\left[ \begin{array}{c}
\mathbf{1} \cdot y \\
x \cdot y
\end{array} \right].
%
\end{align}
$$
The solution for the normal equations is
$$
\begin{align}
a &= \left( \mathbf{A}^{*} \mathbf{A} \right)^{-1} \mathbf{A}^{*} y \\
%
&=
%
\left(
\det \left( \mathbf{A}^{*} \mathbf{A} \right)
\right)^{-1}
%
\left[ \begin{array}{rr}
x \cdot x & -\mathbf{1} \cdot x \\
-\mathbf{1} \cdot x & \mathbf{1} \cdot \mathbf{1}
\end{array} \right]
%
\left[ \begin{array}{c}
\mathbf{1} \cdot y \\
x \cdot y
\end{array} \right].
%
\end{align}
$$
The task is simplified with the a priori constraint that the $x$ values have zero mean:
$$
\mathbf{1} \cdot x = 0.
$$
Now the solution is
$$
\begin{align}
a
%
&=
%
\left(
\det \left( \mathbf{A}^{*} \mathbf{A} \right)
\right)^{-1}
%
\left[ \begin{array}{cc}
x \cdot x & 0 \\
0 & \mathbf{1} \cdot \mathbf{1}
\end{array} \right]
%
\left[ \begin{array}{c}
\mathbf{1} \cdot y \\
x \cdot y
\end{array} \right].
%
\end{align}
$$
Fix the denominator
Turn attention to the denominator
$$
\det
\left( \mathbf{A}^{*} \mathbf{A} \right)
%
=
%
\left( \mathbf{1} \cdot \mathbf{1} \right) \left( x \cdot x \right) -
\left( \mathbf{1} \cdot x \right)
%
=
%
\left( \mathbf{1} \cdot \mathbf{1} \right) \left( x \cdot x \right).
$$
The solution is
$$
a_{LS} = \left( \left( \mathbf{1} \cdot \mathbf{1} \right) \left( x \cdot x \right)\right)^{-1}
%
\left[ \begin{array}{c}
\left( x \cdot x \right) \left( \mathbf{1} \cdot y \right) \\
\left( \mathbf{1} \cdot \mathbf{1} \right) \left( x \cdot y \right)
\end{array} \right].
$$
The explicit form for intercept $a_{0}$ and slope $a_{1}$ makes the constraints apparent:
$$
a_{0} =
\frac{\mathbf{1} \cdot y} {\mathbf{1} \cdot \mathbf{1}}, \quad
%
a_{1} =
\frac{ x \cdot y} { x \cdot x }
%
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1769283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
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
| Plug in $c = \frac1{ab}$, and set the partial derivatives $\frac{\partial}{\partial a}$ and $\frac{\partial}{\partial b}$ equal to zero to find candidates for a minimum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1769608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Check that $B$ is a basis $B= [a,b,c]$ where $a=[-3,2,1]^t$, $b=[-3,2,-1]^t$ and $c=[1,6,-1]^t$
How do I make this into a matrix to find if its a basis of $\mathbb{R}^3$?
I think the matrix will look like this but I'm not sure
$$\begin{bmatrix}
-3&-3&1\\
2&2&6\\
1&-1&1\\
\end{bmatrix}$$
| You could also set
\begin{eqnarray*}
a&=&-3e_1+2e_2+e_3,\\
b&=&-3e_1+2e_2-e_3,\\
c&=&e_1+6e_2+e_3,
\end{eqnarray*}
to indicate the basis change from the old basis $\{e_1,e_2,e_3\}$ to the new one $\{a,b,c\}$.
Visually, the above table allows to find the new components of a vector, say,
$$X=8e_1+199e_2-10e_3,$$
first you solve for the $e_i$
\begin{eqnarray*}
e_1&=&\frac{-1}{5}a+\frac{-1}{10}b+\frac{1}{10}c,\\
e_2&=&\frac{-1}{20}a+\frac{1}{10}b+\frac{3}{20}c,\\
e_3&=&\frac{1}{2}a+\frac{-1}{2}b.
\end{eqnarray*}
With them, subbed in $x$, we're going to give its new components.
This correspond to a matricially multiplication
$$
\left(
\begin{array}{ccc}
-\frac{1}{5} & -\frac{1}{20} & \frac{1}{2} \\
-\frac{1}{10} & \frac{1}{10} & -\frac{1}{2} \\
\frac{1}{10} & \frac{3}{20} & 0 \\
\end{array}
\right)\cdot\left(
\begin{array}{c}
8 \\
199 \\
-10 \\
\end{array}
\right),$$
the $3\times 3$ matrix is the inverse of your $B$.
So, finally
$$X=
-\frac{251}{20}a+\frac{161}{10}b+\frac{373}{20}c.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1769994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is there a systematic way of finding the factorization over the closure of $\mathbb{Z}_2$ of $p(x) = x^{32} - x$? And similar polynomials of the form $x^{p^n} - x$. I know that the degrees of the irreducible monic polynomials that factorize $x^{32} - x$ will have degree $d \vert 5 = 1, 5$. Also, I know that $x$ and $(x-1)$ are factors, so that leaves me with:
$x^5, x^5+x^4, x^5+x^4+x^3, x^5+x^4+x^3+x^2, ..., x^4, x^4+x^3, x^4+x^3+x^2,...$
It seems like I can spend an entire month working by trial and error. Is there some better way?
| It sounds like you know that $x^{32} - x$ is $x(x - 1)$ times all irreducible polynomials of degree $5$. There are $6$ of these out of the $32$ (monic) polynomials of degree $5$. You can rule out most of these by checking for linear factors: first, everything divisible by $x$ is reducible, which leaves $16$ possibilities (constant term must be $1$). Incidentally, this rules out every example you wrote down, some of which aren't even of degree $5$.
Second, everything divisible by $x + 1$ is reducible, which leaves $8$ possibilities (sum of terms must be odd), namely
$$x^5 + x + 1$$
$$x^5 + x^2 + 1$$
$$x^5 + x^3 + 1$$
$$x^5 + x^4 + 1$$
$$x^5 + x^3 + x^2 + x + 1$$
$$x^5 + x^4 + x^2 + x + 1$$
$$x^5 + x^4 + x^3 + x + 1$$
$$x^5 + x^4 + x^3 + x^2 + 1$$
$6$ of these are irreducible, so you want to rule out $2$ more. These $2$ are the ones of the form (irreducible quadratic)(irreducible cubic). There's exactly one irreducible quadratic, namely $x^2 + x + 1$, so the only thing you need to do is check what's divisible by this. $x^2 + x + 1$ divides $x^3 - 1$, so you can reduce the exponents $\bmod 3$ to do this. You get that $x^5 + x + 1$ and $x^5 + x^4 + 1$ are divisible by $x^2 + x + 1$ and that's it; the other $6$ are the irreducibles of degree $5$.
In total this took less than 5 minutes, I think.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1770330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
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.
| Assuming we know $\lim_n \sum_{k=1}^n\frac{1}{k^2} = \frac{\pi^2}{6}$ (see e.g. https://en.wikipedia.org/wiki/Basel_problem) then
$$\lim_n \frac{6}{\pi^2}\sum_{k=1}^n\frac{1}{k^2} =1.$$
Now write
$$\frac{6}{\pi^2}\sum_{k=1}^n\frac{1}{k^2}=1+\frac{6}{\pi^2}\sum_{k=1}^n\frac{1}{k^2}-1 = 1+\frac{1}{\left(\frac{6}{\pi^2}\sum_{k=1}^n\frac{1}{k^2}-1\right)^{-1}}.$$
Denote $a_n := \left(\frac{6}{\pi^2}\sum_{k=1}^n\frac{1}{k^2}-1\right)^{-1}$. Observe that $\lim_n a_n=\infty$. Hence,
$$\lim_n \left(1+\frac{1}{a_n}\right)^{a_n} =: e,$$
by definition.
Thus,
$$\left(\frac{6}{\pi^2}\sum_{k=1}^n \frac{1}{k^2}\right)^n =\left[\left( 1+\frac{1}{a_n}\right)^{a_n}\right]^{n\ a_n^{-1}} .$$
Taking limit
$$\lim_n\left(\frac{6}{\pi^2}\sum_{k=1}^n \frac{1}{k^2}\right)^n = e^{\lim_n n \ a_n^{-1}},$$
where in the last step we used the continuity of the exponential function.
Now, let us compute $n\ a_n^{-1}$.
$$\lim_n \frac{\frac{6}{\pi^2}\sum_{k=1}^n\frac{1}{k^2}-1}{\frac{1}{n}}.$$
The above limit, which is indefinite, can be computed easily by using Stolz-Cesàro Theorem (see e.g. https://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem). If $c_n := \frac{6}{\pi^2}\sum_{k=1}^n\frac{1}{k^2}-1$ and $b_n:= \frac{1}{n}$ then
$$\lim_n = \frac{c_n}{b_n} = \lim_n \frac{c_{n+1}-c_n}{b_{n+1}-b_n} = \frac{\frac{6}{\pi^2}\frac{1}{(n+1)^2}}{\frac{-1}{n(n+1)}}=-\frac{6}{\pi^2}.$$
Altogether,
$$\lim_n \left( \frac{6}{\pi^2}\sum_{k=1}^n\frac{1}{k^2}\right)^n = e^{-\frac{6}{\pi^2}}.$$
| {
"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": 1
} |
Solving without induction show that $a_{n}=2n-1$ Let $a_{1}=1$,and such $$4S_{n}=n(a_{n}+a_{n+1})$$
where $S_{n}=a_{1}+a_{2}+\cdots+a_{n}$
find $a_{n}$
since $a_{2}=3$,and we can easy to prove $a_{n}=2n-1$
Induction Methods
Assume $a_{k}=2k-1$, so we have
$$S_{k}=a_{1}+a_{2}+\cdots+a_{k}=1+3+5+\cdots+(2k-1)=k^2.$$
then we have
$$a_{k+1}=\dfrac{4S_{k}}{k}-a_{k}=4k-(2k-1)=2k+1$$
Question: can we without induction to find this $a_{n}?$
| If we want to solve this without any guessing and without using induction in any way, the best way to do so is to work with generating functions. This method, given that it really makes no assumptions on the final form, takes considerably more work, but can also solve generalizations of this problem.
In particular, the big trick is to consider the 'function' defined as
$$a(x) = a_1+a_2x+a_3x^2+\ldots.$$
We treat this as a formal sum, meaning we ignore any issues of convergence and just treat it algebraically. We then express the recurrence relation as an equation on $a(x)$ and use that to find $a(x)$ and then work backwards to find the coefficients $a_i$. A bit more explaining is necessary for why calculus works, but let's ignore those issues.
So, our equation is
$$4S_n=n(a_n+a_{n+1}).$$
First, we can express the left hand side using another generating function:
$$S(x)=S_1+S_2x+S_3x^2+\ldots.$$
A well known relation lets us write $S(x)$ in terms of $a(x)$:
\begin{align*}S(x)=\frac{a(x)}{1-x}=a(x)\cdot (1+x+x^2+\ldots)=a_1&+a_2x+a_3x^2+\ldots\\&+a_1x + a_2x^2+\ldots \\&\hskip.45in +a_1x^2+\ldots\end{align*}
where we use $\frac{1}{1-x}=1+x+x^2+\ldots$ and the last equality shows the distribution of this over $a(x)$, noting that each column is a partial sum.
For the other side, we have $n$, which hints that we should try differentiating something with coefficients of $a_n+a_{n+1}$. In particular, let $$B(x)=(1+x)a(x)= a_1+(a_1+a_2)x + (a_2+a_3)x^2 + \ldots$$ and note its derivative $$B'(x)=(a_1+a_2) + 2(a_2+a_3)x + 3(a_3+a_4) + \ldots.$$ If we see that he coefficients of $x^{n-1}$ in $B'(x)$ are $n(a_n+a_{n+1})$ and the coefficients of $x^{n-1}$ in $S(x)$ are $S_n$, we get that we want
$$4S(x)=B'(x)$$
Note that the product rule gives $B'(x)=a(x) + (1+x)a'(x)$. Expanding both sides in terms of $a$ gives
$$\frac{4a(x)}{1-x}=a(x) + (1+x)a'(x)$$
which is a differential equation. Then, we can rearrange this equation as
$$\frac{(3+x)a(x)}{1-x^2}=a'(x)$$
$$\frac{3+x}{1-x^2}=\frac{a'(x)}{a(x)}$$
The left side may be pulled apart using partial fractions
$$\frac{1}{1+x}+\frac{2}{1-x}=\frac{a'(x)}{a(x)}.$$
Integrating both sides in $x$ gives
$$\log(1+x) - 2\log(1-x) + C_1 = \log(a(x))$$
Taking exponents of both sides, and letting $C_2=e^{C_1}$ gives
$$C_2\cdot \frac{1+x}{(1-x)^2}=a(x)$$
Moreover, noting that the constant term on the left is just $C_2$, we get $C_2=a_1=1$ thus
$$\frac{1+x}{(1-x)^2}=a(x)$$
Then, we break this up by partial fractions to yield
$$\frac{-1}{1-x}+\frac{2}{(1-x)^2}=a(x).$$
This is convenient. It is well known that we have $\frac{1}{1-x}=1+x+x^2+\ldots$ and $\frac{1}{(1-x)^2}$ can be seen to be $1+2x+3x^2+4x^3+\ldots$ by various methods. Thus, the coefficient of $x^n$ on the left will be $-1 + 2(n+1)=2n+1$. This must equal the coefficient $a_{n+1}$ on the right side so $2n+1=a_{n+1}$ meaning $2n-1=a_n$.
It is worthy of note that, if one applies the proper framework to this, the fact that the algebra checks out is sufficient to prove the correctness of this - that is, we don't need to go back and prove that our system satisfies the original conditions. Moreover, the same framework quickly can solve things like $6S_n=n(a_n+a_{n+1})$, where the answer is harder to guess; in this particular case, we'd get $$a(x)=\frac{(1+x)^2}{(1-x)^3}=\frac{4}{(1-x)^3}- \frac{4}{(1-x)^2} + \frac{1}{1-x}=\sum_{n=0}^{\infty}(2(n+1)(n+2) - 4(n+1) + 1)x^n$$ which gives $a_n=1-2n+2n^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1772266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Understanding steps to obtain derivative of $|x_n|^{\frac{3}{7}}$ I was trying to solve the following derivative
$$|x_n|^{\frac{3}{7}}$$
as follows
$$(|x_n|^{\frac{3}{7}})' \\= \frac{3}{7}(|x_n|^{\frac{3}{7} - 1}) \cdot (|x_n|)' \\= \frac{3}{7}|x_n|^{\frac{-4}{7}} \cdot \frac{x_n}{|x_n|} \\= \frac{3}{7}|x_n|^{\frac{-4}{7} + 1} \cdot x_n \\= \frac{3}{7}|x_n|^{\frac{3}{7}} \cdot x_n$$
which is different from the result that this online calculator gives me, which is
$$\frac{3}{7} x^{\frac{3}{7}} \frac{1}{|x|}$$
What did I do wrong?
| The absolute value just implies you have symmetry about the y-axis.
Assume $x>0$, then
$$\frac{d}{dx} x^{3/7} = \frac{3}{7}x^{-4/7} = \frac{3}{7}x^{3/7-1}=\frac{3}{7}\frac{x^{3/7}}{x}(x>0)$$
If $x<0$ then we just need to reverse the sign of the argument so $-x$ is positive:
$$\frac{d}{dx} (-x)^{3/7} = \frac{-3}{7}(-x)^{-4/7} = \frac{-3}{7}(-x)^{3/7-1}=\frac{-3}{7}\frac{(-x)^{3/7}}{-x}=\frac{3}{7}\frac{(-x)^{3/7}}{x}(x<0)$$
Note that the final result has the same exact form except that the numerator is corrected to always be positive, while the denominator is allowed to vary in sign. We can merge these two results by simply taking the numerator to be $|x|^{3/7}$ and we get the final form:
$$ \frac{3}{7}\frac{|x|^{3/7}}{x} \equiv \frac{3}{7}\frac{x}{|x|^{11/7}}$$
Note that the formula give by your online calculator is not correct: $\frac{3}{7}\frac{x^{3/7}}{|x|}$ is a complex number for $x<0$! (just calculate $(-2)^{3/7}$).
As for your derivation, you made a mistake in going from
$$\frac{3}{7}|x_n|^{\frac{-4}{7}} \cdot \frac{x_n}{|x_n|}$$
to
$$\frac{3}{7}|x_n|^{\frac{-4}{7} + 1} \cdot x_n$$
You distributed the numerator twice (note the +1 in exponent yet $x$ is still there) and lost the denominator.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1772961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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 ?
| What you are looking at is $$\frac{1+x^4+(x+1)^4}{1+x^2+(x+1)^2}$$ where $x=2012$. Expanding and simplifying, this is $$\frac{2+4x+6x^2+4x^3+2x^4}{2+2x+2x^2}=\frac{1+2x+3x^2+2x^3+x^4}{1+x+x^2}$$
$$=\frac{(1+x+x^2)^2}{1+x+x^2}=1+x+x^2.$$
Thus the final answer is $$2012^2+2012+1,$$ and from here you can calculate the number by hand.
| {
"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": 2
} |
Is $f(x) = \sqrt[3]{x}$ continuous on $ [0,\infty)$ and is it uniformly continuous on $[0,\infty)$ Note, there is a similar question here: Show $\sqrt[3]{x}$ is or isn't uniformly continuous.
I am asking my question anyway because the one here does not ask about the Sequential Criterion for Absence of Uniform continuity
$ f(x)=\sqrt[3]{x}$ is continuous on $[0,\infty)$, note
$|f(x)-f(c) | \rightarrow |\sqrt[3]{x}-\sqrt[3]{c}| <\epsilon$
$|\sqrt[3]{x}-\sqrt[3]{c}|= |(x-c)(\sqrt[3]{x^2}+\sqrt[3]{xc}+\sqrt[3]{c^2})| $
If we insist on $\delta < 1$, then we ensure that $x$ is within $(c-1,c+1)$
Then $(\sqrt[3]{x^2}+\sqrt[3]{xc}+\sqrt[3]{c^2})| $ $\geq$ $3\sqrt[3]{c^2}$, giving us $|(x-c)(3\sqrt[3]{c^2})|<\epsilon$.
So $\delta = \dfrac{\epsilon} {3\sqrt[3]{c^2}}$
Is it uniformly continuous on $[0,\infty)$?
Here I use the Sequential Criterion for Absence of Uniform Continuity which states:
A function $f: A \rightarrow \mathbb{R}$ fails to be uniformly continuouson $A$ if and only if $\exists \epsilon_0 >0$ and two sequences $(x_n), (y_n) \in A$ satisfying $|x_n - y_n| \rightarrow 0$ but $|f(x_n) - f(y_n)| \geq \epsilon_0$
I let $x_n=n, y_n=n+\frac{1}{n}$ and observe that $|x_n - y_n| \rightarrow 0$
but
$$|f(x_n) - f(y_n)| = \left| \left( \sqrt[3]{n}-\sqrt[3]{\left(n+\frac{1}{n} \right) }\, \right) \left\{ \frac{\sqrt[3]{n^2} + \sqrt[3]{n\left( n + \frac 1 n \right) + \sqrt[3]{\left( n + \frac 1 n \right)^2}}}{\sqrt[3]{n^2} + \sqrt[3]{ n \left( n + \frac 1 n \right) + \sqrt[3]{\left( n + \frac 1 n \right)^2}}} \right\} \right| $$
$$= \frac{|n-(n + \frac{1}{n})|}{\sqrt[3]{n^2}+\sqrt[3]{n(n+\frac{1}{n})}+\sqrt[3]{(n+\frac{1}{n}})^2},$$
which goes to $0$ and thus $f$ is uniformly continuous on $[0,\infty)$.
| Regarding uniform continuity...
You have shown that there is a single pair of sequences for which $|x_n - y_n| \rightarrow 0$ and not $|f(x_n) - f(y_n)| \geq \epsilon_0$. You have not shown this for any other sequence. Since you need to show this for any possible pair of sequences, you have only begun your task.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1774772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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.
| hint: $\dfrac{x^2+3}{x^6(x^2+1)}= x^{-6}+\dfrac{2}{x^6(x^2+1)}$. At this point, let $x = \dfrac{1}{y}$, then the second part becomes $\dfrac{y^6}{y^2+1}= y^4-\dfrac{y^4}{y^2+1}=y^4-y^2 + \dfrac{y^2}{y^2+1}=y^4-y^2+1 - \dfrac{1}{y^2+1}$,and you can take it from here.
| {
"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": 4
} |
Finding a generating function for $\{(n+2)C_{n+1}\}^\infty_{n=0}$ I'm trying to come up with a generating function for $\{(n+2)C_{n+1}\}^\infty_{n=0}$ where $C_n$ is the $n$th Catalan number.
I know we can write $(n+2)C_{n+1} = 2(2n+1)C_n$. I also tried to follow this post: Simplifying Catalan number recurrence relation
However I am getting somewhat stuck because of the added $(n+2)$ term.
|
Note that $$(n+1)C_n=\binom{2n}{n}\qquad\qquad n\geq 0$$ are the Central Binomial Coefficients with the generating series representation
\begin{align*}
\sum_{n=0}^\infty\binom{2n}{n}x^n=\frac{1}{\sqrt{1-4x}}\qquad\qquad |4x|<1\tag{1}
\end{align*}
A generating function for $\{(n+2)C_{n+1}\}^\infty_{n=0}$ is therefore
\begin{align*}
\sum_{n= 0}^\infty& (n+2)C_{n+1}x^n\\
&=\sum_{n=1}^\infty (n+1)C_{n}x^{n-1}\\
&=\frac{1}{x}\sum_{n=1}^\infty \binom{2n}{n}x^{n}\\
&=\frac{1}{x}\left(\frac{1}{\sqrt{1-4x}}-1\right)
\end{align*}
Hint: The representation (1) is an application of the binomial series
\begin{align*}
(1+x)^{\alpha}=\sum_{n=0}^{\infty}\binom{\alpha}{n}x^n\qquad |x|<1, \alpha\in\mathbb{C}
\end{align*}
and the relation
\begin{align*}
\binom{-\frac{1}{2}}{n}=\frac{(-1)^n}{4^n}\binom{2n}{n}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1778251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that $\tan x +\frac{1}{2}\tan \frac{x}{2}+\dots + \frac{1}{2^n}\tan \frac{x}{2^n} = \frac{1}{2^n}\tan \frac{x}{2^n}-2\cot(2x) $ Show that $$\tan x +\frac{1}{2}\tan \frac{x}{2}+\dots + \frac{1}{2^n}\tan \frac{x}{2^n} = \frac{1}{2^n}\tan \frac{x}{2^n}-2\cot(2x), \quad (n=0,1,\ldots). $$
I tried to prove this using induction but with no result. I'm not sure how to begin to solve this.
| Hint:
$$\tan \beta - \cot \beta= -2\cot 2 \beta$$
Then
$$\tan x +\frac{1}{2}\tan \frac{x}{2}+\dots + \frac{1}{2^n}\tan \frac{x}{2^n} = \frac{1}{2^n}\tan \frac{x}{2^n}-\color{red}{2\cot(2x)} $$
$$\color{red}{\tan x} +\frac{1}{2}\tan \frac{x}{2}+\dots + \frac{1}{2^n}\tan \frac{x}{2^n} = \frac{1}{2^n}\tan \frac{x}{2^n}+\color{red}{\tan x - \cot x} $$
$$\frac{1}{2}\tan \frac{x}{2}+\dots + \frac{1}{2^n}\tan \frac{x}{2^n} = \frac{1}{2^n}\tan \frac{x}{2^n}\color{red}{ - \cot x} $$
$$\tan \frac{x}{2}+\dots + \frac{1}{2^n}\tan \frac{x}{2^n} = \frac{1}{2^n}\tan \frac{x}{2^n}\color{red}{ - 2\cot x} $$
...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1778357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Ellipsoid moment of inertia matrix Some background info: torque $\tau$ is defined as $$\tau = I*d\omega$$
Where $I$ is the moment of inertia matrix and $d\omega$ is an object's rotational acceleration. As I understand it, the inertia matrix acts just like mass in that it counteracts the torque (for example, if an object is spinning around the x axis, a big value of $I_{xx}$ means that the object needs more torque around the x axis in order to spin).
However, angular momentum $M$ can be defined as $$M=I * \omega$$
Where $\omega$ is the rotational velocity. So it seems that torque is the time derivative of angular momentum.
Using these facts, how would I find the moment of inertia matrix for an ellipsoid with uniform density of the form $$\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}≤9$$ with $a≠b≠c≠0$? Would I have to use spherical coordinates somehow? I'm not given any torque or angular velocity information. Any guidance is appreciated.
| In usual notation,
\begin{align*}
\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2} &= 1 \\
I_{ij} &=
\iiint \rho(\mathbf{r}) (r^{2} \delta_{ij}-x_{i} x_{j}) d^{3} \, \mathbf{r} \\
I &= \frac{m}{5}
\begin{pmatrix}
b^2+c^2 & 0 & 0 \\
0 & c^2+a^2 & 0 \\
0 & 0 & a^2+b^2 \\
\end{pmatrix}
\end{align*}
In your case: $$I=\frac{3m}{5}
\begin{pmatrix}
b+c & 0 & 0 \\
0 & c+a & 0 \\
0 & 0 & a+b \\
\end{pmatrix}$$
See also the link here.
P.S.:
By symmetry, $i\neq j \implies I_{ij}=0$
Let $x=a X$, $y=b Y$ and $z=c Z$
\begin{align*}
\iiint_{x^2/a^2+y^2/b^2+z^2/c^2<1} x^{2} dV &=
abc \iiint_{X^2+Y^2+Z^2<1} a^{2} X^{2} dX dY dZ \\
&=
abc\int_{-1}^{1}
\left[
\int_{-\sqrt{1-X^2}}^{\sqrt{1-X^2}}
\left( \int_{-\sqrt{1-X^2-Y^2}}^{\sqrt{1-X^2-Y^2}} dZ \right) dY
\right] a^2X^2 dX \\
&=
abc\int_{-1}^{1}
\left[
\int_{-\sqrt{1-X^2}}^{\sqrt{1-X^2}}
2\sqrt{1-X^2-Y^2} dY
\right] a^2X^2 dX \\
&= abc\int_{-1}^{1} \pi (1-X^2) a^2X^2 dX \\
&= \pi abc \times \frac{4a^2}{15} \\
\rho &= \frac{3m}{4\pi abc} \\
I_{11} &= \rho \iiint (y^2+z^2) dV \\
&= \frac{m}{5} (b^2+c^2)
\end{align*}
Finish the rest by symmetry.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1778627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
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 $y> 1$
| Well, there is probably an easy algebraic method:
$$y=\sqrt{x^2+1}-x$$$$\implies y^2+x^2+2xy=x^2+1$$$$\implies y^2+2xy-1=0$$
to find range of $y$, we reform this equation as
$$0x^2+2xy+y^2-1=0$$
this is quadratic in $x$ and since $x$ is real , $y$ should also be real so discriminent should be greater than zero
$$D=(2y)^2-4(y^2-1)(0)>0$$
and this implies $$y^2>0$$$$\implies y>0$$or$$y<0$$ but $y<0$ is ruled out since
$|x|=\sqrt{x^2}<\sqrt{x^2+1}$, so $x<\sqrt{x^2+1}$ ,therefore $y>0$ for all values of $x$ so $$y\in (0,\infty)$$
| {
"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": 1
} |
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?
| Hint of one possible solution: (assume your derivation is correct)
$$4\cos^2x-4\cos x-2=(2\cos x-1)^2-3\ge -3$$
and on $(0,\pi)$ we have $0<\sin x\le 1$.
Thus the LHS of your last equation should be $\ge -3$ on $(0,\pi)$.
Then you only need to check when the equality is satisfied, which is simple.
| {
"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": 1
} |
General formula for a diagonal parabola. What is the general formula for a diagonal parabola facing a given direction with a given vertex (x,y)?
| The general equation is
$$
y' = mx'^2
$$
where $y'$ and $x'$ are coordinates of a point in a rotated and transalted coordinate system. The good news is that if you're willing to write a point $(x, y)$ as a vector $v = \begin{bmatrix}x \\ y \\ 1\end{bmatrix}$, then the equation of a general parabola is just
$$
v^t M v = 0
$$
where $M$ is the matrix
$$
\frac{1}{2}
\begin{bmatrix}
2d & 0 & 0\\
0 & 0 & 1\\
0 & 1 & 0\\
\end{bmatrix}
$$
The case $d = 1$ corresponds to $y = -x^2$. $d = 2$ corresponds to $y = -(2x)^2$ etc. I went with $ y = -x^2$ to put fewer negative signs in the matrix, and hence give me some hope of getting things right. If you want $y = x^2$ instead, simply rotate by 180 degrees in the formula below (i.e., replace $t$ by $t + 180^\circ$.
If you want to rotate solutions to that equation by an angle $t$ (easier to type than $\theta$), you let
$$
R = \begin{bmatrix}
\cos t & -\sin t & 0\\
\sin t & \cos t & 0\\
0 & 0 & 1\\
\end{bmatrix}
$$
and use the equation
$$
v^t R^t M R v = 0
$$
This will have, as its solutions, the same parabola rotated by angle $t$ clockwise.
Writing $c$ for $\cos t$ and $s$ for $\sin t$, this becomes
$$
\frac{1}{2}
\begin{bmatrix}x & y& 1\end{bmatrix}
\begin{bmatrix}
c & s & 0\\
-s & c & 0\\
0 & 0 & 1\\
\end{bmatrix}
\begin{bmatrix}
2d & 0 & 0\\
0 & 0 & 1\\
0 & 1 & 0\\
\end{bmatrix}
\begin{bmatrix}
c & -s & 0\\
s & c & 0\\
0 & 0 & 1\\
\end{bmatrix}
\begin{bmatrix}x \\ y\\ 1\end{bmatrix} = 0.
$$
Multiplying through by $2$, and doing some matrix multiplies, this becomes
$$
\begin{bmatrix}x & y& 1\end{bmatrix}
\begin{bmatrix}
c & s & 0\\
-s & c & 0\\
0 & 0 & 1\\
\end{bmatrix}
\begin{bmatrix}
2dc & -2ds & 0\\
0 & 0 & 1\\
s & c & 0\\
\end{bmatrix}
\begin{bmatrix}x \\ y\\ 1\end{bmatrix} = 0.
$$
and then
$$
\begin{bmatrix}x & y& 1\end{bmatrix}
\begin{bmatrix}
2dc^2 & -2dcs & s\\
-2dcs & 2ds^2 & c\\
s & c & 0\\
\end{bmatrix}
\begin{bmatrix}x \\ y\\ 1\end{bmatrix} = 0.
$$
Written out without matrices, that's
$$
2dc^2 x^2 - 4dcs xy + 2ds^2 y^2 +2sx + 2cy = 0
$$
and you can substitute in $\cos t$ for $c$ and $\sin t$ for $s$.
To translate that to have its "vertex" at a point $(a, b)$, you could do a similar matrix trick, or something far simpler:
$$
2dc^2 (x-a)^2 - 4dcs (x-a)(y-b) + 2ds^2 (y-b)^2 + 2s(x-a) + 2c(y-b) = 0
$$
And that's your answer. Dividing by 2 makes it a little prettier:
$$
dc^2 (x-a)^2 - 2dcs (x-a)(y-b) + ds^2 (y-b)^2 + s(x-a) + c(y-b) = 0.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1781248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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
| Define $$b_n := \frac{n(n-1)}{a_n}$$
Then your recurrence reduces to $$b_{n+1}-b_n=2n+4$$
So, $$b_n = b_1 + 2\sum_{i=1}^{n-1} i + 4\sum_{i=1}^{n-1} 1 = b_1 + n(n-1)+4(n-1)=b_1+n^2+3n-4$$
$$\prod_{i=1}^n a_i = \prod_{i=1}^n \frac{n(n-1)}{b_i} = \prod_{i=1}^n \frac{n^2-n}{b_1+n^2+3n-4}$$
Now from the value of $b_1=0$ for nonzero $a_1$ so I think there might be some telescoping here.
$$\prod_{i=1}^n \frac{n}{n+4} = \frac{1}{\color{red}{5}}\frac{2}{\color{red}{6}}\frac{3}{\color{red}{7}}\frac{4}{\color{red}{8}}\color{red}{\frac{\color{red}{5}}{9}\frac{{6}}{10}\dots\frac{n-4}{n}}\frac{\color{red}{n-3}}{n+1}\frac{\color{red}{n-2}}{n+2}\frac{\color{red}{n-1}}{n+3}\frac{\color{red}{n}}{n+4}=\frac{4!}{(n+1)(n+2)(n+3)(n+4)}$$
So the limit is (assuming you meant $n^4$ inside the limit)
$$\lim_{n\to\infty} n^4 \frac{4!}{(n+1)(n+2)(n+3)(n+4)} = \fbox{24}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1783377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to demonstrate that $2^{2^n - 2} + 1$ is a nonprime number? This, considering $n ≥ 3$.
I have tried by induction; I suppose that it's true for all n less than or equal to k (and greater than or equal to 3), but then I stride when I go to prove for n = k + 1.
$2^{2^{k+1} - 2 } + 1 = $
$2^{2* 2^k - 2} + 1$
| Multiply by $4$ and look at things mod $5$:
$$4(2^{2^n-2}+1)=2^{2^n}+4=2^{4\cdot2^{n-2}}+4\equiv1+4\equiv0\mod5$$
so $5$ is a prime divisor of $2^{2^n-2}+1$. Note that for $n=2$, $2^{2^2-2}+1=5$, but for $n\ge3$, $2^{2^n-2}+1\gt5$, so $5$ is a proper divisor.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1784395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Limits using derivatives If a differentiable function $f(x)$ satisfies a functional rule
$f(x) + f(x+2) + f(x+4) = 0$ for all $x$ belonging to real numbers,
Then find the value of :
$$\lim \limits_{x \to 0} \frac{\bigl(f(x+12)\bigr)^2- f(x) f(0) - f(x+6) f(18) + \bigl(f(18)\bigr)^2}{x\left({\frac{\pi}{4} - \tan^{-1}{(1-x)}}\right)} $$
what's the method?
The answer is $32$.
| If dont know if the following can help you
\begin{align}
&f(x) + f(x+2) + f(x+4) = 0\hspace{1cm} (I)\\
&f(x+2) + f(x+4) + f(x+6) = 0\hspace{1cm} (II)\\
&f(x+6) + f(x+8) + f(x+10) = 0\hspace{1cm} (III)\\
&f(x+8) + f(x+10) + f(x+12) = 0\hspace{1cm} (IV)\\
&f(x+12) + f(x+14) + f(x+16) = 0\hspace{1cm} (V)\\
&f(x+14) + f(x+16) + f(x+18) = 0\hspace{1cm} (VI)\\
\end{align}
First $(I)-(II)$ gives $f(x)=f(x+6)$ In particular $f(0)=f(6)$
also $(I)-(II)+(III)-(IV)$ gives $f(x)=f(x+12)$ In particular $f(0)=f(12)$
also also $(I)-(II)+(III)-(IV)+(V)-(VI)$ gives $f(x)=f(x+18)$
etc ... etc ..
Noting that $\lim \limits_{x \to 1} x\left({\frac{\pi}{4} - \tan^{-1}{(1-x)}}\right)=\frac{\pi}{4}$. So
$$\lim \limits_{x \to 1} \frac{\bigl(f(x+12)\bigr)^2- f(x) f(0) - f(x+6) f(18) + \bigl(f(18)\bigr)^2}{x\left({\frac{\pi}{4} - \tan^{-1}{(1-x)}}\right)}= \frac{4}{\pi} \lim \limits_{x \to 1} (f(x)+f(0))^{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1786133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The minimum value of $\frac{a^3 + b^3 + c^3 }{\sqrt{a^4 + b^4 + c^4 }}$ . When $a^2 + b^2 + c^2 = 1 $ Asume $a, b, c $ is non-negative real.
I got above equation at this situation ;
$\vec {x}= (a, b, c)$ ,
$\vec {y} = (a^2 , b^2 , c^2 ) $
$$ cos \phi = \frac{ \vec x \cdot \vec {y}} { \Vert {\vec {x} \Vert} \times\Vert {\vec {y} \Vert}} = \frac{ a^3 + b^3 + c^3 } { \sqrt{ a^4 + b^4 + c^4 }}$$
In this case I want to know the Maximum value of $\phi$ .
But I can't compute.
I want your help.
| Let the minimum be $m$. Then we have the homogeneous symmetric inequality
$$(a^3+b^3+c^3)^2 \geqslant m^2(a^2+b^2+c^2)(a^4+b^4+c^4)$$
Using $a+b+c = 3u, ab+bc+ca = 3v^2, abc = w^3$, the above reduces to a quadratic in terms of $w^3$ with positive leading term, so the inequality will hold if it holds for $b=c=1$ and $b=1, c=0 {(\star)}$ . Thus it is enough to have the following satisfied for $a \in \mathbb R^+$:
$$\frac{a^3+2}{\sqrt{(a^2+2)(a^4+2)}} , \frac{a^3+1}{\sqrt{(a^2+1)(a^4+1)}} \geqslant m$$
It turns out the more restrictive condition is the left most, so we find using single variable calculus the minima for $\dfrac{a^3+2}{\sqrt{(a^2+2)(a^4+2)}}$, which is your best $m \approx 0.9576$.
P.S.: There is a third condition $(\star)$ as the quadratic minimum could be at a stationary point, and while it is not restrictive in this case, it is not as easily tractable as the others and is omitted above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1788425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Two real numbers, $x$ and $y$, satisfy the condition $x + y = 2 $. Show $xy(x^2+y^2) \leq 2$
Question: Two real numbers, $x$ and $y$, satisfy the condition $x + y = 2 $.
Show $xy(x^2+y^2) \leq 2$
What I have attempted:
Consider $$x+y=2$$
$$ \Leftrightarrow (x+y)^2 = 2^2 $$
$$ \Leftrightarrow (x+y)^2 = 4 $$
Notice that $$ x^2 + y^2 = (x+y)^2 - 2xy $$
But $(x+y)^2 = 4$
$$ \Longrightarrow x^2 + y^2 = 4-2xy $$
Now we have this inequality; $$xy(x^2+y^2) \leq 2$$
$$\Leftrightarrow xy(4-2xy) \leq 2 $$
$$ \Leftrightarrow 2xy(2-xy) \leq 2 $$
$$ \Leftrightarrow 2(2xy-x^2y^2) \leq 2 $$
Now I am stuck, how should I continue also is there another way to approach this question?
| You have $y=2-x$, so
$$
x^2+y^2=(x+y)^2-2xy=4-2xy=4-2x(2-x).
$$
Then
$$
xy(x^2+y^2)=x(2-x)(4-2x(2-x))
=2r(2-r),
$$
where $r=x(2-x)$.
So consider the parabola $-2r^2+4r$. It's maximum occurs at $r=-4/-4=1$ ("$-b/2a$"), so
$$
2r(2-r)\leq 2(2-1)=2.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1789868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
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.
| Let $\zeta_{\pm}=\sqrt{4\pm 2\sqrt{3}}$. We have:
$$ \frac{1}{\zeta_{+}-\zeta_{-}}=\frac{\zeta_{+}+\zeta_{-}}{\zeta_{+}^2-\zeta_{-}^2}=\sqrt{\frac{\zeta_{+}^2+\zeta_{-}^2+2\zeta_{+}\zeta_{-}}{(\zeta_{+}^2-\zeta_{-}^2)^2}} $$
but $\zeta_{+}\zeta_{-}=2$, $\zeta_{+}^2-\zeta_{-}^2=4\sqrt{3}$ and $\zeta_{+}^2+\zeta_{-}^2=8$, hence:
$$ \frac{1}{\zeta_{+}-\zeta_{-}}=\sqrt{\frac{8+4}{48}}=\color{red}{\frac{1}{2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1790343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
Differential Equation Problem: $2\cot x\frac{\mathrm{d}y}{\mathrm{d}x}=4-y^2$ Solve in terms of $\sec^2 x $$$2\cot x\frac{\mathrm{d}y}{\mathrm{d}x}=4-y^2$$
if $y=0$ when $x=\pi/3$
My attempt $$\int\tan x\,\mathrm{d}x=\int\frac{2}{4-y^2}\,\mathrm{d}y$$
Then by using partial fractions on the RHS
\begin{align}\ln\vert\sec x\vert+C_0&=\frac{1}{2}\left(\int\frac{1}{2-y}+\frac{1}{2+y}\,\mathrm{d}y\right)\\ &=\frac{1}{2}\left(\ln\vert2-y\vert+\ln\vert2+y\vert\right)\\&=\frac{1}{2}\ln\vert4-y^2\vert\end{align}
By taking exponents of both sides
$$\vert\sec^2x\vert+C_1=\vert4-y^2\vert$$
$$\sec^2(\pi/3)+C_1=4\rightarrow C_1=0$$ and the solution is $\sec^2x=\vert4-y^2\vert$
However the actual answer is $\sec^2x=\frac{1}{2}\ln\left\vert\frac{2+y}{2-y}\right\vert+8$
My question is have I gone wrong and if so where? I have tried using WolframAlpha though it does not seem to be very helpful with problems like this. My only thoughts are that maybe rearranging altered the equation since after $y\neq2.$
| $$\int\tan x\,\mathrm{d}x=\int\frac{2}{4-y^2}\,\mathrm{d}y$$
$$\int\tan x\,\mathrm{d}x=\int\frac{\sin x}{\cos x}dx=-\int \frac{d(\cos x)}{\cos x}=-\ln x+C=\ln C_0-\ln \cos x=\ln \frac{C_0}{\cos x}$$
$$2\int \frac{dy}{4-y^2}=\frac12\int\left(\frac1{2-y}+\frac1{2+y}\right)dy=\frac12(\ln|2+y|-\ln|2-y|)=\frac12\ln|\frac{2+y}{2-y}|=$$
$$=\ln \sqrt{\frac{2+y}{2-y}}$$
Then $$\ln \frac{C_0}{\cos x}=\ln \sqrt{\frac{2+y}{2-y}}$$
$$\frac{C_0}{\cos x}= \sqrt{\frac{2+y}{2-y}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1792927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Need help solving $x^4-3x^3-11x^2+3x+10=0$ Solve $x^4-3x^3-11x^2+3x+10=0$
I have tried to solve this equation using 'general formula from roots' from https://en.wikipedia.org/wiki/Quartic_function.
$$ax^4+bx^3+cx^2+dx+e=0$$
$$x_{1,2}=-\frac b{4a}-S \pm 0.5\sqrt{-4S^2-2P+ \frac q S}$$
$$x_{3,4}=-\frac b{4a} + S \pm 0.5\sqrt{-4S^2-2P-\frac q S}$$
$$p=\frac{8ac-3b^2}{8a^2}$$
$$q= \frac{b^3-4abc+8a^2d}{8a^3}$$
$$S=0.5\sqrt{-\frac{2p} 3 + \frac{Q+\Delta_0} Q}{3a}$$
$$Q=\left(\Delta_1+\sqrt{\Delta_1^2-4\Delta_0^3} \cdot 0.5\right)^{1/3}$$
$$\Delta_0=c^2-3bd+12ae$$
$$\Delta_1=2c^3-9bcd+27eb^2+27ad^2-72ace$$
Those formula does not work. I ended up with complex roots.
Please help me to use those formulas to solve quartic equation. Any help will be very much appreciated.
| $$x^4 - 3x^3 - 11x^2 + 3x + 10 = 0$$
$$x^4 - 3x^3 - 10x^2 - x^2 + 3x + 10 = 0$$
$$x^2(x^2 - 3x - 10) - (x^2 - 3x - 10) = 0$$
$$(x^2 - 1)(x^2 - 3x - 10) = 0$$
$$(x - 1)(x + 1)(x - 5)(x + 2) = 0$$
$$x = \{1,-1,5,2\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1796015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Solution of $x^y=y^x$ and $x^2=y^3$ Solve the given set of equations:
$x^y=y^x$ and $x^2=y^3$ where $x,y \in \mathbb{R}$
Would any other solution exist other that $x=y=1$ because I think $x^2=y^3$
will only be true for $x=y=1$ or $x=y=0$
| Following my answer here, we can deduce from the first equation that for some $a \neq 1$, we have
$$
x = a^{1/(a-1)}, \quad y = a^{a/(a-1)}
$$
Thus, the second equation becomes
$$
a^{2/(a-1)} = a^{3a/(a-1)} \implies\\
a = a^{3a/2} \implies\\
1 = a^{3a/2 -1} \implies\\
\left(\frac 32 a - 1\right) \log a = 0
$$
Thus, we get a solution at $a = 2/3$, which is to say
$$
x = (2/3)^{-3} = \frac{27}{8}, \qquad y = (2/3)^{-2} = \frac{9}{4}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1797132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
How to prove the inequality $ \frac{1-e^{-x^2}}{x^2}e^{-(x-n)^2}<\frac{2}{n^2}$ for $0Can somebody prove that this inequality is true for $0<x<n$?
$$ \frac{1-e^{-x^2}}{x^2}e^{-(x-n)^2}<\frac{2}{n^2}$$
I'm pretty much stuck.
| We may notice that $\frac{1-e^{-x^2}}{x^2}\leq \frac{2}{2+x^2}$ for any $x\in\mathbb{R}$ and $e^{-(x-n)^2}\leq \frac{1}{1+(x-n)^2}$ for any $x\in[0,n]$, hence it is enough to prove that:
$$ \forall x\in(0,n),\qquad \frac{2}{2+x^2}\cdot\frac{1}{1+(n-x)^2}<\frac{2}{n^2} \tag{1}$$
or:
$$ \forall x\in(0,n),\qquad (2+x^2)\cdot(1+(n-x)^2) > n^2\tag{2} $$
But due to the Cauchy-Schwarz inequality we have:
$$ \forall x\in(0,n),\qquad (\sqrt{x^2+1}^2+1)(1+(n-x)^2)\geq \left(\sqrt{x^2+1}+n-x\right)^2 \tag{3}$$
so $(2)$ is trivial, $(1)$ follows and the claim is proved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1799542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Hottest and coldest points on a heated circular plate (use Lagrange multipliers) A circular plate given by the relationship $x^2 + y^2 \leq 1$ is heated according to the spatial temperature function $T(x,y) = 2x^2 + y^2-y$. Find the hottest and coldest point on the plate using Lagrange multipliers.
I tried to solve it but I got only one point (0,1) instead of 4 points .. I want to know how can I get 4 points ?? Is there any method to solve this problem?
| Write
$$T (x,y) = 2 x^2 + y^2 - y = 2 x^2 + \left(y - \frac{1}{2}\right)^2 - \frac{1}{4} \geq -\frac{1}{4}$$
Thus, the coldest point is $(0,\frac{1}{2})$, where the temperature is $-\frac{1}{4}$.
The hottest point must be on the circle $x^2 + y^2 = 1$, which is the domain of interest's boundary.
Introduce the Lagrangian
$$\mathcal{L} (x,y,\lambda) = T (x,y) + \lambda \left(x^2 + y^2 - 1\right)$$
Taking the partial derivatives with respect to $x,y,\lambda$ and finding where they vanish, we get
$$(2+\lambda) \, x = 0, \qquad (1+\lambda) \, y = \frac{1}{2}, \qquad x^2+y^2=1$$
If $\lambda = -2$, then we get $y = -\frac{1}{2}$. Since $x^2 + y^2 = 1$, we have the two hot points
$$\left(\pm\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)$$
where the temperature is $\dfrac{9}{4}$. If $\lambda \neq -2$, then
$$x = 0, \qquad y = \frac{1}{2(1+\lambda)}, \qquad x^2+y^2=1$$
Hence, $(1+\lambda)^2 = \frac{1}{4}$, which gives us $\lambda \in \{-\frac{3}{2},-\frac{1}{2}\}$. If
*
*$\lambda = -\frac{3}{2}$, we get the hot point $(0,-1)$, where the temperature is $2$.
*$\lambda = -\frac{1}{2}$, we get the point $(0,1)$, where the temperature is $0$.
Since $2 < \frac{9}{4}$, the hottest points are $\left(\pm\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1800686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How can you prove that $1+ 5+ 9 + \cdots +(4n-3) = 2n^{2} - n$ without using induction? Using mathematical induction, I have proved that
$$1+ 5+ 9 + \cdots +(4n-3) = 2n^{2} - n$$
for every integer $n > 0$.
I would like to know if there is another way of proving this result without using PMI. Is there any geometric solution to prove this problem? Also, are there examples of problems where only PMI is our only option?
Here is the way I have solved this using PMI.
Base Case: since $1 = 2 · 1^2 − 1$, the formula holds for $n = 1$.
Assuming that the
formula holds for some integer $k ≥ 1$, that is,
$$1 + 5 + 9 + \dots + (4k − 3) = 2k^2 − k$$
I show that
$$1 + 5 + 9 + \dots + [4(k + 1) − 3] = 2(k + 1)^2 − (k + 1).$$
Now if I use hypothesis I observe.
$$
\begin{align}
1 + 5 + 9 + \dots + [4(k + 1) − 3]
& = [1 + 5 + 9 + \dots + (4k − 3)] + 4(k + 1) −3 \\
& = (2k^2 − k) + (4k + 1) \\
& = 2k^2 + 3k + 1 \\
& = 2(k + 1)^2 − (k + 1)
\end{align}
$$
$\diamond$
| Follow the rainbow!
$$ \Large \color{PaleVioletRed}1 + \color{DarkViolet}5 + \color{DodgerBlue}9 + \dots + \color{LightCoral}{(4n-3)} = n(2n-1) = 2n^2-n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1802846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "53",
"answer_count": 14,
"answer_id": 7
} |
Evaluate the algebraic limit $\lim_{n\to\infty} \frac{2n}{\sqrt{n+2}(\sqrt{n+1} + \sqrt{n-1})} $
$$\lim_{n\to\infty} \frac{2n}{\sqrt{n+2}(\sqrt{n+1} + \sqrt{n-1})} $$
I know that its value is somewhere in $(0, \infty)$ but I have no idea how to find the exact value.
Update: I've had a mistake in formulation of the limit in the original question so I updated the limit.
| $$\frac{2n}{\sqrt{n+1} + \sqrt{n-1}}$$
$$=\frac{1}{\sqrt{n}}\frac{1}{\sqrt{1+\frac{1}{n}} + \sqrt{1-\frac{1}{n}}}$$
$$=2\sqrt{n} \cdot \frac{1}{1+\frac{1}{2n}-\frac{1}{8n^2} + 1-\frac{1}{2n}-\frac{1}{8n^2} + o\left(\frac{1}{n^2}\right)}$$
$$=2\sqrt{n} \cdot \frac{1}{2-\frac{1}{4n^2}+o\left(\frac{1}{n^2}\right)}$$
$$=\sqrt{n} \cdot \left(1+\frac{1}{8n^2}+o\left(\frac{1}{n^2}\right)\right)\underset{n\to +\infty}{\longrightarrow}\infty$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1805762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Prove $\int_{0}^{\infty}\frac{2x}{x^8+2x^4+1}dx=\frac{\pi}{4}$ $$\int_{0}^{\infty}\frac{2x}{x^8+2x^4+1}dx=\frac{\pi}{4}$$
$u=x^4$ $\rightarrow$ $du=4x^3dx$
$x \rightarrow \infty$, $u\rightarrow \infty$
$x\rightarrow 0$, $u\rightarrow 0$
$$=\int_{0}^{\infty}\frac{2x}{u^2+2u+1}\cdot\frac{du}{4x^3}$$
$$=\frac{1}{2}\int_{0}^{\infty}\frac{1}{u^2+2u+1}\cdot\frac{du}{x^2}$$
$$=\frac{1}{2}\int_{0}^{\infty}\frac{1}{u^2+2u+1}\cdot\frac{du}
{\sqrt{u}}$$
Convert to partial fractions
$$\int_{0}^{\infty}\frac{1}{\sqrt{u}}-\frac{2}{u+1}+\frac{1}{(u+1)^2}du$$
$$=\left.2\sqrt{u}\right|_{0}^{\infty}-\left.2\ln(1+x)\right|_{0}^{\infty} -\left.\frac{1}{1+u}\right|_{0}^{\infty}$$
Where did I went wrong during my calculation?
| Here is an approach using contour integration in case anyone is
interested. Suppose we seek to verify that
$$\int_0^\infty \frac{2x}{x^8+2x^4+1} dx = \frac{\pi}{4}$$
or alternatively
$$\int_0^\infty \frac{x}{x^8+2x^4+1} dx = \frac{\pi}{8}.$$
We use a quarter pizza slice contour with the straight components
$\Gamma_0$ and $\Gamma_1$ on the positive real axis and the positive
imaginary axis and having radius $R$ ($\Gamma_2.$)
The denominator here is
$$(x^4+1)^2$$
so the poles are double and located at
$$\rho_{0,1,2,3} = \exp(\pi i/4 + 2\pi i k/4)
= \exp(\pi i/4 + \pi i k/2)$$
with $k = 0,1,2,3.$ Fortunately we can see by inspection that only the
first pole $\rho_0$ is inside the contour (argument is $\pi/4.$)
For the residue we get
$$\frac{1}{2\pi i}
\int_{|z-\rho_0|=\epsilon} \frac{z}{z^8+2z^4+1} \; dz.$$
Exploiting the symmetry put $w=z\exp(-\pi i/4)$ and $z=w\exp(\pi i/4)$
to get
$$\exp(\pi i/2) \frac{1}{2\pi i}
\int_{|w\exp(\pi i/4)-1|=\epsilon}
\frac{w}{w^8-2w^4+1} \; dw
\\ = \frac{i}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{w}{w^8-2w^4+1} \; dw.$$
The residue is thus given by
$$i\times \lim_{w\rightarrow 1}
\left(\frac{(w-1)^2 w}{w^8-2w^4+1}\right)'
= i\times \lim_{w\rightarrow 1}
\left(\frac{w}{(w+1)^2 (w^2+1)^2}\right)'
\\ = i\times \lim_{w\rightarrow 1}
\left(\frac{1}{(w+1)^2 (w^2+1)^2}
\\ - \frac{w}{(w+1)^4 (w^2+1)^4}
(2(w+1)(w^2+1)^2+(w+1)^2 2(w^2+1) 2w\right).$$
This works out to
$$i \times \left(\frac{1}{16} - \frac{16+32}{256}\right)
= -\frac{i}{8}.$$
Returning to the main computation, on the part of the contour that is
on the positive imaginary axis which is $\Gamma_1$ we obtain
$$\int_{\Gamma_1} \frac{z}{8z^8+2z^4+1} \; dz$$
which yields with $z=\exp(\pi i/2) x$
$$- \int_0^R \frac{\exp(\pi i/2) x}{8x^8+2x^4+1}
\; \exp(\pi i/2) dx
= \int_{\Gamma_0} \frac{z}{8z^8+2z^4+1} \; dz.$$
Finally we have by the ML bound for the circular component
$$\lim_{R\rightarrow\infty}
\left|\int_{\Gamma_2} \frac{z}{8z^8+2z^4+1} \; dz\right|
\le \lim_{R\rightarrow\infty} 2\pi R/4 \times
\frac{R}{8R^8-2R^4+1} = 0.$$
It follows that
$$\int_0^\infty \frac{x}{8x^8+2x^4+1} \; dx
= \frac{1}{2}\times 2\pi i \times -\frac{i}{8}
= \frac{\pi}{8}$$
which is the claim.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1806132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 2
} |
find $c$ and $b$ in terms of $x$ and $a$ I have this geometry problem.
Supose any $\triangle{ABC}$, where $\overline{CE} \perp \overline{AB}$; $\overline{CM}$ is median; $n$ is $proy_{\overline{CM}}\overline{AB}$; $\angle{CMA}$ is obtuse.
Find:
$c^2;b^2;c^2+b^2$ in terms of $a$ and $x$
| For $CME$,
$EM=\sqrt{x^2-h^2}=n$
for $CBE$, we have that
$BE=\sqrt{a^2-h^2}$
We have,$BM=BE+EM$,
Therefore,$BM=\sqrt{a^2-h^2}+\sqrt{x^2-h^2}$
As CM is median,we can say, $c=2BM= 2[\sqrt{a^2-h^2}+\sqrt{x^2-h^2}]$
Also, $b^2 = AC^2 = CE^2+AE^2=h^2+(AM+ME)^2 = h^2+(c/2+n)^2$
therefore, $b^2=h^2+[(\sqrt{a^2-h^2}+\sqrt{x^2-h^2})+\sqrt{x^2-h^2}]^2$
$b^2= h^2+(\sqrt{a^2-h^2}+2\sqrt{x^2-h^2})^2$
You get $b,c$ so you can calculate $b^2+c^2$ also.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1806529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to solve this recurrence relation $ f(n) = \frac{10+7f(n-1)}{2+f(n-1)}$ I had a problem in which I ended up getting the following recurrence relation:
$$\begin{align}
&f(1) = 7\\
&f(n) = \dfrac{10+7f(n-1)}{2+f(n-1)}
\end{align}$$
I haven't solved much recurrence relations, and I am unaware of any general procedure. The only thing I was able to do till now was to get this relation. I have tried putting $n$ as $n+1$, but it soon gets unsolvable.
In a more crude form, $$f(n) = 5+ \dfrac{1}{\frac{1}{f(n-1)}+\frac{1}{2}}$$
| The solutions for
$$x=\frac{10+7x}{2+x}$$
are $x=\frac{5\pm\sqrt{65}}{2}$.
Then,
$$f(n)-\frac{5\pm\sqrt{65}}{2}=\frac{10+7f(n-1)}{2+f(n-1)}-\frac{5\pm\sqrt{65}}{2}=\frac{10\mp 2\sqrt{65}+(9\mp\sqrt{65})f(n-1)}{2(2+f(n-1))}$$
from which we have
$$\begin{align}\frac{f(n)-(5+\sqrt{65})/2}{f(n)-(5-\sqrt{65})/2}&=\frac{{10- 2\sqrt{65}+(9-\sqrt{65})f(n-1)}}{{10+ 2\sqrt{65}+(9+\sqrt{65})f(n-1)}}\\&=\frac{9-\sqrt{65}}{9+\sqrt{65}}\cdot \frac{f(n-1)-(5+\sqrt{65})/2}{f(n-1)-(5-\sqrt{65})/2}\end{align}$$
which can be written as
$$g(n)=\alpha g(n-1)$$
This is a geometric progression
where $$g(n)=\frac{f(n)-(5+\sqrt{65})/2}{f(n)-(5-\sqrt{65})/2},\quad \alpha=\frac{9-\sqrt{65}}{9+\sqrt{65}}$$
Having $g(n)=\alpha^n$, we have
$$\begin{align}f(n)&=\frac{(5+\sqrt{65})/2-\alpha^n\cdot(5-\sqrt{65})/2}{1-\alpha^n}\\&=\color{red}{\frac{(5+\sqrt{65})(9+\sqrt{65})^n-(5-\sqrt{65})(9-\sqrt{65})^n}{2((9+\sqrt{65})^n-(9-\sqrt{65})^n)}}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1809236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Prove using induction on n that: $8\mid5^n+2(3^{n-1})+1$ How can we use induction to prove that $8\mid5^n+2(3^{n-1})+1$ for any natural $n$?
| If it is true for $n$ then for $n+1$ we have
\begin{align}
5^{n+1}+2(3^n) + 1 &= 5(5^n) + 3(3^{n-1} + 1) \\
&= 5(5^n + 2(3^{n-1}) + 1) - 2*2(3^{n-1}) - 4 \\
&= 5(5^n + 2(3^{n-1}) + 1) - 4(3^{n-1} - 1).
\end{align}
By the induction hypothesis the first terms is divisible by 8. The second term is 4 times an even number, hence also divisible by 8.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1810993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Let $\mathbf{X}$ be a Markov chain on a square find $p_{1,1}(n)$ Consider a square like this
$$\begin{array}\\
1 & - & 2\\
| & & |\\
3 & - & 4
\end{array} $$
such that you can go from each state with chance $\tfrac{1}{2}$ to the neighbouring states. So we have the following transition matrix
$$P = \begin{pmatrix}0 & \tfrac{1}{2} & \tfrac{1}{2}& 0\\
\tfrac{1}{2} & 0 & 0 & \tfrac{1}{2}\\
\tfrac{1}{2}&0 & 0 & \tfrac{1}{2}\\
0 & \tfrac{1}{2} & \tfrac{1}{2}& 0 \end{pmatrix}. $$
Let $p_{i,j}(n)$ be the position $(i,j)$ in the matrix $P^n$. I want to find $p_{1,1}(n)$.
This is my approach:
*
*Diagonalize P, the eigevalues of $P$ are $-1,1,0,0$ therefore $P = U D U^{-1} $ where
$$D = \begin{pmatrix}1 & 0 & 0 & 0\\
0 & -1 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\ \end{pmatrix} $$
Thus $P^n = U D^n U^{-1}$.
*We know that $p_{1,1}(0) = 1$ and $p_{1,1}(1) = 0$
I would say that $p_{1,1}(n) = A + (-1)^n B$ for $A,B \in \mathbb{R}$. But this is obviously not true because $A = B = \tfrac{1}{2}$ and therefore
$$p_{1,1}(2) = \frac{1}{2} + \frac{1}{2} = 1. $$
This is not possible.
I intuitively see that there is symmetry, you can't go back to your start position in odd steps. But I want to make this more formal with linear algebra. I don't see why
$$p_{1,1}(n) = \frac{1}{2}A + \frac{1}{2}(-1)^n B = \frac{1}{2} + \frac{1}{2}(-1)^n. $$
I guess this is true because two eigenvalues are zero. But I haven't found something that states this.
| The mistake is in the equation $p_{1,1}(n)=A+(-1)^nB$. Instead, it should be $A+(-1)^nB+0^nC$, where $0^n$ is defined to be $1$ if $n=0$ and $0$ for integers $n>0$.
EDIT: By the way, this equation is derived as follows:
\begin{align*}
p_{1,1}(n)&=e_1^TP^ne_1 \\
&= e_1^TUD^nU^{-1}e_1\\
&= e_1^TU\begin{pmatrix}1^n & 0 & 0 & 0 \\ 0 & (-1)^n & 0 & 0 \\ 0 & 0 & 0^n & 0 \\ 0 & 0 & 0 & 0^n\end{pmatrix}U^{-1}e_1
\end{align*}
where $e_1=(1,0,0,0)^T$. Then, since $e_1$ and $U$ are constant (i.e., not depending on $n$), expanding this out gives an equation of the form $p_{1,1}(n)=A+(-1)^nB+0^nC$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1811430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Obtain the equation of the sphere which passes through the points $(1,0,0),(0,1,0),(0,0,1)$ and has its radius as small as possible. Obtain the equation of the sphere which passes through the points $(1,0,0),(0,1,0),(0,0,1)$ and has its radius as small as possible.
Let the sphere passes through $(x_1,y_1,z_1)$
Then i obtained the equation of sphere as $(x^2+y^2+z^2)(x_1+y_1+z_1)-(x_1^2+y_1^2+z_1^2-1)(x+y+z)+x_1^2+y_1^2+z_1^2-x_1-y_1-z_1=0$
I am stuck here.I do not know how to minimize the radius.The answer given in my book is $3(x^2+y^2+z^2)-2(x+y+z)-1=0$
| Write the equation of a candidate sphere as
$$ (x - x_0)^2 + (y - y_0)^2 + (z - z_0)^2 = r^2. $$
Plugging in the points, we get
$$ (1 - x_0)^2 + y_0^2 + z_0^2 = r^2, \\
x_0^2 + (1 - y_0)^2 + z_0^2 = r^2, \\
x_0^2 + y_0^2 + (1 - z_0)^2 = r^2.
$$
Comparing the first two equations, we get
$$ (1 - x_0)^2 + y_0^2 = x_0^2 + (1 - y_0)^2 \implies 1 - 2x_0 + x_0^2 + y_0^2 = x_0^2 + 1 - 2y_0 + y_0^2 \implies x_0 = y_0. $$
Similarly we see that $x_0 = y_0 = z_0$ and so the center of the sphere must lie on the line $x = y = z$. Denoting the center by $(t,t,t)$ and returning to the first equation, we obtain
$$ 1 - 2t + 3t^2 = r^2 $$
so to minimize $r$ (or $r^2$), we need to minimize the quadratic $1 - 2t + 3t^2$. The minimum $\frac{2}{3}$ is attained at the vertex $t = \frac{1}{3}$ and so the equation is
$$ \left( x - \frac{1}{3} \right)^2 + \left( y - \frac{1}{3} \right)^2 + \left( z - \frac{1}{3} \right)^2 = \frac{2}{3} \iff \\
x^2 + y^2 + z^2 - \frac{2}{3}(x + y + z) + \frac{1}{3} = \frac{2}{3} \iff \\
3(x^2 + y^2 + z^2) - 2(x + y + z) - 1 = 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Find $f'(x)$ at given value of x Find $f'(x)$ at the given value of x
$f(x)=\sqrt{x+2}$
Find $f'(7)$
My question for this one is do I approach this question by trying to find the derivative of the initial equation and then once I have found the derivative do I simply plug in 7 for x and solve?
So once I start solving I would get $\frac{\sqrt{x+h+2}-\sqrt{x+2}}{h}$
$\lim_{h\to0} \frac{\sqrt{x+h+2}-\sqrt{x+2}}{h} = \lim_{h\to0} \frac{\sqrt{x+h+2}-\sqrt{x+2}}{h} \cdot \frac{\sqrt{x+h+2}+\sqrt{x+2}}{\sqrt{x+h+2}+\sqrt{x+2}}$
From that point I would get
$\frac{(x+h+2-x-2)}{h\sqrt{x+h+2}+\sqrt{x+2}}$
| Yes. That is exactly how you do it!
Find $f'(x)$, then plug in $x=7$ into the resulting function and compute.
Here's how:
$$
\begin{align*}
f'(x)&=\lim_{h\to0} \frac{\sqrt{x+h+2}-\sqrt{x+2}}{h} \\
&= \lim_{h\to0} \frac{(\sqrt{x+h+2}-\sqrt{x+2})}{h} \cdot \frac{(\sqrt{x+h+2}+\sqrt{x+2})}{(\sqrt{x+h+2}+\sqrt{x+2})}\\
&=\lim_{h\to0}\frac{x+h+2-x-2}{h(\sqrt{x+h+2}+\sqrt{x+2})}\\
&=\lim_{h\to 0} \frac{1}{\sqrt{x+h+2}+\sqrt{x+2}}\\
f'(x)&=\frac{1}{2\sqrt{x+2}}
\end{align*}
$$
Now plugging in $x=7$ gives: $$f'(7)=\frac{1}{2\sqrt{7+2}}=\frac{1}{6}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Volumes of Revolution question bounded by 2 boundaries and rotated about x and y axes I'm asking this question for a friend who is working on a volumes of revolution question (she is not familiar with stackexchange, and I'm not particularly good at volumes of revolution problems). The question is given as:
Calculate the volume of revolution of $R$ where $R$ is the region bounded by $y=5x-x^{2}$, $y=3+x$ rotated about $a)$ $y=3$, $b)$ $x=3$
Apart from having trouble solving the question, she has trouble visualizing what this question would look like on a graph. Any help would be appreciated!
| First off, we make a sketch of the region $R$.
We can find the intersection of the two curves by setting
$$y=5x-x^2=3+x$$
So that $x^2-4x+3=(x-1)(x-3)=0$ and the points of intersection are $(1,4)$ and $(3,6)$. To find the volume of the solid of revolution about $y=3$, it will be easiest to integrate washers:
$$\begin{align}V&=\int\pi\left(r_{\text{outer}}^2-r_{\text{inner}}^2\right)dh=\int_1^3\pi\left[(5x-x^2-3)^2-(3+x-3)^2\right]dx\\
&=\int_1^3\pi\left[x^4-10x^3+30x^2-30x+9\right]dx=\pi\left[\frac{x^5}5-\frac{5x^4}2+10x^3-15x^2+9x\right]_1^3\\
&=\pi\left[\frac{242}5-200+260-120+18\right]=\frac{32\pi}5\end{align}$$
To find the volume of the solid of revolution about $x=3$, it will be easiest to integrate oatmeal boxes:
$$\begin{align}V&=\int2\pi rh\,dr=\int_1^32\pi(3-x)\left[5x-x^2-3-x\right]dx\\
&=\int_1^32\pi\left[x^3-7x^2+15x-9\right]dx=2\pi\left[\frac{x^4}4-\frac{7x^3}3+\frac{15x^2}2-9x\right]_1^3\\
&=2\pi\left[20-\frac{182}3+60-18\right]=\frac{8\pi}3\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Evaluation of $\int x^8\sqrt{x^2+1}\ dx$
Evaluation of $\displaystyle \int x^8\sqrt{x^2+1}\ dx$
$\bf{My\: Try::}$ Put $x=\tan \theta\;,$ Then $dx = \sec^2 \theta d \theta$
So $$I = \int \sec ^{3}\theta \cdot \tan^{8}\theta d \theta=\int \frac{\sin^{8}\theta}{\cos^{5}\theta }d\theta$$
Now How can I solve it after that, Help Required, Thanks
| Apply the reduction formula
$$\int x^n\sqrt{1+x^2}dx= I_n=\frac{x^{n-1}}{n+2}(1+x^2)^{3/2}-\frac{n-1}{n+2}I_{n-2}
$$
4 times to reduce the integral
\begin{align}
\int x^8\sqrt{1+x^2}\ dx
=7\left(\frac{x^7}{70}-\frac{x^5 }{80}+\frac{x^3 }{96}-\frac{x}{128}\right)(1+x^2)^{3/2}+\frac7{128}I_0
\end{align}
where
$
I_0=\int \sqrt{1+x^2}\ dx
=\frac{x}{2}(1+x^2)^{1/2}+\frac{1}{2}\sinh^{-1}x
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 9,
"answer_id": 4
} |
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.
| Notice that
$$\sin ( \pi \sqrt {n^2 + k^2} - n \pi) = \sin ( \pi \sqrt {n^2 + k^2} ) \cos (n \pi) - \cos ( \pi \sqrt {n^2 + k^2} ) \sin (n \pi) = \\
(-1)^n \sin ( \pi \sqrt {n^2 + k^2} ) ,$$
so
$$\sin ( \pi \sqrt {n^2 + k^2} ) = (-1)^n \sin ( \pi \sqrt {n^2 + k^2} - n \pi) = (-1)^n \sin \frac {\pi^2 (\sqrt {n^2 + k^2})^2 - n^2 \pi^2} {\pi \sqrt {n^2 + k^2} + n \pi} = \\
(-1)^n \sin \frac {\pi^2 k^2} {\pi \sqrt {n^2 + k^2} + n \pi} ,$$
therefore your series is
$$\sum \limits _{n \ge 1} (-1)^n \sin \frac {\pi^2 k^2} {\pi \sqrt {n^2 + k^2} + n \pi} .$$
Notice that the fraction inside the sine decreases to $0$ with respect to $n$, so from a certain $n_0$ onwards it will surely be in $[0, \frac \pi 2]$, an interval on which $\sin$ increases, which means that for $n \ge n_0$ the function $n \mapsto \sin \dfrac {\pi^2 k^2} {\pi \sqrt {n^2 + k^2} + n \pi} $ decreases to $0$ (in particular, this also shows that for $n \ge n_0$ the fraction is positive). Since the series is alternating, Leibniz's test tells us that it is convergent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find all whole number solutions of the following equation While training for a math olympiad(university level) I stumbled upon the following problem. Find all $n, k \in \mathbb{N}$ such that
$${ n \choose 0 } + {n \choose 1}+{n \choose 2} + {n \choose 3} = 2^k.$$
Now I have thought about two ways of solving this. The first is by expanding the left hand side to get a polynomial in $n$, and by this finding restrictions on $n$ and $k$. The other way would be by repeatedly using Pascal's formula and the fact that
$$2^j = \sum_{i=0}^j {j \choose i}$$
for all $j \in \mathbb{N}$. Both have not given me any success. Am I on a right track? Could you give me some hints, or an attempt at a solution? Thanks in advance.
| Assume $n>3$.
Expanding the LHS and simplifying gives
$$\frac{1}{6}(n+1)(n^2-n+6) = 2^k.$$
So except for one factor of $3$, both $n+1$ and $n^2+n-6$ are powers of $2$. If the factor of $3$ is in $n+1$, then $n = 3\cdot 2^r-1$ with $r\ge 1$, so that
$$n^2+n-6 = (3\cdot 2^r-1)^2 - (3\cdot 2^r-1) + 6
= 8 - 3\cdot 2^4 + 9\cdot 2^{2r} - 3\cdot 2^{r+1}
= 8(1 - 2^{r-3} - 2^r + 2^{2r} + 2^{2r-3}).$$
In order for this to be a power of $2$, we must have $r=3$; otherwise the second factor is odd. $r=3$ gives $n = 3\cdot 2^3-1 = 23$.
Alternatively, the factor of $3$ is in $n^2-n+6$, so that $n+1 = 2^r$ is a power of $2$ with $r\ge 1$. But then
\begin{align*}
n^2-n+6 &= (2^r-1)^2 - (2^r-1) + 6 = 2^{2r} - 3\cdot 2^r + 8 \\
&= 9 -3\cdot 2^r + (2^{2r}-1) \\
&= 9 - 3\cdot 2^r + 3(2^{2r-2}+2^{2r-4} + \ldots + 2^2 + 1) \\
&= 12 - 3\cdot 2^r + 3(2^{2r-2}+2^{2r-4}+\ldots + 2^2) \\
&= 3(4-2^r + 2^{2r-2}+2^{2r-4} + \ldots + 2^2) \\
&= 24(1-2^{r-3}+2^{2r-5}+2^{2r-7}+\ldots+ 2),
\end{align*}
and in order for this to be $3$ times a power of $2$, we must again have $r=3$, so that $n = 2^3-1=7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1815121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
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.
| A meta-answer: if you already have the sequence $0,1,1,0,1,1\ldots$, then you may take the power of $-1,$ but you actually don't need exponentiation at all (similarly for other periodical sequences):
\begin{align}
a_n &=1 - 2(n^2 \bmod 3)\\
&=1 - 2n^2 + 6\lfloor n^2/3\rfloor\\
&=1 - 2\cdot \lvert (n-1) {\bmod} 3 -1\rvert\\
&= 1 - 2(\lceil n/3 \rceil - \lfloor n/3 \rfloor) \\
&= 1- 2\cdot \mathbf{1}_{\{n: \, 3\not\mid n\}}\\
&=\ldots
\end{align}
which I think is neater. Also:
$$
a_n= (-1)^{2k+1 \bmod 3}
$$
| {
"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": 10
} |
Each digit of $\dfrac{n(n+1)}{2}$ equals $a$
For which digits $a$ does there exist an integer $n \geq 4$ such that each digit of $\dfrac{n(n+1)}{2}$ equals $a$?
I was first thinking of looking at $\dfrac{n(n+1)}{2} \pmod{100}$, but it doesn't look like it has a small period so it may be hard to arrive at a conclusion. Is there an easier way to solve this question?
| Note that $n(n+1)/2 = z$ has an integer solution iff $1+8z$ is a square.
So the question is whether $1 + \dfrac{8 a}{9} (10^k-1)$ can be a square, or equivalently whether $9 - 8 a + 8 a \cdot 10^k$ can be a square.
You can rule out $a = 2, 3, 4, 7, 8, 9$ this way because $9-8a$ is not a square $\mod (8 a \cdot 10^k)$ for $k=1$ or $2$.
That leaves $a=1$.
EDIT: For $a = 1$, you want $1 + 8 \cdot 10^k$ to be a square with $k \ge 2$. Suppose it is $x^2$. Since $x^2 \equiv 1 \mod 5^k$, we may assume $x \equiv 1 \mod 5^k$.
So let $x = 1 + 5^k y$. Then we have $1 + 2 y \cdot 5^k + y^2 \cdot 5^{2k} = 1 + 8 \cdot 10^k$, or $2 y + y^2 \cdot 5^k = 2^{k+3}$. In particular, $y \mid 2^{k+3}$, so $y$ is a power of $2$. If $y = 2^j$, the equation becomes
$2^{j+1} + 2^{2j} \cdot 5^k = 2^{k+3}$. Considering factors of $2$, two of $j+1, 2j, k+3$ must be equal. But it's easy to rule out all these possibilities.
See also OEIS sequence A045914.
EDIT: It's also interesting to consider other bases than $10$. Here there are some infinite families of solutions, as well as some (such as $52 \dfrac{98^4-1}{98-1} = \dfrac{9944 \cdot 9945}{2}$ or $105 \dfrac{159^6-1}{159-1} = \dfrac{4634175 \cdot 4634176}{2}$) for which I don't see an obvious pattern.
In base $8j+1$, where $j$ is an integer, $j \dfrac{(8j+1)^{2k}-1}{8j+1-1} = \dfrac{n(n+1)}{2}$ where $n = \dfrac{(8j+1)^k-1}2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1820126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
If $f(1-x) + 2f(x) = 3x$, what is $f(0)?$ I did the following:
$$f(1-0) + 2f(0) = 3\cdot 0$$
$$f(1) + 2f(0) = 0$$
This reminds me of the equation of the straight line in the plane, then:
$$\left< \begin{pmatrix}
{1}\\
{2}
\end{pmatrix}
,
\begin{pmatrix}
{f(1)}\\
{f(0)}
\end{pmatrix} \right> =0$$
$(1,2)$ is a normal vector to $(f(1),f(0))$, then It's possible that: $(f(1)=-2,f(0)=1)$ because:
$$\left< \begin{pmatrix}
{1}\\
{2}
\end{pmatrix}
,
\begin{pmatrix}
{-2}\\
{1}
\end{pmatrix}\right> =-2+2=0$$
With generality, It is possible that for all $\alpha$:
$$\left< \begin{pmatrix}
{1}\\
{2}
\end{pmatrix}
,
\begin{pmatrix}
{\alpha\cdot (-2)}\\
{\alpha \cdot 1}
\end{pmatrix}\right> =
\left< \begin{pmatrix}
{1}\\
{2}
\end{pmatrix}
,
\alpha \begin{pmatrix}
{ -2}\\
{ 1}
\end{pmatrix}\right> =1\cdot -2\alpha+2\alpha \cdot 1=-2\alpha + 2 \alpha =0$$
I tagged with "functional equations" because it seems to be related. I don't know if my solution is correct, what I'm saying with this is that there are infinite solutions but I don't know if assuming arbitrary values for $f(1),f(0)$ can be made because of $3x$.
EDIT: I know that It is possible to solve:
$$f(1) + 2f(0) = 0$$
$$f(0) + 2f(1) = 3$$
And get the solution. But why do I have to use the latter instead of the former?
| $$f(1-x)+2f(x)=3x$$
set $1-x=t$
$$f(t)+2f(1-t)=3(1-t)$$
so $f(1-x)+2f(x)=3x$ and $f(x)+2f(1-x)=3(1-x)$
$$f(x)=3x-1$$ $f(0)=-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1820223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Five white balls, five black balls There are 5 white balls and 5 black balls in a box. A visually-challenged man takes 5 balls from the box... What's the probability that all the balls which this individual picked are white if we know for sure that there 3 white balls among those 5 balls he took?
Using the formula $P(A | B) = P(A \cap B)/P(B)$, I get a probability of $1/100$.
| $$\frac{\frac{C\left( \begin{matrix}
5 \\
5 \\
\end{matrix} \right)}{C\left( \begin{matrix}
10 \\
5 \\
\end{matrix} \right)}}{\frac{C\left( \begin{matrix}
5 \\
3 \\
\end{matrix} \right)C\left( \begin{matrix}
5 \\
2 \\
\end{matrix} \right)+C\left( \begin{matrix}
5 \\
4 \\
\end{matrix} \right)C\left( \begin{matrix}
5 \\
1 \\
\end{matrix} \right)+C\left( \begin{matrix}
5 \\
5 \\
\end{matrix} \right)C\left( \begin{matrix}
5 \\
0 \\
\end{matrix} \right)}{C\left( \begin{matrix}
10 \\
5 \\
\end{matrix} \right)}}=\frac{1}{126}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1820383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
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 think it would be better if you first use $t=z-2$. This way, you have $(x^2+y^2+z^2=1)\rightarrow (x^2+y^2+t^2+4t+3=0)$. Then, use the spherical coordinates. Now, your integral would be really simple, and you need to find the proper boundaries for new integral.
$(x^2+y^2+t^2+4t+3=0) \rightarrow \rho^2+4\rho cos\phi +3=0 \rightarrow \rho =-2cos\phi \pm \sqrt{4(cos\phi)^2-3} \rightarrow -2cos\phi - \sqrt{4(cos\phi)^2-3}\leq \rho\leq -2cos\phi + \sqrt{4(cos\phi)^2-3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1821448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\lfloor\frac{n+1}{2}\rfloor+\lfloor\frac{n+2}{4}\rfloor+\lfloor\frac{n+4}{8}\rfloor+\lfloor\frac{n+8}{16}\rfloor+ \dots=n$
Prove $$\left[\dfrac{n+1}{2}\right]+\left[\dfrac{n+2}{4}\right]+\left[\dfrac{n+4}{8}\right]+\left[\dfrac{n+8}{16}\right] + \dots=n$$
where $[x]=\lfloor x\rfloor$
$$$$
It was suggested that somehow I use the identity $[x]=\left[\dfrac x2\right]+\left[\dfrac{x+1}{2}\right]$$$$$After struggling for a while, I realised I wasn't getting anywhere using his hint, probably because I couldn't really understand how I was to use it. Instead I tried to use the Squeeze Theorem by rewriting the $nth$ term of the series (referred to later as S) as$$$$ $$t_n=\left[\dfrac{n+2^k}{2^{k+1}}\right] \text{ where } 0\le k<\infty$$
$$\Rightarrow \dfrac{n+2^k}{2^{k+1}}-1<\left[\dfrac{n+2^k}{2^{k+1}}\right]\le \dfrac{n+2^k}{2^{k+1}}$$
$$$$
$$ \lim_{k\to \infty}(k+1)\left(\dfrac{n+2^k}{2^{k+1}}-1\right)<S\le \lim_{k\to \infty}(k+1)\left( \dfrac{n+2^k}{2^{k+1}}\right)$$
However these bounds are too loose as the limits diverge to $-\infty$ and $\infty$ respectively.
$$$$
Could somebody please show me how to prove the series is equal to $n$, either through the given hint, or through the selection of tighter bounds for the Squeeze Theorem? Many thanks!
| For $x=\frac{n}{2^k}$, the identity you have means:
$$\left\lfloor \frac{n}{2^k}\right\rfloor =\left \lfloor \frac{n}{2^{k+1}}\right\rfloor+\left\lfloor \frac{n+2^k}{2^{k+1}}\right\rfloor$$
So prove by induction for any $m\geq 0$ that:
$$ n = \left\lfloor\frac{n}{2^m}\right\rfloor+\sum_{k=0}^{m-1}\left\lfloor\frac{x+2^{k}}{2^{k+1}}\right\rfloor$$
The pick $m$ so that $2^m>n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1821536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 1
} |
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.