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What are different ways to compute $\int_{0}^{+\infty}\frac{\cos x}{a^2+x^2}dx$? I am interested to compute the following integral
$$I=\int_{0}^{+\infty}\frac{\cos x}{a^2+x^2}dx$$
where $a\in\mathbb{R}^+$. Let me explain my first idea. As the integrand is an even function of $x$ then
$$2I=\int_{-\infty}^{+\infty}\frac{\cos x}{a^2+x^2}dx=\lim_{R\to+\infty}\int_{-R}^{R}\frac{\cos x}{a^2+x^2}dx:=\lim_{R\to+\infty}J$$
So, I first focus on computing the $J$ integral by first modifying it as follows
\begin{align*}
J&=\int_{-R}^{R}\frac{\cos x}{a^2+x^2}dx=\int_{-R}^{R}\frac{\cos x}{a^2+x^2}dx+i\int_{-R}^{R}\frac{\sin x}{a^2+x^2}dx \\
&= \int_{-R}^{R}\frac{(\cos x+i\sin x)}{a^2+x^2}dx = \int_{-R}^{R}\frac{\exp(ix)}{a^2+x^2}dx
\end{align*}
Then, I use the well-known techniques of complex variable theory. First, I replace the real variable $x$ in $J$ with a complex variable $z$ and consider a contour integral over $C=C_1\cup C_2$
$$K:=\int_{C}\frac{\exp(iz)}{a^2+z^2}dz$$
Then, according to the Cauchy's integral theorem and the Residue theorem, I get
\begin{align*}
K=J+\int_{C_2}\frac{\exp(iz)}{a^2+z^2}dz &= \int_{C_3}\frac{\exp(iz)}{a^2+z^2}dz=\int_{C_3}\frac{\exp(iz)}{(z+ia)(z-ia)}dz \\
&=2\pi i \frac{\exp(i^2a)}{2ia}=\frac{\pi}{a}\exp(-a)
\end{align*}
Next, taking the limit $R\to+\infty$ from the above relation, we obtain
$$2I+\lim_{R\to+\infty}\int_{C_2}\frac{\exp(iz)}{a^2+z^2}dz=\frac{\pi}{a}\exp(-a)$$
but, we can show that
$$\lim_{R\to+\infty}\int_{C_2}\frac{\exp(iz)}{a^2+z^2}dz=0$$
and then we can obtain the final result
$$I=\frac{\pi}{2a}\exp(-a)$$
First, please check my steps to see the final result is correct or not. Second, is there any other way to compute $I$?
|
A rather exotic approach leading to a known functional equation, which has an exponential solution.
Consider a function ($a>0$):
$$f(a)=a \int_{-\infty}^\infty \frac{\cos x}{a^2+x^2} dx=\int_{-\infty}^\infty \frac{\cos a x}{1+x^2} dx=\pi e^{-a}$$
Let's square it and change the dummy variable:
$$f^2(a)=\int_{-\infty}^\infty \int_{-\infty}^\infty \frac{\cos a x \cos a y}{(1+x^2)(1+y^2)} dx dy=$$
$$=\frac{1}{2} \int_{-\infty}^\infty \int_{-\infty}^\infty \frac{\cos a (x-y)+ \cos a (x+y)}{(1+x^2)(1+y^2)} dx dy$$
Due to the infinite limits, we can easily make substitutions in the form $x \pm y=t$, which will lead to the following expression under the integral:
$$\frac{\cos a t}{y^2+1} \left(\frac{1}{(y+t)^2+1} +\frac{1}{(y-t)^2+1} \right)$$
We will do partial fraction decomposition to integrate w.r.t. $y$.
$$\frac{1}{((y+t)^2+1)(y^2+1)}=\frac{1}{t (4+t^2)} \left(\frac{2y+3t}{(y+t)^2+1}-\frac{2y-t}{y^2+1} \right)$$
$$\frac{1}{((y-t)^2+1)(y^2+1)}=\frac{1}{t (4+t^2)} \left(\frac{-2y+3t}{(y-t)^2+1}-\frac{-2y-t}{y^2+1} \right)$$
Let's consider separately the 'problematic' integrals, but with finite limits:
$$\int_{-L}^L \frac{2y dy}{(y+t)^2+1}=\int_{-L-t}^{L+t} \frac{2u du}{u^2+1}-2t\int_{-L-t}^{L+t} \frac{du}{u^2+1} $$
The first integral vanishes, the second after taking the limit $L \to \infty$, gives us $-2 \pi t$. In the same way we find the other integral with $(y-t)^2$.
So the two 'problematic' integrals give us:
$$-2 \pi \int_{-\infty}^\infty \frac{\cos at ~dt}{4+t^2}$$
Grouping the other terms we get:
$$\frac{3t}{(y+t)^2+1}+\frac{3t}{(y-t)^2+1}+\frac{2t}{y^2+1}$$
After integration w.r.t. $y$ and adding all the results, we obtain:
$$f^2(a)=2\pi \int_{-\infty}^\infty \frac{\cos at ~dt}{4+t^2}=\pi f(2a)$$
The functional equation:
$$f^2 (a)=\pi f(2a)$$
has a general solution:
$$f(a)=\pi e^{c a}$$
We should have $c<0$, as can be seen by considering the original integral definition and taking the limit $a \to \infty$.
I'm not sure how to prove $c=-1$, but it should be possible.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2734148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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|
Evaluate $\int \frac{x\tan(x) \sec(x)}{(\tan(x)-x)^2}$ Find value of following integral $$\int \frac{x\tan(x) \sec(x)}{(\tan(x)-x)^2}\text{dx}$$
the numerator is $\text{d[sec(x)]}$ but that isnt work due to $x$ in denominator. First we can simplify as $$\int\frac{x\sin(x)}{(\sin(x)-x\cos(x))^2} \text{dx} = \int \frac{x}{(\sin(x)-x\cos(x))} \text{dx}+\int \frac{x^2 \cos(x)}{(\sin(x)-x\cos(x))^2} \text{dx}$$
but again its not manipulative. Suggest a useful substitution or method.
Thanks a lot!
|
\begin{array}{rcl}
\displaystyle \int \dfrac{x \sec x \tan x \,dx}{(\tan x - x)^2} &=& \displaystyle\int\color{red}{\left(-x\csc x\right)}\cdot \color{blue}{\left(\dfrac{-\tan^2 x}{\left(\tan x - x\right)^2}\right)\,dx}\\
&=&\displaystyle \int \color{red}{(-x\csc x)\cdot \color{blue}{d\left(\dfrac{1}{\tan x - x}\right)}}\\
&=& \dfrac{x\csc x}{x-\tan x}-\displaystyle \int \dfrac{d(-x\csc x)}{\tan x - x}\\
&=& \dfrac{x\csc x}{x-\tan x}-\displaystyle \int \dfrac{x\csc x \cot x -\csc x}{\tan x - x}\,dx\\
&=& \dfrac{x\csc x}{x-\tan x}-\displaystyle \int \dfrac{\csc x \cot x\left(x - \tan x\right)}{\tan x - x}\,dx\\
&=& \dfrac{x\csc x}{x-\tan x} - \displaystyle \int \left(-\csc x \cot x\right)\,dx\\
&=& \dfrac{x\csc x}{x -\tan x} - \csc x + C\\
&=& \csc x \left(\dfrac{x}{x - \tan x} - 1\right)+C\\
&=& \dfrac{\csc x\tan x}{x - \tan x} + C\\
&=& \dfrac{\sec x}{x - \tan x} + C
\end{array}
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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|
How does Tom Apostol deduce the inequality $\frac{k^3}{3} + k^2 \lt \frac{(k + 1)^3}{3}$ in section I 1.3 of his proof by induction example? In his book Calculus, Vol. 1, Tom Apostol writes the following proof for proving the following predicate by induction:
$$A(n): 1^2 + 2^2 + \cdots + (n - 1)^2 \lt \frac{n^3}{3}.$$
I have omitted the Base Case due to lack of specific relevance. An excerpt from the book's page:
Assume the assertion has been proved for a specific value of $n$, say $n = k$. That is, assume we have proved
$$A(k): 1^2 + 2^2 + \cdots + (k - 1)^2 \lt \frac{k^3}{3}$$
for a fixed $k \ge 1$. Now using this, we shall deduce the corresponding result for $k + 1:$
$$A(k + 1): 1^2 + 2^2 + \cdots + k^2 \lt \frac{(k + 1)^3}{3}.$$
Start with $A(k)$ and add $k^2$ to both sides. This gives the inequality
$$1^2 + 2^2 + \cdots + k^2 \lt \frac{k^3}{3} + k^2.$$
To obtain $A(k + 1)$ as a consequence of this, it suffices to show that
$$\frac{k^3}{3} + k^2 \lt \frac{(k + 1)^3}{3}.$$
But this follows at once from the equation
$$\frac{(k + 1)^3}{3} = \frac{k^2 + 3k^2 + 3k + 1}{3} = \frac{k^3}{3} + k^2 + k + \frac13.$$
Therefore, we have shown that $A(k + 1)$ directly follows from $A(k)$.
I can not understand the last two steps. Why do the expressions seem to flip on the inequality in step 4 such that $\frac{k^3}{3} + k^2$ is now on the other side of the less than symbol, and how does the final equation prove the assertion?
Thank you.
|
If $a<b, b<c; a<c$
So, as $1^2 + 2^2 + \cdots + k^2 \lt \dfrac{k^3}3+k^2,$
and if we can show $\dfrac{k^3}3+k^2<\dfrac{(k+1)^3}3$
we can safely conclude,
$$1^2 + 2^2 + \cdots + k^2 \lt\dfrac{(k+1)^3}3$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2736318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Equations involving $2 \times 2$ Matrices There are two possible values of A in the solution of the matrix equation
$\left[\begin{matrix}2A+1&-5\\-4&A\end{matrix}\right]^{-1}\cdot \left[\begin{matrix}A-5&B\\2A-2&C\end{matrix}\right]= \left[\begin{matrix}14&D\\E&F\end{matrix}\right]$ where $A,B,C,D,E,F \space all \space \in \mathbb{R}$. The absolute value of the difference of these two solutions is?
Options given :
*
*$\dfrac{8}{3}$
*$\dfrac{11}{3}$
*$\dfrac{1}{3}$
*$\dfrac{19}{3}$
I have first obtained the result of $\left[\begin{matrix}2A+1&-5\\-4&A\end{matrix}\right]^{-1}$
$= \dfrac{1}{(A(2A+1)-20)} \cdot \left[\begin{matrix}A&5\\4&2A+1\end{matrix}\right]$
where $\left[\begin{matrix}A&5\\4&2A+1\end{matrix}\right] =$ adjoint($\left[\begin{matrix}2A+1&-5\\-4&A\end{matrix}\right]$)
Then the next step is to find
$\dfrac{1}{(A(2A+1)-20)} \cdot \left[\begin{matrix}A&5\\4&2A+1\end{matrix}\right] \cdot \left[\begin{matrix}A-5&B\\2A-2&C\end{matrix}\right]= \left[\begin{matrix}14&D\\E&F\end{matrix}\right]$
$\implies \dfrac{1}{(A(2A+1)-20)} \cdot \left[\begin{matrix}A(A-5)+10(A-1)&AB+5C\\4(A-5)+2(A-1)(2A+1)&4B+C(2A+1)\end{matrix}\right]=\left[\begin{matrix}14&D\\E&F\end{matrix}\right]$
$\therefore \dfrac{A^2-5A+10A-10}{2A^2+A-20}=14$[From the above written equation]
$\implies 27A^2+23A-270=0$
whose curve is like this :
and the roots are :
$-10$, $-\dfrac{23}{27}$
The difference between the two is $-\dfrac{247}{27}$
Hence none of the options are matching. Where am I wrong and if Yes, please correct me and give the correct answer please. I request anyone.
|
First keep in mind that if $x_{1,2}$ are the roots of a quadratic $ax^2+bx +c=0,\,a\neq 0$ then
$$|x_1-x_2|={\sqrt{\Delta}\over a}$$
Second you do not need to invert the first matrix just multiply both sides with it so the matrix equation is equivalent to
$$\begin{bmatrix} A-5 & B\\2A-2 & C\end{bmatrix}=\begin{bmatrix}2A+1 & -5\\-4 & A\end{bmatrix}\cdot\begin{bmatrix}14 & D\\E & F\end{bmatrix}$$
This is equivalent to
$$\begin{bmatrix} A-5 & B\\2A-2 & C\end{bmatrix}=\begin{bmatrix}28A+14-5E & (2A+2)D-5F\\-56+AE & -4D+AF\end{bmatrix}$$
We just identify the first column
$$\begin{cases} 5E=27A+19\\5EA=10A+270\end{cases}$$
Substituting $5E$ one gets
$$3A^2+A-30=0$$
This means $\Delta=361=19^2$ and
$$|x_1-x_2|={19\over 3}$$
Your mistake is in the last transformation
$$A^2-5A+10A=28A^2+14A-280\iff 27A^2+9A-270\iff 3A^2+A-30=0$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to solve the cubic $x^3-3x+1=0$? This was a multiple choice question with options being
$$(A)-\cos\frac{2\pi}{9},\cos\frac{8\pi}{9},\cos\frac{14\pi}{9} \\
(B)-2\cos\frac{2\pi}{9},2\cos\frac{8\pi}{9},2\cos\frac{14\pi}{9} \\
(C)-\cos\frac{2\pi}{11},\cos\frac{8\pi}{11},\cos\frac{14\pi}{11} \\
(D)-2\cos\frac{2\pi}{11},2\cos\frac{8\pi}{11},2\cos\frac{14\pi}{11}$$
I tried to eliminate options using the sum and product of roots but I can't figure out if $$\cos\frac{2\pi}{9}+\cos\frac{8\pi}{9}+\cos\frac{14\pi}{9}=0$$
or
$$\cos\frac{2\pi}{11}+\cos\frac{8\pi}{11}+\cos\frac{14\pi}{11}=0$$
|
To follow on from your method, $\cos(a) + \cos(b) = 2 \cos({a+b\over2}) \cos({a-b\over2})$, so $$\cos\frac{2\pi}{9}+\cos\frac{8\pi}{9}=2 \cos\frac{5\pi}{9}\cos\frac{3\pi}9=\cos\frac{5\pi}{9}$$
but $\cos\frac{5\pi}{9} = -\cos\frac{14\pi}{9}$, so the sum $\cos\frac{2\pi}{9}+\cos\frac{8\pi}{9}+\cos\frac{14\pi}{9}=0$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
I want to verify my proofs. 1).Prove: $\lim_{n \to \infty} \frac{3n^2-4}{n^2-4} = 3$.
A proof: there exists an $\epsilon>0$. We define $N=\lceil{\sqrt{\frac{8}{\epsilon}+4}}\rceil$ (notice $N>2$). So:
$$
\begin{aligned}
|\frac{3n^2-4}{n^2-4}-3| < \epsilon \Leftrightarrow |\frac{3n^2-4-3n^2+12}{n^2-4}|<\epsilon \Leftrightarrow |\frac{8}{n^2-4}| < \epsilon
\Leftrightarrow \frac{8}{n^2-4}<\epsilon \Leftrightarrow\\ 8<n^2\epsilon-4\epsilon \Leftrightarrow \frac{8+4\epsilon}{\epsilon}<n^2 \Leftrightarrow \sqrt{\frac{8}{\epsilon}+4}<n.
\end{aligned}
$$
This statement is true for all $n>N$,
because: $$\lceil{\sqrt{\frac{8}{\epsilon}+4}}\rceil \geq \sqrt{\frac{8}{\epsilon}+4}.$$
thus, the limit is 3. $\Box$
2).$a_nb_n=1$. Prove or contradict: If $\lim_{n\to\infty}|a_n|=1$ so $\lim_{n\to\infty}|b_n|=1$.
This statement is true.
A proof: We choose $\epsilon = \frac{1}{2}$, so:
$$
-\epsilon < a_nb_n-1 < \epsilon\Rightarrow -\epsilon+1<a_nb_n<\epsilon+1 \Rightarrow -\frac{1}{2}+1<a_nb_n<\frac{1}{2}+1
\Rightarrow \\ \frac{1}{2}<a_nb_n<1\frac{1}{2}
$$
Also:
$$
||a_n|-1|<\epsilon \Rightarrow \frac{1}{2} <|a_n| <1\frac{1}{2}
$$
It's obvious that $a_n\neq 0$ for almost all n, otherwise $\lim_{n\to\infty}a_n=0$.
If $a_n>0$ for almost all n, also $b_n>0$ for almost all n. we can divide and get:
$$
1<\frac{a_nb_n}{|a_n|}<1 \Rightarrow1<\frac{a_nb_n}{a_n}<1 \Rightarrow 1<b_n<1
$$
If $a_n<0$ for almost all n, also $b_n<0$ for almost all n. we can divide and get:
$$
1<\frac{a_nb_n}{|a_n|}<1 \Rightarrow1<\frac{a_nb_n}{-a_n}<1 \Rightarrow 1<-b_n<1
$$
And from the sandwich theorem we get $\lim_{n\to\infty}|b_n| = 1$. $\Box$
|
As an alternative to simplify note that
$$\lim_{n \to \infty} \frac{3n^2-4}{n^2-4} = \lim_{n \to \infty} \frac{3n^2-12+8}{n^2-4} =\lim_{n \to \infty} 3+\frac{8}{n^2-4}$$
and we can show that eventually $0\le\frac1{n^2-4}\le\frac1n$ and $\frac1n\to 0$.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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|
Short way for upper triangularization
We are given a matrix $$A =
\begin{bmatrix}
3 & 0 & 1 \\
-1 & 4 & -3 \\
-1 & 0 & 5 \\
\end{bmatrix}$$
and we are asked to find a matrix $P$ such that $P^{-1}AP$ is upper triangular.
Here, we first find one eigenvalue as $\lambda= 4$. Then the matrix $$
A-4I =
\begin{bmatrix}
-1 & 0 & 1 \\
-1 & 0 & -3 \\
-1 & 0 & 1 \\
\end{bmatrix}$$
has basis formed from $f_1 = (-1,-1,-1)^T$, $f_2 = (1,-3,1)^T$. We extend this to a basis of the whole space by adjoining $f_3 = (1,0,0)^T$, and so we have base-change matrix
$$ P =
\begin{bmatrix}
-1 & 1 & 1 \\
-1 & -3 & 0 \\
-1 & 1 & 0 \\
\end{bmatrix}.$$
Then, by using some computational tools, we find that
$$ P^{-1}AP =
\begin{bmatrix}
3 & 1 & 1 \\
-1 & 5 & 0 \\
0 & 0 & 4 \\
\end{bmatrix}$$
Now, we need to look at $$B = \begin{bmatrix}
3 & 1 \\
-1 & 5 \\
\end{bmatrix},$$ which has eigenvalue $\lambda = 4$. So we have $$B-4I = \begin{bmatrix}
-1 & 1 \\
-1 & 1 \\
\end{bmatrix}.$$
So basis for the image of this is $(1,1)^T$. We extend this to the basis $(1,1)^T$, $(1,0)^T$ of $\mathbb{R}^2$. Now, going back to $\mathbb{R}^3$, we have the matrix $$Q = \begin{bmatrix}
1 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1 \\
\end{bmatrix}.$$
Then, by using calculation tools, we get
$$Q^{-1}P^{-1}APQ = \begin{bmatrix}
4 & -1 & 0 \\
0 & 4 & 1 \\
0 & 0 & 4 \\
\end{bmatrix},$$
which is in upper triangular form.
Now, what I wanted to ask is that is there a way to directly find the matrix $R = PQ$ such that $R^{-1}AR$ is upper triangular, without going through these steps?
|
Suppose we begin with the matrix $$A =
\begin{bmatrix}
3 & 0 & 1 \\
-1 & 4 & -3 \\
-1 & 0 & 5 \\
\end{bmatrix}$$
and use "double Gaussian elimination". That is, whenever we multiply by an elementary row operation $E$ on the left, we must also multiply by the corresponding column operation $E^{-1}$ on the right. We begin by removing the $-1$ in the bottom left corner by doing a row operation of the form $R_3 \leftarrow R_3 + \lambda R_1$. We must then also perform the column operation $C_1 \leftarrow C_1 - \lambda C_3$. Since we want this to cancel the bottom-left cell, we see that $\lambda$ must satisfy
$$-1 + 3 \lambda - \lambda(5 + \lambda) = 0 \iff \lambda^2+2\lambda+1 = (\lambda+1)^2 = 0$$
So we take $\lambda = -1$, and apply those operations to get
$$A_1 =
\begin{bmatrix}
4 & 0 & 1 \\
-4 & 4 & -3 \\
0 & 0 & 4 \\
\end{bmatrix}$$
Now, this is almost in upper-triangular form: if we perform the paired operations $(R_1 \leftrightarrow R_2, C_1 \leftrightarrow C_2)$ then we end up with
$$A_2 =
\begin{bmatrix}
4 & -4 & -3 \\
0 & 4 & 1 \\
0 & 0 & 4 \\
\end{bmatrix}$$
which is upper-triangular. Let $P$ be the matrix tracking the column operations, so that
$$P =
\begin{bmatrix}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 1 & 1 \\
\end{bmatrix}$$
Then we have $P^{-1} A P = A_2$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Use of principle of inclusion and exclusion in counting Here is the question
Q:- How many bit strings of length 8 contain three consecutive zeros?
I tried to solve this by making the cases that three consecutive zeros start from bit number 1, 2 and so on. I approached to my answer correctly. I obtained 107 as my answer.
But I am confused about, how can I solve this question using inclusion-exclusioin principle? Can anyone help me with this?
|
As N. F. Taussig shows, Inclusion-Exclusion is not easy to apply to this problem. One other method that can be used is generating functions. To count the number of strings with no more than two zeros in succession, we will represent the atomic strings
$$
\begin{array}{}
1&x\\
10&x^2\\
100&x^3
\end{array}\tag1
$$
Since the atomic strings in $(1)$ can only represent strings that start with a one, we need to also represent $0$-$2$ zeros at the start with
$$
\begin{array}{}
-&1\\
0&x\\
00&x^2
\end{array}\tag2
$$
Using $(2)$ once and $(1)$ as many times as needed, we get the generating function
$$
\begin{align}
\left(1+x+x^2\right)\color{#090}{\sum_{k=0}^\infty\left(x+x^2+x^3\right)^k}
&=\frac{1+x+x^2}{\color{#090}{1-x-x^2-x^3}}\\
&=\left(1+x+x^2\right)\color{#090}{\left(1+x+2x^2+\dots\right)}\\[6pt]
&=1+2x+4x^2+\dots\tag3
\end{align}
$$
where the denominator $1-x-x^2-x^3$ says that the coefficients of the generating function, for $n\ge3$, obey the recurrence
$$
a_n=a_{n-1}+a_{n-2}+a_{n-3}\tag4
$$
Note that $(4)$ is the same recursion that N. F. Taussig got in their answer.
Using $(4)$, we can extend $(3)$ as far as we want:
$$
\begin{align}
\frac{1+x+x^2}{1-x-x^2-x^3}
&=1+2x+4x^2+7x^3+13x^4+24x^5+44x^6\\
&+81x^7+\color{#C00}{149}x^8+274x^9+504x^{10}+\dots\tag5
\end{align}
$$
$(5)$ says that the number of strings of length $8$ with no more than two zeros in succession is $149$. Since there are $256$ strings of length $8$, we get that there are $256-149=107$ strings of length $8$ with at least $3$ zeros in succession.
|
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|
How to find mean and standard deviation? An urn contains a large number of cards of which 1/4 have the number 1,1/4 have the number 2 and 1/2 have the number 3.
a) Let $X$ be the number of the card when a card is taken from the urn. Find the mean and standard deviation of $X$.
b) Let $X$ be the sample mean when card samples are taken, compute $\mu_{\overline x}$ and $\sigma_{\overline x}$.
Any hint please I don't know what to do
Edit: By definition $E(x)=\sum xp(x),$ thus $E(x)=9/4$
|
Your probability mass function (pmf) $p$ is
$$
p(x)=P(X=x)=
\begin{cases}
\frac{1}{4} &\mbox{ if } x=1, \\
\frac{1}{4} &\mbox{ if } x=2, \\
\frac{1}{2} &\mbox{ if } x=3. \\
\end{cases}
$$
a.) The expected value (mean) of $X$ is
$$
\mu=\mu_X = E[X] = \sum_{x=1}^{3} xp(x) = 1\left(\frac{1}{4}\right)+ 2\left(\frac{1}{4}\right)+ 3\left(\frac{1}{2}\right) = \frac{9}{4}
$$
and the variance of $X$ is
$$
V(X) = \sum_{x=1}^{3}(x-\mu)^2 p(x)
= \left(1-\frac{9}{4}\right)^2 \frac{1}{4} + \left(2-\frac{9}{4}\right)^2 \frac{1}{4} + \left(3-\frac{9}{4}\right)^2 \frac{1}{2} = \frac{11}{16}.
$$
So the standard deviation of $X$ is
$$
\sigma = \sigma_X = \sqrt{V(X)} = \sqrt{\frac{11}{16}}=\frac{\sqrt{11}}{4}.
$$
b.) We have
$$
\mu_{\overline X} = \mu_X = \frac{9}{4}
$$
while
$$
\sigma_{\overline X} = \frac{\sigma_X}{\sqrt{n}} = \frac{\sqrt{11}}{4\sqrt{n}},
$$
where $n$ is the sample size.
|
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|
Find the value of $ab$ $$5^a = 16, 8^b = 25$$
*
*Find the value of $ab$.
So, this question seems simple to solve. However, the most important thing is to know where to start. That's why I couldn't solve this problem. I've been looking for a method/strategy to solve all kinda questions which involve exponential terms.
Now, rewriting the inequalities.
$$5^a = 16 \implies 5^a = 4^2 \implies 5^a = 2^4$$
$$8^b = 25 \implies 8^b = 5^2 \implies 2^{3b} = 5^2$$
Or what about giving letters like $k$, $t$ or somewhat?
Regards!
|
$5^a = 16$ so $a =\log_5 16= \log_5 2^4 = 4\log_5 2$
$8^b = 25$ so $b = \log_8 25=\log_8 5^2 = 2\log_8 5$
So $ab =4\log_5 2*2\log_8 5=8\log_5 2*\log_8 5$.
That may (or may not) be simplified.
Use $\log_a b = \frac {\log_M b}{\log_M a}$ to convert to a common base.
$\log_5 2 = \frac {\ln 2}{\ln 5}$ and $\log_8 5 = \frac {\ln 5}{\ln 8}=\frac {\ln 5}{3\ln 2}$. (Nothing special about $\ln$. We could just as easily used $\log_{10}$. Or $\log$ base anything.)
So $ab = 8\log_5 2*\log_8 5= 8\frac {\ln 2}{\ln 5}\frac {\ln 5}{3\ln 2} = \frac 83$.
...
Another trick would be to recognize $\log_a b = \frac 1{\log_b a}$ and that $\log_{a^k} b = \frac 1k \log_a b$.
So then:
$\log_5 16*\log_8 25 = \frac {\log_5 16}{\log_25 8} = \frac {4\log_5 2}{3*\frac 12 \log_5 2} = \frac 83$
But that a little subtle and sophisticated and it's not reasonable to expect everyone to see that right away.
....
I suppose if you get really comfortable with exponents you could simply do:
$8^{ab} = (8^b)^a = 25^a = (5^2)^a = (5^a)^2 = 16^2 = 2^2*8^2$
$ab = \log_8 8^{ab} = \log_8 4*8^2 = \log_8 2^2 + \log_8 8^2 = 2\log_8 2 + 2 = 2\frac 23$.
.... Actually that's pretty slick!
....
Actually without logarithms this is okay if you recognize that this are using powers.
$5^a = 2^4$ and $8^b = 2^{3b} = 5^2$.
So we can do either $5^{ab} = 2^{4b}=2^{3b*\frac 43} = (5^2)^{\frac 43}= 4^{\frac 83}$ and $ab = \frac 83$ or we can do $2^{3ab} = 5^{2a} = 2^{8}$ and $ab =\frac 83$.
But that's kind of round about that one simply assumed you knew logarithms:
$a = 4*\log_5 2$ and $3b = 2 \log_2 5$ and so $3ab = 8 \log_5 2\log_2 5$ is the exact same thing.
It's a little be clever to note that $\log_m n *\log_n m = 1$ but if one knows that we can convert bases via $\log_m n = \frac {\log_M n}{\log_M n}$ so $\log_2 5 = \frac {\log_M 5}{\log_M 2}$ and $\log_5 2 = \frac {\log_M 2}{\log_M 2}$ leads to that directly.
|
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|
How to maximise this bivariate function: $\frac{x}{x + y} + \frac{50 - x}{100 - x - y}$ I wish to maximise
$ g(x,y) = \frac{x}{x + y} + \frac{50 - x}{100 - x - y}$
subject to the constraints $x , y \in \mathbb{N}$. $ 1≤ x≤ 50 ,0≤y≤50$
I know $g(x,y) = f(x,y) + f(50-x,50-y)$ if that helps.
I know it is maximised at $x = 1$, $y = 0$, but cannot rigorously show why, I have tried finding the gradient, and then the critical points, but the gradient is just messy, so I don't think this is the correct way.
|
We have $$g(x,y) = \frac{x}{x + y} + \frac{50 - x}{100 - x - y}=2-\frac y{x+y}+\frac{y-50}{100-x-y}$$ so $$\frac{\partial g}{\partial x}=\frac y{(x+y)^2}+\frac{y-50}{(100-x-y)^2}=0$$ gives $$\color{red}{y(100-(x+y))^2=-(y-50)(x+y)^2}\implies200y-4y(x+y)=(x+y)^2\tag{1}$$ and since $$g(x,y) = \frac{x}{x + y} + \frac{50 - x}{100 - x - y}$$ we have $$\frac{\partial g}{\partial y}=-\frac x{(x+y)^2}+\frac{x-50}{(100-x-y)^2}=0$$ so $$\color{blue}{x(100-(x+y))^2=(x-50)(x+y)^2}\implies-200x+4x(x+y)=(x+y)^2\tag{2}$$ Dividing the blue equation by the red gives $$\frac yx=-\frac{y-50}{x-50}\implies x=\frac{25y}{y+25}\tag{3}$$ Equating $(1)$ and $(2)$, $$50(x+y)=(x+y)^2\implies(x+y)(x+y-50)=0\tag{4}$$ Can you continue with $(3)$ and $(4)$?
|
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|
Real roots of an equation A friend gave me this equation which I have trouble finding the real roots. $$x^9+3x^6+3x^3-16x+9=0$$ One can easily see that 1 is a root then with the help of Horner's method this can be simplified. However I am looking for an elegant solution if possible not just to use a computer to do that madness calculation. I reduced it to $$x^3+1=2(2x-1)^{\frac{1}{3}}$$ I don't have any useful attempts.
EDIT: After several tries I think I found something useful. After getting to this form $x^3-2(2x-1)^{\frac{1}{3}}+1=0$ we might notice that it's similar to the equation that you provided me in the answers namely$$(x-1)(x^2 +x-1)=x^3 - 2x+1=0$$ Now in order that this to give me the solution, we must have that $-2(2x-1)^{\frac{1}{3}}=-2x\, $ If we cube both sides we return to $x^3 - 2x+1=0$ (the solutions) Is this a repeatedly loop in the equation? Can this be useful?
|
The polynomial can be factored by considering the product of
$$(x^3 + a x^2 + b x + c) \cdot (x^6 + \alpha x^5 + \beta x^4 + \gamma x^3 + \delta x^2 + \eta x + \mu).$$
With a little bit of effort it can be found that
$$(x - 1) (x^2 + x - 1) (x^6 + 2 x^4 + 2 x^3 + 4 x^2 + 2 x + 9) = 0.$$
The real roots are then obtained as $x=1$ and roots from $x^2 + x -1=0$.
|
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|
the volume of the expanding cube is increasing at the rate of $24 cm^3/min$, how fast is the surface area increasing when surface area is $216cm^2$? If the volume of the expanding cube is increasing at the rate of $24 cm^3/min$, how fast is the surface area increasing when surface area is $216cm^2$?
My Approach :
$V=l^3$
$$\dfrac {dV}{dt}=3l^2\dfrac {dl}{dt}$$
$$24=3l^2\dfrac {dl}{dt}$$
Also,
$S=6l^2$
$$\dfrac {dS}{dt}=12l.\dfrac {dl}{dt}$$
How do I proceed?
|
The volume: $V=x^3$. The surface: $S=6x^2$. Given the surface is $216$, we can find the side:
$$6x^2=216 \Rightarrow x=6.$$
Given the volume increases at the rate $24$, we can find at what rate the side is increases:
$$\frac{dV}{dt}=3x^2\cdot \frac{dx}{dt}=24 \Rightarrow \frac{dx}{dt}=\frac{24}{3\cdot 6^2}=\frac29.$$
Given the volume and surface equation, we can differentiate the volume:
$$V=S\cdot \frac{x}{6} \Rightarrow \frac{dV}{dt}=\frac{dS}{dt}\cdot \frac{x}{6}+S\cdot \frac{1}{6}\cdot \frac{dx}{dt}=24 \Rightarrow \\
\frac{dS}{dt}\cdot \frac{6}{6}+216\cdot \frac{1}{6}\cdot \frac{2}{9}=24 \Rightarrow \\
\frac{dS}{dt}=16. $$
|
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|
Solution of system of equations involving $x_{1},x_{2},x_{3}$
Solve for $x_{1},x_{2},x_{3}$, given
$ax^2_{1}+bx_{1}+c=x_{2}$
$ax^2_{2}+bx_{2}+c=x_{3}$
$ax^2_{3}+bx_{3}+c=x_{1}$
Try: from $(1)$ and $(2)$
$a(x^2_{1}-x^2_{2})+b(x_{1}-x_{2})=(x_{2} - x_{3})$
And from $(2)$ and $(3)$
$a(x^2_{2}-x^2_{3})+b(x_{2}-x_{3})=(x_{3}-x_{1})$
Could some help me to find $x_{1},x_{2},x_{3}$, Thanks
|
I would start with solving $$ax^2_{1}+bx_{1}+c=x_{2}$$
Via the quadratic formula,
$$x_1=\frac{-b\pm\sqrt{b^2-4a(c-x_2)}}{2a}$$
Repeating the same process, we have
$$x_2=\frac{-b\pm\sqrt{b^2-4a(c-x_3)}}{2a}$$
and
$$x_3=\frac{-b\pm\sqrt{b^2-4a(c-x_1)}}{2a}$$
Hence,
\begin{align}
x_1&=\frac{-b\pm\sqrt{b^2-4a(c-x_2)}}{2a}\\
&=\frac{-b\pm\sqrt{b^2-4a\left(c-\frac{-b\pm\sqrt{b^2-4a(c-x_3)}}{2a}\right)}}{2a}\\
&=\frac{-b\pm\sqrt{b^2-4a\left(c-\frac{-b\pm\sqrt{b^2-4a\left(c-\frac{-b\pm\sqrt{b^2-4a(c-x_1)}}{2a}\right)}}{2a}\right)}}{2a}\\
\end{align}
in which $x_1$ can be painfully solved for
to get
$$x_1=\frac{\pm\sqrt{-4 a c + b^2 - 2 b + 1} - b + 1}{2 a}$$
Then just repeat for $x_2$ and $x_3$ for which you will get
$$x_1=x_2=x_3=\frac{\pm\sqrt{-4 a c + b^2 - 2 b + 1} - b + 1}{2 a}$$
|
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$\sqrt[5]{2^4\sqrt[3]{16}} = ? $ $$\sqrt[5]{2^4\sqrt[3]{16}} = ? $$
Rewriting this and we have
$$\sqrt[5\cdot 3]{2^{4\cdot 3}4}$$
$$\sqrt[15]{2^{12}2^2}$$
Finally we get
$$\sqrt[15]{2^{12}2^2} = \sqrt[15]{2^{14}} = 2^{\frac{15}{14}}$$
Am I right?
|
$$2^4\sqrt[3]{16}=2^4\cdot2^{4/3}=2^{16/3}\implies\sqrt[5]{2^4\sqrt{16}}=\left(2^{16/3}\right)^{1/5}=2^{16/15}$$
so no, you are not right...but almost.
|
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A Problem Dealing with putting balls in bin and Expected Value - possible wrong answer Please consider the problem below. Is my answer correct. If is not correct then where did I go wrong?
Problem:
You keep tossing balls into $n$ bins until one of the bins has two balls. For each toss there is a $\frac{1}{n}$ probability that the ball you toss lands in any one of the bins. What is the expected number of tosses?
Answer:
Let $p_i$ be the probability that after $i$ tosses we have at least one bin
with two balls.
\begin{eqnarray*}
p_1 &=& 0 \\
p_2 &=& 1 - ( 1 - \frac{1}{n} ) = \frac{1}{n} \\
p_3 &=& 1 - (1 - \frac{1}{n})(1 - \frac{1}{n-1}) \\
p_3 &=& 1 - (\frac{n-1}{n})(\frac{n-1-1}{n-1}) \\
p_3 &=& 1 - (\frac{n-2}{n}) = \frac{2}{n} \\
p_4 &=& 1 - (1 - \frac{1}{n})(1 - \frac{1}{n-1})(1 - \frac{1}{n-2}) \\
p_4 &=& 1 - ( \frac{n-1}{n} )( \frac{n-2}{n-1} )( \frac{n - 2 -1}{n - 2} ) \\
p_4 &=& 1 - \frac{n-3}{n} = \frac{3}{n} \\
\end{eqnarray*}
Now for $1 <= i <= n$ we have: $p_i = \frac{i-1}{n}$.
\begin{eqnarray*}
E &=& 2p_2 + 3p_3 + 4p_4 + ... (n+1)p_{n+1} \\
E &=& \sum_{i = 1}^{n} \frac{i(i+1)}{n} = \frac{1}{2n} \sum_{i=1}^{n} i^2 + i \\
E &=& \frac{1}{2n}(\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} ) \\
E &=& \frac{n+1}{4n} ( \frac{2n+1}{3} + 1 ) \\
\end{eqnarray*}
Here is an update to my answer:
Let $p_i$ be the probability that after $i$ tosses we have at least one bin
with two balls.
\newline
\begin{eqnarray*}
p_1 &=& 0 \\
p_2 &=& 1 - ( 1 - \frac{1}{n} ) = \frac{1}{n} \\
p_3 &=& 1 - (\frac{n-1}{n})( \frac{n-2}{n}) \\
p_3 &=& 1 - \frac{(n-1)(n-2)}{n^2} = \frac{n^2 - (n^2 - 3n + 2)}{n^2} \\
p_3 &=& \frac{3n-2}{n^2} \\
p_4 &=& 1 - (\frac{n-1}{n})(\frac{n-2}{n})(\frac{n-3}{n}) \\
p_4 &=& 1 - \frac{(n^2-3n+2)(n-3)}{n^3}\\
p_4 &=& 1 - \frac{n^3-3n^2+2n - 3n^2 +9n - 6}{n^3}\\
p_4 &=& \frac{3n^2-2n + 3n^2 - 9n + 6}{n^3}\\
p_4 &=& \frac{3n^2 + 3n^2 - 11n + 6}{n^3}\\
\end{eqnarray*}
\begin{eqnarray*}
E &=& 2p_2 + 3p_3 + 4p_4 + ... (n+1)p_{n+1} \\
\end{eqnarray*}
Now, am on the right track? That is, is what I have so far correct?
Thanks,
Bob
|
The number $n$ of bins is given, and fixed. Denote by $E(j)$ $(0\leq j\leq n)$ the expected number of additional tosses when there are $j$ empty bins and the game is not yet over. Then we have the following recursion:
$$E(j)=1+{j\over n}E(j-1)\quad(n\geq j\geq 1),\qquad E(0)=1\ .$$
This gives
$$\eqalign{
E(n)&=1+{n\over n}E(n-1)\cr
&=1+{n\over n}\left(1+{n-1\over n}E(n-2)\right)\cr
&=1+{n\over n}\left(1+{n-1\over n}\left(1+{n-2\over n}E(n-3)\right)\right)\cr
&=\ldots\cr}$$
and leads to
$$E(n)=1+\sum_{k=1}^n{n(n-1)\cdots\bigl(n-(k-1)\bigr)\over n^k}\ ,$$
as in Joseph Eck's answer.
|
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|
Infinite positive integer solutions of the equation: $x^2+x+1=(y^2+y+1)(n^2+n+1)$ Could anybody help me some ideas on the below problem:
Let $n$ be a positive integer. Prove that there are infinite pairs of positive integer $(x\, y)$ such that
$$x^2+x+1=(y^2+y+1)(n^2+n+1).$$
Thanks in advance.
|
HINT:
Fix $n$. You get a quadratic equation in $x$, $y$. You should reduce it to a Pell equation as follows:
$$(4 x^2 + 4 x + 4 ) = (n^2 + n+1)(4 y^2 + 4 y + 1)\\
(2x+1)^2 + 3 = N \cdot ((2y+1)^2 + 3)\\
(2x+1)^2 - N (2y+1)^2 = 3(N-1)\\
a^2 - N b^2 = 3(N-1)$$
You have a solution for the last equation, $a=2n+1$, $b=1$, coming from the obvious equality $(n^2 + n+1)=(n^2 + n+1)(0^2 + 0+1)$.
Consider now a solution for the Pell equation
$$A^2 - N B^2 =1$$
with $A$, $B$ positive integers ($N = n^2 + n+1$ is odd, not a square). Note that $A$, $B$ have opposite parities. Now
$$(a'+b'\sqrt{N}) = (A+B\sqrt{N})(a+b\sqrt{N})$$
satisfies again
$$a'^2 -N b'^2 = 3(N-1)$$
and $a'$, $b'$ are odd positive integers. Since there are infinitely many such $A$, $B$, we get infinitely many solutions $(a,b)$ odd positive integers.
|
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How to solve this complex limits at infinity with trig? Please consider this limit question
$$\lim_{x\rightarrow\infty}\frac{a\sin\left(\frac{a(x+1)}{2x}\right)\cdot \sin\left(\frac{a}{2}\right)}{x\cdot \sin\left(\frac{a}{2x}\right)}$$
How should I solve this? I have no idea where to start please help.
|
$$\lim_{x\rightarrow\infty}\frac{a\color{#f00}{\sin(\frac{a(x+1)}{2x})}\cdot \sin(\frac{a}{2})}{\color{#00ff}{x\cdot \sin(\frac{a}{2x})}}\tag{1}$$
$\color{#f00}{\sin\left(\frac{a(x+1)}{2x}\right)} = \sin\left(\frac{a}{2}+\frac{a}{2x}\right);\quad$ so $x \to \infty \implies \sin\left(\frac{a}{2}+0\right) = \sin\left(\frac{a}{2}\right)$
L'Hopital:
$\color{#00ff}{x\sin\left(\frac{a}{2x}\right)} = \left(\frac{\sin \left(\frac{a}{2x}\right)}{\frac{1}{x}}\right) =\left(\frac{-\frac{a\cos \left(\frac{a}{2x}\right)}{2x^2}}{-\frac{1}{x^2}}\right)=\left(\frac{a}{2}\cos \left(\frac{a}{2x}\right)\right);\quad$ so $x \to \infty \implies\frac{a}{2}\cos(0) = \frac{a}{2}$
$$\frac{a\color{#f00}{\sin\left(\frac{a}{2}\right)}\cdot \sin(\frac{a}{2})}{\color{#00ff}{\frac{a}{2}}}\tag{2}$$
$$2\sin ^2\left(\frac{a}{2}\right)\tag{3}$$
$$1-\cos\left(a\right)\tag{3}$$
|
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EC Point calculation Following the Guide to Elliptic Curve Cryptography, it provides the following elliptic curve on $E(\mathbb{F}_p)$ with $p=29$ on page 80:
$E: y^2 = x^3 + 4x + 20$
Page 81 provides a list of the points on the curve. For $x=2$ it includes the points $(2,6)$ and $(2,23)$. Let's do the calculation:
\begin{align*}
x &= 2\\
y &= \pm \sqrt{x^3 + 4x + 20} \bmod 29\\
y &= \pm \sqrt{2^3 + 4 \times 2 + 20} \bmod 29\\
y &= \pm \sqrt{8 + 8 + 20} \bmod 29\\
y &= \pm \sqrt{36} \bmod 29\\
y &= \pm 6 \bmod 29\\
y_1 &= +6 \bmod 29 = 6\\
y_2 &= -6 \bmod 29 = 23\\
\end{align*}
Thus, we have the points $(2,6)$ and $(2,23)$.
Doing the same calculation for $x=3$ fails:
\begin{align*}
x &= 3\\
y &= \pm \sqrt{x^3 + 4x + 20} \bmod 29\\
y &= \pm \sqrt{3^3 + 4 \times 3 + 20} \bmod 29\\
y &= \pm \sqrt{27 + 12 + 20} \bmod 29\\
y &= \pm \sqrt{59} \bmod 29\\
y &= \pm 7,68 \bmod 29
\end{align*}
This fails, since it is not an integer.
If I do the mod before the sqrt, I get the correct result for $x=3$:
$$\pm\sqrt{59 \bmod 29} = \pm\sqrt{1} = 1$$
If I do the same for $x=2$, then it fails again:
$$\pm\sqrt{36 \bmod 29} = \pm\sqrt{7} = \pm 2,64$$
Conclusion: Doing mod after sqrt breaks $x=3$, doing mod before sqrt breaks $x=2$.
Question: What do I miss respectively what am i doing wrong? Is there a special sqrt-operation for mod?
[EDIT 1]: Using Wolfram Alpha, I get the correct numbers.
*
*$x=2$: $y^2 = 36 \bmod 29$ $\Rightarrow y = 6$ and $y=23 \Rightarrow (2,6)$ and $(2,23)$
*$x=3$: $y^2 = 59 \bmod 29$ $\Rightarrow y = 1$ and $y=28 \Rightarrow (3,1)$ and $(3,28)$
But how does it work?
[EDIT 2]: Just found Cipolla's algorithm, which solves $x^2\equiv n \pmod{p}$.
|
This is arithmetic in finite fields. You have to take modular square roots. Here the examples with Pari/GP
? x=Mod(36,29)
%1 = Mod(7, 29)
? x^(1/2)
%2 = Mod(6, 29)
? x=Mod(59,29)
%3 = Mod(1, 29)
? x^(1/2)
%4 = Mod(1, 29)
So yes, there is there a special sqrt-operation for mod?
See e.g. Modular square root or the Tonelli-Shanks algorithm. A Q/A on this site is Modular Arithmetic - Find the Square Root.
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Inequality about the most common divergent series We know that:
$1+\frac{1}{2}+\frac{1}{3}+....$ is a divergent series.
I have a small problem about this series.
Show that: $1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2^{2006}} >50$
How to prove this one?
Can you use a basic knowledge to explain for a 14 years old student?
|
Notice $1+\frac{1}{2}+\dots + \frac{1}{n} =\int\limits_{i=1}^{n+1} \frac{1}{\lfloor x \rfloor } \geq \int\limits_{i=1}^{n+1} \frac{1}{ x } = \ln(n+1)$
We conclude your sum is at least $\ln(2^ {2006} +1)$ and the rest is easy.
A basic way to prove it is by noticing
$\frac{1}{2^ n} + \frac{1}{2^n + 1} + \dots + \frac{1}{2^{n+1}-1}> \frac{2^ n}{2^{n+1}} = \frac{1}{2}$ and notice we can split the sum into way more than $100$ of these sums.
|
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British Maths Olympiad (BMO) 2004 Round 1 Question 1 alternative approaches? The questions states:
Solve the simultaneous equations (which I respectively label as $
> \ref{1}, \ref{2}, \ref{3}, \ref{4}$)
$$\begin{align} ab + c + d &= 3 \tag{1} \label{1} \\ bc + d + a &= 5
\tag{2} \label{2} \\ cd + a + b &= 2 \tag{3} \label{3} \\ da + b + c
&= 6 \tag{4} \label{4} \end{align}$$
where $a,b,c,d$ are real numbers.
I solved this system after quite a while by taking
$eqns$ 1 - 3 = $eqns$ 4 - 2
which yields $a + c = 2$
You can then substitute that in and find the other variables
I also noticed that $(a+1)(b+1) + (a+1)(d+1) + (c+1)(b+1) + (c+1)(d+1) = 20$ but that line didnt really help me.
I'm interested in seeing the other approaches people can take with this system.
Additionally, is there a sufficient enough hint to take another route? Did I miss an easy solution?
|
Here's a way after you get $a+c=2$.
Take equation $2$ subtract equation $1$ to get $(b-1)(c-a)=2$.
Likewise, take equation $3$ subtract equation $4$ to get $(d-1)(c-a) = -4$.
Finally, we see that
\begin{align}
\frac{b-1}{d-1}=\frac{(b-1)(c-a)}{(d-1)(c-a)}= -\frac{1}{2} \ \ \implies \ \ 2b+d =3.
\end{align}
Next, take equation $1$ plus equation $2$ to get $b(a+c)+(a+c)+2d=8$ which implies $b+d = 3$ since $a+c=2$.
Now, we see that $b=0$ and $d=3$. Using equation $2$, we have that $a=2$ and $c=0$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding the range of $f(x)=\cos x\left(\sin x+\sqrt{\smash[b]{\frac12+\sin^2x}}\right)$
How do I find the number of integers in the range of
$$f(x)= \cos x\left( \sin x + \sqrt{\dfrac 12 +\sin^2 x} \right)?$$
I set the derivative equal to $0$ but the method isn't efficient here because it gives a very complicated trigonometric equation. What's the proper way to do this then?
|
Since you are asking about integers in the range, we do not need to fully know the end points of the range, and we don't need calculus.
Clearly the integer $0$ is in the range, considering $x=\pi/2$.
Since $f(\pi/4)=\frac{\sqrt{2}}{2}\left(\frac{\sqrt{2}}{2}+1\right)=\frac{1+\sqrt{2}}{2}>1$, the integer $1$ is in the range as well (by the Intermediate Value Theorem).
Is $2$ in the range? If so, then $$
\begin{align}
2&=\cos x\left( \sin x + \sqrt{\dfrac 12 +\sin^2 x} \right)\\
2-\sin(x)\cos(x)&=\cos(x)\sqrt{\dfrac 12 +\sin^2 x}\\
4-4\sin(x)\cos(x)+\sin^2(x)\cos^2(x)&=\cos^2(x)\left(\dfrac 12 +\sin^2 x\right)\\
4-4\sin(x)\cos(x)&=\frac{1}{2}\cos^2(x)\\
4-2\sin(2x)&=\frac{1}{2}\cos^2(x)
\end{align}
$$
This has no solutions since the smallest the left side can be is $2$, and the largest the right side can be is $\frac{1}{2}$.
So $2$ is not in $f$'s range. Again using the Intermediate Value Theorem, no integer larger than $2$ is in the range either, or else $2$ would be in the range.
Lastly, note that $x\mapsto\pi-x$ preserves $\left( \sin x + \sqrt{\dfrac 12 +\sin^2 x} \right)$ but negates $\cos(x)$. So $f(\pi-x)=-f(x)$. And this means that $f$ takes negative values just as it takes positive values. That is, $-1$ is in the range, but not $-2$ or anything smaller.
So the integers in the range of $f$ make up the set $\{-1,0,1\}$.
|
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|
A functional equation problem: $ \frac { f \left( f ( y ) ^ 2 + x y \right) } { f ( y ) } = f ( x ) + y $
Let $ \mathbb R ^ + $ denote the set of the positive real numbers. Find all functions $ f : \mathbb R ^ + \to \mathbb R ^ + $ satisfying
$$ \frac { f \left( f ( y ) ^ 2 + x y \right) } { f ( y ) } = f ( x ) + y $$
for all $ x , y \in \mathbb R ^ + $.
I am very thankful for any solution, please help!
I tried to set $ x = y = 1 $, $ x = y = 2 $, $ x = 1 $, $ y = 2 $, so on, but this problem is more difficult.
|
$ \def \R {\mathbb R ^ +} $
You can show that the only $ f : \R \to \R $ satisfying
$$ f \left( x y + f ( y ) ^ 2 \right) = \big( f ( x ) + y \big) f ( y ) \tag 0 \label 0 $$
for all $ x , y \in \R $ is the identity function. It's straightforward to verify that the identity function is a solution. We try to prove the converse. The messiest part is showing $ f ( 1 ) = 1 $, and after that, everything goes rather smoothly. Before going further, let's make the simple observation that $ f $ must be injective: if $ f ( x ) = f ( y ) $, then by \eqref{0} we have
$$ x = \frac { f \left( y x + f ( x ) ^ 2 \right) } { f ( x ) } - f ( y ) = \frac { f \left( x y + f ( y ) ^ 2 \right) } { f ( y ) } - f ( x ) = y \text . $$
Showing $ f ( 1 ) = 1 $
Letting $ a = f ( 1 ) $ and plugging $ y = 1 $ in \eqref{0} we get
$$ f \left( x + a ^ 2 \right) = a \big( f ( x ) + 1 \big) \text . \tag 1 \label 1 $$
Assuming $ a > 1 $, \eqref{1} gives
$$ f \left( x + n a ^ 2 \right) = a ^ n f ( x ) + \frac { a ^ { n + 1 } - a } { a - 1 } \tag 2 \label 2 $$
by induction on the positive integer $ n $. Setting $ x = y = 1 + n a ^ 2 $ in \eqref{0} and using \eqref{2} we get
$$ f \left( \left( 1 + n a ^ 2 \right) ^ 2 + \left( \frac { a ^ { n + 2 } - a } { a - 1 } \right) ^ 2 \right) = \frac { a ^ { n + 2 } - a } { a - 1 } \left( \frac { a ^ { n + 2 } - a } { a - 1 } + 1 + n a ^ 2 \right) \text . \tag 3 \label 3 $$
Also, letting $ m = 2 n + \left\lfloor n ^ 2 a ^ 2 + \left( \frac { a ^ { n + 1 } - 1 } { a - 1 } \right) ^ 2 \right\rfloor $ and $ b = \left( 1 + n a ^ 2 \right) ^ 2 + \left( \frac { a ^ { n + 2 } - a } { a - 1 } \right) ^ 2 - m a ^ 2 $, we can see that $ m $ is a positive integer and $ b \in \R $, and thus we can use \eqref{2} to get
$$ f \left( \left( 1 + n a ^ 2 \right) ^ 2 + \left( \frac { a ^ { n + 2 } - a } { a - 1 } \right) ^ 2 \right) = a ^ m f ( b ) + \frac { a ^ { m + 1 } - a } { a - 1 } \text . $$
But this means that the right-hand side of \eqref{3} must be greater than $ \frac { a ^ { m + 1 } - a } { a - 1 } $, which can't happen for large enough $ n $. Thus the assumption $ a > 1 $ is absurd, and we must have $ a \le 1 $. Now, if $ a < 1 $, we can set $ x = 1 - a ^ 2 $ and $ y = 1 $ in \eqref{0} to get $ a = a \Big( f \left( 1 - a ^ 2 \right) + 1 \Big) $, which contradicts $ a , f \left( 1 - a ^ 2 \right) \in \R $. Thus the assumption $ a < 1 $ is absurd, too, and we must have $ a = 1 $.
Showing $ f = \operatorname {Id} _ { \R } $
Using \eqref{1} and induction on the positive integer $ n $, we get
$$ f ( x + n ) = f ( x ) + n \text , \tag 4 \label 4 $$
which in particular shows $ f ( n ) = n $ for any positive integer $ n $. Hence, letting $ y = n $ in \eqref{0} we have
$$ f ( n x ) = n f ( x ) \tag 5 \label 5 $$
for any positive integer $ n $. Letting $ x = \frac n y $ in \eqref{0} and using \eqref{4} and \eqref{5}, we get
$$ f \left( f ( y ) ^ 2 \right) - y f ( y ) = n \Bigg( f ( y ) f \left( \frac 1 y \right) - 1 \Bigg) \text , $$
and since the left-hand side doesn't depend on $ n $, both sides must be equal to $ 0 $. In particular, we have
$$ f \left( f ( y ) ^ 2 \right) = y f ( y ) \text . \tag 6 \label 6 $$
Letting $ x = \frac { f ( y ) ^ 2 } y $ in \eqref{0} and using \eqref{5} and \eqref{6} we get
$$ f \left( \frac { f ( y ) ^ 2 } y \right) = y \text . \tag 7 \label 7 $$
Using \eqref{4} and \eqref{7}, we can see that
$$ f \left( \frac { f ( y ) ^ 2 + 2 f ( y ) + 1 } { y + 1 } \right) = f \left( \frac { \big( f ( y ) + 1 \big) ^ 2 } { y + 1 } \right) = f \left( \frac { f ( y + 1 ) ^ 2 } { y + 1 } \right) \\
= y + 1 = f \left( \frac { f ( y ) ^ 2 } y \right) + 1 = f \left( \frac { f ( y ) ^ 2 } y + 1 \right) = f \left( \frac { f ( y ) ^ 2 + y } y \right) \text . $$
By injectivity of $ f $, we have
$$ \frac { f ( y ) ^ 2 + 2 f ( y ) + 1 } { y + 1 } = \frac { f ( y ) ^ 2 + y } y \text , $$
or equivalently
$$ y f ( y ) ^ 2 + 2 y f ( y ) + y = y f ( y ) ^ 2 + y ^ 2 + f ( y ) ^ 2 + y \text , $$
which can be simplified to $ \big( f ( y ) - y \big) ^ 2 = 0 $, and thus $ f $ is the identity function.
|
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|
Finding the residue of $\frac{1}{z^3 \sin z}$ at z = 0 Given the function $f(z) = \frac{1}{z^3 \sin{(z)}}$, what is the residue of $f(z)$ at $z = 0$?
I want to find $b_1$ from the Laurent expansion. So I did the following:
\begin{align*}
\frac{1}{z^3 \sin{(z)}}
&= ( \dots \frac{b_5}{z^5} + \frac{b_4}{z^4} + \frac{b_3}{z^3} + \frac{b_2}{z^2} + \frac{b_1}{z} + a_0 + a_1z + \dots )\\
1& = \Big ( \dots \frac{b_5}{z^5} + \frac{b_4}{z^4} + \frac{b_3}{z^3} + \frac{b_2}{z^2} + \frac{b_1}{z} + a_0 + a_1z + \dots \Big ) \cdot \Big ( z^3 \sin{(z)} \Big )\\
&= \Big ( \dots \frac{b_5}{z^2} + \frac{b_4}{z} + b_3 + b_2z + b_1z^2 + a_0z^3 + a_1z^4 + \dots \Big ) \cdot \Big ( \sin{(z)} \Big )\\
&= \Big ( \dots \frac{b_5}{z^2} + \frac{b_4}{z} + b_3 + b_2z + b_1z^2 + a_0z^3 + a_1z^4 + \dots \Big ) \cdot \Big ( z - \frac{z^3}{3!} + \frac{z^5}{5!} - \dots \Big )\\
\end{align*}
After some more thought...
Is it true to say that because f(z) has a pole of order 4 at $z=0$ that our $b_n$'s only go out to the 4th term? Meaning there are no $b_5$, $b_6$, etc like how wrote previously. That is,
$\frac{1}{z^3 \sin{(z)}}
= \Big ( \frac{b_4}{z^4} + \frac{b_3}{z^3} + \frac{b_2}{z^2} + \frac{b_1}{z} + a_0 + a_1z + \dots \Big )\\$
followed by
\begin{align*}
1
&= \Big ( \frac{b_4}{z} + b_3 + b_2z + b_1z^2 + a_0 + \dots \Big ) \cdot \Big ( z - \frac{z^3}{3!} + \frac{z^5}{5!} - \dots\Big )\\
\end{align*}
Which then when multiplying out $b_1z^2$ with each term from sin(z)'s Laurent expansion will never yield a $\frac{1}{z}$ term, concluding that the coefficient $b_1 = 0$?
|
The function is even so the residue at $z = 0$ is zero, since the residue is the coefficient of ${1 \over z}$ in the Laurent expansion.
|
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|
Generating function for the Catalan numbers I know that generating function $f(x)$ for the Catalan numbers is
\begin{equation}
f(x)=\cfrac{1\pm \sqrt{1-4x}}{2x}\ .
\end{equation}
It is often said that we should choose
\begin{equation}
f(x)=\cfrac{1- \sqrt{1-4x}}{2x}
\end{equation}
because $f(x)$ should be continuous at $x=0$, but I can't understand why $f(x)$ should be continuous.
What is the problem if $f(x)$ is not continuous at $x=0$ ?
|
We choose the negative sign in
\begin{align*}
f(x)=\frac{1\color{blue}{\pm} \sqrt{1-4x}}{2x}
\end{align*}
since we want to expand $f$ in a power series.
According to the binomial series expansion we have for $|x|<\frac{1}{4}$ the following representation at $x=0$
\begin{align*}
\sqrt{1-4x}&=\sum_{n=0}^\infty \binom{\frac{1}{2}}{n}(-4x)^n\\
&=1-2x-2x^2-4x^3-10x^4-\cdots
\end{align*}
so that
\begin{align*}
\frac{1+\sqrt{1-4x}}{2x}=\color{blue}{\frac{1}{x}}-1-x-2x^2-5x^3-14x^4-\cdots\tag{1}
\end{align*}
whereas
\begin{align*}
\frac{1-\sqrt{1-4x}}{2x}=1+x+2x^2+5x^3+14x^4+\cdots\tag{2}
\end{align*}
Note the latter (2) is a power series, while the former (1) is not a power series.
|
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|
Find integer triples $(x,y,z)$ such that $2018^x=y^2+z^2+1$ To find the triples, I had some tries.
First , consider $x\ge 2$, then $2018^x=0\pmod 4$. But for the right side $y^2+z^2+1$. Consider 3 situations.
*
*case 1. Both $y$ and $z$ are even. Then $y^2+z^2+1=1\pmod 4$.
*case 2. One of them are even, the other is odd. Then $y^2+z^2+1=2\pmod 4$.
*case 3. Both $y$ and $z$ are odd. Then $y^2+z^2+1=3\pmod 4$.
Thus for $x\ge 2$, there is no solution. Then $x=0$ or $x=1$.
Now for $x=0$, $2018^0=y^2+z^2+1$ ,then $y^2+z^2=0$, hence we have $y=0$ and $z=0$.
For $x=1$, $2018^1=y^2+z^2+1$, then my question is how to solve $y^2+z^2=2017$.
Also since I didn't learn Number Theory systematically, I don't know if I was proceeding in the right way for this question.
Lastly, Are there more general methods for problems like this?
|
You have solved almost all cases. So let me add the solution for $x=1$, and exclude negative $x$.
Since $2017$ is a prime which is congruent $1$ modulo $4$, it is representable as the sum of two squares. The solutions are, up to sign and order, given by
$$
2017=44^2+9^2,
$$
see here:
$2017$ as the sum of two squares
|
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|
Proof Verification: Using Taylor series to arrive at a certain inequality If $x>0$ show that $|(1+x)^{1/3} - (1 + \frac{1}{3}x - \frac{1}{9}x^2)| \le \frac{5}{81}x^3$.
Proof:
Let $f(x) = (1 +x)^{1/3}$
Then, estimating $f(x)$ at the point $x_o = 0$, we have that:
$f'(x) = 1/3(1+x)^{-2/3}, f"(x) = -2/9(1+x)^{-5/3}, f^{(3)}(x)= 10/27 (1+x)^{-8/3}$
Then,
$P_{3} = 1 + \frac{1}{3}x - \frac{1}{9}x^2 + \frac{5}{81}x^3$
And we have that:
$R_3 = \frac{-10}{243} (1+c)^{-11/3}x^4 <0$ for $c \in (0,x)$
Then this implies that:
$(1+x)^{1/3} < 1+ \frac{1}{3}x -\frac{1}{9}x^2 +\frac{5}{81}x^3$
$\implies (1+x)^{1/3}- (1+ \frac{1}{3}x -\frac{1}{9}x^2 ) <\frac{5}{81}x^3$
I don't understand how to transform the above expression into the one that we're required to show. Can anyone explain why it should be $\le$ instead of $<$. Also, how and why do we need the absolute signs?
Please help. Thank you so much.
|
You've got $(1+x)^{1/3}- (1+ \frac{1}{3}x -\frac{1}{9}x^2 ) <\frac{5}{81}x^3$, but the value on the left side could be a very big negative number, so you actually have to show that the absolute value is smaller than $\frac{5}{81} x^3$.
Working with $P_3, R_3$ is "too much". I suggest to get $P_2,R_2$. Doing that you will get $\leq$ instead of $<$. Feel free to ask me if you don't understand.
|
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|
If $x^2-bx+c=0$ has real roots, then prove that both are greater than $1$ when $c+1>b>2$.
If $ x^2-bx+c=0$ has real roots, prove that both roots are greater than $1$, when $c+1>b>2$.
Working
I tried to prove the given inequality by taking roots greater than $1$.
Let $\alpha$, $\beta$ be the roots of the quadratic equation. So $$\alpha+\beta =b$$
$$\alpha\cdot\beta =c$$
Since $\alpha>1$ and $\beta>1$, it can be deduced that,
$$\alpha+\beta >2\implies b>2$$ $$\alpha\cdot\beta >1\implies c>1\implies c+1>2$$
to combine these two inequalities I need another link between $c$ and $b$. How to proceed? Thanks.
|
Given $x^2-bx+c=0$ we need
*
*$\Delta =b^2-4c\ge0$
and
*
*$x_1=\frac{b-\sqrt{b^2-4c}}{2}>1 \iff b-2> \sqrt{b^2-4c} \stackrel{b>2}\iff b^2-4b+4\ge b^2-4c \iff 4c+4\ge4b \iff c+1\ge b$
|
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|
Integration of $ \int_1^2 \frac{3x^3 + 3x^2 - 5x + 4}{3x^3 - 2x^2 +3x -2} \mathrm dx$ Evaluate $$\int_1^2 \frac{3x^3 + 3x^2 - 5x + 4}{3x^3 - 2x^2 +3x -2}\mathrm dx$$I believe I should use the partial fractions technique to evaluate this integral, but I am not getting anywhere when I try it.
|
Apologies if it's unclear - I'm not sure how to get the polynomial division in Mathjax, if someone could advise me on how to do this it would be appreciated.
From here I'll use Mathjax however:
$$5x^2-8x+6=A(x^2+1)+B(3x-2)$$
Let $A=px+q, B=rx+s$. If $p>0$, this equation transforms into a cubic so $p=0$.
Hence $qx^2+q+(rx+s)(3x-2)=5x^2-8x+6$
Expanding and collecting like terms we get:
$$q+3r=5$$
$$3s-2r=-8$$
$$q-2s=6$$
I instantly checked $q=2$ and $r=1$ (seemed like an obvious first choice given the first statement) and it works with $s=-2$.
Hence: $$\frac{5x^2-8x+6}{(x^2+1)(3x-2)}=1+\frac{2}{3x-2}+\frac{x-2}{x^2+1}$$
Integrate these seperately and sum the results:
$$\int_1^2{1 dx}=\bigg[x\bigg]^2_1=1$$
$$\int_1^2{\frac{2}{3x-2}dx}=2\int_1^2{\frac{1}{3x-2}dx}=2\bigg[\frac 13\ln|3x-2|\bigg]^2_1=\frac 23\ln(4)=\frac 43\ln(2)$$
$$\int_1^2{\frac{x-2}{x^2+1}dx}=\int_1^2{\frac{x}{x^2+1}dx}-\int_1^2{\frac{2}{x^2+1}dx}$$
Note the first of those is of the form $\int{\frac{f'(x)}{2f(x)}dx}$, this integral therefore is $\bigg[\frac 12\ln|x^2+1|\bigg]^2_1=\frac 12(\ln5-\ln2)$
The second half requires $u$-substitution. Let $x=\tan(u)$, then $dx=\sec^2(u) du$. Our integral becomes:
$$2\int_{\tan^{-1}(1)}^{\tan^{-1}(2)}{\frac{1}{sec^2(u)}\cdot \sec^2(u) du}=2\bigg[1\bigg]^{\tan^{-1}(2)}_{\tan^{-1}(1)} \because 1+\tan^2(u)=\sec^2(u)$$
$$=2\tan^{-1}(2)-\frac \pi2\approx 0.6435 \text{(this is also $\tan^{-1}\bigg(\frac 34\bigg)$)}$$
Summing all we get:$$1+\frac 56 \ln(2)+ \frac 12 \ln(5)-2\tan^{-1}(2)+\frac \pi2$$
|
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|
A trigonometric integral (guessed from a combinatorics formula) In class, I defined the binomial coefficient using an integral:
$$\binom{n}{k} = \displaystyle \int_0^{2\pi}\dfrac{dt}{2\pi} e^{-ikt}(1+e^{it})^n.$$
I succeeded in demonstrating many standard properties of the binomial coefficient directly using integration: Pascal's identity, Vandermonde identity, Hockey stick. But I could not show that $$\sum_{k=0}^n \binom{n}{k}=2^n.$$
It turns out I have to show the following:
$$\int_{0}^{2\pi}\dfrac{dt}{2\pi}\dfrac{\sin\left(\dfrac{(n+1)t}{2}\right)}{\sin\dfrac{t}2}\cos^n\dfrac{t}{2}=1$$
I do not know how to perform this integration! I need help. It is better if the solution did not involve contour integration.
|
There is a mistake in my presentation that eludes me, but the essence is the following.
Using
$$\sum_{k=1}^{\infty} p^{k} \, \sin(k \, x) = \frac{p \, \sin(x)}{1 - 2 \, p \, \cos(x) + p^{2}}$$
then let $S_{n}$ be the desired summation to be evaluated to obtain:
\begin{align}
S_{n} &= \sum_{k=0}^{n} \binom{n}{k} \\
&= \frac{1}{2 \pi} \, \int_{0}^{2\pi} \left(2 \, \cos\left(\frac{t}{2}\right)\right)^{n} \, \frac{\sin\left(\frac{(n+1) \, t}{2} \right)}{\sin\left(\frac{t}{2}\right)} \, dt \\
&= \frac{1}{2 \pi} \, \int_{0}^{2\pi} \left(2 \, \cos\left(\frac{t}{2}\right)\right)^{n+1} \, \frac{\sin\left(\frac{(n+1) \, t}{2} \right)}{\sin(t)} \, dt
\end{align}
Now,
\begin{align}
\sum_{n=0}^{\infty} S_{n} &= \frac{1}{2 \pi} \, \int_{0}^{2\pi} \left[ \sum_{n=0}^{\infty} \left(2 \, \cos\left(\frac{t}{2}\right)\right)^{n+1} \, \sin\left(\frac{(n+1) \, t}{2} \right)\right] \, \frac{dt}{\sin(t)} \\
&= \frac{1}{2 \pi} \, \int_{0}^{2\pi} \left[ \sum_{n=1}^{\infty} \left(2 \, \cos\left(\frac{t}{2}\right)\right)^{n} \, \sin\left(\frac{n \, t}{2} \right)\right] \, \frac{dt}{\sin(t)} \\
&= \frac{1}{2 \pi} \, \int_{0}^{2 \pi} \frac{2 \cos\left(\frac{t}{2}\right) \sin\left(\frac{t}{2}\right)}{1 - 4 \cos^{2}\left(\frac{t}{2}\right) + 4 \cos^{2}\left(\frac{t}{2}\right)} \, \frac{dt}{\sin(t)} \\
&= \frac{1}{2\pi} \, \int_{0}^{2 \pi} dt = 1 \\
&= - (-1) = - \frac{1}{1 - 2} = - \sum_{n=0}^{\infty} 2^{n}.
\end{align}
This yields
$$\sum_{n=0}^{\infty} \binom{n}{k} = - 2^{n}.$$
As stated, the mistake that eludes me is the sign error.
|
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"url": "https://math.stackexchange.com/questions/2789309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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|
solving differential equation $\frac{d^3y}{dx^3}+\frac{d^2y}{dx^2}=x^2+2x+2$ How would you solve this third order differential equation:
$$\frac{d^3y}{dx^3}+\frac{d^2y}{dx^2}=x^2+2x+2$$
My first thought was to take a double integral:
$$\iint\frac{d^3y}{dx^3}+\frac{d^2y}{dx^2}dxdx=\iint{x^2+2x+2}dxdx$$
so:
$$y+\frac{dy}{dx}=\frac{x^4}{12}+\frac{x^3}{3}+x^2+c_1x+c_2$$
This is what I got to but I am unsure beyond here
|
Well, there are many ways you can you solve this DE. However, the simplest would be using the characteristic equation and since its a non-homogeneous DE, we need to find a particular solution as well. Finally we need to use the superposition principle to add the solution.
Now, in symbols and numbers
$$
r^3 +r^2 = 0 \\
r = 0\; \text{or}\; r = -1
$$
Note that for $r =0$ it is a double root because every n-polynomial has n roots. Therefore,
$$
y_c = c_1 + c_2x + c_3e^{-x}
$$
now plugging $y_p = Ax^4 + Bx^3 + Cx^2$ in such a way that it satisfies the DE we get
$$
24Ax + B + 12Ax^2 + 6Bx +2C = x^2+2x+2 \\
12Ax^2+ (24A + 6B)x +(B+2C) = x^2 +2x+2
$$
hence $A=\frac{1}{12}$, $B = 0$ and $C=1$ therefore $y_p = \frac{1}{12}x^4 + x^2$ and our general solution would be the sum, so
$$
y = c_1+ +c_2x+ c_3e^{-x} +\frac{1}{12}x^4 +x^2
$$
Hope this helps !
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2791300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Eccentricity of ellipse in terms of eccentric angles of extremities of focal chord?
If $A$ and $B$ are eccentric angles of the extremities of a focal chord of an ellipse then eccentricity of the ellipse is?
The answer given is $\dfrac{\sin A + \sin B}{\sin (A+B)}$.
Now I tried to do this by assuming extremities to be $P(a\cos A, b\sin A)$ and $Q (a\cos B, b\sin B)$ and focus to be $F(ae , 0)$. By equating slopes of $PF$ and $FQ$, I got $e=\dfrac{\sin (A+B)}{\sin A -\sin B}$. I can't find any way to get the required answer from this. Was my approach right? What should I do?
|
The equation of chord joining pairs of point $[A]$ and $[B]$ is:
$$\frac{x}{a} \cos \frac{A+B}{2}+\frac{y}{b} \sin \frac{A+B}{2}
=\cos \frac{A-B}{2}$$
Put $(x,y)=(\pm ae,0)$, then
$$e=\pm \frac{\cos \frac{A-B}{2}}{\cos \frac{A+B}{2}}$$
Also,
\begin{align}
\frac{\sin A+\sin B}{\sin (A+B)} &=
\frac{2\sin \frac{A+B}{2} \cos \frac{A-B}{2}}
{2\sin \frac{A+B}{2} \cos \frac{A+B}{2}} \\
&= \frac{\cos \frac{A-B}{2}}{\cos \frac{A+B}{2}} \\
\frac{\sin (A-B)}{\sin A-\sin B} &=
\frac{2\sin \frac{A-B}{2} \cos \frac{A-B}{2}}
{2\sin \frac{A-B}{2} \cos \frac{A+B}{2}} \\
&= \frac{\cos \frac{A-B}{2}}{\cos \frac{A+B}{2}} \\
\end{align}
In conclusion:
$$\fbox{
$e
=\pm \frac{\cos \frac{A-B}{2}}{\cos \frac{A+B}{2}}
=\pm \frac{\sin A+\sin B}{\sin (A+B)}
=\pm \frac{\sin (A-B)}{\sin A-\sin B}
=\pm \frac{1+\tan \frac{A}{2} \tan \frac{B}{2}}
{1-\tan \frac{A}{2} \tan \frac{B}{2}} \,
$}$$
See also another forum here.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2792481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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|
Finding the area between 4 curves by changing my basis to create a square and integrating The question number $3$ at hand: https://i.imgur.com/DL11Izh.jpg
My work on it: https://i.imgur.com/AzaQKHS.jpg?1
I thought I was doing everything right but it seems to be wrong, the final answer should be $$\frac{6}{35}(b^{\frac{7}{6}}-a^{\frac{7}{6}})(d^{\frac{5}{6}}-c^{\frac{5}{6}})$$
|
Let $(x,y)=(u^3 v^3,u^2 v^4)$
\begin{array}{ccccc}
ax^2<y^3<bx^2 & \implies &
au^6v^6<u^6v^{12}<bu^6v^6 & \implies &
a<v^6<b \\
cy^3<x^4<dy^3 & \implies &
cu^6v^{12}<u^{12}v^{12}<du^6v^{12} & \implies & c<u^6<d
\end{array}
\begin{align*}
\iint_A dx\, dy &=
\int_{\sqrt[6]{a}}^{\sqrt[6]{b}}
\int_{\sqrt[6]{c}}^{\sqrt[6]{d}}
\begin{vmatrix}
x_u & x_v \\
y_u & y_v
\end{vmatrix} du \, dv \\
&=
\int_{\sqrt[6]{a}}^{\sqrt[6]{b}}
\int_{\sqrt[6]{c}}^{\sqrt[6]{d}}
\begin{vmatrix}
3u^2v^3 & 3u^3v^2 \\
2uv^4 & 4u^2v^3
\end{vmatrix} du \, dv \\
&=
\int_{\sqrt[6]{a}}^{\sqrt[6]{b}}
\int_{\sqrt[6]{c}}^{\sqrt[6]{d}}
6u^4v^6 \, du \, dv \\
&= \frac{6}{35}(b^{7/6}-a^{7/6})(d^{5/6}-c^{5/6})
\end{align*}
See also another answer of mine here.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2793197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Evaluation of $\int\frac{1}{(\sin x+a\sec x)^2}dx$
Evaluation of $$\int\frac{1}{(\sin x+a\sec x)^2}dx$$
Try: Let $$I=\int\frac{1}{(\sin x+a\sec x)^2}dx=\int\frac{\sec^2 x}{(\tan x+a\sec^2 x)^2}dx$$
put $\tan x=t$ and $dx=\sec^2 tdt$
So $$I=\int\frac{1}{(a+at^2+t)^2}dt$$
Could some help me how to solve above Integral , thanks in advance.
|
You can write
$$
\frac{1}{(a+t+at^2)^2}=-\frac{1-2a^2+2at+2a^2t^2}{(-1+4a^2)(a+t+at^2)^2}+\frac{2a}{(-1+4a^2)(a+t+at^2)}.
$$
Also is
$$
-\int\frac{1-2a^2+2at+2a^2t^2}{(-1+4a^2)(a+t+at^2)^2}dt =\frac{1+2at}{(-1+4a^2)(a+t+at^2)}
$$
and
$$
\int \frac{2a}{(-1+4a^2)(a+t+at^2)}dt=\frac{4a}{(-1+4a^2)^{3/2}}\arctan\left(\frac{1+2at}{\sqrt{-1+4a^2}}\right).
$$
Hence
$$
\int \frac{1}{(a+t+a t^2)^2}dt=
$$
$$
=\frac{1+2at}{(-1+4a^2)(a+t+a t^2)}+\frac{4a}{(-1+4 a^2)^{3/2}}\arctan\left(\frac{1+2at}{\sqrt{-1+4a^2}}\right).
$$
Where we have used
$$
\int\frac{1}{1+t^2}dt=\arctan(t)+c
$$
In general holds
$$
\int\frac{dt}{(t^2+at+b)^2}=\frac{2t+a}{(-a^2+4b)(t^2+at+b)}+\frac{4\arctan\left(\frac{2t+a}{\sqrt{-a^2+4b}}\right)}{(-a^2+4b)^{3/2}}.
$$
QED
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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|
Show that $ \sqrt{2 + \sqrt{2 + \sqrt{2 + 2 \cos 8\theta}}} = 2 \cos \theta$ Show that:
$$ \sqrt{2 + \sqrt{2 + \sqrt{2 + 2 \cos 8\theta}}} = 2 \cos \theta$$
My try:
As we can see that the LHS is in form of $\cos 8 \theta$ which must be converted into $\cos \theta $ in order to solve the equation.
There are several questions similar on this site but those aren't helping me much regarding this question.
So, I tried converting $\cos 8 \theta$ into $\cos \theta$ and then putting the value in the equation under square root and then further solving it into RHS.
So I got $\cos 8 \theta$ as:
$$ 2\cdot \{ 2\cdot [ 4 \cos ^4 \theta - 4.cos^2 \theta]^2 \}$$
So, I don't know how to solve it further.
Please help, putting the value of $\cos 8 \theta$ in place of that doesn't helps me much.
Thanks in Advance
|
$$\sqrt{2+\sqrt{2+\sqrt{2+2cos8\theta}}}$$
Firstly, $2+2cos8\theta=2+2(2cos^24\theta-1)=4cos^24\theta$
$$\therefore\sqrt{2+\sqrt{2+\sqrt{2+2cos8\theta}}}=\sqrt{2+\sqrt{2+2cos4\theta}}$$
Then, $2+2cos4\theta=2+2(2cos^22\theta-1)=4cos^22\theta$
$$\therefore\sqrt{2+\sqrt{2+2cos4\theta}}=\sqrt{2+2cos2\theta}$$
Finally, $2+2cos2\theta=2+2(2cos^2\theta-1)=4cos^2\theta$
$$\therefore\sqrt{2+\sqrt{2+2cos2\theta}}=\sqrt{4cos^2\theta}=2cos\theta$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2795571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
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|
What is the closed form of :$\int\frac{x+2}{\left(x^2+x+2\right)\left(\sqrt{x^2+2x+3}\right)}\,dx$ over $\mathbb{R}$? I have took many times to get antiderivative of the below integral A in $\mathbb{R}$ , but i didn't succeed however it is a constant from $-\infty \to +\infty$ and it's equal :$1.4389$ , but i'm sure that it has a closed form where inverse symbole calculator didn't give me anything , then Is there any way help me to get it's antiderivative over $\mathbb{R}$ ?
$$A=\int_{\mathbb{R}}\frac{x+2}{\left(x^2+x+2\right)\left(\sqrt{x^2+2x+3}\right)}\,dx$$
|
Using Mathematica with Rubi 4.15.1 we can find very simple form antiderivative:
$$\int \frac{x+2}{\left(x^2+x+2\right) \sqrt{x^2+2 x+3}} \, dx=\\-\sqrt{\frac{1}{14} \left(11+8 \sqrt{2}\right)} \tan ^{-1}\left(\frac{1+2 \sqrt{2}-\left(5+3
\sqrt{2}\right) x}{\sqrt{7 \left(11+8 \sqrt{2}\right)} \sqrt{3+2 x+x^2}}\right)+\sqrt{\frac{1}{14} \left(-11+8 \sqrt{2}\right)} \tanh ^{-1}\left(\frac{1-2
\sqrt{2}-\left(5-3 \sqrt{2}\right) x}{\sqrt{7 \left(-11+8 \sqrt{2}\right)} \sqrt{3+2 x+x^2}}\right)$$
MMA code:
HoldForm[Integrate[(x + 2)/((x^2 + x + 2)*Sqrt[x^2 + 2 x + 3]),
x] == -Sqrt[1/14 (11 + 8 Sqrt[2])] ArcTan[(
1 + 2 Sqrt[2] - (5 + 3 Sqrt[2]) x)/(
Sqrt[7 (11 + 8 Sqrt[2])] Sqrt[3 + 2 x + x^2])] +
Sqrt[1/14 (-11 + 8 Sqrt[2])]
ArcTanh[(1 - 2 Sqrt[2] - (5 - 3 Sqrt[2]) x)/(
Sqrt[7 (-11 + 8 Sqrt[2])] Sqrt[3 + 2 x + x^2])]] // TeXForm
and define integral:
$$\color{blue}{\int_{-\infty }^{\infty } \frac{x+2}{\left(x^2+x+2\right) \sqrt{x^2+2 x+3}} \, dx=\sqrt{\frac{2}{7} \left(11+8 \sqrt{2}\right)} \tan ^{-1}\left(\sqrt{\frac{1}{7}
\left(1+2 \sqrt{2}\right)}\right)-\sqrt{\frac{1}{14} \left(-11+8 \sqrt{2}\right)} \ln \left(1+\frac{1}{\sqrt{2}}+\sqrt{\frac{1}{2}+\sqrt{2}}\right)}$$
MMA code:
HoldForm[Integrate[(x + 2)/((x^2 + x + 2)*Sqrt[x^2 + 2 x + 3]), {x, -Infinity, Infinity}] ==
Sqrt[2/7 (11 + 8 Sqrt[2])] ArcTan[Sqrt[1/7 (1 + 2 Sqrt[2])]] -
Sqrt[1/14 (-11 + 8 Sqrt[2])]Log[1 + 1/Sqrt[2] + Sqrt[1/2 + Sqrt[2]]]] // TeXForm
|
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"url": "https://math.stackexchange.com/questions/2796429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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|
Integral $\int_0^2 \frac{\arctan x}{x^2+2x+2}dx$ I am tring to evaluate
$$I=\int_0^2 \frac{\arctan x}{x^2+2x+2}dx$$
The first thing I did was to notice that
$$\frac{1}{x^2+2x+2}=\frac{1}{(x+1)^2+1}=\frac{d}{dx}\arctan(x+1)$$
So I integrated by parts in order to get
$$I=\arctan 2\arctan 3-\int_0^2\frac{\arctan(x+1)}{1+x^2}dx$$
I let $x=u+1$ but when I do that I get
$$I=\arctan 2\arctan 3+\int_{-1}^1\frac{\arctan(u)}{1+(1+u)^2}du
=\arctan 2\arctan 3$$
Now this is not close to the approximation given by wolfram. What have I done wrong and how to solve this?
|
An elementary solution. Let $I$ denote the integral. Apply the substitution $x=\frac{2t}{t+\sqrt{5}}$ to obtain
$$ I = \int_{0}^{\infty} \frac{\arctan\left(\frac{2t}{t+\sqrt{5}}\right)}{\sqrt{5}+4t+\sqrt{5}t^2} \, dt. \tag{1} $$
Substituting $t \mapsto 1/t$, we find that
$$ I = \int_{0}^{\infty} \frac{\arctan\left(\frac{2}{1+\sqrt{5} t}\right)}{\sqrt{5}+4t+\sqrt{5}t^2} \, dt. \tag{2} $$
But it is easy to check that
$$ \arctan\left(\frac{2t}{t+\sqrt{5}}\right) + \arctan\left(\frac{2}{1+\sqrt{5} t}\right) = \arctan(2) $$
holds, either by utilizing the addition formula for arctan or by differentiating the LHS to check that the LHS is constant and then plugging $t=0$ to determine the value of the constant.
Therefore by averaging $(1)$ and $(2)$, we obtain
$$ I
= \frac{\arctan(2)}{2} \int_{0}^{\infty} \frac{dt}{\sqrt{5}+4t+\sqrt{5}t^2}
= \frac{\arctan(2)\arctan(1/2)}{2}. $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2798550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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|
Integration and hyperbolic function problem According to my cosmology book:$$a\dot{a}^{2}=\frac{\varLambda}{3}a^{3}+K,$$where $a$ is the scale factor, the dot indicates derivative wrt time and $\varLambda$ and $K$ are constants. The author then says “Introducing a new variable $x$ by $a^{3}=x^{2}$ and integrating once more with the initial condition $a\left(0\right)=0$ we obtain”$$a^{3}=\frac{3K}{\varLambda}\sinh^{2}\left(\frac{t}{t_{\varLambda}}\right),$$where $t_{\varLambda}=\frac{2}{\sqrt{3\varLambda}}$. I've tried (with difficulty) integrating the first equation using WolframAlpha but end up with nothing like the second equation. Any suggestions or advice, please?
|
$$a\dot{a}^{2}=\frac{\varLambda}{3}a^{3}+K,$$
$$x^2=a^3 \implies 3a^2 \dot a =2\dot x x \implies 9a^4 \dot a ^2=4x^2\dot x ^2 $$
$$\implies 9a^3(a\dot a^2)=4x^2\dot x^2 \implies (a\dot a ^2)=\frac {4\dot x^2}{9}$$
The equation becomes
$$a\dot{a}^{2}=\frac{\varLambda}{3}a^{3}+K,$$
$$\frac {4\dot x^2}{9}=\frac{\varLambda}{3}x^2+K$$
$$2\dot x=\pm\sqrt {3\varLambda x^2+9K}$$
I integrate for the positive equation...
$$ \int \frac {dx}{\sqrt { x^2+3K/\varLambda }}= \frac{\sqrt {3\varLambda }}{2}\int dt$$
Use a $\sinh(..)$ substitution for the integral and square the result to get $a^3$
$$s^2=3K/\varLambda $$
Substitute $x=s\sinh(m) \implies dx=s\cosh(m)$...for the positive equation we get
$$ \int \frac {dx}{\sqrt { x^2+s^2 }}= \frac{\sqrt {3\varLambda }}{2}\int dt$$
$$ m= \frac{\sqrt {3\varLambda }}{2}t+C$$
$$ arcsinh(x/s)= \frac{\sqrt {3\varLambda }}{2}t+C$$
$$ x= s\sinh(\frac{\sqrt {3\varLambda }}{2}t+C)$$
for the initial condition given $C=0$
$$\boxed{ a^3(t)= \frac {3K}{\varLambda}\sinh^2(\frac{\sqrt {3\varLambda }}{2}t)}$$
Edit for the integral substitute $x=s \sinh(m)$
$$I=\int \frac {dx}{\sqrt { x^2+s^2 }}= \int \frac {s\cosh(m)dm}{\sqrt { s^2\sinh^2(m)+s^2 }}$$
$$I= \int \frac {s\cosh(m)dm}{\sqrt { s^2(\sinh^2(m)+1) }}$$
since you have the equality for hyperbolic functions $\cosh^2(m)-\sinh^2(m)=1$
$$I= \int \frac {s\cosh(m)dm}{\sqrt { s^2\cosh^2(m) }}$$
$$I= \int dm=m+K$$
$$I=arcsinh(x/s)+K$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2799610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Evaluate indefinite integral $\int \tan(\frac{x}{3}) \, dx$ I need help verifying why I am getting an incorrect answer for the question evaluate the integral
$$\int \tan\left(\frac{x}{3}\right) \, dx$$
I simplify the above equation using trig identities to get
$$\int \frac {\sin \left(\frac{x}{3}\right)}{\cos\left(\frac{x}{3}\right)} \, dx$$
I use the substitution method to find
$$ du = -\frac{1}{3} \sin(x/3) \, dx$$ and so $dx = \frac{-3\,du}{\sin\frac{x}{3}}$
I plug the $u$ back into equation
$$ \int \frac {\sin\left(\frac{x}{3}\right)}{u} \cdot\frac {-3\,du}{\sin \left(\frac{x}{3}\right)}$$
I cross out the $\sin \left(\frac{x}{3}\right)$ and (this is where I may be going wrong), I pull out the $-3$ to be in front of the integral sign since it is a constant and solve for
$$-3 \int \frac{1}{u} \, du$$ and get the final answer $$ -3 \biggl|\,\ln \, \cos \frac{x}{3}\biggr| + C $$
But the answer in the back of the book is $ -\frac{1}{3} |\ln \, \cos \frac{x}{3}| + C $
|
Just to make your life a bit simpler without this fraction $\frac{x}{3}$, just use u-sub.
Put: $\frac{x}{3}=u$
Then $\int tan(\frac{x}{3})dx= 3\int tan\ u\ du= 3 $ln |sec u| +C$=$3 ln|sec($\frac{x}{3})|+C$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2804424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
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|
Verify that $\int_{0}^{\pi/2}\tan x\ln(\tan x)\ln(\cos x)\ln^3(\sin x)\mathrm dx=-\frac{3\zeta(6)}{2^8}$ $$\int_{0}^{\pi/2}\tan x\ln(\tan x)\ln(\cos x)\ln^3(\sin x)\mathrm dx=-\frac{3\zeta(6)}{2^8}$$
Any hints how to simplify this integral to more manageable to be solve.
|
Considering $$
\begin{aligned}
J(a, b, c) &=\int_0^{\frac{\pi}{2}} \tan ^a x \cos ^b x \sin ^c x d x \\
&=\int_0^{\frac{\pi}{2}} \sin ^{a+c} x \cos ^{b-a} x d x \\
&=\frac{1}{2} B\left(\frac{a+c+1}{2}, \frac{b-a+1}{2}\right),
\end{aligned}
$$
then $$\boxed{I=\left.\frac{1}{2} \frac{\partial^3}{\partial a \partial b \partial c^3} B\left(\frac{a+c+1}{2}, \frac{b-a+1}{2}\right)\right|_{a=1, b=0=c}}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2805317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
How can I find the sum of $\sum_{n=2}^\infty \frac{(-1)^n}{n^2 - n}$ How can I find the sum of the series:
$$\sum_{n=2}^\infty \frac{(-1)^n}{n^2 - n}$$
I did this:
$$\sum_{n=2}^\infty \left(\frac{(-1)^n}{n-1} + \frac{(-1)^{n-1}}{n}\right)$$
But I am not sure how to proceed.
|
Let us consider a sequence
$$a_n = \sum_{k=n+1}^{2n} \frac{1}{k}$$
Now, let's attempt to get a recursive formula for this.
$$a_n - a_{n-1} = \sum_{k=n+1}^{2n} \frac{1}{k} - \sum_{k=n}^{2n-2} \frac{1}{k} = -\frac{1}{n} + \frac{1}{2n-1} + \frac{1}{2n} = \frac{1}{(2n-1)(2n)}$$
$$a_n = a_{n-1} + \frac{1}{(2n-1)(2n)}$$
Since $a_1 = \frac{1}{2} = \frac{1}{(2\cdot1 - 1)(2\cdot1)}$,
$$a_n = \sum_{k=1}^n \frac{1}{(2k-1)(2k)}$$
Now if we also consider the sequence alternating harmonic series
$$S_n = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{(-1)^{n-1}}{n}$$
We can see that
$$S_{2n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{2n - 1} - \frac{1}{2n}$$
$$S_{2n} = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \cdots + \left(\frac{1}{2n - 1} - \frac{1}{2n}\right)$$
$$S_{2n} = \sum_{k=1}^n \left(\frac{1}{2k-1} - \frac{1}{2k}\right) = \sum_{k=1}^n \frac{1}{(2k-1)(2k)}$$
This means that $S_{2n} = a_n$. If we now look at the expansion of $a_n$
$$a_n = \frac{1}{n+1} + \frac{1}{n+2} + \frac{1}{n+3} + \cdots + \frac{1}{2n}$$
If we factor out $\frac{1}{n}$, we get
$$a_n = \frac{1}{n} \left(\frac{1}{1+\frac{1}{n}} + \frac{1}{1+\frac{2}{n}} + \frac{1}{1+\frac{3}{n}} + \cdots + \frac{1}{1+\frac{n}{n}}\right)$$
As $n$ tends towards infinity, this is the Riemann sum for $\int_0^1 \frac{1}{1+x}~{\rm d}x$.
$$\int_0^1 \frac{1}{1+x}~{\rm d}x = \int_1^2 \frac{1}{x}~{\rm d}x = \ln 2$$
You can then use that to solve your sum.
Alternative: Consider successive derivatives of $\ln(1 + x)$.
$$\frac{{\rm d}}{{\rm d}x} \left[\ln(1 + x)\right] = \frac{1}{1 + x} = \left(1 + x\right)^{-1}$$
$$\frac{{\rm d}^2}{{\rm d}x^2} \left[\ln(1 + x)\right] = -\left(1 + x\right)^{-2}$$
$$\frac{{\rm d}^3}{{\rm d}x^3} \left[\ln(1 + x)\right] = 2\left(1 + x\right)^{-3}$$
$$\frac{{\rm d}^4}{{\rm d}x^4} \left[\ln(1 + x)\right] = -6\left(1 + x\right)^{-4}$$
$$\vdots$$
$$\frac{{\rm d}^n}{{\rm d}x^n} \left[\ln(1 + x)\right] = (-1)^{n-1} \cdot (n-1)! \cdot \left(1 + x\right)^{-n}$$
Using this to create a Taylor series about $x=0$ gives us
$$\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots + \frac{(-1)^{n-1} \cdot x^n}{n} + \cdots = \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \cdot x^n}{n}$$
Plugging in $x=1$ to give the alternating harmonic series gives us
$$\ln 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{(-1)^{n-1}}{n} + \cdots = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}$$
Now knowing that the alternating harmonic series is equal to $\ln 2$, we can solve the sum.
$$\sum_{n=2}^\infty \left(\frac{(-1)^n}{n-1} + \frac{(-1)^{n-1}}{n}\right)$$
$$= \sum_{n=2}^\infty \frac{(-1)^n}{n-1} + \sum_{n=2}^\infty \frac{(-1)^{n-1}}{n}$$
$$= \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n} + \left(-1 + \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}\right)$$
$$= \ln (2) + \ln (2) - 1 = \ln (4) - 1$$
|
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|
Using the limit definition to find a derivative for $\frac{-5x}{2+\sqrt{x+3}}$ I am trying to find the derivative of $\frac{-5x}{2+\sqrt{x+3}}$ using the limit definition of a derivative.
$$\lim_{h \to 0} \frac {f(x+h)-f(x)}{h}$$
What I did is
$$\lim_{h \to 0} \frac{\dfrac{-5(x+h)}{2+\sqrt{x+h+3}}\dfrac{2-\sqrt{x+h+3}}{2-\sqrt{x+h+3}}-\dfrac{-5x}{2+\sqrt{x+3}}\dfrac{2-\sqrt{x+3}}{2-\sqrt{x+3}}}{h}$$
But I am having problem finding the correct answer.
|
You better not rush to multiply by conjugates:
$$\lim_{h \to 0} \frac{\dfrac{-5(x+h)}{2+\sqrt{x+h+3}}-\dfrac{-5x}{2+\sqrt{x+3}}}{h}=\\
\lim_{h \to 0} \frac{\left(\dfrac{-5x}{2+\sqrt{x+h+3}}-\dfrac{-5x}{2+\sqrt{x+3}}\right)+\dfrac{-5h}{2+\sqrt{x+h+3}}}{h}=\\
-5x\cdot \lim_{h \to 0} \frac{\dfrac{1}{2+\sqrt{x+h+3}}-\dfrac{1}{2+\sqrt{x+3}}}{h}-\lim_{h \to 0}\dfrac{5}{2+\sqrt{x+h+3}}=\\
-5x\cdot \lim_{h \to 0} \frac{\sqrt{x+3}-\sqrt{x+h+3}}{h(2+\sqrt{x+h+3})(2+\sqrt{x+3})}-\dfrac{5}{2+\sqrt{x+3}}=\\
-5x\cdot \lim_{h \to 0} \frac{-h}{h(2+\sqrt{x+h+3})(2+\sqrt{x+3})(\sqrt{x+3}+\sqrt{x+h+3})}-\dfrac{5}{2+\sqrt{x+3}}=\\
\frac{15x}{(2+\sqrt{x+3})^2\cdot 2\sqrt{x+3}}-\dfrac{5}{2+\sqrt{x+3}}=\\
\frac{5x+20\sqrt{x+3}+30}{(2+\sqrt{x+3})^2\cdot 2\sqrt{x+3}}.$$
|
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|
$1^n-3^n-6^n+8^n$ is divisible by $10$ Prove that $1^n-3^n-6^n+8^n$ is divisible by $10$ for all $n\in\mathbb{N}$
It is divisible by $2$ and $5$ if we rearrange it will it be enough
$(1^n -3^n)$ and $(6^n -8^n)$ is divisible by $2$.
And
$(1^n-6^n)$ and $(8^n-3^n)$ is divisible by $5$.
Hence $\gcd(2,5)$ is $1$ and it is divisible by $2\cdot5=10$.
Is it correct?
|
Your proof is OK, however, for completeness, you must further show why the binomials are divisible by $2$ and $5$.
If you are familiar with modular arithmetic, it is:
$$1^n-3^n-6^n+8^n\equiv 1-3-6+8\equiv 0 \pmod {10}.$$
|
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|
What is the probability of a symmetric coin fall
The prompt: We have 10 coins and 1 of them is non-symmetric (with probability of head equal $\frac{1}{3}$). We toss a randomly selected coin 6 times, and obtain 3 tails. What is the probability that we tossed a symmetrical coin?
The way I went on about solving the problem was first assuming it to be a win or success if we obtain a Tails, assuming that
NS = Non symmetric coin, S = Symmetric coin, H = heads, T = tails
$$P(S) = \frac{9}{10}; \text{ } P(NS) = \frac{1}{10}; \text{ }
P(T|S) = \frac{1}{2}; \text{ }P(T|NS) = \frac{1}{3}$$
Now Probability of getting a Tail can be obtained using total probability theorem,
$$P(T) = P(T|S)\cdot P(S) + P(T|NS) \cdot P(NS)$$
$$P(T) = \frac{1}{2}\cdot \frac{9}{10} + \frac{1}{3} \cdot \frac{1}{10}
= \frac{29}{60}$$
Using Bernoulli's trials, where winning/success(p) is defined as getting a tails, given by $P(T) = \frac{29}{60}$ and failure(q) is defined by getting a head, $P(H) = \frac{31}{60}$ and n = number of tosses.
$$P(X = x) = {n \choose x} (p)^n \cdot (q)^{n - x}$$
$$P(X = 3) = {6 \choose 3} (\frac{29}{60})^6 \cdot (\frac{31}{60})^3$$
Assuming this to be an event E, we now have $P(E) = {6 \choose 3} (\frac{29}{60})^6 \cdot (\frac{31}{60})^3$,
The desired result is $P(S|E)$, I'm not sure how to go on about finding that, any hint would be much appreciated.
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The probability of getting exactly 3 tails out of 6 tosses is:
a) ${6\choose 3}\left(\frac{1}{2}\right)^6$ in the case of symmetric coin.
b) ${6\choose 3}\left(\frac{1}{3}\right)^3 \left(\frac{2}{3}\right)^3$ in the case of the asymmetric coin.
Then, if you picked your initial coin uniformly at random (probability $1/10$ for each coin), the probability that you picked a symmentric coin is:
$\frac{9\cdot\frac{1}{10}\cdot {6\choose 3}\left(\frac{1}{2}\right)^6}
{9\cdot\frac{1}{10}\cdot {6\choose 3}\left(\frac{1}{2}\right)^6 + \frac{1}{10}{6\choose 3}\left(\frac{1}{3}\right)^3 \left(\frac{2}{3}\right)^3}
= \frac{9\cdot\left(\frac{1}{2}\right)^6}{9\cdot\left(\frac{1}{2}\right)^6 + \left(\frac{1}{3}\right)^3 \left(\frac{2}{3}\right)^3}
=\frac{6561}{6561+512}=\frac{6561}{7073}\approx 92.76\%$
Sanity check: it does make sense that it's slightly better than $90\%$ (the a priori probability) since $3$ heads and $3$ tails is somewhat more in line with what you'd expect from a symmetric coin than from an asymmetric one.
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|
Determine conics by four points and a tangent line. I have small troubles determining conics that go through four points and have a given tangent line.
More specifically $P_1 = (0:1:0), P_2 = (0:0:1), P_3 = (1:0:1), P_4 = (1:-1:0) \in \mathbb{RP}^2$ and $ t: \{x-y+z=0\}$.
I can determine the conics as far as $ax^2+axy+2eyz-axz=0,\ a,e \in \mathbb{R}$ but don't really succeed with incorporating the information of the tangent line.
I read substituting in $z = -x +y$ should work, but i have no clue how to proceed with the obtained result $ax^2+ey^2-exy=0$.
Help is appreciated a lot (also other ways of dealing with the tangent!).
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$\det(\begin{pmatrix}x^2&xy&y^2&xz&yz&z^2\\0&0&1&0&0&0\\0&0&0&0&0&1\\1&0&0&1&0&1\\1&-1&1&0&0&0\\a^2&ab&b^2&ac&bc&c^2\end{pmatrix})=a(c-b-a)yz+bc(-xz+xy+x^2)=0$
Now assume $(a:b:c)$ is the point of tangency making $b=a+c$ and $y=x+z$, substitute and expand:
$2((ac+c^2)x^2-a^2xz-a^2z^2)$
The discriminant reduces to $a^2(2c+a)^2$, so $a=-2c,b=-c,c=c$, which makes the equation $-c^2(8yz-xz+xy+x^2)$, and this with $z=1$ is a hyperbola that actually is tangent to $y=x+1$ at $(-2,-1)$.
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How to solve the given determinant by using properties if possible Let $\alpha,\beta$ be the real roots of the equation $ax^2+bx+c=0$ and let $ {\alpha}^n+{\beta}^n=S_{n}\ for \ n \geq 1 $ then find the value of the determinant of A, where $$A = \left[ \begin{array} { l l l } {\ \ \ \ { 3 }} & { 1 + S _ { 1 } } & { 1 + S _ { 2 } } \\ { 1 + S _ { 1 } } & { 1 + S _ { 2 } } & { 1 + S _ { 3 } } \\ { 1 + S _ { 2 } } & { 1 + S _ { 3 } } & { 1 + S _ { 4 } } \end{array} \right]$$
My try:
$A = \left[ \begin{array} { l l l } { 3+0 } & { 1 + S _ { 1 } } & { 1 + S _ { 2 } } \\ { 1 + S _ { 1 } } & { 1 + S _ { 2 } } & { 1 + S _ { 3 } } \\ { 1 + S _ { 2 } } & { 1 + S _ { 3 } } & { 1 + S _ { 4 } } \end{array} \right]$
$A = \left[ \begin{array} { l l l } { 3 } & { 1 + S_{1} } & { 1 + S _ { 2 } } \\ { S _ { 1 } } & { 1 + S _ { 2 } } & { 1 + S _ { 3 } } \\ {S _ { 2 } } & { 1 + S _ { 3 } } & { 1 + S _ { 4 } } \end{array} \right]+\left[\begin{array} { l l l } { 0 } & { 1 + S _ { 1 } } & { 1 + S _ { 2 } } \\ { 1 } & { 1 + S _ { 2 } } & { 1 + S _ { 3 } } \\ { 1 } & { 1 + S _ { 3 } } & { 1 + S _ { 4 } } \end{array} \right]$
Now solving the determinant second by appling operation $R _ { 2 } \rightarrow R _ { 2 } - R _ { 3 }$ we get
$\left[ \begin{array} { l l l } { 0 } & { 1 + S _ { 1 } } & { 1 + S _ { 2 } } \\ { 0 } & { S _ { 2 } - S _ { 3 } } & { S _ { 3 } - S _ { 4 } } \\ { 1 } & { 1 - S _ { 3 } } & { 1 + S _ { 4 } } \end{array} \right]$
$\Rightarrow ( 1 + S _ { 1 } ) ( S _ { 3 } - S _ { 4 } ) - ( S _ { 2 } - S _ { 3 } ) ( 1 + S _ { 2 } )$
But how to solve other determinant? I am stuck with this question.
Can anyone give me hint to solve this typical determinant?
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$S _ { n } = \alpha ^ { n } + \beta ^ { n } \forall n \geq 1$
$A= \left| \begin{array} { l l l } { 1+1+1 } & { 1+\alpha+\beta } & { 1+\alpha^{2}+\beta^{2} } \\ { 1+\alpha +\beta } & { 1+\alpha^{2}+\beta^{2} } & { 1+\alpha^{3}+\beta^{3}} \\ { 1+\alpha^{2}+\beta^{2} } & { 1+\alpha^{3}+\beta^{3} } & { 1+\alpha^{4}+\beta^{4}} \end{array} \right|$
Now by property of multiplication of determinant we get
$= \left| \begin{array} { l l l } { 1 } & { 1 } & { 1 } \\ { 1 } & { \alpha } & { \beta } \\ { 1 } & { \alpha ^ { 2 } } & { \beta ^ { 2 } } \end{array} \right| \left| \begin{array} { l l l } { 1 } & { 1 } & { 1 } \\ { 1 } & { \alpha } & { \beta } \\ { 1 } & { \alpha ^ { 2 } } & { \beta ^ { 2 } } \end{array} \right|$
$= ( \alpha - \beta ) ^ { 2 } ( \beta - 1 ) ^ { 2 } ( \alpha - 1 ) ^ { 2 }$
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Prove this $\sin(nx)$ identity Without induction How to prove that $$\sin(2nx)=2n\sin x \cos x \prod_{k=1}^{n-1}\left(1-\frac{\sin^2(x)}{\sin^2\frac{k\pi}{2n}}\right)$$
for Natural n
I Tried by several way and the last try is to use euler formula which lead me to
$$\sin(2nx)=\sin x \cos^{2m+1} x \sum_{k=0}^{n-1} \left( (-1)^k\binom{2n}{2k+1}(\cos x \sin x )^{2k}\right)$$
and no idea how to continue or if this method lead to the answer
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Finally it solved ^_^ :
first by use: $\sin^2x=\frac{1-\cos 2x}{2}$
$$\left(1-\frac{\sin^2(x)}{\sin^2\frac{k\pi}{2n}}\right)=\left(\frac{\cos 2x - \cos{\frac{k\pi}{n}}}{2\sin^2\frac{k\pi}{2n}}\right)$$
then by use :$\cos x - \cos y $ formula
I got
$$\left(1-\frac{\sin^2(x)}{\sin^2\frac{k\pi}{2n}}\right)=-\left(\frac{\sin(x+\frac{k\pi}{2n})\sin(x-\frac{k\pi}{2n})}{\sin^2\frac{k\pi}{2n}}\right)$$
so
$$R.H.S=2n\sin x \cos x \prod_{k=1}^{n-1}\left(-\frac{\sin(x+\frac{k\pi}{2n})\sin(x-\frac{k\pi}{2n})}{\sin^2\frac{k\pi}{2n}}\right)$$
but
$$\prod_{k=1}^{n-1}\sin(\frac{k\pi}{2n}-x)= \prod_{k=1}^{n-1}\cos(x+\frac{k\pi}{2n})$$
so
$$R.H.S=2n\sin x \cos x \prod_{k=1}^{n-1}\left(\frac{\sin(2x+\frac{k\pi}{n})}{2\sin^2\frac{k\pi}{2n}}\right)$$
and by This LINK i've got
$$\frac{\sin(2nx)}{2\sin x \cos x 2^{n-1}}=\prod_{k=1}^{n-1}\sin(2x+\frac{k\pi}{n})$$
so
$$R.H.S=\frac{n\sin(2nx)}{2^{2n-2}}\prod_{k=1}^{n-1}\left(\frac{1}{\sin^2\frac{k\pi}{2n}}\right)$$
and Finally from this
$$R.H.S=\frac{n\sin(2nx)}{2^{2n-2}}\left(\frac{1}{n2^{2-2n}}\right)=\sin(2nx)=L.H.S$$
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How to evaluate this integral $\int_{1/3}^{3} \frac{\sin(\frac1x -x)dx}{x}$?
Evaluate
$$\int_{1/3}^{3} \frac{\sin(\frac1x -x)dx}{x}.$$
I am unable to think of any way to solve this.
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$$I=\int_{\frac{1}{3}}^3 \dfrac{\sin \left( \dfrac{1}{x}-x\right)}{x}dx=\underbrace{\int_{\frac{1}{3}}^1\dfrac{\sin \left( \dfrac{1}{x}-x\right)}{x}dx}_J+\underbrace{\int_1^3\dfrac{\sin \left( \dfrac{1}{x}-x\right)}{x}dx}_H$$
$$J=\int_{\frac{1}{3}}^1\dfrac{\sin \left( \dfrac{1}{x}-x\right)}{x}dx\quad t=\dfrac{1}{x}\rightarrow x=\dfrac{1}{t}\rightarrow dx=-\dfrac{1}{t^2}\quad \text{and}\quad \left\{ \begin{array}{lcc}
x=\dfrac{1}{3}\rightarrow t=3 \\
\\ x=1\rightarrow t=1
\end{array}
\right.$$
$$J=\int_3^1\dfrac{\sin \left( t-\dfrac{1}{t}\right)}{\dfrac{1}{t}}\left( -\dfrac{1}{t^2}\right)dt=-\int_1^3\dfrac{\sin \left[ -\left( \dfrac{1}{t}-t\right) \right]}{-t}dt=-\int_1^3 \dfrac{-\sin \left( \dfrac{1}{t}-t\right)}{-t}dt=$$
$$=-\int_1^3\dfrac{\sin \left( \dfrac{1}{t}-t\right)}{t}dt=-\int_1^3\dfrac{\sin \left( \dfrac{1}{x}-x\right)}{x}dx=-H\rightarrow I=-H+H=\boxed{0}$$
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Consider the next succession and prove by induction The exercise says:
Knowing the next succession,
$$a_1=1$$
$$a_{n+1}=\frac{2a_n}{(2n+2)(2n+1)}, n>=1$$
Prove by induction that $a_n=\frac{2^n}{(2n)!}$
What I've done so far is prove for $n=1$
For $n=1$:
$$a_1=\frac{2^1}{(2\times1)!}=\frac{2}{2}=1$$
Which is correct.
Therefore, to prove the induction:
$$a_{n+1}=\frac{2\times a_{n+1}}{(2(n+1)+2)(2n+1)}=$$
$$=\frac{2}{(2(n+1)+2)(2n+1)}\frac{2^{n+1}}{(2(n+1))!}=$$
$$=\frac{2^{n+2}}{(2n+4)(2n+3)}\times\frac{1}{(2(n+1))!}=$$
$$=\frac{2^{n+2}}{4n^2+6n+18n+12}=$$
$$=\frac{2^{n+2}}{4n^2+24n+12}=$$
$$=\frac{2\times2^{n+1}}{2(2n^2+12n+6)}=$$
$$=\frac{2^{n+1}}{2n^2+12n+6}$$
Is this correct? Do I have to simplify even more?
Thanks
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No, it is not correct. What you need to show is that, assuming
$$a_n=\frac{2^n}{(2n)!},$$
then
$$a_{n+1} = \frac{2a_n}{(2n+2)(2n+1)} = \frac{2^{n+1}}{\big(2(n+1)\big)!}.$$
So
$$\frac{2a_n}{(2n+2)(2n+1)} = \frac{2}{(2n+2)(2n+1)}\frac{2^n}{(2n)!}=\dots$$
Try to follow from there.
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Proving the product of four consecutive integers, plus one, is a square I need some help with a Proof:
Let $m\in\mathbb{Z}$. Prove that if $m$ is the product of four consecutive integers, then $m+1$ is a perfect square.
I tried a direct proof where I said:
Assume $m$ is the product of four consecutive integers.
If $m$ is the product of four consecutive integers, then write $m=x(x+1)(x+2)(x+3)$ where $x$ is an integer.
Then $m=x(x+1)(x+2)(x+3)=x^4+6x^3+11x^2 +6x$.
Adding $1$ to both sides gives us:
$m+1=x^4+6x^3+11x^2+6x+1$.
I'm unsure how to proceed. I know I'm supposed to show $m$ is a perfect square, so I should somehow show that $m+1=a^2$ for some $a\in\mathbb{Z}$, but at this point, I can't alter the right hand side of the equation to get anything viable.
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To get a feel for the problem, let's work backwards, starting from a square number. Take some integer $n$.
$n^2 - 1 = (n+1)(n-1).$
OK, so it looks like $m$ has two factors whose difference is 2. Could it be that in our product of consecutive integers, the product of two of them is $n-1$ and of the other two is $n+1$?
Let's throw in a simple example: $1\times2\times3\times4$. Notice how $1\times4=4$ and $2\times3=6$.
What about $2\times3\times4\times5$? This time $2\times5=10$ and $3\times4=12$.
It looks like the product of the "outer" pair is $n-1$ and the product of the "inner pair" is $n+1$.
Now we know how to attack this.
Let $k$ be some integer and $m=k(k+1)(k+2)(k+3)$.
$\begin{align} m &= k(k+1)(k+2)(k+3) \\
&= (k+1)(k+2)\times(k(k+3)) \ \text{ (collecting inner and outer terms)}\\
&= (k^2 + 3k + 2) \times (k^2 + 3k) \\
&= ((k^2 + 3k + 1) + 1) ((k^2 + 3k + 1) - 1) \\
&= (k^2 + 3k + 1) ^ 2 - 1 \qquad \text{ (since }(a+b)(a-b)=a^2-b^2\text{)}.
\end{align}$
Since $k$ is an integer, $k^2 + 3k + 1$ is an integer so $m+1$ is a perfect square.
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|
Different way $\int_{0}^{\infty}e^{-2x}\frac{\ln \cosh(x)}{1+\cosh(x)}dx$ I would like to evaluate this integral, $$I=\int_{0}^{\infty}e^{-2x}\frac{\ln \cosh(x)}{1+\cosh(x)}\mathrm dx$$
An approach might would be:
*
*Remove the $\cosh(x)$
$$I=2\int_{0}^{\infty}e^{-x}\ln\left(\frac{e^x+e^{-x}}{2}\right)\frac{dx}{(e^x+1)^2}$$
*Make a substitution of: $u=e^x$, $du=e^x dx$
$$I=2\int_{1}^{\infty}\frac{\ln(1+u^2)}{(u+u^2)^2}du-2\int_{1}^{\infty}\frac{\ln(2u)}{(u+u^2)^2}du=2(I_1-I_2)$$
*Partial fraction decomp of integral $I_1$ reveals
$$I_1=2\int_{1}^{\infty}\frac{\ln(1+x^2)}{1+x}dx+\int_{1}^{\infty}\frac{\ln(1+x^2)}{(1+x)^2}dx-2\int_{1}^{\infty}\frac{\ln(1+x^2)}{x}dx+\int_{1}^{\infty}\frac{\ln(1+x^2)}{x^2}dx$$
$$I_2=2\int_{1}^{\infty}\frac{\ln(2x)}{1+x}dx+\int_{1}^{\infty}\frac{\ln(2x)}{(1+x)^2}dx-2\int_{1}^{\infty}\frac{\ln(2x)}{x}dx+\int_{1}^{\infty}\frac{\ln(2x)}{x^2}dx$$
abandomed this approach, because too much work
I was thinking of applying integration by parts, $u=\ln\cosh(x)$, $du=\tanh(x) dx$
$dv=\frac{e^{-2x}}{1+\cosh(x)}dx=\frac{2^{-x}}{(e^x+1)^2}dx$
but, when I try to evaluate, it is not possible.
Making a sub: $u=e^x$ and partial fraction decomposition, it is can easily integrate to $$v=4\ln(1+e^x)-\frac{2}{1+e^x}-2e^{-x}-4x$$
This is getting very bored, unfortunately I don't know any other methods. Can anybody evaluate integral $I$ by another approach?
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Let $I$ denote the integral. Then applying the substitution $u = e^{-x}$ followed by integration by parts,
\begin{align*}
I
&= \int_{0}^{1} \frac{2u^2}{(1+u)^2} \log\left(\frac{1+u^2}{2u}\right)\,du \\
&= \int_{0}^{1} 2\left( 1+u - \frac{1}{1+u} - 2\log(1+u)\right)\frac{1-u^2}{u(1+u^2)} \, du \\
&= \underbrace{ \int_{0}^{1} \left( \frac{6}{1+u^2} -\frac{2u}{1+u^2} -2 \right) \, du}_{=(1)} -4 \underbrace{ \int_{0}^{1} \frac{1-u^2}{u(1+u^2)} \log(1+u) \, du }_{=(2)}
\end{align*}
Now $(1)$ is easily computed by direct computation, yielding $(1) = \frac{3\pi}{2} - \log 2 - 2$. For $(2)$, we utilize the identity $\log(1+u)=\int_{0}^{1} \frac{u}{1+uv} \,dv$ to write
$$ (1) = \int_{0}^{1} \int_{0}^{1} \frac{1-u^2}{(1+vu)(1+u^2)} \, dudv. $$
Interchanging the role of $u$ and $v$ does not alter the value of integral, thus we find that
\begin{align*}
(1)
&= \frac{1}{2} \int_{0}^{1} \int_{0}^{1} \frac{1}{1+vu}\left( \frac{1-u^2}{1+u^2} + \frac{1-v^2}{1+v^2} \right) \, dudv \\
&= \int_{0}^{1} \int_{0}^{1} \frac{1-uv}{(1+u^2)(1+v^2)} \, dudv
= \left(\frac{\pi}{4}\right)^2 - \left( \frac{\log 2}{2}\right)^2.
\end{align*}
(The last step easily follows since each term of the double integral factors into the product of two 1-dimensional integrals.) Therefore we have
$$ I = \log^2 2 - \log 2 - \frac{\pi^2}{4} + \frac{3\pi}{2} - 2.$$
|
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|
What tools one should use for inequalities? If $a,b,c>0$ prove that:
$$\frac{1}{a+4b+4c}+\frac{1}{4a+b+4c}+\frac{1}{4a+4b+c}\leq \frac{1}{3\sqrt[3]{abc}}.$$
My first try was the following:
$$\sum_{cyc}\frac{1}{a+4b+4c}\leq\sum_{cyc}\frac{1}{\sqrt[3]{16abc}}=\frac{1}{\sqrt[3]{16abc}}$$
But $\frac{1}{\sqrt[3]{16abc}}\geq \frac{1}{3\sqrt[3]{abc}}$
The I have tried your method from another post:
$$\sum_{cyc}\frac{1}{a+4b+4c}=\sum_{cyc}\frac{1}{a+2b+2(b+2c)}$$
$$\sum_{cyc}\frac{1}{a+2b+2(b+2c)}\leq\sum_{cyc}\frac{1}{9}\left (\frac{1^2}{a+2b}+\frac{2^2}{b+2c} \right )$$
Where I got $$\sum_{cyc}\frac{1}{3}\left ( \frac{1}{a+2b} \right )\leq \frac{1}{3\sqrt[3]{abc}}$$ wich is false.
$$$$
|
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that $f(v^2)\geq0,$ where
$$f(v^2)=16u^3+12uv^2-w^3-24u^2w-3u^2w,$$ which is a linear function.
But a linear function gets a minimal value for an extreme value of $v^2$, which happens for equality case of two variables.
Sines our inequality is homogeneous, we can assume that $abc=1$ and for $b=a$ and $c=1/a^2$ we obtain:
$$\frac{2}{4/a^2+5a}+\frac{1}{\frac{1}{a^2}+8a}\leq\frac{1}{3}$$ or
$$(a-1)^2(40a^4+17a^3-6a^2+8a+4)\geq0,$$ which is obvious.
|
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|
Solving Quasi-Linear Transport Equation with two shockwaves. I want to solve the following Partial Differential Equation by using the method of characteristics. This is the transport equation.
\begin{align}u_t - (1-2u)u_{x}&=0, &-\infty < x < \infty, t >0, \end{align}
\begin{align}u(x,0) &= \begin{cases} \frac{1}{4}, & \mbox{for } x<0, \\ \frac{1}{2}, & \mbox{for } 0 < x < 1, \\ 1, & \mbox{for } 1 < x.\end{cases}\end{align}
By using the method of characteristics, we transform the PDE into the following system of ODE's.
\begin{align}
\frac{d t}{d s} &= 1, &t(s=0)=0, \\
\frac{d x}{d s} &= (1-2u), &x(s=0) = x_0, \\
\frac{d u}{d s} &= 1, &u(s=0) = u(x_0,0).
\end{align}
If we solve these we get the following expressions:
\begin{align}
t(s) &= s, \\
x(s) &= (1-2u)s+x_0.
\end{align}
Thus using $s=t$ we get $x(t) = (1-2u)t+x_0$.
With this we and the given of $u(x_0,0)$ we can make a drawing in the (x,t)-domain of the characteristics. We note that because of $u(x_0,0)$ having three different values that get higher, we will get 2 shock waves. These shockwaves, $x_{s1}(t),x_{s2}(t)$ can be calculated by using the jump in 'q' (which is the 'velocity') divided by the jump in 'u' (which is the density) and then integrating these over t and using the begin conditions. This gives us $x_{s1}(t)=\frac{1}{2}t$ and $x_{s2}(t)=-\frac{1}{2}t+1$.
Now my problem comes in. These shockwaves, they intersect. How can I know their behaviour?
Thanks for your time,
K. Kamal
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Here, we are dealing with the macroscopic traffic-flow model by Lighthill-Witham-Richards (LWR), where $0≤u≤1$ represents the number of cars per unit length. To get an insight about the problem, here is a plot of the characteristic curves $x(t) = (1-2u(x_0,0))\, t + x_0$ in the $x$-$t$ plane:
As written in OP, two shock waves occur, which speeds are given by the Rankine-Hugoniot condition:
\begin{aligned}
s_1 &= \frac{\frac{1}{2}(1-\frac{1}{2}) - \frac{1}{4}(1-\frac{1}{4})}{\frac{1}{2} - \frac{1}{4}} = \frac{1}{4} \\
s_2 &= \frac{1(1-1) - \frac{1}{2}(1-\frac{1}{2})}{1 - \frac{1}{2}} = -\frac{1}{2}
\end{aligned}
Therefore, the solution for small times $t<t^*$ before the interaction is
$$
u(x,t) = \left\lbrace
\begin{aligned}
&1/4 && \textstyle \text{if}\quad x < \frac{1}{4}t \\
&1/2 && \textstyle \text{if}\quad \frac{1}{4}t < x < 1 - \frac{1}{2}t \\
&1 && \textstyle \text{if}\quad 1 - \frac{1}{2}t < x
\end{aligned}
\right.
$$
The shock waves interact at the time $t^*$ such that $\frac{1}{4}t^* = 1-\frac{1}{2}t^*$, i.e. $t^* = 4/3$. The corresponding abscissa is $x^* = 1/3$ (cf. plot of the characteristics). At the interaction, it is as if we would start with new initial conditions $u(x<x^*,t^*) = 1/4$ and $u(x>x^*,t^*) = 1$ at the time $t^*$. A new shock wave is generated with speed
$$
s_3 = \frac{1(1-1) - \frac{1}{4}(1-\frac{1}{4})}{1 - \frac{1}{4}} = -\frac{1}{4}
$$
Therefore, the solution for $t>t^*$ is
$$
u(x,t) = \left\lbrace
\begin{aligned}
&1/4 && \textstyle \text{if}\quad x < \frac{1}{3} - \frac{1}{4}(t-\frac{4}{3}) \\
&1 && \textstyle \text{if}\quad \frac{1}{3} - \frac{1}{4}(t-\frac{4}{3}) < x
\end{aligned}
\right.
$$
|
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|
Find $B=A^2+A$ knowing that $A^3=\begin{bmatrix}4&3\\-3&-2\end{bmatrix}$
Find $B=A^2+A$ knowing that $A^3=\begin{bmatrix}4&3\\-3&-2\end{bmatrix}$
Is there a way to solve this rather than just declaring a matrix $$A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$$ and then trying to solve a system of cubic equations? My attempt:
$$A^3 -I_2 = \begin{bmatrix} 3 & 3\\ -3 & -3\end{bmatrix} = 9 \begin{bmatrix} 1 & 1\\ -1 & -1\end{bmatrix}$$
and
$$B=\frac {A^3-I_2}{A-I_2}-I_2$$
$$(A-I_2)(A^2+A+I_2)=9\begin{bmatrix}1&1\\-1&-1\end{bmatrix}$$
But I get stuck here.
|
Denote the third root of unity as $w=e^{i2\pi/3}$.
Then any of the quantities $\rho=\{w,w^2,w^3=1\}\,$ satisfy $\,\rho^3=1,\,$ i.e. is a cube root of one.
Note that the matrix $X=A^3$ has a single eigenvalue of $\{\lambda=1\}$ with multiplicity two.
Since it is a $2\times 2$ matrix, any analytic function of $X$ (and its first derivative) can be written as linear polynomials
$$\eqalign{
f(X) &= \alpha_1X + \alpha_0I \cr
f'(X) &= \alpha_1I \cr
}$$
Evaluate the cube root function on the eigenvalues to calculate the polynomial coefficients
$$\eqalign{
\alpha_1\lambda + \alpha_0 &= \lambda^{1/3} &\implies \alpha_1+\alpha_0 &= \rho \cr
\alpha_1 &= \tfrac{1}{3}\lambda^{-2/3} &\implies \alpha_1 &= \tfrac{\rho}{3} \cr
}$$
The matrix cube root is therefore
$$\eqalign{
A &= f(X) = \,\,\frac{\rho}{3}X + \frac{2\rho}{3}I \cr
}$$
and the matrix in question is
$$\eqalign{
B &= A^2 + A \cr
&= (\tfrac{\rho}{3}X + \tfrac{2\rho}{3}I)^2 + (\tfrac{\rho}{3}X + \tfrac{2\rho}{3}I) \cr
&= \tfrac{\rho^2}{9}(X^2+4X+4I) + \tfrac{\rho}{3}(X+2I) \cr
&= \frac{1}{9}\Big(\rho^2X^2+(4\rho^2+3\rho)X+(4\rho^2+6\rho)I\Big) \cr\cr
}$$
This gives us 3 possible solutions for the $B$ matrix:
Setting $\rho=w^1=w\,$ yields
$$B_1=\frac{1}{9}\Big(w^2X^2+(w^2-3)X+(2w-4)I\Big)$$
Setting $\rho=w^2\,$ yields
$$B_2=\frac{1}{9}\Big(wX^2+(w-3)X+(2w^2-4)I\Big)$$
Setting $\rho=w^3=1\,$ yields
$$B_3=\frac{1}{9}\Big(X^2+7X+10I\Big)$$
Some properties, peculiar to roots-of-unity, were used to simplify the final expressions. Namely
$$\eqalign{
w &= w^4 \cr
0 &= 1+w+w^2 \cr
0 &= 1+w^2+w^4 \cr\cr
}$$
Interestingly, these properties also mean that $\,\,B_1 + B_2 + B_3 = 0.$
|
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|
show this inequality $ab+bc+ac+\sin{(a-1)}+\sin{(b-1)}+\sin{(c-1)}\ge 3$
let $a,b,c>0$ and such $a+b+c=3abc$, show that
$$ab+bc+ac+\sin{(a-1)}+\sin{(b-1)}+\sin{(c-1)}\ge 3$$
Proposed by wang yong xi
since
$$\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}=3$$
so use Cauchy-Schwarz inequality we have
$$(ab+bc+ac)\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)\ge 9$$
so we have
$$ab+bc+ac\ge 3$$
|
Hint : Prove with Jensen's inequality (apply to the function $f(x)=0.5x^2+\sin(x-1)$ ) this with the obvious restriction $ab+bc+ca\leq6$:
$$ab+bc+ca+\sin(a-1)+\sin(b-1)+\sin(c-1)$$$$\geq$$$$ ab+bc+ca-0.5(a^2+b^2+c^2)+3\sin(\frac{a+b+c}{3}-1)+1.5(\frac{a+b+c}{3})^2$$
And now a last hint prove this (with the condition $3abc=a+b+c$ and $ab+bc+ca\leq6$ ):
$$ab+bc+ca-0.5(a^2+b^2+c^2)+1.5(\frac{a+b+c}{3})^2\geq 3$$
And
$$3\sin(\frac{a+b+c}{3}-1)\geq 0$$
|
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|
The logic in radical symplification I'm having troubles while studying radicals, namely with converting expressions with the form $$\sqrt{a+b\sqrt{c}}$$ to $$a+b\sqrt{c}$$ and vice versa. When I'm dealing with these kind of problem, I usually add and subtract $\sqrt{c}^2$. In example: $$\sqrt{27+10\sqrt{2}}$$ I factor ${27+10\sqrt{2}} = 27+10\sqrt{2}+\sqrt{2}^2-2=\left(\sqrt{2}^2+5\sqrt{2}\right)+\left(5\sqrt{2}+25\right)$ ... until I get to $\sqrt{\left(\sqrt{2}+5\right)^2}$ and finally $\sqrt{2}+5$. However, either I'm missing something, or this does not always work. For instance, I cannot solve $\sqrt{113+8\sqrt{7}}$ using this technique.I would like to fully comprehend the logic behind these conversions so that I can apply it indiscriminately.
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Do not expect integer solutions for problems like that.
For $$\sqrt{27+10\sqrt{2}}= a+b\sqrt{2}$$
We square both sides to get $$27+10\sqrt{2}=a^2+2b^2 +2ab\sqrt 2$$
Thus we solve a system $$ 2ab=10, a^2+2b^2=27$$ and we are lucky to find $$a=5, b=1$$
Therefore $$\sqrt{27+10\sqrt{2}}= 5+\sqrt{2}$$
On the other hand for $$\sqrt{113+8\sqrt{7}} =a+b\sqrt 7$$
We need to solve $$ ab=4, a^2+7b^2=113 $$
We get $a=1$ and $b=4$ thus $$\sqrt{113+8\sqrt{7}} =1+4\sqrt 7$$
|
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|
Using Method of Undetermined coefficients to find maximum value Find the maximum value of ($1$+$x$)$^5$($1$$-$$x$)($1$$-$$2$$x$)$^2$ given that $1/2$$<$$x$$<$$1$.
The solution states the follows
Consider the maximum value of
($A$($1$$+$$x$))$^5$ $*$ ($B$($1$$-$$x$)) $*$ ($C$($2$$x$$-$$1$)$^2$), where $A$, $B$, $C$ are positive integers satisfying $5$$A$$-$$B$$+$$4$$C$$=$$0$, where
($A$($1$$+$$x$))$^5$$=$ ($B$($1$$-$$x$))$=$($C$($2$$x$$-$$1$)$^2$)----------(1)
This then implies that [$B-A$]$/$[$B+A$]$=$[$B+C$]/[$2C+B$], so plugging in $B=5A+4C$ we would have $2(5A-2C)(A+C)=0$-------------(2)
Let ($A,B,C$)=($2,30,5$), from AM-GM inequality we get ($2$($1$$+$$x$))$^5$ $*$ ($30$($1$$-$$x$)) $*$ ($5$($2$$x$$-$$1$)$^2$)$<$ $(15/4)^8$---------------(3)
where the ineqaulity is achieved when $x=7/8$. As a result, the maximum value is $3^7*5^5/$$2$^$22$.
Could someone explain the steps (1), (2), (3) in detail?
I'm not particularly familiar with the method of undetermined coefficients.
|
Since ${\large{\frac{1}{2}}} < x < 1$, it follows that $1+x,\;1-x,\;2x-1$ are positive.
Let $A,B,C$ be fixed positive real numbers, not yet selected.
Applying $\text{AM-GM}$, to $5$ copies of $A(1+x)$, $1$ copy of $B(1-x)$, and $2$ copies of $C(2x-1)$, yields
$$\left(\frac{\;5\bigl(A(1+x)\bigr)+B(1-x)+2\bigl(C(2x-1)\bigr)\!\!}{8}\right)^{\!8}\ge \bigl(A(1+x)\bigr)^5\bigl(B(1-x)\bigr)\bigl(C(2x-1)\bigr)^2\tag{*}$$
with equality if and only if
$$A(1+x)=B(1-x)=C(2x-1)$$
or equivalently (by eliminating $x$ from the above equations), if and only if
$$\frac{B-A}{B+A}=\frac{B+C}{B+2C}\tag{eq1}$$
The idea is to choose $A,B,C$ so that the $\text{LHS}$ of $(\text{*})$ is independent of $x$ (i.e., constant, once $A,B,C$ are selected).
Noting that
$$5A(1+x)+B(1-x)+2C(2x-1)=(5A-B+4C)x + (5A-B+2C)$$
in order to force the $\text{LHS}$ of $(\text{*})$ to be independent of $x$, we'll select $A,B,C$ so that $5A-B+4C=0$, or equivalently, $B=5A+4C$.
Replacing $B$ by $5A+4C$ in $(\text{eq}1)$, we get
\begin{align*}
&\frac{2(A+C)}{3A+2C}=\frac{5(A+C)}{5A+6C}\\[4pt]
\iff\;&\frac{2}{3A+2C}=\frac{5}{5A+6C}\\[4pt]
\iff\;&C=\frac{5A}{2}\\[4pt]
\end{align*}
Hence, by choosing $A=2$, we get $C=5$, and $B=30$.
With these choices, $(\text{*})$ reduces to
\begin{align*}
\left(\frac{\;5\bigl(2(1+x)\bigr)+30(1-x)+2\bigl(5(2x-1)\bigr)\!\!}{8}\right)^{\!8}&\ge \bigl(2(1+x)\bigr)^5\bigl(30(1-x)\bigr)\bigl(5(2x-1)\bigr)^2\\[4pt]
\implies\;\bigl((2^5)(30)(5^2)\bigr)(1+x)^5(1-x)(1-2x)^2&\le \left({\small{\frac{30}{8}}}\right)^8\\[4pt]
\implies\;(1+x)^5(1-x)(1-2x)^2&\le \frac{(3^5)(5^5)}{2^{22}}\\[4pt]
\end{align*}
with equality if and only if
$$2(1+x)=30(1-x)=5(2x-1)$$
or equivalently, if and only if $x={\large{\frac{7}{8}}}$.
|
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|
Prove that $2^{2^5} +1 \equiv 0 \mod 641$
$2^{(2^5)} +1 \equiv 0 \mod 641$
From the totient funcion we have:
$2^{32^{20}} \equiv 1$
Thus:
Either $2^{32} \equiv 1$ or $2^{32} \equiv -1$
But how do I prove that $2^{32} \equiv -1$?
|
You only need $$641=2^4+5^4$$
and $$641=2^7\cdot 5+1$$
This gives you $$2^{32}=(2^7)^4\cdot 2^4\equiv -(2^7)^4\cdot 5^4=-(2^7\cdot 5)^4\equiv -1\mod 641$$
|
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|
For $x^3+px+q=0$, one of the solutions is $\sqrt 3-1$. Find the value of rational number $p$ and $q$, and other two solutions of the equation.
I cannot find any error from my answer. What’s wrong?
|
If $\sqrt{3} -1$ is a root of $x^3+px+q$, it doesn't necessarily follow that $\sqrt{3}+1$ is also a root.
In fact, if $\sqrt{3} -1$ is a root then $\sqrt{3}+1$ cannot be a root, assuming $p,q \in \mathbb{Q}$.
Indeed
$$\alpha = \sqrt{3} + 1 \implies \alpha-1 = \sqrt{3} \implies (\alpha-1)^2 = 3 \implies \alpha^2-2\alpha-2 = 0$$
so $x^2-2x-2$ is the minimal polynomial of $\sqrt{3} + 1$ over $\mathbb{Q}$.
On the other hand
$$\beta = \sqrt{3} - 1 \implies \beta+1 = \sqrt{3} \implies (\beta+1)^2 = 3 \implies \beta^2+2\beta -2 = 0$$
so $x^2+2x-2$ is the minimal polynomial of $\sqrt{3} - 1$ over $\mathbb{Q}$.
If a polynomial $f$ with rational coefficients satisfies $f(\sqrt{3} + 1) = f(\sqrt{3} - 1) = 0$ then $x^2-2x-2$ and $x^2+2x-2$ both divide $f$, and they are irreducible so $\deg f \ge 4$.
However, what you can deduce is that $-\sqrt{3} - 1$ is necessary also a root of $x^3+px+q$. This is because the minimal polynomial of $\sqrt{3}-1$, which is $x^2+2x-2$, must divide $x^3+px+q$. And $-\sqrt{3} - 1$ happens to be a root of $x^2+2x-2$.
Therefore
$$(\sqrt{3}-1)^3 + p(\sqrt{3}-1) + q = 0$$
$$(-\sqrt{3}-1)^3 + p(-\sqrt{3}-1) + q = 0$$
which yields
$$-10+6\sqrt{3} + p(\sqrt{3}-1) + q = 0$$
$$-10+6\sqrt{3} + p(\sqrt{3}-1) + q = 0$$
Subtracting them gives $p = -6$ and adding them gives $q = 4$.
Therefore, your polynomial is $x^3 - 6x + 4 = (x-2)(x^2+2x-2)$ so the third root is $2$.
All of this is unnecessary, of course. The solution from the book is the shortest.
|
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|
Is $\lim_{x\to -\infty} \frac{5x+9}{3x+2-\sqrt{4x^2-7}}=1$?
Evaluate $$\lim_{x\to-\infty} \frac{5x+9}{3x+2-\sqrt{4x^2-7}}$$
My attempt:
$$\lim_{x\to -\infty} \frac{5x+9}{3x+2-\sqrt{4x^2-7}}=\lim_{x\to-\infty}\frac{5x}{-\sqrt{4x^2}}=\frac{-5}{2}$$ According to the answer key, it actually equals $1$.
Thanks in advance.
|
Here's my way :
$$ \lim_{x\to -\infty} \frac {5x+9}{3x+2-\sqrt{4x^2-7}} = \lim_{x\to -\infty} \frac {x (5+ \frac {9}{x})}{3x+2-\sqrt{4x^2(1-\frac{7}{4x^2})}}=\lim_{x\to -\infty}\frac {x (5+ \frac {9}{x})}{3x+2-2|x|\sqrt{1-\frac{7}{4x^2}}}=\lim_{x\to -\infty} \frac {x (5+ \frac {9}{x})}{3x+2+2x\sqrt{1-\frac{7}{4x^2}}} = \lim_{x\to -\infty} \frac {x (5 + \frac {9}{x})}{x (3+\frac{2}{x} +2\sqrt{1-\frac{7}{4x^2}})} = \lim_{x\to -\infty} \frac {5+\frac{9}{x}}{3+\frac{2}{x} +2\sqrt{1-\frac{7}{4x^2}}} = \frac {5}{5} = 1$$
|
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|
Prove $ \min \left(a+b+\frac1a+\frac1b \right) = 3\sqrt{2}\:$ given $a^2+b^2=1$ Prove that
$$ \min\left(a+b+\frac1a+\frac1b\right) = 3\sqrt{2}$$
Given $$a^2+b^2=1 \quad(a,b \in \mathbb R^+)$$
Without using calculus.
$\mathbf {My Attempt}$
I tried the AM-GM, but this gives $\min = 4 $.
I used Cauchy-Schwarz to get $\quad (a+b)^2 \le 2(a^2+b^2) = 2\quad \Rightarrow\quad a+b\le \sqrt{2}$
But using Titu's Lemma I get $\quad \frac1a+\frac1b \ge \frac{4}{a+b}\quad \Rightarrow\quad \frac1a+\frac1b \ge 2\sqrt{2}$
I'm stuck here, any hint?
|
Solution 1
By $H_n \leq G_n \leq A_n \leq Q_n$,
$$a+b+\frac{1}{a}+\frac{1}{b} \ge a+b+\frac{4}{a+b}=(a+b+\frac{2}{a+b})+\frac{2}{a+b} \geq 2\sqrt{2}+\sqrt{\frac{2}{a^2+b^2}} =3\sqrt{2}.$$
Solution 2
$$a+b+\frac{1}{a}+\frac{1}{b} \geq 2(\sqrt{ab}+\frac{1}{2\sqrt{ab}})+\frac{1}{\sqrt{ab}}\geq 2\sqrt{2}+ \sqrt{2}=3\sqrt{2}.$$
|
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|
Convergence of the series $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{x^n}$.
Show that $$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{x^n}$$ converges for every $x>1$.
let $a(x)$ be the sum of the series. does $a$ continious at $x=2$? differentiable?
I guess the first part is with leibniz but I am not sure about it.
|
Let's look at the
partial sums,
and let $y = -1/x$
so
$-1 < y < 0$..
$\begin{array}\\
s_m(y)
&=\sum_{n=1}^{m} (-1)^{n-1}(-y)^n\\
&=\sum_{n=1}^{m} (-1)^{n-1}(-1)^ny^n\\
&=-\sum_{n=1}^{m} y^n\\
&=-y\sum_{n=0}^{m-1} y^n\\
&=-y\dfrac{1-y^m}{1-y}\\
&=\dfrac{-y}{1-y}-\dfrac{-y^{m+1}}{1-y}\\
\end{array}
$
so
$\begin{array}\\
s_m(y)+\dfrac{y}{1-y}
&=\dfrac{y^{m+1}}{1-y}\\
\end{array}
$
Therefore
$\sum_{n=1}^{m} \frac{(-1)^{n-1}}{x^n}+\dfrac{-1/x}{1+1/x}
=\dfrac{1}{(-x)^{m+1}(1+1/x)}
$
or
$\sum_{n=1}^{m} \frac{(-1)^{n-1}}{x^n}-\dfrac{1}{x+1}
=\dfrac{(-1)^{m+1}}{x^{m}(x+1)}
$.
What is needed now is
to show that
$\lim_{m \to \infty} \dfrac{1}{x^{m}(x+1)}
=0$.
(This is from
"What is Mathematics")
Since $x > 1$,
$x = 1+z$
where $z > 0$.
By Bernoulli's inequality,
$x^m
=(1+z)^m
\ge 1+mz
\gt mz
=m(x-1)$,
so
$ \dfrac{1}{x^{m}(x+1)}
\lt \dfrac{1}{m(x-1)(x+1)}
=\dfrac{1}{m(x^2-1)}
$,
so to make
$\dfrac{1}{x^{m}(x+1)}
\lt \epsilon
$
it is enough to take
$m
\gt \dfrac1{\epsilon(x^2-1)}
$.
This is certainly
not the best $m$,
but it is
completely elementary.
|
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|
$\tan(a) = 3/4$ and $\tan (b) = 5/12$, what is $\cos(a+b)$ It is known that
$$\tan(a) = \frac{3}{4}, \:\:\: \tan(b) = \frac{5}{12} $$
with $a,b < \frac{\pi}{2}$.
What is $\cos(a+b)$?
Attempt :
$$ \cos(a+b) = \cos(a) \cos(b) - \sin(a) \sin(b) $$
And we can write $\tan(a) = \sin(a)/\cos(a) = 0.3/0.4 $
and $ \sin(b)/\cos(b) = 0.05/0.12 $
so
$$ \cos(a+b) = (0.4)(0.12) - (0.3) (0.05) = 33/1000$$
Is this correct? Thanks.
|
Solve for sin(a), cos(a), sin(b) and cos(b) (like I said in comments):
$sin(a)=\frac{3}{5}$
$cos(a)=\frac{4}{5}$
$sin(b)=\frac{5}{13}$
$cos(b)=\frac{12}{13}$
Thus $$cos(a+b)=\frac{4}{5}\times\frac{12}{13}-\frac{3}{5}\times\frac{5}{13}=\frac{33}{65}$$
|
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|
If $abc=1$, prove that $\frac{2}{(a+1)^2+b^2+1} + \frac{2}{(b+1)^2+c^2+1} + \frac{2}{(c+1)^2+a^2+1} \le 1$. If $a, b, c \in \mathbb{N}$, and $abc = 1$, prove that:
$$S = \frac{2}{(a+1)^2+b^2+1} + \frac{2}{(b+1)^2+c^2+1} + \frac{2}{(c+1)^2+a^2+1} \le 1$$
Here is my try:
$$\begin{align}\frac{2}{(a+1)^2+(b^2+1)} & \le \frac{1}{\sqrt{(a+1)^2 \cdot (b^2+1)}}\\
& = \frac{1}{(a+1) \cdot \sqrt{b^2+1}}\\
& \le \frac{1}{(a+1) \cdot b}\\
& \le \frac{1}{ab}
\end{align}$$
So,
$$S \le \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} = \frac{a+b+c}{1}$$
Can you help me to continue it, please? Thanks!
|
This is a problem from Olympiad Inequalities, Thomas J. Mildorf, 2005 (short but wonderful reading with many different ideas, a true gem). The idea presented here (replacing $a,b,c$ with $x/y,y/z,z/x$) is often used to normalize the inequality in case of $abc=1$.
|
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|
permutations/combinations with repeated symbols I've been learning some about counting and basic combinatorics. But some scenarios were not explained in my class...
Example problem: You are given 6 tiles. 1 is labeled "1", 2 are labeled "2", and 3 are labeled "3".
Problem 1: How many different ways can you arrange groups of three tiles (order matters)?
Answer 1: 19, {1,2,2} {1,2,3} {1,3,2} {1,3,3} {2,1,2} {2,1,3} {2,2,1} {2,2,3} {2,3,1} {2,3,2} {2,3,3} {3,1,2} {3,1,3} {3,2,1} {3,2,2} {3,2,3} {3,3,1} {3,3,2} {3,3,3}.
Question 1: You can't use the normal $\frac{5!}{\left(5-3\right)!}$ because you have repeated symbols. And you can't divide by the factorials of repeated symbols $\frac{6!}{\left(6-3\right)!\cdot3!\cdot2!}$ like you can when n = k and you make different arrangements of all n of symbols $\frac{n!}{\prod_{i=1}^mn_i}$. What formula would be used to find the answer to a more complex version of this problem?
Problem 2: How many different ways can you make groups of three tiles (order doesn't matter)?
Answer 2: 6, {1,2,2} {1,2,3} {1,3,3} {2,2,3} {2,3,3} {3,3,3}.
Question 2: You can't use the normal $\frac{n!}{\left(n-k\right)!\cdot k!}$ because you have repeated symbols. What formula would be used to find the answer to a more complex version of this problem?
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Not sure this is the best method, but the way I would actually solve such a set of questions would be the following:
Use generating function methods for the second question.
The explicit collection of submultisets of the six tiles is given by the terms of $(1+x)(1+y+y^2)(1+z+z^2+z^3)$ (the $x$ terms correspond to tile 1, $y$ terms to copies of tile 2, and $z$ terms to copies of tile 3), and the numbers of submultisets of a given size are given ignoring the labels of the tiles and just multiplying the polynomials
$$(1+x)(1+x+x^2)(1+x+x^2+x^3)=(1+2x+2x^2+x^3)(1+x+x^2+x^3)$$
$$= 1+(1+2)x+(1+2+2)x^2+(1+2+2+1)x^3+(2+2+1)x^4+(2+1)x^5+x^6$$
$$=1+3x+5x^2+6x^3+5x^4+3x^5+x^6.$$
Looking at the coefficient of $x^3$ in the product tells us that there are 6 submultisets.
Now the question of ordered triples drawn from the multiset is a little more difficult. I think I'd work out the explicit submultisets of size 3 first. I.e. solve question 2, then compute the number of ways to order each submultiset. There are a couple algorithmic methods to compute the submultisets of size 3. The first would be to multiply out the polynomials $(1+x)(1+y+y^2)(1+z+z^2+z^3)$ and compute the degree 3 terms. The others are essentially equivalent, but drop the mention of polynomials, and I find the polynomials are a helpful computational aide.
Now because I only want the degree three terms, I'll only compute those.
We have $$[(1+x)(1+y+y^2)(1+z+z^2+z^3)]_3$$
$$=1[(1+y+y^2)(1+z+z^2+z^3)]_3 + x[(1+y+y^2)(1+z+z^2+z^3)]_2$$
$$=1(z^3+yz^2+y^2z) + x(z^2+yz+y^2)$$
$$=z^3+yz^2+y^2z+xz^2+xyz+xy^2,$$
corresponding to submultisets
$\newcommand{\multset}[1]{\{\{{#1}\}\}}\multset{3,3,3},\multset{2,3,3},\multset{2,2,3},\multset{1,3,3},\multset{1,2,3},$ and $\multset{1,2,2}$.
Now we can work out how many ways there are to order each multiset in the usual manner: the ways to order a multiset of the form $\multset{a,a,a}$ is $3!/3!=1$, $\multset{a,a,b}=3!/2!=3$, and $\multset{a,b,c}=3!=6$. Thus summing the number of ways to order each of our multisets, we get $1+3+3+3+6+3 = 19$ different ordered triples drawn from our multiset.
|
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|
If $AB=BA$, prove that $ A=\begin{bmatrix} a & 0 \\ 0 & a \end{bmatrix} $
Let $A$ a $2\times2$ matrix, if $AB=BA$ for every $B$ of the size $2\times2$, Prove that:
$$
A=\begin{bmatrix} a & 0 \\ 0 & a \end{bmatrix}
$$
$a \in \mathbb{R}$
My attempt:
Let $$
A=\begin{bmatrix} a_1 & b_1 \\ c_1 & d_1 \end{bmatrix}
$$
$$B=\begin{bmatrix} a_2 & b_2 \\ c_2 & d_2 \end{bmatrix}$$
And since $AB=BA$, then
$a_1 a_2 + b_1 c_2 = a_1a_2+b_2 c_1$
So $b_2 c_1=b_1 c_2$
And
$a_1 b_2+b_1 d_2=a_2 b_1+b_2 d_1$
$c_1 a_2+d_1 c_2=c_2 a_1+d_2 c_1$
But what can I do now ?
Thanks :)
|
Note:
$$\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}
\begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix}=
\begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix}
\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \Rightarrow \\
\begin{cases}\require{cancel}\cancel{a_{11}b_{11}}+a_{12}b_{21}=\cancel{b_{11}a_{11}}+b_{12}a_{21}\\
a_{11}b_{12}+a_{12}b_{22}=b_{11}a_{12}+b_{12}a_{22}\\
a_{21}b_{11}+a_{22}b_{21}=b_{21}a_{11}+b_{22}a_{21}\\
a_{21}b_{12}+\cancel{a_{22}b_{22}}=b_{21}a_{12}+\cancel{b_{22}a_{22}}\end{cases}$$
From $(1)$, since $b_{12}$ and $b_{21}$ can be any number, in particular, $b_{12}=0$ and $b_{21}\ne 0$, we get: $a_{12}=0$.
Similarly, for $b_{12}\ne 0$ and $b_{21}=0$, we get $a_{21}=0$.
From $(2)$, since $a_{12}=0$ and $b_{12}$ is an arbitrary number, we get $a_{11}b_{12}=b_{12}a_{22} \Rightarrow a_{11}=a_{22}$.
|
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|
Solve $\tan {(x-\frac{\pi}{4}})=-\tan{(x+\frac{\pi}{2})}$ The question:
Without the use of a calculator, solve for all values of $x$ if $\tan {(x-\frac{\pi}{4}})=-\tan{(x+\frac{\pi}{2})}.$
Using the compound angle formula for solving equations is normally easy, but I stumbled across this problem.
The $LHS$ is easy to expand, but when you apply the compound formula for the $RHS$,
\begin{align}
\tan{(x+\frac{\pi}{2})} & = \frac{\tan{(x)} + \tan{(\frac{\pi}{2})}}{1-\tan{(x)}\cdot\tan{(\frac{\pi}{2})}} \\
\end{align}
You might notice that this is a problem because I cannot evaluate $\tan{(\frac{\pi}{2})}$. So this is what I tried. First I tried writing
\begin{align}
\tan{(x+\frac{\pi}{2})} & = \frac{\sin{(x+\frac{\pi}{2})}}{\cos{(x+\frac{\pi}{2})}} \\
& = \frac{\cos (x)}{\sin (x)}
\end{align}
which I knew was wrong. Anyone know how to get around this?
|
$$\tan \left(x - \frac{\pi}{4}\right) = - \tan \left(x + \frac{\pi}{2}\right)$$
$$\tan \left(x - \frac{\pi}{4}\right) = \tan \left(-x - \frac{\pi}{2}\right)$$
$$x-\frac{\pi}{4}=-x-\frac{\pi}{2}+k\pi,\quad(k\in Z)$$
$$2x=-\frac{\pi}{4}+k\pi$$
$$x=-\frac{\pi}{8}+\frac{k\pi}{2}$$
Valid for any $k\in Z$.
|
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|
What is the next limit equal to? $$\lim_{n\to\infty}(\sqrt{n^2+n}-\sqrt[3]{n^3+1})=?$$
I tried amplifying the hole substraction to form the formula $$a^3-b^3$$ but didn't worked out. Can you help me figure it out?
|
Note that\begin{align}\sqrt{n^2+n}-\sqrt[3]{n^3+1}&=\sqrt[6]{(n^2+n)^3}-\sqrt[6]{(n^3+1)^2}\\&=\frac{(n^2+n)^3-(n^3+1)^2}{\sqrt[6]{(n^2+n)^3}^5+\sqrt[6]{(n^2+n)^3}^4\sqrt[6]{(n^3+1)^2}+\cdots+\sqrt[6]{(n^3+1)^2}^5}\\&=\frac{3n^5+3n^4-n^3-1}{\sqrt[6]{(n^2+n)^3}^5+\sqrt[6]{(n^2+n)^3}^4\sqrt[6]{(n^3+1)^2}+\cdots+\sqrt[6]{(n^3+1)^2}^5}.\end{align}This quotient behaves as $\dfrac{3n^5}{6n^5}$ and therefore its limit is $\dfrac12$.
|
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|
Prove: $\sin\frac{\pi}{20}+\cos\frac{\pi}{20}+\sin\frac{3\pi}{20}-\cos\frac{3\pi}{20}=\frac{\sqrt2}{2}$
Prove: $$\sin\frac{\pi}{20}+\cos\frac{\pi}{20}+\sin\frac{3\pi}{20}-\cos\frac{3\pi}{20}=\frac{\sqrt2}{2}$$
ok, what I saw instantly is that:
$$\sin\frac{\pi}{20}+\sin\frac{3\pi}{20}=2\sin\frac{2\pi}{20}\cos\frac{\pi}{20}$$
and that,
$$\cos\frac{\pi}{20}-\cos\frac{3\pi}{20}=-2\sin\frac{2\pi}{20}\sin\frac{\pi}{20}$$
So,
$$2\sin\frac{2\pi}{20}(\cos\frac{\pi}{20}-\sin\frac{\pi}{20})=\frac{\sqrt2}{2}=\sin\frac{5\pi}{20}$$
Unfortunately, I can't find a way to continue this, any ideas or different ways of proof?
*Taken out of the TAU entry exams (no solutions are offered)
|
$$\sin9^\circ+\cos9^\circ+\sin27^\circ-\cos27^\circ$$
$$=\sin9^\circ+\sin(90-9)^\circ+\sin27^\circ+\sin(27-90)^\circ$$
Observe that $\sin5x=\sin45^\circ$ for $x=9^\circ,81^\circ,27^\circ,-63^\circ$
Now if $\sin5x=\sin45^\circ,$
$5x=180^\circ m+(-1)^m45^\circ$ where $m$ is any integer
$\implies x=36^\circ m+(-1)^m9^\circ$ where $m\equiv0,1,2,3,4\pmod5$
Again, $$\sin5x=5\sin x-20\sin^3x+16\sin^5x$$
$$\implies$$ the roots of $$16s^5-20s^3+5s-\sin45^\circ=0$$
are $x=36^\circ m+(-1)^m9^\circ$ where $m\equiv-1,0,1,2,3\pmod5$
$$\implies\sum_{m=-1}^3\sin(36^\circ m+(-1)^m9^\circ)=\dfrac0{16}$$
$$\implies\sum_{m=0}^3\sin(36^\circ m+(-1)^m9^\circ)=\sin(36^\circ\cdot-1+(-1)^{-1}9^\circ)=-\sin(-45^\circ)=?$$
|
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|
The diophantine equation $5\times 2^{x-4}=3^y-1$ I have this question: can we deduce directly using the Catalan conjecture that the equation
$$5\times 2^{x-4}-3^y=-1$$
has or no solutions, or I must look for a method to solve it. Thank you.
|
From $3^y\equiv1$ mod $5$, we see that $4\mid y$, so writing $y=4z$, we have
$$5\cdot2^{x-4}=3^{4z}-1=(3^{2z}+1)(3^{2z}-1)$$
Since $3^{2z}+1\equiv2$ mod $8$, we can only have $3^{2z}+1=2$ or $10$. We can dismiss the first possibility (since it implies $z=0$, which gives $3^{2z}-1=0$). Thus $3^{2z}+1=10$, so $z=1$ and thus $5\cdot2^{x-4}=10\cdot8$ implies $x-4=4$. We find $(x,y)=(8,4)$ as the only solution.
|
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|
Verifying that $\left[\int_a^b(x-p)\,dx\right]^2\le(b-a)\int_a^b(x-p)^2\,dx$ without Cauchy-Schwarz
For real numbers $a,b,p$, verify that $$\left[\int_a^b(x-p)\,dx\right]^2\le(b-a)\int_a^b(x-p)^2\,dx\tag{1}$$ without using the Cauchy-Schwarz inequality.
Note that this is obviously an immediate consequence of C-S, and I assumed that it had to be shown by integrating directly.
Working backwards, we get $$\left[\frac{b^2-a^2}2-p(b-a)\right]^2\le(b-a)\frac{(b-p)^3-(a-p)^3}3$$ and since $m^3-n^3=(m-n)(m^2+mn+n^2)$, the RHS can be rewritten as $$\frac{(b-a)^2}4(b+a-2p)^2\le\frac{(b-a)^2}3(b^2+ab+b^2+3p^2-3p(b+a))$$ or $$3(b+a-2p)^2\le4((b+a)^2+3p^2-3p(b+a))-4ab$$ which gives $$12p^2-12p(b+a)\le(b-a)^2+12p^2-12p(b+a)$$ or that $$(b-a)^2\ge0$$ which is true.
Any alternative methods?
|
Or note that, if $M:=\frac{1}{b-a}\,\int_a^b\,(x-p)^2\,\text{d}x$, then
$$\int_a^b\,(x-p)^2\,\text{d}x-(b-a)\,M^2=\int_a^b\,\big((x-p)-M\big)^2\,\text{d}x\geq0\,.$$
|
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|
Prove, using first principles, that $\lim\limits_{(x,y)\to(0,0)}{x.y^2=0}$ My try is:
$\left|x.y^2-0\right|<\left|\sqrt{x^2+y^2}.(x^2+y^2)\right|< \delta.\delta^2=\epsilon$
where $0<\sqrt{x^2+y^2}< \delta$ what did I miss here?
|
Writing $f(x,y) = x\cdot y^2$. Let $\varepsilon > 0$, setting $\delta = \varepsilon^{1/3}$, you have that $\sqrt{x^2 + y^2} \leq \delta \implies |f(x,y)- 0| \leq (\sqrt{x^2 + y^2})(\sqrt{x^2 + y^2})^2 \leq \varepsilon^{1/3} \cdot \varepsilon^{2/3} = \varepsilon$.
So, you have proven by constructing it that $\exists \delta, \sqrt{x^2 + y^2} \leq \delta \implies |f(x,y)- 0| \leq \varepsilon$.
|
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|
Evaluating $\frac{1^2}{2^1}+\frac{3^2}{2^2}+\frac{5^2}{2^3}+\cdots$ using sigma notation This question can be solved by method of difference
but I want to solve solve it using sigma notation:
$$\frac{1^2}{2^1}+\frac{3^2}{2^2}+\frac{5^2}{2^3}+\cdots+\frac{(2r +1)^2}{2^r}+\cdots$$
I used the geometric progression summation for the $(1/2)^r$ part, and opened $(2r+1)^2$ to $(4r^2 + 1 + 4r)$. If I now express this in sigma notation, it becomes
$$\frac{4n(n+1)(2n+1)}{6} + n + \frac{4(n\cdot n+1)}{2}$$ but I getting problem while putting upper limit infinity.
Where did I go wrong? Please explain.
Answer = $17$
|
A simple answer to the question
Note that
\begin{multline*}
S\equiv\sum_{n\geq1}\frac{(2n-1)^{2}}{2^{n}}=\sum_{n\geq1}\frac{4n^{2}-4n+1}{2^{n}}=4\sum_{n\geq1}\frac{n^{2}}{2^{n}}-4\sum_{n\geq1}\frac{n}{2^{n}}+\sum_{n\geq1}\frac{1}{2^{n}}\equiv4I_2-4I_1+I_0.
\end{multline*}
since each of the series on the right hand side are convergent.
First, note that $I_0$ is a geometric series with $I_0=1$ (you can find an exposition to geometric series on Wikipedia).
As for $I_1$,
\begin{align*}
I_1 & =\frac{1}{2^{1}}+\frac{2}{2^{2}}+\frac{3}{2^{3}}+\cdots\\
\frac{1}{2}I_1 & =\frac{0}{2^{1}}+\frac{1}{2^{2}}+\frac{2}{2^{3}}+\cdots\\
\frac{1}{2}I_1=I_1-\frac{1}{2}I_1 & =\frac{1}{2^{1}}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+\cdots
\end{align*}
and hence $\frac{1}{2}I_1$ is once again a geometric series with $\frac{1}{2}I_1=1$ so that $I_1=2$.
As for $I_2$, note that
\begin{align*}
I_2 & =\frac{1}{2^{1}}+\frac{4}{2^{2}}+\frac{9}{2^{3}}+\cdots\\
\frac{1}{2}I_2 & =\frac{0}{2^{1}}+\frac{1}{2^{2}}+\frac{4}{2^{3}}+\cdots\\
\frac{1}{2}I_2=I_2-\frac{1}{2}I_2 & =\frac{1}{2^{1}}+\frac{3}{2^{2}}+\frac{5}{2^{3}}+\cdots
\end{align*}
In other words,
$$
\frac{1}{2}I_2=\sum_{n\geq1}\frac{2n-1}{2^{n}}=2I_1-I_0=2\cdot2-1=3.
$$
Putting this all together,
$$
S=4\cdot6-4\cdot2+1=24-8+1=17.
$$
Generalizing the above
We can generalize the approach above. Fix a constant $c\in\mathbb{C}$ with $|c|>1$. For each nonnegative integer $m$, let
$$
\boxed{I_{m}=\sum_{n\geq1}n^{m}c^{-n}}
$$
Note that $I_{0}=(c-1)^{-1}$. If $m>0$, then
\begin{multline*}
\left(c-1\right)c^{-1}I_m=I_m-c^{-1}I_m=\sum_{n\geq1}n^{m}c^{-n}-\sum_{n\geq1}n^{m}c^{-n-1}\\
=\sum_{n\geq1}n^{m}c^{-n}-\sum_{n\geq1}\left(n-1\right)^{m}c^{-n}=\sum_{n\geq1}\left(n^{m}-\left(n-1\right)^{m}\right)c^{-n}\\
=\sum_{n\geq1}\left(n^{m}-\sum_{k=0}^{m}\binom{m}{k}n^{k}\left(-1\right)^{m-k}\right)c^{-n}
=\sum_{n\geq1}\left(\sum_{k=0}^{m-1}\binom{m}{k}n^{k}\left(-1\right)^{m-1-k}\right)c^{-n}\\
=\sum_{k=0}^{m-1}\binom{m}{k}\left(-1\right)^{m-1-k}I_{k}.
\end{multline*}
This yields the recurrence
$$
\boxed{I_{m}=\left(c-1\right)^{-1}c\sum_{k=0}^{m-1}\binom{m}{k}\left(-1\right)^{m-1-k}I_{k}\qquad\text{for }m\geq 1}
$$
Relationship to polylogarithm
Note that $I_{m}$ is related to the polylogarithm:
$$
\operatorname{Li}_{-m}(c^{-1})=\sum_{n\geq1}\frac{\left(c^{-1}\right)^{n}}{n^{-m}}=\sum_{n\geq1}\frac{c^{-n}}{n^{-m}}=\sum_{n\geq1}n^{m}c^{-n}=I_{m}.
$$
Expressing our identity in terms of the polylogarithm,
$$
\boxed{\operatorname{Li}_{-m}(z)=\left(1-z\right)^{-1}\sum_{k=0}^{m-1}\binom{m}{k}\left(-1\right)^{m-1-k}\operatorname{Li}_{-k}(z)\qquad\text{for }m\geq 1\text{ and }0<|z|<1}
$$
There are various other expressions for $\operatorname{Li}_{-m}(z)$
where $m$ is positive and $z$ is complex.
|
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"url": "https://math.stackexchange.com/questions/2864643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Show that $\lim_{m \to \infty} \frac{2^{2m}m!(m+1)!}{(2m+1)!(2m+1)}=0$ After a long way to solve the differential equation $y''+xy'+3y=0$, I arrived at the solution $$y(x) = A_0 \sum_{m=0}^{\infty} (-1)^m \dfrac{2m+1}{2^mm!}x^{2m}+A_1 \sum_{m=0}^{\infty} (-1)^m \dfrac{2^m(m+1)!}{(2m+1)!}x^{2m+1}$$ and now I need to prove that it is convergent for all $x \in \mathbb{R}$ as the book claims.
Applying The Ratio Test, I will need just to show that the two following limits hold: $$\lim_{m \to \infty} \dfrac{2^{2m}m!(m+1)!}{(2m+1)!(2m+1)}=0 \ \ \ \text{and} \ \ \lim_{m \to \infty} \dfrac{(2m+1)!(2m+3)!}{2^{2m+1}[(m+1)!]^2}=0. $$ But after lots of effort I don't know how to prove that the limits are zero and unfortunately I know not much about infinite products as all Real Analysis books teach only infinite sums! So my questions are:
*
*Proving the mentioned limits.
*A self-learn-able book containing lots of theorems and information about infinite products.
--
Added. [after reading @adfriedman's answer]: I have made a mistake regarding the second limit and the correct one is to show that $ \lim_{m \to \infty} \dfrac{(2m+1)!(2m+3)}{2^{2m+1}[(m+1)!]^2}=0$ which holds since $$\dfrac{(2m+1)!(2m+3)}{2^{2m+1}[(m+1)!]^2} = \dfrac{\prod_{k=1}^{2m+3}k}{(2m+2)2^{2m+1}\prod_{k=1}^{m+1}k^2} =\dfrac{\prod_{k=1}^{2m+3}k}{(m+1) \prod_{k=1}^{m+1}(2k)^2} = \dfrac{2m+3}{m+1} \dfrac{1 \times 3 \times \dots \times (2m+1)}{2 \times 4 \times \dots \times (2m+2)} \ge \dfrac{2m+3}{m+1} \dfrac{1}{\prod_{k=1}^{2m+1}(1+1/k)} \to 0$$
|
$y(x) = A_0 \sum_{m=0}^{\infty} (-1)^m \dfrac{2m+1}{2^mm!}x^{2m}+A_1 \sum_{m=0}^{\infty} (-1)^m \dfrac{2^m(m+1)!}{(2m+1)!}x^{2m+1}
$
Use the root test
and Stirling
($n!\approx \sqrt{2\pi n}(n/e)^n$).
$\begin{array}\\
\dfrac{2^m(m+1)!}{(2m+1)!}
&=\dfrac{m+1}{2m+1}\dfrac{2^mm!}{(2m)!}\\
&\approx\dfrac12\dfrac{2^m\sqrt{2\pi m}(m/e)^m}{\sqrt{4\pi m}(2m/e)^{2m}}\\
&=\dfrac12\dfrac1{\sqrt{2}}\left(\dfrac{2m/e}{4m^2/e^2}\right)^m\\
&=\dfrac1{2\sqrt{2}}\left(\dfrac{e}{2m}\right)^m\\
\text{so}\\
\left(\dfrac{2^m(m+1)!}{(2m+1)!}\right)^{1/m}
&\approx\dfrac1{(2\sqrt{2})^{1/m}}\left(\dfrac{e}{2m}\right)\\
&\to 0\\
\end{array}
$
For the first one:
$\begin{array}\\
\dfrac{2m+1}{2^mm!}
&=(2m+1)\dfrac{1}{2^mm!}\\
&\approx(2m+1)\dfrac{1}{2^m\sqrt{2\pi m}(m/e)^m}\\
&=\dfrac{2m+1}{\sqrt{2\pi m}}\dfrac{1}{2^m(m/e)^m}\\
&=\dfrac{2m+1}{\sqrt{2\pi m}}\left(\dfrac{e}{2m}\right)^m\\
\text{so}\\
\left(\dfrac{2m+1}{2^mm!}\right)^{1/m}
&\approx\left(\dfrac{2m+1}{\sqrt{2\pi m}}\right)^{1/m}\left(\dfrac{e}{2m}\right)\\
&\to 0
\qquad\text{since } m^{1/m} \to 1\\
\end{array}
$
|
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|
How to prove $1 + \frac{1}{\sqrt{2}} +... +\frac{1}{\sqrt{n}} <\sqrt{n}\ .\left(2n-1\right)^{1/4} $
Prove that $$1 + \frac{1}{\sqrt{2}} +... +\frac{1}{\sqrt{n}} <\sqrt{n}\ .\Bigl(2n-1\Bigr)^{\frac{1}{4}} $$
My Approach :
I tried by applying Tchebychev's Inequality for two same sets of numbers;
$$1 , \frac{1}{\sqrt{2}} ,... ,\frac{1}{\sqrt{n}}$$
And got , $$\Bigl(1 + \frac{1}{\sqrt{2}} +... +\frac{1}{\sqrt{n}}\Bigr)^2 <n\Bigl(1 + \frac{1}{2} +... +\frac{1}{n}\Bigr) $$
Again I tried by applying Tchebychev's Inequality for another two same sets of numbers;
$$1,\frac{1}{2},...,\frac{1}{n}$$ And got, $$\Bigl(1 + \frac{1}{2} +... +\frac{1}{n}\Bigr)^2 <n\Bigl(1 + \frac{1}{2^2} +... +\frac{1}{n^2}\Bigr)$$
With these two inequities i tried solving further more, but i couldn't. So can you please help me solving this further. And if there is some other approach for this question then please answer that way too.
Thank you.
|
The strict inequality (which only holds for $n\gt1$) can be proved by induction, by showing that
$$\sqrt n\cdot(2n-1)^{1/4}+{1\over\sqrt{n+1}}\lt\sqrt{n+1}\cdot(2n+1)^{1/4}$$
for $n\ge N$ for some $N$ and then checking the base cases up to $N$.
We first rewrite the inductive inequality above as
$$\sqrt{n(n+1)}\cdot(2n-1)^{1/4}\lt(n+1)(2n+1)^{1/4}-1$$
Noting that both sides are positive for $n\ge1$, we can square to the equivalent inequality
$$n(n+1)\sqrt{2n-1}\lt(n+1)^2\sqrt{2n+1}-2(n+1)(2n+1)^{1/4}+1$$
which is certainly true if
$$n(n+1)\sqrt{2n+1}\lt(n+1)^2\sqrt{2n+1}-2(n+1)(2n+1)^{1/4}$$
But this reduces to $2(2n+1)^{1/4}\lt\sqrt{2n+1}$, which simplifies to $16\lt2n+1$. So the inductive inequality holds for $n\ge N=8$. And as luck would have it, the base cases for $n=2$ to $8$ have been checked in Yves Daoust's answer.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Compare $\arcsin (1)$ and $\tan (1)$
Which one is greater: $\arcsin (1)$ or $\tan (1)$?
How to find without using graph or calculator?
I tried using $\sin(\tan1)\leq1$, but how do I eliminate the possibility of an equality without using the calculator?
|
Here is another method. The only real disadvantage is that it requires calculating $e^{-1}$ in decimal form, but only two digits are needed, so this can be done by hand easily. It does not require using very large fractions or remembering the series for $\tan(x)$.
As in another answers, we want to find an upper bound for $\sin(1)$ and a lower bound for $\cos(1)$ so that we can show that $\tan(1) < \frac{\pi}{2} = \sin^{-1}(1)$. From there, given the monotonicity of $\sin^{-1}(x)$, the result follows easily.
$\sin(1)$ can be expanded as
$$ \sin(1) = \frac{1}{2}\left[ \sin(1) + \sinh(1) \right] + \frac{1}{2} \left[ \sin(1) - \sinh(1) \right] $$
The first half represents the positive terms in the Maclaurin series for $\sin$, and the second term represents the negative terms. On the second half, just keep the first term:
$$ \frac{1}{2} \left[ \sin(1) - \sinh(1) \right] = - \sum_{n=0}^\infty \frac{1}{(4n+3)!} \le -\frac{1}{6} $$
Plugging this into the first equation and solving for $\sin(1)$ above, we get
$$ \begin{align} \sin(1) &\le \sinh(1) - \frac{1}{3} \\ &= \frac{e-e^{-1}}{2} - \frac{1}{3} \\ &\le \frac{2.72 - 0.36}{2} - 0.333 \\ &= 0.847 \end{align} $$
Now consider the Maclaurin series for $\cos(1)$. The terms are alternating and decrease in magnitude, so for a lower bound, we can terminate after any negative term. Keep the first four terms:
$$ \begin{align} \cos(1) & \ge 1 - \frac{1}{2} + \frac{1}{24} - \frac{1}{720} \\ & \ge 1 - \frac{1}{2} + \left( \frac{1}{24} - \frac{1}{600} \right) \\ & = 1 - \frac{1}{2} + \frac{1}{25} \\ &= 1 - 0.5 + 0.04 \\ & = 0.54 \end{align} $$
This gives us
$$ \tan(1) \le \frac{0.847}{0.54} < 1.57 < \frac{\pi}{2} = \sin^{-1}(1) $$
Of course, it would have taken me a really long time to come up with this without the use of a calculator for exploration.
|
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"timestamp": "2023-03-29T00:00:00",
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|
order of the splitting field of $x^5 +x^4 +1 $? what is the order of the splitting field of
$x^5 +x^4 +1 = (x^2 +x +1)( x^3 +x+1)$ over $\mathbb{Z_2}$
i thinks it will $6$ because $2.3 = 6$
Pliz help me...
|
Because $x^5 +x^4 +1 = (x^2 +x +1)( x^3 +x+1)$ and the factors are irreducible, the splitting field of $x^5 +x^4 +1$ has degree at least $2 \cdot 3 = 6$.
On the other hand, the splitting field of $x^5 +x^4 +1$ is contained in $\mathbb F_{2^6}$, an extension of degree $6$. Hence this is the splitting field. Indeed, WA tells us that
$$
x^{2^6}-x = x (x + 1) (x^2 + x + 1) (x^3 + x + 1) \cdots
$$
|
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|
Evaluate $\lim_{x\rightarrow0}\left[x\left(\frac{1+3x}{1+2x}\right)^{\frac {1}{x}}\right]$ I want to evaluate the following limit:
$$\lim_{x\rightarrow0}\left[x\left(\frac{1+3x}{1+2x}\right)^{\frac {1}{x}}\right]$$
We have
$$\left(\frac{1+3x}{1+2x}\right)^{\frac 1 x} = e^{\frac{\log\left(\frac{1+3x}{1+2x}\right)}{x}}\equiv e^{h(x)}$$
$$h(x)= {\frac{\log\left(\frac{1+3x}{1+2x}\right)}{x}}$$
The argument of $\log\left(\frac{1+3x}{1+2x}\right)$ tends to 1, therefore:
$$\log\left(\frac{1+3x}{1+2x}\right)\sim\frac{1+3x}{1+2x}-1=\frac{x}{1+2x}\sim\frac {x}{2x}=\frac 1 2$$
This means that
$$h(x)\sim\frac 1 2 \cdot \frac 1 x = \frac 1 {2x}$$
We finally have for $x\rightarrow0$:
$$e^{h(x)}=e^{\frac {1}{2x}}\rightarrow+\infty$$
This should lead to an indeterminate form $[0\cdot\infty]$. Any hints on how to evaluate that limit?
|
Recall that $x\to 0$ implies $1+2x\sim 1$. Hence you should have
$$\log\left(\frac{1+3x}{1+2x}\right)\sim\frac{1+3x}{1+2x}-1=\frac{x}{1+2x}\sim\frac{x}{1}=x.$$
and therefore
$$\lim_{x\rightarrow0}\left(\frac{1+3x}{1+2x}\right)^{\frac {1}{x}}=e^1.$$
|
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|
Prove that every natural number $n>15$ there exist natural numbers $x,y\geqslant1$ which solve the equation. Prove that every natural number $n>15$ exist Natural numbers $x,y\geqslant1$ which solve the equation $3x+5y=n$.
so i try induction.
base case is for $n=16$.
so $\gcd(5,3)=1$, after Euclidean algorithm i found:
$3(32-5t)+5(-16+3t)=16$ so i found that $32-5t>1$c for $x$ and $-16+3t>1$ for $y$ and for $t=6$ $x=2$ and $y=2$ .
now suppose its takes place for $n$.
how i show that for $n+1$?
if there is more elegant way i would love to see.
|
If $x,y$ is a solution, you have $3x+5y=n=n(-3\cdot3+2\cdot5)$ and therefore
$$3(x+3n)= 5(2n-y)$$
As $3,5$ are coprime it exists $k \in \mathbb Z$ such that $x=5k-3n$ and $y= 2n-3k$.
$x >1$means $5k> 1+3n$ and $y>1$ implies $3k<2n-1$.
Which is equivalent to
$$3+9n < 15k < 10n-5 \tag{1}$$
The difference of $10n-5$ and $9n+3$ is equal to $n-8$. For $n-8>15$, i.e $n>23$, finding $k$ satisfying $(1)$ is obviously always possible. Remain the cases $n=16, \dots , 23$ that can be evaluated one by one.
Example of $n=5847$
With this analysis, it is easy to find the solution. We have $$9\cdot 5847+3=52626<15\cdot 3509 =52635 <58465.$$
Hence a solution is $x=5\cdot 3509-3\cdot 5847=4$ and $y=2\cdot 5847-3\cdot 3509=1167$.
We can verify that $3 \cdot 4+ 5\cdot 1167=5847$.
|
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|
Finding n-th term of a matrix I have to find the n-th power of the following matrix $$A=\begin{pmatrix} 2 & 0&1\\ 0 & -3& 0 \\ 1 & 0&2 \end{pmatrix}
$$ I computed a few powers $$A^2=\begin{pmatrix} 4+1& 0&4\\ 0 & 9& 0 \\ 4 & 0&1+4\end{pmatrix}$$
$$A^3 =\begin{pmatrix} 14 & 0&13\\ 0 & -27& 0 \\ 13& 0&14\end{pmatrix}$$
$$A^n=\begin{pmatrix} \frac12(3^n+1) & 0&\frac12(3^n-1)\\ 0 &( -3)^n& 0 \\ \frac12(3^n-1) & 0&\frac12(3^n+1)\end{pmatrix}$$ And is just induction left which is easy, but I found by check and try the n-th term, is it possible to find it in a proper way?
|
You could use Diagonalization.
Let $A$ be a matrix then if you diagonalize $A$ then you get $A=PDP^{-1}$ with $B$ as a diagonal matrix and $P^{-1}$ an invertible matrix. If you try for $A^2$, then it is equal to $A^2=PD^2P^{-1}$. Similarly for $A^n=PD^nP^{-1}$
Here are the steps:
$1.$Find the characteristic polynomial $p(t)$ of $A$.
$2.$Find eigenvalues $λ$ of the matrix $A$ and their algebraic multiplicities from the characteristic polynomial $p(t)$.
$3.$For each eigenvalue $λ$ of $A$, find a basis of the eigenspace $E_λ$.
If there is an eigenvalue $λ$ such that the geometric multiplicity of $λ$, dim$(E_λ)$, is less than the algebraic multiplicity of $λ$, then the matrix $A$ is not diagonalizable. If not, $A$ is diagonalizable, and proceed to the next step.
$4.$If we combine all basis vectors for all eigenspaces, we obtained n linearly independent eigenvectors $v1,v2,…,vn$.
$5.$Define the nonsingular matrix $P=[v_1v_2…v_n]$.
$6.$Define the diagonal matrix $D$, whose $(i,i)$-entry is the eigenvalue λ such that the $i^{th}$ column vector $v_i$ is in the eigenspace $E_λ$
$7.$Then the matrix $A$ is diagonalized as $P^{-1}AP=D$
Eigenvalues $$\begin{vmatrix} 2-\lambda & 0 & 1 \\ 0 & -3-\lambda & 0 \\ 1 & 0 & 2-\lambda \end{vmatrix}=0$$$$\lambda=-3,1,3$$
By using the eigenvalues you need to find the eigenvectors, then we get $$P=\begin{bmatrix} 1 & -1 & 0 \\ 0 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix}$$and $P^{-1}=\begin{bmatrix}\dfrac12&0&\dfrac12\\ -\dfrac12&0&\dfrac12\\ 0&1&0\end{bmatrix}$$$D=P^{-1}AP$$
$$D=\begin{bmatrix}\dfrac12&0&\dfrac12\\ -\dfrac12&0&\dfrac12\\ 0&1&0\end{bmatrix}\begin{bmatrix}2&0&1\\ \:0&-3&0\\ \:1&0&2\end{bmatrix}\begin{bmatrix} 1 & -1 & 0 \\ 0 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix}=\begin{bmatrix}3&0&0\\ 0&1&0\\ 0&0&-3\end{bmatrix}$$Now you can find $$A^n=PD^nP^{-1}$$$$A^n=\begin{bmatrix} \frac12(3^n+1) & 0&\frac12(3^n-1)\\ 0 &( -3)^n& 0 \\ \frac12(3^n-1) & 0&\frac12(3^n+1)\end{bmatrix}$$
|
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|
Differential equation: $\dfrac{dy}{dx}= \dfrac{(x+y)^2}{(x+2)(y-2)}$
The solution of $\dfrac{dy}{dx}= \dfrac{(x+y)^2}{(x+2)(y-2)}$ is given by:
a) $(x+2)^4 (1+\frac{2y}{x})= ke^{\frac{2y}{x}}$
b) $(x+2)^4 (1+ 2\frac{(y-2)}{x+2})= ke^{\frac{2(y-2)}{x+2}}$
c) $(x+2)^3 (1+ 2\frac{(y-2)}{x+2})= ke^{\frac{2(y-2)}{x+2}}$
d) None of these
Attempt:
I have expanded and checked but couldn't spot any exact differentials.
Secondly, it's not a homogeneous equation, so couldn't use $y = vx$.
How do I go about solving this problem?
|
$$\dfrac{dy}{dx}= \dfrac{(x+y)^2}{(x+2)(y-2)}=\left(\dfrac{x+y}{x+2}\right)^2\dfrac{x+2}{y-2}$$
let $w=\dfrac{x+y}{x+2}$ then $y=w(x+2)-x$ and
$$w'(x+2)+w-1=y'=w^2\dfrac{1}{w-1}$$
which is separable
$$\dfrac{w-1}{2w-1}dw=\dfrac{dx}{x+2}$$
with integration the solution will be obtained.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Geometric intuition for the complex shoelace formula The complex shoelace formula for the signed area of a triangle with vertices given by the complex numbers $a, b, c$ is $$\frac{i}{4}
\begin{vmatrix}
1 & 1 & 1 \\
a & b & c \\
\overline{a} & \overline{b} & \overline{c} \\
\end{vmatrix}
$$
I have seen the algebraic proof for this formula using elementary row operations and the multilinear nature of the determinant, however that style of proof appears to imply that the simplicity of the final result (i.e. a determinant involving only conjugates) is a coincidence; furthermore it provides little intuition.
I am looking for a way to understand this formula through a geometric/intuitive argument (not necessarily rigorous) in order to gain a deeper understanding rather than just accepting that the algebra works out.
|
Consider the case $\,a=0\,$, first, where the proposed formula reduces to:
$$
S_{OBC} = \frac{i}{4}
\begin{vmatrix}
1 & 1 & 1 \\
0 & b & c \\
\overline{0} & \overline{b} & \overline{c} \\
\end{vmatrix}
=
\frac{i}{4}
\begin{vmatrix}
b & c \\
\overline{b} & \overline{c} \\
\end{vmatrix}
= \frac{i}{4}\left(b \bar c - \bar b c\right)
= -\,\frac{1}{2}\,\operatorname{Im}(b \bar c) \tag{1}
$$
Let $\,b = |b| e^{i \beta}\,$, $\,c = |c| e^{i\gamma}\,$, then $\,b \bar c = |b|\cdot|c|\cdot e^{i(\beta-\gamma)}=\color{blue}{|b|}\cdot\color{blue}{|c|}\cdot \left(\cos(\beta-\gamma)+ i \color{red}{\sin(\beta-\gamma)}\right)\,$, so $\,(1)\,$ is equivalent to the familiar triangle (signed) area formula $\,S_{OBC} = \frac{1}{2}\,\color{blue}{OB} \cdot \color{blue}{OC} \cdot \color{red}{\sin \widehat{BOC}}\,$.
But area is, of course, invariant under translations, so it follows that in the general case:
$$
S_{ABC} = \frac{1}{2}\,AB \cdot AC \cdot \sin \widehat{BAC} = -\,\frac{1}{2}\,\operatorname{Im}\left((b-a) \overline{(c-a)}\right) \tag{2}
$$
The latter is equivalent to the posted formula via elementary column operations:
$$
\begin{vmatrix}
1 & 1 & 1 \\
a & b & c \\
\overline{a} & \overline{b} & \overline{c} \\
\end{vmatrix}
\;=\;
\begin{vmatrix}
1 & 0 & 0 \\
a & b-a & c-a \\
\overline{a} & \overline{b-a} & \overline{c-a} \\
\end{vmatrix}
$$
|
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|
Proving radius of circle $\dfrac{\triangle}{a}\tan^2\dfrac{A}{2}$
If a circle be drawn touching the inscribed and circumscribed circles of a $\triangle ABC$ and the side $BC$ externally, prove that its radius is: $$r=\dfrac{\triangle}{a}\tan^2\dfrac{A}{2}$$
I tried using triangle formed by circumcenter, incenter and center of above circle as I know all the sides in terms of $r$ to use cosine rule but I don't know any angles.
Please help!
|
The center $J$ of the wanted circle can be constructed by intersecting a line and a parabola, since by setting $s=JL$ we have $JO=R-s$.
If we take $B$ as the origin and $BC$ as the $x$-axis, the equation of the wanted parabola is $y=kx(x-a)$. Since the distance of $O$ from $BC$ is $R\cos A$, the vertex lies at $\left(\frac{a}{2},-\frac{R}{2}(1-\cos A)\right)$ and
$$k=\frac{2R}{a^2}(1-\cos A)=\frac{1-\cos A}{2R\sin^2 A}=\frac{\sin^2\frac{A}{2}}{R\sin^2 A}=\frac{1}{2R\cos^2\frac{A}{2}}.$$
Of course $BL=\frac{a+c-b}{2}$, hence
$$ s = \frac{1}{2R\cos^2\frac{A}{2}}\cdot \frac{a+c-b}{2}\cdot\frac{a-c+b}{2} $$
and I guess you can take it from here, by just recalling that $abc=4R\Delta$ and $2\sin^2\frac{A}{2}=1-\cos A=1-\frac{b^2+c^2-a^2}{2bc}$.
|
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|
Proof: homogeneous second order differential equation This is an homogeneous second order differential equation
$Ay'' + By' + Cy = 0$
Its solution can be retrieved assuming $y = e^{rx}$: in this way, we will have
$y = e^{rx}$
$y' = re^{rx}$
$y'' = r^{2}e^{rx}$
replace:
$Ar^{2}e^{rx} + Bre^{rx} + Ce^{rx} = 0$
$(e^{rx}) (Ar^{2} + Br + C) = 0$
The last is a second degree equation. One can manifest a case in which the roots of this equation are real and coincident. If $r_1 = r_2$, a solution will be the function $y_1 = e^{rx}$ surely. At this point, we must build a linearly independent second solution: we take, for example, $y_2 = xe^{rx}$.
I should prove that $y_2 = xe^{rx}$ is a solution as well: how can I do?
Thank you in advance
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The_lost is right in his comment, we show $xe^{rx}$ solves the equation
$Ay''(x) + By'(x) + Cy(x) = 0, \; A \ne 0, \tag 0$
by simple substitution (here I assume $A \ne 0$ to ensure (0) is a bona-fide second order equation); however, it helps to know something about $r$ first, to wit:
According to the quadratic formula, the roots of
$Ar^2 + Br + C = 0, \; A \ne 0, \tag 1$
are
$r_1, r_2 = \dfrac{-B \pm \sqrt{B^2 - 4AC}}{2A}; \tag 2$
we see that there is one root
$r = r_1 = r_2 = -\dfrac{B}{2A} \tag 3$
precisely when
$B^2 - 4AC = 0, \tag 4$
or, since $A \ne 0$,
$C = \dfrac{B^2}{4A}, \tag 5$
or
$\dfrac{C}{A} = \dfrac{B^2}{4A^2} = \left ( \dfrac{B}{2A} \right )^2; \tag 6$
for then,
$Ar^2 + Br + C = A \left (r^2 + \dfrac{B}{A} r + \dfrac{C}{A} \right )$
$= A \left ( r^2 + \dfrac{B}{A} + \left ( \dfrac{B}{2A} \right )^2 \right ) = A \left ( r + \dfrac{B}{2A} \right )^2; \tag 7$
now if
$y(x) = x e^{rx}, \tag 8$
we calculate:
$y'(x) = e^{rx} + rxe^{rx}; \tag 9$
$y''(x) = re^{rx} + re^{rx} + r^2xe^{rx} = r^2xe^{rx} + 2re^{rx}, \tag{10}$
whence, via (3),
$y''(x) + \dfrac{B}{A}y'(x) + \dfrac{C}{A} y(x) = r^2xe^{rx} + 2re^{rx} + \dfrac{B}{A}(rxe^{rx} + e^{rx}) + \dfrac{C}{A} x e^{rx}$
$= \left (r^2 + \dfrac{B}{A}r + \dfrac{C}{A} \right ) xe^{rx} + \left (2r + \dfrac{B}{A} \right )e^{rx}$
$= \left ( r + \dfrac{B}{2A} \right )^2 xe^{rx} + 2 \left (r + \dfrac{B}{2A} \right ) e^{rx} = 0; \tag{11}$
it then follows that
$Ay''(x) + By'(x) + Cy(x) = A \left (y''(x) + \dfrac{B}{A}y'(x) + \dfrac{C}{A} y(x) \right ) = 0 \tag{12}$
as well.
|
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|
Evaluate $f(x)=x\log{\left|\frac{x+2}{3-x}\right|}$ for $x\rightarrow\infty$ I have the following function:
$$f(x)=x\log{\left|\frac{x+2}{3-x}\right|}$$
I want to find the limit for $x\rightarrow+\infty$.
This is what I do. Since $x>=0$, I can remove the absolute value:
$$f(x)=x\log\left({\frac{x+2}{3-x}}\right)\sim x\left( \frac{x+2}{3-x}-1\right)=x\left(\frac{2x-1}{3-x}\right)=x\left(\frac{2x}{-x}\right)=-2x\rightarrow-\infty$$
The textbook reports that the limit is actually $5$. Why is my solution wrong?
|
Let $x>4$;
$I:= \log\dfrac {|x+2|}{|x-3|}= {\displaystyle \int_{x-3}^{x+2}}(1/t)dt.$
MVT:
$I = (1/r){\displaystyle \int_{x-3}^{x+2}}dt=$
$(1/r)[(x+2)-(x-3)]=5/r,$
where $r \in [x-3,x+2]$.
Hence:
$5\dfrac {x}{x+2} \le xI \le 5\dfrac{x}{x-3}$.
Squeeze :
$\lim_{x \rightarrow \infty} xI =5$.
|
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|
Sketching complex plane Sketch all points in the complex plane such that $Re(\frac{1}{z})=1$
I managed to solve it $(\frac{1}{z})= \frac{|z|}{z|z|}=\frac{(x - iy)}{(x^2+y^2)}$
Meaning that $Re(\frac{1}{z})= \frac{x}{(x^2+y^2)}$
How should I go afterwards?
|
For an alternative approach, use that $\,\operatorname{Re}(w)= \dfrac{1}{2}\left(w + \bar w\right)\,$ for complex $\,w\,$, then (assuming $\,z \ne 0\,$, otherwise $\,\dfrac{1}{z}\,$ is not even defined):
$$
\begin{align}
1=\operatorname{Re}\left(\frac{1}{z}\right)=\frac{1}{2}\left(\frac{1}{z}+\frac{1}{\bar z}\right) = \frac{z+\bar z}{2 z \bar z } \;\;&\iff\;\; z \bar z - \frac{z}{2} - \frac{\bar z}{2} \color{red}{+\frac{1}{4}-\frac{1}{4}} = 0 \\[5px]
&\iff\;\; \left(z-\frac{1}{2}\right)\left(\bar z-\frac{1}{2}\right) = \frac{1}{4} \\[5px]
&\iff\;\; \left|z - \frac{1}{2}\right|^2=\frac{1}{4} \\[5px]
&\iff\;\; \left|z - \frac{1}{2}\right| = \frac{1}{2}
\end{align}
$$
The latter defines the set of points at constant distance $\,\dfrac{1}{2}\,$ from fixed point $\,\dfrac{1}{2}\,$, which is of course the circle of radius $\,\dfrac{1}{2}\,$ centered at $\,\dfrac{1}{2}\,$. However, the origin is excluded because $\,z \ne 0\,$, so the end answer is the same circle but "punctured" at $\,0\,$.
|
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|
Given a triangle whose vertex $A$ is bisecting the base (median), find the angle $X$ Given a triangle whose vertex $A$ is bisecting the base (median), I have to find the angle $X$.
First Approach
From $\Delta AHD$ we can get $AH = HD$. And it will make $45^\circ$ at vertex $A^*$. From $\Delta ABH$, let's Consider angle at $B$ is $\theta$: $\tan \theta = \frac{AH}{BH}$.
From here, I am not able to think forward.
Assumption: If somehow I can prove that $BH = HD$, then $\Delta ABH$ will also be an isosceles triangle. Therefore total angle at vertex $A$ will be $45^\circ + 45^\circ = 90^\circ$.
Possible Second Approach
Assuming angle at $B$ is $Y$, from $\Delta ABD$,
$X + Y + 45^\circ = 180^\circ\\X+Y=135^\circ \qquad(I).$
From $\Delta ABC$: $(X + 15) + Y + 30^\circ = 180^\circ\\X+Y=135^\circ \qquad(II)$
So, from above two equations I am not able to find $X$ and $Y$ (Since, Both are same)
If I can get another linear equation in terms of $X$ and $Y$. By solving those equation, I can get angle at '$A$'
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$\require{cancel}$
Using completely trigonometry
Suppose $AC=q$, let us solve for all the sides in terms of $q$:
$$\frac{\sin(135^\circ)}{q}=\frac{\sin(30^\circ)}{AD}\implies AD=\frac{q}{\sqrt2}$$
$$\frac{\sin(15^\circ)}{CD}=\frac{\sin(135^\circ)}{q}\implies\,CD=\frac{q(\sqrt3-1)}{2}$$
Since $CD=BD$, then $BC=q(\sqrt3-1)$. Then we need to find $AB$:
$$AB^2=q^2+\left(q\left(\sqrt3-1\right)\right)^2-2q^2(\sqrt3-1)\cos(30^\circ)\implies AB=q\sqrt{2-\sqrt{3}}$$
Then we just need to solve for $x$ in:
$$\frac{\sin x}{\frac{q(\sqrt3-1)}2}=\frac{\sin 45^\circ}{q\sqrt{2-\sqrt3}}\implies \sin x=\frac{\sqrt3-1}{2\sqrt{4-2\sqrt3}}$$
The nested radical in the bottom looks ugly, but remember that the radical $\sqrt{a\pm b\sqrt{c}}=\sqrt d\pm\sqrt e$ if and only if $\sqrt{a^2- b^2c}\,\,$ is rational.
Therefore, we have:
$$\sin x=\frac{\cancel{\sqrt3-1}}{2\cancel{(\sqrt3-1)}}=\frac12 \implies$$
$$\bbox[10px, border: 2px black solid]{\therefore x=\arcsin{\frac12}=\frac\pi6=30^\circ}$$
|
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|
Arrangement of $15$ letters with restrictions on which letters are permitted in which positions How many $15$ letter arrangements of $5$ As, $5$ Bs and $5$ Cs have no As in the first $5$ positions, no Bs in the next $5$ positions, and no Cs in the last $5$ positions?
I am not able to frame the equation. I presumed that we fix the first $5$ positions if arrangement is $10 \times 9 \times 8 \times 7 \times 6 \times 5$
I am not able to arrange the middle $5$ positions as some element of C must be overlapping.
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If we choose which of the middle five positions are occupied by $A$s, the others must be occupied by $C$s (since no $B$s are permitted in the middle five positions). If we then choose which of the last five positions are occupied by the remaining $A$s, the remainder must be occupied by $B$s (since no $C$s are permitted in the last five positions). If we then choose which of the first five positions are occupied by the remaining $B$s, the remainder must be occupied by $C$s.
If $k$ of the middle five positions are occupied by $A$s (with the remaining $5 - k$ positions occupied by $C$s), $5 - k$ of the last five positions are occupied by $A$s with the remaining $k$ occupied by $B$s, which means $5 - k$ of the first five positions are occupied by $B$s (and the remaining $k$ are occupied by $C$s). Since choosing which of the middle five positions are occupied by $A$s, which of the last five positions are occupied by $A$s, and which of the first five positions are occupied by $B$s completely determines the sequence, the number of admissible arrangements is
$$\sum_{k = 0}^{5} \binom{5}{k}\binom{5}{5 - k}\binom{5}{5 - k} = \binom{5}{0}\binom{5}{5}^2 + \binom{5}{1}\binom{5}{4}^2 + \binom{5}{2}\binom{5}{3}^2 + \binom{5}{3}\binom{5}{2}^2 + \binom{5}{4}\binom{5}{1}^2 + \binom{5}{5}\binom{5}{0}^2$$
|
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|
What is the radius of the black circle tangent to all three of these circles? The red, blue, and green circles have diameters 3, 4, and 5, respectively.
What is the radius of the black circle tangent to all three of these circles?
I just figured out the radius is exactly $\dfrac{72}{23}$ but I don't know how to do the solution.
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It is simply the Apollonius Problem and solved by the following three quadratic equations using WolframAlpha for center $ (x,y) $ and radius $r$ of the biggest circle, which is tangent to other three circles;
(1) $ x^2 + (y - \frac{3}{2})^2 = (r - \frac{3}{2})^2 $, circle on side of length $ = 3 $,
(2) $ (x - 2)^2 + y ^ 2 = (r - 2)^2 $, circle on side of length $ = 4 $,
(3) $ (x - 2)^2 + (y - \frac{3}{2})^2 = (r - \frac{5}{2})^2 $, circle on side of length $ = 5 $.
The solution is $ x = \frac{36}{23}, y = \frac{24}{23} $ and radius $ r $ is the Answer $ = \frac{72}{23} $.
|
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|
How to calculate $\int_{0}^{\pi/2}\dfrac{1}{\sin^3 x + \cos^3 x}dx$ One way that I tried to solve an integral was to divide integrand by $\cos^3x$ para para obter
$$
\int_{0}^{\pi/2}\dfrac{\sec^3x}{1 + \tan^3x}dx
$$
Through software, I saw that the result involves $\tanh^{-1}\bigl(\frac{1}{\sqrt{2}}\bigr)$, but I do not know how to get there.
|
Let $\tan\frac{x}{2}=t$.
Thus, $$\int\limits_0^{\frac{\pi}{2}}\frac{1}{\sin^3x+\cos^3x}dx=\int\limits_0^1\frac{2}{\left(\left(\frac{2t}{1+t^2}\right)^3+\left(\frac{1-t^2}{1+t^2)}\right)^3\right)(1+t^2)}dt=$$
$$=2\int\limits_0^1\frac{(1+t^2)^2}{(2t+1-t^2)((2t)^2-2t(1-t^2)+(1-t^2)^2)}dt.$$
Now, $$2t+1-t^2=-(t^2-2t+1-2)=-(t-1-\sqrt2)(t-1+\sqrt2)$$ and
$$(2t)^2-2t(1-t^2)+(1-t^2)^2)=t^4+2t^3+2t^2-2t+1=$$
$$=(t^2+t+k)^2-((2k-1)t^2+2(k+1)t+k^2-1)=$$
$$=(t^2+t+2)^2-((2\cdot2-1)t^2+2(2+1)t+2^2-1)=$$
$$=(t^2+t+2)^2-3(t+1)^2=$$
$$=(t^2-(\sqrt3-1)t+2-\sqrt3)(t^2+(1+\sqrt3)t+2+\sqrt3).$$
Can you end it now?
|
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|
Find the general solution (i) $\cot \theta =-\dfrac {1} {\sqrt {3}}$ (ii)$4\cos ^{2}\theta =1$ my attempt for
(i)
$\left. \begin{array} { l } { \cot ( \theta ) = - \frac { 1 \cdot \sqrt { 3 } } { \sqrt { 3 } \sqrt { 3 } } } \\ { 1 \cdot \sqrt { 3 } = \sqrt { 3 } } \end{array} \right.$
$\cot ( \theta ) = - \frac { \sqrt { 3 } } { 3 }$
(ii)
$\left. \begin{array} { l } { \text { Let: } \cos ( \theta ) = u } \\ { 4 u ^ { 2 } = 1 } \end{array} \right.$
$\left. \begin{array} { l } { \text { Divide both sides by } 4 } \\ { \frac { 4 u ^ { 2 } } { 4 } = \frac { 1 } { 4 } } \end{array} \right.$
is it right way to find general solution for these equations?
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We know that $$\cot\left(-\dfrac{\pi}{3}\right)=-\dfrac{\sqrt 3}{3}$$ and $$\cot( \theta+\pi)=\cot \theta$$similarly $$\cos\left(\dfrac{\pi}{3}\right)=\dfrac{1}{2}$$and $$\cos^2(\theta+\pi)=\cos^2\theta$$therefore $$(i)\qquad x=k\pi-\dfrac{\pi}{3}\\(ii)\qquad x=k\pi\pm\dfrac{\pi}{3}$$
|
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|
Finding limit of $\frac{x^2y^2}{x^3+y^3}$ as $(x,y)\to (0,0)$ Here, if we go along $x=-y$ then the function itself is not defined, and limit seems to exist for all other paths of approach
Putting in another way, if we use polar coords, limit is:
$$\lim_{r\to 0} \frac{r^4 \sin^2\theta\cos^2\theta}{r^3(\sin^3\theta + \cos^3\theta)}$$
which is zero only when $(\sin^3\theta + \cos^3\theta)$ is nonzero, and function itself is not defined for values where $(\sin^3\theta + \cos^3\theta) = 0$
So here, what do we say? Does the limit exist? or does it not exist
(Additional Question)
Suppose that limit is being calculated $(x,y) \to (0,0)$. Then can we choose one of the paths as, say $x=0$, although in limit we only let variables to "tend" to a value? Is it allowed to use this path?
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We have that
*
*$x=y=t\to 0 \implies \frac{x^2y^2}{x^3+y^3}=\frac{t^4}{2t^3}=\frac t{2}\to 0$
*$x=t\to 0 \quad y=-t+t^2\to 0 \implies \frac{x^2y^2}{x^3+y^3}=\frac{t^4-2t^5+t^6}{3t^4-3t^5+t^6}=\frac{1-2t+t^2}{3-3t+t^2}\to \frac13$
therefore the limit doesn't exist.
|
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|
Integrate $\frac{x^2-4x+10}{x^2\sqrt x}$
Find the indefinate integral with respect to $x$ of $$\frac{x^2-4x+10}{x^2\sqrt x}$$
For this problem I first made each individual number in the numerator separate from each other for easier integration and then simplified
$$=\int \left(\frac{x^2}{x^2\sqrt x}-\frac{4x}{x^2\sqrt x}+\frac{10}{x^2\sqrt x}\right)dx$$
$$=\int \left(x^2x^{-\frac{3}{2}}-4x^1x^{-\frac{3}{2}}+10x^{-\frac{3}{2}}\right)dx$$
$$=\int \left(x^{\frac{1}{2}}-4x^{-\frac{1}{2}}+10x^{-\frac{3}{2}}\right)dx$$
I then integrated the expression to get
$$\frac{2x^{\frac{3}{2}}}{3}-8x^{\frac{1}{2}}-20x^{-\frac{1}{2}}$$
This is, however, wrong. Any ideas as to why?
Thanks in advance $:)$
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$$\frac{x^2-4x+10}{x^2\sqrt x}$$
$$\frac{1}{\sqrt{x}}-\frac{4}{x\sqrt{x}}+\frac{10}{x^2\sqrt{x}}$$
$$x^{-1/2}-4x^{-3/2}+10x^{-5/2}$$
So it should be $x^{-5/2}$ instead of $x^{-3/2}$
$$I=\int x^{-1/2}-4x^{-3/2}+10x^{-5/2} {dx}=2x^{1/2}+8x^{-1/2}-\frac{20}{3}x^{-3/2}+C$$
|
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|
Prove that $\frac{\sin x}{x}=(\cos\frac{x}{2}) (\cos\frac{x}{4}) (\cos \frac{x}{8})...$ How do I prove this identity:
$$\frac{\sin x}{x}=\left(\cos\frac{x}{2}\right) \left(\cos\frac{x}{4}\right) \left(\cos \frac{x}{8}\right)...$$
My idea is to let
$$y=\frac{\sin x}{x}$$
and
$$xy=\sin x$$
Then use the double angle identity $\sin 2x=2\sin x \cos x$ and its half angle counterparts repeatedly. I see some kind of pattern, but I can't seem to make out the pattern and complete the proof.
|
Note the fact that
$$
\cos \frac{x}{2^k} = \frac12 \cdot \frac{\sin (2^{1-k} x)}{\sin(2^{-k}x)},
$$
and we have
$$
\prod_{k = 1}^n \cos \frac{x}{2^k} = \frac{1}{2^n} \cdot \frac{\sin x}{\sin(2^{-n}x)} = \frac{2^{-n}x}{\sin(2^{-n}x)} \cdot \frac{\sin x}{x}.
$$
For all $x$, as $n \to \infty$, we have
$$
\lim_{n \to \infty} \prod_{k = 1}^n \cos \frac{x}{2^k}= \frac{\sin x}{x} \cdot\lim_{n \to \infty} \frac{2^{-n}x}{\sin(2^{-n}x)} = \frac{\sin x}{x}.
$$
|
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Did I misuse the inductive hypothesis? Note $\log$ is $\log_2$ in this problem.
I am working on the proof of a lemma that my professor used in class. He said we could verify that it works if we want, but I think I used the inductive hypothesis in such a way that I cannot continue my proof. Either that or I am missing a simple algebraic manipulation.
The use of my inductive hypothesis is based on the fact that I am dealing with inequalities, so $lg(n) \leq lg(n+1)$ for all $n$. Bellow is my work so far.
$\sum_{k=2}^{n-1}k\log k \leq \frac{1}{2}n^2 \log n - \frac{1}{8}n^2$
Base step: $2 \leq \frac{9}{2} \log 3 - \frac{9}{8}$
$2 \leq \frac{9}{2} \log 2 - \frac{9}{8} = \frac{27}{8} \leq \frac{9}{2} \log 3 - \frac{9}{8}$
Assume: $\sum_{k=2}^{n-1}k\log k \leq \frac{1}{2}n^2 \log n - \frac{1}{8}n^2$
Show: $\sum_{k=2}^{n}k\log k \leq \frac{1}{2}(n+1)^2 \log (n+1) - \frac{1}{8}(n+1)^2$
$\sum_{k=2}^{n-1}k\log k + n \log n \leq \frac{1}{2}(n+1)^2 \log (n+1) - \frac{1}{8}(n+1)^2$
$\frac{1}{2}n^2 \log n - \frac{1}{8}n^2 + n \log n \leq \frac{1}{2}(n+1)^2 \log (n+1) - \frac{1}{8}(n+1)^2$
$\require{cancel} \cancel{\frac{1}{2}n^2 \log n} - \cancel{\frac{1}{8}n^2} + n \log n \leq \cancel{\frac{1}{2}n^2 \log (n+1)} + n \log (n+1) + \frac{1}{2} \log (n+1) - \cancel{\frac{1}{8}n^2} - \frac{1}{4}n - \frac{1}{8}$
$n \log n \leq n \log (n+1) + \frac{1}{2} \log (n+1) - \frac{1}{4}n - \frac{1}{8}$
At this point I am not sure how to proceed.
|
If you draw some simple diagrams, given some $f(x) > 0$ and $f'(x) > 0,$ you find
$$ \int_{a-1}^b f(x) dx < \sum_{k=a}^b f(k) < \int_{a}^{b+1} f(x) dx $$
You are using $a=2$ and $b=n-1$
This is especially helpful when we know a closed form indefinite integral of $f$
In particular, using logarithm base $e \approx 2.718,$ we have
$$ \int x \log x \; \; dx = \left( \frac{2 x^2 \log x - x^2}{4} \right) + \mbox{constant} $$
NOT SURE: after your edit, I cannot tell which logarithm you are using.
I would say that the constant $2$ matters here, you are using
$$ \frac{\log x}{\log 2} $$
and need to be a little careful about that:
$$ \int \frac{x \log x}{\log 2} \; \; dx = \left( \frac{2 x^2 \log x - x^2}{4 \log 2} \right) + \mbox{constant} $$
|
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"url": "https://math.stackexchange.com/questions/2904471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Determine all integers $x,\ y,\ z$ that satisfy $x+y+z=(x-y)^{2}+(y-z)^{2}+(z-x)^{2}$
Determine all integers $x,\ y,\ z$ that satisfy
$$x+y+z=(x-y)^{2}+(y-z)^{2}+(z-x)^{2}$$
We expand the RHS and sort the equation by $x$, which yields a quadratic equation with the discriminant:
$$D=-12y^{2}-12z^{2}+24yz+12y+12z+1$$
$x$ is an integer, thus $D$ must be square of an integer. We can write similar equalities by solving the original equation for $y$ and $z$. I couldn't get any further.
My other approaches include considering the greatest common divisor of $x,\ y,\ z$ and write $$x=da,\ y=db,\ z=dc$$
where $d=\gcd(x,y,z)$ and $\gcd(a,b,c)=1$ which wasn't much useful.
|
Let $y=u+x$ and $z=v+x$, then $y-z = u-v$ and hence
$$
\begin{align}
(y-u) + y + (y-u+v) &= u^2 + (u-v)^2 + v^2\\
3y &= 2u^2 + 2v^2 - 2uv +2u - v
\end{align}
$$
So we can almost freely choose $u,v$, but we need RHS to be divisible by $3$. By considering the equation modulo $3$, we see that solutions are
$$
(u,v) \equiv (0,0),(0,2),(1,1),(1,2),(2,0),(2,1) \pmod 3
$$
For example, taking $(u,v)\equiv (0,0)\pmod 3$, this gives parametrizations:
$$
\begin{align}
u = 3m, v = 3n &\implies y= 2 m + 6 m^2 - n - 6 m n + 6 n^2\\
&\implies x = -m + 6 m^2 - n - 6 m n + 6 n^2\\
&\implies z = -m + 6 m^2 + 2n - 6 m n + 6 n^2
\end{align}
$$
where we can freely choose $m,n\in\mathbb Z$. The other cases are the same.
Edit 1: One more case: let $(u,v)\equiv (2,1)\pmod 3$, so that
$$
u = 3m+2,\quad v = 3n+1
$$
then
$$
\begin{align}
y &= 3 + 8 m + 6 m^2 - n - 6 m n + 6 n^2\\
x = y-u &= 1 + 5 m + 6 m^2 - n - 6 m n + 6 n^2\\
z =v+x &= 2 + 5 m + 6 m^2 + 2 n - 6 m n + 6 n^2
\end{align}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2905081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Does this matrix sequence always converge? Suppose $a_0, a_1, ... , a_{n-1}$ are real numbers from $(0; 1)$, such that $\sum_{k=0}^{n-1} a_k=1$. Suppose $A = (c_{ij})$ is a $n \times n$ matrix with entries $c_{ij} = a_{(i-j)\%n}$, where $\%$ is modulo operation. Is it always true that $\lim_{m \to \infty} A^m = \frac{1}{n} \begin{pmatrix} 1 & 1 & \cdots & 1 \\1 & 1 & \cdots & 1 \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & \cdots & 1 \end{pmatrix}$?
This statement is true for $n = 2$:
Suppose $A = \begin{pmatrix} a_0 & {1 - a_0} \\ {1 - a_0} & a_0 \end{pmatrix} = \begin{pmatrix} 1 & -1 \\ 1 & 1\end{pmatrix}^{-1}\begin{pmatrix} 1 & 0 \\ 0 & {1-2a_0} \end{pmatrix}\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}$. Because $a_0$ is in $(0; 1)$, $1-2a_0$ is in $(-1;1)$. Thus $$\lim_{m \to \infty} A^m = \begin{pmatrix} 1 & -1 \\ 1 & 1\end{pmatrix}^{-1}\begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix}\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix} = \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$$
However, I do not know how to prove this statement for arbitrary $n$.
Any help will be appreciated.
|
This follows directly from the Perron-Frobenius theorem.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How does the number $3$ help in this (probably Cauchy-based) inequality?
Given $a,b,c>0$ and $a+b+c=3$. Prove that
$$\dfrac{a^2+bc}{b+ca}+\dfrac{b^2+ca}{c+ab}+\dfrac{c^2+ab}{a+bc}\ge3$$
Attempt:
Using Cauchy inequality $(a+b\ge2\sqrt{ab})$:
$\dfrac{a^2}{b+ca}+b+ca\ge 2\sqrt{a^2}=2a$
$\dfrac{b^2}{c+ab}+c+ab\ge 2\sqrt{b^2}=2b$
$\dfrac{c^2}{a+bc}+a+bc\ge 2\sqrt{c^2}=2c$
Adding all the inequalities, we have:
$\dfrac{a^2}{b+ca}+\dfrac{b^2}{c+ab}+\dfrac{c^2}{a+bc}+ab+bc+ca+a+b+c\ge2(a+b+c)$
$\Rightarrow\dfrac{a^2}{b+ca}+\dfrac{b^2}{c+ab}+\dfrac{c^2}{a+bc}+ab+bc+ca\ge 3$, but I don't see how that's relevant and I'm stuck here.
|
By C-S
$$\sum_{cyc}\frac{a^2+bc}{b+ca}=\sum_{cyc}\frac{(a^2+bc)^2}{(a^2+bc)(b+ca)}\geq\frac{\left(\sum\limits_{cyc}(a^2+bc)\right)^2}{\sum\limits_{cyc}(a^2+bc)(b+ca)}.$$
Thus, it's enough to prove that
$$\left(\sum\limits_{cyc}(a^2+ab)\right)^2\geq3\sum\limits_{cyc}(a^2+bc)(b+ca)$$ or
$$\sum_{cyc}(a^4+2a^2b^2+a^2b^2+2a^2bc+2a^3b+2a^3c+2a^2bc)\geq$$
$$\geq3\sum_{cyc}(a^2b+a^3c+a^2b+a^2bc)$$ or
$$\sum_{cyc}(a^4+2a^3b+2a^3c+3a^2b^2+4a^2bc-2a^3b-2a^2b^2-2a^2bc-3a^3c-3a^2bc)\geq0$$ or
$$\sum_{cyc}(a^4-a^3c+a^2b^2-a^2bc)\geq0,$$
which is true by Rearrangement and AM-GM.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Given that the quadratic equation $x^2-px+q=0$ has two roots $r$ and $s,$ Find the quadratic equation that takes $r^3$ and $s^3$ as its roots. Here's what I've tried:
Using Vieta's formulas:
$\;rs = q\;$ and $\,r+s = p$.
Then I cubed $rs$ which is $q^3$ and $(r+s)^3 + rs$ which is $p^3 + pq.$ Thinking that $x^2 - (p^3+pq)x + q^3$ is the answer. Did I do something wrong here?
|
The equation in question would look like
$$(x-r^3)(x-s^3)=x^2-(r^3+s^3)x+r^3s^3=0$$
As you've already identified, the last term is $(rs)^3=q^3$.
The coefficient of $-x$ is
$$\begin{split}
r^3+s^3 &= (r+s)(r^2-rs+s^2) \\
&= p(r^2+2rs+s^2-3rs) \\
&= p\bigg((r+s)^2-3rs\bigg) \\
&= p(p^2-3q)
\end{split}$$
Therefore, the equation is
$$x^2 -p(p^2-3q)x +q^3=0$$
|
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"url": "https://math.stackexchange.com/questions/2909249",
"timestamp": "2023-03-29T00:00:00",
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|
Help: $ |\frac{a+1}{a}- (\frac{xz}{y^2})^k|\leq \frac{1}{b}$ In the paper linear forms in the logarithms of real algebraic numbers close to 1, it is written on page $9$ that
On other hand, a short calculation yields
$$ \left|\frac{a+1}{a}- \left(\frac{xz}{y^2}\right)^k\right|\leq \frac{1}{b}$$
Image of the page :-
Here,
$$\left(\frac{xz}{y^2}\right)^k= \frac{(a+1)(ab^2+1)}{(ab+1)^2}$$ and $ b \geq 2, a\geq 2^{49},k\geq 50 $ (see page $8, 9$).
So, how do we prove the following?
$$ \left|\frac{a+1}{a}- \frac{(a+1)(ab^2+1)}{(ab+1)^2}\right|\leq \frac{1}{b}$$
|
We have, using $b\ge 2$ and $a\ge 2^{49}$,
$$\begin{align}\left|\frac{a+1}{a}- \frac{(a+1)(ab^2+1)}{(ab+1)^2}\right|
&=(a+1)\left|\frac{1}{a}- \frac{(ab^2+1)}{(ab+1)^2}\right|
\\\\&=(a+1)\left|\frac{(ab+1)^2-a(ab^2+1)}{a(ab+1)^2}\right|
\\\\&=(a+1)\left|\frac{1+a(2b-1)}{a(ab+1)^2}\right|
\\\\&=(a+1)\cdot \frac{1+a(2b-1)}{a(ab+1)^2}
\end{align}$$
Here, since
$$ab+1\ge ab$$
we have
$$\frac{1}{ab+1}\color{red}{\le}\frac{1}{ab}$$
Also, we have
$$a+1\le \frac{a^2}{2}\quad\text{and}\quad 1+a(2b-1)\le 2ab$$
Using these gives
$$\left|\frac{a+1}{a}- \frac{(a+1)(ab^2+1)}{(ab+1)^2}\right|
=(a+1)\cdot \frac{1+a(2b-1)}{a(ab+1)^2}\le \frac{a^2}{2}\cdot\frac{2ab}{a(ab)^2}=\frac 1b$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2909494",
"timestamp": "2023-03-29T00:00:00",
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|
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Fractions in Questions and Answers
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