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Find the matrix associated to linear function. Good night, i have a problem when i find the matrix associated to a linear transformation:
$T:P_{1\rightarrow}P_{2}$
$T(p(t))=t(p(t))$
Basis for $P_{1}=\left\{ t,1\right\} $
and for $P_{2}=\left\{ t^{2},t-1,t+1\right\}$
I work in this problem and i made this:
$T(t)=t^{2}=t^{2}+0t+0=\begin{array}{c}
1\\
0\\
0
\end{array}$
$T(1)=t=0t^{2}+t+0=\begin{array}{c}
0\\
1\\
0
\end{array}$
But this matrix is in canonical basis, how i can change this for the polynomial function?
|
As you discovered in the comments, it is easy to solve this one by playing around with linear combinations.
But what about a general strategy?
We know that given a linear transformation $T$ from $V$ to $W$, and two bases $\alpha = (v_1, \ldots, v_n)$ and $\beta = (w_1, \ldots, w_m)$, there is always a unique matrix representing it. The $i$th column of that matrix will be $T(v_i)$ in the $\beta$ coordinate system.
An important special case is when $V = W$ and $T$ is the identity transformation $I$. That's a change of basis matrix.
What you have is three coordinate systems. $P_1 = (t, 1)$. $P_2 = (t^2, t-1, t+1)$. And $P_3 = (t^2, t, 1)$. You have $T$ going from $P_1$ to $P_3$:
$$\begin{pmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
\end{pmatrix}$$
You can also easily find the change of basis matrix from $P_2$ to $P_3$ - just write down $P_2$ in the $P_3$ coordinate system as follows:
$$\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 1 \\
0 & -1 & 1
\end{pmatrix}$$
What you need is the change of basis matrix from $P_3$ to $P_2$. And the way you do that is to just invert the change of basis matrix that you have. Which you can do with row-reduction.
$$
\left[
\begin{array}{ccc|ccc}
1 & 0 & 0 & 1 & 0 & 0\\
0 & 1 & 1 & 0 & 1 & 0\\
0 & -1 & 1 & 0 & 0 & 1
\end{array}
\right]
\to
\left[
\begin{array}{ccc|ccc}
1 & 0 & 0 & 1 & 0 & 0\\
0 & 1 & 1 & 0 & 1 & 0\\
0 & 0 & 2 & 0 & 1 & 1
\end{array}
\right]
\to
\left[
\begin{array}{ccc|ccc}
1 & 0 & 0 & 1 & 0 & 0\\
0 & 1 & 1 & 0 & 1 & 0\\
0 & 0 & 1 & 0 & \frac{1}{2} & \frac{1}{2}
\end{array}
\right]
\to
\left[
\begin{array}{ccc|ccc}
1 & 0 & 0 & 1 & 0 & 0\\
0 & 1 & 0 & 0 & \frac{1}{2} & -\frac{1}{2}\\
0 & 0 & 1 & 0 & \frac{1}{2} & \frac{1}{2}
\end{array}
\right]
$$
And now your answer is simply:
$$
\begin{pmatrix}
1 & 0 & 0\\
0 & 1 & 0
\end{pmatrix}
\quad
\begin{pmatrix}
1 & 0 & 0 \\
0 & \frac{1}{2} & -\frac{1}{2} \\
0 & \frac{1}{2} & \frac{1}{2}
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 & 0 \\
0 & \frac{1}{2} & -\frac{1}{2}\\
\end{pmatrix}
$$
This strategy will always work.
|
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"timestamp": "2023-03-29T00:00:00",
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|
show that $\frac { 1 }{ 1-a } +\frac { 1 }{ 1-b } +\frac { 1 }{ 1-c } \ge \frac { 2 }{ 1+a } +\frac { 2 }{ 1+b } +\frac { 2 }{ 1+c } $ Let $a,b,c$ are positive numbers,if $$a+b+c=1$$
show that $$\frac { 1 }{ 1-a } +\frac { 1 }{ 1-b } +\frac { 1 }{ 1-c } \ge \frac { 2 }{ 1+a } +\frac { 2 }{ 1+b } +\frac { 2 }{ 1+c } $$
I am tried proving it but failed.Any hints will be appreciated.
|
$$ f(x) = \frac{1}{1-x}-\frac{2}{1+x} $$
is not a convex function on $(0,1)$, since:
$$ f''(x) = \frac{2}{(1-x)^3}-\frac{4}{(1+x)^3} \geq 0 $$
is equivalent to:
$$ \frac{1+x}{1-x}\geq \sqrt[3]{2} $$
but if we consider that $f'\left(\frac{1}{3}\right)=\frac{27}{8}$, it is not difficult to prove the algebraic inequality
$$ \forall x\in[0,1],\qquad f(x) \geq \frac{27}{8}\left(x-\frac{1}{3}\right) \tag{1}$$
since $f(x)-\frac{27}{8}\left(x-\frac{1}{3}\right)=0$ is equivalent to $\left(x-\frac{1}{3}\right)^2\left(x+\frac{1}{3}\right)=0$.
Given $(1)$ and $a,b,c\in[0,1]$, $\,a+b+c=1$, it follows that:
$$ f(a)+f(b)+f(c) \geq \frac{27}{8}(a+b+c-1) = 0\tag{2} $$
as wanted.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How can I prove this equation has no real solution? I have an equation
$$x^2-5 x+10+(x-4) \sqrt{1+x}=0. \tag{1}$$
Now I am trying to prove this equation has no real solution.
I tried. Put $t = \sqrt{1+x}$, then, I got
$$t^4+t^3-7 t^2-5 t+16=0. \tag{2}$$
But I can't prove two equations (1) and (2) have not solution.
How can I prove?
|
$$x^2-5x+10-(4-x)\sqrt{x+1}$$
now note: for roots to exist $-1<x<4$ thus $(4-x)>0$ and $\sqrt{x+1}>0$ now using our favourite AM-GM
$$x^2-5x+10-(4-x)\sqrt{x+1}>x^2-5x+10-\frac{(4-x)^2+(x+1)}{2}=\frac{2x^2-10x+20-16-x^2+8x-x-1}{2}=\frac{x^2-3x+3}{2}$$ now discriminant of the quadratic is negative and first coefficient is positive thus no roots exist and quadratic is greater than zero thus your function has no roots
|
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"timestamp": "2023-03-29T00:00:00",
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|
Error in solving $\int \sqrt{1 + e^x} dx$ . I want to solve this integral for $1 + e^x \ge 0$
$$\int \sqrt{1 + e^x} dx$$
I start by parts $$\int \sqrt{1 + e^x} dx = x\sqrt{1 + e^x} - \int x \frac{e^x}{2\sqrt{1 + e^x}} dx $$
Substitute $\sqrt{1 + e^x} = t \implies dt = \frac{e^x}{2\sqrt{1 + e^x}} dx$
So I remain with $$x\sqrt{1 + e^x} - \int \ln((t-1)(t+1)) dt = x\sqrt{1 + e^x} + 2t +(-t+1)\ln(t-1)+ (-t+1)\ln(t+1) $$
that is
$$(x+2)\sqrt{1 + e^x} + (-\sqrt{1 + e^x} + 1) \ln(+\sqrt{1 + e^x} - 1) + (-\sqrt{1 + e^x} + 1) \ln(+\sqrt{1 + e^x} + 1) + C $$
But the solution should be $$2\sqrt{1 + e^x} + \ln(+\sqrt{1 + e^x} - 1) - \ln(+\sqrt{1 + e^x} + 1) $$
Where is the mistake hidden?
|
with $$t=\sqrt{1+e^x}$$ we get $$x=\ln(t^2-1)$$ and from here $$dx=\frac{2t}{t^2-1}dt$$ and our integral will be $$\int\frac{2t^2}{t^2-1}dt$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $2^n+3^n $ is never a perfect square My attempt :
If $n$ is odd, then the square must be 2 (mod 3), which is not possible.
Hence $n =2m$
$2^{2m}+3^{2m}=(2^m+a)^2$
$a^2+2^{m+1}a=3^{2m}$
$a (a+2^{m+1})=3^{2m} $
By fundamental theorem of arithmetic,
$a=3^x $
$3^x +2^{m+1}=3^y $
$2^{m+1}=3^x (3^{y-x}-1) $
Which is not possible by Fundamental theorem of Arithmetic
Is there a better method? I even have a nagging feeling this is wrong as only once is the equality of powers of 2 and 3 $(n) $are used.
|
This is more convoluted than your way, but I already wrote it so I'll just leave it here :-)
$4^m = (a-3^m)(a+3^m) \implies a-3^m = 2^p, a+3^m = 2^q$ with $p+q = 2m$. Note that $q \ge p$
Then $a = 2^p + 3^m = 2^q - 3^m \iff 2^p (2^{q-p}-1) = 2\cdot 3^m$
But then it must be $p=1$, $q = 2m-1$, and we have
$$2^{2(m-1)}-1 = (2^{m-1}-1)(2^{m-1}+1) = 3^m$$
Since only one of the factors in the LHS can be divisible by $3$, it means the other one must be $1$, hence we get $2^{m-1}-1 = 1 \implies m = 2$
But in this case $2^{m-1}+1 = 3 \neq 3^2 $, so this is impossible.
|
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"timestamp": "2023-03-29T00:00:00",
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|
sum of all Distinct solution of the equation $ \sqrt{3}\sec x+\csc x+2(\tan x-\cot x) = 0\;,$
The sum of all Distinct solution of the equation $\displaystyle \sqrt{3}\sec x+\csc x+2(\tan x-\cot x) = 0\;,$
Where $x\in (-\pi,\pi)$ and $\displaystyle x\neq 0,\neq \frac{\pi}{2}.$
$\bf{My\; Try::}$ We can write equation as $$\frac{\sqrt{3}\sin x+\cos x}{\sin x\cos x}-2\frac{\cos 2x}{\sin x\cos x} = 0$$
So we get $$2\frac{\cos \left(x+\frac{\pi}{3}\right)}{\sin x\cos x}-2\frac{\cos 2x}{\sin x\cos x}=0$$
So we get $$\cos (2x) = \cos\left(x+\frac{\pi}{3}\right)$$
So we Get $\displaystyle 2x=2n\pi\pm (x+\frac{\pi}{3})\;,n\in \mathbb{Z}$
So we get $\displaystyle x= 2n\pi+\frac{\pi}{3}$ or $\displaystyle x = \frac{2n\pi-\frac{\pi}{3}}{3}\;,$ Where $n\in \mathbb{Z}$
So we get $$x=\frac{\pi}{3} = \frac{3\pi}{9}\;\;,-\frac{5\pi}{9}\;\;,-\frac{7\pi}{9}$$
So Sum of distinct Roots is $\displaystyle = -\pi$
But it is wrong, Where i have done wrong, Help me
Thanks
|
HINT:
$\cos (a+b) \neq \cos a \cos b + \sin a \sin b $
|
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|
How to see $\cos x \leq \exp(-x^2/2)$ on $x \in [0,\pi/2]$? Can anyone help me with the above inequality? I tried looking at the series expansion and I guess the answer indeed lies there, but I fail to see it.
Thanks
|
Here is another option to prove the inequality by Taylor expansions as the OP suggested. So we are to prove that
$$
\exp(-x^2/2)\ge\cos x,\quad x\in[0,\pi/2].
$$
By doing Taylor expansions at zero with the Lagrange remainders, we can get the estimations
\begin{eqnarray}
\exp(-t)&=&1-t+\frac{t^2}{2!}-\frac{t^3}{3!}+\underbrace{\frac{\exp(-\xi)}{4!}t^4}_{\ge 0}\ge
1-t+\frac{t^2}{2!}-\frac{t^3}{3!},\\
\cos x&=&1-\frac{x^2}{2!}+\frac{x^4}{4!}-\underbrace{\frac{\cos\xi}{6!}x^6}_{\ge 0}\le 1-\frac{x^2}{2!}+\frac{x^4}{4!}.
\end{eqnarray}
Now we can write with $t=x^2/2$
$$
\exp(-x^2/2)\ge 1-\frac{x^2}{2}+\frac{x^4}{8}\color{red}{-\frac{x^6}{48}}\ge
1-\frac{x^2}{2}+\frac{x^4}{8}\color{red}{-\frac{x^4}{12}}=
1-\frac{x^2}{2}+\frac{x^4}{24}\ge\cos x,\quad x\in[0,\pi/2].
$$
Here at the second step we have used the bound
$$
\color{red}{-\frac{x^6}{48}}\ge\color{red}{-\frac{x^4}{12}}\iff x^4(2+x)(2-x)\ge 0
$$
which is true for $0\le x\le\pi/2<2$.
|
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|
Local extrema and minima of the multivariable function $f(x,y) = x^2y+y^2+xy$ Let $f(x,y) = x^2y+y^2+xy$ be a function, I want to find its local extrema an minima.
I easily find that $f$ has 2 critical points: $(x,y)=(0,0)$ and $(x,y) = (-1,0)$.
In order to find its local extrema, I now find the Hessian matrix of the quadratic form: $H_f(a) = \begin{bmatrix}2y&2x+1\\2x+1&2\end{bmatrix}$
Therefore for $(x,y) = (0,0)$ I find $H_f(0,0) = \begin{bmatrix}0&1\\1&2\end{bmatrix}$ and $H_f(-1,0) =\begin{bmatrix}0&-1\\-1&2\end{bmatrix}$
I now have to determine wether those matrices are positive definite, semi-positive definite, semi-negative definite, negative definite or non-definite. In other term if $x \in \mathbb{R^2}$ I have to determine the sign of $x^T.H_f(a).x$
*
*$\begin{bmatrix}x&y\end{bmatrix} \begin{bmatrix}0&1\\1&2\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix} = 2(x+y)^2$ Therefore $H_f(0,0)$ is definite positive, (0,0) is a local extremum.
*$\begin{bmatrix}x&y\end{bmatrix}\begin{bmatrix}0&-1\\-1&2\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix} = -(x-y)^2$ Therefore $H_f(-1,0)$ is definite negative$, (-1, 0) is a local minimum.
Are my calculations correct?
|
The critical points are points $a \in \mathbb R^2$ such that $\mathrm {grad}\, f(a) = 0$. Now $\mathrm{grad}\, f(x,y) = (2xy + y , x^2 + 2y + x)$. You need to solve $$\begin{cases}2xy + y = 0 \\x^2 + 2y + x = 0\end{cases} \iff \begin{cases}y(2x+ 1) = 0\\x(x + 1) + 2y = 0\end{cases} \iff \begin{cases}y= 0 \,\,\text{or}\,\, x = -\frac{1}{2}\\x(x+1) + 2y = 0\end{cases} \iff \ldots$$
Think you can take it from here? From this you should find the third critical point which is
$(x,y) = \left(-\frac{1}{2}, \frac{1}{8}\right)$
|
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|
Laurent series of $\frac{e^z}{z^2+1}$
I cant figure out the laurent series of the following function.
Let $f(z)= \frac{e^z}{z^2+1} $ and $|z|\gt 1$
$$\frac{1}{z^2+1}=\sum_{n=0}^{\infty}(-1)^nz^{-2n-2}$$
and
$$e^z = \sum_{n=0}^{\infty}\frac{z^{n}}{n!}$$
$$e^z*\frac{1}{z^2+1} =\sum_{n=0}^{\infty}\sum_{k=0}^{n}(-1)^nz^{-2n-2}*\frac{z^{n-k}}{(n-k)!}$$
How can I go from here ?
|
Let
\begin{align*}
f(z)&= \frac{e^z}{z^2+1}
\end{align*}
We assume we need a Laurent expansion for all $|z|>1$ with center $z_0=0$.
We obtain
\begin{align*}
f(z)&=\frac{e^z}{z^2+1}\\
&=\frac{e^z}{z^2}\cdot\frac{1}{1+\frac{1}{z^2}}\\
&=\frac{e^z}{z^2}\sum_{k=0}^\infty(-1)^k\frac{1}{z^{2k}}\\
&=\sum_{j=0}^\infty\frac{z^j}{j!}\sum_{k=0}^\infty(-1)^k\frac{1}{z^{2k+2}}\\
&=\sum_{n=-\infty}^\infty\left(\sum_{{j-2k-2=n}\atop{j\geq 0;k\geq 0}}\frac{(-1)^k}{j!}\right)z^n\tag{1}\\
&=\sum_{n=-\infty}^\infty\left(\sum_{{j-2k=n}\atop{j\geq 0;k\geq 1}}\frac{(-1)^{k+1}}{j!}\right)z^n\tag{2}\\
&=\sum_{n=-\infty}^\infty\left(\sum_{k=K}^\infty\frac{(-1)^{k+1}}{(n+2k)!}\right)z^n\tag{3}\\
\end{align*}
Comment:
*
*In (1) we apply the Cauchy product formula
*In (2) we shift the index $k$ by one
*In (3) we replace $j$ with $n+2k$ and since $j=n+2k\geq 0$ and $k\geq 1$ we have to set
\begin{align*}
K=\max\left\{1,-\left\lfloor\frac{n}{2}\right\rfloor\right\}
\end{align*}
|
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|
QR factorization for least squares This is from my textbook
I don't undertand why small errorr in $A^TA$ can lead to large error in cofficient matrix? Because A=QR, so there should be no difference to use A or QR anyway.Could someone give an example? Thank you very much
|
A typical example uses the matrix
$$
\mathbf{A} =
\left(
\begin{array}{cc}
1 & 1 \\
0 & \epsilon \\
\end{array}
\right)
$$
Consider the linear system
$$
\mathbf{A} x= b.
$$
The solution via normal equations is
$$
%
\begin{align}
%
x_{LS} &= \left( \mathbf{A}^{*}\mathbf{A} \right)^{-1}\mathbf{A}^{*}b \\
% x
\left(
\begin{array}{cc}
x_{1} \\
x_{2}
\end{array}
\right)_{LS}
%
&=
%
%
\left(
\begin{array}{cc}
b_{1}-\frac{b_{2}}{\epsilon }\\\frac{b_{2}}{\epsilon }
\end{array}
\right)
%
\end{align}
%
$$
Minute changes in $\epsilon$, for example, $0.001\to0.00001$ create large changes in the solution:
$$
\epsilon = 0.001: \quad x_{LS} =
\left(
\begin{array}{cc}
b_{1}-1000b_{2} \\ 1000 b_{2}
\end{array}
\right)
$$
$$
\epsilon = 0.00001: \quad x_{LS} =
\left(
\begin{array}{cc}
b_{1}-100000b_{2} \\ 100000 b_{2}
\end{array}
\right)
$$
The $\mathbf{QR}$ decomposition is
$$
\mathbf{A} = \mathbf{QR} =
\left(
\begin{array}{cc}
1 & 0 \\
0 & \frac{\epsilon }{\left| \epsilon \right| } \\
\end{array}
\right)
%
\left(
\begin{array}{cc}
1 & 1 \\
0 & \left| \epsilon \right| \\
\end{array}
\right)
$$
|
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|
Proving an algebraic identity Prove:
$$(a + b + c)(ab + bc + ca) - abc = (a + b)(b + c)(c + a)$$
Problem:
I am not sure how to proceed after expanding the brackets on the RHS. I am not sure if I also expanded correctly. My solution is:
|
$$(a+b)(b+c)(c+a)$$
$$=(ab+bc+ca+b^2)(c+a)$$
$$=(ab+bc+ca)(a+c)+ab^2+b^2c$$
Adding $abc$ gives:
$$(ab+bc+ca)(a+c)+ab^2+b^2c+abc$$
$$=(ab+bc+ca)(a+c)+b(ab+bc+ac)$$
$$=(ab+bc+ca)(a+b+c)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the minimum value of expression involving real numbers
Let $n$ positive integer. Find the minimum value of expression:
$$ E=max(\frac {x_1} {1+x_1},\frac {x_2} {1+x_1+x_2}, ... , \frac {x_n}
{1+x_1+..+x_n})$$
where $x_1,x_2, .. , x_n$ are real not negative so that $x_1+x_2+ .. +x_n=1$
My try
For $x_1=1, x_2=0, ...x_n=0$ one gets $E=\frac 1 2$ so $Min \le \frac 1 2$
For $x_1=x_2= ... =x_n=\frac 1 n$ one gets $E=\frac 1 {n+1}$ so $Min \le \frac 1 {n+1}$ (much better)
I suspect the minimum is $\frac 1 {n+1}$ but I cannot prove it (or disprove it).
Update
The minimum cannot be $\frac 1 {n+1}$ because for $n=2, Min=1-\frac{1}{\sqrt2} \lt \frac 1 3$, as @almagest pointed out in a comment.
|
For $n=2$, the answer is $1-\frac{1}{\sqrt{2}}$, when $x_1=\sqrt{2}-1$ and $x_2=2-\sqrt{2}$.
In general, the answer is $1-\frac{1}{\sqrt[n]{2}}$.
Let's assume $a_0=1$, $a_i=1+x_1+x_2+\dots+x_i$ for $1\leqslant i\leqslant n$. Hence, $a_n=2$ and $a_i\leqslant a_{i+1}$. Then, $\frac{x_i}{1+x_1+x_2+\dots+x_i}=\frac{a_i-a_{i-1}}{a_i}$. Thus,
$$E=\max\{1-\frac{a_0}{a_1},1-\frac{a_1}{a_2},\dots,1-\frac{a_{n-1}}{a_n}\}.$$
Moreover, assume $b_i=\frac{a_{i-1}}{a_i}$ for $i=1,2,\dots, n$. Then, we have $b_1b_2\cdots b_n=\frac{a_0}{a_n}=\frac{1}{2}$, and $\frac{1}{2}\leqslant b_i\leqslant{1}$. Assume, $b_m=\min \{b_1,\dots,b_n\}$, then $b_m\leqslant \frac{1}{\sqrt[n]{2}}$.
$$E=\max\{1-b_1,1-b_2,\dots,1-b_n\}=1-b_m\geqslant 1-\frac{1}{\sqrt[n]{2}}.$$
And the equality holds when $b_1=b_2=\dots=b_n=\frac{1}{\sqrt[n]{2}}$. Note that $b_i=\frac{a_{i-1}}{a_i}=\frac{1+x_1+\dots+x_{i-1}}{1+x_1+\dots+x_{i-1}+x_i}$, we can use induction to prove $x_i=\sqrt[n]{2^i}-\sqrt[n]{2^{i-1}}$ in this case.
|
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|
If $a, b, c >0$ prove that $ [(1+a)(1+b)(1+c)]^7 > 7^7a^4b^4c^4 $. I solved it using AM, GM inequalities and reached to $[(1+a)(1+b)(1+c)]^7 > 2^{21}(abc)^\frac72 $ please help how to get $7^7(abc)^4$ in the inequality.
|
Here we have to prove $$(1+a)(1+b)(1+c)>7(abc)^{\frac{4}{7}}$$
So we get $$1+(a+b+c)+(ab+bc+ca)+abc>7(abc)^{\frac{4}{7}}$$
Now Using $\bf{A.M\geq G.M}\;,$ We get $$a+b+c\geq 3(abc)^{\frac{1}{3}}$$
and $$ab+bc+ca\geq 3(abc)^{\frac{2}{3}}$$
So $$1+\sum a+\sum ab+abc\geq 1+3(abc)^{\frac{1}{3}}+3(abc)^{\frac{2}{3}}+abc$$
Now Again Using $\bf{A.M\geq G.M}\;,$ We get
$$\bf{L.H.S}>1+(abc)^{\frac{1}{3}}+(abc)^{\frac{1}{3}}+(abc)^{\frac{1}{3}}+(abc)^{\frac{2}{3}}+(abc)^{\frac{2}{3}}+(abc)^{\frac{2}{3}}+(abc)\geq 7(abc)^{\frac{4}{7}}$$
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|
Condition for three points to lie on a Sphere? If A and B are the points (3,4,5) and (-1,3,-7) respectively then the set of the points P such that $PA^2+PB^2 = K^2$ where K is a constant lie on a proper sphere if K = 1 or K^2 <>= 161/2?
The correct answer is K^2 > 161/2. How?
I know that three non - collinear points always lie on one and only one circle but I don't know similar thing about a sphere. Probably we have to use pythagoras theorem but I'm not sure. Can someone help?
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Note first that the distance between $A$ and $B$ is $\sqrt{161}$. Let $C(1,\frac{7}{2},-1)$ be the midpoint of $AB$ and let $r=\frac{1}{2}\sqrt{2K^2-161}$. Take any point $P(x,y,z)$.
We have $PA^2+PB^2=(x-3)^2+(y-4)^2+(z-5)^2+(x+1)^2+(y-3)^2+(z+7)^2=2x^2+2y^2+2z^2-4x-14y+4z+109=2(x-1)^2+2(y-\frac{7}{2})^2+2(z+1)^2+\frac{161}{2}=2\cdot PC^2+\frac{161}{2}$.
So $PA^2+PB^2=K^2$ iff $K^2=2\cdot PC^2+\frac{161}{2}$ or $PC^2=\frac{1}{2}\sqrt{2K^2-161}=r^2$. In other words, the locus of $P$ is the sphere centre $C$ radius $r$.
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|
Switch from $a\cdot \sin(t) + b \cdot \cos(t)$ to $c \cdot \cos(t+f)$ How could I switch from $a\cdot \sin(t) + b \cdot \cos(t)$ to $c \cdot \cos(t+f)$?
Thank you for your time.
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$$
a\sin t+b\cos t=\sqrt{a^2+b^2}\left[\frac{b}{\sqrt{a^2+b^2}}\cos t +\frac{a}{\sqrt{a^2+b^2}}\sin t\right]$$
$$
=\sqrt{a^2+b^2}\left[\cos \alpha \cos t +\sin\alpha \sin t\right]$$
$$ =\sqrt{a^2+b^2} \cos(t-\alpha). $$
$ (c,f) $ have different symbols.
Useful in combining two waves of same frequency but different amplitude.
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Solve for integers $x, y, z$ such that $x + y = 1 - z$ and $x^3 + y^3 = 1 - z^2$.
Solve for integers $x, y, z$ such that $x + y = 1 - z$ and $x^3 + y^3 = 1 - z^2$.
I think we'll have to use number theory to do it. Simply solving the equations won't do.
If we divide the second equation by the first, we get:
$$x^2 - xy + y^2 = 1 + z.$$
Also, since they are integers $z^2 \ge z \implies -z^2 \le -z$. This implies
$$x + y = 1 - z \ge 1 - z^2 = x^3 + y^3.$$
This shows that atleast one of $x$ and $y$ is negative with the additive inverse of the negative being larger than that of the positive.
I have tried but am not able to proceed further. Can you help me with this?
Thanks.
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A general identity that is nice to know is
$$x^3+y^3=(x+y)(x^2-xy+y^2).$$
Given that $x+y=1-z$ and $x^3+y^3=1-z^2$ we have $z=1-x-y$ and hence
$$x^3+y^3=1-z^2=(1-z)(1+z)=(x+y)(2-x-y).$$
This means that either $x+y=0$, so $y=-x$ and $z=1$, or
$$x^2-xy+y^2=2-x-y.$$
This is a quadratic in $x$, and quadratics are easy. Applying the quadratic formula yields
$$x=\frac{y-1\pm\sqrt{(1-y)^2-4(y^2+y-2)}}{2}=\frac{1}{2}\left(y-1\pm\sqrt{-3(y+3)(y-1)}\right),$$
and for these $x$ to be integers we need $-3(y+3)(y-1)$ to be a square, so certainly this must be nonnegative. It follows that
$$y\in\{-3,-2,-1,0,1\},$$
and these few cases can be checked by hand. For $(x,y,z)$ we find the solutions
$$(-2,-3,6),\ (-3,-2,6),\ (0,-2,3),\ (-2,0,3),\ (1,0,3),\ (0,1,3),$$
and $(x,-x,1)$ for all $x\in\Bbb{Z}$ from before.
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|
How is $ \ \frac{1}{1+\frac{1}{x^2}}=\ \frac{x^2}{1+x^2} = \ 1 - \frac{x^2}{1+x^2}$? How did $$ \ \frac{1}{1+\frac{1}{x^2}}$$ become $$\ \frac{x^2}{1+x^2} = \ 1 - \frac{x^2}{1+x^2}$$
I am trying to figure out the equation.
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$$\frac{1}{1+\frac{1}{x^2}}=\frac{x^2}{x^2}\times\frac{1}{1+\frac{1}{x^2}}=\frac{x^2}{x^2(1+\frac{1}{x^2})}=\frac{x^2}{x^2+1}=\frac{x^2+1-1}{x^2+1}=1-\frac{1}{x^2+1}$$ So, probably a typo somewhere.
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Show that $\frac{1}{x_{1}(x_{1}+1)}+\frac{1}{x_{2}(x_{2}+1)}+\cdots+\frac{1}{x_{n}(x_{n}+1)}\ge\frac{n}{2}$ with $x_{1}x_{2}\cdots x_{n}=1$ Let $x_{1},x_{2},\cdots,x_{n}$ be postive real numbers, and such $x_{1}x_{2}\cdots x_{n}=1$, Show that
$$\dfrac{1}{x_{1}(x_{1}+1)}+\dfrac{1}{x_{2}(x_{2}+1)}+\cdots+\dfrac{1}{x_{n}(x_{n}+1)}\ge\dfrac{n}{2}$$
I try use Cauchy-Schwarz inequality.but I don't see how to make the right estimates.If anyone has an idea how to proceed in that inequality
for $n=3$,even it is hard to prove it
$x_{1}=\dfrac{a}{b},x_{2}=\dfrac{b}{c},x_{3}=\dfrac{c}{a}$
It suceffent to prove
$$\sum_{cyc}\dfrac{b^2}{a(a+b)}\ge\dfrac{3}{2}$$
Use Cauchy-Schwarz inequality we have
$$\sum_{cyc}\dfrac{b^2}{a(a+b)}\sum_{cyc}a(a+b)\ge (a+b+c)^2$$
$$\Longleftrightarrow 2(a+b+c)^2\ge 3\sum_{cyc}(a^2+ab)$$
this inequality is wrong
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Let $x_i=\frac{a_i}{a_{i+1}}$, where $a_i>0$ and $a_{n+1}=a_{1}$. We have
$$\tag{1}\sum_{cyc}\frac{1}{x_i(x_i+1)}=\sum_{cyc}\frac{1}{\frac{a_i}{a_{i+1}}\cdot\left( \frac{a_i}{a_{i+1}}+1\right)}=\sum_{cyc}\frac{a_{i+1}^2}{a_i\cdot\left(a_i+a_{i+1}\right)}$$
Notice that:
$$\sum_{cyc}\frac{a_{i+1}^2}{a_i\cdot (a_i+a_{i+1})}-\sum_{cyc}\frac{a_{i}^2}{a_i\cdot (a_i+a_{i+1})}=\sum_{cyc}\frac{(a_{i+1}-a_i)\cdot(a_{i+1}+a_i)}{a_i\cdot(a_i+a_{i+1})}=\sum_{cyc}\left(\frac{a_{i+1}}{a_i}-1\right)=\sum_{cyc}\frac{a_{i+1}}{a_i}-n\geqslant n\sqrt[n]{\prod_{cyc}\frac{a_{i+1}}{a_i}}-n\geqslant 0$$
Thus
$$\tag{2}\sum_{cyc}\frac{a_{i+1}^2}{a_i\cdot (a_i+a_{i+1})} \geqslant\sum_{cyc}\frac{a_{i}^2}{a_i\cdot (a_i+a_{i+1})}$$
From $(1)$ and $(2)$, we have
$$\sum_{cyc}\frac{a_{i+1}^2}{a_i\cdot(a_i+a_{i+1})}\geqslant \frac12\sum_{cyc}\frac{a_i^2+a_{i+1}^2}{a_i\cdot(a_i+a_{i+1})}\geqslant\frac14\sum_{cyc}\frac{(a_i+a_{i+1})^2}{a_i\cdot(a_i+a_{i+1})}=\frac14\sum_{cyc}\frac{a_i+a_{i+1}}{a_i}=\frac14\sum_{cyc}\left(1+\frac{a_{i+1}}{a_i}\right)=\frac14\left(n+\sum_{cyc}\frac{a_{i+1}}{a_i}\right)\geqslant \frac14\cdot(2n)=\frac{n}{2}$$
QED.
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|
Can someone help me with this question of finding x as exponent? The equation is:
$$6^{x+1} - 6^x = 3^{x+4} - 3^x$$
I need to find x. I forgot how to use logarithm laws. Help would be appreciated. Thanks.
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This means
$$
6^x(6-1) = 3^x(3^4 - 1) \iff \\
5 \cdot 6^x = 80 \cdot 3^x \iff \\
5 \cdot 2^x \cdot 3^x = 80 \cdot 3^x \iff \\
5 \cdot 2^x = 80 \iff \\
2^x = 16 \Rightarrow \\
x = 4
$$
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|
Prove that $\int_{0}^{1}{\ln(x) \over 2-x}\mathrm dx={\ln^2(2)-\zeta(2)\over 2}$ $$I=\int_{0}^{1}{\ln(x) \over 2-x}\,\mathrm{d}x={\ln^2(2)-\zeta(2)\over 2}$$
$$(2-x)^{-1}={1\over2}\left(1-{x\over 2}\right)^{-1}$$
Using binomial series here
$$(2-x)^{-1}={1\over2}\left(1-{x\over 2}\right)^{-1}={1\over2}+{x\over 4}+{x^2\over8}+{x^3\over 16}+\cdots=\sum_{k=0}^{\infty}{x^k\over2^{k+1}}$$
$$I=\sum_{k=0}^{\infty}{1\over 2^{k+1}}\int_{0}^{1}x^k\ln(x)\,\mathrm{d}x$$
$$\int_{0}^{1}x^k\ln(x)\,\mathrm{d}x=-{1\over (1+k)^2}$$
$$I=-\sum_{k=0}^{\infty}{1\over 2^{k+1}(1+k)^2}$$
So this is where I got to, unable to determine this infinite sum. Do anyone know how to prove that,
$$-\sum_{k=0}^{\infty}{1\over 2^{k+1}(1+k)^2}={\ln^2(2)-\zeta(2) \over 2}$$
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start with $$\int_0^1 \ln(1-x) x^{n-1}dx = -\frac{H_{n}}{n}$$
change of variable and expand in power series
$$\int_0^1 \frac{\ln(x)}{2-x}dx = \int_0^1 \frac{\ln(1-x)}{1+x} = \sum_{n=0}^\infty (-1)^n \int_0^1\ln(1-x) x^n dx = \sum_{n=1}^\infty (-1)^n \frac{H_{n}}{n}$$
and use a trick found by Jack D'Aurizio
$$\frac{\ln^2(1-x)}{2} = \int_0^x \frac{-\ln(1-t) }{1-t} dt = \sum_{k=1}^\infty\sum_{n=0}^\infty \int_0^x\frac{t^{n+k}}{k}dt = \sum_{k=1}^\infty\sum_{n=0}^\infty \frac{x^{n+k+1}}{k(n+k+1)}$$ $$ = \sum_{m= 2}^\infty \frac{x^{m}}{m} \sum_{k=1}^{m-1} \frac{1}{k} = \sum_{m =2}^\infty \frac{x^{m}}{m} H_{m-1}$$
hence
$$\frac{\ln^2(2)}{2} =\frac{\ln^2(1-(-1))}{2} = \sum_{m =2}^\infty \frac{(-1)^{m}}{m} H_{m-1} =
-\sum_{m=1}^\infty \frac{(-1)^m}{m^2} +\sum_{m=1}^\infty (-1)^{m}\frac{H_m}{m} $$ $$= \eta(2) +\int_0^1 \frac{\ln(x)}{2-x}dx$$
i.e. with $\eta(2) = (1-2^{1-2})\zeta(2)$
$$\int_0^1 \frac{\ln(x)}{2-x}dx = \frac{\ln^2(2)}{2} - \frac{\zeta(2)}{2}$$
proof of the first formula :
$\displaystyle \frac{1-x^n}{1-x} = \sum_{m=0}^{n-1} x^m\quad$ hence $\quad\displaystyle \int_0^1 \frac{1-x^n}{1-x} dx = \sum_{m=0}^{n-1} \frac{1}{m+1} = H_n\quad$ and integrating by parts $\quad\displaystyle \int_0^1 \frac{1-x^n}{1-x} dx = \lim_{a \to 1^-} (x^n-1)\ln(1-x)\mid_0^a $ $- \int_0^1 n x^{n-1} \ln(1-x)dx = -n\int_0^a x^{n-1} \ln(1-x)dx$
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Differential equation exercise. I am tasked with solving
\begin{cases}
y''(t) &=& \frac{(y(t)')^2}{y} - 2\frac{y'(t)}{y^4(t)} \\
y(0) &=& -1 \\
y'(0) &=& -2
\end{cases}
I proceed by setting $v(s) = y' (y^{-1}(s))$ reducing the problem to
\begin{cases}
v'(s) &=& \frac{v(s)}{s} - \frac{2}{s^4} \\
v(-1) &=& -2
\end{cases}
This is a first order linear ODE, after multiplying by $\frac{1}{|s|}$ (I drop the absolute value and change the sign since $s$ will be in a negative interval). I obtain $$-\frac{v'(s)}{s} + \frac{v}{s^{2}} = \frac{2}{s^5}$$
This gives me $$\frac{v(s)}{s} = \frac{1}{(-2s^4)} + \frac{5}{2} \implies y'(t) = \frac{-1 + 5y(t)^4}{2y(t)^3}$$
And I can't quite manage to integrate the reciprocal of this to get my $y(t)$.
Are my calculations correct? How should I do this? I am utilizing this method because it's the one I am expected to use at the exam (I have already been told that it can make things more difficult).
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\begin{cases}
y''(t) &=& \frac{(y(t)')^2}{y} - 2\frac{y'(t)}{y^4(t)} \\
y(0) &=& -1 \\
y'(0) &=& -2
\end{cases}
This is an ODE of the autonomeous kind. The usual way to reduce the order is the change of function :
$$\frac{dy}{dt}=F(y) \quad\to\quad \frac{d^2y}{dt^2}=\frac{dF}{dy}\frac{dy}{dt}=F\frac{dF}{dy}$$
$$F\frac{dF}{dy}=\frac{F^2}{y} -2\frac{F}{y^4}$$
$$y\frac{dF}{dy}-F =-\frac{2}{y^3}$$
There is no difficulty to solve this linear ODE. You will obtain :
$$F(y)=c_1y+\frac{1}{2y^3}\quad\to\quad \frac{dy}{dt}=c_1y+\frac{1}{2y^3}$$
With conditions $y(0)=-1$ and $y'(0)=-2\,$ then $\,-2=c_1(-1)+\frac{1}{2(-1)^3} \quad\to\quad c_1=\frac{3}{2}$
$$F(y)=\frac{3}{2}y+\frac{1}{2y^3}\quad\to\quad \frac{dy}{dt}=\frac{3}{2}y+\frac{1}{2y^3} = \frac{3y^4+1}{2y^3}$$
$$t=\int \frac{2y^3dy}{3y^4+1}=\frac{1}{6}\ln(3y^4+1)+c_2$$
with conditions : $0=\frac{1}{6}\ln(3(-1)^4+1)+c_2 \quad\to\quad c_2=-\frac{1}{6}\ln(4)$
$$t(y)=\frac{1}{6}\ln(3y^4+1)-\frac{1}{6}\ln(4)$$
$$t(y) =\frac{1}{6}\ln(\frac{3y^4+1}{4})$$
With sign according to condition $y(0)=-1$ , the inversion of $t(y)$ leads to :
$$y(t)=-\left(\frac{4e^{6t}-1}{3} \right)^{1/4}$$
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|
How can I prove this sharp upper bound? Here :
What are sharp lower and upper bounds of the fast growing hierarachy?
Deedlit mentions that for natural $m,n\ge 2$ and natural $k>n+log_2(n)$ , we have $$2\uparrow^{m-1}n<f_m(n)<2\uparrow^{m-1} k$$
The left inequality can be proven by induction. But what about the right inequality ? How can I bound $2\uparrow^{m-1} k$ in a way that induction can be applied ? Or do I need another idea ?
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So we will prove that for $m \ge 2, n \ge 3$, $f_m(n) \le 2\uparrow^{m-1}\lceil n + \log_2(n) + 1\rceil$.
Base case $m=2$: $f_2(n) = n 2^n = 2^{n+\log_2(n)} \le 2 \uparrow \lceil n + \log_2(n) + 1\rceil$.
Inductive step:
We assume that $f_m(n) \le 2\uparrow^{m-1}\lceil n + \log_2(n) + 1 \rceil \le 2 \uparrow^{m-1}(2n)$.
Lemma. For $m,n \ge 1$, $2\uparrow^m (n+2) > 2(2\uparrow^m n) + 2$.
Proof of Lemma: $2\uparrow^m(n+2) = 2 \uparrow^{m-1} (2 \uparrow^{m-1} (2\uparrow^m n)) \ge 2 \uparrow^0 (2\uparrow^0 (2\uparrow^m n)) = 4(2\uparrow^m n) > 2(2\uparrow^m n) + 2$ QED
Let $g(n) = 2 \uparrow^{m-1} n$ and $h(n) = 2 \cdot 2\uparrow^{m-1}(n)$. Then the Lemma is equivalent to $g(n+2) > h(n)+2$.
Lemma 2. $f_m^i(n) < h^i(2n)) < g^i(2n+2)$.
Proof:
$$g^i(2n+2) = g^{i-1}(g(2n+2)) > g^{i-1}(h(2n)+2) > g^{i-2}(h^2(2n)+2) > g^{i-3}(h^3(2n)+2) > \ldots > h^i(2n) > f_m^i(n)$$
QED
Lemma 3. $2n+2 < 2 \uparrow^m \lceil \log_2(n) + 1\rceil$ for $m \ge 2, n \ge 3$.
Proof of Lemma 3: It suffices to prove the case $n = 2^k$, which reduces the lemma to
$$2 \cdot 2^k + 2 < 2 \uparrow^m ( k + 1)$$
Base case $k=2$: $2 \cdot 2^2 + 2 = 10 < 2\uparrow\uparrow 3 \le 2\uparrow^m 3$.
Inductive step:
$$2\uparrow^m(k+2) = 2\uparrow^{m-1}(2\uparrow^m(k+1)) \ge 2(2\uparrow^m(k+1)) > 2(2^{k+1}+2) = 2^{k+2} + 4 > 2^{k+2} + 2$$
QED
So, $$f_{m+1}(n) = f_m^n(n) < g^n(2n+2) <g^n(2\uparrow^m \lceil \log_2(n)+1\rceil) = 2\uparrow^m \lceil n+ \log_2(n) + 1 \rceil$$
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|
Finding $\lim_{x\to 0} \frac{\ln(x+4)-\ln(4)}{x}$ without L'hôpital I have this $\lim_{}$.
$$\lim_{x\to 0} \frac{\ln(x+4)-\ln(4)}{x}$$
Indetermation:
$$\lim_{x\to 0} \frac{\ln(0+4)-\ln(4)}{0}$$
$$\lim_{x\to 0} \frac{0}{0}$$
Then i started solving it:
$$\lim_{x\to 0} \frac{\ln\frac{(x+4)}{4}}{x}$$
$$\lim_{x\to 0} \frac{1}{x}\phantom{2} .\phantom{2} \ln\lim_{x\to 0} \frac{(x+4)}{4}$$
$$\lim_{x\to 0} \frac{1}{x}\phantom{2} .\phantom{2} \ln\lim_{x\to 0} (\frac{x}{4}+1)$$
$$\lim_{x\to 0} \frac{1}{x}\phantom{2} .\phantom{2} \ln\lim_{x\to 0} (1 + \frac{x}{4})$$
$$\ln\lim_{x\to 0} (1 + \frac{x}{4})^\frac{1}{x}$$
Then multiply the power, by 4.
$$\ln\lim_{x\to 0} (1 + \frac{x}{4})^\frac{1.4}{x.4}$$
$$\ln \begin{bmatrix}\lim_{x\to 0} (1 + \frac{x}{4})^\frac{4}{x} \end{bmatrix}^\frac{1}{4} $$
$$ \ln \phantom{2}\mathcal e^\frac{1}{4} = \frac{1}{4}$$
Until here is fine, but someone's telling me that it can be made like this:
$$\lim_{x\to 0} \frac{\ln(x) + \ln(4) -\ln(4)}{x}$$
$$\lim_{x\to 0} \frac{\ln(x)}{x} = 1$$
Alright, it uses logarithmic property to separate the expression, but i said it can't be, because of the indetermination, and $\frac{1}{4} $ is closer of $0$ than $1$. There's another reasonable explanation for this ? Why it can't be or when can it ?
And, there's a trick to way out quickly with this limit, keeping L'Hôspital aside ?
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Consider rewriting the expression as
$\dfrac{\ln(x + 4) - \ln(4)}{x} = \dfrac{\ln\Big(\dfrac{x + 4}{4}\Big)}{x} = \ln\Big(\frac{x}{4} + 1\Big)^{1/x} = \frac{1}{4}\ln\Big(\frac{1}{4/x} + 1\Big)^{4/x}$
Then, we find
$\lim_{x \rightarrow 0} \dfrac{\ln(x + 4) - \ln(4)}{x} = \frac{1}{4}\ln\lim_{x \rightarrow 0}\Big(\frac{1}{4/x} + 1\Big)^{4/x} = \frac{1}{4}\ln(e) = \frac{1}{4}$
|
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|
Problem based on sum of reciprocal of $n^{th}$ roots of unity
Let $1,x_{1},x_{2},x_{3},\ldots,x_{n-1}$ be the $\bf{n^{th}}$ roots of unity. Find: $$\frac{1}{1-x_{1}}+\frac{1}{1-x_{2}}+......+\frac{1}{1-x_{n-1}}$$
$\bf{My\; Try::}$ Given $x=(1)^{\frac{1}{n}}\Rightarrow x^n=1\Rightarrow x^n-1=0$
Now Put $\displaystyle y = \frac{1}{1-x}\Rightarrow 1-x=\frac{1}{y}\Rightarrow x=\frac{y-1}{y}$
Put that value into $\displaystyle x^n-1=0\;,$ We get $\displaystyle \left(\frac{y-1}{y}\right)^n-1=0$
So we get $(y-1)^n-y^n=0\Rightarrow \displaystyle \left\{y^n-\binom{n}{1}y^{n-1}+\binom{n}{2}y^2-......\right\}-y^n=0$
So $\displaystyle \binom{n}{1}y^{n-1}-\binom{n}{2}y^{n-1}+...+(-1)^n=0$ has roots $y=y_{1}\;,y_{2}\;,y_{3}\;,.......,y_{n-1}$
So $$y_{1}+y_{2}+y_{3}+.....+y_{n-1} = \binom{n}{2} =\frac{n(n-1)}{2}\;,$$ Where $\displaystyle y_{i} = \frac{1}{1-x_{i}}\;\forall i\in \left\{1,2,3,4,5,.....,n-1\right\}$
My Question is can we solve it any less complex way, If yes Then plz explain here, Thanks
|
Otherwise you could write $x_i=e^{j\frac {i2\pi} {n}}$ ($j^2=-1$)
Then
$$y_i=\frac {e^{-j\frac {i\pi} {n}}} {e^{-j\frac {i\pi} {n}}-e^{j\frac {i\pi} {n}}}=\frac {e^{-j\frac {i\pi} {n}}}{-2j\sin(\frac {i\pi} {n})}=2j\cot(\frac {i\pi} {n})+\frac 1 2$$
Now consider the fact that, if $x_i$ is an $n^{th}$ root of unity, so is $\bar{x}_i$, so if $x_i\neq \bar x_i$ , your sum contains $y_i+\bar y_i=2\mathcal {Re}(y_i)=1$ exactly $\frac {n-1} {2}$ if $n$ is odd or $\frac {n-2} {2}$ if $n$ is even (but the remaining term is $\frac 1 {1-(-1)}=\frac 1 2$).
So in both cases the result is $\frac {n-1} 2$
I don't know if this can be considered "less complex" but it certainly involves more "down to earth" calculations IMHO.
|
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|
Evaluate the limit $\lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}\right)$
Evaluate the limit $\displaystyle \lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}\right)$, without using a Riemann sum
$\bf{My\; Try:}$ Using the graph of $\displaystyle f(x) = \frac{1}{\sqrt{x}}\;,$ we get
$$\int_{1}^{n+1}\frac{1}{\sqrt{x}}dx <\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}\right)<1+\int_{1}^{n}\frac{1}{\sqrt{x}}dx$$
So we get $$2\sqrt{n+1}-2<\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}\right)<2\sqrt{n}-1$$
So $$\lim_{n\rightarrow \infty}\frac{2\sqrt{n+1}-2}{\sqrt{n}}=2<\lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}\right)<\lim_{n\rightarrow \infty}\frac{2\sqrt{n}-1}{\sqrt{n}} = 2$$
So using the Sandwich Theorem, we get $\displaystyle \lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}\right)=2$
My question is can we solve it by using any other method? If yes then please explain it here.
|
From the Euler-Maclaurin Summation Formula, we have
$$\begin{align}
\frac{1}{\sqrt{n}}\sum_{k=1}^n\frac{1}{\sqrt{k}}&=\frac{1}{\sqrt{n}}\left(1+\int_1^n \frac{1}{\sqrt{x}}\,dx+\frac12\left(\frac{1}{n^{1/2}}-1\right)+O(1)\right)\\\\
&=2+O\left(\frac{1}{n^{1/2}}\right)\\\\
&\to 2\,\,\text{as}\,\,n\to \infty
\end{align}$$
|
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|
Suppose $-a\sin(s) - b\cos(s) = 0$, then $a^2 + b^2 = 1$?
Suppose $-a\sin(s) - b\cos(s) = 0$ and given that $a^2 + b^2 \le 1$, then $a^2 + b^2 = 1$?
I am having trouble getting the above identity. I vaguely recall that
$$a\sin(s) + b\cos(s)=\sqrt{a^2+b^2}\sin(s+\theta)$$
where $$\tan(\theta) = \frac{b}{a}$$
But I still don't see why it's obvious?
|
It makes no sense for a mathematical claim to have a condition of the form "$-P-Q = 0$", so maybe the intended claim is:
Given reals $a,b,t$ such that $1 - a \sin(t) - b \cos(t) = 0$ and $a^2+b^2 \le 1$, then $a^2+b^2 = 1$.
And indeed your method proves this.
For fun, here is an alternative proof that uses weighted AM-GM:
Take reals $a,b,t$ such that $1 - a \sin(t) - b \cos(t) = 0$. Note that $a,b$ are not both zero.
Then $\dfrac1{a^2+b^2} = \dfrac{ a^2 \frac1a \sin(t) + b^2 \frac1b \cos(t) }{a^2+b^2} \le \sqrt{\dfrac{a^2 ( \frac1a \sin(t) )^2 + b^2 ( \frac1b \cos(t) )^2 }{a^2+b^2}} = \dfrac1{\sqrt{a^2+b^2}}$.
Thus $a^2+b^2 \ge 1$.
|
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|
Write the general term of the periodic sequence $1$, $-1$, $-1$, $1$, $-1$, $-1$, $1$, ..., as $(-1)^{g(n)}$ or other closed form How to put mathematically sequence that changes sign like:
$n = 0\quad f = 1$
$n = 1 \quad f = -1$
$n = 2 \quad f = -1$
$n = 3 \quad f = 1$
$n = 4 \quad f = -1$
$n = 5 \quad f = -1$
$n = 6 \quad f = 1$
.....
In the form of (-1)^(something) or similar analytical expression.
|
\begin{align}
a_1 &= 1 \\
a_2 &= -1 \\
a_n &= a_{n-1}a_{n-2}
\end{align}
Using a product recurrence relation, we have
$$
a_n = a_2^{F_{n-1}}a_1^{F_{n-2}}
$$
Where $F_k$ is the $k$'th Fibonacci number. Given a closed form for $F_k$, we can simplify $a_n$.
$$
F_k = \frac{(1 + \sqrt{5})^k - (1 - \sqrt{5})^k}{2^k\sqrt{5}}
$$
$$
a_n = (-1)^{F_{n-1}} = (-1)^{\frac{(1 + \sqrt{5})^{n-1} - (1 - \sqrt{5})^{n-1}}{2^{n-1}\sqrt{5}}}
$$
This was solved so that the starting index of the sequence was $1$. If you want to start at $0$, we simply shift the index by $+1$.
$$
a_n = (-1)^{F_{n}} = (-1)^{\frac{(1 + \sqrt{5})^{n} - (1 - \sqrt{5})^{n}}{2^{n}\sqrt{5}}}
$$
Using Euler's Formula, we can get a far more (or less) elegant solution:
$$
a_n = (-1)^{F_n} = \cos(\pi F_n) + i\sin(\pi F_n) = e^{i\pi F_n}
$$
|
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|
Proof of an inequality involving three numbers $(a,b,c)\gt 0$ If $(a,b,c)\gt 0$
the following inequality holds:
$$\dfrac{a^2}{2a^2+(b+c)^2}+\dfrac{b^2}{2b^2+(c+a)^2}+\dfrac{c^2}{2c^2+(a+b)^2}\lt\dfrac{2}{3}$$
I am stuck to find a proof of it.
Can someone help me? Thanks.
|
WLOG, assume that $a\ge b \ge c$.
We first show that
$$\dfrac{b^2}{2b^2+(c+a)^2}+\dfrac{c^2}{2c^2+(a+b)^2} \le \frac{(b+c)^2}{2(b+c)^2+a^2}.\qquad (1)$$
Indeed, it is equivalent to each of the following inequalities:
$$\dfrac{c^2}{2c^2+(a+b)^2} \le \frac{(b+c)^2}{2(b+c)^2+a^2} - \dfrac{b^2}{2b^2+(c+a)^2}$$
$$\dfrac{c^2}{2c^2+(a+b)^2} \le \frac{(b+c)^2(c+a)^2 - a^2b^2}{(2(b+c)^2+a^2)(2b^2+(c+a)^2)}$$
$$\dfrac{c^2}{2c^2+(a+b)^2} \le \frac{c(a+b+c)(c^2+ca+cb+2ab)}{(2(b+c)^2+a^2)(2b^2+(c+a)^2)}$$
$$\dfrac{c}{2c^2+(a+b)^2} \le \frac{(a+b+c)(c^2+ca+cb+2ab)}{(2(b+c)^2+a^2)(2b^2+(c+a)^2)}$$
$$\dfrac{c}{2c^2+(a+b)^2} \le \frac{c(a+b+c)^2 + 2ab(a+b+c)}{(2(b+c)^2+a^2)(2b^2+(c+a)^2)}$$
It suffices to prove that
$$\dfrac{c}{2c^2+(a+b)^2} \le \frac{c(a+b+c)^2}{(2(b+c)^2+a^2)(2b^2+(c+a)^2)}$$
or equivalently
$$\dfrac{1}{2c^2+(a+b)^2} \le \frac{(a+b+c)^2}{(2(b+c)^2+a^2)(2b^2+(c+a)^2)}$$
The last inequality is true because $2(b+c)^2 + a^2 \le (a+b+c)^2$ and $2b^2+(c+a)^2 \le 2c^2+(a+b)^2$.
Back to the original inequality. Applying $(1)$ it suffices to prove
$$\dfrac{a^2}{2a^2+(b+c)^2}+ \frac{(b+c)^2}{2(b+c)^2+a^2} \lt\dfrac{2}{3}.$$
Denote $x=a^2,y=(b+c)^2$ it is easy to show that the above inequality is equivalent to $(x-y)^2 \ge 0$.
|
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|
Prove $\sqrt{1+x}$ can be represented by a power series I need to show that $\sqrt{1+x}$ can be represented as a power series. I need to prove the equality between the function and its Taylor series, not to prove that the Taylor series of the function, $\sum\limits_{n=0}^{\infty}{\frac{1}{2}\choose n}x^n$ converges in $(-1,1)$.
My lead is to use the Lagrange form of the remainder, by I do not know how to show the remainder tends to zero.
Edit:
The $n$-th derivative is
$$
f^{(n)}(0) = \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\cdot\ldots\cdot\left(\frac{1}{2} - n + 1\right)}{n!}
$$
and hence the Lagrange form of the remainder is
$$
R_n(x) = \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\cdot\ldots\cdot\left(\frac{1}{2} - n + 2\right)}{(n+1)!}(1+c)^{\frac{-1-2n}{2}}x^n
$$
But I do not know how to show this tends to $0$ as $n\to\infty$
|
Start by simplifying the expression for $R_n(x)$. The large fraction becomes, after taking absolute values:
$$
\frac{1}{(n+1)!} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{3}{2} \cdots \frac{2n - 3}{2} = \frac{(2n - 3)(2n - 5) \cdots (3)(1)}{2^{n+1}(n+1)!}
$$
Observe that the numerator is bounded above by $(n + 1)!$, and so the large fraction is smaller than $2^{-n-1}$. So we have:
$$
|R_n(x)| \le \frac{(1+c)^{ \frac{1-2n}{2} } } {2^{n+1}} x^n = \frac{(1+c)^{1/2}}{2} \left( \frac{x}{2(1+c)} \right) ^n
$$
For any $c$ between $0$ and $x$, the right hand side tends to $0$ as $n \to \infty$.
|
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|
Show that $\{a_n\}$ defined by $a_{n+1}=\frac{a_n+2}{a_n+1}$ converges Suppose $a_0$ is an arbitrary positive real number. Define the sequence $\{a_n\}$ by $$a_{n+1}=\frac{a_n+2}{a_n+1}$$ for all $n\geq0$. I have to prove that $\{a_n\}$ converges.
My attempt: If $a=\lim_{n\to\infty}{a_n}$ exists, then it should be a solution to $$a=\frac{a+2}{a+1}$$ which is $\sqrt2$. Thus I need to show that $|\sqrt2 - a_n|$ gets arbitrarily small for large $n$. I tried to prove that $|\sqrt2-a_n|<|\sqrt2-a_{n-1}|$ but couldn't.
|
Note: not obvious, to get the matrix for a composition of Möbius transformations you multiply the matrices, in the same order.
The problem is largely about the Pell equations $ p^2 - 2 q^2 = \pm 1.$ You have an integer matrix
$$
M =
\left(
\begin{array}{rr}
1 & 2 \\
1 & 1
\end{array}
\right)
$$ so
$$
M^2 =
\left(
\begin{array}{rr}
3 & 4 \\
2 & 3
\end{array}
\right)
$$ and
$$
M^3 =
\left(
\begin{array}{rr}
7 & 10 \\
5 & 7
\end{array}
\right)
$$ and
$$
M^4 =
\left(
\begin{array}{rr}
17 & 24 \\
12 & 17
\end{array}
\right)
$$
Let us name
$$ M^n =
\left(
\begin{array}{rr}
p_n & 2 q_n \\
q_n & p_n
\end{array}
\right).
$$
Various things happen, because we are dealing with the continued fraction for $\sqrt 2.$ For example,
$$ p_n^2 - 2 q_n^2 = ( -1)^n. $$
Furthermore
$$ p_{n+2} = 2 p_{n+1} + p_n, $$
$$ q_{n+2} = 2 q_{n+1} + q_n. $$
Here we have
$$ p_1 = 1, p_2 = 3, p_3=7, p_4 = 17, $$
$$ q_1 = 1, q_2 = 2, q_3 = 5, q_4 = 12, $$
Note that both $p,q$ got to infinity, both can be written as a constant times $(1 + \sqrt 2)^n$ plus some other constant times $(1 - \sqrt 2)^n,$ the former larger than $1$ and the latter between $-1$ and $0.$
We do not need much specific other than
$$ \frac{p^2}{q^2} - 2 = \frac{(-1)^n}{q^2} $$
so that the limit of $p/q$ really is $\sqrt 2.$ Here I begin to drop the subscript $n$'s on the $p,q$ numbers.
With initial value $a > 0,$ your fraction is the ratio of the entries in
$$
\left(
\begin{array}{rr}
p_n & 2 q_n \\
q_n & p_n
\end{array}
\right)
\left(
\begin{array}{r}
a \\
1
\end{array}
\right) \mapsto
\frac{a p_n + 2 q_n}{a q_n + p_n},
$$
or
$$
\frac{a p + 2 q}{a q + p}
$$
as an abbreviation.
Not as much trickery needed as I had first thought. From here we get
$$
\frac{a p + 2 q}{a q + p} =
\frac{a \frac{p}{q} + 2 }{a + \frac{p}{q}} =
\frac{a \frac{p}{q} + \frac{p^2}{q^2} }{a + \frac{p}{q}} + \frac{2 - \frac{p^2}{q^2}}{a + \frac{p}{q}} =
\frac{p}{q} + \frac{ \frac{(-1)^{n+1}}{q^2}}{a + \frac{p}{q}}
$$
so that
$$
\frac{a p + 2 q}{a q + p} = \frac{p}{q} \; + \; \frac{ (-1)^{n+1}}{a q^2 + pq}
$$
That finishes it. Note that
$$ \left| \frac{p}{q} - \sqrt 2 \right| < \frac{1}{q^2}. $$
With
$$ q_n = \frac{1}{2 \sqrt 2} \left(1 + \sqrt 2 \right)^n - \frac{1}{2 \sqrt 2} \left(1 - \sqrt 2 \right)^n $$
we see that the distance from $\sqrt 2$ becomes quite small.
Meanwhile
$$ p_n = \frac{1}{2 } \left(1 + \sqrt 2 \right)^n + \frac{1}{2 } \left(1 - \sqrt 2 \right)^n $$
|
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|
Does $8a+5$ ever divide $b^2+8$? For natural $a,b$, does $8a+5$ ever divide $b^2+8$ ?
It doesn't for $b$ up to $10^7$.
Couldn't find congruence obstructions for moduli up to $500$.
$b^2+8$ can be even.
|
You say you couldn't find a congruence obstruction for moduli up to $500$, but there is such an obstruction to having $-8 \equiv b^2 \bmod 8a+5$, revealed by Jacobi reciprocity. Since $-8$ is a unit modulo $8a+5$, so would be $b$, and therefore using Jacobi symbols we would get $(\frac{-8}{8a+5}) = (\frac{b^2}{8a+5}) = 1$. Positivity of $a$ is used in the meaning of the Jacobi symbol that was just written down and, more importantly, in computational rules for the Jacobi symbol that will now be used.
Since $8a+5\equiv 5 \bmod 8$ we have $(\frac{2}{8a+5}) = -1$, and $(\frac{-1}{8a+5}) = (-1)^{((8a+5)-1)/2} = 1$, so $(\frac{-8}{8a+5}) = (-1)^3 = -1$. This is a contradiction.
|
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|
The expansion of $(a+b+c+d)^{20}$ Let us consider the expansion of $$(a+b+c+d)^{20}.$$
Find:
*
*The coefficients of $a^{11}b^6c^2d$ and $a^{11}b^9$,
*The total number of terms of this expansion,
*The sum of all the coefficients.
Thank you for your help.
|
Observe that each term in the expansion of $$(a+b+c+d)^{20}=(a+b+c+d)(a+b+c+d) \cdots (a+b+c+d)$$ is obtained by taking one of the four terms $a,b, c$ or $d$ from each of the $20$ factors. Hence, the coefficient of $a^{11}b^6 c^2 d$ is the number of ways to choose $11$ of the $20$ factors for $a$ (which is ${20 \choose 11}$), times the number of ways to choose $6$ of the remaining $9$ factors for $b$ (which can be done in ${9 \choose 6}$ ways), times the number of ways to choose factors for $c$ (which can be done in ${3 \choose 2}$ ways), times the number of ways to choose the factors for $d$ (which can be done in ${1 \choose 1}$ ways). Hence, the coefficient in question is the product ${20 \choose 11}{9 \choose 6}{3 \choose 2}{1 \choose 1}$, which is equal to the multinomial coefficient $\frac{20!}{11! 6! 2! 1!}$. The coefficient of $a^{11} b^9$ can be obtained similarly.
To obtain the total number of terms in this expansion, observe that one term is of the form $\alpha a^7 b^7 c^6$ for some constant $\alpha$, and another is of the form $\beta a^6 b^6 c^6 d^2$. The former term corresponds to the partition $(7,7,6,0)$ of $20$ into $4$ parts, and the latter term corresponds to the partition $(6,6,6,2)$ into 4 parts. The number of ways to divide $20$ into $4$ parts is ${20+4-1 \choose 3}={23 \choose 3}$. To see why, place $20$ beads in a straight horizontal line, and you want to add three vertical lines (separator bars) to obtain the partition of $20$ into $4$ parts. The number of permutations of $20$ identical beads and $3$ identical bars is $\frac{23!}{3! 20!}$.
The sum of all coefficients in an expansion $\alpha a^7 b^7 c^6+ \beta a^6 b^6 c^6 d^2 \cdots$ is the sum $\alpha + \beta + \cdots$. Observe that the latter is obtained from the former by substituting $1$ for $a,b,c$ and $d$. Hence, we can substitute $1$ for $a,b,c$ and $d$ in the left hand side, ie in $(a+b+c+d)^{20}$, and we get $4^{20}$.
|
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|
How would you find the exact roots of $y=x^3+x^2-2x-1$? My friend asked me what the roots of $y=x^3+x^2-2x-1$ was.
I didn't really know and when I graphed it, it had no integer solutions. So I asked him what the answer was, and he said that the $3$ roots were $2\cos\left(\frac {2\pi}{7}\right), 2\cos\left(\frac {4\pi}{7}\right)$ and $2\cos\left(\frac {8\pi}{7}\right)$.
Question: How would you get the roots without using a computer such as Mathematica? Can other equations have roots in Trigonometric forms?
Anything helps!
|
Consider the equation $$\cos4\theta=\cos3\theta$$ whose roots are $$\theta=n\cdot\frac{2\pi}{7}$$
Representing this as a polynomial in $c=\cos\theta$, we have $$8c^4-4c^3-8c^2+3c+1=0$$
$$\Rightarrow (c-1)(8c^3+4c^2-4c-1)=0$$
Now write $x=2c$ and we see that the polynomial equation $$x^3+x^2-2x-1=0$$ has roots as stated in your question. Note that $$2\cos\frac{6\pi}{7}=2\cos\frac{8\pi}{7}$$
|
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|
Common tangents to circle $x^2+y^2=\frac{1}{2}$ and parabola $y^2=4x$ I'm having trouble with this. What i do is say $\epsilon: y=mx+b$ is the tangent and it meets the circle at $M_1(x_1,y_1)$, i equate the $y$ of the tangent with the circle: $y=\pm \sqrt{1/2-x^2}$ and then the same with the parabola at $M_2(x_2,y_2)$, but i can't reach a result. I've also tried with this version of the tangent to the circle: $xx_1+yy_1=1/2$. By theory we know that a conic section has a tangent at a given point when the discriminant is zero when we equate the two. I'm very confused. I don't know how to solve this. If someone could help i would be very grateful. Thanks in advance.
Edit
Below i add a figure of the graph i made with Matlab
|
Let $P = (h,k)$ be a point on the circle $x^2 + y^2 = \dfrac 12$.
$\left( \text{Then}\; h^2 + k^2 = \dfrac 12 \right)$. The origin, $O = (0,0)$ is the center of the circle. So the equation of the line
$\overleftrightarrow{OP}$ is $kx - hy = 0$
A line tangent to the line $kx - hy = 0$ must have an equation of the form $hx + ky = C$ for some number $C$. Since we want the line to pass through the point $(h,k)$, then $C = h^2 + k^2 = \dfrac 12$. So the equation of the line tangent to the circle $x^2 + y^2 = \dfrac 12$ at the point $(h,k)$ is $hx + ky = \dfrac 12$.
This will intersect the parabola $y^2 = 4x$ when
\begin{align}
y^2 &= 4x \\
hx + ky &= \dfrac 12\\
\hline
\dfrac h4 y^2 + ky &= \dfrac 12\\
hy^2 +4ky-2 &= 0\\
\end{align}
This is a quadratic equation. The line will be tangent to the parabola only if this equation has exactly one solution. That means that the discriminant must be equal to $0$.
\begin{align}
B^2 - 4AC &= 0 \\
16k^2 + 8h &= 0 \\
h = -2k^2
\end{align}
We solve for $h$ and $k$
\begin{align}
h^2 + k^2 &= \dfrac 12 \\
h &= -2k^2 \\
\hline
4k^4 + k^2 &= \dfrac 12\\
k^4 + \dfrac 14 k^2 &= \dfrac 18 \\
k^4 + \dfrac 14 k^2 + \dfrac{1}{64}&= \dfrac{9}{64} \\
k^2 + \dfrac 18 &= \pm \dfrac{3}{8} \\
k &= \pm \dfrac 12
\end{align}
This gives us the four candidates
$(h,k) = \left( \pm \dfrac 12, \pm \dfrac 12\right)$, wich corresponds to the four possible tangent lines $y = \pm x \pm 1$. All four of these lines will be tangent to the circle and will intersect the parabola in only one point. But only two of them are going to be tangent to the parabola.
A quick check show that the lines $y = -x -1$ and $y = x + 1$ are the tangent lines.
|
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|
Prove that the square of an integer $a$ is of the form $a^2=3k$, or $a^2=3k+1$, where $k\in \mathbb{Z} $ Here's my attempt to prove this. I'm not sure about it, but i hope i'm correct.
Let $a=λ, λ\in \mathbb{Z}$. Then $a^2=λ^2=3\frac{λ^2}{3}$.
When $λ^2$ is divided by 3 there are three possible remainders $0,1,2$. So
$$ λ^2=3k \lor λ^2=3k+1 \lor λ^2=3k+2 $$
*
*$ λ^2=3k \rightarrow a^2=3\frac{3k}{3}=3k $
*$ λ^2=3k+1 \rightarrow a^2=3\frac{3k+1}{3}=3k+1 $
*$ λ^2=3k+2 \rightarrow a^2=3\frac{3k+2}{3}=3k+2=3μ, μ\in \mathbb{Z} $
Therefore $$a^2=
\begin{cases}
3k \\
3k+1
\end{cases}
$$
where $k$ is an integer.
Can someone verify this? Any help will be appreciated. Thanks in advance!
|
I don't understand how you eliminated the $3k+2 $ case. It might be easier just to look at the cases $a=3\lambda, a=3\lambda+1, a=3\lambda+2$ and multiply out. as follows:
When $a$ is divided by $3$, there are three possible remainders over an exact multiple of $3$: $0, 1,$ and $2$. Therefore consider the three cases
*
*$a=3\lambda$
*$a=3\lambda+1$
*$a=3\lambda+2$
Then $a^2$ calculates out as follows:
*
*$a=3\lambda \implies a^2 = 9 \lambda^2 = 3(3 \lambda^2)$
*$a=3\lambda+1\implies a^2 = 9 \lambda^2 + 6\lambda + 1 = 3(3 \lambda^2 + 2\lambda)+1$
*$a=3\lambda+2\implies a^2 = 9 \lambda^2 + 12\lambda + 4 = 3(3 \lambda^2 + 6\lambda+1)+1$
giving a suitable value for $k$ in each case.
|
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|
Summation of Binomial Coefficient: $\sum\binom{n+k}{2k} \binom{2k}k \frac{(-1)^l}{k+1}$ I am trying to solve this summation problem .
$$\sum\limits_{k = 0}^\infty {\left( {\begin{array}{*{20}{l}}
{n + k}\\
{2k}
\end{array}} \right)} \left( {\begin{array}{*{20}{l}}
{2k}\\
k
\end{array}} \right)\frac{{{{( - 1)}^k}}}{{k + 1}}$$
It will be grateful if someone could help me !!
|
Suppose we seek to evaluate
$$\sum_{k=0}^n {n+k\choose 2k} {2k\choose k}\frac{(-1)^k}{k+1}
= \sum_{k=0}^n {n+k\choose n-k} {2k\choose k}\frac{(-1)^k}{k+1}.$$
Introduce
$${n+k\choose n-k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n-k+1}} (1+z)^{n+k}\; dz.$$
This vanishes when $k\gt n$ and we may extend the sum to infinity. We
obtain
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} (1+z)^{n}
\sum_{k\ge 0} {2k\choose k}\frac{(-1)^k}{k+1} z^k (1+z)^k
\; dz.$$
We recognize the generating function of the Catalan numbers which is
$$\sum_{k\ge 0} {2k\choose k}\frac{1}{k+1} w^k
= \frac{1-\sqrt{1-4w}}{2w}.$$
This is certainly analytic in $|w|<1/4.$ We therefore fix $\epsilon$
so that $|z(1+z)|<1/4,$ for example take $\epsilon = 1/10.$
This yields for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} (1+z)^{n}
\frac{1-\sqrt{1+4z(1+z)}}{2\times (-1)\times z(1+z)}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+2}} (1+z)^{n-1}
\frac{\sqrt{(1+2z)^2}-1}{2}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+2}} (1+z)^{n-1}
\frac{2z}{2}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} (1+z)^{n-1}
\; dz.$$
This is $$[z^n] (1+z)^{n-1} = 0.$$
Here we have used the branch of the logarithm with the cut on the negative real axis.
|
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|
Characterize the elements of a set of polynomials Given $$M = \{x^3 + ax^2 + bx + c \in\mathbb Z_3[x]\}$$ and given $$A=\{f \in M \mid\overline{1} \text{ is a root of }\ f\}.$$
The exercise asks to characterize (tell) what are the elements of $A$ and how many of them there are. My first attempt at solving this exercise was by using Ruffini's rule:
I divided $$x^3 + ax^2 + bx + c$$ by $$x-1$$ and the result is $$(x^2+\overline2 ax + \overline2 x + b)(x-\overline1) + c \in\mathbb Z_3[x]$$Is this a correct solution for this exercise?
|
You want all polynomials $x^3+ax^2+bx+c$ such that $1+a+b+c=0\bmod3\ \ (*)$. Pick any $a$ and any $b$, then $(*)$ holds iff $c=2-a-b\bmod3$. So we have a total of 9 solutions. If you want them explicitly:
$x^3+2,x^3+x+1,x^3+2x$,
$x^3+x^2+1,x^3+x^2+x,x^3+x^2+2x+2$,
$x^3+2x^2,x^3+2x^2+x+2,x^3+2x^2+2x+1$.
|
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|
how to solve $x^{113}\equiv 2 \pmod{143}$ I need to solve $x^{113} \equiv 2 \pmod{143}$
$$143 = 13 \times 11$$
I know that it equals to $x^{113}\equiv 2 \pmod{13}$ and $x^{113}\equiv 2 \pmod{11}$
By Fermat I got
1) $x^{5} \equiv 2 \pmod{13}$
2) $x^{3} \equiv 2 \pmod{11}$
Now I'm stuck..
|
By the Chinese remainder theorem, you have to solve first
$$\begin{cases}x^{113}\equiv2\mod13\\x^{113}\equiv2\mod11\end{cases}$$
Now Little Fermat says for any $x\not\equiv 0\mod 13\enspace(\text{resp. }11)$, one has $x^{12}\equiv 1\mod 13$, resp. $x^{10}\equiv 1\mod 11$. Hence the system of congruences is equivalent to
$$\begin{cases}x^{113\bmod 12}\equiv x^5\equiv2\mod13,\\x^{113\bmod 10}\equiv x^3\equiv2\mod11.
\end{cases}$$
Let's examine the different possibilities for $x$.
We can eliminate the values $0$ and $1$. Hence,
*
*modulo $13$, let's draw a table, using the fast exponentiation algorithm:
$$\begin{matrix}
x& \pm 2&\pm 3&\pm4 &\pm 5 & \pm 6\\
\hline
x^2&4&-4&3&-1&-3\\
x^4&3&3&-4&1&-4\\
\hline
x^5&\pm6&\pm6&\mp3&\pm5&\pm 2
\end{matrix}$$
Thus the solution is $\;\color{red}{x\equiv6\mod 13}$.
*modulo $11$, we have
$$\begin{matrix}
x& \pm 2&\pm 3&\pm4 &\pm 5\\
\hline
x^2&4&-2&5&-1\\
\hline
x^3&\mp3&\pm5&\mp 2&\mp5
\end{matrix}$$
Here the solution is $\;\color{red}{x\equiv-4\mod 11}$.
To solve this system of simultaneous congruences, we need a Bézout's relation between $13$ and $11$:
$$6\cdot 11-5\cdot13=1.$$
The solutions of the system of congruences is
$$x\equiv\color{red}{6}\cdot 6\cdot 11-(\color{red}{-4})\cdot5\cdot 13\equiv656\equiv \color{red}{84 \mod 143}.$$
|
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|
How to obtain this factorization of $x^4+4$? $x^4 + 4 = (x^2 + 2x +2)(x^2 - 2x +2)$
I am curious how would one obtain this factorization?
Clearly, once the factorization is known it is routine to verify it, however the hard part is how to find the factorization in the first place?
Thanks!
I observed that $(A+B)(A-B)=A^2-B^2$ can be applicable here with $A=x^2+2$, $B=2x$.
Is there any other trick?
|
This is pretty straight forward if you employ complex numbers and this formula: $$(a+b)(a-b)=a^2-b^2$$
Here we go:
\begin{align}
x^4 + 4 & = (x^2)^2 - (-4) \\
& = (x^2)^2 - (2i)^2 \\
& = (x^2 + 2i) \ (x^2 - 2i) \\
& = (x+ i\sqrt{2i}) \ (x- i\sqrt{2i}) \ (x+ \sqrt{2i}) \ (x- \sqrt{2i})
\end{align}
We know that $\sqrt{i}=e^{i{\pi\over4}} = {1+i\over\sqrt{2}}$
Thus,
\begin{align}
x^4 + 4 & = (x+ i\sqrt{2i}) \ (x- i\sqrt{2i}) \ (x+ \sqrt{2i}) \ (x- \sqrt{2i}) \\
& = (x+ 1-i) \ (x- 1+i) \ (x+ 1+i) \ (x- 1-i)\\
&= (x \pm 1 \pm i)
\end{align}
You can also get these roots by using the polar form of complex number, as mentioned by Ross Millikan, or you can also find the roots of the quadratic equations, as written by azc.
|
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|
Finding the action of T on a general polynomial given a basis I am given a question as follows:
Suppose $T: P_{2} \rightarrow M_{2,2}$ is a linear transformation whose action on a basis for $P_{2}$ is
$$T(2x^2+2x+2)=\begin{bmatrix} 2&4 \\ 2&8 \end{bmatrix} \\T(4x^2+2x+2)=\begin{bmatrix}4&6 \\ -2&10 \end{bmatrix} \\T(1)=\begin{bmatrix} 1&1 \\ 2&1 \end{bmatrix}.$$
Find the action of $T$ on a general polynomial with $a,b,c$ as constants.
Assuming that $T$ is a linear transformation, then for any vectors $\vec{v_{1}}, ..., \vec{v_{n}}$ and any scalars $a_{1}, ..., a_{n}$, $T(a_{1}\vec{v_{1}}+...+a_{n}\vec{v_{n}})=a_{1}T(\vec{v_{1}})+...+a_{n}T(\vec{v_{n}})$.
The answer is supposed to be in matrix form.
What I have tried doing so far is taking the basis polynomials and putting them into a matrix, which I then multiplied by a constant vector consisting of $a,b,c$:
$$\begin{bmatrix} 2&4&0 \\ 2&2&0 \\ 2&2&1\end{bmatrix} \begin{bmatrix} a \\ b \\c \end{bmatrix}= \begin{bmatrix} 2a+4b \\ 2a+2b \\ 2a+2b+c \end{bmatrix}$$
Given what I have done so far, I am not sure if this is the right approach, and I am not sure where to proceed from here. Any help would be appreciated!
|
Hint:
Let $p_1 = 2x^2 + 2x + 2$, $p_2 = 4x^2 + 2x + 2$, $p_3 = 1$. Then
$$
-\frac{1}{2}p_1 + \frac{1}{2}p_2 = x^2, \\
p_1 -\frac{1}{2}p_2 - p_3 = x, \\
p_3 = 1.
$$
Thus
$$
ax^2 + bx + c = \left(-\frac{a}{2} + b\right)p_1 + \frac{a-b}{2}p_2 + (c - b)p_3
$$
and as $T$ is a linear operator we get
$$
T(ax^2 + bx + c) = \left(-\frac{a}{2} + b\right)T(p_1) + \frac{a-b}{2}T(p_2) + (c - b)T(p_3) = \\
\left(-\frac{a}{2} + b\right)\begin{bmatrix} 2&4 \\ 2&8 \end{bmatrix} + \frac{a-b}{2}\begin{bmatrix}4&6 \\ -2&10 \end{bmatrix} + (c - b)\begin{bmatrix} 1&1 \\ 2&1 \end{bmatrix} = \\
= \begin{bmatrix} a-b+c&a+c \\ -2a+b+2c&a+2b+c \end{bmatrix}.
$$
|
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|
Evaluation of Irrational Integral
Evaluation of $$\int\frac{x^4}{(1-x^4)^{\frac{3}{2}}}dx$$
$\bf{My\; Try::}$ Let $$I = \int\frac{x^4}{(1-x^4)^{\frac{3}{2}}}dx = -\frac{1}{4}\int x\cdot \frac{-4x^3}{(1-x^{4})^{\frac{3}{2}}}dx$$
Using Integration by parts, We get
$$I =\frac{x}{2(1-x^4)^{\frac{1}{2}}}-\int\frac{1}{(1-x^4)^{\frac{1}{2}}}dx$$
Now How can I solve after that, Help required, Thanks
|
An elliptic integral (of the first kind) is one of the form
$$
F(t\mid k^2)=\int\frac{1}{\sqrt{1-k^2\sin^2t}}\,dt
$$
or by substituting $x=\sin t$
$$
G(x\mid k^2)=\int\frac{1}{\sqrt{(1-x^2)(1-k^2x^2)}}\,dx
$$
In our case, the integral
$$
\int\frac{1}{\sqrt{1-x^4}}\,dx
$$
is $G(x\mid-1)$.
|
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|
If $f(x) = \frac{\cos x + 5\cos 3x + \cos 5x}{\cos 6x + 6\cos4x + 15\cos2x + 10}$then.. If $f(x) = \frac{\cos x + 5\cos 3x + \cos 5x}{\cos 6x + 6\cos4x + 15\cos2x + 10}$ then find the value of $f(0) + f'(0) + f''(0)$.
I tried differentiating the given. But it is getting too long and complicated. So there must be a way to simplify $f(x)$. What is it?
|
Try to simplify the denominator of f(x):
Note that by the identity $\cos(2t)=2\cos^2t-1$
$$
\cos(3t)+6\cos(2t)+15\cos t+10=4\cos^3t+12\cos^2t+12\cos t+4=4(\cos t+1)^3
$$
so
$$
\cos(6x)+6\cos(4x)+15\cos(2x)+10=4(\cos(2x)+1)^3
$$
Therefore
$$
f(x)=\frac{1}{4}\cdot\overbrace{\frac{\cos x}{(\cos(2x)+1)^3}}^{(1)}+\frac{5}{4}\cdot\overbrace{\frac{\cos(3x)}{(\cos(2x)+1)^3}}^{(2)}+\frac{1}{4}\cdot\overbrace{\frac{\cos(5x)}{(\cos(2x)+1)^3}}^{(3)}
$$
Now, in order to compute $f'(x)$ and $f''(x)$ compute $(i)'$ and $(i)''$ for each $i\in\{1,2,3\}$. Then assign $x=0$.
|
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|
Prove $\lim_{(x,y)\to (0,0)} \frac{x^2 y^3}{x^4 + y^4} =0$ without $\varepsilon - \delta$. Unlike Multivariable Delta Epsilon Proof $\lim_{(x,y)\to(0,0)}\frac{x^3y^2}{x^4+y^4}$ --- looking for a hint I would like to avoid the $\varepsilon - \delta$ criterium.
Prove $$\lim_{(x,y)\to (0,0)} \frac{x^2 y^3}{x^4 + y^4} =0 \,.$$
Approaching this limit from $y=0$, $x=0$, $y=x$, $y=x^2$ etcetera all yields 0 as value, so my proposal is that this limit is indeed 0.
I have been able to solve most similar limits so far by finding some convergent upper bound for the absolute limit, but with this one the difference between the numerator and the denominator is so small I can't find anything to fit inbetween. For example, $(x,y)\to(0,0)$,
$$ \left| \frac{x^2 y}{x^2 + y^2} \right| \le \left| \frac{(x^2 + y^2)y}{x^2 + y^2} \right| \to 0 \,. $$
Also, Continuity of $\frac{x^3y^2}{x^4+y^4}$ at $(0,0)$? and Proving $ \frac{x^3y^2}{x^4+y^4}$ is continuous. contain some helpful hints.
|
Hint 1: $$\frac{x^2 y^3}{x^4+y^4}=\frac{y}{(x/y)^2+(y/x)^2}. $$
Hint 2: $f(u)=u^2+u^{-2}$ has an absolute minimum.
|
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|
If $x= m-m^2-2$ then find $x^4+3x^3+2x^2-11x+6$ where m is a cube root of unity If $$x= m-m^2-2$$ then find $$x^4+3x^3+2x^2-11x+6$$ where $m$ is a cube root of unity.
My try:
Since $ m+ m^2+1=0$ the value of $x$ is $-1$.
Let $f(x)=x^4+3x^3+2x^2-11x+6$
then $ f(-1)=5$ So the answer is $5$.
Am I correct? In my book it says that answer is $1$. Please help me to identify my mistake.
|
$$x=m-m^2-2=-(m+m^2+1)+2m-1=2m-1$$
And $2m-1 \neq1$, here is your mistake. Then we can continue our calculations :
$$f(x)=x^4+3x^3+2x^2-11x+6\\=16m^4-32m^3+24m^2-8m+1+24 m^3-36 m^2+18 m-3+8 m^2-8 m+2-22m+11+6\\=16 m^4-8 m^3-4 m^2-20 m+17\\=16m^2(m^2+m+1)-24m^3-20m^2-20m+17\\=-24m^3-20m^2-20m+17=-24m(m^2+m+1)+4m^2+4m+17\\=4m^2+4m+17\\=4(m^2+m+1)+13\\=13$$
|
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|
Finding the maximum value of $n$ What's the maximum value for $n$ for which there is a set of distinct positive integers $k_{1}, k_{2}, ... , k_{n}$ for which $k_{1}^{2} + k_{2}^{2} + ... + k_{n}^{2} = 2002$
|
$1^2+2^2+3^2+4^2+5^2+6^2+7^2+8^2+9^2+10^2+11^2+12^2+14^2+16^2+18^2+24^2=2002$ which uses 16 distinct squares.
$1^2+\dots+18^2=2109>2002$, so any 18 distinct squares have too large a sum. So the only question is whether it can be done with 17.
Note that $1^2+2^2+\dots+17^2=1785=2002-217$ and $18^2-10^2=224>217$, so we must include $1^2+2^2+\dots+10^2=385$ if we have 17 squares.
If we start from $1^2+\dots+17^2=1785$, and add a square larger than $18^2$ we are adding at least $19^2=361$, so we must drop at least $12^2$. That works!
$1^2+\dots+11^2+13^2+14^2+15^2+16^2+17^2+19^2=2002$ with 17 squares.
|
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|
For which $a$ and $b$ is this matrix diagonalizable? For which $a$ and $b$ is this matrix diagonalizable?
$$A=\begin{pmatrix} a & 0 & b \\ 0 & b & 0 \\ b & 0 & a \end{pmatrix}$$
How to get those $a$ and $b$? I calculated eigenvalues and eigenvectors, but don't know what to do next?
|
Calculating the eigenvalues, $b, b+a, a-b,$ one can then easily calculate their respective eigenvectors. For eigenvalue $b,$
\begin{equation}
\begin{pmatrix} a-b&0&b\\0&0&0\\b&0&a-b \end{pmatrix}\begin{pmatrix} 0\\1\\0\end{pmatrix} = \begin{pmatrix} 0\\0\\0\end{pmatrix},
\end{equation}
There may be special cases, such as if $a = 0,$ then the vector $\begin{pmatrix} 1\\0\\1\end{pmatrix}$ works too.
Likewise for $a+b,$
\begin{equation}
\begin{pmatrix} -b&0&b\\0&-a&0\\b&0&-b \end{pmatrix}\begin{pmatrix} 1\\0\\1\end{pmatrix} = \begin{pmatrix} 0\\0\\0\end{pmatrix},
\end{equation}
And last for $a-b,$
\begin{equation}
\begin{pmatrix} b&0&b\\0&2b-a&0\\b&0&b \end{pmatrix}\begin{pmatrix} 1\\0\\-1\end{pmatrix} = \begin{pmatrix} 0\\0\\0\end{pmatrix}.
\end{equation}
Because no assumptions about the values of $a$ and $b$ were made along the way, this operator must be diagonalizable (which we already knew from symmetry), with an eigenbasis \begin{equation} \begin{pmatrix} 0\\1\\0\end{pmatrix}, \begin{pmatrix} 1\\0\\-1\end{pmatrix},
\text{ and }
\begin{pmatrix} 1\\0\\1\end{pmatrix}\end{equation}
Note that if one remembers that matrices of the form $\begin{pmatrix} a & b\\ b& a \end{pmatrix} $ have the eigenbasis $\begin{pmatrix} 1 & 1\\ 1& -1 \end{pmatrix},$
then from a quick glance at the original matrix, you can already "know" the eigenvectors since this $2 \times 2$ is hidden within it.
|
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|
Roots of $y=x^3+x^2-6x-7$ I'm wondering if there is a mathematical way of finding the roots of $y=x^3+x^2-6x-7$?
Supposedly, the roots are $2\cos\left(\frac {4\pi}{19}\right)+2\cos\left(\frac {6\pi}{19}\right)+2\cos\left(\frac {10\pi}{19}\right)$, $2\cos\left(\frac {2\pi}{19}\right)+2\cos\left(\frac {14\pi}{19}\right)+2\cos\left(\frac {16\pi}{19}\right)$ and $2\cos\left(\frac {8\pi}{19}\right)+2\cos\left(\frac {12\pi}{19}\right)+2\cos\left(\frac {18\pi}{19}\right)$.
Anything helps. I don't think substituting $x$ with something like $t+t^{-1}$ will work.
|
We may take $\xi$ as a primitive $19$-th root of unity, $\xi=\exp\left(\frac{2\pi i}{19}\right)$, then check that the elementary symmetric functions of
$$ a = \xi^2+\xi^3+\xi^5+\xi^{14}+\xi^{16}+\xi^{17} $$
$$ b = \xi^4+\xi^6+\xi^9+\xi^{10}+\xi^{13}+\xi^{15} $$
$$ c = \xi^1 + \xi^7 + \xi^8+\xi^{11}+\xi^{12}+\xi^{18} $$
simplify giving the coefficients of $x^3+x^2-6x-7$. That depends on a Kummer sum, the cubic equivalent of a Gaussian period.
For short: the Galois group of the minimal polynomial of $\xi$ is $\mathbb{Z}/(18\mathbb{Z})$, and the polynomial $x^3+x^2-6x-7$ encodes a subgroup of such a group. Notice that $\{1,7,8,11,12,18\}$ are exactly the cubic residues in $\mathbb{Z}/(19\mathbb{Z})^*$, that is generated by $2$, and $\!\!\pmod{19}$
$$2\cdot\{1,7,8,11,12,18\}=\{2,3,5,14,16,17\},$$
$$2^2\cdot\{1,7,8,11,12,18\}=\{4,6,9,10,13,15\},$$
giving the exponents associated with $c,a,b$.
We may perform the same trick with the prime $p=13\equiv 1\pmod{3}$. $\mathbb{Z}/(p\mathbb{Z})^*$ is generated by $2$ and the cubic residues $\!\!\pmod{p}$ are $\{1,5,8,12\}$, hence if we take $\xi=\exp\left(\frac{2\pi i}{13}\right)$ we have that:
$$ x_1 = \xi^1+\xi^5+\xi^8+\xi^{12} = 2\cos\left(\frac{2\pi}{13}\right)+2\cos\left(\frac{10\pi}{13}\right),$$
$$ x_2 = \xi^2+\xi^3+\xi^{10}+\xi^{11} = 2\cos\left(\frac{4\pi}{13}\right)+2\cos\left(\frac{6\pi}{13}\right),$$
$$ x_3 = \xi^4+\xi^6+\xi^7+\xi^{9} = 2\cos\left(\frac{8\pi}{13}\right)+2\cos\left(\frac{12\pi}{13}\right),$$
are conjugated algebraic numbers, roots of the polynomial
$$ p(x) = x^3+x^2-4x+1.$$
|
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|
Find the value of $h$ if $x^2 + y^2 = h$
Consider equation $x^2 + y^2 = h$ that touches the line $y=3x+2$ at some point $P$. Find the value of $h$
I know that $x^2 + y^2 = h$ is a circle with radius $\sqrt{h}$.
Also, since $y = 3x + 2 $ is a tangent, we know that the slope of the radius perpendicular to the tangent is $M_{OP}= -\frac{1}{3}$. I'm not sure how I can determine the value of $h$ though?
|
So we have equation $x^2 + y^2 = r^2$, who’s geometrical representation would be a circle with the radius $r$. Now we also have an equation of a line represented by: $y = mx + c$ that touches the circle at point $P$. To find this point we will use the method of solving simultaneous equation where one is a quadratic and the other is linear so we will take the value of y and replace it in the first equation as follows:
$$x^2 + (3x + 2)^2 = h$$
$$x^2 + 9x^2 + 2*3x*2 + 4 = h$$
$$10x^2 + 12x + 4 = h$$
$$10x^2 + 12x + 4 - h = 0$$
so now we will have something like:
$$x^2 + (mx + c)^2 = r^2$$
$$x^2 + m^2x^2 + 2mcx + c^2 = r^2$$
$$(1 + m^2)x^2 + 2mcx + c^2 - r^2 = 0$$
since this equation is a quadratic equation in x, something like $(ax^2 + bx + c = 0)$ where we have:
$a = 10$; $b = 12$; $c = 4 - h$;
using the values in discriminant $(b^2 - 4ac)$:
$$b^2 - 4ac = (2mc)^2 - 4(1 + m^2)(c^2 - r^2)$$
$$= 4m^2c^2 - 4(c^2 + m^2c^2 - r^2 - m^2r^2)$$
$$ = 4m^2c^2 - 4c^2 - 4m^2c^2 + 4r^2 + 4m^2r^2$$
$$ = -4c^2 + 4r^2 + 4m^2r^2$$
if the discriminant is zero, our equation will have equal roots and the line intersects the circle in one single point which is our case also. So because this line is a tangent to the circle we know:
$$-4c^2 + 4r^2 + 4m^2r^2 = 0$$
$$4r^2 + 4m^2r^2 = 4c^2$$
$$c^2 = r^2 + m^2r^2$$
this will be the condition for a line to be tangent to the circle, so replacing with our values we will have the tangent:
$$(4 - h)^2 = h + 3^2h$$
$$h^2 - 8h + 16 = h + 9h$$
$$h^2 - 18h = -16$$
$$h(h - 18) = -16$$
since:
$$(h-9)^2 = (h-9)(h-9)$$
$$h^2 - 18h + 81$$
$$h(h -18) + 81$$
we can write:
$$(h-9)^2 - 65 = 0$$ so:
$$h = 9 \pm \sqrt65$$
I hope this helps
|
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|
Proving Gerretsen's Inequality Today in class we were shown Gerretsen's inequality:
$$16Rr-5r^2\leq s^2 \leq 4R^2+4Rr+3r^2$$
Where $R$, $r$, and $s$ are the circumradius, in radius, and semiperimeter of a triangle. After some research, I found that this inequality was proven by showing the following:$$IH^2=4R^2+4Rr+3r^2-\frac{1}{4}\left(a+b+c\right)^2$$
$$9IG^2=\frac{1}{4}(a+b+c)^2-16Rr+5r^2$$
Where $IH$ is the distance between the incenter and the orthocenter, $IG$ is the distance between the incenter and the centroid, and $a$, $b$, and $c$ denote the sides of the triangle. How are these two equalities derived? Any help is very much appreciated.
|
Your question is very interesting. Let me give my idea.
Firstly, we will compute $IG$. Note that we have $$\vec{GA} + \vec{GB} + \vec{GC} = \vec{0}.$$ Then, $$3\vec{IG} = \vec{IA} + \vec{IB} + \vec{IC}.$$
$$9IG^2 = \sum_{\circlearrowleft} IA^2 + 2\sum_{\circlearrowleft}\vec{IA}\vec{IB} $$
Recall that $IA^2 = r^2 + (s-a)^2$ (I don't go to this detail, but you can see it easily), and $2\vec{IA}\vec{IB} = IA^2 + IB^2 - AB^2$.
Then, one has: \begin{eqnarray}9IG^2 &=& 3\sum_{\circlearrowleft} IA^2 - \sum_{\circlearrowleft} a^2 = 9r^2 + 3[(s-a)^2 + (s-b)^2 + (s-c)^2] - (a^2+b^2+c^2) \\ &=& 9r^2 + 5s^2 - 4(ab+bc+ca). \ \ \ (1)\end{eqnarray}
Now, use the Heron's formula, we have $$S = sr = \sqrt{s(s-a)(s-b)(s-c)}$$.
So, $$sr^2 = s^3 - s^2(a+b+c) + s(ab+bc+ca) - abc = - s^3 + s(ab+bc+ca) -abc.$$
Note that $abc= 4RS = 4Rrs$. We get $$sr^2 = -s^3 + s(ab+bc+ca) -4Rrs$$ or $$ab+bc+ca = r^2+s^2+4Rr\ \ \ \ (2)$$
Substitute (2) to (1), we get $$9IG^2 = 9r^2 + 5s^2 - 4(r^2 + s^2 + 4Rr) = 5r^2 + s^2 - 16Rr.$$
EDIT: Let me compute the $IH$. I use two lemmas.
Lemma 1 One has $$a\vec{IA} + b\vec{IB} + c\vec{IC} = \vec{0}.$$
Lemma 2 One has $$HA^2 = 4R^2 - a^2; HB^2 = 4R^2 - b^2; HC^2 = 4R^2-c^2.$$
I don't give the detail prove of two above lemmas, but you can see it like this:
*
*lemma 1: the bisector of angle A divides the side $BC$ by ratio $b:c$, and so on with two other bisectors.
*lemma 2: it is easy to use the Euler line, which implies that $HA$ is two times of the distance from the circumcenter to the side $BC$.
Using lemma 1, we have $$2s \vec{HI} = a\vec{HA} + b\vec{HB} + c\vec{HC}.$$
Then, \begin{eqnarray}4s^2 HI^2 &=& \sum_{\circlearrowleft} a^2HA^2 + 2\sum_{\circlearrowleft} ab \vec{HA}\vec{HB}\\
&=& \sum_{\circlearrowleft} a^2(4R^2 - a^2) + \sum_{\circlearrowleft} ab (HA^2 + HB^2 -AB^2)\ \ \ \ \ \mbox{(using the lemma 2)}\\
&=& 4R^2(a^2+b^2+c^2) - (a^4+b^4+c^4) + \sum_{\circlearrowleft} ab (8R^2 - a^2-b^2-c^2) \\
&=& 4R^2(a^2+b^2+c^2 + 2ab +2bc +2ca) - (a^4+b^4+c^4) - (ab+bc+ca)(a^2+b^2+c^2) \\
&=& 16R^2s^2 - (a^4+b^4+c^4) - (ab+bc+ca)(a^2+b^2+c^2).\end{eqnarray}
I leave the part $(a^4+b^4+c^4) + (ab+bc+ca)(a^2+b^2+c^2)$ for you, because of the simple but long computation.
I give the answer, $$4s^2IH^2 = 16R^2s^2 + 16Rrs^2 + 12r^2s^2 - 4s^4.$$
Then, you get the equality.
|
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|
Expressing Tornheim sums in terms of Riemann's Zeta If
$$T(a,b,c)=\sum_{r\geq1}\sum_{s\geq1} \frac{1}{r^as^b(r+s)^c}$$
How to prove that :
$$T(3,1,2)=-\frac13 \zeta(6)+\frac{\zeta^2(3)}{2}$$
I tried some algebraic manipulations but did not work. Can you please help me ?
Any solution will be appreciated
|
The easiest way to do it is to use partial fraction decomposition. Let us choose the variable for the decomposition to be $s$. Then we have:
\begin{equation}
\frac{1}{r^a s^b (s+r)^c} =
\sum\limits_{l_1=1}^b \binom{b+c-1-l_1}{c-1} \frac{(-1)^{b-l_1}}{s^{l_1} r^{b+b+c-l_1}} +
\sum\limits_{l_1-1}^c \binom{b+c-1-l_1}{b-1} \frac{(-1)^b}{(s+r)^{l_1} r^{a+b+c-l_1}}
\end{equation}
Now we sum over $s$ and we get :
\begin{equation}
\sum\limits_{s=1}^\infty \frac{1}{r^a s^b (s+r)^c} =
\sum\limits_{l_1=1}^b \binom{b+c-1-l_1}{c-1} \frac{(-1)^{b-l_1} \zeta(l_1) 1_{l_1 \ge 2}}{ r^{a+b+c-l_1}} +
\sum\limits_{l_1=1}^c \binom{b+c-1-l_1}{b-1} \frac{(-1)^b(\zeta(l_1) 1_{l_1 \ge 2} - H_r^{(l_1)})}{ r^{a+b+c-l_1}}
\end{equation}
I think that this that doesn't require any explanation except that for $l_1=1$ we are getting singular terms. Therefore in that case we simply apply the following identity $H_r^{(1)} = \sum_{s\ge 1}\left( 1/s - 1/(s+r)\right)$. Now we sum over the variable $r$ and we get:
\begin{eqnarray}
&&T(a,b,c)=\sum\limits_{l_1=1}^b \binom{b+c-1-l_1}{c-1} (-1)^{b-l_1} \zeta(l_1) 1_{l_1 \ge 2} \zeta(a+b+c-l_1) +
\sum\limits_{l_1=1}^c \binom{b+c-1-l_1}{b-1} (-1)^b\zeta(l_1) 1_{l_1 \ge 2} \zeta(a+b+c-l_1) - \\
&&
\sum\limits_{l_1=1}^c \binom{b+c-1-l_1}{b-1} (-1)^b {\bf H}^{(l_1)}_{a+b+c-l_1}(+1)
\end{eqnarray}
where ${\bf H}^{(l)}_n(t) := \sum\limits_{r\ge 1} H_r^{(l)}/r^n t^r$.
Now if we take $(a,b,c)=(3,1,2)$ we get:
\begin{eqnarray}
T(3,1,2)&=& 0 + (-1)^1 \zeta(2) \zeta(4) - \sum\limits_{l_1=1}^2 \binom{2-l_1}{0} (-1)^1 {\bf H}^{(l_1)}_{6-l_1}(+1)\\
&=&(-1)^1 \zeta(2) \zeta(4)+ {\bf H}^{(1)}_5(+1) + {\bf H}^{(2)}_4(+1) \\
&=&(-1)^1 \zeta(2) \zeta(4)+\left(-\frac{1}{2} \zeta(3)^2-1/3 \zeta(2) \zeta(4)+\frac{7}{3} \zeta(6)\right)+\left(+1 \zeta(3)^2 + \frac{4}{3} \zeta(2) \zeta(4)-\frac{8}{3} \zeta(6)\right) \\
&=& \frac{1}{2} \zeta(3)^2 - \frac{1}{3} \zeta(6)
\end{eqnarray}
where in the second line from the bottom we used my answer to Calculating alternating Euler sums of odd powers .
|
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|
finding the angle between two vectors If $a,b$ and $c$ be three vectors such that $|\, a \,|=3$, $|\, b \,|=5$ and $|\, c \,|=7$ and $a+b+c=0$. Then find the angle between $a$ and $b$.
I tried by taking $a=-(b+c)$ and $b=-(c+a)$. But couldn't proceed further. Please help.
|
\begin{align*}
\mathbf{a+b+c} &=\mathbf{0} \\
\mathbf{c} &=-(\mathbf{a+b}) \\
\mathbf{c}\cdot \mathbf{c} &= (\mathbf{a+b})\cdot (\mathbf{a+b}) \\
c^2 &= a^2+b^2+2\mathbf{a\cdot b} \\
2\mathbf{a\cdot b} &= c^2-a^2-b^2 \\
\cos \theta_{ab} &= \frac{c^2-a^2-b^2}{2ab} \\
&= \frac{7^2-3^2-5^2}{2(3)(5)} \\
&= \frac{1}{2} \\
\theta_{ab} &= 60^{\circ}
\end{align*}
|
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|
Compute $\int_{0}^{2\pi}\frac{\sin^2\theta}{5+3\cos\theta}\,d\theta$ I''m stuck in a exercise in complex analysis concerning integration of rational trigonometric functions. I have the solution but I don't understand a specific part. Here it goes:
We want to find $\int_{0}^{2\pi}\frac{\sin^2\theta}{5+3\cos\theta}\,d\theta$.
Let $z=e^{i\theta}$ so that $d\theta=dz/iz$, $\cos \theta = \frac12 (z+z^{-1})$ and $\sin \theta = \frac {1}{2i}(z-z^{-1})$. We have
$$\begin{align}
I&=\int_0^{2\pi} \frac{\sin^2\theta}{5+3\cos \theta}d\theta\\\\
&=\oint_C \frac{-\frac14(z-z^{-1})^2}{5+3\frac12 (z+z^{-1})}\frac{dz}{iz}\\\\
&=\oint_C \frac{\frac i4 (z^2-2+z^{-2})}{5z+ \frac32 z^2 + \frac32}dz\\\\
&=\oint_C \frac{\frac i2 (z^2-2+z^{-2})}{3z^2 + 10z + 3}dz\\\\
&=\frac i2 \oint_C \frac{z^4-2z^2+1}{z^2(3z^2 + 10z + 3)}dz \quad (1)
\end{align}$$
where $C$ is the unit circle in the complex $z$-plane.
If I understand the last step correctly we multipliy numerator/denominator by $z^2$ in order to get rid of the negative power of $z$.
Now for the part that I don't understand in the solution. It is said that
$$(1) = \frac i2 \oint_C \frac{z^4-2z^2+1}{z^23(z+3)(z+\frac13)}dz = \frac i6 \oint_C \frac{z^4-2z^2+1}{z^2(z+3)(z+\frac13)}dz$$
Where does this $3$ come from in the denominator? If I factor out $3z^2 + 10z + 3$ I simply get $(z+3)(z+\frac13)$.
What is even stranger to me is that, after the computation of residues, the author is getting the correct solution of $\frac{2\pi}{9}$ (I verified with Wolfram) while I'm getting $\frac{2\pi}{3}$ so I'm missing a factor of $\frac13$ (i.e. I'm missing that $3$ in the denominator).
Can someone help me with this?
|
$$\begin{align*} 3z^2 + 10z + 3
&= 3 \left(z^2 + \frac{10}{3}z + 1\right) \\
&= 3\left(z^2 + 3z + \frac{1}{3}z + 1\right) \\
&= 3\left(z(z + 3) + \frac{1}{3}(z + 3)\right) \\
&= 3\left(z + \frac{1}{3}\right)(z+3). \end{align*}$$
You are missing a factor of $2$ when you go from step $3$ to step $4$. The denominator should be $3z^2 + 10z + 3$.
|
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|
Smallest number whose $\sin(x)$ in radian and degrees is equal Question:
What is the smallest positive real number $x$ with the property that the sine of $x$ degrees is equal to the sine of $x$ radians?
My try: 0. But zero isn't a positive number. How do I even begin to solve it? I tried taking inverse on both sides of $\sin \theta = \sin x$, but that didn't help.
|
We have
$$\sin (x) = \sin (\beta x)$$
where $\beta := \frac{\pi}{180}$. Using
$$\sin (\alpha x) = \frac{e^{i \alpha x} - e^{-i \alpha x}}{2i}$$
we conclude that the equation $\sin (x) = \sin (\beta x)$ can be rewritten as follows
$$2 \, \sin \left( \left(\frac{1-\beta}{2}\right) x\right) \, \cos \left( \left(\frac{1+\beta}{2}\right) x\right) = 0$$
The smallest positive solution is, thus,
$$\min \left( \frac{2\pi}{1-\beta}, \frac{\pi}{1+\beta} \right) = \min \left( \frac{2\pi}{1-\frac{\pi}{180}}, \frac{\pi}{1+\frac{\pi}{180}} \right) = \frac{\pi}{1+\frac{\pi}{180}} = \left(\frac{1}{\pi}+\frac{1}{180}\right)^{-1}$$
|
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|
Finding Laurent and Taylor series I need to find both a Laurent and a Taylor expansion.
$$f(z)=\frac{z}{(z-1)(z-2)} = \frac{-1}{(z-1)}+\frac{2}{(z-2)}$$
If I choose $z_0=0$
$$f(z)=\frac{1}{(1 + z)} - \frac{4}{\left(1 - \frac{z}{4}\right)}$$
$$f(z)=\sum_{n}^{\infty}(-1)^n{z^n} - 4\sum_{n}^{\infty}(\frac{z}{4})^n$$
Which is a Taylor series.
What value of $z_0$ would you pick for a Laurent series?
|
It is sufficient to consider a Laurent expansion around $z=0$. Depending on the region which you then choose you obtain a principal part of a Laurent expansion respectively an expansion as Taylor series.
The function
\begin{align*}
f(z)&= \frac{-1}{z-1}+\frac{2}{z-2}\\
\end{align*}
has two simple poles at $1$ and $2$.
We look at the poles $1$ and $2$ and see they determine three regions.
\begin{align*}
|z|<1,\qquad\quad
1<|z|<2,\qquad\quad
2<|z|
\end{align*}
*
*The first region $ |z|<1$ is a disc with center $0$, radius $1$ and the pole $1$ at the boundary of the disc. In the interior of this disc all two fractions with poles $1$ and $2$ admit a representation as Taylor series at $z=0$.
*The second region $1<|z|<2$ is the annulus with center $0$, inner radius $1$ and outer radius $2$. Here we have a representation of the fraction with pole $1$ as principal part of a Laurent series at $z=0$, while the fraction with pole at $2$ admits a representation as power series.
*The third region $|z|>2$ containing all points outside the disc with center $0$ and radius $2$ admits for all fractions a representation as principal part of a Laurent series at $z=0$.
A Tayor series expansion of $\frac{1}{z+a}$ at $z=0$ is
\begin{align*}
\frac{1}{z+a}&=\frac{1}{a}\cdot\frac{1}{1+\frac{z}{a}}\\
&=\sum_{n=0}^{\infty}\frac{1}{a^{n+1}}(-z)^n
\end{align*}
The principal part of $\frac{1}{z+a}$ at $z=0$ is
\begin{align*}
\frac{1}{z+a}&=\frac{1}{z}\cdot\frac{1}{1+\frac{a}{z}}=\frac{1}{z}\sum_{n=0}^{\infty}\frac{a^n}{(-z)^n}
=-\sum_{n=0}^{\infty}\frac{a^n}{(-z)^{n+1}}\\
&=-\sum_{n=1}^{\infty}\frac{a^{n-1}}{(-z)^n}
\end{align*}
|
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|
Sum and product of analytic functions that is not analytic The function
$$f(x) = \frac{2 + \cos x}{3} (2π - x) + \sin x$$
is the sum/product of analytic functions ($\cos(x)$,$\sin(x)$, linear), but all it's derivatives at $2\pi$ are $0$ ($f^n(2\pi)=0$).
I simply don't understand why.
Thanks,
|
Set $x-2\pi=t$, so $x=t+2\pi$ and $\cos x=\cos t$, $\sin x=\sin t$. Thus you can consider
$$
g(t)=-\frac{2}{3}t-\frac{1}{3}t\cos t+\sin t
$$
and the Taylor coefficients at $0$ of $g$ are the same as the Taylor coefficients at $2\pi$ of $f$. We have
\begin{align}
g(t)&=-\frac{2}{3}t-\frac{1}{3}t\left(1-\frac{1}{2}t^2+\frac{1}{24}t^4+o(t^4)\right)+t-\frac{1}{6}t^3+\frac{1}{120}t^5+o(t^5)
\\[6px]
&=-\frac{2}{3}t-\frac{1}{3}t+\frac{1}{6}t^3-\frac{1}{72}t^5
+t-\frac{1}{6}t^3+\frac{1}{120}t^5+o(t^5)
\\[6px]
&=-\frac{1}{180}t^5+o(t^5)
\end{align}
contradicting your statement.
However, even if all derivatives are zero, there would be no problem: the function would be the constant zero function (in the whole circle of convergence of the Taylor series at $2\pi$).
Since the functions involved are entire, the circle of convergence is the whole line (if you think real) or the whole plane (if you think complex). Thus, in order to check that some derivative is nonzero at $2\pi$ you just have to check that $f(x)\ne0$ for some $x$. And
$$
f(0)=\frac{2+\cos0}{3}(2\pi-0)+\sin 0=2\pi\ne0
$$
|
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|
solve $x(x^2+y^2)^{-1/2}+yy\prime(x^2+2y^2)=0$ Help with this excercise.. :)
$$x(x^2+y^2)^{-1/2}+yy\prime(x^2+2y^2)=0$$
the book says it is an exact differential equation, but how?
$$x(x^2+y^2)^{-1/2}+yy\prime(x^2+2y^2)=0$$
$$x(x^2+y^2)^{-1/2}dx+y(x^2+2y^2)dy=0$$
$M=x(x^2+y^2)^{-1/2}$
$N=y(x^2+2y^2)$
$$\frac{\partial M}{\partial y}=-\frac{xy}{(x^2+y^2)^{3/2}}$$
$$\frac{\partial N}{\partial x}=2xy$$
I cant find the integrating factor,, :(
|
Hint:
$x(x^2+y^2)^{-1/2}+yy'(x^2+2y^2)=0$
$y\dfrac{dy}{dx}(x^2+2y^2)=-\dfrac{x}{\sqrt{x^2+y^2}}$
$\dfrac{dx}{dy}=-\dfrac{y(x^2+2y^2)\sqrt{x^2+y^2}}{x}$
Let $t=y^2$ ,
Then $\dfrac{dx}{dy}=\dfrac{dx}{dt}\dfrac{dt}{dy}=2y\dfrac{dx}{dt}$
$\therefore2y\dfrac{dx}{dt}=-\dfrac{y(x^2+2y^2)\sqrt{x^2+y^2}}{x}$
$\dfrac{dx}{dt}=-\dfrac{(x^2+2y^2)\sqrt{x^2+y^2}}{2x}$
$\dfrac{dx}{dt}=-\dfrac{(x^2+2t)\sqrt{x^2+t}}{2x}$
Let $u=x^2$ ,
Then $\dfrac{du}{dt}=2x\dfrac{dx}{dt}$
$\therefore\dfrac{1}{2x}\dfrac{du}{dt}=-\dfrac{(x^2+2t)\sqrt{x^2+t}}{2x}$
$\dfrac{du}{dt}=-(x^2+2t)\sqrt{x^2+t}$
$\dfrac{du}{dt}=-(u+2t)\sqrt{u+t}$
Let $v=-\sqrt{u+t}$ ,
Then $u=v^2-t$
$\dfrac{du}{dt}=2v\dfrac{dv}{dt}-1$
$\therefore2v\dfrac{dv}{dt}-1=(v^2+t)v$
$2v\dfrac{dv}{dt}=v^3+tv+1$
$(vt+v^3+1)\dfrac{dt}{dv}=2v$
This belongs to an Abel equation of the second kind.
|
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|
Find all the numbers $n$ such that $\frac{4n-5}{60-12n}$ can't be reduced.
Find all the numbers $n$ such that $\frac{4n-5}{60-12n}$ can't be reduced.
Attempt:
$$\gcd(4n-5,60-12n)=(4n-5,-8n+55)=(4n-5,-4n+50)=(4n-5,45)$$
$$n=1: (4-5,45)=1\quad \checkmark\\
n=2: (3,45)=3\quad \times\\
n=3: (7,45)=1\quad \checkmark\\
n=4: (11,45)=1\quad \checkmark\\
n=5: (15,45)=15\quad \times\\
n=6: (19,45)=1\quad \checkmark\\
n=7: (23,45)=1\quad \checkmark\\
n=8: (27,45)=9\quad \times\\
\vdots$$
So the answer is that it can't be reduced for $n=1,3,4,6,7,..$ i.e
$$\bigg\{n\bigg|n\notin \begin{cases}a_1=2\\a_n=a_{n-1}+3\end{cases}\bigg\}$$
I want to verify that my solution is correct
|
I want to verify that my solution is correct
Your solution is not correct.
For $m\not=1$,
$$\frac{4\times\color{red}{5m}-5}{60-12\times\color{red}{5m}}=\frac{5(4m-1)}{5(12-12m)}$$
can be reduced.
If $n$ is of the form $3k+2$, then
$$\frac{4(3k+2)-5}{60-12(3k+2)}=\frac{3(4k+1)}{3(-12k+12)}$$
can be reduced.
If $n$ is not of the form $3k+2$, then since $4n-5$ is not divisible by $3$, we get
$$\gcd(4n-5,12(5-n))=\gcd(4n-5,4(5-n))=\gcd (4n-5,15)=\gcd(4n-5,5)$$
Now note that
$$\gcd(4n-5,5)=1\iff n\not\equiv 0\pmod 5$$
Hence, the answer is that $$n\not\equiv 2\pmod 3\quad\text{and}\quad n\not\equiv 0\pmod 5.$$
|
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|
Proving that $1\cdot 2+2\cdot 3+\cdots+n\left( n+1 \right) =\frac { n\left( n+1 \right) \left( n+2 \right) }{ 3 } $ by induction
Prove that
$$1\cdot 2+2\cdot 3+\cdots+n\left( n+1 \right) =\frac { n\left( n+1 \right) \left( n+2 \right) }{ 3 }. $$
I can get to $1/3(k+1)(k+2) + (k+1)(k+2)$ but then finishing off and rearranging the problem to $1/3(k+2)(k+3)$ is where I always get stuck.
|
Let $ 2 = \frac{1 \cdot 2 \cdot 3}{3} $ be the basis. The inductive step consists of simple distribution of multiplication:
$$
\begin{align}&\frac{n(n+1)(n+2)}{3} + (n+1)(n+2) \\&= \tfrac{1}{3}n^3+2n^2+\tfrac{11}{3}n+2 \\&= \frac{(n + 1)\Big((n+1)+1\Big)\Big((n + 1)+2\Big)}{3}\end{align}
$$
|
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|
Maximize $k=x^2+y^2$ Subject to $x^2-4x+y^2+3=0$ Question
Let $x$ and $y$ be real numbers satisfying the equation $x^2-4x+y^2+3=0$. Find the maximum and minimum values of $x^2+y^2$.
My work
Let $k=x^2+y^2$
Therefore, $x^2-4x+y^2+3=0$ ---> $k-4x+3=0$ .
What do I do next? How do I find an expression in terms of $k$ that I can maximize?
|
Although this kind of problem is usually solved using Lagrange multipliers, this one can be solved using methods from precalculus.
We wish to maximize the value of $x^2+y^2$ on the circle
\begin{equation}
(x-2)^2+y^2=1
\end{equation}
Which is defined on the interval $[1,3]$.
Since $x^2+y^2=4x-3$ the maximum and minimum values of $x^2+y^2$ will be the maximum and minimum values of $4x-3$ on the interval $[1,3]$, namely $1$ and $9$.
|
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|
How to simplify this:$\sqrt{5\sqrt{3}+6\sqrt{2}}$? How to simplyfy this:$\sqrt{5\sqrt{3}+6\sqrt{2}}$.
I know I should use nested radicals formula but which one is $A$ and $B$.Using the fact $A>B^2$ you can find $A$ and $B$.
But $C^2=A-B^2$ isn't a rational number then we have again a nested radical.
What to do?
|
$$\begin{align}
\sqrt{5\sqrt{3}+6\sqrt{2}}
&= \sqrt{5\sqrt{3}+\left(2\sqrt{6}\right)\sqrt{3}} \\
&= \sqrt[4]{3}\cdot\sqrt{5+2\sqrt{6}} \\
&= \sqrt[4]{3}\cdot\sqrt{2+2\sqrt{6} + 3} \\
&= \sqrt[4]{3}\cdot\sqrt{\frac{4+4\sqrt{6}+\left(\sqrt{6}\right)^2}{2}} \\
&= \sqrt[4]{3}\cdot\sqrt{\frac{\left(2+\sqrt{6}\right)^2}{2}} \\
&= \sqrt[4]{3}\cdot\frac{2\sqrt{2}+2\sqrt{3}}{2}\\
&= \sqrt[4]{3}\cdot\left(\sqrt{2} + \sqrt{3}\right).
\end{align}$$
|
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|
Prove $\sum_{n=1}^\infty \frac{6}{n(3n-1)}=9 \ln 3-\sqrt{3} \pi$ This series represents the sum of reciprocal pentagonal numbers (multiplied by $3$). I got its value from Wolfram Alpha:
$$\sum_{n=1}^\infty \frac{6}{n(3n-1)}=9 \ln 3-\sqrt{3} \pi$$
My attempt: Turn the series into an integral:
$$\sum_{n=1}^\infty \frac{x^{3n}}{n}=-\ln(1-x^3)$$
$$\sum_{n=1}^\infty \frac{x^{3n-2}}{n}=-\frac{\ln(1-x^3)}{x^2}$$
$$\sum_{n=1}^\infty \frac{1}{n(3n-1)}=-\int_0^1 \frac{\ln(1-x^3)}{x^2} dx$$
I don't know how to solve this integral.
We can also transform the common term into two parts:
$$\frac{1}{n(3n-1)}=\frac{3}{3n-1}-\frac{1}{n}$$
But this gives us two diverging series, so it doesn't help.
|
We can also continue your way. We have $$-\int\frac{\log\left(1-x^{3}\right)}{x^{2}}dx=\frac{\log\left(1-x^{3}\right)}{x}-\int\frac{3x}{1-x^{3}}dx
$$ $$=\frac{\log\left(1-x^{3}\right)}{x}-\int\frac{1-x}{x^{2}+x+1}dx-\int\frac{1}{x-1}dx
$$ $$=\frac{\log\left(1-x^{3}\right)}{x}-\log\left(x-1\right)-\frac{1}{2}\int\frac{2x+1}{x^{2}+x+1}dx-\frac{3}{2}\int\frac{1}{x^{2}+x+1}dx
$$ $$=\frac{\log\left(1-x^{3}\right)}{x}-\log\left(x-1\right)-\frac{1}{2}\log\left(x^{2}+x+1\right)-\frac{3}{2}\int\frac{1}{\left(x+\frac{1}{2}\right)^{2}+\frac{3}{4}}dx
$$
$$=\color{red}{\frac{\log\left(1-x^{3}\right)}{x}-\log\left(x-1\right)-\frac{1}{2}\log\left(x^{2}+x+1\right)-\sqrt{3}\arctan\left(\frac{2x+1}{\sqrt{3}}\right)+C.}$$
|
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|
Showing that $2^6$ divides $3^{2264}-3^{104}$
Show that $3^{2264}-3^{104}$ is divisible by $2^6$.
My attempt: Let $n=2263$. Since $a^{\phi(n)}\equiv 1 \pmod n$ and
$$\phi(n)=(31-1)(73-1)=2264 -104$$
we conclude that $3^{2264}-3^{104}$ is divisible by $2263$.
I have no idea how to show divisibility by $2^6$.
|
Another approach:
$$\text{Euler's formula: }\quad a^{\phi(n)}\equiv 1 \pmod{n} \text{ when} \gcd(a,n)=1$$
$\implies a^{32}\equiv 1 \pmod{2^6}$
We want $3^{2264}-3^{104}\equiv 0 \pmod{2^6}$
$$\phi(2^6)=2^5=32$$
$$ 3^{2264}-3^{104}=3^{2240}3^{24}-3^{96}3^{8}=(3^{70})^{32}3^{24}-(3^3)^{32}3^8\equiv 3^{24}-3^8\equiv 17^6-17^2\\
\equiv 33^3-33\equiv \color{green}{0 \pmod{2^6}}$$
|
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|
Find the highest point of intersection
Find the highest point of intersection of the sphere $x^2+y^2+z^2=30$ and the cone $x^2+2y^2-z^2=0$.
Am I supposed to use the Lagrange multiplier for this?
EDIT:
So this is what I've tried...
$z^2=-x^2-y^2+30$ and $z^2=x^2+2y^2$. Then $-x^2-y^2+30 = x^2+2y^2$.
This gives us $3y^2=-2x^2+30$, which is then: $y^2=-\frac23x^2+10$. Substitute $y^2$ in to the sphere equation, we have:
$z^2 = -x^2 - (-\frac23x^2+10)+30 = -\frac13x^2+20$, so the maximum $z$ is $\sqrt{20}$...?
If this is true, then calculating the rest of the point, we have $(0,\sqrt{10},\sqrt{20})$....?
|
By using Lagrange multipliers, we need to solve system $$F'_x=0,$$ $$F'_y=0,$$ $$F'_z=0,$$ where is $F(x,y,z,\alpha,\beta)=z+\alpha(x^2+y^2+z^2-30)+\beta(x^2+2y^2-z^2)$,
with conditions $$x^2+y^2+z^2-30=0,$$ $$x^2+2y^2-z^2=0.$$
Stationary points are ($\pm \sqrt{15}$,$0$,$\pm \sqrt{15}$) and ($0$,$\pm\sqrt{10}$, $\pm\sqrt{20}$).
We get that maximal z is $z=\sqrt{20}$ (in point ($0$,$\pm\sqrt{10}$, $\sqrt{20}$)).
|
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|
Find the probability that $1984!$ is divisible by $n$
Let $a,b,c,d$ be a permutation of the numbers $1,9,8,4$ and let $n = (10a+b)^{10c+d}$. Find the probability that $1984!$ is divisible by $n$.
I was told this could be solved by casework on $a$ and using Fermat's Little Theorem. For example, if $a = 1$, there are $6$ possibilities; if $a = 4$, there are $4$ possibilities; and if $a = 8,9$, there are $5$ possibilities.
How do I use Fermat's Little Theorem to get this?
|
The following PARI/GP program determines the permutations for which
$1984!$ divides the given number :
? q=0;x=[1,9,8,4];for(j=1,24,p=numtoperm(4,j);a=x[p[1]];b=x[p[2]];c=x[p[3]];d=x[
p[4]];if(Mod(1984!,(10*a+b)^(10*c+d))==0,q=q+1;print(q," ",a," ",b,"
",c," ",d)))
1 1 9 4 8
2 1 8 9 4
3 1 8 4 9
4 1 4 9 8
5 1 4 8 9
6 9 1 8 4
7 9 1 4 8
8 9 8 1 4
9 9 8 4 1
10 9 4 1 8
11 8 1 9 4
12 8 1 4 9
13 8 9 1 4
14 8 4 1 9
15 8 4 9 1
16 4 9 1 8
17 4 9 8 1
18 4 8 1 9
19 4 8 9 1
20 1 9 8 4
?
The failing permutations are :
? q=0;x=[1,9,8,4];for(j=1,24,p=numtoperm(4,j);a=x[p[1]];b=x[p[2]];c=x[p[3]];d=x[
p[4]];if(Mod(1984!,(10*a+b)^(10*c+d))<>0,q=q+1;print(q," ",a," ",b,"
",c," ",d)))
1 9 4 8 1
2 8 9 4 1
3 4 1 9 8
4 4 1 8 9
?
So, $20$ out of the $24$ permutations are sucessful. Therefore, the probability that $1984!$ divides the number is $\frac{5}{6}$.
|
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|
Closed form of function $f(n) = \frac1n \sum\limits _{k=1}^{n-1} f(k)$? Could anyone help me get to the closed form of the function? Here $\mathbb{N}=\{1,2,3,\ldots\}$.
Find all functions $f:\mathbb{N}\to\mathbb{R}$ such that $f(1)=1$ and
$$f(n) = \frac 1 n \sum _{k = 1}^{n-1}f(k)$$
for every integer $n>1$.
So far, I see that
$$f(2)=\frac{1}{2}f(1)=\frac12,$$
$$f(3)=\frac{1}{3}\left(f(1)+f(2)\right)=\frac{1}{3}\left(1+\frac{1}{2}\right)=\frac12,$$
$$f(4)=\frac14\left(f(1)+f(2)+f(3)\right)=\frac14\left(1+\frac12+\frac12\right)=\frac12.$$
I guess that $f(n)=\frac12$ for every $n\in\mathbb{N}$. How to show this?
|
Start by computing the first few terms to gain an intuition into the problem, we can see that $f(2) = \frac{1}{2} f(1) = \frac{1}{2}$. And then, it follows that $f(3) = \frac{1}{3} \left(1 + \frac{1}{2}\right) = \frac{1}{2} $, etc...
Claim: $f(n) = \frac{1}{2}$ for all $n \geq 2$.
Proof: Assume the above holds, then $$f(n+1) = \frac{1}{n+1} \left(f(1) + \cdots + f(n) \right) = \frac{1}{n+1}\left(1 + \frac{n-1}{2}\right) = \frac{1}{n+1} \cdot \frac{n+1}{2} = \frac{1}{2}$$
and by induction, we are done.
|
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|
Show : $(-1)^{n}n^{-\tan\left(\tfrac{\pi}{4}+\tfrac{1}{n} \right)}=\tfrac{(-1)^{n}}{n}+\mathcal{O}\left(\tfrac{\ln(n)}{n^{2}} \right)$
I would like to show that :
$$(-1)^{n}n^{-\tan\left(\dfrac{\pi}{4}+\dfrac{1}{n} \right)}=\dfrac{(-1)^{n}}{n}+\mathcal{O}\left(\dfrac{\ln(n)}{n^{2}} \right)$$
My proof:
Note that :
\begin{align*}
e^{x}&=1+x+\mathcal{O}\left(x^{2}\right)\quad (x\to 0)\\
\tan(x)&=x+\mathcal{O}\left(x^{3} \right)\\
\tan(a+b)&=\dfrac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}\\
x^{a}&=e^{a\ln(x)}\quad (x>0)
\end{align*}
then:
*
*$-\tan\left(\dfrac{\pi}{4}+\dfrac{1}{n} \right)=\left(1-\tan^{-1}\left(\dfrac{1}{n}\right) \right)^{-1}$
*$\dfrac{1}{n} \underset{ \overset { n \rightarrow +\infty } {} } {\longrightarrow }0$
*$\left(\dfrac{1}{n}+\mathcal{O}\left(\dfrac{1}{n^{3}} \right)\right)^{-1}=\left(\dfrac{1}{n}\right)^{-1}\left(1+\mathcal{O}\left( \dfrac{1}{n^{2}}\right) \right)^{-1}=n\left( 1-\mathcal{O}\left( \dfrac{1}{n^{2}}\right)\right)$
Thus :
\begin{align*}
(-1)^{n}n^{-\tan\left(\dfrac{\pi}{4}+\dfrac{1}{n} \right)}&=(-1)^{n}e^{-\tan\left(\dfrac{\pi}{4}+\dfrac{1}{n} \right)\ln(n)} \\
&=(-1)^{n}\exp\left[-\tan\left(\dfrac{\pi}{4}+\dfrac{1}{n} \right)\ln(n)\right]\\
&=(-1)^{n}\exp\left[\left(1-\tan^{-1}\left(\dfrac{1}{n}\right) \right)^{-1}\ln(n)\right]\\
&=(-1)^{n}\exp\left[\left(1-\left(\dfrac{1}{n}+\mathcal{O}\left(\dfrac{1}{n^{3}} \right)\right)^{-1} \right)^{-1}\ln(n)\right]\\
\end{align*}
I'm stuck
|
First, we note that $\tan\left(\frac{\pi}{4}+\frac1n\right)$ can be expanded as
$$\begin{align}
\tan\left(\frac{\pi}{4}+\frac1n\right)&=\frac{1+\tan(1/n)}{1-\tan(1/n)}\\\\
&=1+\frac2n +\frac2{n^2}+O\left(\frac{1}{n^3}\right)
\end{align}$$
Then,
$$\begin{align}
e^{-\tan\left(\frac{\pi}{4}+\frac1n\right)\log(n)}&=e^{-\left(1+\frac2n +\frac2{n^2}+O\left(\frac{1}{n^3}\right)
\right)\log(n)}\\\\
&=\frac1n\,e^{-\left(\frac{2\log(n)}n +\frac{2\log(n)}{n^2}+O\left(\frac{\log(n)}{n^3}\right)\right)}\\\\
&=\frac1n\,\left(1-2\frac{\log(n)}{n}-2\frac{\log(n)}{n^2}+O\left(\frac{\log^2(n)}{n^2}\right)\right)\\\\
&=\frac1n +O\left(\frac{\log(n)}{n^2}\right)
\end{align}$$
Therefore, we find that
$$(-1)^n\,n^{-\tan\left(\frac{\pi}{4}+\frac1n\right)}=(-1)^n\frac1n +O\left(\frac{\log(n)}{n^2}\right)$$
|
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|
Calculate limit for, $\lim\limits_{x\to 0}\frac{1-cos(x^6)}{x^{12}}$, but in there have suprize. Let's think about this function, $\quad \to f(x)=\dfrac{x^2-1}{x-1}$,
$\lim\limits_{x\to 1}\dfrac{x^2-1}{x-1}=0/0$ ,
First Solution :
$\lim\limits_{x\to 1}\dfrac{x^2-1}{x-1}=\lim\limits_{x\to 1}\dfrac{x+1}{1}$
$=\lim\limits_{x\to 1}\dfrac{x+1}{1}=\dfrac{1+1}{1}=2$
And Second Solution with l'hopital:
$\lim\limits_{x\to 1}\dfrac{x^2-1}{x-1}\overbrace{\longrightarrow}^{l'hopital}\lim\limits_{x\to 1}\dfrac{2x}{1}=2$
But...
Let's we take this function ,$f(x)=\dfrac{1-cos(x^6)}{x^{12}}$
$\lim\limits_{x\to 0}\dfrac{1-cos(x^6)}{x^{12}}=0/0$,
For this funciton couldn't simplification,I have just , graph of function and l'hôpital,
1-L'hôpital;
$\lim\limits_{x\to 0}\dfrac{1-cos(x^6)}{x^{12}}=\lim\limits_{x\to 0}\dfrac{6.x^5.sin(x^6)}{12.x^{11}}=\underbrace{\lim\limits_{x\to 0}\dfrac{sin(x^6)}{x^6}}_1.\dfrac{1}{2}=\dfrac{1}{2}$
2-Graph of function;
Graph telling me ,$\lim\limits_{x\to 0}\dfrac{1-cos(x^6)}{x^{12}}=0$
l'hôpital $\lim\limits_{x\to 0}\dfrac{1-cos(x^6)}{x^{12}}=1/2$
What we do now?
And link of graph , if you can't see clearly,https://www.desmos.com/calculator/jfz4kslm4w
|
Hint:
$$
\begin{align}
\frac{1-\cos\left(x^6\right)}{x^{12}}
&=\frac{1-\cos\left(x^6\right)}{\sin^2\left(x^6\right)}\left(\frac{\sin\left(x^6\right)}{x^6}\right)^2\\
&=\frac{1-\cos\left(x^6\right)}{1-\cos^2\left(x^6\right)}\left(\frac{\sin\left(x^6\right)}{x^6}\right)^2\\
&=\frac1{1+\cos\left(x^6\right)}\left(\frac{\sin\left(x^6\right)}{x^6}\right)^2\\
\end{align}
$$
|
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|
show that $b,c \in \Bbb N$ for $b=c^{\frac{1}{c-1}}$ $\iff$ c = 2. I am trying to show that $b,c \in \Bbb N$ for $b=c^{\frac{1}{c-1}}$ $\iff$ c = 2. The reverse implication is easy, but the forward implication is tricky. It seems essentially to state the the (n-1)th root of a natural number n is not an integer for n>2.
|
Define $f(x) = \dfrac{\ln x}{x-1}$. Then, $f'(x) = -\dfrac{1}{(x-1)^2}\ln x + \dfrac{1}{x-1} \cdot \dfrac{1}{x} = -\dfrac{x(\ln x - 1)+1}{x(x-1)^2}$.
For $x \ge 3$, we have $f'(x) < 0$, so $f$ is decreasing on $[3,\infty)$.
Hence, for all $c \ge 3$, we have $f(c) \le f(3)$, i.e. $\dfrac{\ln c}{c-1} \le \dfrac{\ln 3}{2}$.
Exponentiating both sides to get $c^{\tfrac{1}{c-1}} \le \sqrt{3} < 2$.
Trivially, $c^{\tfrac{1}{c-1}} > 1$ for integers $c \ge 3$. Therefore, $1 < c^{\tfrac{1}{c-1}} < 2$ for all integers $c \ge 3$.
Hence, $c^{\tfrac{1}{c-1}}$ is not an integer for any integer $c \ge 3$.
|
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|
inequality $\sqrt{\cos x}>\cos(\sin x)$ for $x\in(0,\frac{\pi}{4})$ How can I prove the inequality $\sqrt{\cos x}>\cos(\sin x)$ for $x\in(0,\frac{\pi}{4})$ ?
The derivative of $f(x):=\sqrt{\cos x}-\cos(\sin x)$ is very unpleasant, so the standard method is probably not be the right choice...
|
Both sides being positive, we can raise each to the fourth power, to get
$$ \cos^2 x > \cos^4 \sin x $$
and then (as suggested by @Bacon), change variable to $u=\sin x$, which gives us
$$ 1-u^2 > \cos^4 u, \qquad u\in(0,1/\sqrt 2) $$
By Taylor's theorem $\cos u = 1 - \frac12 u^2 + \frac{\cos \xi}{4!}u^4$ for some $\xi\in(0,1/\sqrt2)$, which implies $\cos\xi>0$. Raising this to the fourth power gives us
*
*Terms only involving $1$ and $\frac12u^2$, namely $1 - 4\cdot\frac12u^2 +6\cdot\frac14u^4 - 4\cdot\frac18u^6+\frac1{16}u^8 $
*Further negative terms involving one or three factors of $-\frac12u^2$, which we can ignore
*Positive tems with exactly one $\frac{\cos\xi}{24}u^4$, namely $4\frac{\cos\xi}{24}u^4 + 12\frac{\cos\xi}{24\cdot 4}u^8$
*Fewer than $\binom42\cdot 3^2=54$ terms with two or more factors of $\frac{\cos\xi}{24}u^4$.
Thus we have
$$ \cos^4 u < 1 - 2u^2 + \frac32u^4 + \frac1{16}u^8 + \frac4{24}u^4 + \frac{1}{8} u^8 + \frac{54}{24^2}u^8 $$
As long as $u^2 < \frac12$, this gives
$$ \cos^4 u < 1 - 2u^2 + \frac34u^2 + \frac1{128}u^2 + \frac1{12}u^2 + \frac{1}{64}u^2 + \frac{54}{4068}u^2 $$
But $\frac34+\frac1{128}+\frac1{12}+\frac1{64}+\frac{54}{4068} < \frac{15}{20} + \frac1{20} + \frac2{20} + \frac1{20} + \frac1{20} = 1$, so this gives
$$ \cos^4 u < 1 - u^2 $$
as required.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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|
Rewriting product to a binomial I'm currently researching Wigner matrices. I wanted to calculate the moments of its spectral density. The probability density is
$$\frac{1}{2\pi} \sqrt{4-x^2} \text{ for } x \in [-2,2] $$
I have found an expression for the $k^{th}$ moment by integrating
$$ \int_{-2}^2 \frac{x^k}{2\pi} \sqrt{4-x^2} \, dx.$$
This is $0$ if $k$ is uneven and
$$ \frac{\prod _{i=1}^{\frac{k}{2}} \frac{4k - 8i +12}{k - 2i +4}}{k+1}$$ if $k$ is even.
It is known that the moments are the Catalan numbers for k is even $$ \frac{1}{k+1}\binom{2k}{k}. $$ In Mathematica i found that my solutions are the Catalan numbers. But I can't figure out how to rewrite my product to the expression of the Catalan numbers. Which would be a way more intuitive expression.
Are there any tricks one can use to write products to binomials?
|
Since we have to consider $k$ even, we set $k=2l$ and show
The following is valid for $l\geq 1$:
\begin{align*}
\frac{1}{2l+1}\prod_{i=1}^l\frac{8l-8i+12}{2l-2i+4}=\frac{1}{l+1}\binom{2l}{l}
\end{align*}
with $\frac{1}{l+1}\binom{2l}{l}$ the Catalan numbers.
We obtain
\begin{align*}
\frac{1}{2l+1}\prod_{i=1}^l\frac{8l-8i+12}{2l-2i+4}
&=\frac{1}{2l+1}\prod_{i=0}^{l-1}\frac{8l-8i+4}{2l-2i+2}\tag{1}\\
&=\frac{1}{2l+1}\prod_{i=0}^{l-1}\frac{8i+12}{2i+4}\tag{2}\\
&=\frac{1}{2l+1}\prod_{i=0}^{l-1}\frac{2(2i+3)}{i+2}\\
&=\frac{2^l}{2l+1}\cdot\frac{(2l+1)!!}{(l+1)!}\tag{3}\\
&=\frac{2^l}{l+1}\cdot\frac{(2l)!}{l!\cdot(2l)!!}\tag{4}\\
&=\frac{1}{l+1}\cdot\frac{(2l)!}{l!l!}\tag{5}\\
&=\frac{1}{l+1}\binom{2l}{l}\\
\end{align*}
and the claim follows.
Comment:
*
*In (1) we shift the index $i$ by one
*In (2) we reverse the order of multiplication $i \rightarrow l-1-i$
*In (3) we use factorials and double factorials instead of the product symbol
\begin{align*}
(2l+1)!!=(2l+1)(2l-1)\cdots 5\cdot 3\cdot 1
\end{align*}
*In (4) we use the identity
$
(2l)!=(2l)!!(2l-1)!!
$
*In (5) we use the identity
\begin{align*}
(2l)!!=(2l)(2l-2)\cdots 4\cdot 2=2^l\cdot l!
\end{align*}
Note: In OPs question the case $k$ even should be written as
\begin{align*}
\frac{1}{\frac{k}{2}+1}\binom{k}{\frac{k}{2}}
\end{align*}
in order to be consistent with the stated product expression.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Proof about Pythagorean triples Show that if $(x,y,z)$ is a Pythagorean triple, then $10\mid xyz$
Proof
First, if $x$, $y$, $z$ are all odd, then so are $x^2$, $y^2$, $z^2$, so $x^2+y^2$ is even, which means that $x^2+y^2 \neq z^2 $. Hence, at least one of $x$, $y$, $z$ is even, so $2\mid xyz$ (clear).
Next, for any $n \in Z$, if $5$ doesn't divide $n$, then $(n^2)^2=n^4 \equiv 1\pmod{5}$ (as you can check or quote Euler's theorem), and therefore $n^2 \equiv \pm 1\pmod{5}$. Now, if $5$ doesn't divide $xy$, then $x^2 \equiv \pm 1\pmod{5}$ and $y^2 \equiv \pm 1\pmod{5}$, so
$$x^2+y^2 \equiv -2,0,2 \pmod{5}$$
Therefore if $x^2+y^2=z^2$ and $5$ doesn't divide $xy$ then $z^2=x^2+y^2 \equiv -2,0,2\pmod{5}$, so $5\mid z^2$ (why? it looks weird to me because $z^2$ can be also congruent to $2$ and $-2$) and hence $5\mid z$ (otherwise $z^2 \equiv \pm 1\pmod{5}$). It follows that $x^2+y^2=z^2$, then either $5\mid xy$ or $5\mid z$ (how does this follow? if $5$ doesn't divide $z$ then how can $5$ divide $xy$), so in any case $5\mid xyz$
Finally, we can conclude that if $x^2+y^2=z^2$ then $2\mid xyz$ and $5\mid xyz$, so $10\mid xyz$ (intuitively it looks right but I can't prove it!)
|
No problem with divisibility by $2$, good.
Suppose $5\nmid xyz$, so $5$ divides none of the numbers $x$, $y$ and $z$.
Then $x^2\equiv\pm1\pmod{5}$ and the same for $y$ and $z$. However, it can't be $x^2\equiv1$ and $y^2\equiv-1$, otherwise $z^2\equiv0\pmod{5}$. Similarly it's impossible that $x^2\equiv-1$ and $y^2\equiv1$.
Suppose $x^2\equiv1$ and $y^2\equiv1$. Then $z^2\equiv1+1\pmod{5}$, which is a contradiction. Similarly $x^2\equiv-1\equiv y^2$ can be dismissed.
Hence $5\mid xyz$.
Alternatively, which is what the book seems to be doing, suppose $5\nmid xy$. Then, since $x^2\equiv\pm1$ and $y^2\equiv\pm1$, we can conclude that
\begin{align}
z^2&=x^2+y^2\equiv2\pmod5 &&\text{or} \\[4px]
z^2&=x^2+y^2\equiv-2\pmod5 &&\text{or} \\[4px]
z^2&=x^2+y^2\equiv0\pmod5
\end{align}
The first two cases are impossible, as seen above, so only the third case remains: since $5\mid z^2$, we conclude $5\mid z$.
|
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"url": "https://math.stackexchange.com/questions/1865153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
nature of the series $\sum (-1)^{n}n^{-\tan\left(\tfrac{\pi}{4}+\tfrac{1}{n} \right)}$ I would like to study the nature of the following serie:
$$\sum_{n\geq 0}\ (-1)^{n}n^{-\tan\left(\dfrac{\pi}{4}+\dfrac{1}{n} \right)} $$
we can use simply this question :
Show : $(-1)^{n}n^{-\tan\left(\tfrac{\pi}{4}+\tfrac{1}{n} \right)}=\tfrac{(-1)^{n}}{n}+\mathcal{O}\left(\tfrac{\ln(n)}{n^{2}} \right)$
$$\sum_{n\geq 1} (-1)^{n}n^{-\tan\left(\dfrac{\pi}{4}+\dfrac{1}{n} \right)}=\sum_{n\geq 1}\dfrac{(-1)^{n}}{n}+\sum_{n\geq 1} \mathcal{O}\left(\dfrac{\ln(n)}{n^{2}} \right) $$
Since :
*
*Alternating harmonic series $\sum_{n\geq 1}\dfrac{(-1)^{n}}{n}$ is convergent
*$\sum_{n\geq 1} \mathcal{O}\left(\dfrac{\ln(n)}{n^{2}} \right) $ is convergent since $\left| \mathcal{O}\left(\dfrac{\ln(n)}{n^{2}} \right)\right|\leq \dfrac{\ln(n)}{n^{2}}$
thus $\sum_{n\geq 0}\ (-1)^{n}n^{-\tan\left(\dfrac{\pi}{4}+\dfrac{1}{n} \right)} $ is convergent
My question :
*
*Why we can't take expansion up to $2$ $\frac{1}{1-z}=1+z+O(z^2)$ instead of $\frac{1}{1-z}=1+z+z^2+O(z^3)$
\begin{align}
\tan\left(\frac{\pi}{4}+\frac1n\right)&=\left(1+\tan(1/n)\right)\left(1-\tan(1/n)\right)^{-1}\\
&= \left(1+\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{3}}\right)\right)\left(1-\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{3}}\right)\right) \\
&= \left(1+\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{3}}\right)\right)\left(1-\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{3}}\right)+\mathcal{O}\left(\left(\frac{-1}{n}+\mathcal{O}\left(\frac{1}{n^3}\right)\right)^{2} \right) \right) \\
&= \left(1+\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{3}}\right)\right)\left(1-\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{2}}\right) \right) \\
&=1-\frac{1}{n^2}+\mathcal{O}\left( \frac{1}{n^{2}}\right)
\end{align}
then :
\begin{align}
e^{-\tan\left(\frac{\pi}{4}+\frac1n\right)\log(n)}&=e^{\left(-1-\frac{1}{n^2}+\mathcal{O}\left( \frac{1}{n^{2}}\right)\right)\log(n)}\\\\
&=\frac1n\,e^{\left(-\frac{\log(n)}n^{2}+O\left(\frac{\log(n)}{n^2}\right)\right)}\\\\
&=\frac1n\,\left(1-\frac{\log(n)}{n^{2}}+O\left(\frac{\log(n)}{n^2}\right)+O\left(\frac{\log^{2}(n)}{n^4}\right) \right)\\\\
&=\frac1n +O\left(\frac{\log(n)}{n^3}\right)
\end{align}
$$(-1)^{n}e^{-\tan\left(\frac{\pi}{4}+\frac1n\right)\log(n)}=(-1)^{n}\frac1n +O\left(\frac{(-1)^{n}\log(n)}{n^3}\right) $$
and since $\sum O\left(\frac{\log(n)}{n^3}\right)$ and $(-1)^{n}\frac1n$ are convergent
i prove that the series is conevrgent only by using expansion up to 2 why we can't take it
|
Without digging into asymptotics, the original series is simply convergent by Dirichlet's test, since $\{(-1)^n\}$ has bounded partial sums and
$$ n^{-\tan\left(\frac{\pi}{4}+\frac{1}{n}\right)} = \frac{1}{n^{\frac{1+\tan(1/n)}{1-\tan(1/n)}}}$$
is decreasing towards zero for any $n\geq 3$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to find $ \tan \left(\frac{x}{2}\right) $ knowing that $\cos \left(x\right)+\sin \left(x\right)=\frac{7}{5} $ Good evening to everyone. I don't know how to find $ \tan \left(\frac{x}{2}\right) $ knowing that $$\cos \left(x\right)+\sin \left(x\right)=\frac{7}{5} $$ and x$\in (0,\frac{\pi}{3})$ Here's what I've tried:
$$\tan \left(\frac{x}{2}\right) = \frac{1-\cos \left(x\right)}{\frac{7}{5}- \cos \left(x\right)}$$ But I don't know what to do from here. Can someone explain to me how to solve this? Thanks for any answers.
|
Another approach:
$$\sqrt{2}\cos\left(x + \frac{\pi}{4}\right) = \cos(x) + \sin(x)$$
Thus,
$$\cos(x) + \sin(x) = \frac{7}{5} \rightarrow \sqrt{2}\cos\left(x + \frac{\pi}{4}\right) = \frac{7}{5}$$
And hence
$$x + \frac{\pi}{4} = \arccos\left(\frac{7}{5\sqrt{2}}\right) \rightarrow x = \arccos\left(\frac{7}{5\sqrt{2}}\right) - \frac{\pi}{4}$$
Thus,
$$\tan\left(\frac{x}{2}\right) = \tan\left(\frac{1}{2}\arccos\left(\frac{7}{5\sqrt{2}}\right) - \frac{\pi}{8} \right) $$
Now from here there is some nasty algebra to arrive at the final answer. So, in observing the results as given above, I think it's interesting to note the identify that comes out from my clearly overly complicated approach:
$$\tan\left(\frac{1}{2}\arccos\left(\frac{7}{5\sqrt{2}}\right) - \frac{\pi}{8} \right) = \frac{5 \pm 1}{12}$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
At which points is the following function continuous? From Royden and Fitzpatrick's Real Analysis, Fourth Edition (Chapter 1, Problem 48):
Let $f$ be the function defined by
$$
f(x) = \begin{cases}
x & \text{if $x$ is irrational} \\
p \cdot \sin \frac{1}{q} & \text{if $x = \frac{p}{q}$ in lowest terms}
\end{cases}
$$
At what points is $f$ continuous?
This function is similar to the one considered here.
I was able to show that $f$ is continuous at $0$ and not continuous at any non-zero rational number.
I believe $f$ is continuous on the irrationals, but am having trouble proving this conjecture.
My idea was to pick a small enough neighborhood around $x$ that the only rationals in the neighborhood would have $q$ so large that $\left|1 - q \sin \frac{1}{q}\right|$ was arbitrarily small.
The rest of the proof will follow nicely if I can show such a neighborhood exists, but I am having trouble constructing it.
|
Let us first show that $f$ is continuous at 0.
By construction, $f(0) = 0$.
Now pick any $\epsilon > 0$.
Since $\lim_{n \to \infty} n \sin \dfrac{1}{n} = 1$, there exists an index $N$ such that
$$
n \sin \dfrac{1}{n} < 1+\epsilon
$$
for all $n \geq N$.
Let $\delta = \min\left\{\dfrac{1}{N},\dfrac{\epsilon}{1+\epsilon}\right\}$.
For any non-zero rational number $x = \dfrac{p}{q}$, where $\dfrac{p}{q}$ is in lowest terms, we have
\begin{align*}
\left|x\right| = \left|\frac{p}{q}\right| < \delta \qquad &\Rightarrow \qquad \frac{1}{\delta} \leq \frac{|p|}{\delta} < |q| \\
&\Rightarrow \qquad N \leq |q| \\
&\Rightarrow \qquad \left|q\right| \cdot \left|\sin \frac{1}{q}\right| < 1 + \epsilon \\
&\Rightarrow \qquad |f(x)| = \left|p \cdot \sin \frac{1}{q}\right| \leq \delta \cdot \left|q\right| \cdot \left|\sin \frac{1}{q}\right| < \delta \cdot (1 + \epsilon) \leq \epsilon
\end{align*}
If $x$ is irrational, then
$$
|x| < \delta \qquad \Rightarrow \qquad |f(x)| = |x| < \epsilon
$$
Thus $|x - 0| < \delta$ implies $|f(x) - 0| < \epsilon$, so $f(x)$ is continuous at zero.
Let us now show that $f$ is discontinuous at all non-zero rational numbers.
Let $x = \dfrac{p}{q}$ be a non-zero rational number and consider the sequence
$$
x_n = \frac{p}{q} \cdot \frac{(p\cdot q + 1)^n +q}{(p\cdot q + 1)^n} =: \frac{p_n}{q_n}
$$
By construction, $p_n$ and $q_n$ are relatively prime and $\lim_{n\to \infty} x_n = x$.
We therefore have
$$
\lim_{n\to \infty} f(x_n) = \lim_{n \to \infty} \frac{p_n}{q_n} \cdot q_n \cdot \sin \frac{1}{q_n} = \left(\lim_{n \to \infty} \frac{p_n}{q_n}\right) \cdot \left( \lim_{n \to \infty} q_n \cdot \sin \frac{1}{q_n}\right) = \frac{p}{q}
$$
However
$$
f(x) = p \sin \frac{1}{q} \neq \frac{p}{q}
$$
because the left-hand side is an irrational number for all integers $q$.
Since $\lim_{n\to \infty} x_n = x$ but $\lim_{n\to \infty} f(x_n) \neq f(x)$, $f$ is not continuous at $x$.
Lastly, we can also show that $f$ is continuous at all irrational numbers.
Let $x$ be an irrational number and fix $\epsilon > 0$.
Let $x = [a_0; a_1, a_2, \cdots]$ denote the continued fraction expansion of $x$ and define the sequences
\begin{align*}
h_n = a_n h_{n-1} + h_{n-1} \qquad h_{-1} = 1 \qquad h_{-2} = 0 \\
k_n = a_n k_{n-1} + k_{n-1} \qquad h_{-1} = 0 \qquad h_{-2} = 1
\end{align*}
Since the $a_n \geq 1$ for all $n \geq 1$, $k_n \to \infty$.
It is possible to show $\left\{\dfrac{h_n}{k_n}\right\} \to x$ and
$$
\left|x - \frac{h_n}{k_n}\right| < \left|x - \frac{p}{q}\right|
$$
for any rational $\dfrac{p}{q}$ satisfying $0 < q < k_n$,
This means we can find an index $N$ such that
\begin{gather*}
\left|x - \frac{h_N}{k_N}\right| < \frac{\epsilon \cdot |x|}{\epsilon + 2 \cdot |x|}
\end{gather*}
and
$$
\left|1 - n \cdot \sin \frac{1}{n}\right| < \frac{\epsilon}{2 \cdot |x|}
$$
for all $n \geq k_N$.
Let $\delta = \left|x -\dfrac{h_N}{k_N}\right|$.
Then for any rational $\dfrac{p}{q}$ satisfying $\left|x - \dfrac{p}{q}\right| < \delta$, we must have $q \geq k_N$.
But this implies
\begin{align*}
\left|x - p \cdot \sin \frac{1}{q}\right| &\leq \left|x - \frac{p}{q}\right| + \left| \frac{p}{q}\right| \left| 1 - q \cdot \sin \frac{1}{q}\right| \\[8pt]
&\leq \left|x - \frac{p}{q}\right|\left(1 +\left| 1 - q \cdot \sin \frac{1}{q}\right|\right) + \left|x\right| \left| 1 - q \cdot \sin \frac{1}{q}\right|\\
&< \epsilon
\end{align*}
For any irrational $x'$ satisfying $|x' - x| < \delta$, we have $|x' - x| < \epsilon$ because $\delta < \epsilon$.
Thus $|x' - x| < \delta$ implies $|f(x') - f(x)| < \epsilon$, so $f$ is continuous at $x$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Another way to evaluate $\int\frac{\cos5x+\cos4x}{1-2\cos3x}{dx}$? What I've done is this:$$\int\dfrac{\cos5x+\cos4x}{1-2\cos3x}{dx}$$
$$\int \dfrac{\sin 3x}{\sin 3x}\left[\dfrac{\cos5x+\cos4x}{1-2\cos3x}\right]{dx}$$
$$\dfrac {1}{2}\int\dfrac{\sin 8x -\sin 2x +\sin 7x -\sin x}{\sin 3x - \sin 6x}$$
$$-\dfrac {1}{2}\int\dfrac{ \sin \frac{7x}{2} +\sin \frac{5x}{2} } {\sin \frac{3x}{2} }$$
$$-\int\dfrac{ \sin {3x}\cos \frac{x}{2} } {\sin \frac{3x}{2} }$$
$$-\int\dfrac{2\sin \frac{3x}{2} \cos \frac{3x}{2}\cos \frac{x}{2} } {\sin \frac{3x}{2} }$$
$$-\int {2\cos \frac{3x}{2}\cos \frac{x}{2} }$$
$$ -\left(\frac{\sin 2x}{2} +\sin x \right) +c $$
Is there any other way to do so ? Is it possible to do it by substitution ?
|
Is there any other way to evaluate it ?
Hint. One may observe that
$$
\frac{\cos(5x)+\cos(4x)}{1-2\cos(3x)}=-\cos(2x)-\cos (x)
$$ then the evaluation is easier.
Edit. Here is a way to obtain such a simplification. One may set $u=e^{ix}$ then using De moivre's formula for $\cos(\cdot)$ one gets
$$
\begin{align}
\frac{\cos(5x)+\cos(4x)}{1-2\cos(3x)}&=\frac{\dfrac{u^5+\dfrac1{u^5}}2+\dfrac{u^4+\dfrac1{u^4}}2}{1-\big(u^3+\dfrac1{u^3}\big)}
\\\\&=-\frac12\:\frac{1+u+u^3+u^4}{u^2}
\\\\&=- \dfrac{u^2+\dfrac1{u^2}}2-\dfrac{u+\dfrac1{u}}2
\\\\&=-\cos(2x)-\cos (x).
\end{align}
$$
|
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|
How to prove by induction that $3^{3n}+1$ is divisible by $3^n+1$ for $(n=1,2,...)$ So this is what I've tried:
Checked the statement for $n=1$ - it's valid.
Assume that $3^{3n}+1=k(3^n+1)$ where $k$ is a whole number (for some n). Proving for $n+1$: $$3^{3n+3}+1=3^33^{3n}+1=3^3(3^{3n}+1)-26=3^3k(3^n+1)-26=3^3k(3^n+1)-3^3+1=3^3[k(3^n+1)-1]+1$$ and I'm stuck. Any help please?
|
As an interesting alternative, note that $x^3 +1 = (x+1)(x^2 - x + 1)$, so setting $x = 3^n$ gives $3^{3n} + 1 = (3^n + 1)(3^{2n} - 3^n + 1)$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $\sin x + \sin y = 1$ and $\cos x + \cos y = 0$, solve for $x$ and $y$
*
*$\sin x + \sin y = 1$
*$\cos x + \cos y = 0$
Any valid pair of $(x, y)$ is fine, as the restrictions on the board in the image below are obscured.
I got the question from chapter 26 of a comic called Yamada-kun.
How can I solve this equation?
|
HINT
Squaring both equations you get
$$
\sin^2 x + \sin^2 y + 2\sin x \sin y = 1\\
\cos^2 x + \cos^2 y + 2\cos x \cos y = 0
$$
Now add them together to get
$$
2 + 2 \sin x \sin y + 2 \cos x \cos y = 1
$$
or in other words
$$
\frac{-1}{2} = \cos x \cos y + \sin x \sin y = \cos (x-y)
$$
|
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"url": "https://math.stackexchange.com/questions/1866796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Compute $1 \cdot \frac {1}{2} + 2 \cdot \frac {1}{4} + 3 \cdot \frac {1}{8} + \cdots + n \cdot \frac {1}{2^n} + \cdots $ I have tried to compute the first few terms to try to find a pattern but I got
$$\frac{1}{2}+\frac{1}{2}+\frac{3}{8}+\frac{4}{16}+\frac{5}{32}+\frac{6}{64}$$
but I still don't see any obvious pattern(s). I also tried to look for a pattern in the question, but I cannot see any pattern (possibly because I'm overthinking it?) Please help me with this problem.
|
Let
$$ L = \sum_{n=1}^\infty \frac{n}{2^n} $$
Then,
$$ L = \frac{1}{2} + \sum_{n=2}^\infty \frac{n}{2^n} \\
L = \frac{1}{2} + \sum_{n=1}^\infty \frac{n+1}{2^{n+1}} \\
L = \frac{1}{2} + \frac{1}{2} \sum_{n=1}^\infty \frac{n}{2^n} + \frac{1}{2} \sum_{n=1}^\infty \frac{1}{2^n} \\
L = \frac{1}{2} + \frac{L}{2} + \frac{1}{2} \sum_{n=1}^\infty \frac{1}{2^n} \\
\frac{L}{2} = \frac{1}{2} + \frac{1}{2} \sum_{n=1}^\infty \frac{1}{2^n} \\
\frac{L}{2} = \frac{1}{2} + \frac{1}{2} \\
\boxed{L = 2}
$$
|
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"url": "https://math.stackexchange.com/questions/1867561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Trigonometry Olympiad problem: Evaluate $1\sin 2^{\circ} +2\sin 4^{\circ} + 3\sin 6^{\circ}+\cdots+ 90\sin180^{\circ}$
Find the value of
$$1\sin 2^{\circ} +2\sin 4^{\circ} + 3\sin 6^{\circ}+\cdots+ 90\sin180^{\circ}$$
My attempt
I converted the $\sin$ functions which have arguments greater than $90^\circ$ to $\cos$ but I have gone no where with it!
I also tried using double angle formula for the angles which are even.
|
As a complement to other answers, you should know that the following identities are valid
$$ \begin{align}
2 \sin \frac{1}{2} x \sum_{k=1}^{n} \cos kx &= +\sin(n+\frac{1}{2})x - \sin\frac{1}{2}x \\
2\sin \frac{1}{2} x \sum_{k=1}^{n} \sin kx &= -\cos(n+\frac{1}{2})x + \cos\frac{1}{2}x
\end{align}$$
Thsese formulas can easily be proved by moving $\sin \frac{1}{2} x$ into the summation, using the identities
$$ \begin{align}
2 \sin \frac{1}{2} x \cos kx &= + \sin(k + \frac{1}{2})x - \sin(k - \frac{1}{2})x \\
2 \sin \frac{1}{2} x \sin kx &= - \cos(k + \frac{1}{2})x + \cos(k - \frac{1}{2})x
\end{align}$$
and the telescoping property of partial sums. In your example, we have
$$x=2\frac{\pi}{180}=\frac{\pi}{90}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1868623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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|
Partial fractions and using values not in domain I'm studying partial fraction decomposition of rational expression. In this video the guy decompose this rational expression:
$$ \frac{3x-8}{x^2-4x-5}$$
this becomes:
$$\frac{3x-8}{(x-5)(x+1)} = \frac{A}{x-5} + \frac{B}{x+1} $$
$$[(x-5)(x+1)]\times \frac{3x-8}{(x-5)(x+1)} = \frac{A}{x-5} + \frac{B}{x+1}\times[(x-5)(x+1)]$$
$$3x-8 = A(x+1) + B(x-5)$$
then he substitutes $x$ with $-1$ this is when I feel something is wrong.
because $x=-1$ is not in the domain of the original function and all the statement below the original function hold true only if they comply within the domain of the original function isn't it? so at the time this guy substitute $x=-1$ the statement become false?
I'm newbie in math, please explain to me with easy understanding and step by step.
|
Observe that the equation
\begin{eqnarray}
\frac{3x-8}{(x+1)(x-5)}=\frac{A}{x-5}+\frac{B}{x+1}
\end{eqnarray}
is valid only for $x\in \mathbb{R}\backslash \{-1,5\}$, as you have rightly pointed out. However, the equation
\begin{eqnarray}
3x-8=A(x+1)+B(x-5)
\end{eqnarray}
is valid for all $x\in \mathbb{R}$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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|
Compute trigonometric limit without use of de L'Hospital's rule
$$ \lim_{x\to 0} \frac{(x+c)\sin(x^2)}{1-\cos(x)}, c \in \mathbb{R^+} $$
Using de L'Hospital's rule twice it is possible to show that this limit equals $2c$. However, without the use of de L'Hospital's rule I'm lost with the trigonometric identities.
I can begin by showing
$$
\lim\frac{x\sin (x^2)(1+\cos(x))}{\sin^2x}+\frac{c\sin x^2(1+\cos x)}{\sin^2x}=\lim\frac{\sin (x^2)(1+\cos(x))}{\sin x}+\frac{c\sin x^2(1+\cos x)}{\sin^2x},
$$
and here I'm getting stuck. I will appreciate any help.
|
Using Taylor series make life "easy". Start with $$\sin(x)=x-\frac{x^3}{6}+O\left(x^4\right)$$ $$\sin(x^2)=x^2-\frac{x^6}{6}+O\left(x^8\right)$$ $$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^6\right)$$ So,$$\frac{(x+c)\sin(x^2)}{1-\cos(x)}=\frac{(x+c)\left(x^2-\frac{x^6}{6}+O\left(x^8\right) \right)}{\frac{x^2}{2}-\frac{x^4}{24}+O\left(x^6\right)}$$ Now, long division to get $$\frac{(x+c)\sin(x^2)}{1-\cos(x)}=2 c+2 x+\frac{c x^2}{6}+\frac{x^3}{6}+O\left(x^4\right)$$ which shows the limit and how it is approached.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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|
Evaluation of $\int^{\pi/2}_{0} \frac{x \tan(x)}{\sec(x)+\tan(x)}dx$ Evaluate the given integral:
$$\int^{\pi/2}_0 \frac{x \tan(x)}{\sec(x)+\tan(x)}dx$$
I multiplied and divided by $\sec(x)+\tan(x)$ to get denominator as $1$ but In calculation of integral, $x$ is creating problem. Is there any way to eliminate $x$ here, like it would have been eliminated if upper limit was $\pi$?
|
There might be an easier solution exploiting symmetry but the one follows is what I came up:
\begin{align*}
\int_{0}^{\pi/2} \frac{x\tan x}{\sec x + \tan x} \, {\rm d}x &= \int_{0}^{\pi/2} \frac{x \tan x}{\frac{1}{\cos x} + \frac{\sin x}{\cos x}} \, {\rm d}x \\
&= \int_{0}^{\pi/2} \frac{x \cos x \frac{\sin x}{\cos x}}{1+\sin x} \, {\rm d}x\\
&= \int_{0}^{\pi/2} \frac{x \sin x}{1+ \sin x} \, {\rm d}x \\
&=\int_{0}^{\pi/2} \frac{\left ( \frac{\pi}{2}-x \right ) \cos x}{1+\cos x} \, {\rm d}x \\
&= \frac{\pi}{2} \int_{0}^{\pi/2} \frac{\cos x}{1+ \cos x} \, {\rm d}x - \int_{0}^{\pi/2} \frac{x \cos x}{1+ \cos x} \, {\rm d}x \\
&= \frac{\pi^2-2\pi}{4} + \frac{\pi}{2} - \frac{\pi^2}{8} -\log 2 \\
&= \frac{\pi^2}{8} - \log 2
\end{align*}
The last integrals standing are trivial. If needed I may add a derivation.
Addendum:
For the first integral we have that:
\begin{align*}
\int_{0}^{\pi/2} \frac{\cos x}{1+\cos x} \, {\rm d}x &=\int_{0}^{\pi/2} \frac{\cos x+1 -1}{1+ \cos x} \, {\rm d}x \\
&= \int_{0}^{\pi/2} \left ( 1 - \frac{1}{1+ \cos x} \right ) \, {\rm d}x \\
&= \frac{\pi}{2} - \int_{0}^{\pi/2} \frac{{\rm d}x}{1+ \cos x} \\
&= \frac{\pi}{2} - \int_{0}^{\pi/2} \frac{{\rm d}x}{2\cos^2 \left ( \frac{x}{2} \right )} \\
&= \frac{\pi}{2} - \left [\tan \frac{x}{2} \right ]_0^{\pi/2} \\
&= \frac{\pi}{2} - \tan \frac{\pi}{4} \\
&= \frac{\pi}{2}-1
\end{align*}
and
\begin{align*}
\int_{0}^{\pi/2} \frac{x \cos x}{1+\cos x} \, {\rm d}x &= \int_{0}^{\pi/2} \frac{x (\cos x+1) -x}{1+\cos x} \, {\rm d}x \\
&= \int_{0}^{\pi/2} x \, {\rm d}x - \int_{0}^{\pi/2} \frac{x}{1+ \cos x} \, {\rm d}x\\
&= \left [ \frac{x^2}{2} \right ]_0^{\pi/2} - \int_{0}^{\pi/2} \frac{x}{1+\cos x} \, {\rm d}x\\
&= \frac{\pi^2}{8} - \left [ x \tan \left ( \frac{x}{2} \right ) \right ]_0^{\pi/2} + \int_{0}^{\pi/2} \tan \left ( \frac{x}{2} \right ) \, {\rm d}x \\
&=\frac{\pi^2}{8} - \frac{\pi}{2} - \left [ 2 \log \cos \frac{x}{2} \right ]_0^{\pi/2} \\
&= \frac{\pi^2}{8} - \frac{\pi}{2} - 2 \log \cos \frac{\pi}{4} \\
&= \frac{\pi^2}{8} - \frac{\pi}{2} + \log 2
\end{align*}
since it is well known that $\displaystyle \int \frac{{\rm d}x}{1+ \cos x} = \tan \frac{x}{2}+c, \; c \in \mathbb{R}$ and
$$\int \tan \frac{x}{2} \, {\rm d}x = \int \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}} \, {\rm d}x = -\frac{1}{2}\int \frac{\left ( \cos \frac{x}{2} \right )'}{\cos \frac{x}{2}} \, {\rm d}x = - 2 \log \cos \frac{x}{2} + c , \; c \in \mathbb{R}$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
Linear programming exercise I want some help for the following exercise
I know that the leaving variable is the basic variable associated with the smallest nonnegative ratio with the strictly positive denominator. I can't understand how the current basic variables ($x_5$, $x_6$, $x_7$, $x_8$) are matched with the equations.
|
Let $x_r$ and $x_j$ denote the leaving and entering variables respectively. ($r$ stands for "row"; $j$ runs through columns.)
Write the simplex tableaux.
$$
\require{enclose}
% inner horizontal array of arrays
\begin{array}{cc}
% inner array of minimum values
\begin{array}{r|r|r|r}
\text{basis} & a_j & b & b/a_j \\
\hline
5 & 1 & 4 & 4 \\
6 & 5 & 8 & 8/5 \\
\to7 & 2 & 3 & \enclose{circle}{3/2} \\
8 & * & 0 & -
\end{array}
&
% inner array of minimum values
\begin{array}{r|r|r|r}
\text{basis} & a_j & b & b/a_j \\
\hline
5 & 2 & 4 & 2 \\
6 & * & 8 & - \\
\to7 & 3 & 3 & \enclose{circle}{1} \\
8 & * & 0 & -
\end{array} \\
j=1,r=7 & j=2,r=7 \\
% inner array of minimum values
\begin{array}{r|r|r|r}
\text{basis} & a_j & b & b/a_j \\
\hline
5 & * & 4 & - \\
6 & * & 8 & - \\
7 & * & 3 & - \\
\to8 & 1 & 0 & \enclose{circle}{0}
\end{array}
&
% inner array of minimum values
\begin{array}{r|r|r|r}
\text{basis} & a_j & b & b/a_j \\
\hline
\to5 & 5 & 4 & \enclose{circle}{4/5} \\
6 & 6 & 8 & 4/3 \\
7 & 3 & 3 & 1 \\
8 & * & 0 & -
\end{array} \\
j=3,r=8 & j=4,r=5
\end{array}
$$
Fix $j$. Observe that $r$ is the index which minimises $\{b/a_{rj} \mid a_{rj}>0 \}$. (You may refer to a theoretical explanation for such choice.)
|
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"url": "https://math.stackexchange.com/questions/1876075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Convergence/Divergence of $\sum_{n=1}^{\infty}\frac{n+n^2+\cdots+n^n}{n^{n+2}}$
$$\sum_{n=1}^\infty \frac{n+n^2+\cdots+n^n}{n^{n+2}}$$
$$\sum_{n=1}^\infty \frac{1}{n^{n}}=\sum_{n=1}^{\infty}\frac{n^2}{n^{n+2}}=\sum_{n=1}^\infty \frac{n+n+\cdots+n}{n^{n+2}}\leq\sum_{n=1}^\infty \frac{n+n^2+\cdots+n^n}{n^{n+2}}\leq \sum_{n=1}^{\infty}\frac{n^n+n^n+\cdots+n^n}{n^{n+2}}=\sum_{n=1}^\infty \frac{n^{n+1}}{n^{n+2}}=\sum_{n=1}^\infty \frac{1}{n}$$
But is still does not help to conclude about converges/diverges
|
Easy with equivalents (this requires the general term to have a constant sign) and some computation:
$$n+n^2+\dots+n^n=\frac{n(n^n-1)}{n-1}\sim_\infty n^n$$
so
$$\frac{n+n^2+\cdots+n^n}{n^{n+2}}\sim_\infty\frac{n^n}{n^{n+2}}=\frac1{n^2},$$
which converges.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1878090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Formula for $1^k+2^k+3^k...n^k$ for $n,k \in \mathbb{N}$ So I've been looking for a formula where I can input the parameter $k$ and it will give me a formula for $1^k+2^k+3^k...+ n^k$ with $n,k \in \mathbb{N}$. The result is always a polynomial with $k+1$ as highest power. I've taken the time to calculate the polynomes for $k=1$ to $k=10$ by hand and using the interpolation feature of Wolfram Alpha. Here are the results (I'll only show the coefficients for the sake of clarity. the coefficients are always from $n^{k+1}$ to $n^1$. the constant is always $0$. So $\frac{1}{2},-\frac{1}{2}$ becomes $\frac{1}{2}n^2-\frac{1}{2}n$):
*
*$k=1$ : $\frac{1}{2},-\frac{1}{2}$
*$k=2$ : $\frac{1}{3},\frac{1}{2},\frac{1}{6}$
*$k=3$ : $\frac{1}{4},\frac{1}{2},\frac{1}{4},0$
*$k=4$ : $\frac{1}{5},\frac{1}{2},\frac{1}{3},0,-\frac{1}{30}$
*$k=5$ : $\frac{1}{6},\frac{1}{2},\frac{5}{12},0,-\frac{1}{12},0$
*$k=6$ : $\frac{1}{7},\frac{1}{2},\frac{1}{2},0,-\frac{1}{6},0,\frac{1}{42}$
*$k=7$ : $\frac{1}{8},\frac{1}{2},\frac{7}{12},0,-\frac{7}{24},0,\frac{1}{12},0$
*$k=8$ : $\frac{1}{9},\frac{1}{2},\frac{2}{3},0,-\frac{7}{15},0,\frac{2}{9},0,-\frac{1}{30}$
*$k=9$ : $\frac{1}{10},\frac{1}{2},\frac{3}{4},0,-\frac{7}{10},0,\frac{1}{2},0,-\frac{3}{20},0$
*$k=10$ : $\frac{1}{11},\frac{1}{2},\frac{5}{9},0,1,0,1,0,-\frac{1}{2},0,\frac{5}{66}$
There are a few things i notice: Firstly, the coefficient of the highest power seems to be $\frac{1}{k+1}$. Secondly, the coefficient of the second highest power seems to be $\frac{1}{2}$ with the exception of $k=1$. Thirdly, all coefficients of the fourth, sixth, eight highest power and so on seem to be $0$.
What is the formula that will output the coefficients for any value of $k$?
|
Another way to derive a formula for $$S(k,n)=\sum_{s=1}^{n}{s^k}$$ is to use the binomial expansion. To do this, you can start by changing the summation index $s$ to $t+1$. That is:$$S(k,n)=\sum_{t=0}^{n-1}{(t+1)^k}=1+\sum_{t=1}^{n-1}{(t+1)^k}$$Using binomial expansion we have:$$S(k,n)=1+\sum_{t=1}^{n-1}{\sum_{j=0}^{k}{\binom{k}{j}t^j}}$$Changing the order of summation (notice the independence) gets:$$S(k,n)=1+\sum_{j=0}^{k}{\sum_{t=1}^{n-1}{\binom{k}{j}t^j}}=1+\sum_{j=0}^{k}{\left(\binom{k}{j}\sum_{t=1}^{n-1}{t^j}\right)}$$The inner sum equals $S(j,n-1)$, so: $$S(k,n)=1+\sum_{j=0}^{k}{\binom{k}{j}S(j,n-1)}$$Now, by excluding the last two terms of the summation (the terms obtained by $j=k-1$ and $j=k$) we finally get: $$S(k,n)=1+S(k,n-1)+kS(k-1,n-1)+\sum_{j=0}^{k-2}{\binom{k}{j}S(j,n-1)}$$ Since $S(k,n)-S(k,n-1)=n^k$: $$S(k-1,n-1)=\frac{1}{k}\left(n^k-1-\sum_{j=0}^{k-2}{\binom{k}{j}S(j,n-1)}\right)$$ The formula shows a recursive relation between $S(k-1,n-1)$ and lower sums, i.e., $S(0,n-1),S(1,n-1),\dots,S(k-2,n-1)$. By changing $k$ to $k+1$ and $n$ to $n+1$ the original format emerges: $$S(k,n)=\frac{1}{k+1}\left((n+1)^{k+1}-1-\sum_{j=0}^{k-1}{\binom{k+1}{j}S(j,n)}\right)$$ You can now start from $S(0,n)=n$ and derive $S(k,n)$ for $k=1,2,\dots$. For example: $$S(1,n)=\frac{1}{2}\left( (n+1)^2-1-\binom{2}{0}S(0,n)\right)=\frac{1}{2}\left( n^2+2n+1-1-n \right)=\frac{1}{2}n(n+1)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Can this systems of equations be solved? There is a system of equations as follow:
$$\left[\begin{matrix}x\\ y\\ \end{matrix} \right] = \left[\begin{matrix}a& b\\b & c\\\end{matrix} \right]\left[\begin{matrix}\frac{1}{x}\\ \frac{1}{y}\\ \end{matrix} \right]$$
Given $ \left[\begin{matrix}a& b\\b & c\\\end{matrix} \right]$, can this system of equations be solved for $x$ and $y$?
And how to solve the extended case?
$$\left[\begin{matrix}x_1\\ x_2\\ \vdots \\x_n \end{matrix} \right] =
\left[\begin{matrix}a_{11}& a_{12} & \dots &a_{1n}\\a_{12} & a_{22} & \dots &a_{2n}\\ \vdots & \vdots & &\vdots \\a_{1n} & a_{2n} & \dots &a_{nn}\end{matrix} \right]\left[\begin{matrix}\frac{1}{x_1}\\ \frac{1}{x_2}\\ \vdots \\ \frac{1}{x_n}\\ \end{matrix} \right]$$
|
After pie314271's answer
The equation $$b^2(x^2-a)+c(x^2-a)^2=b^2x^2$$ reduces to $$a\left(a c- b^2\right)-2 a c x^2+c x^4=0$$ which is just a quadratic in $x^2$ and the solutions are given by $$x=\pm\sqrt{a\pm\frac{\sqrt{a}}{\sqrt{c}}\,b}$$ So, the whole set of solutions is
$$\left\{x= -\sqrt{a-\frac{\sqrt{a} b}{\sqrt{c}}},y= \frac{\sqrt{c}
\sqrt{a-\frac{\sqrt{a} b}{\sqrt{c}}}}{\sqrt{a}}\right\},\left\{x=
\sqrt{a-\frac{\sqrt{a} b}{\sqrt{c}}},y= -\frac{\sqrt{c} \sqrt{a-\frac{\sqrt{a}
b}{\sqrt{c}}}}{\sqrt{a}}\right\},\left\{x=-\sqrt{a+\frac{\sqrt{a}
b}{\sqrt{c}}},y= -\frac{\sqrt{c} \sqrt{a+\frac{\sqrt{a}
b}{\sqrt{c}}}}{\sqrt{a}}\right\},\left\{x= \sqrt{a+\frac{\sqrt{a}
b}{\sqrt{c}}},y=\frac{\sqrt{c} \sqrt{a+\frac{\sqrt{a}
b}{\sqrt{c}}}}{\sqrt{a}}\right\}$$
|
{
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"url": "https://math.stackexchange.com/questions/1881765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Computing $\int \frac{1}{(1+x^3)^3}dx$ I tried various methods like trying to break it into partial fractions after factorizing, applying substitutions but couldn't think of any. How will we integrate this?
|
Hint. Another path. Consider a fixed real number $a>0$. From
$$
a^3+x^3=(a+x)(a^2-ax+x^2)
$$ one gets
$$
\begin{align}
\frac{1}{a^3+x^3}&=\frac{1}{3 a^2 (a+x)}+\frac{2 a-x}{3 a^2 \left(a^2-a x+x^2\right)}
\end{align}
$$ integrate the preceding identity to get
$$
\int\frac{dx}{a^3+x^3}=\frac1{3a^2}\ln(a+x)-\frac1{6a^2}\ln(a^2-a x+x^2)+\frac1{a^2\sqrt{3} }\arctan\left(\frac{2x-a}{a\sqrt{3}}\right)
$$ put $a^3 \to a$ and just differentiate twice with respect to $a$.
|
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"url": "https://math.stackexchange.com/questions/1884093",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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|
Integration of trigonometric function $\int\frac{\sin(2x)}{\sin(x)-\cos(x)}dx$
$$\int\frac{\sin(2x)}{\sin(x)-\cos(x)}dx$$
My attempt: Firstly, $\sin(2x)=2\sin(x)\cos(x)$.
After that, eliminate the $\cos(x)$ seen in both the numerator and denominator to get
$$2\int\frac{\sin(x)}{\tan(x)-1}\ dx.$$
From here onwards, should I convert $\sin(x)$, $\tan(x)$ to half-angles and use $\tan(x/2)=t$?
But this would be a time consuming method. Any suggestions?
|
Hint:
$$\int\frac{\sin(2x)}{\sin(x)-\cos(x)}dx = \int\frac{2\sin(x)\cos(x) }{\sin(x)-\cos(x)}dx = \int\frac{\sin(x)\cos(x)+\cos(x)\sin(x) }{\sin(x)-\cos(x)}dx\\
=\int\frac{\sin(x)\cos(x)-\sin^2(x)+\cos(x)\sin(x)-\cos^2(x)+1 }{\sin(x)-\cos(x)}dx \\=\int -\sin(x)dx+\int \cos(x)dx + \int \frac{1}{\sin(x) -\cos(x)}dx \\= \cos(x) + \sin(x) +\int \frac{1}{\sin(x) -\cos(x)}dx $$
For $\int \frac{1}{\sin(x) -\cos(x)}dx$:
Notice that $$\int \frac{1}{\sin(x) -\cos(x)}dx = \int \frac{1}{\sqrt{2} \sin(x-\frac{1}{4} \pi ) }dx$$
Let $u = x-\frac{1}{4} \pi$,
$$ \int \frac{1}{\sqrt{2} \sin\left(x-\frac{1}{4} \pi \right) }dx = \int \frac{1}{\sqrt{2} } \csc(u)du $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1884160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Better way to evaluate $\int \frac{dx}{\left (a +b\cos x \right)^2}$ $$\int \frac{dx}{\left (a +b\cos x \right)^2}$$
$$u=\frac{b +a \cos x}{a +b\cos x }$$
$$du=\frac{\sin x\left(b^2 -a^2\right)}{ \left (a +b.\cos x \right)^2}$$
$$\frac{du}{\sin x\left(b^2 -a^2\right)}=\frac{dx}{\left (a +b\cos x \right)^2}$$
$$\cos x=\frac{au -b}{a - bu} $$
$$\sin x=\sqrt{1-\left(\frac{au -b}{a - bu}\right)^2}$$
It is becoming messy with this .I also tried it using half angle formula but didn't find it good.
|
HINT
Put $$A=\frac{\ sin x}{a+b\cos x} $$now differentiate both sides and after simplification integrate it up to get the desired result.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1887194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
What is $\sqrt{-4}\sqrt{-9}$? I assumed that since $a^c \cdot b^c = (ab)^{c}$, then something like $\sqrt{-4} \cdot \sqrt{-9}$ would be $\sqrt{-4 \cdot -9} = \sqrt{36} = \pm 6$ but according to Wolfram Alpha, it's $-6$?
|
The property $a^c \cdot b^c = (ab)^{c}$ that you mention only holds for integer exponents and nonzero bases. Since $\sqrt{-4} = (-4)^{1/2}$, you cannot use this property here.
Instead, use imaginary numbers to evaluate your expression:
$$
\begin{align*}
\sqrt{-4} \cdot \sqrt{-9} &= (2i)(3i) \\
&= 6i^2 \\
&= \boxed{-6}
\end{align*}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1888773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
How to solve this determinant equation in a simpler way
Question Statement:-
Solve the following equation
$$\begin{vmatrix}
x & 2 & 3 \\
4 & x & 1 \\
x & 2 & 5 \\
\end{vmatrix}=0$$
My Solution:-
$$\begin{vmatrix}
x & 2 & 3 \\
4 & x & 1 \\
x & 2 & 5 \\
\end{vmatrix}=
\begin{vmatrix}
x+5 & 2 & 3 \\
x+5 & x & 1 \\
x+7 & 2 & 5 \\
\end{vmatrix} \tag{$C_1\rightarrow C_1+C_2+C_3$}$$
$$=\begin{vmatrix}
0 & 2 & 3 \\
0 & x & 1 \\
2 & 2 & 5 \\
\end{vmatrix}+
(x+5)\begin{vmatrix}
1 & 2 & 3 \\
1 & x & 1 \\
1 & 2 & 5 \\
\end{vmatrix}\tag{1}$$
On opening the first determinant in the last step above we get $2(2-3x)$.
On simplifying the secind determinant we get,
$$(x+5)\begin{vmatrix}
1 & 2 & 3 \\
1 & x & 1 \\
1 & 2 & 5 \\
\end{vmatrix}=(x+5)\begin{vmatrix}
1 & 2 & 3 \\
0 & x-2 & -2 \\
0 & 0 & 2 \\
\end{vmatrix}
(R_2\rightarrow R_2-R_1)
(R_3\rightarrow R_3-R_1)$$
$=2(x+5)(x-2)$
Substituting the values obtained above in $(1)$, we get
$$=\begin{vmatrix}
0 & 2 & 3 \\
0 & x & 1 \\
2 & 2 & 5 \\
\end{vmatrix}+
(x+5)\begin{vmatrix}
1 & 2 & 3 \\
1 & x & 1 \\
1 & 2 & 5 \\
\end{vmatrix}=2(2-3x)+2(x+5)(x-2)=2(2-3x+x^2+3x-10)=2(x^2-8)$$
Now, as $\begin{vmatrix}
x & 2 & 3 \\
4 & x & 1 \\
x & 2 & 5 \\
\end{vmatrix}=0$, $\therefore 2(x^2-8)=0\implies x=\pm2\sqrt2$
As you can see there was lot of work in my solution so if anyone can provide me with some techniques to solve it faster, or a technique which includes less amount of pen and more thinking.
|
Just expand the determinant! We want
$$
5x^2+2x+24-2x-40-3x^2=0
$$
which simplifies to $2x^2-16=0$.
(In general, finding determinants via row/column relations is faster when a matrix is large. But $3 \times 3$ matrices aren't that large yet...)
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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|
Integration: $\int\frac{dx}{\sqrt{x(1-x)}}$ My teacher wrote this on the blackboard: $$\int_0^z \frac{1}{\pi \sqrt{x(1-x)}}dx= \frac{2}{\pi} \arcsin(\sqrt{z})$$
But when I try to calculate the integral:
\begin{align}
\int \frac{1}{\pi \sqrt{x(1-x)}}dx&=\int \frac{1}{\pi \sqrt{(\frac{1}{2})^2-(x-\frac{1}{2})^2}}dx\\
&= \int \frac{1}{\pi \sqrt{(\frac{1}{2})^2-k^2}}dk=2 \int \frac{1}{\pi \sqrt{1-(2k)^2}}dk\\
&= \int \frac{1}{\pi \sqrt{1-a^2}}da\\
&= \frac{1}{\pi} \arcsin(a)\\
&=\frac{1}{\pi} \arcsin\left(2(x-\frac{1}{2})\right)
\end{align}
So
$$\int_0^z \frac{1}{\pi \sqrt{x(1-x)}}dx= \frac{1}{\pi} \arcsin(2z-1)-\frac{1}{\pi} \arcsin(-1)= \frac{1}{\pi} \arcsin(2z-1) - \frac{1}{2}$$
How I come to the solution of my teacher?
|
if you make the substitution $x=u^2$ the integral becomes:
$$
\int_0^{\sqrt{z}} \frac2{\pi \sqrt{1-u^2}}du
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1891834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
How to evaluate $\int \frac{1+x}{1+\sqrt x} dx$ I picked this exercice from B. Demidovitch and i started solving, but when it's complete, i get only the half of the solution (the first two fractions and the ln, doesn't show up).
$$\int \frac{1+x}{1+\sqrt x} dx$$
The Solution is:
$$2 \left[\frac{\sqrt{x^3}}{3}-\frac{x}{2}+2\sqrt{x}-2ln(1+\sqrt{x})\right] + C$$
|
By setting $\color{red}{x=z^2}$ we just have to compute
$$ I=\int \frac{2z(1+z^2)}{1+z}\,dz $$
and by polynomial division we have $2z(1+z^2)=(2z^2-2z+4)(1+z)-4$, hence
$$ I = \int (2z^2-2z+4)\,dz -4\int\frac{dz}{1+z} = \color{red}{\frac{z^3}{3}-z^2+4-4\log(1+z)+C}. $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1893114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Let a, b, c be positive real numbers. Prove that Let a,b,c be positive real numbers. Prove that
$$\frac{a^3+b^3+c^3}{3}\geq\frac{a^2+bc}{b+c}\cdot\frac{b^2+ca}{c+a}\cdot\frac{c^2+ab}{a+b}\geq abc$$
I will post what I had solved originally, however it is unfortunately incorrect. Please help solve and/or aid in finding my mistakes :]
|
This is what I have although it is incorrect any help would be greatly appreciated.!
$\mathbf{1.)}$ Consider now the expression $\frac{a^3+b^3+c^3}{3}$
The expressions has minimum value given by:
$a\geq b \geq c > 0 \implies a^3 \geq b^3 \geq c^3$
Thus $\frac{a^3+b^3+c^3}{3}$ has minimum value $c^3$
$\mathbf{2.)}$ On the other hand, consider the expression $\frac{a^2+bc}{b+c}$
Again we have $a \geq b \geq c >0 \implies a^2+bc \geq c^2+c\cdot c = 2c^2$
$ a \geq b \geq c > 0 \implies a^2+bc \leq a^2 + a \cdot a = 2a^2$
$ a \geq b \geq c > 0 \implies b+c \geq 2c $
and $ a \geq b \geq c > 0 \implies b+c \leq 2a$
$\mathbf{3.)}$ Finally we know that a positive fraction is maximized when the numerator is as large as possible and the denominator is as small as possible
Therefore we have $ \frac{a^2+bc}{b+c} \leq \frac{2a^2}{2c} = \frac{a^2}{c} \leq a $
Using identical considerations, we have $\frac{b^2+ca}{ca} \leq \frac{a^2}{c} \leq a $ and $\frac{c^2+ab}{a+b}\leq \frac{a^2}{c} \leq a$
So that we have
Hence, $\frac{a^2+bc}{b+c} \cdot \frac{b^2+ca}{c+a} \cdot \frac{c^2+ab}{a+b} \leq a^3$
This is the maximum value of the expression
$\mathbf{4.)}$ The minimum value will be achieved for a fraction when the numerator is as small as possible and denominator is as large as possible
Again, by an overall similar consideration, we have $\frac{a^2+bc}{b+c} \geq \frac{a^2+c^2}{2b}$ is the minimum value of this expression
Similarly $\frac{b^2+ca}{c+a} \geq \frac{b^2+c^2}{2a}$ and $\frac{c^2+ab}{a+b} \geq \frac{c^2+b^2}{2a}$
Hence
But we have $a^2+b^2 \geq 2b^2$, $c^2+b^2 \geq 2c^2$
Hence,
Therefore, $\frac{a^2+bc}{b+c} \cdot \frac{b^2+ca}{c+a} \cdot \frac{c^2+ab}{a+b} \geq \frac{bc^4}{a^2}$
$\mathbf{5.)}$ And finally, we have $a^3 \geq abc$ is the maximum value of $abc$
The minimum value of $\frac{a^3+b^3+c^3}{3}$ is $c^3$ which is greater than $a^3$ which is the maximum value of $\frac{a^2+bc}{b+c}\cdot \frac{b^2+ca}{c+a} \cdot \frac{c^2+ab}{a+b}$
That is,
So that $\frac{a^3+b^3+c^3}{3} \geq \frac{a^2+bc}{b+c}\cdot \frac{b^2+ca}{c+a} \cdot \frac{c^2+ab}{a+b} $
Again we have $\frac{a^2+bc}{b+c}\cdot \frac{b^2+ca}{c+a} \cdot \frac{c^2+ab}{a+b} \geq \frac{bc^4}{a^2}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1893626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
If $a +b $,= 5 and $ab = 6$ . Find $\frac{1}{a}$ + $\frac{1}{b}$ $a +b $ = 5 and $ab = 6$ . Find $\cfrac{1}{a}$ + $\cfrac{1}{b}$
Any Ideas on how to begin?
Many Thanks
|
$$\frac { 1 }{ a } +\frac { 1 }{ b } =\frac { a+b }{ ab } =\frac { 5 }{ 6 } $$
|
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|
Is the series $\sqrt{1} - \sqrt{2} + \sqrt{3} - \sqrt{4} + \dots$ summable? Is the series $\sqrt{1} - \sqrt{2} + \sqrt{3} - \sqrt{4} + \dots$ summable? I think it diverges although:
$$ \sqrt{n+1} - \sqrt{n} \approx \frac{1}{2\sqrt{n}}$$
for example by the Mean Value Theorem $f(x+1)-f(x) \approx f'(x)$ and then I might argue:
$$ \sum_{n \geq 1} (-1)^{n+1} \sqrt{n} = \frac{1}{2}\sum_{m \geq 1} \frac{1}{\sqrt{2m}} = \infty $$
Are these Cesaro summable? For an even number of terms:
$$\sqrt{1} - \sqrt{2} + \sqrt{3} - \sqrt{4} + \dots - \sqrt{2n}
\approx - \frac{1}{2\sqrt{2}}\left( \frac{1}{\sqrt{1}} +
\frac{1}{\sqrt{2}} + \dots + \frac{1}{\sqrt{n}} \right)
\approx \sqrt{\frac{n}{2}}$$
so the Cesaro means tend to infinity. Does any more creative summation method work?
The result is from paper called "The Second Theorem of Consistency for Summable Series" in Vol 6 of the Collected Works of GH Hardy
the series $1 - 1 +1 - 1 \dots$ is summable $(1,k)$ for any $k$ but not summable $(e^n, k)$ for any value of $k$.
The series $\sqrt{1} - \sqrt{2} + \sqrt{3} - \sqrt{4} + \dots$ is summable $(n,1)$ but not $(e^{\sqrt{n}},1)$ and so on...
Here things like $(1,k), (n,1)$ refer to certain averaging procedures, IDK
|
I thought it might be instructive to present a brute force approach. To that end we proceed.
Let $S_n=\sum_{k=1}^n (-1)^{k-1}\sqrt{k}$ be the sequence of interest. Then, we can write the even and odd terms, respectively by
$$\begin{align}
S_{2n}&=\sum_{k=1}^n \left(\sqrt{2k-1}-\sqrt{2k}\right)\\\\
S_{2n+1}&=1+\sum_{k=1}^n \left(\sqrt{2k+1}-\sqrt{2k}\right)
\end{align}$$
Then, the Cesaro Sum is given by
$$\begin{align}
\frac{\sum_{n=0}^N (S_{2n}+S_{2n+1})}{2N+1}&=\frac{1+\sum_{n=1}^N\left(1+ \sum_{k=1}^n \left(\sqrt{2k+1}-2\sqrt{2k}+\sqrt{2k-1}\right)\right)}{2N+1} \tag 1\\\\
&=\frac{N+1}{2N+1}+\frac{1}{2N+1}\sum_{n=1}^N\sum_{k=1}^n \left(\sqrt{2k+1}-2\sqrt{2k}+\sqrt{2k-1}\right)\tag 2\\\\
&=\frac{N+1}{2N+1}+\frac{1}{2N+1}\sum_{k=1}^N\sum_{n=k}^N \left(\sqrt{2k+1}-2\sqrt{2k}+\sqrt{2k-1}\right)\tag 3\\\\
&=\frac{N+1}{2N+1}+\frac{1}{2N+1}\sum_{k=1}^N(N+1-k)\left(\sqrt{2k+1}-2\sqrt{2k}+\sqrt{2k-1}\right) \tag 4\\\\
&=\frac{N+1}{2N+1}+\frac{N+1}{2N+1}\sum_{k=1}^N \left(\sqrt{2k+1}-2\sqrt{2k}+\sqrt{2k-1}\right) \tag 5\\\\
&-\frac{1}{2N+1}\sum_{k=1}^N k\left(\sqrt{2k+1}-2\sqrt{2k}+\sqrt{2k-1}\right)
\end{align}$$
NOTES:
In going from $(1)$ to $(2)$, we simply carried our the trivial sum $\sum_{k=1}^n (1)=N$.
In going from $(2)$ to $(3)$, we interchanged the order of summation.
In going from $(3)$ to $(4)$, we evaluated the inner sum.
In going from $(4)$ to $(5)$, we split the expression into the sum of three terms.
As $N\to \infty$, the first term in $(5)$ approaches $\frac12$ while the second term approaches $\frac12 \sum_{k=1}^\infty \left(\sqrt{2k+1}-2\sqrt{2k}+\sqrt{2k-1}\right)$. The third term in $(5)$ can be shown to approach $0$ since $\sum_{k=1}^N k \left(\sqrt{2k+1}-2\sqrt{2k}+\sqrt{2k-1}\right) =O\left(N^{1/2}\right)$.
Therefore, we find that
$$\lim_{N\to \infty}\frac{\sum_{n=0}^N (S_{2n}+S_{2n+1})}{2N+1}=\frac12+\frac12 \sum_{k=1}^\infty \left(\sqrt{2k+1}-2\sqrt{2k}+\sqrt{2k-1}\right) $$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Solve for x and find an approximate value $15 = \dfrac{((x+3)+(2x-3))h}{2}$
h = $(2x-3) -(x+3) $
& it is also given that $\sqrt{19}$ = 4.36
How can I simplify this & can you help me by explaining the steps
|
$$30 = ((2x-3)+(x+3))((2x-3)-(x+3))=(2x-3)^2-(x+3)^2=3x^2-18x$$
$$x^2-6x=10$$
$$x^2-6x+9=19$$
$$x=3\pm\sqrt {19}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
If $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1$, what can we say about $\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}$?
Suppose that $a,b,c$ are three real numbers such that $\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=1$. What are the possible values for $\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}$?
After clearing the denominators, we have $$a(c+a)(a+b)+b(b+c)(a+b)+c(b+c)(c+a)=(a+b)(b+c)(a+c)\,.$$
That is,
$$a^3+b^3+c^3+abc=0\,.$$
But then I'm stuck. This question is related, but a bit different.
Thank you for your help!
|
Assuming
$$\frac a{b+c}+\frac b{c+a}+\frac c{a+b}=1$$
we also have
$$\frac{a^{2}}{b+c}+\frac {ab}{c+a}+\frac {ac}{a+b}= a$$
as well as
$$\frac{ab}{b+c}+\frac {b^2}{c+a}+\frac {bc}{a+b}= b$$
and
$$\frac{ac}{b+c}+\frac {bc}{c+a}+\frac {c^2}{a+b}= c$$
These three sum together as but all terms without $(.)^2$ on the other side on sorting terms with same denominator you get
$$\frac{a^{2}}{b+c} + \frac {b^2}{c+a} + \frac {c^2}{a+b} =$$
$$ a+b+c -(\frac {ac}{b+c} + \frac{ab}{b+c}) - (\frac {ab}{c+a} + \frac {bc}{c+a}) - (\frac {ac}{a+b} + \frac {bc}{a+b}) = $$
$$ a + b + c - (a) - (b)- (c) = 0$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the last digit of $7^{802} -3^{683}$
Find the last digit of $$7^{802} -3^{683}$$
$$7^{802} -3^{683}=(50 -1)^{401} -3(10 -1)^{341}$$
$$=50k -1 -3(10m -1) \space \text{where k,m are fixed integer } $$
$$=50k -30m +2 $$
So,the last digit will be $2$
Someone just added this question, but deleted it. I solved this question and got this thing. I just want to know if my answer correct or not.
|
Okay, after reading the comments I realize my answer is completely out of the experience level of the OP. But read ahead for a taste of exciting things to come.
=====
Very nice!
I want to introduce you to the idea of Fermat's Little Thereom however.
If $\gcd(a, m) = 1$. And $\phi(m) = $ the number of natural numbers less than $m$ that are relatively prime to $m$ then:
$a^{\phi (m)} \equiv 1 \mod m$
(or in other words $a^{\phi(m)} = km + 1$ for some $k$.)
So... as $1,3,7$ and $9$ are relatively prime to $10$ so $\phi(10) = 4$. And $\gcd(7,10) = \gcd(3,10) = 1$.
So $7^4 = 1 \mod 10$ ($7^4 = 2401$) and $3^4 \equiv 1 \mod 10$ ($3^4 = 81$).
So $7^{802} - 3^{683} = 7^{2 + 4k} - 3^{3 + 4j}$
$= 7^2*(7^4)^k - 3^3*(3^4)^j \equiv 7^2*1^k - 3^3*1^j \mod 10$
$\equiv 49 - 27 \mod 10 \equiv 9 -7 \mod 10 = 2 \mod 10$
So $2$ is the last digit.
.........
Okay, I admit, now that I've typed the whole thing out that seems a LOT more complicated than your elegant fast solution.
But it's useful to know.
If I asked "what is the remainder when $8^{386} - 6^{216}$ is divided by $13$" it might be easier to use Fermat's Little Theorem, rather than "$8^2 = 64 = 5*13 - 1$ and $6^6 = 46656 = 3589*13 - 1$ so $(-1)^{193} - (-1)^36 = -1 -1 =-2$ so $11$ is the remainder"
[$13$ is prime so all numbers less than $13$ are relatively prime so $\phi(13) = 2$. So $8^{386} \equiv 1 \mod 13$ and $6^216 \equiv 1 \mod 13$.
[So $8^{386=2 + 12k} - 6^{216=12j} \equiv 8^2 - 1 \mod 13$
[$\equiv 63 \mod 13 \equiv 11 \mod 13$]
|
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"timestamp": "2023-03-29T00:00:00",
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|
Generating function for cubes of Harmonic numbers By generalizing the approach in Integral involving a dilogarithm versus an Euler sum. meaning by using the integral representation of the harmonic numbers and by computing a three dimensional integral over a unit cube analytically we have found the generating function of cubes of harmonic numbers. We have:
\begin{eqnarray}
&&S^{(3)}(x) := \sum\limits_{n=1}^\infty H_n^3 x^n =
\frac{-18 \text{Li}_3\left(1-\frac{1}{x}\right)+6 \text{Li}_3\left(\frac{1}{x}\right)-18 \text{Li}_3(x)}{6(1-x)}+
\frac{6 \log ^3(1-x)-9 \log (x) \log ^2(1-x)+3 \left(3 \log ^2(x)+\pi ^2\right) \log (1-x)}{6(x-1)}+\frac{-\log (x) \left(2 \log ^2(x)+ 3 i \pi \log (x)+5 \pi ^2\right)}{6 (x-1)}
\end{eqnarray}
Clearly some of the terms on the right hand side are complex even though the whole expression is of course real. The first two terms in the first fraction on the rhs are complex and the middle term in the last fraction is complex. My question is how do I simplify the right hand side to get rid of the complex terms?
|
Here we provide a closed form for another related sum. We have:
\begin{eqnarray}
&&\sum\limits_{n=1}^\infty \frac{H_n^3}{n} \cdot x^n =\\
&&3 \zeta(4)-3 \text{Li}_4(1-x)+3 \text{Li}_3(1-x) \log (1-x)+\log (x) \log ^3(1-x)+\\
&&\text{Li}_2(x){}^2-2 \text{Li}_4(x)-3 \text{Li}_4\left(\frac{x}{x-1}\right)+\frac{3}{2} \left(\text{Li}_2(x)-\frac{\pi ^2}{6}\right) \log ^2(1-x)+2 \text{Li}_3(x) \log (1-x)+\frac{1}{8} \log ^4(1-x)
\end{eqnarray}
We obtained this formula by dividing the right hand side in the question above by $x$ and then integrating.
|
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.