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Very hard inequality: $\frac{a}{\sqrt{a+pb}}+\frac{b}{\sqrt{b+pc}}+\frac{c}{\sqrt{c+pa}} \le k_p \sqrt{a+b+c}.$ Given $p>0$. Find the smallest real number $k_p$ such that the following inequality holds for any non-negative reals $a,b,c$:
$$\frac{a}{\sqrt{a+pb}}+\frac{b}{\sqrt{b+pc}}+\frac{c}{\sqrt{c+pa}} \le k_p \sqrt{a+b+c}.$$
Some particular cases:
*
*$k_p = \sqrt{\frac{3}{p+1}}$ for $0\le p\le \frac{1}{2}$.
*$k_1=\frac{5}{4}$.
*$k_{3/2} = \frac{2\sqrt{6}-3\sqrt{2}}{\sqrt{-1+\sqrt{3}}}+\sqrt{-5+3\sqrt{3}}$.
*$k_2 = \frac{2\sqrt{3}-2}{\sqrt{2\sqrt{3}}}+\sqrt{-1+\frac{2}{\sqrt{3}}}$.
*$k_4 = \frac{17 -\sqrt{33}}{6\sqrt{-1+\sqrt{33}}}+\sqrt{\frac{-5+\sqrt{33}}{12}}$.
|
Consider the vectors $v = \left(\sqrt a,\sqrt b,\sqrt c\right)$ and $u = \left(\frac{\sqrt a}{\sqrt{a + pb}},\frac{\sqrt b}{\sqrt{b + pc}},\frac{\sqrt c}{\sqrt{c + pa}}\right)$
Applying the scalar product inequality, we get:
$$\frac{a}{\sqrt{a+pb}}+\frac{b}{\sqrt{b+pc}}+\frac{c}{\sqrt{c+pa}} = v\cdot u \leq |v||u| = \sqrt{a+b+c}\cdot \sqrt{\frac{a}{a+pb} + \frac{b}{b+pc} + \frac{c}{c+pa}}$$
The equality holds iff the vectors are parallel, that is, iff $\sqrt{a+pb}=\sqrt{b+pc}=\sqrt{c+pa}$
Therefore you must assume $a+pb=b+pc=c+pa\Rightarrow a=b=c$ and the expression on the right side becomes:
$$RHS=\sqrt{a+b+c}\cdot \sqrt{\frac{a}{a+pa} + \frac{a}{a+pa} + \frac{a}{a+pa}} = \sqrt{a+b+c}\cdot \sqrt{\frac{3}{p + 1}}$$
Which is the result you got for $0\leq p \leq \frac{1}{2}$. Actually, for $p=1$, $\sqrt{\frac{3}{p + 1}}$ verifies the inequality and is smaller than $\frac{5}{4}$, so you might want to review that result. In order to finish the problem, it only remains to prove that
$$\sqrt{\frac{a}{a+pb} + \frac{b}{b+pc} + \frac{c}{c+pa}}\leq \sqrt{\frac{3}{p + 1}}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
LU Factorisation (4x4 matrix) - most efficient method I understand how to do LU factorisation but I'm not sure I'm being very efficient. I first find the row echelon form of A, noting the elementary operations $E_i$ in order.
$$
E_1E_2...E_nA = U $$ then $$ L = E_1^{-1}E_2^{-1}...E_n^{-1}
$$
But is this the quickest way for a 4x4 matrix?
I've been given a class problem which (going by every other question) shouldn't take as long as it's taken me. I'd be interested to hear what method you'd be using!
$$
A =
\begin{pmatrix}
1 & -2 & -2 & -3 \\
3 & -9 & 0 & -9 \\
-1 & 2 & 4 & 7 \\
-3 & -6 & 26 & 2
\end{pmatrix}
$$
|
We can use Doolittle's Method:
$$\begin{bmatrix} 1 & 0 & 0 &0\\ l_{21} & 1 & 0 &0 \\ l_{31} & l_{32} & 1 &0 \\ l_{41} & l_{42} & l_{43} & 1 \end{bmatrix} \cdot \begin{bmatrix} u_{11} & u_{12} & u_{13} &u_{14}\\ 0 & u_{22} & u_{23} &u_{24} \\ 0 & 0 & u_{33} &u_{34} \\0 & 0 & 0 & u_{44} \end{bmatrix} = \begin{bmatrix} 1 & -2 & -2 & -3 \\ 3 & -9 & 0 & -9 \\ -1 & 2 & 4 & 7 \\ -3 & -6 & 26 & 2 \end{bmatrix}$$
Solving for each of the variables, in the correct order yields:
*
*$u_{11} = 1, u_{12} = -2, u_{13} = -2, u_{14} = -3$
*$l_{21} = 3, u_{22} = -3, ...$
*$l_{31} = -1, l_{32} = 4, ...$
*$\ldots $
So, we arrive at:
$$A = \begin{bmatrix} 1 & -2 & -2 & -3 \\ 3 & -9 & 0 & -9 \\ -1 & 2 & 4 & 7 \\ -3 & -6 & 26 & 2 \end{bmatrix} = LU = \begin{bmatrix} 1 & 0 & 0 & 0\\ 3 & 1 & 0 & 0 \\ -1 & 0 & 1 & 0 \\ -3 & 4 & -2 & 1\end{bmatrix} \cdot \begin{bmatrix} 1 & -2 & -2 & -3 \\ 0 & -3 & 6 & 0 \\ 0 & 0 & 2 & 4 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$
We could have also used Crout's or Choleski's Method for the $LU$ approach. See: what are pivot numbers in LU decomposition? please explain me in an example
Please note that sometimes an LU decomposition is not possible, and sometimes, when it is, we have to resort to using permutation matrices and other approaches.
|
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|
What is that function? What is the set of the convergence in the reals of the series
$$f(x):= \sum\limits_{n=1}^{\infty} \frac 1 n \sin \left (n+\frac x n\right)?$$ Is the function $f(x)$ bounded?
Edit: more exact title.
|
For any fixed $x$; and any $N\geq 1$
\begin{align*}
\sum_{n=1}^{N} \frac{1}{n} \sin \left (n+\frac x n\right) &=
\sum_{n=1}^{N} \frac{1}{n} \sin n \cos \frac{x}{n} + \sum_{n=1}^{N} \frac{1}{n} \cos n \sin \frac{x}{n}
\end{align*}
Now,
*
*$\frac{\sin n}{n} \cos \frac{x}{n} = \frac{\sin n}{n} - \frac{\sin n}{2n^3}x^2 + o(\frac{1}{n^3})$; the first series converges iff $\sum \frac{\sin n}{n}$ does. It does.
*$\lvert \frac{1}{n} \cos n \sin \frac{x}{n} \rvert \leq \frac{\lvert x\rvert}{n^2}$ (as $\vert \sin t\rvert \leq \lvert t\rvert$): the second series converges as well.
|
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How to solve $x$ in $(x+1)^4+(x-1)^4=16$? I'm trying my hand on these types of expressions. How to solve $x$ in $(x+1)^4+(x-1)^4=16$? Please write any idea you have, and try to keep it simple. Thanks.
|
Let denote by $P(X)$ the following polynomial:
$$P(X)=(X+1)^4+(X-1)^4-16.\quad\quad(1)$$
We immediately see that $X=1$ and $X=-1$ are roots of $P(X)$. Therefore, $P(X)$ can be written as:
$$P(X)=(X-1)(X+1)Q(X), \quad\quad\;\;\;\;\;\;(2)$$
where $Q(X)$ is a polynomial with degree $2$. Hence, $Q(X)=aX^2+bX+c$ for some $a$, $b$, and $c$.
Now, if we develop $P(X)$ for both $(1)$ and $(2)$ we get:
First, with $(1)$:
$$P(X)=2X^4+12X^2-14.\quad\quad(3)$$
Second, with $(2)$:
$$P(X)=(X-1)(X+1)(aX^2+bX+c)=aX^4+bX^3+(c-a)X^2-bX-c.\quad(4)$$
Hence, $(3)=(4)$
$$2X^4+12X^2-14=aX^4+bX^3+(c-a)X^2-bX-c.$$
Then,
$$ \begin{array}{lr}
a=2\\
b=0\\
c=14
\end{array}$$
And finally,
$$P(X)=(X+1)(X-1)(2X^2+14)=2(X+1)(X-1)(X^2+7).$$
Now it is clear how to solve $P(X)=0$ (how to find the two other roots):
$$S_{\mathbb{R}}=\{-1, +1\}.$$
and
$$S_{\mathbb{C}}=\{-1, +1, +i\sqrt{7}, -i\sqrt{7}\}.$$
|
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|
Transformation Matrix $M_B^B$ of $P_3$ for $B = (1,x,x^2,x^3)$. Is that correct? I have the following task and just wanted to check weather this is (written) correct(ly).
Let $V$ be the vector space of all polynomials of grade $\le 3$ and $f: V \rightarrow V, p \rightarrow p'$ an $\mathbb{R}-$linear map. Calculate the transformation matrix using the basis $B=(1,x,x^2,x^3)$.
Well first I get the image of the basis-vectors:
$$f(1) = 0, \ f(x) = 1, \ f(x^2) = 2x, \ f(x^3) = 3x^2 $$
The next step is to express those using the canonical basis in $\mathbb{R}^3$, right? So what I did was:
$$f(1) = 0 = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} = \lambda_1 \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} + \lambda_2 \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix} + \lambda_3 \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix} + \lambda_4 \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} \Rightarrow \lambda_1 = \lambda_2 = \lambda_3 = \lambda_4 = 0$$
$$f(x) = 1 = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} = \lambda_1 \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} + \lambda_2 \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix} + \lambda_3 \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix} + \lambda_4 \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} \Rightarrow \lambda_2 = \lambda_3 = \lambda_4 = 0, \ \lambda_1 = 1 $$
$$f(x^2) = 2x = \begin{pmatrix} 0 \\ 2 \\ 0 \\ 0 \end{pmatrix} = \lambda_1 \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} + \lambda_2 \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix} + \lambda_3 \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix} + \lambda_4 \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} \Rightarrow \lambda_1 = \lambda_3 = \lambda_4 = 0, \ \lambda_2 = 2 $$
$$f(x^3) = 3x^2 = \begin{pmatrix} 0 \\ 0 \\ 3 \\ 0 \end{pmatrix} = \lambda_1 \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} + \lambda_2 \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix} + \lambda_3 \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix} + \lambda_4 \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} \Rightarrow \lambda_2 = \lambda_1 = \lambda_4 = 0, \ \lambda_3 = 3 $$
Which results into the following transformation matrix:
$$M_B^B = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0& 0 & 0 & 3 \\ 0 & 0 & 0 & 0 \end{pmatrix} $$
My problem is that I am not quite sure whether I've placed the resulting images on the right row. E.g. for $f(x^3) = 3x^2$:
$$ 3x^2 = \begin{pmatrix} 0 \\ 0 \\ 3 \\ 0 \end{pmatrix} \text{ or } 3x^2 = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 3 \end{pmatrix}$$
I am quite sure its the first one since one has to order the coefficients in regards to the basis one uses - but I thought it'd be better to check that.
Than you very much for your help.
FunkyPeanut
|
Let $V$ be a vector space and $T$ an operator on $V$. Since for the basis $(v_1,...,v_n)$ of $V$, $T$ maps each $v_k$ as a linear combination $Tv_k =a_{1,k}v_1 \ + \ ...\ + \ a_{n,k}v_n$ and is represented by the matrix
$$\left[\begin{matrix}
a_{1,1}&\cdots& &a_{1,n} \\
\vdots& \ddots & &\vdots\\
a_{n,1}&\cdots & & a_{n,n}
\end{matrix}\right]$$
We see that for the your example (extended to $n+1$ dimensions) $f$ maps each $x^k$ to $kx^{k-1}$, which is the linear combination $0 \cdot 1 \ +\ ... \ + \ kx^{k-1} + \ ... \ + \ 0 x^n$
So the matrix representation would be
$$\left[\begin{matrix}
0 & 1& \cdots& &0 \\
\vdots&&\ddots&&\vdots\\
&&&&n\\
0&&\cdots&&0
\end{matrix}\right]$$
Which is what you have.
|
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|
Show that $\int_0^1 \! \frac{1+x^2}{1+x^4} \, \operatorname d \!x=1+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{9}+\cdots$ As the title states, trying to solve $$\int_0^1 \! \frac{1+x^2}{1+x^4} \, \operatorname d \!x=1+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{9}+\cdots$$
|
Simply express the integrand as the series that converges for $|x| < 1$ and then interchange the order of summation and integration.
Hint: It is straightforward to show that
$$\frac{1 + x^2}{1 + x^4} = 1 + x^2 - x^4 - x^6 + x^8 + x^{10} - \cdots$$
because
$$\frac{1 + x^2}{1 + x^4} = \frac{1}{1 + x^4} + \frac{x^2}{1 + x^4}.$$
Hopefully you know what the geometric series are. Now integrate the series above term-by-term.
|
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|
How to find the coefficient of $x^8$ in $\prod\limits_{i=1}^{10}{\left(x-i\right)}$? How to find the coefficient of $x^8$ in $(x-1) (x-2) . . .(x-10)$.
Is there any general formula to solve this kind of problems?
|
The roots of the polynomial $(x - 1)(x-2)\cdots(x-10) = 0$ are $x = 1,2,\dots,10$.
Apply Vieta's Formulas to deduce that the coefficient of $x^8$ is the sum of all pairwise products of these roots (a.k.a the second elementary symmetric function of the roots):
$$(1\cdot2 + 1\cdot 3 + \dots + 1\cdot 10) + (2\cdot 3 + 2\cdot 4 + \dots + 2\cdot 10) + \dots + (9\cdot 10)\\
= 1320$$
Hence, the coefficient of $x^8$ is $1320$. The Wolf verifies.
|
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If $AD=BD$, $\angle ADC=3\angle CAB$, $AB=\sqrt{2}$, $BC=\sqrt{17}$, $CD=\sqrt{10}$. Find $AC$ In quadrilateral $ABCD$, we have
$$AD=BD,\angle ADC=3\angle CAB,AB=\sqrt{2},$$ $$BC=\sqrt{17},CD=\sqrt{10}$$
Find the $AC=?$
My idea: let $$\angle CAB=x.\angle ADC=3x,\angle ADB=y,$$
then we have
$$\angle CAD=90-\dfrac{y}{2}-x,\angle ACD=\dfrac{y}{2}+90-2x$$then we have
$$\dfrac{\dfrac{\sqrt{2}}{2}}{\sin{\dfrac{y}{2}}}=BD$$
and
$$\dfrac{BD}{\sin{BCD}}=\dfrac{DC}{\sin{DBC}}=\dfrac{BC}{\sin{BDC}}$$
then
$$\dfrac{\sqrt{2}}{\sin{\dfrac{y}{2}}\sin{BCA}}=\dfrac{\sqrt{10}}{\sin{DBC}}=\dfrac{\sqrt{17}}{\sin{(3x-y)}}$$
and in $\Delta ABC$,we have
$$\dfrac{\sqrt{17}}{\sin{x}}=\dfrac{\sqrt{2}}{\sin{ACB}}=\dfrac{AC}{\sin{ABC}}$$
in $\Delta ADC$,we have
$$\dfrac{AC}{\sin{3x}}=\dfrac{\sqrt{10}}{\sin{(90-y/2-x)}}=\dfrac{AD}{\sin{(90+y/2-2x)}}$$then I fell very ugly,and I can't.maybe have other idea. Thank you
|
You only have to solve the 4-equation system below:
\begin{cases}
\overline{AC}^2+\overline{AB}^2-2\overline{AC}\cdot \overline{AB}\cos(x)=\overline{BC}^2 \\
\overline{AD}^2+\overline{CD}^2-2\overline{AD}\cdot \overline{CD}\cos(3x)=\overline{AC}^2 \\
\overline{BD}^2+\overline{CD}^2-2\overline{BD}\cdot \overline{CD}\cos(3x-y)=\overline{BC}^2 \\
\frac{\overline{AD}}{\sin\left(\frac{\pi-y}{2}\right)} = \frac{\overline{AB}}{\sin(y)} \\
\end{cases}
The first 3 equations can be obtained by applying Carnot Theorem in the triangles $\Delta ABC$, $\Delta ACD$ and $\Delta BDC$, while the last is easy to find by applying the Law of Sines in $\Delta ABD$. However there are only 4 variables ($\overline{AC}$,$\overline{AD}$, $x$ and $y$) beacuse $\overline{AD} = \overline{BD}$, so the system becomes easier to solve:
\begin{cases}
\overline{AC}^2+2-2\sqrt{2}\cdot \overline{AC}\cos(x)=17 \\
\overline{AD}^2+10-2\overline{AD}\cdot \sqrt{10}\cos(3x)=\overline{AC}^2 \\
\overline{AD}^2+10-2\overline{AD}\cdot \sqrt{10}\cos(3x-y)=17 \\
\frac{\overline{AD}}{\cos\left(\frac{y}{2}\right)} = \frac{\sqrt{2}}{\sin(y)} \\
\end{cases}
|
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|
Integrating a function with substitution Totally forgot how to integrate.
$$ \int \frac{1}{x^2 \sqrt{x^2+4}}dx$$
Just need a tip, for this what would I use to substitute?
|
Put $x = 2 \tan t $, then $dx = 2 \sec^2 t dt $. and $\sqrt{x^2 +4} = \sqrt{ 4 \tan^2 t + 4 } = 2 \sec t$ hence,
$$ \int \frac{dx}{x^2 \sqrt{x^2+4}} = \int \frac{2 \sec^2 t dt}{4 \tan^2 t 2 \sec t} = \frac{1}{4} \int \frac{ \sec t dt }{\tan^2 t} = \frac{1}{4} \int \frac{\frac{1}{\cos t}}{\frac{\sin^2t}{\cos^2 t}} = \frac{1}{4} \int \frac{\cos t dt}{\sin^2 t} = \frac{1}{4} \int \frac{d ( \sin t)}{\sin^2 t} = \frac{-1}{4}\frac{1}{\sin t} + C$$
|
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If $\sigma(p^m)=2^n$ for prime $p$,then $m=1$ and $n$ is prime Exercise from Beginning Number Theory by Neville Robbins:
Let $\sigma(a)$ denote the sum of divisors of $a$.Then we have to prove that if $\sigma(p^m)=2^n$ for some prime $p$,then $m=1$ and $n$ is a prime.
The result does not hold for $p=2$.Now,$$\dfrac{p^{m+1}-1}{p-1}=2^n$$
$$\implies p^{m+1}-1=2^n(p-1)$$
Then,$$p^{m+1}\equiv 1\pmod {2^n}$$ and $$p^{m+1}\equiv 1\pmod {p-1}$$ I can't go anywhere from here.A hint will be appreciated.
|
Excluding $m = 0$ - when $\sigma(p^0) = 2^0$ - we see that $p$ must be an odd prime, since $\sigma(2^m) \equiv 1 \pmod{2}$. So $\sigma(p^m) = 1 + p + \dotsc + p^m \equiv m+1 \pmod{2}$ forces $m$ to be odd. Then let $k = \frac{m+1}{2}$ in
$$\sigma(p^m) = \frac{p^{2k}-1}{p-1} = (p^k+1)\cdot\frac{p^k-1}{p-1}.$$
If $\sigma(p^m) = 2^n$, then $p^k+1 = 2^r$ and $\dfrac{p^k-1}{p-1} = 2^s$, where $r+s = n$ and $s < r$. Since $\gcd(p^k+1,p^k-1) = \gcd(p^k+1,2) = 2$, it follows that $s \leqslant 1$, and $\frac{p^k-1}{p-1} = 1 + p + \dotsc + p^{k-1} \leqslant 2$ then yields $k = 1$ and hence $m = 1$, so $p = 2^n-1$.
If $2^n-1$ is prime, then $n$ itself must be prime: $2^t-1$ is a nontrivial divisor of $2^{s\cdot t}-1$ for all $s,t > 1$, so $2^n-1$ is composite when $n$ is composite (and often also when $n$ is prime).
|
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Find sequence function and general rule the function $$a_{n+2}=3a_{n+1}-2a_n+2$$
is given, and $$a_0=a_1=1, (a_n)_{n\ge0}$$
multiplying everything by $$/\sum_{n=0}^\infty x^{n+2}$$
also adding $$\sum_{n=0}^\infty (a_{n+2}x^{n+2}+a_1x+a_0)-a_1x-a_0=\sum_{n=0}^\infty (3a_{n+1}x^{n+2}+a_0)-a_0-\sum_{n=0}^\infty 2a_nx^{n+2}+\sum_{n=0}^\infty 2x^{n+2}$$
we get $$R(X)-2x-1=3x(R(X)-1)-2x^2R(X)+\frac{2x^2}{1-x}$$
$$R(X)(1-3x+2x^2)+x-1=\frac{2x^2}{1-x}$$
$$R(X)(1-3x+2x^2)=\frac{3x^2-2x+1}{1-x}$$
$$R(X)=\frac{3x^2-2x+1}{(1-x)^2(1-2x)}=\frac{A}{(1-x)^2}+\frac{B}{1-2x}$$
so $$3x^2-2x+1=A(1-2x)+B(1-x)^2$$
$$3x^2-2x+1=x^2(B)+x(-2A-2B)+A+B$$
B=3
-2A-2B=-2
A+B=1$$\quad\Longrightarrow\quad A=-2, B=3$$
$$R(X)=\frac{-2}{(1-x)^2}+\frac{3}{1-2x}=-2\sum_{n=0}^\infty nx^{n-1}+3\sum_{n=0}^\infty (2x)^2=$$
how to proceed?
|
It is better to work generating functions by multiplying by $z^n$ and sum over $n \ge 0$, which here gives:
$$
\frac{A(z) - a_0 - a_1 z}{z^2}
= 3 \frac{A(z) - a_0}{z}
- 2 A(z)
+ 2 \frac{1}{1 - z}
$$
This results in:
$$
A(z) = \frac{1 - 3 z + 4 z^2}{1 - 4 z + 5 z^2 - 2 z^3}
= \frac{2}{1 - 2 z} + \frac{1}{1 - z} - \frac{2}{(1 - z)^2}
$$
Read off the coefficients:
$$
a_n = 2 \cdot 2^n + 1 - 2 \binom{-2}{n}
= 2^{n + 1} - 2 n - 1
$$
|
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Finding $a$ and $b$ from $a^3+b^3$ and $a^2+b^2$ Question 1
Two numbers are such that the sum of their cubes is 14 and the sum of their squares is 6. Find the sum of the two numbers.
I did
$a^2+b^2=6$ and $a^3+b^3=14$ Find $a$ and $b$, two numbers. but got lost when trying to algebraicly solve it.
Thank you, Any help is appreciated
|
The problem doesn't specify any restriction on the numbers $a$ and $b$, so we're going to assume they are complex. Let $S=a+b$ denote the sum. Then
$$S^2=(a+b)^2=a^2+2ab+b^2=6+2ab$$
and
$$S^3=(a+b)^3=a^3+3a^2b+3ab^2+b^3=14+3abS$$
Rewriting the first of these as $2ab=S^2-6$ and multiplying both sides of the second by $2$ gives
$$2S^3=28+3(2ab)S=28+3(S^2-6)S=28+3S^3-18S$$
or
$$S^3-18S+28=(S-2)(S^2+2S-14)=0$$
The possible values of $S$ are thus $2$, $-1+\sqrt{15}$, and $-1-\sqrt{15}$.
If you care to chase down the actual values of $a$ and $b$, the first two of these give real values while the third gives a pair of complex conjugates. In particular, $S=2$ leads to $1\pm\sqrt2$ for $a$ and $b$.
|
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|
Find value range of $2^x+2^y$ Assume $x,y \in \Bbb{R}$ satisfy $$4^x+4^y = 2^{x+1} + 2^{y+1}$$, Find the value range of $$2^x+2^y$$
I know $x=y=1$ is a solution of $4^x+4^y = 2^{x+1} + 2^{y+1}$ , but I can't go further more. I can only find one solution pair of $4^x+4^y = 2^{x+1} + 2^{y+1}$. It seems very far from solve this question...
|
Let us denote $2^{x}=a,2^{y}=b,$where $a,b\in
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
^{+}$. Then we have $a^{2}+b^{2}=2(a+b)$ which is equivalent to $%
(a-1)^{2}+(b-1)^{2}=2.$So we take the part of the circle with center $M(1,1)
$ and radius $r=\sqrt{2}$ in the first district of the plane as the domain of the
function $f(a,b)=a+b$. We are looking for maxium and minium values of $f$ in
the given domain. Any point $(a,b)$ in the domain is $\sqrt{a^{2}+b^{2}}$
units far from $(0,0)$ which is also in the given circle (though not in
domain). We also know that two points on a circle are at most diameter
length far from each other. So $\sqrt{a^{2}+b^{2}}=2\sqrt{2}$ at most. Thus $%
a^{2}+b^{2}=8.$ Then we have $2(a+b)=8$ and finally $a+b=4$ at maximum. We
look at the edges (because they are closer to (0,0) ) of the domain for
mininum value which is at $(2,0)$ or $(0,2)$. Then $a+b$ converges to $2$ at
minimum. We have the value range $(2,4]$.
|
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|
Showing $\sin{\frac{\pi}{13}} \cdot \sin{\frac{2\pi}{13}} \cdot \sin{\frac{3\pi}{13}} \cdots \sin{\frac{6\pi}{13}} = \frac{\sqrt{13}}{64}$ I would like to show that
$$ \sin{\frac{\pi}{13}} \cdot \sin{\frac{2\pi}{13}} \cdot \sin{\frac{3\pi}{13}} \cdots \sin{\frac{6\pi}{13}} = \frac{\sqrt{13}}{64} $$
I've been working on this for a few days. I've used product-to-sum formulas, writing the sines in their exponential form, etc. When I used the product-to-sum formulas, I'd get a factor of $1/64$, I obtained the same with writing the sines in their exponential form. I'd always get $1/64$ somehow, but never the $\sqrt{13}$.
I've come across this: http://mathworld.wolfram.com/TrigonometryAnglesPi13.html, (look at the 10th equation). It says that this comes from one of Newton's formulas and links to something named "Newton-Girard formulas", which I cannot understand. :(
Thanks in advance.
|
Use this formula (found here, and mentioned recently on MSE here):
$$\prod _{k=1}^{n-1}\,\sin \left({\frac {k\pi }{n}} \right)=\frac{n}{2^{n-1}} .$$
Let $n=13$, which gives $$\left(\sin{\frac{\pi}{13}} \cdot \sin{\frac{2\pi}{13}} \cdot \sin{\frac{3\pi}{13}} \cdots \sin{\frac{6\pi}{13}}\right)\left(\sin{\frac{7\pi}{13}} \cdot \sin{\frac{8\pi}{13}} \cdot \sin{\frac{9\pi}{13}} \cdots \sin{\frac{12\pi}{13}}\right) = \frac{13}{{2^{12}}}.$$
Use the fact that $\sin\dfrac{k\pi}{13}=\sin\dfrac{(13-k)\pi}{13}$ to see that this is the same as
$$\left(\sin{\frac{\pi}{13}} \cdot \sin{\frac{2\pi}{13}} \cdot \sin{\frac{3\pi}{13}} \cdots \sin{\frac{6\pi}{13}}\right)^2 = \frac{13}{2^{12}}$$
and take the square root of both sides to get your answer.
|
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|
Probability of number formed from dice rolls being multiple of 8
A fair 6-sided die is tossed 8 times. The sequence of 8 results is recorded to form an 8-digit number. For example if the tosses give {3, 5, 4, 2, 1, 1, 6, 5}, the resultant number is $35421165$. What is the probability that the number formed is a multiple of 8.
I solved this by listing all possibilities for the last 3 digits that give multiples of 8, and found this to be $\frac{1}{8}$.
The solution key agrees with my answer, but also says that "There are quicker ways to solve the problem using a more advanced knowledge of number theory"
What would be a faster way to solve this using number theory?
|
Here's a generalization:
The sample space for a six-sided die is $S = \{1,2,3,4,5,6\}$.
Theorem: There are exactly $\color{blue}{3}^n$ n-tuples $(x_{n-1}, x_{n-2}, \cdots, x_1, x_0)$ that satisfy $f(n) = 10^{n-1}x_{n-1} + 10^{n-2}x_{n-2} + \cdots + 10x_1 + x_0 \equiv 0 \mod{2^n}$ for $x_i \in S$
Proof by induction:
Base case $n=1$: We have $f(1) = x_0 \equiv 0 \mod{2}$ which is clearly satisfied by exactly $\color{blue}{3}$ 1-tuples $(2), (4)$ and $(6)$.
Inductive step: Assume true for $n$. Now for $n+1$ we have $\begin{array}{l}f(n+1) &= 10^nx_n &+ & 10^{n-1}x_{n-1} + \cdots + 10x_1 + x_0 \equiv 0 \mod{2^{n+1}} &\text{(i)}\\&\Rightarrow 0 &+ & 10^{n-1}x_{n-1} + \cdots + 10x_1 + x_0 \equiv 0 \mod{2^{n}}\\&\Rightarrow & & 10^{n-1}x_{n-1} + \cdots + 10x_1 + x_0 \equiv 0 \mod{2^{n}}& \text{(ii)}\end{array}$
But $\text{(ii)}$ has exactly $\color{blue}{3}^n$ n-tuple solutions from our inductive hypothesis.
For each such n-tuple $(x_{n-1}, x_{n-2}, \cdots, x_1, x_0)$ , there are $\color{blue}{3}$ unique $x_n \in S$ that satisfy $\text{(i)}$ per Lemma below. So, the number of (n+1)-tuple solutions $(x_n, x_{n-1}, \cdots, x_1, x_0)$ is $\color{blue}{3} \cdot \color{blue}{3}^n = \color{blue}{3}^{n+1}$ and this completes the proof by induction.
Answer:
The probability of forming an n-digit number $x_{n-1}x_{n-2}\cdots x_1x_0 = f(n)$ divisible by $2^n$ is thus $\frac{\text{count of n-digit numbers } \equiv 0 \mod{2^n}}{\text{total count of n-digit numbers}} = \frac{3^n}{6^n} = \frac{1}{2^n}$. In your case, $n=3$, giving the probability of $\frac{1}{8}$.
Lemma: $10^n x + m \equiv 0 \mod{2^{n+1}} $ has exactly $\color{blue}{3}$ unique solutions $x\in S$ given that $m \equiv 0 \mod{2^n}$.
Proof: We have $m \equiv 0 \mod{2^n} \implies m = 2^n\ell$ where $\ell \in \mathbb{Z}$.
$\begin{array}{cl} \therefore &10^n x + m &\equiv 0\mod{2^{n+1}}&\\ \iff & 10^n x + 2^n\ell &= 2^{n+1} q & (q \in \mathbb{Z})\\ \iff & 5^nx + \ell &= 2q &(\text{divide by } 2^n)\\\iff & x + \ell &\equiv 0 \mod{2} & (5^n \equiv 1 \mod{2})\end{array}$
And this has exactly $\color{blue}{3}$ unique solutions $x\in S$ since,
$\mod{2}:\begin{cases}\ell \equiv 0 \implies x + 0 \equiv 0 \implies x \equiv 0 &\implies x \in \{2, 4, 6\} \\ \ell \equiv 1 \implies x + 1 \equiv 0 \implies x \equiv 1 &\implies x \in \{1, 3, 5\}\end{cases}$
|
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|
Solve $z^2 - iz = |z - i|$ I have the equation:
$z^2 - iz = |z - i|$
The solutions are $i$, $-\sqrt3/2 + i/2$, $\sqrt3/2 + i/2$
Can someone please walk me through or give me a hint...
|
Here is a solution which is little different, algebraically, though it is "spiritually similar" to the work of Mario Gallegos:
given that
$z^2 - iz = \vert z - i \vert, \tag{1}$
we first take the modulus of each side, noting that
$\vert z^2 - iz \vert = \vert z(z -i) \vert = \vert z \vert \vert z -i \vert; \tag{2}$
we thus obtain
$\vert z \vert \vert z -i\vert = \vert \vert z -i \vert \vert = \vert z -i\vert. \tag{3}$
From (3) we see that $\vert z -i \vert \ne 0$ implies $\vert z \vert = 1$; $z =i$ is clearly a solution as may be seen directly from (1); so we look for solutions with $\vert z \vert = 1$, $z \ne i$. $\vert z \vert = 1$ implies $z = e^{i\theta}$ with $0 \le \theta < 2\pi$; then
$z^2 - iz = e^{2i\theta} - ie^{i\theta} = \cos 2 \theta +i\sin 2\theta -i\cos \theta + \sin \theta. \tag{4}$
We see from (1) that $z^2 -iz$ must be real; this means the imaginary part of (4) must vanish, or
$\sin 2\theta - \cos \theta = 0; \tag{5}$
since $\sin 2 \theta = 2(\sin \theta)(\cos \theta)$ we have
$2(\sin \theta)(\cos \theta) = \cos \theta: \tag{6}$
if $\cos \theta = 0$, then $\theta = \pi / 2, 3\pi/2 \;$ or $z = \pm i$; we have seen that $z = i$ is a solution and $z = -i$ may be ruled out by direct substitution into (1); thus we may assume $\cos \theta \ne 0$, in which case we have $\sin \theta = 1/2$, or $\Im z = 1/2$. We now use the identity $\sin^2 \theta + \cos^2 \theta = 1$ to arrive at $\Re z = \cos \theta = \pm \sqrt 3 / 2$, so that in addition to $z =i$, the other two solutions are
$z = \pm \dfrac{\sqrt 3}{2} + \dfrac{i}{2}; \tag{7}$
all three solution may be verified by direct substitution into (1).
Hope this helps. Cheerio,
and.as always,
Fiat Lux!!!
|
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|
number of solutions to $x_1 + x_2 + x_3 + x_4 + x_5 = 31$ via generating function? I will be very happy to understand how to solve this problem with generating function:
How many solutions are there to the equation
$$x_1 + x_2 + x_3 + x_4 + x_5 = 31$$
where $x_i$ is a nonnegative integer,
and $x_2$ is even and $x_3$ is odd?
thanks!
|
Non-negative $x_1$ corresponds to $\frac{z}{1 - z}$, $x_2$ even gives rise to $\frac{1}{1 - z^2}$, odd $x_3$ means $\frac{z}{1 - z^2}$, non-restriced $x_4$, $x_5$ are $\frac{1}{1 - z}$ each:
\begin{align}
[z^{31}] &\frac{z}{1 - z}
\cdot \frac{1}{1 - z^2}
\cdot \frac{z}{1 - z^2}
\cdot \frac{1}{(1 - z)^2} \\
&= [z^{31}] \frac{z^2}{(1 - z)^3 (1 - z^2)^2} \\
&= [z^{29}] \left(
\frac{5}{64 (1 + z)}
+ \frac{1}{32 (1 + z)^2}
+ \frac{5}{64 (1 - z)}
+ \frac{1}{8 (1 - z)^2}
+ \frac{3}{16 (1 - z)^3}
+ \frac{1}{4 (1 - z)^4}
+ \frac{1}{4 (1 - z)^5}
\right)
\end{align}
Use the fact that $\binom{-n}{k} = (-1)^k \binom{k + n - 1}{n - 1}$:
\begin{align}
[z^{31}] &\frac{z^2}{1 - z}
\cdot \frac{1}{1 - z^2}
\cdot \frac{z}{1 - z^2}
\cdot \frac{1}{(1 - z)^2} \\
&= \frac{5}{64} \binom{-1}{29}
+ \frac{1}{32} \binom{-2}{29}
+ \frac{5}{64} \binom{-1}{29} (-1)^{29}
+ \frac{1}{8} \binom{-2}{29} (-1)^{29}
+ \frac{3}{16} \binom{-3}{29} (-1)^{29}
+ \frac{1}{4} \binom{-4}{29} (-1)^{29}
+ \frac{1}{4} \binom{-5}{29} (-1)^{29} \\
&= 11560
\end{align}
|
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|
One step Gauss Seidel method Apply one step of the Gauss Seidel method to $A\textbf{x} = b$ with
A = $\begin{bmatrix}
4 & 2 & 1 \\
1 & 4 & 1 \\
1 & 2 & 4
\end{bmatrix}$, b = $\begin{bmatrix}
4\\
5\\
8
\end{bmatrix}$ $x_{0}$ = $\begin{bmatrix}
64\\
64\\
-128
\end{bmatrix}$
Does the method converge to the solution?
To answer the question, this is what I did...
$$x_{1} = \frac{4 - 2x_{2} - x_{3}}{4} => x_{1} = \frac{4-2(64)+128}{4} = 1$$
$$x_{2} = \frac{5-x_{1}-x_{3}}{4} = \frac{5-1+128}{4} = 33$$
$$x_{3} = \frac{8-x_{1}-2x_{2}}{4} = \frac{8-1-66}{4} = -59/4$$
Did I do it right? If not I am confused. Can someone help me do this correct? Thanks.
|
Yes, your first iteration is correct. Here are some iterates and details.
Iteration 1:
$$\left(
\begin{array}{cccccccccc}
1. \\ 33. \\ -14.75
\end{array}
\right)$$
Iteration 2:
$$\left(
\begin{array}{cccccccccc}
-11.8125 \\ 7.89063 \\ 1.00781
\end{array}
\right)$$
Iteration 3:
$$\left(
\begin{array}{cccccccccc}
-3.19727 \\ 1.79736 \\ 1.90063
\end{array}
\right)$$
... Iteration 10:
$$\left(
\begin{array}{cccccccccc}
0.208341 \\ 0.812503 \\ 1.54166
\end{array}
\right)$$
You can compare this against the actual result of:
$$x = \left(
\begin{array}{c}
\frac{5}{24} \\
\frac{13}{16} \\
\frac{37}{24} \\
\end{array}
\right) = \left(
\begin{array}{c}
0.208333 \\
0.8125 \\
1.54167 \\
\end{array}
\right)$$
|
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|
Solving $L= \frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}$ priveded $a+b+c=0$
Let $a,b,c$ be such that $a+b+c=0$ and suppose that
$$L= \frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}.$$
Find the value of $L$.
I can only see the symmetry of these function but cannot solve it.
|
Since $a+b+c|(a^3+b^3+c^3-3abc)$, $a+b+c = 0 \implies a^3+b^3+c^3 = 3abc$.
Then Wolfram Alpha says that this gives the sum as $1$.
|
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|
Find The Minimum Value of the quantity Find the minimum value of the quantity $$\frac{(a^2+3a+1)(b^2+3b+1)(c^2+3c+1)}{abc}$$,where $$a,b,c>0$$ and $$ a,b,c\in R $$are positive real numbers.
|
Hint: $$\frac{(a^2+3a+1)(b^2+3b+1)(c^2+3c+1)}{abc} = \left(3+a+\frac1a \right)\left(3+b+\frac1b \right)\left(3+c+\frac1c \right)$$
Now can you show that $x + \dfrac1x$ has a minimum of $2$ for positive real $x$?
|
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|
Number theory - rational number Are there any $x, y$ that fit in below
$\sqrt{4y^2-3x^2}$ such that an rational number is yielded.
Appreciate if explanation is given.
|
We can find infinitely many solutions by considering the Pell equation $y^2-3s^2=1$.
We look for perfect squares $4y^2-3x^2$, where $x$ is even, say $2s$. Then we want $4y^2-12s^2$ to be a perfect square, say $(2t)^2$. So we want to solve the Diophantine equation $y^2-3s^2=t^2$.
There are already infinitely many solutions with $t=1$. For $(2+\sqrt{3})(2-\sqrt{3})=1$. Thus $(2+\sqrt{3})^n(2-\sqrt{3})^n=1$.
But $(2+\sqrt{3})^n=y_n+s_n\sqrt{3}$ for integers $y_n$, $s_n$. It follows that $y_n^2-3s_n^2=1$.
|
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|
Is the modulus of i^n 1 for all n? As the tile says, is $\left | i^{n} \right | = 1$ for all real values of n?
|
We note that $|i| = 1$: $|i| = |0+1i| = \sqrt{0^2+1^2} = \sqrt{1} = 1$.
Suppose that $|i^n| = 1$. Then, $|i^{n+1}| = |i^ni| = |i^n||i| = 1\cdot 1 = 1$. By induction, $|i^n| = 1$ for all positive integers $n$.
Another method:
In general, $|z|^2 = z\overline{z}$.
$$|i^n|^2 = (i^n)\overline{(i^n)} = \underbrace{i\ \overline{i} \cdots i\ \overline{i}}_{n \textrm{ times}} = 1\cdot 1 \cdots 1 = 1.$$
A third method:
$$z \cong \begin{pmatrix} a & -b \\ b & a\end{pmatrix}$$ and $$|z|^2 = \det \begin{pmatrix} a & -b \\ b & a\end{pmatrix}.$$
So $$|i^n|^2 = \det \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}^n = \left[\det \begin{pmatrix} 0 & -1 \\ 1 & 0\end{pmatrix}\right]^n = 1.$$
|
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|
How to find $\int \frac {dx}{(x-1)^2\sqrt{x^2+6x}}$? find the integral of $f(x)=\frac1{(x-1)^2\sqrt{x^2+6x}}$
my attempt =
$(x-1)=a$, $a=x+1$ so the integral'd be
$\int \frac {dx}{(x-1)^2\sqrt{x^2+6x}}=\int\frac{da}{a^2\sqrt{a^2+8a+7}} $
lets say $\sqrt{a^2+8a+7}=(a+1)t$
so $a=\frac{7-t}{t-1}$ and $da=\frac{-6dt}{(t-1)^2}$
$\int\frac{da}{a^2\sqrt{a^2+8a+7}}=\int\frac{\frac{-6dt}{(t-1)^2}}{(\frac{7-t}{t-1})^2\frac{6t}{t-1}}=-\int\frac{(t-1)dt}{(7-t)^2t}$
$\frac{(t-1)}{(7-t)^2t}=\frac{A}{t}+\frac{B}{7-t}+\frac{C}{(7-t)^2}$
then $A=B=\frac{-1}{7}$ and $C=6$
$-\int\frac{(t-1)dt}{(7-t)^2t}=\frac{1}{7}\int\frac{dt}{t}+\frac{1}{7}\int\frac{dt}{7-t}-6\int\frac{dt}{(7-t)^2}=\frac{ln|t|}{7}-\frac{ln|7-t|}{7}+6\frac{1}{7-t}$
finally we substitute $t=\frac{7+a}{a+1}$ and a=x+1
is my solution attempt correct? if it is, is there another simpler way to solve?
edit: a should equal to x-1 and x =a+1
|
Put $$\frac{1}{x-1}=t$$ $\implies$ $$ \frac{dx}{(x-1)^2}=-dt$$ So
$$-I=\int \frac{t \,dt}{\sqrt{7t^2+8t+1}}$$ Put $$ 7t^2+8t+1=z^2$$ $\implies$
$$(7t+4)dt=zdz$$ So
$$ -7I=\int \frac{7t+4-4 \,dt}{\sqrt{7t^2+8t+1}}=\int \frac{7t+4 \,dt}{\sqrt{7t^2+8t+1}}-\int \frac{4 \,dt}{\sqrt{7t^2+8t+1}}$$$$$$ So
$$-7I=z-\int \frac{4 \,dt}{\sqrt{7t^2+8t+1}}=z-\frac{4}{\sqrt{7}}\int \frac{dt}{\sqrt{(t+\frac{4}{7})^2-(\frac{3}{7})^2}}$$ So
$$-7I=\sqrt{7t^2+8t+1}-\frac{4}{\sqrt{7}}Ln\left|(t+\frac{4}{7})+\sqrt{(t+\frac{4}{7})^2-(\frac{3}{7})^2}\right|+Const$$ Finally Replace $t=\frac{1}{x-1}$
|
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|
put numbered balls in four similar boxes of a specific capacity... With how many ways can we put $12$ numbered balls in $4$ similar(not numbered) boxes of capacity $3$ each one?
Is it maybe $3^4$ ?
|
You can choose $3$ out of $12$ balls for the first box, $3$ out of $9$ for the second...and so on...But since the boxes are similar you should divide the result by all the possible combinations of the $4$ boxes ($4!$), so the number requested is:
$$\frac{\binom{12}{3}\cdot\binom{9}{3}\cdot\binom{6}{3}\cdot\binom{3}{3}}{4!}=\frac{\frac{12!}{9!\cdot 3!}\cdot\frac{9!}{6!\cdot3!}\cdot\frac{6!}{3!\cdot3!}\cdot\frac{3!}{3!\cdot0!}}{4!}=\frac{12!}{3!^4\cdot4!}=15400$$
If the boxes weren't similar the solution would be: $$\binom{12}{3}\cdot\binom{9}{3}\cdot\binom{6}{3}\cdot\binom{3}{3}=\frac{12!}{9!\cdot 3!}\cdot\frac{9!}{6!\cdot3!}\cdot\frac{6!}{3!\cdot3!}\cdot\frac{3!}{3!\cdot0!}=\frac{12!}{3!}=369600$$
|
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|
Is the following Markov Chain a martingale? Say I have a finite, ergodic Markov chain with states ${0,1,2,3}$ and with the following transition matrix:
$$\begin{bmatrix}
\frac{7}{10} & \frac{3}{10} & 0 &0\\
\frac{1}{10} & \frac{6}{10} & \frac{3}{10} &0\\
0 & \frac{3}{10} & \frac{6}{10} & \frac{1}{10} \\
0& 0& \frac{3}{10} & \frac{7}{10}\\
\end{bmatrix}$$
If I define $M_n$ to be the value of the state (so either $0,1,2$ or $3$) divided by $5$ at time $n$, is $M_n$ a martingale? I am inclined to believe it is not because the expected value of $M_{n+1}$ given the filtration at $n$ is not equal to $M_n$ (at least by my calculations).
|
For every martingale $(M_n)$, $\mathbb{E}(M_{n+1} | M_n ) = M_n$. But here, $\mathbb{E}(M_{n+1}|M_n=0) = \frac{3}{50}$. Therefore, $(M_n)$ is not a martingale.
|
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|
12 balls in a box to distribute by 2 bags. There is a box with 12 balls inside, 4 green, 5 blue and 3 black.
We want to take out 4 balls for each one of 2 bags (bag A and bag B), and we take note of the number of balls of each color that the bags have.
How many ways are there to distribute the balls?
|
We could start by figuring out the number of ways that $4$ balls do NOT go in either bag (that would also determine the $8$ that do).
The possibilities are:
*
*4 green $\displaystyle \Rightarrow \dbinom{4}{4} = 1$
*3 green, 1 blue $\displaystyle \Rightarrow \dbinom{4}{3} \dbinom{5}{1} = 4\cdot5 = 20$
*3 green, 1 black $\displaystyle \Rightarrow \dbinom{4}{3} \dbinom{3}{1} = 4\cdot3 = 12$
*2 green, 2 blue $\displaystyle \Rightarrow \dbinom{4}{2} \dbinom{5}{2} = 6\cdot10 = 60$
*2 green, 1 blue, 1 black $\displaystyle \Rightarrow \dbinom{4}{2} \dbinom{5}{1} \dbinom{3}{1} = 6\cdot5\cdot3 = 90$
*2 green, 2 black $\displaystyle \Rightarrow \dbinom{4}{2} \dbinom{3}{2} = 6\cdot3 = 18$
*1 green, 3 blue $\displaystyle \Rightarrow \dbinom{4}{3} \dbinom{5}{3} = 4\cdot10 = 40$
*1 green, 2 blue, 1 black $\displaystyle \Rightarrow \dbinom{4}{3} \dbinom{5}{2} \dbinom{3}{1} = 4\cdot10\cdot3 = 120$
*1 green, 1 blue, 2 black $\displaystyle \Rightarrow \dbinom{4}{3} \dbinom{5}{1} \dbinom{3}{2} = 4\cdot5\cdot3 = 60$
*1 green, 3 black $\displaystyle \Rightarrow \dbinom{4}{3} \dbinom{3}{3} = 4\cdot1 = 4$
There are also different ways to place $4$ balls into bag $A$ (which would in turn decide which $4$ go into bag $B$). I leave that for you to do.
|
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|
Why the linear numerator for fractions with irreducible denominators and constant numerators for reducible denominators? For example:
$\Large{\frac{2x^3+5x+1}{(x^2+4)(x^2+x+2)}}$ breaks down to $\Large{\frac{ax+b}{x^2+4}+\frac{cx+d}{x^2+x+2}}$
I have been told that since the denominators are irreducible, the numerators will be either linear or constant.
Now my question is for something like $\Large{\frac{2x^3+5x+1}{x^2-4}}$ you would make it equal $\Large{\frac{A}{x+2}+\frac{B}{x-2}}$, why do we assume that the numerators are constant? Why couldn't they be linear??
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Because remainder of division is always smaller then the degree of the polynomial so we assume it's exactly 1 smaller,and find if the other terms are zero
$$\frac{2x^3+5x+1}{(x+2)(x-2)}=\frac{ax+b}{x+2}+\frac{cx+d}{x-2}=\frac{ax+b-a(x+2)}{x+2}+a+\frac{cx+d-c(x-2)}{x-2}+c=\frac{ax+b-ax-2a}{x+2}+\frac{cx+d-cx+2c}{x-2}+a+c=\frac{b-2a}{x+2}+\frac{2c+d}{x-2}+a+c$$
|
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Why does $\dfrac{2}{\sqrt{-4x^2-4x}}$ simplify into $\dfrac{1}{\sqrt{-x^2-x}}$?
Why does $\dfrac{2}{\sqrt{-4x^2-4x}}$ simplify into $\dfrac{1}{\sqrt{-x^2-x}}$?
What's going on here? How is it being simplified?
|
$$\dfrac{2}{\sqrt{-4x^2-4x}}=\dfrac{2}{\sqrt{4(-x^2-x)}}=\dfrac{2}{\sqrt{4}\sqrt{(-x^2-x)}}\\
=\dfrac{2}{2\sqrt{(-x^2-x)}}=\dfrac{1}{\sqrt{(-x^2-x)}}$$
|
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Finding an explicit formula for $a_n$ defined recursively by The sequence $a_n$ is defined recursively by $a_0=0$, $a_n=4a_{n-1}+1$. I must use generating functions to solve this. $n\geq1$.
I have found a pattern:
$$\sum_{n=1}^\infty(4a_{n-1}+1)x^n = x+5x^2+21x^3+85x^4+341x^5+\ldots$$
If we subtract 1 from each term, respectively (ignoring the first term) we would have:
$$x+4x^2+20x^3+84x^4+340x^5+\ldots$$
As you can see, the pattern is the second term increases by 4, the third term increases by 4^2, the fourth term increases by 4^3, and the fifth term increases by 4^4. I wanted to use the formula for the geometric series, but the common ratio isn't 4, but $4^n$. Is there any way I can use this pattern to help me find the explicit formula, using generating functions?
Using $$\sum_{n=0}^\infty 4^nx^n = \frac{1}{1-4x}$$ Isn't helping me too much. But I have a feeling I can manipulate this generating function to solve this.
However, this might seem helpful:
$$\sum_{n=0}^\infty4^nx^{n+1} = 4^0x+4^1x^2+4^2x^3+4^3x^4+\ldots = x+4x^2+16x^3+64x^4+\ldots$$
After some work I have gotten this:
$$A(x)=\frac{x^2}{(1-x)(x-4)}$$ Where $A(x)$ is our unknown generating function (which we have been looking for). But I don't think this is correct because it does not match the answer down below.
Edit: I have finally figured it out, I will be back with my LaTex code to present it.
Final edit:
Find an explicit formula for $a_n$ using the generation function technique.
\newline We will begin by multiplying each side by $x^n$ and summing over $n\geq1$. We will also let $A(x)=\sum_{n=0}^\infty a_nx^n$ be the unknown generating function which we are looking for.
$$\begin{align*}
\implies \sum_{n=1}^\infty a_nx^n&=\sum_{n=1}^\infty (4a_{n-1}+1)x^n
\end{align*}
$$
Because
$a_0=0$ is given, we can rewrite $A(x)$:
$$
\begin{align*}
A(x)=a_0+\sum_{n=1}^\infty a_nx^n=\sum_{n=1}^\infty a_nx^n
\end{align*}
$$
Now, we continue to manipulate our equality.
$$
\begin{align*}
\implies \sum_{n=1}^\infty a_nx^n&=\sum_{n=1}^\infty (4a_{n-1}+1)x^n \\[2mm]
&=\sum_{n=1}^\infty 4a_{n-1}x^n + x^n \\[2mm]
&=\sum_{n=1}^\infty 4a_{n-1}x^n+\sum_{n=1}^\infty x^n \\[2mm]
&=\sum_{n=1}^\infty 4a_{n-1}x^n+ \frac{x}{1-x} \\[2mm]
&=\sum_{n=0}^\infty 4a_{n}x^{n+1}+ \frac{x}{1-x} \\[2mm]
&=a_0+\sum_{n=1}^\infty 4a_{n}x^{n+1}+ \frac{x}{1-x} \\[2mm]
&=\sum_{n=1}^\infty 4a_{n}x^{n+1}+ \frac{x}{1-x} \\[2mm]
&=4x\sum_{n=1}^\infty a_{n}x^{n}+ \frac{x}{1-x}. \\[2mm]
\end{align*}$$
It is clear now, that we have a multiple of our previously stated $A(x)$ on the left and right hand side.
$$
\begin{align*}
\implies&\sum_{n=1}^\infty a_{n}x^{n}=4x\sum_{n=1}^\infty a_{n}x^{n}+ \frac{x}{1-x} \\[2mm]
\implies&A(x)=4xA(x)+\frac{x}{1-x} \\[2mm]
\implies&A(x)-4xA(x)=\frac{x}{1-x}\\[2mm]
\implies& A(x)(1-4x)=\frac{x}{1-x}\\[2mm]
\implies& A(x)=\frac{x}{(1-x)(1-4x)}=\frac{x}{4x^2-5x+1} \\[2mm]
\end{align*}$$
Thus, the explicit formula for $a_n$ is given by $A(x)=\dfrac{x}{4x^2-5x+1}$ through the use of generating functions.
Finally, it follows that we have $$A(x)=\sum_{n=0}^\infty\frac{4^n-1}{3}x^n=\frac{x}{4x^2-5x+1}.$$
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\on}[1]{\operatorname{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
A particular solution is $\ds{-\,{1 \over 3}}$.
Then
$\ds{\pars{~\mbox{with}\ a_{0} = 0~}}$,
\begin{align}
a_{n} + {1 \over 3} & =
4\pars{a_{n - 1} + {1 \over 3}} =
4^{2}\pars{a_{n - 2} + {1 \over 3}} =
4^{3}\pars{a_{n - 3} + {1 \over 3}}
\\[5mm] & = \cdots =
4^{n}\pars{a_{0} + {1 \over 3}} = {4^{n} \over 3}
\implies
\bbx{a_{n} = {4^{n} - 1 \over 3}} \\ &
\end{align}
|
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|
Show that an integer of the form $8k + 7$ cannot be written as the sum of three squares. I have figured out a (long, and tedious) way to do it. But I was wondering if there is some sort of direct correlation or another path that I completely missed.
My attempt at the program was as follows:
A number of the form, $8k + 7 = 7 (mod 8)$. That is, we are looking for integers a, b, c such that $a^2 + b^2 + c^2 = 7 (mod 8)$.
LONG and TEDIOUS way:
$$(8k)^2 = 0 (mod 8)$$
$$(8k+1)^2 = 1 (mod 8)$$
$$(8k+2)^2 = 4 (mod 8)$$
$$(8k+3)^2 = 1 (mod 8)$$
$$(8k+4)^2 = 0 (mod 8)$$
$$(8k+5)^2 = 1 (mod 8)$$
$$(8k+6)^2 = 4 (mod 8)$$
$$(8k+7)^2 = 1 (mod 8)$$
That is, using three of these modulo there is no way to arrive at
$$a^2 + b^2 + c^2 = 7 (mod 8)$$
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Note that $x^2\equiv (-x)^2\pmod 8$. So the squares mod $8$ are $0^2=0$, $1^2=1$, $2^2=4$ and $3^2=1$. It is evident that three of these numbers cannot add up $7$.
|
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|
The proof of $e^x \leq x + e^{x^2}$ How can we prove the inequality $e^x \le x + e^{x^2}$ for $x\in\mathbb{R}$?
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$$e^{x} = 1 + x + \frac{x^2}{2!} + \dots \le 1 + x + 2 \left(\left( \frac{x}{2} \right )^2+\left( \frac{x}{2} \right )^3 +\dots \right)\le 1 + x+\left( \frac{x}{2} \right )^2 \left( \frac {4}{2 - x}\right)$$
$$x + e^{x^2} \ge x +1 + x^2 + \frac{x^4}{2!} = 1+ x+ \left( \frac{x}{2} \right )^2 2(2+x^2) \ge 1+ x+ 4 \left( \frac{x}{2} \right )^2 $$
The inequality is true for all $\Bbb R \setminus (1,3)$, let $b \in (1,2)$,
Also,
$$e^{(1+b)^2}+1+b \ge e^{(1+b)^2} \ge e^{1+b}$$
hence the inequality holds for all $x\in \Bbb R$
|
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|
Intersection multiplicity
Let $f=y^2-x^3$ and $g=y^3-x^7$. Calculate the intersection multiplicity of $f$ and $g$ at $(0,0)$.
I know the general technique for this (passing to the local ring) but I having difficulty with the fact that $3,7,2$ have no common factors.
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Using the properties of intersection number can save much time for the
calculation of intersection numbers.
If $P$ is a point, two affine plane curves $F$ and $G$ have no common
components, then: (see section 3.3 in Fulton's algebraic curves)
*
*$I(P,F\cap G)=I(F\cap (G+AF))$ for any $A\in k[X,Y]$;
*$I(P,F\cap G)=m_p(F)m_P(G)$, when $F$ and $G$ have not tangent
lines in common at $P$.
*$I(P,F\cap GH)=I(P,F\cap G)+I(P,F\cap H)$ for $H\in k[X,Y]$.
Now, $P=(0,0)$,
$I(P,y^2-x^3\cap y^3-x^7)=I(P, y^2-x^3\cap
((y^3-x^7)-(y^2-x^3)x^4))$
$\quad =I(P,y^2-x^3\cap y^2)+I(P,y^2-x^3\cap y-x^4).$
And $I(P, y^2-x^3\cap y^2)=I(P,x^3\cap y^2)=3\times 2=6$.
$I(P, y^2-x^3\cap y-x^4)=I(P, y^2-x^3\cap
((y-x^4)-(y^2-x^3)x))$
$\quad =I(P,y^2-x^3\cap y(1-xy))=3$.
Thus $9=6+3$ is the
desired number.
$I(P,y^2-x^3\cap y(1-xy))=I(P,y^2-x^3\cap y)+I(P,y^2-x^3\cap (1-xy))$,
$I(P,y^2-x^3\cap y)=I(P,-x^3\cap y)=3$ and $I(P,y^2-x^3\cap (1-xy))=0$ ($P$ is not the zero of $1-xy$).
|
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Find all real $x$ ,such $8x^3-20$ and $2x^5-2$ is perfect square
Find all real numbers $x$,such
$$8x^3-20,2x^5-2$$ is the perfect square of an integer
My idea: First we find the real number $x$ such $$8x^3-20,2x^5-2$$ is postive integer numbers,and second the real numbers such
$$8x^3-20=m^2,2x^5-2=n^5$$
How prove this all real numbers?
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First observe that $x$ is a positive integer. $x^5$ and $x^3$ are rationals; then $x=(x^3)^2/x^5$ is rational; finally $(2x)^5$ is an even positive integer so $2x$ must be an even positive integer.
In order to find a bound for $x$, consider the product of $8x^3−20=m^2$ and $2x^5−2=n^2$ and complete the square.
\begin{align*}
(mn)^2 &= (8x^3-20)(2x^5-2) \\
&= 16x^8 - 40x^5 - 16x^3 + 40 \\
&= (4x^4-5x)^2 -(16x^3+25x^2-40) \\
&= (4x^4-5x-1)^2 +(8x^4-16x^3-25x^2-10x+39).
\end{align*}
If $x\ge4$ then the remaining terms
$16x^3+25x^2-40$ and $8x^4-16x^3-25x^2-10x+39$ are positive and we find
$$
4x^4-5x-1 < mn < 4x^4-5x
$$
that is impossible. Hence, $x\le3$.
For $x=1$ we have $8x^3-20<0$, no solution.
For $x=2$ we have $8x^3-20=44$, no solution.
For $x=3$ we have $8x^3-20=14^2$ and $2x^5-2=22^2$.
|
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|
Polar Coordinates --- Equation of a line
Hey, does anyone know how to tackle this question? I've tried using the formula r=d*sec(theta-alpha)but I'm not sure what each of the variables are equal to. If anyone can offer any help, it would be greatly appreciated!
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The most direct approach, which would be preferred were this an exam question, is to find the equation for the line in Cartesian coordinates in the manner one would in pre-calculus, and then transform that result into polar coordinates. The point $ \ (-11, -7) \ $ (which is on the circle of radius $ \ \sqrt{(-11)^2 + (-7)^2} \ = \ \sqrt{170} \ $ ) is on the radius with a slope of $ \ \frac{-7}{-11} \ \ , $ so the tangent line $ \ L \ $ has slope $ \ -\frac{11}{7} \ $ .
The equation of $ \ L \ $ is then
$$ \ y - (-7) \ = \ -\frac{11}{7} · (x - [-11] ) \ \ \Rightarrow \ \ 7y + 49 \ = \ -11x - 121 \ \ \Rightarrow \ \ 7y + 11x \ = \ -170 \ \ . $$
In polar coordinates, this becomes
$$ 7y + 11x \ = \ -170 \ \ \rightarrow \ \ 7r \ \sin \theta \ + \ 11r \ \cos \theta \ = \ -170 $$
$$ \Rightarrow \ \ r · (7 \ \sin \theta \ + \ 11 \ \cos \theta) \ = \ -170 \ \ \Rightarrow \ \ r \ = \ \frac{-170}{7 \sin \theta \ + \ 11 \cos \theta} \ \ . $$
The negative sign in the numerator of the expression places the tangency in the third quadrant. (The solution given by Biswajit Banerjee works in an equivalent way, but the vector from $ \ O \ $ to $ \ P \ \ , \ \langle -11 \ , \ -7 \ \rangle \ , $ should have been used.)
As for the formula you inquired about, we could also start from the polar straight-line equation $ \ r \ = \ d · \sec(\theta - \alpha) \ \ , $ in which $ \ d \ $ represents the perpendicular (shortest) distance from the origin to the line and $ \ \alpha \ $ is the counter-clockwise angle by which the "vertical" line $ \ r \ = \ d · \sec \theta \ $ is rotated (making it also the angle that our line makes to the $ \ y-$axis.
Now this will mean that $ \ d \ = \ \sqrt{170} \ \ , $ which immediately raises the question of what happened to the radical and where the negative sign comes from. Since $ \ d > 0 \ $ places $ \ r \ = \ d · \sec \theta \ $ in the half-plane "to the right" of the $ \ y-$axis , we will want to use $ \ d \ = \ -\sqrt{170} \ \ . $ Referring to the graph above, we can then express the rotation angle as $ \ \alpha \ = \ \arctan(7/11) \ = \ \arcsin(7/\sqrt{170}) \ = \ \arccos(11/\sqrt{170}) \ \approx \ 0.5667 \ \ . $ (Since $ \ 0 < \alpha < \frac{\pi}{2} \ $ , any of these is valid.) So we could write the line equation as $ \ r \ = \ -\sqrt{170} · \sec(\theta - \alpha) \ \ , $ applying the value determined.
Another way to work with the secant form is to invoke the angle-addition formula for cosine:
$$ r \ = \ d · \sec(\theta - \alpha) \ = \ \frac{d}{\cos(\theta - \alpha)} \ = \ \frac{d}{\cos \theta · \cos \alpha \ + \ \sin \theta · \sin \alpha } \ \ . $$
For the moment, we'll pretend we don't know much about $ \ d \ $ or $ \ \alpha \ \ , $ other than that we'll choose to work with $ \ 0 < \alpha < \frac{\pi}{2} \ \ . $ We can use the Cartesian line equation or even just refer to the graph and make a similarity argument to find the intercepts of the line to be $ \ \left(-\frac{170}{11} \ , \ 0 \right) \ \ \text{and} \ \ \left(0 \ , \ -\frac{170}{7} \right) \ \ . $ Our polar equation then yields for the intercepts
$$ r(\pi) \ = \ \frac{d}{\cos \pi · \cos \alpha \ + \ \sin \pi · \sin \alpha} \ = \ \frac{d}{(-1) · \cos \alpha } \ = \ \ \frac{170}{11} $$
and
$$ r \left(\frac{3\pi}{2} \right) \ = \ \frac{d}{\cos \left(\frac{3\pi}{2} \right) · \cos \alpha \ + \ \sin \left(\frac{3\pi}{2} \right) · \sin \alpha} \ = \ \frac{d}{(-1) · \sin \alpha } \ = \ \ \frac{170}{7} \ \ , $$
since the intercepts lie at a positive radius from the origin in those directions. We conclude that
$$ \cos \alpha \ \ = \ \ -\frac{11}{170} · d \ \ , \ \ \sin \alpha \ \ = \ \ -\frac{7}{170} · d $$
$$ \Rightarrow \ \ r \ = \ \frac{d}{\cos \theta · \left(-\frac{11}{170} · d \right) \ + \ \sin \theta · \left(-\frac{7}{170} · d \right) } \ \ = \ \ \frac{-170}{11 \cos \theta \ + \ 7 \sin \theta } \ \ . $$
We asked earlier why there is no square-root in the numerator. If we insert the actual trigonometric values for $ \ \alpha \ $ taken in the interval $ \ 0 < \alpha < \frac{\pi}{2} \ $ into the earlier version of the polar equation, we would have
$$ r \ = \ \frac{-\sqrt{170}}{\cos \theta · \left(\frac{11}{\sqrt{170}} \right) \ + \ \sin \theta · \left(\frac{7}{\sqrt{170}} \right) } \ \ , $$
which simplifies to our result.
|
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|
Can someone explain the steps of manipulation of this equation for the value of x?
*
*$x = \sum_{j=1}^{n-1} 5j + 7$
*$x = \left(\sum_{j=0}^{n-1} 5j + 7 \right) - 7$
*$x = \left(5\sum_{j=0}^{n-1} j + \sum_{j=0}^{n-1}7 \right) - 7$
*$x = \left(5\sum_{j=0}^{n-1} j + 7n \right) - 7$
*$x = \frac{5n^2}{2} + 7n - 7$
I follow steps 1 to 3.
I understand that $\sum_{j=0}^{n-1}$ is equal to $\frac{n^2}{2}$ in steps 4-5 but I don't understand the following:
How does $\sum_{j=0}^{n-1}$ translate to $n$ in step 4 and what happens to $j$?
Update:
I believe that step 5 should be $x = \frac{5n(n-1)}2 + 7n - 7 = \frac 52 n^2 + \frac 92 n -7.$
|
4: What do you get if you do $7+7+...+7+7~~ n$ times? $~~7n!$
(from $0 $ to $ n-1$ there are n steps.)
For step 5 you have to write: $$x = \frac{5n^2-5n}{2} + 7n - 7$$
as$$\sum_{j=0}^{n-1} j=\sum_{j=1}^{n-1} j=\sum_{j=1}^{n} j-n=\frac{(n+1)n}{2}-n=\frac{n^2+n-2n}{2}=\frac{n^2-n}{2}$$
|
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|
Count subsets whose cardinalities are congruent to 0, 1 and 2 modulo 3 respectively Given a set of N elements, compute the number of subsets whose cardinalities are congruent to 0, 1 and 2 modulo 3 respectively.
Any hints would be appreciated.
Thanks!
|
You want to evaluate
\begin{align}a_N=\binom N0+\binom N3+\binom N6+\binom N9+\cdots,\\b_N=\binom N1+\binom N4+\binom N7+\binom N{10}+\cdots,\\c_N=\binom N2+\binom N5+\binom N8+\binom N{11}+\cdots.\end{align}
Note that, if $x^3=1$, then
\begin{align} (1+x)^N&=\binom N0+\binom N1x+\binom N2x^2+\binom N3x^3+\binom N4x^4+\binom N5x^5+\cdots\\&=\binom N0+\binom N1x+\binom N2x^2+\binom N3+\binom N4x+\binom N5x^2+\cdots\\&=a_N+b_Nx+c_Nx^2.\end{align}
So write down the equation
\begin{align}a_N+xb_N+x^2c_N=(1+x)^N\end{align}
with $x$ replaced in turn with each of the three cube roots of unity, and solve the resulting system of three linear equations for the three unknowns $a_N, b_N, c_N$.
The cube roots of unity are $1$, $\omega=\cos\frac{2\pi}3+i\sin\frac{2\pi}3$, and $\omega^2=\bar\omega$, so the equations are
$$a_N+b_N+c_N=2^N;$$
$$a_N+\omega b_N+\omega^2c_N=(1+\omega)^N;$$
$$a_N+\omega^2b_N+\omega c_N=(1+\bar\omega)^N.$$
To solve for $a_N$ we add all the equations and divide by $3$; since $1+\omega+\omega^2=0$ we get
$$a_N=\frac{2^N+(1+\omega)^N+(1+\bar\omega)^N}3=\frac{2^N+2\Re(1+\omega)^N}3.$$
Since $(1+\omega)^N=(\cos\frac\pi3+i\sin\frac\pi3)^N=\cos\frac{N\pi}3+i\sin\frac{N\pi}3$, this simplifies to
$$a_N=\frac{2^N+2\cos\frac{N\pi}3}3.$$
In like manner, we get
$$b_N=\frac{2^N-\cos\frac{N\pi}3+\sqrt{3}\sin\frac{N\pi}3}3$$
and
$$c_N=\frac{2^N-\cos\frac{N\pi}3-\sqrt{3}\sin\frac{N\pi}3}3.$$
Alternatively, use Pascal's identity $\binom Nk=\binom{N-1}k+\binom{N-1}{k-1}$ to derive the linear recurrences $a_N=a_{N-1}+c_{N-1},b_N=b_{N-1}+a_{N-1},c_N=c_{N-1}+b_{N-1}$ with initial values $a_0=1,b_0=c_0=0$. As an alternative to the matrix method, you could use the guess-and-check method: calculate a few values, observe the simple pattern, and verify by induction.
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block matrix multiplication If $A,B$ are $2 \times 2$ matrices of real or complex numbers, then
$$AB = \left[
\begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right]\cdot
\left[
\begin{array}{cc} b_{11} & b_{12} \\ b_{21} & b_{22} \end{array} \right]
=
\left[
\begin{array}{cc} a_{11}b_{11}+a_{12}b_{21} & a_{11}b_{12}+a_{12}b_{22} \\ a_{21}b_{11}+a_{22}b_{21} & a_{22}b_{12}+a_{22}b_{22} \end{array} \right]
$$
What if the entries $a_{ij}, b_{ij}$ are themselves $2 \times 2$ matrices? Does matrix multiplication hold in some sort of "block" form ?
$$AB = \left[
\begin{array}{c|c} A_{11} & A_{12} \\\hline A_{21} & A_{22} \end{array} \right]\cdot
\left[
\begin{array}{c|c} B_{11} & B_{12} \\\hline B_{21} & B_{22} \end{array} \right]
=
\left[
\begin{array}{c|c} A_{11}B_{11}+A_{12}B_{21} & A_{11}B_{12}+A_{12}B_{22} \\\hline A_{21}B_{11}+A_{22}B_{21} & A_{22}B_{12}+A_{22}B_{22} \end{array} \right]
$$
This identity would be very useful in my research.
|
It is always suspect with a very late answer to a popular question, but I came here looking for what a compatible block partitioning is and did not find it:
For $\mathbf{AB}$ to work by blocks the important part is that the partition along the columns of $\mathbf A$ must match the partition along the rows of $\mathbf{B}$. This is analogous to how, when doing $\mathbf{AB}$ without blocks—which is of course just a partitioning into $1\times 1$ blocks—the number of columns in $\mathbf A$ must match the number of rows in $\mathbf B$.
Example: Let $\mathbf{M}_{mn}$ denote any matrix of $m$ rows and $n$ columns irrespective of contents. We know that $\mathbf{M}_{mn}\mathbf{M}_{nq}$ works and yields a matrix $\mathbf{M}_{mq}$. Split $\mathbf A$ by columns into a block of size $a$ and a block of size $b$, and do the same with $\mathbf B$ by rows. Then split $\mathbf A$ however you wish along its rows, same for $\mathbf B$ along its columns. Now we have
$$
A = \begin{bmatrix}
\mathbf{M}_{ra} & \mathbf{M}_{rb} \\
\mathbf{M}_{sa} & \mathbf{M}_{sb}
\end{bmatrix},
B = \begin{bmatrix}
\mathbf{M}_{at} & \mathbf{M}_{au} \\
\mathbf{M}_{bt} & \mathbf{M}_{bu}
\end{bmatrix},
$$
and
$$
AB = \begin{bmatrix}
\mathbf{M}_{ra}\mathbf{M}_{at} + \mathbf{M}_{rb}\mathbf{M}_{bt}
& \mathbf{M}_{ra}\mathbf{M}_{au} + \mathbf{M}_{rb}\mathbf{M}_{bu} \\
\mathbf{M}_{sa}\mathbf{M}_{at} + \mathbf{M}_{sb}\mathbf{M}_{bt}
& \mathbf{M}_{sa}\mathbf{M}_{au} + \mathbf{M}_{sb}\mathbf{M}_{bu}
\end{bmatrix}
=
\begin{bmatrix}
\mathbf{M}_{rt} & \mathbf{M}_{ru} \\
\mathbf{M}_{st} & \mathbf{M}_{su}
\end{bmatrix}.
$$
All multiplications conform, all sums work out, and the resulting matrix is the size you'd expect. There is nothing special about splitting in two so long as you match any column split of $\mathbf A$ with a row split in $\mathbf B$ (try removing a block row from $\mathbf A$ or further splitting a block column of $\mathbf B$).
The nonworking example from the accepted answer is nonworking because the columns of $\mathbf A$ are split into $(1, 2)$ while the rows of $\mathbf B$ are split into $(2, 1)$.
|
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|
Is my proof correct? (standard algebra, some calculus) I'm teaching myself mathematics and I tried to solve a problem from the STEP exam (Cambridge's admission test). The solution is considerably shorter and less "clumsy" than mine but I would like to know if my proof is at least correct.
Thank you.
We have $$f'(x)=a-\frac{x^4+3x^2}{x^4+2x^2+1}$$ and we want to prove that if $a\geq\frac{9}{8}$ then $f'(x)\geq 0$ for all x. \
If the statement is proven for $a=\frac{9}{8}$, the case of $a > \frac{9}{8}$ follows. Therefore we only need to prove that $$ \frac{9}{8} \geq \frac{x^4+3x^2}{x^4+2x^2+1} $$
The proof:
\begin{gather}
\frac{9}{8} \geq \frac{x^4+3x^2}{x^4+2x^2+1} \\
\frac{9(x^4+2x^2+1)-8(x^4+3x^2)}{8(x^4+2x^2+1)} \geq 0 \\
\frac{x^4-6x^2+9}{8x^4+16x^2+8} \geq 0 \;\; \text{(3)}
\end{gather}
Since $\forall x \; 8x^4+16x^2+8 > 0 $, in fact $ \geq 8 $, then for (3) to be incorrect $ x^4-6x^2+9 < 0 $. I will prove this is impossible for all x.
\begin{gather*}
x^4-6x^2+9 < 0 \\
x^4-6x^2 < -9
\end{gather*}
However the global minimum of $ x^4-6x^2 $ is $-9$:
$\frac{d}{dx}x^4-6x^2=4x^3-12x=0$ when $x = 0, x =\sqrt{3}$ and $x=\sqrt{3}$ ,when plugged back in, yields $ 9-6*3 = -9 $ the global minimum.
It follows that $ x^4-6x^2 \geq -9 $.
Therefore the statement (3) is correct and I have proved that, when $a \geq \frac{9}{8}$ $$ f'(x) \geq 0 $$
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Yes you have done it in the right sense .
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|
$f(\frac1{1234}) + f(\frac3{1234}) + f(\frac5{1234}) + ..... + f(\frac{1231}{1234}) + f(\frac{1233}{1234}) = ?$ If $f(x) = \frac{e^{2x-1}}{1+e^{2x-1}}$, then how to evaluate $$f(\frac1{1234}) + f(\frac3{1234}) + f(\frac5{1234}) + ..... + f(\frac{1231}{1234}) + f(\frac{1233}{1234}) ?$$
I know I'm missing some trick here, what is it?
|
Using the trick you've identified,
\begin{align*}
f(x) + f(1 - x) &= \frac{e^{2x - 1}}{1 + e^{2x - 1}} + \frac{e^{2(1 - x) - 1}}{1 + e^{2(1 - x) - 1}} \\
&= \frac{e^{2x - 1}(1 + e^{1 - 2x}) + e^{1 - 2x}(1 + e^{2x - 1})}{(1 + e^{2x - 1})(1 + e^{1 - 2x})} \\
&= \frac{e^{2x - 1} + 1 + e^{1 - 2x} + 1}{1 + e^{2x - 1} + e^{1 - 2x} + 1} \\
&= 1
\end{align*}
Now count how many terms there are.
|
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|
Is there an elegant way to simplify $\frac{\tan(x+20^{\circ })-\sin(x+20^{\circ })}{\tan(x+20^{\circ })+\sin(x+20^{\circ })}$ I wonder how to solve this equation: $$\frac{\tan(x+20^{\circ })-\sin(x+20^{\circ })}{\tan(x+20^{\circ })+\sin(x+20^{\circ })}=4\sin^{2}\left(\frac{x}{2}+10^{\circ }\right)$$ in an elegant/shorter way.
My way:
$$\frac{\sin(x+20^{\circ })-\sin(x+20^{\circ })\cdot cos(x+20^{\circ})}{\sin(x+20^{\circ })+\sin(x+20^{\circ })\cdot \cos(x+20^{\circ})}=2[1-(\cos(x+20^{\circ}))]$$
so
$$1-\cos(x+20^{\circ})=2[1-\cos(x+20^{\circ})][1+\cos(x+20^{\circ})]$$
which gives
$$\cos(x+20^{\circ})=1\Rightarrow x=-20^{\circ}+360^{\circ}k\\\cos(x+20^{\circ})=-0.5\Rightarrow x=140^{\circ}+360^{\circ}k,\, \, x=-100^{\circ}+360^{\circ}k$$
Thanks.
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$4sin^2(\frac{x + 20^\circ}2) = 4\cdot1/2\cdot(1-\cos(x+20^\circ)) = 2-2\cos(x+20^\circ)$
Applying componendo and dividendo on $\frac{tg(x+20^{\circ })-sin(x+20^{\circ })}{tg(x+20^{\circ })+sin(x+20^{\circ })}=4sin^{2}(\frac{x}{2}+10^{\circ })$ ,
$-\sec(x+20^\circ) = \frac{2-2\cos(x+20^\circ)+1}{2-2\cos(x+20^\circ) -1}$, $cos(x+20^\circ)\ne1/2$
$-\sec(x+20^\circ)\cdot(1-{2\cos(x+20^\circ)}) = {3-2\cos(x+20^\circ)}$
$-\sec(x+20^\circ)+2=3-2\cos(x+20^\circ)$
$2cos(x+20^\circ)-\sec(x+20^\circ)-1=0$
$2cos^2(x+20^\circ)-cos(x+20^\circ)-1=0$
Can you solve it further?
|
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|
Simplifying $\sum_{k=1}^n (n-k)2^{k-1}$ via combinatorics in Aigner, A course in Combinatorics I must simplify $\sum_{k=1}^n (n-k)2^{k-1}$ using the rule of sum. How might I go about doing this?
Thank you and regards.
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$\sum_{k=1}^n(n-k)2^{k-1} = \sum_{k = 1}^n n2^{k-1} - \sum_{k = 1}^n k2^{k-1}$.
In the first sum, we can factor out $n$ from each term so $\sum_{k=1}^n n2^{k-1} = n(1+2+ \cdots + 2^{n-1}) = n(2^n-1)$.
For the second term, we use a standard calculus argument. $1+x+x^2+ \cdots + x^k = \frac{x^{k+1}-1}{x-1}$.
Then differentiating both sides gives $1+2x+ \cdots + kx^{k-1} = \frac{k x^{k+1}-(k+1) x^k+1}{(x-1)^2}$. Let $x = 2$ for our sum in hand. Its easy to finish from here.
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|
A mathematical phenomenon regarding perfect squares.... I was working on identifying perfect squares for one of my programs regarding Pythagorean triplet.
And I found that for every perfect square if we add its digits recursively until we get a single digit number, e.g. 256 -> 13 -> 4 etc. we get the single digit as either 1,4,7 or 9.
Is it an unnoticed phenomenon?
What can be mathematical reason behind this?
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The standard observation is that "add the digits of a number recursively" is essentially just a fancy way to say "find the value of the number modulo 9".
So your question is really asking "which numbers are squares modulo 9?" This is easy to answer with modular arithmetic: since there are only 9 numbers modulo 9, we can use brute force
*
*$0^2 = 0$
*$1^2 = 1$
*$2^2 = 4$
*$3^2 = 9 = 0 \pmod 9$
*$4^2 = 16 = 9 + 7 = 7 \pmod 9$
*$5^2 = 25 = 9\cdot 2 + 7 = 7 \pmod 9$
*$6^2 = 36 = 9\cdot 4 = 0 \pmod 9$
*$7^2 = 49 = 9\cdot 5 + 4 = 4 \pmod 9$
*$8^2 = 64 = 9\cdot 7 + 1 = 1 \pmod 9$
We could have saved some trouble by using $x^2 = (-x)^2$; e.g. $(-3) = 6 \bmod 9$, thus $6^2 = (-3)^2$. We only needed to check $0$ through $4$.
Because of the actual calculation you do, the representatives of the equivalence classes modulo $9$ that you wind up with are the numbers $1$ through $9$, rather than $0$ through $8$. Note $0 = 9 \bmod 9$. Thus, the squares are $1, 4, 7, 9$, as you've observed.
|
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Number theory problems. It is given: $3m+1=$perfect square. Express $m+1$, as the sum of $3$ perfect squares.
I tried to solve the problem by checking for odd and even values perfect squares $4k^2,{(2k+1)}^2$. I got somewhat convincing form of result from $3m+1={(2k+1)}^2$:
$m+1=\dfrac43(k^2)+\dfrac43(k)+1,$ so that $k$ must be a multiple of $3$,
but I could not find the exact value of $k$, so that RHS is a perfect square.
as for the part $3m+1=4k^2$, I am hopeless.
Help please.
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$3m + 1 = n^2$. This means: $n = 3k - 1$ or $n = 3k + 1$. If $n = 3k + 1$ then:
$m = \dfrac{n^2 - 1}{3}$, so $m + 1 = \dfrac{n^2 + 2}{3} = \dfrac{(3k + 1)^2 + 2}{3} = \dfrac{9k^2 + 6k + 3}{3} = 3k^2 + 2k + 1 = k^2 + k^2 + (k+1)^2$.
The other case is similar.
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Continuous functions that satisfies $f(x) + f(1-x) + f(\sqrt{x^2+(1-x)}) = 0$ and $f(\frac12)=0$ $f:[0,1]\rightarrow \mathbb{R}$ is a continuous function which satisfies
$$
f(x) + f(1-x) + f\left(\sqrt{x^2+(1-x)}\right) = 0 \text{ and } f\left(\tfrac12\right)=0.
$$
Can someone give explicit examples of $f$, apart from the trivial solution, $f(x)=0$? And are there infinitely many solutions for $f$?
I can derive some properties like $f(\frac{\sqrt3}2)=0$ or $f(0)=-2f(1)$, but I can't generate particular examples. Continuity seems important here, but I can't see how to use it, for if we take $f(x)=0$, for $x\in (0,1)$, and $f(1)=1$, $f(0)=-2$ also works.
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Here is the set of all possible $f$.
The algebra gets a little gross, so bear with me.
First, let
$f_1: \left[1 - \frac{\sqrt{3}}{2}, 1 \right]
\to \mathbb{R}$ be an arbitrary continuous function such that $f_1(x) = 0$ on the closed interval $\left[ \frac{\sqrt{3}}{2}, \frac{7}{4} - \frac{\sqrt{3}}{2} \right]$,
and $f_1(\tfrac12) = 0$,
and finally,
$f_1(\tfrac{\sqrt{3}}{2}) + f_1(\frac{7}{4} - \frac{\sqrt{3}}{2}) = 0$.
Second, define
$f_2 : \left[1 - \frac{\sqrt{3}}{2}, 1 \right] \to \mathbb{R}$
by
$$
f_2(x) =
\begin{cases}
f_1(x)
& \text{if } 1 - \frac{\sqrt{3}}{2} \le x \le \frac{\sqrt{3}}{2} \\
- f_1\left( \frac{1 - \sqrt{4x^2 - 3}}{2} \right)
- f_1\left( \frac{1 + \sqrt{4x^2 - 3}}{2} \right)
& \text{if }
\frac{\sqrt{3}}{2} \le x \le \frac{7}{4} - \frac{\sqrt{3}}{2} \\
f_1(x)
& \text{if }
\frac{7}{4} - \frac{\sqrt{3}}{2} \le x \le 1 \\
\end{cases}
$$
Verify that
$f_2$ is continuous, and satisfies
$f_2(x) + f_2(1-x) + f_2(\sqrt{x^2 - x + 1}) = 0$
on its domain.
This is a bit gross, so I'm skipping the details for now.
Finally, define
$$
f_3(x) =
\begin{cases}
-f_2(1 - x) - f_2(\sqrt{x^2 - x + 1})
&\text{if } 0 \le x \le 1 - \frac{\sqrt{3}}{2} \\
f_2(x)
&\text{if } 1 - \frac{\sqrt{3}}{2} \le x \le 1 \\
\end{cases}
$$
Verify that $f_3$ is continuous,
and satisfies
$f_2(x) + f_2(1-x) + f_2(\sqrt{x^2 - x + 1}) = 0$
on its domain.
This is not too hard given that we already showed it for $f_2$.
Now just set $f = f_3$.
The end result is a function $f$
such that $f_1 = f$ on
$\left[1 - \frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2} \right]$
and $\left[ \frac{7}{4} - \frac{\sqrt{3}}{2}, 1 \right]$,
and $f$ satisfies the desired property.
So we get to choose $f$ however we want on
$\left[1 - \frac{\sqrt{3}}{2} , \frac{\sqrt{3}}{2} \right]$
and $\left[ \frac{7}{4} - \frac{\sqrt{3}}{2} , 1 \right]$,
and then $f$ is decided for us everywhere else.
|
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|
Find all values of the parameter $a$ so that $f(x)= y$ is always positive Find all values of the parameter $a$ so that $y>0$.
The equation is:
$$y = (a − 1)x^2 − (a + 1)x + a + 1$$
Thanks!
|
First of all, the discriminant should be negative but $a-1>0$.
We get
$$(a+1)^2 - 4(a-1)(a+1) < 0 \Longleftrightarrow a^2+2a+1 - (4a^2 - 4) = -3a^2+2a +5 < 0.$$
By using quadratic formula
$$-3a^2+2a+5 = (-3a +5 )(a + 1) < 0.$$
The formula above will have negative value when $a < -1$ or $a > \frac{5}{3}$.
Because the value of $a$ should be greater than 1, so that
$$a > \frac{5}{3}.$$
|
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Floor function inequality: $\frac{n^2}{x}-\left\lfloor\frac{n^2}{x}\right\rfloor+\frac{2n+1}{x}-\left\lfloor\frac{2n+1}{x}\right\rfloor<1$ I would like to dissect the following inequality to figure out its properties.
$$\frac{n^2}{x}-\left\lfloor\frac{n^2}{x}\right\rfloor+\frac{2n+1}{x}-\left\lfloor\frac{2n+1}{x}\right\rfloor>1$$
where $n>0$ and $0<x<n^2$ and $x$ is an integer.
I would like to know will $\frac{n^2}{x}-\left\lfloor\frac{n^2}{x}\right\rfloor+\frac{2n+1}{x}-\left\lfloor\frac{2n+1}{x}\right\rfloor<1$ for all $x>2n+1$?
|
(Getting this off the unanswered list.)
For any $r\in\Bbb R$, $r-\lfloor r\rfloor$ is the fractional part of $r$, usually written $\{r\}$. Thus, you’re asking about
$$\left\{\frac{n^2}x\right\}-\left\{\frac{2n+1}x\right\}\,.$$
Clearly
$$0\le\left\{\frac{n^2}x\right\},\left\{\frac{2n+1}x\right\}\le\frac{x-1}x=1-\frac1x<1$$
for any positive integer $x$, so
$$\frac1x-1\le\left\{\frac{n^2}x\right\}-\left\{\frac{2n+1}x\right\}\le 1-\frac1x<1$$
for any positive integer $x$.
|
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|
Evaluation of $\int_0^1 \frac{\log^2(1+x)}{x} \ dx$ One of the ways to approach it lies in the area of the dilogarithm, but is it possible to evaluate it
by other means of the real analysis (without using dilogarithm)?
$$\int_0^1 \frac{\log^2(1+x)}{x} \ dx$$
EDIT: maybe you're aware of some easy way to do that. I'd appreciate it!
Some words on the generalization case (by means of the real analysis again)?
$$F(n)=\int_0^1 \frac{\log^n(1+x)}{x} \ dx, \space n\in \mathbb{N}$$
|
On the path of Felix Marin,
\begin{align}J&=\int_0^1 \frac{\ln(1+x)^2}{x}\\
&\overset{y=\frac{1}{1+x}}=\int_{\frac{1}{2}}^1 \frac{\ln^2 x}{x(1-x)}\,dx\\
&=\int_{\frac{1}{2}}^1 \frac{\ln^2 x}{x}\,dx+\int_{\frac{1}{2}}^1 \frac{\ln^2 x}{1-x}\,dx\\
&=\frac{1}{3}\left(\ln^3 (1)-\ln^3\left(\frac{1}{2}\right)\right)+\int_0^1 \frac{\ln^2 x}{1-x}\,dx-\int_0^{\frac{1}{2}} \frac{\ln^2 x}{1-x}\,dx\\
&=\frac{1}{3}\ln^3 2+\int_0^1 \frac{\ln^2 x}{1-x}\,dx-\int_0^{\frac{1}{2}} \frac{\ln^2 x}{1-x}\,dx\\
&\overset{y=\frac{x}{1-x},\text{the 2nd integral}}=\frac{1}{3}\ln^3 2+\int_0^1 \frac{\ln^2 x}{1-x}\,dx-\int_0^1\frac{\ln^2\left(\frac{x}{1+x}\right)}{1+x}\,dx\\
&=\frac{1}{3}\ln^3 2+\int_0^1 \frac{\ln^2 x}{1-x}\,dx-\int_0^1\frac{\ln^2 x}{1+x}\,dx-\int_0^1\frac{\ln^2 (1+x)}{1+x}\,dx+2\int_0^1\frac{\ln(1+x)\ln x}{1+x}\,dx\\
&\overset{IBP}=\frac{1}{3}\ln^3 2+\int_0^1 \frac{\ln^2 x}{1-x}\,dx-\int_0^1\frac{\ln^2 x}{1+x}\,dx-\int_0^1\frac{\ln^2 (1+x)}{1+x}\,dx-J\\
&=\frac{1}{3}\ln^3 2+\int_0^1 \frac{2x\ln^2 x}{1-x}\,dx-\frac{1}{3}\ln^3 2-J\\
&\overset{y=x^2}=\frac{1}{4}\int_0^1 \frac{\ln^2 x}{1-x}\,dx-J\\
J&=\frac{1}{8}\int_0^1 \frac{\ln^2 x}{1-x}\,dx\\
&=\frac{1}{8}\times 2\zeta(3)\\
&=\boxed{\frac{1}{4}\zeta(3)}
\end{align}
NB: i assume that, \begin{align}\int_0^1 \frac{\ln^2 x}{1-x}\,dx=2\zeta(3)\end{align}
(proof: Taylor expansion)
|
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|
Find the minimum value of $P=\frac{1}{4(x-y)^2}+\frac{1}{(x+z)^2}+\frac{1}{(y+z)^2}$ Let $x,y,z$ be real numbers such that $x>y>0, z>0$ and $xy+(x+y)z+z^2=1$.
Find the minimum value of $$P=\frac{1}{4(x-y)^2}+\frac{1}{(x+z)^2}+\frac{1}{(y+z)^2}$$
I tried using some ways, but failed. Please give me an idea. Thank you.
|
From @DeepSea's above, suppose $b\equiv f(\,a\,),\,ab= 1\,\therefore\,{b}'= -\,\dfrac{b}{a},\,(\,a- 2\,b\,)(\,a- 2^{\,(\,\frac{1}{2}\,)}\,)\geqq 0$.
$$a- b> 0 \tag{assume}$$
Let $$W(\,a\,)= \frac{1}{4(\,a- b\,)^{\,2}}+ \frac{1}{a^{\,2}}+ \frac{1}{b^{\,2}}- 3$$
$$\therefore\,{W}'(\,a\,)= -\,\frac{2}{a^{\,3}}+ \frac{{b}'- 1}{2(\,a- b\,)^{\,3}}- \frac{2\,{b}'}{b^{\,3}}$$
$$\therefore\,{W}'(\,a\,)= -\,\frac{2}{a^{\,3}}+ \frac{2}{ab^{\,2}}- \frac{-\,\dfrac{b}{a}- 1}{2(\,a- b\,)^{\,3}}$$
$$\therefore\,{W}'(\,a\,)= \frac{(\,a- 2\,b\,)(\,b- 2\,a\,)(\,a+ b\,)(\,2\,a^{\,2}- 3\,ab+ 2\,b^{\,2})}{2\,a^{\,3}b^{\,2}(\,a- b\,)^{\,2}}$$
$$\therefore\,(\,a- 2\,b\,){W}'(\,a\,)\geqq 0$$
$$\therefore\,(\,a- 2^{\,(\,\frac{1}{2}\,)}\,){W}'(\,a\,)\geqq 0$$
$$\therefore\,W(\,a\,)\geqq W(\,2^{\,(\,\frac{1}{2}\,)}\,)= 3$$
$$\because\,W(\,a\,)- W(\,2^{\,(\,\frac{1}{2}\,)}\,)= (\,a- 2^{\,(\,\frac{1}{2}\,)}\,){W}'(\,a\,) \tag{tangent equation}$$
|
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|
Find the Fourier series for $f(x) := |\sin(x)|$ and the sum of $\sum_{n=1}^{\infty} \frac {(-1)^{n+1}} {4n^2-1}$.
Find the Fourier series for $f(x) := |\sin(x)|$ and the sum of $\sum_{n=1}^{\infty} \frac {(-1)^{n+1}} {4n^2-1}$.
I have computed $$c_n = \frac 1 {2\pi} \int^{\pi}_{-\pi} |\sin(x)|e^{-inx} dx = \frac 1 {2\pi} \int^{\pi}_{0} |\sin(x)|e^{-inx} + |\sin(-x)|e^{-in(-x)} dx = \frac {1} {\pi} \frac {1} {n^2-1} ((-1)^{n+1} -1)$$
However, this doesn't correspond that well with the sum of the series I must find. What have I done wrong ?
|
The exponential Fourier series for this particular $f(x)$ is
\begin{align}
|\sin(ax)| = \sum_{n= -\infty}^{\infty} c_{n} e^{-i nx},
\end{align}
where
\begin{align}
c_{n} = \frac{1}{2\pi} \int_{-\pi}^{\pi} |\sin(ax)| \ e^{-i nx} dx.
\end{align}
This integral can be calculated as follows.
\begin{align}
c_{n} &= \frac{1}{2\pi} \int_{-\pi}^{\pi} |\sin(ax)| \ e^{-i nx} dx \\
&= \frac{1}{2\pi} \int_{-\pi}^{0} |\sin(ax)| \ e^{-i nx} dx + \frac{1}{2\pi} \int_{0}^{\pi}
|\sin(ax)| \ e^{-i nx} dx \\
&= \frac{1}{2\pi} \int_{0}^{\pi} |\sin(-ax)| \ e^{i nx} dx + \frac{1}{2\pi} \int_{0}^{\pi}
|\sin(ax)| \ e^{-i nx} dx \\
&= \frac{1}{\pi} \int_{0}^{\pi} \sin(ax) \cos(nx) dx \\
&= \frac{1}{2\pi} \int_{0}^{\pi} \left[ \sin(a+n)x + \sin(a-n)x \right] dx\\
&= \frac{1}{2\pi} \left[ \frac{2a}{a^{2} - n^{2}} - \frac{\cos(a\pi + n \pi)}{a+n}
- \frac{\cos(a\pi - n \pi)}{a-n} \right] \\
c_{n} &= \frac{a}{\pi (a^{2} - n^{2})} \left[ 1 - (-1)^{n} \cos(a\pi) \right].
\end{align}
From this the Fourier series is
\begin{align}
|\sin(ax)| = \frac{a}{\pi} \sum_{-\infty}^{\infty} \frac{(1 - (-1)^{n} \cos(a\pi))}
{\pi (a^{2} - n^{2})} \ e^{-i nx}.
\end{align}
The summation can be changed and is seen in the following.
\begin{align}
\sum_{-\infty}^{\infty} b_{n} e^{-i nx} &= \sum_{-\infty}^{-1} b_{n} e^{-i nx} + b_{0}
+ \sum_{1}^{\infty} b_{n} e^{-i nx} \\
&= b_{0} + \sum_{n=0}^{\infty} \left( b_{-n} e^{i nx} + b_{n} e^{-i nx} \right).
\end{align}
From this it is seen that
\begin{align}
|\sin(ax)| = \frac{1 - \cos(a\pi)}{a \pi} - \frac{2a}{\pi} \sum_{n=0}^{\infty}
\frac{(1-(-1)^{n} \cos(a\pi))}{n^{2}-a^{2}} \ \cos(nx)
\end{align}
or
\begin{align}
|\sin(ax)| = \frac{\sin^{2}\left(\frac{a\pi}{2}\right)}{2 a \pi} - \frac{2a}{\pi} \sum_{n=0}^{\infty}
\frac{(1-(-1)^{n} \cos(a\pi))}{n^{2}-a^{2}} \ \cos(nx).
\end{align}
Now, when $a=1$ this reduces to
\begin{align}
|\sin(x)| &= \frac{2}{\pi} - \frac{2}{\pi} \sum_{n=0}^{\infty} \frac{1+(-1)^{n}}{n^{2}-1} \ \cos(nx) \\
&= \frac{2}{\pi} - \frac{4}{\pi} \sum_{n=0}^{\infty} \frac{\cos(2nx)}{4n^{2}-1}.
\end{align}
In order to evaluate the series in question let $x = \pi/2$ for which the Fourier series
is reduced to
\begin{align}
|\sin(\pi/2)| &= \frac{2}{\pi} - \frac{4}{\pi} \sum_{n=0}^{\infty} \frac{\cos(n \pi)}{4n^{2}-1} \\
&= \frac{2}{\pi} + \frac{4}{\pi} \sum_{n=0}^{\infty} \frac{(-1)^{n-1}}{4n^{2}-1}.
\end{align}
which leads to
\begin{align}
\sum_{n=0}^{\infty} \frac{(-1)^{n-1}}{4n^{2}-1} = \frac{\pi - 2}{4}.
\end{align}
|
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|
Integral $I:=\int_0^1 \frac{\log^2 x}{x^2-x+1}\mathrm dx=\frac{10\pi^3}{81 \sqrt 3}$ Hi how can we prove this integral below?
$$
I:=\int_0^1 \frac{\log^2 x}{x^2-x+1}\mathrm dx=\frac{10\pi^3}{81 \sqrt 3}
$$
I tried to use
$$
I=\int_0^1 \frac{\log^2x}{1-x(1-x)}\mathrm dx
$$
and now tried changing variables to $y=x(1-x)$ in order to write
$$
I\propto \int_0^1 \sum_{n=0}^\infty y^n
$$
however I do not know how to manipulate the $\log^2 x$ term when doing this procedure when doing this substitution. If we can do this the integral would be trivial from here.
Complex methods are okay also, if you want to use this method we have complex roots at $x=(-1)^{1/3}$. But what contour can we use suitable for the $\log^2x $ term?
Thanks
|
Here is another way to do it. Let $$I=\int_{0}^{1} \frac{(\ln(x))^2}{x^2-x+1} \ dx =\int_{0}^{1} \frac{(\ln(x))^2 (1+x)}{1+x^3} \ dx.$$ By a change of variables $x=\frac{1}{u}, \ dx =-\frac{1}{u^2} \ du,$ we have
$$I=\int_{1}^{\infty} \frac{(\ln(u))^2 (1+u)}{1+u^3} \ du,$$ which implies
$$I=\frac{1}{2} \int_{0}^{\infty} \frac{(\ln(x))^2 (1+x)}{1+x^3} \ dx.$$ Now split the integrand into two terms
$$I=\frac{1}{2} \int_{0}^{\infty} \frac{(\ln(x))^2}{x^3+1} \ dx + \frac{1}{2} \int_{0}^{\infty} \frac{x(\ln(x))^2}{x^3+1} \ dx.$$ The two terms are equal to one another, which can be shown by the same change of variables $x= \frac{1}{u}$ on the first term.Thus,
$$I= \int_{0}^{\infty} \frac{(\ln(x))^2}{x^3+1} \ dx.$$
Now consider the triple integral
$$J=\int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\infty} \frac{xy}{(1+x^2)(1+y^2)(1+x^2y^2z^3)} \ dz \ dy \ dx.$$
Make the change of variables $z=\frac{u}{(xy)^{\frac{2}{3}}}, \ dz = \frac{1}{xy^{\frac{2}{3}}} \ du,$ to see
$$J=\int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\infty} \frac{(xy)^{\frac{1}{3}}}{(1+x^2)(1+y^2)(1+u^3)} \ du \ dy \ dx.$$ Apply the well known formula
$$\int_{0}^{\infty} \frac{t^m}{1+t^n} \ dt= \frac{\pi}{n} \csc \left( \frac{\pi(m+1)}{n} \right)$$ thrice to find $$J=\frac{2 \pi^3}{9\sqrt{3}}.$$
Now reverse the order of integration as such. $$J=\int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\infty} \frac{xy}{(1+x^2)(1+y^2)(1+x^2y^2z^3)} \ dx \ dz \ dy.$$ Applying partial fractions or Mathematica gives us that
$$J=\int_{0}^{\infty}\int_{0}^{\infty}\frac{y\ln(y)}{(1+y^2)(-1+y^2z^3)} \ dz \ dy+ \frac{3}{2} \int_{0}^{\infty}\int_{0}^{\infty}\frac{y\ln(z)}{(1+y^2)(-1+y^2z^3)} \ dz \ dy=J_1+J_2$$ Reverse the order of integration on $J_2$ and use partial fractions to see
$$J_2=\frac{9}{4}I.$$
Now we focus on the first term $J_1$ Apply partial fractions or use Mathematica to see that
$$J_1=-\frac{\pi}{3 \sqrt{3}} \int_{0}^{\infty} \frac{y^{\frac{1}{3}} \ln(y)}{y^2+1} \ dy.$$
We expand $$J_1 =\frac{\pi}{3 \sqrt{3}} \int_{0}^{\infty} \int_{0}^{\infty} \frac{ty^{\frac{1}{3}}}{(1-t^2y^2)(1+t^2)} \ dt \ dy.$$ This equality is true by using partial fractions to evaluate the inner integral. Reversing the order of integration, and using partial fractions again, we get that
$$J_1=\frac{-\pi^2}{18} \int_{0}^{\infty} \frac{t^{\frac{-1}{3}}}{1+t^2} \ dt= \frac{-\pi^3}{18\sqrt{3}}$$ by our aforementioned well-known integral formula.
Since $$J=J_1+ \frac{9}{4}I,$$ and $J=\frac{2 \pi ^3}{9\sqrt{3}}$ and $J_1= \frac{-\pi^3}{18\sqrt{3}},$ we have that $$I=\frac{4}{9}\left(\frac{2 \pi ^3}{9\sqrt{3}}+\frac{\pi^3}{18\sqrt{3}}\right)=\frac{10 \pi ^3}{81\sqrt{3}}.$$
|
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|
Inequality $\sum\frac{x}{(x + n^2)^2}<\frac{1}{2} \sum \frac{1}{x + n^2} $
$x\geq0$, then, we have
$$\sum_{n=1}^{\infty}\frac{x}{(x + n^2)^2}<\frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{x + n^2} $$
The problem is not easy, even $x=1$. Any help will be appreciated
|
I think this approach is also valid, and simpler.
A term-wise subtraction (LHS - RHS) gives:
$$
\frac{x}{(x+n^2)^2} - \frac{1}{2(x+n^2)} = \frac{2x - (x+n^2)}{2(x+n^2)^2} = \frac{x-n^2}{2(x+n^2)^2}
$$
for sufficiently large number of terms, this can be proved to be less than zero.
|
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|
Trigonometry substitution Given the expansion of $\cos(5x) = 16\cos^5(x)-20\cos^3(x)+5\cos(x)$. How can I find the find the value of $\cos(\frac{\pi}{10})$ and $\cos(\frac{7\pi}{10})$ using $\cos(5x) = 0$ ? I know that I can change $\cos (5x)$ as a quartic equation and can get $\cos(x) = \pm\sqrt \frac{5 \pm\sqrt 5}{8}$. Which means $\cos(\frac{5\pi}{10}) =\cos(\frac{\pi}{2})= 0$ but how I am going to decide about the $\pm$ signs? and what about $\cos(\frac{7\pi}{10})$ ?
|
First of all, $$0<\frac\pi{10}<\frac\pi2<\frac{7\pi}{10}$$
So, $\displaystyle\cos\frac\pi{10}>0;\cos\frac{7\pi}{10}<0$
$\displaystyle\implies-\cos\frac{7\pi}{10}=\cos\left(\pi-\frac{7\pi}{10}\right)=\cos\frac{3\pi}{10}>0$
and using Prosthaphaeresis Formulas, $\displaystyle\cos\frac\pi{10}-\cos\frac{3\pi}{10}=2\sin\frac\pi5\sin\frac{2\pi}5>0$
$\displaystyle\implies\cos\frac\pi{10}>\cos\frac{3\pi}{10}$
Can you determine the required values from here?
|
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|
How can I find the following product? $ \tan 20^\circ \cdot \tan 40^\circ \cdot \tan 80^\circ.$ How can I find the following product using elementary trigonometry?
$$ \tan 20^\circ \cdot \tan 40^\circ \cdot \tan 80^\circ.$$
I have tried using a substitution, but nothing has worked.
|
tan 20 tan 40 tan 80 = (sin 20 sin 40 sin 80) / (cos 20 cos 40 cos 80)
Solving for numerator :
sin 20 sin 40 sin 80 = (sin 80 sin 20)(sin 60 sin 40) / (sin 60)
= (1/2)(cos 60 - cos 100)(1/2)(cos 20 - cos 100) / (sin 60) ... (product rule)
= (1/4)(1/2 + sin 10)(cos 20 + sin10) / (sin 60) ... (cos 60 = 1/2 & cos 100 = -sin 10)
= [(1/8)cos 20 + (1/8)sin 10 + (1/4)sin 10 cos 20 + (1/4) sin^2 10] / (sin 60)
= [(1/8)cos 20 + (1/8)sin 10 + (1/8)(sin -10 + sin 30) + (1/4)(1/2)(1 - cos 20)] / (sin 60)
= [(1/8(1/2) + 1/8] / (√3/2) ... (sin 30 = 1/2)
= √3/8
Solving for denominator :
sin 160
= 2sin 80 cos 80
= 4sin 40 cos 40 cos 80
= 8 sin 20 cos 20 cos 40 cos 80
But, sin 160 = sin 20
So, sin 20 = 8 sin 20 cos 20 cos 40 cos 80
Cancel sin 20. Therefore, cos 20 cos 40 cos 80 = 1/8
So, tan 20 tan 40 tan 80 = numerator/denominator = (√3/8) / (1/8) = √3 = tan 60
Thus, x = 60º. (This is just the principal value, being one of an infinite other values) `enter preformatted text here`
|
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|
Prove that $A^k = 0 $ iff $A^2 = 0$ Let $A$ be a $ 2 \times 2 $ matrix and a positive integer $k \geq 2$. Prove that $A^k = 0 $ iff $A^2 = 0$.
I can make it to do this exercise if I have $ \det (A^k) = (\det A)^k $. But this question comes before this.
Thank you very much for your help!
|
A^k=0 iff A^2=0 k positive interger >2 you can write k=2+n and n>=1. Using induction: A^2+n=0 Then A^2=0. Take A^3=A^2A=0. And prove A^2+n+1=0 =>A^2=0 and A^2+nA=0. Then A^2=0.
First direction A^2=0 then A^k=0.
You can take A and B as a matrix.. Lets represent both matrix . We have A=[a b over c d] and B=[b1 b2 over b3 b4] then we have B times A=0.
(multiplying) we have
BA= (b1a+b2c and b1b+ b2d over b3a+b4c and b3b+b4d)
It was the matrix. Then
b1a+b2c=0 b1b+ b2d=0 b3a+b4c=0 b3b+b4d=0
Where we have
(ad-bc)b1=0 (ad-bc)b2=0 (ad-bc)b3=0 and
(ad-bc)b4=0
Then we have 2 possibilities:
b1=b2=b3=b4=0 or ad-bc=0 this is B=0 or ad=bc
Obviously if A^2=0 then A^k =0 for every k bigger than 2.
A second direction A^k=0 then A^2=0 :
Now let's prove A^k=0 for k>2 with k as positive integer. Then A^2=0.
Watch first k>2, then you can write k=2+n with k positive integer n>=1.
Let's prove using induction that:
if A^2+n=0 then A^2=0
For n=1 we have A^3=A^2.A=0.
Previously we worked on matrix we can say A^2=0 or ad=bc.
Let's represent A^3 as matrix:
A^3=[(a^2+bc)a+(ab+bd)c and (a^2+bc)b+(ab+bd)d over (ca+dc)a+(bc+d^2)c and (ca+dc)b+(bc+d^2)d)]
It was the matrix.
A^3=0 then
(a^2+bc)(a+d)=0 (ab+bc)(a+d)=0
(ca+dc)(a+d)=0 (bc+d^2)(a+d)=0
Then let's represent the matrix as : A^2=[a^2+bc and ab+bd over
ca+dc and bc+d^2]=[0 and 0 over 0. And 0]
Then
a=-d but we have ad=bc then a^2+bc=0
bc+d^2=0 ab+bd=0 and ca+dc=0 then A^2=0
And prove that A^2+n+1=0 then A^2=0
If A^2+n.A=0 then A^2+n.A=0. Because previous work A^2+n=0 or ad=bc
(induction) we conclude A^2+n=0 and
Because the induction hypothesis we have A^2=0 finally we have A^k=0 then A^2=0. Done
|
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|
Linear Algebra determinant reduction Prove, without expanding, that
\begin{vmatrix}
1 &a &a^2-bc \\
1 &b &b^2-ca \\
1 &c &c^2-ab
\end{vmatrix} vanishes.
Any hints ?
|
Simply add a multiple of the first column to the last
$$\begin{vmatrix} 1 & a & a^2-bc\\1&b&b^2-ac\\1&c&c^2-ab\end{vmatrix}
=\begin{vmatrix} 1 & a & a^2-bc+(1)(ab+ac+bc)\\1&b&b^2-ac+(1)(ab+ac+bc)\\1&c&c^2-ab+(1)(ab+ac+bc)\end{vmatrix}
= \begin{vmatrix} 1 & a & a(a+b+c)\\1&b&b(a+b+c)\\1&c&c(a+b+c)\end{vmatrix}
= 0$$
The final step follows because the second and third column are linearly dependent.
|
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|
Show that the equation $2a^4+2a^2b^2+b^4=c^2$ hasn't an integer solution Question:
Prove that: if $a,b,c\in Z$,and $a\neq 0$ show that: the equation
$$2a^4+2a^2b^2+b^4=c^2$$ has no solution.
My idea:
since
$$a^4+(a^2+b^2)^2=c^2$$
and I want use Pythagorean triple
and let $$a^2=m^2-n^2. a^2+b^2=2mn, c=m^2+n^2,(m,n)=1$$
$$\Longrightarrow m^2-n^2+b^2=2mn$$
$$\Longrightarrow (m-n)^2+b^2=2n^2$$
maybe my idea is wrong?
maybe the David H idea is usefull?
since
$$2a^2(a^2+b^2)=(c-b^2)(c+b^2)$$
then $b$ and $c$ are either both even or both odd,and $c>b^2>b$
can usefull
But I can't continue.
|
Let $(a,b,c)$ be a solution of
$$ 2a^4 + 2a^2b^2 + b^4 = c^2.$$
By evaluating the different values of $a,b,c$ modulo $4$ (Edit: $8$), one can easily show that $a$ must be even. If $b$ would also be even, then $(\frac a2, \frac b2, \frac c4)$ is also a solution to the equation. Hence if $a\neq 0$ we may assume that $b$ is odd and thus $c$.
Now we use, what you already wrote about Pythagorean triples. Note that automatically
$$ a^2 = 2mn\quad\text{and}\quad a^2 + b^2 = m^2-n^2$$
because $a$ is even. We also deduce the equations
$$ 2mn +b^2 = m^2 - n^2 \iff (m+n)^2 + b^2 = 2m^2.$$
So $m$ odd because $b$ is odd. The equation is also equivialent to
$$\left(\frac{m+n-b}2\right)^2+\left(\frac{m+n+b}2\right)^2 = m^2.$$
Therefore there must be $x$ and $y$ with
$$ m= x^2+y^2,\quad n= 2y(x-y),\quad b= y^2-x^2+2xy.$$
Therefore
$$2mn = 4(x^2+y^2)y(x-y)$$
must be the square $a^2$. It should be easy to check that the three factors are pairwise coprime, thus
$$ x^2+y^2 = z^2,\quad y = w^2, \quad x-y = t^2.$$
Putting these equations together we get
$$ z^2 = (t^2 + w^2) + w^4 = 2w^4 + 2w^2t^2 + t^4.$$
Hence $(w,t,z)$ is also a solution of the above equation, but $|w|<|a|$. By infinite decent, there is no solution $(a,b,c)$ with $a\neq 0$.
(Note that one needs to pay more attention on coprimeness at a few steps which I hope is correct or can be worked around.)
|
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|
How to solve $\int \frac{x^4 + 1 }{x^6 + 1}$?
How to solve $\int \frac{x^4 + 1 }{x^6 + 1}$ ?
The numerator is a irreducible polynomial so I can't use partial fractions.
I tried the substitutions $t = x^2, t=x^4$ and for the formula $\int u\,dv = uv - \int v\,du$ I tried using: $u=\frac{x^4 + 1 }{x^6 + 1} , \,dv=\,dx \\ u=\frac{1}{x^6 + 1} , \,dv= (x^4 + 1) \,dx \\u=x^4 + 1 , \,dv=\frac{\,dx}{x^6 + 1}$
But I always get more complicated integrals.
Any hints are appreciated!
|
Hint: Decompose the denominator using $a^3+b^3=(a+b)(a^2-ab+b^2)$. Then add $\&$ subtract $x^2$ in the numerator.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/811911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
}
|
Compute $\int_0^{\pi/4}\frac{(1-x^2)\ln(1+x^2)+(1+x^2)-(1-x^2)\ln(1-x^2)}{(1-x^4)(1+x^2)} x\exp(\frac{x^2-1}{x^2+1}) dx$
Compute the following integral
\begin{equation}
\int_0^{\Large\frac{\pi}{4}}\left[\frac{(1-x^2)\ln(1+x^2)+(1+x^2)-(1-x^2)\ln(1-x^2)}{(1-x^4)(1+x^2)}\right] x\, \exp\left[\frac{x^2-1}{x^2+1}\right]\, dx
\end{equation}
I was given two integral questions by my teacher. I can answer this one although it took a lot of time to compute it. I want to share this problem to the other users here and I would love to see how Mathematics SE users compute this monster. Thank you.
|
From the numerator, collect the logarithmic terms first.
$$\displaystyle \int_0^{\pi/4} x\frac{(1+x^2)+(x^2-1)\ln\left(\frac{1-x^2}{1+x^2}\right)}{(1-x^4)(1+x^2)}\exp\left(\frac{x^2-1}{x^2+1}\right)\,dx$$
Rewrite $(1-x^4)=(1-x^2)(1+x^2)$ and divide the numerator by $(1+x^2)$.
$$\displaystyle \int_0^{\pi/4} \frac{x}{(1-x^2)(1+x^2)}\left(1+\frac{x^2-1}{x^2+1}\ln\left(\frac{1-x^2}{1+x^2}\right)\right)\exp\left(\frac{x^2-1}{x^2+1}\right)\,dx $$
Use the substitution $\displaystyle \frac{x^2-1}{x^2+1}=t \Rightarrow \frac{4x}{(1+x^2)^2}\,dx=dt$ to get:
$$\displaystyle \frac{1}{4}\int_{a}^{-1}\frac{e^t}{t}\left(1+t\ln(-t)\right)\,dt= \frac{1}{4}\int_{a}^{-1}e^t\left(\frac{1}{t}+\ln(-t)\right)\,dt$$
where $\displaystyle a=\frac{\pi^2/16-1}{\pi^2/16+1}$
Since $\displaystyle \int e^x(f'(x)+f(x))\,dx=e^xf(x)+C $, the above definite integral is:
$$\displaystyle \frac{1}{4}\left(e^t \ln(-t) \right|_{a}^{-1}=-\frac{1}{4}e^a\ln(-a) \approx \boxed{0.284007} $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/815863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "116",
"answer_count": 2,
"answer_id": 0
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|
Quick method for finding eigenvalues and eigenvectors in a symmetric $5 \times 5$ matrix? The matrix $B$:
$B =
\pmatrix{
0 & 0 & 0 & 0 & 0 \cr
0 & 8 & 0 & -8 & 0 \cr
0 & 0 & 8 & 0 & -8 \cr
0 & -8 & 0 & 8 & 0 \cr
0 & 0 & -8 & 0 & 8 \cr
}$
Which has nonzero eigenvalues $\lambda_1=16$ and $\lambda_2=16$ and corresponding eigenvectors:
v$_1 =
\pmatrix{
0\cr
\frac{1}{2} \sqrt2 \cr
0 \cr
-\frac{1}{2} \sqrt2 \cr
0\cr
}$ and v$_2 =
\pmatrix{
0\cr
0\cr
\frac{1}{2} \sqrt2 \cr
0 \cr
-\frac{1}{2} \sqrt2 \cr
}$
What is the method for obtaining these eigenvalues and corresponding eigenvectors?
It's a large matrix and I'm hoping there's some kind of easy trick to it. From what I can remember of eigen decomposition, normally I'd do:
$Ax = \lambda x \implies|A-\lambda I|x = 0$
$\implies
\det \pmatrix{
0-\lambda & 0 & 0 & 0 & 0 \cr
0 & 8-\lambda & 0 & -8 & 0 \cr
0 & 0 & 8-\lambda & 0 & -8 \cr
0 & -8 & 0 & 8-\lambda & 0 \cr
0 & 0 & -8 & 0 & 8-\lambda \cr
}$ $\pmatrix{
x_1 \cr
x_2 \cr
x_3 \cr
x_4 \cr
x_5 \cr
}$ = $\pmatrix{
0 \cr
0 \cr
0 \cr
0 \cr
0 \cr
}$
So the determinant is
$\implies -\lambda \det \pmatrix{
8-\lambda & 0 & -8 & 0 \cr
0 & 8-\lambda & 0 & -8 \cr
-8 & 0 & 8-\lambda & 0 \cr
0 & -8 & 0 & 8-\lambda \cr
}$
$\implies -\lambda * [ (8- \lambda)\det \pmatrix{
8-\lambda & 0 & -8 \cr
0 & 8-\lambda & 0 \cr
-8 & 0 & 8-\lambda \cr
}-8 \det \pmatrix{
0 & 8-\lambda & -8 \cr
-8 & 0 & 0 \cr
0 & -8 & 8-\lambda \cr
}]$
etc.
There's got to be an easier way?
|
Hint: if $A$ and $B$ are square of the same order, $$\det\begin{pmatrix} A & B \\ B & A\end{pmatrix}=\det(A-B)\det(A+B)$$
So setting
$$A=\begin{pmatrix}8-\lambda & 0 \\ 0 & 8-\lambda\end{pmatrix}$$
and
$$B=\begin{pmatrix}-8 & 0\\0&-8\end{pmatrix}$$
shows that the determinant of your $4\times 4$ matrix is $(16-\lambda)^2\lambda^2$, and therefore the determinant of your original matrix is $$\boxed{-\lambda^3(16-\lambda)^2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/816087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 1
}
|
Can One Integrate $\frac{1}{i} \int \frac{e^{ix}-e^{-ix}}{e^{ax}+e^{-ax}+e^{ix}+e^{-ix}}dx$ I'd like to integrate
$$I(a)=\int \frac{\sin(x)}{\cosh(ax)+\cos(x)}dx$$
by changing $sin$ etc... into their exponential representation.
Using $e^{ix} = \cos(x) + i \sin(x)$ and $e^{ax} = \cosh(ax)+\sinh(ax)$ we have
$$\frac{1}{i} \int \frac{e^{ix}-e^{-ix}}{e^{ax}+e^{-ax}+e^{ix}+e^{-ix}}dx =I(a)$$
If one substitutes $e^x = y$ we get $y^i$, can we do this? Can one get this method to work? A very similar integral has an answer:
http://integralsandseries.prophpbb.com/topic397.html?sid=77128c640169d5c07a9b32a5e7c35bc2
|
Consider the integral
\begin{align}
I(a) = \int \frac{\sin(x) }{ \cos(x) + \cos(ax) } \ dx.
\end{align}
This integral can be readily evaluated as is given by
\begin{align}
I(a) = \frac{1}{a^{2}-1} \ \left[ (a+1) \ln\left( \cos\left(\frac{(1-a)x}{2}\right)\right) - (a-1) \ln\left( \cos\left(\frac{(1+a)x}{2}\right)\right) \right].
\end{align}
Now letting $a \rightarrow i a$ leads to
\begin{align}
I(ia) &= \frac{-1}{a^{2}+1} \ \left[ (1+ia) \ln\left( \cos\left(\frac{(1-ia)x}{2}\right)\right) + (1-ia) \ln\left( \cos\left(\frac{(1+ia)x}{2}\right)\right) \right] \\
&= \frac{-1}{a^{2}+1} \ \left[ \ln\left(\cos\left(\frac{(1-ia)x}{2}\right) \cos\left(\frac{(1+ia)x}{2}\right) \right) + ia \ \ln\left(\frac{\cos\left(\frac{(1-ia)x}{2}\right)}{\cos\left(\frac{(1+ia)x}{2}\right)}\right) \ \right] \\
I(ia) &= \frac{-1}{a^{2}+1} \ \left[ \ln\left(\frac{\cos(x) + \cosh(ax)}{2}\right) + ia \
\ln\left( \frac{1+i \tan(x/2) \tanh(ax/2)}{1-i \tan(x/2) \tanh(ax/2)}\right) \right].
\end{align}
The second term can be reduced as follows. It is known that
\begin{align}
x+iy = \sqrt{x^{2}+y^{2}} \ e^{i \tan^{-1}(y/x)}
\end{align}
which helps lead to
\begin{align}
\ln(x+iy) = \frac{1}{2} \ \ln(x^{2}+y^{2}) + i \tan^{-1}(y/x).
\end{align}
From this result it is seen that
\begin{align}
\ln\left( \frac{1+i \tan(x/2) \tanh(ax/2)}{1-i \tan(x/2) \tanh(ax/2)}\right) = 2 i \ \tan^{-1}\left(\tan\left(\frac{x}{2}\right) \ \tanh^{-1}\left( \frac{ax}{2} \right) \right)
\end{align}
and now $I(ia)$ becomes
\begin{align}
I(ia) &= \frac{2a}{a^{2}+1} \ \tan^{-1}\left(\tan\left(\frac{x}{2}\right) \ \tanh^{-1}\left( \frac{ax}{2} \right) \right) - \frac{1}{a^{2}+1} \ \ln\left(\frac{\cos(x) + \cosh(ax)}{2}\right).
\end{align}
Hence the desired integral value is
\begin{align}
\int \frac{\sin(x) }{ \cos(x) + \cosh(ax) } \ dx = \frac{2a}{a^{2}+1} \ \tan^{-1}\left(\tan\left(\frac{x}{2}\right) \ \tanh^{-1}\left( \frac{ax}{2} \right) \right) - \frac{1}{a^{2}+1} \ \ln\left(\frac{\cos(x) + \cosh(ax)}{2}\right).
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/817519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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|
Simple algebra formula for which I can't find the right answer I have the formula $y + (z + 1) = \frac{1}{2} \cdot (z + 1) \cdot (z + 2)$, and I should work to $y = \frac{1}{2}\cdot z \cdot (z + 1)$.
Somebody showed me how it's done:
$y + (z + 1) = \frac{1}{2} \cdot (z + 1) \cdot (z + 2)$
$y + (z + 1) = \frac{1}{2} \cdot ((z + 1) \cdot (z + 2))$
$y + (z + 1) = \frac{1}{2} \cdot (z^2 + 3z + 2)$
$y + (z + 1) = \frac{1}{2}(z^2) + \frac{1}{2}(3z) + \frac{1}{2}(2)$
$y + (z + 1) = \frac{1}{2}(z^2) + 1\frac{1}{2}z + 1$
$y = \frac{1}{2}(z^2) + 1\frac{1}{2}z + 1$ - z - 1
$y = \frac{1}{2}(z^2) + \frac{1}{2}z$
$y = \frac{1}{2}z(z + 1)$
Great! But, my try went completely wrong, and I don't understand what I'm doing wrong:
$y + (z + 1) = \frac{1}{2} \cdot (z + 1) \cdot (z + 2)$
$y + (z + 1) = \frac{1}{2} \cdot ((z + 1) \cdot (z + 2))$
$y + (z + 1) = \frac{1}{2} \cdot (z^2 + 3z + 2)$
$y = \frac{1}{2} \cdot z^2 + 3z + 2 - z - 1$
$y = \frac{1}{2} \cdot z^2 + 2z + 1$
$y = \frac{1}{2} \cdot (z^2 + 2z + 1)$
$y = \frac{1}{2}(z^2) + \frac{1}{2}(2z) + \frac{1}{2}1$
$y = \frac{1}{2}(z^2) + z + \frac{1}{2}$
But from this last step, I can't get anywhere near $y = \frac{1}{2}z(z + 1)$, and I do not understand what I did wrong.
|
There is an error in the fourth line of your second set of equations:-
$$y = \frac{1}{2} \cdot z^2 + 3z + 2 - z - 1$$
should be
$$y = \frac{1}{2} \cdot (z^2 + 3z + 2 \color{red}{- 2z - 2})\\\Rightarrow y = \frac{1}{2} \cdot (z^2 + z)=\frac{1}{2}\cdot z(z+1)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/818475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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|
Linear algebra, power of matrices $P^{-1}AP =
\begin{pmatrix}
-1 & 1 & 0 & 0 \\
0 & -1 & 1 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & 2 \\
\end{pmatrix}
$
with
$P=
\begin{pmatrix}
-1 & 1 & 0 & 0 \\
0 & -1 & 1 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}
$
and $P^{-1}$ is the inverse of $P$
Find $A$, $A^{100}$?
I found $A$, $P^{-1}AP=B$ and multiplied by $P$ and $P^{-1}$. How can I find $A^{100}$? I have to use eigenvalues/vectors.
|
Because of the way that $P$ and $A$ are constructed, it is not hard to see that $PA=AP$. Consequently, $P^{-1}AP=P^{-1}PA=A$. So that makes life easier. You can then write
$$
A =
\left[\begin{array}{cccc}
-1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & 2
\end{array}\right]+
\left[\begin{array}{cccc}
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{array}\right] = D+N,
$$
where $D$ is the diagonal matrix and $N$ is nilpotent of order $3$ (meaning that $N^{3}=0$, but $N^{2} \ne 0$.) You have $DN=ND=-N$ and
$$
N^{2} = \left[\begin{array}{cccc}
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{array}\right],\;\;\; N^{3}=0.
$$
This is how the Jordan canonical form works. $N$ works on the same part of the space corresponding to the columns with $-1$ entries. So $DN=ND=-N$. And $DN^{2}=N^{2}D=-N^{2}$. So,
$$
(D+N)^{100} = D^{100}+(100)D^{99}N+(50)(99)D^{98}N^{2}=D^{100}-100N+50(99)N^{2}.
$$
You can write down the matrix $A^{100}$ from this. Nothing further is needed.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/819222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
$\sum_{k=1}^{\infty} \frac{a_1 a_2 \cdots a_{k-1}}{(x+a_1) \cdots (x+a_k)}$ Hey guys I was reading Alfred van der Poortens paper regarding Apery's constant and I came across this pretty equality. For all $a_1, a_2, \ldots$
\begin{align}
\large\sum_{k=1}^{\infty}\frac{a_1a_2\cdots a_{k-1}}{(x+a_1)\cdots (x+a_k)} = \frac 1x
\end{align}
I tried fiddling around with the the summands by trying to use a partial fraction decomposition but it just got messy.
link to paper: http://maths.mq.edu.au/~alf/Humid%20Summer/45.pdf
|
$$(x+a_1)(x+a_2)(x+a_3)(x+a_4)(x+a_5)... =A(x)\tag 1$$
$$\frac{A(x)}{A(x)}=1=x\frac{(x+a_2)(x+a_3)(x+a_4)(x+a_5)...}{(x+a_1)(x+a_2)(x+a_3)(x+a_4)(x+a_5)... }+a_1\frac{(x+a_2)(x+a_3)(x+a_4)(x+a_5)...}{(x+a_1)(x+a_2)(x+a_3)(x+a_4)(x+a_5)... } \tag 2$$
$$1=\frac{x}{(x+a_1)}+a_1\frac{(x+a_2)(x+a_3)(x+a_4)(x+a_5)...}{(x+a_1)(x+a_2)(x+a_3)(x+a_4)(x+a_5)... } \tag 3$$
$$1=\frac{x}{(x+a_1)}+a_1\frac{x(x+a_3)(x+a_4)(x+a_5)...+a_2(x+a_3)(x+a_4)(x+a_5)...}{(x+a_1)(x+a_2)(x+a_3)(x+a_4)(x+a_5)... } \tag 4$$
$$1=\frac{x}{(x+a_1)}+\frac{a_1x}{(x+a_1)(x+a_2)}+a_1a_2\frac{(x+a_3)(x+a_4)(x+a_5)...}{(x+a_1)(x+a_2)(x+a_3)(x+a_4)(x+a_5)... } \tag 5$$
$$1=\frac{x}{(x+a_1)}+\frac{a_1x}{(x+a_1)(x+a_2)}+a_1a_2\frac{x(x+a_4)(x+a_5)...}{(x1+a_1)(x+a_2)(x+a_3)(x+a_4)(x+a_5)... }+a_1a_2\frac{a_3(x+a_4)(x+a_5)...}{(x+a_1)(x+a_2)(x+a_3)(x+a_4)(x+a_5)... } \tag 6$$
$$1=\frac{x}{(x+a_1)}+\frac{a_1x}{(x+a_1)(x+a_2)}+\frac{a_1a_2x}{(x+a_1)(x+a_2)(x+a_3)}+a_1a_2a_3\frac{(x+a_4)(x+a_5)...}{(x+a_1)(x+a_2)(x+a_3)(x+a_4)(x+a_5)... } \tag 7$$
If we continue in that way to create series then the last term will be :
$$\frac{a_1a_2a_3a_4a_5....}{x(x+a_1)(x+a_2)(x+a_3)(x+a_4)(x+a_5)... }\tag 8$$
The last term must go to zero otherwise the relation I wrote above must be as shown below
$$\frac{1}{x}-\frac{\prod\limits_{j=1}^{\infty}a_j}{x\prod\limits_{j=1}^{\infty} (x+a_j)}=\frac{1}{x+a_1}+\frac{a_1}{(x+a_1)(x+a_2)}+\frac{a_1a_2}{(x+a_1)(x+a_2)(x+a_3)}+\frac{a_1a_2a_3}{(x+a_1)(x+a_2)(x+a_3)(x+a_4)}+... \tag 9$$
$$\frac{1}{x}-\frac{\prod\limits_{j=1}^{\infty}a_j}{x\prod\limits_{j=1}^{\infty} (x+a_j)}=\frac{1}{x+a_1}+\sum\limits_{k=1}^ \infty\frac{\prod\limits_{j=1}^{k}a_j}{\prod\limits_{j=1}^{k+1} (x+a_j)} \tag {10}$$
If $\frac{\prod\limits_{j=1}^{\infty}a_j}{x\prod\limits_{j=1}^{\infty} (x+a_j)}=0$ then the relation will be
$$\frac{1}{x}=\frac{1}{x+a_1}+\sum\limits_{k=1}^ \infty\frac{\prod\limits_{j=1}^{k}a_j}{\prod\limits_{j=1}^{k+1} (x+a_j)} \tag {11}$$
Also The general formula can be written for n terms.
$$\frac{1}{x}-\frac{\prod\limits_{j=1}^{n}a_j}{x\prod\limits_{j=1}^{n} (x+a_j)}=\frac{1}{x+a_1}+\sum\limits_{k=1}^ {n-1} \frac{\prod\limits_{j=1}^{k}a_j}{\prod\limits_{j=1}^{k+1} (x+a_j)} \tag {12}$$
I had similiar question about that relation for $x=1$. If you want to see my question and answer, the link
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/821338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
Stuck on finding the value of $\sum_{n=0}^\infty {n(n+1) \over 3^n}$ I am trying to find the value of the series
$$
\sum_{n=0}^\infty {n(n+1) \over 3^n}
$$
Here's what I have done so far:
$$
\sum_{n=0}^\infty {n(n+1) \over 3^n}=\sum_{n=0}^\infty {n^2 \over 3^n}+\sum_{n=0}^\infty {n \over 3^n}
$$
Determining the values of the both terms separately:
\begin{eqnarray}
S_1 = \sum_{n=0}^\infty {n \over 3^n} = {1 \over 3^1}+{2 \over 3^2}+{3 \over 3^3}+ \dots \\
= {1 \over 3^1} + {1 \over 3^2} + {1 \over 3^3} + \dots \\ +{1 \over 3^2} + {1 \over 3^3} + \dots \\ +{1 \over 3^3} + \dots \\
= {1 \over 3^1} + {1 \over 3^2} + {1 \over 3^3} + \dots \\
+{1 \over 3^1}\left( {1 \over 3^1}+{1 \over 3^2}+\dots\right) \\
+ {1 \over 3^2}\left( {1 \over 3^1}+\dots\right)\\
+ \dots\\
= \sum_{n=1}^\infty{1 \over 3^n} \cdot \sum_{n=0}^\infty {1 \over 3^n}
\end{eqnarray}
Next evaluating $\sum_{n=1}^\infty{1 \over 3^n}$:
\begin{align*}
3S_1 - S_1 = \sum_{n=1}^\infty{1 \over 3^{n-1}} - \sum_{n=1}^\infty{1 \over 3^n} = \\
1 - \require{enclose}\enclose{updiagonalstrike}{1 \over 3^1} + \enclose{updiagonalstrike}{1 \over 3^1} - \enclose{updiagonalstrike}{1 \over 3^2} + \enclose{updiagonalstrike}{1 \over 3^2} - \enclose{updiagonalstrike}{1 \over 3^3} + \enclose{updiagonalstrike}{1 \over 3^3} - \dots - {1 \over 3^n} \\
= 1-{1 \over 3^n} = 2S_1;
\end{align*}
$$
S_1=\sum_{n=1}^\infty{1 \over 3^n} = \frac 12 \lim_{n \to \infty} \left( 1-{1 \over 3^n} \right) = \frac 12
$$
And thus we receive
$$
\sum_{n=1}^\infty{n \over 3^n} = \frac 14;
$$
I was trying to evaluate the second term $\sum_{n=0}^\infty {n^2 \over 3^n}$, however it wasn't a success. Could you give me a hint, how I can proceed with this series?
|
$|x|<1:\\\sum_{n=0}^{\infty }x^{n}=\frac{1}{1-x}\Rightarrow 1+\sum_{n=1}^{\infty }x^{n}=\frac{1}{1-x}\\\Rightarrow \sum_{n=1}^{\infty }x^{n}=\frac{x}{1-x}\\\frac{d}{dx}\sum_{n=0}^{\infty }x^{n}=\frac{d}{dx}\frac{1}{1-x}\Rightarrow \sum_{n=1}^{\infty }nx^{n-1}=\frac{1}{(1-x)^2}\Rightarrow \sum_{n=1}^{\infty }nx^{n}=\frac{x}{(1-x)^2}\\\sum_{n=1}^{\infty }(n+1)x^n=\frac{x}{(1-x)}+\frac{x}{(1-x)^2}\\$
$
\begin{array}{l}
\left( {\sum\limits_{n = 0}^{ + \infty } {\left( {n + 1} \right)x^n } } \right)^\prime =\sum\limits_{n = 1}^{ + \infty } {n\left( {n + 1} \right)x^{n - 1} } = \frac{1}{{\left( {1 - x} \right)^2 }} + \frac{{1 + x}}{{\left( {1 - x} \right)^3 }} \\
\Rightarrow \sum\limits_{n = 0}^{ + \infty } {n\left( {n + 1} \right)x^n } = \frac{x}{{\left( {1 - x} \right)^2 }} + \frac{{x^2 + x}}{{\left( {1 - x} \right)^3 }} \\
x = \frac{1}{3} \Rightarrow \sum\limits_{n = 0}^{ + \infty } {n\left( {n + 1} \right)\left( {\frac{1}{3}} \right)^n } = \frac{{\left( {\frac{1}{3}} \right)}}{{\left( {1 - \left( {\frac{1}{3}} \right)} \right)^2 }} + \frac{{\left( {\frac{1}{3}} \right)^2 + \left( {\frac{1}{3}} \right)}}{{\left( {1 - \left( {\frac{1}{3}} \right)} \right)^3 }} \\
\Rightarrow \sum\limits_{n = 0}^{ + \infty } {n\left( {n + 1} \right)\left( {\frac{1}{3}} \right)^n } = \frac{9}{4} \\
\end{array}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/822059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Can $A, B$ fail to commute if $e^A=e^B=e^{A+B}=id$? Consider the real $n \times n$-matrices $A$ and $B$.
Can $A, B$ fail to commute if $e^A=e^B=e^{A+B}=id$ ?
|
Yes, $A$ and $B$ can fail to commute. Consider
$$
A = \left[\begin{array}{@{}rrr@{}}
0 & 1 & 0 \\
-1 & 0 & 0 \\
0 & 0 & 0
\end{array}\right],\qquad
B = \left[\begin{array}{@{}ccc@{}}
0 & -\tfrac{1}{2} & \tfrac{\sqrt{3}}{2} \\
\tfrac{1}{2} & 0 & 0 \\
-\tfrac{\sqrt{3}}{2} & 0 & 0
\end{array}\right],
$$
so that
$$
A + B = \left[\begin{array}{@{}ccc@{}}
0 & \tfrac{1}{2} & \tfrac{\sqrt{3}}{2} \\
-\tfrac{1}{2} & 0 & 0 \\
-\tfrac{\sqrt{3}}{2} & 0 & 0
\end{array}\right].
$$
Since each of $A$, $B$, and $A + B$ has Frobenius norm equal to $\sqrt{2}$, we have
$$
\exp(2\pi A) = \exp(2\pi B) = \exp\bigl(2\pi (A + B)\bigr) = I.
$$
However, it's easy to check $AB$ is not symmetric, so
$$
BA = (-B^{T})(-A^{T}) = B^{T}A^{T} = (AB)^{T} \neq AB.
$$
It follows, of course, that $2\pi A$ and $2\pi B$ don't commute, either.
|
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|
Two ways to show that $\sin x -x +\frac {x^3}{3!}-\frac {x^5}{5!}< 0$ Show that: $\large \sin x -x +\frac {x^3}{3!}-\frac {x^5}{5!}< 0$ on: $0<x<\frac {\pi}2$
I tried to solve it in two ways and got a little stuck:
One way is to use Cauchy's MVT, define $f,g$ such that $f(x)=\sin x -x +\frac {x^3}{3!}$ and $g(x)=\frac {x^5}{5!}$ so $\frac {f(x)}{g(x)}<1$ and both are continuous and differentiable on $x\in (0,\frac{\pi}2)$ so
$$\large \frac {f(x)}{g(x)}=\frac{f(x)-f(0)}{g(x)-g(0)}=\frac {f'(c)}{g'(c)}\Rightarrow
\frac{\sin x -x +\frac {x^3}{3!}} {\frac {x^5}{5!}}=\frac {\cos y -1 +\frac{y^2}{2}}{\frac{y^4}{24}}$$
Such that $y\in (0,x)$.
Now what ? I need to show that $\frac {f'(y)}{g'(y)}<1$ but I tried to input $\frac{\pi} 2$ as a value to $f'/g'$ but it was larger than 1.
The other way: With Taylor expansion.
$\large \sin x=x-\frac {x^3}{3!}+\frac {x^5}{5!}+R_{5,0}(x)\Rightarrow\\
\large \sin x-x+\frac {x^3}{3!}-\frac {x^5}{5!}=R_{5,0}(x)$
Now we're left with showing that $R_{5,0}(x)<0$
So $\large R_{5,0}(x)=\frac{-\sin(c)c^7}{7!}$ for $c\in (0,x)$, now can I say that because it's negative and $(\sin(c)c^7>0)$ for all $c\in(0,x)$, $R_{5,0}(x)<0$ ?
Also, would it be correct to say that: $\large R_{5,0}(x)=\frac{cos(c)c^6}{6!}$?
|
Using the Mean value theorem is the right idea. You need the MVT to prove the following useful lemma:
If $f'(x)>g'(x)$ for all $a\ge x>0$ and $f(0)=g(0)$, then $f(x)>g(x)$ for all $a\ge x>0$.
Letting $g(x)=\sin(x)$ and $f(x)=x-\frac{x^3}{3!} +\frac{x^5}{5!}$ and then iterating this lemma will give you your answer.
Proof of the lemma:
Suppose we are given $f$ and $g$ such that $f'(x)>g'(x)$ for all $a\ge x>0$ and $f(0)=g(0)$. Let $h(x)=f(x)-g(x)$, then $h(0)=0$ and $h'(x)>0$. By MVT there is an $x_0\in [0,x]$, for any $0<x\le a$, such that
$$h'(x_0)=\frac{f(x)-g(x)-(f(0)-g(0))}{x}=\frac{f(x)-g(x)}{x}>0.$$
Since $x>0$, we know $f(x)-g(x)>0$ for any $x\in [0,a]$.
|
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|
Writing $2$ roots of a cubic in terms of the third root Let $\theta$ be a root of $x^3-3x+1$. Since the discriminant is a square, the splitting field of this polynomial is just $\mathbb Q(\theta)$. Now, I want to write the other roots as linear combinations of $1$, $\theta$, and $\theta^2$.. If we let $\alpha$ and $\beta$ be other roots, then $x^3-3x+1=(x-\theta)(x-\alpha)(x-\beta)$, from which we get that $\alpha+\beta=-\theta$ and $\alpha\beta=\theta^2-3$. Thus, $\alpha$ and $\beta$ are the roots of the quadratic $x^2+\theta x+(\theta^2-3)=0$. The discriminant of this quadratic is $12-3\theta^2$, so I'll be done if I can find rational $a, b$, and $c$ such that $(a+b\theta+c\theta^2)^2=12-3\theta^2$. But when I set up the equations, there was no obvious way to solve them, and even Wolfram Alpha ran out of computation time. Is there a less bashy way of doing this problem, or am I missing a trick in solving the resulting system of equations?
|
$(a+b\theta+c\theta^2)^2 = 12-3\theta^2$ and $\theta^3 = 3\theta-1$ (from the initial polynomial, and the fact that plugging in $\theta$ will give zero).
Multiplying out gives
$\begin{align*}a^2+2ab\theta+(2ac+b^2)\theta^2+2bc\theta^3+c^2\theta^4 & = a^2+2ab\theta+(2ac+b^2)\theta^2+2bc(3\theta-1)+c^2(3\theta-1)\theta \\ & = a^2-2bc + (2ab+6bc-c^2)\theta+(2ac+b^2+3c^2)\theta^2\end{align*}$
From here, equate coefficients. $a^2-2bc = 12, 2ab+6bc-c^2=0, 2ac+b^2+3c^2 = -3$
Wolframalpha comes up with $a=4,b=-1,c=-2$ or $a=-4,b=1,c=2$. Checking these out, both solutions satisfy the system of equations.
|
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|
Help with Evaluating a Logarithm A precalculus text asks us to evaluate $\log_{8}\dfrac{\sqrt{2}\cdot\sqrt[3]{256}}{\sqrt[6]{32}}$
I do the following:
$\log_{8}\dfrac{\sqrt{2}\cdot\sqrt[3]{(2^2)^3\cdot 2^2}}{\sqrt[6]{2^3\cdot 2^2}}$
$\equiv \log_{8}\dfrac{\sqrt{2}\cdot 2^2\cdot\sqrt[3]{2^2}}{\sqrt{2}\cdot\sqrt[6]{2^2}}$
$\equiv \log_{8}\dfrac{2^2\cdot\sqrt[3]{2^2}}{\sqrt[3]{2}}$
and then I'm stumped.
Hints?
|
$$ \log_{8}\left(\frac{2^2\cdot\sqrt[3]{2^2}}{\sqrt[3]{2}}\right) = \log_8 \left(2^2\cdot \sqrt[3]2\right) = \log_8 2^{7/3} = \frac 73 \log_8(2)= \frac 73 \cdot \frac 13 = \frac 79$$
$$8^{1/3} = 2 \iff \log_8 2 = \frac 13$$
|
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|
Help Evaluating $\lim_{x \to \infty} \sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x}$ Does anyone know how to evaluate the following limit?
$$
\lim_{x \to \infty} \sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x}
$$
The answer is $\frac{1}{2}$, but I want to see a step by step solution if possible.
|
$$\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}=\frac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}=\frac{\sqrt{1+\frac{1}{\sqrt{x}}}}{\sqrt{1+\sqrt{\frac{1}{x}+\frac{1}{x^{\frac{3}{2}}}}}+1}$$
I got the above by first multiplying by the conjugate to $\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}$ then dividing both the numerator and denominator by $\frac{1}{\sqrt{x}}$. Now take $x$ to infinity.
|
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|
How does Wolfram get from the first form to the second alternate form? So, I was trying to compute an integral but I couldn't actually manage getting anywhere with it in its initial form. So, I inserted the function in Wolfram Alpha and I really got a nicer form (second alternate form). But I want to understand how that was done, I really need some help here.
Input:
$\dfrac{1}{(e^t + 1)(e^t + 2)(e^t + 3)}$
Alternate forms:
$\dfrac{1}{11e^t + 6e^{2t} + e^{3t} + 6}\\
-\dfrac{1}{e^t + 2} + \dfrac{1}{2(e^t + 3)} + \dfrac{1}{2(e^t + 1)}
$
|
$\dfrac{1}{(e^t + 1)(e^t + 2)(e^t + 3)} = \dfrac{1}{(x + 1)(x + 2)(x + 3)}$, where $x = e^t$.
Let $\dfrac{1}{(x + 1)(x + 2)(x + 3)} = \dfrac{A}{x + 1} + \dfrac{B}{x + 2} + \dfrac{C}{x + 3}$.
Then $A(x + 2)(x + 3) + B(x + 1)(x + 3) + C(x + 1)(x + 2) = 1$.
For $x = -1$:
$A(-1 + 2)(-1 + 3) = 1 \Rightarrow 2A = 1 \Rightarrow A = 1/2$
For $x = -2$:
$B(-2 + 1)(-2 + 3) = 1 \Rightarrow -B = 1 \Rightarrow B = -1$
For $x = -3$:
$C(-3 + 1)(-3 + 2) = 1 \Rightarrow 2C = 1 \Rightarrow C = 1/2$
$$\boxed{\dfrac{1}{(e^t + 1)(e^t + 2)(e^t + 3)} = \dfrac{1}{2(e^t + 1)} - \dfrac{1}{e^t + 2} + \dfrac{1}{2(e^t + 3)}}$$
Alternative Method
Now, I'll tell you an easier method (which has the same logic as above, and only looks different). Using this method, you can just look at the fraction and write down the partial fractions right away.
Write:
$\dfrac{1}{(x + 1)(x + 2)(x + 3)} = \dfrac{}{x + 1} + \dfrac{}{x + 2} + \dfrac{}{x + 3}$
Ignore (or cover with your finger) the first factor in the denominator:
$\dfrac{1}{\boxed{\color{white}{(x + 1)}}(x + 2)(x + 3)}$
Substitute $x = -1$: $\dfrac{1}{(-1 + 2)(-1 + 3)} = \dfrac{1}{2}$.
Write this down as the coefficient of the first fraction:
$\dfrac{1}{(x + 1)(x + 2)(x + 3)} = \dfrac{1}{2(x + 1)} + \dfrac{}{x + 2} + \dfrac{}{x + 3}$
Similarly for the second factor: $\dfrac{1}{(x + 1)\boxed{\color{white}{(x + 2)}}(x + 3)}$, with $x = -3$ gives $\dfrac{1}{(-2 + 1)(-2 + 3)} = -1$.
And for the third factor: $\dfrac{1}{(x + 1)(x + 2)\boxed{\color{white}{(x + 3)}}}$, with $x = -2$ gives $\dfrac{1}{(-3 + 1)(-3 + 2)} = \dfrac{1}{2}$.
In this manner you get $\boxed{\dfrac{1}{(x + 1)(x + 2)(x + 3)} = \dfrac{1}{2(x + 1)} + \dfrac{-1}{x + 2} + \dfrac{1}{2(x + 3)}}$ in one step.
|
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|
A function satisfying $f(\frac1{x+1})\cdot x=f(x)-1$ and $f(1)=1$? $f:[0,\infty)\to\mathbb{R}$ is a continuous function which satisfies $f(1)=1$ and:
$$f(\frac1{x+1})\cdot x=f(x)-1$$
Does there exist such a function, if they do, are there infinitely many? And is there any explicit example? I have been able to derive these:
*
*$f(1)\cdot0=f(0)-1$, which means $f(0)=1$.
*$f(\frac1{2})=f(1)-1=0$
*Similarly, $f(\frac23)=-2$ and then $f(\frac35)=\frac{-9}2$ and so on.
This is going on without any pattern I can see. Is there anything special in the sequence $(1,\frac12,\frac23,...,a_n,...)$ where $a_n=\frac1{a_{n-1}+1}$? I have not been able to use continuity as well. Please help!
|
Such a function doesn't exist.
Forget $f(1) = 1$. Just assume $f(x)$ satisfy the functional equation on $[0,\infty)$. We have
$$
\begin{align}
f(x) &= 1 + x f\left(\frac{1}{x+1}\right)\\
&= 1 + x \left[ 1 + \frac{1}{x+1}f\left(\frac{x+1}{x+2}\right)\right]\\
&= 1 + x \left[ 1 + \frac{1}{x+1} + \frac{1}{x+2} f\left(\frac{x+2}{2x+3}\right) \right]\\
&\;\vdots\\
&= 1 + x \left[ 1 + \left( \sum_{k=2}^{p-1} \frac{1}{F_{k-1} x + F_k} \right) + \frac{1}{F_{p-1} x + F_p}f\left(\frac{F_{p-1} x + F_p}{F_p x + F_{p+1}}\right)\right]\\
&\;\vdots
\end{align}
$$
where $F_k$ is the $k^{th}$ Fibonacci number. Notice
$$\lim_{p\to\infty} \frac{F_{p+1}}{F_p} = \phi = \frac{1+\sqrt{5}}{2}
\quad\implies\quad
\lim_{p\to\infty} \frac{F_{p-1} x + F_p}{F_p x + F_{p+1}} = \frac{1}{\phi}
$$
If we further assume $f(x)$ is continuous at a single point $\displaystyle\;\frac{1}{\phi}$,
the last expression converges as $p \to \infty$ and we get:
$$f(x) = 1 + x + \sum_{k=2}^\infty \frac{x}{F_{k-1} x + F_k}$$
Substitute $x$ by $1$, we get
$$f(1) = 1 + 1 + \sum_{k=2}^\infty\frac{1}{F_{k-1}+F_k} = \frac{1}{F_1} + \frac{1}{F_2} + \sum_{k=3}^\infty \frac{1}{F_k}
= \sum_{k=1}^\infty \frac{1}{F_k}$$
The number at the RHS is known as the
Reciprocal Fibonacci constant
$\psi$ with a value $$\approx 3.359885666243177553172011302918927179688905133731$$
differs from $1$.
|
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|
Help in Solving a Trigonometric Equation Solve the equation $$\left(\sin x + \cos x\right)^{1+\sin(2x)} = 2$$ when $-\pi \le x \le \pi $ .
I have tried to use $\sin (2x) = 2\sin x \cos x$ identity but I this doesn't lead me to a conclusion.
I will appreciate the help.
|
To solve $(\sin x+\cos x)^{1+\sin 2x}=2$ when $−\pi\leq x\leq \pi$.
Let $z=\sin x+\cos x$, then $z^2=1+\sin 2x$. Hence the equation is equivalent to $z^{z^2}=2$ which has the only possible solution $z=\sqrt{2}$.
Hence $\sin x+\cos x=\sqrt{2}\sin(x+\frac{\pi}{4})=\sqrt{2}$.
Thus $\sin(x+\frac{\pi}{4})=1$, which implies $x+\frac{\pi}{4}=\frac{\pi}{2}$. Hence $x=\frac{\pi}4$.
Or using $z^2=1+\sin 2x$ with $z=\sqrt{2}$ gives $1+\sin 2x=2$
Hence $\sin 2x=1$ implies $2x=\frac{\pi}{2}$. Hence $x=\frac{\pi}4$.
|
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|
Range of the function $f(x) = \frac{x^2+14x+9}{x^2+2x+3}\;,$ where $x\in \mathbb{R}$ Calculation of Range of the function $\displaystyle f(x) = \frac{x^2+14x+9}{x^2+2x+3}\;,$ where $x\in \mathbb{R}$
(Can we solve it Using $\bf{A.M\geq G.M}$) Inequality.
$\bf{My\; Try::}$ Let $\displaystyle y = f(x) = \frac{x^2+14x+9}{x^2+2x+3} = \frac{(x+1)^2+12(x+1)-4}{(x+1)^2+1}$
Now Let $(x+1) = t\;,$ where $t\in \mathbb{R}$
So $\displaystyle y=\frac{t^2+12t-9}{t^2+1} = 1+\frac{12t-10}{t^2+1}$
Now I did not understand How can I solve after that
Help me
Thanks
|
Hint:
try to find a line $y=k$ such that
$$
\frac{x^2+14x+9}{x^2+2x+3} - k = 0
$$
has only one solution, that $k$ is a max (above the max there is no solution, below there are two solutions) or a min (below the min there is no solution, abore there are two solutions).
Thus
$$
\frac{[1-k]x^2+2[7-k]x+[9-3k]}{x^2+2x+3} = 0
$$
Set discriminant to $0$ and find $k$...
So we get$$[7-k]^2 - [1-k][0-3k] = 0,$$ whence $$-2k^2-2k+40=0,$$ so $$k^2+k-20=0,$$ thus $$[k+5][k-4]=0,$$ so the min is $-5$ and the max is $+4$...
|
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|
Problem of quadratic equation If $\alpha$ be a root of $ 4x^2 +2x -1 = 0 $ , prove that the other root is $4\alpha^3 - 3\alpha$ . I have tried to do it but of no success.[$4\alpha^3 -2\alpha$ = $\dfrac {-1}{2}$ and $4\alpha^4 - 3\alpha^2$ = $\dfrac {-1}{4}$ ] .How to prove it?
|
If $\alpha$ is a root then
$$
4 \alpha^2 + 2 \alpha - 1 = 0.
$$
Therefore
$$
\begin{eqnarray}
4 \Big( -\tfrac{1}{2} - \alpha \Big)^2 + 2 \Big( -\tfrac{1}{2} - \alpha \Big) - 1 &=&
4 \Big( \tfrac{1}{4} + \alpha + \alpha^2 \Big)
- 2 \Big( \tfrac{1}{2} + \alpha \Big) - 1\\
&=& 4 \alpha^2 + 2 \alpha - 1\\
&=& 0.
\end{eqnarray}
$$
So if $\alpha$ is a root, then $- \tfrac{1}{2} - \alpha$ is also a root, but
$$
\begin{eqnarray}
4 \alpha^3 - 3 \alpha &=& \alpha \Big( 4\alpha^2 - 3 \Big)\\
&=& \alpha \Big( 1 - 2 \alpha - 3 \Big)\\
&=& - \Big( 2 \alpha^2 + 2 \alpha \Big)\\
&=& - \Big( \frac{1}{2} - \alpha + 2 \alpha \Big)\\
&=& - \frac{1}{2} - \alpha,
\end{eqnarray}
$$
So $ 4 \alpha^3 - 3\alpha$ is also a root.
|
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|
Algebraic inequality:$\frac a{b^2} + \frac b{c^2} +\frac c{a^2} \geq \frac1a+\frac1b+ \frac1c$ Prove that :
If a,b,c $\in \mathbb{R^+}$
$$\frac a{b^2} + \frac b{c^2} +\frac c{a^2} \geq \frac1a+\frac1b+ \frac1c$$
My attempt :
We know that the sequence {a,b,c} and {$\frac1{a^2},\frac1{b^2},\frac1{c^2}$} are oppositely ordered thus from rearrenegement inequality we conclude -
$$\frac a{b^2} + \frac b{c^2} +\frac c{a^2} \geq \frac{a}{a^2}+\frac{b}{b^2}+ \frac{c}{c^2}$$
Is this correct?
|
$\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(\dfrac{a}{b^{2}}+\dfrac{b}{c^{2}}+\dfrac{c}{a^{2}}\right)\geq\left(\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{a}\right)^{2}$ : Cauchy-Schwarz
$\therefore$ $\dfrac{a}{b^{2}}+\dfrac{b}{c^{2}}+\dfrac{c}{a^{2}}\geq \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$
|
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|
using residue for integration Hi how do u calculate the integral which have square root ? for example for this integral (because of branches points I always baffle) :
$$\int_0^1 \frac{(1-x)^{1/4}\, x^{3/4}}{5-x}\, dx$$
|
Consider $$f(z) = \frac{z^{3/4}(z-1)^{1/4}}{5-z} = \frac{|z|^{3/4}e^{3i/4 \arg(z)} |z-1|^{1/4} e^{i/4 \arg(z-1)} }{5-z}$$
where $ 0 \le \arg(z), \arg(z-1) < 2 \pi$.
By omitting the line segment $[0,1]$, $f(z)$ is well-defined on the complex plane and real-valued on the real axis for $x > 1$.
Then integrating clockwise around a dog-bone contour,
$$ \int_{0}^{1} \frac{x^{3/4} e^{3i/4 (0)} (1-x) e^{i/4(\pi)}}{5-x} \ dx + \int_{1}^{0} \frac{x^{3/4} e^{3i/4(2 \pi)}(1-x)^{1/4} e^{i/4(\pi)}}{5-x} \ dx$$
$$ = e^{i \pi /4}\int_{0}^{1} \frac{x^{3/4}(1-x)^{1/4}}{5-x} \ dx- e^{- i \pi /4}\int_{0}^{1} \frac{x^{3/4}(1-x)^{1/4}}{5-x} \ dx$$
$$ = \frac{2 i}{\sqrt{2}} \int_{0}^{1} \frac{x^{3/4}(1-x)^{1/4}}{5-x} \ dx$$
$$ = 2 \pi i \Big(\text{Res}[f(z),5] + \text{Res}[f(z),\infty] \Big)$$
where
$$ \begin{align} \text{Res}[f(z), 5] &= - \lim_{z \to 5} \Big( |z|^{3/4}e^{3i/4 \arg(z)} |z-1|^{1/4} e^{i/4 \arg(z-1)} \Big) \\ &= - 5^{3/4} \sqrt{2} \end{align}$$
and
$$ \text{Res}[f(z), \infty] = \frac{19}{4}$$
since when we expand $f(z)$ in a Laurent series at $\infty$ we get
$$\begin{align} f(z) &= \frac{z^{3/4} z^{1/4} (1-\frac{1}{z})^{1/4}}{-z(1-\frac{5}{z})} \\ &= -\frac{(1-\frac{1}{z})^{1/4}}{1-\frac{5}{z}} \\ &= -\left(1- \frac{1}{4z} + \ldots \right)\left(1+\frac{5}{z} + \ldots\right) \\ &= - 1 - \frac{19}{4z} + \ldots \end{align}$$
Therefore,
$$ \frac{2i}{\sqrt{2}} \int_{0}^{1} \frac{x^{3/4}(1-x)^{1/4}}{5-x} \ dx = 2 \pi i \left(-5^{3/4} \sqrt{2} + \frac{19}{4} \right) $$
which implies
$$\begin{align} \int_{0}^{1} \frac{x^{3/4}(1-x)^{1/4}}{5-x} \ dx &= \pi \sqrt{2}\left(-5^{3/4} \sqrt{2} + \frac{19}{4} \right) \\ &\approx 0.0945976635 \end{align}$$
|
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}
|
What's the integral of $\int_0^\infty \frac{dx}{(x^4+1)^5}$ $$\int_0^\infty \frac{dx}{(x^4+1)^5}$$
My answer would be : $\dfrac{\Gamma(\tfrac{1}{4})\Gamma(\tfrac{19}{4})}{4\Gamma(5)}$
Solution: You can use this technique. – Mhenni Benghorbal
|
Let $x = \sqrt{\tan \theta}$. Then, $dx = \dfrac{1}{2\sqrt{\tan \theta}}\cdot \sec^2 \theta d\theta =\dfrac{d\theta}{2\sin^{1/2}\theta\cos^{3/2} \theta}$. If $x = 0$, $\theta = 0$ and if $x \to \infty \ \Rightarrow \ \theta = \pi/2$.
$I =\int_{0}^{\infty}\dfrac{dx}{(x^4 + 1)^5} = \dfrac{1}{2}\int_{0}^{\pi/2}\dfrac{\cos^{10}\theta d\theta}{\sin^{1/2}\theta\cos^{3/2}\theta} = \dfrac{1}{2}\int_{0}^{\pi/2}\sin^{-1/2}\theta\cos^{17/2}\theta d\theta$
Beta function: $B(x,y) = 2\int_{0}^{\pi/2}\sin^{2x - 1}\theta \cos^{2y - 1}\theta d\theta $
Thus, $x = 1/4$ and $y = 19/4$ and $I = \dfrac{1}{2}\cdot \dfrac{1}{2}B(1/4,19/4) = \dfrac{1}{4}\dfrac{\Gamma(1/4)\Gamma(19/4)}{\Gamma(5)}$
|
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|
what is wrong in my answer and what will be the correct solution for this probability question ? A and B play a game where each is asked to select a number from 1 to 5. If
the two numbers match, both of them win a prize. The probability that they
will not win a prize in a single trial is -
a)1/25 b) 24/25 c)2/25 d)23/25
My proposed solution - Total number of ways in which both of them can select a number each:
=5×5=25
total number of ways in which two numbers match that is (1,1),(2,2),(3,3),(4,4),(5,5) = 5
probability that they will not win a prize = 1 - 5/25 = 4/5.
Tell me what is wrong in my answer and what will be the correct solution ?
|
The rationale behind the solution is this:
The game conductor has a particular number from 1 to 5 that he wants A and B to pick. Let us say "2". Then
P(A picking 2) $= \frac{1}{5}$
P(B picking 2) $= \frac{1}{5}$
P(A picking other than 2) $= \frac{4}{5}$
P(B picking other than 2) $= \frac{4}{5}$
P(Both picking 2) $= \frac{1}{5}\cdot\frac{1}{5}$
P(A picks 2, B picks other than 2)$ = \frac{1}{5}\cdot\frac{4}{5}$
P(A picks other than 2, B picks 2) $= \frac{4}{5}\cdot\frac{1}{5}$
P(B picks other than 2, B picks other than 2)$ = \frac{4}{5}\cdot\frac{4}{5}$
They don't win the game when the last three scenarios happen $= \frac{1}{5}\cdot\frac{4}{5}+\frac{4}{5}\cdot\frac{1}{5}+\frac{4}{5}\cdot\frac{4}{5} = \frac{24}{25}$
They win the prize $= \frac{1}{5}\cdot\frac{1}{5} = \frac{1}{25}$
Obviously, the question has been worded badly. Your answer is correct for the wording of the question.
Thanks
Satish
|
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Find A,B,C,D consecutive numbers based on written addition formula $A, B, C, D$ are consecutive digits: $B$ is greater by $1$ than $A$, $C$ is greater by $1$ than $B$, $D$ is greater by one than $D$.
Four $X$s are digits $A,B,C,D$ in unknown order. Find $A,B,C,D$
\begin{matrix}
& A & B & C & D\\
& D & C & B & A\\
+& X & X & X & X\\
\hline
1& 2 & 3 & 0 & 0\\
\end{matrix}
The only
thing I can think about is assign $A = y, B = y+1, C = y+2, D=y+3$ and then insert it above.
Not homework.. I try to prepare myself.
EDIT: after assignment I get:
\begin{matrix}
& 2y+3 & 2y+3 & 2y+3 & 2y+3 \\
+ & X & X & X & X \\
\hline
1 & 2 & 3 & 0 & 0\\
\end{matrix}
So:
\begin{matrix}
& 2A+3 & 2A+3 & 2A+3 & 2A+3\\
+ & X & X & X & X\\
\hline
1 & 2 & 3 & 0 & 0\\
\end{matrix}
|
$ABCD+DCBA = 1111\cdot(2A+3)$
If $A=1$, $12300-5\cdot1111=6745$.
If $A=2$, $12300-7\cdot1111=4523$.
If $A=3$, $12300-9\cdot1111=2301$.
The next three possibilities will be too small, as we need a four digit number. So $A=2$.
|
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|
numeric solutions on quadric surfaces Maybe it's a trivial thing, but I can't seem to find solution I'm looking for. I need to find a parametric solution to the following equation ($\mathbf{A}$ is positive definite):
$$
\mathbf{x}^T\mathbf{A}\mathbf{x}+\mathbf{b}^T\mathbf{x}+c=0
$$
How to find $x$, then, in some parametric form? That is, I want to generate some $n-1$ dimensional parameter (something like spherical coördinates) and get one of the solutions in the original space.
|
Well you can factor the problem by completing the square as follows,
\begin{align}
x^T A x + b^T x + c &= (x^T A x + b^T x + \frac{b^TA^{-1}b}{4}) + c - \frac{b^TA^{-1}b}{4} \\
&= (x^T A^{1/2} + \frac{1}{2}b^T A^{-1/2})(A^{1/2} x + \frac{1}{2}A^{-1/2} b) + c - \frac{b^TA^{-1}b}{4} \\
&= (x + \frac{1}{2}A^{-1} b)^T A (x + \frac{1}{2}A^{-1} b) + c - \frac{b^TA^{-1}b}{4},
\end{align}
or
$$x^T A x + b^T x + c = (x-p)^T A (x-p) - q$$
where $p := -\frac{1}{2}A^{-1} b$ and $q := \frac{b^TA^{-1}b}{4} - c$. So, an equivalent formulation of your equation is,
$$(x-p)^T A (x-p) = q.$$
So, the solution set is an ellipsoid centered at $p$ with axes given by the singular vectors $u_i$ of $A$ and axis lengths $\frac{\sqrt{q}}{\sigma_i}$.
Edit: from the comment below for completeness, a numerical scheme for drawing samples from the solution set is,
*
*Draw a collection of gaussian random vectors and put them as the columns of the matrix $X$.
*Normalize the columns of $X$ to have length $\sqrt{q}$.
*Compute a collection of random points on the ellipse as $S = A^{-1/2} X + p \mathbb{1}^T$ ($\mathbb{1}^T$ is the vector of all ones).
|
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|
Maximizing Area of Triangle in Circle I was playing around with another example that I made up where I am trying to maximize the area of a triangle inscribed in a circle of radius. I want to do the problem using the method of Lagrange Multipliers.
Attempt: Consider the circle of radius 2 given by $x^2+y^2=4$. The function we are trying to maximize is $A(x,y) = \frac{1}{2}xy$. Let $h(x,y)=4-x^2-y^2$.
Then, $\nabla h(x,y) = [-2x \ \ -2y ]$ and $\nabla A(x,y) = [\frac{y}{2} \ \ \frac{x}{2}]$. We want, $ [\frac{y}{2} \ \ \frac{x}{2}]= \lambda[-2x \ \ -2y ] $. So, $[x \ \ y]=[-4\lambda y \ \ -4\lambda x]$.
Hence, $(-4\lambda y)^2+(-4\lambda x)^2 = 4 \Rightarrow 4\lambda^2(y^2+x^2)=1 \Rightarrow x^2+y^2 = \frac{1}{4\lambda^2} \Rightarrow \lambda = \frac{1}{4}$. It then follows that $[x \ \ y] = [-0.5y \ \ -0.5x]$. And, $h(-0.5y,y) = 4-\frac{5}{4}y^2 = 0 \Rightarrow [x \ \ y] = [\frac{-4}{\sqrt{5}} \ \ \frac{4}{\sqrt{5}}]$. But then $A(x,y)<0$.
Question: What am I doing wrong?
|
The area is wrong. For a triangle inscribed in a circle (WLOG), you can assume one axis is parallel to the $x$ axis, say at height $y$. Then, the three points are:
$$\left(\sqrt{4-y^2},y\right),\ \left(-\sqrt{4-y^2},y\right), (x_3, y_3)$$
The area is therefore: $$A=(y_3-y)\sqrt{4-y^2}$$
|
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|
If $ x^2+y^2+z^2 =1$ for $x,y,z \in \mathbb{R}$, then find maximum value of $ x^3+y^3+z^3-3xyz $. If $ x^2+y^2+z^2 =1$, for $x,y,z \in \mathbb{R}$, what is the maximum of
$ x^3+y^3+z^3-3xyz $ ?
I factorize it... Then put the maximum values of $x+y+z$ and min value of $xy+yz+zx$...
But it is wrong as they don't hold simultaneously!
Also can it be solved using partial differentiation ?
And plz provide a solution without it.!
|
because we find the maximum,so let $A>0$
let $$x^3+y^3+z^3-3xyz=A>0\Longrightarrow (x+y+z)(x^2+y^2+z^2-xy-yz-xz)=A$$
so
$$x^2+y^2+z^2=\dfrac{A}{x+y+z}+xy+yz+xz=\dfrac{A}{x+y+z}+\dfrac{1}{2}[(x+y+z)^2-x^2-y^2-z^2]$$
By AM-GM inequality
\begin{align*}\Longrightarrow \dfrac{3}{2}(x^2+y^2+z^2)&=\dfrac{A}{x+y+z}+\dfrac{1}{2}(x+y+z)^2\\
&=\dfrac{A}{2(x+y+z)}+\dfrac{A}{2(x+y+z)}+\dfrac{1}{2}(x+y+z)^2\\
&\ge \dfrac{3}{2}\sqrt[3]{A^2}
\end{align*}
so
$$A\le 1$$
|
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|
A series converging (or not) to $\ln 2$ I have come across the following series, which I suspect converges to $\ln 2$:
$$\sum_{k=1}^\infty \frac{1}{4^k(2k)}\binom{2k}{k}.$$
I could not derive this series from some of the standard expressions for $\ln 2$. The sum of the first $100 000$ terms agrees with $\ln 2$ only up to two digits.
Does the series converge to $\ln 2$?
|
Hint: (or outline -- a lot of details and justifications must be done where there are $(\star)$'s)
*
*for $\lvert x\rvert < \frac{1}{4}$,
\begin{align*}
f(x)&\stackrel{\rm def}{=}\frac{1}{2}\sum_{k=1}^{\infty} \binom{2k}{k}\frac{x^k}{k} = \frac{1}{2}\sum_{k=1}^{\infty} \binom{2k}{k}\int_0^x t^{k-1}dt \\
&\stackrel{(\star)}{=} \frac{1}{2} \int_0^x \left(\sum_{k=1}^{\infty} \binom{2k}{k} t^{k-1}\right)dt = \frac{1}{2} \int_0^x \frac{1}{t} \left(\sum_{k=1}^{\infty} \binom{2k}{k} t^{k}\right)dt \\
&\stackrel{(\star)}{=} \frac{1}{2} \int_0^x \frac{1}{t} \left(\frac{1}{\sqrt{1-4t}}-1\right)dt \\
&= \frac{1}{2} 2\ln\frac{2}{\sqrt{1-4x}+1}
\end{align*}
*"so" ($\star$)
\begin{align*}
f(x)\xrightarrow[t\to\frac{1}{4}^-]{}\ln 2 \tag{$\star$}
\end{align*}
|
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|
Nested radicals and n-th roots There are many beautiful infinite radical equations,
some relatively straightforward, some much more subtle:
$$
x = \sqrt{ x \sqrt{ x \sqrt{ x \sqrt{ \cdots } } } }
$$
$$
\sqrt{2} = \sqrt{ 2/2 + \sqrt{ 2/2^2 + \sqrt{ 2/2^4 + + \sqrt{ 2/2^8 + \sqrt{ \cdots}}}}}
$$
$$
3 = \sqrt{1 + 2\sqrt{1 + 3\sqrt{ 1 + 4\sqrt{ \cdots }}}}
$$
But I have seen far fewer analogous equations for $n$-roots.
Here is one:
$$
2 = \sqrt[3]{6 + \sqrt[3]{6 + \sqrt[3]{6 + \sqrt[3]{\cdots}}}}
$$
My question is:
Q. Are there truly "more" beautiful infinite radical equations,
or is it just our natural gravitation toward the simpler square-root equations that
leads to collections emphasizing radicals?
I am aware this question is vague, but perhaps some nevertheless have insights.
|
Here's a nested radical for 4th roots that is not so well known. Given the tetranacci numbers (an analogue of the fibonacci numbers). Let $y$ be the tetranacci constant, or the positive real root of,
$$y^4-y^3-y^2-y-1=0$$
Then,
$$y(3-y) = \sqrt[4]{41-11\sqrt[4]{41-11\sqrt[4]{41-11\sqrt[4]{41-\dots}}}} = 2.06719\dots$$
The family can be found in this MSE post.
|
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|
Proving an identity involving binomial coefficients and fractions I've been trying to prove the following formula (for $n > 1$ natural, $a, b$ non-zero reals), but I don't know where to start.
$$\sum_{j=1}^n \binom{n-1}{j-1} \left( \frac{a-j+1}{b-n+1} \right) \left( \frac{a}{b} \right)^{j-1} \left( \frac{b-a}{b} \right)^{n-j} = \frac{a}{b}$$
Wolfram says it's right, but so far I've been unable to give a proof. I wonder if there's a combinatorial proof to it. Any help, hint or reference will be much appreciated.
|
Replace $j$ by $j+1$ and $n$ by $n+1$ rewrite as Greg Martin suggested:
\begin{align*}
\sum_{j=0}^n \binom nj \left( \frac{a-j}{b-n} \right) \left( \frac{a}{b} \right)^j \left( \frac{b-a}{b} \right)^{n-j} \tag I
\end{align*}
Let $p=\dfrac{a}{b}$ and $q=1-p$ and consider:
\begin{align*}
\left(q+px\right)^n &= \sum_{j=0}^{n}\binom nj p^j q^{n-j} x^j\tag 1 \\
npx\, \left(q+px\right)^{n-1} &= \sum_{j=0}^{n}j \binom nj p^j q^{n-j} x^j\tag 2 \\
\end{align*}
$(1)$ is by binomial expansion, $(2)$ is by differentiating w.r.t x on both sides.
Now substituting $x=1$, $(1)$ and $(2)$ become
\begin{align*}
1 &= \sum_{j=0}^{n}\binom nj p^j q^{n-j}\tag 3 \\
np &= \sum_{j=0}^{n}j \binom nj p^j q^{n-j} \tag 4 \\
\end{align*}
From $(\mathrm{I})$,$(3)$ and $(4)$,
\begin{align*}
\sum_{j=0}^n \binom nj \left( \frac{a-j}{b-n} \right) \left( \frac{a}{b} \right)^j \left( \frac{b-a}{b} \right)^{n-j} &= \frac{a}{b-n}\cdot 1 - \frac{1}{b-n}\cdot np \\
&= \frac{a\left(b-n\right)}{b-n} \\
&= \frac{a}{b}
\end{align*}
|
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Proof of the inequality $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} \geq \frac{3}{2}$ I am currently attempting to prove the following inequality
$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \geq \dfrac{3}{2}$ for all $ a,b,c>0$
My instinctive plan of attack is to use the AM/GM inequality with $x_1=\dfrac{a}{b+c}$ etc.
Using that I get this
$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \geq 3\times \sqrt[3]{\dfrac{abc}{(a+b)(b+c)(a+c)}}$
From here, I used the fact that $(a+b)(b+c)(a+c)\geq 8abc$, which can be easily proven by considering that $a+b\geq 2\sqrt{ab}$
But by using this, I get the following...
$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \geq 3\times \sqrt[3]{\dfrac{abc}{(a+b)(b+c)(a+c)}} \leq 3 \times \sqrt[3]{\dfrac{abc}{8abc}} = \dfrac{3}{2}$
Everything seems so perfect because I get the value $\dfrac{3}{2}$ as required, but this method isn't valid due to the change in direction! What is going on?
Is there a way of proving this inequality otherwise then?
|
We can also use rearrangement inequality to prove it.
$\displaystyle\frac a{b+c}+\frac b{a+c}+\frac c{a+b}$
$\displaystyle=\frac12\left(\left(\frac a{b+c}+\frac b{a+c}+\frac c{a+b}\right)+\left(\frac a{b+c}+\frac b{a+c}+\frac c{a+b}\right)\right)$
$\displaystyle\geq\frac12\left(\left(\frac a{a+c}+\frac b{a+b}+\frac c{b+c}\right)+\left(\frac a{a+b}+\frac b{b+c}+\frac c{a+c}\right)\right)$
$=\dfrac32$
|
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|
Equation of parabola, tangent at vertex Two tangents on a parabola are $x-y=0$ and $x+y=0$. If $(2,3)$ is the focus of the parabola, then find the equation of tangent at the vertex.
Thanks.
My thoughts:
Can't figure out anything :(
|
We are given a pair of perpendicular tangents. The point of intersection of tangents lie on directrix of parabola. Clearly, these tangents intersect at origin. Let the equation of directrix be $y=mx$, then the equation of parabola is
$$(x-2)^2+(y-3)^2=\frac{(y-mx)^2}{1+m^2}$$
Since the parabola is tangent to $y=x$, we get the following quadratic:
$$(x-2)^2+(x-3)^2=\frac{x^2(m-1)^2}{1+m^2}=x^2\left(1-\frac{2m}{1+m^2}\right)$$
The discriminant of the above quadratic must be zero. With this condition, $m=2/3$ and $m=3/2$. Obviously, the second solution is not possible i.e $m=2/3$.
The equation of directrix is hence $y=\dfrac{2}{3}x$. The equation of axis of parabola is perpendicular to directrix and passes through focus. Its equation is $3x-2y-12=0$. The point of intersection of directrix and axis is: $\left(\dfrac{36}{5},\dfrac{24}{5}\right)$.
The vertex is midpoint of the above point and focus i.e $\left(\dfrac{23}{5},\dfrac{39}{10}\right)$. Hence, the tangent at vertex is:
$$y-\frac{39}{10}=\frac{2}{3}\left(x-\frac{23}{5}\right) \Rightarrow \boxed{4x-6y+5=0}$$
|
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|
inequality method of solution Im looking for an efficent method of solving the following inequality: $$\left(\frac{x-3}{x+1}\right)^2-7 \left|\frac{x-3}{x+1}\right|+ 10 <0$$
I've tried first determining when the absolute value will be positive or negative etc, and than giving it the signing in accordance to range it is in, bur it turned out to be quite complex and apperently also wrong. Are there any other ways?
|
$$\left(\frac{x-3}{x+1}\right)^2-7 \left|\frac{x-3}{x+1}\right|+ 10 \lt 0 \Rightarrow \\ \left(\left|\frac{x-3}{x+1}\right|\right)^2-7 \left|\frac{x-3}{x+1}\right|+ 10 \lt 0 $$
$$\Delta=\left(-7\right)^2-4 \cdot 10=49-40=9$$
$$\left|\frac{x-3}{x+1}\right|_{1,2}=\frac{-\left(-7\right) \pm \sqrt{\Delta}}{2}=\frac{7 \pm \sqrt{9}}{2}=\frac{7 \pm 3}{2}$$
$\left(\left|\frac{x-3}{x+1}\right|\right)^2-7 \left|\frac{x-3}{x+1}\right|+ 10 \lt 0 \Rightarrow \left|\frac{x-3}{x+1}\right| \in \left(\frac{7-3}{2},\frac{7+3}{2}\right)=\left(2,5\right)$
$$\left|\frac{x-3}{x+1}\right|\gt 2 \Rightarrow \frac{x-3}{x+1}\lt -2 \text{ OR } \frac{x-3}{x+1}\gt 2 $$
and
$$\left|\frac{x-3}{x+1}\right|\lt 5 \Rightarrow -5\lt \frac{x-3}{x+1}\lt 5 $$
Solve at each case for $x$.
Can you continue?
|
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|
Finding the derivative of sinus and cosinus. Trigonometric identities How can we see that $$\sin(x+h)-\sin(x)=2\sin\left(\frac h2\right)\cos\left(x+\frac h2\right)$$
How can we see that $$\cos(x+h)-\cos(x)=-2\sin\left(\frac h2\right)\sin\left(x+\frac h2\right)$$
Do these identities have a name?
|
You asked for a name: Looks like an application of the reverse of the Prosthaphaeresis identities, or sum-to-product identities.
$$
\sin a − \sin b = 2 \cos\frac{a + b}{2} \sin\frac{a − b}{2} \\
\cos a − \cos b = −2 \sin\frac{a + b}{2} \sin\frac{a − b}{2}
$$
using $a = x + h$ and $b = x$.
Another link here.
I usually use the complex definitions of $\sin$ and $\cos$ to proof, but you probably want an elementary one.
Note there are four identities in total.
$$
\sin a + \sin b = 2 \sin\frac{a + b}{2} \cos\frac{a − b}{2} \\
\cos a + \cos b = 2 \cos\frac{a + b}{2} \cos\frac{a − b}{2}
$$
which would yield
$$
\sin(x+h) + \sin(x) =
2 \sin\left(x+\frac{h}{2}\right) \cos\left(\frac{h}{2}\right) \\
\cos(x+h) + \cos(x) =
2 \cos\left(x+\frac{h}{2}\right) \cos\left(\frac{h}{2}\right)
$$
|
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|
Supremum $\sup_x \left\{x-\ln\left(1+\frac{\sin x-\cos x}{2}\right)\right\}$ I would like to know the supremum
$$\sup_x \left\{x-\ln\left(1+\frac{\sin x-\cos x}{2}\right)\right\}$$
when $0\leq x \leq \frac{\pi}{4}$.
After the derivative is easily found that the function is increasing. But, is there a way that does not involve the derivative to find this sup?
Thanks!
|
Let
$$ f(x)=\frac{e^x}{1+\frac{\sin x-\cos x}{2}}. $$
We only have to show $f(x)$ is increasing in $[0,\frac{\pi}{4}]$. In fract, for $0<x_1<x_1+\alpha<\frac{\pi}{4}$ ($\alpha\in(0,\frac{\pi}{4})$),
\begin{eqnarray*}
\frac{f(x_1+\alpha)}{f(x_1)}&=&e^{\alpha}\frac{2+\sin x_1-\cos x_1}{2+\sin (x_1+\alpha)-\cos (x_1+\alpha)}=e^{\alpha}\frac{\sqrt{2}+\sin(x_1-\frac{\pi}{4})}{\sqrt{2}+\sin(x_1+\alpha-\frac{\pi}{4})}.
\end{eqnarray*}
It is not hard to check
$$ \sin(x_1+\alpha-\frac{\pi}{4})-\sin(x_1-\frac{\pi}{4})\le \alpha. $$
So
$$\sin(x_1+\alpha-\frac{\pi}{4})\le \alpha+\sin(x_1-\frac{\pi}{4})$$
and hence
\begin{eqnarray*}
\frac{f(x_1+\alpha)}{f(x_1)}\ge e^{\alpha}\frac{\sqrt{2}+\sin(x_1-\frac{\pi}{4})}{\sqrt{2}+\alpha+\sin(x_1-\frac{\pi}{4})}=e^{\alpha}\frac{\sqrt{2}-\sin(\frac{\pi}{4}-x_1)}{\sqrt{2}+\alpha-\sin(\frac{\pi}{4}-x_1)}.
\end{eqnarray*}
Note the following
$$ g(x)=\frac{\sqrt{2}-x}{\sqrt{2}+\alpha-x}, x\in[-1,0] $$
is decreasing. Since $-1<\sin(\frac{\pi}{4}-x_1)\le 0$, we have
\begin{eqnarray*}
\frac{f(x_1+\alpha)}{f(x_1)}&\ge& e^{\alpha}\frac{\sqrt{2}-\sin(\frac{\pi}{4}-x_1)}{\sqrt{2}+\alpha-\sin(\frac{\pi}{4}-x_1)}\\
&=&e^\alpha g(\sin(\frac{\pi}{4}-x_1))\ge e^\alpha g(0)\\
&=&e^\alpha\frac{\sqrt{2}}{\sqrt{2}+\alpha}\\
&\ge&(1+\alpha)\frac{1}{1+\frac{\alpha}{\sqrt{2}}}\ge 1.
\end{eqnarray*}
This implies that $f(x)$ is increasing and so is the original function. So the original function reaches its max at $x=\pi/4$.
|
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|
How to solve $\frac{2x+1}{2x-3}+\frac{7x\:}{9-4x^2}=1+\frac{x-4}{2x+3}$ for $x$? Can somebody explain me this one!
$\frac{2x+1}{2x-3}+\frac{7x\:}{9-4x^2}=1+\frac{x-4}{2x+3}$
My book says the answer is $x_1 = 0$; $x_2 = 6$.
I tried to solve it and got stuck somewhere in:
$4x^2+x+3/(2x-3)(3x-1) = 0$
I can't find my mistake.
|
Note that $$9 - 4x^2 = -(4x^2 - 9) = -(2x+3)(2x-3)$$
$$\frac{2x+1}{2x-3}+\frac{7x\:}{9-4x^2}=1+\frac{x-4}{2x+3}\frac{2x+1}{2x-3}-\frac{7x\:}{(2x+3)(2x-3)}=1+\frac{x-4}{2x+3}$$
Now, multiply both sides of the equation by $(2x - 3)(2x+ 3)$:
$$(2x+1)(2x+3) - 7x = (2x+3)(2x-3) +(x-4)(2x-3)$$
Simplify by expanding as needed, and forming a quadratic equation (all terms on left, with $0$ on the right).
$$\begin{align} &\quad 4x^2 +8x+3 - 7x -(4x^2 -9) - (2x^2 - 11x +12) = 0\\
& \iff -2x^2 +12x = 0 \\ &\iff -2x(x -6) = 0 \\ &\implies x = 0\, \text{ or }\,x = 6\end{align}$$
|
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|
Maclaurin series expansion Write the Maclaurin series for: $6\cos{5x^2}$
Find the first five coefficients.
For this question: I repeatedly come to the answer of: $6-75 x^4+\frac{625 x^8}{4}+O(x^9)$
With the coefficients being $6, -75, \frac{625}{4}$ and $0$. These answers are wrong though.
Any suggestions on how to get the 2nd, 3rd, 4th, and fifth coeffient? I have coefficient #1 correct.
|
$$
\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} - \ldots
$$
Thus,
$$
6\cos(5x^2) = 6\biggl[1 - \frac{(5x^2)^2}{2} + \frac{(5x^2)^4}{24} - \frac{(5x^2)^6}{720} + \frac{(5x^2)^8}{40320} - \ldots \biggr]
$$
|
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|
Prove $\int_0^{\infty} \left(\sqrt{1+x^{4}}-x^{2}\right)\ dx=\frac{\Gamma^{2}\left(\frac{1}{4}\right)}{6\sqrt{\pi}}$ I have in trouble for evaluating following integral
$$\int_0^{\infty} \left(\sqrt{1+x^{4}}-x^{2}\right)\ dx=\frac{\Gamma^{2}\left(\frac{1}{4}\right)}{6\sqrt{\pi}}$$
It seems really easy, but I don't know how to handle it at all.
(The results are well known, here I tried to evaluate it but I failed)
I tried to use the relation
$$\sqrt{1+x^{4}}-x^{2}=\frac{1}{\sqrt{1+x^{4}}+x^{2}}$$
but I couldn't find the desired results.
|
Let $x = \sqrt{\tan \theta}$. Note that, $x^4 = \tan^2 \theta$ and $dx = \dfrac{\sec^2\theta d\theta}{2\sqrt{\tan\theta}}$. Like this,
$$
I = \int_{0}^{\infty}( \sqrt{1 + x^4} - x^2)dx = \int_{0}^{\pi/2} \dfrac{(\sec \theta + \tan \theta)\sec^2\theta d\theta}{2\sqrt{\tan \theta}} = \dfrac{1}{2}\int_{0}^{\pi/2} \dfrac{\cos^{1/2}\theta (1 - \sin \theta)d\theta}{\sin^{1/2}\theta \cos^3 \theta}
$$
$$
= \dfrac{1}{2}\int_{0}^{\pi/2}\cos^{-5/2}\theta\sin^{-1/2}\theta d\theta - \dfrac{1}{2}\int_{0}^{\pi/2}\cos^{-5/2}\theta\sin^{1/2}\theta d\theta = I_1 + I_2
$$
Note that, $I_1 = \dfrac{1}{2}B(-3/4,1/4)$, because
$$
(2m-1=-5/2 \ \Rightarrow \ m = -3/4 \ \text{and} \ 2n-1 = -1/2 \ \Rightarrow \ n=1/4)
$$
But, $B(m,n) = \dfrac{1}{2}\dfrac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}$. Thus,
$$
I_1 = \dfrac{1}{4}\dfrac{\Gamma(-3/4)\Gamma(1/4)}{\Gamma(-1/2)} = \dfrac{1}{4}\cdot \dfrac{(-4)}{3}\dfrac{\Gamma(1/4)\cdot \Gamma(1/4)}{(-2)\sqrt{\pi}} = \dfrac{\Gamma(1/4)^2}{6\sqrt{\pi}}
$$
If $x \to 0$, $\Gamma(x) \to \infty$, so that $I_2 = 0$. Furthermore, if $x < 0, x \neq -1,-2,\ldots$, define $\Gamma(x) = \Gamma(x + 1)/x$.
|
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|
The area of circle The question is to prove that area of a circle with radius $r$ is $\pi r^2$ using integral. I tried to write $$A=\int\limits_{-r}^{r}2\sqrt{r^2-x^2}\ dx$$
but I don't know what to do next.
|
Another interesting solution, not yet mentioned by the other answers, is to use this substitution:
$$
x = r\cos\theta \;\therefore\; \frac{dx}{d\theta} = -r\sin\theta
\;\therefore\; dx = -r\sin\theta \; d\theta
$$
Substituting the integral limits, we have
$$x = -r \implies -r = r\cos\theta \implies \cos\theta = -1 \implies \theta = \pi$$
$$x = r \implies r = r\cos\theta \implies \cos\theta = 1 \implies \theta = 0$$
With this we can do:
$$
A=-r \int\limits_{\pi}^{0}2\sqrt{r^2-(r\cos\theta)^2} \sin\theta \; d\theta =
-2r \int\limits_{\pi}^{0}\sqrt{r^2(1 - \cos^2\theta)} \sin\theta \; d\theta =
$$
$$
-2r \int\limits_{\pi}^{0}\sqrt{r^2\sin^2\theta} \sin\theta \; d\theta =
-2r \int\limits_{\pi}^{0}r\sin^2\theta \; d\theta =
-2r^2 \int\limits_{\pi}^{0}\sin^2\theta \; d\theta =
$$
$$
-2r^2 \int\limits_{\pi}^{0}\frac{1}{2}(1 - \cos(2\theta)) \; d\theta =
$$
Now we substitute $u = 2\theta \implies \frac{du}{d\theta} = 2 \implies d\theta = \frac{du}{2}$, and limits range from $2\pi$ to $0$:
$$
\frac{-r^2}{2} \int\limits_{2\pi}^{0}(1 - \cos u) \; du =
\frac{-r^2}{2} (u - \sin u)\Big{|}_{2\pi}^{0} =
$$
$$
\frac{-r^2}{2} [(0 - \sin 0) - (2\pi - \sin2\pi)] =
\frac{-r^2}{2} [-2\pi] = \pi r^2
$$
|
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|
Is $7^{8}+8^{9}+9^{7}+1$ a prime? (no computer usage allowed)
Prove or disprove that $$7^{8}+8^{9}+9^{7}+1$$ is a prime number, without using a computer.
I tried to transform $n^{n+1}+(n+1)^{n+2}+(n+2)^{n}+1$, unsuccessfully, no useful conclusion.
|
Trial division mod $47$, as per the hint.
Firstly, for positive integer $a$, observe that $$a \equiv \overbrace{a \mod 50}^{\text{reduced residue}}+3\lfloor a/50\rfloor \pmod {47}.$$ This makes taking mod $47$s much easier.
*
*We compute $7^2=49 \equiv 2 \pmod {47}$. So $7^8 \equiv 2^4 = 16 \pmod {47}$.
*We compute $8^2=64 \equiv 14+3 = 17 \pmod {47}$. So $8^4 \equiv 17^2 = 289 \equiv 39+3 \times 5=54 \equiv 4+3=7 \pmod {47}.$ So $8^8 \equiv 7^2 = 49 \equiv 2 \pmod 7 \pmod {47}.$ So $8^9 \equiv 2 \times 8 = 16 \pmod {47}$.
*I happen to have memorized that $9^3=729$ (I used to set my alarm to 7:29am because it is equal to $3^6$). So $9^3 \equiv 729 = 29+3 \times 14=71 \equiv 21+3=24 \pmod {47}$. So $9^6 \equiv 24^2=576 \equiv 26+3 \times 11=59 \equiv 9+3=12 \pmod {47}$. So $9^7=12 \times 9=108=8+3 \times 2=14 \pmod {47}$.
Finally $16+16+14+1=47 \equiv 0 \pmod {47}$.
|
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|
Prove this polylogarithmic integral has the stated closed form value Question. Prove the following polylogarithmic integral has the stated value:
$$I:=\int_{0}^{1}\frac{\operatorname{Li}_2{(1-x)}\log^2{(1-x)}}{x}\mathrm{d}x=-11\zeta{(5)}+6\zeta{(3)}\zeta{(2)}.$$
I was able to arrive at the proposed value for the integral through a combination of educated guessing and after-the-fact numerical checking (the value of the integral was approximately $I\approx0.457621...$), but I'm at a loss for how to derive it rigorously. Thoughts?
|
Since no answers have been posted, I'll expand on my comment above.
There is a general formula that states $$\sum_{n=1}^{\infty}\frac{H_{n}^{(r)}}{n^{q}}=\zeta(r)\zeta(q)-\frac{(-1)^{r-1}}{(r-1)!}\int_{0}^{1}\frac{\text{Li}_{q}(x) \log^{r-1}(x) }{1-x}dx $$ where $$ H_{n}^{(r)} = \sum_{k=1}^{n} \frac{1}{k^{r}} .$$
A proof can be found here.
Making the substitution $u = 1-x$, $$ \sum_{n=1}^{\infty}\frac{H_{n}^{(r)}}{n^{q}}=\zeta(r)\zeta(q)-\frac{(-1)^{r-1}}{(r-1)!}\int_{0}^{1}\frac{\text{Li}_{q}(1-u) \log^{r-1}(1-u)}{u}dx .$$
Therefore, $$ \int_{0}^{1} \frac{\text{Li}_{2}(1-x)\log^{2}(1-x)}{x} \ dx = 2 \zeta (3) \zeta (2) - 2 \sum_{n=1}^{\infty} \frac{H_{n}^{(3)}}{n^{2}} .$$
To simplify the evaluation of that Euler sum slightly, I'm first going to evaluate $ \displaystyle \sum_{n=1}^{\infty} \frac{H_{n}^{(2)}}{n^{3}}$ and then use the identity $$\sum_{n=1}^{\infty} \frac{H_{n}^{(r)}}{n^{q}} + \sum_{n=1}^{\infty} \frac{H_{n}^{(q)}}{n^{r}} = \zeta(r) \zeta(q) + \zeta(r+q) . \tag{1} $$
Consider $$f(z) = \frac{ \pi \cot (\pi z) \ \psi_{1}(-z)}{z^{3}} $$ where $\psi_{1}(z)$ is the trigamma function.
The function $f(z)$ has poles of order $3$ at the positive integers, a pole of order 6 at the origin, and simple poles at the negative integers.
On the sides of a square with vertices at $\pm \left( N+ \frac{1}{2} \right) \pm i \left( N+ \frac{1}{2} \right)$, $\cot (\pi z)$ is uniformly bounded.
And when $z$ is large in magnitude and not on the positive real axis, $\psi_{1}(-z)$ is approximately $ \displaystyle - \frac{1}{z}$.
So as $N \to \infty$ through the integers, $ \displaystyle \int f(z) \ dz$ will vanish on all four sides of the square.
Therefore, $$ \sum_{n=-\infty}^{\infty} \text{Res} [f(z), n] = 0.$$
The Laurent expansion of $\psi_{1}(-z)$ at the positive integers (including $0$) is $$ \psi_{1}(-z) = \frac{1}{(z-n)^{2}} + \sum_{m=0}^{\infty} (m+1) \left( (-1)^{m} H_{n}^{(m+2)} + \zeta(m+2) \right) (z-n)^{m} . $$
And the Laurent expansion of $\pi \cot \pi z$ at the integers is
$$ \pi \cot (\pi z) = \frac{1}{z-n} - 2 \sum_{m=1}^{\infty} \zeta(2m) (z-n)^{2m-1} .$$
So at the positive integers, $$f(z) = \frac{1}{z^{3}} \left(\frac{1}{(z-n)^{3}} + \frac{H_{n}^{(2)} - \zeta(2)}{(z-n)} + \mathcal{O}(1) \right) $$
which implies
$$ \begin{align} \text{Res} [f(z), n] &= \text{Res} \left[\frac{1}{z^{3}(z-n)^{3}}, n \right] + \text{Res} \left[\frac{H_{n}^{(2)}-\zeta(2)}{z^{3}(z-n)}, n \right] \\ &= \frac{6}{n^{3}} + \frac{H_{n}^{(2)}}{n^{3}} -\frac{\zeta(2)}{n^{3}} . \end{align} $$
At the negative integers,
$$\text{Res} [f(z), -n] = - \frac{\psi_{1}(n)}{n^{3}} = \frac{H_{n-1}^{(2)} - \zeta(2)}{n^{3}} = \frac{H_{n}^{(2)}}{n^{3}} - \frac{1}{n^{5}} - \frac{\zeta(2)}{n^{3}} .$$
And at the origin,
$$ f(z) = \frac{1}{z^{6}} - \frac{\zeta(2)}{z^{4}} + \frac{2 \zeta(3)}{z^{3}} + \left(\zeta(4) - 2 \zeta^{2}(2) \right) \frac{1}{z^{2}} + \Big(4 \zeta(5) - 4 \zeta(3) \zeta(2) \Big) \frac{1}{z} + \mathcal{O}(1)$$
while implies
$$ \text{Res}[f(z),0] = 4 \zeta(5) - 4 \zeta(3) \zeta(2) .$$
Summing up all the residues,
$$6 \zeta(5) + \sum_{n=1}^{\infty} \frac{H_{n}^{(2)}}{n^{3}} - \zeta(3) \zeta(2) + \sum_{n=1}^{\infty} \frac{H_{n}^{(2)}}{n^{3}} - \zeta(5) - \zeta(3) \zeta(2) + 4 \zeta(5) - 4 \zeta(3) \zeta(2) = 0$$
which implies
$$ \sum_{n=1}^{\infty} \frac{H_{n}^{(2)}}{n^{3}} = 3 \zeta(3) \zeta(2) - \frac{9}{2} \zeta(5) .$$
Then using $(1)$,
$$\sum_{n=1}^{\infty} \frac{H_{n}^{(3)}}{n^{2}} = \zeta(3) \zeta(2) + \zeta(5) - 3 \zeta(3) \zeta(2) + \frac{9}{2} \zeta(5) = - 2 \zeta(3) \zeta(2) + \frac{11}{2} \zeta(5). $$
So finally we have
$$ \begin{align} \int_{0}^{1} \frac{\text{Li}_{2}(1-x)\log^{2}(1-x)}{x} \ dx &= 2 \zeta (3) \zeta (2) - 2 \Big(- 2 \zeta(3) \zeta(2) + \frac{11}{2} \zeta(5) \Big) \\ &= 6 \zeta(3) \zeta(2) - 11 \zeta(5) . \end{align}$$
|
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|
maximum area of a rectangle inscribed in a semi - circle with radius r.
A rectangle is inscribed in a semi circle with radius $r$ with one of its sides at the diameter of the semi circle. Find the dimensions of the rectangle so that its area is a maximum.
My Try:
Let length of the side be $x$,
Then the length of the other side is $2\sqrt{r^2 -x^2}$, as shown in the image.
Then the area function is
$$A(x) = 2x\sqrt{r^2-x^2}$$
$$\begin{align}A'(x) &= 2\sqrt{r^2-x^2}-\frac{4x}{\sqrt{r^2-x^2}}\\
&=\frac{2}{\sqrt{r^2-x^2}} (r^2 - 2x -x^2)\end{align}$$
setting $A'(x) = 0$,
$$\implies x^2 +2x -r^2 = 0$$
Solving, I obtained:
$$x = -1 \pm \sqrt{1+r^2}$$
That however is not the correct answer, I cannot see where I've gone wrong? Can someone point out any errors and guide me the correct direction. I have a feeling that I have erred in the differentiation.
Also how do I show that area obtained is a maximum, because the double derivative test here is long and tedious.
Thanks!
|
Equation of circle: $x^2+y^2=r^2$
$x = \pm (r^2-y^2) $
Thus, length of rectangle is $x-(-x)=2x$ and height is $y$.
Area $A = 2xy$. Maximizing A is equivalent to maximizing $ A^2$
$A^2 = 4x^2y^2 = 4(r^2-y^2)y^2=4r^2y^2-4y^4$
Let, $f(z)= 4r^2z-4z^2 $
Then, $f'(z) = 4r^2-8z$
Equating, $f'(z)=0$ we get,
$\mathbf{z=\frac12r^2}$
Therefore, max. $A^2$ or max. $f(z) = 4r^2\times \frac12r^2-4\times(\frac12r^2)^2 = 2r^4-r^4=r^4$
Hence, $\mathbf{max. A = \sqrt{r^4} = r^2}$
|
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|
Find odd numbers $(o_1,o_2,o_3,o_4)$ such that $o_1^2-o_2^2=2(o_3^2-o_4^2)$ such that $o_1>o_2$ and $o_3>o_4$ I am working on a graph labeling problem and am stuck at the following problem on odd numbers.
Find (all) odd numbers $(o_1,o_2,o_3,o_4)$ such that $o_1^2-o_2^2=2(o_3^2-o_4^2)$ such that $o_1>o_2$ and $o_3>o_4$?
Ideally I would like to prove that no such 4-tupule $(o_1,o_2,o_3,o_4)$ exists. However if they exist, I want to count, for given $n$, how many such 4-tupules exists such that the largest odd number i.e. $o_1 \leq n$.
My approach: Say the 4-tupule is $(2p+1,2q+1,2r+1,2s+1)$. The above equation simplifies to $(p-q)(p+q+1) = 2(r-s)(r+s+1)$. I am stuck here..
Thanks for going through this.
|
Describing all solutions is quite intricate. However, with composite $n \equiv 3 \pmod 8$ where all prime factors of $n$ are $1,3 \pmod 8,$ then there are multiple expressions as $u^2 + 2 v^2;$ with two solutions, you can arrange in your pattern. The predictable kind are illustrated below: if $n$ is the product of $r$ distinct primes, an odd number of which are $3 \bmod 8$ and the others $1 \bmod 8,$ then there will be $2^{r-1}$ different expressions $n=u^2 + 2 v^2$ with positive integers $u,v.$
here are some
51 == 3 = 3 * 17
123 == 3 = 3 * 41
187 == 3 = 11 * 17
219 == 3 = 3 * 73
267 == 3 = 3 * 89
291 == 3 = 3 * 97
323 == 3 = 17 * 19
So
$$ 51 = 7^2 + 2 \cdot 1^2 = 1^2 + 2 \cdot 5^2 $$
$$ 123 = 11^2 + 2 \cdot 1^2 = 5^2 + 2 \cdot 7^2 $$
$$ 187 = 13^2 + 2 \cdot 3^2 = 5^2 + 2 \cdot 9^2 $$
.........
$$ 627 = 25^2 + 2 \cdot 1^2 = 23^2 + 2 \cdot 7^2 = 17^2 + 2 \cdot 13^2 = 7^2 + 2 \cdot 17^2 $$
...........
$$ 2091 = 43^2 + 2 \cdot 11^2 = 37^2 + 2 \cdot 19^2 = 29^2 + 2 \cdot 25^2 = 13^2 + 2 \cdot 31^2 $$
|
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|
Value of $\cos^2\alpha-\sin^2\alpha$ My problem is from Israel Gelfand's Trigonometry textbook.
Page 48. Exercise 8: b) If $\tan\alpha=r$, write an expression in terms of $r$ that represents the value of $\cos^2\alpha-\sin^2\alpha$.
The attempt at a solution:
Well I solved a) which was similar, question was: a) If $\tan\alpha=2/5$, find the numerical value of $\cos^2\alpha-\sin^2\alpha$. Solution was this, since $\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{2}{5}$, then $\cos^2\alpha-\sin^2\alpha=5^2-2^2=21$. b) part is more general version of a) I guess, but I can't seem to find identity that would allow me to solve it. I would appreciate some hints, thank you in advance.
|
My Approach:
$\tan \alpha=r$
$\implies \tan^2 \alpha=r^2$
$\implies \large \frac{\sin^2 \alpha}{\cos^2 \alpha}=r^2$
Apply Componendo and Dividendo
$\implies \large \frac{\sin^2 \alpha+\cos^2 \alpha}{\sin^2 \alpha-\cos^2 \alpha}=\large \frac{r^2+1}{r^2-1}$
$\implies \large \frac{1}{\sin^2 \alpha - \cos ^2 \alpha}=\large \frac{r^2+1}{r^2-1}$
$\implies {\sin^2 \alpha - \cos ^2 \alpha}=\large \frac{r^2-1}{r^2+1}$
Multiply both sides by -1
$\implies {\cos ^2 \alpha-\sin^2 \alpha}=\large \frac{1-r^2}{1+r^2}$
|
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|
If the circle $x^2+y^2+4x+22y+c=0$ bisects the circmuference of the circle $x^2+y^2-2x+8y-d=0$ the... Problem :
If the circle $x^2+y^2+4x+22y+c=0$ bisects the circmuference of the circle $x^2+y^2-2x+8y-d=0$ then c +d equals
(a) 60
(b) 50
(c) 40
(d) 30
Solution :
Equation of common chord of the circles is given by $S-S'=0$ where $S = x^2+y^2+4x+22y+c=0 ; S' = x^2+y^2-2x+8y-d=0$
$\Rightarrow 4x+22y+c-(2x+8y-d)=0$
$\Rightarrow 4x+22y+c-2x-8y+d =0$
$\Rightarrow c+d = -(2x+14y)$
Now how to get the value of c +d , please suggest thanks..
|
you had an error:
$$S-S'=4x+22y+c-(-2x+8y-d)\\=6x+14y+c+d=0$$
so
$$c+d=-6x-14y$$
because $S$ besects $S'$, so the center of $S'$ is on the line $S-S'$. The center is $(1,-4)$, so
$$c+d=-6*1-14*(-4)=50$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/873506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.