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How can I simplify this expression
I know that I cannot just do $\frac{x^{-1}}{x^{-3}}$ and $\frac{y^{-1}}{y^{-3}}$ and get $x^{2}+y^{2}$ but how can i factor either the numerator or denominator to get something to cancel out?
|
$\dfrac{x^{-1}+y^{-1}}{x^{-3}+y^{-3}}=\dfrac{x^3y^3(x^{-1}+y^{-1})}{x^3y^3(x^{-3}+y^{-3})}= \dfrac{x^{2}y^3+x^3y^{2}}{x^{3}+y^{3}}=\dfrac{(x^2y^2)(x+y)}{x^{3}+y^{3}}=\dfrac{(x^2y^2)(x+y)}{(x+y)(x^2-xy+y^2)}=\dfrac{(x^2y^2)}{(x^2-xy+y^2)}$
|
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|
Solving Diophantine equation without brute force I am having trouble trying to solve this Diophantine equations:
$$
^3=4^2+4−3
$$
I was wondering if anyone could help me find the integer solutions to these and any advice of techniques to use?
Thank you!
|
Another way: You can rewrite the equation:
$$x^3+8=4y^2+4y+5$$
Or $$(x+2)(x^2-2x+4)=4y^2+4y+5$$
Let $(x+2;x^2-2x+4)=d (d\in N^+)$
Note that : $x^2-2x+4=x(x+2)-4(x+2)+12\Rightarrow d\mid 12$
We will prove $4y^2+4y+5\not\mid12$. Assume $4y^2+4y+5\mid12$, then we have $4y^2+4y+5=12k(k\in Z)$
$\Leftrightarrow (2y+1)^2=12k-4=4(3k-1)$
But $(2y+1)^2\equiv0;1\pmod{3}$ and $4(3k-1)\equiv2\pmod{3}$
Can you continue?
|
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|
prove that the triangle is isosceles
In a $\triangle ABC$, If
$\begin{vmatrix}
1 & \;\;1\;\;&\;\; 1\;\;\\\\
\displaystyle \cot \frac{A}{2} & \displaystyle \cot \frac{B}{2} & \displaystyle \cot \frac{C}{2} \\\\
\displaystyle \tan\frac{B}{2}+\tan \frac{C}{2} &\;\;\displaystyle \tan \frac{C}{2}+\tan\frac{A}{2} & \;\;\displaystyle\tan \frac{A}{2}+\tan \frac{B}{2}\end{vmatrix}=0$
Then prove that the triangle is isosceles.
Try: Let $\displaystyle \tan \frac{A}{2}=p\;\;,\tan \frac{B}{2}=q\;\;,\tan \frac{C}{2}=r$
Using $$\tan\bigg(\frac{A}{2}+\frac{B}{2}\bigg)=\tan\bigg(\frac{\pi}{2}-C\bigg)$$
So $$\sum \tan\frac{A}{2}\tan\frac{B}{2}=1\Rightarrow pq+qr+rp=1$$
So $$\begin{vmatrix}1& 1& 1\\
\displaystyle \frac{1}{p}& \displaystyle \frac{1}{q}& \displaystyle \frac{1}{r}\\
q+r & r+p& p+q\end{vmatrix}=0$$
So $$\frac{1}{pqr}\begin{vmatrix}p& q& r\\
\displaystyle 1 & \displaystyle 1 & \displaystyle 1 \\
p(q+r) & q(r+p)& r(p+q)\end{vmatrix}=0$$
So $$\frac{1}{pqr}\begin{vmatrix}p& q& r\\
\displaystyle 1 & \displaystyle 1 & \displaystyle 1 \\
1-qr & 1-rp& 1-pq\end{vmatrix}=0$$
So $$-\begin{vmatrix}p^2& q^2& r^2\\
\displaystyle p & \displaystyle q & \displaystyle r \\
1 & 1& 1\end{vmatrix}=0$$
So we have $(p-q)(q-r)(r-p)=0.$
So either $p=q$ or $q=r$ and $r=p.$
Could some help me some short way to solve it? Thanks.
|
With transformation $R_3 \longrightarrow R_3- \left(\tan\left(\dfrac{A}{2}\right)+\tan\left(\dfrac{B}{2}\right)+\tan\left(\dfrac{C}{2}\right)\right)R_1$
we get $\begin{vmatrix}
1 & \;\;1\;\;&\;\; 1\;\;\\\\
\displaystyle \cot \frac{A}{2} & \displaystyle \cot \frac{B}{2} & \displaystyle \cot \frac{C}{2} \\\\
\displaystyle -\tan\frac{A}{2} &\;\;\displaystyle -\tan \frac{B}{2} & \;\;\displaystyle -\tan \frac{C}{2}\end{vmatrix}$
Now factor out $(-1)$ from Row $3$ and multiply the columns by $\displaystyle \tan\frac{A}{2}, \displaystyle \tan\frac{B}{2}, \displaystyle \tan\frac{C}{2}$ respectively to obtain
$-\begin{vmatrix}
\displaystyle \tan\frac{A}{2} &\;\;\displaystyle \tan \frac{B}{2} & \;\;\displaystyle\tan \frac{C}{2}\;\;\\\\
1 & \;\;1\;\;&\;\; 1 \\\\
\displaystyle \tan^2\frac{A}{2} &\;\;\displaystyle \tan^2 \frac{B}{2} & \;\;\displaystyle\tan^2 \frac{C}{2}\end{vmatrix}$ which can be simplified to the Vandermonde Determinant and hence it evaluates to $\displaystyle \left(\tan\frac{A}{2} - \tan\frac{B}{2}\right)\left(\tan\frac{B}{2} - \tan\frac{C}{2}\right)\left(\tan\frac{C}{2} - \tan\frac{A}{2}\right)$
|
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|
Confusion with Summation notation I need to compute the value of this:
$$\frac{1}{3}\left(\sum_{n=1}^{3} \frac{z_{n}}{f} + N(2,2)\right)$$
the $N$ is a gaussian noise with mean=2 and standard deviation=2.
The question:
is this equivalent to:
$$
\frac{1}{3}\left(\frac{z_{1}}{f} + N(2,2)+\frac{z_{2}}{f} + N(2,2)+\frac{z_{3}}{f} + N(2,2)\right)
$$
or
$$
\frac{1}{3}\left(\left(\frac{z_{1}}{f} +\frac{z_{2}}{f} +\frac{z_{3}}{f}\right) + N(2,2)\right)
$$
|
It could be either one. It is more likely to be the latter; at least I myself would interpret it as
$$\frac{1}{3}\left(\sum_{n=1}^{3} \frac{z_{n}}{f} + N(2,2)\right)=\frac{1}{3}\left(\left(\frac{z_{1}}{f} +\frac{z_{2}}{f} +\frac{z_{3}}{f}\right) + N(2,2)\right),$$
because the former would more likely be written with extra braces, i.e.
$$\frac{1}{3}\left(\frac{z_{1}}{f} + N(2,2)+\frac{z_{2}}{f} + N(2,2)+\frac{z_{3}}{f} + N(2,2)\right)=\frac{1}{3}\left(\sum_{n=1}^{3}\left(\frac{z_{n}}{f} + N(2,2)\right)\right),$$
but not everyone is as generous with the braces, and some may omit them to avoid clutter.
Your best chances at interpreting the expression properly is to inform with whomever gave it to you.
|
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|
Determinant of block matrix with equal diagonals
Let $$Q=\begin{bmatrix}A&B\\-B&A\end{bmatrix}$$ where $A,B\in \mathbb{R}^{n\times n}$. Prove that $$\det(Q)=\det(A^2+B^2)$$
Since $A$ and $B$ do not commute, I cannot use Schur's formula. Also, $A$ and $B$ may not be invertible. Any comment or response is appreciated.
|
The trick, if $A$ and $B$ were commutative, is to notice that
$$\begin{bmatrix}A&B\\-B&A\end{bmatrix} \cdot \begin{bmatrix}A&-B\\B&A\end{bmatrix}=\begin{bmatrix}A^2+B^2&0\\0&A^2+B^2\end{bmatrix},$$
which I will refer to as $Q\cdot Q^\top = D$. It is true that $\det(M)=\det(M^\top)$ for any real matrix, so we obtain $\det(D)=\det(Q)^2$. All one has to show now is $\det(D)=\det(A^2+B^2)^2$, which is easy because of the zero block matrices. We conclude $\det(Q)=\pm\det(A^2+B^2)$.
If $A$ and $B$ do not commute, this does not hold in general. A counterexample is given by
$$A=\begin{bmatrix}1&0\\1&1\end{bmatrix},\quad B=\begin{bmatrix}1&1\\0&1\end{bmatrix}.$$
We have $AB=\begin{bmatrix}1&1\\1&2\end{bmatrix}$, $BA=\begin{bmatrix}2&1\\1&1\end{bmatrix}$, $A^2+B^2=\begin{bmatrix}2&2\\2&2\end{bmatrix}$ which yields $\det(A^2+B^2)=0$, but
$$
\det(Q)
=\det\left(\begin{bmatrix}A&B\\-B&A\end{bmatrix}\right)
=\det\left(\begin{bmatrix}1&0&1&1\\1&1&0&1\\-1&-1&1&0\\0&-1&1&1\end{bmatrix}\right)=1.
$$
|
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|
If nine coins are tossed, what is the probability that the number of heads is even? If nine coins are tossed, what is the probability that the number of heads is even?
So there can either be 0 heads, 2 heads, 4 heads, 6 heads, or 8 heads.
We have $n = 9$ trials, find the probability of each $k$ for $k = 0, 2, 4, 6, 8$
$n = 9, k = 0$
$$\binom{9}{0}\bigg(\frac{1}{2}\bigg)^0\bigg(\frac{1}{2}\bigg)^{9}$$
$n = 9, k = 2$
$$\binom{9}{2}\bigg(\frac{1}{2}\bigg)^2\bigg(\frac{1}{2}\bigg)^{7}$$
$n = 9, k = 4$
$$\binom{9}{4}\bigg(\frac{1}{2}\bigg)^4\bigg(\frac{1}{2}\bigg)^{5}$$
$n = 9, k = 6$
$$\binom{9}{6}\bigg(\frac{1}{2}\bigg)^6\bigg(\frac{1}{2}\bigg)^{3}$$
$n = 9, k = 8$
$$\binom{9}{8}\bigg(\frac{1}{2}\bigg)^8\bigg(\frac{1}{2}\bigg)^{1}$$
Add all of these up:
$$=.64$$ so there's a 64% chance of probability?
|
Your approach is good also, you probably made a mistake in calculations. The number of favorable outcomes is
$$\binom{9}{0}+\binom{9}{2}+\binom{9}{4}+\binom{9}{6}+\binom{9}{8}=1+36+126+84+9=256$$
The number of all possible outcomes is $512$ thus the probability to get even number of heads is $0.5$.
|
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|
Circle problem in which i wanted to find value of $PQ$ Find a value of $PQ$:
Let the radius of bigger circle be $R$, and that of the smaller circles be $r_1$ and $r_2$.
Hence we can find value of $R = 10$.
I don't know how to proceed further to find the Value of $r_1$ and $r_2$.
|
Using the power of the point $O$
with respect to circle $O_1$
we have
\begin{align}
|OT_2|\cdot|OT_1|&=|OP|^2
,\\
\text{or }\quad
R(R-2r_1)&=(R-(|AD|-r_1))^2
,\\
r_1&=
\sqrt{2R(2R-|AD|)}-(2R-|AD|)
\tag{1}\label{1}
.
\end{align}
Similarly,
\begin{align}
r_2&=\sqrt{2R\cdot|AD|}-|AD|
\tag{2}\label{2}
.
\end{align}
Note that given
$a=16=4\cdot4$,
$b=12=4\cdot3$
means that
$\triangle ABC$
is a scaled version of the famous $3-4-5$ triangle,
so $R$ and $|AD|$ can be easily calculated,
\begin{align}
R&=4\cdot\frac 52=10
,\\
|AD|&=4\cdot3\cdot\frac 35=\frac{36}5
,\\
r_1&=\frac{16}5
,\\
r_2&=\frac{24}5
,\\
|PQ|&=8
.
\end{align}
|
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|
Calculate the sum $S_n = \sum\limits_{k=1}^{\infty}\left\lfloor \frac{n}{2^k} + \frac{1}{2}\right\rfloor $ I am doing tasks from Concrete Mathematics by Knuth, Graham, Patashnik for trainning, but there are a lot of really tricky sums like that:
Calculate sum $$S_n = \sum_{k=1}^{\infty} \left\lfloor \frac{n}{2^k} + \frac{1}{2} \right\rfloor $$
My idea
I had the idea to check when $$\frac{n}{2^k} < \frac{1}{2}$$
because then $$ \forall_{k_0 \le k} \left\lfloor \frac{n}{2^k} + \frac{1}{2} \right\rfloor=0$$
It should be $$ k_0 = \log_2(2n) $$ but I don't know how it helps me with this task (because I need not only "stop moment" but also sum of considered elements
Book idea
Let $$S_n = \sum_{k=1}^{\infty} \left\lfloor \frac{n}{2^k} + \frac{1}{2} \right\rfloor $$
then $$ S_n-S_{n-1} = 1$$
and then solve this recursion. But I write $S_n - S_{n-1}$ and I don't see how it can be $1$ , especially that is an infinite sum.
|
Consider how $\left\lfloor \frac{n}{2^k} + \frac{1}{2} \right\rfloor$ depends on the binary representation of $n$. Dividing by $2^k$ shifts the digits to the right by $k$. There are then two cases:
*
*If the $k^{th}$ binary digit of $n$ is $0$, then adding $1/2$ will not cause a carry, so $\left\lfloor \frac{n}{2^k} + \frac{1}{2} \right\rfloor$ is just equal to the truncation of $n/2^k$ after the decimal point.
*If the $k^{th}$ binary digit of $n$ is $1$, then adding $1/2$ will cause a carry, so $\left\lfloor \frac{n}{2^k} + \frac{1}{2} \right\rfloor$ is just equal to this truncation plus one.
Now, letting $f_k(n)=\left\lfloor \frac{n}{2^k} + \frac{1}{2} \right\rfloor$ for brevity, consider the difference between $f_k(n)$ and $f_k(n-1)$. The difference between the binary representations of $n$ and $n-1$ is summarized as follows; if $n$ has $t$ trailing zeroes, then $n$ ends with $100\dots0$, while $n-1$ ends with $011\dots1$. Otherwise, the representations are the same.
*
*If $k>t+1$, then $f_k(n)=f_k(n-1)$, since the binary representations of $n$ and $n-1$ are equal after the first $t+1$ places, and the first $t+1$ places do not affect the computation of $f_k(n)$ and $f_k(n-1)$.
*If $k=t+1$, then $f_k(n)=f_k(n-1)+1$. The former has a carry, while the latter does not.
*If $k<t+1$, then $f_k(n)=f_k(n-1)$. The latter will have a carry, and the former will not. You can verify this on your own with some examples.
Therefore, in
$$
S_n-S_{n-1}=\sum_{k=1}^\infty \left\lfloor \frac{n}{2^k} + \frac{1}{2} \right\rfloor-\left\lfloor \frac{n-1}{2^k} + \frac{1}{2} \right\rfloor
$$
exactly one summand is equal to $1$ and the rest are zero, so $S_n-S_{n-1}=1$.
|
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|
Find range of $x$ satisfying $\left \lfloor \frac{3}{x} \right \rfloor+\left \lfloor \frac{4}{x} \right \rfloor=5$ Find range of $x$ satisfying $$\left \lfloor \frac{3}{x} \right \rfloor +\left \lfloor \frac{4}{x} \right \rfloor=5$$ Where $\lfloor\cdot\rfloor$ is the floor function
My try:
As far as domain of LHS is concerned we have $x \ne 0$ and since RHS is positive, we have $x \gt 0$
Now since LHS is sum of two positive integers, let us suppose:
$$\left \lfloor \frac{3}{x} \right \rfloor=m$$ and
$$\left \lfloor \frac{4}{x} \right \rfloor=5-m$$
Thus we have:
$$ m \le \frac{3}{x} \lt m+1$$
$$5-m \le \frac{4}{x} \lt 6-m$$
Adding both we get:
$$5 \le \frac{7}{x} \lt 7$$
$\implies$
$$1 \lt x \le \frac{7}{5}$$
Hence $$x \in (1, 1.4]$$
But answer in book is given as $$x \in (1,\frac{4}{3})$$
What went wrong?
|
The inequality after "Adding both we get:" is true, but it is not the whole story. You have lost information here, which means that not every solution to this inequality is a solution to both of the constituent inequalities.
Here is a simpler example: suppose we seek the range of solutions to
$$0<x<2$$ and $$1<x<3$$
Obviously the answer is $1<x<2$; but by your method, adding both gives
$$1<2x<5$$
which has a wider range of solutions. Your addition has lost the information that $x<2$ and $1<x$.
|
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|
Evaluate using Riemann sums $\int_{\pi/2}^{3\pi/2}(4\sin 3x - 3 \cos 4x)dx$ I have to evaluate this integral using Riemann sums. I've already tried:
$$\int_{\pi/2}^{3\pi/2}(4\sin 3x - 3\cos 4x)dx = [\delta x = \frac{\pi}{n}, x_i = \frac{\pi}{2} + \frac{i\pi}{n}] =$$$$= \lim_{n \rightarrow \infty}\sum_{i = 1}^n(4\sin(3(\frac{\pi}{2} + \frac{i\pi}{n})) - 3\cos(4(\frac{\pi}{2} + \frac{i\pi}{n})))\frac{\pi}{n}
= \lim_{n \rightarrow \infty}(\sum_{i = 1}^n(4\sin\frac{3(\pi n + 2\pi i)}{2n} - 3\sum_{i = 1}^n\cos \frac{4(\pi n + 2\pi i)}{2n})\frac{\pi}{n}
= \lim_{n \rightarrow \infty} (4\Re[\sum_{k = 1}^ne^{\frac{4i(\pi n + 2\pi k)}{2n}}] - 3\Re[\sum_{k = 1}^ne^{\frac{3i(\pi n + 2\pi k)}{2n}}])\frac{\pi}{n}
$$
But what can i do with this $e^{.....}$ expressions ? Or maybe here exists a simpler way to evaluate this using Riemann sums.
|
To evaluate the Riemann sums use the identities:
$$\sum_{k=1}^n \sin kx = \frac{\sin \frac{nx}{2} \sin \frac{(n+1)x}{2}}{\sin \frac{x}{2}}, \quad \sum_{k=1}^n \cos kx = \frac{\sin \frac{nx}{2} \cos \frac{(n+1)x}{2}}{\sin \frac{x}{2}}, $$
which can be derived as the imaginary and real parts of the geometric sum $\sum_{k=1}^n e^{ikx}$.
Thus,
$$\begin{align}\int_{\pi/2}^{3\pi/2} 4 \sin 3x \, dx &= \lim_{n \to \infty}\frac{4\pi}{n}\sum_{k=1}^n \sin\left(\frac{3\pi}{2} + k\frac{3\pi}{n} \right) \\ &= \lim_{n \to \infty}-\frac{4\pi}{n}\sum_{k=1}^n \cos\left(k\frac{3\pi}{n} \right) \\ &= \lim_{n \to \infty}-\frac{4\pi}{n}\frac{\sin \frac{3\pi n}{2n} \cos \frac{3\pi(n+1)}{2n}}{\sin \frac{3 \pi }{2n}} \\ &= \lim_{n \to \infty}-\frac{3}{8}\frac{\frac{3\pi}{2n}}{\sin \frac{3 \pi }{2n}}\sin \frac{3\pi}{2} \cos \left(\frac{3\pi}{2}(1 + 1/n)\right) \\ &= -\frac{3}{8} \cdot 1 \cdot (-1)\cdot 0 = 0\end{align}$$
The integral of $3 \cos 4x$ can be handled in a similar way.
|
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|
Solve a system equation in $\mathbb{R}$ - $\sqrt{x+y}+\sqrt{x+3}=\frac{1}{x}\left(y-3\right)$ how to solve a system equation with radical
$$\sqrt{x+y}+\sqrt{x+3}=\frac{1}{x}\left(y-3\right)$$
And $$\sqrt{x+y}+\sqrt{x}=x+3$$
This system has $1$ root is $x=1;y=8$,but i have no idea which is more clearly to solve it. I tried substituting and squaring to find the factor but failed.
|
Let $\sqrt{x+y}=a\ge0,\sqrt{x+3}=b\ge0$
$$(b^2-3)(a+b)=a^2-b^2$$
If $a+b=0,a=b=0$
Else $b^2-3=a-b\iff a=b^2+b-3=x+b\iff x=a-b\ \ \ \ (1)$
Squaring we get
$$x+y=x^2+x+3+2bx$$
$$\iff y=x^2+3+2x\sqrt{x+3}$$
The value of $x,y$ must satisfy $(1)$
|
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|
Algebraic equation I mean if we change interval of "x" to:
$k>1$ and $x\in(\sqrt[3]{k},\sqrt[5]{k^3})$.
Define the $B(x)$ as a polynomial as follows;
$B(x)=-2x^7-3x^6k+2k^2x^4-k^3x^3+2k^3x^2+3xk^4-2xk^3+k^4$
I want to check that whether $B(x)$ is positive?
if not, for which values of $k$ it is positive?
Do we have to repeat the method separately for:
$$ y:=\sqrt[3]k $$
then plot B(x,y) and find "y"
and also by defining
$$ z:=\sqrt[5]{k^3} $$
and plot B(x,z)
and find "z"?
And finally finding intersection of the solution in terms of "k?"
|
Let $y:=\sqrt[3]k$ and solve the equation
$$x^7+3x^6-2x^5-\left(y^3+1\right)x^4+\left(2y^3-3\right)x^2+1=0.$$
We easily draw
$$y=\sqrt[3]{\frac{x^7+3x^6-2x^5-x^4-3x^2+1}{x^4-2x^2}}$$ which is the zero set.
That curve is asymptotic to $x=y$ and $x=\sqrt2$, and there are two roots which satisfy $1<x<y$ for every value of $y$ above the minimum. The minimum is obtained by canceling the derivative, which amounts to $$3x^9+6x^8-12x^7-24x^6+12x^5+10x^4-4x^2+4=0.$$
We have one root $x\approx1.77368776,y\approx2.97774722$, then $k\approx26.4036205$.
There is also the isolated solution $x=1,y=1,k=1$.
|
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|
Prime numbers $pFind the prime numbers $p<q<r$ such that $r^2-q^2-p^2$ is a perfect square.
I think the only solution is (2,3,7) but i cannot prove it.
The equation would be $r^2-q^2-p^2=k^2$ equivalently
$q^2+p^2+k^2=r^2$ which is somehow a classical diophantine equation (of the pythagorean quadruples) which have a parametrization but how do i use it? The problem is from a magazine at 8 th grade and i don't think the solution is supposed to use the general solution.
Actually the general solution of $q^2+p^2+k^2=r^2$ is given by:
$ q=a^2+b^2-c^2-d^2$
$p=2(mq+np)$
$k=2(nq-mp)$
$r=a^2+b^2+c^2+d^2$
IT would follow that one of the numbers q or p is even and prime, so it is 2.
|
Your guess is correct: $(2,3,7)$ is the unique solution. There is a simple proof using two basic and easy to verify facts:
*
*if $m$ is odd, then $m^2\equiv 1\pmod 4$;
*if $m$ is not divisible by $3$, then $m^2\equiv 1\pmod 3$.
We proceed as follows.
Write $r^2-p^2-q^2=k^2$ with an integer $k$.
If we had $p>2$, this would imply $p^2\equiv q^2\equiv r^2\equiv 1\pmod 4$, leading to $k^2\equiv r^2-p^2-q^2\equiv 3\pmod 4$, which contradicts the first basic fact above. Thus $p=2$, whence $r^2=q^2+k^2+4$.
If we had $q>3$, this would imply $r^2\equiv q^2\equiv 1\pmod 3$ leading to $k^2\equiv 2\pmod 3$, which contradicts the second basic fact above. Thus $q=3$, and, as a result, $r^2=13+k^2$. Consequently, $(r-k)(r+k)=13$, which is only possible if $r-k=1$ and $r+k=13$. This yields $r=7$, $k=6$.
To summarize, the only solution is $p=2$, $q=3$, $r=7$.
|
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"url": "https://math.stackexchange.com/questions/3141520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Prove closed form for $\sum_{n\in\Bbb N}\frac1{5n(5n-1)}$ While looking for solutions to a difficult geometric problem, I encountered this sum:
$$
\sum_{n\in\Bbb N}\frac1{5n(5n-1)} = \frac1{4\cdot 5} + \frac1{9\cdot 10} + \frac1{14\cdot 15} + \ldots
$$
A bit of numerical exploring has convinced me that the answer is
$$
\sum_{n=1}^\infty\frac1{5n(5n-1)} = \frac14\log(5) + \frac{\sqrt{5}}{10}\log(\rho) -\frac{\pi}{10}\cot\left(\frac{\pi}{5} \right)
$$
where $\rho$ is the "Golden Ratio" $\rho = \frac{1+\sqrt{5}}{2}$.
But I can't find a way to prove it.
Prove
$$
\sum_{n=1}^\infty\frac1{5n(5n-1)} = \frac14\log(5) + \frac{\sqrt{5}}{10}\log(\rho) -\frac{\pi}{10}\cot\left(\frac{\pi}{5} \right)
$$
|
Is easy to prove the last step of Andreas:
$$
\sum_{n=1}^\infty\frac1{5n(5n-1)} =\\
\frac{\log 10}{5} + \frac{\pi}{10}\cot\left(\frac{4\pi}{5}\right) -\frac1{20} (\log(16/5) + \sqrt 5 (\log(5 - \sqrt 5) - \log(5 + \sqrt 5)))\\
\simeq 0.07756
$$
which is exactly as given in the question: $\frac14\log(5) + \frac{\sqrt{5}}{10}\log(\rho) -\frac{\pi}{10}\cot\left(\frac{\pi}{5} \right) \simeq 0.07756$
You just have to check these three adds:
a) $$ \frac{\log 10}{5} -\frac1{20} \log(16/5) = \frac14\log(5) $$
steps:
$$ \frac{\log 10}{5} -\frac1{20} \log(16/5) =\\ \frac{\log 2}{5}+\frac{\log 5}{5}-\frac{\log 16}{20}+\frac{\log 5}{20} =\\
\frac{\log 2}{5}+\frac{\log 5}{5}-\frac{4\log 2}{20}+\frac{\log 5}{20} =\\
(\frac{1}{5}-\frac{4}{20})\log2+(\frac{1}{5}+\frac{1}{20})\log5 =\\
\frac14\log(5)
$$
b) is trivial: $$ \frac{\pi}{10}\cot\left(\frac{4\pi}{5}\right) = -\frac{\pi}{10}\cot\left(\frac{\pi}{5} \right) $$
c) $$ -\frac1{20} (\sqrt 5 (\log(5 - \sqrt 5) - \log(5 + \sqrt 5))) = \frac{\sqrt{5}}{10}\log(\rho) $$
steps:
$$ -\frac1{2} (\log(5 - \sqrt 5) - \log(5 + \sqrt 5)) = \log(\rho) $$
$$ \log(5 + \sqrt 5) - \log(5 - \sqrt 5) = 2\log(\rho) $$
$$ \log \frac {5 + \sqrt 5} {5 - \sqrt 5} = \log(\rho^2) $$
$$ \frac {5 + \sqrt 5} {5 - \sqrt 5} = \rho^2 $$
$$ 5 + \sqrt 5 = (5 - \sqrt 5)\rho^2 $$
$$ 5 + \sqrt 5 = (5 - \sqrt 5)\frac{3+\sqrt 5}{2} $$
$$ 5 + \sqrt 5 = 5 + \sqrt 5 $$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Evaluating $\lim\limits_{x \to 0} \frac{(1+x)^{1/x} - e + \frac{1}{2}ex}{x^2}$ without expansions in limits
Evaluate $\lim\limits_{x \to 0} \frac{(1+x)^{1/x} - e + \frac{1}{2}ex}{x^2}$
One way that I can immediately think of is expanding each of the terms and solving like,
$$(1+x)^{1/x} = e^{\log_e (1+x)^{1/x}} = e^{\frac{1}{x} (x-\frac{x^2}{2} -\frac{x^3}{3}+...)}$$
and then after complete expansion of each and every and substuting into back to limit and solving I get $\frac{11e}{24}$ as an answer.
Now, this is a relatively long and complicated way to solve as you can see. I want to know if there is an easier way to solve this problem. Please help. Thank you!
|
Write $f(x) = \frac{1}{x}\log(1+x)$ and $f(0) = 1$. We know that $f$ so defined is analytics near $0$. Now, by the L'Hospital's rule applied twice,
\begin{align*}
\lim_{x\to0} \frac{e^{f(x)} - e + \frac{e}{2}x}{x^2}
&= \lim_{x\to0} \frac{e^{f(x)}f'(x) + \frac{e}{2}}{2x} \\
&= \lim_{x\to0} \frac{e^{f(x)}f''(x) + e^{f(x)}f'(x)^2}{2} \\
&= \frac{e}{2}f''(0) + \frac{e}{2}f'(0)^2.
\end{align*}
Since $ f(x) = 1 - \frac{1}{2}x + \frac{1}{3}x^2 + \cdots $ near $0$, it follows that $f'(0) = -\frac{1}{2}$ and $f''(0) = \frac{2}{3}$. Therefore the limit equals
$$ \frac{e}{2}\cdot\frac{2}{3} + \frac{e}{2}\left(-\frac{1}{2}\right)^2
= \frac{11}{24}e. $$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3144560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Proof that there exist only 17 Wallpaper Groups (Tilings of the plane) I had a professor who once introduced us to Wallpaper Groups. There are many references that exist to understand what they are (example Wiki, Wallpaper group).
The punchline is
$$There \,\, are \,\, exactly \,\, 17 \,\, wallpaper \,\, groups \,\,(17 \,\, ways \,\, to \,\, tile \,\, the \,\, plane)$$
My question is $2$-fold:
*
*Can someone sketch out the proof or at least give some high level ideas of why this may be true?
*Can someone refer me to a website or a textbook that develops the proof in detail?
|
Sketch of the proof: Let $\Gamma \le {\rm Iso}(\Bbb R^2)$ be a wallpaper group. Then $\Gamma$ has a normal subgroup isomorphic
to $\Bbb Z^2$ with finite quotient $F$. This finite group acts on the lattice $\Bbb Z^2$ by conjugation.
We obtain a faithful representation
$$
F \hookrightarrow {\rm Aut}(\Bbb Z^2)\cong GL_2(\Bbb Z).
$$
The group $GL_2(\Bbb Z)$ has exactly $13$ different conjugacy classes of finite subgroups,
called arithmetic ornament classes:
\begin{align*}
C_1 & \cong \left\langle \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right\rangle,\;
C_2 \cong \left\langle \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \right\rangle,\;
C_3 \cong \left\langle \begin{pmatrix} 0 & 1 \\ -1 & -1 \end{pmatrix} \right\rangle, \\
C_4 & \cong \left\langle \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \right\rangle, \;
C_6 \cong \left\langle \begin{pmatrix} 0 & 1 \\ -1 & 1 \end{pmatrix} \right\rangle, \;
D_1 \cong \left\langle \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \right\rangle, \\
D_1 & \cong \left\langle \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \right\rangle, \;
D_2 \cong \left\langle \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix},\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}
\right\rangle, \\
D_2 & \cong \left\langle \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}
\right\rangle, \; D_3 \cong \left\langle \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},
\begin{pmatrix} 0 & 1 \\ -1 & -1 \end{pmatrix}\right\rangle, \\
D_3 & \cong \left\langle \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix},\begin{pmatrix} 0 & 1 \\ -1 & -1 \end{pmatrix}
\right\rangle,\\
D_4 & \cong \left\langle \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}
\right\rangle, \; D_6\cong \left\langle \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},
\begin{pmatrix} 0 & 1 \\ -1 & 1 \end{pmatrix}\right\rangle.
\end{align*}
This is an easy computation. Here $C_1,C_2,C_3,C_4,C_6$ are cyclic groups and $D_1,D_2,D_3,D_4,D_6$ are dihedral groups. We use here, that the order $n$ of a subgroup must satisfy $\phi(n)=deg(\Phi_n)\mid 2$, so that $n=1,2,3,4,6$. This is called the crystallographic condition. The wallpaper groups arise from these $13$ classes by equivalence classes of extensions
$$
1\rightarrow \Bbb Z^2\rightarrow \Gamma\rightarrow F\rightarrow 1,
$$
determined by $H^2(F,\Bbb Z^2)$.
By computing $H^2(F,\Bbb Z^2)$ in each case we obtain $18$ inequivalent extensions, because in $13$ cases the cohomology is trivial, and in three cases we get $C_2,C_2$ and $C_2\times C_2$, i.e., $5$ additional possibilities, so that $13+5=18$.
This yields $17$ different groups, because two of them turn out to be isomorphic.
|
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"timestamp": "2023-03-29T00:00:00",
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|
What is the value of $\lim_{x \to 1^+} \frac{\sqrt{x^2-1}+x-1}{ \sqrt{4x-4}+x^2-1}$? $$\lim_{x \to 1^+} \frac{\sqrt{x^2-1}+x-1}{ \sqrt{4x-4}+x^2-1} = ?$$
I have done these steps to find the answer:
*
*$x^2-1=(x+1)(x-1)$
*$\sqrt{4x-4}=2\sqrt{x-1}$
*$\displaystyle\lim_{x \to 1^+} \frac{\sqrt{x^2-1}+x-1}{ \sqrt{4x-4}+x^2-1} = \lim_{x \to 1^+} \frac{\sqrt{ (x+1)(x-1) }+x-1}{ 2\sqrt{x-1} + (x+1)(x-1) }$
So how do I remove what causes the hole function not to become $\frac{0}{0}$ and solve the limit?
|
Set $x-1=t^2$, which you can because the limit is for $x\to1^+$. Then the limit becomes
$$
\lim_{t\to0^+}\frac{t\sqrt{t^2+2}+t^2}{2t+t^2(t^2+2)}=
\lim_{t\to0^+}\frac{\sqrt{t^2+2}+t}{2+t(t^2+2)}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3150178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Prove that $\Bbb Q(\sqrt{2},3^{1/3})=\Bbb Q(\sqrt{2}+3^{1/3})$
Prove that $\Bbb Q(\sqrt{2},3^{1/3})=\Bbb Q(\sqrt{2}+3^{1/3})$
My attempt:
Firstly, since, $\Bbb Q(\sqrt{2}+3^{1/3}) \subseteq \Bbb Q(\sqrt{2},3^{1/3})$ , I computed $(\sqrt{2}+3^{1/3})^{-1} = 6-4\sqrt{2}+4(3^{1/3})+3(3^{2/3})-3(\sqrt{2}3^{1/3})-2(\sqrt{2}3^{2/3})$ and also tried some naive manipulation by considering powers of $(\sqrt{2}+3^{1/3})$ and tried to eliminate terms ( imitating the proof for $\Bbb Q(\sqrt{2},\sqrt{3})=\Bbb Q(\sqrt{2}+\sqrt{3})$ ) but it gets very messy.
Secondly, I am trying this problem by Galois theory but since, $\Bbb Q(\sqrt{2},3^{1/3})|\Bbb Q$ is not Galois,I don't get how to proceed.
Thanks in advance for help!
|
Let $\alpha = \sqrt2+ \sqrt[3]{3}$.
First isolate $\sqrt[3]{3}$ and raise everything to the third power:
$$\sqrt[3]{3} = \alpha - \sqrt{2} \implies 3 = (\alpha - \sqrt{2})^3 = \alpha^3-3\alpha^2\sqrt{2}+6\alpha - 2\sqrt{2}$$
Then isolate $\sqrt{2}$
$$\sqrt{2}(3\alpha^2+2) = \alpha^3+6\alpha-3$$
so $$\sqrt{2} = \frac{\alpha^3+6\alpha-3}{3\alpha^2+2} \in \mathbb{Q}(\alpha)$$
Then also $\sqrt[3]{3} = \alpha - \sqrt{2} \in \mathbb{Q}(\alpha)$ so $\mathbb{Q}(\alpha) = \mathbb{Q}(\sqrt2, \sqrt[3]{3})$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3151252",
"timestamp": "2023-03-29T00:00:00",
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|
Lower Bound on the Sum of Reciprocal of LCM While reading online, I encountered this post which the author claims that
\begin{align}
S(N, 1):=\sum_{1\le i, j \le N} \frac{1}{\text{lcm}(i, j)} \geq 3H_N-2
\end{align}
and $S(N, 1) \geq H_N^2$ where $H_N$ is the partial harmonic sum.
The prove of the latter inequality is already given in the post. For the former, we see that
\begin{align}
S(N, 1)
=&\ 2\sum^N_{j=2}\sum^{j-1}_{i=1}\frac{1}{\text{lcm}(i, j)} +H_N\\
\geq&\ 2\sum^N_{j=2} \frac{1}{\text{lcm}(1, j)}+H_N=\ 3H_N -2.
\end{align}
My question is whether there exists $C$, independent of $N$, such that the following bound holds
\begin{align}
s(N):=\sum_{N \leq i, j \leq 2N} \frac{1}{\text{lcm}(i, j)} \geq C \log N.
\end{align}
Actually, I know for a fact that this is true by $(1.7)$ in this paper but couldn't prove it.
Another question: How does one show
\begin{align}
\sum_{N^2/4< k< N^2/3} (\#\left\{N/3<q<N/2 \big|\ q\mid k\right\})^2\sim \sum_{N/3<q, q'<N/2} \frac{N^2}{\text{lcm}(q,q')}.
\end{align}
I started by
\begin{align}
\sum_{N^2/4< k< N^2/3} (\#\left\{N/3<q<N/2 \big|\ q\mid k\right\})^2 =&\ \sum_{N^2/4< k< N^2/3}\left( \sum_{\substack{N/3<q<N/2\\ q\mid k}}1 \right)^2\\
=&\ \sum_{N^2/4< k< N^2/3} \sum_{\substack{N/3<q,\ q'<N/2\\ q\mid k,\ q'\mid k}} 1.
\end{align}
Here I know I should interchange the order of summation, but I don't know how to get the lcm to show up.
Any help/suggestion is highly welcome.
|
Sketch proof for 1: assume that there is $c>0$ absolute constant s.t for each $1 \leq a \leq N^{\frac{1}{4}}$ ($N$ high enough), there are at least $c\frac{N^2}{a^2}$ pairs $N \leq k,l \leq 2N$ with $gcd(k,l)=a$; then as $lcm(k,l)gcd(k,l)=kl$,
$s(N):=\sum_{N \leq k, l \leq 2N} \frac{1}{\text{lcm}(k, l)} =\sum_{N \leq k, l \leq 2N} \frac{gcd(k,l)}{kl} \geq \frac{1}{4N^2}\sum_{N \leq k, l \leq 2N}gcd(k,l) \geq \frac{1}{4N^2}\sum_{1 \leq a \leq N^{\frac{1}{4}}}a(c\frac{N^2}{a^2}) \geq \frac{c}{16}\log N $
Let now $1 \leq a \leq N^{\frac{1}{4}}$, considering the first number $N \leq k(a) < N+a$ which is divisble by $a$, it follows that $k(a)+qa$ is divisble by $a$ and within our range as long as $(q+1)a \leq N$, so there are at least $[\frac{N}{a}-1]$ such, so at least $[\frac{N}{a}-1]^2$ pairs with gcd divisible by $a$; same reasoning give at most $[\frac{N}{a}+1]^2$ such pairs, so applying this with $2a,3a...2Na$ (since the gcd is at most $2N$ in our interval) we get that the number of pairs with gcd precisely $a$ is at least: $\begin{align}\frac{N^2}{a^2}(1-\frac{1}{4}-\frac{1}{9}-.. )- 2N(\log 2N+1) \geq \frac{N^2}{a^2}(1-(\frac{\pi^2}{6}-1))-2N(\log 2N+1) \end{align}$ $\geq \frac{N^2}{6a^2}-2N(\log 2N+1)$, and that is $ \geq \frac{N^2}{12a^2}$ for say $N \geq 40000$ since $a \leq N^{\frac{1}{4}}, \frac{N^2}{12a^2} \geq \frac{N^2}{12\sqrt N} \geq 2N(\log 2N+1)$, for $N$ large as noted so we are done.
For the second claim, note that switching the sum, each ${N^2/4< k< N^2/3}$ appears for a pair $(q,q')$ precisely when it is a multiple of the lcm of the pair and it is fairly clear that for any $m \geq 1$ in a ~$N^2$ interval there are ~$\frac{N^2}{m}$ multiples of it, so the claim follows
Edit: as requested a more detailed explanation of the estimate above for the number of pairs with gcd $a$:
because the number of multiples of $a$ between $N, 2N$ is between $\frac{N}{a}-1, \frac{N}{a}+1$, this means that we have at least $(\frac{N}{a}-1)^2=\frac{N^2}{a^2}-2\frac{N}{a}+1$ pairs with gcd a multiple of $a$, but if the gcd is not $a$, it means it is $2a, 3a,...2Na$ (this last is very crude and we can do much better but we do not need), but for $2a$ we have at most $(\frac{N}{2a}+1)^2=\frac{N^2}{4a^2}+2\frac{N}{2a}+1$ pairs with gcd that (actually less since some will have gcd $4a$ etc but we again use a rough estimate as that is enough); putting things together we get by subtracting all those cases with gcd $2a,3a...2Na$, the main term is $\frac{N^2}{a^2}(1-\frac{1}{4}-\frac{1}{9}-.. )$ which we estimate by $\frac{N^2}{6a^2}$ with a simple computation; the remainder is $(-2\frac{N}{a}+1)+(-2\frac{N}{2a}-1)+(-2\frac{N}{3a}-1)+...$ (at most $2N$ terms) and that clearly estimates by $-2N\log 2N - 2N$, so we get the estimate above and use that $a \leq N^{\frac{1}{4}}$ to conclude as noted for large enough $N$ since obviously $\frac{N^2}{12a^2} \geq \frac{N^2}{12\sqrt N}$ will dominate $2N\log 2N + 2N$, allowing us to keep another $\frac{N^2}{12a^2}$ from the main term
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve the Diophantine equation $24x^4-5y^4=z^2$ I want to solve $24x^4-5y^4=z^2$ in integers not all zero, and to fix ideas, I want to find them such that x,y are coprime.
I've tried plugging in small values of $x$ and $y$ and they don't return a square value for $24x^4-5y^4$. So I guess that no (nontrivial) solutions exist.
How would I prove this? I can't seem to derive a contradiction in $\mathbb{R}$ nor in $\mathbb{Q}_p$. So I guess that this equation violates Hasse's local-global principle. That is, I have to derive a contradiction by some other method than arguing in $\mathbb{R}$ or mod $p$ for some prime $p$.
The only example I know of a method for solving an equation that violates Hasse's principle is for the example $2a^2+17b^4=c^4$. I shall now describe the method, and my question is, does this method generalize to my equation $24x^4-5y^4=z^2$? (And if not, how else would I go about solving it?)
Here is the method I know for showing that no integers solutions exist to $2a^2+17b^4=c^4$, where $a,b,c$ are not all zero, and $(b,c)=1$. For any prime $p$ dividing $a$, since $17b^4 \equiv c^4$ modulo $p$, we get that $17$ is a quadratic residue mod $p$. By quadratic reciprocity, we get that $p$ is a quadratic residue mod $17$, where $p$ was any prime factor of $a$.
This implies that $a$ is a quadratic residue mod $17$. Write $a \equiv x^2$ mod $17$. Then $2x^4 \equiv c^4$ mod $17$, and then $2$ is a fourth power mod $17$. This is a contradiction.
|
Here's a rudimentary proof. Suppose $z^2=24x^4-5y^4$. Then looking modulo $4$ gives
\begin{align}
z^2 &= 24x^4-5y^4\\
&\equiv -y^4\mod 4
\end{align}
Since the only squares modulo $4$ are $0$ and $1$, this implies $z^2\equiv y^4\equiv 0\mod 4$, so $2|z,y$. Then setting $\overline{y}=\frac{y}{2}$ and $\overline{z}=\frac{z}{2}$, our original equation becomes
\begin{align}
4\overline{z}^2 &= 24x^4-16\cdot5\overline{y}^4\\
\overline{z}^2 &= 6x^4-4\cdot5\overline{y}^4\\
\end{align}
This then immediately implies that $2|\overline{z}$, so we can again set $w=\frac{\overline{z}}{2}$ and find
\begin{align}
4w^2 &= 6x^4-4\cdot5\overline{y}^4\\
2w^2 &= 3x^4-2\cdot5\overline{y}^4\\
\end{align}
Now this gives $2|x$, so we finally set $\overline{x}=\frac{x}{2}$ and get
\begin{align}
2w^2 &= 16\cdot3\overline{x}^4-2\cdot5\overline{y}^4\\
w^2 &= 24\overline{x}^4-5\overline{y}^4.
\end{align}
Thus we get a new solution with every term strictly smaller, so by infinite descent no solutions can exist.
|
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|
How to compute $\gcd(d^{\large 671}\! +\! 1, d^{\large 610}\! −\!1),\ d = \gcd(51^{\large 610}\! +\! 1, 51^{\large 671}\! −\!1)$ Let $(a,b)$ denote the greatest common divisor of $a$ and $b$.
With $ \ d = (51^{\large 610}\! + 1,\, 51^{\large 671}\! −1)$
and $\ \ x \,=\, (d^{\large 671} + 1,\, \ d^{\large 610} −1 )$
find $\ X = (x\bmod 10)$
I used $y=51^{61}$ to reduce $d$ to $d=(y^{10}+1,y^{11}-1) = (y^{10}+1,y+1)$.
What should I do now?
|
First note: $\gcd(a^m \pm 1, a+1)=\gcd((a^{m}\pm 1)-(a^{m}+a^{m-1}),a+1) = \gcd(a^{m-1}\mp 1, a+1)$
And via induction $\gcd(a^{m}+1, a+1) = \gcd(2, a+1)$ if $m$ is even. $\gcd(a^{m} - 1,a+1) =\gcd(0, a+1) = a+1$ if $m$ is even. (And the opposite results if $m$ is odd).
so if we let $a= 51^{61}$.
Then $\gcd(51^{610} + 1, 51^{671} - 1)=$
$\gcd(a^{11}-1,a^{10} + 1)=$
$\gcd((a^{11} -1)-(a^{11} + a), a^{10} + 1) =$
$\gcd(a^{10} + 1, a+ 1) = 2$
...
Let $b = 2^{61}$ and so
$\gcd(2^{671}+1, 2^{610} -1)= \gcd (b^{11} + 1, b^{10} -1)=$
$\gcd((b^{11}+1)-(b^{11}-b), b^{10}-1)=\gcd(b^{10}-1,b+1)=b+1= 2^{61}+1$
|
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|
Prove that $\frac{1}{ab}+\frac{1}{cd} \geq \frac{a^2+b^2+c^2+d^2}{2}$ Let $a,b,c,d$ be positive real numbers such that $a+b+c+d=4$. Prove that
$$\frac{1}{ab}+\frac{1}{cd} \geq \frac{a^2+b^2+c^2+d^2}{2}$$
I write
$$a^2+b^2+c^2+d^2=16-2\left(ab+cd+\left(a+b\right)\left(c+d\right)\right)$$. Then the inequality is equivalent to
$$\frac{1}{ab}+\frac{1}{cd} +ab+cd+\left(a+b\right)\left(c+d\right) \geq 8$$
But now I don't know how to change $ab,cd$ into the forms of $a+b$ and $c+d$. Moreover, from the last inequality, we have $\frac{1}{ab}+\frac{1}{cd} \geq 4$, whereas $\left(a+b\right)\left(c+d\right) \leq 4$. I cannot handle this. Please help me.
|
$ \frac{2}{a^2+b^2} \leq \frac{1}{ab} , \frac{2}{c^2+d^2} \leq \frac{1}{cd}$
$ \Rightarrow \frac{2(a^2+b^2+c^2+d^2)}{(a^2+b^2)(c^2+d^2)} \leq \frac{1}{ab} + \frac{1}{cd}$
$ \Rightarrow \frac{a^2+b^2+c^2+d^2}{2} \cdot (\frac{4}{(a^2+b^2)(c^2+d^2)} - 1)\leq \frac{1}{ab} + \frac{1}{cd} - \frac{a^2+b^2+c^2+d^2}{2}$
also, $ \frac{4}{(a^2+b^2)(c^2+d^2)} \geq \frac{1}{abcd} \geq 1 (\because 1 = (\frac{a+b+c+d}{4})^4 \geq abcd)$
$ \therefore 0 \leq \frac{a^2+b^2+c^2+d^2}{2} \cdot (\frac{4}{(a^2+b^2)(c^2+d^2)} - 1)\leq \frac{1}{ab} + \frac{1}{cd} - \frac{a^2+b^2+c^2+d^2}{2} , \frac{1}{ab} + \frac{1}{cd} \geq \frac{a^2+b^2+c^2+d^2}{2} $
|
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"timestamp": "2023-03-29T00:00:00",
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|
Determine the number of ordered triple $(x, y, z)$ of integer numbers (negatives and positives) satisfying $|x| + |y| + |z| \le 6$ Determine the number of ordered triple $(x, y, z)$ of integer numbers (negatives and positives) satisfying $|x| + |y| + |z| \le 6$
I know that final answer is 377, but how?
Edit:
Drawing from David K's answer:
One way to count the ways is to first do the ways for which $x > 0,$ $y > 0,$ and $z > 0.$
That's the number of ways to put $6$ or fewer indistinguishable balls into $3$ numbered bins under the constraint that every bin must have at least one ball,
which is the number of ways to put $3$ or fewer indistinguishable balls in $3$ numbered bins without that constraint, which is the number of ways of putting exactly $3$ indistinguishable balls in $4$ numbered bins.
Multiply by $8$ to take into account all the cases where $x < 0$ or $y < 0$ or $z < 0.$
$x,y,z >0 => x + y + z <= 3, n = Cr(3+4-1,3) * (8) = 160$
Now let $x = 0,$ $y > 0,$ and $z > 0.$
Count the number of ways to put up to $6$ balls in $2$ bins if each bin must contain at least one ball. Multiply by four to account for all the cases were $y < 0$ or $z < 0.$ Multiply that result by $3$ to account for the fact that we could have chosen $y= 0$ or $z=0$ instead of $x = 0.$
$x=0, x,y>0 => y+z <=6, n = Cr(4+3-1, 4) * 3 * 14 = 180$
Now let $x = y = 0$ and $z > 0.$ There are $6$ ways for that to happen. Multiply by $2$ to account for $z < 0,$ then by $3$ to account for the other choices of which variables are zeros.
$x,y = 0, z>0 => z<=6, n = 6 * 2 * 3 = 36$
$x,y,z=0, n = 1$
Sum of all of them is : 160 + 180 + 36 + 1 = 377
|
Here is a brute force solution. Letters $x,y,z$ stand for nonzero integers.
Case 1. Permutations of $(x,x,x)$.
$(1,1,1)$
$(2,2,2)$
$2$ combinations, one permutation of each, $2^3=8$ ways to assign $\pm$ signs, so $2\cdot1\cdot8=\boxed{16}$ solutions.
Case 2. Permutations of $(x,x,y)$, $x\ne y$.
$(1,1,2)$
$(1,1,3)$
$(1,1,4)$
$(2,2,1)$
$4$ combos, $3$ perms each, $2^3=8$ sign choices, so $4\cdot3\cdot8=\boxed{96}$ solutions.
Case 3. Perms of $(x,y,z)$ with $x,y,z$ all different.
$(1,2,3)$
Just $1$ combo, $6$ perms, $8$ sign choices, so $1\cdot6\cdot8=\boxed{48}$ solutions.
Case 4. Perms of $(x,x,0)$.
$(1,1,0)$
$(2,2,0)$
$(3,3,0)$
$3$ combos, $3$ perms each, $2^2=4$ sign choices, so $3\cdot3\cdot4=\boxed{36}$ solutions.
Case 5. Perms of $(x,y,0)$, $x\ne y$.
$(1,2,0)$
$(1,3,0)$
$(1,4,0)$
$(1,5,0)$
$(2,3,0)$
$(2,4,0)$
$6$ combos, $6$ perms each, $4$ sign choices, so $6\cdot6\cdot4=\boxed{144}$ solutions.
Case 6. Perms of $(x,0,0)$.
$(1,0,0)$
$(2,0,0)$
$(3,0,0)$
$(4,0,0)$
$(5,0,0)$
$(6,0,0)$
$6$ combos, $3$ perms each, $2$ sign choices, so $6\cdot3\cdot2=\boxed{36}$ solutions.
Case 7. Perms of $(0,0,0)$.
$(0,0,0$.
Just $1\cdot1\cdot1=\boxed{1}$ solution
The final answer is
$$16+96+48+36+144+36+1=\boxed{377}$$
provided $0$'s are allowed. However, if you really mean "negatives and positives" (no $0$'s) then you only get
$$16+96+48=\boxed{160}$$.
|
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|
Prove: if $c^2+8 \equiv 0$ mod $p$ then $c^3-7c^2-8c$ is a quadratic residue mod $p$. I want to show:
If $c^2+8 \equiv 0$ mod $p$ for prime $p>3$, then $c^3-7c^2-8c$ is a quadratic residue mod $p$.
I have calculated that $c^3-7c^2-8c \equiv -7c^2-16c \equiv 56- 16c \equiv 8(7-2c) \equiv c^2 (2c -7)$, so it should be enough to see that $2c -7$ is a quadratic residue. What now?
|
$$2c-7\equiv2c-7+c^2+8\pmod p\equiv(c+1)^2$$
|
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|
Trigonometric residue integral The trigonometric complex integral is
$$
\int_0^{2\pi} \frac 1{(2+\sin \theta)^2} d\theta.
$$
My answer is below, and it is correct because I verified this numerically on WolframAlpha: https://www.wolframalpha.com/input/?i=%5Cint_0%5E(2pi)+1%2F%7B(2%2B%5Csin+%5Ctheta)%5E2%7D+d%5Ctheta
|
I applied the substitution $z=e^{i\theta}$ and then used the residue theorem to obtain
\begin{align*}
\int_0^{2\pi} \frac 1{(2+\sin \theta)^2} d\theta &= \int_{|z|=1} \frac 1{(2+\frac 1{2i}(z-\frac 1z))^2} \frac{dz}{iz} \\
&= 4i\int_{|z|=1} \frac z{(z^2+4iz-1)^2} dz \\
&= 4i \cdot 2\pi i \operatorname{Res}_{z=(-2+\sqrt 3)i} \frac z{(z^2+4iz-1)^2} \\
&= -8\pi \frac d{dz}\left.\left((z-(-2+\sqrt 3)i)^2 \frac z{(z^2+4iz-1)^2} \right)\right\vert_{z=(-2+\sqrt 3)i} \\
&= -8\pi \frac d{dz}\left.\left(\frac z{(z-(-2-\sqrt 3)i)^2}\right)\right\vert_{z=(-2+\sqrt 3)i} \\
&= -8\pi\left.\left(\frac{-z^2-(2+\sqrt 3)^2}{(z+(2+\sqrt 3)i)^4}\right)\right\vert_{z=(-2+\sqrt 3)i} \\
&= \boxed{\frac{4\pi\sqrt 3}9}
\end{align*}
|
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|
Complex sum of sine and cosine functions
By using the complex representations of sine and cosine, show that $$\sum_{m=0}^n\sin m\theta =\frac{\sin\frac{n}{2}\theta\sin\frac{n+1}{2}\theta}{\sin\frac{1}{2}\theta}$$
So I am not too sure how to go about this proof. I tried to substitute sine for its complex representation but I can't see how to evaluate the sum?
|
Based upon the finite geometric series formula and
$\sin \theta=\frac{1}{2i}\left(\exp(i\theta)-\exp(-i\theta)\right)$
we obtain:
\begin{align*}
&\color{blue}{\frac{\sin\frac{n}{2}\theta\sin\frac{n+1}{2}\theta}{\sin\frac{1}{2}\theta}}\\
&\qquad=\frac{\frac{1}{2i}\left(\exp\left(i\frac{n}{2}\theta\right)-\exp\left(-i\frac{n}{2}\theta\right)\right)
\frac{1}{2i}\left(\exp\left(i\frac{(n+1)}{2}\theta\right)-\exp\left(-i\frac{(n+1)}{2}\theta\right)\right)}
{\frac{1}{2i}\left(\exp\left(i\frac{\theta}{2}\right)-\exp\left(-i\frac{\theta}{2}\right)\right)}\\
&\qquad=\frac{\exp\left(i\left(n+\frac{1}{2}\right)\theta\right)-\exp\left(i\frac{\theta}{2}\right)
-\exp\left(-i\frac{\theta}{2}\right)+\exp\left(-i\left(n+\frac{1}{2}\right)\theta\right)}
{\frac{1}{2i}\left(\exp\left(i\frac{\theta}{2}\right)-\exp\left(-i\frac{\theta}{2}\right)\right)}\\
&\qquad=\frac{\exp\left(-i\frac{\theta}{2}\right)\left(\exp\left(i(n+1)\theta\right)-1\right)}{2i\exp\left(-i\frac{\theta}{2}\right)\left(\exp\left(i\theta\right)-1\right)}
-\frac{\exp\left(i\frac{\theta}{2}\right)\left(1-\exp\left(-i(n+1)\theta\right)\right)}{2i\exp\left(i\frac{\theta}{2}\right)\left(1-\exp\left(-i\theta\right)\right)}\\
&\qquad=\frac{1}{2i}\sum_{m=0}^n\exp\left(im\theta\right)-\frac{1}{2i}\sum_{m=0}^n\exp\left(-im\theta\right)\\
&\qquad=\sum_{m=0}^n\frac{1}{2i}\left(\exp\left(im\theta\right)-\exp\left(-im\theta\right)\right)\\
&\qquad\,\,\color{blue}{=\sum_{m=0}^n\sin m\theta}
\end{align*}
and the claim follows.
|
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|
Proof for $\frac{\cos^6 x}{\cos 6x}=1$ I tried some transformations but nothing is bringing me to 1.
$$\frac{\cos^6 x}{\cos 6x} = 1$$
I tried something like $\left(\cos^2 x\right)^3$, then $\left(2\cos^2x-1\right)^3$. And, for $\cos 6x$, to transform into $\cos(2(3x))$ or $\cos(3 (2x))$. But it's not going well.
|
Let $\cos^2x=t$.
Thus, since $$\cos3\alpha=4\cos^2\alpha-3\cos\alpha,$$ we obtain:
$$\cos^6x=\cos6x$$ or
$$t^3=2(4\cos^3x-3\cos x)^2-1$$ or
$$t^3=2t(4t-3)^2-1$$ or
$$31t^3-48t^2+18t-1=0$$ or
$$31t^3-31t^2-17t^2+17t+t-1=0$$ or
$$(t-1)(31t^2-17t+1)=0.$$
1. $t=1$.
Thus, $$\cos^2x=1$$ or $$\sin{x}=0$$ or $$x=\pi k,$$ where $k\in\mathbb Z$.
*$t=\frac{17+\sqrt{17^2-4\cdot31}}{62}=\frac{17+\sqrt{165}}{62}.$
Here, $$\frac{1+\cos2x}{2}=\frac{17+\sqrt{165}}{62}$$ or
$$\cos2x=\frac{\sqrt{165}-14}{31},$$ which gives
$$x=\pm\frac{1}{2}\arccos\frac{\sqrt{165}-14}{31}+\pi k.$$
3. $t=\frac{17-\sqrt{165}}{62}.$
Here, $$\cos2x=\frac{-14-\sqrt{165}}{31},$$ which gives
$$x=\pm\frac{1}{2}\arccos\frac{-14-\sqrt{165}}{31}+\pi k.$$
|
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|
Convergence of $\sum_{n=1}^\infty\frac{k^2}{k^2-2k+5}$, $\sum_{n=1}^\infty\frac{6\cdot 2^{2n-1}}{3^n}$, and $\sum_{n=1}^\infty\frac{2^n+4^n}{e^n}$ I have three series questions:
1) $$\sum_{n=1}^{\infty} \frac{k^2}{k^2 - 2k + 5}$$
I'm going to use this theorem:
So $$\lim_{x\to \infty} \frac{x^2}{x^2-2x+5}$$
So $$\lim_{x\to \infty} \frac{2x}{2x-2}$$
So $$\lim_{x\to \infty} \frac{2}{2} = 1$$
Is that right?
2) $$\sum_{n=1}^{\infty} \frac{6 \cdot 2^{2n-1}}{3^n}$$
I'm a tad stuck. I can get to here:
$$\sum_{n=1}^{\infty} \frac{2 \cdot 3 \cdot 2^{2n-1}}{3 \cdot 3^{n-1}}$$
$$\sum_{n=1}^{\infty} \frac{2^{2n}}{3^{n-1}}$$
Is this valid:
$$\sum_{n=1}^{+\infty} \frac{2^{n} \cdot 2^{n}}{3^{n-1}}$$
$$\sum_{n=1}^{+\infty} \frac{2^{n} \cdot 2 \cdot 2^{n-1}}{3^{n-1}}$$
$$\sum_{n=1}^{+\infty} 2^{n+1} \frac{2^{n-1}}{3^{n-1}}$$
So using this theorem:
it looks like this could be:
$$a_n = \frac{1}{1-r} = \frac{2^{n+1}}{\frac{1}{3}}$$
which diverges. Is this right? Is there a better way?
3) $$\sum_{n=1}^{\infty} \frac{2^n + 4^n}{e^n}$$
I have no idea where to even start.
|
In both cases you can use the necessary condition for convergence of a series:
If
$$\sum_{n=0}^\infty a_n$$
converges, then
$$a_n\to 0$$
(the same is true if the sum starts at some integer different from zero).
This implies that if $a_n\to 0$ is not true (whether because the limit is a non zero number, or infinity, or it doesn't exist), then
$$\sum_{n=0}^\infty a_n$$
does not converge.
In the first case, you have
$$\frac{n^2}{n^2-2n+5}=\frac{n^2}{n^2}\cdot\frac{1}{1-\tfrac2n+\tfrac5{n^2}}\to 1$$
as $n\to \infty$.
So
$$\sum_{n=0}^{\infty}\frac{n^2}{n^2-2n+5}$$
does not converge.
You can use a similar argument in the second example, since
$$\frac{6\cdot2^{2n-1}}{3^n}=\frac{6\cdot 2^{2n}\cdot 2^{-1}}{3^n}=\frac{6\cdot (2^2)^n}{2\cdot 3^n}=3\left(\frac43\right)^n\to \infty$$
as $n\to\infty$, since $\frac43>1$.
|
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|
Proving difference between two functions is small when $x$ is small... I'm really struggling to prove the following claim and I was wondering whether anyone could help me.
Claim:$$ 0<|x| \leq 10^{-4} \implies \bigg{|} \frac{x}{e^x-1} - \frac{1}{\sum_{k=1}^3 \frac{x^{k-1}}{k!}} \bigg{|} \leq 10^{-12}$$
My attempt at a proof: I have already proven the following inequality,
$$\bigg{|} \frac{e^x-1}{x} - \sum_{k=1}^n \frac{x^{k-1}}{k!} \bigg{|} \leq \frac{|x|^ne^{|x|}}{(n+1)!}$$
So let $n=3$, then
\begin{align*}
\bigg{|} \frac{x}{e^x-1} - \frac{1}{\sum_{k=1}^3 \frac{x^{k-1}}{k!}} \bigg{|} &= \bigg{|} \frac{x}{(e^x-1)(\sum_{k=1}^3 \frac{x^{k-1}}{k!})} \cdot \left( \frac{e^x-1}{x} - \sum_{k=1}^3 \frac{x^{k-1}}{k!} \right) \bigg{|} \\
&= \bigg{|} \frac{x}{(e^x-1)(\sum_{k=1}^3 \frac{x^{k-1}}{k!})} \bigg{|} \cdot \bigg{|} \frac{e^x-1}{x} - \sum_{k=1}^3 \frac{x^{k-1}}{k!} \bigg{|} \\
&\leq \bigg{|} \frac{x}{(e^x-1)(\sum_{k=1}^3 \frac{x^{k-1}}{k!})} \bigg{|} \cdot \frac{|x|^3e^{|x|}}{(3+1)!} \\
&\leq \bigg{|} \frac{x}{(e^x-1)(\sum_{k=1}^3 \frac{x^{k-1}}{k!})} \bigg{|} \cdot \frac{(10^{-4})^3 e^{10^{-4}}}{4!} \\
&= 10^{-12} \cdot \bigg{|} \frac{x}{(e^x-1)(\sum_{k=1}^3 \frac{x^{k-1}}{k!})} \bigg{|} \cdot \frac{ e^{10^{-4}}}{4!}
\end{align*} Hence, it suffices to show that for $0 < |x| \leq 10^{-4}$,
\begin{align*}
f(x) := \bigg{|} \frac{x}{(e^x-1)(\sum_{k=1}^3 \frac{x^{k-1}}{k!})} \bigg{|} < \frac{4!}{e^{10^{-4}}}
\end{align*}
Since $f(x)$ is monotonically decreasing on the intervals $[-10^{-4}, 0)$ and $(0, 10^{-4}]$, we know that for $0 <|x| \leq 10^{-4}$,
\begin{align*}
f(x) \leq f \left( -10^{-4} \right) \approx 1.0001 < \frac{4!}{e^{10^{-4}}}
\end{align*} Therefore, for $0 < |x| \leq 10^{-4}$,
\begin{align*}
\bigg{|} \frac{x}{e^x-1} - \frac{1}{\sum_{k=1}^3 \frac{x^{k-1}}{k!}} \bigg{|} \leq 10^{-12}
\end{align*} $\square$
My big problem here is that I don't actually know how to prove my claim that $f(x)$ is monotonically decreasing. Does anyone know how to prove this?
Or even better, does anyone know of a more direct proof that avoids this problem altogether?
Thanks!
|
For $|x|<\epsilon=10^{-4}$, by the mean value theorem $$\left|\frac{e^x-1}{x}\right|=e^y\ge e^{-\epsilon}\ge1-\epsilon,$$ and $$1+\frac{x}{2}+\frac{x^2}{6}\ge1-\frac{\epsilon}{2},$$ so it follows that $$\left|\frac{x}{e^x-1}\frac{1}{1+\frac{x}{2}+\frac{x^2}{6}}\right|\le\frac{1}{(1-\epsilon)^2}\le2\le\frac{4!}{e^\epsilon}$$
|
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|
Determining a marginal of a joint categorical distribution I have a set of $n$ discrete variables $\mathcal{A} = \{ a_{1}, \dots, a_{n} \}$, where every $a_{i} \in \mathcal{A}$ can take on the values in $ \{ 0, 1, \dots, m \}$ and $n,m \in \mathbb{N}$. I wish to create a joint measure -- an unnormalised probability distribution -- over all joint events described by the variables and then determine a marginal probability distribution over some $a_{i} \in \mathcal{A}$. Each event must obey the following rule: $a_{i} \neq a_{j}$, if $a_{i} = k$ and $k > 0.$ That is to say, multiple variables may equal zero in any event, but every non-zero value must be unique. Here is an example for $n = 3$ and $m = 2$:
\begin{array}{|ccc|c|} \hline
a_{1} & a_{2} & a_{3} & \phi(a_{1}, a_{2}, a_{3}) \\ \hline
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 1 \\
0 & 0 & 2 & 1 \\
0 & 1 & 0 & 1 \\
0 & 1 & 2 & 1 \\
0 & 2 & 0 & 1 \\
0 & 2 & 1 & 1 \\
1 & 0 & 0 & 1 \\
1 & 0 & 2 & 1 \\
1 & 2 & 0 & 1 \\
2 & 0 & 0 & 1 \\
2 & 0 & 1 & 1 \\
2 & 1 & 0 & 1 \\ \hline
& \text{Elsewhere} & & 0\\ \hline
\end{array}
I wish to create a similar joint measure for an arbitrary $n$ and $m$ and then determine the marginal distribution, $$P(a_{i}) = \frac{1}{K} \sum\limits_{\mathcal{A} - \{a_{i}\} } \phi(\mathcal{A}) \text{,}$$ where $$ K = \sum_{\mathcal{A}} \phi (\mathcal{A}) = \sum_{k=0}^{\min(m, n)} k! \binom{n}{k} \binom{m}{k} $$ is a normalising constant and $\mathcal{A} - \{a_{i}\}$ denotes the set difference. Quite obviously, all marginal distributions over every $a_{j} \in \mathcal{A}$ are equivalent, as the joint measure is symmetrical. Is it possible to find an expression for this marginal distribution, as I have done for the normalising constant?
Edit. I believe the marginal distribution is given by $$ P(a_{i} = 0) = \frac{1}{K} \sum\limits_{k=0}^{\min(m, n-1)} k! \binom{n-1}{k} \binom{m}{k} $$ and $$ P(a_{i} \neq 0) = \frac{1}{K} \sum\limits_{k=0}^{\min(m-1, n-1)} k! \binom{n-1}{k} \binom{m-1}{k} \text{;}$$ however, I would like to show that $$\sum_{j=0}^{m} P(a_{i} = j) = 1 \text{.}$$
|
Here's how you prove this is a probability distribution. I have removed the upper limits of the summations; this is allowed since all subsequent terms are zero.
\begin{align}
\sum_{j=0}^m P(a_i=j)
&=P(a_i=0)+mP(a_i\neq 0)
\\&=\frac{1}{K} \sum_{k\ge 0} k! \binom{n-1}{k} \binom{m}{k}+m\cdot\frac1K\sum_{k\ge 0}k!\binom{n-1}k\binom{m-1}k.
\end{align}
Multiplying by $K$,
\begin{align}
K\sum_{j=0}^m P(a_i=j)
&= \sum_{k\ge 0} k! \binom{n-1}{k} \binom{m}{k}+\sum_{k\ge 0}k!\binom{n-1}k\cdot m\cdot \binom{m-1}k
\\ &= \sum_{k\ge 0} k! \binom{n-1}{k} \binom{m}{k}+\sum_{k\ge 0}k!\binom{n-1}k\cdot (k+1)\binom{m}{k+1}
\\ &= \sum_{k\ge 0} k! \binom{n-1}{k} \binom{m}{k}+\sum_{k\ge 0}(k+1)!\binom{n-1}k\binom{m}{k+1}
\\ &= \sum_{k\ge 0} k! \binom{n-1}{k} \binom{m}{k}+\sum_{k\ge 1}k!\binom{n-1}{k-1}\binom{m}{k}
\\ &= 0!\binom{n-1}0\binom{m}0+\sum_{k\ge 1} k! \left[\binom{n-1}{k}+ \binom{n-1}{k-1}\right]\binom{m}{k}
\\ &= 0!\binom{n-1}0\binom{m}0+\sum_{k\ge 1} k! \binom{n}{k}\binom{m}{k}
\\ &= \sum_{k\ge 0} k! \binom{n}{k}\binom{m}{k}
\\&=K.
\end{align}
Now divide by $K$.
|
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|
Find all values of $c$ for the Mean Value Theorem's $f'(c)$ of the function $f(x) = 2x^3-6x^2-90x+6$ of the range $[-5, 8]$ I need to find all values of $c$ for the $f'(c)$ of the function $f(x) = 2x^3-6x^2-90x+6$ of the range $[-5, 8]$ such that $f'(c) = -10$. I have already found that $f'(c) = \frac{-130}{13}=-10$.
I thought I could find the values of $c$ by differentiating $f(x)$ which becomes $6x^2-12x-90$, and applying the Mean Value Theorem $f'(c) = \frac{f(b)-f(a)}{b - a}$ again.
$f(b) = 6(-5)^2-12(-5)-90 = 150+60-90 = 120$
$f(a) = 6(8)^2-12(8)-90 = 384-186 = 198$
$f'(c) = \frac{120-198}{-5-8} = \frac{78}{13} = 6$
Then I took $6x^2-12x-90$ again, replaced $x$ with $c$, and set it equal to $6$ before solving for $c$.
$6c^2-12c-90 = 6 \to 6(c^2-2c-16) = 0$
$\frac{2 \pm \sqrt{2^2-4(1)(-17)}}{2(1)} = \frac{2 \pm \sqrt{74}}{2}$ or $c = \frac{2 - \sqrt{74}}{2}, \frac{2 + \sqrt{74}}{2}$
But this answer is wrong.
How can I find the values for $c$ here?
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You are overcomplicating things; there is no need to apply the mean value theorem to $f'$.
To find all values of $c\in[-5,8]$ for which $f'(c)=-10$ you indeed want to first compute $f'$. You correctly found that
$$f'(c)=6c^2-12c-90,$$
and setting this equal to $-10$ yields
$$6c^2-12c-90=-10.$$
This is a quadratic equation in $c$, which you can solve yourself, judging from your attempted solution.
Caution: The quadratic equation above has (at most) two solutions. Make sure to check whether the solutions are in fact in the interval $[-5,8]$.
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|
$ \lim\limits_{x\rightarrow +\infty} \left(\frac{2}{\pi} \arctan x \right)^x$ and $\lim_{x\rightarrow 3^+} \frac{\cos x \ln(x-3)}{\ln(e^x-e^3)}$?
I got stuck on two exercises below
$$
\lim\limits_{x\rightarrow +\infty} \left(\frac{2}{\pi} \arctan x \right)^x \\
\lim_{x\rightarrow 3^+} \frac{\cos x \ln(x-3)}{\ln(e^x-e^3)}
$$
For the first one , let $y=(\frac{2}{\pi} \arctan x )^x $, so $\ln y =x\ln (\frac{2}{\pi} \arctan x )$, the right part is $\infty \cdot 0$ type, but seemly, the L 'hopital's rule is useless. PS: I know the $\infty \cdot 0$ can be become to $\frac{\infty}{\infty}$ or $\frac{0}{0}$. But when I use the L 'hopital's rule to the $\frac{\infty}{\infty}$ or $\frac{0}{0}$ the calculation is complex and useless.
For the second one , it is $\frac{\infty}{\infty}$ type, also useless the L 'hopital's rule is. How to calculate it ?
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Without L'Hospital
$$y=\left(\frac{2}{\pi} \arctan (x) \right)^x\implies \log(y)=x \log\left(\frac{2}{\pi} \arctan (x) \right) $$
Now, by Taylor for large values of $x$
$$\arctan (x)=\frac{\pi }{2}-\frac{1}{x}+\frac{1}{3 x^3}+O\left(\frac{1}{x^4}\right)$$
$$\frac{2}{\pi} \arctan (x) =1-\frac{2}{\pi x}+\frac{2}{3 \pi x^3}+O\left(\frac{1}{x^4}\right)$$ Taylor again
$$\log\left(\frac{2}{\pi} \arctan (x) \right)= -\frac{2}{\pi x}-\frac{2}{\pi ^2 x^2}+O\left(\frac{1}{x^3}\right)$$
$$\log(y)=x\log\left(\frac{2}{\pi} \arctan (x) \right)= -\frac{2}{\pi }-\frac{2}{\pi ^2 x}+O\left(\frac{1}{x^2}\right)$$ Just continue with Taylor using $y=e^{\log(y)}$ if you want to see not only the limit but also how it is approached
|
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|
Why does $\sin(x) - \sin(y)=2 \cos(\frac{x+y}{2}) \sin(\frac{x-y}{2})$? Why does this equality hold?
$\sin x - \sin y = 2 \cos(\frac{x+y}{2}) \sin(\frac{x-y}{2})$.
My professor was saying that since
(i) $\sin(A+B)=\sin A \cos B+ \sin B \cos A$
and
(ii) $\sin(A-B) = \sin A \cos B - \sin B \cos A$
we just let $A=\frac{x+y}{2}$ and $B=\frac{x-y}{2}$. But I tried to write this out and could not figure it out. Any help would be appreciated
|
Following your professor's advice, let $A=\frac{x+y}{2}$, $B=\frac{x-y}{2}$. Then $$x=A+B\\y=A-B$$So the LHS of your equation becomes $$\sin(A+B)-\sin(A-B)$$Now you just use the usual addition/subtraction trigonometric identities (i) and (ii) listed to evaluate this. It should give $2\cos A\sin B$ as required.
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|
Nice way of showing the locus of $\left|z-\frac{1}{z}\right| = 2$ is the union of 2 intersecting circles. For visualisation, the following code was run in Mathematica.
ContourPlot[{Abs[x + I y - 1/(x + I y)] == 2}, {x, -2.42, 2.42}, {y, -2.42, 2.42}, Axes -> True]
Here is the locus plot generated by the above code.
I do not see a nice way of showing the locus is what is is apart from brute force algebraic factorisation, which basically requires knowing the form of the locus a priori.
A less cumbersome approach is desired, if such a method exists.
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$$\begin{align}
\left|z-{1\over z}\right|=2
&\iff|z-1||z+1|=2|z|\\
&\iff((x-1)^2+y^2)((x+1)^2+y^2)=4(x^2+y^2)\\
&\iff(x^2+y^2+1-2x)(x^2+y^2+1+2x)=4(x^2+y^2)\\
&\iff(x^2+y^2+1)^2-4x^2=4(x^2+y^2)\\
&\iff(x^2+y^2-1)^2-4x^2=0
\end{align}$$
On the other hand, the equation for the locus of two circles of radius $r$, centered at $(x,y)=(\pm c,0)$, is
$$((x-c)^2+y^2-r^2)((x+c)^2+y^2-r^2)=0$$
This can be rewritten as
$$(x^2+y^2+c^2-r^2)^2-4c^2x^2=0$$
All we need is to let $c=1$ and $r=\sqrt2$.
|
{
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|
Sum the first n^3 how provide with using mathematical induction? I having next formula and need solving with helps mathematical induction.
$\sum_{i=1}^n i^3 = \biggl(\frac{n(n+1)}{2}\biggl)^2$
My next steps in solving the task:
a)I check the validity of this assertion for n = 1
$\sum_{i=1}^1 1^3 = \biggl(\frac{1(1+1)}{2}\biggl)^2$ = $1$
b)Next, I try to prove the following assertion for n + 1
$\sum_{i=1}^n i^3 = \biggl(\frac{n(n+1)}{2}\biggl)^2$ = $\sum_{i=1}^n i^3$ + $(i + 1)^3$ = $\biggl(\frac{n(n+1)}{2}\biggl)^2$ + $(n+1)^3$ = $\biggl(\frac{n(n+1)}{2}\biggl)^2$ + $(n+1)^3$ = $\biggl(\frac{n(n+1)}{2}\biggl)^2$ + $(n+1)^3$
c)I right solving ? Or have problems ?
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You need to show that
$$
\sum_{i=1}^{n+1} i^3 = \left( \frac{(n+1)(n+2)}{2}\right)^2,
$$
using as an hypothesis that
$$
\sum_{i=1}^{n} i^3 = \left( \frac{n(n+1)}{2}\right)^2.
$$
So,
\begin{align*}
\sum_{i=1}^{n+1} i^3=& \sum_{i=1}^n i^3 + (n+1)^3= \left( \frac{n(n+1)}{2}\right)^2 + (n+1)^3\\
=& (n+1)^2\left( \frac{n^2}{4}+n+1\right)=\frac{(n+1)^2}{4}(n^2+4n+4)\\
=&\frac{(n+1)^2(n+2)^2}{4} = \left( \frac{(n+1)(n+2)}{2}\right)^2
\end{align*}
|
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|
Find $\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \left(e^{(1+\frac{k}{n})^2} - \frac{3e^{(1 + \frac{3k}{n})}}{2\sqrt{1 + \frac{3k}{n}}}\right).$ Find
$$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \left(e^{(1+\frac{k}{n})^2} - \frac{3e^{(1 + \frac{3k}{n})}}{2\sqrt{1 + \frac{3k}{n}}}\right)$$
Choices:
|
1
B: $\frac{1}{2}$
C: $\frac{1}{4}$
D: 0
My attempt is:
The question is directly equal to
$$\int_0^1\left(e^{(1+x)^2} - \frac{3e^{(1+3x)}}{2\sqrt{1+3x}}\right)dx$$
|
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|
Need help for integration I need help with finding the following integral
$$\int_0^{+\infty} \frac{x^a\,dx}{(1+x)^2(A+x^a)}, $$ where $A>0$ and $a \in \mathbb{R}\,.$
This is my first trial.
\begin{align}
\int_0^\infty \dfrac{x^a\,{\rm d}x}{(1+x)^2\,(A+ x^a)} \\
&= \int_0^1 \dfrac{x^a\,{\rm d}x}{(1+x)^2\,(A+ x^a)}
+ \int_1^\infty \dfrac{x^a\,{\rm d}x}{(1+x)^2\,(A+ x^a)} =: I_1 + I_2\,. \notag
\end{align}
After applying binomial expansion and interchanging the infinite summation and integral I got the (possible) solution for $I_1$ as
\begin{align} \label{2.1}
I_1 &= \int_0^1 \sum_{k \geq 0} \binom{-2}k \dfrac{x^{a+k}\, {\rm d}x}{A+ x^a}
= \dfrac1a \sum_{k \geq 0} (-1)^k \dfrac{(2)_k}{k!} \int_0^1 \dfrac{t^{\frac{k+1}a}\,
{\rm d}t}{A+t} \notag \\
&= \dfrac1A \sum_{k \geq 0} (-1)^k \dfrac{(2)_k}{k!\, (a+k+1)}\,
{}_2F_1\left(1, \frac{k+1}a+1; \frac{k+1}a+2; -\dfrac1A\right)\notag \\
&= \dfrac1A \sum_{k \geq 0} (-1)^k \dfrac{k+1}{a+k+1}\,
{}_2F_1\left(1, \frac{k+1}a+1; \frac{k+1}a+2; -\dfrac1A\right)\,.
\end{align}
My doubt and question is did I employ interchanging the infinite summation and integral on appropriate way? Is it possible or not?
|
I give you a hint to solve your integral analytical:
First you split the integral in two parts:
$$\int_0^{\infty } \frac{x^a}{(1+x)^2 \left(A+x^a\right)} \, \
dx=\int_0^{\infty } \frac{1}{(1+x)^2} \, dx-\int_0^{\infty } \
\frac{A}{(1+x)^2 \left(A+x^a\right)}\, dx$$
The first integral is trivial:
$$\int_0^{\infty } \frac{x^a}{(1+x)^2 \left(A+x^a\right)} \, dx=1-A \
\int_0^{\infty } \frac{1}{(1+x)^2 \left(A+x^a\right)} \, dx$$
Now you have to do two times partial integration to reduce the problem on an integral including the term $\text{log}[x+1]$:
$$\int_0^{\infty } \partial _x\left(-\frac{1}{(1+x) \left(A+x^a\right)}\right) \, dx=\int_0^{\infty } \frac{a x^{-1+a}}{(1+x) \
\left(A+x^a\right)^2}\, dx+\int_0^{\infty } \frac{1}{(1+x)^2 \left(A+x^a\right)} \, dx$$
$$\text{F6}(A,a)=1+A \left(-\frac{1}{A}+\int_0^{\infty } \frac{a \
x^{-1+a}}{(1+x) \left(A+x^a\right)^2} \, dx\right)$$
Take the integral on the right side to repeat the partial integration:
$$\int_0^{\infty } \partial _x\frac{\text{Log}[1+x] \left(a x^{-1+a}\right)}{\left(A+x^a\right)^2} \, dx=\int_0^{\infty } \frac{(-1+a) a \
x^{-2+a}\text{Log}[1+x]}{\left(A+x^a\right)^2} \, dx-\int_0^{\infty } \frac{2 a^2 x^{-2+2 a} \text{Log}[1+x]}{\left(A+x^a\right)^3} \, \
dx+\int_0^{\infty} \frac{a x^{-1+a}}{(1+x) \left(A+x^a\right)^2} \, dx$$
This leads to the integrals:
$$\text{F11}[(A,a)=\kappa [a,A] \int_0^{\infty } \frac{x^{2 (a-1)} \text{Log}[1+x]}{\left(A+x^a\right)^3} \, dx-\lambda [a,A] \
\int_0^{\infty } \frac{x^{-2+a}\text{Log}[1+x]}{\left(A+x^a\right)^2} \, dx$$
with the abbreviations:
$$\kappa (A,a)=2 a^2 A$$
and
$$\lambda (A,a)=A a (-1+a)$$
The first and second integral on the right hand side in a similar form are already solved by Sasha using the
Plancherel theorem.
For a similar derivation to [Sasha] we start with the introduction of k:
$$\text{F12}(k,A,a)=\int_0^{\infty } \frac{t^{a (k+1)} \text{Log}[1+t]}{\left(A+t^a\right)^{k+2} t^2} \, dt$$
to write
$$\text{F13}(A,a)=\kappa (A,a) \text{F12}(1,A,a)-\lambda (A,a) \text{F12}(0,A,a)$$
Now we only have to calculate the $F12(k,A,a)$. For the calculation of $F12(k,A,a)$ we use the calculation of [Sasha] and take over the derivation of the $G_{1}\left( t\right)$, the only difference between our calculation and [Sacha] is the function $G_{2}\left( t\right)$
Mathematica has some problems to get the Inverse Mellin - Transform for general values of A and a. So we calculate for some different values of a and A:
$$\text{G21}(k,A,a)=\int_0^{\infty } \frac{t^{a (k+1)} \
t^{s-2}}{\left(A+t^a\right)^{k+2}} \, dt$$
$$\text{H21}(k,A,a)=\frac{\text{Gamma}[1-s] \text{Gamma}[s]^2 \text{G21}\left(k,A,\frac{1}{\alpha }\right)}{\text{Gamma}[s+1]}$$
$$\text{h21}(k,A,a)=\text{InverseMellinTransform}[\text{H21}(A,\lambda ),s,t]$$
$$\text{h21}(3,2)=\frac{2^{1+k} \text{MeijerG}\left[\left\{\left\{-1,-\frac{1}{2},0\right\},\{1\}\right\},\left\{\left\{0,0,\frac{1}{2} (-1+k),\frac{k}{2}\right\},\{\}\right\},\frac{t}{9}\right]}{27\pi \text{Gamma}[2+k]}$$
other values are calculated in the same way. Final we get:
$$F14\left( A,k,a\right) =\frac{1}{a~A^{\frac{1}{a}+1}\left( k+1\right) !}
H_{3,3}^{3,2}\left( \frac{1}{A^{\frac{1}{a}}}\left\vert
\begin{array}{c}
\left( -\frac{1}{a},\frac{1}{a}\right) ,\left( 0,1\right) ,\left( 1,1\right)
\\
\left( 0,1\right) ,\left( 0,1\right) ,\left( k-\left( \frac{1}{a}-1\right) ,%
\frac{1}{a}\right)
\end{array}\right. \right) $$
For $a=\frac{1}{2}$ we get:
$$F14\left( A,k,\frac{1}{2}\right) =\frac{2^{k+1}}{A^{3}\pi \left( k+1\right) !}%
G_{4,4}^{4,3}\left( \frac{1}{A^{2}}\left\vert
\begin{array}{c}
-1,-\frac{1}{2},0,1 \\
0,0,\frac{k-1}{2},\frac{k}{2}%
\end{array}%
\right. \right) $$
Finally we get:
$$\int_0^{\infty } \frac{x^a}{(1+x)^2 \left(A+x^a\right)} \, \
dx=\kappa (a,A) \text{F14}(A,1,a)-\lambda (a,A) \text{F14}(A,0,a)$$
So the general solution of integral can be expressed in the form of two H-Fox-Functions, which are discussed in detail in
Mathai
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|
Analytical solution to $\sqrt{a^2-x^2} + \sqrt{b^2-x^2} = \sqrt{a^2-x^2} \cdot \sqrt{b^2-x^2}$ Can't find beauty analytical solution to such equation :
$$\sqrt{a^2-x^2} + \sqrt{b^2-x^2} = \sqrt{a^2-x^2}\cdot\sqrt{b^2-x^2}$$
Assuming $a,\ b \in \mathbb{N},\ a \le b,\ x \in \mathbb{R}$
Is it possible to find a solution for general case?
|
So, we have $$\dfrac1{\sqrt{a^2-x^2}}+\dfrac1{\sqrt{b^2-x^2}}=1$$
WLOG $a^2-x^2=\sin^4t, b^2-x^2=\cos^4t$
$\implies b^2-a^2=\cos2t$
$$4a^2-4x^2=(2\sin^2t)^2=(1+a^2-b^2)^2$$
$$\iff4x^2=?$$
|
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|
Find all integral solutions to $xy^2 + (x^2 + 1)y + x^4 + 1 = 0$. I have an equation $$xy^2 + (x^2 + 1)y + x^4 + 1 = 0$$ and I need to find a set of solutions for it in integers. I tried to solve it using Maple with the isolve () command, but it doesn’t give anything. This suggests that either the Maple algorithms cannot solve it or that there are no solutions. However, I know that he is resolved. Maybe you know special algorithms for solving this equation or other methods?
|
This answer makes use of the rational root theorem a lot. Let $x$ and $y$ be integers such that
$$xy^2 + (x^2 + 1)y + x^4 + 1 = 0.\tag{1}$$
We first make a few simple observations.
*
*If $x=0$ then $y=-1$, and if $y=-1$ then $x=0$.
*There are no other integral solutions with $|x|\leq2$.
*Equation $(1)$ shows that $y$ divides $x^4+1$.
Lemma 1: If $(x,y)$ is an integral solution to $(1)$, then there exists $u\in\Bbb{Z}$ such that
$$y=ux^4-2x^3-x^2-x-1.$$
Proof. If $x=0$ then this follows from the first observation above. So suppose $x\neq0$. Viewing $(1)$ as a polynomial in $x$ we see that
$$x^4+yx^2+y^2x+y+1=0,$$
and hence that $x$ divides $y+1$. Let $z=\tfrac{y+1}{x}\in\Bbb{Z}$ so that plugging in $y=zx-1$ into $(1)$ yields
$$x^3+x^2z^2+x^2z-2xz-x+z+1=0.$$
this in turn shows that $x$ divides $z+1$. Let $w=\tfrac{z+1}{x}\in\Bbb{Z}$ so that plugging in $z=wx-1$ yields
$$w^2x^3-wx^2-2wx+w+x^2+1=0,$$
which in turn shows that $x$ divides $w+1$. Let $v=\tfrac{w+1}{x}\in\Bbb{Z}$ so that plugging in $w=vx-1$ yields
$$v^2x^4-2vx^3-vx^2-2vx+v+x^2+2x+2=0.$$
which in turn shows that $x$ divides $v+2$. Let $u=\tfrac{v+2}{x}\in\Bbb{Z}$ so that
\begin{eqnarray*}
y&=&zx-1=(wx-1)x-1=((vx-1)x-1)x-1\\
&=&(((ux-2)x-1)x-1)x-1\\
&=&ux^4-2x^3-x^2-x-1.\hspace{100pt}\square
\end{eqnarray*}
Lemma 2: If $x\neq0$ then $|u|\leq1$.
Proof. If $|u|>1$ then for all $x\notin\{-1,0,1,2\}$ we have
$$|y|=|ux^4-2x^3-x^2-x-1|>x^4+1,$$
contradicting the fact that $y$ divides $x^4+1$. As there are no integral solutions with $x\in\{-1,1,2\}$ it follows that $|u|\leq1$ if $x\neq0$.$\hspace{20pt}\square$
At this point we can simply try the three possible values of $u$ to get three polynomials in $x$, and see whether they have integral roots:
If $u=0$ then $y=-(2x^3+x^2+x+1)$. Plugging this back into $(1)$ yields
$$x^4(4x^3+4x^2+3x+6)=0,$$
and the rational root theorem shows that $x=0$ is the unique solution.
If $u=1$ then $y=x^4-2x^3-x^2-x-1$. Plugging this back into $(1)$ yields
$$x^4(x^2+1)(x^3-4x^2+x+7)=0,$$
which again has $x=0$ as its unique integral solution by the rational root theorem.
If $u=-1$ then $y=-x^4-2x^3-x^2-x-1$. Plugging this back into $(1)$ yields
$$x^4(x^5+4x^4+6x^3+5x^2+5x+5)=0,$$
which again has $x=0$ as its unique integral solution by the rational root theorem.
This shows that $(x,y)=(0,-1)$ is the unique integral solution to $(1)$.
|
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|
Prove that for all $x\in \mathbb R$ $, \arctan x=\frac{\pi}{2}-\arccos(\frac{x}{\sqrt{1+x^{2}}})$
Prove that for all $x\in \mathbb R$, $$\arctan x=\frac{\pi}{2}-\arccos \left(\frac{x}{\sqrt{1+x^{2}}}\right)$$
From Lagrange form of Taylor's theorem I have:$$\arctan x+\arccos\left(\frac{x}{\sqrt{1+x^{2}}}\right)=x-\frac{c_{1}x^{2}}{\sqrt{1+c_{1}^{2}}}+\frac{\pi}{2}-x-\frac{c_{2}x^{2}}{2\sqrt{(1-c_{2}^{2})^{3}}}=-\frac{c_{1}x^{2}}{\sqrt{1+c_{1}^{2}}}+\frac{\pi}{2}-\frac{c_{2}x^{2}}{2\sqrt{(1-c_{2}^{2})^{3}}}$$I know that: $$\left(-\frac{c_{1}x^{2}}{\sqrt{1+c_{1}^{2}}}-\frac{c_{2}x^{2}}{2\sqrt{(1-c_{2}^{2})^{3}}}\right) \rightarrow 0$$But I don't know how to show that: $$\left(-\frac{c_{1}x^{2}}{\sqrt{1+c_{1}^{2}}}-\frac{c_{2}x^{2}}{2\sqrt{(1-c_{2}^{2})^{3}}}\right) = 0$$ Can you help me?
|
Draw a right triangle whose tangent is $x$. Label the opposite side $x$ and the other leg $1$. Now find the length of the hypotenuse; it's $\sqrt{x^2 + 1}$.
Then, compute the cosine of the complementary angle. You are done.
|
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|
Why is this limit evaluated like so? Question:
If
$$\lim_{x \to 0}{\frac{-1 + \sqrt{(\tan x - \sin x) + \sqrt{(\tan x - \sin x) + \sqrt{(\tan x - \sin x) + \cdots \infty}}}}{-1 + \sqrt{x^3 + \sqrt{x^3 + \sqrt{x^3 + \cdots \infty}}}}} = \frac{1}{k}$$
Then find the value of $k$.
The way I approached the problem was by substituting $x = 0$ in the limit:
$$\frac{-1 + \sqrt{(\tan 0 - \sin 0) + \sqrt{(\tan 0 - \sin 0) + \sqrt{(\tan 0 - \sin 0) + \cdots \infty}}}}{-1 + \sqrt{0^3 + \sqrt{0^3 + \sqrt{0^3 + \cdots \infty}}}} = \frac{1}{k}$$
$$\implies \frac{-1 + \sqrt{0 + \sqrt{0 + \sqrt{0 + \cdots \infty}}}}{-1 + \sqrt{0 + \sqrt{0 + \sqrt{0 + \cdots \infty}}}} = \frac{1}{k}$$
$$\implies \frac{-1}{-1} = \frac{1}{k}$$
$$\implies \frac{1}{1} = \frac{1}{k}$$
$$\implies k = 1$$
But according to the given solution, the answer is 2.
I did find this question, but I do not understand why I cannot just put $x = 0$ in the limit.
Any help is appreciated.
|
Okay. when we look at these continued roots....lets look at the one in the numerator, the logic will hold for the one in the denominator.
First the roots will not be defined as real functions if $x<0$
$\lim_\limits{x\to 0} \tan x - \sin x = 0$
$\tan x - \sin x > 0$ for all $0<x<\frac {\pi}2$
When x is small and positive then $\tan x - \sin x$ will also be small and positive.
$\tan x - \sin x<\sqrt{\tan x - \sin x}$
We are looking at an infinite sequence.
$s_{n+1} = \tan x - \sin x + \sqrt {s_{n}}$
Where we are taking small positive numbers and adding something on, and since this sequence is infinite, it will eventually be something of non trivial magnitude no matter how small $\tan x - \sin x$ may have been at the start.
Eventually, it approaches 1, but it can't get more than trivally larger than one.
Because if $s_n > 1, 1<\sqrt {s_n} < s_n$
This is why you can't treat it like you do continuous functions and just plug zero.
|
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|
Showing matrix $\left[\begin{smallmatrix} 4 & 1 & 1\\1 & 2 & -1 \\ 1 & -1 & 3 \end{smallmatrix}\right]$ is positive definite $$\begin{bmatrix}
x_1 & x_2 & x_3
\end{bmatrix}\begin{bmatrix}
4 & 1 & 1\\1 & 2 & -1 \\ 1 & -1 & 3
\end{bmatrix}\begin{bmatrix}
x_1\\x_2\\x_3
\end{bmatrix} = \\4x_1^2 + x_1x_2 + x_3x_1 + x_2x_1 + 2x_2^2 -x_2x_3 + x_1x_3-x_2x_3+3x_3^2$$
which I cannot put in a form that is $>0$. So how can this matrix be positive definite?
|
Note that we can show that a matrix is positive definite by looking at its $n$ upper left determinants. Note that for the matrix
$$\begin{bmatrix}
4 & 1 & 1\\1 & 2 & -1 \\ 1 & -1 & 3
\end{bmatrix}$$
We have
$$\begin{vmatrix}
4
\end{vmatrix} = 4$$
and
$$\begin{vmatrix}
4 & 1\\
1 & 2
\end{vmatrix} = 7$$
and lastly
$$\begin{vmatrix}
4 & 1 & 1\\
1 & 2 & -1\\
1 & -1 & 3
\end{vmatrix} = 13$$
Since $4,7,13 > 0$ the matrix is positive definite.
|
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|
$\int \frac{1}{{1-2x-x^2}} \, \mathrm{d}x $ substitution I have this integral.
$$\int \frac{1}{{1-2x-x^2}} \, \mathrm{d}x $$
But I am unable to do it right and I just don't know where is the problem in my steps.
My steps:
Complete the square
$$\int \frac{1}{{2-(x+1)^2}} \, \mathrm{d}x $$
$$\frac{1}{2}*\int \frac{1}{{-(x+1)^2}} \, \mathrm{d}x $$
Substitute $$t=x+1$$
$$\frac{1}{2}*\int \frac{1}{{-t^2}} \, \mathrm{d}x $$
$$\frac{1}{2}*\int {{-t^{-2}}} \, \mathrm{d}x $$
$$\frac{1}{2}* \frac{-t^{{-1}}}{{-1}}+c $$
$$ \frac{1}{{2(x+1)}}+c $$
But the result is different.
I will be thankful for any help.
|
Use the u-substation $u=\frac{x+1}{\sqrt{2}}$ and then do a partial fraction decomposition on the fraction that you will get:
$$
\int\frac{1}{1-2x-x^2}\,dx=\int\frac{1}{2-(x+1)^2}\,dx=\\
\frac{\sqrt{2}}{2}\int\frac{1}{1-\left(\frac{x+1}{\sqrt{2}}\right)^2}\frac{d}{dx}\left(\frac{x+1}{\sqrt{2}}\right)\,dx=
\frac{\sqrt{2}}{2}\int\frac{1}{1-u^2}\,du=\\
\frac{\sqrt{2}}{2}\int\frac{1}{(1-u)(1+u)}\,du=
\frac{\sqrt{2}}{2}\int\left(\frac{1}{2}\frac{1}{1-u}-\frac{1}{2}\frac{1}{1+u}\right)\,du=\\
\frac{\sqrt{2}}{4}\left(-\ln{|1-u|-\ln{|1+u|}}\right)+C=\\
\frac{\sqrt{2}}{4}\ln{\left|\frac{1+u}{1-u}\right|}+C.
$$
And then do a back-substitution.
|
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|
Why do I get different results when testing for increasing/decreasing intervals of a function here? I have the function $f(x)=6+\frac{6}{x}+\frac{6}{x^2}$ and I want to find the intervals where it increases or decreases. The problem is that when I find $f'(x)=0$, which becomes $x=-2$. Once I put $-2$ on a number line, and find whether the numbers higher and lower than $-2$ produce positive or negative numbers when plugged into $f'(x)$, I get both positives and negatives for different numbers when $x > -2$. For example, $f'(-1) > 0$ and $f'(5) < 0$.
I am not sure if this is because the function has a vertical asymptote at $x=0$ and a horizontal attribute at $y=6$. If so, I'd like to know what I need to do to handle them.
How I found the first derivative:
$\frac{d}{dx} \frac{6}{x}=\frac{[0] - [6\cdot1]}{x^2}=\frac{-6}{x^2}$
$\frac{d}{dx}\frac{6}{x^2}=\frac{[0]-[6 \cdot 2x]}{(x^2)^2} =\frac{-12}{x^3}$
$f'(x)=\frac{-6}{x^2}-\frac{-12}{x^3}$
How I found the asymptotes:
When you combine the fractions of $f(x)$, $f(x)=\frac{6x^2+6x+6}{x^2}$. When set equal to zero, $x^2$ has an x-value of zero. Therefore, $f(x)$ has a vertical asymptote at $x=0$.
Once the fractions are combined, and when everything but the coefficients of the leading terms on the numerator and denominator are removed, $f(x)=\frac{6}{1}$. So $f(x)$ has a horizontal asymptote at $y=6$.
How I found the value of $f'(x)=0$:
$\frac{-6}{x^2}-\frac{12}{x^3}=0$
Add $\frac{12}{x^3}$ to both sides
$\frac{6}{x^2}=\frac{-12}{x^3}$
Multiply both sides by $x^3$.
$6x=-12$
Divide both sides by $6$.
$x = \frac{-12}{6}=-2$
|
Notice that
$$ f^\prime(x)=-\frac{6}{x^2} -\frac{12}{x^3}=-\frac{6}{x^3}(x+2)$$
So $f^\prime$ is undefined at $x=0$ and $f$ has slope $0$ at $x=-2$.
You should therefore check for increasing/decreasing on the intervals $(-\infty,-2),\,(-2,0)$ and $(0,\infty)$
|
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|
Find the partial sum formula of $\sum_{i=1}^n \frac{x^{2^{i-1}}}{1-x^{2^i}}$ Given next series:
$$\frac{x}{1 - x^2} + \frac{x^2}{1 - x^4} + \frac{x^4}{1 - x^8} + \frac{x^8}{1 - x^{16}} + \frac{x^{16}}{1 - x^{32}} + ... $$
and $|x| < 1$. Need to derive $S_n$ formula from series partial sums.
I could only find that $S_{k+1}=\frac{S_k}{1-x^{2^k}} + \frac{x^{2^k}}{1-x^{2^{k+1}}}$. But this is incorrect answer, of course, but I don't know what to do next...
Thank you in advance!
|
Hint:
Use method of cancellation
$$\frac {x^2} {1-x^4} = \left[\frac 1 {1-x^2} - \frac 1 {1-x^4}\right]$$
$$\frac {x^4} {1-x^8} = \left[\frac 1 {1-x^4} - \frac 1 {1-x^8}\right]$$
|
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|
Prove the inequality $\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+xy} \geq \frac{3}{1+\frac{(x+y)^2}{4}}$ when $x^2+y^2=1$ I have to prove the inequality
$$
\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+xy} \geq \frac{3}{1+\frac{(x+y)^2}{4}}
$$
when $x^2+y^2=1$, using Cauchy-Schwarz Inequality.
The RHS is equal to $\frac{12}{5+2xy}$.
I can prove, using C-S, that the RHS is $\geq \frac{9}{4+xy}$ or $\geq \frac{12}{5+4xy}$ but I can't go further.
I can prove the inequality only using A.M.-G.M. inequality proving that $xy\leq\frac{1}{2}$ and simplifying the expression all together, but this is not what I want.
Thanks
|
For $xy\geq0$ by C-S $$\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+xy}=\frac{3}{2+x^2y^2}+\frac{1}{1+xy}=$$
$$=\frac{4}{\frac{4(2+x^2y^2)}{3}}+\frac{1}{1+xy}\geq\frac{(2+1)^2}{\frac{4(2+x^2y^2)}{3}+1+xy}.$$
Id est, it's enough to prove that
$$\frac{9}{\frac{4(2+x^2y^2)}{3}+1+xy}\geq\frac{12}{5+2xy}$$ or
$$(1-2xy)(1+8xy)\geq0,$$ which is true by C-S again:
$$2=(x^2+y^2)(1^2+1^2)\geq(x+y)^2=1+2xy,$$ which gives $1-2xy\geq0.$
For $xy\leq0$ by C-S again we obtain
$$\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+xy}=\frac{1}{\frac{2+x^2y^2}{3}}+\frac{1}{1+xy}\geq\frac{4}{\frac{2+x^2y^2}{3}+1+xy}.$$
Thus, it's enough to prove that
$$\frac{4}{\frac{2+x^2y^2}{3}+1+xy}\geq\frac{12}{5+2xy}$$ or
$$x^2y^2+xy-2\leq0$$ or
$$(1-xy)(2+xy)\geq0,$$ which is obvious.
|
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|
Show that $\sum_{k=1}^{n}(5^{2k}-5^{2k-1})=\frac{5^{2n+1}-5}{6}$, for n = 1,2,3, ... Show that
$\sum_{k=1}^{n}(5^{2k}-5^{2k-1})=\frac{5^{2n+1}-5}{6}$, for n = 1,2,3, ...
n = 1
$LHS = 5^2 - 5 = 20$
$RHS =\frac{ 5^{3} - 5}{6} = \frac{120}{6} = 20$
n = p
$LHS_{p} = (5^{2(1)}-5^{2(1)-1}) + (5^{2(2)}-5^{2(2)-1})+....+ (5^{2(p)}-5^{2(p)-1}) $
$RHS_{p} =\frac{ 5^{2(p)+1} - 5}{6} $
n = p + 1
$LHS_{p+1} = (5^{2(1)}-5^{2(1)-1}) + (5^{2(2)}-5^{2(2)-1})+....+ (5^{2(p)}-5^{2(p)-1})+ (5^{2(p+1)}-5^{2(p+1)-1}$
$RHS_{p+1} =\frac{ 5^{2(p+1)+1} - 5}{6} $
Show that: $ RHS_{p+1} = RHS_{p} + 5^{2(p+1)}-5^{2(p+1)-1} $
$\frac{ 5^{2(p+1)+1} - 5}{6} = \frac{ 5^{2(p)+1} - 5}{6} + 5^{2(p+1)}-5^{2(p+1)-1} $
And here's where I am stuck :)
|
Hint:
$$5^{2k}-5^{2k-1}=5^{2k-1}(5-1)=4\cdot5^{2k-1}$$
So, we have
$$\sum_{k=1}^n4\cdot5^{2k-1}$$ which is a geometric Series
Use Induction proof dealing with geometric series
|
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|
Lyapunov's theory for differential equations Use Lyapunov's theory to discuss stability of the origin for
$$x' = -x^3 - xy^2 + xy^5$$
$$y' = -y^3 + x^2y - 3x^2y^4$$
$$L(x,y) = ax^2 + by^2$$
In class, we define $L(x,y)$ as being positive definite. Then we determine if the origin is a critical point. The we take the derivative of $L$ and determine if it's a negative definite or negative semidefinite.
I can get to the point where we take the derivative. Simplifying and determining the negative definite or negative semidefinite is the problem.
|
Given the system
$\dot x = -x^3 - xy^2 + xy^5, \tag 1$
$\dot y = -y^3 + x^2y - 3x^2y^4, \tag 2$
it is easily seen that $(0, 0)$ is a critical point; we seek a Lyapunov function of the form
$L(x,y) = ax^2 + by^2; \tag 3$
we have
$\dot L = 2ax \dot x+ 2by \dot y$
$= 2ax(-x^3 - xy^2 + xy^5) + 2by(-y^3 + x^2y - 3x^2y^4)$
$= -2ax^4 -2ax^2y^2 + 2ax^2y^5 -2by^4 +2bx^2y^2 -6bx^2y^5$
$= -2ax^4 - 2by^4 + 2(b - a)x^2y^2 + (2a - 6b)x^2y^5; \tag 4$
we may exploit our freedom in the selection of $a$ and $b$ to ensure that
$\dot L(x, y) < 0 \tag 5$
along the trajectories of (1)-(2) in a neighborhood of the origin: we choose
$0 < a = 3b; \tag 6$
then
$2a - 6b = 0, \tag 7$
$b - a = -2b; \tag 8$
such a choice of $a$ and $b$ eliminates the complicating term $(2a - 6b)x^2y^5$ from (4); we are left with an expression containing only even powers of $x$ and $y$, viz.
$\dot L = -6bx^4 - 2by^4 - 2bx^2y^2 < 0 \tag 9$
whenever at least one of $x$, $y$ does not vanish; thus
$L = 3bx^2 + by^2 \tag{10}$
is a suitable Lyapunov function for (1)-(2) in the vicinity of $(0, 0)$, and in conclude that the origin is an asymptotically stable critical point of (1)-(2).
|
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|
Showing that the simple continued fraction of $\sqrt{d}$ has period length 1 iff $d=a^2+1$ Given that I know if $d$ is an integer that $\sqrt{d}=[\alpha_0,\bar{\alpha_1},...\bar{\alpha_n},\bar{2\alpha_0}]$. I want to show that $\sqrt{d}$ has period length 1 if and only if $d=a^2+1$, for some natural number a.
Here is what I think I have to do , could someone please verify it for me ?
($\Leftarrow$) suppose $d=a^2+1$, then $\sqrt{d}=\sqrt{a^2+1}$.
$(a-1)^2\leq a^2+1, \Rightarrow (a-1)\leq \sqrt{a^2+1}$
So the floor of $\sqrt{a^2+1}$ is $a-1=\alpha_0$.
Subtracting this and inverting gives:
$\tfrac{1}{\sqrt{a^2+1}-(a-1)}=\tfrac{\sqrt{a^2+1}+(a-1)}{2a}$, which has floor 1.
So we know $\alpha_1=1$, subtracting this and inverting gives:
$\tfrac{2a}{\sqrt{a^2+1}-(a-1)}=\tfrac{2a(\sqrt{a^2+1}+(a-1)}{2a}=\sqrt{a^2+1}+(a-1)$, which has floor $2a-2$, subtracting and inverting gives,
$\tfrac{1}{\sqrt{a^2+1}-(a-1)}$, so we're back to the start , and so the continued fraction has period length 1.
$(\Rightarrow) $ suppose $\sqrt{d}$ has period length 1. Then $\sqrt{d}=[\alpha_0,\bar{\alpha_1}]=[\alpha_0,\bar{\alpha_1},\bar{2\alpha_0}]$.
So then we can write :
$\sqrt{d}=\alpha_0+\tfrac{1}{\alpha_1+\tfrac{1}{2\alpha_0+\tfrac{1}{...}}}=\alpha_0+\tfrac{1}{\alpha_1+\tfrac{1}{2\sqrt{d}}}$.
I haven't got this to work out yet, but does showing $d=a^2+1$, come down to reaaranging what I've just written ?
|
$$\sqrt{a^2}\lt\sqrt{a^2+1}\lt\sqrt{a^2+2a+1}$$
$$\therefore a\lt\sqrt{a^2+1}\lt a+1\implies\left\lfloor\sqrt{a^2+1}\right\rfloor=a$$
Hence,
$$\begin{align}
\sqrt{d}
&=\sqrt{a^2+1}\\
&=a+\sqrt{a^2+1}-a\\
&=a+\frac1{\left(\frac1{\sqrt{a^2+1}-a}\right)}\\
&=a+\frac1{a+\sqrt{a^2+1}}\\
&=a+\frac1{a+\sqrt{d}}\\
&=a+\frac1{2a+\frac1{a+\sqrt{d}}}\\
&=a+\frac1{2a+\frac1{2a+\dots}}\\
\end{align}$$
You should be able to prove the if and only if part yourself.
|
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|
Inscribed circle in a right-angled triangle
In right-angled $\triangle ABC$ with catheti $a = 11\,\text{cm}, b=7\,\text{cm}$ a circle has been inscribed. Find the radius and altitude from $C$ to the hypotenuse.
I found that the hypotenuse is $c = \sqrt{170}$ and the radius is $r = \frac{a+b-c}{2}=\frac {18 - \sqrt{170}}{2}$. I think that the altitude $CH=$ the sum of radii of circles inscribed in $\triangle ABC, \triangle AHC, \triangle HBC$ but I don't understand how I should calculate them. Thank you in advance!
|
I assume you mean that $CH$ is the altitude from $C$ to the hypotenuse.
Since $\triangle ABC$ is a right triangle, and since
$\triangle AHC$ and $\triangle HBC$ each share one angle with $\triangle ABC$,
the three triangles are similar. We have
$$
\frac{CH}{AC} = \frac{BC}{AB}
$$
and (equivalently)
$$
\frac{CH}{BC} = \frac{AC}{AB}.
$$
The radii of the three incircles are proportional to $AB,$ $AC,$ and $BC,$
so their sum is
\begin{align}
\frac{a+b-c}{2} + \frac{a(a+b-c)}{2c} + \frac{b(a+b-c)}{2c}
&= \frac{(a+b+c)(a+b-c)}{2c}
\\
&= \frac{(a+b)^2 - c^2}{2c} \\
&= \frac{ab}{c},
\end{align}
which is indeed equal to the altitude.
|
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|
Intriguing Limit Prove that:
$$L=\lim_{n\to\infty} \frac {\sqrt 2 n^{\left(n-\frac 12\right)}}{n!}\left(\frac {(2\sqrt[n] {n} -1)^n}{n^2}\right)^{ \frac {n\left(n-\frac 12\right)}{\ln^2 n}}=\sqrt {\frac {e}{\pi}}$$
My method:
Properties I am going to use :
1)Stirling's approximation:$$n!\sim\sqrt {2\pi n} \left(\frac ne\right)^n$$
2)Property 2 : $$\sqrt[n] {n}\sim 1+\frac {\ln n}{n}+\frac{\ln^2 n}{2n^2}$$
3)Property 3: For all continuous and differentiable functions $f,g$ (In their domain respectively), if $\lim_{x\to\infty} g(x)=0$ then for large enough $x$ we have $$(1+g(x))^{f(x)}\sim e^{f(x)\cdot g(x)}$$
Using Stirling's approximation we get $$L=\lim_{n\to\infty} \frac {e^n}{\sqrt{\pi} n}\left(\frac {\displaystyle (2\sqrt[n]{n} -1)^n}{n^2}\right)^{\frac {n\left(n-\frac 12\right)}{\ln^2n}}$$
Using Property 2 we get $$L=\lim_{n\to\infty} \frac {e^n}{\sqrt {\pi} n}\left(\frac { \left(1+\frac {2\ln n}{n}+\frac{\ln^2n}{n^2}\right)^n}{n^2}\right)^{\frac {n\left(n-\frac 12\right)}{\ln^2n}}$$
And using the property 3 we get $$L=\lim_{n\to\infty} \frac {e^n}{\sqrt{\pi} n} \displaystyle \frac {e^{\frac {n(2n-1)}{\ln n}}\cdot e^{ \left(n-\frac 12\right)}}{ n^{\frac {n(2n-1)}{\ln^2 n}}}$$
Using that $$n^{\frac {n(2n-1)}{\ln^2 n}}=e^{\frac {n(2n-1)}{\ln n}}$$
Using this alongwith previous results we get $$L=\lim_{n\to\infty} \frac {e^n}{\sqrt{\pi} n} \displaystyle \frac {e^{\frac {n(2n-1)}{\ln n}}}{e^{ \frac {n(2n-1)}{\ln n}}}\cdot e^{ \left(n-\frac 12\right)}=\lim_{n\to\infty} \frac {e^{2n}}{\sqrt{e\pi} n}$$
Which clearly doesn't converge.Can someone please point out my mistake in above working. Also some new suggestions to solve this question will be quite beneficial.
|
We can start with
$$
\log\left(n^{1/n}\right)=\frac{\log(n)}n
$$
and use the power series for $e^x$ to get
$$
2n^{1/n}-1=1+2\frac{\log(n)}n+\frac{\log(n)^2}{n^2}+\frac{\log(n)^3}{3n^3}+O\!\left(\frac{\log(n)^4}{n^4}\right)
$$
The power series for $\log(1+x)$ yields
$$
\begin{align}
\log\left(2n^{1/n}-1\right)
&=\overbrace{2\frac{\log(n)}n+\frac{\log(n)^2}{n^2}+\frac{\log(n)^3}{3n^3}}^x\overbrace{-2\frac{\log(n)^2}{n^2}-2\frac{\log(n)^3}{n^3}}^{-x^2/2}\overbrace{+\frac83\frac{\log(n)^3}{n^3}}^{x^3/3}\\
&=2\frac{\log(n)}n-\frac{\log(n)^2}{n^2}+\frac{\log(n)^3}{n^3}+O\!\left(\frac{\log(n)^4}{n^4}\right)
\end{align}
$$
Multiply by $n$ and use the power series for $e^x$ where $x=-\frac{\log(n)^2}n+\frac{\log(n)^3}{n^2}+O\!\left(\frac{\log(n)^4}{n^3}\right)$:
$$
\left(2n^{1/n}-1\right)^n=n^2\left(1-\frac{\log(n)^2}n+\frac{\log(n)^4+2\log(n)^3}{2n^2}+O\!\left(\frac{\log(n)^6}{n^3}\right)\right)
$$
Divide by $n^2$ and use the power series for $\log(1+x)$:
$$
\log\left(\frac{\left(2n^{1/n}-1\right)^n}{n^2}\right)=-\frac{\log(n)^2}n+\frac{\log(n)^3}{n^2}+O\!\left(\frac{\log(n)^6}{n^3}\right)
$$
Multiply by $\frac{n\left(n-\frac12\right)}{\log^2(n)}$, use the power series for $e^x$, and apply Stirling's Formula to get
$$
\begin{align}
\frac{\sqrt2n^{\left(n-\frac 12\right)}}{n!}\left(\frac{\left(2n^{1/n}-1\right)^n}{n^2}\right)^{\frac{n\left(n-\frac12\right)}{\log^2(n)}}
&=\frac{\sqrt2n^{\left(n-\frac 12\right)}}{\sqrt{2\pi n}\,n^ne^{-n}}ne^{\frac12-n}\left(1+O\!\left(\frac{\log(n)^4}n\right)\right)\\
&=\sqrt{\frac e\pi}\left(1+O\!\left(\frac{\log(n)^4}n\right)\right)
\end{align}
$$
Therefore,
$$
\lim_{n\to\infty}\frac{\sqrt2n^{\left(n-\frac 12\right)}}{n!}\left(\frac{\left(2n^{1/n}-1\right)^n}{n^2}\right)^{\frac{n\left(n-\frac12\right)}{\log^2(n)}}=\sqrt{\frac e\pi}
$$
|
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|
How to find $\sum_{n=1}^{\infty}\frac{H_nH_{2n}}{n^2}$ using real analysis and in an elegant way? I have already evaluated this sum:
\begin{equation*}
\sum_{n=1}^{\infty}\frac{H_nH_{2n}}{n^2}=4\operatorname{Li_4}\left( \frac12\right)+\frac{13}{8}\zeta(4)+\frac72\ln2\zeta(3)-\ln^22\zeta(2)+\frac16\ln^42
\end{equation*}
using the identity
$\displaystyle\frac{1}{1-x^2}\ln\left(\frac{1-x}{1+x}\right)=\sum_{n=1}^{\infty}\left(H_n-2H_{2n}\right)x^{2n-1}$
but kind of lengthy. any other approaches?
|
Outstanding solution due to Cornel Valean.
Recall the generating function $\displaystyle-\ln(1+x)\ln(1-x)=\sum_{n=1}^\infty x^{2n}\frac{H_{2n}-H_n}{n}+\frac12\sum_{n=1}^\infty\frac{x^{2n}}{n^2},$ where if we multiply both sides by $\ln(1+x)/x$ and use the simple fact $\displaystyle\int_0^1x^{2n-1}\ln(1+x)\ dx=\frac{H_{2n}-H_n}{2n}$ then the Au-Yeung eries result, $\displaystyle\sum_{n=1}^\infty\left(\frac{H_n}{n}\right)^2=\frac{17}{4}\zeta(4),$ and $\displaystyle\sum_{n=1}^\infty\frac{H_n}{n^3}=\frac54\zeta(4)$, we have
$$\small{\sum_{n=1}^\infty\frac{H_{2n}H_n}{n^2}-2\sum_{n=1}^\infty\frac{H_{2n}^2}{(2n)^2}-2\sum_{n=1}^\infty\frac{H_n}{(2n)^3}=\frac{29}{16}\zeta(4)-\int_0^1\frac{\ln(1-x)\ln^2(1+x)}{x}\ dx=\frac{23}{16}\zeta(4)}\tag{1}$$
where $\displaystyle\int_0^1\frac{\ln(1-x)\ln^2(1+x)}{x}\ dx=-\frac38\zeta(4)$ is an already famous integral elementary to evaluate using the algebraic identity, $\displaystyle6a^2b=(a+b)^3-(a-b)^3-2b^3$. since $\displaystyle\sum_{n=1}^\infty(-1)^{n-1}\frac{H_n}{n^3}=\frac{11}4\zeta(4)-\frac74\ln(2)\zeta(3)+\frac12\ln^22\zeta(2)-\frac1{12}\ln^42-2\operatorname{Li}_4\left(\frac12\right)$ and $\displaystyle\sum_{n=1}^\infty(-1)^{n-1}\frac{H_n^2}{n^2}=\frac{41}{16}\zeta(4)-\frac74\ln(2)\zeta(3)+\frac12\ln^22\zeta(2)-\frac1{12}\ln^42-2\operatorname{Li}_4\left(\frac12\right)$, if using for the last two series in (1) that $2\sum_{n=1}^\infty a_{2n}=\sum_{n=1}^\infty a_n-\sum_{n=1}^\infty (-1)^{n-1}a_n$, we conclude that $\displaystyle\sum_{n=1}^\infty\frac{H_nH_{2n}}{n^2}=\frac{13}8\zeta(4)+\frac72\ln(2)\zeta(3)-\ln^22\zeta(2)+\frac1{6}\ln^42+4\operatorname{Li}_4\left(\frac12\right)$ and the solution is complete.
|
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|
Solving $3x^2 - 4x -2 = 0$ by completing the square I can't understand the solution from the textbook (Stroud & Booth's "Engineering Mathematics" on a problem that involves solving a quadratic equation by completing the square.
The equation is this:
$$
\begin{align}
3x^2 - 4x -2 = 0 \\
3x^2 - 4x = 2
\end{align}
$$
Now, divide both sides by three:
$$x^2 - \frac{4}{3}x = \frac{2}{3}$$
Next, the authors add to both sides the square of the coefficient of $x$, completing the square on the LHS:
$$x^2 - \frac{4}{3}x + \left(\frac{2}{3}\right)^2 = \frac{2}{3} + \left(\frac{2}{3}\right)^2$$
Now, the next two steps (especially the second step) baffle me. I understand the right-hand side of the first quation (how they get the value of $\frac{10}{9}$), but the last step is a complete mystery to me:
$$
\begin{align}
x^2 - \frac{4}{3}x + \frac{4}{9} = \frac{10}{9} \\
\left(x - \frac{2}{3}\right)^2 = \frac{10}{9}
\end{align}
$$
Can anyone please explain how they went from the first step to the second step?
|
Maybe it would be easier with prettier numbers? If you were to solve the equation $$x^2+200x+10000=0,$$ you’d recognize that $$(x+100)^2 = x^2+200x+10000$$ which would simplify things greatly, quickly leading to the solution $x=-100$.
If, however, your equation does not match the $(x+a)^2 = x^2 + 2ax + a^2$ formula perfectly, you'll have to add something to make it match. For example, to solve $$x^2+200x+9999=0,$$ you’ll have to transform it like this:
$$x^2 + 200x+9999+1-1=0$$
$$x^2 + 200x+10000-1=0$$
$$(x-100)^2-1=0$$
Note that $x^2+200x$ can only be completed by 10000 and not by any other number.
|
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|
Minimum possible values of $\frac{x^2+y^2+z^2+1}{xy+yz+z}$ and $\frac{x^2+y^2+z^2+1}{xy+y+z}$ Let $k$ and $m$ be the minimum possible values of $$\frac{x^2+y^2+z^2+1}{xy+yz+z} \quad \text{and} \quad \frac{x^2+y^2+z^2+1}{xy+y+z}$$ respectively where $x,y,z$ are non-negative real numbers. What is the value of $km+k+m$?
I used the AM-GM inequality to get a minimum value for $x^2+y^2+z^2+1$ but the problem is with getting an upper bound for $xy+yz+z$ and $xy+y+z$.
This problem is from India IMC 2017 team contest.
|
Let $U = [0,\infty)^3$ and $V = [0,\infty)^3 \times (0,\infty)$.
We will parameterize $U$ by elements in $V$ through following map:
$$V \in (X,Y,Z,W) \mapsto (x,y,z) = \left(\frac{X}{W},\frac{Y}{W},\frac{Z}{W}\right) \in U$$
The minimum $k$ we seek equals to
$$k = \min_{(x,y,z)\in U}\left\{\frac{x^2+y^2+z^2+1}{xy+yz+z}\right\}$$
Notice
$$\frac{1}{k} = \max_{(x,y,z)\in U}\left\{\frac{xy+yz+z}{x^2+y^2+z^2+1}\right\}
= \max_{(X,Y,Z,W)\in V}\left\{\frac{XY+YZ+ZW}{X^2+Y^2+Z^2+W^2}\right\}\\
= \max_{(X,Y,Z,W)\in V\cap S^3}\{ XY+YZ + ZW\}
= \frac12 \max_{u \in V\cap S^3}\{ u^T\Lambda u \}
$$
where $u^T = (X,Y,Z,W)$ and
$\Lambda$ is the matrix $\begin{bmatrix}
0 & 1 & 0 & 0\\
1 & 0 & 1 & 0\\
0 & 1 & 0 & 1\\
0 & 0 & 1 & 0\end{bmatrix}$
Since $\Lambda$ is real symmetric, it can be diagonalized by orthogonal matrices and its eigenvalues are real. Let $\lambda_1 \ge \lambda_2 \ge \lambda_3 \ge \lambda_4$ be the eigenvalues of $\Lambda$ and $v_1, v_2, v_3, v_4$ be a corresponding set of orthonormal eigenvectors.
For any $u \in S^3$, we have $u = \sum_{k=1}^4 \alpha_k v_k$ where $\alpha_k = u^T v_k$. Furthermore,
$$u^T \Lambda u = \sum_{k=1}^4 \lambda_k \alpha_k^2 \le \lambda_1\sum_{k=1}^4 \alpha_k^2 = \lambda_1 u^T u = \lambda_1$$
This implies
$$\max_{u \in V\cap S^3}\{ u^T\Lambda u \}
\le \max_{u \in S^3}\{ u^T \Lambda u \} \le \lambda_1$$
Since $\Lambda$ is a non-negative irreducible,
Perron-Frobenius theorem tells us $\lambda_1$ is simple and $v_1$ can be chosen to belong to $(0,\infty)^4 \subset V$. This means
$$\max_{u \in V\cap S^3}\{ u^T\Lambda u \} \ge v_1^T \Lambda v_1 = \lambda_1 v_1^T v_1 = \lambda_1$$
and hence $\max\limits_{u \in V\cap S^3}\{ u^T\Lambda u \} = \lambda_1$.
As a corollary, we find
$$k = \frac{2}{\lambda_1}$$
It is not hard to work out the eigenvalues/eigenvalues of $\Lambda$. They are
$$\lambda_k = 2\cos(\frac{k\pi}{5})
\quad \text{ and } \quad v_k^T \propto \left( \sin\frac{k\pi}{5}, \sin\frac{2k\pi}{5}, \sin\frac{3k\pi}{5}, \sin\frac{4k\pi}{5}\right)$$
In particular, $\lambda_1 = 2\cos\frac{\pi}{5} = \varphi$ and $v_1^T \propto (1, \varphi, \varphi, 1)$ where $\varphi = \frac{1+\sqrt{5}}{2}$ is the golden ratio.
From this, we can conclude $$k = \frac{2}{\varphi}$$
and the minimum value of $k$ is achieved at $(x,y,z) = (1,\varphi,\varphi)$.
For minimum $m$, the derivation is similar, we have
$$m = \min_{(x,y,z)\in U}\left\{\frac{x^2+y^2+z^2+1}{xy+y+z}\right\} = \frac{2}{\varphi}$$
and the minimum value is achieved at $(x,y,z) = (\frac{1}{\varphi},1,\frac{1}{\varphi})$.
Combine all these, we obtain
$$mk + m + k = \left(\frac{2}{\varphi}\right)^2 + 2 \left(\frac{2}{\varphi}\right) = \frac{4}{\varphi^2}(\varphi + 1) = 4$$
|
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|
How prove with using mathematical induction $\prod_{i=0}^n 1+q^{2^i} = \frac{1-q^{2^{n+1}}}{1-q}$?
Prove the identity
\begin{align}
\prod_{i=0}^n \left(1+q^{2^i}\right) = \frac{1-q^{2^{n+1}}}{1-q}
\end{align}
for each nonnegative integer $n$.
To begin with, I cannot verify the equality itself.
$\prod_{i=0}^n 1+q^{2^i} = \frac{1-q^{2^{n+1}}}{1-q}$
If $n=1$ then I have next expressions
$(1+q)(1+q^2) = \frac{1-q^{2^{1+1}}}{1-q} = $
$ = (1+q)(1+q^2) = \frac{1-q^{4}}{1-q}$
Well played and calculated =(.Speak me where i s mistaked.
|
We will start off with the basis step using the base value of $n = 0$.
Basis step
$\prod_{i=0}^{0} (1 + q^{2^i}) = \dfrac{1 - q^{2^{n + 1}}}{1 - q}$
=> $1 + q^{2^0} = \dfrac{1 - q^{2^{0 + 1}}}{1 - q}$
=> $1 + q = \dfrac{1 - q^2}{1 - q}$
=> $(1 + q)(1 - q) = 1 - q^2$
=> $1 - q^2 = 1 - q^2$
The basis step is hereby proved by equality and we move forward to the inductive step.
Inductive step
If it is the case that $\prod_{i=0}^{n} (1 + q^{2^i}) = \dfrac{1 - q^{2^{n + 1}}}{1 - q}$ is true, then is $\prod_{i=0}^{k} (1 + q^{2^i}) = \dfrac{1 - q^{2^{k + 1}}}{1 - q}$ also true for any $0 \leq k$, which forms our hypothesis.
For this equality to be true, our hypothesis must also be true for $k + 1$.
$\prod_{i=0}^{k} (1 + q^{2^i})(1 + q^{2^{k + 1}}) = \dfrac{1 - q^{2^{k + 2}}}{1 - q}$
We use our hypothesis for a substitution:
$(\dfrac{1 - q^{2^{k + 1}}}{1 - q})(1 + q^{2^{k + 1}}) = \dfrac{(1 - q^{2^{k + 1}})(1 + q^{2^{k + 1}})}{1 - q} = \dfrac{1 - (q^{2^{k + 1}})^2}{1 - q} = \dfrac{1 - q^{2^{k + 1} * 2}}{1 - q} = \dfrac{1 - q^{2^{k + 2}}}{1 - q}$, which completes our proof.
|
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|
Minimum value of $\sqrt{x^2+25}+\sqrt{y^2+16}$ if $x+y=12$ If $x,y\in\mathbb R^+$, $x+y=12$, what is the minimum value of $\sqrt{x^2+25}+\sqrt{y^2+16}$?
I got the question from a mathematical olympiad competition (of China) for secondary 2 student, so I don't expect an "analysis" answer.
The answer should be $15$, when $x=\frac{20}{3}$ and $ y=\frac{16}{3}$ (just answer, no solution :< ).
I try to use AM-GM inequality, but I couldn't manage to get the answer.
$$\sqrt{x^2+25}+\sqrt{y^2+16}=\sqrt{x^2+25}+\sqrt{(x-12)^2+16}=\sqrt{x^2+25}+\sqrt{x^2-24x+160}$$
Can this help?
I also tried to plot the graph, see here.
Any help would be appreciated. Thx!
|
$\sqrt{x^2+25}$ is the distance between $(x,0)$ and $(0,5)$, while $\sqrt{(x-12)^2+16}$ is the distance between $(x,0)$ and $(12,-4)$. Hence we need to find $x$ such that the sum of these two distances reaches the minimum. That is exactly the point that the line through $(12,-4)$ and $(0,5)$ meets the $x$-axis. Therefore we get $x = \frac{20}{3}$ and so $y = 12-\frac{20}{3} = \frac{16}{3}$.
|
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|
Uniform convergence of $\sum_{n=0} ^\infty \frac{(-1)^n}{2n+1} \left(\frac{2z}{1-z^2}\right)^{2n+1}$
I am given:
$$ f(z)=\sum_{n=0} ^\infty \frac{(-1)^n}{2n+1} \left(\frac{2z}{1-z^2}\right)^{2n+1}$$
And asked to show that this complex function series converges uniformly for $|z|\leq \frac{1}{3}$. I am also asked to determine its complex derivative in simplified form.
$1)$ I try:
$$ f'(z)=\frac{2z^2 +2}{(1-z^2)^2} \cdot \sum_{n=1} ^\infty \frac{(-1)^n}{2n+1} \cdot (2n+1) \left(\frac{2z}{1-z^2}\right)^{2n} $$
Here I pulled out the factor from the chain rule (+quotient rule) as it does not depend on $n$.
$$ f'(z)= \frac{2z^2 +2}{(1-z^2)^2} \cdot \sum_{n=1} ^\infty (-1)^n \left(\frac{2z}{1-z^2}\right)^{2n} $$
However, I seem to now be missing the factor of $2n$ in the enumerator as I wanted it to be a complex cosine, what went wrong here? I also do not have the right index yet, I suspect this may be related to my problem.
$2)$ For convergence I recognise this as a power series so I can use one of the many tests, I will use the ratio test:
$$ \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|=\lim_{n \to \infty} \left|-\frac{2n+1}{2n+3} \left(\frac{2z}{1-z^2}\right)^2\right|=\frac{4 |z|^2}{|1-z^2|^2}$$
For convergence we need $\frac{4 |z|^2}{|1-z^2|^2} <1$
Observe by reverse triangle inequality $$\frac{2 |z|}{|z^2-1|} \leq \frac{2 |z|}{||z|^2-1|}$$
Now since $|z|\leq 1/3$, we have $ 1\geq ||z|^2 -1| \geq \frac{8}{9} $ or $1\leq \frac{1}{||z|^2 -1|} \leq \frac{9}{8} $ therefore
$$\frac{2 |z|}{|z^2-1|} \leq 2 \cdot 1/3 \cdot 9/8 =3/4$$
The square of this is expression is what we are interested in, this is $9/16<1$
|
Observe for $|z|\leq \frac{1}{3}$ We have:
$$|f_n|=\left| \frac{(-1)^n}{2n+1} \left(\frac{2z}{1-z^2}\right)^{2n+1} \right| =
\frac{1}{2n+1} \left|\left(\frac{2z}{1-z^2}\right)\right|^{2n+1} $$
By reverse triangle inequality we get:
$$|f_n|\leq \left|\left(\frac{2|z|}{|1|-|z^2|}\right)^{2n+1} \right|=\left|\left(\frac{2|z|}{|1|-|z|^2}\right)^{2n+1} \right| \leq \left(\frac{2 / 3}{1-1/9}\right)^{2n+1}=(\frac 3 4)^{2n+1}
= \frac{3}{4} \cdot \left(\frac{9}{16}\right)^n= M_n $$
By the Weiserstrass M-test $\sum M_n$ is a convergent infinite geometric series hence the original sum is convergent in the ball of radius $\frac 1 3$.
|
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|
For which $k\in \Bbb N$ does $x^2+x+1$ divide $ (x + 1)^{2k}+x^{2k} + 1$? I tried many attempt to solve this problem but failed every time.
The polynomial $x^{2k} + 1 + (x + 1)^{2k}$ is not divisible by $x^2 + x + 1$. Find the value of $k \in \mathbb{N}$
My question is:
can this be solved with using remainder/factor theorem only?
I haven't learned Number Theory yet so maybe I don't know sufficient number of lemmas.
|
You said you haven't learned number theory yet.
I don't know if you have learned about complex roots of unity.
Here is a proof where you just need to know the binomial expansion and mathematical induction.
Let $f_n(x)=(x+1)^n+x^n+1.$
I claim $f_{6s}(x)=q_1(x)(x^2+x+1)+3,$
$ f_{6s+2}(x)=q_2(x)(x^2+x+1),$ and
$f_{6s+4}(x)=q_3(x)(x^2+x+1).$
Base case $(s=0)$:
$f_0(x)=(x+1)^0+x^0+1=1+1+1=3.$
$f_2(x)=(x+1)^2+x^2+1=x^2+2x+1+x^2+1=2x^2+2x+2=2(x^2+x+1)$
$f_4(x)=(x+1)^4+x^4+1=x^4+4x^3+6x^2+4x+1+x^4+1=2x^4+4x^3+6x^2+4x+2$
$=(2x^2+2x+2)(x^2+x+1).$
Induction step:
$f_{n+6}(x)-f_n(x)=(x+1)^{n+6}-(x+1)^n+x^{n+6}-x^n$
$=(x+1)^n((x+1)^6-1)+x^n(x^6-1).$
Note that $x^6-1=(x^3+1)(x^3-1)=(x^3+1)(x-1)(x^2+x+1)$ is a multiple of $x^2+x+1$.
Therefore also $(x+1)^6-1=((x+1)^3+1)((x+1)-1)((x+1)^2+(x+1)+1)$ is
a multiple of $x^2+x+1$ [since $(x+1)^3+1=x^3+3x^2+3x+2=(x+2)(x^2+x+1)],$
and we are basically done.
|
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|
What is the solution of the equation $xyp^2 + (3x^2 - 2y^2)p - 6xy=0$, where $p = \frac{dx}{dy}$ What is the solution of the equation $xyp^2 + (3x^2 - 2y^2)p - 6xy=0$, where $p = \frac{dx}{dy}$
I was trying to solve it by dividing the whole equation by $xy$ and then integrate it
$\frac{dy}{dx}[\frac{dy}{dx} + (\frac{3x}{y} - \frac{2y}{x})] = 6$, but still this equation is non separable.
Please tell me how to solve it.
|
Using quadratic formula, we have
$$p=\frac{2y^2 - 3x^2\pm\sqrt{(3x^2-2y^2)^2+24x^2y^2}}{2xy}$$
$$=\frac{2y^2 - 3x^2\pm(3x^2+2y^2)}{2xy}. $$
That is:
$$\frac{dy}{dx}=p=\frac{2y}{x} \text{ or } \frac{-3x}{y}. $$
In both cases, we can easily solve the ODE.
|
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|
not every matrix is sum of two idempotent matrices i'm reading the proof of theory show that the matrix ring $M_{n}(R)$ over $R$ contains an element which cannot be written as sum of two idempotents. ($R$ is a ring with $1$, and $n$ is a positive integer greater than $1$)
Lemma: Let $a$ be an element of $R$ with $a^{2} = 0$. if $a = e + f$ for idempotents $e, f$ then $4a = 0$.
theory's proof: We write $M_{n}(R) = \sum_{i,j=1}^{n}Re_{ij} $ where $e_{ij}$ are matrix units. Suppose,
to the contrary, that every element of $M_{n}(R)$ is the sum of two idempotents. Then, by Lemma, $4e_{12} = 0 $ and so $4R = 0$. Consider the element $a = e_{11} + e_{12} + e_{21}$, and choose idempotents $e = \sum r_{ij}e_{ij}$ and $f$ such that $a = e + f$. Since $a - e = f = f^{2} = a^{2} — ae — ea + e$ , we get $a^{2}= a + ae + ea — 2e$ . Comparing the coefficients of $e_{11}$, $e_{12}$ and $e_{21}$ on both sides, we get $1 = r_{12} + r_{21}$, $0 = r_{11} + r_{22} - r_{12}$ and $0 = r_{11} + r_{22} - r_{21}$ therefore $1 = 2r_{12}$. Then $4R = 0$ implies that $1 = 4r_{12}^{2} = 0$, which is a contradiction.
Please explain me why we got $4R=0$? And compare the coefficients of $e_{11}$, $e_{12}$ and $e_{21}$?
(Sorry, English is not my native language)
|
Why we got $4R=0$
Because $\begin{bmatrix}0&4r\\0&0\end{bmatrix}=r4\begin{bmatrix}0&1\\0&0\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$ for any $r$, so $4R=\{0\}$.
And compare the coefficients of $e_{11}$, $e_{12}$ and $e_{21}$?
Just write out $a^{2}= a + ae + ea — 2e$:
$\begin{bmatrix}2&1\\ 1&1\end{bmatrix}= \begin{bmatrix}1&1\\1&0\end{bmatrix}^2=\begin{bmatrix}1&1\\1&0\end{bmatrix}+\begin{bmatrix}1&1\\1&0\end{bmatrix}\begin{bmatrix}r_{11}&r_{12}\\r_{21}&r_{22}\end{bmatrix}+\begin{bmatrix}r_{11}&r_{12}\\r_{21}&r_{22}\end{bmatrix}\begin{bmatrix}1&1\\1&0\end{bmatrix}-2\begin{bmatrix}r_{11}&r_{12}\\r_{21}&r_{22}\end{bmatrix}$
$\begin{bmatrix}2&1\\ 1&1\end{bmatrix}= \begin{bmatrix}1&1\\1&0\end{bmatrix}+\begin{bmatrix}r_{11}+r_{21}&r_{12}+r_{22}\\r_{11}&r_{12}\end{bmatrix}+\begin{bmatrix}r_{11}+r_{12}&r_{11}\\r_{21}+r_{22}&r_{21}\end{bmatrix}-\begin{bmatrix}2r_{11}&2r_{12}\\2r_{21}&2r_{22}\end{bmatrix}$
$\begin{bmatrix}2&1\\ 1&1\end{bmatrix}= \begin{bmatrix}1+r_{21}+r_{12}&1+r_{22}+r_{11}-r_{12}\\1+r_{11}+r_{22}-r_{21}&r_{12}+r_{21}-2r_{22}\end{bmatrix}$
|
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"url": "https://math.stackexchange.com/questions/3220317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Computing volume inside a ball and outside a cylinder I try to calculate the volume inside the ball $\ x^2 + y^2 + z ^2 = 4 $ the outside the cylinder $\ x^2+y^2=2x $ using double integral.
Since the shape is symmetric I chose $\ z = 0 $ as bottom limit and the ball is the upper limit. I try to convert to polar form and so I get
$$\ x^2 + y^2 + z ^2 = 4 \Rightarrow z = \sqrt{4-r^2} \\x^2 + y^2 = 2x \Rightarrow r = 2 \cos \theta $$
therefore the integral should be
$$\ 4 \cdot \int_0^{\pi/2} \int_0^{2\cos\theta} (\sqrt{4-r^2}) \ r \ dr \ d \theta$$
but this integral is way to messy for me to calculate. I mean first step of calculating integral for $\ r $ is okay but then the value I get and calculating integral of $\ \theta $ is beyond me and I believe it is beyond the scope of the course. I guess I'm missing something in the process here?
|
Let $I = \int_0^{\pi/2} \int_0^{2\cos\theta} (\sqrt{4-r^2}) \ r \ dr \ d \theta$
$I = \frac{-1}{2}\int_0^{\pi/2} \int_0^{2\cos\theta} (\sqrt{4-r^2}) \ d(4-r^2) \ d \theta$
$I = \frac{-1}{2}.\frac{2}{3}\int_0^{\pi/2}\big[(4-r^2)^{3/2}\big]^{2cos\theta}_0d\theta$
$I = \frac{-1}{2}.\frac{2}{3}\int_0^{\pi/2}\big[(4-4cos^2\theta)^{3/2} - 4^{3/2}\big]d\theta$
$I = \frac{-1}{3}.2^3\int_0^{\pi/2}\big[(sin^2\theta)^{3/2} - 1\big]d\theta$
$I = \frac{-1}{3}.2^3\int_0^{\pi/2}\big[(sin^3\theta)- 1\big]d\theta$
$\bigg[ \int^{\pi/2}_0sin^mxdx = \frac{m-1}{m}.\frac{m-3}{m-2}...1 \bigg]$, if m is odd
$I = \frac{-2^3}{3}\bigg[\big[\frac{2}{3}.1\big] -\frac{\pi}{2}\bigg]$
Thus, $I = \frac{8}{18}[3\pi - 4] = \frac{4}{9}[3\pi - 4]$
Can you proceed now?
|
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"url": "https://math.stackexchange.com/questions/3220705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Exact value of $100!\times\left(1+\sum_{n=1}^{100} \frac{(-1)^n (n^2+n+1)}{n!}\right)$ Problem
Find the exact value of $$100!\times\left(1+\sum_{n=1}^{100} \frac{(-1)^n (n^2+n+1)}{n!}\right)$$
What I tried :
multiply $(n-1)$ to sumnation's numerator and denominator then it changed to
$$100!\times\left(1+\sum_{n=1}^{100} \frac{(-1)^n (n^3-1)}{n(n-2)!}\right)$$
But this didn't give me any clues to solve this.
I just guess telescoping is the way to solve because this is finite-sum, but I don't have any idea.
|
Write the sum like
$$\sum_{n=1}^{100} \frac{(-1)^n n^2}{n!}+\frac{(-1)^n n}{n!} + \frac{(-1)^n}{n!}$$
$$=\sum_{n=1}^{100} \frac{(-1)^n n^2}{n!}+\sum_{n=1}^{100}\frac{(-1)^n n}{n!} +\sum_{n=1}^{100} \frac{(-1)^n}{n!}$$
$$=\sum_{n=1}^{100} \frac{(-1)^n n^2}{n!}+\sum_{n=0}^{99}\frac{(-1)^{n+1}}{n!} +\sum_{n=1}^{100} \frac{(-1)^n}{n!}$$
$$=\sum_{n=1}^{100} \frac{(-1)^n n^2}{n!}+\sum_{n=0}^{99}\frac{(-1)^{n+1}}{n!} +\sum_{n=1}^{100} \frac{(-1)^n}{n!}$$
$$=\sum_{n=1}^{100} \frac{(-1)^n n^2}{n!}-1+\frac{1}{100!}.$$
Because only the zeroth term of the second sum and the 100th term of the third sum survive. Now your expression is
$$100! \times \left(1+\sum_{n=1}^{100} \frac{(-1)^n n^2}{n!}-1+\frac{1}{100!}\right)$$
$$= 100!\sum_{n=1}^{100} \frac{(-1)^n n^2}{n!} +1.$$
If $n$ is odd and at least three, then two successive terms of the remaining sum are
$$\frac{-n}{(n-1)!}+\frac{n+1}{n!} = \frac{-1}{(n-2)!}+\frac{1}{n!}.$$ So the sum from 3 to 100 telescopes to $-1 + 1/99!.$ Tacking on the $n=1$ and $n=2$ terms gives us
$$100!\left(\frac{-1}{0!} +\frac{2}{1!} -1 +\frac{1}{99!}\right) +1 = 101.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show that the characteristic polynomial is the same as the minimal polynomial
Let $$A =\begin{pmatrix}0 & 0 & c \\1 & 0 & b \\ 0& 1 & a\end{pmatrix}$$
Show that the characteristic and minimal polynomials of $A$ are the same.
I have already computated the characteristic polynomial
$$p_A(x)=x^3-ax^2-bx-c$$
and I know from here that if I could show that the eigenspaces of $A$ all have dimension $1$, I would be done. The problem is that solving for the eigenvalues of this (very general) cubic equation is difficult (albeit possible), meaning it would be difficult to find bases for the eigenspaces.
A hint would be appreciated.
|
Compute:
$$A^2 = \begin{pmatrix} 0 & c & ac \\ 0 & b & c + ab \\ 1 & a & b + a^2\end{pmatrix}.$$
So, we just need to show that $A^2, A, I$ are linearly independent. Clearly $A$ is not a multiple of $I$, so we just need to show there is no solution to the equation
$$A^2 = pA + qI \iff \begin{pmatrix} 0 & c & ac \\ 0 & b & c + ab \\ 1 & a & b + a^2\end{pmatrix} = p\begin{pmatrix} 0 & 0 & c \\ 1 & 0 & b \\ 0 & 1 & a\end{pmatrix} + q\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}$$
for $p$ and $q$. In particular, if you examine the entries in the left column, bottom row, we get
$$1 = 0p + 0q,$$
which means there is indeed no solution. Hence $I, A, A^2$ are linearly independent, so no quadratic of $A$ will be equal to the $0$ matrix. Thus, the minimal polynomial must be (at least) a cubic, and equal to the characteristic polynomial
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $\sin(18^\circ)=\frac{a + \sqrt{b}}{c}$, then what is $a+b+c$? If $\sin(18)=\frac{a + \sqrt{b}}{c}$ in the simplest form, then what is $a+b+c$?
$$ $$
Attempt:
$\sin(18)$ in a right triangle with sides $x$ (in front of corner with angle $18$ degrees), $y$, and hypotenuse $z$, is actually just $\frac{x}{z}$, then $x = a + \sqrt{b}, z = c$. We can find $y$ as
$$ y = \sqrt{c^{2}- (a + \sqrt{b})^{2}} $$
so we have
$$ \cos(18) = \frac{y}{z} = \frac{\sqrt{c^{2}- (a + \sqrt{b})^{2}}}{c}$$
I also found out that
$$b = (c \sin(18) - a)^{2} = c^{2} \sin^{2}(18) - 2ac \sin(18) + a^{2}$$
I got no clue after this.
The solution says that $$ \sin(18) = \frac{-1 + \sqrt{5}}{4} $$
I gotta intuition that we must find $A,B,C$ such that
$$ A \sin(18)^{2} + B \sin(18) + C = 0 $$
then $\sin(18)$ is a root iof $Ax^{2} + Bx + C$, and $a = -B, b = B^{2} - 4AC, c = 2A$.
Totally different. This question is not asking to prove that $sin(18)=(-1+\sqrt{5})/4$, that is just part of the solution.
|
Edit: This is now obsolete - it applies to the original version of the question, without the condition "in simplest form":
It's clear that $a+b+c$ is not determined, since $$\frac{2a+\sqrt{4b}}{2c}=\frac{a+\sqrt b}c$$but $$2a+4b+2c\ne a+b+c.$$
|
{
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"url": "https://math.stackexchange.com/questions/3230730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
real solution of equation $(x^2+6x+7)^2+6(x^2+6x+7)+7=x$ is
Number of real solution of equation $(x^2+6x+7)^2+6(x^2+6x+7)+7=x$ is
Plan
Put $x^2+6x+7=f(x)$. Then i have $f(f(x))=x$
For $f(x)=x$
$x^2+5x+7=0$ no real value of $x$
For $f(x)=-x$
$x^2+8x+7=0$
$x=-7,x=-1$
Solution given is all real solution
Help me please
|
Let $x^2+6x+7=y$.
Thus, $y^2+6y+7=x$, which gives
$$x^2+6x+7-(y^2+6y+7)=y-x$$ or
$$(x-y)(x+y+7)=0.$$
Thus, $x=y,$ which does not give real solutions, or
$$y=-x-7,$$ which gives
$$x^2+6x+7=-x-7.$$
Can you end it now?
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show that $2 - \sqrt{2}$ is irrational I suppose $2 - \sqrt{2} $ is rational.
so $$2- \sqrt{2} = {a/b} $$
where a,b are integers and gcd(a,b) = 1.
$$\text{Step 1. } 2 = (a/b)^2 \text{ //squared both sides }$$
$$\text{Step 2. } 2b^2 = a^2 \text{ //We see $a^2$ is even }$$
$$\text{Step 3. }2b^2 = (2k)^2$$
Step 3 since $a^2$ is even and so $a$ is even too. Also, k is an integer.
$$\text{Step 4. }b^2 = (2k)^2 \text{ //We see $b^2$ is even }$$
Step 4 since $b^2$ is even and so $b$ is even too.
We see $ a$ and $b$ are even. We see $gcd(a,b) \neq 1$
Is this proof correct ?
|
Yet another proof: $0<2-\sqrt{2}<1$, so $(2-\sqrt{2})^n\to 0$. Note that $(2-\sqrt{2})^n=a_n-b_n\sqrt{2}$ for some integers $a_n,b_n$ by binomial theorem, so $$0<(2-\sqrt{2})^n=(a_n-2b_n)+b_n(2-\sqrt{2}).\tag{1}$$
If $2-\sqrt{2}$ were rational, say $=p/q$, $p,q\in\mathbb{N}$, then by equation (1) $(2-\sqrt{2})^n$ must be at least $1/q$, contradicting $(2-\sqrt{2})^n\to 0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3235941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Why does $ \lim_{x \to 2} \frac{x^2-4}{x-2} =4 $ if x cannot be 2? I know that $ \lim_{x \to 2} \frac{x^2-4}{x-2} $ is evaluated as follows :-
$$ \lim_{x \to 2} \frac{x^2-4}{x-2} \\ = \lim_{x \to 2} \frac{(x+2)(x-2)}{x-2} \\ = \lim_{x \to 2} x+2 \\ = 2+2 \\ = 4 $$
By looking at the function $ \frac{x^2 - 4}{x - 2} $, I can see that 2 is not in its domain. Therefore, I am not able to understand how $ \lim_{x \to 2} x+2 = 2+2 $.
|
Limit doesn't have to do with the existence of function at that point. In the context of the given problem,
$$f(x) = \frac{x^2-4}{x-2}= \frac{(x-2)(x+2)}{x-2}$$
We study the behavior of $f$ as $x$ approaches a certain point. For example let's study the RHL by considering the value of $f$ at some points:
$$f(2.01) = 4.01$$
$$f(2.001) = 4.001$$
$$f(2.0001) = 4.0001$$
You see how when we move $x$ closer to $2$, $f(x)$ gets closer to $4$.
Similarly for the LHL,
$$f(1.99) = 3.99$$
$$f(1.999) = 3.999$$
$$f(1.9999) = 3.9999$$
The same thing happens when we approach $x$ closer to $2$, $f(x)$ gets closer to $4$.
Now in short hand we can do it like this
$$\lim_{x \to 2} \frac{(x-2)(x+2)}{x-2}$$
Now since $ x \neq 2$, $x - 2 \neq 0$. Therefore we can cancel the factor in the numerator and denominator above. Hence we get
$$\lim_{x \to 2} \frac{(x-2)(x+2)}{x-2}$$
$$=\lim_{x \to 2} x+2$$
$$=4$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3238436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
positive root of the equation $x^2+x-3-\sqrt{3}=0$ $x^2+x-3-\sqrt{3}=0$
using the quadratic formula we get
$$x=\frac{-1+\sqrt{13+4\sqrt{3}}}{2}$$ for the positive root
but the actual answer is simply $x=\sqrt3$
I am unable to perform the simplification any help would we helpful
|
Note that $(1+ 2 \sqrt3)^2 = 1 + 4 \sqrt 3 + 4 \cdot \sqrt 3^2 = 13 + 4 \sqrt 3$. The expression is now easy to simplify.
|
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|
prove $\sum_{n=1}^\infty \frac{H_n^2}{n^4}=\frac{97}{24}\zeta(6)-2\zeta^2(3)$ this series was evaluated by Cornel Valean here using series manipulation.
I took a different path as follows:
using the identity:$$\frac{\ln^2(1-x)}{1-x}=\sum_{n=1}^\infty x^n\left(H_n^2-H_n^{(2)}\right)$$
multiply both sides by $\ln^3x/x$ then integrate
$$-6\sum_{n=1}^\infty \frac{H_n^2-H_n^{(2)}}{n^4}=\int_0^1\frac{\ln^2(1-x)\ln^3x}{x(1-x)}\ dx$$
I was able here to find
\begin{align}
\sum_{k=1}^\infty\frac{H_k^{(2)}}{k^4}&=\frac43\zeta^2(3)-\frac23\sum_{k=1}^\infty\frac{H_k^{(3)}}{k^3}\\
&=\zeta^2(3)-\frac13\zeta(6)
\end{align}
as for the integral, it seems very tedious to calculate it using the derivative of beta function.
can we find it with or without using beta function?
|
For a slight variation on a theme.
As
$$\int_0^1 x^{n - 1} \ln^2 (1 - x) \, dx = \frac{H^2_n}{n} + \frac{H^{(2)}_n}{n},$$
for a proof of this result, see here, we can write the sum as
\begin{align}
\sum_{n = 1}^\infty \frac{H^2_n}{n^4} &= \sum_{n = 1}^\infty \frac{1}{n^3} \cdot \frac{H^2_n}{n}\\
&= - \sum_{n = 1}^\infty \frac{H^{(2)}_n}{n^4} + \int_0^1 \frac{\ln^2 (1 - x)}{x} \sum_{n = 1}^\infty \frac{x^n}{n^3} \, dx\\
&= - \sum_{n = 1}^\infty \frac{H^{(2)}_n}{n^4} + \int_0^1 \frac{\ln^2 (1 - x) \operatorname{Li}_3 (x)}{x} \, dx.\tag1
\end{align}
Making use of the following Maclaurin series expansion for $\ln^2 (1 - x)$, namely
$$\ln^2 (1 - x) = 2 \sum_{n = 1}^\infty \frac{H_n x^{n + 1}}{n + 1},$$
the integral in (1) can be re-written as
\begin{align}
\sum_{n = 1}^\infty \frac{H^2_n}{n^4} &= - \sum_{n = 1}^\infty \frac{H^{(2)}_n}{n^4} + 2 \sum_{n = 1}^\infty \frac{H_n}{n + 1} \underbrace{\int_0^1 x^n \operatorname{Li}_3 (x) \, dx}_{\text{IBP 3 times}}\\
&= - \sum_{n = 1}^\infty \frac{H^{(2)}_n}{n^4} + 2 \sum_{n = 1}^\infty \frac{H_n}{n+ 1} \left [\frac{\zeta (3)}{n + 1} - \frac{\zeta (2)}{(n + 1)^2} + \frac{H_{n + 1}}{(n + 1)^3} \right ]\\
&= - \sum_{n = 1}^\infty \frac{H^{(2)}_n}{n^4} + 2 \zeta (3) \underbrace{\sum_{n = 1}^\infty \frac{H_n}{(n + 1)^2}}_{n \, \mapsto \, n - 1} -2 \zeta (2) \underbrace{\sum_{n = 1}^\infty \frac{H_n}{(n + 1)^3}}_{n \, \mapsto \, n - 1} + 2 \underbrace{\sum_{n = 1}^\infty \frac{H_n H_{n + 1}}{(n + 1)^2}}_{n \, \mapsto \, n - 1}\\
&= - \sum_{n = 1}^\infty \frac{H^{(2)}_n}{n^4} + 2 \zeta (3) \sum_{n = 1}^\infty \frac{1}{n^2} \left (H_n - \frac{1}{n} \right ) - 2 \zeta (2) \sum_{n = 1}^\infty \frac{1}{n^3} \left (H_n - \frac{1}{n} \right )\\
& \qquad + 2 \sum_{n = 1}^\infty \frac{H_n}{n^4} \left (H_n - \frac{1}{n} \right )\\
&= - \sum_{n = 1}^\infty \frac{H^{(2)}_n}{n^4} + 2 \zeta (3) \sum_{n = 1}^\infty \frac{H_n}{n^2} - 2 \zeta^2 (3) - 2 \zeta (2) \sum_{n = 1}^\infty \frac{H_n}{n^3} + 2 \zeta (2) \zeta (4)\\
& \qquad + 2 \sum_{n = 1}^\infty \frac{H^2_n}{n^4} - 2 \sum_{n = 1}^\infty \frac{H_n}{n^5}\\
\Rightarrow \sum_{n = 1}^\infty \frac{H^2_n}{n^4} &= \sum_{n = 1}^\infty \frac{H^{(2)}_n}{n^4} - 2 \zeta (3) \sum_{n = 1}^\infty \frac{H_n}{n^2} + 2 \zeta (2) \sum_{n = 1}^\infty \frac{H_n}{n^3} + 2 \sum_{n = 1}^\infty \frac{H_n}{n^5}\\
& \qquad + 2 \zeta^2 (3) - 2 \zeta (2) \zeta (4).\tag2
\end{align}
Making use of the following results:
\begin{align}
\sum_{n = 1}^\infty \frac{H_n}{n^2} &= 2 \zeta (3)\\
\sum_{n = 1}^\infty \frac{H_n}{n^3} &= \frac{5}{4} \zeta (4)\\
\sum_{n = 1}^\infty \frac{H_n}{n^5} &= -\frac{1}{2} \zeta^2 (3) + \frac{7}{4} \zeta (6)\\
\sum_{n = 1}^\infty \frac{H^{(2)}_n}{n^4} &= \zeta^2 (3) - \frac{1}{3} \zeta (6)\\
\zeta (2) \zeta (4) &= \frac{7}{6} \zeta (6)
\end{align}
substituting into (2) leads to
$$\sum_{n = 1}^\infty \frac{H^2_n}{n^4} = \frac{97}{24} \zeta (6) - 2 \zeta^2 (3),$$
as desired.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Quickest way to find characteristic polynomial from a given matrix Find the Rational form of
$$
A=\begin{pmatrix} 1&2&0&4\\ \:\:\:4&1&2&0\\ \:\:\:0&4&1&2\\ \:\:\:2&0&4&1\end{pmatrix}
$$
I don't wanna the solution, instead I would like to know a quickest way to calculate $\det (\lambda I-A)$.
$$
\begin{vmatrix}
x-1 &-2 &0 &-4 \\
-4&x-1 &-2 &0 \\
0&-4 &x-1 & -2 \\
-2&0 &-4 &x-1
\end{vmatrix}
$$
here I could see the polynomial but the procedure is quite long.
https://www.symbolab.com/solver/matrix-eigenvectors-calculator/eigenvectors%20%5Cbegin%7Bpmatrix%7D1%262%260%264%5C%5C%20%20%20%20%204%261%262%260%5C%5C%20%20%20%20%200%264%261%262%5C%5C%20%20%20%20%202%260%264%261%5Cend%7Bpmatrix%7D
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Add columns $\,2,3,4\,$ to the first one (this is a circulant matrix), and get:
$$\det(t I-A)=\begin{vmatrix}t-7&-2&0&-4\\
t-7&t-1&-2&0\\
t-7&-4&t-1&-2\\
t-7&0&-4&t-1\end{vmatrix}\stackrel{R_i-R_1}=\begin{vmatrix}t-7&-2&0&-4\\
0&t+1&-2&4\\
0&-2&t-1&2\\
0&2&-4&t+3\end{vmatrix}=$$$${}$$
$$=(t-7)\begin{vmatrix}
t+1&-2&4\\
-2&t-1&2\\
2&-4&t+3\end{vmatrix}=(t-7)\left[(t^2-1)(t+3)+24+8t+8-4t-12-8t+8\right]=$$
$$=(t-7)\left[(t^2-1)(t+3)-4t+28\right]=(t-7)\left[t^3+3t^2-5t+25\right]=$$
$$=(t-7)(t+5)(t^2-2t+5)=(t+t)(t+5)(t-(1-2i))(t-(1+2i))$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find Maxima and Minima of $f( \theta) = a \sin^2 \theta + b \sin \theta \cos \theta + c \cos^2 \theta$ Show that, whatever the value of $\theta$, the expression
$$a \sin^2 \theta + b \sin \theta \cos \theta + c \cos^2 \theta\ $$
Lies between
$$\dfrac{a+c}{2} \pm \dfrac 12\sqrt{ b^2 + (a-c)^2} $$
My try:
The given expression can be reduced as sum of sine functions as:
$$(a-c) \sin^2 \theta + \dfrac b2 \sin 2 \theta + c \tag{*} $$
Now, there is one way to take everything as a function of $\theta$ and get the expression in the form of $ a \sin \theta + b \cos \theta = c$ and dividing it by $ \sqrt{ a^2 + c^2} $ both sides, but square in sine function is a big problem, also both have different arguments.
Other way, I can think of is taking $ \tan \dfrac \theta 2 = t$ and getting sine and cosine function as $ \sin \theta = \dfrac{ 2t}{1+t^2} $ while cosine function as $ \dfrac{ 1-t^2} {1+t^2}$ solving. So getting $(*)$ as a function of $t$, and simplifying we get,
$$ f(t) = \dfrac{2 Rt + 2 R t^3 + R_0 t - R_0 t^3}{1+t^4 + 2t^2} + c\tag{1}$$
For $R_0 = 2b, R = (a-c)$ , but this is where the problem kicks in!, The Range of given fraction seems $ (-\infty,+ \infty)$ and is not bounded!
So what's the problem here? Can it be solved?
Thanks :)
Edit : I'd like to thank @kaviramamurthy for pointing out that as $t \rightarrow \pm \infty, f(t) \rightarrow c$. That's a mistake here.
|
It's not actually a full answer but, in my view, interesting generalization.
Note that the problem is a special case of the following statement.
Proposition. Let $\beta$ be a symmetric bilinear form in $n$-dimensional euclidean space $\left(V, \langle\cdot, \cdot\rangle\right)$ and $\lambda_1\geq \lambda_2\geq\ldots\geq\lambda_n$ are eigenvalues of linear operator $A$, which is defined by equality $\beta(x, y)=\langle Ax, y\rangle$. For vector subspace $L\subset V$ define
$$
\underline{\lambda}(L):=\min\{\beta(v,v)| v\in L,\|v\|=1\},
\\
\overline{\lambda}(L):=\max\{\beta(v,v)| v\in L,\|v\|=1\}.
$$
Then, for all $1\leq k\leq n$ the following equlaity holds
$$
\lambda_k=\max\{\underline{\lambda}(L)|\dim L= k\}=\min\{\overline{\lambda}(L)|\dim L= n+1-k\}.
$$
In our case $n=2$ and operator $A$ (and symmetric bilinear form $\beta$) has matrix
\begin{pmatrix}
a & b/2 \\
b/2 & c
\end{pmatrix}
Characteristic equation of this matrix is $(a-\lambda)(c-\lambda)-\frac{b^2}{4}=0$, so we have eigenvalues $\lambda_1=\dfrac{a+c}{2}+\dfrac{1}{2}\sqrt{b^2+(a-c)^2}$ and $\lambda_2=\dfrac{a+c}{2}-\dfrac{1}{2}\sqrt{b^2+(a-c)^2}$.
Using proposition above we obtain that
$$
\min_{\|v\|=1}\beta(v, v)=\lambda_2, \max_{\|v\|=1}\beta(v, v)=\lambda_1.
$$
Finally, note that $\|v\|=1$ is equivalent to $v=(\sin\theta, \cos\theta)^T$ for some $\theta\in [0,2\pi)$. Thus,
$$
\min_{\theta} (a \sin^2 \theta + b \sin \theta \cos \theta + c \cos^2 \theta) = \lambda_2=\dfrac{a+c}{2}-\dfrac{1}{2}\sqrt{b^2+(a-c)^2},
\\
\max_{\theta} (a \sin^2 \theta + b \sin \theta \cos \theta + c \cos^2 \theta) = \lambda_1=\dfrac{a+c}{2}+\dfrac{1}{2}\sqrt{b^2+(a-c)^2},
$$
as desired.
|
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|
Determining the coefficients of $(1 + x + x^2 +\cdots+x^n)^{n-1}$ Suppose we have the following polynomials:
$$f_1(x)=(1 + x + x^2)$$
$$f_2(x)=(1 + x + x^2 + x^3)^2$$
$$f_3(x)=(1 + x + x^2 + x^3 + x^4)^3$$
$$f_4(x)=(1 + x + x^2 + x^3 + x^4 + x^5)^4$$
$$\vdots$$
$$f_{n-1}(x)=(1 + x + x^2 + x^3 +x^4+ x^5+\cdots+x^n)^{n-1}$$
upon expanding them we get:
$$f_1(x)=1 + x + x^2$$
$$f_2(x)=1 + 2 x + 3 x^2 + 4 x^3 + 3 x^4 + 2 x^5 + x^6$$
$$f_3(x)=1 + 3 x + 6 x^2 + 10 x^3 + 15 x^4 + 18 x^5 + 19 x^6 + 18 x^7 +
15 x^8 + 10 x^9 + 6 x^{10} + 3 x^{11} + x^{12}$$
$$f_4(x)=1 + 4 x + 10 x^2 + 20 x^3 + 35 x^4 + 56 x^5 + 80 x^6 + 104 x^7 +
125 x^8 + 140 x^9 + 146 x^{10} + 140 x^{11} + 125 x^{12} + 104 x^{13} +
80 x^{14} + 56 x^{15} + 35 x^{16} + 20 x^{17} + 10 x^{18} + 4 x^{19} + x^{20}$$
$$\vdots$$
$$f_{n-1}(x)=1 + ?x + ?x^2 + ?x^3 +?x^4+ ?x^5+\cdots+?x^{n(n-1)}$$
I'm wondering how to determine the coefficients for the n-th order? I can observe that the coefficients are symmetric.
|
Hint
$$p(x)=(1 + x + x^2 +\cdots+x^n)^{n-1}=\left(\frac{x^{n+1}-1}{x-1}\right)^{n-1}$$ $$ p(x)(x-1)^{n-1}\equiv(x^{n+1}-1)^{n-1}$$
and remember that $p(x)$ has degree equal to $(n+1)(n-1)-(n-1)=n^2-n$. So,
$$p(x)=a_0+a_1x+...+a_{n^2-n}x^{n^2-n}$$
|
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|
Solve $x^2+5x+6 \equiv 0 \pmod{\!11\cdot 17}$ Solve $x^2+5x+6 \equiv 187 \mod 187$
Solution
$$x^2+5x+6 \equiv 187 \mod 187$$
$$ (x+\frac{5}{2})^2 \equiv \frac{1}{4}$$
$$ 4(x+\frac{5}{2})^2 \equiv 1$$
$$ y:= x+\frac{5}{2} $$
$$ 4y^2 \equiv 1 \mod 11 \wedge 4y^2 \equiv 1 \mod 17 $$
$$ ( 2y \equiv 1 \mod 11 \vee 2y \equiv 10 \mod 11 ) \wedge ( 2y \equiv 1 \mod 17 \vee 2y \equiv 13 \mod 17) $$
$$ ( y \equiv 6 \mod 11 \vee y \equiv 5 \mod 11 ) \wedge ( y \equiv 9 \mod 17 \vee y \equiv 15 \mod 17) $$
Combining that from CRT I got:
$$ y \in \left\{49, 60,83,94 \right\} $$
and for example:
$$ x+\frac{5}{2} \equiv 94 \mod 187$$
$$ 2x \equiv 183 \mod 187$$
some calculus and get...
$$x \equiv 185 $$
And the same thing for each other case.
Question
Is there any faster (or smarter) way to solve equations like that?
|
Since $x^2+5x+6=(x+2)(x+3)$,\begin{align}x^2+5x+6\equiv187\pmod{187}&\iff(x+2)(x+3)\equiv0\pmod{187}\\&\iff(11\mid x+2\vee11\mid x+3)\wedge(17\mid x+2\vee17\mid x+3).\end{align}
|
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|
Interesting inequality $\frac{x^{m+1}+1}{x^m+1} \ge \sqrt[2m+1]{\frac{1+x^{2m+1}}{2}}$
Prove that $$\frac{x^{m+1}+1}{x^m+1} \ge \sqrt[k]{\frac{1+x^k}{2}}$$ for all real $x \ge 1$ and for all positive integers $m$ and $k \le 2m+1$.
My work. If $k \le 2m+1$ then $$\sqrt[k]{\frac{1+x^k}{2}}\le \sqrt[2m+1]{\frac{1+x^{2m+1}}{2}}.$$ I need prove that for all real $x \ge 1$ and for all positive integers $m$ the inequality $$\frac{x^{m+1}+1}{x^m+1} \ge \sqrt[2m+1]{\frac{1+x^{2m+1}}{2}}$$ holds.
|
The result can be proved using AM $\geq$ GM.
\begin{align}
\sqrt[2m+1]{\frac{1+x^{2m+1}}{2}}&=\sqrt[2m+1]{\prod_{i=0}^{2m} \frac{1+x^{i+1}}{1+x^{i}}}\leq\frac{1}{2m+1} \sum_{i=0}^{2m} \frac{1+x^{i+1}}{1+x^{i}} \\
&=\frac{1}{2m+1} \left[\sum_{i=0}^{m-1} \left(\frac{1+x^{i+1}}{1+x^{i}}+\frac{1+x^{2m-i+1}}{1+x^{2m-i}} \right)+\frac{1+x^{m+1}}{1+x^m}\right]\\
&=\frac{1}{2m+1}\left[ \sum_{i=0}^{m-1} \left( 2x- \frac{x-1}{1+x^{i}} -\frac{x-1}{1+x^{2m-i}}\right)+ \frac{1+x^{m+1}}{1+x^{m}}\right]\\
&=\frac{1}{2m+1}\left[ \sum_{i=0}^{m-1}\left( 2x- \frac{x-1}{1+x^m}\left[\frac{1+x^m}{1+x^{i}}+\frac{1+x^m}{1+x^{2m-i}}\right]\right)+ \frac{1+x^{m+1}}{1+x^{m}}\right]\\
&=\frac{1}{2m+1}\left[ \sum_{i=0}^{m-1}\left( 2x- \frac{x-1}{1+x^m}\left[2+\frac{x^i(x^m-1)(x^{m-i}-1)^2}{(1+x^i)(1+x^{2m-i})}\right]\right)+ \frac{1+x^{m+1}}{1+x^{m}}\right]\\
&\leq \frac{1}{2m+1}\left[ \sum_{i=0}^{m-1}\left(2x-(x-1)\frac{2}{1+x^{m}}\right)+ \frac{1+x^{m+1}}{1+x^{m}}\right]\\
&=\frac{1+x^{m+1}}{1+x^{m}}.
\end{align}
|
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|
$\int_{0}^{\pi /2} (\cos x)^{2.5} \mathrm dx$ $$\int_{0}^{\pi /2} (\cos x)^{2.5} \mathrm dx$$
I tried changing to sine and substituting, but it's getting complex. Help.
|
It is a difficult problem because you face elliptic integrals.
$$\int\cos ^{\frac{5}{2}}(x)\,dx=\frac{2}{5} \left(3 E\left(\left.\frac{x}{2}\right|2\right)+\sin (x) \cos^{\frac{3}{2}}(x)\right)$$ making
$$\int_0^{\frac \pi 2}\cos ^{\frac{5}{2}}(x)\,dx=\frac{3}{5} \sqrt{2} \left(2
E\left(\frac{1}{2}\right)-K\left(\frac{1}{2}\right)\right)=\frac{\sqrt{\pi } \,\Gamma \left(\frac{7}{4}\right)}{2\, \Gamma
\left(\frac{9}{4}\right)}$$
What you could do is to use the Taylor series expansion of $\cos(x)$ and use the binomial theorem to get
$$\cos ^{\frac{5}{2}}(x)=1-\frac{5 x^2}{4}+\frac{55 x^4}{96}-\frac{139 x^6}{1152}+\frac{1709
x^8}{129024}-\frac{22201 x^{10}}{23224320}+\frac{233771
x^{12}}{6131220480}+O\left(x^{13}\right)$$ and integtermwise. Using this expansion and the given bounds, you should arrive to $\approx 0.719173$ while the exact value should be $\approx 0.718884$.
In fact, integrals of the type
$$\int\cos ^{a}(x)\,dx=-\frac{\sin (x) \cos ^{a+1}(x)}{(a+1)
\sqrt{\sin ^2(x)}} \,
_2F_1\left(\frac{1}{2},\frac{a+1}{2};\frac{a+3}{2};\cos ^2(x)\right)$$ where appears the gaussian hypergeometric function which simplifies only if $a$ is integer.
$$\int_0^{\frac \pi 2}\cos ^{a}(x)\,dx=\frac{\sqrt{\pi }\,\Gamma \left(\frac{a+1}{2}\right)}{2\, \Gamma\left(\frac{a+2}{2}\right)}$$
|
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|
Compute $\lim\limits_{n\to \infty} \left(n^3 \int_{n} ^{2n}\frac{x dx} {1+x^5}\right) $ Compute $\lim\limits_{n\to \infty} \left(n^3 \int_{n} ^{2n}\frac{x dx} {1+x^5}\right) $. I tried to apply the first mean value theorem for definite integrals and then apply the squeeze theorem, but it didn't work. The answer given by the book is $\frac{7}{24}$,but I can't see how to get to it.
|
We should start by noting that
$$
0 \leq \int_n^{2n}\frac{x}{1+x^5}dx\leq \int_n^{2n} \frac{x}{1+x^4} dx=\left[\frac 12 \arctan(x^2)\right]_n^{2n}=\frac 12 (\arctan(2n)-\arctan n) \to 0.
$$
This justifies that we can apply L'Hospital's rule getting
$$
\lim n^3 \int_n^{2n}\frac{x}{1+x^5}\,dx = \lim \frac{\int_n^{2n}\frac{x}{1+x^5}\,dx}{\frac{1}{n^3}} = \lim \frac{2 \cdot \frac{2n}{1+(2n)^5}-\frac{n}{1+n^5}}{\frac{-3}{n^4}} = \frac{4}{-3 \cdot 32}- \frac{1}{-3 \cdot 1}=\frac{7}{24}.
$$
|
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|
What property was used in this sine transformation? I have this expression:
$$
ψ(χ) = A\sin^3(\frac{πχ}{α})
$$
And somehow the book i read equalizes the previous equation to this one:
$$
ψ(χ) = \frac{A}{4}[3\sin(\frac{πχ}{α}) - \sin(\frac{3πχ}{α})]
$$
What trigonometric identity was used to make this possible?
|
By DeMoivre's identity,
$\cos(3x)+i\sin(3x)=(\cos(x)+i\sin(x))^3=$
$\cos^3(x)+3i\cos^2(x)\sin(x)-3\cos(x)\sin^2(x)-i\sin^3(x)$.
Grouping the imaginary terms together, we see that $\sin(3x)=3\cos^2(x)\sin(x)-\sin^3(x)$.
We can rewrite $3\cos^2(x)\sin(x)$ as $3\sin(x)(1-\sin^2(x))$, which gives us $\sin(3x)=3\sin(x)-4\sin^3(x)$.
Finally, we can write $\sin^3(x)=\frac{1}{4}(3\sin(x)-\sin(3x))$.
|
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|
Calculate the maximum value of $\sum_{cyc}\frac{1}{\sqrt{a^2 + b^2}}$ where $a, b, c > 0$ and $abc = a + b + c + 2$.
$a$, $b$ and $c$ are positives such that $abc = a + b + c + 2$. Caculate the maximum value of $$\large \frac{1}{\sqrt{a^2 + b^2}} + \frac{1}{\sqrt{b^2 + c^2}} + \frac{1}{\sqrt{c^2 + a^2}}$$
This problem is adapted from a recent competition. Here's what I got, and I am pretty proud of it. (Actually not.)
$$abc = a + b + c + 2$$
$$\iff abc + (ab + bc + ca) + (a + b + c) + 1 = ab + bc + ca + 2(a + b + c) + 3$$
$$\iff (a + 1)(b + 1)(c + 1) = (a + 1)(b + 1) + (b + 1)(c + 1) + (c + 1)(a + 1)$$
$$\frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c + 1} = 1$$
And I'm done.
|
So, by Cauchy-Schwarz, we have $a^2+b^2 \geq (a+b)^2/2$, so we just have to prove that
$\displaystyle \sum_{cyc} \frac{1}{a+b} \leq 3/4$ (It's obvious that this maximum occurs when $a=b=c=2$, I will prove it.)
Letting, $a=2k,b=2l,c=2m$ we have to prove that $\displaystyle \sum_{cyc} \frac{1}{k+l} \leq 3/2$ subject to the condition $4klm=k+l+m+1$.
By expanding and letting $k+l+m=p, kl+lm+mk, klm=r$ we have to prove that $3pq-3r \geq 2p^2+2q$.
Substituting $r=(p+1)/4$, we have to prove that $q(3p-2) \geq \frac{8p^2+3p+3}{4}$.
Now, it is known that $q^2 \geq 3pr=\frac{3p^2-3p}{4}$ so it suffices to show that $\sqrt{3p^2+3p}\cdot (3p-2) \geq \frac{8p^2+3p+3}{2}$ which is equivalent to :
$(p-3)(44p^3+48p^2-9p+3)$.
By AM-GM, $p+1=4r \leq 4p^3/27$ so easily $p \geq 3$.
So, $p-3 \geq 0$ and $44p^3+48p^2-9p+3 >48p^2-9p=p(48p-9)>0$ and the problem is solved.
|
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|
Minimal Rook Difference Grids In the below grid all 18 orthogonal differences are distinct, with a difference of 18 missing.
Could the highest number be 18? The resulting graph would have valence 4, making it an Eulerian Graceful graph with edges(mod 4)=2. Rosa (1967) proved Eulerian Graceful graphs must have edges(mod 4)=0 or 3, so 18 is impossible.
Thus the minimal $3\times3$ rook difference grid has $rdg(3,3)=19$.
For $rdg(1,n)$ see Golomb Ruler.
$rdg(2,3)=9$ and $rdg(2,4)=16$, as shown below.
What are values for larger grids?
|
I found a new optimal solution for the $3 \times 5$ rook graph:
\begin{array}{|c|c|c|c|c|}
\hline
5 & 46 & 45 & 28 & 0 \\
\hline
31 & 7 & 1 & 44 & 11 \\
\hline
40 & 21 & 13 & 6 & 42 \\
\hline
\end{array}
Also I investigated the queen graph.
$qdg(2,2)=6$, so the $2 \times 2$ queen graph is graceful:
\begin{array}{|c|c|}
\hline
6 & 5 \\
\hline
2 & 0 \\
\hline
\end{array}
$qdg(2,3)=13$, so the $2 \times 3$ queen graph is graceful:
\begin{array}{|c|c|c|}
\hline
0 & 13 & 6 \\
\hline
10 & 2 & 1 \\
\hline
\end{array}
$qdg(2,4)=22$, so the $2 \times 4$ queen graph is graceful:
\begin{array}{|c|c|c|c|}
\hline
1 & 13 & 22 & 0 \\
\hline
21 & 3 & 6 & 17 \\
\hline
\end{array}
$qdg(2,5) \leq 34$. The $2 \times 5$ queen graph has 33 edges:
\begin{array}{|c|c|c|c|c|}
\hline
9 & 0 & 20 & 30 & 3 \\
\hline
34 & 32 & 1 & 16 & 8 \\
\hline
\end{array}
$qdg(3,3) \leq 29$. The $3 \times 3$ queen graph has 28 edges:
\begin{array}{|c|c|c|}
\hline
2 & 19 & 5 \\
\hline
28 & 29 & 13 \\
\hline
7 & 0 & 25 \\
\hline
\end{array}
And for a bit of fun I found that $qdg(4,4) \leq 97$. The $4 \times 4$ queen graph has 76 edges:
\begin{array}{|c|c|c|c|}
\hline
2 & 91 & 23 & 47 \\
\hline
58 & 97 & 1 & 88 \\
\hline
5 & 0 & 72 & 84 \\
\hline
78 & 59 & 86 & 36 \\
\hline
\end{array}
|
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|
Evaluate $\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n}{(2n+1)^3}$ How to prove that
$$\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n}{(2n+1)^3}=\frac{7\pi}{16}\zeta(3)+\frac{\pi^3}{16}\ln2+\frac{\pi^4}{32}-\frac1{256}\psi^{(3)}\left(\frac14\right)$$
where $H_n=1+\frac1{2}+\frac1{3}+...+\frac1{n}$ is the $n$th harmonic number.
This sum was proposed by Cornel and I solved it using integration but can we solve it using series manipulation?
The integral representation of the sum is $\ \displaystyle\frac12\int_0^1\frac{\ln^2x\ln(1+x^2)}{1+x^2}\ dx$ in case it is needed.
|
A nice generalization proved here:
$$\sum_{n=1}^\infty\frac{(-1)^{n}H_n}{(2n+1)^{2a+1}}=(2a+1)\beta(2a+2)-\frac{\ln(2)|E_{2a}|}{(2a)!}\left(\frac{\pi}{2}\right)^{2a+1}$$
$$-\frac12\left(\frac{\pi}{2}\right)^{2a+1}\sum_{k=1}^{a} \frac{|E_{2a-2k}|}{(2a-2k)!}{\pi^{-2k}}(2^{2k+1}-1)\zeta(2k+1),$$
where $E_r$ is the Euler numbers.
Set $a=1,2,3 $ we get
$$\sum_{n=1}^\infty\frac{(-1)^n H_n}{(2n + 1)^3}=3\beta(4)-\frac{7\pi}{16}\zeta(3)-\frac{\pi^3}{16}\ln(2)$$
$$\sum_{n=1}^\infty\frac{(-1)^n H_n}{(2n + 1)^5}=5\beta(6)-\frac{31\pi}{64}\zeta(5)-\frac{7\pi^3}{128}\zeta(3)-\frac{5\pi^5}{768}\ln(2)$$
$$\sum_{n=1}^\infty\frac{(-1)^n H_n}{(2n + 1)^7}=7\beta(8)-\frac{127\pi}{256}\zeta(7)-\frac{31\pi^3}{512}\zeta(5)-\frac{35\pi^5}{6144}\zeta(3)-\frac{61\pi^7}{92160}\ln(2)$$
|
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|
How to prove this inequality for $a,b,c>0$? How to prove the inequality for $a,b,c>0$ :
$$\frac{2a-b-c}{2(b+c)^2}+\frac{2b-a-c}{2(a+c)^2}+\frac{2c-b-a}{2(b+a)^2}\geq 0$$
?
|
Also, we can use SOS:
$$\sum_{cyc}\frac{2a-b-c}{(b+c)^2}=\sum_{cyc}\frac{a-b-(c-a)}{(b+c)^2}=$$
$$=\sum_{cyc}(a-b)\left(\frac{1}{(b+c)^2}-\frac{1}{(a+c)^2}\right)=\sum_{cyc}\frac{(a-b)^2(a+b+2c)}{(a+c)^2(b+c)^2}\geq0.$$
|
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|
Integral $\int_{-\infty}^{\infty} \arctan(e^x) \arctan(e^{-x})dx=\frac{7}{4}\zeta(3)$
How to prove that $$\int_{-\infty}^{\infty} \arctan(e^x) \arctan(e^{-x})\rm dx=\frac{7}{4}\zeta(3)?$$
This integral appeared while attempting to solve an unanswered question, namely:
Closed form of :$ \int_{-\infty}^{\infty}\arctan\left(e^{-x^2 \text{erf}(x)}\right)\,\arctan\left(e^{x^2\text{erf(x)}}\right)\,dx $
I happened to find that
$$\int_{-\infty}^{\infty} \arctan(e^x) ~\arctan(e^{-x})~dx=\frac{7}{4}\zeta(3).$$
How can you work out the answer on the right by hand?
I guess that there could be multiple approaches to solve this one.
|
Here is another slight variation on a theme.
Let
$$I = \int_{-\infty}^\infty \arctan (e^x) \arctan (e^{-x}) \, dx.$$
Setting $u = e^x$ gives
\begin{align}
I &= \int_0^\infty \arctan u \arctan \left (\frac{1}{u} \right ) \frac{du}{u}\\
&= \int_0^1 \arctan u \arctan \left (\frac{1}{u} \right ) \frac{du}{u} + \int_1^\infty \arctan u \arctan \left (\frac{1}{u} \right ) \frac{du}{u}\\
&= 2 \int_0^1 \arctan x \arctan \left (\frac{1}{x} \right ) \frac{dx}{x},
\end{align}
where in the second of the integrals a substitution of $x \mapsto 1/x$ has been enforced.
As
$$\arctan \left (\frac{1}{x} \right ) = \frac{\pi}{2} - \arctan x, \qquad x > 0,$$
we can write the above integral as
$$I = \pi \int_0^1 \frac{\arctan x}{x} \, dx - 2 \int_0^1 \frac{\arctan^2 x}{x} \, dx = \pi I_1 - 2 I_2.$$
For the first integral $I_1$, we have
\begin{align}
I_1 &= \int_0^1 \frac{\arctan x}{x} \, dx\\
&= \sum_{n = 0}^\infty \frac{(-1)^n}{2n + 1} \int_0^1 x^{2n} \, dx\\
&= \sum_{n = 0}^\infty \frac{(-1)^n}{(2n + 1)^2}\\
&= \mathbf{G},
\end{align}
where $\mathbf{G}$ denotes Catalan's constant.
For the second integral $I_2$, integrating by parts gives
$$I_2 = -2 \int_0^1 \frac{\ln x \arctan x}{1 + x^2} \, dx.$$
Enforcing a substituition of $x \mapsto \arctan x$ leads to
$$I_2 = -2 \int_0^{\frac{\pi}{4}} x \ln (\tan x) \, dx.$$
The integral appearing above has been evalauted elsewhere on this site (see, for example, here). The result is:
$$\int_0^{\frac{\pi}{4}} x \ln (\tan x) \, dx = \frac{7}{16} \zeta (3) - \frac{\pi \mathbf{G}}{4}.$$
Thus
$$I_2 = \frac{\pi \mathbf{G}}{2} - \frac{7}{8} \zeta (3),$$
giving
$$I = \pi \mathbf{G} - 2 \left (\frac{\pi \mathbf{G}}{2} - \frac{7}{8} \zeta (3) \right ),$$
or
$$\int_{-\infty}^\infty \arctan (e^x) \arctan (e^{-x}) \, dx = \frac{7}{4} \zeta (3),$$
as required.
|
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|
What am I doing wrong solving $\sqrt{x^2+1}- 2x+1>0$? First, I started looking at where $\sqrt{x^2+1}$ is defined: $\sqrt{x^2+1}>0$ is defined everywhere. Next I
$\sqrt{x^2+1}- 2x+1>0$
$\sqrt{x^2+1}> 2x-1$
$0> 3x^2-4x$
$0> x(3x-4)$ and I solve this for $x\in (0,\frac{4}{3}]$.
I know that this solution is wrong, because I went and drew this graph. The correct result is $x\in (-\infty,\frac{4}{3}]$.
I'm totally confused on how to solve irrational inequalities now, because the official solving in my textbook looks like this:
$\sqrt{x^2+1}> 2x-1$. This inequality is fulfilled , if the right side is negative, therefore $x<\frac{1}{2}$. If $x\geq\frac{1}{2}$, the right side is positive or equals to $0$ and we get $0> 3x^2-4x$ which is true for $x\in (0,\frac{4}{3}]$. Now with previous condition $x\geq\frac{1}{2}$ we get the solution $[\frac{1}{2},\frac{4}{3})$.The complete solution set is $x\in (-\infty,\frac{4}{3}]$.
I never solved this type of inequality in this way, because it's messy- Why would I look at the conditions $x<\frac{1}{2}$ and $x\geq\frac{1}{2}$ when I can immediately tell where irrational part is defined and where not?
In the end, this textbook solution process only confused me. Could anyone please explain it why the correct solution is $(-\infty,\frac{4}{3}]$ or more concretely: Where did I miss the part of the solution $(-\infty,0]$?
|
The OP mentioned the solution given in the book, so here we express that logic.
$\quad \{ x \in \Bbb R \, | \, \sqrt{x^2+1} - 2x+ 1 \gt 0\} =$
$\quad\quad\quad \{ x \in (-\infty, \frac{1}{2}] \; | \; \sqrt{x^2+1} - 2x+ 1 \gt 0\} \; \bigcup $
$\quad\quad\quad\{ x \in [\frac{1}{2},\frac{4}{3}] \; | \; \sqrt{x^2+1} - 2x+ 1 \gt 0\} \; \;\;\;\;\,\bigcup$
$\quad\quad\quad \{ x \in [\frac{4}{3},+\infty) \; | \; \sqrt{x^2+1} - 2x+ 1 \gt 0\} \;=$
$\quad\quad\quad(-\infty, \frac{1}{2}] \; \bigcup \; [\frac{1}{2} , \frac{4}{3}) \; \bigcup \; \emptyset = (-\infty, \frac{4}{3}) $
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Finding the sum $\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+\cdots$ The sum of series is
$${1\over(3\times5)}+{1\over(5\times7)}+{1\over(7\times9)}+\cdots$$
I used to solve this problem as its $n$th term
$${1\over (2n+1)(2n+3)}$$
Now how can I proceed??
|
Hint : The $n$ th term of your series is $$ a_n= \dfrac{1}{2}\left ( \dfrac{1}{2n+1}- \dfrac{1}{2n+3}\right)$$
expanding your series you get $$ \dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}...\right)$$
as you see the terms cancel out each other the only term left is $\dfrac{1}{3}$ with a factor of $\dfrac{1}{2}$ outside, giving your sum of the series as $\dfrac{1}{6}$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Proving inequality for induction proof: $\frac1{(n+1)^2} + \frac1{n+1} < \frac1n$ In an induction proof for $\sum_{k = 1}^n \frac{1}{k^2} \leq 2 - \frac{1}{n}$ (for $n \geq 1$), I was required to prove the inequality $\frac{1}{(n+1)^2} + \frac{1}{n+1} < \frac{1}{n}$.
This is my attempt:
$$ \begin{align} \frac{1}{(n+1)^2} + \frac{1}{n+1} &= \frac{1+(n+1)}{(n+1)^2} \\
&=\frac{n+2}{(n+1)^2} \\
&=\frac{n+2}{n^2 + 2n + 1} \\
&<\frac{n+2}{n^2 + 2n} (\text{since } n \geq 1)\\
&=\frac{n+2}{n(n+2)}\\
&=\frac{1}{n} \end{align} $$
I am just wondering if there is a simpler way of doing this.
|
To prove the inequality you were required to prove, note
$\dfrac1{n+1}+\dfrac1{(n+1)^2}<\dfrac1{n+1}+\dfrac1{(n+1)^2}+\dfrac1{(n+1)^3}+...=\dfrac{\dfrac1{1+n}}{1-\dfrac1{n+1}}=\dfrac1n$
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
If $u=(1+\cos t)(1+\cos 2t)-\sin t\sin 2t$ and $v=\sin t(1+\cos 2t)+\sin 2t(1+\cos t)$, then $u^2+v^2=4(1+\cos t)(1+\cos 2t)$
If
$$u = (1+\cos\theta)(1+\cos2\theta) - \sin\theta \sin 2\theta \qquad v = \sin\theta (1+\cos2\theta) + \sin2\theta(1+\cos\theta)$$
then show that
$$u^2 + v^2 = 4(1+\cos\theta)(1+\cos2\theta)$$
I have simplified the values of $u$ and $v$ and got:
$$u = 2\cos\theta (1+\cos2\theta) \qquad v=\sin2\theta(1+\cos\theta)$$
Then I tried to square both $u$ and $v$ individually before doing summation.
Still could not prove the statement.
|
Assigning $a$, $b$, $c$, $d$ as
$$a := 1+\cos\theta \qquad b := 1+\cos2\theta \qquad c := \sin\theta \qquad d := \sin 2\theta$$
we can write
$$u = ab - c d \qquad v = bc + ad$$
When squaring and adding, the $-2abcd$ will cancel with $2abcd$, leaving
$$u^2 + v^2 = a^2 b^2 + c^2 d^2 + b^2 c^2 + a^2 d^2 = \left(a^2+c^2\right)\left(b^2+d^2\right)$$
Now,
$$\begin{align}
a^2 + c^2 &= \left(1 + 2 \cos\theta + \cos^2\theta\right) + \sin^2\theta = 2\left(1 + \cos\theta\right) \\[4pt]
b^2 + d^2 &= \left(1 + 2 \cos 2\theta + \cos^22\theta\right) + \sin^22\theta = 2\left(1 + \cos 2\theta\right)
\end{align}$$
and the result follows. $\square$
Note. The sum reduces even further to
$$16\cos^2\frac{\theta}{2}\cos^2\theta$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove the inequality $b\le \sqrt{a^2 + b^2}$ How do i prove $b\le \sqrt{a^2 + b^2}$ ?
What i think is :
$\begin{align}
a^2& \gt 0,\, b^2 \gt 0 \\
b &= \sqrt{b^2}\\
&\le\sqrt{a^2+b^2} \\
\therefore b &\le \sqrt{a^2 + b^2}
\end{align}$
Am i right? Please do a correction.
|
It is not true that $b =\sqrt {b^{2}}$. You have to say $b \leq |b| =\sqrt {b^{2}} \leq \sqrt {a^{2}+b^{2}}$.
|
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|
Expanding the Polynomial Using Taylor Series? Expand the polynomial
$$f(x) = x^3-2x^2-3x+5$$
In power of $(x-2)$
This might be a simple question but all we have to do to solve this question is to expand using the Taylor series with $x =2 $ , as opposed to the Maclaurin at $x=0$.
$f(x) = f(a) + \frac{f'(a)(x-a)^1}{1!}+\frac{f''(a)(x-a)^2}{2!}+ \cdot \cdot \cdot$
$f(2)= -1+1(x-2)+\frac{8(x-2)^2}{2!}+\frac{6(x-2)^3}{3!}$
$f(2) = -1 +(x-2) +4(x-2)^2+(x-2)^3$
So in the end this would be the solution correct?
|
Just like for the change of basis for numbers, you can use Horner's algorithm:
divide the polynomial by $x-2$, then divide again the quotient by $x-2$, and so on until you obtain a constant quotient. The coefficients of the successive powers of $x-2$ are the successive remainders (including the final quotient).
Illustration:
\begin{array}{rrrrr}
& 1 &-2 & -3 & 5 \\
+&\downarrow&2 & 0 & -6 \\
\hline
\times2 &1 & 0 &-3 & \color{red}{-1} \\
+&\downarrow & 2&4\\
\times 2&1&2&\color{red}{1}\\ \hline
+&\downarrow & 2\\
\times 2&\color{red}1 & \color{red}4
\end{array}
So the expansion of the polynomial in powers of $x-2$ is
$$f(x)=-1+(x-2)+4(x-2)^2+(x-2)^3.$$
|
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|
If distinct integers $a,b,c,d$ are consecutive terms of an arithmetic progression such that . . .
If distinct integers $a,b,c,d$ are consecutive terms of an arithmetic progression such that
$$a^2 + b^2 + c^2=d$$
then $a+b+c+d$ equals?
$$(A)\ 0 \quad (B)\ 1 \quad (C)\ 2 \quad (D)\ 4$$
Here is what I tried: assume the common difference is $x$.
Thus,
$$a^2 + a^2 + 2ax + x^2 +a^2 +4ax +4x^2=a + 3x$$
That is,
$$3a^2 + 6ax + 2x^2=a + 3x$$
First I assumed it as a quadratic in $a$ and solved for it.
$$a=\frac{(1-6x) \pm \sqrt{12x^2-48x +1}}{6}$$ but to no avail.
Then I assumed it as a quadratic in $x$ and got:
$$x=\frac{(3-6a) \pm \sqrt{12a^2 -36a +81}}{6}$$
And I noticed that I'm stuck. Please help. Thanks in advance!
Edit: I made a calculation error as pointed out by multiple answers, and so the equation should have been $$3a^2 + 6ax + 5x^2=a + 3x$$ which leads to different solutions.
|
You made a slight error, this needs to be $5x^2$ and not $2x^2$.
So you have this equation $$3a^2+a(6x-1)+(5x^2-3x)=0$$
This equation has discriminant $\Delta=-24x^2+24x+1$
The way to solve this for Diophantine equations is to introduce an integer $m$ such that $\Delta=m^2$ (because we need the square root to be an integer for the solution $a$ to have a chance to be an integer).
Now we are left to solve $$24x^2-24x+(m^2-1)=0$$
Once again since $x$ is an integer, the discriminant $\delta=672-96m^2=96(7-m^2)$ should also be a perfect square.
In this case notice that $m$ can only take values $\ 0,1,2\ $ since we need $\delta\ge 0$
$96=2^5\times 3$ so we need to multiply by $6$ to have perfect square and the solution is $m=1$.
Finally $x=\dfrac{24\pm\sqrt{96\times 6}}{2\times 24}\in\{0,1\}$
The last step is to verify the $x$ found satisfy the original problem.
*
*$x=0$
Then $a=b=c=d\implies d=3a^2=a\implies a=\frac 13$ which is not an integer. So this is not a valid solution.
*
*$x=1$
With this value we arrive to $$3a^2+5a+2=(a+1)(3a+2)=0$$
And only $a=-1$ is a valid integer.
Conclusion: $a=-1,\ b=0,\ c=1,\ d=2$
We can verify $a^2+b^2+c^2=1+0+1=2=d$ and $a+b+c+d=-1+0+1+2=2$.
|
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|
$\lim_{n\to \infty} \ \frac{1}{n} \Bigl[(a+\frac{1}{n})^2+(a+\frac{2}{n})^2+\cdots+(a+\frac{n-1}{n})^2\Bigr]$ without L'Hopital Find limit:
$$a_n = \lim_{n\to \infty} \ \frac{1}{n} \Bigl[\Bigl(a+\frac{1}{n}\Bigr)^2+\Bigl(a+\frac{2}{n}\Bigr)^2+\cdots+\Bigl(a+\frac{n-1}{n}\Bigr)^2\Bigr]$$
I tried limiting it with $$n\cdot\frac{1}{n}\Bigl(a+\frac{1}{n}\Bigr)^2\leqslant a_n \leqslant n \cdot \frac{1}{n}\Bigl(a+\frac{n-1}{n}\Bigr)^2$$
but that got me to the answer that left side limits to $a^2$ and right side to $(a+1)^2$
and so I didn't squeeze it the right way. Help, and keep it simple.
|
Converting this
to a Riemann sum,
$\begin{array}\\
\frac{1}{n}\sum_{k=1}^n (a+\frac{k}{n})^2
&\to \int_0^1 (a+x)^2dx\\
&= \int_a^{a+1} x^2dx\\
&=\dfrac{x^3}{3}|_a^{a+1}\\
&=\dfrac{a^3+3a^2+3a+1-a^3}{3}\\
&=\dfrac{3a^2+3a+1}{3}\\
&=a^2+a+\frac13\\
\end{array}
$
|
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|
Any suggestions on a closed form of $\sum_{a=1}^{N} \left({\lfloor{\frac{N+1}{a}}\rfloor+\lfloor{\frac{N-1}{a}}\rfloor}\right)$ Now the sum is approximated by $$\sum_{a=1}^{N} \left({\left\lfloor{\frac{N+1}{a}}\right\rfloor + \left\lfloor{\frac{N-1}{a}}\right\rfloor}\right) \approx 2 \sum_{a=1}^{N} \left\lfloor{\frac{N}{a}}\right\rfloor$$.
when $N$ is large. This part of my on going research in polynomial factoring with constraints over the coefficients. I am looking for a possible closed form solution or if I can combine these two sum terms into one term.
|
A first useful point to consider is that
$$
\eqalign{
& \left\lfloor {{{N + 1} \over a}} \right\rfloor + \left\lfloor {{{N - 1} \over a}} \right\rfloor = \cr
& = {{N + 1} \over a} + {{N - 1} \over a} - \left\{ {{{N + 1} \over a}} \right\} - \left\{ {{{N - 1} \over a}} \right\} = \cr
& = {{2N} \over a} - \left( {\left\{ {{{N + 1} \over a}} \right\} + \left\{ {{{N - 1} \over a}} \right\}} \right) \cr}
$$
And a second one is that
$$
\eqalign{
& \left\{ x \right\} + \left\{ y \right\} = \cr
& = \left\{ {x + y} \right\} + \left\lfloor {\left\{ x \right\} + \left\{ y \right\}} \right\rfloor = \cr
& = \left\{ {x + y} \right\} + \left[ {1 \le \left\{ x \right\} + \left\{ y \right\}} \right] \cr}
$$
where $[P]$ denotes the Iverson bracket
From these, you can get a rough estimate, considering that the fractional part is less than one,
and on avg. it can be take to equal $1/2$.
Then it depends on the approximation you need to reach.
A further point is that
$$
\eqalign{
& \left\lfloor {x + y} \right\rfloor = \cr
& = \left\lfloor x \right\rfloor + \left\lfloor y \right\rfloor + \left\lfloor {\left\{ x \right\} + \left\{ y \right\}} \right\rfloor = \cr
& = \left\lfloor x \right\rfloor + \left\lfloor y \right\rfloor + \left[ {1 - \left\{ x \right\} \le \left\{ y \right\}} \right]\quad \Rightarrow \cr
& \Rightarrow \;\quad \left\lfloor {{{N + 1} \over a}} \right\rfloor
= \left\lfloor {{{N - 1} \over a}} \right\rfloor + \left\lfloor {{2 \over a}} \right\rfloor + \left[ {1 - \left\{ {{2 \over a}} \right\} \le \left\{ {{{N - 1} \over a}} \right\}} \right] \cr}
$$
which for $a=1$ and for $a=2$ gives simple results, and for $3 \le a$ becomes
$$
\eqalign{
& \left\lfloor {{{N + 1} \over a}} \right\rfloor \quad \left| {\;3 \le a} \right.\quad = \cr
& = \left\lfloor {{{N - 1} \over a}} \right\rfloor + \left[ {{{a - 2} \over a} \le \left\{ {{{N - 1} \over a}} \right\}} \right] = \cr
& = \left\lfloor {{{N - 1} \over a}} \right\rfloor + \left[ {a - 2 \le \left( {N - 1} \right)\bmod a} \right] = \cr
& = \left\lfloor {{{N - 1} \over a}} \right\rfloor + \left[ {a - 2 = \left( {N - 1} \right)\bmod a} \right] + \left[ {a - 1 = \left( {N - 1} \right)\bmod a} \right] = \cr
& = \left\lfloor {{{N - 1} \over a}} \right\rfloor + \left[ {0 = \left( {N + 1} \right)\bmod a} \right] + \left[ {0 = N\bmod a} \right] = \cr
& = \left\lfloor {{{N - 1} \over a}} \right\rfloor + \left[ {a\backslash N} \right] + \left[ {a\backslash \left( {N + 1} \right)} \right] \cr}
$$
so that the sum of the two terms differ for the number of divisors
of $N$ and $N+1$.
|
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|
Show that $3^{7^n}+5^{7^n}\equiv 1 \pmod{7^{n+1}}$ Show that $3^{7^n}+5^{7^n}\equiv 1 \pmod{7^{n+1}}$
I tried to induct on n:
For $n = 0$ we have $3+5 = 8$ and $8 \equiv 1 \pmod{7^{n+1}}$.
Suppose it is true for $n = k$:
$$3^{7^k}+5^{7^k}\equiv 1 \pmod{7^{k+1}}$$
so $3^{7^k}+5^{7^k}=7^{k+1}*q_1+1$
For $n = k+1$:
$$3^{7^{k+1}}+5^{7^{k+1}}\equiv r \pmod{7^{k+2}}$$
so $3^{7^{k+1}}+5^{7^{k+1}}=7^{k+2}*q_2+r$
Now if we subtract them we get that:
$$3^{7^{k+1}}+5^{7^{k+1}}-3^{7^k}-5^{7^k}=7^{k+1}(7q_2-q1)+r-1$$
From Euler's theorem we know that $a^{\phi(7^{k+1})}\equiv 1 \pmod{7^{k+1}}$
and $\phi(7^{k+1}) = 6\cdot7^{k}$ so $3^{6\cdot7^{k}}\equiv 1 \pmod{7^{k+1}}$ and $5^{6 \cdot 7^{k}}\equiv 1 \pmod{7^{k+1}}$.
So $3^{7^{k+1}}+5^{7^{k+1}}-3^{7^k}-5^{7^k}\equiv3^{7^k}\cdot3^{6\cdot{7^{k}}}+5^{7^k}\cdot5^{6\cdot{7^{k}}}-3^{7^k}-5^{7^k}\equiv 0 \pmod{7^{k+1}}$
Finally I get that $r\equiv1\pmod{7^{k+1}}$.
Here I got stuck. I don't know how to show that $r=1$.
|
Quick solution using Lifting the Exponent Lemma:
Note that $3^{7^n} + 5^{7^n} - 1 \equiv -4^{7^n} - 2^{7^n} - 1 \equiv -(4^{7^n} + 2^{7^n} + 1) \equiv -\dfrac{8^n-1}{2^n-1} \pmod{7^{n+1}}$.
Now, by LTE, we have $\nu_7(8^{7^n} - 1) = \nu_7(8-1) + \nu_7(7^n) = n+1$, and $2^{7^n} \equiv 2^{7^n \pmod{6}} \equiv 2^{1^n} \equiv 2 \pmod{7}$, so $7 \nmid 2^{7^n} - 1$. Thus, we have
$$\frac{8^{7^n}-1}{2^{7^n} - 1} \equiv 0 \pmod{7^{n+1}}$$
so we're done.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find : $\int_0^{b}\frac{\ln x}{x^2-1}dx$ where $b>1$ Evaluate :
$I=\int_0^{b}\frac{\ln x}{x^2-1}dx$ where $b>1$
My attempt :
I use integral by part : $u=\ln x$ $dv=\frac{1}{x^2-1}$
Then : $u'=\frac{1}{x}$ and $v=\frac{\ln \frac{x-1}{x+1}}{2}$
So :
$I=[\frac{\ln x\ln\frac{x-1}{x+1}}{2}]_0^{b}$
$+\int_0^{b}\frac{\ln (1+x)-\ln (x-1)}{2x}dx$
My problem here is limit of integral because $b>1$ ?
And I can't complete the last integral !
|
$
\begin{align}
\int_0^b\dfrac{\ln x}{x^2-1}~dx&=\int_0^1\dfrac{\ln x}{x^2-1}~dx+\int_1^b\dfrac{\ln x}{x^2-1}~dx
\\&=\dfrac{\pi^2}{8}+\int_1^b\frac{\ln x}{x^2\left(1-\dfrac{1}{x^2}\right)}~dx
\\&=\dfrac{\pi^2}{8}+\sum\limits_{n=0}^\infty\int_1^b\frac{\ln x}{x^{2n+2}}~dx
\\&=\dfrac{\pi^2}{8}-\sum\limits_{n=0}^\infty\int_1^b\ln x~d\left(\dfrac{1}{(2n+1)x^{2n+1}}\right)
\\&=\dfrac{\pi^2}{8}-\left[\dfrac{\ln x}{(2n+1)x^{2n+1}}\right]_1^b+\sum\limits_{n=0}^\infty\int_1^b\dfrac{1}{(2n+1)x^{2n+1}}~d(\ln x)
\\&=\dfrac{\pi^2}{8}-\dfrac{\ln b}{(2n+1)b^{2n+1}}+\sum\limits_{n=0}^\infty\int_1^b\dfrac{1}{(2n+1)x^{2n+2}}~dx
\\&=\dfrac{\pi^2}{8}-\dfrac{\ln b}{(2n+1)b^{2n+1}}-\sum\limits_{n=0}^\infty\left[\dfrac{1}{(2n+1)^2x^{2n+1}}\right]_1^b
\\&=\dfrac{\pi^2}{8}-\dfrac{\ln b}{(2n+1)b^{2n+1}}+\sum\limits_{n=0}^\infty\dfrac{1}{(2n+1)^2}-\sum\limits_{n=0}^\infty\dfrac{1}{(2n+1)^2b^{2n+1}}
\\&=\dfrac{\pi^2}{4}-\dfrac{\ln b}{(2n+1)b^{2n+1}}-\sum\limits_{n=0}^\infty\dfrac{1}{(2n+1)^2b^{2n+1}}
\end{align}
$
|
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|
LU Decomposition of $A$ I am currently working with a practice problem in which we are asked to find the $LU$ factorization of some matrix $A$, given below:
$$
A= \begin{pmatrix}
1 & 2 & 3 \\
2 & 4 & 7 \\
3 & 6 & 10 \\
3 & 6 & 10 \\
\end{pmatrix}
$$
In the first step, I added $-2$ of the first row to the second, $-3$ of the first row to the third, and $-3$ of the first row to the forth to give the matrix:
$$
\begin{pmatrix}
1 & 2 & 3 \\
0 & 0 & 1 \\
0 & 0 & 1 \\
0 & 0 & 1 \\
\end{pmatrix}
$$
In the second step, I added $-1$ of the second row to the third row, and $-1$ of the second row to the fourth row, to give the matrix $U$:
$$
U = \begin{pmatrix}
1 & 2 & 3 \\
0 & 0 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{pmatrix}
$$
Given these steps, I determined the matrix $L$ to be:
$$
L= \begin{pmatrix}
1 & 0 & 0 & 0 \\
-2 & 1 & 0 & 0 \\
-3 & -1 & 1 & 0 \\
-3 & -1 & 0 & 1 \\
\end{pmatrix}
$$
However, although the answer $U$ was found to be correct, the answer $L$ was given the following form:
$$
L= \begin{pmatrix}
1 & 0 & 0 & 0 \\
2 & 1 & 0 & 0 \\
3 & 1 & 1 & 0 \\
3 & 1 & 0 & 1 \\
\end{pmatrix}
$$
However, I don't understand why all of the terms that I found to be negative are positive in the given answer. Can anyone help me figure it out? Thanks!
(Note: this is not homework, but independent study.)
|
Using notation from here, you obtained that
$$L_{24}(-1)L_{23}(-1)L_{14}(-3)L_{13}(-3)L_{12}(-2)A = U$$
so
$$A = L_{12}(-2)^{-1}L_{13}(-3)^{-1}L_{14}(-3)^{-1}L_{23}(-1)^{-1}L_{24}(-1)^{-1}U = L_{12}(2)L_{13}(3)L_{14}(3)L_{23}(1)L_{24}(1)U$$
where $L = L_{12}(2)L_{13}(3)L_{14}(3)L_{23}(1)L_{24}(1)$ is lower triangular with unit diagonal.
Applying these transformations in order to the identity matrix $I$ yields the correct $L$.
|
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|
Prove $\sum_\text{cyc}\frac{1}{3-ab}\le\frac{3}{2}$
Let $a,b,c>0$ such that $a^2+b^2+c^2=1$. Prove that $$\frac{1}{3-ab}+\frac{1}{3-bc}+\frac{1}{3-ca}\le\frac{3}{2}$$
By Am/Gm we have: $$ {1\over 3-ab}\leq {2\over 6-a^2-b^2} = {2\over 5+c^2}<{2
\over 5}$$
so
$$\sum_{cyc} {2\over 5+a^2} < {6\over 5} <{3\over 2}$$
So I get much better bound. Where did I go wrong?
Edit: If nothing is wrong then what is a smallest constant $m$ such $$\frac{1}{3-ab}+\frac{1}{3-bc}+\frac{1}{3-ca}\le m$$ For $a=b=c=\sqrt{3}/3$ we get $m\geq {9\over 8}$.
|
Hint.-Because of $a^2+b^2+c^2=1$, each $a,b,c$ must be in the interval $[-1,1]$ then one has for positive values
$$\frac13\le\frac{1}{3-ab}\le\frac12\\\frac13\le\frac{1}{3-ac}\le\frac12\\\frac13\le\frac{1}{3-bc}\le\frac12$$ Thus the sum is less or equal than $\dfrac32$.
If some of $a,b,c$ is negative, the fraction is less than $\dfrac13$ then one has the same majorant $\dfrac32$ for the sum.
|
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|
How to solve $\log_{3}(x) = \log_{\frac{1}{3}}(x) + 8$ I'm trying to solve $\log_{3}(x) = \log_{\frac{1}{3}}(x) + 8$. I am getting x = 4 but the book gets x = 81. What am I doing wrong?
\begin{align*}
\log_{3}(x) & = \log_{\frac{1}{3}}(x) + 8\\
\log_{3}(x) & = \frac{\log_{3}(x)}{\log_{3}(\frac{1}{3})} + 8\\
\log_{3}(x) & = \frac{\log_{3}(x)}{-\log_{3}(3)} + 8\\
\log_{3}(x) & = \frac{\log_{3}(x)}{-1} + 8\\
\log_{3}(x) & = -\log_{3}(x) + 8\\
2\log_{3}(x) & = 8\\
\log_{3}(x) & = 4\\
x & = 4
\end{align*}
What am I doing wrong?
|
When you want to find the value of $~x~$ inside a logarithm function, first you have to make both side in term of logarithm first (convert four into a logarithm term) and then inverse it to find $~x~$ :
$\log_{3}(x) = 4(1)$ $\Leftarrow$ convert $~1~$ into logarithm base $~3~$
$\log_{3}(x) = 4.\log_{3}(3)$ $\Leftarrow$ raise $~4~$ as exponent
$\log_{3}(x) = \log_{3}(3^{4})$ $\Leftarrow$ inverse it using exponent function
$3^{\log_{3}(x)} = 3^{\log_{3}(3^{4})}$
$ x = 3^{4} = 81$
|
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|
Is every prime contained in the set $\,\{n\in\mathbb{N}:n|10^k-1\}\cup\{2,5\}\,$ for $k=1,2,3,...$?
Is every prime contained in the set
$\,\{n\in\mathbb{N}:n|10^k-1\}\cup\{2,5\}\,$ for $k=1,2,3,...$?
Let $p\notin\{2,5\}$ be a prime number. Then $\frac{1}{p}$ has a decimal period of length at most $p-1$. Denote $\frac{1}{p}$ by $$\frac{1}{p}=0.\overline{d_1d_2...d_r}$$ where $1\leq r \lt p$ and $d_i\in\{0,1,2,...,9\}$.
Let $d=0.d_1d_2...d_r\cdot10^r\;$ (e.g., if $p=7$, then $\frac{1}{7}=0.\overline{142857}$, $r=6$, and $d=142857$).
Let's consider $\frac{1}{10^m-1}$ for some $m\in\mathbb{N}$:
$$
\frac{1}{10^m-1}=\frac{1}{\underbrace{99...9}_{\text{$m$ times}}}=0.\overline{\underbrace{00...0}_{m-1 \\ \text{times}}1}=\sum_{i=1}^\infty \frac{1}{10^{i\cdot m}}
$$
So
\begin{align}
\frac{1}{p}&=0.\overline{d_1d_2...d_r}=d\cdot \left(\frac{1}{10^r}\right)+d\cdot \left(\frac{1}{10^{2r}}\right)+d\cdot \left(\frac{1}{10^{3r}}\right)+\cdots \\ &=d\cdot\left(\frac{1}{10^r}+\frac{1}{10^{2r}}+\frac{1}{10^{3r}}+\cdots\right) \\ &=d\cdot\sum_{i=1}^\infty \frac{1}{10^{i\cdot r}} \\ &=\frac{d}{10^r-1}
\end{align}
Rearranging terms, we have $\,10^r-1=p\cdot d \, \Longrightarrow \, p|10^r-1$.
Does this prove that the set of primes is contained in the set $\{n\in\mathbb{N}:n|10^k-1\}\cup\{2,5\}$?
Clearly, $2$ and $5$ are missing from $\{n\in\mathbb{N}:n|10^k-1\}$. Are there any other primes not contained in this set?
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A prime number $p$ divides $10^k-1$ if and only if $10^k \equiv 1 \bmod p$. But if $p \nmid 10$ then $10^{p-1} \equiv 1 \bmod p$ by Fermat's little theorem.
Therefore $p$ divides $10^{p-1}-1$ for all prime numbers $p \not\in \{ 2,5 \}$.
|
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|
Minimizing distance between an ellipse and a point Problem
An ellipse has the formula:
$$\frac{x^2}{9} + \frac{y^2}{4} = 1$$
What is the shortest distance from the ellipse to the point $P = (a,0)?$
Attempted solution:
I have previously solved the same problem with P = (1,0) and P = (2,0). For P = (1,0), the minimum where x = 1.8 and y = 1.6 and a distance of $\frac{4\sqrt{5}}{5}$.
For P = (2,0), the local minimum was larger than the distance between P and the ellipse at y = 0., which was equal to 1. So there is some threshold for $a$ for which the local minimum is not truly the smallest distance.
Basic strategy:
1) Solve the ellipse formula for y and put that into the distance formula for the distance between the ellipse and the point (Pythagorean theorem).
2) Take the derivative.
3) Set the derivative to 0 and solve for x.
4) Use the ellipse formula to solve for y.
5) Put in x and y in the distance formula and get the distance.
However, at some threshold the distance between the point and the ellipse where y = 0 is even shorter. So the shortest distance will have one of two cases, one between 0 and just below the threshold and one for the threshold value and above.
I figured that I could apply the same method I did for the previous cases of P = (1,0) and P = (2,0) and see if I can find some way to find this threshold.
Use the ellipse formula and solve for y:
$$\frac{x^2}{9} + \frac{y^2}{4} = 1 \Rightarrow y = \sqrt{4\Big(1-\frac{x^2}{9}\Big)}$$
Write down distance formula
$$D = \sqrt{(x-a)^2 + y^2}$$
Put in y from the ellipse formula
$$D = \sqrt{(x-a)^2 + \Big(\sqrt{4\Big(1-\frac{x^2}{9}\Big)}\Big)^2}$$
Simplify:
$$D = \sqrt{\frac{5}{9} \cdot x^2 - 2ax +4}$$
Take the derivative:
$$D' = \frac{\frac{10}{9}x-2a}{2\sqrt{\frac{5}{9}x^2 - 2ax +4}}$$
Set the derivative to zero and solve for x:
$$\frac{10}{18}x -a = 0 \Rightarrow x = 1.8a$$
Use ellipse function and solve for y:
$$y = \sqrt{4\Big(1-\frac{(1.8a)^2}{9}\Big)} = a\sqrt{0.76}$$
Putting in x and y to the distance formula
$$D = \sqrt{(1.8a-a)^2 + (a\sqrt{0.76})^2} = \sqrt{0.64a^2 + 0.76a^2} = \sqrt{1.4a^2} = a\sqrt{1.4}$$
...but I feel like I am not getting anywhere. This is not the correct distance and the threshold is nowhere to be seen.
Expected answer is:
$$\sqrt{4-\frac{4}{5}a^2}$$
for $0 < a < \frac{5}{3}$ and
$$|a-3|$$
if $a \geq \frac{5}{3}$
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Let $(x-a)^2+y^2=k^2,$ where $k\geq0.$
Thus, the equation $$(x-a)^2+4\left(1-\frac{x^2}{9}\right)=k^2$$ has real roots.
We have
$$\frac{5}{9}x^2-2ax+a^2+4-k^2=0,$$ which gives
$$a^2-\frac{5}{9}(a^2+4-k^2)\geq0$$
and $$k\geq2\sqrt{1-\frac{1}{5}a^2},$$ as your result.
|
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|
Why do the digits of a number squared follow a similar quotient? I realized that, any number $k$ with $n$ digits, and when $k$ is squared, i.e $k^2$ will have $2n-1$ or $2n$ digits.
Per example, let $k = 583$, thus $n = 3$ and the digits of $583^2$ are $2n$.
But I started thinking, how can I narrow the result to be more precise?
But, I asked myself. How can I determine when I will have $2n-1$ or $2n$ digits?
What I did was the following:
$1)$ The last number of $1$ digit whose square has $2n-1$ digits, is $3$ since $3^2 = 9$, and the next number $4^2 = 16$ have $2n$ digits. And the quotient between the max numbers of $1$ digit($9$) and the last number of $1$ digit to the pow of $2$ with $2n-1$ digits(3) is: $9/3 = 3$
$2)$ The last number of $2$ digits whose square has $2n-1$ digits, is $31$, since $31^2 = 961$. The quotient here is $3.19354839$
$3)$ The last number of $3$ digits whose square has $2n-1$ digits, is $316$ since $316^2 = 99856$. The quotient here is $3.16139241$
$4)$ The last number of $8$ digits whose square has $2n-1$ digits, is $31622776$, since $31622776^2 = 9.9999996\cdot10^{14}$. The quotient here is $3.16227768871$
The quotient is each time smaller and closer to $3.16$
$i)$ Why does the quotient between the largest number with $n$ digits and the last number with $n$ digits squared that have $2n-1$ follow this "pattern" closer and closer to $3.16$?
$i)$ With this I can assure for all numbers that: If i have a number, per example $k = 7558$ and $k$ have $4$ digits and the quotient between $9999/7558 < 3.2$, then $k$ have $2n = 8$ digits?
That more generally, I can assure you this?:
If i have a number $k$ with $n$ digits, this number have
$2n$ digits if $\frac{10^{n}-1}{k} \leq 3.2$ otherwise it will have $2n-1$
digits
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If $k^2$ has $2n$ digits, then it is true that $10^{2n-1} \leq k^2 < 10^{2n}$, so we have $10^{n-1}\sqrt{10} = \sqrt{10^{2n-1}} \leq k < \sqrt{10^{2n}} = 10^n$.
If $k^2$ has $2n-1$ digits, then it is true that $10^{2n-2} \leq k^2 < 10^{2n-1}$, so we have $10^{n-1} = \sqrt{10^{2n-2}} \leq k < \sqrt{10^{2n-1}} = 10^{n-1}\sqrt{10}$.
So the cutoff you've observed is exactly at $\sqrt{10} \approx 3.162277660168379332$, times powers of 10.
|
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|
Find the number of roots of the polynomial $P(z)=z^5+2z^3+3$ in the closed unit disk $\{z\colon |z|\le 1\}$.
Find the number of roots of the polynomial $P(z)=z^5+2z^3+3$ in the closed unit disk $\{z\colon |z|\le 1\}$.
My solution:
Set $f(z)=3$.
For every $0<\delta<1$ and $|z|=\delta$, we have $|P(z)-f(z)|\le \delta^5+2\delta^3<3\delta<3=|f(z)|$. Hence $P(z)$ and $f(z)=3$ has the same number of zeros, which is $0$, inside the open unit disk.
Now consider the behavior of $P(z)$ when $|z|=1$. If we have $P(z)=0$, then $z^5=-2z^3-3$, and $1=|z^5|=|-2z^3-3|$ when $|z|=1$ which forces $-2z^3=2$ and $z^3=-1$.
Clearly, $z=-1$ is a zero of $P(z)$.
For the other two roots of $z^3=-1$, note that: $$ P(z)=z^2\cdot z^3+2z^3+3=z^2-2+3=z^2+1=0\implies z=\pm 1 $$
Therefore we conclude that there is only one root, $z=-1$, of $P(z)$ in the closed unit disk.
However, I am asking for other solutions, such as applying the symmetric version of Rouché''s theorem like in this post. Thank you.
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$f(z)=z^5+2z^3+3 \Rightarrow f'(z)=5z^4+6z^2>0 \Rightarrow f(z)$ is an increasing function. So $f(z)=0$ can have at most oone real root and it is $z=-1.$
Take $$-3=z^5+2z^3 \Rightarrow 3=|z^5+2z^3|\le |z|^5+2|z|^3 \le 3.$$
This proves that at least one root of this equation is s.t. |z|=1.
Other roots cannot not satisfy $|z|<1$, so they will lie outside the unit disk i.e.,$z>1$
In fact, one can numerically check that apart from -1, this equation has two pairs non-real complex conjugate roots roots such that $|z|>1.$
|
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|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.