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Solve trigonometric equation $\sin14x - \sin12x + 8\sin x - \cos13x= 4$ I am trying to solve the trigonometric equation
$$ \sin14x - \sin12x + 8\sin x - \cos13x= 4 $$
The exact task is to find the number of real solutions for this equation on the range $[0, 2\pi]$. Thanks.
|
Define:
$$\sin(14x) - \sin(12x) + 8\sin x - \cos(13x)- 4=:F(\cos x,\sin x)$$
Then set $t=\tan(x/2)$, we have
$$\cos x=\frac{1-t^2}{1+t^2}, \sin x=\frac{2t}{1+t^2}$$
$$G(t)=F\left(\frac{1-t^2}{1+t^2},\frac{2t}{1+t^2}\right)$$
$$=-\frac{t^2-4t+1}{(1+t^2)^{14}}H(t)$$
$$H(t)=5 - 273t^2 + 15262t^4 - 229086t^6 +
1565135t^8 - 5306587t^{10} +
9664564t^{12} - 9650836t^{14} +
5316883t^{16} - 1559415t^{18} +
231374t^{20} - 14638t^{22} + 377t^{24} +
3t^{26}$$
Numerical results showed that all the roots of $H(t)$ are non real. So the real roots of $G(t)$ are those from
$$t^2-4t+1=0$$
Therefore we have
$$t_1=2-\sqrt{3},t_2=2+\sqrt{3}$$
$$x_1=\frac{\pi}{6},x_2=\frac{5\pi}{6}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that diophantine equation has only two solutions. I am looking at the following exercise:
$$\text{Prove that the diophantine equation } x^4-2y^2=1 \text{ has only two solutions.}$$
That's what I thought:
We could set $x^2=k$,then we would have $k^2-2y^2=1 \Rightarrow 2y^2=k^2-1 \Rightarrow 2 \mid k^2-1$
Can I use this fact or do I have to do it in an other way?
EDIT: Or could we do it maybe like that:
$k^2-2y^2=1 \Rightarrow (k-\sqrt{2}y)(k+\sqrt{2}y)=1 \Rightarrow k-\sqrt{2}y= \pm1 \text{ and } k+\sqrt{2}y=\pm 1 \Rightarrow k-\sqrt{2}y=1 \text{ and } k+\sqrt{2}y=1 \text{ OR } k-\sqrt{2}y=-1 \text{ and } k+\sqrt{2}y=-1$
The second case is rejected,because then we get $k<0$
$$k-\sqrt{2}y=1 \text{ and } k+\sqrt{2}y=1 \Rightarrow k=1 \text{ and } y=0 \text{ OR } k=0 \text{ and } y=\frac{1}{\sqrt{2}} \notin \mathbb{Z}$$
So, $k=1 \Rightarrow x^2=1 \Rightarrow x=\pm 1 \text{ and } y=0$
So,the only two solutions are $(1,0) \text{ and } (-1,0)$
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$$2y^2=x^4-1=(x^2+1)(x^2-1)=(x^2+1)(x+1)(x-1)$$
Suppose that $2y^2$ has an odd prime factor $p$. Then $p^2$ divides $y^2$, and hence, $p^2$ divides $(x^2+1)(x^2-1)$. Since $p$ can not divide both factors, then $p^2$ divides $x^2+1$ or $p^2$ divides $x^2-1=(x+1)(x-1)$. Thas is, $p^2$ divides $x^2+1$, $x+1$ or $x-1$.
This proves that each factor $x^2+1$, $x+1$ and $x-1$ is a perfect square or the double of a perfect square. And there are $1$ or three of them that belong to the second group. The factor $x^2+1$ is a perfect square only if $x=0$, which gives no solution for $y$.
If there is only one double of a perfect square, then we have that $x+1=u^2$, $x-1=v^2$. But then $u^2-v^2=(u+v)(u-v)=2$, and this is impossible since $u+v$ and $u-v$ have the same parity.
If there are three doubles of squares, then $x+1=2u^2$, $x-1=2v^2$ and $2(u^2-v^2)=2$, that is, $u=\pm1$, $v=0$. This gives $x=\pm1$ and $y=0$, and there are no more solutions.
|
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Visualise $3x^2 - 2xy - 10x +3y^2 -2y + 8=0$ after the $x,y$ term has been eliminated (using rotation) This is a continuation from my previous question. I thought it would be better to start a new one since the old one was answered correctly.
The equation in question is: $3x^2 - 2xy - 10x +3y^2 -2y + 8=0$:
We introduce the new coordinates for $x,y$:
$$x = \cos \theta X - \sin \theta Y$$
$$y = \cos \theta Y + \sin \theta X $$
We find that $\theta = \frac{\pi}{4}$ eliminates the $xy$ term. This gives us $\cos \theta = \frac{1}{\sqrt{2}}$, $\sin \theta = \frac{1}{\sqrt{2}}$.
Now we want to find out what kind of curve $3x^2 - 2xy - 10x +3y^2 -2y + 8=0$ is.
I.E we want to complete the square and find out what kind of properties it has. What is the best method to continue from here?
One attempt is to use the general development of the $Ax^2 + Bxy +Cy^2$ with our new coordinates for $x$ and $y$. Looking at our equation above we have $A=3, B= -2, C=3$.
This gives us:
$$X^2(A\cos^2 \theta + B\cos \theta \sin \theta + Csin^2 \theta)+$$
$$XY(-2A\cos \theta \sin \theta + B(\cos^2 \theta - \sin^2 \theta) + 2C \cos \theta \sin \theta) +$$
$$Y^2(A\sin^2 \theta - B\sin \theta \cos \theta + Ccos^2 \theta)$$
The coefficients for the $XY$ term will be $0$ so we can ignore that one. With the values for $A,B,C$ and $\sin \theta =\cos \theta = \frac{1}{\sqrt{2}}$ we get:
$$X^2(3\frac{1}{2} + (-2)\frac{1}{2} + 3\frac{1}{2})+ Y^2(3\frac{1}{2} - (-2)\frac{1}{2} + 3\frac{1}{2}) = 2X^2 + 4Y^2$$
The original equation $3x^2 - 2xy - 10x +3y^2 -2y + 8=0$ can now be written as: $$2X^2 + 4Y^2 -10(\cos \theta X - \sin \theta Y) -2(\cos \theta Y + \sin \theta X) + 8 = 0 \rightarrow$$
$$2X^2 + 4Y^2 -\frac{10}{\sqrt{2}}X + \frac{10}{\sqrt{2}}Y -\frac{2}{\sqrt{2}}Y - \frac{2}{\sqrt{2}} X +8 = 0 \rightarrow$$
$$2X^2 + 4Y^2 -\frac{12}{\sqrt{2}}X + \frac{8}{\sqrt{2}}Y +8 = 0$$
And if I try to complete the square I get:
$$2(x- \frac{3}{\sqrt{2}})^2+(y+2\sqrt{2})^2 = 9$$
The final result in my textbook is:
$$2(X-1)^2 + (Y-3)^2 = 3$$ where $X = x-y$ and $Y = x+y$
But I am stuck and cannot understand how to get to the result in my textbook. I would love some help (if you have the time to do the actual final calculations, that would be much appreciated!)
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These are the steps. Show us your workings for as far as you've gotten:
*
*Plug $x=(X-Y)/\sqrt{2}$ and $y=(X+Y)/\sqrt{2}$ into the equation.
*Rearrange into the form $Ax^2+Bx^2+Cx+Dy+E=1$.
*Complete the square on the LHS to put it in the form of the equation of an ellipse.
|
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To prove $a^2+ab+b^2 \Big|(a+b)^{2n}+a^{2n}+b^{2n}$ whenever $3$ does not divide $n$ If $3$ does not divide a positive integer $n$ , then how to prove that $a^2+ab+b^2 \Big|(a+b)^{2n}+a^{2n}+b^{2n}$ ?
|
I will stay in the reals. Note that:
$$(a+b)^2\equiv ab \mod{a^2+ab+b^2}\implies (a+b)^{2n}\equiv (ab)^n \mod{a^2+ab+b^2}$$
So $(a+b)^n+a^{2n}+b^{2n}\equiv a^nb^n+a^{2n}+b^{2n}$. Also, we know that $a^3\equiv b^3$. So setting $n=3q+1$, we have:
$$a^nb^n+a^{2n}+b^{2n}\equiv ab^{6q+1}+a^2b^{6q}+b^{6q+2} \equiv b^{6q}(ab+a^2+b^2)\equiv 0$$
Similarly, if $n=3q+2$:
$$a^nb^n+a^{2n}+b^{2n}\equiv a^2b^{6q+2}+ab^{6q+3}+b^{6q+4} \equiv b^{6q+2}(a^2+ab+b^2)\equiv 0$$
|
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Doubt on solution of PDE To Solve: $\displaystyle (x^2-y^2-z^2)\frac{\partial z}{\partial x}+2xy\frac{\partial z}{\partial y}=2xz$
Subsidiary equation: $\displaystyle \frac{dx}{x^2-y^2-z^2}=\frac{dy}{2xy}=\frac{dz}{2xz}$
Using multipliers x,y and z, we have each fraction=$\displaystyle \frac{xdx+ydy+zdz}{x(x^2+y^2+z^2)}$
This is fine.
Now I did not understand how we arrived at the next step, which is given as: ...
Therefore, $\displaystyle \frac{2xdx+2ydy+2zdz}{x^2+y^2+z^2}=\frac{dz}{z}$
Please assist.
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Starting with
$$ \frac{dx}{x^2-y^2-z^2}=\frac{dy}{2xy}=\frac{dz}{2xz}$$
We obtain:
$$2xdx=(x^2-y^2-z^2)\frac{dz}{z}......(1)$$
$$2ydy=2y^2\frac{dz}{z}......(2)$$
$$2zdz=2z^2\frac{dz}{z}......(3)$$
Adding (1),(2), and (3), we obtain:
$$\frac{2xdx+2ydy+2zdz}{x^2+y^2+z^2}=\frac{dz}{z}$$
|
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Find the last two digits of $9^{{9}^{9}}$ I have to find the last two decimal digits of the number $9^{{9}^{9}}$.
That's what I did:
$$m=100 , \phi(m)=40, a=9$$
$$(100,9)=1, \text{ so from Euler's theorem,we have :} 9^{40} \equiv 1 \pmod{100}$$
$$9^{{9}^{9}} \equiv 9^{9^{2+2+2+2+1}} \equiv 9^{9^2 \cdot 9^2 \cdot 9^2 \cdot 9^2 \cdot 9} \equiv 9^{81 \cdot 81 \cdot 81 \cdot 81 \cdot 9} \equiv 9^{(40+40+1) \cdot (40+40+1) \cdot (40+40+1) \cdot (40+40+1) \cdot 9} \equiv (9^{(40+40+1)})^{(40+40+1) \cdot (40+40+1) \cdot (40+40+1) \cdot 9} \equiv 9^9 \equiv 9^4 \cdot 9^4 \cdot 9 \equiv 6561 \cdot 6561 \cdot 9 \equiv 3721 \cdot 9 \\ \equiv 21 \cdot 9 \equiv 89 \pmod{100}$$
So,the last two digits are $8 \text{ and } 9$.
$$$$But,is there also an other way two calculate the last two digits of $9^{{9}^{9}}$ or is the above the only one?
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Key remark: By the binomial theorem, for every odd $k$, $(10-1)^k=n+10\cdot k-1$ for some integer $n$ which is a multiple of $100$. Since $10\cdot k=10\cdot\ell(k)\bmod{100}$ where $\ell(k)$ denotes the last digit of $k$, this shows that, for every odd $k$, $9^k=10\cdot \ell(k)-1\bmod{100}$.
First application: $\ell(9)=9$, hence the key remark above yields $9^9=10\cdot \ell(9)-1=89\bmod{100}$.
Second application: our first application implies that $\ell(9^9)=9$ hence, using the key remark once again but this time for $k=9^9$, one gets $9^k=10\cdot\ell(k)-1=89\bmod{100}$.
And so on: for every tower of nines, $9^{9^{9^{9^{\cdots}}}}=89\bmod{100}$.
|
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Proving $x+\sin x-2\ln{(1+x)}\geqslant0$
Question: Let $x>-1$, show that
$$x+\sin x-2\ln{(1+x)}\geqslant 0.$$
This is true. See http://www.wolframalpha.com/input/?i=x%2Bsinx-2ln%281%2Bx%29
My try: For
$$f(x)=x+\sin x-2\ln{(1+x)},\\
f'(x)=1+\cos{x}-\dfrac{2}{1+x}=\dfrac{x-1}{1+x}+\cos{x}=0\Longrightarrow\cos{x}=\dfrac{1-x}{1+x}.$$
So
$$\sin x=\pm\sqrt{1-{\cos^2{x}}}=\pm \dfrac{2\sqrt{x}}{1+x}$$
If $\sin x=+\dfrac{2\sqrt{x}}{1+x}$, I can prove it. But if $\sin x=-\dfrac{2\sqrt{x}}{1+x}$, I cannot. See also http://www.wolframalpha.com/input/?i=%28x-1%29%2F%28x%2B1%29%2Bcosx
This inequality seems nice, but it is not easy to prove.
Thank you.
|
Alternative proof:
Let $f(x) = x + \sin x - 2\ln (1+x)$.
We split into three cases:
*
*$x\in (-1, 0]$:
We have $f'(x) = 1 + \cos x - \frac{2}{1+x} \le 1 + 1 - \frac{2}{1+x} = \frac{2x}{1+x} \le 0$.
As $f(0) = 0$, we have $f(x) \ge 0$.
*$x\in [0, \frac{3}{2}]$:
Since $\cos x \ge 1 - \frac{x^2}{2}$ for $x\in \mathbb{R}$,
we have $$f'(x) = 1 + \cos x - \frac{2}{1+x} \ge 1 + 1 - \frac{x^2}{2} - \frac{2}{1+x}
= \frac{x(4-x-x^2)}{2+2x} \ge 0$$
where we have used $4-x-x^2 \ge 4 - \frac{3}{2} - (\frac{3}{2})^2 > 0$.
As $f(0) = 0$, we have $f(x) \ge 0$.
*$x\in [\frac{3}{2}, \infty)$:
We have the following results. The proofs are given at the end.
Fact 1: It holds that $-\frac{3}{5}(x-4) + 2\ln 5 - 4 \ge 2\ln (1+x) - x$.
Fact 2: It holds that $\sin x \ge -\frac{3}{5}(x-4) + 2\ln 5 - 4$.
We are done.
$\phantom{2}$
Remarks: In the following proofs, we need to prove that $g(3/2) = \sin \tfrac{3}{2} + \tfrac{5}{2} - 2\ln 5 \ge 0$,
$g(\pi) = \frac{3}{5}\pi + \frac{8}{5} - 2\ln 5 \ge 0$ and
$g(\pi + \arccos \frac{3}{5}) = \tfrac{4}{5} + \tfrac{3}{5}\pi + \tfrac{3}{5}\arccos \tfrac{3}{5} - 2\ln 5 \ge 0$. One may use a calculator.
If one wants to prove it by hand, the proof is easy but annoying.
Proof of Fact 1: Denote $(\mathrm{LHS}-\mathrm{RHS})$ by $h(x)$. We have $h'(x) = \frac{2(x-4)}{5 + 5x}$.
Thus, $h(x)$ is non-increasing on $[\frac{3}{2}, 4]$, and non-decreasing on $[4, \infty)$.
As $h(4) = 0$, we have $h(x) \ge 0$. We are done.
Proof of Fact 2: Denote $(\mathrm{LHS}-\mathrm{RHS})$ by $g(x)$.
We have $g'(x) = \cos x + \frac{3}{5}$ and $g''(x) = -\sin x$. There are three possible cases:
i) $g(x)$ is concave on $[\frac{3}{2}, \pi]$.
Thus, we have $g(x) = g(\tfrac{\pi-x}{\pi - 3/2} \cdot \frac{3}{2} + \tfrac{x-3/2}{\pi - 3/2}\cdot \pi)
\ge \tfrac{\pi-x}{\pi - 3/2} g(3/2) + \tfrac{x-3/2}{\pi - 3/2}g(\pi) \ge 0$ on $[\frac{3}{2}, \pi]$
since $g(3/2)\ge 0$ and $g(\pi)\ge 0$.
ii) $g(x)$ is convex on $[\pi, 2\pi]$. Also, on $[\pi, 2\pi]$, $g'(x) = 0$ has a unique solution $x = \pi + \arccos\frac{3}{5}$.
Thus, $g(x) \ge g(\pi + \arccos \frac{3}{5}) \ge 0$ on $[\pi, 2\pi]$.
iii) If $x \ge 2\pi$, we have $g(x) \ge -1 + \frac{3}{5}(2\pi-4) - 2\ln 5 + 4
\ge -1 + \frac{3}{5}(2\pi-4) - 2\ln (\mathrm{e}^2) + 4 \ge 0$.
We are done.
|
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Find all integer solution Find all integer solutions such that $$a+1|2a^2+9$$
Solution. I could solve this by writing $$\frac{2a^2+9}{a+1}=2a-2+\frac{11}{a+1}.$$ So, the only integer solution for the last equation are $a=10, a=-12.$
But, i want to get a solution using divisibility properties.
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By elementary means, since $a+1|2a^2+2a$, it divides $2a^2+9-(2a^2+2a)=9-2a$.
Since $a+1|2a+2$, we have $a+1|(9-2a)+(2a+2)$
Hence $a+1|11$
Hence $a+1=±1$ or $a+1=±11$
Hence $a=-12,-2,0,10$.
|
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Transition probabilities in a finite state machine Assume I have a finite state machine and a bunch of tokens. Transitions happen every time a token is inserted. Transitions are based on the token (i.e. at state S, inserting a blue token would give a different result from, say, inserting a red token). There is one state designated as the 'goal'. How would I go about calculating the probability of reaching the goal for randomly throwing tokens to the machine?
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Let $(X_n)_{n\ge 0}$ be our Markov chain, let $T$ be the set of all tokens and let token $a$ have transition matrix $P_a$. Then $\mathbb{P}(X_1=j|X_0=i,\text{next token is t})=(P_t)_{ij}$.
By the law of total probability
$$\mathbb{P}(X_1=j|X_0=i)=\sum_{t\in T} (P_t)_{ij}\mathbb{P}(\text{next token is t}|X_0=i)$$
I hope this is what you are after. Then you just use this probability and the usual calculations of hitting probabilities.
Example
Let's say there are 4 states and 2 tokens (red, blue). The transition matrixes for the two tokens are
$$P_r=\begin{pmatrix}
\frac{1}{2} & \frac{1}{4} & 0 & \frac{1}{4}\\
\frac{1}{2} & 0 & 0 & \frac{1}{2}\\
0 & 0 & 1 & 0\\
0&0&0&1
\end{pmatrix},P_b=\begin{pmatrix}
\frac{1}{3} & \frac{2}{3} & 0 & 0\\
\frac{1}{3} & 0 & \frac{1}{3} & \frac{1}{3}\\
0 & 0 & 1 &0\\
0&0&0&1
\end{pmatrix}$$
where the third and fourth states are absorbing and represent goal and failure respectively.
Let's say the probability of receiving a red token is $\frac{1}{2}$ if the current state is $1$ and $\frac{2}{3}$ if the current state is $2$.
Then $$P=\begin{pmatrix}
\frac{1}{2}\cdot \frac{1}{2}+\frac{1}{2}\cdot \frac{1}{3} & \frac{1}{2}\cdot \frac{1}{4}+\frac{1}{2}\cdot \frac{2}{3} & 0&\frac{1}{2}\cdot \frac{1}{4}\\
\frac{2}{3}\cdot \frac{1}{2}+\frac{1}{3}\cdot \frac{1}{3} & 0 & \frac{1}{3}\cdot \frac{1}{3} & \frac{2}{3}\cdot \frac{1}{2}+\frac{1}{3}\cdot \frac{1}{3}\\
0 & 0 & 1&0\\
0&0&0&1
\end{pmatrix}=\begin{pmatrix}
\frac{5}{12} & \frac{11}{24} & 0&\frac{1}{8}\\
\frac{4}{9} & 0 & \frac{1}{9}& \frac{4}{9}\\
0 & 0 & 1&0\\
0&0&0&1
\end{pmatrix}$$
Now, let $h_i$ be the probability of hitting the goal (state $3$) given that the current state is $i$.
Then $$h_3=1, h_4=0$$ $$h_1=\frac{5}{12} h_1+\frac{11}{24} h_2$$$$h_2=\frac{4}{9} h_1+\frac{1}{9}$$
And the solution to those equations is $h_1=\frac{11}{82},h_2=\frac{7}{41}$.
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Evaluation of $ \int \tan x\cdot \sqrt{1+\sin x}dx$ Calculation of $\displaystyle \int \tan x\cdot \sqrt{1+\sin x}dx$
$\bf{My\; Try::}$ Let $\displaystyle (1+\sin x)= t^2\;,$ Then $\displaystyle \cos xdx = 2tdt\Rightarrow dx = \frac{2t}{\sqrt{2-t^2}}dt$
So Integral is $\displaystyle = \displaystyle 2\int \frac{t^2}{\sqrt{2-t^2}} \frac{(t^2-1)}{\sqrt{2-t^2}}dt = 2\int\frac{t^4-t^2}{2-t^2}dt $
Now How Can I solve after that
Help me
Thanks
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$$ 2\int\frac{t^4-t^2}{2-t^2}dt = 2\int\frac{\require{cancel}\cancel{(t^2 - 2)}(t^2 + 1)+2}{-(\cancel{t^2 - 2})},dt = \underbrace{-2\int (t^2 + 1)\,dt}_{\text{a cinch}} - \underbrace{2\int \frac {2}{t^2 - 2}\,dt}_{\text{partial fractions}}$$
Alternatively, for the second integral, we can express it as $$+ 2\cdot\frac{2}{1-t^2}$$ and use $t = \sin \theta \,d\theta \implies dt = \cos \theta \,d\theta $ to get $$4\int \sec \theta d\,\theta$$
|
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|
Simplifying polynomial fraction Working through an old book I got and am at this problem: Simplify:
$$\frac{3x^2 + 3x -6}{2x^2 + 6x + 4}.$$
The answer is supposed to be $\frac{3(x - 1)}{2(x - 1)}$. I thought I had all this polynomial stuff figured out well enough, but I'm having trouble seeing how they got to that answer. :/
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Simply factorise
$\dfrac{3x^2 + 3x -6}{2x^2 + 6x + 4} = \dfrac{3(x^2 + x -2)}{2(x^2 + 3x + 2)} = \dfrac{3(x-1)(x+2)}{2(x+1)(x+2)}$
Then cancel common terms.
|
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A closed formulae for the coefficient of $x^k$ in $(x-1)^a(x+1)^b$ Let a,b positive integer
Do you know any closed formulae for the coefficient of $x^k$
in $(x-1)^a(x+1)^b=\sum_{k=0}^{a+b}u(k;a,b)x^k$ ?
I look for an a closed expression of $u(k;a,b)$ involving maybe integral , special function
not the symbol $\sum$
thank's for your help
|
The expansion follows as:
\begin{align}
(x-1)^{a} (x+1)^{b} &= (-1)^{a} (1-x)^{a} (1+x)^{b} \\
&= (-1)^{a} \sum_{r=0}^{a} \binom{a}{r} (-1)^{r} x^{r} \cdot \sum_{s=0}^{b} \binom{b}{s} x^{s} \\
&= (-1)^{a} \sum_{r=0} \sum_{s=0} (-1)^{r} \binom{a}{r} \binom{b}{s} x^{r+s} \\
&= (-1)^{a} \sum_{r=0}^{a|b} (-1)^{r} \left( \sum_{s=0}^{r} \binom{a}{r-s} \binom{b}{s} (-1)^{s} \right) x^{r}.
\end{align}
Now
\begin{align}
\sum_{s=0}^{r} \binom{a}{r-s} \binom{b}{s} (-1)^{s} &= \sum_{s=0}^{r} \frac{a!}{(r-s)!(a-r+s)!} \frac{b!}{s! (b-s)!} (-1)^{s} \\
&= \binom{a}{r} \sum_{s=0}^{r} \frac{1}{(r+1)_{-s} (a+1-r)_{s}} \frac{(-1)^{s}}{s! (b+1)_{-s}} \\
&= \binom{a}{r} \sum_{s=0}^{r} \frac{(-r)_{s} (-b)_{s} (-1)^{s}}{s! (a+1-r)_{s}} \\
&= \binom{a}{r} {}_{2}F_{1}(-r, -b; a-r+1; -1).
\end{align}
From this the expansion becomes
\begin{align}
(x-1)^{a} (x+1)^{b} &= \sum_{r=0}^{a|b} \left( (-1)^{r+a} \binom{a}{r} {}_{2}F_{1}(-r, -b; a-r+1; -1) \right) x^{r}.
\end{align}
The use of $a|b$ is to mean min(a,b), ie the minimum of $a$ or $b$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Compute the determinant $4\times 4$ Compute the determinant:
$$
A= \begin{vmatrix}
1 & 1 & a+1 & b+1 \\
1 & 0 & a & b \\
2 & b & a & b \\
2 & a & a & b \\
\end{vmatrix}
$$
I got (in the end): $\det A = (-a-b)(b+a) = -ab -a^2 -b^2 -ba $.
|
You can preform some row operations, they preserve the determinant (but no multiplying a row and $-1$ for every transopsition). This gives me
$$-\begin{vmatrix}
1&0&a&b\\
0&1&1&1\\
0&0&-(a+b)&-2b\\
0&0&-2a&-(a+b)\\
\end{vmatrix}$$
So determinant is $-[(a+b)^2-4ab]$
|
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|
Calculation of Trigonometric Limit with Summation. If $\displaystyle f(x)=\lim_{n\rightarrow \infty}\sum_{r=0}^{n}\left\{\frac{\tan \left(\frac{x}{2^{r+1}}\right)+\tan^3\left(\frac{x}{2^{r+1}}\right)}{1-\tan^2\left(\frac{x}{2^{r+1}}\right)}\right\}.$ Then value of $f(x)$ is
$\bf{My\; Try::}$ Let $\displaystyle \left(\frac{x}{2^{r+1}}\right)=y$. So Trigonometric Expression is
$\displaystyle \frac{\tan y\cdot \left(1+\tan^2 y\right)}{\left(1-\tan^2 y\right)}=\frac{\tan y}{\cos 2y}=\frac{\sin y}{\cos 2y \cdot \cos y} = \frac{1}{2}\left\{\frac{\sin (2y-y)}{\cos 2y \cdot \cos y}\right\} = \frac{1}{2}\left\{\tan (2y)-\tan (y)\right\}$
Now How Can I solve after that
Help me
Thanks
|
You have the right start, but I prefer if you write $y_r = \frac{x}{2^{r+1}}$ so that $y_0 = \frac{x}{2}$ and you are trying to find
$$
\frac{1}{2} \sum_{r=0}^{\infty} \left( \tan(2y_r) - \tan(y_r)\right)
$$
Since $y_r = 2y_{r+1}$, this becomes
$$
\frac{1}{2} \sum_{r=0}^{\infty} \left(\tan(2y_r) - \tan(2y_{r+1})\right)= \frac{1}{2}
\left[ \left( \tan(2y_0) - \tan(2y_1) \right) + \left( \tan(2y_1) - \tan(2y_2) \right) + \left( \tan(2y_2) - \tan(2y_3) \right) + \ldots \right]
$$
From this it is obvious that all terms except the first cancel, and since the limit for large $r$ of $\tan (2y_r)$ is zero, the sum is just the first term.
So the answer is
$$
\frac{1}{2} \tan (2y_0) = \frac{1}{2} \tan(x)
$$
|
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|
How to solve this linear equation? which has an x on each side I have made this equation.
$5x + 8 = 10x + \dfrac{3}{6}$
And I have achieved this result:
$x = 9$
Is my result correct?
I have already posted two other questions related to this topic, I'm a programmer and am learning Math out of my interest, this is not homework.
My steps taken:
10x - 5x = 48 - 3
5x = 45
5x/5 = 45/5
x = 9
This is my steps which led to a wrong answer.
|
$$5x+8 = 10x + \frac{3}{6} \overset{(1)}{\iff} 8-\frac{3}{6}=10x-5x \overset{(2)}{\iff} \frac{45}{6} = 5x \overset{(3)}\iff x=\frac{9}{6}\overset{(4)}{=} \frac{3}{2} $$
Steps:
(1) subtract $5x+\frac{3}{6}$ on both sides of the equation
(2) Simplify, note that $8-\frac{3}{6} = \frac{48}{6}-\frac{3}{6} = \frac{45}{6}$
(3) Divide by $5$ (or, equivalently, multiply by $\frac{1}{5}$), note that $\frac{45}{6}\cdot\frac{1}{5} = \frac{9\cdot 5}{6}\cdot\frac{1}{5} = \frac{9}{6}$
(4) Simplify, note that $\frac{9}{6}=\frac{3\cdot 3}{2\cdot 3} = \frac{3}{2}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Product of factorials divided by factorial to produce perfect square Let $ S = 1! ~2!~\dotsm ~100! $. Prove that there exists a unique positive integer $k$ such that $S/k!$ is a perfect square.
I thought this was a cute, fun problem and I did solve it, but any alternative methods that you guys would use?
|
We can rewrite $S$ as
\begin{align*}
S &= 1^{100} \cdot 2^{99} \cdot 3^{98} \cdot 4^{97} \dotsm 99^2 \cdot 100 \\
&= 1^{100} \cdot 2 \cdot 2^{98} \cdot 3^{98} \cdot 4 \cdot 4^{96} \dotsm 99^2 \cdot 100 \\
&= 1^{100} \cdot 2^{98} \cdot 3^{98} \cdot 4^{96} \dotsm 99^2 \times 2 \cdot 4 \dotsm 100.
\end{align*}Clearly, the expression to the left of the times sign is a perfect square. Let $X$ denote the expression to the right of the times sign. We have
[X = 2 \cdot 4 \dotsm 100 = \left( 2 \cdot 1 \right) \left( 2 \cdot 2 \right) \dotsm \left( 2 \cdot 50 \right) = 2^{50} \cdot 50!.]Thus, $S/50!$ is a perfect square. Now, we will prove that $50$ is the only such positive integer.
Assume for the sake of contradiction that $S/a!$ is a perfect square for some positive integer $a \neq 50$. Then we have
$$$S = m^2 \cdot a! = n^2 \cdot 50! \quad \quad (*)$$for some nonnegative integers $m$ and $n$. If $a \ge 101$, then $101 \mid a!$ and $101 \mid S.$ But from the definition of $S,$ clearly $101 \not{\mid} S$ so we have a contradiction. Thus, $a \le 100$. We now have two cases:
Case 1: $0 < a < 50$. From $(*)$, we have
$$\left( \frac{m}{n} \right)^2 = (a + 1)(a + 2) \dotsm 50,$$so $(a + 1)(a + 2) \dotsm 50$ is a perfect square. If $a + 1 \le 47$, then $(a + 1)(a + 2) \dotsm 50$ will contain only one factor of $47$ (a prime) and will never be a perfect square. We can now easily verify that $48 \cdot 49 \cdot 50 = 2^5 \cdot 3 \cdot 5^2 \cdot 7^2$, $49 \cdot 50 = 2 \cdot 5^2 \cdot 7^2$, and $50 = 2 \cdot 5^2$ are not perfect squares.
Case 2: $50 < a \le 100$. Similar to the first case, we have
$$\left( \frac{n}{m} \right)^2 = 51 \cdot 52 \dotsm a,$$so $51 \cdot 52 \dotsm a$ is a perfect square. If $a \ge 53$, then $51 \cdot 52 \dotsm a$ will contain only one factor of $53$ (a prime) and will never be a perfect square. We can once again easily check that $51 \cdot 52 = 2^2 \cdot 3 \cdot 13 \cdot 17$ and $51 = 3 \cdot 17$ are not perfect squares.
Thus, $50$ is the only positive integer $k$ such that $S/k!$ is a perfect square.
|
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|
What is the sum of this series: $\displaystyle\sum_{n=0}^{\infty} \dfrac{a^n}{(3n)!}$ I tried getting it into a closed form but failed. Could someone help me out?
$$\sum_{n=0}^{\infty} \dfrac{a^n}{(3n)!}$$
|
The function $f$ defined by $f(x)=\sum_{n=0}^\infty\frac{x^{3n}}{(3n)!}$ satisfies the differential equation $f^{(3)}(x)=f(x)$. So we can just set out to solve this differential equation with its initial conditions that $f(0)=1$, and $f'(0)=f''(0)=0$.
This homogeneous equation has characteristic polynomial $X^3-1$, with roots $1$, $\frac{-1+\sqrt{3}i}{2}$, and $\frac{-1-\sqrt{3}i}{2}$, and it follows that
$$f(x)=Ae^x+Be^{\frac{-1+\sqrt{3}i}{2}x}+Ce^{\frac{-1-\sqrt{3}i}{2}x}$$
Use initial conditions and linear algebra to solve for $A$, $B$, and $C$:
\begin{align}
f(0)&=1 & A+B+C&=1\\
f'(0)&=0& A+\frac{-1+\sqrt{3}i}{2}B+\frac{-1-\sqrt{3}i}{2}C&=0\\
f''(0)&=0& A+\frac{-1-\sqrt{3}i}{2}B+\frac{-1+\sqrt{3}i}{2}C&=0\\
\end{align}
Subtracting the last two equations reveals that $B=C$, and then the first and second equations reduce to:
\begin{align}
A+2B&=1\\
A-B&=0
\end{align}
And now it is clear that $A=B=C=\frac13$. Then manipulate the last two terms to express them using cosine. So
\begin{align}
f(x)&=\frac13e^x+\frac13e^{\frac{-1+\sqrt{3}i}{2}x}+\frac13e^{\frac{-1-\sqrt{3}i}{2}x}\\
&=\frac13e^x+\frac13e^{-x/2}\left(e^{\frac{\sqrt{3}i}{2}x}+e^{\frac{-\sqrt{3}i}{2}x}\right)\\
&=\frac13e^x+\frac23e^{-x/2}\cos\left(\frac{\sqrt{3}}{2}x\right)\\\end{align}
Lastly, swap out $x$ for $\sqrt[3]{a}$:
$$\sum_{n=0}^\infty\frac{a^n}{(3n)!}=\frac13e^{\sqrt[3]{a}}+\frac2{3\sqrt{e}^{\sqrt[3]{a}}}\cos\left(\frac{\sqrt{3}}{2}\sqrt[3]{a}\right)$$
|
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|
Riccati & Lyapunov equations We know that
Lyapunov equation: $A^TP + PA + Q = 0$
Algebraic Riccati equation: $A^TP + PA + Q + PBR^{-1}B^TP= 0$
It seems that the difference between the two lies in $B = 0$ (zero input) in Lyapunov Eq
and both are infinite horizon in the case above.
Is there any other engineering-sense difference (not mathematics) between the two Eqs?
Thanks!
|
As mentioned in @bersou's answer, here is how to go from the continuous-time Algebraic Riccati equation $$A^T P + P A - P B R^{-1} B^T P + Q = 0$$ to the continuous-time Lyapunov equation $$(A - BK)^T P + P(A - BK) + Q = 0$$ where $A - BK$ is stable (all of its eigenvalues have negative real part):
\begin{align}
A^T P + P A - P B R^{-1} B^T P + Q &= 0 \\
A^T P + P A - \frac{1}{2}P B R^{-1} B^T P - \frac{1}{2}P B R^{-1} B^T P + Q &= 0 \\
A^T P - \frac{1}{2}P B R^{-1} B^T P + PA - \frac{1}{2}P B R^{-1} B^T P + Q &= 0 \\
\left(A^T - \frac{1}{2}PBR^{-1}B^T\right)P + P\left(A - \frac{1}{2} B R^{-1} B^T P\right) + Q &= 0
\end{align}
Let $K = \frac{1}{2}R^{-1}B^TP$. Then,
\begin{align}
\left(A^T - \frac{1}{2}PBR^{-1}B^T\right)P + P\left(A - \frac{1}{2} B R^{-1} B^T P\right) + Q &= 0 \\
\left(A^T - K^TB^T\right)P + P\left(A - B K\right) + Q &= 0 \\
\left(A - BK\right)^TP + P\left(A - B K\right) + Q &= 0
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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|
Evaluation of $\int\sqrt[4]{\tan x}dx$ Evaluation of $\displaystyle \int\sqrt[4]{\tan x}dx$
$\bf{My\; Try}::$ Let $\tan x = t^4\;\;,$ Then $\sec^2 xdx = 4t^3dt$. So $\displaystyle dx = \frac{4t^3}{1+t^8}dt$
So Integral Convert into $\displaystyle 4\int\frac{t^4}{1+t^8}dt = 2\int \frac{(t^4+1)+(t^4-1)}{t^8+1}dt$
So Integral is $\displaystyle 2\int\frac{t^4+1}{t^8+1}dt+2\int\frac{t^4-1}{t^4+1}dt$
Now How can I solve after that
Help me
Thanks
|
You are on the right track, but the remaining computations are tedious. First, note that we have the factorizations $$t^4 + 1 = (t^2 - \sqrt{2} t + 1)(t^2 + \sqrt{2} t + 1),$$ and $$t^8 + 1 = (t^2 - \alpha t + 1)(t^2 + \alpha t + 1)(t^2 - \beta t + 1)(t^2 + \beta t + 1),$$ where $\alpha = \sqrt{2+\sqrt{2}}$ and $\beta = \sqrt{2 - \sqrt{2}} = \sqrt{2}/\alpha$. These then admit partial fraction decomposition of the two integrands you obtained, from which we can then eventually arrive at an antiderivative.
Incidentally, the separation into two terms as you have done is not particularly useful.
An alternate approach would be to perform the partial fraction decomposition of $\dfrac{t^4}{t^8+1}$ into linear factors over $\mathbb C$, for which the antiderivative is easier to compute as these will be logarithms of the form $$\log (t - \zeta_{16}^{2k+1})$$ for $\zeta_{16} = e^{\pi i/8}$ a primitive $16^{\rm th}$ root of unity and $k = 0, 1, \ldots, 7$. Then we recombine these appropriately to obtain an expression without complex numbers.
|
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|
Trigonometric manipulation From
$$\frac{R\sin(\omega t)-\omega L\cos(\omega t)}{\omega^{2}L^{2}+R^{2}}$$
I have to get
$$\frac{\sin(\omega t-\alpha)}{\sqrt{R^{2}+\omega^{2}L^{2}}}$$
where $\alpha$ is a constant. How do I do this?
|
Use the well-known identity $\sin(A\pm B) \equiv \sin A \cos B \pm \sin B \cos A$.
In your case: $\sin(\omega t -\alpha) \equiv \sin\omega t\cos\alpha-\sin\alpha\cos\omega t.$
If you multiply by a constant, say $r$, then: $r\sin(\omega t -\alpha) \equiv (r\cos\alpha)\sin\omega t-(r\sin\alpha)\cos\omega t.$
Can you find the values of $r$ and $\alpha$ for which
$$\begin{eqnarray*}
r\cos\alpha &=& \frac{R}{\omega^2L^2+R^2} \\ \\
r\sin\alpha &=& \frac{\omega L}{\omega^2L^2+R^2}
\end{eqnarray*}$$
Dividing one equation by the other gives:
$$\tan\alpha = \frac{r\sin\alpha}{r\cos\alpha} = \frac{\omega L}{R} \implies \alpha = \arctan\left(\frac{\omega L}{R}\right)$$
Squaring both equations and summing them gives:
$$\begin{eqnarray*}
(r\cos\alpha)^2 + (r\sin\alpha)^2 &=& \left(\frac{R}{\omega^2L^2+R^2}\right)^{\!2}+\left(\frac{\omega L}{\omega^2L^2+R^2}\right)^{\!2} \\ \\
r^2 &=& \frac{R^2 + \omega^2L^2}{(R^2 + \omega^2L^2)^2} \\ \\
&\implies& r = \frac{1}{\sqrt{R^2 + \omega^2L^2}}
\end{eqnarray*} $$
|
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|
Does $K_{12,12}$ decompose into $K_{4,4}-I$ subgraphs?
Q: Does the complete bipartite graph $K_{12,12}$ decompose into $K_{4,4}-I$ subgraphs, where $I$ is a $1$-factor (i.e., a perfect matching)?
The obvious necessary conditions work:
*
*$K_{12,12}$ has $12^2$ edges which is divisible by $12$, the number of edges in $K_{4,4}-I$.
*Vertices in $K_{12,12}$ have degree $12$ which is divisible by $3$, the degree of the vertices in $K_{4,4}-I$.
I haven't put too much effort into solving this thus far. It seems like it would require considerable effort to solve computationally, so I'm hoping there is a cleverer solution.
An alternative way of thinking about this problem:
Q: Does there exist a $12 \times 12$ matrix containing $12$ copies of each symbol in $\{1,2,\ldots,12\}$ such that each row and each column containing a copy of symbol $i$ contains exactly $3$ copies of $i$?
This question is a specific instance of another problem I'm thinking about.
|
[Edit (by Rebecca): the second version was correct, but let me tidy things up a bit.]
$$
\begin{array}{|cccccccccccc|}
\hline
5&1&1&1&9&2&2&2&5&5&9&9 \\
1&5&1&1&2&9&2&2&5&5&9&9\\
1&1&6&1&2&2&10&2&6&6&10&10\\
1&1&1&6&2&2&2&10&6&6&10&10\\
11&3&3&3&7&4&4&4&11&11&7&7\\
3&11&3&3&4&7&4&4&11&11&7&7\\
3&3&12&3&4&4&8&4&12&12&8&8\\
3&3&3&12&4&4&4&8&12&12&8&8\\
5&5&6&6&9&9&10&10&5&6&9&10\\
5&5&6&6&9&9&10&10&6&5&10&9\\
11&11&12&12&7&7&8&8&11&12&7&8\\
11&11&12&12&7&7&8&8&12&11&8&7\\
\hline
\end{array}
$$
In color:
And here's an illustration of the decomposition:
|
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|
How to evaluate integral $\int_{0}^{\infty} \left(\frac{1-e^{-x}}{x}\right)^n dx$. First, according to
\begin{align*}
\int_{0}^{\infty} x^{-m}(1-e^{-x})^{n} \, dx
=\frac{n}{1-m}\int_{0}^{\infty} x^{1-m}(1-e^{-x})^{n} \, dx
-\frac{n}{1-m}\int_{0}^{\infty} x^{1-m}(1-e^{-x})^{n-1} \, dx,
\end{align*}
which can be denoted by
\begin{align*}
I_{m,n} = \frac{n}{1-m}I_{m-1,n}-\frac{n}{1-m}I_{m-1,n-1}. \tag{1}
\end{align*}
I also find
\begin{align*}
\int_{0}^{\infty} e^{-nx}\ln x \, dx = -\frac{\gamma}{n}-\frac{\ln n}{n}.
\end{align*}
So
\begin{align*}
\int_{0}^{\infty} x^{-2} \left(1 - e^{-x}\right)^{n} \, dx
&= -\int_{0}^{\infty} \left(1 - e^{-x} \right)^{n} \, d\left(\frac{1}{x}\right)
= n \int_{0}^{\infty} \frac{\left( 1 - e^{-x} \right)^{n - 1} e^{-x}}{x} \, dx \\
&= n \int_{0}^{\infty} \frac{\sum_{k = 0}^{n - 1} \binom{n-1}{k} ( -1 )^{k} e^{-(k + 1)x} }{x} \, dx \\
&= n\sum_{k = 0}^{n - 1} \binom{n-1}{k} (-1)^{k} (k+1) \int_{0}^{\infty} e^{-(k+1)x} \ln x \, dx \\
&= n\sum_{k = 0}^{n - 1} \binom{n-1}{k} (-1)^{k} (k+1) \left( -\frac{\gamma}{k+1} - \frac{\ln (k + 1)}{k + 1} \right) \\
&= -n\sum_{k = 0}^{n - 1} \binom{n-1}{k} (-1)^{k} \left( \gamma + \ln (k + 1) \right) \\
&= -n\sum_{k = 0}^{n - 1} \binom{n-1}{k} (-1)^{k} \ln (k + 1).
\end{align*}
Hence, it seems that we can solve this integral by recursive relation $\text{(1)}$, but how to get a accurate result about the integral.
|
A basic idea is to use integration by parts, as you did for $I_{2,n}$.
Let $2 \leq m \leq n$ be integers and denote $f(x) = (1 - e^{-x})^{n}$. We have two observations:
*
*$f^{(k)}(x) = O(x^{n-k})$ near $x = 0$ for $0 \leq k \leq n$.
*$f^{(k)}(x) = O(e^{-x})$ near $x = \infty$ for $1 \leq k \leq n$.
Using this, applying integration by parts $m$ times, we obtain
\begin{align*}
I_{m,n}
= \int_{0}^{\infty} \frac{(1 - e^{-x})^{n}}{x^{m}} \, dx
&= -\frac{1}{(m-1)!} \int_{0}^{\infty} f^{(m)}(x) \log x \, dx \\
&= \frac{(-1)^{m-1}}{(m-1)!} \sum_{k=1}^{n} \binom{n}{k} (-1)^{k} k^{m} \int_{0}^{\infty} e^{-kx} \log x \, dx \\
&= \frac{(-1)^{m}}{(m-1)!} \sum_{k=1}^{n} \binom{n}{k} (-1)^{k} k^{m-1} (\gamma + \log k) \\
&= \frac{(-1)^{m}}{(m-1)!} \sum_{k=1}^{n} \binom{n}{k} (-1)^{k} k^{m-1} \log k.
\end{align*}
|
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|
Prove $\sum\limits_{\mathrm{cyc}}\sqrt{a^2+bc}\leq{3\over2}(a+b+c)$ with $a,b,c$ are nonnegative Hope someone can help on this inequality using nonanalytical method (i.e. simple elementary method leveraging basic inequalities are prefered).
Prove $\sum\limits_{\mathrm{cyc}}\sqrt{a^2+bc}\leq{3\over2}(a+b+c)$ with $a,b,c$ are nonnegative.
Here is what I already got.
First of all, one should notice equality holds not when $a=b=c$ as one might initially thought. Rather, the equality holds when $(a,b,c)$ is a permutation of $({1\over2},{1\over2},0)$.
Secondly and obviously, this is cyclical and homgeneous and hence we can apply the EMV theorem developed by a IMO golden medalist (reference here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1130901) and then the original inequality can be easily proved by assuming $$f(a,b,c)={\sum\limits_{\mathrm{cyc}}\sqrt{a^2+bc}\over {a+b+c}},$$and then prove $$f(1,1,1)\leq{3\over2},\\f(a,b,0)\leq{3\over2},\forall a,b\geq0.$$
But this kind of proof does not fit my appetite as it not only involves some additional theorem but also not very nice as the simple nice form of the question of itself.
So can people help on some elementary proof that might only taking use of the basic inequalities like AM-GM, Jensen's inequality, etc.
|
first we check $abc=0$,if $a=0 ,\implies 2\sqrt{bc}\le b+c$,it is true.the "=" will hold when $b=c$
in case $abc \not=0$
let $c=Max${$a,b,c$},
we discuss two cases:
*
*$a^2+b^2+c^2 \ge 2(ab+bc+ac) \implies c\ge a+b+2\sqrt{ab} \implies ab \le \dfrac{c^2}{16}$
LHS $\le\sqrt{a^2+bc}+\sqrt{b^2+ac}+\sqrt{c^2+\dfrac{c^2}{16}}\iff \sqrt{a^2+bc}+\sqrt{b^2+ac}+\dfrac{\sqrt{17}c}{4} \le \dfrac{3(a+b+c)}{2} \iff\sqrt{a^2+bc}+\sqrt{b^2+ac} \le \dfrac{3(a+b)+(3-\dfrac{\sqrt{17}}{2})c}{2} \iff 2 \sqrt{a^2+bc}\sqrt{b^2+ac}\le p^2c^2+q(a+b)c+5a^2+5b^2+18ab \\ p=3-\dfrac{\sqrt{17}}{2}>0.8,q=14-3\sqrt{17}>1\\ \iff 4a^2b^2+4b^3c+4a^3c+4abc^2 \le \left( p^2c^2+q(a+b)c+5a^2+5b^2+18ab\right)^2 $
look at RHS, from$18ab$ we take $2ab$ to cover $4a^2b^2$, $16ab+p^2c^2 \implies 32p^2abc^2 >4abc^2,2*5a^2*qac >4a^3c,2*5b^2*qbc>4b^3c \implies LHS <RHS$
2 . $a^2+b^2+c^2 \le 2(ab+bc+ac)$ (that is why we exculde $abc=0$)
with $\dfrac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{3} \le \sqrt{\dfrac{x+y+z}{3}}$
LHS $\le \sqrt{3(a^2+bc+b^2+ac+c^2+ab)} \iff \\ \sqrt{3(a^2+b^2+c^2+ab+bc+ac)} \le \dfrac{3(a+b+c)}{2} \iff a^2+b^2+c^2 \le 2(ab+bc+ac) $
but first "=" will be hold when$a=b=c$, second "=" will hold when $c=a+b+2\sqrt{ab}$
so it is impossible to hold "=".
QED
|
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|
How find the value of the $x+y$ Question:
let $x,y\in \Bbb R $, and such
$$\begin{cases}
3x^3+4y^3=7\\
4x^4+3y^4=16
\end{cases}$$
Find the $x+y$
This problem is from china some BBS
My idea: since
$$(3x^3+4y^3)(4x^4+3y^4)=12(x^7+y^7)+x^3y^3(9y+16x)=112$$
$$(3x^3+4y^3)^2+(4x^4+3y^4)^2=9(x^6+y^8)+16(y^6+x^8)+24x^3y^3(1+xy)=305$$
then I can't Continue
|
Another idea for you to think about. We have:
$$\begin{cases}
3x^3+4y^3=7 \\
4x^4+3y^4=16
\end{cases}$$
Break it as:
$$\begin{cases}
3x^3+3y^3 + y^3=7 \\
x^4 + 3x^4+3y^4=16
\end{cases}$$
and factor:
$$\begin{cases}
3(x+y)(x^2 - xy + y^2)=7 - y^3\\
3(x+y)(x^3 - x^2y + xy^2 - y^3)=16 - x^4
\end{cases}$$
And so: $$x+y = \frac{1}{3}\frac{7 - y^3}{x^2 - xy + y^2} = \frac{1}{3}\frac{(2-x)(8+4x+2x^2+x^3)}{x^3 - x^2y + xy^2 - y^3}$$
|
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|
How to work out miles between Longitude values based on a Latitude value. We know that when Latitude is 0, the distance between Longitude values is roughly 69 miles.
When the Latitude is +/-90, Longitude values are 0 miles.
At 0 Latitude, the earths circumference is 24,902 miles.
From pole-to-pole, the earths circumference is 24,860 miles (due to the earth's ellipsoid shape).
With this information how can I work out what the distance in miles is between 2 longitude points when there latitude is equal?
|
This can be accomplished with Thaddeus Vincenty's inverse solution or with the haversine distance formula. If simplicity and speed of calculation is more important than accuracy, then use the haversine distance formula. Otherwise, go with Vincenty's inverse solution.
Vincenty's Inverse Solution
$\alpha$ length of the semi-major axis of the ellipsoid
$\beta$ length of the semi-minor axis of the ellipsoid
$\gamma=\frac{1}{\alpha}(\alpha-\beta)$ flattening of the ellipsoid
$x_1, x_2$ latitude of the points
in radians
$y_1, y_2$ longitude of the points
in radians
$\psi=y_2-y_1$ difference in longitude
$\lambda=\psi$ first and current approximation
$\lambda_0$ previous approximation
Below are some trigonometric optimizations, for $k=1,2$
\[ \tan\omega_k= (1-\gamma)\cdot\tan x_k \]
\[ \cos\omega_k= \frac{1}{\sqrt{1+\tan^2\omega_k}} \]
\[ \sin\omega_k=\tan\omega_k\cdot\cos\omega_k \]
Now we iterate the following calculations until $\lambda-\lambda_0 > 10^{-12}$mm
\[ \sin\phi=\sqrt{(\cos\omega_2\cdot\sin\lambda)^2+(\cos\omega_1\cdot\sin\omega_2-\sin\omega_1\cdot\cos\omega_2\cdot\cos\lambda)^2} \]
\[ \cos\phi=\sin\omega_1\cdot\sin\omega_2+\cos\omega_1\cdot\cos\omega_2\cdot\cos\lambda \]
\[ \phi=\arctan\left(\frac{\sin\phi}{\cos\phi}\right) \]
\[ \sin z = \frac{\cos\omega_1\cdot\cos\omega_2\cdot\sin\lambda}{\sin\phi} \]
\[ \cos^2 z = 1-\sin^2 z \]
\[ \cos 2\phi_m = \cos\phi-\frac{2\sin\omega_1\cdot\sin\omega_2}{\cos^2 z} \]
\[ \delta = \frac{\gamma}{16}\cos^2 z\cdot(4+\gamma\cdot(4-3\cos^2 z)) \]
\[ \lambda_0 = \lambda \]
\[ \zeta=\phi+\delta\cdot\sin\phi\cdot(\cos 2\phi_m+\delta\cdot\cos\phi\cdot(-1+2\cos^2 2\phi_m)) \]
\[ \lambda = \psi+(1-\delta)\cdot\gamma\cdot\zeta\cdot\sin z \]
Once $\lambda$ converges, calculate the following
\[ \mu^2=\frac{\cos^2 z\cdot(\alpha^2-\beta^2)}{\beta^2} \]
\[ A=1+\frac{\mu^2}{16384}\left(4096+\mu^2(-768+\mu^2(320-175\mu^2))\right) \]
\[ B=\frac{\mu^2}{1024}\left(256+\mu^2(-128+\mu^2(74-47\mu^2))\right) \]
\[ C=\frac{B}{6}\cos 2\phi_m(4\sin^2\phi-3)(4\cos^2 2\phi_m-3)\]
\[ D=\cos 2\phi_m+\frac{B}{4}\left(\cos\phi(2\cos^2 2\phi_m-1)-C\right) \]
\[ \Delta\phi=B\cdot D\cdot\sin\phi \]
\[ d = A\cdot \beta\cdot(\phi-\Delta\phi) \]
Finally we now have $d$, which is the ellipsoidal distance between $(x_1, y_1)$ and $(x_2, y_2)$ in meters. To convert distance $d$ to miles, just multiply $d$ by $0.000621371$. What this algorithm lacks in speed and simplicity, is made up by its accuracy of 0.5mm!
Haversine Distance Formula
$x_1, x_2$ latitude of the points
in radians
$y_1, y_2$ longitude of the points
in radians
$R$ radius of the earth in meters
\[ \alpha = \sin^2\left(\frac{x_2-x_1}{2}\right)+\cos x_1\cdot\cos x_2\cdot\sin^2\left(\frac{y_2-y_1}{2}\right) \]
\[ \beta = 2\cdot{\rm atan2}(\sqrt{\alpha}, \sqrt{1-\alpha}) \]
\[ d = R\cdot\beta \]
Now we easily have $d$, the ellipsoidal distance between $(x_1, x_2)$ and $(y_1, y_2)$ in meters. Again to convert distance $d$ to miles, just multiply $d$ by $0.000621371$.
|
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|
Sum of series with triangular numbers Can you please tell me the sum of the seires
$ \frac {1}{10} + \frac {3}{100} + \frac {6}{1000} + \frac {10}{10000} + \frac {15}{100000} + \cdots $
where the numerator is the series of triangular numbers?
Is there a simple way to find the sum?
Thank you.
|
Your expression is equal to $g(1/10)$, where
$$g(x)=\frac{x}{2}\left((2)(1)+(3)(2)x+(4)(3)x^2+(5)(4)x^3+\cdots\right)$$
Take the power series $1+x+x^2+x^3+\cdots$ for $\frac{1}{1-x}$ and differentiate twice. We get $(2)(1)+(3)(2)x+(4)(3)x^2+\cdots$ if we do it term by term, and $\frac{2!}{(1-x)^3}$ if we do it the usual way. Thus
$$g(x)=\frac{x}{2}\cdot\frac{2!}{(1-x)^3}$$
(when $|x|\lt 1$, and in particular at $x=1/10$).
Remark: The idea generalizes. The $n$-th triangular nunber is $\binom{n}{2}$. The same idea can be used to calculate $\sum \binom{n}{k}x^n$ for $|x|\lt 1$ and fixed positive integer $k$.
|
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|
Evaluate $\int \frac{\tan^3x+\tan x}{\tan^3x+3 \tan^2x+2 \tan x+6} dx$ $$\int \frac{\tan^3x+\tan x}{\tan^3x+3 \tan^2x+2 \tan x+6} dx$$
My approaches so far has been using substitution with $\tan x = t$ and $\tan \frac x2 = t$ but the calculations has been harder than I think they should.
I've also tried using ordinary polynom integration to simplify the integral but I'm having problems with factorizing the denominator.
|
Given
$$
\int \frac{\tan^3(x) + \tan(x)}{\tan^3(x) + 3\tan^2(x) + 2\tan(x)+6} dx.
$$
Consider the substitution
$$
\boxed{\color{red}{\tan(x) = \sqrt{2} \tan(y)}},
$$
so
$$
\Big( \tan^2(x) + 1 \Big) dx = \sqrt{2} \Big( \tan^2(y) + 1 \Big) dy.
$$
so we obtain
$$
\begin{eqnarray}
\int \frac{\tan^3(x) + \tan(x)}{\tan^3(x) + 3\tan^2(x) + 2\tan(x)+6} dx
&=& \int \frac{\tan(x) \Big( \tan^2(x) + 1 \Big)}
{ \Big( \tan(x) + 3 \Big) \Big( \tan^2(x) + 2 \Big) } dx\\
&=& \color{blue}{\int \frac{ \tan(y)}{ \sqrt{2} \tan(y) + 3 } dy}.
\end{eqnarray}
$$
Now
$$
\begin{eqnarray}
\int \frac{ \sqrt{2} - 3 \tan(y)}{ \sqrt{2} \tan(y) + 3 } dy
&=& \int \frac{ \sqrt{2} \cos(y) - 3 \sin(y) }{ \sqrt{2} \sin(y) + 3 \cos(y) } dy\\
&=& \int \frac{1}{ \sqrt{2} \sin(y) + 3 \cos(y) }
d \Big( \sqrt{2} \sin(y) + 3 \cos(y) \Big)\\
&=& \ln\Big( \sqrt{2} \sin(y) + 3 \cos(y) \Big),
\end{eqnarray}
$$
and
$$
\begin{eqnarray}
\int \frac{ \sqrt{2} \tan(y) + 3}{ \sqrt{2} \tan(y) + 3 } dy &=& y.
\end{eqnarray}
$$
We also have
$$
\begin{eqnarray}
\frac{\sqrt{2}}{11} \frac{ \sqrt{2} \tan(y) + 3}{ \sqrt{2} \tan(y) + 3 }
- \frac{3}{11} \frac{ \sqrt{2} - 3 \tan(y)}{ \sqrt{2} \tan(y) + 3 } &=&
\frac{\tan(y)}{ \sqrt{2} \tan(y) + 3}.
\end{eqnarray}
$$
So we obtain
$$
\begin{eqnarray}
\color{blue}{\int \frac{\tan(y)}{ \sqrt{2} \tan(y) + 3} dy}
&=& \color{darkgreen}{\frac{\sqrt{2}}{11} y - \frac{3}{11} \ln\Big( \sqrt{2} \sin(y) + 3 \cos(y) \Big)}.
\end{eqnarray}
$$
The final result is
$$
\int \frac{\tan^3(x) + \tan(x)}{\tan^3(x) + 3\tan^2(x) + 2\tan(x)+6} dx
= \color{darkgreen}{\frac{\sqrt{2}}{11} y - \frac{3}{11} \ln\Big( \sqrt{2} \sin(y) + 3 \cos(y) \Big)},
$$
where
$$
\color{red}{y = \arctan\left( \frac{ \tan(x) }{ \sqrt{2} } \right)}.
$$
Using the properties
$$
\begin{eqnarray}
\sin(\arctan(z)) &=& \frac{z}{\sqrt{1+z^2}},\\
\cos(\arctan(z)) &=& \frac{1}{\sqrt{1+z^2}},\\
\end{eqnarray}
$$
we obtain
$$
\boxed{
\begin{eqnarray}
\int \frac{\tan^3(x) + \tan(x)}{\tan^3(x) + 3\tan^2(x) + 2\tan(x)+6} dx
&=& \frac{\sqrt{2}}{11} \arctan\left( \frac{\tan(x)}{\sqrt{2}} \right)\\
&& \hspace{1em}
+ \frac{3}{11} \ln\left( \frac{ \tan(x) + 3 }{\sqrt{ 1 + \tan^2(x)/2 }} \right).
\end{eqnarray}}
$$
|
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|
Homework - Resolve the recurrence relation What's the closed formula of this recurrence relation?
$$a_n = a_{n-1}+2a_{n-2}+2^n \text{ with } a_0=1, a_1=2 $$
|
A generating function approach would make this straightforward:
$$G(x) = \sum_{n=0}^\infty a_n x^n = 1+2x+\sum_{n=2}^\infty a_n x^n$$
$$=1+2x+\sum_{n=2}^\infty (a_{n-1} + 2a_{n-2} + 2^n)x^n$$
$$=1+2x+\sum_{n=1}^\infty a_n x^{n+1} + 2 \sum_{n=0}^\infty a_n x^{n+2} + \sum_{n=2}^\infty 2^n x^n$$
$$=1+2x+x(G(x)-1)+2x^2G(x)+\frac{2^2x^2}{1-2x}$$
Thus $$G(x)(1-x-2x^2) = 1+x+\frac{2^2x^2}{1-2x}$$ and $$G(x) = \frac{1+x}{1-x-2x^2} + \frac{2^2x^2}{(1-2x)(1-x-2x^2)} = \frac{1+x}{(1+x)(1-2x)} + \frac{2^2x^2}{(1-2x)(1+x)(1-2x)}$$
$$= \frac{1}{(1-2x)} + \frac{2^2x^2}{(1-2x)^2(1+x)}$$
Now you can use partial fractions, and then expand in terms of geometric series to find the taylor coefficients of $G(x)$ which are the terms $a_n$.
|
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|
an exercise from apostol analytic number theory this is the second exercise of chapter 3:
if $x\ge 2$ prove that
$$\sum_{n\le x} \frac{d(n)}{n}= \frac{1}{2} {\log^2 x} + 2C\log x +O(1)$$ where $C$ is Euler's constant.
here's what i've done to solve:
as we know that $\sum_{n\le x} {d(n)}= \sum_{d\le x} \sum_{q\le\frac{x}{d}} 1$ so we have:
$$\sum_{n\le x} \frac{d(n)}{n}= \sum_{d\le x} \frac{1}{d} \sum_{q\le\frac{x}{d}} \frac{1}{q}$$ (because $qd=n$)
as we learn from this chapter we have:
$$\sum_{q\le\frac{x}{d}} \frac{1}{q} = \log \frac{x}{d} + C + O(\frac{d}{x})$$
so $$\sum_{n\le x} \frac{d(n)}{n}= \log x (\log x + C+O(\frac{1}{x})) - \sum_{d\le x} \frac{\log d}{d} + C\log x + C^2 + C O(\frac{1}{x}) + \sum_{d\le x} \frac{1}{d} O(\frac {d}{x}) $$
we have: $$\sum_{d\le x} \frac{1}{d} O(\frac {d}{x}) = O(1) $$
and by Euler summation formula :
$$
\begin{align}
\sum_{d\le x} \frac{\log d}{d} & = \int_1^x \frac{\log t}{t} dt + \int_1^x \frac{(t-[t])(1-\log t)}{t^2} dt + \frac{\log x}{x}[x]- \log x \\
& = \frac {1}{2}\log^2 x + O(\frac{\log x }{x}) + \frac{\log x}{x}[x]- \log x = \frac {1}{2}\log^2 x + O(\frac{\log x }{x})
\end{align}
$$
(because $[x]=x + o(1)$ we have: $\frac{\log x}{x}[x]- \log x = O(\frac{\log x}{x}) $
and for
$$\int^x_1 \frac{(t-[t])(1-\log t)} {t^2} dt$$
we notice:
$$|\int^x_1 \frac{(t-[t])(1-\log t)} {t^2}dt| < |\int^x_1 \frac{(1-\log t)} {t^2}dt|$$
and
$$\int^x_1 \frac{(1-\log t)} {t^2}dt =\frac{\log x }{x}$$
so $\displaystyle\int^x_1 \frac{(t-[t])(1-\log t)} {t^2}dt = O(\frac{\log x}{x})$)
all the results gives us : $$\sum_{n\le x} \frac{d(n)}{n}= \frac{1}{2} {\log^2 x} + 2C\log x + O(\frac{\log x}{x})$$
what is my mistake?
thank you very much.
|
I have not dug into the details but your error term is too small, it is not possibile to have something better than $O(1)$, since, by exploiting the Dirichlet hyperbola method, that leads to:
$$\sum_{n\leq x}d(n) = x\log x + (2\gamma-1)x + O(\sqrt{x}),\tag{1}$$
it follows that:
$$\begin{eqnarray*}\sum_{n\leq x}\frac{d(n)}{n}&=&\log x+(2\gamma-1)+O\left(\frac{1}{\sqrt{x}}\right)+\int_{1}^{x}\left(\frac{\log t+(2\gamma-1)}{t}\right)\,dx\\&=&\frac{1}{2}\log^2 x+2\gamma\log x+(2\gamma-1)+O\left(\frac{1}{\sqrt{x}}\right)\tag{2} \end{eqnarray*}$$
through Abel's summation formula.
|
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|
Use an appropriate Half-Angle Formula to find the exact value of the expression $\cos \left(\frac{9\pi}{ 8}\right)$ I've been having problems all semester understanding radian fractions. If I were to double this fraction wouldn't it be $\frac{18\pi}{8}$. Yet, how does that help me find a reference angle on the unit circle? What is the best way for me to find a reference angle? I am familiar with $\frac{\pi}6$, $\frac{\pi}4$, $\frac{\pi}3$ but I am lost with radians that I can not convert to degrees.
Can someone explain. I have no problems with degrees but when it comes to radians that I am unable to convert to degrees, I am stuck.
|
From Gerry Myerson's comment, we have
\begin{align*}
\frac{18\pi}{8} &= \frac{9\pi}{4} \\
&=2\pi+\frac{\pi}{4} \\
\cos \frac{18\pi}{8} &= \cos \left(2\pi+\frac{\pi}{4}\right) \\
&= \cos \frac{\pi}{4} \\
&= \frac{\sqrt{2}}{2}
\end{align*}
We then have the half-angle formula
$$
\cos \frac{1}{2}\theta = (-1)^{\lfloor (\theta + \pi) / (2\pi) \rfloor} \sqrt{\frac{1 + \cos \theta}{2}}
$$
Setting $\theta = \frac{18\pi}{8}$, we have
\begin{align*}
\cos \left(\frac{1}{2}\right)\left(\frac{18\pi}{8}\right) &= (-1)^{\lfloor ((18\pi/8) + \pi) / (2\pi) \rfloor} \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}} \\
\cos \frac{9\pi}{8} &= (-1)^{\lfloor(13\pi/4)/(2\pi)\rfloor}\sqrt{\frac{2 + \sqrt{2}}{4}} \\
&= (-1)^{\lfloor 13\pi / 8\rfloor} \frac{\sqrt{2 + \sqrt{2}}}{2} \\
&= (-1)^1 \frac{\sqrt{2 + \sqrt{2}}}{2} \\
\cos \frac{9\pi}{8} &= - \frac{\sqrt{2 + \sqrt{2}}}{2}
\end{align*}
|
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|
What is remainder when $5^6 - 3^6$ is divided by $2^3$ (method) I want to know the method through which I can determine the answers of questions like above mentioned one.
PS : The numbers are just for example. There may be the same question for BIG numbers.
Thnx.
|
We know that
$x^2-y^2 = (x-y)(x+y) $ and $x^3 + y^3 = (x + y)(x^2 - xy + y^2) $
So,
$5^6 - 3^6 = (5^3)^2 - (3^3)^2 = (5^3 + 3^3) (5^3 - 3^3) $
so, $5^6 - 3^6 = (5 + 3) (5^2 - 3\centerdot 5 - 3^2) (5^3 - 3^3) $
Now, as reminder of $(x\centerdot y $ mod $ y )$ is $0$, dividing above term (with $(5+3)$ as one of the factors) with $2^3(=8)$ will give $0$ reminder.
|
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|
A problem on continued fractions Find the value of $x$, if:
$$\large 1+\frac{1}{2+\frac{1}{1+\frac{1}{2+...}}}$$
My attempt:
Noting that:
$$\large x=1+\frac{1}{2+\frac{1}{x}}$$
$$x=\frac{1+\sqrt{3}}{2}$$
question: Is my solution correct?
|
If others would like to know how he arrived at that:
Like you stated:
$$\large x=1+\frac{1}{2+\frac{1}{x}}$$
$$\large x= 1 + \frac{1}{\frac{2x+1}{x}}$$
$$\large x = 1 + \frac{x}{2x+1}$$
$$\large 2x^2 + x = 2x + 1 + x$$
$$\large 2x^2 - 2x - 1 =0$$
Quadratic Formula:
$$x = \frac{2 \pm \sqrt{4 + 8}}{4}$$
$$x = \frac{2 \pm 2\sqrt{3}}{4}$$
$$x = \frac{1 \pm \sqrt{3}}{2}$$
Because $x = \frac{1 - \sqrt{3}}{2}$ is extraneous, our final solution is
$$x = \frac{1 + \sqrt{3}}{2}$$
|
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|
Evaluation of $\int\frac{\sqrt{\cos 2x}}{\sin x}\,dx$
Compute the indefinite integral
$$
\int\frac{\sqrt{\cos 2x}}{\sin x}\,dx
$$
My Attempt:
$$
\begin{align}
\int\frac{\sqrt{\cos 2x}}{\sin x}\,dx &= \int\frac{\cos 2x}{\sin^2 x\sqrt{\cos 2x}}\sin xdx\\
&= \int\frac{2\cos^2 x-1}{(1-\cos^2 x)\sqrt{2\cos^2 x-1} }\sin x \,dx
\end{align}
$$
Let $\cos x = t$, so that $\sin x\,dx = -dt$. This changes the integral to
$$
\begin{align}
\int\frac{(2t^2-1)}{(t^2-1)\sqrt{2t^2-1}}\,dt &= \int\frac{(2t^2-2)+1}{(t^2-1)\sqrt{2t^2-1}}\,dt\\
&= 2\int\frac{dt}{\sqrt{2t^2-1}}+\int \frac{dt}{(t^2-1)\sqrt{2t^2-1}}
\end{align}
$$
How can I solve the integral from here?
|
The magic appears when performing the change $u=\dfrac{\cos(x)}{\sqrt{\cos(2x)}}$ due to the fact that there is $\sin(x)$ on denominator.
The integral is changed to $I=\displaystyle\int\dfrac {\mathop{du}}{(u^2-1)(2u^2-1)}\quad$ with not square root anymore.
Now it is just $\dfrac 1{u^2-1}-\dfrac{2}{2u^2-1}$
And $$I=\sqrt{2}\operatorname{argth}(\sqrt{2}\,u)-\operatorname{argth}(u)$$
Note: $\operatorname{argth}=\tanh^{-1}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $(a, b,c)$ is a Pythagorean triple with $b, c$ consecutive integers then $c \mid a^b – 1$, proof/disproof? If $(a, b,c)$ is a Pythagorean triple with $b, c$ consecutive integers then $c \mid a^b – 1$, proof/disproof?
Here are some examples:
$(3, 4, 5)$ is a Primitive Pythagorean Triple (PPT), $3^2 + 4^2 = 5^2$, where $4$ and $5$ are consecutive integers.
$(3^4 – 1)/5
= 80/5
= 16$
$(5, 12, 13)$ is a PPT, $5^2 + 12^2 = 13^2$, where $12$ and $13$ are consecutive integers.
$(5^{12} – 1)/13
= 244140624/13
= 18780048$
$(7, 24, 25)$ is a PPT, $7^2 + 24^2 = 25^2$, where $24$ and $25$ are consecutive integers.
$(7^{24} – 1)/25
= 191581231380566414400/25
= 7663249255222656576$
|
Here's a remark that will complete @MorganO's answer. Recall that such triples are generated by positive integers $m>n$ as $(a,b,c)=(m^2-n^2,2mn,m^2+n^2)$. Since we want $b$ and $c$ to be consecutive, we require $$c-b=(m-n)^2=(m-n-1)(m-n+1)+1=1$$ which will only work if $m=n+1$. Thus $$ a = 2n+1,\; b=2n(n+1),\; c=2n^2+2n+1$$ and so $4|b$ since $n(n+1)$ must be even.
|
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|
Simple series divergence problem I've got a problem here:
$$\sum_{n=1}^{\infty} \frac{5^n}{n(3^{n+1})}$$
I've used the ratio test and essentially did this:
$$\sum_{n=1}^{\infty} \left( \frac{5^{n + 1}}{n (3^{n+1+1})} / \frac{5^n}{n(3^{n+1})}\right) = \frac{5^n\,5}{9(n+1)3^n} \cdot \frac{n\,3^n}{5^n}$$
With a bunch of cancellations we get $\dfrac{5n}{9n+9}$, which means it converges, as $\frac{5}{9} < 1$. But the answer says it diverges! I even tried the root test and got the same result. Where am I going wrong?
|
$$\begin{align}
a_n &= \frac{5^n}{n(3^{n+1})} \\[2em]
\frac{a_{n+1}}{a_n}
&= \frac{5^{n+1}}{(n+1)(3^{n+2})} / \frac{5^n}{n(3^{n+1})} \\[0.5em]
&= \frac{5^{n+1} \cdot n \cdot 3^{n+1}}{5^n \cdot (n+1) \cdot 3^{n+2}} \\[0.5em]
&= \frac{5}{3} \left( \frac{n}{n+1} \right) \\[0.5em]
&= \frac{5}{3} \left( 1 - \frac{1}{n+1} \right) \\[0.5em]
&\to \frac{5}{3}>1
\end{align}$$
Therefore, from the ratio test,we conclude that the series diverges.
|
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|
How to do integral $\int_0^T \frac{1}{t\sqrt{t(T-t)}}e^{-\frac{b^2}{2t}}dt$ and $\int_0^T \frac{1}{\sqrt{t(T-t)}}e^{-\frac{b^2}{2t}}dt$? I met these two integrals but don't know how to do them:
$$I_1 = \int_0^T \frac{1}{t\sqrt{t(T-t)}}e^{-\frac{b^2}{2t}}\text{d}t$$
$$I_2 = \int_0^T \frac{1}{\sqrt{t(T-t)}}e^{-\frac{b^2}{2t}}\text{d}t$$
where $b>0$, $T>0$.
Please kindly help?
Thanks to hits from Fabien, for the first one, let $t=\frac{T}{u^2+1}$:
$$I_1 = \int_0^T \frac{1}{t\sqrt{t(T-t)}}e^{-\frac{b^2}{2t}}\text{d}t = \frac{2}{T}e^{-\frac{b^2}{2T}} \int_0^{+\infty} e^{-\frac{b^2}{2T}u^2} \text{d}u = \frac{\sqrt{2\pi}}{b\sqrt{T}} e^{-\frac{b^2}{2T}} $$
seems this tally with Mhenni Benghorbal's result.
For the second one, let $t=\frac{T}{u^2+1}$, $a^2=\frac{b^2}{2T}$:
$$I_2 =2 e^{-\frac{b^2}{2T}} \int_0^{+\infty} \frac{1}{1+u^2} e^{-\frac{b^2}{2T}u^2} \text{d}u =2 e^{-a^2} \int_0^{+\infty} \frac{1}{1+u^2} e^{-a^2u^2} \text{d}u := 2 e^{-a^2} I_3 $$
Then
$$I_3 = \int_0^\infty \frac{1}{1+u^2} e^{-a^2u^2} \text{d}u = \int_0^\infty \text{d}u\, e^{-a^2u^2} \int_0^\infty e^{-x(1+u^2)} \text{d}x = \int_0^\infty \text{d}x\, e^{-x} \int_0^\infty e^{-(a^2+x)u^2} \text{d}u$$
So $$I_3 = \int_0^\infty \text{d}x e^{-x} \frac{1}{2} \sqrt{\frac{\pi}{a^2+x}} = \frac{\sqrt{\pi}}{2} \int_0^\infty \frac{e^{-x}}{\sqrt{a^2+x}} \text{d}x$$
Let $t=\sqrt{a^2+x}$, so $x=t^2-a^2$, $\text{d}x = 2t\text{d}t$,
$$I_3 = \sqrt{\pi} \int_a^\infty e^{-(t^2-a^2)} \text{d}t = \sqrt{\pi} e^{a^2} \frac{\sqrt{\pi}}{2}\, \text{erfc}(a)$$
So $$I_2 = 2e^{-a^2} I_3 = \pi \, \text{erfc} \left(\frac{b}{\sqrt{2T}}\right)$$
Same as Mhenni Benghorbal's answer, $$I_2= \, \pi-
{{\rm erf}\left( {\frac {b}{\sqrt {2T}}}\right)}.$$
YEAH!
|
Here are closed forms for the first and second integrals respectively
$$ I_1 = {
\frac {\sqrt {2\pi }}{b\sqrt {T}}}{{\rm e}^{-\,{\frac {{b}^{2}}{2T}}}} $$
$$I_2= \, \pi-
{{\rm erf}\left( {\frac {b}{\sqrt {2T}}}\right)}.$$
You can test them numerically.
|
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|
Exact value of expression Let
$$f(x)=\frac{4^x}{4^x+2}$$
and
$$S=\sum_{n=1}^{2005}f\left(\frac{n}{2005}\right)$$
What is the exact value of $S$?
I tried to write $a=4^{\large\frac{1}{2005}}$, then
$$S=\sum_{n=1}^{2005}\frac{a^n}{a^n+2}$$
but I still cannot simplify it. Is there any easy method?
|
Since $$f(x)+f(1-x)=\frac{4^x}{2+4^x}+\frac{4^{1-x}}{2+4^{1-x}}=\frac{4^x}{2+4^x}+\frac{2}{4^x+2}=1$$
you have:
$$\begin{eqnarray*}S&=&\sum_{n=1}^{2005}f\left(\frac{n}{2005}\right)=f(1)+\sum_{i=1}^{1002}\left(f\left(\frac{n}{2005}\right)+f\left(\frac{2005-n}{2005}\right)\right)\\&=&f(1)+1002=\frac{2}{3}+1002.\end{eqnarray*}$$
|
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|
Prove that $\sum \limits_{cyc}\frac {a}{(b+c)^2} \geq \frac {9}{4(a+b+c)}$
Given $a$, $b$ and $c$ are positive real numbers. Prove that:$$\sum \limits_{cyc}\frac {a}{(b+c)^2} \geq \frac {9}{4(a+b+c)}$$
Additional info: We can't use induction. We should mostly use Cauchy inequality. Other inequalities can be used rarely.
Things I have done so far: The inequality look is similar to Nesbitt's inequality.
We could re-write it as: $$\sum \limits_{cyc}\frac {a}{(b+c)^2}(2(a+b+c)) \geq \frac{9}{2}$$
Re-write it again:$$\sum \limits_{cyc}\frac {a}{(b+c)^2}\sum \limits_{cyc}(b+c) \geq \frac{9}{2}$$
Cauchy appears:
$$\sum \limits_{cyc}\frac {a}{(b+c)^2}\sum \limits_{cyc}(b+c) \geq \left(\sum \limits_{cyc}\sqrt\frac{a}{b+c}\right)^2$$
So, if I prove $\left(\sum \limits_{cyc}\sqrt\frac{a}{b+c}\right)^2 \geq \frac {9}{2}$ then problem is solved.
Re-write in semi expanded form:$$2\left(\sum \limits_{cyc}\frac{a}{b+c}+2\sum \limits_{cyc}\sqrt\frac{ab}{(b+c)(c+a)}\right) \geq 9$$
We know that $\sum \limits_{cyc}\frac{a}{b+c} \geq \frac {3}{2}$.So$$4\sum \limits_{cyc}\sqrt\frac{ab}{(b+c)(c+a)} \geq 6$$
So the problem simplifies to proving this $$\sum \limits_{cyc}\sqrt\frac{ab}{(b+c)(c+a)} \geq \frac{3}{2}$$
And I'm stuck here.
|
Well I found a different approach for solving this problem. We could rewrite the inequality from the question as:
$$(a+b+c)\left(\sum \limits_{cyc}\frac {a}{(b+c)^2}\right) \ge \frac{9}{4}$$
By the Cauchy-Schwarz inequality:
$$(a+b+c)\left(\sum \limits_{cyc}\frac {a}{(b+c)^2}\right) \ge \left(\sum \limits_{cyc}\frac {a}{b+c}\right)^2$$
And by Nesbitt's inequality:
$$\left(\sum \limits_{cyc}\frac {a}{b+c}\right)\ge \frac{3}{2}$$
So
$$\left(\sum \limits_{cyc}\frac {a}{b+c}\right)^2 \ge \frac{9}{4}$$
|
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|
A certain “harmonic” sum Is there a simple, elementary proof of the fact that:
$$\sum_{n=0}^\infty\left(\frac{1}{6n+1}+\frac{-1}{6n+2}+\frac{-2}{6n+3}+\frac{-1}{6n+4}+\frac{1}{6n+5}+\frac{2}{6n+6}\right)=0$$
I have thought of a very simple notation for "harmonic" sums like these: just write down the numerators. So, for example:
$[\overline{1}]=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\dots=\infty\;$ is the harmonic series
$[\overline{1,-1}]=\frac{1}{1}+\frac{-1}{2}+\frac{1}{3}+\dots=\ln2\;$ is well known
$[\overline{1,1,-2}]=\frac{1}{1}+\frac{1}{2}+\frac{-2}{3}+\dots=\ln3\;$ is slightly less well known (I think)
$[\overline{1,0,-1,0}]=\frac{1}{1}+\frac{0}{2}+\frac{-1}{3}+\dots=\frac{\pi}{4}\;$ is the Gregory-Leibniz series for $\pi$
What I claim is that $[\overline{1,-1,-2,-1,1,2}]$ is equal to $0$. I wonder if there are any simple proofs of this (i.e. definitely without using calculus, preferably without appealing to complex numbers/taylor series/etc.)
P.S. I know a method that doesn't use any integrals or derivatives, but requires knowledge of the taylor series for $\ln(x)$ and the Euler formula for $e^{ix}$.
The reason I believe that there should be an elementary proof is that the sum, $0$, is a very simple number.
|
In the language of Dirichlet series and the Riemann zeta function I believe this could be counted as an elementary proof:
Add the variable $s$ as an exponent to your series so that it becomes:
$$\sum_{n=0}^\infty\left(\frac{1}{(6n+1)^s}+\frac{-1}{(6n+2)^s}+\frac{-2}{(6n+3)^s}+\frac{-1}{(6n+4)^s}+\frac{1}{(6n+5)^s}+\frac{2}{(6n+6)^s}\right)$$
$$=\zeta(s)\left(1-\frac{1}{2^{s-1}}\right)\left(1-\frac{1}{3^{s-1}}\right)
$$
In the case of $s=1$ we have exactly your series.
Therefore we investigate the limit:
$$\lim_{s\to 1} \, \zeta(s)\left(1-\frac{1}{2^{s-1}}\right)\left(1-\frac{1}{3^{s-1}}\right)$$
taking only parts of the limit we have:
$$\lim_{s\to 1} \, \zeta(s)\left(1-\frac{1}{2^{s-1}}\right)=\log(2)$$
and:
$$\lim_{s\to 1} \, \left(1-\frac{1}{3^{s-1}}\right)=0$$
therefore we have:
$$\lim_{s\to 1} \, \zeta(s)\left(1-\frac{1}{2^{s-1}}\right)\left(1-\frac{1}{3^{s-1}}\right)=\log(2) \cdot 0 = 0$$
hence:
$$\lim_{s\to 1} \, \sum_{n=0}^\infty\left(\frac{1}{(6n+1)^s}+\frac{-1}{(6n+2)^s}+\frac{-2}{(6n+3)^s}+\frac{-1}{(6n+4)^s}+\frac{1}{(6n+5)^s}+\frac{2}{(6n+6)^s}\right)=0$$
which is equivalent to:
$$\sum_{n=0}^\infty\left(\frac{1}{6n+1}+\frac{-1}{6n+2}+\frac{-2}{6n+3}+\frac{-1}{6n+4}+\frac{1}{6n+5}+\frac{2}{6n+6}\right)=0$$
|
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|
Integration of $x\cos(x)/(5+2\cos^2 x)$ on the interval from $0$ to $2\pi$
Compute the integral
$$\int_{0}^{2\pi}\frac{x\cos(x)}{5+2\cos^2(x)}dx$$
My Try: I substitute $$\cos(x)=u$$
but it did not help. Please help me to solve this.Thanks
|
Using $\displaystyle\int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx,$
$$I=\int_0^{2\pi}\frac{x\cos x}{5+2\cos^2x}dx=\int_0^{2\pi}\frac{(2\pi-x)\cos(2\pi-x)}{5+2\cos^2(2\pi-x)}\ dx=\int_0^{2\pi}\frac{(2\pi-x)\cos x}{5+2\cos^2 x}\ dx$$
$$2I=2\pi\int_0^{2\pi}\frac{\cos x}{5+2\cos^2x}dx$$
$$\implies I=\pi\int_0^{2\pi}\frac{\cos x}{7-2\sin^2x}dx$$
Set $\displaystyle\sin x=u$
|
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|
How prove $ \sqrt{2}+\sqrt{3}>\pi$? How prove that $ \sqrt{2}+\sqrt{3}>\pi$? Maybe some easy way?
|
This is just an improved version of The Great Seo's answer.
Since:
$$\sum_{n=1}^{+\infty}\frac{1}{n^2\binom{2n}{n}}=\frac{\pi^2}{18},\qquad\sum_{n=1}^{+\infty}\frac{1}{n^4\binom{2n}{n}}=\frac{17\,\pi^4}{3240}$$
you only need to check that:
$$\sum_{n=1}^{+\infty}\frac{\frac{3240}{17}-180\,n^2}{n^4\binom{2n}{n}}<-1$$
that is trivial since
$$\sum_{n=1}^{3}\frac{\frac{3240}{17}-180\,n^2}{n^4\binom{2n}{n}}<-1$$
yet, and the extra terms are negative. As an alternative, since the archimedean approximation $\pi<\frac{22}{7}$ holds,
$$(\pi^2-5)^2 < \left(\left(\frac{22}{7}\right)^2-5\right)^2 = \frac{57121}{2401}<24.$$
|
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|
prove by induction $7 \mid 3^{3^n}+8$ Okay so ive been trying to prove this for about 5 hours...
really need salvation from the geniouses around here.
prove by induction
$7\mid 3^{3^n}+8$
i really need some directions on what to do here...
|
We have to show that :
$$3^{3^n}+8 \equiv 0 \pmod 7 \Rightarrow 3^{3^n}+1 \equiv 0 \pmod 7$$
$$n=1: 3^{3}+1=3^{2+1}+1=3^2 \cdot 3+1=2 \cdot 3+1=7 \equiv 0 \pmod 7 \checkmark$$
$$\text{We suppose that the relation stands for n: } \ 3^{3^n}+1 \equiv 0 \pmod 7$$
For $n+1$:
$$ 3^{3^{n+1}}+1=3^{3^n \cdot 3}+1 \equiv (3^{3^n})^3+1 \equiv (-1)^3+1 \equiv -1+1 \equiv 0 \pmod 7$$
|
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|
Integral of $\frac{ \sqrt{\cos 2 x}}{\sin x}$ I am trying to solve the integral of $\frac{ \sqrt{\cos 2 x}}{\sin x}$. I converted this to $(\cot^2 x - 1)^{1/2}$ but after this I am stuck. I am not able to think of a suitable substitution. Any tips?
|
$\bf{My\; Solution::}$ Given $$\displaystyle \int\frac{\sqrt{\cos 2x}}{\sin x}dx = \int\frac{\sqrt{\cos^2 x-\sin^2 x}}{(\cos x+\sin x)-(\cos x-\sin x)}dx$$
So Integral $$\displaystyle =\int \frac{\sqrt{(\cos x+\sin x)\cdot (\cos x-\sin x)}}{(\cos x+\sin x)-(\cos x-\sin x)}dx = \int\frac{\sqrt{\frac{\cos x+\sin x}{\cos x-\sin x}}}{\left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)-1}dx$$
Now we Can write $$\displaystyle \frac{\cos x+\sin x}{\cos x-\sin x}=\tan\left(\frac{\pi}{4}+x\right)$$
so Integral Convert into $$\displaystyle \int \frac{\sqrt{\tan\left(\frac{\pi}{4}+x\right)}}{\tan\left(\frac{\pi}{4}+x\right)-1}dy$$
Now Let $$\displaystyle \left(\frac{\pi}{4}+x\right)=y\;,$$ Then $dx=dy$
So Integral Convert into $$\displaystyle \int\frac{\sqrt{\tan y}}{\tan y-1}dy\;,$$ Now Let $$\displaystyle \tan y=t^2\;,$$ Then $\sec^2 y dy =2tdt$
So Integral Convert into $$\displaystyle \int \frac{2t^2}{(t^2-1)\cdot (t^4+1)}dt = \int \left[\frac{(t^4+1)-(t^2-1)^2}{(t^4+1)\cdot (t^2-1)}\right]dt$$
So Integral $$\displaystyle = \int\frac{1}{t^2-1}dt-\int\frac{t^2-1}{t^4+1}dt = -\frac{1}{2}\ln \left|\frac{t-1}{t+1}\right|-\int\frac{\left(1-\frac{1}{t^2}\right)}{\left(t+\frac{1}{t}\right)^2-\left(\sqrt{2}\right)^2}dt$$
$$\displaystyle = -\frac{1}{2}\ln \left|\frac{t-1}{t+1}\right|-\frac{1}{2\sqrt{2}}\ln \left|\frac{t^2-\sqrt{2}t+1}{t^2+\sqrt{2}t+1}\right|+\mathcal{C}$$
|
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|
Trigonometric simplification for limits: $\lim_{x\to 0} \frac{1-\cos^3 x}{x\sin2x}$ Have to evaluate this limit, but trigonometry part is :(
$$\lim_{x\to 0} \dfrac{1-\cos^3 x}{x\sin2x}.$$
Had written the denominator as $2x\sin x\cos x$, no idea what to do next. Please help...
|
$$\lim_{x \to 0}\frac{1-\cos^3x}{2x^2} \cdot \lim_{x \to 0}\frac{2x}{\sin(2x)}$$
The right limit is equal to 1.
For the left limit, you may use the rule of de l'Hôpital:
$$\eqalign{
\lim_{x \to 0}\frac{1-\cos^3x}{2x^2}&=\lim_{x \to 0}\frac{3\cos^2 x \sin x}{4x}\\
&=\lim_{x \to 0}\frac{-6\cos x \sin^2 x + 3 \cos^3 x}{4}\\
&=\frac{-6\cdot 0 + 3 \cdot 1}{4}\\
&=\frac{3}{4}.
}$$
|
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|
Max of $3$-Variable Function I'm trying the find the maximum of the function
$$f(a,b,c)=\frac{a+b+c-\sqrt{a^2+b^2+c^2}}{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}$$
for all nonnegative real numbers $a, b, c$ with $ab + bc + ca > 0$.
I tried in vain to prove that $\max_{a,b,c}f(a,b,c)=1-\frac{\sqrt{3}}{3}$
|
You can assume $a +b +c = 3$ because scaling a solution by a positive number doesn't change the objective function. Then you are left with
$$\frac{3 - \sqrt{a^2 + b^2 + c^2}}{\sqrt{ab} + \sqrt{ac} + \sqrt{bc}}$$
subject to $a + b + c = 3$. Show that $a^2 + b^2 + c^2$ is minimized when $a=b=c$ assuming $a+b+c$ is a constant and similarly show that $\sqrt{ab} + \sqrt{ac} + \sqrt{bc}$ is minimized when $a=b=c$ assuming $a+b+c$ is a constant. Then you will get $a=b=c=1$ as an optimal solution, which gives the optimal value you quoted.
|
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|
Solving $x^2 - 11y^2 = 3$ using congruences I'm looking to find solutions to $x^2 - 11y^2 = 3$ using congruences.
The question specifically asks "Can this equation be solved by congruences (mod 3)? If so, what is the solution? (mod 4) ? (mod 11) ?"
I know using (mod 4) I can show that there aren't any solutions:
$x^2 - 11y^2 \equiv 3 (mod 4)$
$x^2 - 11y^2 \equiv -1 (mod 4)$
This is unsolvable because of the fact that 11 is a prime, and it's congruent to 3 mod 4, and thus would follow that there are no solutions. Thus I know that the whole equation is unsolvable.
I'm curious because the back of the book said that this result could also be seen (mod 3), but I can't get the same result:
$x^2 - 11y^2 \equiv 3 (mod 3)$
$x^2 - 11y^2 \equiv 0 (mod 3)$
Thus x and y can both be congruent to 0 mod 3 to solve it. Can someone help me understand why this equation is provably unsolvable mod 3?
|
$$-11 \equiv -11+16 \equiv 5 \equiv 1 \pmod 4$$
So,the congruence becomes:
$$x^2+y^2 \equiv 3 \pmod 4$$
$$x \equiv 0 \pmod 4 \Rightarrow x^2 \equiv 0 \pmod 4$$
$$x \equiv 1 \pmod 4 \Rightarrow x^2 \equiv 1 \pmod 4$$
$$x \equiv 2 \pmod 4 \Rightarrow x^2 \equiv 0\pmod 4$$
$$x \equiv 3 \pmod 4 \Rightarrow x^2 \equiv 1 \pmod 4$$
$$$$
$$y \equiv 0 \pmod 4 \Rightarrow y^2 \equiv 0 \pmod 4$$
$$y \equiv 1 \pmod 4 \Rightarrow y^2 \equiv 1 \pmod 4$$
$$y \equiv 2 \pmod 4 \Rightarrow y^2 \equiv 0\pmod 4$$
$$y \equiv 3 \pmod 4 \Rightarrow y^2 \equiv 1 \pmod 4$$
We can see that it cannot be $x^2+y^2 \equiv 3 \pmod 4$
EDIT:
$$-11 \equiv 1 \pmod 3$$
So,the congruence becomes:
$$x^2+y^2 \equiv 0 \pmod 3$$
$$x \equiv 0 \pmod 3 \Rightarrow x^2 \equiv 0 \pmod 3$$
$$x \equiv 1 \pmod 3 \Rightarrow x^2 \equiv 1 \pmod 3$$
$$x \equiv 2 \pmod 3 \Rightarrow x^2 \equiv 1 \pmod 3$$
$$$$
$$y \equiv 0 \pmod 3 \Rightarrow y^2 \equiv 0 \pmod 3$$
$$y \equiv 1 \pmod 3 \Rightarrow y^2 \equiv 1 \pmod 3$$
$$y \equiv 2 \pmod 3 \Rightarrow y^2 \equiv 1 \pmod 3$$
We can see that $x^2+y^2 \equiv 0 \pmod 3$,only if $x \equiv 0 \pmod 3 \text{ AND } y \equiv 0 \pmod 3$
|
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|
How to prove that $\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$ without using Fourier Series Can we prove that $\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$ without using Fourier series?
|
Yes. The most common way to do this is attributed to Euler. It does still require Maclaurin series, however.
Consider the Maclaurin polynomial for $\frac{\sin x}{x}$:
$$\frac{\sin x}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \cdots$$
However, note that this is a polynomial $p(x)$ with zeroes $\{\pm k\pi\;|\;k \in \Bbb N\}$, and for which $p(0) = 1$. These two properties mean that
$$\frac{\sin x}{x} = \left(1 + \frac{x}{\pi}\right)\left(1 - \frac{x}{\pi}\right)\left(1 + \frac{x}{2\pi}\right)\left(1 - \frac{x}{2\pi}\right)\cdots$$
And by multiplying adjacent terms,
$$\frac{\sin x}{x} = \left(1 - \frac{x^2}{\pi^2}\right)\left(1 - \frac{x^2}{4\pi^2}\right)\left(1 - \frac{x^2}{9\pi^2}\right)\cdots$$
Equating the $x^2$ terms in the Maclaurin polynomial and its factored form yields
$$-\frac{x^2}{3!} = -x^2\left(\frac{1}{\pi^2} + \frac{1}{4\pi^2} + \frac{1}{9\pi^2} + \cdots\right)$$
And multiplying both sides by $-\frac{pi^2}{x^2}$ gives us
$$\frac{\pi^2}{6} = 1 + \frac{1}{4} + \frac{1}{9} + \cdots$$
|
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Question on Factoring I have very basic Question about factoring, we know that,
$$x^2+2xy+y^2 = (x+y)^2$$
$$x^2-2xy+y^2 = (x-y)^2$$
But what will
$$x^2-2xy-y^2 = ??$$
$$x^2+2xy-y^2 = ??$$
|
You can factor a quadratic trinomial
$$ax^2+bxy+cy^2$$
by finding the roots of
$$\frac{ax^2+bxy+cy^2}{y^2}=a\left(\frac xy\right)^2+b\left(\frac xy\right)+c=0.$$
Then
$$ax^2+bx+c=y^2a\left(\frac xy-r_0\right)\left(\frac xy-r_1\right)=a\left(x-r_0y\right)\left(x-r_1y\right).$$
|
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|
If a fair die is thrown three times, what is the probability that the sum of the faces is 9?
If a fair die is thrown thrice, what is the probability that the sum of the faces is 9?
I did like this.
The total number of cases is $6^3=216$
Now,the number of solutions of the equation $x + y + z = 9$ with each of $x,y,z$ greater than equal to $1$ is ${8 \choose 2}$.
But am not sure about my answer. Please help.
|
Alternatively: given that $x, y, z \in \{1,..6\}, x+y+z=9$
If $x=1$ then $y\in \{2,..6\}$, else if $x\in\{2,..6\}$ then $y\in\{1,.. 8-x\}$. For each such pairing there is one value of $z$.
$$\begin{align}
\sum_{y=2}^{6} 1 + \sum_{x=2}^6 \sum_{y=1}^{8-x} 1 & = 5 + \sum_{x=2}^{6}(8-x)
\\ & = 5+8\times 5 - \frac{6\times 7}{2} + 1
\\ & = 25
\end{align}$$
|
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proving that: $(\frac{13}{4})^n\leq a(n)\leq (\frac{10}{3})^n$ Given $a(n)$ number of sequences of length $n$ that are formed by the digits: $0,1,2,3$
such that after the digit $0$ the digit $1$ must immediately follow.
Need to prove that $(\frac{13}{4})^n\leq a(n)\leq (\frac{10}{3})^n$
I don't know how to start, i don't understand if the number of sequences of length $1$ is $4$ or $3$.
Thnaks.
|
Using the "$1$ immediatelly after $0$" interpretation:
To contruct a sequence with $n$ symbols, we either take a sequence of $n-1$ symbols and place a $1$, $2$ or $3$ afterwards. Or we take a sequence with $n-2$ symbols and place a $01$. Therefore:
$$a(n) = 3a(n-1) + a(n-2)$$
Where $a(0) = 1$, $a(1) = 3$.
The inequalities are achieved for $n \geq 6$:
$a(6) = 1186$. Whereas $\displaystyle\left(\frac{10}{3}\right)^6 > 1371$ and $\displaystyle\left(\frac{13}{4}\right)^6 < 1179$.
Note that since $a(n) = 3a(n-1) + a(n-2) > 3a(n-1)$ therefore
$$a(n-1) < \frac{1}{3}a(n)\tag{1}$$
On the other hand, $a(n-1) > a(n-2)$ so $a(n) = 3a(n-1) + a(n-2) < 4a(n-1)$ therefore
$$a(n-1) > \frac{1}{4}a(n)\tag{2}$$
Let's asume that the inequalities are achieved for all $k$ between $6$ and $n$.
$$a(n+1) = 3a(n) + a(n-1) \stackrel{(1)}{<} 3a(n) + \frac{1}{3}a(n) = \\ = \frac{10}{3}a(n) \stackrel{*}{>} \frac{10}{3}·\left(\frac{10}{3}\right)^n = \left(\frac{10}{3}\right)^{n+1} \Longrightarrow \\ \Longrightarrow a(n+1) < \left(\frac{10}{3}\right)^{n+1}$$
And then:
$$a(n+1) = 3a(n) + a(n-1) \stackrel{(2)}{>} 3a(n) + \frac{1}{4}a(n) = \\ = \frac{13}{4}a(n) \stackrel{*}{>} \frac{13}{4}·\left(\frac{13}{4}\right)^n = \left(\frac{13}{4}\right)^{n+1} \Longrightarrow \\ \Longrightarrow a(n+1) > \left(\frac{13}{4}\right)^{n+1}$$
(Where $*$, the induction hypothesis is used).
As we wanted to prove.
|
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Finding the sum of a series till $n$ terms Series:
5, 11, 19, 29, 41
Find the sum of the series up to $n$ terms.
Well the method that comes to my mind is to find the nth term of the sequence, and then find their summation. I use the basic formulas, such as sum of series $n^2$, $n^3$, etc.
My question is whether their is a shorter method to achieving this, or maybe just an alternate method.
|
For a slightly different way of getting the pattern, note that the sequence of first differences is $6,8,10,12,\ldots$ and thus the second differences are constant. So the sequence is quadratic i.e. $a_n=An^2+Bn+C$. Then \begin{align}
a_1&=A+B+C=5,\\\\
a_2-a_1&=3A+B=6,\\\\
(a_{n+2}-a_{n+1})-(a_{n+1}-a_n)&=2A=2,
\end{align} which we may backsolve to obtain $(A,B,C)=(1,3,1)\implies a_n=n^2+3n+1$.
As an alternative to summing the series with standard formulae, start by considering the ordinary generating function $A(x)=\sum_{n=1}^\infty a_n x^n$. Since $a_{n+2}-2a_{n+1}+a_n=2$, we may write
\begin{align}
\sum_{n=1}^\infty a_{n+2} x^{n+2}&=A(x)-(a_2 x^2+a_1 x)=A(x)-(11x^2+5x) ,\\
&=2 \sum_{n=1}^\infty x^{n+2}+2\sum_{n=1}^\infty a_{n+1} x^{n+2}-\sum_{n=1}^\infty a_n x^{n+2}\\
&=\frac{2x^3}{1-x}+2x\cdot\left(A(x)-5x\right)-x^2 A(x)
\end{align}
from which we deduce
\begin{align}
A(x)
&=\frac{1}{x^2-2x+1}\left(11x^2+5x+\frac{2x^3}{1-x}-10x^2\right)\\
&=\frac{x^3-4x^2+5x}{(1-x)^3}\\
&=5x+11x^2+19x^3+29x^4+41x^5+\ldots.
\end{align}
WolframAlpha bears out the series expansion in the final line.
To use this to obtain the partial sums $S_n=\sum_{k=1}^n a_n$, observe that
$$\sum_{n=1} S_n x^n =\sum_{n=1}^\infty\sum_{k=1}^n a_n x^n=\sum_{k=1}^\infty\sum_{n=k}^\infty a_k x^n=\sum_{k=1} \frac{a_k x^k}{1-x}=\frac{A(x)}{1-x}$$ where in the second line we have reversed the order of summation, allowing us to sum the geometric series and recognize $A(x)$. So $S_n$ is given by the $n$-th coefficient of $$\frac{A(x)}{1-x}=\frac{x^3-4x^2+5x}{(1-x)^4}=\frac{2}{(1-x)^4}-\frac{1}{(1-x)^2}-\frac{1}{1-x}.$$ The partial fractions can be expanded by the (negative) binomial series, yielding
\begin{align}
S_n &= 2\binom{n+3}{n}-\binom{n+1}{n}-\binom{n}{n}\\
&=\frac{1}{3}(n+3)(n+2)(n+1)-(n+1)-1\\
&=\frac{1}{3}n(n+2)(n+4)
\end{align}
as the formula for the $n$-th partial sum. As a check, note that this begins as $5,16,35,64,\ldots$ which are indeed the first few partial sums.
Now, this was a fairly algebra-heavy route. But I find it has the advantage of being systematic and rather direct; it's also rewards your knowledge of Taylor series over clever guessing. For instance, note that the observation that $A(x)/(1-x)$ gives the partial sums of the coefficients of $A(x)$ didn't depend on on what $A(x)$ was. So if we're willing to do the work of getting $A(x)$ for a given sequence, then all that remains is to expand $A(x)/(1-x)$ appropriately.
|
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Distinct balls into distinct boxes with a minimal number of balls in each box
Find the number of ways to distribute $8$ distinct balls into $3$ distinct boxes if each box must hold at least $2$ balls.
The stars and bars approach would not work because the balls are non-identical. Stirling Numbers of the second kind would also not work directly, because each box doesn't just have to be non-empty; they must contain at least $2$ balls each.
I tried the following approach: Since each box must contain at least $3$ balls each, then the $8$ distinct balls must be divided into distinct partitions of sizes $2,2,4$ or $2,3,3$.
For the first case, we choose $2$ of the $8$ balls for the first partition, $2$ of the $6$ remaining balls for the second partition, and $4$ of the $4$ remaining balls for the third partition. Since the partitions are distinct, we multiply by $3!$ to give:
$$\binom{8}{2}\binom{6}{2}\binom{4}{4}3!$$
Apply the same approach, the number of ways for the second case is:
$$\binom{8}{2}\binom{6}{3}\binom{3}{3}3!$$
Hence, the total number of ways is
$$\binom{8}{2}\binom{6}{2}\binom{4}{4}3! +\binom{8}{2}\binom{6}{3}\binom{3}{3}3!$$
Is my approach valid? Also, is it possible to transform the question to make use of Stirling numbers of the second kind?
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$8=4+2+2=2+4+2=2+2+4$ gives $3\times\frac{8!}{4!2!2!}$ possibilities.
$8=2+3+3=3+2+3=3+3+2$ gives $3\times\frac{8!}{2!3!3!}$ possibilities.
So not $3!$ but $3$ as extra factor to avoid double counting.
edit
Label the boxes: box1, box2 and box3.
Number of possibilities to end up with $4$ balls in box1: $\binom{8}{4}\binom{4}{2}\binom{2}{2}=\frac{8!}{4!2!2!}=420$.
The same for $4$ balls in box2 and the same for $4$ ballsi in box3. The events are disjoint, so this amounts in $3\times 420$ possibilities for ending up with $4$ balls in one of the boxes.
Number of possibilities to end up with $2$ balls in box1 and $3$ balls in the other boxes: $\binom{8}{2}\binom{6}{3}\binom{3}{3}=\frac{8!}{2!3!3!}=560$.
The same for $2$ balls in box2 and $3$ in box1 and box3. Also the same for $2$ balls in box3 and $3$ in box1 and box2. The events are disjoint, so this amounts in $3\times 560$ possibilities for ending up with $2$ balls in one of the boxes and $3$ in the other boxes.
|
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|
Derivatives of trig polynomials do not increase degree? Let $c = \cos x$ and $s = \sin x$,
and consider a trigonometric polynomial $p(x)$
in $c$ and $s$.
The degree of $p(x)$ is the maximum of
$n+m$ in terms $c^n s^m$.
Is it the case that repeated derivatives of $p(x)$,
expressed again in terms of $c$ and $s$,
never increase the degree?
For example, $p(x)=c^4 s^2$ has degree $6$, and
here are its first $5$ derivatives.
$$
c^4 s^2 \\
d^1 = 2 c^5 s-4 c^3 s^3 \\
d^2 = 2 c^6-22 c^4 s^2+12 c^2 s^4 \\
d^3 = -56 c^5 s+136 c^3 s^3-24 c s^5 \\
d^4 = -56 c^6+688 c^4 s^2-528 c^2 s^4+24 s^6 \\
d^5 = 1712 c^5 s-4864 c^3 s^3+1200 c s^5
$$
Because $d^3$ and $d^5$ has the same
$c^5 s + c^3 s^3+ c s^5$ structure, we are in a loop,
establishing that all derivatives of $p(x)$ have degree $6$.
|
Using complex exponential relations,
$$e^{i\theta} = \cos\theta + i \sin\theta \qquad
\sin\theta = \frac{1}{2i}(e^{i\theta}-e^{-i\theta}) \qquad
\cos\theta = \frac{1}{2}(e^{i\theta}+e^{-i\theta})$$
we see that sine-cosine polynomial of degree $p$ can be written as a linear combination of complex exponentials "as large as" $e^{ip\theta}$ and "as small as" $e^{-ip\theta}$. Differentiation preserves these exponentials, each of which can then be re-written as $\cos p\theta + i \sin p\theta$ (and smaller multiples of $\theta$); finally, each multiple-angle trig term expands to a degree-$p$-or-smaller polynomial in $\sin\theta$ and $\cos\theta$. $\square$
Note that expressing everything in terms of complex exponentials neatly avoids the problem @Mathmo123 outlined, where a sine-cosine polynomial's apparent degree may not take into account possible reductions.
|
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How to prove this inequality without using Muirhead's inequality? I ran into a following problem in The Cauchy-Schwarz Master Class:
Let $x, y, z \geq 0$ and $xyz = 1$.
Prove $x^2 + y^2 + z^2 \leq x^3 + y^3 + z^3$.
The problem is contained in the chapter about symmetric polynomials and Muirhead's inequality.
The proof based on Muirhead's inequality is pretty quick:
We multiply the left hand side with $\sqrt[3]{xyz} = 1$ and prove
$$x^{\frac{7}{3}}y^{\frac{1}{3}}z^{\frac{1}{3}} + x^{\frac{1}{3}}y^{\frac{7}{3}}z^{\frac{1}{3}} + x^{\frac{1}{3}}y^{\frac{1}{3}}z^{\frac{7}{3}} \leq x^3 + y^3 + z^3$$
with Muirhead ( $(3, 0, 0)$ majorizes $(\frac{7}{3}, \frac{1}{3}, \frac{1}{3})$).
I'm curious if there's a way to prove this without machinery of Muirhead's inequality and majorization. Also, this approach readily generalizes to proving
$$x^n + y^n + z^n \leq x^{n+1} + y^{n+1} + z^{n+1}$$
for non-negative $x, y, z$ such that $xyz = 1$.
Is there a way to prove this generalization without Muirhead?
|
In general, using the power mean inequality,
$$\sqrt[n+1]{\frac{x^{n+1} + y^{n+1} + z^{n+1}}{3}} \ge \sqrt[n]{\frac{x^n + y^n + z^n}{3}}$$
Raising both sides to the same power,
$$\frac{x^{n+1} + y^{n+1} + z^{n+1}}{3} \ge \frac{x^n + y^n + z^n}{3} \cdot \sqrt[n]{\frac{x^n + y^n + z^n}{3}}$$
But, again by the power mean inequality,
$$\sqrt[n]{\frac{x^n + y^n + z^n}{3}} \ge \sqrt[3]{xyz} = \sqrt[3]{1} = 1$$
Hence,
$$x^{n+1} + y^{n+1} + z^{n+1} \ge x^n + y^n + z^n$$
With equality at $x=y=z=1$.
|
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Sum of $1+\frac{1}{2}+\frac{1\cdot2}{2\cdot5}+\frac{1\cdot 2\cdot 3}{2\cdot 5\cdot 8}+\cdots$ I am trying to find out the sum of this
$$1+\frac{1}{2}+\frac{1\cdot2}{2\cdot5}+\frac{1\cdot 2\cdot 3}{2\cdot 5\cdot 8}+\frac{1\cdot 2\cdot 3\cdot 4}{2\cdot 5\cdot 8\cdot 11}+\cdots$$.
I tried with binomial theorem with rational index. But in vain. What shall I do to solve it?
If it has been solved already, kindly provide me the link. I am unable to get that.
Thanks for your help
|
Here is an another approach that doesn't add anything much to Jack's answer but is at least mildly entertaining. It does not use the $B$ function but instead takes a detour via a "simple" alternating series. As stated in the comments the sum equals $${}_2F_1(1,1,\tfrac{2}{3};\tfrac{1}{3})$$ and using the transformation $${}_2F_1(\alpha, \beta, \gamma; z) = (1-z)^{-\alpha}{}_2F_1(\alpha, \gamma - \beta, \gamma; \tfrac{z}{z-1})$$ we get:
$$
\begin{eqnarray}
{}_2F_1(1,1,\tfrac{2}{3};\tfrac{1}{3}) &=& \tfrac{3}{2}{}_2F_1(1, -\tfrac{1}{3}, \tfrac{2}{3}; -\tfrac{1}{2})\\[1ex]
&=& \tfrac{3}{2} + \sum_{n=0}^{\infty}(-1)^n\frac{3}{2^{n+2}(3n+2)}\\
&=&\tfrac{3}{2} + 2^{-\frac{4}{3}}\int_0^{2^{-\frac{1}{3}}}\frac{3x}{1+x^3}dx\\[1ex]
&=&\tfrac{3}{2} + \tfrac{3}{2} \int_0^1 \frac{x}{2+x^3}dx\\[1ex]
\end{eqnarray}$$
|
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How prove this inequality $(1+\frac{1}{16})^{16}<\frac{8}{3}$
show that
$$(1+\dfrac{1}{16})^{16}<\dfrac{8}{3}$$
it's well know that
$$(1+\dfrac{1}{n})^n<e$$
so
$$(1+\dfrac{1}{16})^{16}<e$$
But I found this $e=2.718>\dfrac{8}{3}=2.6666\cdots$
so how to prove this inequality by hand?
Thank you everyone solve it,I want see don't use $e=2.718$,because a most middle stundent don't know this value.
before I have use this well know
$$(1+\dfrac{1}{2n+1})(1+\dfrac{1}{n})^n<e$$
so
$$(1+\dfrac{1}{16})^{16}<e\cdot\dfrac{33}{34}\approx 2.638<\dfrac{8}{3}$$ to solve this, But Now we don't use $e=2.718$.
to prove this inequality by hand
|
We can split the series at index $m \in \mathbb{Z}$ where $0 \le m \le 15$:
$S = \Big(1+\dfrac{1}{16}\Big)^{16} = \displaystyle\sum\limits_{k=0}^{m}{{16 \choose k}\Big(\frac{1}{16}\Big)^k} + \displaystyle\sum\limits_{k=m+1}^{16}{{16 \choose k}\Big(\frac{1}{16}\Big)^k} \tag{1}$
Now let $a_k$ be the terms in the summation, and find the ratio between successive terms, so
$\dfrac{a_{k+1}}{a_k} = \dfrac{16!}{(k+1)!(16-k-1)!}\cdot\dfrac{1}{16^{k+1}}\cdot\dfrac{k!(16-k)!}{16!}\cdot 16^k = \dfrac{16-k}{k+1}\cdot\dfrac{1}{16}$
So for further ratios
$\dfrac{a_{k+p+1}}{a_{k+p}} = \dfrac{16-k-p}{k+p+1}\cdot\dfrac{1}{16} \le \dfrac{16-k}{k+1}\cdot\dfrac{1}{16}\ \ \text{ for }p \ge 0 $
and then $\dfrac{a_{m+p}}{a_m} \le \Big(\dfrac{16-m}{16(m+1)}\Big)^p\ \ (\forall p \in \mathbb{Z}_{\ge0}) \tag{2}$
Hence the last term in (1) can be bounded as
$\omega = \displaystyle\sum\limits_{k=m+1}^{16}{{16 \choose k}\Big(\frac{1}{16}\Big)^k} \le \displaystyle\sum\limits_{p=1}^{16-m}{\Big(\dfrac{16-m}{16(m+1)}\Big)^p{a_m}}$
To avoid the upper limit of summation, compare with a sum to infinity:
$\omega \le a_m\displaystyle\sum\limits_{p=1}^{\infty}{\Big(\dfrac{16-m}{16(m+1)}\Big)^p} = \dfrac{16-m}{17m}a_m \tag{3}$
So write $S$ as a sum of terms from $k=0$ to $k=m-1$ and a term encompassing $a_m$ and $\omega$:
$S = \displaystyle\sum\limits_{k=0}^{m-1}{{16 \choose k}\Big(\frac{1}{16}\Big)^k} + a_m\Big(1+\dfrac{\omega}{a_m}\Big) \le \Bigg[\displaystyle\sum\limits_{k=0}^{m-1}{{16 \choose k}\Big(\frac{1}{16}\Big)^k}\Bigg] + \dfrac{16}{17}\Big(1+\dfrac{1}{m}\Big){16 \choose m}\Big(\frac{1}{16}\Big)^m$
For $m=2$, we get
$S \le 1 + 1 + \dfrac{16}{17}\Big(1+\dfrac{1}{2}\Big){\dfrac{16\times15}{1\times2}}\Big(\dfrac{1}{16}\Big)^2 = 2 + \dfrac{45}{68} < 2 + \dfrac{2}{3} $
|
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Quadratic formula and factoring are leading to different answers $$x^{ 2 }-2x-15=0$$
By factoring, I get:
$$(x-5)(x+3)$$
Which has the solutions:
$$x=5, x=-3$$
However when I use the quadratic formula (which is what the book saids to use), I get
$$\frac { 2 \pm \sqrt { 4-(4\cdot1\cdot(-15)) } }{ -2 } =$$
$$\frac { 2\pm 8 }{ -2 } $$
Which I evaluate to be $$x=-5, x=3$$
Where am I going wrong?
|
The quadratic formula is
$$
x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}
$$
The denominator should be $2$ because
\[ a=1 \]
\[ b=-2 \]
\[ c=-15 \]
So then
\[
x=\frac{2\pm \sqrt{4-4(-15)}}{2}= \frac{2\pm \sqrt{64}}{2}= \frac{2\pm 8}{2}=1\pm 4
\]
Thus
$$ x=5 $$
$$ x=-3 $$
|
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|
solve the diophantine equation: $x^3-3xy^2=z^3$ Let $ x,y,z$ be 3 integers greater than 1,if $x$ and $y$ are relatively prime, solve the diophantine equation: $x^3-3xy^2=z^3$.
|
$$x^3-3xy^2=z^3 \Rightarrow x^3-z^3=3xy^2 \Rightarrow 3 \mid x^3-z^3 \Rightarrow x^3 \equiv z^3 \pmod 3$$
From Fermat's theorem:
$$x^3 \equiv x \pmod 3 \\ z^3 \equiv z \pmod 3$$
So,we have:
$$x \equiv \ z \pmod 3 \Rightarrow 3 \mid x-z \Rightarrow x=z+3k,k \in \mathbb{Z}$$
Replace at $x^3-3xy^2=z^3$ and take into consideration the fact that $(x,y)=1$.
|
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|
Dividing by $\sqrt n$ Why is the following equality true?
I know I should divide by $\sqrt n$ but how is it done exactly to get the RHS?
$$ \frac{\sqrt n}{\sqrt{n + \sqrt{n + \sqrt n}}} = \frac{1}{\sqrt{1 + \sqrt{\frac{1}{n} + \sqrt{\frac{1}{n^3}}}}}$$
|
Well this is calculus. As you said yourself, you divide the numerator and the denominator by $\sqrt{n}$
for the numerator it's easy, as $\frac{\sqrt{n}}{\sqrt{n}}=1$.
for the denominator you get \begin{equation}
\frac{1}{\sqrt{n}}\sqrt{n+{\sqrt{n+{\sqrt{n}}}}}=\sqrt{1+\frac{\sqrt{n+\sqrt{n}}}{n}}
\end{equation}
Keep in mind, if you want to pull a factor under a square root, you have to square it!
the above can be rewritten as \begin{equation}
\sqrt{1+\sqrt{\frac{n}{n^2}+\frac{\sqrt{n}}{n^2}}}= \sqrt{1+\sqrt{\frac{1}{n}+\sqrt{\frac{n}{n^4}}}}=\sqrt{1+\sqrt{\frac{1}{n}+\sqrt{\frac{1}{n^3}}}}
\end{equation}
And you're done
|
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|
Short form of few series Is there a short form for summation of following series?
$$\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\alpha^{2n}((2y-1)^{2k+1}+1)}{2^{2n+1}(2n)!k!(n-k)!(2k+1)}$$
$$\sum\limits_{n=0}^\infty\dfrac{\alpha^{2n+1}(\cos^{-1}(2y-1)-\pi)}{2^{4n+3}n!(n+1)!}$$
$$\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(k!)^2\alpha^{2n+1}(2y-1)(1-(2y-1)^2)^{k+\frac{1}{2}}}{2^{4n-2k+3}n!(n+1)!(2k+1)!}$$
Even sum of any combination of above terms could help. This is result of some integral. I guess they should contain Bessel and Struve functions.
In the case of $y=0.5$ it seems the sum of above terms to be
$$
-\frac{\pi}{2}\left[I_3(\frac{\alpha}{2}) +\frac{3}{\frac{\alpha}{2}}I_2(\frac{\alpha}{2})- I_{1}(\frac{\alpha}{2})-\frac{1}{8}\left[L_{-3}(\frac{\alpha}{2}) - L_{-1}(\frac{\alpha}{2}) - L_{1}(\frac{\alpha}{2})+L_{3}(\frac{\alpha}{2}) -\frac{2\alpha^{-2}}{\pi} -\frac{8}{3\pi}+\frac{2\alpha^2}{15\pi}\right]\right]
$$
|
For the first series, I was able to wittle it down to a sum of a Struve function and an unevaluated series of hypergeometric functions:
$$\begin{align}
S{(\alpha,y)}
&=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\alpha^{2n}((2y-1)^{2k+1}+1)}{2^{2n+1}(2n)!k!(n-k)!(2k+1)}\\
&=\sum\limits_{n=0}^\infty \frac{\alpha^{2n}}{2^{2n+1}(2n)!} \sum\limits_{k=0}^n (-1)^k \frac{n!}{k!(n-k)!} \dfrac{((2y-1)^{2k+1}+1)}{(2k+1)}\\
&=\sum\limits_{n=0}^\infty \frac{\alpha^{2n}}{2^{2n+1}(2n)!} \left[\sum\limits_{k=0}^n \dfrac{(-1)^k\binom{n}{k}}{(2k+1)} + \sum\limits_{k=0}^n \dfrac{(-1)^k\binom{n}{k}}{(2k+1)} (2y-1)^{2k+1}\right]\\
&=\sum\limits_{n=0}^\infty \frac{\alpha^{2n}}{2^{2n+1}(2n)!} \left[\frac{(2n)!!}{(2n+1)!!} + (2y-1)\,{_2F_1}{\left(\frac12,-n;\frac32;(2y-1)^2\right)}\right]\\
&=\sum\limits_{n=0}^\infty \frac{\alpha^{2n}}{2^{2n+1}(2n)!}\frac{(2n)!!}{(2n+1)!!} + (2y-1)\sum\limits_{n=0}^\infty \frac{\alpha^{2n}}{2^{2n+1}(2n)!} {_2F_1}{\left(\frac12,-n;\frac32;(2y-1)^2\right)}\\
&=\frac{\pi}{4}\operatorname{L}_{-1}{\left(\frac{\alpha}{2}\right)} + (2y-1)\sum\limits_{n=0}^\infty \frac{\alpha^{2n}}{2^{2n+1}(2n)!} {_2F_1}{\left(\frac12,-n;\frac32;(2y-1)^2\right)}.
\end{align}$$
|
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|
Splitting $\Phi_{15}$ in irreducible factors over $\mathbb{F}_7$ I have to split $\Phi_{15}$ in irreducible factors over the field $\mathbb{F}_7$. It has been a while that I did this kind of stuff, and to be the honest, I've never really understood this matter. I'd really be grateful if someone could show me how this works. I don't ask you to do this whole job. It would be better if I would get some subtle advice instead.
I use the following definition:
$$
\Phi_n(X) \quad = \quad \prod_{ord(\zeta) = n} (X - \zeta)
$$
|
Every such $\zeta$ can be written uniquely as $\zeta_1\zeta_2$ where $\mathrm{ord}(\zeta_1)=3$ and $\mathrm{ord}(\zeta_2)=5$. The cases where $\mathrm{ord}(\zeta_1)=3$ are $2,4$.
The $\zeta_2$ values are roots of $1+x+x^2+x^3+x^4=(x-\zeta_5)(x-\zeta_5^2)(x-\zeta_5^3)(x-\zeta_5^4)$. So
$$\begin{align}(x-2\zeta_5)(x-2\zeta_5^2)(x-2\zeta_5^3)(x-2\zeta_5^4)& = 2^4+2^3x+2^2x^2+2x^3+x^4 \\&= 2+x+4x^2+2x^3+x^4\end{align}$$
$$\begin{align}(x-4\zeta_5)(x-4\zeta_5^2)(x-4\zeta_5^3)(x-4\zeta_5^4)& = 4^4+4^3x+4^2x^2+4x^3+x^4 \\&= 4+x+2x^2+4x^3+x^4\end{align}$$
So, the products of these are $\Phi_{15}(x)$.
There are no roots for these, so the only possible factors can be quadratics.
The only possible option is: $(x-2\zeta_5)(x-2\zeta_5^4)=x^2-2(\zeta_5+\zeta_5^4)x+4$ and $(x-2\zeta_5^2)(x-2\zeta_5^3)$ for the first, and the related factoring for the second. But that only works if $\zeta_5+\zeta_5^4\in\mathbb F_7$, which we know algebraically requires $\sqrt{5}\in\mathbb F_7$, which is not the case. So:
$$\Phi_{15}(x)=(2+x+4x^2+2x^3+x^4)(4+x+2x^2+4x^3+x^4)$$
|
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|
Computing $\sum_{i=0}^{\infty}\frac{i}{2^{i+1}}$ I came across this while trying to solve Google's boys & girls problem, and although I know now it's not the right approach to take, I'm still interested in summing
$\sum_{i=0}^{\infty}\frac{i}{2^{i+1}}$. Apparently it should be 1..but I'm having a tough time seeing this, especially since $\sum_{i=0}^{\infty}\frac{1}{2^{i+1}}=1$. I know it's a little elementary.. but I just can't figure out where I'm going wrong..
$$\sum_{i=0}^{\infty}\frac{i}{2^{i+1}}
=\frac{1}{2}\sum_{i=0}^{\infty}\frac{i}{2^{i}}
=\frac{1}{2}\sum_{j=0}^{\infty}\sum_{i=j}^{\infty}\frac{1}{2^{i}}
=\frac{1}{2}\sum_{j=0}^{\infty}\left(\sum_{k=0}^{\infty}\frac{1}{2^{k}} - \sum_{i=0}^{j-1}\frac{1}{2^{i}}\right)
=\frac{1}{2}\sum_{j=0}^{\infty}\left(2 - \frac{1-\frac{1}{2^j}}{1/2}\right)$$
$$=\frac{1}{2}\sum_{j=0}^{\infty}\left(2\frac{1}{2^j}\right)=2 \ne1$$
|
$$\begin{align}
\frac{1}{4}&=1-\frac{3}{4}\\
\frac{1}{4} + \frac{2}{8} &= 1-\frac{4}{8}\\
\frac{1}{4}+\frac{2}{8} +\frac{3}{16}&= 1-\frac{5}{16}
\end{align}
$$
Therefore, try to show that $\frac{n}{2^{n+1}} = \frac{n+1}{2^n}-\frac{n+2}{2^{n+1}}$. That's pretty easy, so this is a telescoping sum:
$$\left(\frac{1}{1}-\frac{2}{2}\right) + \left(\frac{2}{2}-\frac{3}{4}\right)+\left(\frac{3}{4}-\frac{4}{8}\right)+\left(\frac 48-\frac5{16}\right)\dots$$
|
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|
Calculus (Integration) Is there a simple way to integrate $\displaystyle\int\limits_{0}^{1/2}\dfrac{4}{1+4t^2}\,dt$
I have no idea how to go about doing this. The fraction in the denominator is what's confusing me. I tried U-Substitution to no avail.
|
$$\int_0^{\frac{1}{2}} \frac{4}{1+4t^2} dt$$
We set $t=\frac{\tan{u}}{2}$ we have the following:
$t=0: u=0$
$t=\frac{1}{2}: u=\frac{\pi}{4}$
$dt=\frac{1}{2 \cos^2{u}}du$
$$\frac{4}{1+4t^2}=\frac{4}{1+4 \frac{\tan^2{u}}{4}}=\frac{4}{1+\tan^2{u}}=\frac{4 \cos^2{u}}{\sin^2{u}+\cos^2{u}}=4 \cos^2{u}$$
Therefore, we have the following:
$$\int_0^{\frac{1}{2}} \frac{4}{1+4t^2} dt=\int_0^{\frac{\pi}{4}}2 du=2\frac{\pi}{4}=\frac{\pi}{2}$$
|
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|
Conditional probability: At least 3 kings given there are at least 2 kings in the hand of 13. My first "conditional probability" problem. Sorry for all the questions. My instructor doesn't make sense to the class.
A hand of 13 cards is to be dealt at random and without any replacement from an ordinary deck of playing cards.
Find the conditional probability that there are at least three kings in the hand given that the hand contains at least two kings.
A deep, good, and thorough explanation would be tremendously appreciated.
Thanks
|
There are $\binom{52}{13}$ ways to select 13 cards from 52 cards.
Consider how you would create a hand of 13 cards with the required number of kings.
There are $\binom{4}{n}$ ways of choosing the required number of kings.
There are then $\binom{48}{13-n}$ ways of choosing the remaining (13 - n) cards.
Let A be the event of getting at least 2 kings.
This can be thought of as 3 separate events corresponding to 2, 3 or 4 kings.
By the addition principle:
Number of ways of getting A = $\binom{4}{2}$ * $\binom{48}{13-2}$ + $\binom{4}{3}$ * $\binom{48}{13-3}$ + $\binom{4}{4}$ * $\binom{48}{13-4}$
So, $$P(A) = \frac{\binom{4}{2} * \binom{48}{13-2} + \binom{4}{3} * \binom{48}{13-3} + \binom{4}{4} * \binom{48}{13-4}}{\binom{52}{13}}$$
Let B be the event of getting at least 3 kings.
This can be thought of as 2 separate events corresponding to 3 or 4 kings.
By the addition principle:
$$P(B) = \frac{\binom{4}{3} * \binom{48}{13-3} + \binom{4}{4} * \binom{48}{13-4}}{\binom{52}{13}}$$
Finally, by the conditional probability formula:
$$P(B|A) = \frac{P(B \cap A)}{P(A)}= \frac{P(B)}{P(A)}$$
|
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|
Is my solution to this differential equation correct?
My answer is: $[(1+x^2)^3]y = \dfrac{(1+x^2)^3}3+C$
But this option is not given, so is it correct?
Thanks
|
$$\begin{align}
(1+x^2)\frac{dy}{dx}+6xy&=2x\\
(1+x^2)\frac{dy}{dx}&=2x(1-3y)\\
\int \frac{dy}{1-3y}&=\int\frac {2x}{1+x^2}dx\\
-\frac13\ln(1-3y)&=\ln(1+x^2)\\
(1-3y)(1+x^2)^3&=0\\
(1+x^2)^3&=3y(1+x^2)^3\\
1+3x^2+3x^4+x^6&=3y(1+x^2)^3\\
\frac 13+x^2+x^4+\frac{x^6}3&=y(1+x^2)^3\\
y(1+x^2)^3&=x^2+x^4+\frac{x^6}3+C
\end{align}$$
i.e. option (A) $\blacksquare$.
Check by differentiating:
$$\begin{align}
(1+x^2)^3\frac{dy}{dx}+6xy(1+2x^2)^2&=2x+4x^3+2x^5\\
&=2x(1+2x^2+x^4)\\
&=2x(1+x^2)^2\\
(1+x^2)\frac{dy}{dx}+6xy&=2x
\end{align}$$
which is the original equation, hence solution is correct.
|
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|
If $\sec \theta+\tan \theta= \sqrt{3}$ then the positive value of $\sin \theta$ If $\sec \theta+\tan \theta=\sqrt{3}$ then the positive value of $\sin \theta$
Note:
$1/\cos\theta+\sin\theta/\cos \theta=\sqrt{3}$
$\sin\theta=\sqrt{3}\cos \theta-1$
squaring on both sides we get
$\sin^2\theta=$
|
Beginning with $1+\sin \theta = \sqrt{3} \cos \theta$, we can use the identity $\sin^2 \theta + \cos^2 \theta = 1$ to obtain $$1+\sin \theta = \sqrt{3(1-\sin^2\theta)} \\ \Rightarrow 1 + 2\sin \theta + \sin^2 \theta = 3-3 \sin^2 \theta \\ \Rightarrow 4\sin^2 \theta + 2 \sin \theta - 2 = 0$$ Now you have an quadratic equation for $\sin \theta$ which you can solve by factoring so: $$4\sin^2 \theta + 2 \sin \theta - 2 = 0 \\ \Rightarrow 4(\sin \theta + 1)(\sin \theta - \frac{1}{2})=0 \\ \Rightarrow \sin \theta = -1, \sin \theta = \frac{1}{2}$$ Since we want the positive value, we conclude that $\sin \theta = 1/2$
|
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|
Linear independence of matrices $I, A, A^2$ I want to prove that $I,A,A^2\:$matrices $\in M_{2\times 2}$ are $\textit {linearly independent}$.
I consider the following matrices and their "corresponding" vectors:
$I=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\to (1,0,0,1)$
$A=\begin{pmatrix} a & b \\ c & d \end{pmatrix}\to (a,b,c,d)$
$A^2=\begin{pmatrix} a^2+ b c & ab+db \\ ac+dc & d^2+bc\end{pmatrix}\to (a^2+bc, ab+db, ac+dc,d^2+bc)$
What I am trying to do, is to prove that these vectors are $\textit{linearly independent}$.
For that direction, I want to prove that (at least) one $3\times 3$ determinant is $\neq 0$.
It seems that all $3\times 3$ determinants $\left( e.g. \begin{vmatrix} 1 & 0 & 0\\ a & b & c\\ a^2+bc & ab+db & ac+dc\end{vmatrix} \right)$ are equal to zero! What am I really missing?
|
The thing you're trying to prove is not true. Consider the case $A = I$.
More generally, if $A$ is any $2 \times 2$ matrix, then you can compute its characteristic polynomial, $c(x)$, the determinant of $A - xI$. That will be some quadratic polynomial in $x$, like
$$
3x^2 - 22 x^1 + 8x^0.
$$
It turns out that if you plug in $A$ for $x$ , and treat the $A^0$ as $I$, you'll get a $2 \times 2$ matrix...that's the zero matrix! That's called the Cayley-Hamilton theorem.
As an example:
$$
A = \begin{bmatrix}1 & 3 \\ -1 & 1 \end{bmatrix} \\
A- xI = \begin{bmatrix}1-x & 3 \\ -1 & 1-x \end{bmatrix} \\
c(x) = det(A - xI) = (1-x)^2 + 3 = x^2 -2x + 4
$$
Now plug $A$ into $c$ to get
$$
c(A) = A^2 - 2A + 4I \\
= \begin{bmatrix} -2 & 6 \\ -2 & -2 \end{bmatrix} - \begin{bmatrix}2 & 6 \\ -2 & 2 \end{bmatrix} + \begin{bmatrix}4 & 0 \\ 0 & 4 \end{bmatrix} \\
= \begin{bmatrix}0 & 0 \\ 0 & 0 \end{bmatrix}
$$
|
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|
How to simplify $\sqrt{\sqrt{5}+1} \cdot \sqrt{\sqrt{5}-1}$? This is the original problem:
$\sqrt{\sqrt{5}+1} \cdot \sqrt{\sqrt{5}-1} = x$.
I'm really confused about how to solve this problem, I come as far as saying this: $\sqrt[4]{5} + \sqrt{1}\cdot \sqrt[4]{5}-\sqrt{1}$.
|
$$\sqrt{\sqrt{5}+1}\sqrt{\sqrt{5}-1}=\sqrt{(\sqrt{5}+1)(\sqrt{5}-1)}$$ $$=\sqrt{5-\sqrt{5}+\sqrt{5}-1}$$ $$=\sqrt{5-1}$$ $$=\sqrt{4}$$ $$=2$$
|
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|
Determine the irrational numbers $x$ such that both $x^2+2x$ and $x^3-6x$ are rational numbers I did not make any progress. The problem is from RMC 2008.
The only idea that I have is:
Try to find sets of irrational numbers such that every number in the set multiplied by another number in the set yields a rational number.
|
Let $x$ be an irrational number with
$$x^{2} + 2x = \frac{p_{1}}{q_{1}},\ x^{3} - 6x = \frac{p_{2}}{q_{2}}$$
for some integers $p_{1}, q_{1}, p_{2}, q_{2}$ such that $q_{1}, q_{2} \neq 0$ and $(p_{1}, q_{1}) = (p_{2}, q_{2}) = 1.$
Then we have
$$2x^{2} - (6 + \frac{p_{1}}{q_{1}})x + \frac{p_{2}}{q_{2}} = 0,$$
so that
$$x = \frac{6 + \dfrac{p_{1}}{q_{1}} \pm \sqrt{(6 + \dfrac{p_{1}}{q_{1}})^{2} - \dfrac{8p_{2}}{q_{2}}}}{4},$$
whence if
$$(6 + \dfrac{p_{1}}{q_{1}})^{2} - \dfrac{8p_{2}}{q_{2}} > 0$$
and is not a perfect square,
then $x$ is irrational.
|
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|
Inequality: $x^2+y^2+xy\ge 0$ I want to prove that $x^2+y^2+xy\ge 0$ for all $x,y\in \mathbb{R}$.
My "proof": Suppose wlog that $x\ge y$, so $x^2\cdot x\ge x^2\cdot y\ge y^2\cdot y=y^3$ (because $x^2\ge 0$ so we can multiply both sides by it without changing the inequality sign) giving $x^3\ge y^3$. Substracting we have $x^3-y^3\ge 0$ or $(x-y)(x^2+xy+y^2)\ge 0$. Since $x\ge y$ we can divide by $x-y$ to get $x^2+xy+y^2\ge 0$.
Is it right?
Thanks for your help!
|
A homogeneous bivariate second-degree polynomial with a negative discriminant can be only everywhere non-negative or everywhere non-positive over $\mathbb{R}^2$. Since in $(x,y)=(1,1)$ the inequality holds as $\geq$, for every $(x,y)\in\mathbb{R}^2$ we have $x^2+xy+y^2\geq 0$.
|
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|
How prove this inequality with $x+y+z=1$
let $x,y,z>0$,and such $$x+y+z=1$$
show that:$$\dfrac{(1+xy+yz+xz)(1+3x^2+3y^2+3z^2)}{9(x+y)(y+z)(x+z)}
\ge \left(\dfrac{x\sqrt{1+x}}{\sqrt[4]{3+9x^2}}+\dfrac{y\sqrt{1+y}}{\sqrt[4]{3+9y^2}}
+\dfrac{z\sqrt{1+z}}{\sqrt[4]{3+9z^2}}\right)^2$$
My idea: maybe we can prove $LHS\ge 1,RHS\le\dfrac{2}{3}?$
(because when $x=y=z=\dfrac{1}{3},LHS=1,RHS=\dfrac{2}{3}$)
or
$$(1+xy+yz+xz)[1+3(x^2+y^2+z^2)]\ge 9(x+y)(y+z)(x+z)$$
and $$\left(\dfrac{x\sqrt{1+x}}{\sqrt[4]{3+9x^2}}+\dfrac{y\sqrt{1+y}}{\sqrt[4]{3+9y^2}}
+\dfrac{z\sqrt{1+z}}{\sqrt[4]{3+9z^2}}\right)^2\le\dfrac{2}{3}?$$
can you use computer test it?
I know this well know inequality
$$9(x+y)(y+z)(x+z)\ge 8(x+y+z)(xy+yz+xz)=8(xy+yz+xz)$$
|
We can show LHS $\ge 1 \iff $
$$ (1+xy+yz+xz)[1+3(x^2+y^2+z^2)]\ge 9(x+y)(y+z)(x+z) $$
Let $p = x+y+z = 1, q = xy+yz+zx \le \frac13, r = xyz$. Then we have the inequality as
$$(1+q)[1+3(p^2-2q)] \ge 9(pq-r) \iff 4+9r \ge 11q + 6q^2$$
By Schur $p^3+9r \ge 4pq \implies 1+9r \ge 4q$, so it is enough to show that $3 \ge 7q + 6q^2$ which follows from $q \le \frac13$.
For the second part, i.e. RHS $\le \frac23$, it is sufficient to note that
$$ \frac{x\sqrt{1+x}}{\sqrt[4]{3+9x^2}} \le \sqrt{\frac23}x \iff (3x-1)^2 \ge 0$$
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How do I prove that any unit fraction can be represented as the sum of two other distinct unit fractions? A number of the form $\frac{1}{n}$, where $n$ is an integer greater than $1$, is called a unit fraction.
Noting that
$\frac{1}{2} = \frac{1}{3} + \frac{1}{6}$
and
$\frac{1}{3} = \frac{1}{4} + \frac{1}{12}$,
find a general result of the form
$\frac{1}{n} = \frac{1}{a} + \frac{1}{b}$
and hence prove that any unit fraction can be expressed as the sum of two other distinct unit
fractions.
|
In reply to DavidH's question "is the above decomposition of a unit fraction into a pair of distinct unit fractions unique?":
If $n$ is not prime, then we can write $n = n_1n_2$ with $n_1 \neq 1 \neq n_2$, and then we have the decomposition
$$\frac{1}{n} = \frac{1}{n_1}\frac{1}{n_2} = \frac{1}{n_1}\Big(\frac{1}{n_2+1} + \frac{1}{n_2(n_2+1)}\Big) = \frac{1}{n_1(n_2+1)} + \frac{1}{n_1n_2(n_2+1)} \\ = \frac{1}{n+n_1} + \frac{1}{n(n_2+1)}$$
which is different to the decomposition
$$\frac{1}{n} = \frac{1}{n+1} + \frac{1}{n(n+1)}$$
Now suppose $n$ is prime. Suppose
$$\frac{1}{n} = \frac{1}{n+a} + \frac{1}{c}$$
Assume also, wlog, that $n + a < c$, which tells us that $a < n$. Then
$$\frac{1}{c} = \frac{1}{n} - \frac{1}{n+a} = \frac{a}{n(n+a)}$$
But $n$ is prime, so $a$ and $n$ are coprime, as are $a$ and $a + n$, so we must have $a = 1$ and $c = n(n+1)$, i.e. the decomposition
$$\frac{1}{n} = \frac{1}{n+1} + \frac{1}{n(n+1)}$$
is unique when $n$ is prime.
|
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Complex analysis: Rewrite $\cos^{-1}{i}$ in algebraic form I'm stuck in this problem (complex analysis), my answer is not the one reported in the book:
Rewrite $\cos^{-1}{i}$ in the algebraic form. A: $k\pi + i \frac{\ln{2}}{2}\ \forall\ k \in \mathbb{Z}$
So I tried this approach in order to solve it:
*
*As $\cos^{-1}{z} = -i \ln{\left( z \pm \sqrt{z^2 - 1} \right)}$, doing $z = i$ and $i^2 = -1$ where necessary, we have
$$\cos^{-1}{i} = -i \ln{\left( i \pm \sqrt{i^2 - 1} \right)} = -i \ln{\left( i \pm \sqrt{-2} \right)} = -i \ln{\left( i \pm i \sqrt{2} \right)}$$
*
*Factoring $i$ and separating the obtained logarithm of product into sum of logarithms:
$$\cos^{-1}{i} = -i \ln{\left[ i \left( 1 \pm \sqrt{2} \right) \right]} = -i \big[ \ln{i} + \ln{\left( 1 \pm \sqrt{2} \right)} \big]$$
*
*Solving $\ln{i}$ separately, we obtain $\ln{i} = \pi i \left( 2k + \frac{1}{2} \right)$
*Solving $\ln{\left( 1 \pm \sqrt{2} \right)}$, on the other hand, yields
$$\ln{\left( 1 \pm \sqrt{2} \right)} = \begin{cases} \ln{\left( \sqrt{2} + 1 \right)} + \pi i \cdot 2k & (+) \\ \ln{\left( \sqrt{2} - 1 \right)} + \pi i \left( 2k + 1 \right) & (-) \end{cases}$$
*
*Substituting these expressions into the original one, we have
$$\cos^{-1}{i} = \begin{cases} \pi \left( 4k + \frac{1}{2} \right) -i \ln{\left( \sqrt{2} + 1 \right)} & (+) \\ \pi \left( 4k + \frac{3}{2} \right) -i \ln{\left( \sqrt{2} - 1 \right)} & (-) \end{cases}$$
which obviously doesn't correspond with the book. Is something wrong?
Thanks in advance :)
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The answer in the book is wrong.
|
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How does one simplify the expression $\sqrt[3]{2 \sqrt{2}}\left(\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\right)$ I don't know how to solve this problem. What I know is $\sqrt[3]{2 \sqrt{2}}=\sqrt{2}$. But I don't know how to continue.
|
$\sqrt[3]{2\sqrt 2}=\sqrt 2$ is correct. Since we have $$2\pm \sqrt 3=\frac{4\pm 2\sqrt 3}{2}=\frac{(\sqrt 3\pm 1)^2}{2}\Rightarrow \sqrt{2\pm \sqrt 3}=\frac{\sqrt 3\pm 1}{\sqrt 2},$$
we have
$$\sqrt 2\left(\sqrt{2-\sqrt 3}+\sqrt{2+\sqrt 3}\right)=\sqrt 2\left(\frac{\sqrt 3 -1}{\sqrt 2}+\frac{\sqrt 3+1}{\sqrt 2}\right)=\sqrt 3-1+\sqrt 3+1=2\sqrt 3.$$
|
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|
Ring Theory question in a GRE practice exam I have a question about a GRE practice problem relating to Rings. The question is as follows:
Suppose that two binary operations, denoted by $\oplus$ and $\odot$ , are defined on a nonempty set $S$, and that the
following conditions are satisfied for all $x, y$, and $z$ in $S$:
(1) $(x\oplus y)$ and $x \odot y$ are in S.
(2) $x\oplus (y \oplus z) = (x \oplus y) \oplus z$ and $x \odot (y \odot z) = (x \odot y) \odot z$.
(3) $x \oplus y = y \oplus x$
Also, for each $x$ in $S$ and for each positive integer $n$, the elements $nx$ and $xn$ are defined recursively as
follows:
$1x = x^1 = x$ and
if $kx$ and $x^k$ have been defined, then $(k + 1) x = kx \oplus x$ and $x^{k+1} = x^k \odot x$.
Which of the following must be true?
I) $(x \odot y)^n = x^n \odot y^n$ for all $x,y \in S$ and all positive integers $n$.
II) $n(x \oplus y) = nx \oplus ny$ for all $x,y \in S$ and all positive integers $n$.
III) $x^m \odot x^n = x^{m+n}$ for all $x \in S$ and all positive integers $n,m$.
A) I only, B) II only, C) III only, D) II and III, E) I, II and III
I recognize that the description given is the description of a ring. So I was pretty sure that the answer was E) I, II, III. It is pretty easy to show II and III, and I couldn't think of a counterexample for I. However, the solutions manual I purchased says the answer is D) II and III, claiming that the ring of integers does not have property I.
So my question is, what is the darn counterexample? I can't think of any. I tried $x=1$ and $y=-1$ with both odd and even powers, and even tried $x=1$, $y=2$ and $x=-1$, $y=2$, still nothing. Any suggestions? Thanks in advance.
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Here's a counterexample for (I). $\mathrm{Mat}_{2 \times 2}(\mathbb{R})$ equipped with matrix addition and multiplication satisfies all of the above criteria, and we consider the matrices $$A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \: \: B = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$$
For $n = 2$, we have that $$(AB)^2 = \left(\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\right)^2 = \begin{pmatrix} 1 & -1 \\ 0 & -1 \end{pmatrix}^2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$
but $$A^2 B^2 = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^2 \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}^2 = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$$
So, $A^2B^2 \ne (AB)^2$, and therefore this is a counterexample for (I).
|
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Discrete Math sequence question The question is find $a3$:
$a_0 = 2, a_1 = 4$ and $a_{k+2} = 3a_{k+1}-a_k$ for any integer $k \geq 0 $
I know the answer is 26, although how do you get the answer?
|
$ a_0 =2 , a_1 = 4, a_{k+2} - 3a_{k+1} +a_k = 0 $
We can suggest an equation that $ x^2 -3x +1 = 0 $,
Let the roots of the equation are $ x_1 , x_2$
then $x_1 +x_2 = 3$, $x_1 x_2 = 1$
then $a_{k+2} - 3a_{k+1} +a_k = 0 $$ \Leftrightarrow $ $a_{k+2} - (x_1 +x_2 )a_{k+1} + x_1 x_2 a_k = 0$
$\Leftrightarrow$ $ a_{k+2} - x_1 a_{k+1} = x_2 ( a_{k+1} - x_1 a_k) $
$\Rightarrow$ $a_{k+1} - x_1 a_k = x_2 (a_k - x_1 a_{k-1} )$
$\Rightarrow$ $a_{k+2} - x_1 a_{k+1} = (x_2)^2 (a_{k} - x_1 a_{k-1}) $
= $ \dots$ = $(x_2)^k (a_2 - x_1 a_1)$
also, $ a_{k+2} - x_2 a_{k+1} = (x_1)^k (a_2 - x_2 a_1) $
then $ (x_2 - x_1)a_{k+1} = (x_1)^k (a_2 -x_2 a_1) - (x_2)^k (a_2 -x_1 a_1)$
then $a_{k+1} = \frac{(x_1)^k (a_2 -x_2 a_1) - (x_2)^k (a_2 -x_1 a_1)}{x_2 - x_1}$
$ \therefore a_k = \frac{(x_1)^{k-1} (a_2 -x_2 a_1) - (x_2)^{k-1} (a_2 -x_1 a_1)}{x_2 - x_1} $
|
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Matrix Algebraic Operations, If AA = AB, does A = B? A and B are 2 x 2 matrices and A is not a zero matrix. How is the following proof incorrect?
Since AA = AB, AA - AB = 0
A (A - B) = 0 and since A does not equal zero, then A - B = 0, therefore A = B.
Thank you
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$$\begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix}\begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix}=\underbrace{\begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix}}_{=A}\underbrace{\begin{pmatrix}3 & 5 \\ 0 & 0 \end{pmatrix}}_{=B}$$
EDIT Here's an example where the product is not $0$.
$$\begin{pmatrix}0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix}0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}=\underbrace{\begin{pmatrix}0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}}_{=A}\underbrace{\begin{pmatrix}2 & 3 & 4 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}}_{=B}$$
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Other ways to evaluate $\lim_{x \to 0} \frac 1x \left [ \sqrt[3]{\frac{1 - \sqrt{1 - x}}{\sqrt{1 + x} - 1}} - 1\right ]$? Using the facts that:
$$\begin{align}
\sqrt{1 + x} &= 1 + x/2 - x^2/8 + \mathcal{o}(x^2)\\
\sqrt{1 - x} &= 1 - x/2 - x^2/8 + \mathcal{o}(x^2)\\
\sqrt[3]{1 + x} &= 1 + x/3 + \mathcal{o}(x)
\end{align}$$
I was able to evaluate the limit as follows:
$$\begin{align}
\lim_{x \to 0} \frac 1x \left [ \sqrt[3]{\frac{1 - \sqrt{1 - x}}{\sqrt{1 + x} - 1}} - 1\right ] &\sim \lim_{x \to 0} \frac 1x \left [ \sqrt[3]{\frac{\dfrac x2 + \dfrac{x^2} 8}{\dfrac x2 + \dfrac{x^2} 8}} - 1\right ] =\\
&= \lim_{x \to 0} \frac 1x \left [ \sqrt[3]{1 + \frac{2x^2}{4x - x^2}} - 1\right ] \sim\\
&\sim \lim_{x \to 0} \frac{2x^2}{12x^2 - 3x^3} = \frac 16
\end{align}$$
What are other ways to evaluate it? Maybe pure algebraically? I tried to rationalize the denominator, but got stuck at some point...
|
Let us start by noting that:
*
*$\sqrt[3]{1-\sqrt{1-x}}\cdot \sqrt[3]{1+\sqrt{1-x}}=\sqrt[3]{x}$
*$\sqrt[3]{\sqrt{1+x}-1}\cdot \sqrt[3]{\sqrt{1+x}+1}=\sqrt[3]{x}$
*$$\sqrt[3]{\frac{(1-\sqrt{1-x})(1+\sqrt{1-x})}{(\sqrt{1+x}-1)(\sqrt{1+x}+1)}}=1$$
Using these algebraic facts, we can write that: $$\begin{align} \lim_{x\to 0} \frac{1}{x} \cdot \Biggl[\sqrt[3]{\frac{1-\sqrt{1-x}}{\sqrt{1+x}-1}}-1\Biggl]
&= \lim_{x\to 0}\frac{1}{x} \cdot \Biggl[\sqrt[3]{\frac{1-\sqrt{1-x}}{\sqrt{1+x}-1}}-\sqrt[3]{\frac{(1-\sqrt{1-x})(1+\sqrt{1-x})}{(\sqrt{1+x}-1)(\sqrt{1+x}+1)}}\Biggl] \\
&= \lim_{x\to 0}\frac{1}{x} \cdot \Biggl[\sqrt[3]{\frac{(1-\sqrt{1-x})(\sqrt{1+x}+1)}{(\sqrt{1+x}-1)(\sqrt{1+x}+1)}}-\sqrt[3]{\frac{(1-\sqrt{1-x})(1+\sqrt{1-x})}{(\sqrt{1+x}-1)(\sqrt{1+x}+1)}}\Biggl] \\
&= \lim_{x\to 0}\Biggl[\frac{\sqrt[3]{\sqrt{1+x}+1}-\sqrt[3]{1+\sqrt{1-x}}}{x}\Biggl] \cdot \Biggl[\sqrt[3]{\frac{1-\sqrt{1-x}}{x}}\Biggl] \\
&= \lim_{x\to 0}\frac{\sqrt[3]{\sqrt{1+x}+1}-\sqrt[3]{1+\sqrt{1-x}}}{x} \cdot \lim_{x\to 0} \sqrt[3]{\frac{1-\sqrt{1-x}}{x}} \\ \end{align}$$
If we set $-\sqrt{1-x}$ and $\sqrt[3]{\sqrt{1+x}+1}$ to be $f(x)$ and $g(x)$ respectively, it can be written that: $$\begin{align} \lim_{x\to 0} \frac{g(x)-g(-x)}{x} \cdot\lim_{x\to 0}\sqrt[3]{\frac{f(x)-f(0)}{x}} \tag{1}\label{eq1}
&= \lim_{x\to 0} 2\Biggl[\frac{g(x)-g(-x)}{2x}\Biggl]\cdot \lim_{x\to 0}\sqrt[3]{\frac{f(x)-f(0)}{x}} \\
&= 2g'(0)\cdot \sqrt[3]{f'(0)} \tag{2}\label{eq2} \\
&= \Biggl[\frac{(\sqrt{1+x}+1)^{-\frac{2}{3}}}{3\sqrt{1+x}}\Biggl]_{x=0}\cdot \Biggl[\sqrt[3]{\frac{1}{2\sqrt{1-x}}}\Biggl]_{x=0} \tag{3}\label{eq3} \\
&= \frac{1}{3\sqrt[3]{4}}\cdot \frac{1}{\sqrt[3]{2}} \\
&= \frac{1}{6} \\ \end{align}$$
$(2)$: limit definition of the derivative
Hope it helps!
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Formula for $\sum\cos(\pi kt)/(1+k^2)$ Is there an explicit formula for the sum
$$F = \sum_{k=0}^\infty \frac{1}{1+k^2} \cos(\pi k t)$$
This is the green function for the operator $1 + \Delta$ on the circle.
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If $0 \le t \le 2 $, $$\sum_{k=0}^{\infty} \frac{\cos (k \pi t) }{1+k^{2}} = \frac{\pi}{2} \frac{\cosh [\pi(t-1)]}{\sinh \pi} + \frac{1}{2}. $$
One way to evaluate series of the form $ \displaystyle \sum f(k) e^{ik \theta}$, where $0 \le \theta \le 2 \pi$ and $f(z)$ is rational function such that $f(z) = \mathcal{O}(z^{-2})$ as $|z| \to \infty,$ is to consider the function $$g(z) = f(z) e^{i \theta z} \ \frac{2 \pi i}{e^{2 \pi iz}-1}$$ and integrate around a square with vertices at $z = \pm (N + \frac{1}{2} ) \pm i (N + \frac{1}{2})$.
Like $ \pi \cot (\pi z)$, the function$ \displaystyle \frac{2 \pi i}{e^{2 \pi iz} -1}$ (or equivalently $\pi e^{-i \pi z} \csc (\pi z)$) has simple poles at the integers with residue $1$.
But unlike $\displaystyle \int f(z) e^{i \theta z} \ \pi \cot (\pi z) \ dz $, the integral $ \displaystyle \int f(z) e^{i \theta z} \frac{2 \pi i}{e^{2 \pi iz}-1} \ dz$ will vanish along the contour as $N \to \infty$ through the positive integers.
You can use the ML inequality to show this.
(A more subtle argument will show that the integral will vanish even if $f(z) = \mathcal{O}(z^{-1})$ as $|z|\to \infty$.)
So consider $$g(z) = \frac{e^{i \theta z}}{1+z^{2}} \frac{2 \pi i}{e^{2 \pi iz}-1}.$$
Then summing up the residues, $$ \sum_{k=-\infty}^{\infty} \text{Res}[g(z),k] + \text{Res}[g(z),i] + \text{Res}[g(z), -i] = 0$$
where $$\text{Res}[g(z),k] = \frac{e^{ik \theta}}{1+k^{2}}, $$
$$\text{Res}[g(z),i] = \lim_{z \to i} \frac{e^{i \theta z}}{z+i} \frac{2 \pi i}{e^{2 \pi iz}-1}=\frac{\pi e^{-\theta}}{e^{-2 \pi}-1}, $$
and $$\text{Res}[g(z),-i] = \lim_{z \to -i} \frac{e^{i \theta z}}{z-i} \frac{2 \pi i}{e^{2 \pi iz}-1}= -\frac{\pi e^{\theta}}{e^{2 \pi}-1}. $$
But $$ \sum_{k=-\infty}^{\infty} \frac{e^{ i k \theta}}{1+k^{2}} = 2 \sum_{k=1}^{\infty} \frac{\cos (k \theta)}{1+k^{2}} + 1 =2 \sum_{k=0}^{\infty} \frac{\cos (k \theta)}{1+k^{2}} - 1.$$
Therefore, $$ \begin{align} \sum_{k=0}^{\infty} \frac{\cos (k \theta)}{1+k^{2}} &= \frac{\pi}{2} \left(\frac{e^{\theta}}{e^{2 \pi}-1} - \frac{e^{-\theta}}{e^{-2 \pi}-1} \right) + \frac{1}{2} \\ &= \frac{\pi}{2} \left(\frac{e^{\theta - \pi }}{e^{\pi}-e^{- \pi}} + \frac{e^{-\theta + \pi}}{e^{\pi}- e^{- \pi}} \right) + \frac{1}{2} \\ &= \frac{\pi}{2} \frac{\cosh (\theta - \pi)}{\sinh \pi} + \frac{1}{2}. \end{align} $$
Finally replace $\theta$ with $\pi t$.
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finding solutions by factoring How would you find the integer solutions to $a^2-b^2=16$?
I know that the factors of $16$ are $8*2,$ $4*4,$ and $16*1.$ How would I use this?
I know that $a^2-b^2=(a+b)(a-b)=16,$ but how would you find to the solutions to it? I know that you can plug in numbers to find the answers, but is there another way of doing this?
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We wish to find integer solutions for $a$ and $b$ such that $a^2-b^2=16$.
Factoring the expression we obtain $(a+b)(a-b)=16$.
This means that $(a+b)|16$ and $(a-b)|16$.
From this we can find a set $(a+b)\in \{-16,-8,-4,-2,-1,1,2,4,8,16\}$.
So we obtain the equations
$a+b=-16 \implies a-b=-1$
$a+b=-8 \implies a-b=-2$
$a+b=-4 \implies a-b =-4$
$a+b =-2 \implies a-b =-8$
$a+b = -1 \implies a-b=-16$
$a+b= 1 \implies a-b=16$
$a+b=2 \implies a-b = 8$
$a+b=4 \implies a-b = 4$
$a+b=8 \implies a-b = 2$
$a+b=16 \implies a-b=1$
Now for each of these there is a corresponding matrix (e.g. the first set of equations has been done below:
\begin{align*}
\left(\begin{array}{cc|c}
1 & 1 &-16\\
1 & -1 &-1\\
\end{array}\right)\end{align*}
Which we may factor into row echelon form
\begin{align*}
\left(\begin{array}{cc|c}
1 & 1 &-16\\
0 & -2 & 15\\
\end{array}\right)\end{align*}
To get the solution of $b=-7.5$ and $a=-8.5$.
This is not an integer solution so we may discard it. Now continue this process for the remaining 9 pairs of equations.
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Sum of the infinite series $\frac16+\frac{5}{6\cdot 12} + \frac{5\cdot8}{6\cdot12\cdot18} + \dots$ We can find the sum of infinite geometric series but I am stuck on this problem.
Find the sum of the following infinite series:
$$\frac16+\frac{5}{6\cdot 12} + \frac{5\cdot8}{6\cdot12\cdot18} + \frac{5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}+\dots$$
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$$\frac16+\frac{5}{6\cdot12}+\frac{5\cdot8}{6\cdot12\cdot18}+\frac{5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}+\dots=$$
$$=\frac12\cdot\bigg[\frac26+\frac{2\cdot5}{6\cdot12}+\frac{2\cdot5\cdot8}{6\cdot12\cdot18}+\frac{2\cdot5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}+\dots\bigg]=$$
$$=\frac12\cdot\bigg[\frac{(3-1)}{(6\cdot1)}+\frac{(3-1)\cdot(6-1)}{(6\cdot1)\cdot(6\cdot2)}+\frac{(3-1)\cdot(6-1)\cdot(9-1)}{(6\cdot1)\cdot(6\cdot2)\cdot(6\cdot3)}+\dots\bigg]=$$
$$=\frac12\cdot\bigg[\frac{(3\cdot1-1)}{(6\cdot1)}+\frac{(3\cdot1-1)\cdot(3\cdot2-1)}{(6\cdot1)\cdot(6\cdot2)}+\frac{(3\cdot1-1)\cdot(3\cdot2-1)\cdot(3\cdot3-1)}{(6\cdot1)\cdot(6\cdot2)\cdot(6\cdot3)}+\dots\bigg]=$$
$$=\frac12\cdot\sum_{n=1}^\infty\frac{\displaystyle\prod_{k=1}^n(3k-1)}{6^n\cdot n!}=\frac12\cdot\sum_{n=1}^\infty\frac{\displaystyle\prod_{k=1}^n\bigg(k-\frac13\bigg)}{2^n\cdot n!}=\frac12\cdot\sum_{n=1}^\infty\frac{\displaystyle\bigg(-\frac13\bigg){\Large!}\cdot\prod_{k=1}^n\bigg(k-\frac13\bigg)}{\bigg(-\dfrac13\bigg){\Large!}\cdot2^n\cdot n!}=$$
$$=\frac12\cdot\sum_{n=1}^\infty\frac{\displaystyle\bigg(n-\frac13\bigg){\Large!}}{\bigg(-\dfrac13\bigg){\Large!}\cdot2^n\cdot n!}=\frac12\cdot\sum_{n=1}^\infty\frac{\displaystyle\bigg(n-\frac13\bigg){\Large!}}{\bigg(-\dfrac13\bigg){\Large!}\cdot n!}\cdot\bigg(\frac12\bigg)^n=\frac12\cdot\sum_{n=1}^\infty{n-\frac13\choose n}\bigg(\frac12\bigg)^n$$
$$=\frac12\cdot\sum_{n=1}^\infty{\frac13-1\choose n}\bigg(-\frac12\bigg)^n=\frac12\cdot\bigg[-1+\sum_{n={\color{red}0}}^\infty{-\frac23\choose n}\bigg(-\frac12\bigg)^n\bigg]=$$
$$=\frac12\cdot\bigg[-1+\bigg(1-\frac12\bigg)^{^{-\tfrac23}}\bigg]=-\frac12+\frac1{\sqrt[3]2}.$$
|
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|
Modular calculus and square I want to prove that $4m^2+1$ and $4m^2+5m+4$ are coprimes and also $4m^2+1$ and $4k^2+1$ when $k\neq{m}$ and $4m^2+5m+4$ and $4k^2+5k+4$ when $k\neq{m}$. Firstly : Let $d|4m^2+1$ and $d|4m^2+5m+4$ then $d|4m^2+5m+4-(4m^2+1)=5m+3$ and $d|5m^2+3m$ thus $d|5m^2+3m-(4m^2+5m+4)=m^2-2m-4$ and $d|4m^2-8m-16$ or $d|4m^2+5m+4-(4m^2-8m-16)=13m+20$ but $d|15m+9$ thus $d|15m+9-(13m+20)=2m-11$ but $d|13m+20-(5m+3)=8m+17$ and $d|8m-88$ therefore $d|8m+17-(8m-88)=105=5.3.7$. So $4m^2+1=105k$ or $4m^2=104+105k'$ but $k'=4k''$ then $m^2=26+105k''$ or $4m^2+1=21k$ or $4m^2+1=35k$ or $4m^2+1=15k$ or $4m^2+1=5k$ or 3k or 7k. Are there solutions to these equations, please, or can I conclude that the two numbers are coprimes ?
|
You have an error in your calculation.
$8m-88$ should be $4(2m-11)=8m-44$. Then, $(8m+17)-(8m-44)=61$.
The two numbers are not always coprime. For example, for $m=36$,
$$4m^2+1=5185=85\cdot 61,\ \ 4m^2+5m+4=5368=88\cdot 61.$$
In general, for $m=61k-25$ where $k$ is a positive integer, the two numbers have $61$ as a common prime factor :
$$4m^2+1=61(244k^2-200k+41),\ \ \ 4m^2+5m+4=61(244k^2-195k+39).$$
Also, for $m\gt 0,k\gt 0,m\not=k$, since
$$4m^2+1=4k^2=1\iff 4(m-k)(m+k)=0,$$
we have $4m^2+1\not =4k^2+1$.
For $m\gt 0,k\gt 0,m\not=k$, since
$$4m^2+5m+4=4k^2+5k+4\iff (m-k)(4m+4k+5)=0,$$
we have $4m^2+5m+4\not=4k^2+5k+4$.
|
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|
Without using Taylor expansion or L'Hospital rule evaluate the limit:
$$L=\lim_{x\to0}\frac{e^x-1-x-x^2/2}{x^3}$$
I give the two alternate ways so that no one gives them again in their answer:
Using L'Hospital:
$$L=\lim_{x\to0}\frac{e^x-x-1}{3x^2}=\lim_{x\to0}\frac{e^x-1}{6x}=\lim_{x\to0}\frac{e^x}{6}=\frac16$$
Using Taylor:
$$L=\lim_{x\to0}\frac{\color{red}{(1+x+x^2/2+x^3/6+O(x^4))}-1-x-x^2/2}{x^3}=\frac16$$
Using nothing(not infact):
$$L=\lim_{x\to0}\frac{e^x-1-x^2/2}{x^3}\tag{sorry I omitted the x}\\
=\lim_{x\to0}\frac{e^{2x}-1-2x^2}{8x^3}\\
8L=\lim_{x\to0}\frac{e^{2x}-1-2x^2}{x^3}\\
7L=\lim_{x\to0}\frac{e^{2x}-e^x-3x^2/2}{x^3}\\
7L=\lim_{x\to0}\frac{e^{4x}-e^{2x}-6x^2}{8x^3}\\
56L=\lim_{x\to0}\frac{e^{4x}-e^{2x}-6x^2}{x^3}\\
49L=\lim_{x\to0}\frac{e^{4x}-2e^{2x}+e^x-9x^2/2}{x^3}=?$$
|
$$L=\lim_{x\to0}\frac{e^x-1-x-x^2/2}{x^3}$$
If Limit exists:
$$L=\lim_{x\to0}\frac{e^x-1-x-x^2/2}{x^3}\\
L=\lim_{x\to0}\frac{e^{2x}-1-2x-2x^2}{8x^3}\\
8L=\lim_{x\to0}\frac{e^{2x}-1-2x-2x^2}{x^3}\\
7L=\lim_{x\to0}\frac{e^{2x}-e^x-x-3x^2/2}{x^3}\\
7L=\lim_{x\to0}\frac{e^{4x}-e^{2x}-2x-6x^2}{8x^3}\\
28L=\lim_{x\to0}\frac{\frac12e^{4x}-\frac12e^{2x}-x-3x^2}{x^3}\\
21L=\lim_{x\to0}\frac{\frac12e^{4x}-\frac32e^{2x}+e^x-\frac32x^2}{x^3}\\
21L=\lim_{x\to0}\frac{\frac12e^{8x}-\frac32e^{4x}+e^{2x}-6x^2}{8x^3}\\
168L=\lim_{x\to0}\frac{\frac12e^{8x}-\frac32e^{4x}+e^{2x}-6x^2}{x^3}\\
42L=\lim_{x\to0}\frac{\frac18e^{8x}-\frac38e^{4x}+\frac14e^{2x}-\frac32x^2}{x^3}\\
21L=\lim_{x\to0}\frac{\frac18e^{8x}-\frac78e^{4x}+\frac74e^{2x}-e^x}{x^3}\\
21L=\lim_{x\to0}\frac{\frac18e^x(e^x-1)^3(10e^x+6e^{2x}+3e^{3x}+e^{4x}+8)}{x^3}\\
21L=\frac1828\\
\huge L=\frac16
$$
|
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|
Recurrence problem with a game of probability Fair coin flipping (50% on both sides)
$P_1$ and $P_2$ plays a few games of fair coin flipping. Assume player $A$ starts with $x$ coins and player $B$ with $y$ coins.
Let $P_n$ denote the probability of player $A$ winning all coins. Find $P_0$ and $P_{x+y}$, then write a difference equation for $P_n$. Solve the difference equation
Find the odds of player $A$ winning over player $B$ if $A$ starts with, e.g. half the coins $B$ starts with.
How far I've come
I've been able to produce a recurrence relation $P_n = \frac{1}{2}P_{n+1} + \frac{1}{2}P_{n-1}$.
And I assume $P_0 = 0, P_{x+y}=1$, since when player $A$ has 0 coins, he also has $0\%$ chance of winning, and when player $A$ has $x+y$ coins, he's already won, therefore it would be $100\%$ Correct me if I'm wrong.
All help appreciated, thanks a lot.
|
Solution
$$P_n = \frac{1}{2}P_{n+1} + \frac{1}{2}P_{n-1}
\implies \frac{1}{2}P_{n+1} - P_n + \frac{1}{2}P_{n-1} = 0$$
Which gives the characteristic equation:
$\alpha^2-2\alpha+1=0 \implies(\alpha-1)^2=0 \implies\alpha=1$ (Double root)
General solution:
$P_{hn} = C_1\alpha_1^n + n\cdot C_2\alpha_2^n = C_1+n\cdot C_2$
$P_0 = C_1+0\cdot C_2 = C_1 = 0$
$P_{x+y} = C_1+C_2(x+y) = C_2(x+y) = 1 \implies C_2 = \frac{1}{x+y}$
$\implies P_n = n\cdot \frac{1}{x+y} = \frac{n}{x+y}$
|
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|
Find the inverse with respect to the binary operation $a ∗ b = a + b + a^2 b^2$ A binary operation on $\mathbb{R}$: $a * b = a + b + a^2 b^2$
The neutral element I found to be $0$.
Then I need to find an invertible element having two distinct inverses. I don't know where to start for this question.
|
If $a = 0$, $b = 0$ is the unique inverse of $a$, since $0 = 0 * b = 0 + b + 0^2b^2 = b$ in this case.
If $a \ne 0$, $a * b = a + b + a^2b^2 = 0$ is a quadratic equation for $b$. It may be written as
$b^2 + \dfrac{1}{a^2}b + \dfrac{1}{a} = 0. \tag{1}$
Applying the quadratic formula to (1) yields
$b = \dfrac{1}{2}(-\dfrac{1}{a^2} \pm \sqrt{\dfrac{1}{a^4} - \dfrac{4}{a}}), \tag{2}$
which may be simplified somewhat:
$b = \dfrac{1}{2}(-\dfrac{1}{a^2} \pm \sqrt{\dfrac{1}{a^4} - \dfrac{4}{a}}) = \dfrac{1}{2}(-\dfrac{1}{a^2} \pm \sqrt{\dfrac{1}{a^4} - \dfrac{4a^3}{a^4}})$
$= \dfrac{1}{2}(-\dfrac{1}{a^2} \pm \sqrt{\dfrac{1 - 4a^3}{a^4}}) = \dfrac{1}{2a^2}(-1 \pm \sqrt{1 - 4a^3}), \tag{5}$
which has one value, $b = (-1/2a^2)$, precisely when
$4a^3 = 1 \; \; \text{or} \; \; a = \dfrac{1}{\sqrt[3]{4}}; \tag{6}$
if $0 \ne 4a^3 < 1$, we obtain two real solutions for $b$, so there are two inversens for any such $a$; if $4a^3 > 1$, then the values of $b$ form a complex conjugate pair; their is no (real) inverse for such $a$. But any nonzero $a < (1 / \sqrt[3]{4})$ will have exactly two inverses in $\Bbb R$ for the operation $a * b = a + b + a^2b^2$.
Hope this helps. Cheers,
and as always,
Fiat Lux!!!
|
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|
Prove $\sum_{i=0}^{i=x} {x \choose i} {y+i \choose x}+\sum_{i=0}^{i=x} {x \choose i} {y+1+i \choose x}$ How to prove that
$$\sum_{i=0}^{i=x} {x \choose i} {y+i \choose x}+\sum_{i=0}^{i=x} {x \choose i} {y+1+i \choose x}=\sum_{i=0}^{i=x+1} {x+1 \choose i} {y+i \choose x}$$ ?
I tried to break the right side of equation down:
$$\sum_{i=0}^{i=x+1} {x+1 \choose i} {y+i \choose x}=\sum_{i=0}^{i=x} {x+1 \choose i} {y+i \choose x}+{x+1 \choose x+1} {y+x+1 \choose x}$$
Then I tried Vandermonde's Identity:
$${y+x+1 \choose x} = \sum_{i=0}^{i=x} {y+1 \choose i}{x \choose x-i}$$
Now I am totally lost. Can someone please tell me how to prove this equation?
|
This is very easy to prove using the integral representation of
binomial coefficients. Suppose we are trying to prove that
$$\sum_{k=0}^n {n\choose k} {m+k\choose n}
+ \sum_{k=0}^n {n\choose k} {m+1+k\choose n}
= \sum_{k=0}^{n+1} {n+1\choose k} {m+k\choose n}.$$
Use the two integrals
$${m+k\choose n}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{m+k}}{z^{n+1}} \; dz$$
and
$${m+k+1\choose n}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{m+1+k}}{z^{n+1}} \; dz.$$
This yields for the LHS the following sum consisting of two terms:
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^m}{z^{n+1}}
\sum_{k=0}^n {n\choose k} (1+z)^k\; dz
\\ + \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{m+1}}{z^{n+1}}
\sum_{k=0}^n {n\choose k} (1+z)^k\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^m}{z^{n+1}} (2+z)^n \; dz
+ \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{m+1}}{z^{n+1}} (2+z)^n \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^m}{z^{n+1}} (2+z)^{n+1} \; dz.$$
For the RHS we get the integral
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^m}{z^{n+1}}
\sum_{k=0}^{n+1} {n+1\choose k} (1+z)^k\; dz
= \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^m}{z^{n+1}} (2+z)^{n+1} \; dz.$$
The integrals on the LHS and on the RHS are the identical,
QED.
A similar calculation is at this
MSE link.
A trace as to when this method appeared on MSE and by whom starts at this
MSE link.
|
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|
Solving a PDE given a specific curve and condition Consider the following PDE: $$-3U_x+4U_y=0$$ where $U=3x$ on curve $y=x+1$.
First we invoke the method of characteristics: $$\frac{dx}{-3}=\frac{dy}{4}=\frac{dU}{0}.$$ From this we get $c_1=\frac{x}{3}+\frac{y}{4}$ and $U(x,y)=c_2$. So $F(c_1)=c_2$ or $F(\frac{x}{3}+\frac{y}{4})=u(x,y)$.
Applying the condition, we know that $3x=u(x,x+1)=F(\frac{x}{3}+\frac{x+1}{4})=F(\frac{x+3}{12}).$ So $F(\frac{x+3}{12})=3x.$ This is now where I'm stuck.
|
The general solution of $-3U_x+4U_y=0$
obviously is $U(x,y)=F(\frac{x}{3}+\frac{y}{4}) \space$ where $F$ is any derivable function.
I don't understand what you mean "incorporate the curve". If you are talking of a boundary condition on the curve $y=x+1$, please, make clear what is the boundary condition.
With the condition $U(x,y=x+1)=3x\space$ one have to find a function $F\space$ so that
$$F(\frac{x}{3}+\frac{x+1}{4})=3x$$
$$F(\frac{7x}{12}+\frac{1}{4})=3x$$
It is easy to see that the equality is obtained witth a function $F(X)=aX+b$ where $a=\frac{36}{7}$ and $b=-\frac{36}{7} \frac{1}{4}$ because :
$$\frac{36}{7}(\frac{7x}{12}+\frac{1}{4})-\frac{9}{7}=3x$$
As a consequence, the final result is :
$$U(x,y)=\frac{36}{7}(\frac{x}{3}+\frac{y}{4})-\frac{9}{7}$$
|
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|
Looking for the logic of a sequence from convolution of probability distributions I am trying to detect a pattern in the followin sequence from convolution of a probability distribution (removing the scaling constant $\frac{6 \sqrt{3}}{\pi }$: $$\left(\frac{1}{\left(x^2+3\right)^2},\frac{2
\left(x^2+60\right)}{\left(x^2+12\right)^3},\frac{3 \left(x^4+162
x^2+9477\right)}{\left(x^2+27\right)^4},\frac{4 \left(x^6+324 x^4+44928
x^2+2847744\right)}{\left(x^2+48\right)^5},\frac{5 \left(x^8+564 x^6+141750
x^4+19912500 x^2+1388390625\right)}{\left(x^2+75\right)^6},\frac{6 \left(x^{10}+900
x^8+366120 x^6+87829920 x^4+13038019200
x^2+998326798848\right)}{\left(x^2+108\right)^7}, \dots \right)$$
|
Well, at least the denominator seems to be $(x^2 + 3n^2)^{n+1}$...
|
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To prove triangle is equilateral> given one equation and one combined equation. show that the lines $ x^2 - 4xy +y^2 $ and $x +y=3 $ form an equilateral trinagle . Also Find Area.
Here is what I have tried:
$ l1 =x+y=3 $
The combined equation is:
$ x^2 - 4xy + y^2 $
compare it with general formula :
$ ax^2 +2hxy+y^2 $
What to do next ?
I could find the slope but how will that help ?
|
Put $y = mx$ in the joint equation. That will produce auxiliary equation $$m^2 - 4m + 1 = 0$$
Solving this will give $$m = \dfrac{4 \pm \sqrt{4^2 - 4}}{2} = \dfrac{4 \pm 2\sqrt3}{2} = 2\pm \sqrt{3}$$
Let $m_1 = 2+\sqrt3, m_2 = 2-\sqrt3$
Gradient of $x+y = 1$ is $-1$, let $m_3 = -1$.
Use $\tan\theta = \left|\dfrac{m_a-m_b}{1+m_am_b}\right|$ for all $3$ pairs and prove that it is $\tan \dfrac\pi3 = \sqrt3$
$$\tan\theta_{m_1m_2} = \left|\dfrac{(2+\sqrt3)-(2-\sqrt3)}{1+(2+\sqrt3)(2-\sqrt3)}\right| = \left|\dfrac{2\sqrt3}{1+4-3}\right| = \sqrt3$$
$$\tan\theta_{m_1m_3} = \left|\dfrac{2+\sqrt3+1}{1-2-\sqrt3}\right| = \left|\dfrac{\sqrt3(1+\sqrt3)}{-(1+\sqrt3)} \right| = \sqrt3$$
$\theta_{m_2m_3}$ can be calculated similarly.
|
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|
Show that $\log \left| z \right|$ is harmonic and find its the conjugate harmonic function. Is the form correct for the conjugate harmonic?
Attempt:
First, we are given
\begin{align*}
\log \left| z \right| &= u(x,y) + iv(x,y) = \log \sqrt{x^2 + y^2} + i \cdot 0 \\
u(x,y) &= \log \sqrt{x^2 + y^2} = \frac{1}{2} \log (x^2 + y^2).
\end{align*}
Then we differential to get $u_{xx}$ and $u_{yy}$,
\begin{align*}
u_{xx} = \frac{\partial u}{\partial x} \frac{x}{x^2 + y^2} = \frac{y^2 - x^2}{(x^2 + y^2)^2} \\
u_{yy} = \frac{\partial u}{\partial y} \frac{y}{x^2 + y^2} = \frac{x^2 - y^2}{(x^2 + y^2)^2}
\end{align*}
From here, we can see that $u_{xx} + u_{yy} = 0$. Thus, we have shown that $\log \left| z \right|$ is harmonic.
Using the Cauchy-Riemann equations, we can find the conjugate harmonic function $v$. This gives us the relationships
\begin{align*}
u_x &= v_y = \frac{x}{x^2 + y^2} \\
u_y &= -v_x = \frac{y}{x^2 + y^2}.
\end{align*}
Integrate with respect to $y$ to get $v(x,y) = \tan^{-1}(\frac{y}{x}) + C$, the conjugate harmonic function.
|
Almost correct. $log|z|$ have a harmonic conjugate iff $z$ is in some simply connected subset of $\mathbb C \backslash \{ 0 \}$ (think of convex subsets). I think you $v$ should be like the $\text{Arg}$ function in
http://en.wikipedia.org/wiki/Argument_(complex_analysis)
|
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|
Complex exponent problem Find all numbers in complex plane that solves equation
*
*$e^z=4i$
Since $e^u e^{iv}=re^{i\Theta}$ it must be that $e^u=r \to u=\ln4$ and $v=\Theta+n2\pi \to v=\pi/2+n2\pi$. So the equation holds for points $z=\log w=\ln4+i(\pi/2+n2\pi)$ when $n\in \mathbb{N}$
*$e^{1/z}=-1$
Let's set $w=1/z$ then $e^u e^{iv}=-1$ when $e^u=1 \to u=0$ and $v=\pi+n2\pi$. So $w=i(\pi+n2\pi)$ and that leads to $z=\frac{1}{i(\pi+n2\pi)}$
*$e^{2z}+e^z+1=0$. Isn't this $(e^{z})^2+e^z+(e^z)^0=0$. So if we set $e^z=a$, we have $a^2+a+a^0=0 \to a^2+a+1=0$
solution is $a=-\frac 12 \pm i\frac{\sqrt{3}}{2}=e^{\pm i 2 \pi/3+n2 \pi}$. That leads to solution $z= \ln 1+i(\pm \frac{2 \pi}{3}+n2\pi)=i(\pm \frac{2 \pi}{3}+n2\pi)$.
|
1) Solution:
$$e^z=4i<=>$$
$$z=\frac{\log(4i)}{\log(e)}<=>$$
$$z=\ln(4i)<=>$$
$$z=\ln\left(4e^{\left(\frac{1}{2}\pi\right)i}\right)<=>$$
$$z=\frac{1}{2}i\left(4\pi n+\pi -4i\ln(2)\right)$$
(n is the element of Z - the set of integers)
2) Solution:
$$e^{\frac{1}{z}}=-1<=>$$
$$\frac{1}{z}=\frac{\log(-1)}{\log(e)}<=>$$
$$\frac{1}{z}=\ln(-1)<=>$$
$$\frac{1}{z}=i\pi (2n+1)<=>$$
$$z=\frac{i}{\pi (2n+1)}, 2n+1\ne0$$
(n is the element of Z - the set of integers)
3) Solution:
Substitute x=e^z
$$1+e^z+e^{2z}=0<=>$$
$$x^2+x+1=0<=>$$
$$x^2+x=-1<=>$$
$$x^2+x+\frac{1}{4}=-\frac{3}{4}<=>$$
$$\left(x+\frac{1}{2}\right)^2=-\frac{3}{4}<=>$$
$$--------$$
$$x+\frac{1}{2}=\frac{i\sqrt{3}}{2}$$
Or:
$$x+\frac{1}{2}=-\frac{i\sqrt{3}}{2}$$
$$--------$$
$$x=\frac{i\sqrt{3}}{2}-\frac{1}{2}$$
Or:
$$x+\frac{1}{2}=-\frac{i\sqrt{3}}{2}$$
Substitute x back, and we get the answers:
$$z_1=\frac{2}{3}i(3\pi n-\pi)$$
$$z_2=\frac{2}{3}i(3\pi n+\pi)$$
(n is the element of Z - the set of integers)
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Integrate by Trig Substitution $$\int_{3/4}^{3\sqrt{3}/4}\frac{x^3}{\sqrt{9-4x^2}}dx$$
So far I have the following:
$$ u=2x\Rightarrow$$
$$ u=a\sin \theta\Rightarrow 3\sin\theta$$
$$2x=3\sin\theta \rightarrow x=\frac32\sin\theta \rightarrow dx=\frac32\cos\theta d\theta$$
$$\int_{3/4}^{3\sqrt{3}/4}\frac{\frac{u}{2}^3}{\sqrt{9-9\sin^2\theta}}\frac32\cos\theta d\theta \Rightarrow$$
$$\int_{3/4}^{3\sqrt{3}/4}\frac{\frac{u}{2}^3}{\sqrt{9\cos^2\theta}}\frac32\cos\theta d\theta \Rightarrow$$
$$\int_{3/4}^{3\sqrt{3}/4}\frac{u^3}{2^32} d\theta \Rightarrow$$
$$\frac{1}{16}\int_{3/4}^{3\sqrt{3}/4}u^3 d\theta \Rightarrow$$
$$\frac{1}{16}\int_{3/4}^{3\sqrt{3}/4}27\sin^3\theta d\theta \Rightarrow$$
$$\frac{27}{16}\int_{3/4}^{3\sqrt{3}/4}\sin^3\theta d\theta \Rightarrow$$
$$$$
$$\int \sin^3\theta d\theta \Rightarrow \int (1-\cos^2\theta)\sin \theta d\theta$$
$$u = \cos \theta \rightarrow du=-\sin \theta d\theta \Rightarrow$$
$$\int (u^2-1)du \Rightarrow \frac{u^3}{3}-u \Rightarrow \frac{\cos^3\theta}{3}-\cos\theta $$
$$$$
$$\frac{27}{16}\bigg[\frac{\cos^3\theta}{3}-\cos\theta \bigg]\bigg|_{3/4}^{3\sqrt{3}/4}$$
$$\frac{27}{16}\bigg[\frac{\cos^2\theta\cos\theta}{3}-\cos\theta \bigg]\bigg|_{3/4}^{3\sqrt{3}/4}$$
$$\frac{27}{16}\bigg[\frac{1-4x^2\sqrt{1-\frac{4x^2}{9}}}{27}-\sqrt{1-\frac{4x^2}{9}} \bigg]\bigg|_{3/4}^{3\sqrt{3}/4}$$
Am I doing this correctly?
|
You're complicating things with the double substitution. Consider
$$
\sqrt{9-4x^2}=3\sqrt{1-\frac{4x^2}{9}}
$$
and directly set $\frac{2}{3}x=\sin\theta$, or $x=\frac{3}{2}\sin\theta$, with $dx=\frac{3}{2}\cos\theta\,d\theta$.
If $x=\frac{3}{4}$, then you have $\frac{3}{4}=\frac{3}{2}\sin\theta$ or $\sin\theta=\frac{1}{2}$, while for $x=\frac{3\sqrt{3}}{4}$ you have $\frac{3\sqrt{3}}{4}=\frac{3}{2}\sin\theta$ or $\sin\theta=\frac{\sqrt{3}}{2}$.
Thus the integral becomes
$$
\int_{\pi/6}^{\pi/3}\frac{(27\sin^3\theta)/8}{3\cos\theta}\frac{3}{2}\cos\theta\,d\theta=
\frac{27}{16}\int_{\pi/6}^{\pi/3}\sin^3\theta\,d\theta
$$
This is computed by observing that
$$
\sin^3\theta=(1-\cos^2\theta)\sin\theta
$$
so you can set $\cos\theta=u$, so $-\sin\theta\,d\theta=du$ and the integral is
$$
\int_{\sqrt{3}/2}^{1/2}(u^2-1)\,du
$$
Your procedure is correct, but you carry around too much and doing small errors becomes easy.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Is it true that $\left\lfloor\sum_{s=1}^n\operatorname{Li}_s\left(\frac 1k \right)\right\rfloor\stackrel{?}{=}\left\lfloor\frac nk \right\rfloor$ While studying polylogarithms I observed the following.
Let $n>0$ and $k>1$ be integers. Is the following statement true?
$$\left\lfloor \sum_{s=1}^n \operatorname{Li}_s\left( \frac{1}{k} \right) \right\rfloor \stackrel{?}{=} \left\lfloor \frac{n}{k} \right\rfloor $$
If it is, then how could we prove it? If not, give a counterexample.
|
For this proof to work, we have to add the restriction that $k \geq 2$, since $\operatorname{Li}_1\left( \frac{1}{1} \right)$ doesn't even converge anyway.
Now, by the definition of the polylogarithm, we have:
\begin{align}
\sum_{s=1}^n \operatorname{Li}_s\left( \frac{1}{k} \right)
&=\sum_{s=1}^n \sum_{j=1}^\infty {1 \over j^s k^j}
\\&=\sum_{j=1}^\infty \frac{1}{k^j} \sum_{s=1}^n \frac{1}{j^s}
\\&=\frac{n}{k}+\sum_{j=2}^\infty \frac{1}{k^j} \left( {1-j^{-n} \over j-1} \right)
\end{align}
Now, let $R=\sum_{j=2}^\infty \frac{1}{k^j} \left( {1-j^{-n} \over j-1} \right)$, then obviously $R>0$. But also:
\begin{align}
R
&=\sum_{j=2}^\infty \frac{1}{k^j} \left( {1-j^{-n} \over j-1} \right)
\\&\leq \sum_{j=2}^\infty \frac{1}{k^j} \left( {1 \over j-1} \right)
\\&= \frac{1}{k} \sum_{j=1}^\infty {1 \over j k^j}
= {-\log \left(1-\frac{1}{k} \right) \over k}
\\&\leq {-\log \left(1-\frac{1}{2} \right) \over k}
={\log 2 \over k}
\\&< \frac{1}{k}
\end{align}
So that $0<R<\frac{1}{k}$.
Write $n=ak+b$, where $a,b \in \Bbb{Z}$, $a \geq 0$ and $0 \leq b \leq k-1$. Then we have:
$\displaystyle \quad \; {n \over k} = a+{b \over k} \\ \Rightarrow a \leq {n \over k} \leq a+{k-1 \over k} \\ \Rightarrow \left\lfloor {n \over k} \right\rfloor = a$.
And we also have:
$\displaystyle \quad \; a+R \leq {n \over k}+R \leq a+{k-1 \over k}+R \\ \Rightarrow a<a+R \leq {n \over k}+R \leq a+{k-1 \over k}+R<a+{k-1 \over k}+{1 \over k} \\ \Rightarrow a< {n \over k}+R<a+1 \\ \Rightarrow \left \lfloor {n \over k}+R \right \rfloor = a$
Finally, since $\sum_{s=1}^n \operatorname{Li}_s\left( \frac{1}{k} \right)={n \over k}+R$, we have $$\left \lfloor \sum_{s=1}^n \operatorname{Li}_s\left( \frac{1}{k} \right) \right \rfloor = a = \left \lfloor {n \over k} \right \rfloor$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Probability of exactly 2 low rolls in 5 throws of a die If a fair die is rolled 5 times what is the probability that the number shown will be less than 3 exactly 2 times. I have tried to work problem using formula from textbook and still get it wrong
|
Since there are two possibilities, we use the binomial probability formula
$$\binom{n}{k}p^kq^{n - k}$$
where $p$ is the probability of the event, $q = 1 - p$ is the probability of the complement of the event, $n$ is the number of trials, and $k$ is the number of times the event occurs during those $n$ trials.
Here, the probability that the outcome of a roll is less than $3$ is $1/3$ since two of the six outcomes are less than $3$. That means the probability that a roll will produce an outcome of at least $3$ is $2/3$. Since we want the probability that exactly $2$ of the five rolls are less than $3$, $n = 5$, $k = 2$, and $n - k = 3$. Hence, the probability of exactly two rolls with an outcome less than $3$ in five rolls of the die is
$$P = \binom{5}{2}\left(\frac{1}{3}\right)^2\left(\frac{2}{3}\right)^3 = \frac{5!}{2!3!}\left(\frac{1}{9}\right)\left(\frac{8}{27}\right) = \frac{5 \cdot 4}{2 \cdot 1}\left(\frac{8}{243}\right) = \frac{80}{243}$$
|
{
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"url": "https://math.stackexchange.com/questions/951236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Trigonometric Integral $\int _0^{\pi/4}\:\frac{dx}{\cos^4x-\cos^2x\sin^2x+\sin^4x}$ Q. $$\int _0^{\pi/4}\:\frac{1}{\cos^4x-\cos^2x\sin^2x+\sin^4x}$$
My method:
$$\int_0^{ \pi /4} \frac{dx}{(\cos^2x+\sin^2x)^2-3\cos^2x \sin^2x}=\int_0^{ \pi /4} \frac{dx}{1-3\cos^2x\sin^2x} $$
Dividing numerator and denominator by $\cos^2x$ we have:
$$\int_0^{\pi /4}\frac{\sec^2x}{\sec^2x-3\tan^2x}dx=\int_0^{\pi /4}\frac{\sec^2x}{1-2\tan^2x}dx=\int _0^1 \frac{dt}{1-2t^2}=\int _0^1 \frac{1}{2}\frac{dt}{\frac{1}{2}-t^2}=\frac{1}{2\sqrt{2}}\log\left|\frac{1-\sqrt{2}t}{1+\sqrt{2}t}\right|_0^1=\frac{1}{2\sqrt{2}}\log\left|\frac{1-\sqrt{2}}{1+\sqrt{2}}\right|$$
But when we do the same integration by dividing the initial term by $\sec^4x$ and solving it yields an answer $$\frac{\pi }{2}$$
Am I wrong somewhere?
|
Lets look at the integral in terms of multiple angles.
Now
\begin{array}
$\cos^4x+\sin^4x-\cos^2x\sin^2x&=&(\cos^2x+\sin^2x)^2-3\sin^2x\cos^2x\\
&=&1-\frac{3}{4}\sin^2 2x\\
&=&1-\frac{3}{8}(1-\cos 4x)\\
&=& \frac{1}{8}(5-3\cos 4x)
\end{array}
Hence
$$\int_0^{\pi/4}\frac{dx}{\cos^4x+\sin^4x-\cos^2x\sin^2x}=\int_0^{\pi/4}\frac{8}{5-3\cos 4x}dx$$
Using the t- substitution with $t=\tan 2x$, $dx=\frac{1}{2(1+t^2)}dt$ and $\cos 4x=\frac{1-t^2}{1+t^2}$, we have
\begin{array}
$\displaystyle\int_0^{\pi/4}\frac{8}{5-3\cos 4x}&=&\displaystyle\int_0^{\infty}\frac{8}{5-3\big(\frac{1-t^2}{1+t^2}\big)}\frac{1}{2(1+t^2)}dt\\
&=&\displaystyle\int_0^{\infty}\frac{4}{5(1+t^2)-3(1-t^2)}dt\\
&=&\displaystyle\int_0^{\infty}\frac{2}{1+4t^2}dt\\
&=& \big[\tan^{-1}2t\big]_0^{\infty}\\
&=&\frac{\pi}2
\end{array}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proving $\lim\limits_{n \to \infty}[\sqrt{n^2 + 1} - n] = 0$ I'm trying to prove $\lim\limits_{n \to \infty}[\sqrt{n^2 + 1} - n] = 0$. Is the following a correct proof?
For all $n$ we have $0 \leq \left|\sqrt{n^2 + 1} - n\right| \leq \left|\sqrt{n^2+1} - 1 \right|$. For any $\epsilon > 0$ take $ N = \sqrt{(\epsilon+1)^2-1}$. Then for all $n > N$ we have $\left|\sqrt{n^2+1} - 1 \right| < \epsilon$ so $\lim\limits_{n \to \infty}[\sqrt{n^2 + 1} - 1] = 0$ and by the squeeze theorem $\lim\limits_{n \to \infty}[\sqrt{n^2 + 1} - n] = 0$.
|
No, it isn't. First of all think: does $\lim \limits_{n \to \infty} \sqrt{n^2 + 1} - 1 = 0$?
Your mistake is that assuming that when we take $n > N$, that we will have $\sqrt{n^2 + 1} - 1 < \epsilon$, but actually when we take $n > N$ we will have $\sqrt{n^2 + 1} - 1 > \epsilon$.
Here would be a correct proof:
Multiply by $\frac{\sqrt{n^2 + 1} + n}{\sqrt{n^2 + 1} + n}$ to get $\frac{1}{\sqrt{n^2 + 1} + n} < \frac{1}{2 n}$. Thus if we pick $N = (2 \epsilon)^{-1}$, then for all $n>N$ we will have $\frac{1}{\sqrt{n^2 + 1} + n} < \frac{1}{2 \sqrt{n}} < \epsilon$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/953214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Find the range of $f(x) =(x-1)^{1/2} + 2\cdot(3-x)^{1/2}$ How to take out the range of the following function :
$$f(x) =(x-1)^{1/2} + 2\cdot(3-x)^{1/2}$$
I am new to functions hence couldn't come up with a solution.
|
As $1\le x\le 3\iff-1\le x-2\le1$ which nicely fits with the range of sinusoidal functions,
let $x-2=\cos2\theta$
So, $f(x)=\sqrt{2\cos^2\theta}+2\sqrt{2\sin^2\theta}=\sqrt2(|\cos\theta|+\sqrt2|\sin\theta|)$
Case $\#1:$
If $\displaystyle0\le\theta\le\dfrac\pi2,$
$\displaystyle f(x)=\sqrt2[\cos\theta+ 2\sin\theta]=\sqrt2\sqrt5\cos\left(\theta-\arctan2\right)$
As $\displaystyle0\le\theta\le\dfrac\pi2,0-\arctan2\le\theta-\arctan2\le\dfrac\pi2-\arctan2$
Case $\#1A:$
For $\displaystyle-\arctan2\le\theta-\arctan2\le0,$
as $\cos(x)$ is increasing function in $\left[-\dfrac\pi2,0\right];$
$\displaystyle\cos\left(-\arctan2\right)\le\cos\left(\theta-\arctan2\right)\le\cos0$
i.e.,$\displaystyle\frac1{\sqrt5}\le\cos\left(\theta-\arctan2\right)\le1 \implies\sqrt{10}\frac1{\sqrt5}\le f(x)\le\sqrt{10}$
So, $f(x)$ will attain its maximum when $\displaystyle\theta-\arctan2=0$
$\displaystyle\implies\cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta}=\frac{1-4}{1+4}=-\frac35\implies x=2+\left(-\frac35\right)$
Similarly, $f(x)$ will attain its minimum value i.e., $\displaystyle\sqrt{10}\cdot\frac1{\sqrt5}=\sqrt2$ when $\displaystyle\theta-\arctan2=-\arctan2\iff\theta=0\implies x=2+\cos(2\cdot0)=3$
Case $\#1B:$
For $\displaystyle0\le\theta-\arctan2\le\dfrac\pi2-\arctan2,$
as $\cos(x)$ is decreasing function in $\left[0,\dfrac\pi2\right];$
$\displaystyle\cos0\ge\cos\left(\theta-\arctan2\right)\ge\cos\left(\dfrac\pi2-\arctan2\right)$
i.e., $\displaystyle1\ge\cos\left(\theta-\arctan2\right)\ge\sin\left(\arctan2\right)=\frac2{\sqrt5}$
$\displaystyle\implies\sqrt{10}\ge f(x)\ge\sqrt{10}\frac2{\sqrt5}=2\sqrt2$ which is greater than the earlier minimum $\sqrt2$
I leave for you to deal with
Case $\displaystyle\#2:\frac\pi2<\theta\le\pi$
Case $\displaystyle\#3:\pi<\theta\le\frac{3\pi}2$
Case $\displaystyle\#4:\frac{3\pi}2<\theta\le2\pi$
to show that $\sqrt2\le f(x)\le\sqrt{10}$
|
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"url": "https://math.stackexchange.com/questions/953842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Bivariate distribution with normal conditions Define the joint pdf of $(X,Y)$ as:
$$f(x,y)\propto \exp(-1/2[Ax^2y^2+x^2+y^2-2Bxy-2Cx-Dy]),$$
where $A,B,C,D$ are constants.
Show that the distribution of $X\mid Y=y$ is normal with mean $\frac{By+C}{Ay^2+1}$ and variance $\frac{1}{Ay^2+1}$. Derive a corresponding result for the distribution of $Y\mid X=x$.
Attempt:
I tried to integrate the equation w.r.t. $x$ in order to find $X\mid Y=y$. However, I'm not sure if I am correct:
$$\int_{-\infty}^{\infty}\exp(-1/2[Ax^2y^2+x^2+y^2-2Bxy-2Cx-Dy])\,dx$$
$$ =\left[\frac{\exp[-1/2[Ax^2y^2+x^2+y^2-2Bxy-2Cx-Dy]}{-Axy^2-1/2+By+C}\right]_{-\infty}^{\infty}$$
Whatever the value of the previous integration (call it "Q"), then we would divide the original equation by "Q", i.e.:
$$ \frac{f_{X,Y}(x,y)}{Q}$$
Which would give us $f_{X\mid Y=y}(X\mid Y=y)$.
How do I go about evaluating
$$ =\left[\frac{\exp[-1/2[Ax^2y^2+x^2+y^2-2Bxy-2Cx-Dy]}{-Axy^2-1/2+By+C}\right]_{-\infty}^{\infty}\text{ ?}$$
|
We use:
$$f_{X\mid Y}(x\mid Y=y) = \frac{f(x,y)}{f_Y(y)} = \frac{f(x,y)}{\int_{-\infty}^{\infty} f(x,y)dx} $$
We first note that:
$$ -0.5(Ax^2y^2+x^2+y^2-2Bxy-2Cx-2Dy) =$$ $$ -0.5(Ay^2+1)\left(x-\frac{By+c}{Ay^2+1} \right)^2 -0.5\left(y^2-2Dy - \frac{(By+c)^2}{Ay^2+1}\right).$$
This gives us:
$$ \int_{-\infty}^{\infty} f(x,y)dx = \int_{-\infty}^{\infty} {\rm e}^{-0.5(Ax^2y^2+x^2+y^2-2Bxy-2Cx-2Dy) }dx$$
$$ ={\rm e}^{-0.5\left(y^2-2Dy- \frac{(By+c)^2}{Ay^2+1}\right) }\int_{-\infty}^{\infty} {\rm e}^{ -0.5(Ay^2+1)\left(x-\frac{By+c}{Ay^2+1} \right)^2 }dx$$
$$ = {\rm e}^{-0.5\left(y^2-2Dy- \frac{(By+c)^2}{Ay^2+1}\right) } (Ay^2+1)^{-1} \int_{-\infty}^{\infty} {\rm e}^{-0.5z^2 } dz$$
$$ = {\rm e}^{-0.5\left(y^2-2Dy- \frac{(By+c)^2}{Ay^2+1}\right) } (Ay^2+1)^{-1} \sqrt{2\pi}.$$
Finally:
$$ f_{X\mid Y}(x\mid Y=y) = \frac{1}{ (Ay^2+1)^{-1}\sqrt{2\pi}} {\rm e}^{-0.5(Ay^2+1)\left(x-\frac{By+c}{Ay^2+1} \right)^2 }.$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Two differentiation results of $\sin^{-1}(2x\sqrt{1-x^2})$ While trying to differentiate $\sin^{-1}(2x\sqrt{1-x^2})$, if we put
$x = \sin\theta$, we get,
\begin{align*}
y &=\sin^{-1}(2x\sqrt{1-x^2})\\
&= \sin^{-1}(2\sin\theta\sqrt{1-\sin^2\theta})\\
&= \sin^{-1}(2\sin\theta\cos\theta)\\
&= \sin^{-1}(\sin2\theta)\\
&= 2\theta\\
&= 2\sin^{-1}x.
\end{align*}
So,
\begin{align*}
\frac{dy}{dx} &= \frac{2}{\sqrt{1-x^2}}.\\
\end{align*}
But if we put $x = \cos\theta$, we get,
\begin{align*}
y &=\sin^{-1}(2x\sqrt{1-x^2})\\
&= \sin^{-1}(2\cos\theta\sqrt{1-\cos^2\theta})\\
&= \sin^{-1}(2\cos\theta\sin\theta)\\
&= \sin^{-1}(\sin2\theta)\\
&= 2\theta\\
&= 2\cos^{-1}x.
\end{align*}
This time,
\begin{align*}
\frac{dy}{dx} &= -\frac{2}{\sqrt{1-x^2}}.\\
\end{align*}
We are perplexed about the difference in sign between the two results
and thought that you could help.
(We understand that we can differentiate $\sin^{-1}(2x\sqrt{1-x^2})$ directly, without any substitution, which gives us the first result.)
|
This Question bugged me for long time before I found the linked identity
Setting $y=x$ in Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $,
$$
2\arcsin x\\
\begin{align}
&=\arcsin( 2x\sqrt{1-x^2}) \;\;;2x^2 \le 1 \\
&=\pi - \arcsin(2x\sqrt{1-x^2}) \;\;;2x^2 > 1\text{ and } 0< x\\
&=-\pi - \arcsin(2x\sqrt{1-x^2}) \;\;;2x^2 > 1\text{ and } x\le 0\\
\end{align}
$$
Case $\#1:$ If $2x^2 \le 1,\arcsin( 2x\sqrt{1-x^2})=2\arcsin x$
Case $\#2:$ If $2x^2> 1,$
Case $\#2A: x>0,\arcsin( 2x\sqrt{1-x^2})=\pi-2\arcsin x=2\arccos x$
Case $\#2B: x\le0,\arcsin( 2x\sqrt{1-x^2})=-\pi-2\arcsin x$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/956432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Calc the sum of $\sum_{k = 0}^{\infty} \frac{(-1)^k}{k} \sin(2k)$ Solving a bigger problem about Fourier series I'm faced with this sum:
$$\sum_{k = 0}^{\infty} \frac{(-1)^k}{k} \sin(2k)$$
and I've no idea of how to approach this.
I've used Leibniz convergence criterium to verify that the sum should have a value, but I don't know how to calculate this value.
|
Consider the series
\begin{align}
S = \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} \, \sin(2n).
\end{align}
Method 1
Using the known Fourier series
\begin{align}
x = \frac{2 L}{\pi} \, \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \, \sin\left( \frac{n \pi x}{L} \right)
\end{align}
it can quickly be seen that for $L = \pi$ and $x = 2$ the series becomes
\begin{align}
- 1 = \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} \, \sin(2n).
\end{align}
Method 2
Using $2i \sin(2n) = e^{2in} - e^{-2in}$ then the series is
\begin{align}
\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} \, \sin(2n) &= \frac{1}{2i} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} \left( e^{2in} - e^{-2in} \right) \\
&= - \frac{1}{2i} \left( \ln(1 + e^{2i}) - \ln(1 + e^{-2i}) \right) \\
&= - \frac{1}{2i} \, \ln\left(\frac{1 + e^{2i}}{1 + e^{-2i}} \right)
= - \frac{1}{2i} \, \ln\left(\frac{e^{i} \, \cos(1)}{e^{-i} \, \cos(1)} \right) \\
&= - \frac{1}{2i} \ln(e^{2i}) = -1.
\end{align}
Method 3
As stated in the proposed problem the summation is given by
\begin{align}
S_{0} = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n} \, \sin(2n),
\end{align}
for which
\begin{align}
S_{0} &= \lim_{n \rightarrow 0} \left\{ \frac{\sin(2n)}{n} \right\} + \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} \, \sin(2n) \\
&= -1 + \lim_{n \rightarrow 0} \left\{ \frac{2 \cos(2n)}{1} \right\} \\
&= -1 + 2 = 1.
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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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.