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$\begin{cases} x^4+x^2y^2+y^4=21 \\ x^2+xy+y^2=3 \end{cases}$
I should solve the following system: $$\begin{cases} x^4+x^2y^2+y^4=21 \\
x^2+xy+y^2=3 \end{cases}$$
by reducing the system to a system of second degree.
What can I look for in such situations? What is the way to solve this kind of systems? The only thing I see here is that we can factor:
$$\begin{cases} x^2(x^2+y^2)+y^4=21 \\
x(x+y)+y^2=3 \end{cases}$$
|
You can substitute $u=x+y$ and $v=xy$. Then
$$x^2+xy+y^2=(x+y)^2-xy=u^2-v $$
and
$$x^4+x^2y^2+y^4=(x^2+y^2)^2-x^2y^2=(u^2-2v)^2-v^2 $$
You obtain the system
$$\begin{cases}
u^4-4u^2v+3v^2=21\\ u^2-v=3
\end{cases}$$
From the second equation you get $v=u^2-3$. Substitute that in the first equation, and you'll immediately find $u^2=1$. After that, you can find $u$ and $v$, and I assume you can finish.
|
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|
Help with inequality problem
Given $a$ , $b$ , $c \ge 0$ show that
$$\frac{a^2}{(a+b)(a+c)} + \frac{b^2}{(b+a)(b+c)}+ \frac{c^2}{(c+a)(c+b)} \ge \frac{3}{4}.$$
I tried using Titu's lemma on it, resulting in
$$\frac{a^2}{(a+b)(a+c)}+\frac{b^2}{(b+a)(b+c)}+ \frac{c^2}{(c+a)(c+b)}\ge \frac{(a+b+c)^2}{a^2+b^2+c^2 + 3(ab + bc + ca)} $$
And I am stuck here.
|
Hint.
As the inequality is homogeneous, making the substitutions $b=\lambda a, c=\mu a$ we have
$$\frac{a^2}{(a+b)(a+c)} + \frac{b^2}{(b+a)(b+c)}+ \frac{c^2}{(c+a)(c+b)} = \frac{\lambda ^2 \mu +\lambda ^2+\lambda \mu ^2+\lambda +\mu ^2+\mu }{(\lambda +1) (\mu +1) (\lambda +\mu )} = f(\lambda,\mu)
$$
with stationary points as the solutions $(\lambda^*,\mu^*)$ for
$$
\nabla f = 0\Rightarrow \cases{\lambda ^2 \mu -\mu ^2 = 0\\ \lambda ^2-\lambda \mu ^2=0}
$$
now discarding the solution $\lambda=0,\mu = 0$ we have to demonstrate that $f(\lambda^*,\mu^*)\ge \frac 34$.
|
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|
Solving logarithm leaving in terms of $p$ and $q$ I would like to check the steps if Part a) is done correctly.
For Part b), how do I continue from below? I seem to stuck for $\log_{10}(5)$…
Here is the problem:
Given that $p = \log_{10} 2$ and $q = \log_{10} 7$, express the following in terms of $p$ and $q$.
a) $\log_{7} 4 = \frac{\log_{10} 4}{\log_{10} 7} = \frac{2 \log_{10} 2}{\log_{10} 7} = \frac{2p}{q}$
b) $\log_{10} \sqrt[3]{\frac{25}{49}} = \log_{10}5^\frac{2}{3} - \log_{10}7^\frac{2}{3} = \frac{2}{3}\log_{10}5 - \frac{2q}{3}$
[Source]
|
Hint: Write $5 = 10/2$. ${}{}$
|
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|
Element of order $7$ in $GL(4,2)$ Find an element of order $7$ in $GL(4,2)$, the group of all invertible $4 \times 4$ matrices with entries in $\mathbb{F}_2$.
I'd like a more a constructive way to find the required element in $GL(4,2)$ in lieu of explicity going through elements of $GL(4,2)$ and computing their orders aimlessly.
First, I know that if $A$ is the desired element, we know $A$ satisfies $A^7 = I$ $\Rightarrow$ $A$ satisfies the polynomial $p(x) = x^7 - 1$ $\Rightarrow$ $p(x) = x^7-1$ is an annihilating polynomial for $A$. This means that the minimal polynomial of $A$ must divide $p(x) = x^7 - 1$. Over $\mathbb{F}_2$, I found that this polynomial factors completely as $p(x) = (x+1)(x^3 + x^2 + 1)(x^3 + x + 1)$.
Now, the minimal polynomial of $A$ cannot equal $p(x)$, since $p(x)$ has degree greater than $4$. Similarly, the minimal polynomial of $A$ cannot involve both factors $(x^3 + x^2 + 1)$ and $(x^3 + x + 1)$. Furthermore, the minimal polynomial cannot simply be $(x+1)$ alone, as this would mean that $A$ is equal to the identity matrix multiplied by $-1$, which has order $2$. Thus, for degree reasons, it must be that the minimal polynomial is either $m_A(x) = (x+1)(x^3 + x^2 + 1)$ or $m_A(x) = (x+1)(x^3 + x + 1)$. In either case, the minimal polynomial coincides with the characteristic polynomial, since the characteristic polynomial must have degree $4$.
Thus, the characteristic polynomial of $A$ is either $p_A(x) = (x+1)(x^3 + x + 1)$ or $p_A(x) = (x+1)(x^3 + x^2 + 1)$. Then, once can build the companion matrices corresponding to each of these characteristic polynomials, and check if either of them has order $7$. It seems, a prior, one of them must be of order $7$.
However, neither of the corresponding companion matrices is working for me as a solution. Where did I make a mis-step in my logic ? Is there an even better method to do this ?
Thanks!
|
If $\lambda\in\mathbb{F}_{2^3}$ is such that $\lambda^3=\lambda+1 $ then $\lambda^6=\lambda^2+1 $ and $\lambda^7 = \lambda^3+\lambda = 1 $, so the companion matrix of $x^3-x-1$ (as an element of $\text{GL}(3,\mathbb{F}_2)$) is a $3\times 3$ matrix with order $7$, which can be easily completed to a $4\times 4$ matrix with order $7$. Given $M=\left(\begin{smallmatrix}0&0&1\\1&0&1\\0&1&0\end{smallmatrix}\right)$ we have
$M^2=\left(\begin{smallmatrix}0&0&1\\1&0&1\\0&1&0\end{smallmatrix}\right)\quad$ $M^3=\left(\begin{smallmatrix}1&0&1\\1&1&1\\0&1&1\end{smallmatrix}\right)\quad$
$M^4=\left(\begin{smallmatrix}0&1&1\\1&1&2\\1&1&1\end{smallmatrix}\right)\quad$
$M^5=\left(\begin{smallmatrix}1&1&1\\1&2&2\\1&1&2\end{smallmatrix}\right)\quad$ $M^6=\left(\begin{smallmatrix}1&1&2\\2&2&3\\1&2&2\end{smallmatrix}\right)\quad$
$M^7=\left(\begin{smallmatrix}1&2&2\\2&3&4\\2&2&3\end{smallmatrix}\right)\equiv I\pmod{2}.$
|
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|
Correlation between two matrices what is the correlation/relationship between two matrices
A = \begin{bmatrix}4&0&0&0\\0&3&0&0\\0&0&2&0\end{bmatrix}
B = \begin{bmatrix}1/4&0&0\\0&1/3&0\\0&0&1/2\\0&0&0\end{bmatrix}
where A*B = \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}
A cannot be an inverse of B because it's not square. So is there a word that describes the relation between this two matrices A and B?
EDIT1: Changed dimensions of A from 4X4 to 3X4
|
You may be reaching for the Moore-Penrose pseudoinverse, usually denoted as $A^+$, which is also defined for rectangular and singular matrices.
In this case:
$$B^+ = \begin{bmatrix}1/4&0&0\\0&1/3&0\\0&0&1/2\\0&0&0\end{bmatrix}^+
= \begin{bmatrix}4&0&0&0\\0&3&0&0\\0&0&2&0\end{bmatrix}$$
And:
$$B^+\cdot B=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$$
$$B\cdot B^+=\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&0\end{bmatrix}$$
|
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|
Solve $y' + y^2 = \frac{1}{x^2}$ by introducing $z = xy$ as a new function. Question:
Solve the equation:
$$y' + y^2 = \frac{1}{x^2},~~~~~~~~~x > 0$$
by introducing $z = xy$
Attempted answer:
$z = xy \Rightarrow y = \frac{z}{x}$
Taking the derivative of $y$ with respect to $x$ using he product rule:
$$y' = \frac{z'}{x} - \frac{z}{x}$$
Adding this into the original equation:
$$\frac{z'}{x} - \frac{z}{x^2} + \frac{z^2}{x^2} = \frac{1}{x^2}$$
Putting $z$ and $x$ on each side:
$$\frac{z'}{1-z^2 + z} = \frac{1}{x}$$
Integrating on each side:
$$\int \frac{1}{1-z^2 + z}dz = \int \frac{1}{x} dx$$
This produces
$$z(x) = \frac{(1+\sqrt{5})c_1x^{\sqrt{5}}+1-\sqrt{5}}{(2c_1 x^{\sqrt{5}}+2)}$$
Substituting back to $y$:
$$y(x) = \frac{(1+\sqrt{5})c_1x^{\sqrt{5}}+1-\sqrt{5}}{x(2c_1 x^{\sqrt{5}}+2)}$$
This seems all and well, but the expected answer contains yet another solution:
$$y = \frac{1\pm\sqrt{5}}{2x}$$
How does this second solution arise? Since there is an $x^2$ in the question, I think that a solution might have been dropped at some point.
|
Because of the division that must be different from zero as GSofer pointed in his good answer but you can also deduce that solution by other methods:
$$y' + y^2 = \frac{1}{x^2}$$
By inspection $y_p=\frac A x$
$$-\frac A {x^2}+\frac {A^2}{x^2}=\frac 1 {x^2}$$
$$ \implies A^2-A-1=0$$
$$\implies A=\frac 12 \pm\frac {\sqrt 5}{2} \implies y =\frac {1\pm \sqrt 5}{2x}$$
|
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|
We roll a fair die until a $5$ appears. What is the expected value of the minimum value rolled?
Question: Given a fair dice, we roll until we get a $5.$ What is the expected value of the minimum value rolled?
Answer is $\frac{137}{60}.$
There is a similar question asked in MSE but I do not understand the method used by Henry.
In particular, if we let $X$ be the minimum value rolled up to and including $5$, then
$$E(X) = \sum_{x=1}^5 xP(X=x) = 1 \times \frac12 + 2 \times \frac16 + 3 \times \frac1{12}+4 \times \frac{1}{20}+5 \times \frac15 = \frac{137}{60}.$$
It seems that we are using the fact that
$$P(X=x) = \frac{1}{x(x+1)}.$$
I do not understand how to obtain the equation above.
|
Assume that a $5$ is first seen on roll $n$.
$5$ is the lowest seen with $n-1$ $6$s then one $5$.
$4$ is the lowest seen with $n-1$ $4$s and $6$s, but not all $6$s then one $5$.
$3$ is the lowest seen with $n-1$ $3$s, $4$s, and $6$s, but not all $4$s and $6$s then one $5$.
$2$ is the lowest seen with $n-1$ $2$s, $3$s, $4$s, and $6$s, but not all $3$s, $4$s, and $6$s then one $5$.
$1$ is the lowest seen with $n-1$ $1$s, $2$s, $3$s, $4$s, and $6$s, but not all $2$s, $3$s, $4$s, and $6$s then one $5$.
$$
\begin{array}{c|l|l}
\text{lowest}&\text{chance with $n$ rolls}&\text{sum over $n$}\\
\hline
5&\,\left(\frac16\right)^{n-1}\frac16&\frac15\\
4&\,\left[\left(\frac26\right)^{n-1}-\left(\frac16\right)^{n-1}\right]\frac16&\frac1{20}\\
3&\,\left[\left(\frac36\right)^{n-1}-\left(\frac26\right)^{n-1}\right]\frac16&\frac1{12}\\
2&\,\left[\left(\frac46\right)^{n-1}-\left(\frac36\right)^{n-1}\right]\frac16&\frac16\\
1&\,\left[\left(\frac56\right)^{n-1}-\left(\frac46\right)^{n-1}\right]\frac16&\frac12
\end{array}
$$
Expected value $=5\cdot\frac15+4\cdot\frac1{20}+3\cdot\frac1{12}+2\cdot\frac16+1\cdot\frac12=\frac{137}{60}$
|
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|
Solve $2^m=7n^2+1$ Solve $2^m=7n^2+1$ with $(m,n)\in \mathbb{N}^2$
Here is what I did:
First try, I have seen first that the obvious solutions are $n=1$ and $m=3$ , and $n=3$ and $m=6$, then I proved by simple congruences that $m$ must be divisible by $3$ so $m=3k$, If we add $27$ to the equation we will have $2^{3k}+3^3=7(n^2+2^2)$, but unfortunately I tried to do something with Legendre symbol or the multiplicative order but I found nothing interesting.
Second try,I let $n=2k+1$ then I worked in $\mathbb{Z}\left[ \frac{-1+\sqrt{-7}}{2} \right] $ and the equation becomes $7\times 2^{m-2}=\left( 7k+4+\frac{-1+\sqrt{-7}}{2} \right) \left( 7k+3-\frac{-1+\sqrt{-7}}{2} \right) $ but I didn't find something interesting because the two factors are not coprime.
|
HINT
$2^m\equiv 1$ mod $7$ and so $m=3k$. For $n>0$, we now have
$$2^k-1=au^2, 2^{2k}+2^k+1=bv^2$$
where either $\{a,b\}=\{1,7\}$ or $\{a,b\}=\{3,21\}.$
Each of the four possibilities gives an elliptic curve $$bv^2=3+3au^2+a^2u^4.$$
Of these, the case $a=7,b=1$ is impossible modulo $7$.
|
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|
Easy double sum trouble : $\sum_{1 \leq i,j \leq n} (i+j)^2$ There's a mistake somewhere but I really can't see where.
$\sum_{1 \leq i,j \leq n} (i+j)^2 = \sum_{i=1}^n \sum_{j=1}^n \left( i^2+2ij+j^2\right) = \sum_{i=1}^n \left( \sum_{j=1}^n i^2 + 2i\sum_{j=1}^n j + \sum_{j=1}^n j^2\right) $
$= \sum_{i=1}^n\left( ni^2 + 2i\frac{n(n+1)}{2} + \frac{n(n+1)(2n+1)}{6}\right) $
$ = n \sum_{i=1}^n i^2 + 2\frac{n(n+1)}{2} \sum_{i=1}^n i + \frac{n^2(n+1)(2n+1)}{6} = 2\frac{n^2(n+1)(2n+1)}{6} + 2\left( \frac{n(n+1)}{2}\right) ^2$
When I evaluate for $n=3$, analytically it gives me $156$, and manually : $4 + 9 + 9 +16+16+16+25+25+81 = 201$.
Thanks for the help !
|
The last number is $(3+3)^2$ which is $36$.
And here is analytic result $$\frac{1}{6} \left(7 n^4+12 n^3+5 n^2\right)$$
|
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|
$\int\sqrt{x^2\sqrt{x^3\sqrt{x^4\sqrt{x^5\sqrt{x^6\sqrt{x^7\sqrt{x^8\ldots}}}}}}}\,dx$ I was attempting to solve an MIT integration bee problem (1) when I misread the integral and wrote (2) instead.
$$\int\sqrt{x\cdot \sqrt[3]{x\cdot \sqrt[4]{x\cdot\sqrt[5]{x\ldots } }}}\,dx\tag{1}$$
$$\int\sqrt{x^2\sqrt{x^3\sqrt{x^4\sqrt{x^5\sqrt{x^6\sqrt{x^7\sqrt{x^8\ldots}}}}}}}\,dx\tag{2}$$
I was able to solve (1), as the integrand simplifies to $x^{e-2}$, however, I'm struggling with solving (2).
If we rewrite the roots as powers, we get:
$$\int x^\frac{2}{2}\cdot x^\frac{3}{4}\cdot x^\frac{4}{8}\cdot x^\frac{5}{16}\ldots\,dx$$
combining the powers we get:
$$\int x^{\frac{2}{2}+\frac{3}{4}+\frac{4}{8}+\frac{5}{16}+\ldots}$$
the exponent is the infinite sum
$$\sum^{\infty}_{n=1}\frac{n+1}{2^n}\tag{3} $$
we can split this into:
$$\sum^{\infty}_{n=1}\frac{n}{2^n}+\sum^{\infty}_{n=1}\frac{1}{2^n} $$
The right sum is well known except here the sum begins at $n=1$, meaning that the right sum evaluates to 1. Messing around with desmos, the integrand appears to be $x^3,x>0$ implying that (3) converges to 3 and the $\sum^{\infty}_{n=1}\frac{n}{2^n}$ converges to 2.
Which is part I'm struggling with. Any ideas?
$$\sum^{\infty}_{n=1}\frac{n}{2^n}$$
|
Hint:
If $$f(x)=\sum_{n=1}^{\infty}x^n=\frac{x}{1-x}$$ $\forall$ $\vert x\vert \lt 1$. Then what is $$\frac{f'\left(\frac{1}{2}\right)}{2}$$
|
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|
How to solve following differential equation $y = \frac{xy'}{2} + \frac{y'^2}{x^2}$? Given the following differential equation:
$$y = \frac{xy'}{2} + \frac{y'^2}{x^2}$$
I tried to solve it downgrading and as a quadratic equation for unknown $y'$ but it did not bring any results
UPD Possible solution:
$$y = \frac{xy'}{2} + \frac{y'^2}{x^2}$$
Let $y' = p$:
$$y = \frac{xp}{2} + \frac{p^2}{x^2}$$
Now multiply both side by 2:
$$2y = xp + 2\frac{p^2}{x^2}$$
$$d(2y) = d(xp) + d(2\frac{p^2}{x^2})$$
$$2dy = xdp+pdx + 2\frac{x^2d(p^2)-p^2d(x^2)}{x^4}$$
$$2pdx = xdp+pdx + 2 \frac{x(x2pdp - p^22dx)}{x^4}$$
Then multiply both side by $x^3$:
$$x^3pdx = x^4dp + 4xpdp - 4p^2dx$$
$$(x^3p + 4p^2)dx = (x^4 + 4xp)dp$$
$$p(x^3 + 4p)dx = x(x^3 + 4p)dp$$
$$(x^3 + 4p)(pdx - xdp) = 0$$
Now solve $(x^3 + 4p) = 0$ and $(pdx -xdp) =0 $ we got $p = \frac{-x^3}{4}$ and $p = xConst$
Solution is:
$y = -\frac{x^4}{16}$ and $y = \frac{x^2Const}{2} + Const^2$
|
Hint. If we differentiate the equation
$$
y = \frac{xy'}{2} + \frac{y'^2}{x^2}
$$
we obtain
$$
y'=\frac{y'}{2}+\frac{xy''}{2}+\frac{2y'y''}{x^2}-\frac{2y'^2}{x^3}
$$
or
$$
-\frac{y'}{2}+\frac{xy''}{2}+\frac{2y'y''}{x^2}-\frac{2y'^2}{x^3}=0.
$$
or
$$
(xy''-y')\left(\frac{1}{2}+\frac{2y'}{x^3}\right)=0.
$$
|
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|
if $mx+3|x+4|-2=0$ has no solutions, solve for $m$
If $mx+3|x+4|-2=0$ has no solutions, which of the following value could be $m$?
(A)5
(B)$-\frac{1}{2}$
(C)-3
(D)-6
(E)$\frac{10}{3}$
my attempt:
$$mx-2=-3|x+4| \\ m^2x^2-4mx+4=9x^2+72x+144 \\ (9-m^2)x^2+(4m+72)x+140=0$$
because the equation has no solutions, therefore
$$(4m+72)^2-4(9-m^2)140<0 \\ 576m^2+576m+144<0 \\ 4m^2+4m+1<0 \\ (2m+1)^2<0$$
maybe I made a mistake but I couldn't find it
|
The hint.
For $x\geq-4$ we obtain:
$$mx+3x+12-2=0,$$ which gives a value $m=-3$.
For $x\leq-4$ we obtain $$mx-3x-12-2=0,$$ which gives a value $m=3.$
Now, check that for $m=3$ our equation has root, while for $m=-3$ our equation has no roots.
|
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|
Linear differential equations, integrating factor Solve the following differential equation:
$$ dr+(2r \cot\theta+\sin 2\theta)d\theta=0$$
I have tried like this:
$$ \frac{dr}{d\theta}+2r\cot\theta=-\sin{2\theta}$$
\begin{align}
I.F. &=e^{\int{2\cot\theta d\theta}}\\
& =e^{-2\log\sin\theta}\\
& =\frac 1{\sin^2\theta}\\
\end{align}
$$\therefore\frac r{\sin^2\theta}=-\int\frac{\sin 2\theta d\theta}{\sin^2\theta}\\
\implies \frac r{\sin^2\theta}=-2\int{\cot\theta d\theta}\\
\implies \frac r{\sin^2\theta}=2\log\sin\theta+c $$
But in my book the answer is:
$$2r{\sin^2\theta}+{\sin^4\theta}=c$$
Please check out which is correct..
|
$$dr+(2r \cot\theta+\sin 2\theta)d\theta=0$$
$$\sin \theta dr+(2r \cos\theta+2\sin^2 \theta\cos\theta )d\theta=0$$
Multiply by $2\sin \theta$:
$$2\sin^2 \theta dr+(4r \cos\theta\sin \theta+4\sin^3 \theta\cos\theta )d\theta=0$$
$$2\sin^2 \theta dr+2r d\sin^2 \theta+ d\sin^4 \theta=0$$
$$2(\sin^2 \theta dr+r d\sin^2 \theta)+ d\sin^4 \theta=0$$
$$2 dr\sin^2 \theta+ d\sin^4 \theta=0$$
Integrate:
$$2 r\sin^2 \theta+ \sin^4 \theta=C$$
|
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|
Inverse of $I + A$ I am trying to solve the following exercise in Artin, without breaking into cases for even and odd $k$.
A square matrix $A$ is called nilpotent if $A^k = 0$ for some $k > 0$. Prove that if $A$ is nilpotent, then $I + A$ is invertible.
Here's my attempt.
I claim that the inverse of $I + A$ is
$$(I + A)^{-1} = \sum\limits_{n=0}^{k-1} (-1)^n A^n.$$
We prove this is a right inverse.
\begin{align*}
(I + A)\sum\limits_{n=0}^{k-1} (-1)^n A^n & = \sum\limits_{n=0}^{k-1} (-1)^n (I + A)A^n \\
& = \sum\limits_{n=0}^{k-1} (-1)^n (A^n + A^{n+1})
\end{align*}
If $n$ is even, then $n + 1$ is odd and vice-versa. Hence, the series telescopes:
\begin{align*}
& = (A^0 + A^{1}) - (A^1 + A^2) + (A^2 + A^3) + \ldots \pm (A^{k-1} + A^k) \\
& = I \pm A^k \\
& = I
\end{align*}
It seems that the sign of the final term is dependent on whether $k$ is even or odd. However, it shouldn't matter, because if $A^k = 0$, then $-A^k = 0$.
Is there a better way to formalize this fact?
|
You could use
$$
\sum\limits_{n=0}^{k-1} (-1)^n (A^n + A^{n+1}) =\sum\limits_{n=0}^{k-1} (-1)^n A^n +\sum\limits_{n=0}^{k-1} (-1)^n A^{n+1} = \sum\limits_{n=0}^{k-1} (-1)^n A^n -\sum\limits_{n=0}^{k-1} (-1)^{n+1} A^{n+1} =\sum\limits_{n=0}^{k-1} (-1)^n A^n -\sum\limits_{n=1}^{k} (-1)^n A^n = I - (-1)^kA^k
$$
|
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|
How to find: $\lim_{x\to 0}\frac{\ln(e+x^2)-\cos x}{e^{x^2}-\cos x}$
Evaluate without L'Hospital's rule: $\displaystyle\lim_{x\to 0}\frac{\ln(e+x^2)-\cos x}{e^{x^2}-\cos x}$
My attempt:
$$e^{x^2}-\cos x=e^{x^2}-1+1-\cos x=x^2\cdot\left(\frac{e^{x^2}-1}{x^2}+\frac{1-\cos x}{x^2}\right)$$
$$\frac{\ln(e+x^2)-\cos x}{x^2}=\frac{\ln(e+x^2)-1+1-\cos x}{x^2}=\frac{\ln\left(1+\frac{x^2}{e}\right)}{\frac{x^2}{e}\cdot e}+\frac{1-\cos x}{x^2}$$
$$\lim_{x\to 0}\frac{\ln(e+x^2)-\cos x}{e^{x^2}-\cos x}=\lim_{x\to 0}\frac{\frac{\ln\left(1+\frac{x^2}{e}\right)}{\frac{x^2}{e}}\cdot\frac{1}{e}+\frac{1-\cos x}{x^2}}{\frac{e^{x^2}-1}{x^2}+\frac{1-\cos x}{x^2}}=\frac{\frac{1}{e}+\frac{1}{2}}{\frac{3}{2}}=\frac{2(2+e)}{6e}$$
|
$$L=\lim_{x \rightarrow 0} \frac{\ln (e+x^2)-\cos x}{e^{x^2}-\cos x}$$ Let us use
$$e^{x^2}=1+x^2+O(x^4), \cos x= 1+ x^2/2+O(x^4), \ln(1+z)=z-z^2/2+O(z^4),$$
then $$L=\lim_{x \rightarrow 0}\frac{1+x^2/e-1+x^2/2+O(x^4)}{3x^2/2+O(x^4)}=\frac{2+e}{3e}$$
|
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|
A generalization of the (in)famous IMO 1988 problem 6: If $\frac{a^2 + b^2 - abc}{ab + 1}$ is a positive integer then it is a square. This question is motivated by the famous IMO $1988$ problem $6$. Is the following true?
Let $a,b$ be positive integers and $c \ge 0$ be a non-negative integer. If $\dfrac{a^2 + b^2 - abc}{ab + 1}$ is a positive integer then it is a square.
|
We will adapt the proof for $c=0$. So let us consider $c\ge 0$ to be fixed.
Assume we have a solution $(a,b)$ of the problem, so that the number
$$
k =\frac{a^2+b^2-abc}{ab+1}\tag{$1$}
$$
is an integer, $k\in \Bbb Z$, and it is positive, $k>0$.
We can and do assume that $a>b\ge 0$. (The case $a=b$ is easily eliminated, since then in $a^2(2-c)/(a^2+1)$ the factor $a^2$ of the numerator, and the denominator are relatively prime, so $2-c\in \{0,1,2\}$ is a multiple of the denominator. The case $c=2$ leads to $k=0$, not allowed. If $c=1$, we obtain $a=b=0$ as unique case of an integer $k$, but then again $k=0$. It remains $c=0$, the OIM case, then $a=b=1$. We observe that in this case there is also the solution $(1,0)$ near $(1,1)$, a solution which makes the sum $a+b$ smaller.)
We furthermore can and do assume, that $(a,b)$ with $a>b\ge 0$ is a solution that minimizes
$$
a+b\ .
$$
Let us show that $b=0$.
Assume the contrary, $a>b>0$.
We formally substitute $a$ in the equation $(1)$ with an indeterminate $X$ and write the corresponding equation of degree two in $X$ explicitly:
$$
\underbrace{X^2-b(c+k)X + b^2-k}_{f(X)} = 0\ .
$$
One solution is known, $x_1=a\in \Bbb Z$, there is by Vieta an other solution, making their sum $b(c+k)$, so the second solution is
$x_2b(c+k)-a$. Let us show that
$0\le x_2< x_1=a$. We first compute
$$
\begin{aligned}
f(a+1)
&=a^2+2a+1\ -\ a(bc+bk)\ +\ (bc+bk)\ +\ b^2-k\\
&=2a+1 \ - \ (bc+bk)\ ,\\
af(a+1)
&=2a^2+a \ -\ a(bc+bk) \\
&=2a^2+a \ -\ (a^2+b^2-k) \\
&=2a^2-b^2+a +k \\
&>0\ .
\end{aligned}
$$
Since $x_1=a$ is a root, and the other root $x_2$ is an integer, and (on $\Bbb Z$) in the interval $I$ between the roots $x_1,x_2$ (in the correct order) the function $f$ takes values $\le 0$, and outside $I$ values $>0$, the second root $x_2=bc+bk-a$ is $\le x_1=a$. (Or use $f(a+1)>0$ combined with the above computation of $f(a+1)$.)
The initial discussion about the case $a=b$ shows that this case ($x_1=x_2)$ is here excluded. So we have the strict inequality
$$ x_1 = bc+bk-a<a=x_2 \ .\tag{$2$}$$
Let us see that $0\le x_2=bc+bk-a$. (And the case of equality is of course possible.) Assume this is not the case, so that $\color{blue}{x_2<0}$. In the equality
$$
\frac{x_2^2+b^2-x_2bc}{x_2b+1}=k>0
$$
the denominator is then $<0$, so the numerator $x_2^2+b^2-x_2bc$ is also strictly negative. This implies $c>0$, and the expression $x_2/b$ is then between the two roots of the equation $Y^2-cY+1$, which have the same sign (Vieta product is one), and thus positive (Vieta sum is $c> 0$). We get $\color{red}{x_2>0}$. Contradiction with the assumption made in blue. So we have indeed $x_2\ge 0$.
It remains to see that the solution $(x_2,b)$ is breaking the chosen minimality of $(a,b)=(x_1, b)$, since $x_2+b<x_1+b$.
Contradiction with the assumption $b>0$. So $b=0$, which gives $k=a^2$, a square.
$\square$
This is an adaptation of the solution given in https://en.wikipedia.org/wiki/Vieta_jumping#Standard_Vieta_jumping for the case $c=0$.
|
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|
How to multiply $\sqrt{5x^2+2x+1}$ with $\frac{1}{x}$? I believe the answer says, $\sqrt{5+\frac{2}{x}+\frac{1}{x^2}}$, but I don't see how they obtained it. I'm quite rusty with radicals.
|
If $x>0$ then $x=\sqrt{x^2}$. So if $x>0$, then
$$\begin{align*}
\frac{1}{x}\cdot\sqrt{5x^2+2x+1} &= \frac{\sqrt{5x^2+2x+1}}{x} \\
&= \frac{\sqrt{5x^2+2x+1}}{\sqrt{x^2}} \\
&= \sqrt{\frac{5x^2+2x+1}{x^2}} \\
&= \sqrt{\frac{5x^2}{x^2}+\frac{2x}{x^2}+\frac{1}{x^2}} \\
&= \sqrt{5+\frac{2}{x}+\frac{1}{x^2}} \\
\end{align*}$$
But if $x<0$, then $x=-\sqrt{x^2}$. So in this case we would get that
$$\begin{align*}
\frac{1}{x}\cdot\sqrt{5x^2+2x+1} &= \frac{\sqrt{5x^2+2x+1}}{x} \\
&= \frac{\sqrt{5x^2+2x+1}}{-\sqrt{x^2}} \\
&= -\sqrt{\frac{5x^2+2x+1}{x^2}} \\
&= -\sqrt{\frac{5x^2}{x^2}+\frac{2x}{x^2}+\frac{1}{x^2}} \\
&= -\sqrt{5+\frac{2}{x}+\frac{1}{x^2}} \\
\end{align*}$$
|
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|
Why does't quadratic formula work to factor polynomial when $a \ne 1$? $$2x^2 + 3x + 1$$
applying quadratic formula:
$$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$$
$$a=2, b=3, c=1$$
$$x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot1}}{2\cdot2}$$
$$x = \frac{-3 \pm \sqrt{9-8}}{4}$$
$$x = \frac{1}{4}[-3 + 1],~~~x=\frac{1}{4}[-3-1]$$
$$x_1 = -1/2,~~~x_2 = -1$$
therefore:
$$2x^2 + 3x + 1 = (x + 1/2)(x+1)$$
Now I check it:
$$2x^2 + 3x + 1 = (x + \frac{1}{2})(x+1) = x \cdot x + x \cdot 1 + \frac{1}{2}\cdot x + \frac{1}{2}\cdot 1$$
$$2x^2 + 3x + 1 = (x + \frac{1}{2})(x+1) = x^2 + x + \frac{1}{2}x + \frac{1}{2}$$
$$2x^2 + 3x + 1 = (x + \frac{1}{2})(x+1) = x^2 + x + \frac{1}{2}x + \frac{1}{2}$$
$$2x^2 + 3x + 1 = (x + \frac{1}{2})(x+1) = x^2 + \frac{3}{2}x + \frac{1}{2}$$
but:
$$2x^2 + 3x + 1 \ne x^2 + \frac{3}{2}x + \frac{1}{2}$$
Why does't quadratic formula work when $a \ne 1$?
however, I can pull out the 1/2.
$$2x^2 + 3x + 1 \ne \frac{1}{2}[2x^2 + 3x + 1]$$
I feel that this is something I must have missed in grade school...
Does this mean you need to multiple by "a" if $a \ne 1$?**
|
Algebra students learn that the quadratic formula says
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$
When is this true? It's helpful to remember that the quadratic formula comes from the following statement, which is true for any complex numbers $a$, $b$, $c$, and $x$ (as long as $a\ne0$):
$$\text{If }ax^2+bx+c=0,\quad\text{then }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$
So the quadratic formula is good for getting the roots of a quadratic polynomial. Now if your goal is to factor a quadratic polynomial, then it's helpful to know the following:
$$\text{If }p,\,q\text{ are the roots of }ax^2+bx+c,\quad\text{then }ax^2+bx+c=a(x-p)(x-q).$$
Hence, if you want to factor $2x^2+3x+1$, then you could note that
$$2x^2+3x+1=2(x-p)(x-q)$$
where $p$, $q$ are the roots of $2x^2+3x+1$. Since the quadratic formula says that the roots of $2x^2+3x+1$ are $-\frac{1}{2}$ and $-1$, we have that
$$2x^2+3x+1=2\left(x+\frac{1}{2}\right)\left(x+1\right).$$
|
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|
Prove that $A^2=a_1^2+a_2^2-a_3^2-a_4^2$ for all integers $A$.
Prove that:
For every $A\in\mathbb Z$, there exist infinitely many $\{a_1,a_2,a_3,a_4\}\subset\mathbb Z$ given $a_m\neq a_n $ such that $$A^2=a_1^2+a_2^2-a_3^2-a_4^2$$
After many hours, I found a general formula satisfying the statement for all $A$ and $B$. $$A^2=(3A+B)^2+(9A+2B)^2-(5A+B)^2-(8A+2B)^2$$ This was derived by noticing the following pattern, to which I used the first equation below as a seed and multiplied through $A^2$. I was then able to find a second paramater $B$ due to the arithmetic progression of the pattern, in order to prove the infinitude of $\{a_n\}_{n=1}^{4}$. $$\begin{align}1^2+5^2+8^2 &= 3^2+9^2 \\ 1^2+6^2+10^2 &= 4^2+11^2 \\ 1^2+7^2+12^2 &= 5^2+13^2 \\ &\vdots\end{align}$$
My question is, is there a way one could prove the statement without the involvement of such curious patterns surrounding square numbers? Apologies if this question is somewhat vague.
Edit:
Fun fact, it appears there is also a general formula for the equation $$A=a_1^2+a_2^2+a_3^2-a_4^2+a_5^2$$ That is, $$A^2=(A+B)^2+(A+3B)^2+(A+8B)^2-(A+5B)^2-(A+7B)^2$$
Edit 2: It appears that the first general equation at which I arrived in this question is actually part of an even more general equation $$(pq +s)^2=\big\{p(3q+r)+3s\big\}^2+\big\{p(9q+r)+3s\big\}^2-\big\{p(5q+r)+s\big\}^2-\big\{2p(4q+r)+4s\big\}^2+4pqs$$ where $(p,q,r,s)=(1,A,B,0)$. Note that an interesting fact implies when $p$, $q$ and $s$ are square numbers.
|
Given any non-negative integer $A$, let $a_1=A+2n+1$ for any of $n=1,2,3,\cdots$.
For convenience let $b=a_1^2-A^2 = (A+2n+1)^2-A^2= (4n+2)A+4n^2+4n+1$. Thus $b$ is odd and $\geq9$.
Now let $a_2=(b-1)/2$ and $a_3=(b+1)/2$, implying that $a_2^2-a_3^2=-b$, and let $a_4=0$. Then:
$$a_1^2+a_2^2-a_3^2-a_4^2=a_1^2+(-b)-0=b+A^2-b=A^2$$
Also $a_3=a_2+1>a_2$, $a_1=A+2n+1>0=a_4$, and: $$a_2=(b-1)/2=(2n+1)A+2n^2+2n>A+2n+1=a_1$$ so the four terms are all unequal.
For negative $A$, let $a_1=|A|+2n+1$ and proceed in the same way.
|
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|
If $\cos^{-1}(\frac{x}{a})+\cos^{-1}(\frac{y}{b})=\theta$, then $\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{2xy} {ab}\cos(\theta)$ If $\cos^{-1}(\frac{x}{a})+\cos^{-1}(\frac{y}{b})=\theta$, prove
that $\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{2xy}{ab}\cos(\theta)=\sin^2(\theta)$
My trial:Let:
$\cos^{-1}(\frac{x}{a})=\alpha$, and $\cos^{-1}(\frac{x}{a})=\beta$
$\sin(\theta)= \sin(\alpha) \cos(\beta)+\cos(\alpha)\sin(\beta)$
$\sin (\theta)=\sqrt{1-\frac{y^2}{b^2}}\bigl(\frac{x}{a}\big)+\sqrt{1-\frac{x^2}{a^2}}\big(\frac{y}{b}\bigr)$
$\sin^2(\theta)=\bigl(1-\frac{y^2}{b^2}\bigr)\frac{x^2}{a^2}+\bigl(1-\frac{x^2}{a^2}\bigr)\frac{y^2}{b^2}+\frac{2xy}{ab}\sqrt{(1-\frac{y^2}{b^2})((1-\frac{x^2}{a^2})}$
$\sin^2(\theta)=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{2xy}{ab}\sqrt{(1-\frac{y^2}{b^2})((1-\frac{x^2}{a^2})}-\frac{2x^2y^2}{a^2b^2}$
I could not get the final solution, any idea, maybe there is a better algorithm?
|
Hint
Using $\cos(A+B)$ formula
$$\cos\theta-\dfrac{xy}{ab}=-\sqrt{\left(1-\dfrac{x^2}{a^2}\right)(\cdots)}$$
Take square in both sides
Alternatively
Let $\cos^{-1}\dfrac xa=A,\cos A=?$ etc.
$$\cos\theta=\cos(A+B)$$
Rearrange and square both sides
$$(\cos\theta-\cos A\cos B)^2=(-\sin A\sin B)^2=(1-\cos^2A)(1-\cos^2B)$$
Replace the values of $\cos A,\cos B$ to eliminate foreign elements $A,B$
|
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|
First order PDE problem $yu_{x} - xu_{y} = x^2$ I am a beginner to PDE, trying to solve $$yu_{x} - xu_{y} = x^2$$ using characteristic line, first I have \begin{align*}
\frac{dx}{y} &= -\frac{dy}{x} = \frac{du}{x^2}
\end{align*}
Then I get the characteristic line is given by $$C = \frac{1}{2}y^2 - \frac{1}{2}x^2$$
Next, I solve the first term and third term, I have \begin{align*}
\frac{du}{x^2} &= \frac{dx}{y} \\
du &= \frac{x^2}{y}dx
\end{align*}
Here is my problem, that is we can not integrate right-hand side without eliminating the variable $y$, but if we try to replace $y$ in terms of $x$ and $C$, the result does not look integrable and really messy.
Can someone help me, please? Thanks!
|
$$yu_x-xu_y=x^2$$
Charpit-Lagrange system of characteristic ODEs:
$$\frac{dx}{y}=\frac{dy}{-x}=\frac{du}{x^2}$$
A first characteristic equation comes from $\frac{dx}{y}=\frac{dy}{-x}$
$$x^2+y^2=c_1$$
A second characteristic equation comes from $\frac{dx}{\sqrt{c_1-x^2}}=\frac{du}{x^2}$
$du=\frac{x^2dx}{\sqrt{c_1-x^2}}\quad\implies\quad u=\int\frac{x^2dx}{\sqrt{c_1-x^2}}=-\frac{xy}{2}+\frac{x^2+y^2}{2}\tan^{-1}\left(\frac{y}{x}\right)+c_2$
$$u+\frac{xy}{2}-\frac{x^2+y^2}{2}\tan^{-1}\left(\frac{y}{x}\right) =c_2$$
The solution of the PDE expressed on the form of implicit equation $c_2=F(c_1)$ is :
$$u+\frac{xy}{2}-\frac{x^2+y^2}{2}\tan^{-1}\left(\frac{y}{x}\right) =F(x^2+y^2)$$
$$\boxed{u(x,y)=-\frac{xy}{2}+\frac{x^2+y^2}{2}\tan^{-1}\left(\frac{y}{x}\right) +F(x^2+y^2)}$$
$F$ is an arbitrary function.
The function $F$ has to be determined according to some boundary condition which is missing in the wording of the question.
|
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|
For △ABC, prove $\frac a{h_a} + \frac b{h_b} + \frac c{h_c} \ge 2 (\tan\frac{\alpha}2+ \tan\frac{\beta}2 + \tan\frac{\gamma}2)$
Given $\triangle ABC$, (using the main parameters and notation), prove that $$ \frac{a}{h_a} + \frac{b}{h_b} + \frac{c}{h_c} \ge 2 \cdot \left(\tan\frac{\alpha}{2} + \tan\frac{\beta}{2} + \tan\frac{\gamma}{2}\right)$$
Great, another problem that already has a solution. We have many ways of expressing the area of a triangle, for example $$2 \cdot [ABC] = \frac{abc}{2R} = \sqrt{2(a + b + c) \cdot \sum_{cyc}\frac{c + a - b}{2}} = ah_a = bh_b = ch_c$$
The above equations are used in the solution I have provided below.
I would be greatly appreciated if you could come up with any other solutions.
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In the standard notation we need to prove that:
$$\sum_{cyc}\frac{a}{\dfrac{2S}{a}}\geq2\sum_{cyc}\sqrt{\frac{1-\frac{b^2+c^2-a^2}{2bc}}{1+\frac{b^2+c^2-a^2}{2bc}}}$$ or
$$\sum_{cyc}a^2\geq4S\sum_{cyc}\sqrt{\frac{(a+b-c)(a+c-b)}{(b+c-a)(a+b+c)}}$$ or
$$\sum_{cyc}a^2\geq\sum_{cyc}(a^2-(b-c)^2),$$ which is obvious.
|
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|
Derivative of $\dfrac{\sqrt{3-x^2}}{3+x}$ I am trying to find the derivative of this function
$f(x)=\dfrac{\sqrt{3-x^2}}{3+x}$
$f'(x)=\dfrac{\dfrac{1}{2}(3-x^2)^{-\frac{1}{2}}\frac{d}{dx}(3-x)(3+x)-\sqrt{3-x^2}}{(3+x)^2}$
$=\dfrac{\dfrac{-2x(3+x)}{2\sqrt{3-x^2}}-\sqrt{3-x^2}}{(3+x)^2}$
$=\dfrac{\dfrac{-x(3+x)}{\sqrt{3-x^2}}-\sqrt{3-x^2}}{(3+x)^2}$
$\dfrac{-x(3+x)}{\sqrt{3-x^2}(3+x)^2}-\dfrac{\sqrt{3-x^2}}{(3+x)^2}$
$\dfrac{-x}{\sqrt{3-x^2}(3+x)}-\dfrac{\sqrt{3-x^2}}{(3+x)^2}$
At this point, I want to transform this derivative into the form of $\dfrac{3(x+1)}{(3+x)^2\sqrt{3-x^2}}$
How do I do this? This form is given by Wolfram:
https://www.wolframalpha.com/input/?i=derivative+%283-x%5E2%29%5E%281%2F2%29%2F%283%2Bx%29
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Wow. There are many good solutions here.
@James Warthington, I'd like to add another approach but using a more methodical application of the Chain Rule.
$y=(3+x)^{-1}(3-x^2)^ \frac{1}{2}$
First variable substitution, let $\mu=3+x$, and note(for later) that $\frac{d}{dx}\mu=1$
Second variable substitution, let $\omega=3-x^2$, and note(for later) that $\frac{d}{dx}\omega=-2x$
Our equation is now $y=\mu^{-1}\omega^ \frac{1}{2}$. Now let's apply the Differential Operator to both sides...
$\frac{d}{dx}y=\frac{d}{dx}[\mu^{-1}\omega^ \frac{1}{2}]$. $\space\space\space$ Apply the multiplicative operation of derivatives...
$y'=\mu^{-1}\frac{d}{dx}\omega^ \frac{1}{2}+\omega^ \frac{1}{2}\frac{d}{dx}\mu^{-1}$
You cannot $\frac{d}{dx}\mu$ or $\frac{d}{dx}\omega$ directly. You must apply the Chain Rule! This rule essentially says you can change the $x$ to $\mu$ but only if you multiply by $\frac{d\mu}{dx}$. Same thing applies to $\omega$.
$y'=\mu^{-1}\frac{d}{d\omega}\omega^ \frac{1}{2}\frac{d\omega}{dx}+\omega^ \frac{1}{2}\frac{d}{d\mu}\mu^{-1}\frac{d\mu}{dx}$
Now, this is very straight-forward stuff.
$\frac{d}{d\omega}\omega^ \frac{1}{2}$ is $\frac{1}{2\sqrt\omega}$ $\space\space\space$...and...$\space\space\space$ $\frac{d}{d\mu}\mu^{-1}$ is $\frac{-1}{\mu^{2}}$. $\space\space\space$Now let's substitute everything...
$y'=\left(\mu^{-1}\right)\left[\frac{1}{2\sqrt\omega}\right]\left(-2x\right)+ \omega^ \frac{1}{2} \left[ \frac{-1}{\mu^{2}} \right] \left(1\right)$. $\space\space\space\space\space$Simplifying...
$y'=\frac{-2x}{2\mu\sqrt\omega}$ + $\frac{-1\sqrt\omega}{\mu^2}$ $\space\space\space$...and factor out $\frac{-1}{\mu}$ ...$\space\space\space$ $y'=\left(\frac{-1}{\mu}\right)\left[\frac{x}{\sqrt\omega}+\frac{\sqrt\omega}{\mu}\right]$
Get a common denominator in the square brackets and multiply through...
$y'=-\frac{x\mu+\omega}{\mu^2\sqrt\omega}$. $\space\space\space\space\space$Now re-substitute variables...
$y'=-\frac{x(3+x)+\left(3-x^2\right)}{(3+x)^2\sqrt{3-x^2}}$
Expand and factor the numerator...
$y'=-\frac{3x+x^2+3-x^2}{(3+x)^2\sqrt{3-x^2}}$
$y'=-\frac{3(x+1)}{(3+x)^2\sqrt{3-x^2}}$
The technique used by @Claude Leibovici, logarithmic differentiation, is very interesting. I've never come across this approach and will certainly study it and add it to my bag of tricks!
|
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|
Evaluate $\int_{0}^{\infty} \frac{x^2+x+1}{x^6+x^4+1} dx$ $$\int_{0}^{\infty} \frac{x^2+x+1}{x^6+x^4+1} dx $$
Wolfram says $1.80276\ldots $
Calculating this seems complicated to me because the residues are pretty hard to find and I have tried the infinite residue method but it didn't work out.
|
Without complex analysis.
From a formal point of view, there is closed form solution.
Let $(a,b,c)$ to be the roots of $y^3+y^2+1=0$; one is real and negative (say $a$) and the other two $(b,c)$ are complex conjugate.
$$\frac{x^2+x+1}{x^6+x^4+1}=\frac{x^2+x+1}{(x^2-a)(x^2-b)(x^2-c)}$$ Now, using partial fraction decomposition, the rhs write
$$-\frac{1}{(a-b) (a-c) (b-c)}\left(\frac{(x+a+1) (c-b)}{x^2-a}+\frac{(b-a) (x+c+1)}{x^2-c}+\frac{(a-c) (x+b+1)}{x^2-b} \right)$$ So, we face three integrands of the form
$$\frac{A+B x}{x^2-d}=\frac{A}{x^2-d}+\frac B 2\frac {2x}{x^2-d }$$
$$\int_0^t \frac{A+B x}{x^2-d}\,dx=-\frac{A}{\sqrt{d}}\tanh ^{-1}\left(\frac{t}{\sqrt{d}}\right)+\frac B 2 \log \left(1-\frac{t^2}{d}\right)$$ Now, the problem is to make $t\to \infty$ and to simplify.
The real solution is given by
$$a=-\frac{1}{3} \left(1+2 \cosh \left(\frac{1}{3} \cosh
^{-1}\left(\frac{29}{2}\right)\right)\right)$$ I prefer to avoid typing the explicit solutions of$(b,c)$.
|
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|
Difference between sinc and $\cos$ can be expressed using Bessel function $J_{3/2}$ I've tried to prove this property of Bessel function but I don't seem to be going anywhere
$$\sqrt{\frac 12 \pi x} J_\frac 32 (x) = \cfrac{\sin x}{x} - \cos x$$
I have tried substituting $\frac 32$ for $J_n (x)$ and then manipulating with the product but it doesn't seem to give me something similar with the series on my LHS. I don't know if there is another different approach which I must follow.
|
Using the series expansion of $J_{3/2}(x)$ and the Legendre duplication formula for the gamma function, we find
$$
\sqrt {\frac{{\pi x}}{2}} J_{3/2} (x) = \sqrt {\frac{{\pi x}}{2}} \left( {\tfrac{1}{2}x} \right)^{3/2} \sum\limits_{n = 0}^\infty {( - 1)^n \frac{{\left( {\frac{1}{4}x^2 } \right)^n }}{{n!\Gamma \left( {n + \frac{5}{2}} \right)}}}
\\
= \sqrt \pi \sum\limits_{n = 0}^\infty {( - 1)^n \frac{{x^{2n + 2} }}{{2^{2n + 2} n!\Gamma \left( {n + \frac{5}{2}} \right)}}} = \sum\limits_{n = 0}^\infty {( - 1)^n \frac{{(2n + 2)}}{{(2n + 3)!}}x^{2n + 2} }
\\
= \sum\limits_{n = 1}^\infty {( - 1)^{n + 1} \frac{{2n}}{{(2n + 1)!}}x^{2n} } = \sum\limits_{n = 1}^\infty {\frac{{( - 1)^n }}{{(2n + 1)!}}x^{2n} } - \sum\limits_{n = 1}^\infty {( - 1)^n \frac{{x^{2n} }}{{(2n)!}}}
\\
= \sum\limits_{n = 0}^\infty {\frac{{( - 1)^n }}{{(2n + 1)!}}x^{2n} } - \sum\limits_{n = 0}^\infty {( - 1)^n \frac{{x^{2n} }}{{(2n)!}}} = \frac{{\sin x}}{x} - \cos x.
$$
|
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|
integration of. $\int \frac{x}{x^3-3x+2}$ I am trying to integrate :
$\Large \int \frac{x}{x^3-3x+2}dx$
I decomposed the fraction and got :
$\Large \frac{x}{x^3-3x+2} = \frac {x}{(x-1)^2(x+2)}$
Then I tried to get two different fractions:
$ \Large \frac {x}{(x-1)^2(x+2)} = \frac {Ax}{(x-1)^2} + \frac {B}{(x+2)}$
Well I got A=1/3 and B=-1/3. as possible values. but that was not correct. I guess I missed something.
|
You don't have all the components in the decomposition. It should be
$$\frac {x}{(x-1)^2(x+2)} = \frac {Ax+C}{(x-1)^2} + \frac {B}{x+2}
=\frac {\frac29x+\frac19}{(x-1)^2} - \frac {\frac29}{x+2}$$
|
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If $\frac{1}{a+3}+\frac{1}{b+4}+\frac{1}{c+5}=\frac{7}{12}$ for positive integer $a$, $b$, $c$, then find $\frac{a}{a+3}+\frac{b}{b+4}+\frac{c}{c+5}$
Given $a, b, c$ are positive integers, and that
$$\frac{1}{a + 3} + \frac{1}{b+4} + \frac{1}{c+5} = \frac{7}{12},$$
compute:
$$\frac{a}{a+3} + \frac{b}{b+4} + \frac{c}{c+5}$$
Source (Romanian Math Magazine, Gazeta Matematica S:E19.333, this problem marked as targetting 6th graders)
I tried writing the terms as $1 - \frac{1}{a+3}$ and similarly for the others, but I get $3 - \frac{3}{a+3} - \frac{4}{b + 4} - \frac{5}{c + 5}$ which I don't find very helpful.
Many thanks in advance!
|
It is given that $a,b,c \geq 1$. If $a \geq 2$. Then:
$$\frac{1}{a+3}+\frac{1}{b+4}+\frac{1}{c+5} \leq \frac{1}{5}+\frac{1}{5}+\frac{1}{6}=\frac{17}{30} < \frac{7}{12}$$
So $a=1$ and
$$\frac{1}{b+4}+\frac{1}{c+5}=\frac{1}{3}$$
If $b \geq 3$, we get:
$$\frac{1}{b+4}+\frac{1}{c+5} \leq \frac{1}{7}+\frac{1}{6} = \frac{13}{42} < \frac{1}{3}$$
So $b\in \{1,2\}$. Checking $b=1$, we don't get an integer $c$, so $b=2$ and $c=1$. The final answer is:
$$\frac{a}{a+3}+\frac{b}{b+4}+\frac{c}{c+5} = \frac{3}{4}$$
|
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|
Minimize $\frac{2}{1-a}+\frac{75}{10-b}$
Let $a,b>0$ and satisfy $a^2+\dfrac{b^2}{45}=1$. Find the minimum
value of $\dfrac{2}{1-a}+\dfrac{75}{10-b}.$
WA gives the result that $\min\left(\dfrac{2}{1-a}+\dfrac{75}{10-b}\right)=21$ with $a=\dfrac{2}{3},b=5$.
Consider making a transformation like this
$$\dfrac{2}{1-a}+\dfrac{75}{10-b}=\frac{2\cdot (ma)^2}{ma\cdot ma\cdot (1-a)}+\frac{75\cdot(nb)^2}{nb\cdot nb\cdot(10-b)}.$$
If $2m=2n=1$, $ma=1-a$, and $ nb=10-b$ can all hold, we can apply AM-GM inequality as follows
$$\frac{2\cdot (ma)^2}{ma\cdot ma\cdot (1-a)}+\frac{75\cdot(nb)^2}{nb\cdot nb\cdot(10-b)}\geq \frac{2\cdot (ma)^2}{\left(\frac{ma
+ma+(1-a)}{3}\right)^3}+\frac{75\cdot(nb)^2}{\left(\frac{nb+nb+(10-b)}{3}\right)^3}=\cdots$$
But this is invalid, since $n=\frac{1}{2},nb=10-b$ can not satisfy $b=5$.
How to solve it?
|
I come up with a solution which is based on the clue I posted above.
\begin{align*}
\frac{2}{1-a}+\frac{75}{10-b}&=\frac{2}{1-a}+\frac{15}{2}\cdot\frac{b}{10-b}+\frac{15}{2}\\
&=\frac{2\cdot \frac{a}{2}\cdot \frac{a}{2}}{(1-a)\cdot \frac{a}{2}\cdot\frac{a}{2}}+\frac{15}{2}\cdot\frac{b\cdot b}{(10-b)\cdot b}+\frac{15}{2}\\
&\ge\frac{\frac{a^2}{2}}{\left(\frac{(1-a)+\frac{a}{2}+\frac{a}{2}}{3}\right)^3}+\frac{15}{2}\cdot\frac{b^2}{\left(\frac{(10-b)+ b}{2}\right)^2}+\frac{15}{2}\\
&\ge \frac{27}{2}\left(a^2+\frac{b^2}{45}\right)+\frac{15}{2}\\
&=\frac{27}{2}\cdot1+\frac{15}{2}\\
&=21,
\end{align*}
which hold as equality iff $1-a=\dfrac{a}{2},10-b=b,$ namely $a=\dfrac{3}{2},b=5$.
|
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|
Finding an Explicit Formula for Probability Given the Recursive Formula I am attempting to solve a problem asking for the formula to find a probability, $p(n)$, $n$ being 3, 4, 5,... I was able to determine that $p(3) = 1/4$ and the $p(n)$ given not $p(n-1)= \frac{n-2}{2(n-1)}$. From that, I can write the recursive formula:
$$p(n) = p(n-1) + \frac{(1-p(n-1))(n-2)}{2(n-1)}$$
How can I turn this into an explicit formula for $p(n)$? Thanks.
|
I will post a solution by @robjohn
Let $t(n) = 1 - p(n)$, then we rewrite the equation as
$$ t(n) = \frac{n}{2(n-1)} t(n-1) = \frac{n}{2(n-1)}\cdot\frac{n-1}{2(n-2)}\cdot t(n-2)$$
you proceed to a generic step $k$
$$t(n) = \frac{n}{2(n-1)}\cdot\frac{n-1}{2(n-2)}\cdot\ldots\cdot\frac{n-k+1}{2(n-k)}\cdot t(n-k) = \frac{n}{2^k (n-k)}\cdot t(n-k)$$
Now, substitute $k = n-3$:
$$t(n) = \frac{n}{2^{n-3} \cdot 3}\cdot t(3) = \frac{n}{2^{n-3} \cdot 3} \cdot \frac{3}{4} = \frac{n}{2^{n-1}}$$
Therefore, $$p(n) = \frac{2^{n-1}-n}{2^{n-1}}.$$
|
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|
Find $\int_0^{2\pi} \frac{1}{3 + \cos x} dx$. I have to find the integral
$$\int_0^{2\pi} \dfrac{1}{3 + \cos x} dx$$
I tried using the Weierstrass subtitution, but replacing the bounds, I get:
$$t_1 = \tan \dfrac{0}{2} = \tan 0 = 0$$
$$t_2 = \tan \dfrac{2 \pi}{2} = \tan \pi = 0$$
Resulting in the integral:
$$\int_0^0 \dfrac{1}{3 + \dfrac{1 - t^2}{1 + t^2}} \cdot \dfrac{2}{1 + t^2} dt$$
which obviously equals $0$ since the bounds are the same. Is this correct? It feels wrong.
|
By the tangent half-angle substitution we obtain:
\begin{align}2\int_0^{\pi} \dfrac{1}{3 + \cos x} \,dx&=2\int_0^{\infty}\frac{1}{3+\frac{1-t^2}{1+t^2}}\frac{2}{1+t^2}\,dt\\&=
2\int_0^{\infty}\frac{1}{t^2+2}\,dt\\&=
2\left(\frac{1}{\sqrt{2}}\arctan\left(\frac{x}{\sqrt2{}}\right)\right)\Bigg|_0^{\infty}
\\&=2\left(\frac{1}{\sqrt{2}}\frac{\pi}{2}-0\right)\\&=
\frac{\pi}{\sqrt{2}}
\end{align}
|
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|
Find all values of $m$ such that the equation $ mx^4 + x^3 + (8m - 1)x^2 + 4x + 16m = 0$ has nonnegative roots.
Find all values of $m$ such that the equation $$\large mx^4 + x^3 + (8m - 1)x^2 + 4x + 16m = 0$$ has nonnegative roots.
For an equation to have nonnegative roots, it mustn't only have negative roots.
Let $y = x^2 - x + 4$ $(y > 0)$, we have that $$mx^4 + x^3 + (8m - 1)x^2 + 4x + 16m = mx^2 + (2m - 1)xy + my^2$$
For the equation $$mx^2 + (2m - 1)xy + my^2 = 0$$ to have only negative roots, it must be satisfied that $\left\{ \begin{align} [(2m - 1)y]^2 - 4m^2y^2 \ge 0\\ (1 - 2m)y < 0\\ my^2 > 0 \end{align} \right.$ $\implies \left\{ \begin{align} (1 - 4m)y^2 \ge 0\\ 1 - 2m < 0\\ m > 0 \end{align} \right.$ $\implies \left\{ \begin{align} 1 - 4m \ge 0\\ m > \dfrac{1}{2}\\ m > 0 \end{align} \right.$ $\iff \left\{ \begin{align} m \le \frac{1}{4}\\ m > \dfrac{1}{2} \end{align} \right. \implies m \in \varnothing$.
Thus for $\forall m \in \mathbb R$, the equation $$mx^4 + x^3 + (8m - 1)x^2 + 4x + 16m = 0$$ has nonnegative roots.
Is the above solution correct? And should my attempt was inaccurate and yours is (hopefully) helpful, please post an answer.
|
Like Quadratic substitution question: applying substitution $p=x+\frac1x$ to $2x^4+x^3-6x^2+x+2=0$
divide both sides by $x^2$
$$0=m(x^2+16/x^2)+x+4/x+8m-1=m(x+4/x)^2+x+4/x-1$$
Now as $x\ge0,$ $$\dfrac{x+4/x}2\ge\sqrt{x\cdot4/x}=?$$
|
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|
Find the equation of tangents for $x^3+y^3-3xy=0$ at $x=0,y=0$ $$x^3+y^3-3xy=0$$
Find the equation of tangents at $x=0,y=0$
My attempt is as follows:-
Attempt $1$:
$$3x^2+3y^2\dfrac{dy}{dx}-3\left(x\dfrac{dy}{dx}+y\right)=0$$
$$\dfrac{dy}{dx}(y^2-x)=y-x^2$$
$$\dfrac{dy}{dx}=\dfrac{y-x^2}{y^2-x}$$
but when placing $x=0,y=0$ we are getting undefined quantity.
But actual answer is $xy=0$, how can we proceed here?
Attempt $2$: (Parametric method)
$$\dfrac{y}{x}=t$$
$$y=xt$$
Putting this in the original equation
$$x^3+x^3t^3-3x^2t=0$$
$$x^2(x+xt-t)=0$$
$$x=\dfrac{t}{t+1}$$
$$y=\dfrac{t^2}{t+1}$$
$$\dfrac{dy}{dx}=\dfrac{\dfrac{2t(t+1)-t^2}{(t+1)^2}}{\dfrac{t+1-t}{(t+1)^2}}$$
$$\dfrac{dy}{dx}=t^2+2t$$
$x=0$, then $\dfrac{t}{t+1}=0 \implies t=0$
$y=0$, then $\dfrac{t^2}{t+1}=0$ also $\implies t=0$
So we are getting slope as $0$, hence $y=0$ can be the answer, but actual answer is $xy=0$
|
Continue with what you obtained
$$y'= \frac{y-x^2}{y^2-x} $$
and note that the near the origin, $y' = \frac yx$, or
$$\frac{y-x^2}{y^2-x} = \frac yx \implies x^3+y^3 = 2xy$$
Then, substitute $x^3+y^3 - 3xy=0$ to obtain the equation near the origin,
$$xy=0$$
which represents the $x$- and the $y$-axes and also the tangent lines of the given curve at the origin.
|
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Solving the sequence $a_{n+1}=a_{n}a_{n-1}+\sqrt{(a_n^2-1)(a_{n-1}^2-1)}$: proving that $2+2a_n$ is a perfect square Question:
Let $a_1=a_2=97$ and $a_{n+1}=a_{n}a_{n-1}+\sqrt{(a_n^2-1)(a_{n-1}^2-1)}$ for $n>1$. Prove that
(a) $2+2a_n$ is a perfect square, and (b) $2+\sqrt{2+2a_n}$ is a perfect square.
I changed the given recursive formula by squaring, and the result was as follows:
$$(a_{n+1}^2+a_n^2+a_{n-1}^2)-2a_{n+1}a_na_{n-1}-1=0$$
$\Rightarrow$ $$(a_{n+1}+a_n+a_{n-1})^2-2(a_{n+1}a_n+a_na_{n-1}+a_{n-1}a_{n+1}+a_{n+1}a_na_{n-1})-1=0$$
$\Rightarrow$ $$(1+a_{n+1})^2+(1+a_n)^2+(1+a_{n-1})^2-2(a_{n+1}+a_n+a_{n-1}+a_{n+1}a_n+a_na_{n-1}+a_{n-1}a_{n+1}+a_{n+1}a_na_{n-1}+1)-2=0$$
$\Rightarrow$ $$(1+a_{n+1})^2+(1+a_n)^2+(1+a_{n-1})^2-2(1+a_{n+1})(1+a_n)(1+a_{n-1})-2=0$$
And consequently, I lost the way :(
I thought of proving in a inductive way, that is :
$$2+2a_1=196=14^2$$
When we set $2+2a_n=k^2$, $2+2a_{n-1}=l^2$,
$$ 2+2_{n+1}=\frac{(k^2-2)(m^2-2)}{4}+\sqrt{\left(\left(\frac{k^2-2}{2}\right)^2-1\right)\left(\left(\frac{m^2-2}{2}\right)^2-1\right)}$$
As a result, I again lost the way :/
I think I sill have not got the main points. Could you give me some clues about this problem? Thanks.
|
Simply redoint the argument of @Martin R: so I understand it better.
If $a = \frac{1}{2}(s +1/s)$ then $a^2-1= \left(\frac{1}{2}(s - 1/s)\right)^2$. Now for
$a = \frac{1}{2}(s + 1/s)$, $b = \frac{1}{2}(t + 1/t)$ ( $s$, $t> 1$) we get
$$a b + \sqrt{(a^2-1)(b^2-1)}= \frac{1}{4} (s+1/s)(t+ 1/t) + \frac{1}{4} (s-1/s)(t- 1/t)=\\ = \frac{1}{2}( s t + 1/(s t) )$$
|
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|
Integer solutions to the product of four consecutive integers $ \bullet \textbf{Question} $
The product of four consecutive integers $ x, x + 1, x + 2, x + 3 $ can be written as the product of two consecutive integers, find all integer solutions for $ x $.
$ \bullet \textbf{Rephrasing} $
I decided to name the other integer as $ y $ so that this equation is a plot on a graph,
$$ y(y + 1) = x(x + 1)(x + 2)(x + 3) \tag{1} $$
$ \bullet \textbf{Attempt} $
To solve for $ y $ I did these simple steps,
$$ y^2 + y = x(x + 1)(x + 2)(x + 3) \tag{2} $$
$$ y^2 + y - x(x + 1)(x + 2)(x + 3) = 0 \tag{3} $$
$$ y = \frac{-1 \pm{\sqrt{1 + 4x(x + 1)(x + 2)(x + 3)}}}{2} \tag{4} $$
I then figured that when,
$$ -1 \pm{\sqrt{1 + 4x(x + 1)(x + 2)(x + 3)}} \equiv 0\pmod{2} \tag{5} $$
then $ y $ will be a integer, so this is just solving for all $ x $, so $ y $ will be a integer but not for only integer $ x $'s.
This then means that
$$ \pm{\sqrt{1 + 4x(x + 1)(x + 2)(x + 3)}} \equiv 1\pmod{2} \tag{6} $$
$$ {\sqrt{1 + 4x(x + 1)(x + 2)(x + 3)}} \equiv 1\pmod{2} \tag{7} $$
This is the end of my knowledge, if you try to square it will include the integer solution for $ y $ but also include non-integer solutions for $ y $.
Am I going in the correct direction, or is this a dead end and I need to apply a new approach?
|
We have $$y^2+y=(x^2+3x)(x^2+3x+2)$$ or$$y^2+y+1=(x^2+3x+1)^2$$ or
$$(2y+1)^2+3=(2x^2+6x+2)^2$$ or
$$(2x^2+6x-2y+1)(2x^2+6x+2y+3)=3.$$
Now, solve a number of systems.
|
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Solve $xy''-y'-x^3y=0$ I want to solve $xy''-y'-x^3y=0$. My solution:
$y = z\cdot \exp(kx^2)$
$y' = z'\cdot \exp(kx^2) + z\cdot 2kx\exp(kx^2)$
$y'' = z''\cdot \exp(kx^2) + z'\cdot 4kx\exp(kx^2) + z\cdot (2k+4k^2x^2)\exp(kx^2)$
Plug in:
$z''\exp(kx^2)+z'(4kx-\frac{1}{x})\exp(kx^2)+z(2k+4k^2x^2-\frac{1}{x}\cdot2kx-x^2)\exp(kx^2)=0
$
$2k+4k^2x^2-\frac{1}{x}\cdot2kx-x^2=0\iff k=\frac{1}{2}$
Plug in:
$z''+z'(2x-x^{-1})=0$
$z'=:u$
$u'+u(2x-x^{-1})=0$
$\frac{\mathrm{d}u}{u}=-(2x-x^{-1})\mathrm{d}x$
$\ln|u|=-(x^2-\ln|x|)+c_1$
$u=\frac{c_1x}{e^{x^2}}$
Substitute back:
$z'=c_1xe^{-x^2}$
$z=-\tfrac{1}{2}c_1e^{-x^2}+c_2$
$y(x)=(-\tfrac{1}{2}c_1e^{-x^2}+c_2)e^{\frac{1}{2}x^2}$
$y(x)=c_2e^{\frac{1}{2}x^2}-\tfrac{1}{2}c_1e^{-\frac{1}{2}x^2}$
Right? Is there another way solving this ode? Thanks!
edit: @Lutz Lehmann: Sorry there was a square missing. Fixed it.
|
We can really smoke this with the substitution $u=x^2$:
$$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=2x\frac{dy}{du}$$
$$\frac{d^2y}{dx^2}=\frac{d}{dx}\left(2x\frac{dy}{du}\right)=2\frac{dy}{du}+4x^2\frac{d^2y}{du^2}$$
So
$$x\frac{d^2y}{dx^2}-\frac{dy}{dx}-x^3y=4x^3\frac{d^2y}{du^2}-x^3y=4x^3\left(\frac{d^2y}{du^2}-\frac14y\right)=0$$
With solution
$$y=c_1e^{\frac12u}+c_2e^{-\frac12u}=c_1e^{\frac12x^2}+c_2e^{-\frac12x^2}$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
If $h(x) = \dfrac{f(x)}{g(x)}$, then find the local minimum value of $h(x)$ Let $f(x) = x^2 + \dfrac{1}{x^2}$ and $g(x) = x – \dfrac{1}{x}$
, $x\in R – \{–1, 0, 1\}$. If $h(x) = \dfrac{f(x)}{g(x)}$, then the local minimum value
of $h(x)$ is :
My attempt is as follows:-
$$h(x)=\dfrac{x^2+\dfrac{1}{x^2}}{x-\dfrac{1}{x}}$$
$$h(x)=\dfrac{x^2+x^2-2+2}{x-\dfrac{1}{x}}$$
$$h(x)=x-\dfrac{1}{x}+\dfrac{2}{x-\dfrac{1}{x}}$$
Case $1$: $x-\dfrac{1}{x}>0$
$$\dfrac{x^2-1}{x}>0$$
$$x\in(-1,0) \cup (1,\infty)$$
$$AM\ge GM$$
$$\dfrac{x-\dfrac{1}{x}+\dfrac{2}{x-\dfrac{1}{x}}}{2}>\sqrt{2}$$
$$h(x)\ge 2\sqrt{2}$$
$$x-\dfrac{1}{x}=\dfrac{2}{x-\dfrac{1}{x}}$$
$$x^2+\dfrac{1}{x^2}-2=2$$
$$x^2+\dfrac{1}{x^2}=4$$
$$x^4-4x^2+1=0$$
$$x^2=2\pm\sqrt{3}$$
Only $x=\sqrt{2+\sqrt{3}},-\sqrt{2-\sqrt{3}}$ are the valid solutions.
Case $2$: $x-\dfrac{1}{x}<0$
$$x\in(-\infty,-1) \cup (0,1)$$
$$h(x)=-\left(\dfrac{1}{x}-x+\dfrac{2}{\dfrac{1}{x}-x}\right)$$
By $AM\ge GM$, $h(x)\ge-2\sqrt{2}$
We will get this minimum value at $-\sqrt{2+\sqrt{3}},\sqrt{2-\sqrt{3}}$
So answer should have been $-2\sqrt{2}$ but actual answer is $2\sqrt{2}$. What am I missing here.
|
Note that
$$\frac{f(x)}{g(x)}=\frac{(x-1/x)^2+2}{x-1/x}=x-\frac1x+2\frac{1}{x-\frac1x}.
$$
Letting $u=x-1/x$ we take the derivative of $u+2/u$ to get $u'-\frac{2}{u^2}u'=u'(1-2/u^2)=0$. As $u'$ is always positive we must have $2=u^2=(x-1/x)^2$ which is easily solved; $x=\pm\sqrt{2\pm\sqrt3}$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Trouble with trig substitution I have a few questions and a request for an explanation.
I worked this problem for a quite a while last night. I posted it here.
Here is the original question: $$\int\frac{-7 x^2}{\sqrt{4x-x^2}} dx$$
And here is the work that I did on it:
Help with trig sub integral
Sorry that the negative sometimes gets cut off in the photo, and yes I know it's not fully simplified there.
\begin{align}
-7 \int\frac{x^2}{\sqrt{4x-x^2}} dx
&= -7 \int\frac{x^{3/2}}{\sqrt{4-x}} dx \\
&= -7 \int\frac{8\sin^3\theta\ 2\cos\theta}
{\sqrt{4-4\sin^2\theta}} d\theta
&& \begin{array}{c}
2\sin\theta = \sqrt x \\
2\cos\theta = dx \\
(2\sin\theta)^3 = x^{3/2}
\end{array} \\
&= -7 \cdot 8 \int\frac{\sin^3\theta\ 2\cos\theta}
{2\sqrt{1 - \sin^2\theta}} d\theta \\
&= -56 \int \sin^3\theta\, d\theta \\
&= -56 \int (1 - \cos^2\theta) \sin\theta\, d\theta \\
&= -56 \int(\sin\theta - \cos^2\theta\sin\theta)\,d\theta \\
&= 56 \cos\theta - 56 \int u^2\, du &&
\begin{array}{c}
u = \cos\theta \\
du = \sin\theta
\end{array}\\
&= 56 \cos\theta - 56 \frac{\cos^3\theta}{3} + C \\
&= 56 \left(\frac{\sqrt{4-x}}{2}
- \frac13 \left( \frac{\sqrt{4-x}}{2}\right)^3 \right)+ C
&& \cos\theta = \frac{\sqrt{4-x}}{2}\\
&= 56 \frac{\sqrt{4-x}}{2}
\left(1 - \frac{4-x}{12}\right) + C\\
\end{align}
My first question is the more involved one: Is the algebra in my original work sound? If it is, why doesn't it work in this instance?
My second question is: is this a correct solution?
$$14\left(\frac{\sqrt{4x-x^2}(x-2)}{2}-2\sqrt{4x-x^2}-3\arcsin\left(\frac{x-2}{2}\right)\right)+C$$
It is for webwork, and I used two out of three chances. I'd prefer to keep my perfect webwork mark, obviously ;p
Finally, I was kind of impressed with Ans4's square completion and had to run it through to see that it was correct. That's such a useful skill. Do you have some specific advice about how I could improve my math tricks to that point?
|
Disclaimer: Since you are saying that it is your last attempt on Webwork, so I would suggest that you should read through the steps carefully to see if there is any typo or mistake on my part.The approach is along the same lines however.
Since $4x-x^2=-(x^2-4x+2^2-2^2) =-(x-2)^2+2^2 $
So
$\int \frac{x^2}{\sqrt{4x-x^2}}dx =\int \frac{x^2}{\sqrt{2^2-(x-2)^2}}dx$
Let $x-2 = 2 \sin u$ then $dx = 2 \cos u du$
Now back to integral
$\int \frac{x^2}{\sqrt{2^2-(x-2)^2}}dx = \int \frac{(2+2\sin u)^2}{\sqrt{2^2-(2\sin u)^2}}2\cos udu$
$= \int \frac{(2+2\sin u)^2}{\sqrt{2^2(1-\sin^2 u)}}2\cos udu$
$= \int \frac{(2+2\sin u)^2}{\sqrt{2^2(\cos^2 u)}}2\cos udu$
$= \int \frac{(2+2\sin u)^2}{2(\cos u)}2\cos udu$
$= \int (2+2\sin u)^2du$
$= 4(\int (1+\sin^2 u+2\sin u ))du$
$= 4(\int 1 du + \int \sin^2 u du +2 \int \sin u) du$
$= 4(u + \int \sin^2 u du -2 \cos u )$
$= 4(u + \int \sin^2 u du -2 \cos u )$
$=4(u + \frac{1}{2}\left(\int (1-\cos 2u) du \right) -2 \cos u) $
$=4( u + \frac{1}{2}\left(u- \frac{\sin 2u}{2} \right) -2 \cos u )$
$= 4(u + \frac{1}{2}\left(u- \frac{2\sin u\cos u}{2} \right) -2 \cos u )$
$=4( u + \frac{1}{2}\left(u- {\sin u\cos u} \right) -2 \cos u )$
$=4( \frac{3u}{2} - \frac{1}{2}{\sin u\cos u} -2 \cos u )$
$= 6u - 2\sin u\cos u -8 \cos u $
Since $\sin u = \frac{x -2}{2}$ So
$\cos u = \sqrt{1-\sin^2 u} = \sqrt{1-(\frac{x -2}{2})^2} = \sqrt{\frac{4 -x^2 +4x -4}{4}} =\frac{\sqrt{4x -x^2}}{2}$
and
$u = \arcsin(\frac{x -2}{2})$
which leads to
$\int \frac{x^2}{\sqrt{4x-x^2}}dx = 6 \arcsin(\frac{x -2}{2}) - 2 \frac{x -2}{2}\frac{\sqrt{4x -x^2}}{2} -8 \frac{\sqrt{4x -x^2}}{2} $
$= 6 \arcsin(\frac{x -2}{2}) - \frac{(x -2)\sqrt{4x -x^2}}{2} -4 \sqrt{4x -x^2}$
Now you can combine your original integral and the integral constant $C$ as
$\int \frac{-7x^2}{\sqrt{4x-x^2}}dx = -7\left( 6 \arcsin(\frac{x -2}{2}) - \frac{(x -2)\sqrt{4x -x^2}}{2} -4 \sqrt{4x -x^2} \right)+C$
|
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|
An approximation related to Euler's constant and the Harmonic number Let's consider Euler's constant $\gamma$, i.e.,
$$\gamma=\lim_{n\to \infty} \sum_{k=1}^n\frac{1}{k}-\ln(n).$$
Prove the following approximation:
$$\sum_{k=1}^{m-1}\frac{1}{k}-\ln(m)+\frac{1}{2m}+\frac{1}{12m^2}\approx \gamma.$$
The above approximation can be found in many places, e.g. John D. Cook's blog and appears back in Concrete Mathematics asymptotics chapter as a non-trivial exercise of Euler's summation formula. While there are more efficient algorithms that estimates Euler's constant, this approximation allows also one way to look at large values of the Harmonic number (as mentioned in John's blog).
|
We have
\begin{align}
\sum\limits_{k = 1}^m {\frac{1}{k}} - \log m &= \sum\limits_{k = 1}^m {\frac{1}{k}} - \log \prod\limits_{k = 2}^m {\frac{k}{{k - 1}}}
\\
&= \sum\limits_{k = 1}^m {\frac{1}{k}} - \sum\limits_{k = 2}^m {\log \frac{k}{{k - 1}}} \\&= 1 + \sum\limits_{k = 2}^m {\left[ {\frac{1}{k} - \log \frac{k}{{k - 1}}} \right]} \\ &= 1 + \sum\limits_{k = 2}^\infty {\left[ {\frac{1}{k} - \log \frac{k}{{k - 1}}} \right]} - \sum\limits_{k = m + 1}^\infty {\left[ {\frac{1}{k} - \log \frac{k}{{k - 1}}} \right]}
\\
&= 1 + \sum\limits_{k = 2}^\infty {\left[ {\frac{1}{k} + \log \left( {1 - \frac{1}{k}} \right)} \right]} - \sum\limits_{k = m + 1}^\infty {\left[ {\frac{1}{k} + \log \left( {1 - \frac{1}{k}} \right)} \right]} .
\end{align}
By Taylor's theorem
$$
\frac{1}{k} + \log \left( {1 - \frac{1}{k}} \right) = - \frac{1}{{2k^2 }} + \mathcal{O}\!\left( {\frac{1}{{k^3 }}} \right),
$$
whence the infinite series is convergent and we can write
$$
\sum\limits_{k = 1}^m {\frac{1}{k}} - \log m = \gamma - \sum\limits_{k = m + 1}^\infty {\left[ {\frac{1}{k} + \log \left( {1 - \frac{1}{k}} \right)} \right]} ,
$$
with some constant $\gamma$. By Taylor's formula,
$$
\frac{1}{k} + \log \left( {1 - \frac{1}{k}} \right) = - \sum\limits_{j = 2}^\infty {\frac{1}{{jk^j }}} ,
$$
hence
\begin{align}
\sum\limits_{k = 1}^m {\frac{1}{k}} - \log m - \gamma = \sum\limits_{k = m + 1}^\infty {\sum\limits_{j = 2}^\infty {\frac{1}{{jk^j }}} } = \sum\limits_{j = 2}^\infty {\frac{1}{j}\sum\limits_{k = m + 1}^\infty {\frac{1}{{k^j }}} } = \sum\limits_{j = 2}^\infty {\frac{1}{{j!}}\sum\limits_{k = m + 1}^\infty {\frac{{(j - 1)!}}{{k^j }}} } .
\end{align}
By the Euler integral
$$
\frac{{(j - 1)!}}{{k^j }} = \int_0^{ + \infty } {e^{ - kt} t^{j - 1} dt} ,
$$
whence, using the geometric series and the Taylor series of the exponential function,
\begin{align}
\sum\limits_{k = 1}^m {\frac{1}{k}} - \log m - \gamma &= \sum\limits_{j = 2}^\infty {\frac{1}{{j!}}\sum\limits_{k = m + 1}^\infty {\int_0^{ + \infty } {e^{ - kt} t^{j - 1} dt} } } \\& = \sum\limits_{j = 2}^\infty {\frac{1}{{j!}}\int_0^{ + \infty } {\frac{{e^{ - (m + 1)t} }}{{1 - e^{ - t} }}t^{j - 1} dt} }
\\ &= \int_0^{ + \infty } {\frac{{e^{ - (m + 1)t} }}{{1 - e^{ - t} }}\frac{1}{t}\sum\limits_{j = 2}^\infty {\frac{{t^j }}{{j!}}} dt} \\ &= \int_0^{ + \infty } {\frac{{e^{ - mt} }}{{e^t - 1}}\frac{{e^t - t - 1}}{t}dt} \\ & = \int_0^{ + \infty } {e^{ - mt} \left( {1 - \frac{t}{{e^t - 1}}} \right)\frac{1}{t}dt} .
\end{align}
Now for $0<t<2\pi$,
$$
\left( {1 - \frac{t}{{e^t - 1}}} \right)\frac{1}{t} = \frac{1}{2} - \sum\limits_{n = 1}^\infty {\frac{{B_{2n} }}{{(2n)!}}t^{2n - 1} } ,
$$
with $B_n$ being the Bernoulli numbers. Noting that our function tends to zero at infinity and employing Taylor's theorem, we have that
$$
\left| {\left( {1 - \frac{t}{{e^t - 1}}} \right)\frac{1}{t} - \left( {\frac{1}{2} - \sum\limits_{n = 1}^{N - 1} {\frac{{B_{2n} }}{{(2n)!}}t^{2n - 1} } } \right)} \right| \le C_N t^{2N - 1}
$$
for $t>0$ and each positive $N$ with a suitable positive constant $C_N$. Therefore, using the Euler integral,
\begin{align}
\sum\limits_{k = 1}^m {\frac{1}{k}} - \log m - \gamma &= \int_0^{ + \infty } {e^{ - mt} \left( {\frac{1}{2} - \sum\limits_{n = 1}^{N - 1} {\frac{{B_{2n} }}{{(2n)!}}t^{2n - 1} } } \right)dt} + \mathcal{O}(1)\int_0^{ + \infty } {e^{ - mt} t^{2N - 1} dt}
\\ &= \frac{1}{2}\int_0^{ + \infty } {e^{ - mt} dt} - \sum\limits_{n = 1}^{N - 1} {\frac{{B_{2n} }}{{(2n)!}}\int_0^{ + \infty } {e^{ - mt} t^{2n - 1} dt} } \\ &\quad \, + \mathcal{O}(1)\int_0^{ + \infty } {e^{ - mt} t^{2N - 1} dt}
\\ &= \frac{1}{{2m}} - \sum\limits_{n = 1}^{N - 1} {\frac{{B_{2n} }}{{2n}}\frac{1}{{m^{2n} }}} + \mathcal{O}\! \left( {\frac{1}{{m^{2N} }}} \right).
\end{align}
Re-arranging and subtracting $1/m$ from both sides gives
$$
\sum\limits_{k = 1}^{m - 1} {\frac{1}{k}} = \log m + \gamma - \frac{1}{{2m}} - \sum\limits_{n = 1}^{N - 1} {\frac{{B_{2n} }}{{2n}}\frac{1}{{m^{2n} }}} + \mathcal{O}\!\left( {\frac{1}{{m^{2N} }}} \right).
$$
Taking $N=2$ yields your approximation.
|
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|
Integer solutions of $\frac{(a+b)(a+c)(b+c)}{2} + (a+b+c)^3 = 1 -abc $ (AusPol 1994) Find all integer solutions of
$$\frac{(a+b)(a+c)(b+c)}{2} + (a+b+c)^3 = 1 -abc $$
Attempt:
I noticed that $a,b,c>0$ or $a,b,c<0$ can't happen. Besides, if one of them is zero, we can find some solutions. Suppose $c=0$, we get
$$(a+b)\frac{ab}{2} + (a+b)^3 = 1 $$
$$\implies (a+b)(ab+ 2(a+b)^2) = 2 $$
So, we have $(a+b) \in \{\pm 1,\pm 2\}$. Just $(a+b) = 1$ has solution, with $a= 1$ and $b= 0$ or $a= 0$ and $b=1$.
So, due to the symmetry of the equation, just the case $a,b>0$ and $c<0$ is missing. Besides, $\gcd(a,b,c) =1$.
|
Rewriting, we have:
$$(a+b)(b+c)(c+a)+2(a+b+c)^3+2abc=2$$
The case of $c=0$ was solved using the fact that the LHS was divisible by $a+b$. Then, we could say that it was divisible by $(a+b+kc)$ since $kc=c=0$. Since the equation is symmetric, we can guess:
$$(a+b)(b+c)(c+a)+2(a+b+c)^3+2abc=m(a+b+kc)(a+kb+c)(ka+b+c)$$
which turns out to be true for $m=1$ and $k=2$. Now, we have:
$$(a+b+2c)(a+2b+c)(2a+b+c)=2$$
If WLOG $a+b+2c=a+2b+c$ (assuming two of the factors are equal), then we have $b=c$ which gives:
$$(a+3b)^2(2a+2b)=2 \implies (a+3b)^2(a+b)=1$$
We then have $a+b=1$ and $a+3b= \pm 1$ which gives $(a,b)=(1,0),(2,-1)$.
If none of the three factors were equal, it is easy to see that the factors have to be $-2,-1,1$. WLOG, set:
$$a+b+2c=-2,a+2b+c=-1,2a+b+c=1 \implies 4(a+b+c)=-2$$
which is clearly impossible. Thus, the set of all solutions are:
$$(a,b,c)=(1,0,0),(0,1,0),(0,0,1),(2,-1,-1),(-1,2,-1),(-1,-1,2)$$
|
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|
Proving that $\prod_{k=1}^{n}\tan\left(\frac{\pi k}{2n+1}\right)=\sqrt{2n+1}$ using geometry I have to prove that the following holds. A hint to use complex numbers has been given. I have tried to make a start but not to any result.
$$\prod_{k=1}^{n}\tan\left(\frac{\pi k}{2n+1}\right)=\sqrt{2n+1}$$
My Attempt:
Let us consider $(2n+1)^{\text{th}}$ roots of unity, $z_k=\exp\left(\frac{2k\pi i}{2n+1}\right)$. We can rewrite the product in terms of $\arg(z_k)$ as $\prod_{k=1}^{n}\tan\left(\frac{1}{2}\arg(z_k)\right)$. Or equivalently so, if we consider $(4n+2)^{\text{th}}$ roots of unity, we get this product as $\prod_{k=1}^{n}\tan(\arg(\zeta _k))$, where $\zeta_k= \exp\left(\frac{2k\pi i}{4n+2}\right)$.
I know it can be proved by proving the expression for $\prod\sin(\frac{1}{2}\arg(z_k))$. But I was wondering, is there a way to use telescopic products or purely the geometry of the complex roots of unity to arrive at this
|
Building on @WETutorialSchool's hint, start from the identity$$i\tan\frac{\theta}{2}=\frac{2i\sin\frac{\theta}{2}\exp\frac{i\theta}{2}}{2\cos\frac{\theta}{2}\exp\frac{i\theta}{2}}=\frac{\exp i\theta-1}{\exp i\theta+1}.$$Define $z:=\exp\frac{2i\pi}{2n+1}$ so$$\frac{z^k-1}{z^k+1}=i\tan\frac{k\pi}{2n+1},\,\frac{z^{2n+1-k}-1}{z^{2n+1-k}+1}=\frac{z^{-k}-1}{z^{-k}+1}=\frac{1-z^k}{1+z^k}=-i\tan\frac{k\pi}{2n+1}.$$Since $\prod_{k=1}^n\tan\frac{k\pi}{2n+1}$ is a product of the positive tangents of $n$ acute angles,$$\prod_{k=1}^n\tan\frac{k\pi}{2n+1}=\sqrt{\prod_{k=1}^ni\tan\frac{k\pi}{2n+1}\cdot-i\tan\frac{k\pi}{2n+1}}=\sqrt{\prod_{k=1}^n\frac{z^k-1}{z^k+1}\frac{z^{2n+1-k}-1}{z^{2n+1-k}+1}}=\sqrt{\prod_{k=1}^{2n}\frac{z^k-1}{z^k+1}}.$$To prove $\prod_{k=1}^{2n}\frac{z^k-1}{z^k+1}=2n+1$, consider the values of $z^k+1,\,0\le k\le 2n$. They are the roots of $(w-1)^{2n+1}-1$, so their product is $-1$ times this polynomial's constant term, i.e. $2$. In other words, $\prod_{k=0}^{2n}(z^k+1)=2$ and $\prod_{k=1}^{2n}(z^k+1)=1$. Similarly, $\prod_{k=1}^{2n}(z^k-1)$ is the product of the roots of $\frac{(w+1)^{2n+1}-1}{w}$, so is equal to its constant term, $\binom{2n+1}{1}$.
|
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|
A and B commute. Delete the same row and column from each. Do they still commute? Suppose that two matrices $A$ and $B$ commute. If I delete the $i$th row and $i$th column from each, do they necessarily commute? Does the answer change if we guarantee that $A$ and $B$ are Hermitian?
For example, consider the matrices
$A=
\begin{bmatrix}
\frac{4}{3} & \frac{1}{3\sqrt{2}} & -\frac{1}{\sqrt{6}} \\
\frac{1}{3\sqrt{2}} & \frac{7}{6} & -\frac{1}{2\sqrt{3}}\\
-\frac{1}{\sqrt{6}} & -\frac{1}{2\sqrt{3}} & \frac{3}{2}
\end{bmatrix}
$
and
$B=
\begin{bmatrix}
\frac{5}{3} & \frac{\sqrt{2}}{3} & -\frac{\sqrt{2}}{\sqrt{3}} \\
\frac{\sqrt{2}}{3} & \frac{4}{3} & -\frac{1}{\sqrt{3}}\\
-\frac{\sqrt{2}}{\sqrt{3}} & -\frac{1}{\sqrt{3}} & 2
\end{bmatrix}
$.
A patient calculation shows that $AB - BA = 0$. Let's construct the truncated matrices $A_r$ and $B_r$ formed by deleting the third column and third row.
Then
$A_r=
\begin{bmatrix}
\frac{4}{3} & \frac{1}{3\sqrt{2}} \\
\frac{1}{3\sqrt{2}} & \frac{7}{6}
\end{bmatrix}
$
and
$B_r=
\begin{bmatrix}
\frac{5}{3} & \frac{\sqrt{2}}{3} \\
\frac{\sqrt{2}}{3} & \frac{4}{3}
\end{bmatrix}
$.
Another patient, but faster calculation shows that $A_r B_r - B_r A_r=0$. Then for these particular matrices, the answer to the question is yes.
A similar calculation can be done for the non-Hermitian $A=
\begin{bmatrix}
\frac{11}{7} & 0 & -\frac{4}{7} \\
-\frac{1}{7} & 1 & \frac{1}{7}\\
-\frac{3}{7} & 0 & \frac{10}{7}
\end{bmatrix}$ and $B=
\begin{bmatrix}
\frac{15}{7} & 0 & -\frac{8}{7} \\
-\frac{2}{7} & 1 & \frac{2}{7}\\
-\frac{6}{7} & 0 & \frac{13}{7}
\end{bmatrix}$, again affirming yes for this particular example.
This question has a physical motivation from quantum mechanics, where whether or not two Hermitian matrices commute determines whether or not they are simultaneously observable. I'm imagining the case where we are unable to access a certain state, which would correspond to deleting the row and column corresponding to that state.
|
The answer is no in general. Consider
$$
A=
\begin{bmatrix}
6 & 1 & 8 \\
7 & 5 & 3 \\
2 & 9 & 4
\end{bmatrix}
\quad\text{and}\quad
B=
\begin{bmatrix}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{bmatrix}
.$$
Then $AB$ computes the row sums of $A$ while $BA$ computes the column sums of $A$. Since this $A$ is a magic square (!), the answers match:
$$
AB=BA=
\begin{bmatrix}
15 & 15 & 15 \\
15 & 15 & 15 \\
15 & 15 & 15
\end{bmatrix}
.$$
However, removing any row and any column breaks the magic: for example, removing the last rows and columns yields
$$
A_r=
\begin{bmatrix}
6 & 1 \\
7 & 5
\end{bmatrix}
\quad\text{and}\quad
B_r=
\begin{bmatrix}
1 & 1 \\
1 & 1
\end{bmatrix}
,$$
for which
$$
A_rB_r=
\begin{bmatrix}
7 & 7 \\
12 & 12
\end{bmatrix}
\quad\text{while}\quad
B_rA_r=
\begin{bmatrix}
13 & 6 \\
13 & 6
\end{bmatrix}
.$$
|
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|
What is the locus of $z^2+\bar{z}^2=2$? I have to prove that it's the equation of an equilateral hyperbola
$$z^2+z^{-2}=2$$
I try this
$z^2+z^{-2} +2 = 2+2$
$(z^2+1/z^2 +2 ) = 4$
$(z+1/z)^2=4$
$ z+1/z= 2 $
$ x+yi + 1/(x+yi) = 2 $
$ ((x+yi)^2+1)/(x+yi)=2$
$ x^2+2xyi-y^2+1=2x+2yi$
$ x^2-2x+1 +2xyi -y^2=2yi $
$(x-1)^2 +2xyi-y^2= 2yi$
I don't know how to end it.
|
The locus of $$z^2+(\bar z)^2=2 \implies (x+iy)^2+(x-iy)^2=2 \implies x^2-y^2=1$$ which is a hyperbola.
|
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|
How to find the div(curl $v$)?
let $v= ( v_1 ,v_2,v_3)$ be a vector field on $\mathbb{R}^3$ where $v_1 = \sqrt {1+x^2+y^2} , v_2= \sqrt {1+z^2} $and $v_3= \sqrt{ 1 +x^2y^2z^2}$. Evaluate div(curl $v$)
My attempt : $$Cur v= \begin{bmatrix} i &j&k\\ \frac{dv_1}{dx}& \frac{dv_2}{dy} & \frac{dv_3}{dz} \\ \sqrt {1+x^2+y^2} &\sqrt {1+z^2} & \sqrt{ 1 +x^2y^2z^2} \end{bmatrix}$$
$$Cur v= \begin{bmatrix} i &j&k\\ \frac{x}{\sqrt {1+x^2+y^2}}& 0 & \frac{zx^2y^2}{ \sqrt{ 1 +x^2y^2z^2}} \\ \sqrt {1+x^2+y^2} &\sqrt {1+z^2} & \sqrt{ 1 +x^2y^2z^2} \end{bmatrix}$$
= $i(\frac{2zx^2y^2}{ \sqrt{ 1 +x^2y^2z^2}})(\sqrt {1+z^2}) - j((\sqrt {1+x^2+y^2})(\frac{2zx^2y^2}{ \sqrt{ 1 +x^2y^2z^2}})-(\frac{x}{\sqrt {1+x^2+y^2}})\sqrt{ 1 +x^2y^2z^2})) + k(\sqrt {1+z^2} )(\frac{x}{\sqrt {1+x^2+y^2}})$
After that im not able to proceed further
|
It is a general fact that $\operatorname{div}(\operatorname{curl}(F))=0$. You can verify this by expanding. Let $F=(P,Q,R)$. Then
$$\operatorname{curl}(F)=(R_y-Q_z)i-(R_x-P_z)j+(Q_x-P_y)k$$
by definition. Then
$$ \operatorname{div}(\operatorname{curl}(F))=R_{xy}-Q_{xz}-R_{yx}+P_{yz}+Q_{zx}-P_{zy}$$
$$=(R_{xy}-R_{yx})+(P_{yz}-P_{zy})+(Q_{zx}-Q_{xz})=0+0+0.$$
Here, I used the fact that mixed second partials agree for $\mathscr{C}^2$ functions. The only thing that one should be careful about here is whether or not your functions are $\mathscr{C}^2$, but I am fairly certain they are by inspection.
|
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|
Prove that $\binom{n}{0} + 2\binom{n}{1} + ...(n+1)\binom{n}{n} = (n + 2)2^{n-1}$. How can I prove the following identity?
$$\binom{n}{0} + 2\binom{n}{1} + ...(n+1)\binom{n}{n} = (n +
2)2^{n-1}$$
I thought about differentiating this:
$$(1 + x) ^ n = \sum_{k = 0} ^ {n} \binom{n}{k}x^k$$
and then evaluating it at $x = 1$, but I didn't get to my desired result. I just kept finding that $\displaystyle\sum_{k = 0}^n k \binom{n}{k} = n2^{n-1}$.
|
Here's an alternative proof that does not depend on derivatives:
\begin{align}
\sum_k (k+1) \binom{n}{k}
&= \sum_k k\binom{n}{k} + \sum_k \binom{n}{k} \\
&= \sum_k n\binom{n-1}{k-1} + \sum_k \binom{n}{k} \\
&= n 2^{n-1} + 2^n \\
&= (n+2) 2^{n-1}
\end{align}
|
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|
How $\sum_1^6\sin x \equiv2\sin\frac{7}{2}(\cos\frac{5}{2}+\cos\frac{3}{2}+\cos\frac{1}{2})$? I just learned this cool trick but I can't figure out why it works, obviously I know how it's done:
1 + 2 + 3 + ... + n ( I know the no of terms have to be even for it to work in pairs)
$\sin$(1)+ $\sin$(2) + $\sin$(3)+... = [$\sin$(1)+ $\sin$(n)] + [$\sin$(2) + $\sin$(n - 1)] + ...
upto halfway through the series and stop since it repeats the pairs in reverse basically doubling the sum.
then use sum-product identities,
$\sin\alpha+\sin\beta\equiv2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}$
$\cos\alpha+\cos\beta\equiv2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}$
$\sum_1^6\sin x \equiv2\sin\frac{7}{2}(\cos\frac{5}{2}+\cos\frac{3}{2}+\cos\frac{1}{2})$
or
$\sum_1^4\cos x \equiv2\cos\frac{5}{2}(\cos\frac{3}{2}+\cos\frac{1}{2})$
But why in $2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}$ does
when n = 6 , $\sum_1^6\sin x$
$\sin\frac{\alpha+\beta}{2}$ takes the first (1 + n) while $\cos\frac{\alpha-\beta}{2}$ does (1 - n) + (2 - (n - 1)) + (3 -(n - 2))
Why $\sum_x^n\tan x$ doesn't work using $\tan\alpha+\tan\beta\equiv\frac{\sin(\alpha+\beta)}{\cos\alpha\cos\beta}$ ?
|
For the first question, you just need to look at the values of $\alpha$ and $\beta$. As you have noted, they are paired so the sum is always $n+1$. The first few $(\alpha,\beta)$ pairs are $$(n,1)\\(n-1,2)\\(n-2,3)\\\vdots$$ The sums are always $n+1$ and the differences are $$n-1\\n-3\\n-5\\\vdots$$
Symbolically, we have $$\alpha=n-k\\\beta=k+1\\\alpha+\beta=n+1\\\alpha-\beta=n-2k-1$$
As to why a similar formula doesn't work for tangent, it's because you don't have a common denominator.
|
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|
How to prove $xy>x+y$ if $x,y \in (2,\infty)\subseteq \Bbb R$ I am new to proofs. This is my first proof-based mathematics class, and it is a hard transition from my high school classes. I am a second semester freshman.
My original question was:
Prove that there is a real number with the property that for any two
larger numbers, there is another real number that is larger than the
sum of the two numbers and less than their product.
I have attempted this by a constructive proof (which is possible according to my distinguished professor) and am very unwilling to attempt a nonconstructive proof until I have proven it by a constructive proof. I said:
"Let $x,y,z,$ and $t \in \Bbb R$, with $y>x$ and $z>x$. Because $x,y,z,$ and $t$ are real numbers, if we choose a pair of $y$ and $z$ such that $yz>(y+z)$ , the following always holds $\exists t(yz>t>(y+z))$. Let $x=2$ and let $\epsilon$ be some arbitrary infinitesimal positive quantity. The smallest product of $y$ and $z$ would then be the product of $y=2+\epsilon$ and $z=2+\epsilon$ and the product would be $yz=4+2\epsilon + \epsilon^2$. Similarly, the smallest sum of $y$ and $z$ would be the sum if $y=2+\epsilon$ and $z=2+\epsilon$ which is $(y+z)=4+2\epsilon$. We see that $yz>(y+z)$ which is equivalent to $4+2\epsilon + \epsilon^2 > 4+2\epsilon$..."
And this is where I get stuck. I don't think I am near the end of the proof, I feel like I need a lemma to complete this. I want the lemma to be the proof of:
Prove that for any pairs of real numbers $x$ and $y$, where $x>2$ and
$y>2$, the following always holds: $xy>(x+y)$
I thought of proving this lemma by approaching using the AM-GM inequality which I learned in competitive mathematics during high school, since one side is a product and one side is a summation, but I have not started it. I feel like I am being a little redundant in the proof, going in circles, or writing without much thinking.
Disregarding the original question (although I appreciate hints), how would I approach the proof of the lemma?
|
If $x>z$ and $y > z$ then $a=x-z> 0$ and $b=y-z>0$.
And $x+y = (z+a) + (z+b) = 2z +(a+b)$. And $xy = (z+a)(z+b) = z^2 + (a+b)z + ab$.
We want to find a $z$ so that $2z + (a+b) < z^2 +(a+b)z + ab$ for all possible positive $ab$.
This will happen if and only if $0 < (z^2-2z) + (a+b)(z-1) + ab=z(z-2) + (a+b)(z-1) + ab$ for all positive $a,b$
We know that $ab > 0$ and $(a+b)>0$ for all positive $a,b$ so if we can have $z(z-2)\ge 0$ and $z-1\ge 0$ that will be good enough.
If $z = 2$ and $x>2$ and $y > 2$ and $a= x-2>0;b=y-2>0$ then $xy=(2+a)(2+b) = 4+2(a+b)+ab> 4+a+b = x+y$.
So $z=2$ is just such a number.
And if $x+y < xy$ then $w=\frac {(x+y) + xy}2$ will be that $x+y < w < xy$ and $w$ is such a number as we need.
Note: We haven't proven that $z=2$ is the least such possible value but we weren't asked to prove that.
But we can prove that.
If $z < 2$ we just have to prove there exist any $x,y> z$ where this fails.
If $x = y = 1$ then $x+y = 2 >1=xy$ so this will fail if $z< 1$. ANd if $1\le z< 2$ then if we let $x+y = \frac {z+2}2$. Then $x+y = z+ 2$ while $xy= \frac{z^2 + 4z + 4}{4} = \frac {z^2}4 + z + 1= z + (\frac {z^2}4 +1)$. But $1\le x < 2$ so $z+(\frac {z^2}4 +1) < z + (1+1)=z+2 =x+y$. So that is a counter example.
So $z$ is the smallest such number. (Similar for any $z > 2$ this will hold. After all, if $x,y > z > 2$ then $x,y > 2$ and we know that in that case $x+y < xy$.)
|
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|
Calculus 2: Integration by Parts Stuck on Integral of Product With ArcTan Inside I'm stuck on the following problem:
$$\int_0^{1/3} y \tan^{-1}(3y)\,dy$$
I think my last line in my work below is correct but I don't know what to do beyond that.
A step through of the problem would be appreciated.
|
Integrate by parts $\left(f = \tan^{-1}3y, g' = y \Rightarrow f' = \frac{3}{9y^2 + 1}, g = \frac{y^2}{2}\right)$:
$$\int y \tan^{-1}(3y) dy$$
$$= \frac{y^2 \tan^{-1}(3y)}{2} - \int \frac{3y^2}{2(9y^2 + 1)} dy$$
$$= \frac{y^2 \tan^{-1}(3y)}{2} - \frac{3}{2} \int \frac{y^2}{9y^2 + 1} dy$$
Write $y^2$ as $\frac{1}{9}(9y^2+1) - \frac{1}{9}$ then we get:
$$\frac{y^2 \tan^{-1}(3y)}{2} - \frac{3}{2} \int \frac{\frac{1}{9}(9y^2+1) - \frac{1}{9}}{9y^2 + 1} dy$$
$$= \frac{y^2 \tan^{-1}(3y)}{2} - \frac{3}{2} \int \frac{\frac{1}{9}(9y^2+1)}{9y^2 + 1} - \frac{\frac{1}{9}}{9y^2 + 1}dy$$
$$= \frac{y^2 \tan^{-1}(3y)}{2} - \frac{3}{2} \int \frac{1}{9} - \frac{1}{9} \frac{1}{9y^2 + 1}dy$$
$$= \frac{y^2 \tan^{-1}(3y)}{2} - \frac{1}{6} \int 1 - \frac{1}{9y^2 + 1}dy$$
$$= \frac{y^2 \tan^{-1}(3y)}{2} - \frac{1}{6} \left(\int 1\,dy - \int \frac{1}{9y^2 + 1}dy\right)$$
Using substitution $u=3y$, we get:
$$= \frac{y^2 \tan^{-1}(3y)}{2} - \frac{1}{6} \left(\int 1\,dy - \frac{1}{3} \int \frac{1}{u^2 + 1}du\right)$$
$$= \frac{y^2 \tan^{-1}(3y)}{2} - \frac{1}{6} \left(y - \frac{1}{3} \tan^{-1}u + C\right)$$
$$= \frac{y^2 \tan^{-1}(3y)}{2} - \frac{1}{6} y + \frac{1}{18} \tan^{-1}3y + C$$
$$= \frac{(9y^2 + 1) \tan^{-1}(3y) - 3y}{18} + C$$
Now you can evaluate the integral:
$$\int_{0}^{\frac{1}{3}} y \tan^{-1}(3y) \, dy = \frac{\pi - 2}{36}$$
|
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|
Evaluate $\lim_{n\to\infty}\sum_{k=1}^n\frac{k^3}{\sqrt{n^8+k}}$ & $\lim_{n\to\infty}n\bigg(\sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}-\frac{1}{4}\bigg)$. This is a homework question. I have to find two limits:
i. $$\lim_{n\to \infty} \sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}$$
ii. $$\lim_{n\to \infty} n\bigg(\sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}-\frac{1}{4}\bigg)$$
I was able to solve the first one:
$\sqrt{n^8+1}\le \sqrt{n^8+k}\le \sqrt{n^8+n}$
$\implies \displaystyle\dfrac{1}{\sqrt{n^8+1}}\ge \dfrac{1}{\sqrt{n^8+k}}\ge \dfrac{1}{\sqrt{n^8+n}}$
$\implies \displaystyle\dfrac{k^3}{\sqrt{n^8+1}}\ge \dfrac{k^3}{\sqrt{n^8+k}}\ge \dfrac{k^3}{\sqrt{n^8+n}}$
$\implies \displaystyle\sum_{k=1}^n\dfrac{k^3}{\sqrt{n^8+1}}\ge \sum_{k=1}^n\dfrac{k^3}{\sqrt{n^8+k}}\ge \sum_{k=1}^n\dfrac{k^3}{\sqrt{n^8+n}}$
$\implies \displaystyle\dfrac{n^2(n+1)^2}{4\sqrt{n^8+1}}\ge \sum_{k=1}^n\dfrac{k^3}{\sqrt{n^8+k}}\ge \dfrac{n^2(n+1)^2}{4\sqrt{n^8+n}}$
The upper fraction goes to $\frac{1}{4}$ and the lower as well. From the Squeeze theorem:
$$\lim\limits_{n\to \infty} \sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}} = \frac{1}{4}$$
The same method did not work for the second limit. Can I get a clue or a hint, please?
|
The second limit can be squeezed as well with :
$$n^4 \leq \sqrt{n^8+k} \leq n^4+1$$
Thus
$$n\left(\sum_{k=1}^n \frac{k^3}{n^4+1}-\frac{1}{4}\right)\leq n\left(\sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}-\frac{1}{4}\right)\leq n\left(\sum_{k=1}^n \frac{k^3}{n^4}-\frac{1}{4}\right)$$
or
$$n\left[\frac{n^2(n+1)^2}{4(n^4+1)}-\frac{1}{4}\right]\leq n\left(\sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}-\frac{1}{4}\right)\leq n\left[\frac{n^2(n+1)^2}{4n^4}-\frac{1}{4}\right]$$
or
$$n\cdot \frac{2n^3+n^2-1}{4(n^4+1)}\leq n\left(\sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}-\frac{1}{4}\right)\leq n\cdot \frac{2n^3+n^2}{4n^4}$$
and squeezing
$$\lim_{n\to \infty} n\left(\sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}-\frac{1}{4}\right)=\frac{1}{2}$$
|
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|
How to prove the asymptotic expansion $\overline{H}_n \sim \log(2) -(-1)^n\left (\frac{1}{2n}-\frac{1}{4 n^2} +\frac{1}{8n^4}\mp\ldots\right)$? It is well-known that the harmonic sum $H_{n}= \sum_{k=1}^{n}\frac{ 1}{k}$ has the following asymptotic expansion for $n\to\infty$
$$H_n = \sum_{k=1}^{n}\frac{1}{k}\sim \gamma+\log \left(n\right)+\frac{1}{2 n}-\frac{1}{12 n^2}+\frac{1}{120 n^4}-\frac{1}{252 n^6}\pm \ldots\tag{1}$$
The alternating harmonic sum is defined as
$$\overline {H}_{n} = \sum_{k=1}^{n}\frac{(-1)^{k+1}}{k}\tag{2}$$
and we ask for its asymptotic expansion.
At first I tried to use the representation
$$\overline{H}_{n} =\log (2)+ (-1)^{n+1} \Phi (-1,1,n+1)\tag{3}$$
where $ \Phi (z,s,a)=\sum_{k=0}^{\infty} \frac{z^k}{(k+a)^s}$ is a special function called Lerch transcendent (https://en.wikipedia.org/wiki/Lerch_zeta_function) which is just the tail of the expansion of $\log(2)$ starting at the $(n+1)$st term. But I couldn't find the asymptotics of $\Phi$. Also Mathematica wouldn't do it.
So I came up with another idea and found
$$\overline{H}_{n} \sim \log(2) -(-1)^n \left(\frac{1}{2n}-\frac{1}{4 n^2} +\frac{1}{8n^4} - \frac{1}{4n^6}+\ldots\right)\tag{4}$$
I have looked up possibly related proofs. This reference contains two of them.
Asymptotic expansion at order 2 of $\int_0^1 \frac{x^n}{1+x} \, dx$
But mine was still different.
What would be your proof?
|
My idea was to express $\overline{H}_k$ by $H_k$ and then use the asyptotic expansion of $H_k$.
Indeed, $\overline{H}_n$ can be expressed as follows ($m=1,2,3,\ldots$}
$$\overline{H}_{2m} = H_{2m} -H_{m}\tag{5a}$$
$$\overline{H}_{2m+1} = H_{2m+1} -H_{m}\tag{5b}$$
The (simple) proof is left as an exercise to the reader.
For the asymptotic expressions of the even version we find from $(1)$
$$\overline{H}_{2m}\overset{m\to\infty,m->\frac{n}{2}} = \log (2) \\-\frac{1}{2 n}+\frac{1}{4 n^2}-\frac{1}{8 n^4}+\frac{1}{4 n^6} -\frac{17}{16 n^8}\pm\ldots\tag{6a}$$
For the odd version we have, to begin with,
$$\overline{H}_{2m+1}\overset{m\to\infty, m->\frac{n-1}{2}}=\log (2)
\\
+\frac{1}{2 (n-1)}-\frac{3}{4 (n-1)^2}+\frac{1}{(n-1)^3}-\frac{9}{8 (n-1)^4}+\frac{1}{(n-1)^5}-\frac{3}{4 (n-1)^6}
\\
+\frac{1}{(n-1)^7}-\frac{33}{16 (n-1)^8}+\frac{1}{(n-1)^9}\mp\ldots$$
Taking the asymptotics of this in turn we get
$$\overline{H}_{2m+1}\overset{m\to\infty, m->\frac{n-1}{2}}=\log (2)\\+ \frac{1}{2 n}-\frac{1}{4 n^2}+\frac{1}{8 n^4}-\frac{1}{4 n^6}+\frac{17}{16 n^8}\mp\ldots\tag{6b}$$
Finally, combining $(6a)$ and $(6b)$ gives the expression $(4)$ of the OP.
Combining this with $(3)$ we have also derived the asymptotics of the Lerch $\Phi$ function from that of the harmonic number.
|
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|
Integral $\int \arcsin \left(\sqrt{\frac{x}{1-x}}\right)dx$ $$\int \arcsin\left(\sqrt{\frac{x}{1-x}}\right)dx$$
I'm trying to solve this integral from GN Berman's Problems on a course of mathematical analysis (Question number 1845)
I tried substituting $x$ for $t^2$:
$$2\int t\arcsin\left(\frac{t}{\sqrt{1-t^2}}\right) dt$$
And then evaluating it by parts, which made it very complicated.
So how do I go about solving this?
|
Here is a nice solution that utilizes symmetry instead of $u$ susbsitution. Notice that the domain of this function is at most $\left[0,\frac{1}{2}\right]$, so for the moment let's define
$$F(x) \equiv \int_0^x \arcsin\left(\sqrt{\frac{t}{1-t}}\right)\:dt$$
We could do this integral directly, or since the integrand has a fixed point at $0$, note that the integral is the same as the area of the box $x \cdot \arcsin\left(\sqrt{\frac{x}{1-x}}\right)$ minus the integral of the inverse function:
$$F(x) = x \arcsin\left(\sqrt{\frac{x}{1-x}}\right) - \int_0^{\arcsin\left(\sqrt{\frac{x}{1-x}}\right)} 1 - \frac{1}{1+\sin^2 y}\:dy$$
$$= (x-1) \arcsin\left(\sqrt{\frac{x}{1-x}}\right) + \int_0^{\arcsin\left(\sqrt{\frac{x}{1-x}}\right)}\frac{1}{\cos^2y+2\sin^2 y}\:dy$$
The integral reduces to
$$\int_0^{\arcsin\left(\sqrt{\frac{x}{1-x}}\right)}\frac{\sec^2y}{1+2\tan^2 y}\:dy = \frac{1}{\sqrt{2}}\arctan\left(\sqrt{2}\tan y\right)\Biggr|_0^{\arcsin\left(\sqrt{\frac{x}{1-x}}\right)}$$
$$ = \frac{1}{\sqrt{2}}\arctan\left(\sqrt{\frac{2x}{1-2x}}\right)$$
which gives us a final answer of
$$F(x) = (x-1) \arcsin\left(\sqrt{\frac{x}{1-x}}\right)+\frac{1}{\sqrt{2}}\arctan\left(\sqrt{\frac{2x}{1-2x}}\right)$$
making the most general antiderivative
$$\int \arcsin\left(\sqrt{\frac{x}{1-x}}\right)\:dx = (x-1) \arcsin\left(\sqrt{\frac{x}{1-x}}\right)+\frac{1}{\sqrt{2}}\arctan\left(\sqrt{\frac{2x}{1-2x}}\right) + C$$
This solution has the benefit of not having complicated substitutions to undo, like the other solutions have, but we got lucky since this technique will not always work (what are the odds of the inverse function being more easily integrable than the original?)
|
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$f(x)=\frac{\sin x}{x}$, prove that $|f^{(n)}(x)|\le \frac{1}{n+1}$ This is a homework question.
Let $f: (0, \infty) \to \mathbb{R}$ with $f(x)=\dfrac{\sin x}{x}$. I
have to prove that
$$|f^{(n)}(x)|\le \frac{1}{n+1}$$
where $f^{(n)}$ is the nth derivative of $f$.
I started to do a few derivatives:
$f^{(1)}(x)=\dfrac{1}{x^2}(x \cos x-\sin x)$
$f^{(2)}(x)=\dfrac{1}{x^3}((2-x^2) \sin x-2x\cos x)$
$f^{(3)}(x)=\dfrac{1}{x^4}(3(x^2-2) \sin x-x(x^2-6)\cos x)$
$f^{(4)}(x)=\dfrac{1}{x^5}(4x(x^2-6)\cos x + (x^4-12x^2+24)\cos x)$
I noticed that in denominator, there is always $x^{n+1}$ (because there is $x$ in denominator of $f$), but I couldn't spot a pattern for the numerator that would help me prove the inequality. Can I get a hint or a clue, please?
|
I will assume (since it's not specified) that $n$ is a non-negative integer.
Claim: For $f:(0,\infty)\to \mathbb{R}$, defined by $f(x)=\dfrac{\sin x}{x}$ and any non-negative integer $n$, we have:
$$f^{(n)}(x)=\frac{1}{x^{n+1}}\int_0^xu^n\cos\left(u+\frac{n\pi}{2}\right)\,du$$
Proof: I will prove this by induction. The case $n=0$ is obvious:
$$f^{(0)}(dx)=f(x)=\frac{\sin x}{x}=\frac{1}{x}\int_0^x\cos u\,du$$
Now assume it is true for some positive integer $n \geq 1$. Then, integrating by parts:
$$
\begin{aligned}
f^{(n+1)}(x)&=(f^{(n)})'(x)\\
&=\frac{1}{x^{n+2}}\left(-(n+1)\int_0^xu^n\cos\left(u+\frac{n\pi}{2}\right)+x\cdot x^n\cos\left(x+\frac{n\pi}{2}\right)\right)\\
&=\frac{1}{x^{n+2}}\left[-u^{n+1}\cos(u+\frac{n\pi}{2})\bigg|_0^x+\int_0^xu^{n+1}(\cos\left(u+\frac{n\pi}{2}\right))'\,du+x^{n+1}\cos(x+\frac{n\pi}{2})\right] \\
&=\frac{1}{x^{n+2}}\int_0^xu^{n+1}\cos\left(u+\frac{(n+1)\pi}{2}\right)\,du
\end{aligned}
$$
Claim proved. Now, the inequality follows immediately:
$$|f^{(n)}(x)|=\left|\frac{1}{x^{n+1}}\int_0^xu^n\cos\left(u+\frac{n\pi}{2}\right)\,du\right|\leq \frac{1}{x^{n+1}}\int_0^xu^n\,du=\frac{1}{n+1}$$
|
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|
Why does $3x^4 + 16x^3 + 20x^2 - 9x - 18$ = $(x+3)(x+2)(\frac{-1 \mp \sqrt 37 }{6})3$? $$3x^4 + 16x^3 + 20x^2 - 9x - 18 $$
When simplified I arrive to:
$$ (x+3)(x+2)(\frac{-1 \mp \sqrt 37 }{6}) $$
But the math book wrote:
$$ (x+3)(x+2)(\frac{-1 \mp \sqrt 37 }{6})3 $$
with that extra 3 at the end. The graph calculator seem to agree with that extra 3 as well. what did I do wrong?
|
For any real $k$ we have
$$3x^4+16x^3+20x^2-9x-18=3\left(x^4+\frac{16}{3}x^3+\frac{20}{3}x^2-3x-6\right)=$$
$$=3\left(\left(x^2+\frac{8}{3}x+k\right)^2-\frac{64}{9}x^2-\frac{16k}{3}x-2kx^2-k^2+\frac{20}{3}x^2-3x-6\right)=$$
$$=3\left(\left(x^2+\frac{8}{3}x+k\right)^2-\left(\left(2k+\frac{4}{9}\right)x^2+\left(\frac{16k}{3}+3\right)x+k^2+6\right)\right).$$
Now, we'll choose $k$ such that $$2k+\frac{4}{9}>0$$ and
$$\left(\frac{16k}{3}+3\right)^2-4\left(2k+\frac{4}{9}\right)(k^2+6)=0,$$
which gives $k=\frac{5}{2}.$
Id est, $$3x^4+16x^3+20x^2-9x-18=3\left(\left(x^2+\frac{8}{3}x+\frac{5}{2}\right)^2-\left(\frac{49}{9}x^2+\frac{49}{2}x+\frac{49}{4}\right)\right)=$$
$$=3\left(\left(x^2+\frac{8}{3}x+\frac{5}{2}\right)^2-\left(\frac{7}{3}x+\frac{7}{2}\right)^2\right)=3(x^2+5x+6)\left(x^2+\frac{x}{3}-1\right)=$$
$$=3(x+2)(x+3)\left(x-\frac{-\frac{1}{3}+\sqrt{\frac{37}{9}}}{2}\right)\left(x-\frac{-\frac{1}{3}-\sqrt{\frac{37}{9}}}{2}\right)=$$
$$=3(x+2)(x+3)\left(x-\frac{-1+\sqrt{37}}{6}\right)\left(x-\frac{-1-\sqrt{37}}{6}\right).$$
|
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|
solve for $(\cos x)(\cos2x)(\cos3x)=\frac{1}{4}$ solve for $(\cos x)(\cos2x)(\cos3x)=\frac{1}{4}$. what is the general solution for $x$.
I wrote $\cos2x$ as $2\cos^2x-1$ and $\cos3x$ as $4\cos^3x-3\cos x$. Then the expression gets reduced to $\cos x(2\cos^2x-1)(4\cos^3x-3\cos x)=\frac{1}{4}$. substituted $\cos x=t$ and got
$t(2t^2-1)(4t^3-3t)=\frac{1}{4}$. how do i proceed further?
|
Hint:
Clearly, $\sin x\ne0$
$$1=4\cos x\cos2x\cos3x$$
$$\implies\sin x=2(\sin2x)\cos2x\cos3x=\sin4x\cos3x$$
$$\implies2\sin x=2\sin4x\cos3x=\sin7x+\sin x$$
$$\implies0=\sin7x-\sin x=2\sin3x\cos4x$$
|
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|
Integral $\int_{0}^{\frac{\pi}{4}}x\ln(\cos x)dx$ I would like to learn more about this integral:
$$\int_{0}^{\frac{\pi}{4}}x\ln(\cos x)dx=\frac{1}{128}\Big(16\pi \operatorname{C}-21\zeta(3)-4\pi^2\ln(2)\Big)$$
where $\operatorname{C}$ is the Catalan's constant.
I have tried integration by parts , substitution $y=\cos x$, etc. But, it reveals nothing. WA gives the anti-derivative in polylogarithmic functions that I don't know about.
I think it is a hard integral. If you have an elementary method to guide me, it would be nice.
Thanks a lot!
|
By the Fourier series of $\ln (\cos x)$
$$
\ln (\cos x)=-\ln 2+\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} \cos (2 k x),
$$
we have $$
x \ln (\cos x)=-x \ln 2+\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} x \cos (2 k x)
$$
$$
\begin{aligned}
\int_{0}^{\frac{\pi}{4}} x \ln (\cos x) d x&=-\ln 2 \int_{0}^{\frac{\pi}{4}} x d x+\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} \int_{0}^{\frac{\pi}{4}} x \cos (2 k x) d x \\
&=-\frac{\pi^{2} \ln 2}{8}+\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} \int_{0}^{\frac{\pi}{4}} x \cos (2 k x) d x
\end{aligned}
$$
$$
\begin{aligned}
\int_{0}^{\frac{\pi}{4}} x \cos (2 k x) d x &=\frac{1}{2 k} \int_{0}^{\frac{\pi}{4}} x d(\sin (2 k x)) \\
&=\left[\frac{x \sin (2 k x)}{2 k}\right]_{0}^{\frac{\pi}{4}}-\frac{1}{2 k} \int_{0}^{\frac{\pi}{4}} \sin (2 k x) d x \\ \\
&= \frac{\pi}{8 k} \sin \left(\frac{k \pi}{2}\right)+\frac{1}{4 k^{2}}\left(\cos \frac{k \pi}{2}-1\right)
\end{aligned}$$
$$
\begin{aligned}
\int_{0}^{\frac{\pi}{4}} x \ln (\cos x) d x&= -\frac{\pi^{2} \ln 2}{32}+\frac{\pi}{8} \sum_{k=1}^{\infty} \frac{(-1)^{k+1} \sin \left(\frac{k \pi}{2}\right)}{k^{2}}+ \frac{1}{4} \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^{3}}\left(\cos \frac{k \pi}{2}-1\right)\\&=-\frac{\pi^{2} \ln 2}{32}+\frac{\pi G}{8}+\frac{1}{4}S,
\end{aligned}$$
where $G$ is the Catalan’s Constant.
$$\begin{aligned}S&=2 \sum_{k=1}^{\infty} \frac{(-1)^{k}}{k^{3}}\left(\sin ^{2} \frac{k \pi}{4}\right)\\&
=2\left[-\frac{1}{2} \sum_{k=0}^{\infty} \frac{1}{(2 k+1)^{3}}+\frac{1}{2^{3}} \sum_{k=1}^{\infty} \frac{1}{(2 k+1)^{3}}\right] \\
&=2\left[-\frac{3}{8} \sum_{k=0}^{\infty} \frac{1}{(2 k+1)^{3}}\right] \\
&=-\frac{3}{4}-\frac{7}{8}\zeta(3) \\
&=-\frac{21}{32} \zeta(3) \end{aligned}$$
Now we can conclude that$$
\boxed{\int_{0}^{\frac{\pi}{4}} x \ln (\cos x) d x= -\frac{\pi^{2} \ln 2}{32}+\frac{\pi G}{8}-\frac{21}{128} \zeta(3)} $$
|
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|
If $x^4+12x-5$ has roots $x_1,x_2,x_3,x_4$ find polynomial with roots $x_1+x_2,x_1+x_3,x_1+x_4,x_2+x_3,x_2+x_4,x_3+x_4$ I have the polynomial $x^4+12x-5$ with the roots $x_1,x_2,x_3,x_4$ and I want to find the polynomial whose roots are $x_1+x_2,x_1+x_3,x_1+x_4,x_2+x_3,x_2+x_4,x_3+x_4$.
I found the roots $x_1=-1+\sqrt{2},x_2=-1-\sqrt{2},x_3=1-2i,x_4=1+2i$. And after long computations the polynomial is $x^6+20x^2-144$. Are there clever way to find it?
|
Eliminate $x$ out of the system
$$s^4 + 12 s - 15 = 0 \\
t^4 + 12 t - 15 =0 \\
x- (s+t) = 0$$
and get an equation of degree $10$ in $x$
$$x^{10} + 96 x^7 - 60 x^6 - 144 x^4 + 1920 x^3 - 1600 x^2 - 13824 x + 11520=0 \ \ (*)$$
Now, some of these solutions ($4$ of them) of the above are in fact solution of the equation in $x$ obtained from
$$s^4 + 12 s - 15 = 0\\
x - (s+s) = 0$$
that is
$$x^4 + 96 x - 80=0 \ \ (**)$$
Divide the polynomials $(*)$ by $(**)$ and get
$$ x^6 + 20 x^2 - 144=(x-2)(x+2)(x^4 + 4 x^2 + 36) $$
|
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|
Divisibility by $33^{33}$ Let $P_n=(19+92)(19^2+92^2)\cdots(19^n+92^n)$ for each positive integer $n$. Determine , with proof the least positive integer $n$, if it exists , for which $P_n$ is divisible by $33^{33}$.
I have made no progress concrete enough to show .
|
COMMENT:
$19^n+92^n ≡ 8^n+4^n \ mod (11)$
⇒ $4^n(2^n+1)=11 k$
One solution of this equation is $n=5$, $k=3 \times 1024$, we have:
$4^5(2^5+1)=1024\times 33$
Therefor for $n=5$ and all powers such as $5(2t+1)$ , $19^n+92^n ≡ 0 \mod (33)$ due to geometric progression:
$(19^5)^{2t+1}+(92^5)^{2t+1}=(19^5+92^5)[(19^5)^{2t}-(19^5)^{2t-1}(92^5) + . . . + (92^5)^{2t}]$
We can see that $19^5+92^5$ is divisible by $11^2$ :
$19^5+92^5= 121\times 54490011 $
so we need to have $33/2=16.5$ factors of such kind; we may guess $n=33\times 5=165$. But for this we have 165 times 3 and 17 times power of 5(from 5 to 165) so the product contains a factor like:
$P=3^{165 }\times 11^{2\times 17\times 5=170}$
So it seems $n=165$ is sufficient. The result from brute force , i.e $n=155=31\times 5$, indicates that there must be two hidden power for 33 such that we can get equal powers for 3 and 11.That is two of 5th powers must be divisible by $11^3$.If so power of 11 will be:
$(17-2=15)\times 2+1)\times 5=155$
|
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|
Prove $ \frac{1}{201} < \int^{100}_{0} \frac{e^{-x}}{100+x} dx < \frac{1}{100}$ Prove: $\displaystyle \frac{1}{201} < \int^{100}_{0} \frac{e^{-x}}{100+x} dx < \frac{1}{100}$
What i was trying to do:
On the right side:
$$
\int^{100}_{0} \frac{e^{-x}}{100+x} dx < \int^{100}_{0} \frac{1}{100+x} dx = ln(100+x) |^{100}_0 = ln(200) - ln(100) = ln(2)
$$
But
$$\frac{1}{100} < ln(2)$$
$$
\int^{100}_{0} \frac{e^{-x}}{100+x} dx < \int^{100}_{0} \frac{1}{e^x} dx = -\frac{1}{(e^x)^2} |^{100}_0 = 1 - \frac{1}{e^{99}}
$$
But as the previous:
$$
\frac{1}{100} < 1 - \frac{1}{e^{99}}
$$
For the left side i dont realy have an idea.
So i thought maybe my way of proving is not the right direction, can someone give me a hint?
|
Using the estimation as already given in @BrianMoehring's comment and noting that
*
*$\int_0^{100}e^{-x}dx = 1- \frac{1}{e^{100}}$ you have
$$\frac 1{200}\left(1- \frac{1}{e^{100}}\right) < \int^{100}_{0} \frac{e^{-x}}{100+x} dx < \frac 1{100}\left(1- \frac{1}{e^{100}}\right) < \frac 1{100}$$
The lower bound seems a bit artificial but can be obtained by a further rough estimate using $e^x > 1+x+ \frac{x^2}{2}$ for $x>0$. It is enough to show that
$$\frac 1{201}< \frac 1{200}\left(1- \frac{1}{e^{100}}\right)$$
$$\Leftrightarrow 1-\frac{1}{201}< 1- \frac{1}{e^{100}}$$
$$\Leftrightarrow \frac{1}{201} >\frac{1}{e^{100}}$$
which is true since
$$\frac{1}{e^{100}} < \frac{1}{1+100+\frac{100^2}{2}}< \frac 1{201}$$
|
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|
Remainder when $2^{55}+1$ is divided by $33$ There's this problem I encountered in a math Olympiad for my country.
Find the remainder when $2^{55}+1$ is divided by $33$.
My approach was to consider $2^{55}$ as the sum of numbers in the 56th row of Pascal's triangle. Then I showed that apart from 1 and 55, all other numbers have at least one factor of 11 and 3 to spare, so one would consider only 1 and 55 for the answer. Doing this leads to $113 \mod 33=14$. But the choices for that question were between 0 and 5.
Is my reasoning wrong? Or am I missing something?
|
Sticking to Pascal's triangle, you can just start calculating $2^k \pmod{33}$ row by row until you see a 'pattern' and can then get lazy.
$\quad 2^0 \equiv \;\,1 \pmod{33}$
$\quad 2^1 \equiv \;\,2 \pmod{33}$
$\quad 2^2 \equiv \;\,4 \pmod{33}$
$\quad 2^3 \equiv \;\,8 \pmod{33}$
$\quad 2^4 \equiv 16 \pmod{33}$
$\quad 2^5 \equiv 32 \pmod{33}$
$\quad 2^6 \equiv 31 \pmod{33}$
$\quad 2^7 \equiv 29 \pmod{33}$
$\quad 2^8 \equiv 25 \pmod{33}$
$\quad 2^9 \equiv 17 \pmod{33}$
and then repeats since
$\quad 2^{10} \equiv \;\,1 \pmod{33}$
So the pattern repeats after a cycle of length $10$.
We want to get to the $56^{th}$ row ($0 \le k \le 55$), and since $56 = 5 \times 10 + 6$, we use the $6^{th}$ entry of the cycle, $2^5 \equiv 32 \pmod{33}$, so that also
$\tag 1 2^{55} \equiv 32 \pmod{33}$
and adding $1$ we can write the answer,
$\tag {ANS} 2^{55}+1 \equiv 0 \pmod{33}$
|
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|
Compute $S_n=\sum_{k=1}^{n} {n \choose k} k^2$ Compute $S_n=\sum_{k=1}^{n} {n \choose k} k^2$.
This is took from Arthur Engel’s book, from the enumerative combinatorics chapter. I can’t understand the author’s explanation. He says the sum represents the number of ways to choose a comittee, a chairman, and its secretary (possible the same person) from a set with n elements. I don’t understant why this happens.
|
Here is the algebric way to prove it :
Let $ n $ be a positive integer greater than $ 1 $, know : $$ \left(\forall k\in\left[\!\left[1,n-1\right]\!\right]\right),\ k^{2}\displaystyle\binom{n-1}{k}=k\left(n-k\right)\displaystyle\binom{n-1}{k-1}$$
Thus, \begin{aligned} \displaystyle\sum_{k=1}^{n-1}{k^{2}\displaystyle\binom{n-1}{k}}&=\displaystyle\sum_{k=1}^{n-1}{k\left(n-k\right)\displaystyle\binom{n-1}{k-1}} \\ &=n\displaystyle\sum_{k=1}^{n-1}{k\displaystyle\binom{n-1}{k-1}}-\displaystyle\sum_{k=1}^{n-1}{k^{2}\displaystyle\binom{n-1}{k-1}}\\ \displaystyle\sum_{k=1}^{n-1}{k^{2}\displaystyle\binom{n-1}{k}}&=n\displaystyle\sum_{k=0}^{n-2}{\left(k+1\right)\displaystyle\binom{n-1}{k}}-\displaystyle\sum_{k=1}^{n-1}{k^{2}\displaystyle\binom{n-1}{k-1}}\\ \iff \displaystyle\sum_{k=1}^{n-1}{k^{2}\left[\displaystyle\binom{n-1}{k}+\displaystyle\binom{n-1}{k-1}\right]}&=n\displaystyle\sum_{k=0}^{n-2}{k\displaystyle\binom{n-1}{k}}+n\displaystyle\sum_{k=0}^{n-2}{\displaystyle\binom{n-1}{k}}\\ \iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \displaystyle\sum_{k=1}^{n}{k^{2}\displaystyle\binom{n}{k}}&=n\displaystyle\sum_{k=0}^{n-1}{k\displaystyle\binom{n-1}{k}}+n\displaystyle\sum_{k=0}^{n-1}{\displaystyle\binom{n-1}{k}}\end{aligned}
(In the fourth line, we added $ \sum\limits_{k=1}^{n-1}{k^{2}\binom{n-1}{k-1}} $ to both sides of the equation, and in the last line we used pascal's triangle formula, then we added $ n^{2} $ to both sides of the equation)
Using an index change by symmetry, we have $ \sum\limits_{k=0}^{n-1}{k\binom{n-1}{k}}=\sum\limits_{k=0}^{n-1}{\left(n-1-k\right)\binom{n-1}{k}} $, thus $ \sum\limits_{k=0}^{n-1}{k\binom{n-1}{k}}=\frac{n-1}{2}\sum\limits_{k=0}^{n-1}{\binom{n-1}{k}} $, hence : $$ \fbox{$
\begin{array}{rcl}\displaystyle\sum_{k=0}^{n}{k^{2}\displaystyle\binom{n}{k}}&=\displaystyle\frac{n\left(n+1\right)}{2}\displaystyle\sum_{k=0}^{n-1}{\displaystyle\binom{n-1}{k}}\end{array}
$} $$
Which means the final result would be : $ 2^{n-2}n\left(n+1\right) $
|
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|
Finding xy+yz+zx such that the given determinant = 0
$x≠y≠z$
$\begin{vmatrix}x&x^3&x^4-1\\y&y^3&y^4-1\\z&z^3&z^4-1\end{vmatrix} = 0$
Then xy+yz+zx = | A. x+y+z | B. $xyz$ | C. $xyz\over(x+y+z)$ | D. $(x+y+z)\over xyz$ |
Given Ans - D
What I did first was R1->R1-R3 & R2->R2-R3 and throwing (x-z) and (y-z) to the 0.....but this way the opened determinant is still too complex
What I did second was putting values of x and y but with that I was only able to eliminate option A & B
I need help with the correct approach (the correct row transformation) or any other method I can try.
|
We can take advantage of this being a multiple choice question.
Expanding out the determinant $$\begin{vmatrix}x&x^3&x^4-1\\y&y^3&y^4-1\\z&z^3&z^4-1\end{vmatrix}$$ will give products that look like $xy^3(z^4-1)$, which give us six degree-$8$ terms (such as $xy^3z^4$) and six degree-$4$ terms (such as $-xy^3$).
We know the determinant is divisible by $(x-y)(y-z)(x-z)$, because if any two of $x,y,z$ are equal, the determinant is $0$. Factoring that out, we should get the difference of a degree-$5$ polynomial and a degree-$1$ polynomial.
The four answers are predicting that the polynomial we have left is a multiple of:
\begin{array}{cc}
(A) & xy + yz + zx - x - y - z \\
(B) & xy + yz + zx - xyz \\
(C) & (xy + yz + zx)(x + y + z) - xyz \\
(D) & (xy + yz + zx)(xyz) - x - y - z
\end{array}
Only option (D) is the difference of a degree-$5$ polynomial and a degree-$1$ polynomial, so it is the only possibility.
|
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|
Prove that in acute triangle : $\sin A\cos (B-C)=\frac{\sin (2B)+\sin (2C)}{2}$ Im going to prove this identity in acute triangle :
Let $ABC$ acute triangle , $A,B,C$ are angles then :
$$\sin A\cos (B-C)=\frac{\sin (2B)+\sin (2C)}{2}$$
I know that $A,B,C<\frac{π}{2}$
$$\sin A\cos (B-C)=\sin A(\cos B\cos C+\sin A\sin B\sin C)$$
And also :
$$\sin (2B)+\sin (2C)=\sin B\cos B +\sin B\cos B+\sin C\cos C +\sin C\cos C$$
But then I don't know how ?
|
First, we need to derive this little identity:
$\sin (x +y) = \sin x\cos y - \cos x\sin y\\
\sin (x - y) = \sin x\cos y + \cos x\sin y\\
\sin (x +y) + \sin(x-y) = 2\sin x\cos y\\
\sin x\cos y = \frac{\sin (x +y) + \sin(x-y)}{2}$
Now we can apply it with $x = A, y = B-C$
And while we are at it, we can derive similar identities to show that
$\cos x\cos y = \frac {\cos (x+y) + \cos (x-y)}{2}\\
\sin x\sin y = \frac {\cos (x-y) - \cos (x+y)}{2}$
|
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|
Comparing $2$ infinite continued fractions
$A = 1 +\dfrac{1}{1 + \frac{1}{1 + \frac{1}{\ddots}}} \\
B = 2 +\dfrac{1}{2 + \frac{1}{2 + \frac{1}{\ddots}}}$
Given the two infinite continued fractions $A$ and $B$ above, which is larger, $2A$ or $B?$
I used the golden ratio on the $2$ and came up with:
$A = 1 + \dfrac{1}{A} \\
B = 2 + \dfrac{1}{B}$
Converting to quadratic equations:
$A^2 - A - 1 = 0 \\
B^2 -2B - 1 = 0$
Resulting to:
$2A = 1 + \sqrt{5} > 1 + \sqrt{2} = B$
My Question is:
Are there any more ways to solve this type of problem?
|
You can work out a continued fraction form for $2A$ and compare it with $B$.
Start with
$A=1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}...}}}$
Then we multiply by 2, by multiplying the initial $1$ by $2$ and dividing the denominator by $2$ at the first layer:
$2A=2+\dfrac{1}{(1/2)(1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}...}})}$
Now do the indicated multiplication by $1/2$ in the analogous way:
$2A=2+\dfrac{1}{(1/2)+\dfrac{1}{\color{blue}{(2)(1+\dfrac{1}{1+\dfrac{1}...}})}}$
The blue expression just gives us back our original $2A$, so this pattern repeats giving us:
$2A=\color{blue}{2}+\dfrac{1}{\color{brown}{(1/2)}+\dfrac{1}{\color{blue}{2}+\dfrac{1}{\color{brown}{(1/2)}+\dfrac{1}{...}}}}$
Versus:
$B=\color{blue}{2}+\dfrac{1}{\color{brown}{2}+\dfrac{1}{\color{blue}{2}+\dfrac{1}{\color{brown}{2}+\dfrac{1}{...}}}}$
Now note the coloring I gave to the summands. By considering the convergent one can see that when the numerators are all $1$, the value of a continued fraction increases when the blue entries are increased but decreases when the brown entries are increased, provided these entries are all positive. Here the blue entries for $2A$ and $B$ are the same but the brown entries are greater for $B$, therefore $2A>B$.
|
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|
how to calculate $\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{dx}{1+\cos^2x}$? $$
\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}{\frac{1}{1+\cos ^2x}\rm{dx}}=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}{\frac{1}{1+\cos ^2x}\rm{dx}+\int_{\frac{\pi }{2}}^{\frac{3\pi}{4}}{\frac{1}{1+\cos ^2x}\rm{dx}}}
$$
I want to split this integral,$
\int_{\frac{\pi }{4}}^{\frac{3\pi }{4}}{\frac{1}{1+\cos ^2x}\rm{dx}}
$
but I don't know how to calculate the two integrals after splitting?
|
A possible, maybe more convenient way is as follows:
*
*First shift the integral using $x=\frac{\pi}{2} + u$:
$$\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{dx}{1+\cos^2x}= \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{du}{1+\sin^2u}= 2\int_{0}^{\frac{\pi}{4}}\frac{du}{1+\sin^2u}$$
*Now, use $t=\tan u$ and note that $\sin^2u = \frac{t^2}{1+t^2}$:
$$2\int_{0}^{\frac{\pi}{4}}\frac{du}{1+\sin^2u}=2\int_{0}^{1}\frac{dt}{1+2t^2}= \sqrt 2\arctan(\sqrt 2)$$
|
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|
Find the sum of all positive integers $k$ for which $5x^2-2kx+1<0$ has exactly one integral solution.
Find the sum of all positive integers $k$ for which $5x^2-2kx+1<0$ has exactly one integral solution.
My attempt is as follows:
$$\left(x-\dfrac{2k-\sqrt{4k^2-20}}{10}\right)\left(x-\dfrac{2k+\sqrt{4k^2-20}}{10}\right)<0$$
$$\left(x-\dfrac{k-\sqrt{k^2-5}}{5}\right)\left(x-\dfrac{k+\sqrt{k^2-5}}{5}\right)<0$$
$$x\in\left(\dfrac{k-\sqrt{k^2-5}}{5},\dfrac{k+\sqrt{k^2-5}}{5}\right)$$
As it is given that it has got only one integral solution, so there must be exactly one integer between $\dfrac{k-\sqrt{k^2-5}}{5}$ and $\dfrac{k+\sqrt{k^2-5}}{5}$
Let $x_1=\dfrac{k-\sqrt{k^2-5}}{5}$ and $x_2=\dfrac{k+\sqrt{k^2-5}}{5}$ , then $[x_2]-[x_1]=1$ where [] is a greater integer function.
But from here, how to proceed? Please help me in this.
|
Your idea is good; you want to find all positive integers $k$ for which there is precisely on integer between the roots of
$$5x^2-2kx+1=0.$$
Then the distance between the roots can be at most $2$, where the distance between the roots is precisely
$$\frac{1}{5}\sqrt{(-2k)^2-4\cdot1\cdot5}=\frac25\sqrt{k^2-5},$$
as you already found. This is at most $2$ if and only if $\sqrt{k^2-5}\leq5$, or equivalently $k\leq5$. This leaves only $5$ values of $k$ to check.
|
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|
Probability Two People are Born on the Same Date (alternative way) What is the probability two people (individuals) will have the same exact birthday?
There are 365 days in a year and I assume that any person can be born on any random day, so uniformly.
I like to use a slots method when I look at combinations/permutations
Slots of possibilities person 1 was born on (e.g. born on Jan 2)
$\{ \text{Person 1} \} = \frac{ \quad 0 \quad }{1} \frac{ \quad 1 \quad }{2} \frac{ \quad 0 \quad }{3} \quad \cdot \cdot \cdot \cdot \cdot \cdot \quad \frac{ \quad 0 \quad }{364} \frac{ \quad 0 \quad }{365} $
Slots of possibilities person 2 was born on (e.g. born on Jan 3)
$\{ \text{Person 2} \} = \frac{ \quad 0 \quad }{1} \frac{ \quad 0 \quad }{2} \frac{ \quad 1 \quad }{3} \quad \cdot \cdot \cdot \cdot \cdot \cdot \quad \frac{ \quad 0 \quad }{364} \frac{ \quad 0 \quad }{365} $
Add these two slots and you get these two slot possibilities
The two people are not born the same date
$\{ \text{Slot 1 + Slot 2} \} = \frac{ \quad 0 \quad }{1} \frac{ \quad 1 \quad }{2} \frac{ \quad 1 \quad }{3} \quad \cdot \cdot \cdot \cdot \cdot \cdot \quad \frac{ \quad 0 \quad }{364} \frac{ \quad 0 \quad }{365} $
There $365 \choose 2$ ways of arranging two $1$'s and three hundred sixty three $0$'s
OR
Both people are born on the same day (e.g. Jan 3)
$\{ \text{Slot 1 + Slot 2} \} = \frac{ \quad 0 \quad }{1} \frac{ \quad 0 \quad }{2} \frac{ \quad 2 \quad }{3} \quad \cdot \cdot \cdot \cdot \cdot \cdot \quad \frac{ \quad 0 \quad }{364} \frac{ \quad 0 \quad }{365} $
There are 365 ways of arranging one 2 and three hundred sixty four $0$'s
So the probability of two people have matching birthdays
$$\text{P}(\text{matching birthday})=\frac{365 }{{365 \choose 2} +365 } \approx 0.005$$
But this answer is incorrect. I know what the correct answer is and I know how to do it another way. My question is why is the method I laid out not correct? I am less concerned with the answer, where am I wrong in my thinking in looking at all the possibilities? Thank you
|
$\{ \text{Slot 1 + Slot 2} \} = \frac{ \quad 0 \quad }{1} \frac{
\quad 1 \quad }{2} \frac{ \quad 1 \quad }{3} \quad \cdot \cdot
\cdot \cdot \cdot \cdot \quad \frac{ \quad 0 \quad }{364} \frac{
\quad 0 \quad }{365} $
There $365 \choose 2$ ways of arranging two $1$'s and three hundred
sixty three $0$'s
Here is the flaw. You have two different days of birthdays (a,b). So basically you arrange 365 elements. 363 have the same label $x$. Then you have additionally two different labels. So you are looking for the number of ways to arrange the following elements
$$\underbrace{xx...xx}_{=363}ab$$
So you have three different types of elements. Here you use the multinomial coefficient.
$$\binom{365}{363,1,1}=\frac{365!}{363!\cdot 1!\cdot 1!}=365\cdot 364$$
There are 365 ways of arranging one 2 and three hundred sixty four 0's
This is right. Therefore
$$\text{P}(\text{matching birthday})=\frac{365\ }{365\cdot 364 +365 } =\frac{1}{365}$$
|
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|
Let $a,b,c\in R$. If $f(x)=ax^2+bx+c$ is such that $a+b+c=3$ and $f(x+y)=f(x)+f(y)+xy$ for all $x,y\in R$...
Let $a,b,c\in R$. If $f(x)=ax^2+bx+c$ is such that $a+b+c=3$ and $f(x+y)=f(x)+f(y)+xy$ for all $x,y\in R$. Then $\sum_{n=1}^{10} f(n)$ is
From the first equation
$$a(x^2+y^2+2xy)+b(x+y)+c=ax^2+bx+c+ay^2+by+c+xy$$
$$2axy=c+xy$$
Also, the summation will be
$$a(1^2+2^2+3^2...10^2)+b(1+2+3+4...+10)+10c$$
$$=385a+55b+10c$$
$$375a+45b+30$$
I don’t know what to do with the x and y terms. How do I proceed?
|
From your third step you get $c=0$ and $a=1/2, \implies 1/2+b+0=3 \implies b=5/2$.
Next, $f(n)=n^2/2+5n/2$, then
$$S_m=\sum_{n=1}^{m} (n^2/2+5n/2)= \frac{m(m+1)(2m+1)}{12}+\frac{5m(m+1)}{4}$$
$$\implies S_m=\frac{m(m+1)(m+8)}{6}$$
then $$S_{10}=10.11.18/6=330$$
|
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|
Show that $(n+1)^{2/3} -n^{2/3} <\frac{2}{3} n^{-1/3}$ for all positive integers
I am trying to prove
$$(n+1)^{2/3} -n^{2/3} <\frac{2}{3} n^{-1/3}$$
for all positive integers.
My attempts so far have been to Taylor expand the left hand side:
$$(n+1)^{2/3} -n^{2/3}\\
=n^{2/3}\big((1+1/n)^{2/3} -1\big)\\
=n^{2/3}\left(\sum_{\alpha=0}^\infty\frac{\frac{2}{3}(\frac{2}{3}-1)\cdots(\frac{2}{3}-\alpha+1)}{\alpha!} n^\alpha-1\right)\\
=n^{2/3}\sum_{\alpha=1}^\infty\frac{\frac{2}{3}(\frac{2}{3}-1)\cdots(\frac{2}{3}-\alpha+1)}{\alpha!} n^\alpha$$
I also tried proof by induction. Assume that it's true for n=k, so that
$$(n+1)^{2/3} -n^{2/3} < \frac{2}{3}n^{-1/3}\\
n^{2/3}\big((1+1/n)^{2/3} -1\big)<\frac{2}{3} n^{-1/3}\\
(1+1/n)^{2/3} -1<\frac{2}{3} n^{-1}$$
Then I want to prove that $(n+2)^{2/3} -(n+1)^{2/3} < \frac{2}{3}(n+1)^{-1/3}$. The left hand side is:
$$(n+2)^{2/3} -(n+1)^{2/3}\\
=(n+1)^{2/3}\left[\left(1+\frac{1}{n+1}\right)^{2/3}-1^{2/3}\right]\\
<(n+1)^{2/3}\cdot \frac{2}{3} n^{-1}\\
=\frac{2}{3}\frac{(n+1)^{2/3}}{n^{-1}}$$
But this is bigger than $\frac{2}{3}(n+1)^{-1/3}$, so I am stumped!
|
If you are allowed to integrate,
then, for $a > 0$,
$\begin{array}\\
\int_n^{n+1} x^{-a} dx
&= \dfrac{x^{-a+1}}{-a+1}|_n^{n+1}\\
&= \dfrac{(n+1)^{-a+1}-n^{-a+1}}{-a+1}\\
\text{and}\\
\int_n^{n+1} x^{-a} dx
&\lt n^{-a}
\qquad\text{since } x^{-a} \text{ is decreasing}\\
\end{array}
$
Putting $a = \frac13$.
this becomes
$n^{-1/3}
\gt \dfrac{(n+1)^{2/3}-n^{2/3}}{2/3}
= \dfrac32((n+1)^{2/3}-n^{2/3})
$.
Note that,
since
$\int_n^{n+1} x^{-a} dx
\gt (n+1)^{-a}
$
we get
$(n+1)^{-1/3}
\lt \dfrac32((n+1)^{2/3}-n^{2/3})
$.
|
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|
3 circles of radii 1,2 and 3 and centers A, B and C respectively, touch each other.......
3 circles of radii 1,2 and 3 and centers A, B and C respectively, touch each other. Another circle whose center is P touches all 3 externally and has radius $r$. Also $\angle PAB=\theta$ and $\angle PAC=\alpha$. Prove that
$\cos \theta= \frac{3-r}{3(1+r)}$
$\cos \alpha = \frac{2-r}{2(1+r)}$
$r=\frac {6}{23}$
I imagine the circle $P$ to be in the center of the three circles. The $\Delta ABC$ is a right angled at A, so $\theta+\alpha = 90$
I know the lengths $PA, PB$ and $PC$ in terms of $r$
All the angles of the triangle are also well known.
If we consider $\Delta ABP$, then
$$\cos \theta = \frac{PA^2+AB^2-PB^2}{2PA.PB}$$
Here
$$PA=1+r$$
$$AB=3$$
$$PB=2+r$$
so
$$\cos \theta = \frac{9+(1+r^2)-(2+r)^2}{(2)(3)(1+r)}$$
which ends up being 1. I feel my approach is fundamentally wrong, so please correct me.
|
It should be $$\cos\theta=\frac{9+(1+r)^2-(2+r)^2}{2\cdot3\cdot(1+r)}=\frac{6-2r}{6(1+r)}=\frac{3-r}{3(1+r)}.$$
Also, $$\cos\alpha=\frac{16+(1+r)^2-(3+r)^2}{2\cdot4\cdot(1+r)}=\frac{2-r}{2(1+r)}$$ and since $$\alpha+\theta=90^{\circ},$$ we obtain:
$$\left(\frac{3-r}{3(1+r)}\right)^2+\left(\frac{2-r}{2(1+r)}\right)^2=1$$ or
$$4(9-6r+r^2)+9(4-4r+r^2)=36(1+2r+r^2)$$ or
$$(r+6)(23r-6)=0,$$ which gives $$r=\frac{6}{23}.$$
|
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|
For acute $\triangle ABC$, prove $(\cos A+\cos B)^2+(\cos A+\cos C)^2+(\cos B+\cos C)^2\leq3$
Prove that, in an acute $\triangle ABC$,
$$(\cos A+\cos B)^2+(\cos A+\cos C)^2+(\cos B+\cos C)^2\leq3$$
I tried this, but I can't to this.
I used $AM\geq GM$ and got $$3\geq\cos(A-B)+\cos(A-C)+\cos(B-C)$$
But I can't see how to do this question.
|
Let $a^2+b^2-c^2=z$, $a^2+c^2-b^2=y$ and $b^2+c^2-a^2=x$.
Thus, we need to prove that
$$\sum_{cyc}\left(\frac{b^2+c^2-a^2}{2bc}+\frac{a^2+c^2-b^2}{2ac}\right)^2\leq3$$ or
$$\sum_{cyc}\left(\frac{x}{2\cdot\sqrt{\frac{x+z}{2}}\cdot\sqrt{\frac{x+y}{2}}}+\frac{y}{2\cdot\sqrt{\frac{y+z}{2}}\cdot\sqrt{\frac{x+y}{2}}}\right)^2\leq3$$ or
$$\sum_{cyc}\left(\frac{x}{\sqrt{(x+y)(x+z)}}+\frac{y}{\sqrt{(x+y)(y+z)}}\right)^2\leq3$$ or
$$\sum_{cyc}(x\sqrt{y+z}+y\sqrt{x+z})^2\leq3\prod_{cyc}(x+y)$$ or
$$\sum_{cyc}\left(2x^2y+2x^2z+2xy\sqrt{(x+z)(y+z)}\right)\leq\sum_{cyc}(3x^2y+3x^2z+2xyz)$$ or
$$2\sum_{cyc}xy\sqrt{(x+z)(y+z)}\leq\sum_{cyc}(x^2y+x^2z+2xyz),$$ which is true by AM-GM:
$$2\sum_{cyc}xy\sqrt{(x+z)(y+z)}\leq\sum_{cyc}xy(x+z+y+z)=\sum_{cyc}(x^2y+x^2z+2xyz).$$
|
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|
Find the area under the graph of the function $f(x)=(1-2x)^2$ between $x=-2$ and $x=2$ 5.4
Can somebody verify this solution for me?
Find the area under the graph of the function $f(x)=(1-2x)^2$ between $x=-2$ and $x=2$
The area under the graph of $f(x)$ between $x=-2$ and $x=2$ is exactly equal to:
$\int_{-2}^2 (1-2x)^2dx$
$=\int_{-2}^2 1-4x+4x^2dx$
$=(x-\frac{4}{2}x^2+\frac{4}{3}x^3) |_{-2}^2$
$=(x-2x^2+\frac{4}{3}x^3) |_{-2}^2$
$=(2-2(2)^2+\frac{4}{3}(2)^3)-(-2-2(-2)^2+\frac{4}{3}(-2)^3)$
$=(2-8+\frac{32}{3})-(-2-8+\frac{-32}{3})$
$=2-8+\frac{32}{3}+2+8+\frac{32}{3})$
$=4+\frac{64}{3}$
$\frac{12}{3}+\frac{64}{3}$
$=\frac{76}{3}$
|
$$
\int_{-2}^{2}{(1-2x)^{2}dx}\\=\int_{-2}^{2}{1-4x+4x^2dx}\\=\int_{-2}^{2}{1dx}+\int_{-2}^{2}{4xdx}+\int_{-2}^{2}{4x^2dx}
$$
Now lets solve those integrals.
$$
\int_{-2}^{2}{1dx}=4\\
$$
since it has only a constant
$$
\int_{-2}^{2}{4xdx}=0\\
$$
since it's an even function.
$$
\int_{-2}^{2}{4x^2}\\
=4\times\int_{-2}^{2}{x^2}\\
=4[\frac{x^3}{3}]_{-2}^{2}\\
=4\times(\frac{8}{3}-\frac{-8}{3})\\
=4\times\frac{16}{3}\\
=\frac{64}{3}
$$
so
$$
=4-0+\frac{64}{3}
=\frac{76}{3}
$$
|
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|
Given that $a+b+c=0$, show that $2(a^4+b^4+c^4)$ is a perfect square Given that $a+b+c=0$. Show that: $2(a^4+b^4+c^4)$ is a perfect square
MY ATTEMPTS:
I found that when $a+b+c=0$, $a^3+b^3+c^3=3abc$
So I did:
$(a^3+b^3+c^3)(a+b+c)$ -- $a^4+b^4+c^4=-(a^3c+a^3b+ab^3+b^3c+c^3a+c^3b)$
And then I tried to substitute $a^4+b^4+c^4$, but I found nothing that I thought relevant to the question
|
Square $a+b+c=0$
\begin{eqnarray*}
a^2+b^2+c^2=-2(ab+bc+ca).
\end{eqnarray*}
Square this
\begin{eqnarray*}
a^4+b^4+c^4+2(a^2b^2+b^2c^2+c^2a^2)=4(a^2b^2+b^2c^2+c^2a^2)+8abc(a+b+c)
\end{eqnarray*}
The last term is zero ... rearrange
\begin{eqnarray*}
2(a^4+b^4+c^4)=4(a^2b^2+b^2c^2+c^2a^2)=(a^2+b^2+c^2)^2.
\end{eqnarray*}
the last equality follows from the first equation.
|
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|
Prove by mathematical induction that $3^n>2n^3$ I'm having trouble with this question:
"Prove by mathematical induction that for all integers $n\ge 6$, $3^n>2n^3$".
I got to $P(k)=2k^3<3^k$ and $P(k+1)=2(k+1)^3<3^{k+1}=2k^3+6k^2+6k+2<3^k*3$,
but I dont know how I can get $P(k+1)$ from $P(K)$...
Thanks
|
Suppose $2k^3<3^k$. Then
\begin{align}
2(k+1)^3&=2(k^3+3k^2+3k+1)\\
&=2k^3+6k^2+6k+2\\
&<3^k+6k^2+6k+2\\
&< 3^k+k^3+k^2+k\\
&<3^k+4k^3\\
&<3^k+2\cdot3^k\\
&=3^{k+1}
\end{align}
Note that we use in the middle that $6\le k$.
|
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|
Minimizing cubic over elliptic disk without using derivatives Given $x, y$ satisfy $(x-4)^2 + (y-4)^2 + 2xy \leq 32$. Find the minimum value of:
$$P = x^3 + y^3 +3(xy-1)(x+y-2)$$
My attempt:
From $(x-4)^2 + (y-4)^2 + 2xy \leq 32$, I get:
$$(x+y)^2 - 8(x+y) \leq 0$$
$$\implies 0 \leq x+y \leq 8$$
From $P = x^3 + y^3 +3(xy-1)(x+y-2)$, I get:
$$P = (x+y)^3 - 6xy - 3(x+y) + 6 \geq (x+y)^3 - 3(x+y)^2 - 3(x+y) + 6$$
I can use derivatives to find the minimum value but my teacher told me to try to solve this without derivatives. Is there a way to do that?
Edit: There was a mistake in my solving. I was using AM-GM inequality to derive this:
$$- 6xy \geq - \frac{3(x+y)^2}{2}$$
But I forgot that $x$ and $y$ are both real and can be negative.
So now I fixed it. Instead of $\frac{3(x+y)^2}{2}$, it should be $3(x+y)^2$ since $2xy \leq (x + y)^2$
|
Yes, we can.
Let $x+y=2u$ and $xy=v^2$, where $v^2$ can be negative.
Thus, the condition gives $$(x+y)^2-8(x+y)\leq0$$ or
$$0\leq u\leq4.$$
Now, since $u^2\geq v^2$ it's $(x-y)^2\geq0$, we obtain: $$P=8u^3-6uv^2+3(v^2-1)(2u-2)=8u^3-6v^2-6u+6\geq$$
$$\geq8u^3-6u^2-6u+6=\left(2u-\frac{1}{2}\right)^3-\frac{3}{2}u+\frac{1}{8}-6u+6=$$
$$=\left(2u-\frac{1}{2}\right)^3-\frac{15}{4}\left(2u-\frac{1}{2}\right)+\frac{17}{4}=$$
$$=\left(2u-\frac{1}{2}\right)^3+2\cdot\frac{\sqrt{125}}{8}-\frac{15}{4}\left(2u-\frac{1}{2}\right)+\frac{17}{4}-\frac{5\sqrt5}{4}\geq$$
$$\geq 3\sqrt[3]{\left(2u-\frac{1}{2}\right)^3\cdot\left(\frac{\sqrt{125}}{8}\right)^2}-\frac{15}{4}\left(2u-\frac{1}{2}\right)+\frac{17}{4}-\frac{5\sqrt5}{4}=\frac{17-5\sqrt5}{4}.$$
We used the following inequality.
$$\left(2u-\frac{1}{2}\right)^3+\frac{\sqrt{125}}{4}\geq\frac{15}{4}\left(2u-\frac{1}{2}\right),$$ which is
$$\left(u-\frac{1+\sqrt5}{4}\right)^2\left(u+\frac{2\sqrt5-1}{4}\right)\geq0,$$ which is true for $u\geq0.$
The equality occurs for $x=y=\frac{1+\sqrt5}{4},$ which says that we got a mninimal value.
|
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|
Evaluate $\sum _{n=1}^{\infty } \frac{1}{n^5 2^n \binom{3 n}{n}}$ in terms of elementary constants How can we evaluate
$$\sum _{n=1}^{\infty } \frac{1}{n^5 2^n \binom{3 n}{n}}$$
Related: $\sum _{n=1}^{\infty } \frac{1}{n^4 2^n \binom{3 n}{n}}$ is solved recently (see here for the solution) by elementary method, so I wonder if a generalization to weight $5$ can be made.
|
Beta+IBP $3$ times+log factorization yields
*
*$S=\sum _{n=1}^{\infty } \frac{1}{n^5 2^n \binom{3 n}{n}}=\int_0^1 \frac2x \text{Li}_4\left(\frac{x^2(1-x)}2\right) dx\\ =\int_0^1 -\frac{2\left((3 x-2) \text{Li}_3\left(\frac{1}{2} (1-x) x^2\right)\right) \log (x)}{(x-1) x} dx\\ =\int_0^1 \frac{2\left((3 x-2) \text{Li}_2\left(\frac{1}{2} (1-x) x^2\right)\right) \left(\log ^2(x)-\text{Li}_2(1-x)\right)}{(x-1) x} dx\\ =\int_0^1 \left(\frac{2}{x}-\frac{1}{1-x}\right) f(x) \left(\log \left((1-x)^2+1\right)+\log (x+1)-\log (2)\right) dx$
Where
*
*$\small f(x)=2\left(-\text{Li}_3(1-x)+2 \text{Li}_3(x)-2 \text{Li}_2(1-x) \log (x)-2 \text{Li}_2(x) \log (x)+\frac{2 \log ^3(x)}{3}-\log (1-x) \log ^2(x)\right)$
Apply reflection one obtain
*
*$S=\int_0^1 \left(\frac{2 f(1-x) \left(\log \left(x^2+1\right)-\log (2)\right)}{1-x}-\frac{f(1-x) \log \left(x^2+1\right)}{x}+\frac{2 f(x) \log (x+1)}{x}-\frac{f(x) (\log (x+1)-\log (2))}{1-x}\right) \, dx$
Which has $4$-admmisible integrand thus solvable via MZVs of level $4$ (see arXiv $2007.03957$)
$$\sum _{n=1}^{\infty } \frac{1}{n^5 2^n \binom{3 n}{n}}=3 \pi \beta (4)+4 \pi \operatorname{Im} \left(\text{Li}_4\left(\frac{1}{2}+\frac{i}{2}\right)\right)-\frac{51 \text{Li}_5\left(\frac{1}{2}\right)}{2}-15 \text{Li}_4\left(\frac{1}{2}\right) \log (2)+\frac{\pi ^2 \zeta (3)}{4}+\frac{9 \zeta (5)}{2}-3 \zeta (3) \log ^2(2)-\frac{97 \log ^5(2)}{240}+\frac{41}{144} \pi ^2 \log ^3(2)-\frac{61}{960} \pi ^4 \log (2)$$
|
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|
Finding number of integral solutions to an equation.
Find the number of integral solutions to:
$$x^2+y^2-6x-8y=0.$$
My attempt:
The equation can be rewritten as:
$$x^2+y^2-6x-8y+9+16=25,$$
basically adding 25 to both sides, or equivalently,
$$(x-3)^2+(y-4)^2=25.$$
This is a Pythagorean triplet. The only triplet of this form would be $3$, $4$, $5$, so the possibilities for $x - 3$ and $y - 4$ are:
1.) $3$, $4$
2.) $-3$, $-4$
3.) $-3$, $4$
4.) $3$, $-4$
For each of these pairs, there are two different pairs $(x,y)$. For instance, in the first pair we could have $x - 3 = 3$ and $y - 4 = 4$, or $x - 3 = 4$ and $y - 4 = 3$. So there are $8$ solutions.
But, shouldn't we also consider the pair $(0,5)$? Because
$$0^2+5^2=25.$$
Then we have two new pairs:
1.) $0$, $5$
2.) $0$, $-5$
This would give $4$ new solutions, so the total number of solutions should be $8+4=12$.
How is the answer just $8$ then? What am I missing?
|
In the non-negatives,
$$n^2+m^2=25$$ is indeed solved by
$$(0,5),(3,4),(4,3),(5,0),$$ as you can check by exhaustive search on $n$.
Then the signed solutions are
$$(0,\pm5),(\pm3,\pm4),(\pm4,\pm3),(\pm5,0).$$
There are $2+2^2+2^2+2$ of them, that you shift by $(3,4)$ to get $(x,y)$.
You can even minimize the "effort" by noticing that $n$ and $m$ play symmetric roles, and you can solve for $n\le m$ only, giving
$$(0,5),(3,4).$$
Then by playing with the signs and swapping, you get $2\cdot2+2\cdot2^2$ distinct solutions.
|
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|
If $\log_8 3 = P$ and $\log_3 5 = Q$, express $\log_{10} 5$ in terms of $P$ and $Q$. If $\log_8 3 = P$ and $\log_3 5 = Q$, express $\log_{10} 5$ in terms of $P$ and $Q$. Your answer should no longer include any logarithms.
I noted that $\log_5 10=\frac{1}{\log_{10} 5}.$
I also noted that $\log_{5} 10=\log_5 2+\log_5 5=\log_5 2+1.$ I don't know how to continue, how do I finish this problem using my strategy?
|
$$ \log_{10} 5 = \log_{8} 5 \times \log_{10}{8}$$
$$\log_{10}{8}=\frac{1}{\log_{8}{10}} = \frac{1}{\log_{8} 2+\log_8 5}$$
$$\log_8 5 = \log_8 3\times \log_3 5 = PQ, \log_{8} 2=\frac{1}{3}$$
$$ \log_{10} 5 = PQ\times \frac{1}{\frac{1}{3}+PQ}$$
|
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|
Application of lagrange mean value theorem Let $f$ be continuous on $[a,b]$ , differentiable on ($a,b$), and its given $f(a)=a, f(b)=b$. We need to prove the following:
*
*$f'(c1)+f'(c2)=2,$ , for some $c1,c2 \in (a,b)$, and $c1\neq c2$.
*$\frac{1}{f'(c1)}+ \frac{1}{f'(c2)}=2,$ , for some $c1,c2 \in (a,b)$, and $c1\neq c2$.
*$ f'(c)=\frac{1}{b-c} + \frac{1}{a-c}$ , for some $c\in (a,b).$
I got the insight for the 1st one: since we need to apply the theorem twice, it is reasonable to split the intervals into two sub intervals$(a,\frac{a+b}{2})$ and $(\frac{a+b}{2},b)$. After applying the Lagrange mean value theorem on each of these intervals and adding, we easily prove 1.
I thought of a similar argument for 2, but the reciprocals make things messy.
I am absolutely clueless about 3.
Edit: option 3 seems similar to cauchy mean value theorem, but I am still confused on how to proceed.
|
For $2$, since $f(a) = a$ and $f(b) = b$, with $a \lt b$, since $f$ is continuous, note the intermediate value theorem says there's a point, call it $a \lt d \lt b$, where
$$f(d) = \frac{a + b}{2} \tag{1}\label{eq1A}$$
Now, use the Lagrange mean value theorem on the sub-intervals $(a, d)$ and $(d, b)$ to get for that some $c_1 \in (a, d)$ you have
$$f'(c_1) = \frac{f(d) - f(a)}{d - a} \implies \frac{1}{f'(c_1)} = \frac{d - a}{f(d) - f(a)} = \frac{d - a}{\frac{b + a}{2} - a} = \frac{2(d - a)}{b - a} \tag{2}\label{eq2A}$$
and for some $c_2 \in (d, b)$ you have
$$f'(c_2) = \frac{f(b) - f(d)}{b - d} \implies \frac{1}{f'(c_2)} = \frac{b - d}{f(b) - f(d)} = \frac{b - d}{b - \frac{b + a}{2}} = \frac{2(b - d)}{b - a} \tag{3}\label{eq3A}$$
Thus, you then get
$$\begin{equation}\begin{aligned}
\frac{1}{f'(c_1)}+ \frac{1}{f'(c_2)} & = \frac{2(d - a)}{b - a} + \frac{2(b - d)}{b - a} \\
& = \frac{2d - 2a + 2b - 2d}{b - a} \\
& = \frac{2(b - a)}{b - a} \\
& = 2
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
Of course, you also have that $c_1 \neq c_2$.
For $3$, consider
$$g(x) = \ln(b - x) + \ln(x - a) + f(x), \; x \in (a,b) \tag{5}\label{eq5A}$$
which is continuous and differentiable on $(a,b)$. You also have
$$\begin{equation}\begin{aligned}
g'(x) & = -\frac{1}{b - x} + \frac{1}{x - a} + f'(x) \\
& = -\frac{1}{b - x} - \frac{1}{a - x} + f'(x)
\end{aligned}\end{equation}\tag{6}\label{eq6A}$$
Note since $f(x)$ is continuous on a closed interval, it's bounded. This means that as $x \to a^{+}$, you have $g(x) \to -\infty$ and as $x \to b^{-}$, you have $g(x) \to -\infty$. From \eqref{eq5A}, you have
$$\begin{equation}\begin{aligned}
e_m & = g\left(\frac{b+a}{2}\right) \\
& = \ln\left(b - \frac{b+a}{2}\right) + \ln\left(\frac{b+a}{2} - a\right) + f\left(\frac{b+a}{2}\right) \\
& = 2\ln\left(\frac{b-a}{2}\right) + f\left(\frac{b+a}{2}\right)
\end{aligned}\end{equation}\tag{7}\label{eq7A}$$
Consider any $e \lt e_m$. Thus, since $g(x) \to -\infty$ as $x \to a^{+}$, by the intermediate value theorem, there's a point $d_1 \in \left(a, \frac{b+a}{2}\right)$ where $f(d_1) = e$, and likewise (since $g(x) \to -\infty$ as $x \to b^{-}$) there's a point $d_2 \in \left(\frac{b+a}{2},b\right)$ where $f(d_2) = e$. Thus, by Lagrange's mean value theorem, there's a $c \in (d_1,d_2)$ such that
$$g'(c) = \frac{f(d_2) - f(d_1)}{d_2 - d_1} = \frac{e - e}{d_2 - d_1} = 0 \tag{8}\label{eq8A}$$
Thus, from \eqref{eq6A}, you get
$$g'(c) = 0 = -\frac{1}{b - c} - \frac{1}{a - c} + f'(c) \implies f'(c) = \frac{1}{b - c} + \frac{1}{a - c} \tag{9}\label{eq9A}$$
|
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If a,b,c be roots of $2x^3+x^2+x-1=0$ show that some expression is equal to 16. If $a,b,c$ are roots of $2x^3+x^2+x-1=0,$ show that
$$\bigg(\frac{1}{b^3}+\frac{1}{c^3}-\frac{1}{a^3}\bigg)\bigg(\frac{1}{c^3}+\frac{1}{a^3}-\frac{1}{b^3}\bigg)\bigg(\frac{1}{a^3}+\frac{1}{b^3}-\frac{1}{c^3}\bigg)=16$$
My attempt:
Let $\frac{1}{a}=p,\frac{1}{b}=q,\frac{1}{c}=r$
$p+q+r=1$
$pqr=2$
$$pq+qr+rp=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}$$
$$=\frac{ab+bc+b^2}{(abc)^2}$$
$$=4\bigg(\frac{1}{pq}+\frac{1}{qr}+\frac{1}{q^2}\bigg)$$
$$pq+qr+rp=4\bigg(\frac{pq+qr+rp}{pq^2r}\bigg)$$
$$pq^2r=4$$
$$\implies q=2 \implies b=\frac{1}{2}$$
So p, r are roots of $x^2+x+1=0$
$\implies p^3=q^3=1$
But this condition gives a different value of the required expression, so what am I doing wrong? Please tell me the right solution.
|
I think the answer is not equal to $16$.
Let $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3u$, $\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}=3v^2$ and $\frac{1}{abc}=w^3$.
Thus, $u=\frac{1}{3},$ $v^2=-\frac{1}{3}$ and $w^3=2$.
Thus, $$\prod_{cyc}\left(\frac{1}{a^3}+\frac{1}{b^3}-\frac{1}{c^3}\right)=\prod_{cyc}\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}-\frac{2}{c^3}\right)=$$
$$=(27u^3-27uv^2+3w^3)^3-2(27u^3-27uv^2+3w^3)^3+$$
$$+4(27u^3-27uv^2+3w^3)(27v^6-27uv^2w^3+3w^6)-8w^9=-384.$$
|
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|
Find the minimum value of $a^2+b^2+c^2+2abc$ when $a+b+c=3$ and $a,b,c\geq0$.
Given $a,b,c\geq0$ such that $a+b+c=3$, find the minimum value of $$P=a^2+b^2+c^2+2abc.$$
It seems like the minimum value of $P$ is $5$ when $a=b=c=1$, but I can find at least one example where $P<5$.
My attempt:
Without loss of generality, I can suppose that $a\geq b\geq c$ and so $a\geq 1$. Therefore I have:
$$P\geq a^2+b^2+c^2+2bc=a^2+(b+c)^2\geq \frac{(a+b+c)^2}{2}=\frac{9}{2}.$$
The issue with this is that equality doesn't occur with this method.
What's your take on the problem?
|
Your attempt already contains very good observations, and is quite close to a complete solution. All that remains is an analysis of when both inequalities are equalities. You use the following two inequalities:
\begin{eqnarray*}
a^2+b^2+c^2+2abc&\geq&a^2+b^2+c^2+2bc\tag{1}\\
a^2+(b+c)^2&\geq&\frac{(a+b+c)^2}{2}\tag{2}
\end{eqnarray*}
The first inequality is an equality if and only if $abc=bc$, i.e. if and only if either $a=1$ or $bc=0$.
The second inequality is an equality if and only if $a=b+c$. Then from $a+b+c=3$ it follows that $a=\tfrac32$. Then for the first inequality to be an equality we must have $bc=0$, and hence from $b\geq c$ it follows that then $c=0$. Because $a+b+c=0$ it finally follows that $b=\tfrac32$, and so $$(a,b,c)=(\tfrac32,\tfrac32,0).$$
This shows that the minimum of $P$ is indeed $\tfrac92$, and that it is attained precisely at the points
$$(a,b,c)=(0,\tfrac32,\tfrac32),\qquad(a,b,c)=(\tfrac32,0,\tfrac32),\qquad(a,b,c)=(\tfrac32,\tfrac32,0).$$
Original answer, by brute force and elementary methods:
Plugging in $c=3-a-b$, we want the minimum of
$$P=2a^2+2b^2+9-6a-6b+8ab-2a^2b-2ab^2,$$
with the restriction that $a,b\geq0$ and $a+b\leq3$. At the boundary points we either have $a=0$ or $b=0$ or $a+b=3$, and hence correspondingly either
\begin{eqnarray*}
P&=&2b^2-6b+9,\\
P&=&2a^2-6a+9,\\
P&=&2a^2+2(3-a)^2+9-6a-6(3-a)+8a(3-a)+2a^2(3-a)+2a(3-a)^2\\
&=&2a^2-6a+9,
\end{eqnarray*}
where in each case $0\leq a,b\leq3$. From here the local minima on the boundary are easily determined to be at
$$(a,b)=(0,\tfrac32),\qquad(a,b)=(\tfrac32,0),\qquad(a,b)=(\tfrac32,\tfrac32)$$
each with value $\tfrac92$. For the extrema of $P$ on the interior we compute the derivatives of $P$ w.r.t. $a$ and $b$, which shows that
$$4a-6+8b-4ab-2b^2=0\qquad\text{ and }\qquad 4b-6+8a-2a^2-4ab=0,$$
at any interior extremum $(a,b)$ of $P$. Taking the difference shows that
$$0=2(a^2-b^2)-4(a-b)=2(a-b)(a+b-2)$$
so either $a=b$ or $a+b=2$. If $a=b$ then the quadratics above both become
$$0=4a-6+8b-4a^2-2a^2=6(a-1)^2,$$
which shows that $a=b=1$, and then $P=5$, which is not minimal. If $a+b=2$ then
$$0=4(2-a)-6+8a-2a^2-4a(2-a)=2(a-1)^2,$$
which again yields $a=b=1$, again not yielding a minimum.
In conclusion the minimum of $P$ is $\tfrac92$, and it is attained precisely at the points
$$(a,b,c)=(0,\tfrac32,\tfrac32),\qquad(a,b,c)=(\tfrac32,0,\tfrac32),\qquad(a,b,c)=(\tfrac32,\tfrac32,0).$$
|
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|
Question about $\lim\limits_{x \to2} \frac{\sqrt[3]{x} - \sqrt[3]{2}}{\sqrt{x} - \sqrt{2}}$ $\lim\limits_{x \to2} \frac{\sqrt[3]{x} - \sqrt[3]{2}}{\sqrt{x} - \sqrt{2}}$
Let $x=t^6$, then it becomes $\lim\frac{t^2 - (\sqrt[6]{2})^2}{t^3 - (\sqrt[6]{2})^3}$=$\lim\frac{(t+\sqrt[6]{2})(t-\sqrt[6]{2})}{(t-\sqrt[6]{2})(t^2+t\sqrt[6]{2}+(\sqrt[6]{2})^2}$=$\lim\frac{(t+\sqrt[6]{2})}{(t^2+t\sqrt[6]{2}+(\sqrt[6]{2})^2}$
But when $x\to 2$, t can go to $\sqrt[6]{2}$ or -$\sqrt[6]{2}$, which gives two different limits, where I was wrong? Thanks!
|
The common method is to complete:
$$\lim\limits_{x \to2} \frac{\sqrt[3]{x} - \sqrt[3]{2}}{\sqrt{x} - \sqrt{2}}\cdot \frac{\sqrt[3]{x^2}+\sqrt[3]{2x}+\sqrt[3]{2^2}}{\sqrt[3]{x^2}+\sqrt[3]{2x}+\sqrt[3]{2^2}}\cdot \frac{\sqrt{x} + \sqrt{2}}{\sqrt{x} + \sqrt{2}}=\\
\lim\limits_{x \to2} \frac{\sqrt{x} + \sqrt{2}}{\sqrt[3]{x^2}+\sqrt[3]{2x}+\sqrt[3]{2^2}}=\\
\frac{2\sqrt{2}}{3\sqrt[3]{4}}=\frac13\cdot 2^{5/6}.$$
|
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|
show $\int_{1}^{\infty} \frac{1}{\sqrt{x^4-x}} dx$ converges Show $\int_{1}^{\infty} \frac{1}{\sqrt{x^4-x}} dx$ converges.
I was able to show $\int_{2}^{\infty} \frac{1}{\sqrt{x^4-x}} dx$ converges, comparing it with the function $\frac{1}{x^{3/2}}$. I have trouble showing that $\int_{1}^{2} \frac{1}{\sqrt{x^4-x}} dx$ converges due to the fact that the function is not continuous at 1. Not sure how to do this, since I can't evaluate the integral directly.
|
The problem seeming to be at $x=1$, make the Taylor series
$$\frac{1}{\sqrt{x^{4} - x}} =\frac{1}{\sqrt{3} \sqrt{x-1}}-\frac{\sqrt{x-1}}{\sqrt{3}}+\frac{5 (x-1)^{3/2}}{6
\sqrt{3}}-\frac{2 (x-1)^{5/2}}{3 \sqrt{3}}+\frac{13 (x-1)^{7/2}}{24
\sqrt{3}}+O\left((x-1)^{9/2}\right)$$ Integrate termwise to get
$$\frac{2 \sqrt{x-1}}{\sqrt{3}}-\frac{2 (x-1)^{3/2}}{3 \sqrt{3}}+\frac{(x-1)^{5/2}}{3
\sqrt{3}}-\frac{4 (x-1)^{7/2}}{21 \sqrt{3}}+\frac{13 (x-1)^{9/2}}{108
\sqrt{3}}+O\left((x-1)^{11/2}\right)$$
Make $x=2$ and obtain
$$\frac {1207 \sqrt 3 } { 2268}\approx 0.9218$$ More terms you will add to the expansion and closer and closer you will approach to the true value which is $\approx 0.8969$.
|
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|
Computing the Fourier transform of a complex-valued function I have the following function: $e^{-(a+bi)|x|^2}$.
While trying to compute the fourier transform of the following function, I know that fourier transform of the real part remains the same and the result would be $\sqrt{\frac{\pi}{a}}e^{{-\xi^2}/4a}$. I'm not quite sure on how to proceed with the imaginary part.
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Completing the square in the exponent and the well known integral of $e^{-\pi x^2}$ are tools you can use:
$\begin{align}{F}\left\{e^{-(a+bi)|x|^2}\right\} &= {F}\left\{e^{-ax^2}e^{-ibx^2}\right\}\\
\\
&= \dfrac{1}{2\pi}{F} \left\{e^{-ax^2}\right\} * {F}\left\{e^{-ibx^2}\right\}\\
\\
&= \dfrac{1}{2\pi}\left( \sqrt{\frac{\pi}{a}}e^{{-\xi^2}/4a} \right)* \int_{-\infty}^{\infty}e^{-ibx^2}e^{-i\xi x} dx\\
\\
&= \dfrac{1}{2\pi}\left( \sqrt{\frac{\pi}{a}}e^{{-\xi^2}/4a} \right)* \int_{-\infty}^{\infty} e^{-ib\left(x^2+\frac{\xi}{b}x+\frac{\xi^2}{4b^2}\right)}e^{ib\frac{\xi^2}{4b^2}}dx\\
\\
&= \dfrac{1}{2\pi}\left( \sqrt{\frac{\pi}{a}}e^{{-\xi^2}/4a} \right)*\left(e^{{-\xi^2}/{4bi}}\int_{-\infty}^{\infty} e^{-ib\left(x+\frac{\xi}{2{b}}\right)^2}dx \right)\\
\\
&= \dfrac{1}{2\pi}\left( \sqrt{\frac{\pi}{a}}e^{{-\xi^2}/4a} \right)*\left(e^{{-\xi^2}/{4bi}}\int_{-\infty}^{\infty} e^{-\pi\left(\frac{1+i}{\sqrt{2}}\sqrt{\frac{b}{\pi}}x+\frac{1+i}{\sqrt{2}}\frac{\xi}{2\sqrt{\pi b}}\right)^2}dx \right)\\
\\
& \quad\quad\quad\quad u = \frac{1+i}{\sqrt{2}}\sqrt{\frac{b}{\pi}}x+\frac{1+i}{\sqrt{2}}\frac{\xi}{2\sqrt{\pi b}}\\
\\
& \quad\quad\quad\quad \frac{1-i}{\sqrt{2}}\sqrt{\frac{\pi}{b}}du = dx\\
\\
&= \dfrac{1}{2\pi}\left( \sqrt{\frac{\pi}{a}}e^{{-\xi^2}/4a} \right)*\left(\dfrac{1-i}{\sqrt{2}}\sqrt{\dfrac{\pi}{b}}e^{{-\xi^2}/{4bi}}\int_{-\infty+\frac{i}{\sqrt{2}}\frac{\xi}{2\sqrt{\pi b}}}^{\infty+\frac{i}{\sqrt{2}}\frac{\xi}{2\sqrt{\pi b}}} e^{-\pi u^2}du \right)\\
\\
&= \dfrac{1}{2\pi}\left( \sqrt{\frac{\pi}{a}}e^{{-\xi^2}/4a} \right)*\left(\dfrac{1-i}{\sqrt{2}}\sqrt{\dfrac{\pi}{b}}e^{{-\xi^2}/{4bi}}\right)\\
\\
&= \dfrac{1}{2\pi}\left( \frac{\sqrt{\pi}}{\sqrt{a}}e^{{-\xi^2}/4a} \right)*\left(\dfrac{\sqrt{\pi}}{\sqrt{b}e^{i\frac{\pi}{4}}}e^{{-\xi^2}/{4bi}}\right)\\
\\
&= \left[\dfrac{1}{2\sqrt{ab}e^{i\frac{\pi}{4}}}\right]\left[\left( e^{{-\xi^2}/4a} \right)*\left(e^{{-\xi^2}/{4bi}}\right)\right]\\
\end{align}$
where $*$ indicates convolution. I'll also note that $\left(e^{i\frac{\pi}{4}}\right)^2 = i$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find $f^{(80)}(27)$ where $f(x)=(x+3)^{\frac{1}{3}}\cdot (x-27)$
Suppose that $f(x)=(x+3)^{\frac{1}{3}}\cdot (x-27)$. Use a Taylor series expansion to find $f^{(80)}(27)$.
I tried the following:
\begin{align}
f'(x) &= (x+3)^{\frac{1}{3}}\cdot 1+(x-27)\cdot \frac{1}{3}(x+3)^{\frac{-2}{3}}\\
%
f''(x)
&= \frac{1}{3}(x+3)^{-\frac{2}{3}}+\frac{1}{3}(x-27)\cdot -\frac{2}{3}(x+3)^{-\frac{5}{3}}+\frac{1}{3}(x+3)^{-\frac{2}{3}} \\
&= \frac{2}{3}(x+3)^{-\frac{2}{3}}-\frac{1\cdot 2}{3^2}(x+3)^{-\frac{5}{3}}(x-27)\\
%
f'''(x) &= -\frac{2^2}{3^3}(x+3)^{-\frac{5}{3}}-\frac{1\cdot 2}{3^2}(x+3)^{-\frac{5}{3}}-\frac{1\cdot 2\cdot 5}{3^3}(x+3)^{-\frac{8}{3}}
\end{align}
This is very nasty, please help me to solve in some easy way.
|
Let $g(x)=f(x-3)$. Then $g(x)=x^{\frac13}(x+30)$
The Taylor Series of $x^\frac13$ is
$$\begin{split}x^{\frac13}&=a^{\frac13}+\sum^\infty_{n=1}\frac{a^{\frac13-n}\prod^n_{m=1}\big(\frac43-n\big)}{n!}(x-a)^n\\&=a^\frac13+\frac13a^{-\frac23}(x-a)+...+\frac{a^{-\frac{236}3}\prod^{79}_{n=1}\big(\frac43-n\big)}{79!}(x-a)^{79}+ \frac{a^{-\frac{239}3}\prod^{80}_{n=1}\big(\frac43-n\big)}{80!}(x-a)^{80}+...\end{split}$$
So, we put $a=30$ the degree $80$ term of the Taylor series of $g(x)$ is
$$\begin{split}\frac{(x-30)^{80}}{80!}g^{(80)}(30)&=\deg_{80}\big(x^{\frac13}(x-30+60)\big)\\&
=\frac1{80!}\Bigg(79{\bigg(30^{-\frac{236}3}\bigg)\prod^{79}_{n=1}\bigg(\frac4{3}-n\bigg)}(x-30)^{80}+60\bigg({30^{-\frac{239}3}\bigg)\prod^{80}_{n=1}\bigg(\frac4{3}-n\bigg)}(x-30)^{80}\Bigg)
\\&=\frac1{80!}\Bigg(79{\bigg(30^{-\frac{236}3}\bigg)\prod^{79}_{n=1}\bigg(\frac4{3}-n\bigg)}(x-30)^{80}-\frac{472}3\bigg({30^{-\frac{236}3}\bigg)\prod^{79}_{n=1}\bigg(\frac4{3}-n\bigg)}(x-30)^{80}\Bigg)
\\&=\frac{(x-30)^{80}}{80!}\Bigg(-\frac{235}3\bigg({30^{-\frac{236}3}\bigg)\prod^{79}_{n=1}\bigg(\frac4{3}-n\bigg)}\Bigg)\end{split}$$
So,
$$f^{(80)}(27)=g^{(80)}(30)=-\frac{235}3\bigg({30^{-\frac{236}3}\bigg)\prod^{79}_{n=1}\bigg(\frac4{3}-n\bigg)}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding $\lim_{n\to\infty}\prod_{n=1}^{\infty}\left(1-\frac{1}{n(n+1)}\right)$ I have a trouble with this limit of the infinite product:
$$\lim _{n \to\infty}\left(1-\frac{1}{1 \cdot 2}\right)\left(1-\frac{1}{2 \cdot 3}\right) \cdots\left(1-\frac{1}{n(n+1)}\right)$$
My attempt:
We have
$$\prod_{n=1}^{\infty}\left(1-\frac{1}{n(n+1)}\right)=\prod_{n=1}^{\infty}\left(\frac{n^{2}+n-1}{n^{2}+n}\right)=\prod_{n=1}^{\infty} \frac{\left(n-a_{1}\right)\left(n-a_{2}\right)}{n \left(n+1\right)},$$
where $a_{1}=\dfrac{-1+\sqrt{5}}{2}$, $a_{2}=\dfrac{-1-\sqrt{5}}{2}$.
So I would just like a hint as to how to proceed. Any help would be appreciated.
|
I continue from where you left. The infinite product representation of the sine function can be used to finish the calculation:
\begin{align*}
& \mathop {\lim }\limits_{N \to + \infty } \prod\limits_{n = 1}^N {\left( {1 - \frac{{a_1 }}{n}} \right)\left( {1 + \frac{{ - 1 - a_2 }}{{n + 1}}} \right)} = \mathop {\lim }\limits_{N \to + \infty } \prod\limits_{n = 1}^N {\left( {1 - \frac{{a_1 }}{n}} \right)\left( {1 + \frac{{a_1 }}{{n + 1}}} \right)}
\\ & = \mathop {\lim }\limits_{N \to + \infty } \frac{{1 + \frac{{a_1 }}{{N + 1}}}}{{1 + a_1 }}\prod\limits_{n = 1}^N {\left( {1 - \frac{{a_1 }}{n}} \right)\left( {1 + \frac{{a_1 }}{n}} \right)} = \frac{1}{{1 + a_1 }}\prod\limits_{n = 1}^\infty {\bigg( 1 - \frac{{a_1^2 }}{{n^2 }} \bigg)} \\ & = \frac{1}{{1 + a_1 }}\frac{{\sin (\pi a_1 )}}{{\pi a_1 }} = \frac{{\sin (\pi a_1 )}}{{\pi }}.
\end{align*}
|
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|
What is $\frac{a b \sin x}{2\sqrt{(a^2 + b^2 + 2 ab \cos x ) \cdot (a+b - \sqrt{a^2+b^2+ 2 ab \cos x})}}$ for $x \rightarrow 0$? I am trying to find the limit
$\lim_{x \rightarrow0} \frac{a b \sin x}{2\sqrt{(a^2 + b^2 + 2 ab \cos x ) \cdot (a+b - \sqrt{a^2+b^2+ 2 ab \cos x})}}$
for $a,b \in \rm{I\!R}_{+}$. Applying L'Hospital's rule leads to
$\lim_{x \rightarrow0}\frac{\cos x \cdot \sqrt{(a^2 + b^2 + 2 ab \cos x ) \cdot (a+b - \sqrt{a^2+b^2+ 2 ab \cos x})}}{-2 \sin x \cdot (a+b - \sqrt{a^2+b^2+ 2 ab \cos x}) + \frac{\sin x \cdot (a^2 + b^2 + 2 ab \cos x )}{\sqrt{a^2+b^2+ 2 ab \cos x}} }$.
However, this remains with both, cosine and sine. Maybe one could use a trigonometric identity, which I cannot find.
|
Fix $a,b > 0$, and let
$$
f(x)
=
\frac{ab\,\sin x}{2\sqrt{(a^2+b^2+2ab\cos x)(a+b-\sqrt{a^2+b^2+2ab\cos x})}}
$$
As $x$ approaches zero from the right, we have $f(x) > 0$, so the limit from the right, if it exists, $L$ say, must be nonnegative, in which case, since $f$ is an odd function, the limit from the left will be equal to $-L$.
Looking ahead, we'll find that $L > 0$, hence the two-sided limit doesn't exist.
Let's find $L$ . . .
Computing $L^2$, we get
\begin{align*}
L^2&=\lim_{x\to 0} f(x)^2
\\[4pt]
&=\lim_{x\to 0}\, \frac{(ab\,\sin x)^2}{4(a^2+b^2+2ab\cos x)(a+b-\sqrt{a^2+b^2+2ab\cos x})}
\\[4pt]
&=\frac{(ab)^2}{4(a+b)^2}\lim_{x\to 0}\, \frac{\sin^2 x}{a+b-\sqrt{a^2+b^2+2ab\cos x}}
\\[4pt]
&=\frac{(ab)^2}{4(a+b)^2}\lim_{x\to 0}\, \frac{\left(x\left(\!{\Large{\frac{\sin x}{x}}}\!\right)\right)^2}{a+b-\sqrt{a^2+b^2+2ab\cos x}}
\\[4pt]
&=\frac{(ab)^2}{4(a+b)^2}\lim_{x\to 0}\, \frac{x^2}{a+b-\sqrt{a^2+b^2+2ab\cos x}}
\\[4pt]
&=\frac{(ab)^2}{4(a+b)^2}\lim_{x\to 0}\, \frac{x^2}{a+b-\sqrt{a^2+b^2+2ab\cos x}}
\cdot \frac{a+b+\sqrt{a^2+b^2+2ab\cos x}}{a+b+\sqrt{a^2+b^2+2ab\cos x}}
\\[4pt]
&=\frac{(ab)^2}{4(a+b)^2}\lim_{x\to 0}\, \frac{x^q}{a+b-\sqrt{a^2+b^2+2ab\cos x}}
\cdot \frac{2(a+b)}{a+b+\sqrt{a^2+b^2+2ab\cos x}}
\\[4pt]
&=\frac{(ab)^2}{2(a+b)}\lim_{x\to 0}\, \frac{x^2}{2ab-2ab\cos x}
\\[4pt]
&=\frac{ab}{4(a+b)}\lim_{x\to 0}\, \frac{x^2}{1-\cos x}
\\[4pt]
&=\frac{ab}{4(a+b)}\lim_{x\to 0}\, \frac{2x}{\sin x}
\\[4pt]
&=\frac{ab}{4(a+b)}\lim_{x\to 0}\, 2
\\[4pt]
&=\frac{ab}{2(a+b)}
\\[8pt]
\text{hence}\;L&=\sqrt{\frac{ab}{2(a+b)}}
\\[4pt]
\end{align*}
|
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|
Technique for solving $\frac{ax+b}{cx+d}=\frac{px+q}{rx+s}$ where the sum of numerators equals the sum of denominators I was looking up some shortcuts to solve quadratic equations. I got a technique that applies when the sum of the numerators and denominators are equal, but I am unable to understand the reasoning behind it. Here I'm showing an example:
$$ \frac{3x + 4}{6x + 7} = \frac{5x + 6}{2x + 3} $$
The solution goes as follows:
"Minute observation of the question helps us to identify that this question falls in a special category of quadratic equations, where the sum of the numerators (N) and the sum of the denominators (D) are found to be equal to 8x + 10.''
For the first root,
$ N_1 + N_2 = D_1 + D_2 = 0$
or, $ 8x + 10 = 0 $
or, $ x = -5/4 $
For the second root
$ N_1 - D_1 = N_2 - D_2 = 0 $
or, $ 3x + 3 = 0 $
or, $ x = -1 $
Can someone explain the reasoning/proof behind this?
|
Componendo and dividendo (Brilliant) is another method.
Using the third rule with $k=1$, we have:
$$\frac{3x+4+(6x+7)}{3x+4-(6x+7)} = \frac{5x+6+(2x+3)}{5x+6-(2x+3)}$$
$$\Rightarrow \frac{9x+11}{-3x-3} = \frac{7x+9}{3x-3}$$
$$\Rightarrow -9x-11 = 7x+9$$
$$\Rightarrow x = -\frac{5}{4}$$
which is true in general, when we have:
$$\frac{N_1 + D_1}{N_1 - D_1} = \frac{N_2 + D_2}{N_2 - D_2}$$
and $N_1 - D_1 = N_1 + N_2 - D_1 - N_2 = D_1 + D_2 - D_1 - N_2 = D_2 - N_2$, so $N_1 - D_2 = -(N_2 - D_2)$.
Using the fourth rule with $k=1$, we have:
$$\frac{3x+4+(5x+6)}{3x+4+(2x+3)} = \frac{5x+6+(3x+4)}{5x+6+(6x+7)}$$
$$\Rightarrow \frac{8x+10}{5x+5} = \frac{8x+10}{11x+13}$$
$$\Rightarrow 11x+13=5x+5$$
$$\Rightarrow x = -1$$
This can be proven by cross multiplying as well.
|
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|
Prove that if $\sqrt{a}+\sqrt{b}+\sqrt{c}=3$ then $ \frac{a^{2}}{a+2 b^{2}}+\frac{b^{2}}{b+2 c^{2}}+\frac{c^{2}}{c+2 a^{2}} \geq 1 $ Question -
Let $a, b, c$ be positive real numbers such that $\sqrt{a}+\sqrt{b}+\sqrt{c}=3$ . Prove that
$$
\frac{a^{2}}{a+2 b^{2}}+\frac{b^{2}}{b+2 c^{2}}+\frac{c^{2}}{c+2 a^{2}} \geq 1
$$
My try
$$
\frac{a^{2}}{a+2 b^{2}}=a-\frac{2 a b^{2}}{a+2 b^{2}} \geq a-\frac{2 a b^{2}}{3 \sqrt[3]{a b^{4}}}=a-\frac{2(a b)^{2 / 3}}{3}
$$
which implies that
$$
\sum_{c y c} \frac{a^{2}}{a+2 b^{2}} \geq \sum_{c y c} a-\frac{2}{3} \sum_{c y c}(a b)^{\frac{2}{3}}
$$
It suffices to prove that
$$
(a b)^{2 / 3}+( b c)^{2 / 3}+\left (c a)^{2 / 3} \leq 3\right.
$$
beacuse we can easily get that $\sum a \ge 3$
but i am not able to prove it..
note that we have to prove this using only am-gm or weighted am-gm or power mean or any kind of means inequality because author did not introduce any advance inequality yet...
any hints ???
thankyou
|
Elaborating on the above solution:
We first wish to prove that the last inequality holds. Schur's inequality in its simplest form states that: For non-negative real numbers $x,y,z$, we have $x^3+y^3+z^3+3xyz \geq xy(x+y)+xz(x+z)+yz(y+z)$.
Now, expanding the R.H.S., we obtain: $(X+Y+Z)^3+9XYZ$
$=X^3+3X^2Y+3X^2Z+3XY^2+6XYZ+3XZ^2+Y^3+3Y^2Z+3YZ^2+Z^3+9XYZ$
$=X^3+Y^3+Z^3+3XYZ+12XYZ+3X^2Y+3X^2Z+3XY^2+3XZ^2+3Y^2Z+3YZ^2$
$\geq XY(X+Y) +XZ(X+Z) + YZ(Y+Z)+12XYZ+3X^2Y+3X^2Z+3XY^2+3XZ^2+3Y^2Z+3YZ^2$,
$=X^2Y+XY^2+X^2Z+XZ^2+Y^2Z+YZ^2+12XYZ+3X^2Y+3X^2Z+3XY^2+3XZ^2+3Y^2Z+3YZ^2$
$=4X^2Y+4XY^2+4X^2Z+4XZ^2+4Y^2Z+4YZ^2+12XYZ$
$=4(X^2Y+XYZ+XY^2+X^2Z+XYZ+XZ^2+Y^2Z+XYZ+YZ^2)$
$=4[XY(X+Y+Z)+XZ(X+Y+Z)+YZ(X+Y+Z)]$
$=4(X+Y+Z)(XY+YZ+ZX)$
Simply flipping the inequality sign gives us: $4(X+Y+Z)(XY+YZ+ZX) \leq (X+Y+Z)^3+9XYZ$, so we are done.
Finally, we observe that this inequality implies the second last inequality. With $X+Y+Z=3$, we have:
$4(X+Y+Z)(XY+YZ+ZX) \leq (X+Y+Z)^3+9XYZ$
$\Rightarrow 4(XY+YZ+ZX) \leq \ (X+Y+Z)^2+3XYZ$
$\Rightarrow 4(XY+YZ+ZX) -3XYZ \leq \ (X+Y+Z)^2 = 3^2 =9$
(Q.E.D.)
|
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|
Dot product with a scalar function Example 5: Find $\nabla\cdot\left(\frac{\boldsymbol r}{r^2}\right)$ where $\boldsymbol r(x,y,z)=x\boldsymbol i+y\boldsymbol j+z\boldsymbol k$, $r=\sqrt{x^2+y^2+z^2}$.
In this case, why can we not bring the scalar function $r^{-2}$ outside the dot product?
When I leave the function in place, I calculate a result of $\frac{1}{r^{2}}$. When I pull it out, I find $\frac{3}{r^{2}}$.
|
$$\nabla \cdot \left( \frac{\vec{r}}{r^2} \right) = \nabla \cdot \left( \frac{x}{x^2+y^2+z^2}\vec{i} + \frac{y}{x^2+y^2+z^2}\vec{j} + \frac{z}{x^2+y^2+z^2}\vec{k} \right) \\ = \frac{y^2+z^2-x^2}{(x^2+y^2+z^2)^2} + \frac{x^2+z^2-y^2}{(x^2+y^2+z^2)^2} + \frac{x^2+y^2-z^2}{(x^2+y^2+z^2)^2} = \frac{x^2+y^2+z^2}{(x^2+y^2+z^2)^2} \\ = \frac{1}{x^2+y^2+z^2} = \frac{1}{r^2}.$$
|
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|
How to find Laurent-series? I am trying to find the Laurent series for the function $\frac{1}{ z (2i - z)}$.
I already obtained for...
(1) ... $2 < | z |$:
$\frac{1}{ z (2i - z)} = \frac{1}{2 i} \left( \frac{1}{z} + \frac{1}{2i - z} \right)
= \frac{1}{2 i} \left( \frac{1}{z} - \frac{1}{z} \cdot \frac{1}{1 - \frac{2i}{z} } \right)
= \frac{1}{2 i} \left[ \frac{1}{z} - \frac{1}{z} \cdot \sum_{n \geq 0} \left( \frac{2i}{z} \right)^n \right] \\
= \frac{1}{2 i} \left[ \frac{1}{z} - \sum_{n \geq 0} \frac{(2i)^n}{z^{n+1}} \right]
= \frac{1}{2 i} \left[ \frac{1}{z} - \sum_{n \geq 0} \frac{(2i)^n}{z^{n+1}} \right]$
(2) ... $0 < | z | < 2$:
$\frac{1}{ z (2i - z)} = \frac{1}{2 i} \left( \frac{1}{z} + \frac{1}{2i - z} \right)
= \frac{1}{2 i} \left( \frac{1}{z} + \frac{1}{2i} \cdot \frac{1}{1 - \frac{z}{2i} } \right)
= \frac{1}{2 i} \left( \frac{1}{z} + \frac{1}{2i} \cdot \sum_{n \geq 0} \left( \frac{z}{2i} \right)^n \right)
= \frac{1}{2 i} \left( \frac{1}{z} + \sum_{n \geq 0} \ \frac{z^n}{(2i)^{n+1}} \right)$
I am confused about the complex number $i$ in the series ... are my calculations correct?
|
Don't worry about the $i$, just treat it as you would any other complex constant. Both series you wrote look correct, but generally it is more clear to write them in the following forms.
For $|z|>2$ we can write:
$$
\frac{1}{z(2i-z)}=-\frac{1}{z^2}\left(\frac{1}{1-\frac{2i}{z}}\right)=-\frac{1}{z^2}\sum_{n=0}^{\infty}\frac{(2i)^n}{z^n}\\=\sum_{n=0}^{\infty} \frac{-(2i)^n}{z^{n+2}}
=\sum_{n=2}^{\infty} \frac{-(2i)^{n-2}}{z^n}
$$
And in the annulus $0<|z|<2$ we can similarly write:
$$
\frac{1}{z(2i-z)}=\frac{1}{2iz}\left(\frac{1}{1-\frac{z}{2i}}\right)=\frac{1}{2i}\frac{1}{z}\sum_{n=0}^{\infty} \frac{z^n}{(2i)^n} \\
=\sum_{n=0}^{\infty} \frac{z^{n-1}}{(2i)^{n+1}} =\frac{1}{2i}\frac{1}{z}+\sum_{n=0}^{\infty} \frac{z^{n}}{(2i)^{n+2}}
$$
In particular, we can see the residue at $z=0$ is $\frac{1}{2i}$.
|
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|
Solve $(y-u)u_x + (u-x)u_y = x-y$
Solve
$$(y-u)u_x + (u-x)u_y = x-y, \qquad u=0 \text{ when } xy=1.$$
I tried to solve the equation above using characteristic method
$$\begin{cases}
x'=y-u \\
y' = u - x \\
u' = x-y
\end{cases}. \tag{1}$$
Adding together the first and the second equation in $(1)$ we get that $x'+y'=y-x$. Thus (using third equation in $(1)$)
$$u = x+ y + C \implies C = u - x- y.$$
Solving the first and the second equation yields to
$$\frac{1}{2} y^2 - uy = ux - \frac{1}{2}x^2 + \tilde{C} \implies \tilde{C} = \frac{1}{2}(y^2 +x^2)-u(x+y).$$
That gives us
$$F\big(u-x-y, \frac{1}{2}(y^2 +x^2)-u(x+y) \big) = 0.$$
How can I finish this example?
|
$$(y-u)u_x+(u-x)u_y=x-y$$
System of characteristic ODEs (Charpit-Lagrange) :
$$\frac{dx}{y-u}=\frac{dy}{u-x}=\frac{du}{x-y}$$
A first characteristic equation comes from :
$\frac{dx}{y-u}=\frac{dy}{u-x}=\frac{du}{x-y}=\frac{dx+dy+du}{(y-u)+(u-x)+(x-y)}=\frac{dx+dy+du}{0} \quad\implies\quad dx+dy+du=0$
$$u+x+y=c_1$$
A second characteristic equation comes from :
$\frac{dx}{y-u}=\frac{dy}{u-x}=\frac{du}{x-y}=\frac{xdx+ydy+udu}{x(y-u)+y(u-x)+u(x-y)}=\frac{xdx+ydy+udu}{0} \quad\implies\quad xdx+ydy+udu=0$
$$x^2+y^2+u^2=c_2$$
General solution of the PDE expressed on the form of implicit equation $\Phi(c_1,c_2)=0$ or equivalently $c_2=F(c_1)$ :
$$x^2+y^2+u^2=F(x+y+u)$$
$\Phi$ or $F$ are undetermined functions until no condition is specified.
With condition $u=0$ when $xy=1$ :
Then $x^2+y^2+0=F(x+y+0)=x^2+\frac{1}{x^2}=F(x+\frac{1}{x})$
Let $X=x+\frac{1}{x} \quad\implies\quad x=\frac12\left(X\pm\sqrt{X^2-4}\right)$
$F(X)=\frac14\left(X\pm\sqrt{X^2-4}\right)^2+\frac{4}{\left(X\pm\sqrt{X^2-4}\right)^2}$ and after simplification :
$$F(X)=X^2-2$$
Now $F(X)$ is determined. We put it into the above general solution where $X=x+y+u$
Then $F(x+y+u)=(x+y+u)^2-2$ .
$$x^2+y^2+u^2=(x+y+u)^2-2$$
Solving for $u$ leads to :
$$\boxed{u(x,y)=\frac{1-xy}{x+y}}$$
NOTE:
Another method consists in the change of variables
$\zeta=x+y$ ; $\eta=xy$ which leads to a separable PDE.
|
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|
Prove that for all positives $a, b$ and $c$, $(\sum_{cyc}\frac{c + a}{b})^2 \ge 4(\sum_{cyc}ca)(\sum_{cyc}\frac{1}{b^2})$.
Prove that for all positives $a, b$ and $c$, $$\left(\frac{b + c}{a} + \frac{c + a}{b} + \frac{a + b}{c}\right)^2 \ge 4(bc + ca + ab) \cdot \left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\right)$$
Let $ca + ab = m$, $ab + bc = n$ and $bc + ca = p$, we have that $$\left(\frac{m}{a^2} + \frac{n}{b^2} + \frac{p}{c^2}\right)^2 \ge 2(m + n + p) \cdot \left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\right)$$
$$\iff \left(\frac{m}{a^2} + \frac{n}{b^2} + \frac{p}{c^2} - 1\right)^2 \ge 2 \cdot \sum_{cyc((m, n, p), (a, b, c))}\left[n \cdot \left(\frac{1}{c^2} + \frac{1}{a^2}\right)\right] + 1$$
Expanding $\displaystyle \sum_{cyc((m, n, p), (a, b, c))}\left[n \cdot \left(\frac{1}{c^2} + \frac{1}{a^2}\right)\right]$ gives $$2 \cdot \sum_{cyc}\frac{ca}{b^2} + \left(\frac{b + c}{a} + \frac{c + a}{b} + \frac{a + b}{c}\right)$$
Let $\dfrac{b + c}{a} = m'$, $\dfrac{c + a}{b} = n'$ and $\dfrac{a + b}{c} = p'$, we have that $$(m' + n' + p' - 1)^2 \ge 2 \cdot \left[2 \cdot \sum_{cyc}\frac{ca}{b^2} + (m' + n' + p')\right] + 1$$
Moreover, $$(m')^2 + (n')^2 + (p')^2 = \sum_{cyc}\left[\left(\frac{c + a}{b}\right)^2\right] \ge 2 \cdot \sum_{cyc}\frac{ca}{b^2}$$
$$\implies (m' + n' + p' - 1)^2 \ge 2 \cdot \left[(m')^2 + (n')^2 + (p')^2 + m' + n' + p'\right] + 1$$
$$\iff -[(m')^2 + (n')^2 + (p')^2] + 2(m'n' + n'p' + p'm') - 4(m' + n' + p') \ge 0$$, which is definitely not correct.
Another attempt, let $(0 <) \ a \le b \le c \implies ab \le ca \le bc \iff ca + ab \le ab + bc \le bc + ca$
$\iff m \le n \le p$ and $a^2 \le b^2 \le c^2 \iff \dfrac{1}{a^2} \ge \dfrac{1}{b^2} \ge \dfrac{1}{c^2}$.
By the Chebyshev inequality, we have that $$3 \cdot \left(\frac{m}{a^2} + \frac{n}{b^2} + \frac{p}{c^2}\right) \le (m + n + p) \cdot \left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\right)$$
Any help would be appreciated.
|
We use the standard pqr method.
Let $p = a + b + c, q = ab+bc+ca, r = abc$. The inequality becomes
$$\left(\frac{pq}{r} - 3\right)^2 \ge 4q \left(\left(\frac{q}{r}\right)^2 - 2\frac{p}{r}\right)$$
or (after clearing the denominators)
$$p^2q^2 + 2pqr - 4q^3 + 9r^2 \ge 0.$$
We split into two cases:
1) $p^2 \ge 4q$: Since $p^2q^2 \ge 4q^3$, the inequality is true.
2) $p^2 < 4q$: By Schur's inequality $a^2(a-b)(a-c) + b^2(b-c)(b-a) + c^2(c-a)(c-b) \ge 0$
which is written as $6pr - (4q-p^2)(p^2-q) \ge 0$, we have $r \ge \frac{(4q-p^2)(p^2-q)}{6p}$.
As $p^2 \ge 3q$, we have $r \ge \frac{(4q-p^2)(p^2-q)}{6p}\ge 0$.
Thus, we have
\begin{align}
&p^2q^2 + 2pqr - 4q^3 + 9r^2\\
\ge\ & p^2q^2 + 2pq \cdot \frac{(4q-p^2)(p^2-q)}{6p} - 4q^3 + 9\left(\frac{(4q-p^2)(p^2-q)}{6p}\right)^2\\
= \ & \frac{(p^2-3q)(3p^2-q)(p^2-4q)^2}{12p^2}\\
\ge \ & 0.
\end{align}
We are done.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Let $x, y \in \mathbb R$ such that $x^2+y^2=2x-2y+2$. Find the largest possible value of $x^2+y^2$
Let $x, y \in \mathbb R$ s. t . $x^2+y^2=2x-2y+2$. Find the largest possible value of $x^2+y^2-\sqrt{32}$
I know this is a duplicate of another question, but that question has solutions involving calculus and geometry, while I want a solution relying on algebra and basic inequalities only to solve this problem.
|
Let $ z=x+iy$. Then, we have $ |z-(1+i)| =2$ from $x^2+y^2=2x-2y+2$, and
$$|z|= |z-(1+i)+(1+i )| \le | z-(1+i)|+ |1+i |= 2+ \sqrt 2=|z|_{max}$$
Thus, the largest possible value is
$$x^2+y^2= |z|_{max}^2 = (2+\sqrt 2)^2=6+4\sqrt2$$
|
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|
Prove using Jensen's inequality that if $abcd=1$ then $\frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}+\frac{1}{(1+c)^{2}}+\frac{1}{(1+d)^{2}} \geq 1$ Question -
Let $a, b, c, d$ be positive real numbers such that abcd $=1 .$ Prove that
$$
\begin{array}{c}
\frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}+\frac{1}{(1+c)^{2}}+\frac{1}{(1+d)^{2}} \geq 1 \\
\text { (China TST 2004) }
\end{array}
$$
I solved it using CS inequality, and also seen a solution using the fact that $\frac{1}{(1+x)^{2}}+\frac{1}{(1+y)^{2}} \geq \frac{1}{1+x y}$
but, i think the inequality is very much likely to be also solved using jensens but i am not able to find proof, i tried many functions such as $x^2 , 1/x^2 , 1/(x+1)^2$
but none of them working...
can anyone solve it using Jensen inequality ???
|
We can use the Vasc's RCF Theorem. It's like Jensen, but it's not Jensen.
Also, since $f(x)=\frac{1}{(1+e^x)^2}$ has an unique inflection point, we can use Jensen with Karamata, but it's not so nice solution. I am ready to show, if you want.
Indeed, $$f''(x)=\frac{4e^x\left(e^x-\frac{1}{2}\right)}{(1+e^x)^4}.$$
Thus, $f$ is a convex function on $[-\ln2,+\infty)$ and a concave function on $(-\infty,-\ln2]$.
We need to prove that $$\sum_{cyc}f(x)\geq0,$$ where $x+y+z+t=0$.
Now, let $x\geq y\geq z\geq t.$
We'll consider the following cases.
*
*$x\geq y\geq z\geq t\geq-\ln2.$
Thus, by Jensen $$\sum_{cyc}f(x)\geq4f\left(\frac{x+y+z+t}{4}\right)=4f(0)=1.$$
*$x\geq y\geq z\geq-\ln2\geq t$.
Thus, by Jensen again:
$$\sum_{cyc}f(x)\geq3f\left(\frac{x+y+z}{3}\right)+f(t)=3f\left(\frac{-t}{3}\right)+f(t).$$
Thus, it's enough to prove that
$$3f\left(\frac{-t}{3}\right)+f(t)\geq0,$$ which is $$\sum_{cyc}\frac{1}{(1+a)^2}\geq1,$$ where $b=c=a$ and $d=\frac{1}{a^3}$ or
$$\frac{3}{(1+a)^2}+\frac{1}{\left(1+\frac{1}{a^3}\right)^2}\geq1$$ or
$$(a-1)^2(3a^2-2a+2)\geq0,$$ which is obvious.
*$x\geq y\geq-\ln2\geq z\geq t$.
Thus, by Jensen again we have:
$$f(x)+f(y)\geq2f\left(\frac{x+y}{2}\right)^2=\frac{2}{\left(1+e^{\frac{x+y}{2}}\right)^2}=\frac{2}{(1+\sqrt{ab})^2}.$$
Also, since $$(-\ln2,\ln2+z+t)\succ(z,t),$$ by Karamata we obtain:
$$f(z)+f(t)\geq f(-\ln2)+f(\ln2+z+t)=$$
$$=\frac{1}{\left(1+e^{-\ln2}\right)^2}+\frac{1}{\left(1+e^{\ln+z+t}\right)^2}=\frac{4}{9}+\frac{1}{(1+2cd)^2}.$$
Let $\sqrt{ab}=u$.
Thus, it's enough to prove in this case that
$$\frac{4}{9}+\frac{1}{\left(1+\frac{2}{u^2}\right)^2}+\frac{2}{(1+u)^2}\geq1$$ and since $$\left(\frac{1}{\left(1+\frac{2}{u^2}\right)^2}+\frac{2}{(1+u)^2}\right)'=\frac{4(u^3-2)(u^3+6u^2+4)}{(u+1)^3(u^2+2)^3},$$ it's enough to prove the last inequality for $u=\sqrt[3]2,$ which gives
$$\frac{4}{9}+\frac{1}{\left(1+\frac{2}{u^2}\right)^2}+\frac{2}{(1+u)^2}=\frac{4}{9}+\frac{1}{(1+\sqrt[3]2)^2}+\frac{2}{(1+\sqrt[3]2)^2}>\frac{4}{9}+\frac{3}{(1+1.3)^2}>1.$$
*$x\geq-\ln2\geq y\geq z\geq t$.
Thus, since $$\left(-\ln2,-\ln2,2\ln2+y+z+t\right)\succ(y,z,t),$$ by Karamata again we obtain:
$$f(y)+f(z)+f(t)\geq2f(-\ln2)+f(2\ln2+y+z+t)=$$
$$=\frac{8}{9}+\frac{1}{\left(1+e^{2\ln2+y+z+t}\right)^2}=\frac{8}{9}+\frac{1}{(1+4bcd)^2}.$$
Id est, it's enough to prove that:
$$\frac{1}{(1+a)^2}+\frac{8}{9}+\frac{1}{\left(1+\frac{4}{a}\right)^2}\geq1$$ or
$$8a^4+8a^3-15a^2+32a+128\geq0,$$ which is obvious.
The case $-\ln2\geq x\geq y\geq z\geq t$ is impossible and we are done!
|
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|
Show $|a\sqrt{b}-\sqrt{c}|$ equal to zero or larger than $\frac{1}{2}10^{-3}$ when $a, b$ and $c$ are natural numbers strictly less than 100 I need to show that $|a\sqrt{b}-\sqrt{c}|$ is equal to zero or larger than $\frac{1}{2}10^{-3}$ when $a, b$ and $c$ are natural numbers strictly less than 100.
I see why it can be equal to zero. It is the other part that is causing my trouble. I tried to consider $(a\sqrt{b}-\sqrt{c})(a\sqrt{b}+\sqrt{c})=a^2b-c$, but I really don't know if it helps. Any hints would be appreciated.
Thanks!
|
If $a\sqrt{b} - \sqrt{c} \neq 0$, then $|a^2b - c| \geq 1$. We see that:
$$
a\sqrt{b} + \sqrt{c} < 100(10) + 10 = 1010
$$
Therefore:
$$
|a\sqrt{b} - \sqrt{c}| = \frac{|a^2b - c|}{a\sqrt{b} + \sqrt{c}} > \frac{1}{1010} > \frac{1}{2000}
$$
|
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|
$f(t) = \min\{1,t\}$ not operator monotone I want to show that the function on $\mathbb{R}^+$ $f(t) = \min \{1,t\}$ is not operator monotone on the complex $2\times 2$ matrices. My plan is to find matrices $A$ and $B$, $B\geq A$, such that the spectrum of $A$ is in $[0,\infty]$ and the spectrum of $B$ contains elements greater than 1. Then applying $f$ should 'make $B$ smaller' such that $f(B) \not \geq f(A)$. However, I struggle to find a concrete example. Can someone help me finding such $A$ and $B$?
|
You are on the right track. Actually any pair of matrices such that
*
*$A$ has spectrum on $[0, 1]$,
*$B$ has an eigenvalue on $(1, \infty)$, and
*$B - A$ is of rank $1$
*$A$ and $B$ do not commute
will work as an example.
Indeed, $f(t) \leq t$ together with 2. imply that $B \neq f(B) \leq B$. Since also $f(A) = A$,
$$
f(B) - f(A) = f(B) - A \leq B - A.
$$
Since the RHS is of rank $1$, and the second inequality is not equality, the only way the LHS can be positive is that $f(B) - f(A) = c(B - A)$ for some $0 \leq c < 1$, or
$$
A = \frac{1}{1 - c} \left(f(B) - c B\right).
$$
But this implies that $A$ commutes with $B$, contradiction.
As a concrete example one may for instance take
$$
A = \begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix},\;\;
B = A + \begin{bmatrix}
\frac12 & \frac12 \\
\frac12 & \frac12
\end{bmatrix}
= \begin{bmatrix}
\frac32 & \frac12 \\
\frac12 & \frac12
\end{bmatrix}.
$$
Again $f(A) = A$. One may calculate $f(B)$ using the definitions directly but the task might be slightly easier using the following observation: if $f$ and $g$ agree on the spectrum of $B$, then $f(B) = g(B)$. One may check that the eigenvalues of $B$ are $1 \pm \frac{1}{\sqrt{2}}$, so it suffices to find a polynomial $p$ with
$$
p\left(1 - \frac{1}{\sqrt{2}}\right) = 1 - \frac{1}{\sqrt{2}}\; \text{ and } p\left(1 + \frac{1}{\sqrt{2}}\right) = 1;
$$
then $f(B) = p(B)$. Degree $1$ example $p(x) = \frac{1}{2} x + \frac{1}{2} - \frac{1}{2 \sqrt{2}}$ will do, and thus
$$
f(B) = p(B) =
\frac{1}{2}
\begin{bmatrix}
\frac32 & \frac12 \\
\frac12 & \frac12
\end{bmatrix}
+ \left(\frac{1}{2} - \frac{1}{2 \sqrt{2}} \right)
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix} \\
= \begin{bmatrix}
\frac54 - \frac{1}{2 \sqrt{2}} & \frac{1}{4} \\
\frac{1}{4} & \frac34 - \frac{1}{2 \sqrt{2}}
\end{bmatrix}.
$$
Finally,
$$
f(B) - f(A) = \begin{bmatrix}
\frac14 - \frac{1}{2 \sqrt{2}} & \frac{1}{4} \\
\frac{1}{4} & \frac34 - \frac{1}{2 \sqrt{2}}
\end{bmatrix} \not\geq 0,
$$
as the determinant of the matrix is $\frac{1}{4} \left(1 - \sqrt{2}\right) < 0$.
|
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|
Is $\sum\limits_{n=2}^{\infty} \ln \left[1+\frac{(-1)^{n}}{n^{p}}\right](p>0)$ convergent? Is $S=\sum\limits_{n=2}^{\infty} \ln \left[1+\dfrac{(-1)^{n}}{n^{p}}\right](p>0)$ convergent?
I can prove the case when $p\geq 1$: If $p>1$, $\sum\limits_{n=2}^{\infty} \left|\ln \left[1+\dfrac{(-1)^{n}}{n^{p}}\right]\right|\leq \sum\limits_{n=2}^{\infty} \left|\dfrac{(-1)^{n}}{n^{p}}\right|=\sum\limits_{n=2}^{\infty}\dfrac{1}{n^{p}}$, which is convergent, so $S$ converges absolutely. If $p=1$, $\sum\limits_{n=2}^{\infty} \ln \left[\dfrac{n+(-1)^{n}}{n}\right]=\sum\limits_{n=2}^{\infty} \ln(n+(-1)^n)-\ln(n)=0 \text{ or } \ln(\dfrac{2n+1}{2n})$ converges. How to deal with $p<1$?
Any suggestions will be greatly appreciated.
|
1) If $\frac{1}{2} < p$, the series $\sum_{n=2}^\infty \ln (1 + \tfrac{(-1)^n}{n^p})$ is convergent.
It is easy to prove that $0 \le x - \ln (1 + x) \le 2x^2$ for $x > -\frac{3}{4}$.
Note that $\frac{(-1)^n}{n^p} \ge - \frac{1}{3^p} > - \frac{3}{4}$ for $n\ge 2$.
Thus, we have, for $n\ge 2$,
$$0 \le \frac{(-1)^n}{n^p} - \ln \left(1 + \frac{(-1)^n}{n^p}\right) \le \frac{2}{n^{2p}}.$$
Since $\sum_{n=2}^\infty \frac{2}{n^{2p}}$ is convergent, by comparison test,
$\sum_{n=2}^\infty \left[\frac{(-1)^n}{n^p} - \ln \left(1 + \frac{(-1)^n}{n^p}\right)\right]$
is convergent. Since $\sum_{n=2}^\infty \frac{(-1)^n}{n^p}$ is convergent (alternating series test),
$\sum_{n=2}^\infty \ln (1 + \tfrac{(-1)^n}{n^p})$ is convergent.
$\phantom{2}$
2) If $0 < p \le \frac{1}{2}$, the series $\sum_{n=1}^\infty \ln (1 + \tfrac{(-1)^n}{n^p})$ is divergent.
It is easy to prove that $\ln (1+x) \le x - \frac{x^2}{4}$ for $- 1 < x < 1$.
Note that $-1 < \frac{(-1)^n}{n^p} < 1$ for $n \ge 2$.
Thus, we have, for $n\ge 2$,
$$\ln \left(1 + \frac{(-1)^n}{n^p}\right) \le \frac{(-1)^n}{n^p} - \frac{1}{4n^{2p}}.$$
Denote $S_N = \sum_{n=2}^N \ln (1 + \tfrac{(-1)^n}{n^p})$. We have
$$S_N \le \sum_{n=2}^N \frac{(-1)^n}{n^p} - \sum_{n=2}^N \frac{1}{4n^{2p}}.$$
Note that $\sum_{n=2}^\infty \frac{(-1)^n}{n^p}$ is convergent.
Also, $\lim_{N\to \infty} \sum_{n=2}^N \frac{1}{4n^{2p}} = \infty$.
Thus,
$$\lim_{N\to \infty} \left(\sum_{n=2}^N \frac{(-1)^n}{n^p} - \sum_{n=2}^N \frac{1}{4n^{2p}}\right)
= -\infty.$$
Thus, $\lim_{N\to\infty} S_N = -\infty$.
|
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|
Solving this trigonometric equation I want to solve this equation
$$
\arccos(x)+\arcsin(x^2-x+1)=\pi/2
$$
I write this: for all $x\in [-1,1]$
$\arcsin(x^2-x+1)=\pi/2-\arccos(x)$ then $x^2-x+1=\sin(\pi/2-\arccos(x))=\cos(\arccos(x))=x$
$x^2-x+1=x\Rightarrow x^2-2x+1=0\Rightarrow (x-1)^2=0\Rightarrow x=1 $
Is it true ?
|
Yes true indeed, in fact doubly true ( double root) at $ x=1$
Compare terms of two equations with identity
$$ \arccos(x)+\arcsin(x^2-x+1)=\pi/2 $$
$$ \arccos(x)+\arcsin(x)=\pi/2 $$
so we must have
$$ x^2-x+1 = x,\, x^2-2x+1 = 0,\,(x-1)^2=0,\,x= (1,1)$$
Solution can be verified by their graphs at $(x=1)$ their average value is $y = \pi/4$ at $ x=1$ has a tangent point:
$$ \dfrac{\arccos(x)+\arcsin(x^2-x+1)}{2}=\pi/4 $$
It can be seen that $\arccos(x) $ has domain $(-1,1).$
However, $\arcsin(x^2-x+1)$ and hence the sum $\arccos(x) + \arcsin(x^2-x+1)$ must have a restricted domain $(0,1)$.
When $x<0,\,\arcsin(x^2-x+1)$ is imaginary !
|
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|
Prove $\frac{a^2}{(b+c)^2}+\frac{b^2}{(c+a)^2}+\frac{c^2}{(a+b)^2}+\frac{1}{4}\ge \frac{a^2+b^2+c^2}{ab+bc+ca}$ For $a,b,c>0$. Prove that: $$\frac{a^2}{(b+c)^2}+\frac{b^2}{(c+a)^2}+\frac{c^2}{(a+b)^2}+\frac{1}{4}\geqq \frac{a^2+b^2+c^2}{ab+bc+ca}$$
NguyenHuyen gave the following expression$:$
$$\sum \frac12\, \left( 8\,{a}^{3}b+{a}^{3}c+8\,{a}^{2}{b}^{2}+11\,{a}^{2}bc+7\,a
{b}^{3}+13\,a{b}^{2}c+3\,ab{c}^{2}+3\,b{c}^{3}+2\,{c}^{4} \right)
\left( a+b \right) ^{2} \left( a-b \right) ^{2} \geqslant 0$$
My work with Titu's Lemma and Maple and by lucky!
By Titu's Lemma$,$ we have$:$
$$\text{LHS} \geqq \frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc} a^2(b+c)^2} +\frac{1}{4}\geqq \text{RHS}$$
Last inequality equivalent to$:$
$$\,{\frac {\sum\limits_{cyc} \left( a-b \right) ^{2}\Big[bc \left( 2\,{a}^{2}+ab+ca+{c}^{
2} \right) +2\,ac \left( a-c \right) ^{2}+2\,ab \left( {a}^{2}+{b}^{2}
\right) +c \left( b-c \right) ^{2} \left( a+2\,b \right)\Big]}{ 8\left( {
a}^{2}{b}^{2}+{a}^{2}bc+{a}^{2}{c}^{2}+a{b}^{2}c+ab{c}^{2}+{b}^{2}{c}^
{2} \right) \left( ab+ca+bc \right) }}\geqq 0$$
However$,$ it's very hard with me to find a nice SOS if not have Maple$,$ who have a simple proof for it?
Without $\it{uvw}$ and Buffalo Way if you can!
Thanks a lot!
$\lceil$You can also see here. $\rfloor$
|
Also, we can use SOS after the following C-S:
$$\sum_{cyc}\frac{a^2}{(b+c)^2}\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}a^2(b+c)^2}$$ and it's remains to prove that
$$\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}a^2(b+c)^2}+\frac{1}{4}\geq\frac{a^2+b^2+c^2}{ab+ac+bc},$$ which is sixth degree already and it's obvious by $uvw$, which you don't want.
The last inequality we can prove also by the following reasoning.
Since $$(ab+ac+bc)\sum_{cyc}(a^2+ab)\geq3\sum_{cyc}(a^2b^2+a^2bc)$$ it's $$\sum_{cyc}ab(a-b)^2\geq0,$$ we obtain:
$$\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}a^2(b+c)^2}=\frac{(a^2+b^2+c^2)^2}{2\sum\limits_{cyc}(a^2b^2+a^2bc)}\geq\frac{(a^2+b^2+c^2)^2}{\frac{2}{3}(ab+ac+bc)\sum\limits_{cyc}(a^2+ab)}.$$
Thus, it's enough to prove that
$$\frac{(a^2+b^2+c^2)^2}{\frac{2}{3}(ab+ac+bc)\sum\limits_{cyc}(a^2+ab)}+\frac{1}{4}\geq\frac{a^2+b^2+c^2}{ab+ac+bc}.$$
Now, let $a^2+b^2+c^2=k(ab+ac+bc).$
Thus, $k\geq1$ and we need to prove that
$$\frac{k^2}{\frac{2}{3}(k+1)}+\frac{1}{4}\geq k$$ or
$$(k-1)(2k-1)\geq0$$ and we are done!
|
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|
Finding a function $f$ such that $f(3n)=1$, $f(3n+1)=5/3$, $f(3n+2)=8/3$, where $n$ is an integer
Determine a function $f(x)$ such that :
*
*$f(3n)=1$
*$f(3n+1)=5/3$
*$f(3n+2)=8/3$
where $n$ is an integer
How can you determine it? Thanks for the help!
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I guess, topicstarter wants some closed form description of his function like
$$f(k)=\frac{9}{2}(1*(k+1-3[\frac{k+1}{3}])(k+2-3[\frac{k+2}{3}])+\frac{5}{3}*(k-3[\frac{k}{3}])(k+1-3[\frac{k+1}{3}])+\frac{8}{3}*(k+2-3[\frac{k+2}{3}])(k-3[\frac{k}{3}]).$$
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"url": "https://math.stackexchange.com/questions/3671046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Geometric series in proof of Stirling's Formula I am working through a proof of Stirling's Formula in Feller's An Introduction to Probability Theory and it's Applications and am stuck at equation 9.10, where he make a comparison with a geometric series. For full context, he states:
And using the expansion we get:
$$d_n - d_{n+1} = \frac{1}{3(2n+1)^2} + \frac{1}{5(2n+1)^4}+ \dots\tag{9.9}\label{9.9}$$
By comparison of the right side with a geometric series with ratio $(2n+1)^{-2}$ one sees that:
$$0 < d_n - d_{n+1} < \frac{1}{3[(2n+1)^2 - 1]} = \frac{1}{12n} - \frac{1}{12(n+1)}\tag{9.10}\label{9.10}$$
I am struggling to make the jump from equation 9.9 to 9.10. The geometric series with ratio $(2n+1)^{-2}$ would be (based on wikipedia article):
$$\frac{1}{(2n+1)^2} + \frac{1}{(2n+1)^4} + \frac{1}{(2n+1)^6} + \dots = \frac{1}{1 - \frac{1}{(2n+1)^2}}$$
Which leaves me trying to make a comparison between:
$$
d_n - d_{n+1} = \frac{1}{3(2n+1)^2} + \frac{1}{5(2n+1)^4}+ \dots
< \frac{1}{1 - \frac{1}{(2n+1)^2}}
$$
Which intuitively makes sense seeing as all terms on the left hand side have an additional factor in the denominator, ensuring it is less than the right hand side. However, I am still stuck trying to figure out how Feller arrives at equation 9.10.
Any help or input on where I am going wrong is greatly appreciated.
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Your formula for the sum of a geometric series is slightly off. First of all your formula would be the sum of a geometric series with ratio $(2n+1)^{-1}$, not $(2n+1)^{-2}$. More importantly however, the expression $\frac1{1-r}$ is the sum of $1+r+r^2+...$, however in the current case we're lacking the first term, we instead have the "decapitated" geometric series $r+r^2+r^3...$ with sum
$$S=\frac{1}{1-r}-1=\frac{r}{1-r}=\frac1{(2n+1)^2-1}=\frac1{4n^2+4n}=\frac1{4n(n+1)}$$
That last fraction can be rewritten using partial fractions
$$S=\frac{1}{4n}+\frac{-1}{4(n+1)}$$
Each term in the original series is a term from the geometric series multiplied by a value less than or equal to $\frac13$, therefore the sum of the series is at most
$$\frac13S=\frac{1}{12n}-\frac{1}{12(n+1)}$$
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"language": "en",
"url": "https://math.stackexchange.com/questions/3671295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
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