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Find maximum of function $A=\sum _{cyc}\frac{1}{a^2+2}$ Let $a,b,c\in R^+$ such that $ab+bc+ca=1$. Find the maximum value of $$A=\frac{1}{a^2+2}+\frac{1}{b^2+2}+\frac{1}{c^2+2}$$
I will prove $A\le \dfrac{9}{7}$ and the equality occurs when $a=b=c=\dfrac{1}{\sqrt3 }$
$$\frac{\sum _{cyc}\left(b^2+2\right)\left(c^2+2\right)}{\Pi _{cyc}\left(a^2+2\right)}\le \frac{9}{7}$$
$$\Leftrightarrow 7\sum _{cyc}a^2b^2+28\sum _{cyc}a^2+84\le 9a^2b^2c^2+18\sum _{cyc}a^2b^2+36\sum _{cyc}a^2+72 (1)$$
Let $a+b+c=3u;ab+bc+ca=3v^2=1;abc=w^3$ then we need to prove
$$9w^6+11\left(9v^4-6uw^3\right)+8\left(9u^2-6v^2\right)-12\ge 0$$
$$\Leftrightarrow 9w^6+11\left(27v^6-18uv^2w^3\right)+8\left(81u^2v^4-54v^6\right)-324v^6\ge 0$$
$$\Leftrightarrow 9w^6-198uv^2w^3-459v^6+648u^2v^4\ge 0$$
We have: $$f'\left(w^3\right)=18\left(w^3-11uv^2\right)\le 0$$
So $f$ is a decreasing function of $w^3$ it's enough to prove it for an equality case of two variables. Assume $a=b\rightarrow c=\dfrac{1-a^2}{2a}$
$$(1)\Leftrightarrow \frac{(a^2+2)(a^2+4)(3a^2-1)^2}{4a^2}\ge0$$
Please check my solution. It's the first time i use $u,v,w$, if i have some mistakes, pls fix for me. Thanks!
| Your solution is right, but if you want to get all points for this problem, you need to explain, why it's enough to prove the starting inequality for equality case of two variables. It's not enough to say that it's true by uvw.
After words: $f$ decreases you can write the following.
Thus, it's enough to prove our inequality for a maximal value of $w^3$.
Now, $a$, $b$ and $c$ are roots of the equation
$$(x-a)(x-b)(x-c)=0$$ or
$$x^3-3ux^2+3v^2x-w^3=0$$ or
$$g(x)=w^3,$$ where
$$g(x)=x^3-3ux^2+3v^2x.$$
We see that
$$g'(x)=3x^2-6ux+3v^2=3\left(x-(u-\sqrt{u^2-v^2})\right)\left(x-(u+\sqrt{u^2-v^2})\right),$$
which says that $g$ gets a local maximum for $x=u-\sqrt{u^2-v^2}$ and the equation
$$g(x)=w^3$$ has three positive roots for $$\max\left\{0,g\left(u+\sqrt{u^2+v^2}\right)\right\}<w^3\leq g\left(u-\sqrt{u^2-v^2}\right)$$ and $w^3$ gets a maximal value, when the graph of $y=w^3$ is a tangent line to the graph of $g$, which happens for equality cases of two variables.
We got even that these variables should be equal to $u-\sqrt{u^2-v^2}.$
The rest you wrote already.
| {
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How to factorize the polynomial in the ring $\mathbb{Z}_5[x]$ I need to factor the polynomial $x^5+3x^4+x^3+x^2+3$ into the product of the irreducible ones in the ring $\mathbb{Z}_5[x]$.
The problem is I don't see any whole roots (I tried every possible divisor of 3) in the given polynomial. Does it mean that it is already irreducible?
| You need to make it more aggressively.
Check that $2$, $3$ and $4$ are roots, which gives the factorization:
$$x^5+3x^4+x^3+x^2+3\equiv x^5-2x^4+x^3+x^2-12=$$
$$=x^5-2x^4+x^3-2x^2+3x^2-6x+6x-12=(x-2)(x^4+x^2+3x+6)\equiv$$
$$\equiv(x-2)(x^4-9x^2-2x+6)=(x-2)(x^2(x-3)(x+3)-2(x-3))=$$
$$=(x-2)(x-3)(x^3+3x^2-2)\equiv(x-2)(x-3)(x^3-2x^2-32)=$$
$$=(x-2)(x-3)(x^3-4x^2+2x^2-32)=(x-2)(x-3)(x-4)(x^2+2x+8)=$$
$$=(x-2)(x-3)(x-4)((x+1)^2-3).$$
Easy to see that the quadratic factor is irreducible.
| {
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Different methods of solving a linear system of first order DE?
$$ x'(t)=\begin{pmatrix} 1&\frac{-2}3\\3&4\end{pmatrix}x(t).$$
This is the method described in my book:
*
*finding the eigenvalues of matrix $A = \begin{pmatrix} 1&\frac{-2}3\\3&4\end{pmatrix}$:
$$ \begin{vmatrix} 1-\lambda&\frac{-2}3 \\3&4-\lambda\end{vmatrix}=0\iff \lambda=3, \lambda=2.$$
*eigenvector corresponding to $\lambda=3$: $\begin{pmatrix} -1/3 \\ 1\end{pmatrix}$, eigenvector corresponding to $\lambda=2$: $\begin{pmatrix} -2/3 \\ 1\end{pmatrix}$.
*define $C = \begin{pmatrix} -2/3 & -1/3 \\ 1&1\end{pmatrix}$, then $C^{-1} = \begin{pmatrix} -3 & 3 \\ -1&2\end{pmatrix}$, and $$ A = C\operatorname{diag}(\lambda_1,\lambda_2)C^{-1} = \begin{pmatrix} -2/3 & -1/3 \\ 1&1\end{pmatrix} \begin{pmatrix} 2&0\\0&3\end{pmatrix}\begin{pmatrix} -3 & 3 \\ -1&2\end{pmatrix}.$$
*calculate $e^{tA} $. Using previous expression for $A$ we get $$ e^{tA}=\begin{pmatrix} -2/3 & -1/3 \\ 1&1\end{pmatrix} \begin{pmatrix} e^{2t}&0\\0&e^{3t}\end{pmatrix}\begin{pmatrix} -3 & 3 \\ -1&2\end{pmatrix} = \begin{pmatrix} 2e^{2t}+\frac13e^{3t} & -2e^{2t}-\frac23e^{3t} \\ -3e^{2t}-e^{3t} & 3e^{2t}+2e^{3t}\end{pmatrix}.$$
*the solution of the given system, considering the initial condition $x_0=(x_1,x_2)^t$, equals to $e^{tA}x_0$: $$ e^{tA}x_0 = \dots = x_1\begin{pmatrix}2e^{2t}+\frac13e^{3t} \\-3e^{2t}-e^{3t} \end{pmatrix}+x_2\begin{pmatrix} -2e^{2t}-\frac23e^{3t} \\3e^{2t}+2e^{3t}\end{pmatrix}$$
Now, I have looked up some extra exercises online, but these seem to solve such systems in a shorter way: the solution of the system above with be given by $c_1e^{2t}\begin{pmatrix} -2/3 \\ 1 \end{pmatrix} + c_2e^{3t}\begin{pmatrix} -1/3 \\ 1\end{pmatrix}$.
What is the difference between both approaches? The method used by my book seems to have more coefficients (and is therefore maybe a little more detailed/exact?). Are these solution methods equivalent? Which one would you use?
Thanks.
| The advantage of the exponential matrix method is that you do not have to solve for $c_1$ and $c_2$ to satisfy the initial conditions.
You just multiply your exponential matrix by the vector of initial condition and you get your solution.
$$ x(t)=e^{tA}x_0 = \dots = x_1\begin{pmatrix}2e^{2t}+\frac13e^{3t} \\-3e^{2t}-e^{3t} \end{pmatrix}+x_2\begin{pmatrix} -2e^{2t}-\frac23e^{3t} \\3e^{2t}+2e^{3t}\end{pmatrix}$$
There are other simpler ways of finding the exponential matrix using Cayley-Hamilton theorem which only uses the eigenvalues of the original matrix.
The other advantage of exponential matrix is that finding its inverse is almost trivial, so in case of solving an inhomogeneous system,by the method of variation of constants, it is very convenience.
| {
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How to solve this quadratic congruent equation by inspection I found a systematic way (c.f. How to solve this quadratic congruence equation) to solve all congruent equations of the form of $ax^2+bx+c=0\pmod{p}$, or to determine that they have no solution.
But I wonder if there is some easy way to find solutions of simple quadratic congruent equations by inspection, for example, for this equation $$x^2+x+47=0\pmod{7}.$$
My textbook gave the solutions $x\equiv 5\pmod{7}$ and $x\equiv 1\pmod{7}$ directly, without any proof. So I think there must be some easier way to inspect the solutions.
Any help will be appreciated.
| The good old quadratic equation works just fine if $p\neq2$. If $a\not\equiv 0\pmod{p}$ and $b^2-4ac$ is a quadratic residue mod $p$, then the solutions to the quadratic congruence
$$ax^2+bx+c\equiv0\pmod{p},$$
are precisely
$$-\frac{b\pm\sqrt{b^2-4ac}}{2a}.$$
In this particular case we have $p=7$ and $a\equiv b\equiv1\pmod{7}$ and $c\equiv47\equiv5\pmod{7}$. Then
$$b^2-4ac\equiv2\equiv3^2\pmod{7},$$
so the congruence has the two solutions
$$-\frac{1+3}{2}=5\qquad\text{ and }\qquad-\frac{1-3}{2}=1.$$
On the other hand, if you want to solve it purely by inspection, note that there are only $7$ possible solutions to check. Clearly $x=0$ is not a solution, and plugging in $x=1$ yields the first solution. The product of the solutions is congruent to $47\equiv5\pmod{7}$, so the other solution is $x=5$.
| {
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$1/(51 +u^2) + 1/(51 +v^2) + 1/(51 +w^2)$ inequality I need to prove that $$1/(51 +u^2) + 1/(51 +v^2) + 1/(51 +w^2) \leq 1/20$$ given $u + v + w = 9$ and $u,v,w$ positive reals.
Using AM-HM inequality with $(51 +u^2, 51 +v^2, 51 +w^2)$ I arrive at
$(51 +u^2 + 51 +v^2 + 51 +w^2)/3 \geq 3/((1/(51 +u^2) +1/(51 +v^2) 1/(51 +w^2))$, $$or,$$
$1/(51 +u^2) +1/(51 +v^2)+1/(51 +w^2) \leq 9/(51 +u^2 + 51 +v^2 + 51 +w^2)$
In order to take the next step, I need to estimate the RHS with an upper bound. Here is where I face the problem. With $(u+v+w)^2 \leq 3(u^2 + v^2 + w^2)$ from C-S inequality and $u+v+w =9$ we get $27 \leq (u^2 + v^2 + w^2)$ which does not provide an upper bound. I tried several other inequalities, but each lead to a bound on the "wrong" side. Am I at all looking in the right direction?
| You wrote the following:
"Using AM-HM inequality with $(51 +u^2, 51 +v^2, 51 +w^2)$ I arrive at
$(51 +u^2 + 51 +v^2 + 51 +w^2)/3 \geq 3/((1/(51 +u^2) +1/(51 +v^2) 1/(51 +w^2))$, $$or,$$
$1/(51 +u^2) +1/(51 +v^2)+1/(51 +w^2) \leq 9/(51 +u^2 + 51 +v^2 + 51 +w^2)$"
See please better, the last line should be:
$1/(51 +u^2) +1/(51 +v^2)+1/(51 +w^2) \geq 9/(51 +u^2 + 51 +v^2 + 51 +w^2)$
You proved that
$$\sum_{cyc}\frac{1}{51+u^2}\geq\frac{9}{\sum\limits_{cyc}(51+u^2)},$$ which does not help for the proof of your inequality.
By the way, the Tangent Line method helps.
Indeed, we need to prove that
$$\frac{1}{20}-\sum_{cyc}\frac{1}{51+u^2}\geq0$$ or
$$\sum_{cyc}\left(\frac{1}{60}-\frac{1}{51+u^2}\right)\geq0$$ or
$$\sum_{cyc}\frac{(u-3)(u+3)}{51+u^2}\geq0$$ or
$$\sum_{cyc}\left(\frac{(u-3)(u+3)}{51+u^2}-\frac{u-3}{10}\right)\geq0$$ or
$$\sum_{cyc}\frac{(u-3)^2(7-u)}{51+u^2}\geq0,$$ which is true for $\max\{u,v,w\}\leq7.$
Let $u>7$.
Thus, $$\sum_{cyc}\frac{1}{51+u^2}<\frac{1}{51+7^2}+\frac{2}{51}<\frac{1}{20}$$
and we are done!
| {
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Different methods give different answers. Let A,B,C be three angles such that $ A=\frac{\pi}{4} $ and $ \tan B \tan C=p $. Let $A,B,C$ be three angles such that $ A=\frac{\pi}{4} $ and $ \tan B \tan C=p $. Find all possible values of $p$ such that $A,B$ and $C$ are angles of a triangle.
case 1- discriminant
We can rewrite the following equation
$ f(x) = x^2 - (p-1)x + p $
As we know the sum and product of $ \tan C $ and $ \tan B $
Settings discriminant greater than equal to zero.
$ { (p-1)}^2 - 4p \ge 0 $
This gives $ p \le 3 - 2\sqrt2 $. Or $ p \ge 3 + 2\sqrt2 $
solving both equation
$ A + B + C = \pi $
$ C + B + \frac{\pi}{4} = \pi $
$ C + B = \frac{3\pi}{4} $
Using this to solve both the equation give $ p \in $ real
I found this on Quora.
https://www.quora.com/Let-A-B-C-be-three-angles-such-that-A-frac-pi-4-and-tan-B-tan-C-p-What-are-all-the-possible-value-of-p-such-that-A-B-C-are-the-angles-of-the-triangle
the right method
$ 0 \lt B , C \lt \frac{3\pi}{4} $
Converting tan into sin and cos gives
$ \dfrac {\sin B \sin C}{\cos B \cos C} = p $
Now using componendo and dividendo
$ \frac{\cos (B-C) }{- \cos(B+C) } = \frac{p+1}{p-1} $
We know $ \cos (B+C) = 1/\sqrt2 $
We know the range of $B$ and $C$ $(0, 3π/4)$
Thus the range of $B - C$. $(0, 3π/4 )$
Thus range of $\cos(B+C)$ is $ \frac{ -1}{\sqrt2} $ to $1$
Thus using this to find range gives
$ P \lt 0 $ or $ p \ge 3+ 2\sqrt2 $
| For $A,B,C$ to be the angles of a triangle, not only it shall be $A+B+C = \pi$,
but also $0 \le A,B,C$, or strictly greater than $0$ if you exclude the degenerate case.
So, the correct formulation of the problem is
$$
\left\{ \matrix{
A = \pi /4 \hfill \cr
\tan B\tan C = p \hfill \cr
A + B + C = \pi \hfill \cr
0 \le A,B,C \hfill \cr} \right.\quad \Rightarrow \quad \left\{ \matrix{
\tan B\tan C = p \hfill \cr
B + C = 3\pi /4 \hfill \cr
0 \le B,C\left( { \le 3\pi /4} \right) \hfill \cr} \right.
$$
Taking care of this further restiction, whatever approach you follow (correctly) you will get to the
unique right solution.
For instance, given $A=\pi /4$, we may start from the symmetrical case , isosceles triangle
$B=C=3 \pi /8$ and put
$$
\eqalign{
& \left\{ \matrix{
B = 3\pi /8 + D \hfill \cr
C = 3\pi /8 - D \hfill \cr
- 3\pi /8 \le D \le 3\pi /8 \hfill \cr
\tan \left( {3\pi /8 + D} \right)\tan \left( {3\pi /8 - D} \right) = p \hfill \cr} \right.\quad \Rightarrow \cr
& \Rightarrow \quad \left\{ \matrix{
B = 3\pi /8 + D \hfill \cr
C = 3\pi /8 - D \hfill \cr
- 3\pi /8 \le D \le 3\pi /8 \hfill \cr
t = \tan \left( {3\pi /8} \right) = \sqrt 2 + 1 \hfill \cr
- t \le x = \tan D \le t \hfill \cr
{{t^2 - x^2 } \over {1 - t^2 x^2 }} = p \hfill \cr} \right. \cr
& \Rightarrow \quad \left\{ \matrix{
t^2 = 3 + 2\sqrt 2 \hfill \cr
0 \le x^2 \le t^2 \hfill \cr
p(x) = {{t^2 - x^2 } \over {1 - t^2 x^2 }} \hfill \cr} \right.\quad \Rightarrow \cr
& \Rightarrow \quad p \in \left[ {t^2 = 3 + 2\sqrt 2 , + \infty } \right) \cup \left( { - \infty ,0} \right] \cr}
$$
| {
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How to solve an equation with 6 degree polynomial? Does anyone have an idea to solve the following equation if it is possible? It's the best to get analytic solution, but if you can help me to show when equation has all real root with certain conditions of parameters, it is also alright.
Variable is $\Omega$, and parameters are $M_1 >0$ and $r >0, r \neq 1$.
I am trying to find when it has all real roots.
$f(\Omega)=(r-1)\Omega^6 - (r+1) \Omega^5 - \frac{(r-1)}{4} \Omega^4 + \left(\frac{(r+1)}{2}+ \frac{4}{M_1^2}\right)\Omega^3 -\frac{(r-1)}{16} \Omega^2 + \left(\frac{1}{M_1^2}- \frac{r+1}{16}\right)\Omega + \frac{(r-1)}{64}=0.$
Thank you very much for any help!
| The following is a summary and expansion of some of my comments above, and by no means a complete answer to the question.
If $r<1$ then by Descartes' rule of signs $f$ has either $2$ or $0$ positive real roots, and either $4$, $2$ or $0$ negative real roots, and clearly $\Omega=0$ is not a root. It is not true in general that $f$ has at least $4$ roots.
If $r\neq1$ then setting $t:=\frac{1}{16}+\tfrac{1}{8(r-1)}$ and $s=\tfrac{1}{M_1^2}$ we might as well find the roots of
\begin{eqnarray*}
\tfrac{f(x)}{r-1}&=&x^6-\left(1+\tfrac{2}{r-1}\right)x^5-\tfrac14x^4
+\left(\tfrac12+\tfrac{1}{r-1}+\tfrac{4}{M_1^2}\right)x^3-\tfrac{1}{16}x^2 + \left(\tfrac{1}{M_1^2}-\tfrac{1}{16}-\tfrac{1}{8(r-1)}\right)x+\tfrac{1}{16}\\
&=&x^6-16tx^5-\tfrac14x^4+(8t+4s)x^3-\tfrac{1}{16}x^2+(s-t)x+\tfrac{1}{16},
\end{eqnarray*}
with the conditions that $s>0$ and $t\notin[-\tfrac{1}{16},\tfrac{1}{16}]$, or even the nicer looking
$$\frac{64f(\tfrac x2)}{r-1}=x^6-32tx^5-x^4+32(s+2t)x^3-x^2+32(s-t)x+4.$$
Then if $t<0$ and $s>-t$ then most coefficients are positive; setting the above equal to $0$ yields
$$x^6+32(-t)x^5+32(s+2t)x^3+32(s+(-t))x+4=x^4+x^2,$$
where the terms are sorted left and right so that all coefficients are positive. But it is not hard to check that for all real $x$ we have
$$x^6+4>x^4+x^2,$$
so $f$ has no positive real roots if $t<0$ and $s>-t$, or equivalently $r<1$ and
$$\frac{1}{M_1^2}>-\left(\frac{1}{16}+\frac{1}{8(r-1)}\right),$$
which is in turn equivalent to $M_1^2>16\frac{1-r}{1+r}$.
Similarly, if $t>0$ and $s<t$ then all coefficients in the equation
$$(-x)^6+32t(-x)^5+32(s-t)(-x)+4=(-x)^4+32(s+2t)(-x)^3+(-x)^2,$$
are positive. If $c<1.94316$ then for all real $x$ we have
$$x^6+4>x^4+cx^3+x^2,$$
so if $32(s+2t)<1.94316$ then the above has no real solutions, so $f$ has no negative real roots. In particular, by Descartes' rule of signs it has at most $2$ real roots.
EDIT: The inequality $32(s+2t)<1.94316$ is not satisfied as $t>0$ implies $t>\tfrac{1}{16}$. This shows that $f$ has at least two negative real roots if $r>1$.
| {
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Evaluate $(\sum_{k=1}^{7} \tan^2(\frac{k\pi}{16})) - \tan^2(\frac{4\pi}{16})$ The question is:
Evaluate
$$\displaystyle \left(\sum_{k=1}^{7} \tan^2\left(\frac{k\pi}{16}\right)\right) - \left(\tan^2\frac{4\pi}{16}\right)$$
The given answer:$34$
What I've tried:
$$\displaystyle \left(\sum_{k=1}^{7} \tan^2\left(\frac{k\pi}{16}\right)\right) - \left(\tan^2\frac{4\pi}{16}\right)$$
$$=\tan^2\left(\frac{\pi}{16}\right)+\tan^2\left(\frac{2\pi}{16}\right)+\tan^2\left(\frac{3\pi}{16}\right)+\tan^2\left(\frac{5\pi}{16}\right)+\tan^2\left(\frac{6\pi}{16}\right)+\tan^2\left(\frac{7\pi}{16}\right)$$
$$=\tan^2\left(\frac{\pi}{16}\right)+\tan^2\left(\frac{2\pi}{16}\right)+...+\tan^2\left(\frac{\pi}{2}-\frac{2\pi}{16}\right)+\tan^2\left(\frac{\pi}{2}-\frac{\pi}{16}\right)$$
$$=\tan^2\left(\frac{\pi}{16}\right)+\cot^2\left(\frac{\pi}{16}\right)+\tan^2\left(\frac{2\pi}{16}\right)+\cot^2\left(\frac{2\pi}{16}\right)+\tan^2\left(\frac{3\pi}{16}\right)+\cot^2\left(\frac{3\pi}{16}\right)$$
How do I proceed from here?
| Using Expansion of $\tan(nx)$ in powers of $\tan(x)$,
$\tan\dfrac{k\pi}{2n},0\le k<2n,k\ne n$ are the roots of
$$\binom{2n}1t-\binom{2n}3t^3+\cdots+(-1)^{n-1}\binom{2n}{2n-1}t^{2n-1}=0$$
$$\implies \binom{2n}{2n-1}t^{2n-2}-\binom{2n}{2n-3}t^{2n-3}+\cdots+(-1)^{n-1}\binom{2n}1=0$$ will have the roots $\tan\dfrac{k\pi}n, 1\le k<2n;k\ne n$
So, the roots of $$2nc^{n-1}-\binom{2n}3c^{n-2}+\cdots=0$$ will be $c_k=\tan^2\dfrac{k\pi}n,1\le k<n$
$$\sum_{k=1}^{n-1}c_k=\dfrac{\binom{2n}3}{2n}=?$$
Here $2n=16$
| {
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Prove that $\frac{1}{AM \cdot BN} + \frac{1}{BN \cdot CP} + \frac{1}{CP \cdot AM} \le \frac{4}{3(R - OI)^2}$.
$(O, R)$ is the circumscribed circle of $\triangle ABC$. $I \in \triangle ABC$. $AI$, $BI$ and $CI$ intersects $AB$, $BC$ and $CA$ respectively at $M$, $N$ and $P$. Prove that $$\large \frac{1}{AM \cdot BN} + \frac{1}{BN \cdot CP} + \frac{1}{CP \cdot AM} \le \frac{4}{3(R - OI)^2}$$
I have provided my own solution and I would be greatly appreciated if there are any other solutions, perhaps one involving trigonometry. I deeply apologise for the misunderstanding.
| Firstly,
$$\frac{IM}{AM} + \frac{IN}{BN} + \frac{IP}{CP} = \frac{A_{CIB}}{A_{CAB}} + \frac{A_{AIC}}{A_{ABC}} + \frac{A_{BIA}}{A_{BCA}} = 1$$
where $A_m$ denotes the area of shape $m$.
$$\frac{AI}{AM} + \frac{BI}{BN} + \frac{CI}{CP} = 3 - \left(\frac{IM}{AM} + \frac{IN}{BN} + \frac{IP}{CP}\right) = 2$$
We have that $$\frac{1}{AM \cdot BN} + \frac{1}{BN \cdot CP} + \frac{1}{CP \cdot AM} \le \frac{1}{3}\left(\frac{1}{AM} + \frac{1}{BN} + \frac{1}{CP}\right)^2$$
$$\le \frac{1}{3}\left[\left(\frac{AI}{AM}\right)^2 + \left(\frac{BI}{BN}\right)^2 + \left(\frac{CI}{CP}\right)^2\right]\left(\frac{1}{AI^2} + \frac{1}{BI^2} + \frac{1}{CI^2}\right)$$
(according to the Cauchy-Schwarz inequality)
However
$$\left(\frac{AI}{AM}\right)^2 + \left(\frac{BI}{BN}\right)^2 + \left(\frac{CI}{CP}\right)^2 \ge \dfrac{1}{3}\left(\frac{AI}{AM} + \frac{BI}{BN} + \frac{CI}{CP}\right)^2 = \frac{4}{3}$$
Now I just need to prove that
$$\frac{1}{3}\left(\frac{1}{AI^2} + \frac{1}{BI^2} + \frac{1}{CI^2}\right) \le \frac{1}{(R - OI)^2}$$
which can be easily proven because
$$\left\{ \begin{align} AI \ge |AO - OI|\\ BI \ge |BO - OI|\\ CI \ge |CO - OI|\\ \end{align} \right. = |R - OI|$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3227215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Extracting integral solutions from a quartic equation The equation
\begin{equation*}
y^{4} + 4y^{3} + 10y^{2} + 12y - 27 = 0
\end{equation*}
has two integral roots. Without resorting to the quartic formula, how would one extract the roots from it?
| Here's one way we might proceed:
There is an old and simple result:
Let $R$ be a commutative unital ring and let
$p(x) = \displaystyle \sum_0^n p_i x^i \in R[x]; \tag 1$
then $1$ is a root of $p(x)$ if and only if the sum of the coefficients of $p(x)$ is $0$; that is,
$p(1) = 0 \Longleftrightarrow \displaystyle \sum_0^n p_i = 0; \tag 2$
this is an immediate consequence of
$p(1) = \displaystyle \sum_0^n p_i 1^i = \sum_0^n p_i, \tag 3$
as is easy to see.
We can apply this observation to
$q(y) = y^4 + 4y^3 + 10y^2 + 12y - 27, \tag 4$
and immediately conclude $1$ is a zero of $q(y)$. Next we synthetically divide $q(y)$ by $y - 1$, and obtain
$q(y) = y^4 + 4y^3 + 10y^2 + 12y - 27 = (y - 1)(y^3 + 5y^2 + 15y + 27); \tag 5$
We Check:
$(y - 1)(y^3 + 5y^2 + 15y + 27) = y^4 + 5y^3 + 15y^2 + 27y - y^3 - 5y^2 - 15y - 27$
$= y^4+ 4y^3 + 10y^2 + 12y - 27 = q(y); \tag 6$
setting
$s(y) = y^3 + 5y^2 + 15y + 27, \tag 7$
we now invoke the rational root theorem to limit the integer candidates for a zero of this cubic polynomial to the (integer) divisors of the constant term, that is, to $\pm 1$, $\pm 3$, $\pm 9$, and $\pm 27$; we then make a (hopefully) intelligent guess that
$s(-3) = 0 \tag 8$
might hold; and indeed we see that
$s(-3) = -27 + 45 - 45 + 27 = 0; \tag 9$
thus $-3$ is also a root of $q(y)$; now we (again, synthetically) divide $s(y)$ by $y + 3$:
$s(y) = y^3 + 5y^2 + 15y + 27 = (y + 3)(y^2 + 2y + 9), \tag{10}$
and again we may check:
$(y + 3)(y^2 + 2y + 9) = y^3 + 2y^2 + 9y + 3y^2 + 6y + 27$
$= y^3 + 5y^2 + 15y + 27 = s(y); \tag{11}$
it follows that
$q(y) = (y - 1)(y + 3)(y^2 + 2y + 9); \tag{12}$
we have now extracted two integral roots of $q(y)$; the remaining zeroes satisfy the quadratic equation
$y^2 + 2y + 9 = 0; \tag{13}$
the "formula" yields
$y = \dfrac{-2 \pm \sqrt{-32}}{2} = \dfrac{-2 \pm 4 \sqrt{-2}}{2} = -1 \pm 2i\sqrt 2; \tag{14}$
we see the remaining two roots of $q(y)$ form a complex conjugate pair, not integers! We have thus verified that $q(y)$ has precisely two integral zeroes, $1$ and $-3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3233358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
If $\space$ $\forall$ $x \in \Bbb R$, $\space$ $f(f(x))=x^2-x+1$. Find the value of $f(0)$. If $\space$ $\forall$ $x \in \Bbb R$, $\space$ $f(f(x))=x^2-x+1$. Find the value of $f(0)$.
I thought that making $f(x)=0$ implies that $f(0)= 0^2 - 0 + 1 = 1$, but i think that this isn't correct, because the $x$ in $f(f(x))$ isn't equal to $f(x)$ .
Any hints?
| We first consider $f(1)$. Note that $f(f(1)) = 1$, so by substituting $x=f(1)$ we obtain $f(1) = f(f(f(1))) = f(1)^2 - f(1) + 1$. We therefore obtain the quadratic equation $f(1)^2 - 2f(1) + 1 = 0$ which implies $f(1) = 1$.
Also note that $f(f(0)) = 1$, so we have $f(1) = f(f(f(0))) = f(0)^2 - f(0) + 1$. So we obtain $f(0)^2 - f(0) = 0$ giving $f(0) = 0$ or $f(0) = 1$. Note that if $f(0) = 0$ then $f(f(0)) = f(0) = 0 \neq 1$, yielding a contradiction, so we conclude $f(0) = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3235288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Fourier Series Representation for piecewise function I've been posed the following question:
$$
f(x)=
\begin{cases}
1-x^2, & 0 \leqslant|x|<1,\\
0, & 1\leqslant|x|<2\\
\end{cases}
$$
I'm trying to find the Fourier series representation but I am having trouble with $ a_n$ and $b_n$ coefficients.
For reference, I obtained $a_0 = \frac{4}{3}$ and I'm pretty sure this is correct.
For $a_n$, I have tried solving the following
$$\frac{1}{2} \int_{-2}^{2} (1-x^2)\cos \left(\frac{\pi m x}{2} \right) dx$$
to obtain
$$\frac{16}{\pi^2 m^2} \cos(2 \pi m) \cos \left(\frac{\pi n x}{2} \right)$$
but am fairly certain I did not obtain the correct answer.
For $b_n$, the function $1-x^2$ is even and hence $b_n = 0$. However, I'm not sure if this is right either.
Any help will be much appreciated!
| All the odd terms disappear.
Note that $f$ doesn't take value of $1-x^2$ from $1$ to $2$.
\begin{align}
\frac12 \int_{-2}^2 f(x) \cos \left( \frac{m\pi x}{2}\right) \, dx &= \int_0^1 (1-x^2) \cos \left( \frac{m\pi x}{2}\right) \, dx\\
&= \frac{2}{m \pi}\sin \left( \frac{m \pi x}{2} \right)(1-x^2)\big|_0^1 +\frac{4}{m \pi} \int_0^1 x\sin \left(\frac{m \pi x}2 \right) \, dx \\
&= \frac{4}{m \pi} \int_0^1 x\sin \left(\frac{m \pi x}2 \right) \, dx \\
&=\frac4{m \pi} \left[-\frac{2x}{m \pi} \cos \left( \frac{m \pi x}2\right)\big|_0^1 + \int_0^1 \, \frac2{m \pi}\cos \left(\frac{m \pi x}2 \right) dx\right]\\
&=\frac4{m \pi} \left[\frac{-2}{m \pi}\cos\left(\frac{m \pi}{2} \right) + \int_0^1 \, \frac2{m \pi}\cos \left(\frac{m \pi x}2 \right) dx\right]\\
&= \frac4{m \pi} \left[\frac{-2}{m \pi}\cos\left(\frac{m \pi}{2} \right) + \left(\frac2{m \pi}\right)^2\sin \left(\frac{m \pi }2 \right) \right]\\
&=\begin{cases}
\frac{16}{((4k-3)\pi)^3} & ,m=4k-3\\
\frac{8}{((4k-2)\pi)^2}& ,m = 4k-2\\
-\frac{16}{((4k-3)\pi)^3}& , m = 4k-1\\
-\frac{8}{(4k\pi)^2} & ,m=4k
\end{cases}
\end{align}
Also,
$$a_0 = \frac12 \int_{-1}^1 1-x^2\, dx = \int_0^11-x^2 \, dx =1-\frac13=\frac23$$
The fourier series is
\begin{align}\frac13 + \sum_{k=1}^\infty&\left[\frac{16}{((4k-3)\pi)^3} \cos\left( \frac{(4k-3)\pi x}{2}\right) + \frac{8}{((4k-2)\pi)^2} \cos\left( \frac{(4k-2)\pi x}{2}\right)\right. \\&\left.-\frac{16}{((4k-3)\pi)^3}\cos\left( \frac{(4k-1)\pi x}{2}\right)-\frac{8}{(4k\pi)^2}\cos\left( \frac{(4k)\pi x}{2}\right) \right] \end{align}
Here is a second order approximation:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3238965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$x_1,x_2,x_3$ the roots of $x^3+mx+n=0$ Find determinant of $A^{2}$ Let $m,n\in\mathbb{R}; x_1,x_2,x_3 $ the roots of $x^3+mx+n=0$ and the matrix $A=\begin{pmatrix}
1 & 1 &1 \\
x_1 &x_2 &x_3 \\
x^{2}_1 & x^{2}_2 & x^{2}_3
\end{pmatrix}$
I need to find determinant of $A^2$ which is $det(A)\cdot det(A)$
I got $det(A)=(x_2-x_1)(x_3-x_1)(x_3-x_2)$.I know that $x_1+x_2+x_3=0$, $x_1x_2+x_1x_3+x_2x_3=m$, $x_1x_2x_3=-n$.
I expanded the determinant and I tried to factorize but I can't use Vieta formula because I don't get a sum or a product.
Also I tried to find $det(A\cdot A^T)$ but the calculations are very heavy.
How to approach the exercise?
| An alternative method!
You found:
$$\det(A)=(x_2-x_1)(x_3-x_1)(x_3-x_2), \\
x_1+x_2+x_3=0, x_1x_2+x_1x_3+x_2x_3=m, x_1x_2x_3=-n$$
Note:
$$x_1+x_2+x_3=0 \Rightarrow x_1+x_2=-x_3 \Rightarrow x_1^2+x_2^2=x_3^2-2x_1x_2 \ \ (1)\\
$$
Also note the famous formula:
$$x^3+y^3+z^3=3xyz+(x+y+z)[(x+y+z)^2-3(xy+yz+zx)] \ (2) \ \ \text{or}\\
(xy)^3+(yz)^3+(zx)^3=3(xyz)^2+(xy+yz+zx)[(xy+yz+zx)^2-3xyz(x+y+z)] $$
You need to find:
$$\begin{align}\det(A)^2&=(x_2-x_1)^2(x_3-x_1)^2(x_3-x_2)^2=\\
&=(x_1^2+x_2^2-2x_1x_2)(x_1^2+x_3^2-2x_1x_3)(x_2^2+x_3^2-2x_2x_3)\stackrel{(1)}=\\
&=(x_3^2-4x_1x_2)(x_2^2-4x_1x_3)(x_1^2-4x_2x_3)=\\
&=-63x_1^2x_2^2x_3^2-4[(x_1x_2)^3+(x_1x_3)^3+(x_2x_3)^3]+16x_1x_2x_3(x_1^3+x_2^3+x_3^3)\stackrel{(2)}=\\
&=-63n^2-4[3n^2+m(m^2-0)]-16n(0-3n)=\\
&=-27n^2-4m^3.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3239161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Inverse of an upper bidiagonal Toeplitz matrix I have a matrix with the following structure
$$\left[\begin{array}{cccccc|c}
-1 & 1-b & 0 & \dots & 0 & 0 & b \\
0 & -1 & 1-b & \dots & 0 & 0 & b \\
\cdots \\
0 & 0 & 0 & \dots &-1 & 1-b & b \\
0 & 0 & 0 & \dots & 0 & -1 & 1 \\ \hline
0 & 0 & 0 & \dots & 0 & 0 & -1
\end{array}\right]$$
I have to find the inverse of this matrix. I begin by using the block matrix inversion formula
$$\begin{bmatrix}\mathbf {A} &\mathbf {B} \\\mathbf {C} &\mathbf {D} \end{bmatrix}^{-1}=\begin{bmatrix}\left(\mathbf {A} -\mathbf {BD} ^{-1}\mathbf {C} \right)^{-1}&-\left(\mathbf {A} -\mathbf {BD} ^{-1}\mathbf {C} \right)^{-1}\mathbf {BD} ^{-1}\\-\mathbf {D} ^{-1}\mathbf {C} \left(\mathbf {A} -\mathbf {BD} ^{-1}\mathbf {C} \right)^{-1}&\quad \mathbf {D} ^{-1}+\mathbf {D} ^{-1}\mathbf {C} \left(\mathbf {A} -\mathbf {BD} ^{-1}\mathbf {C} \right)^{-1}\mathbf {BD} ^{-1}\end{bmatrix}$$
where
$$\mathbf{A} = \begin{bmatrix}
-1 & 1-b & 0 & \dots & 0 & 0 \\
0 & -1 & 1-b & \dots & 0 & 0 \\
\cdots \\
0 & 0 & 0 & \dots &-1 & 1-b \\
0 & 0 & 0 & \dots & 0 & -1
\end{bmatrix}$$
$$\mathbf{B} = \begin{bmatrix} b \\ b \\ \vdots \\ b \\ 1 \end{bmatrix}$$
$$\mathbf{C} = \begin{bmatrix} 0 & 0 & 0 & \dots & 0 & 0 \end{bmatrix}$$
$$\mathbf{D} = \begin{bmatrix} -1 \end{bmatrix}$$
Given that $\mathbf{C}$ and $\mathbf{D}$ I can simplify the formula as
$$\begin{bmatrix}\mathbf {A} &\mathbf {B} \\\mathbf {C} &\mathbf {D} \end{bmatrix}^{-1}=\begin{bmatrix}\mathbf{A}^{-1} & \mathbf {A}^{-1} \mathbf {B}\\ \mathbf{0} & -1\end{bmatrix}$$
The only part left is to solve for $\mathbf{A}$ which I believe can be called an upper bidiagonal Toeplitz matrix. Unfortunately, I have not been able to find a formula to compute the inverse for the same.
| You can notice that $A=-I + (1-b)N$ where $N$ is the standard nilpotent matrix, so
$$A^{-1} = - \left( I + (1-b)N + (1-b)^2 N^2 + \cdots + (1-b)^{n-1} N^{n-1} \right)$$
and the powers of $N$ are easily computed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3239967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
prove $\ln(1+x^2)\arctan x=-2\sum_{n=1}^\infty \frac{(-1)^n H_{2n}}{2n+1}x^{2n+1}$ I was able to prove the above identity using 1) Cauchy Product of Power series and 2) integration but the point of posting it here is to use it as a reference in our solutions.
other approaches would be appreciated.
| I think this is a much simpler proof.
\begin{align}
\tanh^{-1}x\ln(1-x^2)&=\frac12\{\ln(1+x)-\ln(1-x)\}\{\ln(1+x)+\ln(1-x)\}\tag1\\
&=\frac12\ln^2(1+x)-\frac12\ln^2(1-x)\tag2\\
&=\sum_{n=1}^\infty(-1)^n\frac{H_{n-1}}{n}x^n-\sum_{n=1}^\infty\frac{H_{n-1}}{n}x^n\tag3\\
&=-2\sum_{n=1}^\infty\frac{H_{2n-2}}{2n-1}x^{2n-1}\tag4\\
&=-2\sum_{n=1}^\infty\frac{H_{2n}}{2n+1}x^{2n+1}\tag5
\end{align}
Thus $$\tanh^{-1}x\ln(1-x^2)=-2\sum_{n=1}^\infty\frac{H_{2n}}{2n+1}x^{2n+1}\tag6$$
Replace $x$ with $ix$ we get
$$\tan^{-1}x\ln(1+x^2)=-2\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{2n+1}x^{2n+1}\tag7$$
Explanation:
$(1)$ $\tanh^{-1}x=\frac12\ln\left(\frac{1+x}{1-x}\right)$.
$(2)$ Difference of two squares.
$(3)$ $\frac12\ln^2(1-x)=\sum_{n=1}^\infty\frac{H_n}{n+1}x^{n+1}=\sum_{n=1}^\infty\frac{H_{n-1}}{n}x^n$
$(4)$ $\sum_{n=1}^\infty ((-1)^n-1)a_{n}=-2\sum_{n=1}^\infty a_{2n-1}$
$(5)$ Reindex.
Bonus:
If we differentiate both sides of $(7)$ we obtain another useful identity
$$\frac{\arctan x}{1+x^2}=\frac12\sum_{n=1}^\infty(-1)^n\left(H_n-2H_{2n}\right)x^{2n-1}\tag8$$
another identity follows from integrating both sides of $(8)$:
$$\arctan^2x=\frac12\sum_{n=1}^\infty\frac{(-1)^n\left(H_n-2H_{2n}\right)}{n}x^{2n}\tag9$$
replace $x$ with $ix$ in $(9)$
$$\text{arctanh}^2x=-\frac12\sum_{n=1}^\infty\frac{\left(H_n-2H_{2n}\right)}{n}x^{2n}\tag{10}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3240779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
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I need help with this rate of change problem The solid shown in the figure below consists of a cylinder of the radius (r) and height (h)
and a hemispherical void of the radius (r). The dimensions at a given instant (t) are:
$$h(t)=3t^2+2;r(t)=8-\frac{t^2}{4}$$
Find the rate of change of the volume (V) and surface area (S) of the solid at $t = 4$
seconds. State whether (V) and (S) are increasing, decreasing or neither.
Note:
The volume of Sphere =$\frac{4}{3}πr^3$
$(Ans: \frac{dV}{dt} = - 1105.84 (Deceasing))$
I tried this:
$$\frac{dV}{dt}=\frac{\partial{V}}{\partial{r}}*\frac{dr}{dt}$$
and said:
$$t=4$$
$$r(4)=4$$
$$\frac{\partial{V}}{\partial{r}}=4πr^2;\frac{dr}{dt}=\frac{-t}{2}$$
so:$$\frac{dV}{dt}=(4πr^2)*(\frac{-t}{2})$$
and the final result I get:
$$\frac{dV}{dt}=128π$$
What am I doing wrong?
| Volume-->
$$V=\pi r^2h-\frac{2}{3}\pi r^3$$
$$V=\pi (8-\frac{t^2}{4})^2(3t^2+2)-\frac{2}{3}\pi (8-\frac{t^2}{4})^3$$
$$\frac{dV}{dt}=\pi[(3t^2+2)*2(8-\frac{t^2}{4})*(\frac{-t}{2})+(8-\frac{t^2}{4})^2*6t]-[\frac{2}{3}\pi *3(8-\frac{t^2}{4})^2*(\frac{-t}{2})]$$
$$\frac{dV}{dt}=\pi[\frac{19}{16}t^5-\frac{103}{2}t^3+432t]$$
At $t=4s$,
$$\frac{dV}{dt}=\pi[\frac{19}{16}*4^5-\frac{103}{2}*4^3+432*4]=-352\pi\approx-1105.84$$
Surface Area-->
$$A=2\pi rh+3\pi r^2$$
$$A=2\pi(8-\frac{t^2}{4})(3t^2+2)+3\pi(8-\frac{t^2}{4})^2$$
$$\frac{dA}{dt}=\pi[2(3t^2+2)(\frac{-t}{2})+3*2(8-\frac{t^2}{4})(\frac{-t}{2})]$$
$$\frac{dA}{dt}=\pi[\frac{-9}{4}t^3-26t]$$
At $t=4s$,
$$\frac{dA}{dt}=\pi[\frac{-9}{4}*4^3-26*4]=-248\pi\approx-779.115$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3243043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Non-linear system of equations with three unknowns I have the following non-linear system of equations
\begin{cases}
\tau + e^{(m+\frac{s}{2})} &= a\,\quad(1)\\
\tau^2 + 2\tau\,e^{(m+\frac{s}{2})}+e^{(2m+2s)} &= b.\quad(2)\\
\tau^3+3\tau^2\, e^{(m+\frac{s}{2})}+3\tau\,e^{(2m+2s)}+e^{(3m+\frac{9s}{2})}&= c\quad (3)
\end{cases}
and would like to solve for m, s and $\tau$.
I have tried the following:
Letting $x = e^{m}$ and $y=e^{s/2}$ so as to get
\begin{cases}
\tau + xy &= a\,\quad(3)\\
\tau^2 + 2\tau xy+ x^2y^4&= b.\quad(4)\\
\tau^3 + 3\tau^2xy+ 3\tau\,x^2y^4+x^3y^9 &= c \quad(5)
\end{cases}
Then solving for x and y and in turns m and s.
However I'm struggling to get $\tau$ in terms of the unknowns only.
Any suggestions would be very much appreciated.
| \begin{cases}
t + xy &= a\,\quad(3)\\
t^2 + 2t xy+ x^2y^4 &= b\quad(4)\\
t^3 + 3t^2xy+ 3t\,x^2y^4+x^3y^9 &= c \quad(5)
\end{cases}
$xy=a-t\quad$ that we put into $(4)$ and $(5)$.
$\begin{cases}
t^2 + 2t (a-t)+ (a-t)^2y^2 &= b\quad(6)\\
\color{red}{t^3} + 3t^2(a-t)+ 3t\,(a-t)^2y^2+(a-t)^3y^6 &= c \quad(7)
\end{cases}$
from $(6)$ :
$$y^2=\frac{b-t^2 - 2t (a-t)}{(a-t)^2}=\frac{b+t^2 - 2at}{(a-t)^2} \quad(8)$$
We put $(8)$ into $(7)$ :
$$\color{red}{t^3} + 3t^2(a-t)+ 3t\,(a-t)^2\frac{b+t^2 - 2at}{3t\,(a-t)^2}+(a-t)^3\left(\frac{b+t^2 - 2at}{3t\,(a-t)^2} \right)^3=c$$
After simplification :
$$\alpha t^3+\beta t^2 +\gamma t+\delta = 0\quad
\begin{cases}
\alpha=c-3ab+2a^3 \\
\beta =3(b^2+a^2b-a^4-ac) \\
\gamma=3a(a^2b-2b^2+ac\\
\delta=b^3-a^3c
\end{cases}\tag 9$$
Then, solve the cubic equation for $t$.
Put $t$ into $(8)$. This gives $y$.
Then with $t$ and $y$ obtained, $\quad x=\frac{a-t}{y}$
Obviously the formulas for $t$, $y$ and $x$ will be very complicated.
| {
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"url": "https://math.stackexchange.com/questions/3243241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Difference in my and wolfram's integration. Calculate
$$\int \frac{\sin ^3(x)+1}{\cos (x)+1} \, dx$$
Let $$u = \tan(x/2)$$
$\int \frac{\sin ^3(x)+1}{\cos (x)+1} \, dx = \int \frac{2\left(\frac{8u^3}{(u^2+1)^3}+1 \right)}{(u^2+1)\left( \frac{1-u^2}{u^2+1} + 1 \right)} \, du = \int \frac{8u^3}{(u^2+1)^3 } + 1 \, du$
$$ s := u^2 \wedge ds = 2u \, du $$
$ u + \int \frac{8su}{(s+1)^3 } (2u)^{-1} \, ds = u + 4\int \frac{s}{(s+1)^3 } \, ds = $
$ u + 4\int \frac{1}{(s+1)^2 } - \frac{1}{(s+1)^3 } \, ds = u + 4\left(\frac{-1}{s+1} +\frac{1}{2(s+1)^2} \right) + C =$
$ u + 4\left(\frac{-1}{u^2+1} +\frac{1}{2(u^2+1)^2} \right) + C = $
$ \tan(x/2) + 4\left(\frac{-1}{(\tan(x/2))^2+1} +\frac{1}{2((\tan(x/2))^2+1)^2} \right) + C := F(x)$
$$F(\pi/2) - F(-\pi/2) = 2 $$
Wolfram tells that result ($=2$) is okay, but my integral is different from wolfram result:
$$ \frac{1}{8} \sec \left(\frac{x}{2}\right) \left(8 \sin \left(\frac{x}{2}\right)-4 \cos \left(\frac{x}{2}\right)-3 \cos \left(\frac{3 x}{2}\right)+\cos \left(\frac{5 x}{2}\right)\right) $$
Where I failed?
| Use $$\frac{\sin^3x+1}{\cos{x}+1}=\frac{8\sin^3\frac{x}{2}\cos^3\frac{x}{2}+1}{2\cos^2\frac{x}{2}}=4\sin^3\frac{x}{2}\cos\frac{x}{2}+\frac{1}{2\cos^2\frac{x}{2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3243878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
How to prove : $\cos^32\theta + 3\cos2\theta = 4(\cos^6 \theta -\sin^6 \theta)$ How to prove : $\cos^32\theta + 3\cos2\theta = 4(\cos^6 \theta -\sin^6 \theta)$
| Write $\cos2\theta=x$. We have
$$x=2\cos^2\theta-1=1-2\sin^2\theta$$
$$\frac{1+x}2=\cos^2\theta$$
$$\frac{1-x}2=\sin^2\theta$$
Thus the RHS becomes
$$4\left(\frac{(1+x)^3}8-\frac{(1-x)^3}8\right)=\frac12(1+3x+3x^2+x^3-(1-3x+3x^2-x^3))=x^3+3x=LHS$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3244039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find all numbers x which satisfy $|x^2 + 2| = |x^2 − 11|$. This question is taken from book: Exercises and Problems in Calculus, by
John M. Erdman, available online, from chapter 1.1, question $4$.
Request help, as not clear if my approach is correct.
(4) Find all numbers x which satisfy $|x^2 + 2| = |x^2 − 11|$.
Have two conditions, leading to three intervals;
(i) $x \lt \sqrt{-2}$
(ii) $\sqrt{-2} \le x \lt \sqrt{11}$
(iii) $x \ge \sqrt{11}$
(i) no soln.
(ii) $2x^2 = 9 \implies x = \pm\sqrt{\frac 92}$
(iii) no soln.
Verifying:
First, the two solutions must satisfy that they lie in the given interval,
this means $\sqrt{-2} \le x \lt \sqrt{11}\implies \sqrt{2}i \le x \lt \sqrt{11}$.
Am not clear, as the lower bound is not a real one. So, given the two real values of $x = \pm\sqrt{\frac 92}$ need only check with the upper bound.
For further verification, substitute in values of $x$, for interval (ii):
a) For $x = \sqrt{\frac 92}$, $|x^2 + 2| = |x^2 − 11|$ leads to
$\frac 92 + 2 = -\frac 92 + 11\implies 2\frac 92 = 9$.
b) For $x = -\sqrt{\frac 92}$, $|x^2 + 2| = |x^2 − 11|$ leads to
$\frac 92 + 2 = -\frac 92 + 11\implies 2\frac 92 = 9$.
| $\mathrm{dist}(x^2,-2)=|x^2+2|=|x^2-11|=\mathrm{dist}(x^2,11)$. Therefore you want $x^2$ to be equidistant from $-2$ and $11$, i.e. $x^2=11-\frac{\mathrm{dist}(-2,11)}{2}=-2+\frac{\mathrm{dist}(-2,11)}{2}=\frac{9}{2}$. It follows that $x=\pm \frac{3}{\sqrt{2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3247025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Find $9$'th derivative of $\frac{x^3 e^{2x^2}}{(1-x^2)^2}$ How can I find $9$'th derivative at $0$ of $\displaystyle \frac{x^3 e^{2x^2}}{(1-x^2)^2}$. Is there any tricky way to do that?
This exercise comes from discrete mathematic's exam, so I think that tools like taylor can't be used there.
| We know that
$$\frac{x}{(1-x)^2}=\sum_{n=1}^{\infty}nx^{n}$$
$x\rightarrow x^2$
$$\frac{x^2}{(1-x^2)^2}=\sum_{n=1}^{\infty}nx^{2n}$$
multiply by $x$
$$\frac{x^3}{(1-x^2)^2}=\sum_{n=1}^{\infty}nx^{2n+1}\tag1$$
$$e^x=\sum_{m=0}^{\infty}\frac{x^m}{m!}$$
$x\rightarrow 2x^2$
$$e^{2x^2}=\sum_{m=0}^{\infty}\frac{2^mx^{2m}}{m!}\tag2$$
so
$$y=\frac{x^3 e^{2x^2}}{(1-x^2)^2}=\sum_{n=1}^{\infty}\sum_{m=0}^{\infty}\frac{n2^mx^{2n+2m+1}}{m!}$$
to get the coefficient of $x^9$ ,we should find the solutions of $2m+2n+1=9$ which are
$$(m,n)=(0,4),(1,3),(2,2),(3,1)$$
so the coefficient of $x^9$ is $\frac{46}{3}$
the 9 'th derivative at $x=0$ will be
$$y(0)^{(9)}=\frac{46}{3}*9!=5564160$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3247700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Trigonometric series Cosecant^4 How to prove the following identity
$$
\text{cosec}^4{\frac{π}{n}}+\text{cosec}^4{\frac{2π}{n}}+\text{cosec}^4{\frac{3π}{n}} + \ldots+ \text{cosec}^4{\frac{(n-1)π}{n}} = \frac{(n^2+11)(n^2-1)}{45}
$$
I tried to simplify the sum but I finally get stuck with the same question I don't know how to start the problem
| The best way is probably appealing to Chebyshev polynomials and Vieta's formula.
Note that $x=\sin(k\pi/n)$, $k=0,\dots,2n-1$, are the roots to $\cos(n\arccos(1-2x^2))=1$, i.e., $T_n(1-2x^2)=1$, where $T_n$ is the Chebyshev polynomial of the first kind.
Now using
$$
T_n(y)=n\sum_{k=0}^n(-2)^k\frac{(n+k-1)!}{(n-k)!(2k)!}(1-y)^k
$$
we get
$$
0=T_n(1-2x^2)-1=-1+n\sum_{k=0}^n(-2)^k\frac{(n+k-1)!}{(n-k)!(2k)!}(2x^2)^k
$$
Hence
$$
0=-2n^2x^2+\frac23n^2(n^2-1)x^4-\frac4{45}n^2(n^4-5n^2+4)x^6+\dots
$$
Excluding the two zeros $k=0,n$ (which gives $x=0$), we have
$$
0=-2n^2+\frac23n^2(n^2-1)x^2-\frac4{45}n^2(n^4-5n^2+4)x^4+\dots
$$
So Vieta's formula gives
\begin{align*}
\sum_{k=1}^{n-1}\frac{1}{\sin^2(k\pi/n)}&=\frac13(n^2-1),\\
\sum_{1\leq k<l<n}\frac1{\sin^2(k\pi/n)\sin^2(l\pi/n)}&=\frac2{45}(n^4-5n^2+4)
\end{align*}
Hence
\begin{align*}
\sum_k\operatorname{cosec}^4(k\pi/n)&=\sum_k\frac{1}{\sin^4(k\pi/n)}\\
&=\left(\sum_k\frac{1}{\sin^2(k\pi/n)}\right)^2-2\sum_{k<l}\frac1{\sin^2(k\pi/n)\sin^2(l\pi/n)}\\
&=\left(\frac13(n^2-1)\right)^2-\frac4{45}(n^4-5n^2+4)\\
&=\frac1{45}(n^2+11)(n^2-1).
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3249415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $\lim_{x\to\infty}1+2x^2+2x\sqrt{1+x^2}$ Consider the function
$$f(x)=1+2x^2+2x\sqrt{1+x^2}$$
I want to find the limit $f(x\rightarrow-\infty)$
We can start by saying that $\sqrt{1+x^2}$ tends to $|x|$ when $x\rightarrow-\infty$, and so we have that
$$\lim_{x\rightarrow-\infty}{(1+2x^2+2x|x|)}=\lim_{x\rightarrow-\infty}(1+2x^2-2x^2)=1$$
However, if you plot the function in Desmos or you do it with a calculator, you will find that $f(x\rightarrow-\infty)=0$
What am I missing?
|
We can start by saying that $\sqrt{1+x^2}$ tends to $|x|$ when $x\rightarrow-\infty$,
Not exactly. Both $\sqrt{1+x^2}$ and $|x|$ tend to $\infty$ as $x \to -\infty$. I think what you mean that each is asymptotic to the other, because their ratio tends to $1$ as $x\to-\infty$. But that doesn't mean you can replace one with the other, as we can see here.
As $x\to -\infty$, this limit takes the form $\infty - \infty$, which is an indeterminate form. One way to evaluate such a limit is to rewrite it in a form which is not indeterminate. For convenience, let's first write it as
$$
\lim_{x\to\infty} \left(1 + 2x^2 - 2 x\sqrt{1+x^2}\right)
$$
Here we are just substituting $-x$ for $x$; it doesn't change the sign of $x^2$ or $\sqrt{1+x^2}$, but it does change the sign of $x$. The typical algebraic trick for dealing with radicals such as this is to multiply and divide by the conjugate radical:
\begin{align*}
1 + 2x^2 - 2 x\sqrt{1+x^2}
&= \left(1 + 2x^2 - 2 x\sqrt{1+x^2}\right) \frac{1 + 2x^2 + 2 x\sqrt{1+x^2}}{1 + 2x^2 + 2 x\sqrt{1+x^2}} \\
&= \frac{\left((1 + 2x^2) - 2 x\sqrt{1+x^2}\right)\left((1 + 2x^2) + 2 x\sqrt{1+x^2}\right)}{1 + 2x^2 + 2 x\sqrt{1+x^2}} \\
&= \frac{(1+2x^2)^2 - \left(2x \sqrt{1+x^2}\right)^2}{1 + 2x^2 + 2 x\sqrt{1+x^2}} \\
&= \frac{(1+4x^2+4x^4) - 4x^2(1+x^2)}{1 + 2x^2 + 2 x\sqrt{1+x^2}} \\
&= \frac{1}{1 + 2x^2 + 2 x\sqrt{1+x^2}} \\
\end{align*}
Now as $x\to\infty$, the denominator tends to $\infty$ while the numerator is a constant $1$. Therefore the quotient, and the original expression, tends to zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3249693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
A hard integral, can't seem to prove it How to show that,
$$\int_{-\infty}^{+\infty}\frac{(x-1)^2}{(2x-1)^2+\left(ax^2-x-\frac{(a-1)^2}{2(2a-1)}\right)^2}\mathrm dx=\pi$$
Let assume $a\ge1$
I have tried substitution but it leads to a more complicated integral.
| Noting that
\begin{eqnarray}
&&(2x-1)^2+\left(ax^2-x-\frac{(a-1)^2}{2(2a-1)}\right)^2\\
&=&\left(ax^2-x-\frac{(a-1)^2}{2(2a-1)}+(2x-1)i\right)\left(ax^2-x-\frac{(a-1)^2}{2(2a-1)}-(2x-1)i\right) \\
&=&\left(ax^2+(2i-1)x-\frac{(a-1)^2}{2(2a-1)}-i\right)\left(ax^2-(2i+1)x-\frac{(a-1)^2}{2(2a-1)}+i\right)
\end{eqnarray}
so the equation
$$(2x-1)^2+\left(ax^2-x-\frac{(a-1)^2}{2(2a-1)}\right)^2=0$$
has two roots in the upper half plane
$$ x_1=\frac{(1+2 i)-\sqrt{\frac{2 a^3-(4+8 i) a^2-(4-12 i) a+(3-4 i)}{2
a-1}}}{2 a},x_2=\frac{(1+2 i)+\sqrt{\frac{2 a^3-(4+8 i) a^2-(4-12 i) a+(3-4 i)}{2
a-1}}}{2 a}.$$
Thus after long calculation, one has
\begin{eqnarray}
&&\int_{-\infty}^{+\infty}\frac{(x-1)^2}{(2x-1)^2+\left(ax^2-x-\frac{(a-1)^2}{2(2a-1)}\right)^2}\mathrm dx\\
&=&2\pi i\bigg[\text{Res}_{x=x_1}\bigg(\frac{(x-1)^2}{(2x-1)^2+\left(ax^2-x-\frac{(a-1)^2}{2(2a-1)}\right)^2}\bigg)\\
&&+\text{Res}_{x=x_2}\bigg(\frac{(x-1)^2}{(2x-1)^2+\left(ax^2-x-\frac{(a-1)^2}{2(2a-1)}\right)^2}\bigg)\bigg]\\
&=&2\pi i\bigg[\frac{(x_1^2-1)^2}{4(2x_1-1)+2\left(ax_1^2-x_1-\frac{(a-1)^2}{2(2a-1)}\right)(2ax_1-1)}\\
&&+\frac{(x_2^2-1)^2}{4(2x_2-1)+2\left(ax_2^2-x_2-\frac{(a-1)^2}{2(2a-1)}\right)(2ax_2-1)}\bigg]\\
&=&2\pi i(-\frac{i}{2})\\
&=&\pi.
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3250011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to calculate limit as $n$ tends to infinity of $\frac{(n+1)^{n^2+n+1}}{n! (n+2)^{n^2+1}}$? This question stems from and old revision of this question, in which an upper bound for $n!$ was asked for.
The original bound was incorrect. In fact, I want to show that the given expression divided by $n!$ goes to $0$ as $n$ tends to $\infty$.
I thus want to show:
$$\lim_{n\to\infty}\frac{(n+1)^{n^2+n+1}}{n!(n+2)^{n^2+1}}=0.$$
Using Stirling's approximation, I found that this is equivalent to showing that
$$\lim_{n\to\infty} \frac{\exp(n)}{\sqrt n}\cdot\left(\frac{n+1}{n+2}\right)^{n^2+1}\cdot\left(\frac{n+1}{n}\right)^n=0.$$
However, I don't see how to prove the latter equation.
EDIT: It would already be enough to determine the limit of $$\exp(n)\left(\frac{n+1}{n+2}\right)^{(n^2)}\left(\frac{n+1}{n}\right)^n$$ as $n$ goes to $\infty$.
| The easy part first: We have by basic analysis (where $x\in\Bbb R$)
\begin{equation}\tag 1\label 1\lim_{x\to\infty} \left(\frac{x+1}x\right)^x=\lim_{x\to\infty}(1+1/x)^x=e.\end{equation}
Now comes the harder part:
Note that
\begin{equation}\label 2\tag 2
\lim_{x\to\infty} e^x \left(\frac{x+1}{x+2}\right)^{(x^2)}
= \lim_{x\to\infty} \exp\left(x+x^2\ln\left({x+1\over x+2}\right)\right).
\end{equation}
We now have by Taylor expansion of $\ln(1-y)$ (for $x$ big enough):
\begin{align}\tag 3\label 3
x+x^2
\ln\left(1-\frac1{x+2}\right)
&=x-\sum_{k=1}^\infty\frac1k\frac{x^2}{(x+2)^k}
\\ &=x-\frac{x^2}{x+2}-\frac{x^2}{2(x^2+4x+4)}-\overbrace{\sum_{k=3}^\infty \frac{x^2}{k(x+2)^k}}^{\xrightarrow{x\to\infty} 0}.
\end{align}
The latter sum converges to $0$ since (for $x> 1$), $$\displaystyle\sum_{k=3}^\infty \frac{x^2}{k(x+2)^k}\le\sum_{k=3}^\infty \frac{x^2}{x^k}=\sum_{k=1}^\infty x^{-k}=\frac1x\frac{x}{x-1}=\frac1{x-1}.$$
Thus, by additivity of the limit,
\begin{align}\label 4\tag 4
\lim_{x\to\infty} x+x^2\ln(1-\frac1{x+2}) = \lim_{x\to\infty} \overbrace{x-\frac{x^2}{x+2}}^2-\overbrace{\frac{x^2}{2(x^2+4x+4)}}^\frac12=\frac32.
\end{align}
We can now use continuity of the exponential function and \eqref{2} to find that
\begin{align}\tag 5\label 5
\lim_{x\to\infty} e^x \left(\frac{x+1}{x+2}\right)^{(x^2)}
&= \exp\left(\lim_{x\to\infty} x+x^2\ln\left({x+1\over x+2}\right)\right) \\
&= e^{3/2}.&\eqref 4
\end{align}
We can thus finally assert, using multiplicity of the limit, that your limit equals $0$:
\begin{equation}
\bbox[5px,border:2px solid #C0A000]{
\lim_{x\to\infty} \color{orange}{\frac{x+1}{(x+2)\sqrt x}}
\color{blue}{e^x \left(\frac{x+1}{x+2}\right)^{(x^2)}}
\color{green}{\left(\frac{x+1}x\right)^x}
= \color{orange}0 \cdot \color{blue}{e^{3/2}} \cdot \color{green}e = 0.
}
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3250902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Asymptotic behavior of $a_{n+1}=\frac{a_n^2+1}{2}$ Define a sequence as follows:
$$a_0=0$$
$$a_{n+1}=\frac{a_n^2+1}{2}$$
I would like to know the asymptotic behavior of $a_n$. I already know (by roughly approximating $a_n$ with a differential equation) that
$$a_n\sim 1-\frac{2}{n}$$
as $n\to\infty$. However, my approximation is very crude. Can anyone find a couple more terms? I expect (from numerical data) that the next term is something like $\frac{\log(n)}{n^2}$.
In case anyone wants context for this problem, I am trying to find the optimal strategy for a game with the following rules:
There are $n$ offers of money whose amounts are hidden from you, and whose quantities are random (independently and uniformly distributed from $0$ to $1$). One at a time, the offers are shown to you, and as you view each offer, you may either accept it or reject it. Once you accept an offer, the game is over and you may accept no more offers.
It turns out that $a_n$ is the minimum value of the first offer for which you should accept that offer, in a game with $n+1$ offers. This is why I am interested in the asymptotic behavior of $a_n$.
| Write $a_n = 1 - \epsilon_n$ and notice that $(\epsilon_n)$ solves
$$ \epsilon_{n+1} = \epsilon_n - \frac{\epsilon_n^2}{2}.$$
For the purpose of future use, we allow $\epsilon_0$ to take any value in $(0, 1]$. This type of sequence is well-studied, and here is a method of extracting asymptotic forms up to certain order in a bootstrapping manner.
*
*Since $0 \leq \epsilon_n \leq 1$ and $(\epsilon_n)$ is monotone decreasing, we have $\epsilon_n \to 0$ as $n\to\infty$.
*We have $ \frac{1}{\epsilon_{n+1}} - \frac{1}{\epsilon_n} = \frac{1}{2-\epsilon_n} $. So by Stolz–Cesàro theorem (a.k.a. L'Hospital's theorem for sequence),
$$ \lim_{n\to\infty} \frac{1/\epsilon_n}{n} = \lim_{n\to\infty} \left(\frac{1}{\epsilon_{n+1}} - \frac{1}{\epsilon_n} \right) = 2.$$
This shows that
$$\epsilon_n = (1 + o(1))\frac{2}{n}. $$
*Pushing this idea further, we may utilize $ \frac{1}{\epsilon_{n+1}} - \frac{1}{\epsilon_n} = \frac{1}{2} + \frac{\epsilon_n}{2(2-\epsilon_n)} $. Again, by Stolz–Cesàro theorem,
$$ \lim_{n\to\infty} \frac{\frac{1}{\epsilon_n} - \frac{n}{2}}{\log n}
= \lim_{n\to\infty} \frac{\frac{1}{\epsilon_{n+1}} - \frac{1}{\epsilon_n} - \frac{1}{2}}{\log (n+1) - \log n}
= \frac{1}{2}, $$
and so, $\frac{1}{\epsilon_n} = \frac{n}{2} + \frac{1 + o(1)}{2}\log n$. This in turn implies
$$ \epsilon_n = \frac{2}{n} - (2 + o(1)) \frac{\log n}{n^2}. $$
*Finally, by writing
\begin{align*}
\frac{1}{\epsilon_{n+1}} - \frac{1}{\epsilon_n}
&= \frac{1}{2} + \frac{1}{2(n+1)} + \underbrace{\left( \frac{\epsilon_n}{2(2-\epsilon_n)} - \frac{1}{2(n+1)} \right)}_{=\mathcal{O}(n^{-2})}, \\
\end{align*}
we obtain
\begin{align*}\frac{1}{\epsilon_n}
&= \frac{1}{\epsilon_0} + \frac{n}{2} + \frac{H_n}{2} + \sum_{k=0}^{n-1} \left( \frac{\epsilon_k}{2(2-\epsilon_k)} - \frac{1}{2(k+1)} \right) \\
&= \frac{n}{2} + \frac{H_n}{2} + C\left(\epsilon_0\right) + \mathcal{O}_{\epsilon_0}\left(\frac{1}{n}\right). \tag{*}
\end{align*}
Here, $H_n = \sum_{k=1}^{n} \frac{1}{k}$ is the harmonic number and $C(\epsilon_0)$ is a constant depending only on $\epsilon_0$. Also, the implicit bound of $\mathcal{O}_{\epsilon_0}$ depends only on $\epsilon_0$. Moreover, $C(\cdot)$ admits a nice property. Let $f(x) = x-x^2/2$ so that $\epsilon_{n+1} = f(\epsilon_n)$. Then
$$ C(x) = \frac{1}{\epsilon_0} + \sum_{k=0}^{\infty} \left( \frac{f^{\circ k}(x)}{2(2-f^{\circ k}(x))} - \frac{1}{2(k+1)} \right), $$
where $f^{\circ k}$ denotes the $k$-fold function composition of $f$ together with $f^{\circ 0} = \mathrm{id}$. Then $\text{(*)}$ can be recast to
$$ \frac{1}{f^{\circ n}(x)} = \frac{n}{2} + \frac{H_n}{2} + C(x) + \mathcal{O}_{x}\left(\frac{1}{n}\right). $$
Then it follows that
\begin{align*}
0
&= \frac{1}{f^{\circ (n+1)}(x)} - \frac{1}{f^{\circ (n+1)}(x)} \\
&= \left( \frac{n+1}{2} + \frac{H_{n+1}}{2} + C(x) + \mathcal{O}\left(\frac{1}{n}\right) \right) - \left( \frac{n}{2} + \frac{H_n}{2} + C(f(x)) + \mathcal{O}\left(\frac{1}{n}\right) \right),
\end{align*}
and letting $n\to\infty$ gives
$$ C(f(x)) = C(x) + \frac{1}{2}. $$
Summarizing,
Let $(\epsilon_n)$ be a sequence which solves $\epsilon_{n+1} = f(\epsilon_n)$, where $\epsilon_0 \in (0, 1]$ and $f(x) = x - x^2/2$. Then there exists a function $C : (0, 1] \to \mathbb{R}$ such that
$$ \frac{1}{\epsilon_n} = \frac{n}{2} + \frac{H_n}{2} + C(x) + \mathcal{O}_x\left(\frac{1}{n}\right), $$
and so,
$$ \epsilon_n = \frac{2}{n} - \frac{2H_n}{n^2} - \frac{4C(x)}{n^2} + \mathcal{O}_x\left( \frac{\log^2 n}{n^3} \right). $$
Moreover, $C$ solves $C(f(x)) = C(x) + \frac{1}{2}$.
| {
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"url": "https://math.stackexchange.com/questions/3252295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
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Given $f(x)=ax^3-ax^2+bx+4$ Find the Value of $a+b$ Let $f(x)=ax^3-ax^2+bx+4$. If $f(x)$ divided by $x^2+1$ then the remainder is $0$. If $f(x)$ divided by $x-4$ then the remainder is $51$. What is the value of $a+b$?
From the problem I know that $f(4)=51$.
Using long division, I found that remainder of $\frac{ax^3-ax^2+bx+4}{x^2+1}$ is $a+b+x(b-a)$.
Then
$$a+b+x(b-a)=0$$
I can't proceed any further so I'm guessing the other factor of $f(x)$ is $ax+4$.
Then
$$f(x)=(ax+4)(x^2+1)=ax^3+4x^2+ax+4=ax^3-ax^2+bx+4$$
I found that $a=-4$ and $b=a=-4$. Then $f(x)=-4x^3+4x^2-4x+4$. But I doesn't satisfy $f(4)=51$
| One more way to see the conditions imposed on $f(x)$ are inconsistent/impossible:
Since $x^2 + 1$ divides $f(x)$ with remainder $0$, $f(x)$ factors as
$ax^3 - ax^2 + bx + 4 = (x^2 + 1)(cx + d) = cx^3 + dx^2 + cx + d; \tag 1$
comparing coefficients:
$a = c = -d, \tag 2$
$b = c, \tag 3$
$ d = 4; \tag 4$
thus,
$a = b = c = -4; \tag 5$
thus,
$cx + d = -4x + 4, \tag 6$
and
$f(x) = -4x^3 + 4x^2 - 4x + 4; \tag 7$
then clearly $f(4)$ is even, so
$f(4) \ne 51. \tag 8$
If we choose to ignore the condition
$f(4) = 51, \tag 9$
we may still salvage the inference
$a + b = -8. \tag{10}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3252433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Evaluating $\sum_{k=1}^{n} \frac{1}{k(k+1)}$ I have just started learning sums. I need to evaluate the following sum:
$$S_{n} = \sum_{k=1}^{n} \frac{1}{k(k+1)}$$
$$a_{1} = \frac{1}{2}, a_{2} = \frac{1}{6}, a_{3} = \frac{1}{12}, a_{4} = \frac{1}{20}, a_{5} = \frac{1}{30}, a_{6} = \frac{1}{42}, ...$$
I tried splitting it to partial fractions:
$$\frac{1}{k(k+1)} = \frac{A}{k} + \frac{B}{k+1}$$
$$1 = A(k+1) + Bk$$
$$1 = Ak + A + Bk$$
$$\begin{cases} A = 1 \\ 0 = A + B \quad \Rightarrow B = -1\end{cases}$$
$$\Longrightarrow S_{n} = \sum_{k=1}^{n} \Big( \frac{1}{k} - \frac{1}{k+1} \Big)$$
$$S_{n} = \sum_{k=1}^{n} \frac{1}{k} - \sum_{k=1}^{n} \frac{1}{k+1}$$
Wolframalpha gives answer $S_{n} = \frac{n}{n+1}$
Help's appreciated. P.S: Are the tags okay?
| $$S_{n} = \sum_{k=1}^{n} \frac{1}{k}-\frac{1}{k+1}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{n}+\frac{1}{n}-\frac{1}{n+1}=1-\frac{1}{n+1}$$
Check this: https://en.wikipedia.org/wiki/Telescoping_series
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Five different positive non-integer rational numbers such that increased by 1 the product of any two numbers is a square of some rational number Find at least one example of five rational numbers $x_1, \; x_2, ..., \; x_5$ such that
i) $x_k > 0$ for all $k=1,2,...,5$;
ii) $x_k$ is not an integer for all $k=1,2,...,5$;
iii) if $k \ne m$ then $x_k \ne x_m$ (for all $k=1,2,...,5; \; m=1,2,...,5$);
iv) if $k \ne m$ then $x_k x_m+1$ is a square of some rational number (for all $k=1,2,...,5; \; m=1,2,...,5$).
My work. I found infinitely many examples of four rational numbers. Let $a$ and $b$ are rational numbers. Let $$x_1=a \\ x_2=ab^2+2b \\ x_3=a(b+1)^2+2(b+1) \\ x_4=a \left( 2b(b+1)a+4b+2 \right)^2+2\left( 2b(b+1)a+4b+2 \right).$$ Then $$x_1x_2+1=\left( ab+1\right)^2 \\ x_1x_3+1=\left( a(b+1)+1\right)^2 \\ x_1x_4+1=\left(a \left( 2b(b+1)a+4b+2 \right)+1\right)^2 \\ x_2x_3+1=\left( ab(b+1)+2b+1\right)^2 \\ x_2x_4+1=\left( 2a^2b^2(b+1)+2ab(3b+1)+4b+1\right)^2 \\ x_3x_4+1=\left( 2a^2b(b+1)^2+2a(b+1)(3b+1)+6b+3\right)^2. $$
I also have one idea. Can see that $x_4=x_2 \left(2a(b+1)+2 \right)^2+2 \left(2a(b+1)+2 \right)$. Let $x_5= x_2 \left(2a(b+1)+2 \pm 1\right)^2+2 \left(2a(b+1)+2 \pm 1\right)$. Then $$x_4x_5+1=\left(x_2 \left(2a(b+1)+2 \right) \left(2a(b+1)+2 \pm 1\right)+2\left(2a(b+1)+2 \right) \pm 1 \right)^2 \\ x_2x_5+1=\left( x_2\left(2a(b+1)+2 \pm 1 \right)+1\right)^2.$$
It remains to find the rational numbers $a$ and $b$ such that the numbers $x_1x_5+1$ and $x_3x_5+1$ are squares of some rational numbers.
| See OEIS A192629 for
$$1,3,8,120,\frac {777480}{8288641}$$
and OEIS A030063 (which adds $0$ and deletes the fraction) for more information.
| {
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"source": "stackexchange",
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Find $\int_0^1 \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx$ I came across this integral while i was working on a tough series.
a friend was able to evaluate it giving:
$$\int_0^1 \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx=\frac{\pi^3}{16}\ln2-\frac{7\pi}{64}\zeta(3)-\frac{\pi^4}{96}+\frac1{768}\psi^{(3)}\left(\frac14\right)$$ using integral manipulation. other approaches are appreciated.
| solution by Kartick Betal.
\begin{align}
I&=\int_0^1 \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx=\int_0^\infty \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx-\underbrace{\int_1^\infty \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx}_{\displaystyle x\mapsto 1/x}\\
&=\int_0^\infty \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx-\int_0^1 \frac{x\ln^2x\left(\frac{\pi}{2}-\arctan x\right)}{1+x^2}\ dx\\
&=\int_0^\infty \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx-\frac{\pi}{2}\int_0^1 \frac{x\ln^2x}{1+x^2}\ dx+\int_0^1 \frac{x\ln^2x\arctan x}{1+x^2}\ dx\\
&\small{=\int_0^\infty \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx-\frac{\pi}{2}\cdot\frac3{16}\zeta(3)+\int_0^1 \left(\frac1x-\frac1{x(1+x^2)}\right)\ln^2x\arctan xdx}\\
&=\int_0^\infty \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx-\frac{3\pi}{32}\zeta(3)+\int_0^1 \frac{\ln^2x\arctan x}{x}\ dx-I\\
2I&=\int_0^\infty \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx-\frac{3\pi}{32}\zeta(3)+2\beta(4)\tag1\\
\end{align}
using $\ \displaystyle\arctan x=\int_0^1\frac{x}{1+x^2y^2}\ dy\ $, we get
\begin{align}
K&=\int_0^\infty \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx=\int_0^\infty \frac{\ln^2x}{x(1+x^2)}\left(\int_0^1\frac{x}{1+x^2y^2}\ dy\right)\ dx\\
&=\int_0^1\frac{1}{1-y^2}\left(\int_0^\infty\frac{\ln^2x}{1+x^2}\ dx-\int_0^\infty\frac{y^2\ln^2x}{1+x^2y^2}\ dx\right)\ dy\\
&=\int_0^1\frac{1}{1-y^2}\left(\frac{\pi^3}{8}-\frac{y\pi^3}{8}-\frac{y\pi\ln^2y}{2}\right)\ dy\\
&=\frac{\pi^3}{8}\int_0^1\frac{1-y}{1-y^2}\ dy-\frac{\pi}2\int_0^1\frac{y\ln^2y}{1-y^2}\ dy\\
&=\frac{\pi^3}{8}\int_0^1\frac{1}{1+y}\ dy-\frac{\pi}{16}\int_0^1\frac{\ln^2y}{1-y}\ dy\\
&=\frac{\pi^3}{8}\ln2-\frac{\pi}{8}\zeta(3)\tag{2}
\end{align}
plugging $(2)$ in $(1)$, we get
$$I=\frac{\pi^3}{16}\ln2-\frac{7\pi}{32}\zeta(3)+\beta(4)$$
plugging $\ \displaystyle\beta(4)=\frac1{768}\left(\psi^{(3)}\left(\frac14\right)-8\pi^4\right)\ $ from here, we get the closed form of $\ I$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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maximum and minimum value of $\frac{x^2+y^2}{x^2+xy+4y^2}$
If $x,y\in\mathbb{R}$ and $x^2+y^2>0.$ Then maximum and minimum value of $\displaystyle \frac{x^2+y^2}{x^2+xy+4y^2}$
Plan
Let $$K=\frac{x^2+y^2}{x^2+xy+4y^2}$$
$$Kx^2+Kxy+4Ky^2=x^2+y^2\Rightarrow (4K-1)y^2+Kxy+(K-1)x^2=0$$
put $y/x=t$ and equation is $(4K-1)t^2+Kt+(K-1)=0$
How do i solve it Help me plesse
| Let $v = (x,y)$,
$$
A = \begin{bmatrix} 1 & \frac{1}{2} \\ \frac{1}{2} & 4 \end{bmatrix}
$$
and define the Rayleigh cuocient.
$$
R_A(v) = \frac{v^T A v}{v^Tv}.
$$
Then, it is clear that
$$
\frac{x^2+y^2}{x^2+xy+4y^2} = \frac{1}{R_A(v)}.
$$
It is well-known (Linear Algebra result) that $\operatorname{min}_{v \neq 0} R_A(v) = \lambda_{\operatorname{min}}$ and $\operatorname{max}_{v \neq 0} R_A(v) = \lambda_{\operatorname{max}}$, where $\lambda_{\operatorname{min}}$ and $\lambda_{\operatorname{max}}$ are the smallest and largest eigenvalues of $A$, respectively.
In this case $\lambda_{\operatorname{min}} = \frac{1}{2} \left(5-\sqrt{10}\right)$ and $\lambda_{\operatorname{max}} = \frac{1}{2} \left(5+\sqrt{10}\right)$. Therefore,
$$
\min_{x,y\neq 0} \frac{x^2+y^2}{x^2+xy+4y^2} = \frac{1}{\max_{v\neq 0} R_A(v)} = \frac{1}{\frac{1}{2} \left(5+\sqrt{10}\right)} = \frac{2}{15}(5+\sqrt{10})
$$
and
$$
\max_{x,y\neq 0} \frac{x^2+y^2}{x^2+xy+4y^2} = \frac{1}{\min_{v\neq 0} R_A(v)} = \frac{1}{\frac{1}{2} \left(5-\sqrt{10}\right)} = \frac{2}{15}(5-\sqrt{10}).
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Calculate local maximum and minimum of $f(x)=(2+\sin x)(5-\sin x)$ I am working on my scholarship exam practice which assumes high school or pre-university math knowledge. Could you please have a look on my approach?
The minimum of the function $f(x)=(2+\sin x)(5-\sin x)$ is ......
First, I began with some basic approach.
We know that $-1\leq\sin x\leq1$, so I just test both values $-1$ and $1$ in the function.
$f(-1)=(2+(-1))(5-(-1))=6$
$f(1)=(2+1)(5-1)=12$
Since $-1$ is the minimum of function $\sin x$, I conclude that $f(-1)=6$, which is correct when checked with the value of answer key provided. Please let me know if my approach is not always true or can apply on other similar problems.
But in exam I may be uncertain if my answer is right so I tried calculus approach to check my answer.
$f(x)=(2+\sin x)(5-\sin x)=10+3\sin x-\sin^2 x$
$f'(x)=3\cos x-2\sin x\cos x=0$
$2\sin x\cos x=3\cos x$
$2\sin x=3$
$\sin x=\frac{3}{2}=1.5>1$
Since 1.5 exceeds the range of $\sin x$, then I cannot use this approach. I am wondering why this is the case. Why can't we use this method to find the minimum?
| You cannot divide by $\cos x$ since it may be equal to zero.
It is true that
$$f(x) = (2 + \sin x)(5 - \sin x) = 10 + 3\sin x - \sin^2x$$
Therefore,
$$f'(x) = 3\cos x - 2\sin x\cos x$$
Setting the derivative equal to zero yields
\begin{align*}
f'(x) & = 0\\
3\cos x - 2\sin x\cos x & = 0\\
\cos x(3 - 2\sin x) & = 0
\end{align*}
Hence,
\begin{align*}
\cos x & = 0 & 3 - 2\sin x & = 0\\
x & = \frac{\pi}{2} + n\pi, n \in \mathbb{Z} & -2\sin x & = -3\\
& & \sin x & = \frac{3}{2}
\end{align*}
The equation $\sin x = \frac{3}{2}$ has no real-valued solutions. Notice that we can express
$$x = \frac{\pi}{2} + n\pi, n \in \mathbb{Z}$$
as
$$x = \begin{cases}
\dfrac{\pi}{2} + 2k\pi, k \in \mathbb{Z}\\
-\dfrac{\pi}{2} + 2m\pi, m \in \mathbb{Z}
\end{cases}
$$
which will be useful since the original function has sines and the value sine takes at $\frac{\pi}{2}$ is different from the value it takes at $-\frac{\pi}{2}$.
Notice that since $3 - 2\sin x > 0$ for each $x \in \mathbb{R}$, the sign of the derivative is determined by $\cos x$, which is positive in the intervals $$(-\frac{\pi}{2} + 2n\pi, \frac{\pi}{2} + 2n\pi), n \in \mathbb{Z}$$ and negative in the interval $$(\frac{\pi}{2} + 2n\pi, \frac{3\pi}{2} + 2n\pi), n \in \mathbb{Z}$$ By the First Derivative Test, $f$ has relative maxima at $$x = \frac{\pi}{2} + 2n\pi$$ and relative minima at $$x = -\frac{\pi}{2} + 2\pi$$
If $x = \frac{\pi}{2} + 2k\pi, k \in \mathbb{Z}$, $\sin x = 1$. Thus, the relative maximum value $f$ assumes is
$$f(\frac{\pi}{2} + 2k\pi) = f\left(\frac{\pi}{2}\right) = (2 + 1)(5 - 1) = 3 \cdot 4 =12$$
If $x = -\frac{\pi}{2} + 2m\pi, m \in \mathbb{Z}$, $\sin x = -1$. Thus, the relative minimum value $f$ assumes is
$$f(-\frac{\pi}{2} + 2k\pi) = f\left(\frac{3\pi}{2}\right) = (2 - 1)(5 + 1) = 1 \cdot 6 = 6$$
which should look familiar.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
If $\tan x=3$, then what is the value of ${3\cos{2x}-2\sin{2x}\over4\sin{2x}+5\cos{2x}}$?
If $\tan x=3$, then what is the value of
$${3\cos{2x}-2\sin{2x}\over4\sin{2x}+5\cos{2x}}$$
So what I did is change all the $\sin{2x}$ and $cos{2x}$ with double angle formulas, getting
$${3\cos^2{x}-3\sin^2{x}-4\sin{x}\cos{x}\over5\cos^2{x}-5\sin^2{x}+8\sin{x}\cos{x}}$$
Now I thought of changing the top part to $\sin{x}$ and bottom part to $\cos{x}$ hoping to somehow get $\tan{x}$ in this way, but I ultimately got just
$${3-6\sin^2{x}-4\sin{x}\cos{x}\over-5+10\cos^2{x}+8\sin{x}\cos{x}}$$
Had really no ideas what to either do after this, seems pretty unusable to me. Was there possibly a mistake I made in the transformation or maybe another way of solving this?
| $$\tan2x=\dfrac{2\tan x}{1-\tan^2x}=-\dfrac34$$ then
$${3\cos{2x}-2\sin{2x}\over4\sin{2x}+5\cos{2x}}=\dfrac{3-2\tan 2x}{4\tan2x+5}=\dfrac{9}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3273674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Radical equation solve $\sqrt{3x+7}-\sqrt{x+2}=1$. Cannot arrive at solution $x=-2$ I am to solve $\sqrt{3x+7}-\sqrt{x+2}=1$ and the solution is provided as -2.
Since this is a radical equation with 2 radicals, I followed suggested textbook steps of isolating each radical and squaring:
$\sqrt{3x+7}-\sqrt{x+2}=1$
$(3x+7=(1-\sqrt{x+2})^2$ # square both sides
(Use perfect square formula on right hand side $a^2-2ab+b^2$)
$3x+7=1^2-2(1)(-\sqrt{x+2})+x+2$ # lhs radical is removed, rhs use perfect square formula
$3x+7=1+2(\sqrt{x+2})+x+2$ # simplify
$3x+7=x+3+2\sqrt{x+2}$ # keep simplifying
$2x+4=2\sqrt{x+2}$ # simplify across both sides
$(2x+4)^2=(2\sqrt{x+2})^2$
$4x^2+16x+16=4(x+2)$ # now that radical on rhs is isolated, square both sides again
$4x^2+12x+14=0$ # a quadratic formula I can use to solve for x
For use int he quadratic function, my parameters are: a=4, b=12 and c=14:
$x=\frac{-12\pm\sqrt{12^2-(4)(4)(14)}}{2(4)}$
$x=\frac{-12\pm{\sqrt{(144-224)}}}{8}$
$x=\frac{-12\pm{\sqrt{-80}}}{8}$
$x=\frac{-12\pm{i\sqrt{16}*i\sqrt{5}}}{8}$
$x=\frac{-12\pm{4i*i\sqrt{5}}}{8}$
$x=\frac{-12\pm{-4\sqrt{5}}}{8}$ #since $4i*i\sqrt{5}$ and i^2 is -1
This is as far as I get:
$\frac{-12}{8}\pm\frac{4\sqrt{5}}{8}$
I must have gone of course somewhere further up since the solution is provided as x=-2.
How can I arrive at -2?
| There are actually two solutions: $x = -1; x = -2$ when you continue with the ``Can you finish?'' step of Dr. S.
| {
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"question_score": "3",
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Simplifying $\sqrt\frac{\left(a^2\cos^2t+b^2\sin^2t\right)^3}{\left(b^2\cos^2t+a^2\sin^2t\right)^3}$ I am looking to simplify these term [ I forgot the 3 :( ]
$$\sqrt\frac{\left(a^2\cos^2t+b^2\sin^2t\right)^3}{\left(b^2\cos^2t+a^2\sin^2t\right)^3}$$
where $a$ and $b$ are two non-negative reals.
(This is not homework. I am just trying to make my expression easy, but I didn't find a way.)
Thanks for your help.
| Divide both the numerator and denominator by $a^2+b^2$, then use $\frac{a^2}{a^2+b^2}=\sin^2 u$. Your expression will become $$\sqrt{\frac{\sin^2 u\cos^2 t+\cos^2u\sin^2t}{\cos^2 u\cos^2 t+\sin^2u\sin^2t}}$$
Now use $\sin^2u+\cos^2 u=1$
$$\sqrt{\frac{\sin^2 u\cos^2 t+\cos^2u\sin^2t}{\cos^2 u\cos^2 t+\sin^2u\sin^2t}}=\sqrt{\frac{(1-\cos^2 u)\cos^2 t+(1-\sin^2u)\sin^2t}{\cos^2 u\cos^2 t+\sin^2u\sin^2t}}\\=\sqrt{\frac{(\cos^2 t+\sin^2 t)-(\cos^2 u\cos^2 t+\sin^2u\sin^2t)}{\cos^2 u\cos^2 t+\sin^2u\sin^2t}}\\=\sqrt{\frac{1}{\cos^2 u\cos^2 t+\sin^2u\sin^2t}-1}$$
| {
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"url": "https://math.stackexchange.com/questions/3277124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Problems with functions. I have been working on the following problem relating to functions:
"Consider a line $y=kx+b$ ($k<0$, $b>0$) which is tangent to the parabola $y=x^2−4x+4$.
Find the maximal value of the region bounded by the lines $y=kx+b$, $y=0$ and $x=0$".
The greatest region I found that followed such restrictions had an area of $2.25$. It was a right triangle with sides $3$ and $1.5$. However, I am uncertain whether this is the greatest value. Can anyone confirm?
| Let the parabola be $f(x) = x^2 -4x + 4$.
Let's let the point of the tangent point of the parabola be $(x, y)$
The slope of the tangent line is $k = f'(x) = 2x - 4$.
Given a right triangle made be the lines $y=kx + b$ and $y = 0$ and $x=0$ the base of the triangle will be from $(0,0)$ to the $x$-intercept of the line $y=kx+b$ which is $-\frac bk$ (assuming $k\ne 0$[*]). The height of the triangle will be from $(0,0)$ to the $y$-intecept of the line $y=kx+b$ which is $b$.
So the Area is $\frac 12 (-\frac bk)b= -\frac {b^2}{2k}$. (Note a negative value will just mean that the triangle is under the $x$ axis or to the left of the $y$ axis.)
If $(x,y)$ is the point of tangency then $(x,y)$ is on the line and on the parabola so $y = kx + b = x^2 -4x + 4 = (x-2)^2$.
The slope of the tangent line is $y' = k = 2x -4 = 2(x-2)$.
Thus $y = 2(x-2)x + b = (x-2)^2$ and we get $b = (x-2)^2 - 2(x-2)x = (x-2)(x-2 - 2x)= -(x-2)(x+2)$ or if you prefer the long way: $y = 2x^2-4x + b = x^2 -4x + 4$ so $b=-x^2 +4$.
So for that tangent point we have the area of the triangle is:
$A(x) = -\frac {b^2}{2k} = -\frac {(x-2)^2(x+2)^2}{4(x-2)} = -\frac 14(x-2)(x+2)^2$.
([*] Note: that the factor of $x-2$ in the denominator factored out. At $x = 2$ it's interesting to note that the slope of the tangent line is $0$-- This is the cusp of the parabola. But it doesn't affect the are area of our triangle as it is a removable singularity. If the parabola didn't have a cusp at $(x, 0)$ then the "triangle" would be an open "slice" and have infinite area. As it is at $x=2$ the area is $0$ because the tangent line is $x=0$ and our "triangle" is a line segment.)
To find the max/min of this area set $A'(x) = 0$ and solve.
$A'(x) = -\frac 14[2(x-2)(x+2) + (x+2)^2]=-\frac 14(x+2)[2(x-2) + (x+2)] = -\frac 14(x+2)(3x -2) = 0$
So either $x+2 = 0$ and $x = -2$ or $3x -2 = 0$ and $x =\frac 23$.
$x = -2$ gives as an area of $A(-2)= -\frac 14(x-2)(x+2)^2 = 0$.
(This is is because at $x = -2$ the tangent line $y = kx + b = y = 2(-2-2)x + (-(-2)^2 + 4) = 8x$ passes through the origin and the "triangle" is just a point. This is a minimum.)
(Worth noting for $x < -2$ the triangle will occur to the left of the $y$-axis and below the $x$ axis for n positive area.[**])
$x = \frac 23$ gives us an area of $A(\frac 23) = -\frac 14(-\frac 43)(\frac 83)^2 = \frac {64}{27}$.
That's the maximum.
([*]It's worth noting that at $x = 2$ we get an area of $A(x) =0$. If $x = 2$ then $y = 0$ and this is the cusp of the parabola. This is not a minimum because as $x > 2$ we will have a positive slope and the triangle will be below the $x$-axis and the area will take on negative values.)
([**] Actually there is no global maximum; just a localmaximum at $x = \frac 23$. The triangles resulting from $x < -2$ will result in increasingly large triangles in the fourth quadrant. There is probably something in the text that rules out us having triangles in different quadrants than te parabola. Or maybe you were told $x$ had to be positive.)
I'm embarrassed to admit my not getting $x =2$ as a minimal which I knew had to have area $0$ and not getting why I got the area of $0$ at $x=-2$ instead when it didn't seem to make sense that would have a triangle of $0$ area, took me a long time. We never outgrow making mistakes.
| {
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Evaluate $\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x^2}dx$
How to prove $$\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x^2}\ dx=\text{Im}\left(\operatorname{Li}_3(1+i)\right)-\frac{\pi^3}{32}-G\ln2 \ ?$$
where $\operatorname{Li}_3(x)=\sum\limits_{n=1}^\infty\frac{x^n}{n^3}$ is the trilogarithm and $G=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^2}$ is Catalan's constant
Trying the algebraic identity $\ 4ab=(a+b)^2-(a-b)^2\ $ where $\ a=\ln(1-x)$ and $b=\ln(1+x)\ $is not helpful here and the integral will be more complicated.
Also, applying IBP or substituting $x=\frac{1-y}{1+y}$ is not that useful either.
All approaches are appreciated.
| lets start with $\displaystyle\int_0^\infty\frac{\ln^2(1+x)}{1+x^2}\ dx=2\Im\operatorname{Li}_3(1+i)\quad$ (proved here)
\begin{align}
\int_0^\infty\frac{\ln^2(1+x)}{1+x^2}\ dx&=\int_0^1\frac{\ln^2(1+x)}{1+x^2}\ d+\underbrace{\int_1^\infty\frac{\ln^2(1+x)}{1+x^2}\ dx}_{\small\displaystyle x\mapsto1/x}\\
2\Im\operatorname{Li}_3(1+i)&=2\int_0^1\frac{\ln^2(1+x)}{1+x^2}\ dx-2\int_0^1\frac{\ln x\ln(1+x)}{1+x^2}\ dx+\underbrace{\int_0^1\frac{\ln^2x}{1+x^2}\ dx}_{2\beta(3)}
\end{align}
then
$$\int_0^1\frac{\ln^2(1+x)-\ln x\ln(1+x)}{1+x^2}\ dx=\Im\operatorname{Li}_3(1+i)-\beta(3)\tag{1}$$
now lets start with $\ I=\displaystyle\int_0^1\frac{\ln x\ln(1-x)}{1+x^2}\ dx$ and by setting $x=\frac{1-y}{1+y}$, we get
$$I=\displaystyle\int_0^1\frac{\ln^2(1+x)-\ln x\ln(1+x)}{1+x^2}-\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x^2}\ dx+\ln2\underbrace{\int_0^1\frac{\ln\left(\frac{1-x}{1+x}\right)}{1+x^2}\ dx}_{x=(1-y)/(1+y)}+I$$
then
\begin{align}
\int_0^1\frac{\ln^2(1+x)-\ln x\ln(1+x)}{1+x^2}=\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x^2}\ dx-\ln2\underbrace{\int_0^1\frac{\ln x}{1+x^2}}_{-G}\tag{2}
\end{align}
from $(1)$ and $(2)$ and substituting $\displaystyle\beta(3)=\frac{\pi^3}{32}\ $, the result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3278573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $S=\sum_{n=2}^\infty\frac{_nC_2}{(n+1)!}.$
Prove that $$S=\sum_{n=2}^\infty\frac{_nC_2}{(n+1)!}=\frac{e}{2}-1.$$
$$
S=\sum_{n=2}^\infty\frac{_nC_2}{(n+1)!}=\sum_{n=2}^\infty\frac{n!}{2(n-2)!(n+1)!}=\sum_{n=2}^\infty\frac{1}{2(n+1)(n-2)!}\\
=\frac{1}{2}\bigg[\frac{1}{3.0!}+\frac{1}{4.1!}+\frac{1}{5.2!}+\frac{1}{6.3!}+\dots\bigg]=\frac{1}{2}\bigg[\Big(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\dots\Big)-\Big(\frac{2}{3.0!}+\frac{5}{4.1!}+\dots\Big)\bigg]\\=\frac{e}{2}-\frac{1}{2}\Big(\frac{2}{3.0!}+\frac{5}{4.1!}+\dots\Big)
$$
How do I proceed further as I am stuck with the last infinite series.
| $$S=\sum_{n=2}^\infty\frac{1}{2(n+1)(n-2)!}=\sum_{n=0}^\infty\frac{1}{2(n+3)(n)!}$$
$$=\frac{1}{2}\sum_{n=0}^\infty\int_{0}^{1}\frac{x^{n+2}}{n!}dx=\frac{1}{2}\int_{0}^{1}{x^2e^{x}}dx=\frac{e}{2}-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3280661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate $\int_0^1\arctan x\left(\frac{3\ln(1+x^2)}{1+x}-\frac{2\ln(1+x)}{x}\right)\ dx$ Prove that
$$\int_0^1\arctan x\left(\frac{3\ln(1+x^2)}{1+x}-\frac{2\ln(1+x)}{x}\right)\ dx=\frac{3\pi}{8}\ln^22-\frac{\pi^3}{32}$$
I managed to prove the above equality using integral manipulation (solution be posted soon) , but is it possible to do it in different ways and specifically by harmonic series?
The interesting thing about this problem is that we don't see any imaginary part which is usually involved in such integrals.
Note: the second integral $$\int_0^1 \frac{\arctan x\ln(1+x)}{x}\ dx=\frac{3\pi^3}{32}+\frac{3\pi}{16}\ln^22+\frac32G\ln2+3\text{Im}\operatorname{Li}_3(1-i)$$ was already evaluated here but I calculated the original problem without separating the two integrals.
| @Kemono Chen elegantly proved here
$$\int_0^y\frac{\ln(1+yx)}{1+x^2}dx=\frac12 \tan^{-1}(y)\ln(1+y^2)$$
By integration by parts we have
$$\int_0^y\frac{y\tan^{-1}(x)}{1+yx}dx=\frac12 \tan^{-1}(y)\ln(1+y^2)$$
divide both sides by $1+y$ and integrate from $y=0$ to $y=1$, we get,
\begin{align}
\frac12I&=\frac12\int_0^1\frac{\arctan y\ln(1+y^2)}{1+y}\ dy=\int_0^1\int_0^y\frac{y\arctan x}{(1-yx)(1+y)}\ dy\ dx\\
&=\int_0^1\arctan x\left(\int_x^1\frac{y}{(1-yx)(1+y)}\ dy\right)\ dx\\
&=\int_0^1\arctan x\left(\frac{\ln(1+x)}{x}-\frac{\ln(1+x^2)}{x}+\frac{2\ln(1+x)-\ln(1+x^2)-\ln2}{1-x}\right)\ dx\\
&=\underbrace{\int_0^1\frac{\arctan x\ln(1+x)}{x}\ dx}_{J}-\underbrace{\int_0^1\frac{\arctan x\ln(1+x^2)}{x}\ dx}_{K}\\
&\quad+\underbrace{\int_0^1\frac{\arctan x}{1-x}\left(2\ln(1+x)-\ln(1+x^2)-\ln2\right)\ dx}_{\large{x=(1-y)/(1+y)}}\\
&=J-K-\int_0^1\frac{\left(\frac{\pi}{4}-\arctan x\right)}{x(1+x)}\ln(1+x^2)\ dx\\
&=J-K-\frac{\pi}{4}\int_0^1\frac{\ln(1+x^2)}{x(1+x)}\ dx+K-I\\
\frac32I&=J-\frac{\pi}{4}\left(\frac{3\pi^2}{48}-\frac34\ln^22\right)\\
3I-2J&=\frac{3\pi}{8}\ln^22-\frac{\pi^3}{32}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3282123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Finding zero divisors in a polynomial quotient ring
Is $x^2+x+I$ a zero divisor in $\mathbb{Z}_7[x]/I$, where $I=(x^3+5x^2+2x+5)$?
I know that $\gcd(x^3+5x^2+2x+5,x^2+x)=x+1$, and that it means that $x^2+x$ is indeed a zero divisor.
What I struggle with is finding $g(x)$ such that $$(f(x)+\langle x^3+5x^2+2x+5\rangle)(g(x)+\langle x^3+5x^2+2x+5\rangle)=\langle x^3+5x^2+2x+5\rangle$$ in the quotient ring.
I know how to find an inverse, is this any similar? How is it done?
| Hint:
Note that $6$ is a root of given polynomial. That is, in $\Bbb Z_7$, $$x^3+5x^2+2x+5=(x-6)(x^2+4x+5)=(x+1)(x^2+4x+5)$$ So take $f(x)=x+1$ and $g(x)=x^2+4x+5$ and let $I=\langle x^3+5x^2+2x+5 \rangle$. Then $f+I \neq I$ and $g+I \neq I$ but $$(f+I)(g+I)=fg+I=I$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3283085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Prove that $(abc+xyz) \left(\frac{1}{ay}+\frac{1}{bz}+\frac{1}{cx}\right)\geq3$ Let $a,b,c,x,y,z\in\mathbb{R}_+$ such that $a+x=b+y=c+z=1.$ Prove the inequality
$$(abc+xyz)\left(\frac1{ay}+\frac1{bz}+\frac1{cx}\right)\geq3$$
I tried using AM-HM to get
$$(abc+xyz)\left(\frac1{ay}+\frac1{bz}+\frac1{cx}\right)\geq9\frac{abc+xyz}{ay+bz+cx}$$
and wrote it as $$\frac{abc+xyz}{ay+bz+cx}=\frac{1-(a+b+c)+(ab+bc+ca)}{(a+b+c)-(ab+bc+ca)}=\frac1{a+b+c-ab-bc-ca}-1$$That is, I have to prove that
$$a+b+c-ab-bc-ca\leq\frac34$$
I tried using Cauchy-Schwarz after this as $$(a(1-b)+b(1-c)+c(1-a))\leq\sqrt{(a^2+b^2+c^2)((1-a)^2+(1-b)^2+(1-c)^2)}$$
but didn't get any idea to simplify it further.
Any help would be appreciated!
| By AM-GM
$$(abc+xyz)\sum_{cyc}\frac{1}{ay}=\sum_{cyc}\left(\frac{zx}{a}+\frac{bc}{y}\right)=\sum_{cyc}\left(\frac{xy}{b}+\frac{ab}{x}\right)=$$
$$=\sum_{cyc}\left(\frac{x(1-b)}{b}+\frac{(1-x)b}{x}\right)=\sum_{cyc}\left(\frac{x}{b}+\frac{b}{x}-x-b\right)=$$
$$=\sum_{cyc}\left(\frac{x}{b}+\frac{b}{x}\right)-3\geq\sum_{cyc}2-3=3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3283687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $N \in \mathbb{N}$ so for $n \geq N$: $ \frac{n}{n-1} \frac{\log(n)}{\sqrt{2^n}} \leq \frac{4}{25}\frac{\log(n)^2}{n^{1/5}}. $ Show that there exists some $N \in \mathbb{N}$ so that for all $n \geq N$:
$$
\frac{n}{n-1} \frac{\log(n)}{\sqrt{2^n}} \leq \frac{4}{25}\frac{\log(n)^2}{n^{1/5}}.
$$
Numerically I know that the statement is true for $N = 6$ but if you can show it for some $N \geq 6$, that would be helpful too.
I used that $\frac{n}{n-1} \leq \frac{6}{5}$ for all $n \geq 6$ and that $\sqrt{2^n} \geq n^{1/5}$ to get
$$
\frac{n}{n-1} \frac{\log(n)}{\sqrt{2^n}} \leq \frac{6}{5}\frac{\log (n)}{n^{1/5}}
$$
but I don't know how to proceed from here. I think maybe I've been too rough in my estimation.
Any ideas?
| We have
$$\frac{n}{n-1} \frac{\log(n)}{\sqrt{2^n}} \frac{25}4\frac{n^{1/5}}{\log(n)^2}
\sim\frac{25}4\frac{n^{1/5}}{\log(n)}2^{-n/2}\xrightarrow{n\to\infty}0$$
hence, by definition of limit, there exists $N\in\Bbb N$ such that
$$\frac{n}{n-1} \frac{\log(n)}{\sqrt{2^n}} \frac{25}4\frac{n^{1/5}}{\log(n)^2}\leq 1$$
for every $n\geq N$.
To prove that we can choose $N=6$ consider for $x\geq 6$:
\begin{align}
0\leq\frac{x}{x-1} \frac{\log(x)}{\sqrt{2^x}} \frac{25}4\frac{x^{1/5}}{\log(x)^2}
&\leq \frac{6}{5} \frac{\log(x)}{\sqrt{2^x}} \frac{25}4\frac{x^{1/5}}{\log(x)^2}\\
&=\frac{15}{2}\frac{1}{\sqrt{2^x}}\frac{x^{1/5}}{\log(x)}\\
&=f(x)
\end{align}
do that
\begin{align}
\frac{f'(x)}{f(x)}
&=\frac 1{5x}-\frac 1{x\log(x)}-\frac 12\log(2)\\
&\leq\frac 1{30}-\frac 12\log(2)<0
\end{align}
so that $f'(x)<0$ for $x\geq 6$, hence $f(x)\leq f(6)\leq 1$ for $x\geq 6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3283917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
On the determinant of a Toeplitz-Hessenberg matrix I am having trouble proving that
$$\det
\begin{pmatrix}
\dfrac{1}{1!} & 1 & 0 & 0 & \cdots & 0 \\
\dfrac{1}{2!} & \dfrac{1}{1!} & 1 & 0 & \cdots & 0 \\
\dfrac{1}{3!} & \dfrac{1}{2!} & \dfrac{1}{1!} & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \ddots & \vdots \\
\dfrac{1}{(n-1)!} & \dfrac{1}{(n-2)!} & \dfrac{1}{(n-3)!} & \cdots & \dfrac{1}{1!} &1\\
\dfrac{1}{n!} & \dfrac{1}{(n-1)!} & \dfrac{1}{(n-2)!} & \dfrac{1}{(n-3)!} & \cdots & \dfrac{1}{1!}
\end{pmatrix}
=\dfrac{1}{n!}.
$$
| Hint. In general, let $d_0=d_1=1$ and let $(a_k)_{k=1,2,\ldots}$ be any sequence of numbers. For every $n\ge2$, denote by $d_n$ the determinant of the $n\times n$ Toeplitz-Hessenberg matrix
$$
\begin{pmatrix}
a_1 &1 &0 &0 &\cdots &0\\
a_2 &a_1 &1 &0 &\cdots &0\\
a_3 &a_2 &a_1 &1 &\cdots &0\\
\vdots &\vdots &\vdots &\ddots &\ddots &\vdots\\
a_{n-1} &a_{n-2} &a_{n-3} &\cdots &a_1 &1\\
a_n &a_{n-1} &a_{n-2} &a_{n-3} &\cdots &a_1
\end{pmatrix}.
$$
If one expands the determinant by the first column, one obtains
$$
d_n=-\sum_{k=1}^n(-1)^ka_kd_{n-k}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3288246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Depression of a car
A person on the summit of mountain observes that the angles of depression of a car moving on a straight road at three consecutive mile stones are x, y and z respectively. Prove that the height of the mountain is $\sqrt{\frac{2}{\cot^2x-2\cot^2y+\cot^2z}}$
I am getting the height as $\frac{1}{\cot y-\cot z}$ or as $\frac{1}{\cot x - \cot y}$. Not getting the desired expression.
|
Let the distance between the foot of the mountain and the first mile stone be $a$ . Then the distances between the foot of the mountain and the consecutive milestones are $a+1$ and $a+2$.
Now $\cot x = \frac{a}{h} \ , \cot y = \frac{a+1}{h} \ , \cot z = \frac{a+2}{h}$.
$$\frac{2}{\cot^2x - 2\cot^2y + \cot^2z} = \frac{2}{\frac{a^2}{h^2}-2\frac{a^2+2a+1}{h^2}+\frac{a^2+4a+4}{h^2}} = \frac{2h^2}{a^2-2a^2-4a-2+a^2+4a+4} = \frac{2h^2}{2}$$
Thus,
$$\sqrt{\frac{2}{\cot^2x - 2\cot^2y + \cot^2z}} = h$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3289036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
The limit of $(1+x^2+y^2)^{1\over x^2 + y^2 +xy^2}$ as $(x,y)$ approaches $(0.0)$ Im having trouble solving the following limit problem:
Whats the limit of $(1+x^2+y^2)^{1\over x^2 + y^2 +xy^2}$ as $(x,y)$ approaches $(0.0)$.
I know that the first step is to change $(x,y)$ to polar coordinates. However after I've done that and simplified the expression I'm left with the following:
$$
(1+r^2)^{1\over r^2+r^2\cos(α) \sin^2(α)}.
$$
How do i prove that the expression goes to $e$ when $r→0$?
Thanks in advance!
| Let $\epsilon>0$. Then for $|x|<\epsilon$:
$$
(1+x^2+y^2)^{\frac{1}{x^2+y^2+\epsilon y^2+\epsilon x^2}} \leq
(1+x^2+y^2)^{\frac{1}{x^2+y^2+x y^2}} \leq
(1+x^2+y^2)^{\frac{1}{x^2+y^2-\epsilon y^2-\epsilon y^x}}.
$$
Since
$$
(1+x^2+y^2)^{\frac{1}{x^2+y^2-\epsilon y^2-\epsilon y^x}}=
\Big((1+x^2+y^2)^{\frac{1}{x^2+y^2}}\Big)^{\frac{1}{(1-\epsilon)}}
\rightarrow e^{\frac{1}{(1-\epsilon)}} \text{ as }(x,y)\rightarrow 0
$$
and
$$
(1+x^2+y^2)^{\frac{1}{x^2+y^2+\epsilon y^2+\epsilon y^x}}=
\Big((1+x^2+y^2)^{\frac{1}{x^2+y^2}}\Big)^{\frac{1}{(1+\epsilon)}}
\rightarrow e^{\frac{1}{(1+\epsilon)}} \text{ as }(x,y)\rightarrow 0
$$
it follows that
$$ e^{\frac{1}{(1+\epsilon)}} \leq \lim_{(x,y)\rightarrow 0} (1+x^2+y^2)^{\frac{1}{x^2+y^2+x y^2}} \leq e^{\frac{1}{(1-\epsilon)}}.$$
The above is true for all $\epsilon>0$. Therefore we must have
$$ \lim_{(x,y)\rightarrow 0} (1+x^2+y^2)^{\frac{1}{x^2+y^2+x y^2}} = e^1 = e.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3291695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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minimum value of of $(5+x)(5+y)$
If $x^2+xy+y^2=3$ and $x,y\in \mathbb{R}.$
Then find the minimum value of $(5+x)(5+y)$
What I try
$$(5+x)(5+y)=25+5(x+y)+xy$$
$x^2+xy+y^2=3\Rightarrow (x+y)^2-3=xy$
I am finding $f(x,y)=22+5(x+y)+(x+y)^2$
How do I solve it? Help me please
| Using Rotation of axes
if $t$ is the rotation of axes, $\cot2t=0\iff\cos2t=0$
Set $2t=\dfrac\pi2$
$\sqrt2u=x+y,\sqrt2v=y-x$
$$3=\dfrac{(u-v)^2+(u+v)^2}2=u^2+v^2$$
WLOG $u=\sqrt3\cos t, v=\sqrt3\sin t$
$$(5+x)(5+y)=25+\sqrt2u+\dfrac{2(u^2-v^2)}4=\dfrac{50+2\sqrt2u+u^2-v^2}2$$
$2\sqrt2u+u^2-v^2=2\sqrt2(\sqrt3\cos t)+3(\cos^2t-\sin^2t)$
$=6\cos^2t+2\sqrt6\cos t-3=6\left(\cos t+\dfrac1{\sqrt6}\right)^2-3-6\left(\dfrac1{\sqrt6}\right)^2$
Now $-1\le\cos t\le1\iff -1+\dfrac1{\sqrt6}\le\cos t+\dfrac1{\sqrt6}\le1+\dfrac1{\sqrt6}$
$\implies\left(\cos t+\dfrac1{\sqrt6}\right)^2\le\left(1+\dfrac1{\sqrt6}\right)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3291916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
How to isolate y in a potential function The following is a potential function:
$Φ(x,y)=x^4y^4+x^2e^xy^2-3lnx-\frac{3}{4}e^2=0$
I would like to isolate $y$ to give $y(x)=…$.
An online calculator gave $y=\pm \frac{\sqrt{-\sqrt{x^4(e^2x+12ln(x)+3e^2}+e^x}x^2}{\sqrt{2}}$, yet I would like to learn how to go about this manually.
Thank you.
| As suggested in the comments, the function $\Phi(x,y)$ is quadratic in $y^2$ as
$$\Phi(x,y)=x^4(y^2)^2+x^2e^xy^2-3\ln x-\tfrac{3}{4}e^2=0,$$
and so if $x\neq0$ the quadratic formula gives
\begin{eqnarray*}
y^2
&=&\frac{-x^2e^x\pm\sqrt{(x^2e^x)^2-4x^4(-3\ln x-\tfrac34e^2)}}{2x^4}\\
&=&\frac{-x^2e^x\pm\sqrt{x^4(e^{2x}+12\ln x+3e^2)}}{2x^4}.\\
\end{eqnarray*}
Note that the denominator is positive, so if we have the $-$-sign for the $\pm$-sign then the numerator is negative. But this is impossible because $y^2$ is nonnegative. So we must have the $+$-sign. Then taking square roots shows that
\begin{eqnarray*}
y
&=&\pm\sqrt{\frac{-x^2e^x+\sqrt{x^4(e^{2x}+12\ln x+3e^2)}}{2x^4}}.\\
&=&\pm\frac{\sqrt{-x^2e^x+\sqrt{x^4(e^{2x}+12\ln x+3e^2)}}}{\sqrt{2x^4}}\\
&=&\pm\frac{\sqrt{-x^2e^x+\sqrt{x^4(e^{2x}+12\ln x+3e^2)}}}{x^2\sqrt{2}},
\end{eqnarray*}
which is close to the result you found, but not the same; the denominator is off by a factor $x^2$, and the expression under the outer square root in the numerator is the negative of what you found.
The above can be further simplified to
\begin{eqnarray*}
y
&=&\pm\frac{\sqrt{-x^2e^x+\sqrt{x^4(e^{2x}+12\ln x+3e^2)}}}{x^2\sqrt{2}}\\
&=&\pm\frac{\sqrt{-x^2e^x+x^2\sqrt{e^{2x}+12\ln x+3e^2}}}{x^2\sqrt{2}}\\
&=&\pm\frac{x\sqrt{-e^x+\sqrt{e^{2x}+12\ln x+3e^2}}}{x^2\sqrt{2}}\\
&=&\pm\frac{\sqrt{-e^x+\sqrt{e^{2x}+12\ln x+3e^2}}}{x\sqrt{2}}.
\end{eqnarray*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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2019 AIME I Problems/Problem 5 Regarding the AIME Problem:
Problem Link
Tried to solve backward - going from 4,4 and moving towards the origin and recording the number of ways to reach different points- then find the number of ways to each point of the coordinate axis and calculate the probability of hitting the origin. This approach seems to be clearly wrong as the answer in the source is different. Can somebody point out the fault in my approach.
My Trial Solution
| If you had used probabilities rather than counts, it would have worked.
For example, the probability of reaching position $(3,4)$ is ${\large{\frac{1}{3}}}$, and the same for position $(4,3)$.
But the probability of reaching position $(3,3)$ is
$${\small{\frac{1}{3}}}+{\small{\frac{2}{9}}}={\small{\frac{5}{9}}}$$
which has nothing to do with the count you showed for that position.
Counts are not applicable here since not all paths from $(4,4)$ to $(0,0)$ are equally likely.
To see that using probabilities rather than counts would have worked, let $P(a,b)$ be the probability of reaching position $(a,b)$ from the initial position $(4,4)$.
The goal is to find $P(0,0)$.
By symmetry, we have $P(b,a)=P(a,b)$, so it suffices to compute $P(a,b)$ for $a\le b$.
Starting from $(4,4)$ and working towards $(0,0)$ we get
\begin{align*}
P(4,4)&=1\\[6pt]
P(3,4)&=\frac{1}{3}P(4,4)=\frac{1}{3}\\[6pt]
P(3,3)&=\frac{1}{3}P(4,3)+\frac{1}{3}P(3,4)+\frac{1}{3}P(4,3)=\frac{5}{3^2}\\[6pt]
P(2,4)&=\frac{1}{3}P(3,4)=\frac{1}{3^2}\\[6pt]
P(2,3)&=\frac{1}{3}P(3,4)+\frac{1}{3}P(2,4)+\frac{1}{3}P(3,3)=\frac{1}{3}\\[6pt]
P(1,4)&=\frac{1}{3}P(2,4)=\frac{1}{3^3}\\[6pt]
P(2,2)&=\frac{1}{3}P(3,3)+\frac{1}{3}P(2,3)+\frac{1}{3}P(3,2)=\frac{11}{3^3}\\[6pt]
P(1,3)&=\frac{1}{3}P(2,4)+\frac{1}{3}P(1,4)+\frac{1}{3}P(2,3)=\frac{13}{3^4}\\[6pt]
P(1,2)&=\frac{1}{3}P(2,3)+\frac{1}{3}P(1,3)+\frac{1}{3}P(2,2)=\frac{73}{3^5}\\[6pt]
P(1,1)&=\frac{1}{3}P(2,2)+\frac{1}{3}P(1,2)+\frac{1}{3}P(2,1)=\frac{245}{3^6}\\[6pt]
P(0,0)&=\frac{1}{3}P(1,1)=\frac{245}{3^7}\\[6pt]\\[6pt]
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3297517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How do you go from $\sum_{n \geq 3} \frac{1}{3} \big(\frac{1}{2}\big)^{n-2}s^n$ to $\frac{1}{6} s^3 \sum_{n \geq 0}\big(\frac{s}{2}\big)^{n}$? I'm summing:
$$\sum_{n \geq 3} \frac{1}{3} \bigg(\frac{1}{2}\bigg)^{n-2}s^n$$
Which is just an infinite series, so if I can get it in the form of $\sum_{n\geq 0}$, I can use $s_{\infty}=\frac{a}{1-r}$.
So I've multiplied it by $\frac{2^2}{2^2}$ so that $1/2$ and $s$ are raised to the same power, i.e. $\big(\frac{1}{2}\big)^{n-2}s^n=\frac{2^2}{2^2}\big(\frac{1}{2}\big)^{n-2}s^n=2^2\big(\frac{s}{2}\big)^n$. Then I've summed from $n=0$ to $\infty$ and subtracted off the extra terms from $n=0$ to $2$, due to summing from $n=0$ instead of $n=3$:
$$\frac{2^2}{3} \bigg\{ \sum_{n\geq 0} \bigg(\frac{s}{2}\bigg)^2 -\bigg(\frac{s}{2}\bigg)^0 - \bigg(\frac{s}{2}\bigg)^1-\bigg(\frac{s}{2}\bigg)^2 \bigg\}$$
$$=\frac{4}{3} \bigg\{ \frac{1}{1-\frac{s}{2}} - \frac{s^2+2s+4}{4} \bigg\}$$
$$=\frac{4}{3} \bigg\{ \frac{2}{2-s} - \frac{s^2+2s+4}{4} \bigg\}$$
$$=\frac{4}{3}\bigg\{ \frac{(2)(4)-(2-s)(s^2+2s+4)}{4(2-s)} \bigg\}$$
$$=\frac{1}{3}\bigg\{ \frac{8-(8-s^3)}{(2-s)} \bigg\}$$
$$=\frac{s^3}{3(2-s)}$$
That was all kind of messy. In one fell swoop, my professor just simply wrote down:
$$\sum_{n \geq 3} \frac{1}{3} \bigg(\frac{1}{2}\bigg)^{n-2}s^n = \frac{1}{6} s^3 \sum_{n \geq 0}\bigg(\frac{s}{2}\bigg)^{n}=\frac{s^3}{3(2-s)}$$
This is certainly a cleaner way than my way, however, I'm not even sure what the thought process is here. How exactly can this be reasoned through? How exactly do you go from $\sum_{n \geq 3} \frac{1}{3} \big(\frac{1}{2}\big)^{n-2}s^n$ to $\frac{1}{6} s^3 \sum_{n \geq 0}\big(\frac{s}{2}\big)^{n}$?
Suggested format for an answer is how I've explained the reasoning behind my messy approach. But feel free to explain in any way you see fit.
| Hint : $$ \bigg(\frac{1}{2}\bigg)^{n-2}s^n = \bigg(\frac{1}{2}\bigg)^{1}\bigg(\frac{1}{2}\bigg)^{n-3} s^3 s^{n-3}$$ then apply the change of variables in the summand.
| {
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"url": "https://math.stackexchange.com/questions/3299067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int_{0}^{\infty} \frac{\sin x-x\cos x}{x^2+\sin^2x } dx$ The integral $$\int_{0}^{\infty} \frac{\sin x-x\cos x}{x^2+\sin^2x } dx$$ admits
a nice closed form. The question is: How to evaluate it by hand.
| Note that
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}\log(x-i\sin(x))
&=\frac{1-i\cos(x)}{x-i\sin(x)}\\
&=\frac{x+\sin(x)\cos(x)}{x^2+\sin^2(x)}+i\frac{\sin(x)-x\cos(x)}{x^2+\sin^2(x)}
\end{align}
$$
Taking the imaginary part of both sides
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}\frac1{2i}\log\left(\frac{x-i\sin(x)}{x+i\sin(x)}\right)
&=\frac{\sin(x)-x\cos(x)}{x^2+\sin^2(x)}
\end{align}
$$
Thus,
$$
\int\frac{\sin(x)-x\cos(x)}{x^2+\sin^2(x)}\,\mathrm{d}x
=\frac1{2i}\log\left(\frac{x-i\sin(x)}{x+i\sin(x)}\right)+C
$$
and
$$
\begin{align}
\int_0^\infty\frac{\sin(x)-x\cos(x)}{x^2+\sin^2(x)}\,\mathrm{d}x
&=-\frac1{2i}\log\left(\frac{1-i}{1+i}\right)\\
&=\frac\pi4
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3299202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving $\frac{\sqrt{k(k+1)} }{k-1} \leq1 + \frac{2}{k} + \frac{3}{4k^2}$ for $k \geq 3$. Could you please give me a hint on how to prove the inequality below for $k \geq 3$?
$$\frac{\sqrt{k(k+1)} }{k-1} \leq1 + \frac{2}{k} + \frac{3}{4k^2} $$
Thank you in advance.
| $$\dfrac{\sqrt{k(k+1)} }{k-1} \leq1 + \dfrac{2}{k} + \dfrac{3}{4k^2}=\dfrac{(2k+1)(2k+3)}{4k^2}$$
$$\sqrt{k(k+1)}\le\dfrac{(2k+1)(2k+3)(k-1)}{4k^2}=\frac{2k+1}{2}\times\dfrac{(2k+3)(k-1)}{2k^2}$$ We know that $\sqrt{k(k+1)}\le\dfrac{k+(k+1)}{2}$ (AM–GM inequality) and $\dfrac{(2k+3)(k-1)}{2k^2}=\dfrac{2k^2+k-3}{2k^2}$ is clearly greater than $1$ (because $k\ge3$).
This proves the inequality.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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1025th term of the sequence $ 1,2,2,4,4,4,4,8,8,8,8,8,8,8,8, ... $ Consider the following sequence - $$ 1,2,2,4,4,4,4,8,8,8,8,8,8,8,8, ... $$
In this sequence, what will be the $ 1025^{th}\, term $
So, when we write down the sequence and then write the value of $ n $ (Here, $n$ stands for the number of the below term) above it We can observe the following -
$1 - 1$
$2 - 2 $
$3 - 2$
$4 - 4$
$5 - 4$
. . .
$8 - 8$
$9 - 8$
. . .
We can notice that $ 4^{th}$ term is 4 and similarly, the $ 8^{th}$ term is 8.
So the $ 1025^{th}$ term must be 1024 as $ 1024^{th} $ term starts with 1024.
So the value of $ 1025^{th}$ term is $ 2^{10} $ .
Is there any other method to solve this question?
| $$ 2^0 + 2^1 + 2^2 + 2^3 + ... + 2^n < 1025 = 2^{10}+ 2^0 $$
$$ 2^1 + 2^2 + 2^3 + ... + 2^n < 2^{10} $$
$$ \frac{1}{2^9} + \frac{1}{2^8} + ... + \frac{1}{2^{10-n}} < 1 $$
In the left-hand side of the above inequality, we've acquired a geometrical sequence. The summation formula for geometrical series is;
$$ S = \frac{a(1-r^n)}{1-r} $$
So this summation must be smaller than 1, with this condition stated we can set n = 9.
$$ 2^0 + 2^1 + 2^2 + ... + 2^9 = 1023 $$
We are now able to conclude that the term at place 1024 and 1025 (at the same time) is 1024.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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Finding the number of solutions to $\cos^4(2x)+2\sin^2(2x)=17(1+\sin 2x)^4$ for $x\in(0,2\pi)$
Number of solution of the equation
$\cos^4(2x)+2\sin^2(2x)=17(1+\sin 2x)^4\; \forall $ $x\in(0,2\pi)$
what i try
$\cos^4(2x)+2\sin^2 2x=17(1+\sin^2(2x)+2\sin 2x)^2$
$1+\sin^4 (2x)=17(1+\sin^4 2x+2\sin^2 2x+4\sin^24x+4\sin 2x(1+\sin^2 2x))$
$16\sin^4 (2x)+68\sin^3 2x+34\sin^2 2x+68\sin 2x+68\sin^2 4x+16=0$
How do i solve it Help me please
| Hint: Your equation is equivalent to
$$2 (\sin (2 x)+2) (2 \sin (2 x)+1) (-7 \sin (2 x)+2 \cos
(4 x)-6)=0$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Distance between two circles in a sphere I have a sphere with radius $R$ and $O$ is the origin.
Inside sphere there are 3 circles. The small circle in black colour is fixed with it's position defined by and $\alpha$ angle and among other two circles one is great circle and another small circle can rotate by maintaining same distance and they are always parallel to each other. Both blue circles intersect the black great circle at point A and B. The rectilinear distance of A to B can be calculated using this relation:
$AB = 2RSin {\theta\over2}$
Orientation of upper blue great circle is defined as $\beta$ which is the angle starting from $z$ axis.The range of $\beta $ can vary from $0^0 -360^0$ and it is a known value. Though my both blue coloured circle is parallel to each other, is it possible to get the lower circle's orientation angle from the centre of sphere that respect $\theta$ as it is the rectilinear distance? We always assume that both blue circle intersect the black circle.
| The black circle is the intersection of a plane $P\cdot (0,0,1)=a$ (I don't think you've specified $a$) and the sphere $P\cdot P =1$.
The blue great circle is the intersection of $P\cdot (\cos \beta, 0, \sin \beta) = 0$ with the sphere, and the blue small circle is the intersection of $P\cdot (\cos \beta, 0, \sin \beta) = b$ with the sphere.
The line of intersection between $P\cdot (0,0,1)=a$ and $P\cdot (\cos \beta, 0, \sin \beta) = b$ is fairly obviously $$P(t) = (\frac{b - a\sin\beta}{\cos\beta}, t, a)$$. Intersection with the sphere gives $$t^2 = 1 - a^2 - \left( \frac{b - a\sin\beta}{\cos\beta} \right)^2$$
The angle between two points on the unit sphere is the arccos of their dot product, so the equation to solve is $$\frac{0 - a\sin\beta}{\cos\beta} \frac{b - a\sin\beta}{\cos\beta} + \sqrt{ \left( 1 - a^2 - \left( \frac{0 - a\sin\beta}{\cos\beta} \right)^2 \right) \left( 1 - a^2 - \left( \frac{b - a\sin\beta}{\cos\beta} \right)^2 \right) } + a^2 = \cos \theta$$ which can be simplified to
$$\frac{- a\sin\beta(b - a\sin\beta) + \sqrt{(\cos^2\beta - a^2)(\cos^2\beta + 2ab\sin\beta - a^2 -b^2)}}{\cos^2 \beta} + a^2 = \cos \theta$$
which can be further rearranged into a quartic in $\sin\beta$ (WARNING: I may have made some errors while rearranging):
$$(1 - \cos^2\theta)\sin^4\beta - 2ab(1 - \cos \theta)(\sin^3\beta - \sin\beta) + (\cos^2\theta -k - 2) \sin^2\beta + ( k + 1) = 0$$ where $k = a^2 b^2 - 2a^2 - b^2 + 2a^2 \cos \theta - \cos^2\theta$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3303052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to show that $\sqrt{2+\sqrt{3}} = \dfrac{\sqrt{6}+\sqrt{2}}{2}$
$\sqrt{2+\sqrt{3}} = \dfrac{\sqrt{6}+\sqrt{2}}{2}$
How to change $\sqrt{2+\sqrt{3}}$ into $\dfrac{\sqrt{6}+\sqrt{2}}{2}$
| You can use the identity
\begin{equation}
\sqrt{a+\sqrt{b}} = \sqrt{\frac{a+\sqrt{a^2-b}}{2}} + \sqrt{\frac{a-\sqrt{a^2-b}}{2}}.
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3303949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
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$a, b$ are the integer part and decimal fraction of $\sqrt7$ find integer part of $\frac{a}{b}$
$a, b$ are the integer part and decimal fraction of $\sqrt7$
find integer part of $\frac{a}{b}$
using calculator :
$\sqrt 7$ = 2.645
$\frac{a}{b} = \frac{2}{0.6} = \frac{20}{6} = 3.333$
integer part of $\frac{a}{b} = 3$
how to do it without calculator? if i don't know $\sqrt7$ part
| $2.6^2=6.76<7<7.29=2.7^2.$
So $a=2$ and $0.6<b<0.7.$
So $4>10/3=2/0.6>a/b>2/0.7=20/7>2.$
So the integer part of $a/b$ is $2$ or $3.$
Now $$3\le a/b\iff 3b\le a\iff 3b\le 2 \iff b\le 2/3 \iff$$ $$\iff \sqrt 7=a+b=2+b\le 2+2/3=8/3\iff$$ $$\iff 7\le (8/3)^2=64/9 =7+1/9.$$ Since we used "$\iff$" throughout, and since it is true that $7\le (8/3)^2,$ therefore $3\le a/b.$ And we already have $a/b<4$. So we have $$3\le a/b<4.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3304536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Infinite Series $\sum_{n=1}^{\infty}\frac{4^nH_n}{n^2{2n\choose n}}$ I am trying to find a closed form for this infinite series:
$$ S=\sum_{n=1}^{\infty}\frac{4^nH_n}{n^2{2n\choose n}}$$
Whith $H_n=\sum\limits_{k=1}^{n}\frac{1}{k}$ the harmonic numbers.
I found this integral representation of S:
$$S=2\int_{0}^{1}\frac{x}{1-x^2}\left(\frac{\pi^2}{2}-2\arcsin^2(x)\right)dx$$
Sketch of a proof:
Recall the integral representation of the harmonic numbers: $H_n=\displaystyle\int_{0}^{1}\frac{1-x^n}{1-x}dx$
By plugging it into the definition of S and interchanging the order of summation between $\displaystyle\sum$ and $\displaystyle\int$ (justified by the uniform convergence of the function series $\displaystyle\sum\left(x\to\frac{4^n}{n^2{2n\choose n}}\frac{1-x^n}{1-x}\right)$, because $\forall x\in[0,1],\frac{1-x^n}{1-x}<n$), we get:
$$S=\int_{0}^{1}\frac{1}{1-x}\sum\limits_{n=1}^{\infty}\frac{4^n(1-x^n)}{n^2{2n\choose n}}dx$$
$$=\int_{0}^{1}\frac{1}{1-x}\left(\sum\limits_{n=1}^{\infty}\frac{4^n}{n^2{2n\choose n}}-\sum\limits_{n=1}^{\infty}\frac{(4x)^n}{n^2{2n\choose n}}\right)dx$$
$$=\int_{0}^{1}\frac{1}{1-x}\left(\frac{\pi^2}{2}-\sum\limits_{n=1}^{\infty}\frac{(4x)^n}{n^2{2n\choose n}}\right)dx$$
Using the result $\displaystyle\sum\limits_{n=1}^{\infty}\frac{4^n}{n^2{2n\choose n}}=\frac{\pi^2}{2}$.
At that point, we will rely on the taylor series expansion of $\arcsin^2$:
$$\arcsin^2(x)=\frac{1}{2}\sum\limits_{n=1}^{\infty}\frac{4^n}{n^2{2n\choose n}}x^{2n}, |x|<1$$
Out of which we get $\displaystyle\sum\limits_{n=1}^{\infty}\frac{(4x)^n}{n^2{2n\choose n}}=2\arcsin^2\left(\sqrt{x}\right)$
So,
$$S=\int_{0}^{1}\frac{1}{1-x}\left(\frac{\pi^2}{2}-2\arcsin^2\left(\sqrt{x}\right)\right)dx$$
Which, through the substitution $u=\sqrt{x}$, gives the integral representation above.
But beyond that, nothing so far. I tried to use the integral representation of $\frac{H_n}{n}$ to switch the order of summation, but it didn't lead anywhere. Any suggestion?
| $$S=2\int_{0}^{1}\frac{x}{1-x^2}\left(\frac{\pi^2}{2}-2\arcsin^2(x)\right)dx\overset{IBP}=-4\int_0^1 \frac{\arcsin x\ln(1-x^2)}{\sqrt{1-x^2}}dx$$
$$\overset{x=\sin t}=-8\int_0^\frac{\pi}{2} t \ln(\cos t)dt=8 \ln 2 \int_0^\frac{\pi}{2}t dt+8\sum_{n=1}^\infty \frac{(-1)^n}{n}\int_0^\frac{\pi}{2} t\cos(2n t)dt$$
$$={\pi^2}\ln 2+2\sum_{n=1}^\infty \frac{1-(-1)^n}{n^3}=\boxed{\pi^2 \ln 2 +\frac72 \zeta(3)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3306742",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "7",
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Proof verification induction on sequence I am doing problems that involve inequalities. My understanding is that you through a string of inequalities show that one is less than the other. Kind of like the transitive property. For example:
$2n+1 \lt 2^n$ for $n=3,4,...$
Assume this is true for P(k):
Base case $k = 3$
$LHS: 7 \space \space \space RHS: 8$
$LHS \space \space \lt \space \space RHS$
This holds true for $(k)$
Prove for P(k+1):
$2(n+1)+1 \lt 2^{n+1}$
my understanding is that if I can find something that is greater than $2(n+1)+1$ and obviously below $2^{n+1}$ then the inequality holds true
$2k+3 \lt 2^{k+1}$
$(2k + 1) + 2 \lt 2^k2$
$(2k+1) + 2 \lt 2^k + 2$ by our hypothesis on $k$
It is very obvious that $2^k + 2 \lt 2^k2$ and doesn't need explanation so its safe to assume
$2k+3 = (2k+1)+2 \lt 2^k + 2 \lt 2^k2$
Thus $(2k+3) \lt 2^k2$
This seems perfectly logical to me since we are dealing with integers. These inequalities would not hold for $\mathbb{R}$.
I am having a hard time find a string of inequalities that proves $n^2 \leq 2^n+1$
Assume this holds true for $k$. $P(k) = k^2 \leq 2^k+1$ for $n = 1,2...$
Base case $P(1):$ $LHS: 1 \space \space \space RHS: 3$
$LHS \space \space \leq \space \space RHS$
holds true for $P(k)$
$P(k+1)$:
$(k+1)^2 \leq 2^k + 1$
$k^2 + 2k + 1 \leq 2^k2+1$
$k^2 + 2k + 1 \leq 2^k +1 + 2k +1 \leq 2^k2+1$
$k^2 + 2k + 1 \leq 2^k + 2k + 2 \leq 22^k+1$ by our hypothesis on $k$
I do not know where to go from here
| You can not use induction on $\mathbb R$ as the induction step will prove if it is true for $x$ then it is true for $x+1$ but there is nothing that will assure that it is true for any $k; x < k < x+1$.
Are you trying to prove $2x + 1 < 2^x$ for all real $x\ge 1$? If so do the following.
Use induction to prove that for $n\in \mathbb N;n\ge 3$ that $2n+1 < 2^n$.
Then prove that for any $x \in \mathbb R$; that if $n < x < n+1$ then $2n+1 < 2x+1 < 2(n+1)+1$ and $2^n < 2^x < 2^{n+1}$ so then only way that $2^x \le 2x+1$ could be possible is if $2(n+1)> 2^n$.
Why can prove that is impossible by induction by proving that $2(n+1) \le 2^n; n\ge 3$ and $n\in \mathbb N$.
So.....
Claim 1: $2(n+1) \le 2^n$ if $n\in \mathbb N; n\ge 3$.
Proof by Induction:
Base case: $k=3$ then $2(3+1) = 2^3$.
Induction case: If $2(k+1) \le 2^k$ then
$2(k+2)= 2(k+1) + 2 \le 2^k+2 \le 2^k + 2^k = 2^{k+1}$.
Thus we have proven so for all natural $n> 3$ that $2(n+1)\le 2^n$
So as $2a+1 < 2(n+1)=2n + 2\le 2^n$ we have proven $2n+1 < 2^n$ for all natural $n \ge 3$.
Now if $x\in \mathbb R; x \ge 3$ then there is an $n$ so that $n < x < n+1$ then $2n+1 < 2x + 1 < 2n+2 \le 2^n$. And $2^x = 2^n*2^{x-n}$. As $x-n >0$ we know $2^{x-n} > 1$ so $2^n < 2^n*2^{x-n} = 2^x$. And so $2x+1 < 2^x$. And that's that.
Then you sweitch gears and ask how to prove $n^2 \le 2^n + 1;n \ge 3$. (Was that part of the question? A different question?
Well, base case is easy: $3^2 = 2^3 + 1$.
Induction follows:
If $k^2 \le 2^k + 1$ then
$(k+1)^2 = k^2 + 2k + 1 \le 2^k + 2k + 1 = 2^k + 2(k+1)$. Above we proved that $2(n+1) \le 2^k$ if $k \ge 3$ so $2^k + 2(k+1) \le 2^k + 2^k = 2^{k+1}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to get the value of $A + B ?$ I have this statement:
If $\frac{x+6}{x^2-x-6} = \frac{A}{x-3} + \frac{B}{x+2}$, what is the value of $A+B$ ?
My attempt was:
$\frac{x+6}{(x-3)(x+2)} = \frac{A(x+2) + B(x-3)}{(x-3)(x+2)}$
$x+6=(x+2)A + B(x-3)$: But from here, I don't know how to get $A + B$, any hint is appreciated.
| If $x+6=(x+2)A+B(x-3)$ then $$x+6=x(A+B)+(2A-3B)\implies$$ $$A+B=1\text{ and } 2A-3B=6$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the point located on an offset ellipse I am attempting to find either the x & y coordinates of a point located on an offset ellipse. I have the following information to go on:
*
*The angle of a line which intersects the point from the origin
*The radii of the ellipse
*The offset of the ellipse
This image visually demonstrates the problem (The use of symbols do not match up with the workings provided below).
I start with the equation for an offset ellipse:
$${(x - h)^2 \over a^2} + {(y-k)^2 \over b^2} = 1$$
My y-offset $k == 0$, so I get rid of that straight away...
$$\therefore {(x - h)^2 \over a^2} + {(y)^2 \over b^2} = 1$$
I can substitute $y$ for a simple $y = {mx}$ equation, and $m = \tan\theta$ so:
$$\therefore {(x - h)^2 \over a^2} + {(x\tan\theta)^2 \over b^2} = 1$$
Then I proceed to expand and simplify:
$$b^2{(x - h)^2} + a^2{(x\tan\theta)^2} = a^2b^2$$
$${(x - h)^2} + {a^2x^2\tan^2\theta\over b^2} - a^2 = 0$$
$$x^2(1 + {a^2\tan^2\theta\over b^2})-2hx + h^2 - a^2 = 0$$
From here I use
$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$
to figure out my possible $x$ values, and $y = x\tan\theta$ to get the matching $y$ values.
Does this seem the correct approach, or can anyone spot where I may be going wrong? I get more-or-less correct values once I code it up in C#, but there seems to be a slight margin of error. I'm not confident in my maths so most of my doubts are here!
Thank you for your help.
Update: An image with matching symbols
| The intersections of
$$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$$ and $$y=mx$$
are given by the solutions of
$$b^2(x-h)^2+a^2(mx-k)^2-a^2b^2=0.$$
The coefficients are
$$\begin{cases}A=b^2+a^2m^2,\\B=-2b^2h-2a^2mk,\\C=b^2h^2+a^2k^2-a^2b^2.\end{cases}$$
Then the discriminant
$$\Delta'=a^2b^2(a^2m^2+b^2-(mh-k)^2)$$
and the roots,
$$x=\frac{b^2h+a^2mk\pm\sqrt{\Delta'}}{a^2m^2+b^2},\\y=mx.$$
| {
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In the exapnsion of $(1+x+x^3+x^4)^{10}$, find the coefficient of $x^4$ In the exapnsion of $(1+x+x^3+x^4)^{10}$, find the coefficient of $x^4$.
What's the strategy to approach such problems. Writing expansion seems tedious here.
| You can use the multinomial expansion:
$$
(a+b+c+d)^{10}=\sum_{p+q+r+s=10}\binom{10}{p\ q\ r\ s}a^pb^qc^rd^s
$$
where
$$
\binom{10}{p\ q\ r\ s}=\frac{10!}{p!q!r!s!}
$$
With $a=1$, $b=x$, $c=x^3$, $d=x^4$, we have
$$
a^pb^qc^rd^s=x^{q+3r+4s}
$$
We get $q+3r+4s=4$ if and only if $q=4$, $r=0$, $s=0$ or $q=1$, $r=1$, $s=0$ or $q=0$, $r=0$, $s=1$. Thus the coefficient is
$$
\binom{10}{6\ 4\ 0\ 0}+\binom{10}{8\ 1\ 1\ 0}+\binom{10}{9\ 0\ 0\ 1}
=\frac{10!}{6!4!}+\frac{10!}{8!1!1!}+\frac{10!}{9!1!}=210+90+10=310
$$
| {
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Modular arithmetic $(2n+1)x \equiv -7 \pmod 9$ Find a solution $(2n+1)x \equiv -7 \pmod 9$
I’m sure this is trivial but I still have doubts about it.
I know the equation has solution for certain $n \in \mathbb {Z}$. Actually I have tried a few and got a similar results (with Diophantine equations ). I wonder if there’s general solution for the equation without changing the n for an integer.
Thanks in advance.
| \begin{align}
(2n+1)x &\equiv 2 \pmod 9 \\
5(2n+1)x &\equiv 1 \pmod 9 \\
(10n + 5)x &\equiv 1 \pmod 9 \\
(n-4)x &\equiv 1 \pmod 9 \\
n-4 &\equiv x^{-1} \pmod 9 \\
n &\equiv 4 + x^{-1}
\end{align}
\begin{array}{c}
x & n \equiv x^{-1} + 4 \\
\hline
1 & 5 \\
2 & 0 \\
3 & \text{No solution.} \\
4 & 2 \\
&\text{etc.}
\end{array}
| {
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Computation Of Integrals Computer the Integral: $$\int\frac{2x+1}{(x-1)(x-2)}dx$$
Now using partial fraction we can write $$\frac{2x+1}{(x-1)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2}$$, So we get $$\frac{2x+1}{(x-1)(x-2)}=\frac{A(x-2)+B(x-1)}{(x-1)(x-2)}$$ Now for all $x$ not equal to $1, 2$ we can cancel out the denominator to get $$2x+1=A(x-2)+B(x-1)$$ Now to find $A$ and $B$ how can we put $x=1$ and $x=2$ in this identity as this identity is valid if and only if $x$ is not equal to $1, 2$
| While $\frac{2x+1}{(x-1)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2}$ is only valid for $x\not\in\{1,\,2\}$ for the values of $A,\,B$ we seek, $2x+1=A(x-2)+B(x-1)$ will be valid for all $x\in\Bbb R$ because its validity at other values implies validity on $x\in\{1,\,2\}$ by continuity. You could also find $A,\,B$ from simultaneous equations obtained at other values of $x$, but using $x\in\{1,\,2\}$ is especially convenient, viz. $3=-A,\,5=B$. Or you could just equate coefficients of $x^1=x,\,x^0=1$, viz. $A+B=2,\,-2A-B=1$. That approach will serve you well for other kinds of coefficient-inferring problems.
| {
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Inequality $\frac{\ln(7a+b)}{7a+b}+\frac{\ln(7b+c)}{7b+c}+\frac{\ln(7c+a)}{7c+a}\leq \frac{3\ln(8\sqrt{3})}{8\sqrt{3}}$ I'm interested by the following problem :
Let $a,b,c>0$ such that $abc=a+b+c$ then we have :
$$\frac{\ln(7a+b)}{7a+b}+\frac{\ln(7b+c)}{7b+c}+\frac{\ln(7c+a)}{7c+a}\leq \frac{3\ln(8\sqrt{3})}{8\sqrt{3}}$$
I have tried to use convexity and Jensen but the result is weaker . I try also Karamata's inequality but it fails totaly so I'm a bit lost .
If you have a hint it would be great .
Thank you .
| As pointed out by River Li, first show that for every $x>0$ holds the inequality
$$
\log x\leq\frac{x}{8\sqrt{3}}+\log\left(\frac{8\sqrt{3}}{e}\right).\tag 1
$$
The equality in $(1)$ for holds only for $x=8\sqrt{3}$.
From $(1)$ we have
$$
\sum_{cyc}\frac{\log(7a+b)}{7a+b}\leq \frac{\sqrt{3}}{8}+\sum_{cyc}\frac{1}{7a+b}\log\left(\frac{8\sqrt{3}}{e}\right).
$$
Set
$$
\Pi(x,y,z):=\sum_{cyc}\frac{1}{7x+y},
$$
Consider the inequality
$$
\frac{1}{kx+ly}\leq \frac{A}{kx}+\frac{B}{ly},
$$
with $k,l,x,y>0$, $k=7$, $l=1$ and assume that
$$
\frac{A}{k}+\frac{B}{l}=\frac{1}{k+l}.
$$
Hence $B=\frac{1}{56}(7-8A)$. But for this value of $B$ it is
$$
\frac{1}{kx+ly}-\frac{A}{7x}-\frac{B}{y}=-\frac{(x-y)(49x-56Ax-8Ay)}{56xy(7x+y)}.
$$
Using $49-56A=8A\Leftrightarrow A=\frac{49}{24}$, we get
$$
\frac{1}{7x+y}-\frac{7}{64x}-\frac{1}{64y}=-\frac{7(x-y)^2}{64xy(7x+y)}\tag 2
$$
Summing all three equalities $(2)$ (for the pairs $(x,y),(y,z),(z,x)$) we get
$$
\sum_{cyc}\frac{1}{7x+y}\leq\frac{7}{64}\sum_{cyc}\frac{1}{x}+\frac{1}{64}\sum_{cyc}\frac{1}{x}
$$
and consequently
$$
\Pi(x,y,z)\leq\frac{1}{8}\sum_{cyc}\frac{1}{x}.
$$
This is a ''bad'' estimate, since
$$
\sum_{cyc}\frac{1}{x}\geq\sqrt{3}.
$$
However, I think, someone can work better with $(2)$ and obtain a ''good'' estimate.
| {
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If $\sec A-\cos A=1$, then determine the value of $\tan^2\frac A2$ This is what I tried
$\sec A=\frac{1}{\cos A}$, so the equation becomes
$1-\cos^2A=\cos A$
If we solve the above quadratic equation, we the values of $\cos A$ as $\frac{-1\pm \sqrt5}{2}$
Therefore, $\tan\frac A2$ becomes
$$\sqrt \frac{3-\sqrt 5}{1+\sqrt 5}$$
Squaring that value, the answer remains meaningless
The options are
A) $\sqrt 5+ 2$
B) $\sqrt 5-2$
C) $2-\sqrt5$
D) $0$
Since the options are not matching, where am I going wrong?
| Just notice that
$$
\frac{3-\sqrt 5}{1+\sqrt 5}=\frac{(3-\sqrt 5)(1-\sqrt{5})}{(1+\sqrt 5)(1-\sqrt{5})}=\frac{-4\sqrt{5}+8}{-4}=\sqrt{5}-2.
$$
| {
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Prove that number of partitions $n^2$ on $n$ distinct parts $=$ partitions of $\binom{n}{2}$ on at most $n$ parts
Prove that number of partitions $n^2$ on $n$ distinct parts $=$ partitions of $\binom{n}{2}$ on at most $n$ parts
I tried to find bijection.
We can noticed that
$$ \binom{n}{2} = \frac{n^2-n}{2} $$
which is kind of hint there...
Let find bijection in use of example. Let $n=8$ and $n^2 = 64$
let $\pi$ be $$ 1+2+3+4+5+6+7+8+9+19 $$
Now rearrange them: Odd first, even later:
$$ 2+4+6+8+1+3+5+7+9+19$$
now we do $$n^2 \rightarrow n^2-n$$:
$$ 1+3+5+7+0+2+4+6+8+18$$
Now do $/2$:
$$ 0+1+2+3+0+1+2+3+4+9$$
OK but there are two problems
*
*how can I come back (due to lost $2 \times 0$ at different steps of process)$
*In that way, my second partitions has from $n-2$ to $n$ parts. So it can't be bijection because In should get also partitions with $1$, $2$, ... $n-3$ elements...
| Example for $n=3$ (rather easily generalized):
Take $9$ dots and arrange in three rows so that each row has more dots than the one below it. For instance,
$$
\begin{array}{ccccc}
\bullet&\bullet&\bullet&\bullet&\bullet\\
\bullet&\bullet&\bullet\\
\bullet
\end{array}
$$
Each partition of $9$ into $3$ distinct parts corresponds to the rows in a such array of dots, each partition corresponds to a unique array, and each array corresponds to a unique partition.
Now remove $3$ dots from the top row, $2$ dots from the middle row and one dot from the bottom row (this should be proven to always be possible, also in the general case). You get
$$
\begin{array}{ccc}
\bullet&\bullet\\
\bullet\\
\phantom\bullet
\end{array}
$$
Now read the rows as a partition of the remaining $\binom 32=3$ dots with at most three dots per row (it should be proven that the rows are still sorted from largest to smallest, so that dot arrangements still bijectively correspond to partitions).
| {
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Find "A" in this equation. $$ \sqrt[3] {A-15√3} + \sqrt[3] {A+15√3} = 4 $$
Find "A" ?
The way of exponentiation took too much time, is there any easier method?
| Let $x=A-15\sqrt 3$ and $y=A+15\sqrt 3,$ then the equation becomes $$x^{1/3}+y^{1/3}=4.$$ Taking cubes of both sides gives $$x+y+3(xy)^{1/3}(x^{1/3}+y^{1/3})=4^3.$$ Now since $x+y=2A,$ and $x^{1/3}+y^{1/3}=4,$ the equation becomes $$2A+3(xy)^{1/3}(4)=4^3,$$ which gives $$6(xy)^{1/3}=32-A.$$ Now, cubing and substituting for $xy=A^2-15^2\cdot 3$ gives $$6^3(A^2-15^2\cdot 3)=(32-A)^3.$$ This is a cubic in $A,$ which in general is not so easy to deal with.
| {
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Determine points on line with specific distance from plane There is a line $$p: \dfrac{x-1}{2} = \dfrac{y+1}{3} = \dfrac{z+1}{-1}$$
and plane $$\pi : x+y+2z-3=0.$$
I need to find points $T_1, T_2 \in p$. Requirement when finding those points are that they have a set distance from plane $\pi$, with distance being $\sqrt{6}$, and that the points are on the line.
The solution for this problem is
$T_1=\left( \dfrac 13, -2, -\dfrac 23\right)$
$T_2 = \left( \dfrac{25}{3}, 10, -\dfrac{14}{3} \right)$
Problem is I don't have idea how to get there, so I will apreciate any advice/guidance for solving this problem.
EDIT1
Equation for distance between point and plane is:
$d(pi,T_\pi')=\dfrac{|AX_0+BY_0+CZ_0+D|}{\sqrt{(A^2+B^2+C^2)}})$
so I get:
$\sqrt6=\dfrac{|X_0+Y_0+2Z_0-3|}{\sqrt6}$
$6=|X_0+Y_0+2Z_0-3|$
since the requirement for parallel planes is: $\dfrac{A_1}{A_2}=\dfrac{B_1}{B_2}=\dfrac{C_1}{C_2}$ I deduced that $A_2=B_2=2C_2$
returning to where I left:
$6=|4X_0-3|$
$4X_0-3=6$
and
$4X_0-3=-6$
$X_01=\dfrac{9}{4}$
and
$X_02=\dfrac{-3}{4}$
Finally I have two planes but I will focus on one because I messed up something:
$\pi .. \dfrac{9}{4}X+\dfrac{9}{4}Y+\dfrac{9}{2}-3=0$
If I fit vector of line direction into this equation we get that $t=\dfrac{10}{9}$
Now if try to fit $t$ back in the equation I will get wrong results since x will be $\dfrac{29}{9}$.
What am I currently missing?
| If you are going to work with parallel planes, you can use the fact that they have the same normal vector. This means you only need to change the $d$ in the plane equation. So you want to find a plane $x+y+2z+d=0$ such that its distance to $\pi$ is $\sqrt{6}$.
The distance between two planes with the same normal vector is:
$ d=\frac{|d_1-d_2|}{\sqrt{A^2+B^2+C^2}}$
So, plugging in the values for this specific problem you get:
$ \sqrt{6}=\frac{|d_1-(-3)|}{\sqrt{6}}=\frac{|d_1+3|}{\sqrt{6}}$
So either $d=-9$ or $d=3$
So, for example, your first plane would be:
$x+y+2z-9=0$
Then, fitting the vector of line direction into this equation we get that $t=\frac{11}{3}$
Then you get $x=\frac{25}{3}$
| {
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if $2^n+1=xy$, show $2^a|(x-1) \iff 2^a|(y-1)$
Suppose that $2^n+1=xy$, where $x$ and $y$ are integers greater than $1$ and $n>0$. Show that $2^a|(x-1)$ if and only if $2^a|(y-1)$.
What I noticed is that $2^n+1$ is an odd number for any $n$, so $x$ and $y$ are both odd numbers, and hence $x-1$ and $y-1$ will always be even. So here is what I did,
Let $x-1=2^bd_{1}$ and $y-1=2^cd_{2}$, where $b,c>0$ and $d_{1},d_{2}$ are odd natural numbers. So,
$\begin{align*}
2^n+1 & = (2^bd_{1}+1)(2^cd_{2}+1) \\
& = 2^{b+c}d_{1}d_{2}+2^bd_{1}+2^cd_{2}+1 \\
2^n & = 2^{b+c}d_{1}d_{2}+2^bd_{1}+2^cd_{2}
\end{align*} $
if $2^a|(x-1)$ for some $a$, then $b\ge a$, so in the above equation, $2^a|2^n$,$2^a|2^{b+c}d_{1}d_{2}$ and $2^a|2^bd_{1}$, this implies $2^a|2^cd_{2} \implies c\ge a$. Hence $2^a|(y-1)$. The other implication can be proved in a similar way.
Is this approach correct? Am I missing something? Is there any other way to solve this problem?
| You can rewrite $$2^n+1=xy$$ as $$2^n=(x-1)(y-1)+(x-1)+(y-1)$$
With this equation, we can easily see that for every integer $a$ with $\ 1\le a\le n\ $ , $\ x-1\ $ is divisible by $\ 2^a\ $ if and only if $\ y-1\ $ is divisble by $\ 2^a\ $ , which is what has to be proven.
With some additional considerations, we can also show the claim for other non-negative integers $\ a\ $ , which I will leave as an exercise for the reader.
| {
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Computing $\int\frac {du}{\sqrt{u^2 + s^2}} =\log \lvert (u + \sqrt{u^2 + s^2}) \rvert$ with a substitution Can someone please show me where I am going wrong? It seems there is a contradiction in the formula for the the following integral.
$$\int\frac {du}{\sqrt{u^2 + s^2}} = \log \bigl\lvert u + \sqrt{u^2 + s^2}\bigr\rvert$$
Now, can't we rewrite the integral as follows using the substitution z = -u?
\begin{align}
\int\frac {du}{\sqrt{u^2 + s^2}} &= \int\frac {du}{\sqrt{(-u)^2 + s^2}} = \int\frac {-dz}{\sqrt{z^2 + s^2}}\\
& = -\log \bigl\lvert z + \sqrt{z^2 + s^2}\bigr\rvert = -\log \bigl\lvert (-u + \sqrt{u^2 + s^2})\bigr\rvert
\end{align}
But, clearly $\;\log \bigl\lvert u + \sqrt{u^2 + s^2}\bigr\rvert \ne -\log \bigl\lvert -u + \sqrt{u^2 + s^2} \bigr\rvert$
Also, the quantity inside $\log \bigl\lvert u + \sqrt{u^2 + s^2})\bigr\rvert$ is clearly non-negative for any choice of $u$ and $s$, so why do we need the absolute value of the quantity in our integral?
| Both are not equal . But you'll get a term which will get added to the arbitrary constant and get cancelled in definite integral.
$$I=-\log|-u+\sqrt{u^2+s^2}|+c = \log\bigg\vert\frac{1}{-u+\sqrt{u^2+s^2}}\bigg\vert+c$$
$$I = \log\bigg\vert\frac{u+\sqrt{u^2+s^2}}{-u^2+u^2+s^2}\bigg\vert+c=\log\vert u+\sqrt{u^2+s^2}\vert-\log s^2+c = \log\vert u+\sqrt{u^2+s^2}\vert +k$$
where $k=c-\log s^2$
And this gets cancelled in a definite integral, so both are equivalent.
| {
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Prove that for $n\in\mathbb{N}$, $\sum_{k=1}^{\infty}\left[\frac{n}{5^{k}}\right]=15\iff\left[\frac{n}{5}\right]=13.$ How to show that the following relation? : for $n\in\mathbb{N}$, $$\sum_{k=1}^{\infty}\left[\frac{n}{5^{k}}\right]=15\iff\left[\frac{n}{5}\right]=13.$$ It's not obvious to me. Can anyone help me? Thank you!
| Observe that $\sum_{k=1}^{\infty}\left\lfloor\frac{n}{5^k}\right\rfloor=\nu_5(n!)$, where $\nu_5(n!)$ is the exponent of the largest power of $5$ that divides $n!$ (if you are not familiar with this, then see Legendre's formula).
Since this sum is $15$, thus the highest power of $5$ that divides $n!$ is $15$.
Consider $n!=(1\cdot 2 \cdot 3 \cdot 4 \cdot \color{red}{5})(6\cdot 7 \cdot 8 \cdot 9 \cdot \color{red}{10})\dotsb ((n-4) \cdot (n-3) \cdot (n-2) \cdot (n-1) \cdot \color{red}{n}).$
Each group of $((a+1)\cdot (a+2) \cdot (a+3) \cdot (a+4) \cdot (a+5))$ contributes a single power of $5$, until we reach $25$, where we get a contribution of $2$ towards the exponent of $5$. Likewise until we reach $50$ each such group of five numbers contribute only a single power of $5$ and so on. Thus the number $n$ must be such that
$$n!=\underbrace{(1\cdot 2 \cdot 3 \cdot 4 \cdot \color{red}{5})}_{5^1}\underbrace{(6\cdot 7 \cdot 8 \cdot 9 \cdot \color{red}{10})}_{5^1}\dotsb \underbrace{(21\cdot 22 \cdot 23 \cdot 24 \cdot \color{red}{25})}_{\color{green}{\boxed{5^2}}}\dotsb \underbrace{(61\cdot 62 \cdot 63 \cdot 64 \cdot \color{red}{65})}_{5^1} \dotsb ()$$
When $n=65$, that is the first time we will have $\nu_5(n!)=15$. Thus $65 \leq n < 70$ which implies $\left\lfloor\frac{n}{5}\right\rfloor=13.$
| {
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equation simplification. $(5y-1)/3 + 4 =(-8y+4)/6$ Simplification of this equation gives two answers when approched by two different methods.
Method 1 Using L.C.M( least common multiple)
$(5y-1)/3 + 4 =(-8y+4)/6$
$(5y-1+12)/3 = (-8y+4)/6$
$5y-11 = (-8y+4)/2$
$(5y-11)2= (-8y+4)$
$10y-22 = -8y+4$
$18y=26$
$y = 26/18=13/9$
Method 2 multiplying every term by 3
$3(5y-1)/3 + 4*3 = 3(-8y+4)/6$
$5y-1 + 12 = (-8y+4)/2$
$2(5y-1 + 12) = -8y+4$
$10y-2+24 = -8y+4$
$18y + 22 = 4$
$18y = -18$
$y = -1$
The correct method is method 2 and the correct answer is y = -1
Why is method 1 is incorrect? Could anyone explain why the answer is wrong when using the L.C.M( method 1)?
| When you do two algebraic manipulations simultaneously, like here:
$$(5y-1+12)/3 = (-8y+4)/6$$
$$5y-11 = (-8y+4)/2$$
i.e. multiply through by $3$ AND simplify the left bracket, you have confusion. It should be $5y+11$. And you could always multiply through by $6$ instead of $3$ ($6=\operatorname{lcm}(3,6)$).
I would:
$$(5y-1)/3 + 4 =(-8y+4)/6$$
Multiply both sides by $6$:
$$2(5y-1)+24=-8y+4$$
Expand brackets:
$$10y-2+24=-8y+4$$
Simplify $-2+24$:
$$10y+22=-8y+4$$
Add $8y$ to both sides:
$$10y+22+8y=-8y+4+8y$$
Simplify:
$$18y+22=4$$
Add $-22$ to both sides
$$18y+22-22=4-22$$
Simplify:
$$18y=-18$$
Divide both sides by $18$:
$$y=-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3323817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Uniform convergence of $f_n(x) = \cos^n(x/ \sqrt{n})$ . I'm studying the uniform convergence of the following sequence :
\begin{equation*}
f_n(x) =
\left\{
\begin{split}
& \ \cos^n\left(\frac{x}{\sqrt{n}}\right) \ \ \textrm{if} \ x \in \left[0, \frac{\sqrt{n} \pi}{2} \right] \\
& 0 \ \ \textrm{if} \ x \geq \frac{\sqrt{n} \pi}{2}
\end{split}
\right.
\end{equation*}
to $f : x \longmapsto e^{-x^2/2} $ .
I computed this to see that it converges uniformly but I don't get to a proof. I tried to study the difference $f - f_n$ analytically but it doesn't work.
Thanks for help.
| Too long for comments.
Since composition of Taylor series is one of my hobbies, continuing your work, we have
$$\cos^n\left(\frac{x}{\sqrt{n}}\right)=e^{-\frac{x^2}{2}}\left(1-\frac{x^4}{12 n}+\frac{x^6 \left(5 x^2-32\right)}{1440 n^2}+O\left(\frac{1}{n^3}\right) \right)$$ We have also the "nice"
$$\text{sinc}^n\left(\frac{x}{\sqrt{n}}\right)=e^{-\frac{x^2}{6}}\left(1-\frac{x^4}{180 n}+\frac{x^6 \left(7 x^2-160\right)}{453600 n^2}+O\left(\frac{1}{n^3}\right) \right)$$
$$\text{tanc}^n\left(\frac{x}{\sqrt{n}}\right)=e^{+\frac{x^2}{3}}\left(1+\frac{7 x^4}{90 n}+\frac{x^6 \left(343 x^2+2480\right)}{113400 n^2}+O\left(\frac{1}{n^3}\right) \right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3325964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Is a (short and) beautiful proof for symmetric inequality must exist always? There are several and almost similar inequalities in MSE that some of them can be proved in long page. some of these questions listed below:
*
*For $abc=1$ prove that $\sum\limits_{cyc}\frac{a}{a^{11}+1}\leq\frac{3}{2}.$
*For positive $a$, $b$, $c$ with $abc=1$, show $\sum_{cyc} \left(\frac{a}{a^7+1}\right)^7\leq \sum_{cyc}\left(\frac{a}{a^{11}+1}\right)^7$
*Inequality $\frac{x}{x^{10}+1}+\frac{y}{y^{10}+1}+\frac{z}{z^{10}+1}\leq \frac{3}{2}$
*Inequality: $(a^3+a+1)(b^3+b+1)(c^3+c+1) \leq 27$
*If $abc=1$ so $\sum\limits_{cyc}\frac{a}{a^2+b^2+4}\leq\frac{1}{2}$
*For $abc=1$ prove that $\sum\limits_\text{cyc}\frac{1}{a+3}\geq\sum\limits_\text{cyc}\frac{a}{a^2+3}$
*If $abc=1$ so $\sum\limits_{cyc}\sqrt{\frac{a}{4a+2b+3}}\leq1$.
and so on. One cane pose many many similar question in this way:
Let $f(x)$ be a continuous (and maybe with a special property) then prove that $\sum_{x\in\{a,b,c\}}f(x)\leq 3f(1)$ whenever $abc=1$. or one can generalize this for arbitrary number of variables: $\sum_{cyc}f(x)\leq nf(1)$ whenever $\prod_{i=1}^n x_i=1$.
My argument based on what I read in Problem-Solving Through Problems by Loren C. Larson that
principle of insufficient reason, which can be stated
briefly as follows: "Where there is no sufficient reason to distinguish, there
can be no distinction."
So my question is
Is a (short and) beautiful proof for similar inequalities must exist always as the OPs want for desired answer?
| It not always exists, but very very wanted that we can find a nice solution.
I'll give one example.
In 1988 Walther Janous proposed the following problem (Crux 1366).
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\geq\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt2}.$$
This problem was an open issue until 2005 before Peter Scholze came and found the following.
We need to prove that:
$$\sum_{cyc}\frac{a^2}{\sqrt{a^2+b^2}}\geq\frac{a+b+c}{\sqrt2}$$ or
$$\sum_{cyc}\left(\frac{a^4}{a^2+b^2}+\frac{2a^2b^2}{\sqrt{(a^2+b^2)(b^2+c^2)}}\right)\geq\frac{1}{2}\sum_{cyc}(a^2+2bc),$$ which is true by Rearrangement:
$$\sum_{cyc}\left(\frac{a^4}{a^2+b^2}+\frac{2a^2b^2}{\sqrt{(a^2+b^2)(b^2+c^2)}}\right)-\frac{1}{2}\sum_{cyc}(a^2+2bc)=$$
$$=\sum_{cyc}\left(\frac{1}{2}\frac{a^4+b^4}{a^2+b^2}+\frac{2a^2b^2}{\sqrt{a^2+b^2}}\cdot\frac{1}{\sqrt{b^2+c^2}}\right)-\frac{1}{2}\sum_{cyc}(a^2+2bc)\geq$$
$$\geq\sum_{cyc}\left(\frac{1}{2}\frac{a^4+b^4}{a^2+b^2}+\frac{2a^2b^2}{\sqrt{a^2+b^2}}\cdot\frac{1}{\sqrt{a^2+b^2}}\right)-\frac{1}{2}\sum_{cyc}(a^2+2bc)=\frac{1}{2}\sum_{cyc}\frac{(a-b)^4}{a^2+b^2}\geq0.$$
If there is so beautiful solution for so hard problem, why we can not try to find something for another problems?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3326085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Inductively simplify specific Vandermonde determinant From Serge Lang's Linear Algebra:
Let $x_1$, $x_2$, $x_3$ be numbers. Show that:
$$\begin{vmatrix} 1 & x_1 & x_1^2\\ 1 &x_2 & x_2^2\\ 1 & x_3 &
x_3^2 \end{vmatrix}=(x_2-x_1)(x_3-x_1)(x_3-x_2)$$
The matrix presented above seems to be the specific case of Vandermonde determinant:
$$
\begin{vmatrix}
1 & x_1 & ... & x_1^{n-1}\\
1 &x_2 & ... & x_2^{n-1}\\
... & ... & ... & ...\\
1 & x_n & ... & x_n^{n-1}
\end{vmatrix}=\prod_{i, j}(x_i - x_j), \forall (1 \leq i \leq n) \land (1 \leq j \leq n)
$$
I'm trying to prove the specific case to then generalize it for arbitrary Vandermonde matrices.
My incomplete "proof"
Since determinant is a multilinear alternating function, it can be seen that adding a scalar multiple of one column (resp. row) to other column (resp. row) does not change the value (I omitted the proof to avoid too much text).
Thus considering that $x_1$ is a scalar, we can multiply each column but the last one of our specific Vandermonde matrix by $x_1$ and then starting from right to left subtract $n-1$th column from $n$:
$$\begin{vmatrix} 1 & x_1 & x_1^2\\ 1 &x_2 & x_2^2\\ 1 & x_3 &
x_3^2 \end{vmatrix}=\begin{vmatrix}
x_1 & 0 & 0 \\
x_1 & x_2 - x_1 & x^{2}_2 - x^{2}_1\\
x_1 & x_3 - x_1 & x^{2}_3 - x^{2}_1
\end{vmatrix}$$
Then using the expansion rule along the first row (since all the elements in it but $x_1$ are zero):
$$... =x_1\begin{vmatrix}
x_2 - x_1 & x^{2}_2 - x^{2}_1\\
x_3 - x_1 & x^{2}_3 - x^{2}_1
\end{vmatrix}=(x_1x_2-x^2_1)(x^2_{3}-x^2_1)-(x^{2}_2x_1 - x^{3}_1)(x_3x_1 - x^2_1)$$
The first expansion seems interesting because it contains $x_2 - x_1$ and $x_3 - x_1$ (which are first two factors of specific Vandermonde matrix), but further expansion does not give satisfying results.
Question:
Is this a good simple start of inductively "proving" relation between Vandermonde matrix and its factors? If so what does it lack to show the complete result? Did I make mistake during evaluation?
Thank you!
| "Since determinant is a multilinear alternating function, it can be seen that adding a scalar multiple of one column (resp. row) to other column (resp. row) does not change the value (I omitted the proof to avoid too much text)
" is right. But
$$
\begin{vmatrix} 1 & x_1 & x_1^2\\ 1 &x_2 & x_2^2\\ 1 & x_3 &
x_3^2 \end{vmatrix}
\neq
\begin{vmatrix}
x_1 & 0 & 0 \\
x_1 & x_2 - x_1 & x^{2}_2 - x^{2}_1\\
x_1 & x_3 - x_1 & x^{2}_3 - x^{2}_1
\end{vmatrix}
\neq
(x_1x_2-x^2_1)(x^2_{3}-x^2_1)-(x^{2}_2x_1 - x^{3}_1)(x_3x_1 - x^2_1)
$$
Remember that when you multiply a row or a column by $\lambda$, the determinant is multiplied by $\lambda$. And be careful when distributing $x_1$. We have
\begin{align}
\begin{vmatrix} 1 & x_1 & x_1^2\\ 1 &x_2 & x_2^2\\ 1 & x_3 &
x_3^2 \end{vmatrix}
&=
x_1
\begin{vmatrix}
x_1 & 0 & 0 \\
x_1 & x_2 - x_1 & x^{2}_2 - x^{2}_1\\
x_1 & x_3 - x_1 & x^{2}_3 - x^{2}_1
\end{vmatrix}\\
&= x_1^2
\begin{vmatrix}
x_2 - x_1 & x^{2}_2 - x^{2}_1\\
x_3 - x_1 & x^{2}_3 - x^{2}_1
\end{vmatrix}\\
&=
x_1^2((x_2 - x_1)(x^{2}_3 - x^{2}_1) - (x^{2}_2 - x^{2}_1)(x_3 - x_1))\\
&\neq (x_1x_2-x^2_1)(x^2_{3}-x^2_1)-(x^{2}_2x_1 - x^{3}_1)(x_3x_1 - x^2_1)
\end{align}
Keep in mind that we are trying to have the simplest possible factors.
Here, you can do
\begin{align}
\begin{vmatrix} 1 & x_1 & x_1^2\\ 1 &x_2 & x_2^2\\ 1 & x_3 &
x_3^2 \end{vmatrix}&=_{L_3 \leftarrow L_3 - L_2 \text{ and } L_2 \leftarrow L_2 - L_1}
\begin{vmatrix} 1 & x_1 & x_1^2\\ 0 &x_2 -x_1& (x_2 - x_1)(x_2+x_1)\\ 0 & x_3 - x_2 &
(x_3 - x_2)(x_3+x_2) \end{vmatrix}\\
&=_{L_3 \leftarrow L_3 - L_2} (x_2 - x_1)(x_3-x_2)
\begin{vmatrix} 1 & x_1 & x_1^2\\ 0 &1& x_2 + x_1\\ 0 & 0 &
x_3 -x_1 \end{vmatrix}\\
&=(x_2 - x_1)(x_3-x_2)(x_3-x_1)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3326310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
find min of $a+b$ given sum of $a\geq b\geq c\geq d$ is 9 and square sum is 21 Suppose $a\geq b\geq c\geq d>0$ and all are real numbers, and $a+b+c+d=9,a^2+b^2+c^2+d^2=21$, how to find the minmum of $a+b$?
What I attempted:
I can show $b\geq 1.5$ and $a\leq 3$.for $r\geq 0$, I consider $\sum(a-r)^2=\sum a^2-2r\sum a+4r^2=21-18r+4r^2$, then $a\leq \sqrt{21-18r+4r^2}+r$ which implies $a\leq 3$. Also I guess the min shoul be 5 and there are two solutions: 3,2,2,2 and $2.5,2.5,2.5,1.5$. But I cannot prove it.
New attempts:
*
*$6-2a=(a-3)^2+\sum(b-2)^2$
*$2d-3=\sum(a-2.5)^2+(d-1.5)^2$
then $2(d-a)+3\geq 1$ which implies that $d\leq c\leq b\leq a\leq d+1$
| Set $c+d=x,c^2+d^2=y$ then we have
$$\begin{cases}c=\frac{x+\sqrt{2y-x^2}}{2}\\ d=\frac{x-\sqrt{2y-x^2}}{2}\end{cases}$$
Also it is easy to find out that
$$b=\frac{9-x-\sqrt{42-2y-81+18x-x^2}}{2},$$
Note that $b\geq c$ we have
$$(9-2x)^2\geq 18x-2x^2-39+2\sqrt{(18x-x^2-2y-39)(2y-x^2)}$$
which also implie that
$$6x^2-54x+120\geq 2\sqrt{(18x-x^2-2y-39)(2y-x^2)}\geq 0$$
so $x\geq 5$ or $x\leq 4$ and since $c+d\leq 9/4$ so $x\leq 4$ which means
$$a+b=9-x\geq 5.$$
And "$=$" happens when $(a,b,c,d)=(3,2,2,2) \text{ or } (2.5,2.5,2.5,1.5)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3327246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Range of $f(x,y)=\frac{4x^2+(y+2)^2}{x^2+y^2+1}$ I am trying to find the range of this function:
$$f(x,y)=\frac{4x^2+(y+2)^2}{x^2+y^2+1}$$
So I think that means I have to find minima and maxima. Using partial derivatives gets messy, so I was wondering if I could do some change of variables to make it easier computationally. But no change of coordinates that I can think of have really simplified it much. If I set $2w=y+2$, then I get a problem below. Am I thinking of the right strategy, or is there something better I could do?
| For
$$f(x,y)=\frac{4x^2+(y+2)^2}{x^2+y^2+1}$$
partial derivatives are not so terrifying:
\begin{align}
D_1f(x,y)&=\frac{8x(x^2+y^2+1)-2x(4x^2+(y+2)^2)}{(x^2+y^2+1)^2}\\[4px]
&=\frac{2xy(3y-4)}{(x^2+y^2+1)^2}
\\[8px]
D_2f(x,y)&=\frac{2(y+2)(x^2+y^2+1)-2y(4x^2+(y+2)^2)}{(x^2+y^2+1)^2}\\[4px]
&=\frac{2(-3x^2y+2x^2-2y^2-3y+2)}{(x^2+y^2+1)^2}
\end{align}
The first derivative vanishes for $x=0$, $y=0$ or $y=4/3$.
We have (the denominator is irrelevant as it doesn't vanish)
\begin{align}
D_2f(0,y)&=\frac{-2(2y^2+3y-2)}{\dots} \\[4px]
D_2f(x,0)&=\frac{4(x^2+1)}{\dots} \\[4px]
D_2f(x,4/3)&=\frac{-4(x^2+19/3)}{\dots}
\end{align}
Thus we only get critical points for $x=0$ and $2y^2+3y-2=0$, that is, $y=-2$ or $y=1/2$.
Since $f(0,-2)=0$ and $f(0,1/2)=5$, the minimum is $0$ and the maximum is $5$.
Well, how do we know that $0$ is the minimum? Because obviously the function only takes on nonnegative values. Why $(0,5)$ is a point of maximum and not a saddle point? Because $f(x,y)$ is bounded:
$$
f(x,y)\le\frac{4x^2+4(y+2)^2}{x^2+y^2+1}=4+\frac{16y+12}{x^2+y^2+1}\le16+\frac{16y}{x^2+y^2+1}
$$
Now note that $|y|\le x^2+y^2+1$, because $|y|^2-|y|+1\ge0$. Thus
$$
|f(x,y)|\le 32
$$
Hence the function must have an absolute maximum which has to be at a critical point. Checking with the Hessian would be indeed a nuisance.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3329898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Evaluate $\int _0^1\int _0^1\int _0^1\int _0^1\sqrt{(z-w)^2+(x-y)^2} \, dw \, dz \, dy \, dx$ This page contains an interesting identity$$\int _0^1\int _0^1\int _0^1\int _0^1\sqrt{(z-w)^2+(x-y)^2} \, dw \, dz \, dy \, dx=\frac{1}{15} \left(\sqrt{2}+2+5 \log \left(\sqrt{2}+1\right)\right)$$
Which calculates the average distance between two random points in the unit square. I'm wondering if there's an elementary solution? Any help will be appreciated.
| By symmetry:
$$I=4\int _0^1\int _0^1\int _0^x\int _0^z\sqrt{(z-w)^2+(x-y)^2} \, dw \, dy \, dx dz$$
Substituting $w \to z w, \quad y \to x y$:
$$I=4\int _0^1\int _0^1\int _0^1\int _0^1\sqrt{z^2(1-w)^2+x^2(1-y)^2}~ x z \, dw \, dy \, dx dz$$
Substituting $w \to 1-w, \quad y \to 1-y$:
$$I=4\int _0^1\int _0^1\int _0^1\int _0^1\sqrt{z^2 w^2+x^2 y^2}~ x z \, dw \, dy \, dx dz$$
Substituting $x^2 = u, \quad z^2 = v$:
$$I=\int _0^1\int _0^1\int _0^1\int _0^1\sqrt{v w^2+u y^2}~ \, dw \, dy \, du dv$$
Substituting $v w^2=p, \quad u y^2=q$:
$$I=\int _0^1\int _0^1\int _0^{y^2}\int _0^{w^2} \sqrt{p+q}~ \, dp dq \frac{dy dw}{y^2 w^2}$$
$$I=\frac{2}{3} \int _0^1\int _0^1\int _0^{y^2}\left((q+w^2)^{3/2}-q^{3/2} \right) dq \frac{dy dw}{y^2 w^2}$$
$$I=\frac{4}{15} \int _0^1\int _0^1\left((y^2+w^2)^{5/2}-y^5-w^5 \right) \frac{dy dw}{y^2 w^2}$$
By symmetry:
$$I=\frac{8}{15} \int _0^1\int _0^w\left((y^2+w^2)^{5/2}-y^5-w^5 \right) \frac{dy dw}{y^2 w^2}$$
Substituting: $y = w s$:
$$I=\frac{8}{15} \int _0^1\int _0^1 w^2 \left((1+s^2)^{5/2}-s^5-1 \right) \frac{ds dw}{s^2 }$$
$$I=\frac{8}{45} \int _0^1 \left((1+s^2)^{5/2}-s^5-1 \right) \frac{ds }{s^2 }$$
The last single integral can be evaluated by different methods, and the result agrees with the one from the OP.
One good method here is integration by parts. I leave this as an exercise.
| {
"language": "en",
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"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
Non-graphical solution to $5\log_{4}a\ + 48\log_{a}4 = \frac{a}{8}$ It is required to solve
$$5 \log_{4}a\ + 48\log_{a}4 = \frac{a}{8}$$
Here is my attempt,
Let $$ x = \log_{4}a$$, then $$a = 2^{2x}$$
And our equation becomes
$$ 5x^{2} - x\cdot 2^{2x-3} + 48 = 0$$
But this is as far as I can go. I've tried several substitutions but no progress.
| To localize the solution x=4 you can write
$$
5x^2 - x \cdot 2^{2x - 3} + 48 = 0
$$
as
$$
5x + \frac{{48}}
{x} = 2^{2x - 3}
$$
This tell you that x must be positive. Now, first search for integer solutions. It is obvious that $$x=1$$ is not a solution and therefore you can imagine that $$x \geq 2$$. In this case the RHS is integer and therefore x must be a divisor of 48. Among these numbers you have that only x=4 and x=12 let the LHS a power of 2. Since x=12 is not a solution while x=4 does you have that the only integer solution is x=4.
Now let be
$$
f(x) = 2^{2x - 3} - 5x
$$
and
$$g(x)=\frac{48}{x}$$
It is g(x) <12 if x>4 and g(x)>12 if $$0<x<4$$. We will prove that f(x)>12 if x>4 and f(x)<12 per $$0<x<4$$. Namely let be x=4+t with t>0. It is
$$
\begin{gathered}
2^{2(4 + t) - 3} - 5(4 + t) > 12 \Leftrightarrow \hfill \\
2^{5 + 2t} - 32 - 5t > 0 \Leftrightarrow \hfill \\
32 \cdot 4^t - 32 - 5t > 0 \hfill \\
\end{gathered}
$$
But with t>0 it is
$$4^t>1+t$$
therefore
$$32 \cdot 4^t - 32 - 5t > 32\left( {1 + t} \right) - 32 - 5t = 27t > 0
$$
Hence there is no solutions if x>4. If $$0<x<4$$ we can write $$x=4-t$$ with $$0<t<4$$. It is
$$
\begin{gathered}
2^{2(4 - t) - 3} - 5(4 - t) < 12 \Leftrightarrow \hfill \\
2^{5 - 2t} - 32 + 5t < 0 \Leftrightarrow \hfill \\
32 \cdot \frac{1}
{{4^t }} - 32 + 5t < 0 \hfill \\
\end{gathered}
$$
We have that
$$
\begin{gathered}
32 \cdot \frac{1}
{{4^t }} - 32 + 5t < 32 \cdot \frac{1}
{{1 + t}} - 32 + 5t = \hfill \\
= \frac{{ - 32t}}
{{1 + t}} + 5t = \frac{{ - 27t + 5t^2 }}
{{1 + t}} = t\frac{{5t - 27}}
{{1 + t}} < 0 \hfill \\
\end{gathered}
$$
which is negative for $$0<t<4$$.
Therefore there are no solutions with $$0<x<4$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3334918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Regarding perfect squares Is there any positive integer $n > 2$ such that $(n - 1)(5n - 1)$ is a perfect square? It is observed that $(n - 1)(5n - 1)$ is of the form $4k$ or $4k+ 1$.
Affirmative answers were given by Pspl and Mindlack (by providing some examples).
Now my question is the following:
Is there any characterization of positive integer $n$ such that $(n - 1)(5n - 1)$ is a perfect square?
| As $(n-1)(5n-1)=m^2$ for some $m\in\mathbb{N}$, then
$$\begin{split}(3n-1)^2-(2n)^2&=m^2\\m^2+(2n)^2&=(3n-1)^2\end{split}$$
By the formula for Pythagorean Triple, we can use it.
Let $m=x^2-y^2, 2n=2xy, 3n-1=x^2+y^2$ and assume $x>y>0$. Then
$$\begin{cases}
(x + y)^2 =5n-1 \\
(x-y)^2 = n-1
\end{cases}$$
By the assumption, $x+y,x-y>0$. So $x=\frac{\sqrt{5n-1}+\sqrt{n-1}}{2},y=\frac{\sqrt{5n-1}-\sqrt{n-1}}{2}$.
$\because x,y\in\mathbb{N}$
$\therefore n-1,5n-1$ must be perfect squares in order to have $(n-1)(5n-1)$ be a perfect square.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Simple sum with $i = 2$ step I need to calculate the following sum:
$$S_{n} = \sum_{2 \leq i \leq n } (3i - 2)$$
(two more conditions for the above sum: $n$ is even and $i$ with step $2$ (not sure how to do multi-line))
Adding image of the task with multi-line visible:
I wrote down the first few terms and it looks like this:
$4, 10, 16, 22, ...$
I re-arranged a bit and obtained such sum:
$$S_{n} = \sum_{i = 0}^{n} 4+6i$$
and its terms are the same: $4, 10, 16, 22, ...$
Is the top boundary correct?
Proceeding with the solution:
$$S_{n} = \sum_{i = 0}^{n} 4+6i$$
$$S_{n} = 4 \sum_{i = 0}^{n} + 6 \sum_{i = 0}^{n} i$$
$$S_{n} = 4n + 4 + 6 \sum_{i = 0}^{n} i$$
$$S_{n} = 4n + 4 + 6 \frac{(1+n)n}{2}$$
$$S_{n} = 4n + 4 + 3 (n^2 + n)$$
$$S_{n} = 3n^2 + 7n + 4$$
Testing for $n = 3$:
$$S_{3} = 3(3^2) + 21 + 4 = 52 = \Bigg(4 + 10 + 16 + 22 \Bigg)$$
So the result seems correct. But is it for sure? Plugging $3$ into $n$ should take into account only first $3$ or first $4$ elements? (works fine for first $4$ elements, but seems kinda wrong for me)
| Make the substitution $j=i-2$: $$S_n=\sum_{i=2}^{n}(3i-2) = \sum_{j=0}^{n-2}[3(j+2)-2] = \sum_{j=0}^{n-2}(3j+4) = 3 \cdot \sum_{j=0}^{n-2} j + 4 \cdot \sum_{j=0}^{n-2} 1 .$$ Now use the well known formula $\sum_{j=0}^{m} j = \frac{m(m+1)}{2}$ and get
$$ S_n = 3 \cdot \frac{(n-2)(n-1)}{2} + 4 \cdot (n-1) = \frac{3n^2-9n+6+8n-8}{2} = \frac{3n^2-n-2}{2} = \frac{(3n+2)(n-1)}{2}.$$
| {
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Inverse of a skew-symmetric matrix For $a, x, y, z \in \mathbb R$, let $$M= \left(
\begin{array}{cccc}
\cos(a) & \sin(a) \, x & \sin(a)\, y & \sin(a) \, z \\
-\sin(a) \, x & \cos(a) & \sin(a) \,z & -\sin(a)\, y \\
-\sin(a) \, y & -\sin(a) \, z & \cos(a) & \sin(a) \, x \\
-\sin(a) \, z & \sin(a) \, y & -\sin(a)\, x & \cos(a)
\end{array}
\right).$$
Without doing much calculation, why the matrix $M-I_4$ is invertible and why its inverse is given $$(M-I_4)^{-1}=\,\frac{-1}{2} I_{4} - \frac{\cot(\sqrt{x^2+y^2+z^2} )}{2\sqrt{x^2+y^2+z^2}} A,$$
where $A$ is the skew-symmetric matrix given by
$$A=\left(
\begin{array}{cccc}
0 & x & y & z \\
-x & 0 & z & -y \\
-y & -z & 0 & x \\
-z & y & -x & 0
\end{array}
\right).$$
Thank you in advance
| Here's an approach which is relatively light on computation.
As in the comments, we observe that $A^2 = -(x^2 + y^2 + z^2)I$. It follows that any rational function of $A$ can be written in the form $p I + q A$ for some $p,q$. In other words, there necessarily exist numbers $p,q$ such that
$$
(M-I)^{-1} = ([\cos(a) - 1]I + \sin(a)A)^{-1} = p I + q A
$$
Now, expand the product $(M-I)(pI + qA),$ and solve for the $p$ and $q$ that cause this product to equal $I$. In particular, we have
$$
(M - I)(pI + qA) = ([\cos(a) - 1]I + \sin(a)A)(pI + qA)\\
= p(\cos(a) - 1)I + (q[\cos(a) - 1] + p\sin(a))A + \sin(a)qA^2\\
= \left[p(\cos(a) - 1) - \sin(a)(x^2 + y^2 + z^2)q \right]I + \left[q(\cos(a) - 1) + p\sin(a)\right]A.
$$
Thus, it suffices to find $p,q$ that solve the system of equations
$$
\begin{array}{ccccccc}
(\cos(a) - 1) &p &- &\sin(a)(x^2 + y^2 + z^2)&q &= &1\\
\sin(a) &p &+ &(\cos(a) - 1)&q &= &0.
\end{array}
$$
Solving this system yields
$$
p = \frac{1 - \cos(a)}{D}, \quad q = \frac{\sin(a)}{D}
$$
where $D$ is the determinant of the coefficient matrix of this system, that is
$$
D = \sin^2(a)(x^2 + y^2 + z^2) + (\cos(a) - 1)^2.
$$
I see no general way of simplifying the resulting expression.
| {
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Quartic polynomial roots The equation
$x^4 - x^3-1=0$
has roots α, β, γ, δ. By using the substitution $y=x^3$ find the exact value of $α^6+β^6+γ^6+δ^6$ .
The solution is
$x=y$ (1/3)
$y^4=(1+y)^3$
$y^4 -y^3 -3y^2-3y^2-1=0$
$S$N+4 $=$ $S$N + $S$N+3
$S$-1 = $\frac {0}{1} =0$
$S$2 = $ 1^2 -2*0 =1$
$S$3 = $0+1 =1$
$S$4 = $1+4 =5$
$S$5 = $5+1=6$
$S$6 = $6+1 =7$
Therefore, $α^6+β^6+γ^6+δ^6 = 7$
Can someone explain what is this $S$n is and how did they work out solution from that.
| Multiply each term by $x^n$
$\pmb S_n =x^n$
Then it becomes
$S$N+4 $=$ $S$N + $S$N+3
| {
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Big Factorials And Rules With Their Digits This is the issue I have.
Given that $34! = 295,232,799,**cd**9,604,140,847,618,609,643,5**ab**,000,000$ determine the digits $a$, $b$, $c$, $d$.
As far as I've gotten is $34!$ contains these multiples of $5$. $5$, $10$, $15$, $20$, $25$, $30$. So $34!$ is divisible by $5^7$ ($25$ has two powers of $5$), and since there's at least that power of $2$, the last $7$ digits are $0$, so b is $0$.
| Yes, $b=0$ for the reason you stated ($34!$ is divisible by $10^7$ but not $10^8$ so $b=0$ but $a\neq0$). That's step $1$.
Step $2$. Now say you divided $34!$ by the $10^7$ so you take away seven products of $2$ and seven product of $5$ (speaking in terms of prime factors). What you're left with has at least three products of $2$ (since every other integer up to $34$ is even so there's at least seventeen products of $2$). If a number is divisible by $8$ then it's last three digits is also divisible by $8$ since $1000$ (and any multiple of it) is divisible by $8$. Therefore $35a$ is divisible by $8$.
$35a = 300 + 50 + a = (4 + 2 + a) ($mod $ 8)$. $4+2+a$ can only equal $8$ (not $16$ since $a$ would have to be $10$), there $a=2$.
Step $3$. Now, $34!$ is also divisible by $9$ (which is relatively prime to $10$) so it's digits must add up to a multiple of $9$. The sum of the digits (including $a$ and $b$) comes to $141+c+d$. $141+c+d = (6+c+d)($mod $9)$. $6+c+d=9$ or $18$.
So $c+d=3$ or $12$. Let's call this equation $(1)$.
Step $4$. Now, $34!$ is also divisible by $11$ (which is relatively prime with $10$ and $9$) so (according to a proof I'll address in a second) the sum of the odd-positioned numbers subtracted from the evenly-positioned numbers is a multiple of $11$. This gives the equation $80 - 61 + d - c = (8 + d - c)($mod $11) = 0$ $ ($mod $11)$. So $8+d-c=0$ or $11$ .
So $d-c=-8$ or $3$. Let's call this equation $(2)$.
$(1)+(2)$ gives: $2d=$ $...$
This must be the sum of two evens or two odds for $d$ to be an integer.
So the two potential pairs give the following:
$(c,d)=(10,2)$ or $(0,3)$.
Finally, $(a,b,c,d)=(2,0,0,3)$.
The proof for step $4$ is as follows. Basically, $10=(-1)($mod $11)$ and $100=10^2=1$ $($mod $11)$. Every odd power of ten gives $(-1)($mod $11)$ and every even power gives $1$ $($mod $11)$. So the sum of the evenly-positioned digits (are multiplied by an odd power of $10$ and therefore are essentially multiplied by $-1$. Equally the sum of the oddly-positioned digits (are multiplied by an even power of $10$ and therefore are essentially multiplied by $1$.
| {
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A matrix of order 8 over $\mathbb{F}_3$ What is an example of an invertible matrix of size 2x2 with coefficients in $\mathbb{F}_3$ that has exact order 8?
I have found by computation that the condition that the 8th power of a matrix
$\begin{bmatrix}a & b\\c & d\end{bmatrix}$ is the identity is
$$
b c (a + d)^2 (a^2 + 2 b c + d^2)^2 + ((a^2 + b c)^2 +
b c (a + d)^2)^2=1, \qquad b (a + d) (a^2 + 2 b c + d^2) (a^4 + 4 a^2 b c + 2 b^2 c^2 +
4 a b c d + 4 b c d^2 + d^4)=0, \qquad c (a + d) (a^2 + 2 b c + d^2) (a^4 + 4 a^2 b c + 2 b^2 c^2 +
4 a b c d + 4 b c d^2 + d^4)=0, \qquad b c (a + d)^2 (a^2 + 2 b c +
d^2)^2 + (b c (a + d)^2 + (b c + d^2)^2)^2=1
$$
and the condition for invertibility is $ad\neq bc$. If the 4th power is not the identity, then no power that is not a multiple of 8 is not the identity (because we could cancel out to either get that the first power is the identity or that the second power is the identity, both lead to contradiction). That is another cumbersome condition to write out.
I hope somebody can suggest a nicer way.
| Using a similar strategy to Travis, we can actually answer this question quite nicely:
$$A \text{ has order 8 if and only if }\det(A) = 2 \text{ and } \text{tr}(A) \neq 0.$$ We can do this by looking at the characteristic polynomial $p(x)$ of $A$, with which we can utilize the fact that $p(A) = 0$ as shown below.
Since $A$ is a $2\times 2$ matrix, this takes the simple form $p(x) = x^2 - tx + d$ where $t$ is the trace of $A$ and $d$ the determinant (which cannot be 0 since $A$ is invertible). If the trace is 0, then $A^2 = dI$ and so has order $2$ if $d = 1$ and 4 if $d = 2$. Thus if we are looking for an invertible matrix $A$ with order 8, we know that $d$ and $t$ must be nonzero.
We can now calculate powers of $A$ quite readily through successive squaring, use two key observations to simplify the calculations: 1)in $\mathbb{F}_3$, 2 is the same as $-1$ and so we can replace any appearance of 2 with a negative sign and 2) since $t$ and $d$ are both either $1$ or $2 = -1$, it is always true that $t^2 = d^2 = 1$. The calculations in $\mathbb{F}_3$ are as follows, using the fact that $p(A) = 0$:
$$A^2 = tA-dI$$
$$ A^4 = (tA-dI)^2 = A^2+tdA + I = t(1+d)A + (1-d)I,$$
$$ A^8 = (1+d)^2A^2+(1-d)^2I = -(d+1)A^2 + (d-1)I$$
$$\implies A^8 = [d(d+1)+(d-1)]I - t(d+1)A = -dI - t(d+1)A.$$
We can therefore conclude that the order of $A$ is 8 if and only if $-d = 1$ and $t(d+1) = 0$, which since $t \neq 0$ are both equivalent to $d = 2$. Therefore, an invertible matrix $A \in GL(2,\mathbb{F}_3)$ has order 8 iff $\det(A) = 2$ and tr$(A) \neq 0$ (which we showed is necessary before). You'll notice that this agrees with the concrete examples given by Travis and Chris Culter.
| {
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For $z \in \mathbb{C}$, if $(z^a,z^c,z^b \bar{z}^d,z^{b+d})=(1,1,1,1)$, then $z=1$ under certain gcd conditions for $a,b,c,d$.
Let $z \in \mathbb{C}$. I would like to show that if $(z^a,z^c,z^b \bar{z}^d,z^{b+d})=(1,1,1,1)$ and gcd$(a,b,c,d)=1$ and gcd$(a^2-c^2,b^2-d^2)=4$, then $z=1$.
Note that I am not sure whether this is true or not, but I believe it is. My argument starts as follows:
Assume that $z$ is a primitive $n$th root of unity, that is, $z^n=1$ and $n$ is the smallest such positive integer. Then
$z^a=1 \implies n |a$
$z^c=1 \implies n |c$
$z^b \bar{z}^d=1 \implies z^b=z^d \implies n |(b-d)$
$z^{b+d}=1 \implies n|(b+d)$.
Using the division conditions that n divides $b-d$ and $b+d$ it follows that $n|2b$ and $n|2d$. Moreover, since $n$ divides $a$ and $n$ divides $c$ it follows that $n$ divides $2a, 2b,2c$ and $2d$. Hence $n=1$ or $n=2$ because gcd$(2a,2b,2c,2d)=2$. If $n=2$ then since $2|a$ and $2|c$, then $4$ divides $a^2-c^2$. Also, since $2$ divides $(b-d)$ and $(b+d)$, we have that $4$ divides $b^2-d^2$.
Now, this does not seem to give me a contradiction, so I am stuck.
| The given condition is false. Take $z = -1$, and $a=4,c=2$, with $b=3,d=1$. Then, of course $\mbox{gcd}(a,b,c,d) = 1$, and $a^2-c^2 = 12$ with $b^2-d^2 = 8$ whose $\gcd$ is $4$. Furthermore, $\bar z = z$, so $z^a = z^c= 1$ and $z^b\bar z^d = z^bz^d=z^{b+d} = z^4=1$.
| {
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If $a$, $b$, $c$ are sides of a triangle, prove $2(a+b+c)(a^2+b^2+c^2)\geq3(a^3+b^3+c^3+3abc)$
$a$, $b$, $c$ are sides of a triangle, prove:
$$2(a+b+c)(a^2+b^2+c^2)\geq3(a^3+b^3+c^3+3abc)$$
What I have tried:
$$
⇔2\sum (a+b)ab\geq \sum a^3+9abc
$$
so I can't use
$$\sum a^3+3abc\geq \sum ab(a+b)$$
| Also, we can use SOS here.
Indeed, by your work we need to prove that
$$\sum_{cyc}(2a^2b+2a^2c-a^3-3abc)\geq0.$$
Now, let $a\geq b\geq c$.
Thus, since $$3b-a-c=b+c-a+2b-2c>0,$$ we obtain:
$$\sum_{cyc}(2a^2b+2a^2c-a^3-3abc)=\sum_{cyc}(2a^2b+2a^2c-4abc-a^3+abc)=$$
$$=\sum_{cyc}\left(2c(a-b)^2-\frac{1}{2}(a+b+c)(a-b)^2\right)=\frac{1}{2}\sum_{cyc}(a-b)^2(3c-a-b)\geq$$
$$\geq\frac{1}{2}((a-b)^2(3c-a-b)+(a-c)^2(3b-a-c))\geq$$
$$\geq\frac{1}{2}((a-b)^2(3c-a-b)+(a-b)^2(3b-a-c))=(a-b)^2(b+c-a)\geq0.$$
| {
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Integral solutions of $a^{2} + a = b^{3} + b$ Find all pairs of coprime positive integers $(a,b)$ such that
$b<a$ and $a^2+a=b^3+b$
My approach:
$a(a+1)=b(b^2+1)$ so $a|(b^2+1)$ and $b|(a+1)$
Now after this I am not able to do anything.pls help.
| As you already note, if $a$ and $b$ are coprime and
$$a(a+1)=a^2+a=b^3+b=b(b^2+1),$$
then it follows that $b$ divides $a+1$. Then $a=bc-1$ for some integer $c$, where $c>1$ because $a>b$. Then
$$b^3+b=a^2+a=(bc-1)^2+(bc-1)=c^2b^2-cb,$$
and since $b$ is positive we can divide both sides by $b$ and rearrange to get the quadratic
$$b^2-c^2b+c+1=0,$$
in $b$.
This shows that the integer $b$ is a root of a quadratic equation with discriminant
$$\Delta=(-c^2)^2-4\cdot1\cdot(c+1)=c^4-4c-4.$$
In particular this means $c^4-4c-4$ is a perfect square. Of course $c^4$ is itself a perfect square, and the previous one is
$$(c^2-1)^2=c^4-2c^2+1,$$
which shows that $-4c-4\leq-2c^2+1$, or equivalently
$$2c^2-4c-5\leq0.$$
A quick check shows that this implies $c<3$, so $c=2$. Then this plugging back in yields
$$b^3+b=a^2+a=(2b-1)^2+(2b-1)=4b^2-2b,$$
which we can rearrange to get the cubic
$$b^3-4b^2+3b=0\qquad\text{ and hence }\qquad b^2-4b+3=0.$$
Then either $b=1$ or $b=3$, corresponding to $a=1$ and $a=5$, respectively. Hence the only solution with $a>b$ is $(a,b)=(5,3)$.
| {
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Prove that $0<\sum_{k=1}^n \frac{g(k)}{k} - \frac{2n}{3} < \frac{2}{3}$ Prove
$$0<\sum_{k=1}^n \frac{g(k)}{k} - \frac{2n}{3} < \frac{2}{3}$$
where $g(k)$ is the greatest odd divisor of k
Please Find Holes in my Proof.
Let $k=2m+1$ if we show that the right hand side of the equation is true for odd numbers, then it is true for even numbers since there is a net total of $1/3$ since $g(k)/k = 1$ for odd numbers.
$$\sum_{k=1}^n\frac{g(k)}{k} <\frac{2n}{3} + \frac{2}{3}$$
$$\sum_{k=1}^n\frac{g(k)}{k} <\frac{2(2m+1)}{3} + \frac{2}{3}$$
$$\sum_{k=1}^n\frac{g(k)}{k} <\frac{4m}{3} + \frac{4}{3}$$
From $1$ to $2m+1$ there are $m$ even numbers and $m+1$ even numbers, the value of $\frac{g(k)}{k}$ for even numbers is $\frac{1}{2^{V2(k)}}$ where $V2(k)$ is the exponent of 2 in the factorization of k
$$\sum_{k=1}^n\frac{g(k)}{k} = m+1 + (m)\frac{1}{2} -((\lfloor{\frac{m}{2}\rfloor \frac{1}{4}) +(\lfloor{\frac{m}{4}\rfloor \frac{1}{8}}})...)$$
$$\leq m+1 + (m)\frac{1}{2} - (\frac{m}{8} + \frac{m}{32}...) = \frac{4m}{3} + 1 <\frac{2(2m+1)}{3} + \frac{2}{3}$$
Solving for the left hand side of the inequality
$$0<\sum_{k=1}^n \frac{g(k)}{k} - \frac{2n}{3}$$ Let $k = 2m$ with the same reasoning as above
$$\sum_{k=1}^n\frac{g(k)}{k} = m + (m)\frac{1}{2} -((\lfloor{\frac{m}{2}\rfloor \frac{1}{4}) +(\lfloor{\frac{m}{4}\rfloor \frac{1}{8}}})...)$$
$$\geq m + (m)\frac{1}{2} - (\frac{m}{8} + \frac{m}{32}...) = \frac{4m}{3} + \frac{1}{3} >\frac{2(2m)}{3}$$
I'm not sure about the solution since it was a strict inequality to begin with. Is this a correct solution? Any other solutions are welcome
| Let me prove a slightly sharper inequality. While it looks slightly more complicated it gives a better insight:
$$ \frac 1{3\cdot 2^{\lfloor \log_2(n)\rfloor}}\,\ \le
\,\ \sum_{k=1}^n\frac{g(k)}k\ -\ \frac 23\cdot n\,\ \le
\,\ \frac 23 - \frac 1{3\cdot 2^{\lfloor \log_2(n)\rfloor}}
$$
Even better, the above inequality follows instantly from the
following exact formula:
$$ \sum_{k=1}^n\frac{g(k)}k\ -\ \frac 23\cdot n\,\ =
\,\ \frac 13\cdot \sum_{k\in E_n} 2^{-k}
\qquad\qquad (*) $$
where $\ E_n\subseteq\mathbb N\ $ is such that
$$ n\,\ =\ \sum_{k\in E_n}2^k $$
In order to see (*), it takes two simple lemmas (exercises):
*
*formula (*) holds for $\ n=2^K\ $ for arbitrary non-negative integer K (then $\ E_n=\{K\}$), namely
$$ \sum_{k=1}^{2^K}\frac{g(k)}k\ -\ \frac 23\cdot 2^K\,\ =
\,\ \frac 13\cdot 2^{-K}
\qquad\qquad (\%) $$
*if $\ m\ =\ 2^M+n,\ $ where $\ 2^M>n\ $ then
$$ \sum_{k=2^M+1}^m \frac{g(k)}k\,\ =
\,\ \sum_{k=1}^n \frac{g(k)}k $$
It may help to see that
$$ k\ =\ 2^s\cdot(2\cdot t-1)\qquad \Longrightarrow\qquad
\frac{g(k)}k\ =\ 2^{-s} $$
That's all.
| {
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proof based on cauchy schwartz inequality If a,b,c are positive real numbers,prove that
$$ \frac{a}{b+2c} + \frac{b}{c+2a} + \frac{c}{a+2b} \ge 1 $$
I tried solving and i have no idea how to proceed I mechanically simplified it it looks promising but im still stuck. This is from the excersice on Cauchy Schwartz Inequality.
| As we have, $2+a^3=a^3+1+1\geqslant 3a$, $b^2+1\geqslant 2b$, thus$$\dfrac{a}{a^3+b^2+c}=\frac{a}{3+a^3+b^2-a-b}\leqslant\frac{a}{3a+2b-a-b}=\frac{a}{2a+b}.$$Similarly, we can get $$\dfrac{b}{b^3+c^2+a}\leqslant\frac{b}{2b+c},\,\,\,\,\,\dfrac{c}{c^3+a^2+b}\leqslant\frac{c}{2c+a}.$$It suffices to show$$\frac{a}{2a+b}+\frac{b}{2b+c}+\frac{c}{2c+a}\leqslant 1\iff \frac{b}{2a+b}+\frac{c}{2b+c}+\frac{a}{2c+a}\geqslant 1$$
By Cauchy's inequality, we get$$(b(2a+b)+c(2b+c)+x(2c+a))\left(\frac{b}{2a+b}+\frac{c}{2b+c}+\frac{a}{2c+a}\right)\geqslant (a+b+c)^2.$$ As $b(2a+y)+z(2b+c)+a(2c+a)=(a+b+c)^2$,
So,
$$\frac{b}{2a+b}+\frac{c}{2b+c}+\frac{a}{2c+a}\geqslant 1$$
| {
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The sum of infinite series $(1/2)(1/5)^2 + (2/3)(1/5)^3 + (3/4)(1/5)^4..................$ Initially, I broke this series as $1/2 = 1-1/2, 2/3 = 1-1/3, 3/4 = 1-1/4$, then I got two series as it is
$$= (1/5)^2 + (1/5)^3 + (1/5)^4 +.......-[1/2(1/5)^2 + 1/3(1/5)^3 + 1/4(1/5)^4 +.....]$$
After solving the first part, we reached at this point
$$= 1/20 - [1/2(1/5)^2 + 1/3(1/5)^3 + 1/4(1/5)^4 +.....]$$
Please help someone in solving this question. I am very grateful to you.
| Thanks all who have given the answers. Some people use "Lerch Transenden" methods to solve some uses "integration", "convergence" etc. But I was looking to solve it as using 10+2 level math techniques.
Finally I got success in solving it. See my solution below -
= (1/2)(1/5)^2 + (2/3)(1/5)^3 + (3/4)(1/5)^4 +.........
= (1- 1/2)(1/5)^2 + (1- 1/3)(1/5)^3 + (1- 1/4)(1/5)^4 +.........
= (1/5)^2+(1/5)^3+(1/5)^4+.......−{1/2(1/5)^2+1/3(1/5)^3+1/4(1/5)^4+.....}
= (1/5)+(1/5)^2+(1/5)^3+(1/5)^4+.......−{(1/5)+1/2(1/5)^2+1/3(1/5)^3+1/4(1/5)^4+.....}
= (1/5)/(1- 1/5) - {-log(1- 1/5)}
= (1/5)/(4/5) + log(4/5)
= 1/4 + log(4/5)
Here log(1-x) = -x -x^2/2 -x^3/3 -..........
Once again thank you all of you.
| {
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"url": "https://math.stackexchange.com/questions/3350958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How many $14$-digit even numbers can be formed using $0,1,1,2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5$? What is a good way to approach this question?
How many $14$-digit even numbers can be formed using $0,1,1,2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5$?
I know that I can list the possible outcomes with different ending numbers and then add up all the outcomes. However, is there a way to solve the question without categorizing adding different types together?
| Case 1: The number ends in $0$. Considering all possible orderings of the other $13$ numbers we get $$\frac{13!}{2!2!3!3!3!}$$ possible outcomes (since we have two of $1$ and $2$ each and three of $3,4,5$ each, we divide by $2!2!3!3!3!$) to remove duplicates.
Case 2: The number does not end in $0$. We have $5$ possible choices for the last digit, which leaves us with $12$ possible choices for the leading digit, $12$ possible choices for the second digit (include zero), $11$ choices for the third and so on. Basically,
$$\frac{5\cdot 12\cdot 12!}{2!2!3!3!3!}.$$
Again, we divide to avoid duplicates.
Hence, we get a total of $$\frac{13! + 5\cdot 12\cdot 12!}{2!2!3!3!3!} = 40471200.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3351314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Relate to Dirichlet's theorem Diophantine approximation I have a problem:
Let $a \in \mathbb{Z}$, $a \geq 3$, and set $\xi= \sum_{n=0}^\infty
10^{-a^{2n}}>0$. Then the inequality
$$\Big|\,\xi - \dfrac{x}{y}\,\Big| \leq \dfrac{1}{y^a}$$ has infinitely many solutions with $x,y \in \mathbb{Z}$, $y>0$ and $\gcd(x,y)=1$.
I would imitate the proof of Dirichlet's theorem as follow, I claim my lemma
Lemma: Let $\xi=\sum_{n=0}^\infty 10^{-a^{2n}}$, for every integer $Q \geq 2$, there are integers $x,y$ which are not both equal to $0$, such that
$$|x - \xi y| \leq \dfrac{1}{Q^{a-1}}$$
with $0<y \leq Q$ and $\gcd(x,y) =1$
*
*Try to prove this lemma:
Partition the interval $[0,1]$ into $Q^{a-1}$ subintervals of length $\dfrac{1}{Q^{a-1}}$. Consider $Q^{a-1}+1$ numbers $\xi-[\xi]$,..,$Q^{a-1}\xi-[Q^{a-1}\xi]$ and $1$. By the Dirichlet principle, two among these numbers must lie in the same subinterval of length $\dfrac{1}{Q^{a-1}}$. Hence we can find $x,y \in \mathbb{z}$ such that $|x-y \xi| \leq \frac{1}{Q^{a-1}}$. But now my trouble is $y$ is not smaller than $Q$.
Does anyone have other ideas?
| With regards to other ideas, just straight attack. Noting $y=10^{a^{2n}}$ then
$$\sum\limits_{k=0}^{n}\frac{1}{10^{a^{2k}}}=\frac{\sum\limits_{k=0}^{n}10^{a^{2n}-a^{2k}}}{y}$$
where $x=\sum\limits_{k=0}^{n}10^{a^{2n}-a^{2k}}=\sum\limits_{k=0}^{\color{red}{n-1}}10^{a^{2n}-a^{2k}}\color{red}{+1}=10\cdot Q\color{red}{+1}$ and $\gcd(x,y)=1$.
Now
$$\left|\xi -\frac{x}{y}\right|=
\sum\limits_{k=n+1}^{\infty}\frac{1}{10^{a^{2k}}}=
\frac{1}{10^{a^{2n+2}}}\left(\sum\limits_{k=n+1}^{\infty}\frac{1}{10^{a^{2k}-a^{2n+2}}}\right)< \\
\frac{1}{10^{a^{2n+2}}}\left(\sum\limits_{k=0}^{\infty}\frac{1}{10^{k}}\right)=
\frac{1}{10^{a^{2n+2}}}\left(\frac{1}{1-\frac{1}{10}}\right)=\\
\frac{1}{10^{a^{2n}\cdot a^2}}\cdot \frac{10}{9}=
\frac{1}{y^{a^2}}\cdot \frac{10}{9}<
\frac{1}{y^{a+1}}\cdot \frac{10}{9}=\frac{1}{y^a}\cdot \frac{10}{y\cdot9}<\frac{1}{y^a}$$
for infinitely $n$ and because $y\geq10, \forall n\geq0$ and $a^2 > a+1, \forall a\geq3$.
| {
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"url": "https://math.stackexchange.com/questions/3352604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Is my summation result correct? $\sum_{i=0}^{n-1} \sum_{j=0}^{i-1} (i+j) =\cdots= \frac{n(n-1)^2}{2}$ Is this result correct?
$$\begin{align}
\sum_{i=0}^{n-1} \sum_{j=0}^{i-1} (i+j)
&= \sum_{i=0}^{n-1}\left( i^2 + \frac{i(i-1)}{2} \right)\\
&= \frac{3}{2}\sum_{i=0}^{n-1}i^2 - \frac{1}{2}\sum_{i=0}^{n-1}i \\
&= \frac{3}{2}\frac{n(n-1)(2n-1)}{2} + \frac{n(n-1)}{4} \\
&= \frac{n(n-1)^2}{2}.
\end{align}$$
| You are almost correct. It should be
\begin{align}
\sum_{i=0}^{n-1} \sum_{j=0}^{i-1} (i+j)
&= \frac{3}{2}\sum_{i=0}^{n-1}i^2 - \frac{1}{2}\sum_{i=0}^{n-1}i \\
&= \frac{3}{2}\cdot \frac{n(n-1)(2n-1)}{\color{red}{6}} -\frac{1}{2}\cdot \frac{n(n-1)}{2}\\
&=\frac{n(n-1)((2n-1)-1)}{4}=\frac{n(n-1)^2}{2}.
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/3352845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Olympiad Inequality $\frac{ab+bc+ca}{a+b+c}(\sum_{cyc}\frac{1}{2a+b})\leq 1$ with proof (long post) It takes me a lot a time to solve this :
Let $a,b,c>0$ with $a\geq b\geq c$ then we have :
$$\frac{ab+bc+ca}{a+b+c}(\sum_{cyc}\frac{1}{2a+b})\leq 1$$
My proof :
We need some lemmas to begin :
First lemma :
Let $x,y>0$ then we have :
$$\frac{x}{x+\frac{1}{2}}+\frac{y}{y+\frac{1}{2}}=\frac{\frac{x+y}{2}}{\frac{x+y}{2}+\frac{1}{2}}+\frac{p}{p+\frac{1}{2}}$$
Where $p$ is equal to:
$$\frac{\frac{x}{(x+\frac{1}{2})^2}+\frac{y}{(y+\frac{1}{2})^2}}{\frac{1}{(x+\frac{1}{2})^2}+\frac{1}{(y+\frac{1}{2})^2}}$$
the proof is easy it's just calculus .
Second lemma :
With the notation of the first lemma we have :
$$\frac{\frac{x+y}{2}}{\frac{x+y}{2}+\frac{1}{2}}\geq \frac{p}{p+\frac{1}{2}}$$
Proof :
With the first lemma and the concavity of $\frac{x}{x+\frac{1}{2}}$ and the use of Jensen's inequality :
$$\frac{x}{x+\frac{1}{2}}+\frac{y}{y+\frac{1}{2}}-\frac{\frac{x+y}{2}}{\frac{x+y}{2}+\frac{1}{2}}=\frac{p}{p+\frac{1}{2}}\leq \frac{\frac{x+y}{2}}{\frac{x+y}{2}+\frac{1}{2}} $$
Now we attack the problem :
We prove a stronger result :
We have with $x=\frac{1}{a}$ and $y=\frac{1}{b}$ and $z=\frac{1}{c}$:
$$\sum_{cyc}\frac{1}{2a+b}+\sum_{cyc}\frac{1}{2b+a}=\sum_{cyc}\frac{xy}{2x+y}+\sum_{cyc}\frac{xy}{2y+x}$$
But :
$$\frac{xy}{2x+y}+\frac{yz}{2z+y}=y(\frac{\frac{1}{2}\frac{x}{y}}{\frac{x}{y}+\frac{1}{2}}+\frac{\frac{1}{2}\frac{z}{y}}{\frac{z}{y}+\frac{1}{2}})$$
Now we apply the first lemma to get :
$$y(\frac{\frac{1}{2}\frac{x}{y}}{\frac{x}{y}+\frac{1}{2}}+\frac{\frac{1}{2}\frac{z}{y}}{\frac{z}{y}+\frac{1}{2}})=\frac{\frac{x+z}{4}}{\frac{\frac{x}{y}+\frac{x}{y}}{2}+\frac{1}{2}}+Q$$
Finally we have :
$$\sum_{cyc}\frac{xy}{2x+y}+\sum_{cyc}\frac{xy}{2y+x}=\sum_{cyc}\frac{\frac{x+z}{4}}{\frac{\frac{x}{y}+\frac{x}{y}}{2}+\frac{1}{2}}+Q+R+T$$
We multiply by $\frac{x+y+z}{xy+yz+zx}$ :
$$\frac{x+y+z}{xy+yz+zx}(\sum_{cyc}\frac{xy}{2x+y}+\sum_{cyc}\frac{xy}{2y+x})=\frac{x+y+z}{xy+yz+zx}(\sum_{cyc}\frac{\frac{x+z}{4}}{\frac{\frac{x}{y}+\frac{x}{y}}{2}+\frac{1}{2}})+\frac{x+y+z}{xy+yz+zx}(Q+R+T)$$
But :
$$\frac{x+y+z}{xy+yz+zx}(\sum_{cyc}\frac{\frac{x+z}{4}}{\frac{\frac{x}{y}+\frac{x}{y}}{2}+\frac{1}{2}})=1$$
And with the second lemma we have :
$$\frac{x+y+z}{xy+yz+zx}(\sum_{cyc}\frac{\frac{x+z}{4}}{\frac{\frac{x}{y}+\frac{x}{y}}{2}+\frac{1}{2}})=1\geq\frac{x+y+z}{xy+yz+zx}(Q+R+T)$$
So we prove :
$$\frac{x+y+z}{xy+yz+zx}(\sum_{cyc}\frac{xy}{2x+y}+\sum_{cyc}\frac{xy}{2y+x})\leq 2$$
Or
$$\frac{ab+bc+ca}{a+b+c}(\sum_{cyc}\frac{1}{2a+b}+\sum_{cyc}\frac{1}{2b+a})\leq 2$$
But :
$$\sum_{cyc}\frac{1}{2a+b}-\sum_{cyc}\frac{1}{2b+a}= -\frac{(2 (a - b) (a - c) (b - c) (2 a^2 + 7 a b + 7 a c + 2 b^2 + 7 b c + 2 c^2))}{((2 a + b) (a + 2 b) (2 a + c) (a + 2 c) (2 b + c) (b + 2 c))}$$
And with the condition $a\geq b\geq c$ it's negative so :
$$2\frac{ab+bc+ca}{a+b+c}(\sum_{cyc}\frac{1}{2a+b})\leq\frac{ab+bc+ca}{a+b+c}(\sum_{cyc}\frac{1}{2a+b}+\sum_{cyc}\frac{1}{2b+a})\leq 2$$
Done !
My question is have you an alternative proof ?
Thanks a lot for your time !
| After full expanding we need to prove that
$$\sum_{cyc}(2a^3b-a^2b^2-a^2bc)\geq0$$ or
$$\sum_{cyc}(a^3b+a^3c-a^2b^2-a^2bc)\geq\sum_{cyc}(a^3c-a^3b)$$ or
$$\sum_{cyc}(a^3b+a^3c-a^2b^2-a^2bc)\geq(a-b)(b-c)(c-a)(a+b+c),$$ which is true because by Muirhead
$$\sum_{cyc}(a^3b+a^3c-a^2b^2-a^2bc)\geq0$$ and by the given $$(a-b)(b-c)(c-a)(a+b+c)\leq0$$
| {
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"url": "https://math.stackexchange.com/questions/3354020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Determining The $I^{th}$ Number In The Sequence I've recently come across a sequence of numbers but I can't determine any sensible way to say what the $I^{th}$ number in the pattern will be. The pattern is as follows:
$\frac{1}{2}, \\\frac{1}{4}, \frac{3}{4}, \\\frac{1}{8}, \frac{5}{8}, \frac{3}{8}, \frac{7}{8}, \\\frac{1}{16}, \frac{9}{16}, \frac{5}{16}, \frac{13}{16}, \frac{3}{16}, \frac{11}{16}, \frac{7}{16}, \frac{15}{16},\\ \frac{1}{32}, \frac{17}{32}, \frac{9}{32}, \frac{25}{32}, \frac{5}{32}, \frac{21}{32}, \frac{13}{32}, \frac{29}{32}, \frac{3}{32}, \frac{19}{32}, \frac{11}{32}, \frac{27}{32}, \frac{7}{32}, \frac{23}{32}, \frac{15}{32}, \frac{31}{32}, \\ \textrm{and so on. }$
The denominator is easy to determine, and it's also clear that once a numerator appears, it will also be in subsequent levels, but how the "new numerators" at each level are determined is unclear to me. If I find this I may be able to determine an expression for the $I^{th}$ number.
| Ignore the denominator. For the $i^\text{th}$ number, convert that number to binary, reverse the order of the digits, and then convert that back to a base 10 number.
| {
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"timestamp": "2023-03-29T00:00:00",
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Hard inequality :$\Big(\frac{1}{a^2+b^2}\Big)^2+\Big(\frac{1}{b^2+c^2}\Big)^2+\Big(\frac{1}{c^2+a^2}\Big)^2\geq \frac{3}{4}$ I have a hard problem this is it :
Let $a,b,c>0$ such that $a^ab^bc^c=1$ then we have :
$$\Big(\frac{1}{a^2+b^2}\Big)^2+\Big(\frac{1}{b^2+c^2}\Big)^2+\Big(\frac{1}{c^2+a^2}\Big)^2\geq \frac{3}{4}$$
I try to use Jensen's ienquality applied to the function $f(x)=\frac{1}{x^2}$ but it doesn't works for all the values .
if we apply Am-Gm it's like Jensen's inequality so I forget this ways .
Maybe If we apply Karamata's inequality but I didn't found the right majorization .
I try also to use Muirhead inequality but without success .
So I'm a bit lost with that if you have a hint or a full answer I will be happy to read your work .
| Define: $$F(a,b,c)=\left(\frac{1}{a^2+b^2}\right)^2+\left(\frac{1}{b^2+c^2}\right)^2+\left(\frac{1}{c^2+a^2}\right)^2$$
By the method of Lagrange multipliers, the minimizer of $F$ under the constraint $a^ab^bc^c=1$ must be a critical point of the Lagrangian
$$L(a,b,c,\lambda)=\left(\frac{1}{a^2+b^2}\right)^2+\left(\frac{1}{b^2+c^2}\right)^2+\left(\frac{1}{c^2+a^2}\right)^2+\lambda(a^ab^bc^c-1)$$
To find the critical points we solve $\nabla L=0$:
$$\frac{\partial L}{\partial\lambda}=a^ab^bc^c-1=0$$
$$\frac{\partial L}{\partial a}=-\frac{4a}{(a^2+b^2)^3}-\frac{4a}{(a^2+c^2)^3}+\lambda(a^ab^bc^c)(1+\log a)=0$$
and similarly find the equation for $\frac{\partial L}{\partial b}=0$ and $\frac{\partial L}{\partial c}=0$. We got a system of 4 (non-linear) equations with 4 variables.
Let $a,b,c,\lambda$ be a solution. Since $a^ab^bc^c=1$, at least one of $a,b,c$ is $\leq 1$ and at least one is $\geq 1$, so WLOG we can treat two cases: $a\leq b\leq 1\leq c$ and $a\leq 1\leq b\leq c$ .
*
*First case $a\leq b\leq 1\leq c$: Isolating $\lambda$ in the equations $\frac{\partial L}{\partial a}=0$ and $\frac{\partial L}{\partial b}=0$ leads to
$$\lambda=\left[\frac{1}{(a^2+b^2)^3}+\frac{1}{(a^2+c^2)^3}\right]\cdot\frac{a}{1+\log a}=\left[\frac{1}{(a^2+b^2)^3}+\frac{1}{(b^2+c^2)^3}\right]\cdot\frac{b}{1+\log b}$$
If $a<b$ then
$$\left[\frac{1}{(a^2+b^2)^3}+\frac{1}{(a^2+c^2)^3}\right]>\left[\frac{1}{(a^2+b^2)^3}+\frac{1}{(b^2+c^2)^3}\right]$$
and
$$\frac{a}{1+\log a}>\frac{b}{1+\log b}$$
by contradiction to the equality above. Therefore we conclude $a=b$, and we may reduce our problem to 2 dimensions: Minimize
$$F(a,a,c)=\frac{1}{4a^4}+2\left(\frac{1}{a^2+c^2}\right)^2$$
Among all $a,c$ s.t. $a^{2a}c^c=1$. Using the square super root function we put $c=\text{ssrt}(a^{-2a})$ in the above expression and get
$$G(a)=\frac{1}{4a^4}+2\left(\frac{1}{a^2+\text{ssrt}(a^{-2a})^2}\right)^2$$
Convince yourself that $G$ attains its minimum at $a=1$ (I used a computer, maybe possible to do it analytically), so to minimize $G$ we must take $a=1$. We conclude $a=b=1$, and then $c=\text{ssrt}(1)=1$ as well.
*Second case $a\leq 1\leq b\leq c$: Isolating $\lambda$ in the equations $\frac{\partial L}{\partial b}=0$ and $\frac{\partial L}{\partial c}=0$ leads to
$$\lambda=\left[\frac{1}{(b^2+a^2)^3}+\frac{1}{(b^2+c^2)^3}\right]\cdot\frac{b}{1+\log b}=\left[\frac{1}{(c^2+a^2)^3}+\frac{1}{(c^2+b^2)^3}\right]\cdot\frac{c}{1+\log c}$$
If $b<c$ then
$$\frac{1}{(b^2+a^2)^3}\frac{b}{1+\log b}>\frac{1}{(c^2+a^2)^3}\frac{c}{1+\log c}$$
and
$$\frac{1}{(b^2+c^2)^3}\frac{b}{1+\log b}>\frac{1}{(c^2+b^2)^3}\frac{c}{1+\log c}$$
by contradiction to the equality above. Therefore we conclude $b=c$, and continue as in the first case to show $a=b=c=1$.
To conclude:
$$\min\{F(a,b,c);a^ab^bc^c=1\}=F(1,1,1)=\frac{3}{4}$$
Remark: I may have over-complicated things. If you can show in a more simple way that $a=b=c=\lambda=1$ is the only solution to $\nabla L=0$ then you are done.
| {
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"timestamp": "2023-03-29T00:00:00",
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partial fraction decomposition in integral I have a question that asks to calculate the following integral:
$$ \int_0^\infty {\frac{w\cdot \sin w}{4a^2+w^2}dw} $$
In the official solution they used partial fraction decomposition in order to later use Plancherel's identity:
$$ \frac{w\cdot \sin w}{4a^2+w^2} = $$
$$ \frac{\sin w}{w}\cdot\frac{w^2}{w^2+4a^2} = $$
$$ \frac{\sin w}{w}(1-\frac{4a^2}{w^2+4a^2}) = $$
$$ \frac{\sin w}{w} - \frac{\sin w}{w} \cdot \frac{4a^2}{w^2+4a^2} $$
And then used Plancherel's identity.
But I didn't understand how to expand to partial fraction and in particular I didn't understand this equation:
$$ \frac{\sin w}{w}\cdot\frac{w^2}{w^2+4a^2} = \frac{\sin w}{w}(1-\frac{4a^2}{w^2+4a^2}) $$
Can you please explain how to expand the integrand into partial fraction?
| Consider the function:
$$I(a)=\int_0^\infty {\frac{w\cdot \sin (w)}{w^2+4\cdot a^2}\text{d}w},\space a>0$$
Rewrite the integrand as follows:
$$\frac{w\cdot \sin (w)}{w^2+4\cdot a^2}=\frac{w}{w}\cdot\frac{w\cdot \sin (w)}{w^2+4\cdot a^2}=\frac{\sin(w)}{w}\cdot\frac{w^2}{w^2+4\cdot a^2}=\frac{\sin(w)}{w}\cdot\frac{w^2+4\cdot a^2-4\cdot a^2}{w^2+4\cdot a^2}=\frac{\sin(w)}{w}\cdot\left(\frac{w^2+4\cdot a^2}{w^2+4\cdot a^2}-\frac{4\cdot a^2}{w^2+4\cdot a^2}\right)=\frac{\sin(w)}{w}\cdot\left(1-\frac{4\cdot a^2}{w^2+4\cdot a^2}\right)=\frac{\sin(w)}{w}-\frac{\sin(w)}{w}\cdot\frac{4\cdot a^2}{w^2+4\cdot a^2}$$
Then:
$$I(a)=\int_0^\infty {\frac{\sin (w)}{w}\text{d}w}-\int_0^\infty {\frac{\sin(w)}{w}\cdot\frac{4\cdot a^2}{w^2+4\cdot a^2}\text{d}w}$$
The left-hand integral is known as a Dirichlet integral and it can be derived that it evaluates to $\frac{\pi}{2}$:
$$I(a)=\frac{\pi}{2}-\int_0^\infty {\frac{\sin(w)}{w}\cdot\frac{4\cdot a^2}{w^2+4\cdot a^2}\text{d}w}$$
Let $w\mapsto 2\cdot a\cdot w$:
$$I(a)=\frac{\pi}{2}-\int_0^\infty {\frac{\sin(2\cdot a\cdot w)}{2\cdot a\cdot w}\cdot\frac{4\cdot a^2}{(2\cdot a\cdot w)^2+4\cdot a^2}\cdot(2\cdot a\space\text{d}w)}=\frac{\pi}{2}-\int_0^\infty {\frac{\sin(2\cdot a\cdot w)}{ w}\cdot\frac{1}{w^2+1}\text{d}w}$$
Recognize that the integrand is a continuous and continuously differentiable function and differentiate with respect to $a$ under the integral sign:
$$I'(a)=\frac{\text{d}}{\text{d}w}\left[\frac{\pi}{2}-\int_0^\infty {\frac{\sin(2\cdot a\cdot w)}{ w}\cdot\frac{1}{w^2+1}\text{d}w}\right]=-\int_0^\infty {\frac{\partial}{\partial a}\frac{\sin(2\cdot a\cdot w)}{ w}\cdot\frac{1}{w^2+1}\text{d}w}=-2\cdot\int_0^\infty {\frac{w}{w}\cdot\frac{\cos(2\cdot a\cdot w)}{w^2+1}\text{d}w}=-2\cdot\int_0^\infty {\frac{\cos(2\cdot a\cdot w)}{w^2+1}\text{d}w}$$
Recognize that the integrand is a continuous and continuously differentiable function and differentiate with respect to $a$ under the integral sign:
$$I''(a)=-2\cdot\frac{\text{d}}{\text{d}a}\int_0^\infty {\frac{\cos(2\cdot a\cdot w)}{w^2+1}\text{d}w}=-2\cdot\int_0^\infty {\frac{\partial}{\partial a}\frac{\cos(2\cdot a\cdot w)}{w^2+1}\text{d}w}=4\cdot\int_0^\infty {\frac{w\cdot\sin(2\cdot a\cdot w)}{w^2+1}\text{d}w}$$
Consider the original expression for $I(a)$:
$$I(a)=\int_0^\infty {\frac{w\cdot \sin (w)}{w^2+4\cdot a^2}\text{d}w}$$
Let $w\mapsto 2\cdot a\cdot w$:
$$I(a)=\int_0^\infty {\frac{2\cdot a\cdot w\cdot \sin(2\cdot a\cdot w)}{(2\cdot a\cdot w)^2+4\cdot a^2}\cdot(2\cdot a\space\text{d}w)}=\int_0^\infty {\frac{w\cdot \sin(2\cdot a\cdot w)}{w^2+1}\text{d}w}$$
Recognize that
$$I''(a)=4\cdot I(a)\Rightarrow I''(a)-4\cdot I(a)=0$$
Solving the differential equation yields
$$I(a) = \text{c}_{1}\cdot e^{2\cdot a} + \text{c}_{2}\cdot e^{-2\cdot a}$$
Differentiate with respect to $a$ on both sides:
$$I'(a) = 2\cdot\left(\text{c}_{1}\cdot e^{2\cdot a} - \text{c}_{2}\cdot e^{-2\cdot a}\right)$$
According to the closed form of $I(a)$, as $a$ approaches $0$, $I(a\rightarrow 0)=\text{c}_{1}+\text{c}_{2}$.
According to the integral form of $I(a)$, as $a$ approaches $0$, $I(a\rightarrow 0)=\frac{\pi}{2}-\int_0^\infty {0\space\text{d}w}=\frac{\pi}{2}$.
According to the closed form of $I'(a)$, as $a$ approaches $0$, $I'(a\rightarrow 0)=2\cdot(\text{c}_{1}-\text{c}_{2})$.
According to the integral form of $I'(a)$, as $a$ approaches $0$, $I'(a\rightarrow 0)=-2\cdot\int_0^\infty {\frac{1}{w^2+1}\text{d}w}=-2\cdot\frac{\pi}{2}=-\pi$.
It can be derived that $\text{c}_{1}=0$ and $\text{c}_{2}=\frac{\pi}{2}$.
Then,
$$I(a)=\int_0^\infty {\frac{w\cdot \sin (w)}{w^2+4\cdot a^2}\text{d}w}=\frac{\pi}{2}\cdot e^{-2\cdot a},\space a>0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3356858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How many triplets $(p,q,r)$ of integers between $1$ and $1000$ satisfy $p^2\sin^2x+q\sin2x+r^2\cos^2x>1$, if $\sin x$ and $\cos x$ are non-zero?
We have an inequality of the form
$$p^2\sin^2 x + q\sin2x + r^2\cos^2x > 1$$
for integers $p$, $q$, $r$ with $1\leq p,q,r \leq 1000$, and such that $x$ cannot be any value that makes $\sin x$ or $\cos x$ equal to $0$.
How many triplets of the form $(p,q,r)$ satisfy the above inequality?
Note. I actually need a pseudo code for this, but I was stuck in the mathematical approach itself so I came here.
My approach:
I tried finding an upper bound by breaking $q\sin2x$ as $2q\sin x\cos x$ and clubbing it with the remaining two terms to get something of the form of
$$A\sin x + B\cos x > 1$$
The maximum value of LHS can $\sqrt{A^2+ B^2}$.
But I haven't made any progress after this.
| Divide both sides by $\cos^2x$ and writing $\tan x=t$
$$p^2t^2+2qt+r^2>1+t^2$$
$$t^2(p^2-1)+2qt+r^2-1>0$$
Now the roots of $$t^2(p^2-1)+2qt+r^2-1=0$$ are $t=\dfrac{-q\pm\sqrt{q^2-(p^2-1)(r^2-1)}}{p^2-1}$
So, either $t>$max $\left(\dfrac{-q+\sqrt{q^2-(p^2-1)(r^2-1)}}{p^2-1},\dfrac{-q-\sqrt{q^2-(p^2-1)(r^2-1)}}{p^2-1}\right)$
or either $t<$ min $\left(\dfrac{-q+\sqrt{q^2-(p^2-1)(r^2-1)}}{p^2-1},\dfrac{-q-\sqrt{q^2-(p^2-1)(r^2-1)}}{p^2-1}\right)$
As $\sin x\cos x\ne0,t$ must be non-zero and finite
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3357097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\frac{a}{b^{2}+1} + \frac{b}{c^{2}+1} + \frac{c}{a^{2} + 1} \ge \frac{3}{2}$ $a,b,c > 0$ and $a+b+c=3$, prove
$$ \frac{a}{b^{2} + 1} + \frac{b}{c^{2}+1} + \frac{c}{a^{2}+1} \ge 3/2 $$
Attempt:
Notice that by AM-Gm
$$\frac{a}{b^{2} + 1} + \frac{b}{c^{2}+1} + \frac{c}{a^{2}+1} \ge 3\frac{\sqrt[3]{abc}}{\sqrt[3]{(a^{2}+1)(b^{2}+1)(c^{2}+1)}} $$
Now, AM-GM again
$$ a^{2}+b^{2}+c^{2} + 3 \ge 3 \sqrt[3]{(a^{2}+1)(b^{2}+1)(c^{2}+1)} ... (1)$$
Then $a+b+c = 3 \ge 3 \sqrt[3]{abc} \implies 1 \ge \sqrt[3]{abc}$. Also
$$a^{2} + b^{2} + c^{2} \ge 3 \sqrt[3]{(abc)^{2}}$$
multiply by $1 \ge \sqrt[3]{abc}$ and will get
$$ a^{2} + b^{2} + c^{2} \ge 3 abc ... (2)$$
subtract $(1)$ with $(2)$ and get
$$ 3 \ge 3 \sqrt[3]{(a^{2}+1)(b^{2}+1)(c^{2}+1)} - 3 abc$$
$$ 3 + 3 abc \ge \sqrt[3]{(a^{2}+1)(b^{2}+1)(c^{2}+1)} $$
$$ \frac{3abc}{\sqrt[3]{(a^{2}+1)(b^{2}+1)(c^{2}+1)}} \ge 1 - \frac{3}{\sqrt[3]{(a^{2}+1)(b^{2}+1)(c^{2}+1)}} $$
How to continue..?
| Since for $x>0$ we have (just draw a graph for ${1\over 1+x^2}$ and a tangent at $x=1$) $${1\over 1+x^2}\geq -{1\over 2}x+1$$ it is enough to check if $$-{1\over 2}(ab+bc+ca)+3\geq {3\over 2}$$ i.e. $$3\geq ab+bc+ca$$ is true?
Since $$a^2+b^2+c^2\geq ab+bc+ca$$ that is easy to verify. :)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3360763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
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