Challenge Math Challenge - July 2019

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QuantumQuest

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I may have overthought about problem 2.
Yes indeed. The solution you give in post #19 using simple calculus is much faster and simpler. Well done.

I noticed that P4 required the use of calculus? :eek: What, precisely, does that mean? If my approach doesn't qualify, it's fine.
There are various ways that can be used but I ask for the use of calculus as it is a fast and simple approach. Your first solution for question ##4## gets unnecessarily complex but the second is fine.
 

QuantumQuest

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Let ##f(x) = arctan(x).##

Since ##arctan(x)## is differentiable for all x, by the Mean Value Theorem we know
$$
\frac{f(b) - f(a)}{b - a} = f'(c)
$$
for some c on the interval ##a < c < b##.

##f'(x) = \frac{1}{1 + x^2}##, which is decreasing for ##x > 0##, so for ##0 < a < c < b##, ##f'(b) < f'(c) < f'(a)##.

Substituting, we have
$$
f'(b) < \frac{f(b) - f(a)}{b - a} < f'(a)
$$
$$
\frac{1}{1 + b^2} < \frac{arctan(b) - arctan(a)}{b - a} < \frac{1}{1 + a^2}
$$
$$
\frac{b - a}{1 + b^2} < arctan(b) - arctan(a) < \frac{b - a}{1 + a^2}
$$

Let ##a = 1, b = 4/3##.

$$
\frac{\frac{4}{3} - 1}{1 + (\frac{4}{3})^2} < arctan(4/3) - arctan(1) < \frac{\frac{4}{3} - 1}{1 + 1^2}
$$
$$
\frac{3}{25} < arctan(4/3) - \frac{\pi}{4} < 1/6
$$
$$
\frac{\pi}{4} + \frac{3}{25} < arctan(4/3) < \frac{\pi}{4} + \frac{1}{6}
$$
Well done @Flatlanderr
 
My solution to 11.
View attachment 246120

By the way i wanted to know whether can we solve the same integral but indefinite.If we can please provide a solution.
How you change ##x \to \frac{\pi}{2}-x## without changing interval of integration?
 

Math_QED

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How you change ##x \to \frac{\pi}{2}-x## without changing interval of integration?
The new integration bounds are from ##\pi/2## to ##0## and we get a minus sign for the integral as a result of the substitution. Switching the integral bounds removes the minus sign.
 
How you change ##x \to \frac{\pi}{2}-x## without changing interval of integration?
This is a propertie of definite integral
##\int_0^a f(x) \, dx =## ##\int_0^a f(0+a-x) \, dx ##
 

fresh_42

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Oh, come on ...!

No graphics, handwritings and even more, no uploads on external servers please!
Presentation is part of what should be learnt with these riddles. To know a solution is far from being able to convince others that your solution is one.
 
I am having a lot of problem with latex. I am not that bad at latex in AoPS or Stack exchange but I am not able to use it properly here :(. Example, my first line of answer should be this-

By AM-GM

$$ \frac{(u+(1-v)+v+(1-w)+w+(1-u))}{6} \geq
(u(1-v)v(1-w)w(1-u))^{\frac{1}{6}$$

but the Latex is not working. What am I doing wrong?
 

fresh_42

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I am having a lot of problem with latex. I am not that bad at latex in AoPS or Stack exchange but I am not able to use it properly here :(. Example, my first line of answer should be this-

By AM-GM

$$ \frac{(u+(1-v)+v+(1-w)+w+(1-u))}{6} \geq
(u(1-v)v(1-w)w(1-u))^{\frac{1}{6}$$

but the Latex is not working. What am I doing wrong?
This is due to your laziness. If you copy and paste instead of actually writing it, you also copy hidden control sequences as colors or fonts. This doesn't work in LaTeX here.

If it helps you, you can download a script program, e.g. AutoHotkey, which allows you to abbreviate certain key sequences by keyboard shortcuts. For example I have \frac{}{} on Ctrl+F or \ begin{bmatrix} \ end{bmatrix} on Alt+M (without the blanks).
 
@fresh_42 Thanks for telling my mistake! Now I can latex again 😁
By AM-GM

$$ \frac {(u)+(1-v)+(v)+(1-w)+(w)+(1-u)} {6} \geq [(u)(1-v)(v)(1-w)(w)(1-u)]^{\frac{1}{6}}$$

Implies

$$ \frac {1} {2} \geq [(u)(1-v)(v)(1-w)(w)(1-u)]^{\frac{1}{6}}$$

Squaring both sides (works in this inequality as both sides are positive)

$$ \frac {1} {4} \geq [(u)(1-v)(v)(1-w)(w)(1-u)]^{\frac{1}{3}}$$

Now using GM-HM

$$ \frac {1} {4} \geq [(u)(1-v)(v)(1-w)(w)(1-u)]^{\frac{1}{3}} \geq \frac {3} {\frac{1}{u(1-v)}+\frac{1}{v(1-w)} + \frac{1}{w(1-u)}}$$

Implies

$$ \frac{1}{u(1-v)} + \frac{1}{v(1-w)} + \frac{1}{w(1-u)} \geq 12 $$

W.L.O.G assume ## u(1-v) \leq v(1-w) \leq w(1-u) ##

then ## \frac{3}{u(1-v)} \geq 12 ##

implies $$\frac{1}{4} \geq u(1-v) $$

Hence Proved
 

fresh_42

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@fresh_42 Thanks for telling my mistake! Now I can latex again 😁
By AM-GM

$$ \frac {(u)+(1-v)+(v)+(1-w)+(w)+(1-u)} {6} \geq [(u)(1-v)(v)(1-w)(w)(1-u)]^{\frac{1}{6}}$$

Implies




$$ \frac {1} {2} \geq [(u)(1-v)(v)(1-w)(w)(1-u)]^{\frac{1}{6}}$$




Squaring both sides (works in this inequality as both sides are positive)

$$ \frac {1} {4} \geq [(u)(1-v)(v)(1-w)(w)(1-u)]^{\frac{1}{3}}$$

Now using GM-HM

$$ \frac {1} {4} \geq [(u)(1-v)(v)(1-w)(w)(1-u)]^{\frac{1}{3}} \geq \frac {3} {\frac{1}{u(1-v)}+\frac{1}{v(1-w)} + \frac{1}{w(1-u)}}$$

Implies

$$ \frac{1}{u(1-v)} + \frac{1}{v(1-w)} + \frac{1}{w(1-u)} \geq 12 $$

W.L.O.G assume ## u(1-v) \leq v(1-w) \leq w(1-u) ##

then ## \frac{3}{u(1-v)} \geq 12 ##

implies $$\frac{1}{4} \geq u(1-v) $$

Hence Proved
Well done and far better written than your first attempts, and I do not mean LaTeX, I mean the structure of your proof!

You should consider to download this little helper. It saves so much time, that I can almost type at a normal speed without all these special characters. I even use keys for \alpha , \beta, \omega etc.

If anyone still wants to try: there is another proof using a binomial formula.
 
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BWV

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#7
Isnt #7 just the variance of W?
Var(W)=E[W^2]-E^2[W]
E[W]=E^2[W]=0
So
var(w)=E[W^2]=t
 
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fresh_42

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Well, from science advisors and such experienced members of PF as you are we can certainly expect that they can solve problems meant for the kids still in school.
 

hilbert2

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For anyone learning proof writing and presentation skills in mathematics, it's important to remember that a proof never becomes any better by making it look like something esoteric is taking place there. Most "normal" people won't understand it anyway, even when it's made "as simple as possible but no simpler" (a quote from Einstein). When a professional mathematician writes a new long proof of some theorem, it's difficult enough to follow the reasoning in it even when it contains all the necessary detail and the reader is a professional of equal standing.
 
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Great, I am not normal :cry:
 
Set ##A=1+i, B=1-i, C=-1-i, D=-1+i## as the vertices of the square.
Our circle passes through ##A, D## and ##\frac{C+B}{2}=M=-i##
The center lies on the intersection of y-axis and perpendicular bisector of ##DM##.
equating slopes (or equations lines, depends on you), we get the center ##K= \frac{i}{4}##
Using distance formula we find circle intersects the side DC at point ##L=(-1,-0.5)##.
So we get, the circle divides the side DC into the ratio 3:1
 

Math_QED

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#7
Isnt #7 just the variance of W?
Var(W)=E[W^2]-E^2[W]
E[W]=E^2[W]=0
So
var(w)=E[W^2]=t
It is not. Why do you think so?

Note that the question is changed to calculating ##\mathbb{E}[X]##, which is a lot easier.
 
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283
I know P13 was solved, but I like the exercise so I tried something, too.
For every [itex]0<x<1[/itex] it holds [itex]x(1-x)\leq 1/4[/itex] (find the vertex of the parabola). Assume [itex]u(1-v) > 1/4[/itex] and [itex]v(1-w) > 1/4[/itex]. Need to show [itex]w(1-u) \leq 1/4[/itex].

If [itex]1-u \leq 1-w[/itex], then
[tex]w(1-u) \leq w(1-w) \leq 1/4.[/tex]

Assume [itex]1-u>1-w[/itex]. From
[tex]
v(1-u) > v(1-w) > 1/4 \geq w(1-w)
[/tex]
we further obtain [itex]u<w<v[/itex]. But now
[tex]
1/4 < u(1-v) < u(1-u) \leq 1/4,
[/tex]
which is impossible. Thus [itex]1-u \leq 1-w[/itex] must hold.
 
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168
Problem 9
I recognize this problem as Cholesky decomposition - Wikipedia

The proof of uniqueness is by construction, and this construction uses the Cholesky-Banachiewicz-Crout algorithm, as it might be called. One must find lower-triangular matrix L for matrix A such that they suffer ## \mathbf{A} = \mathbf{L} \mathbf{L}^T ##. It is easy to show that taking the transpose of both sides yields this equation again. Here is that algorithm, for A having size n*n:

For j = 1 to n do:
$$ L_{jj} = \sqrt{ A_{jj} - \sum_{k=1}^{j-1} L_{jk}^2 } $$
For i = (j+1) to n do:
$$ L_{ij} = \frac{1}{L_{jj}} \left( A_{ij} - \sum_{k=1}^{j-1} L_{ik} L_{jk} \right) $$
Next i
Next j

Each new L component depends only on an A component and on previously-calculated L components, so after one pass, the calculation is complete.

I must also calculate L for
$$ A = \begin{pmatrix} 4 & 2 & 4 & 4 \\ 2 & 10 & 17 & 11 \\ 4 & 17 & 33 & 29 \\ 4 & 11 & 29 & 39 \end{pmatrix} $$

I used Mathematica's CholeskyDecomposition[] function and took the result's transpose, since that function calculates an upper triangular matrix. I then verified that that result is correct. It is
$$ L = \begin{pmatrix} 2 & 0 & 0 & 0 \\ 1 & 3 & 0 & 0 \\ 2 & 5 & 2 & 0 \\ 2 & 3 & 5 & 1 \end{pmatrix} $$
 
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Problem 8, partial solution
It is necessary to find the Galois group for the splitting field of ##x^4 - 2x^2 - 2## over Q.

The roots of that equation are ##\pm \sqrt{1 \pm \sqrt{3} } ##. Their four symmetry operations are (identity), (reverse outer square-root sign), (reverse inner square-root sign), and (reverse both square-root signs). The group of these operations is rather obviously ##Z_2 \times Z_2##.
 

Math_QED

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Problem 8, partial solution
It is necessary to find the Galois group for the splitting field of ##x^4 - 2x^2 - 2## over Q.

The roots of that equation are ##\pm \sqrt{1 \pm \sqrt{3} } ##. Their four symmetry operations are (identity), (reverse outer square-root sign), (reverse inner square-root sign), and (reverse both square-root signs). The group of these operations is rather obviously ##Z_2 \times Z_2##.
You are right about the roots, but that's about it. The Galoisgroup is not the Klein-group, nor has it order 4.
 
HI! I just came across those questions and thought if it would be okay if I could send you some math questions for the August math challenge 2019!
 
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Not really my area, but here goes nothing. Based on Brownian motion characterisation under section titled "Mathematics"
Put [itex]X := \int _0^t W_s^2 ds[/itex]. Based on conditions 1 and 4 of the characterisation we have [itex]W_s \sim \mathcal N(0,s)[/itex], where [itex]s\geq 0[/itex]. Therefore
[tex]
s = \mbox{var}(W_s) = \mathbb E (W_s - \mathbb EW_s)^2 = \mathbb E(W_s^2).
[/tex]
Fubini allows us to change order of integration, so we get
[tex]
\mathbb EX = \mathbb E \left ( \int _0^t W_s^2ds\right ) = \int _0^t \mathbb E(W_s^2)ds = \int_0^t sds = \frac{t^2}{2}
[/tex]
Initially, P7 asked for variance, which turns out to be
[tex]
\mbox{var}(X) = \mathbb E (X - \mathbb EX)^2 = \mathbb E\left (X - \frac{t^2}{2}\right )^2 = \mathbb E(X^2) - \frac{t^4}{4}
[/tex]
So we need to calculate the second moment of [itex]X[/itex]. I'm not sure at the moment what's happening here.
 
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Math_QED

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Not really my area, but here goes nothing. Based on Brownian motion characterisation under section titled "Mathematics"
Put [itex]X := \int _0^t W_s^2 ds[/itex]. Based on conditions 1 and 4 of the characterisation we have [itex]W_s \sim \mathcal N(0,s)[/itex], where [itex]s\geq 0[/itex]. Therefore
[tex]
s = \mbox{var}(W_s) = \mathbb E (W_s - \mathbb EW_s)^2 = \mathbb E(W_s^2).
[/tex]
Fubini allows us to change order of integration, so we get
[tex]
\mathbb EX = \mathbb E \left ( \int _0^t W_s^2ds\right ) = \int _0^t \mathbb E(W_s^2)ds = \int_0^t sds = \frac{t^2}{2}
[/tex]
Initially, P7 asked for variance, which turns out to be
[tex]
\mbox{var}(X) = \mathbb E (X - \mathbb EX)^2 = \mathbb E\left (X - \frac{t^2}{2}\right )^2 = \mathbb E(X^2) - \frac{t^4}{4}
[/tex]
So we need to calculate the second moment of [itex]X[/itex]. I'm not sure at the moment what's happening here.
Correct! It should be noted that is non-trivial that the map ##(t,\omega) \mapsto W_t(\omega)## is jointly measurable (i.e. measurable w.r.t. the product sigma algebra), which is necessary to apply Fubini. But you could use this without a proof.

The calculation of ##\mathbb{E}[X^2]## is harder. If you want to attempt it anyway, here is a hint to get you (or someone else who wants to try) started:

$$X^2 = \left(\int_0^t W_u^2 du\right) \left(\int_0^t W_v^2 dv\right)=\int_0^t \int_0^t W_u^2 W_v^2 dudv$$

Now, taking expectations on both sides you can use Fubini on a triple integral and the problem reduces to finding ##\mathbb{E}[W_u^2 W_v^2]##.
 
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It gets weird, there has to be some kind of algebraic trick involved, which I can't think of.
We could try
[tex]
\mathbb E(W_uW_v)^2 = \mbox{var}(W_uW_v) + \mathbb E^2 (W_uW_v)
[/tex]
This brings more complications, though. By linearity we get
[tex]
v\leq u \implies u-v=\mbox{var}(W_u-W_v) = \mathbb E(W_u-W_v)^2 = u - 2\mathbb E(W_uW_v) + v
[/tex]
from which
[tex]
v\leq u \implies \mathbb E(W_uW_v) = v
[/tex]
not sure how that's helpful, though.
 

Math_QED

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It gets weird, there has to be some kind of algebraic trick involved, which I can't think of.
We could try
[tex]
\mathbb E(W_uW_v)^2 = \mbox{var}(W_uW_v) + \mathbb E^2 (W_uW_v)
[/tex]
This brings more complications, though. By linearity we get
[tex]
v\leq u \implies u-v=\mbox{var}(W_u-W_v) = \mathbb E(W_u-W_v)^2 = u - 2\mathbb E(W_uW_v) + v
[/tex]
from which
[tex]
v\leq u \implies \mathbb E(W_uW_v) = v
[/tex]
not sure how that's helpful, though.
Your attempt shows in fact that ##E(W_u W_v) = \min\{u,v\}##, which is correct. But you don't need it here. What you need is ##E(W_u^2 W_v^2)##. Here is a hint:

Suppose ##u \geq v##. Then

$$E(W_u^2 W_v^2) = E[(\{W_u-W_v\}+W_v)^2 W_v^2]$$

and ##W_u- W_v, W_v## are independent variables. What do you know about the expectation of the product of two independent random variables?
 

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