Basic Math Challenge - May 2018

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It's time for a basic math challenge! For more advanced problems you can check our other intermediate level math challenge thread!

RULES:
1) In order for a solution to count, a full derivation or proof must be given. Answers with no proof will be ignored.
2) It is fine to use nontrivial results without proof as long as you cite them and as long as it is "common knowledge to all mathematicians". Whether the latter is satisfied will be decided on a case-by-case basis.
3) If you have seen the problem before and remember the solution, you cannot participate in the solution to that problem.
4) You are allowed to use google, wolframalpha or any other resource. However, you are not allowed to search the question directly. So if the question was to solve an integral, you are allowed to obtain numerical answers from software, you are allowed to search for useful integration techniques, but you cannot type in the integral in wolframalpha to see its solution.
5) Mentors, advisors and homework helpers are kindly requested not to post solutions, not even in spoiler tags, for the challenge problems, until 16th of each month. This gives the opportunity to other people including but not limited to students to feel more comfortable in dealing with / solving the challenge problems. In case of an inadvertent posting of a solution the post will be deleted by @fresh_42

##1.## Finite Field ##\mathbb{F}_8##
a) (solved by @lpetrich ) Find a minimal polynomial to determine the factor ring which is isomorphic to ## \mathbb{F} _8## .
b) (resolved in post #83) From there determine a basis of ## \mathbb{F}_8## over ## \mathbb{F}_2## and write down its multiplication and addition laws.
c) (solved by @lpetrich ) Why does the algebraic closure of a finite field have to be infinite?##\space## (by @fresh_42)

##2.## (solved by @nuuskur ) If ##a## is an odd integer show that ##a^{2^n} \equiv 1(\mod{2^{n+2}})## for all ##n \in \mathbb{N} - \{0\}## ##\space## (by @QuantumQuest)

##3.## (solved by @Zafa Pi ) A carnival has a 3 sided coin with outcomes ## \big \{\text{Heads, Tails, Other}\big\}## and respective probabilities of ## \big\{\frac{2}{5}, \frac{2}{5}, \frac{1}{5}\big\}##.

Rules of play:

Tails: ##+3## as a payoff
Other: game over, lose all accrued winnings.

Otherwise the player may stop at any time and keep the accrued winnings.

question:
How much should a risk neutral player be willing to pay in order to play this game?

(For avoidance of doubt, this refers to playing one 'full' game, which is complete upon termination, and termination occurs on the first occurrence of (a) result of coin toss equals ##\text{Other}## or (b) the player elects to stop.)

optional:
How many rounds would it take on average for the game to terminate? (You may assume a mild preference for shorter vs longer games in the event of any tie breaking concerns.)

Now suppose the player doesn't care about the score and just loves flipping coins -- how long will the game take to terminate, on average, in this case? ##\space## (by @StoneTemplePython)

##4.## (solved by @Hiero ) Calculate the volume ##\mu(A)## of

## A =\{(x,y,z)\in \mathbb{R}^3\,: \,x,y,z \ge 0\; , \;x+y+z \leq \sqrt{2}\; , \;x^2+y^2 \leq 1\,\}## ##\space## (by @fresh_42)

##5.## (solved by @nuuskur ) Determine the open balls with radius ##3## around ## (2,1) \in \mathbb{R}^2## w.r.t.
a) the French Railway metric with Paris at the origin ##P## and Reims at ##R=(2,1)## .
b) the Manhattan metric.
c) the maximum metric. ##\space## (by @fresh_42)

##6.##(solved by @Zafa Pi ) Two urns contain the same total number of balls, some black and some white in each. From each urn are drawn ## \big(n \geq 3\big)## balls with replacement. Find the number of drawings and the composition of the two urns so that the probability that all white balls are drawn from the first urn is equal to the probability that the drawing from the second is either all white or all black. ##\space## (by @StoneTemplePython)

##7.## (solved by @lpetrich ) Show that if for some complex number ##z##, ##{\lvert \sin z\rvert}^2 + {\lvert \cos z\rvert}^2 = 1## then ##z \in \mathbb{R}## ##\space## (by @QuantumQuest)

##8.## (solved by @dRic2 ) Show that of all triangles with given base and given area, the one with the least perimeter is isosceles ##\space## (by @QuantumQuest)

##9.## (solved by @lpetrich ) Solve ##\int_{\Gamma} \omega## with the curve ##\Gamma = \gamma([0,1])## given by

##\gamma : \mathbb{R} \longrightarrow \mathbb{R}^3\; , \;\gamma(t)=(t^2,2t,1)\text{ and }\omega = z^2 dx +2ydy+xzdz##

and compute the exterior derivative ## \nu = d\omega## . As such, the result is an exact ##2-## form. Is it also closed? Show this by direct calculation. ##\space## (by @fresh_42)

##10.## (solved by @lpetrich ) Prove that ##\lim_{n\to\infty} (n!e - \lfloor n!e \rfloor) = 0## ##\space## (by @QuantumQuest)

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member 587159, nuuskur, Helly123 and 4 others

jedishrfu
Mentor
Nice set of problems. I like the spread even if I don't know how to do some of them.

QuantumQuest
Wow... if this is basic then I must be really poorly versed in math, because I don’t even understand the language of many of these questions.

$$\int_0^1 \int_0^{\sqrt{1-x^2}}(\sqrt2-x-y)dydx = \sqrt2\pi/4-2/3$$
?

(I did do it by hand, I just don’t want to type the intermediate steps.)

lekh2003, epenguin and Greg Bernhardt
fresh_42
Mentor
Wow... if this is basic then I must be really poorly versed in math, because I don’t even understand the language of many of these questions.

$$\int_0^1 \int_0^{\sqrt{1-x^2}}(\sqrt2-x-y)dydx = \sqrt2\pi/4-2/3$$
?

(I did do it by hand, I just don’t want to type the intermediate steps.)
Yes, but the procedure belongs to the answer. You should at least roughly tell us what you did.

Yes, but the procedure belongs to the answer. You should at least roughly tell us what you did.
The condition x,y,z ≥ 0 means we are in the positive octant. The next two conditions describe (under) a plane and (inside) a cylinder . So the integral should be of a quarter-cylinder under a plane.

Volume is given by (one way at least) ∫∫z dydx, so I put in the z from the plane equation and chose the integral boundaries to describe the quarter-circle and voila.
(The integral never gets messy, the worst part is the integral of cos^2(a)da which can be done by parts.)

I am sorry I know this is a weak explanation but it’s more of a visual thought-process than a verbal thought-process.

QuantumQuest and fresh_42
fresh_42
Mentor
I am sorry I know this is a weak explanation but it’s more of a visual thought-process than a verbal thought-process.
Your visual thought is right and it is exactly what we do formally. The process is called reduction of variables:
1.) ##A =\{(x,y,z)\in \mathbb{R}^3\,: \,x,y,z \ge 0\; , \;x+y+z \leq \sqrt{2}\; , \;x^2+y^2 \leq 1\,\}##
2.) ##A'=\{\{(x,y)\in \mathbb{R}^2\,: \,x,y \ge 0\; , \;x^2+y^2 \leq 1\,\}\; ; \; A_{(x,y)}= [0,\sqrt{2}-x-y]##
3.) ##A''=[0,1]\; ; \;A'_x=[0,\sqrt{1-x^2}]##
We thus have
\begin{align*}
\mu(A)&=\int_A d\mu \\
&= \int_{A'} \left( \, \int_0^{\sqrt{2}-x-y} \,dz \,\right) \,d\mu_{A'} \\
&= \int_{A''} \left( \,\int_0^{\sqrt{1-x^2}}\,\left( \, \int_0^{\sqrt{2}-x-y} \,dz \,\right)\,dy \,\right) \,d\mu_{A''}\\
&= \int_{0}^1 \left( \,\int_0^{\sqrt{1-x^2}}\,\left( \, \int_0^{\sqrt{2}-x-y} \,dz \,\right)\,dy \,\right) \,dx \\
\end{align*}

Hiero
Solution to problem #7:
Set z = x + i*y where x and y are real. Then,
##\cos z = \cos x \cosh y - i \sin x \sinh y##
and
##\sin z = \sin x \cosh y + i \cos x \sinh y##.
Substituting in,
##|\cos z|^2 + |\sin z|^2 = \cos^2 x \cosh^2 y + \sin^2 x \sinh^2 y + \sin^2 x \cosh^2 y + \cos^2 x \sinh^2 y##
The trigonometric functions in x drop out because of a familiar trig identity, giving
##\cosh^2 y + \sinh^2 y = 2 \sinh^2 y + 1##
Since y is real, this quantity only equals 1 if y = 0, meaning that z must be real.

Greg Bernhardt and QuantumQuest
Gold Member
@lpetrich

Your first two identities i.e. ##\cos z = \cos x \cosh y - i \sin x \sinh y## and ##\sin z = \sin x \cosh y + i \cos x \sinh y## are taught as a proposition so they need to be proved.

OK, I will prove them.
First, Euler's formula for real x: ##e^{i x} = \cos x + i \sin x##
Next, the definitions of the hyperbolic functions ##\cosh x = \frac12 (e^x + e^{-x})## and ##\sinh x = \frac12 (e^x - e^{-x})##.
One can easily rearrange those definitions to get ##e^x = \cosh x + \sinh x## and ##e^{-x} = \cosh x - \sinh x##.

Generalizing Euler's formula to complex args gives ##\cos z = \frac12 (e^{iz} + e^{-iz})## and ##\sin z = \frac{1}{2i} (e^{iz} - e^{-iz})##.

Setting z = x + i*y with both x and y real,
##\cos z = \frac12 (e^{ix -y} + e^{-ix+y}) = \frac12[(\cos x + i \sin x) (\cosh y - \sinh y) + (\cos x - i \sin x) (\cosh y + \sinh y)] = \cos x \cosh y - i \sin x \sinh y##
##\sin z = \frac1{2i} (e^{ix -y} - e^{-ix+y}) = \frac1{2i}[(\cos x + i \sin x) (\cosh y - \sinh y) - (\cos x - i \sin x) (\cosh y + \sinh y)] = \sin x \cosh y + i \cos x \sinh y##

Results that I'd used earlier.

Greg Bernhardt and QuantumQuest
dRic2
Gold Member
I think I came up with the worst proof one could ever imagine for problem #8
If base and area are given then also the height is given (##h = 2A/b##). Now consider the following image:

Now I can express the other two sides (##l_1## and ##l_2##) as a function of ##x## and the two constants (##b## and ##h##). So:

##l_1 = \sqrt{x^2+h^2}##
##l_2 = \sqrt{(b-x)^2+h^2}##

Then the perimeter is

##P = b + l_1 + l_2 = b + \sqrt{x^2+h^2} + \sqrt{(b-x)^2+h^2}##

Then I'm going to find the minimum for this function:

##\dot P = \frac x {\sqrt{x^2+h^2}} + \frac {x-b}{\sqrt{(b-x)^2+h^2}} = 0##

It's evident that ##x = \frac b 2## is a solution (and it is also the minimum). I don't know how to prove it rigorously, though. If you need a rigorous proof I will try to upload it tomorrow (it's gettin late here).

Finally we observe that for ##x = b/2## it happens ##l_1 = l_2##.

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I have the proof for the 8)

As the second derivative of P is too long to do it, It's easier to proof that slope of the curve p(x) with x < b/2 is negative and the slope of the curve
with x > b/2 is positive. I have do it and there is a minimum in x = b/2

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I will now take on problem #8. I will try to do it in a simpler fashion.
Consider a triangle ABC where AB is the base, with length 2b, and C is at perpendicular distance h and perpendicular-line intersection location at point D. That is, D is on AB and CD is perpendicular to AB. Let C have distance x toward A.

The area A = (1/2)*((2b)*h) = b*h. Since by the problem statement, both A and b are fixed, that means that only h is fixed.

The perimeter ##L = 2b + \sqrt{(b-x)^2 + h^2} + \sqrt{(b+x)^2 + h^2}##. One finds the minimum by taking the derivative with respect to the remaining free parameter, x:
$$\frac{dL}{dx} = - \frac{b-x}{\sqrt{(b-x)^2 + h^2}} + \frac{b+x}{\sqrt{(b+x)^2 + h^2}}$$
For solving (dL/dx) = 0, move the right term to the opposite side of the equation:
$$- \frac{b-x}{\sqrt{(b-x)^2 + h^2}} = - \frac{b+x}{\sqrt{(b+x)^2 + h^2}}$$
Square it and take the reciprocal:
$$\frac{(b-x)^2 + h^2}{(b-x)^2} = \frac{(b+x)^2 + h^2}{(b+x)^2}$$
Subtract 1 and take the reciprocal again:
$$\frac{(b - x)^2}{h^2} = \frac{(b + x)^2}{h^2}$$
Multiply by h^2 and subtract the right term from the left term, giving -4*b*x = 0. This has solution x = 0 as its only solution.

Now consider the curvature of L at x = 0. We must take the derivative again:
$$\frac{d^2 L}{dx^2} = \frac{h^2}{(\sqrt{(b-x)^2 + h^2})^3} + \frac{h^2}{(\sqrt{(b+x)^2 + h^2})^3}$$
This is always positive, meaning that the zero-slope point x = 0 gives the minimum length of perimeter.

For x = 0, the distances AD = DB = b, making the triangle isosceles.

QuantumQuest, Greg Bernhardt and Sergio Rodriguez
Gold Member
Gold Member
@lpetrich well done for problem ##8## but the solution of dRic2 came first ;)

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I was going to post 8 and 10 but I was late :c

Question 10:
The limit problem simply asks to prove that the fractional part of n!e tends to equal zero.
##n! e = n! (\frac{1}{1!} + \frac{1}{2!} ...) ##

Thus the fractional part call it ##K ## equals
## K = \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)(n+3)} + ...##

If you take n as a common factor (from each () ) you can show that the limit of this series tends to zero, Thus proving the limit in the question

I am interested in the probabilities questions but they same a bit vague for me. In Question 3, A risk neutral player would want to pay as much as he expects to earn?

StoneTemplePython
Gold Member
I am interested in the probabilities questions but they same a bit vague for me. In Question 3, A risk neutral player would want to pay as much as he expects to earn?

Right. Or equivalently, at what pricing would it be a "fair game"?

Gold Member
@Biker

I can see the points you make but it is not a full solution. In order to have a full solution you must show what the problem asks to be proved, using things that explicitly lead there.

lekh2003
Gold Member
If this is basic math, I am absolutely useless at math. Can we have an extra basic math problem set which us non-undergrads might be able to do?

fresh_42
Mentor
If this is basic math, I am absolutely useless at math. Can we have an extra basic math problem set which us non-undergrads might be able to do?
Yes. We are learning, too. The answers themselves are not so difficult, more their setting - and some of these concepts can be looked up on Wiki, nLab, WolframAlpha etc.

But, you're right. The more as it makes our lives easier to find good questions. And because you support my personal opinion, that it might be a good idea to reverse the normal order: instead of answering randomly posed questions, let us take a subject and ask some exemplary questions around it, e.g. continuity, convergence, series, etc.

As said, we're learning, too.

lekh2003
It is basic in the sense it is mostly routine i.e what we've been doing in various tutorials of algebra, analysis etc. At the same time, it is instructive. Definitions can be looked up by doing some search around the web.

question 10 solution guys,
Using principle of mathematical induction,
when n=1

limn->inf(e-e)=0
Proved true for n=1
Now assume true for n=k,

limn->inf(k!e-⌊k!e⌋)=0 ------1
for n=k+1
limn->inf((k+1)!e-⌊(k+1)!e⌋)
limn->inf(k!e+e-⌊k!e⌋)-⌊(K+1)e⌋)
limn->inf(k!e+e-⌊k!e⌋-⌊(k-1)!e⌋-⌊ke⌋-⌊e⌋)
limn->inf(e-⌊(k!e⌋-2) using equation 1
again rearrange equation 1 and apply above to get,
limn->inf(e-k!e-2)
I am stuck here now

fresh_42
Mentor
question 10 solution guys,
Using principle of mathematical induction,
when n=1

limn->inf(e-e)=0
Proved true for n=1
Now assume true for n=k,

limn->inf(k!e-⌊k!e⌋)=0 ------1
for n=k+1
limn->inf((k+1)!e-⌊(k+1)!e⌋)
limn->inf(k!e+e-⌊k!e⌋)-⌊(K+1)e⌋)
limn->inf(k!e+e-⌊k!e⌋-⌊(k-1)!e⌋-⌊ke⌋-⌊e⌋)
limn->inf(e-⌊(k!e⌋-2) using equation 1
again rearrange equation 1 and apply above to get,
limn->inf(e-k!e-2)
I am stuck here now
Your induction anchor, in case you separate the two occurrences and roles of ##n##, should be ##\lim_{n \to \infty} (1! \cdot e - \lfloor 1! \cdot e \rfloor) =0 ## which is wrong. However, we can't separate the two, so the induction in this form doesn't make sense. @Biker in post #15 has been on the right track, he just didn't work out his idea.

Suyash Singh
Your induction anchor, in case you separate the two occurrences and roles of ##n##, should be ##\lim_{n \to \infty} (1! \cdot e - \lfloor 1! \cdot e \rfloor) =0 ## which is wrong. However, we can't separate the two, so the induction in this form doesn't make sense. @Biker in post #15 has been on the right track, he just didn't work out his idea.
I guess i need to study "exponential and logarithmic series" again:(
I forgot everything

fresh_42
Mentor
I guess i need to study "exponential and logarithmic series" again:(
I forgot everything
I'm not quite sure, whether @Biker's method actually works, I haven't checked the solution to this problem, but I guess the series for ##e## and maybe a good remainder term estimation of the Taylor series could be helpful.

Suyash Singh
Can someone translate this into English? Just kidding but this is not basic unless I know less than I thought. Is the answer... 4? Just kidding but this is like Greek to me so I need to do some more studying then