Consolidated math problems thread

[Zadius]

Almost forgot the second part of the problem.

Equality occurs when C = π/3 and (a - b)² = 0. Therefore, a = b and the triangle must be equilateral.

[Kuroneko]

OK, here's my attempt at solving problem 3:...

Very nice solution. Interestingly, your ending inequality is actually intimately connected to both fnord's inequality and Jaepheth's problem (a simpler solution than my previous spoiler). Both of them can be solved with the same tool that happens to be an easy generalization of that trivial observation to more variables. (After all, just because a problem's been solved doesn't mean it's not instructive to solve it a different way.)

And just because it's also an excellent problem, Grog's deserves some emphasis:

It is well known that the sum of two stochastic variables with binomial distributions with the same "p-value" (I'm not sure what the correct term is here) is a stochastic variable with binomial distribution with the same "p-value". Prove the converse, that is that if a sum of two stocastic [independent] variables [taking only non-negative integer values] gives a binomial distribution then the two stochastic variables have a binomial distribution.

With minor modification so as to not leave open the door for trivial counter-examples (being independent is also important, and wasn't mentioned before).

[Jaepheth]

Ooh, that is a fun one. Spoiler:
Let r_k be the price per mass ("specific price") of flour at Friday k, so Bob pays r_k*(1 kg) money each time while Sue gets ($1)/r_k of flour each time. Hence Bob's average price per mass over n weeks is the arithmetic mean of the r_k's, while Sue's is $1 times the harmonic mean. Hence Sue gets a deal that's no worse than Bob's. One way to prove that the harmonic mean is no greater than the arithmetic mean is to observe that the map f(x) = 1/x is convex on x>0 and apply Jensen's inequality. I'll try to find a more elementary proof later.

I like that one. The proof I saw, if I remember correctly, used Cauchy-Schwarz inequality to show the ratio of Bob/Sue >= 1

[Kuroneko]

I like that one. The proof I saw, if I remember correctly, used Cauchy-Schwarz inequality to show the ratio of Bob/Sue >= 1

Yes, that's the result I was referring to. Explicitly, the connection between (a-b)²≥0 and that result is simply that
For reals vectors u,v with magnitudes U=|u|,V = |v|, we have the dot product <u/U,v/V> = Sum[ u_k/U v_k/V ] ≤ (1/2)Sum [ (u_k/U)² + (v_k/V)² ] = 1, using 2ab≤a²+b². Hence the Cauchy-Schwarz inequality for R^n: <u,v> ≤ UV.
For your problem, take u = (...1/√r_k...) and v = (...√r_k...), where r_k>0 is the price per mass of rice on Friday k. Substituted, squared, and rearranged, it's exactly the harmonic mean <= arithmetic mean inequality.
Now, for fnord's inequality, take u = (a²,b²,c²), v = (b²,c²,a²).

[Grog]

For the triangle problem I found this solution:
Ar(2/sinA+2/sinB+2/sinC)=ab+bc+ac≤a^2+b^2+c^2.
In the above we used Cauchy-Schwarz inequality or something like that and Ar is the area of the triangle. The second derivative of 2/sinx is nonnegative in this interval so 2/sinx is convex
4sqrt(3)=6/sin((A+B+C)/3)≤2/sinA+2/sinB+2/sinC.
Hopefully I have not made some simple mistake this time.

[Grog]

Oh and my solution also shows that we have equality only when all sides are equal.

[Zadius]

OK, here's my attempt at solving problem 3:...

Very nice solution. Interestingly, your ending inequality is actually intimately connected to both fnord's inequality and Jaepheth's problem (a simpler solution than my previous spoiler). Both of them can be solved with the same tool that happens to be an easy generalization of that trivial observation to more variables. (After all, just because a problem's been solved doesn't mean it's not instructive to solve it a different way.)

OK, I know what you mean. I just figured out the following proof that I think uses the generalization to more variables that you're talking about.
Plus, it doesn't require any derivatives be taken like my first proof:

a² + b² + c² ≥ 2absin(C)√3
(a² + b² + c²)² ≥ 12a²b²sin²(C)
(a² + b² + c²)² ≥ 12a²b²[1-cos²(C)]
(a² + b² + c²)² ≥ 12a²b²[1-(a² + b² - c²)²/(4a²b²)]
(a² + b² + c²)² ≥ 12a²b² - 3(a² + b² - c²)²
(a² + b² + c²)² + 3(a² + b² - c²)² - 12a²b² ≥ 0
Now expand and collect:
4a^4 - 4a²b² +4b^4 - 4a²c² - 4b²c² + 4c^4 ≥ 0
2a^4 - 2a²b² +2b^4 - 2a²c² - 2b²c² + 2c^4 ≥ 0
(a^4 - 2a²b² + b^4) + (a^4 - 2a²c² + c^4) + (b^4 - 2b²c² +c^4) ≥ 0
(a² - b²)² + (a² - c²)² + (b² - c²)² ≥ 0

[Grog]

I think I figured out the general case for the birthday problem:

If the probability that a person is born on day i is p_i then the probability that no two of n persons are born on the same day is sum product p_k=f(p_1,p_2,...,p_366) where the sum is over all n sequences S in [366] with no repeating terms and the product is over all elements in S.
Using lagrange multipliers (the condition is p_1+p_2+...+p_366-1=0) we get that all sum product p_k where the sum is over all sequences n-1 S' not containing j and the product is over all the elements in S' are equal for extremums of f, it is now easy to verify that all p_k must be equal too.
Some details are left out but I think this approach should work.

[Kuroneko]

... Plus, it doesn't require any derivatives be taken like my first proof: ...

Very straightforward. The result I was actually thinking of is a bit more general (not necessary for this problem, as your proof aptly demonstrates, but necessary to be also applicable to Jaepheth's), and it can itself be proven by the bound 2ab≤(a²+b²).

For the triangle problem I found this solution: ...
Hopefully I have not made some simple mistake this time.

It's all quite correct. Jensen's inequality is an extremely versatile tool.

I think I figured out the general case for the birthday problem: ...
Some details are left out but I think this approach should work.

I do believe that works. To try to fill in some details to make sure...
The sum-of-products you're referring to is the elementary symmetric polynomial e_n({p_k}), which has the property ∂e_n/∂x_j = e_{n-1}(¬j), where ¬j is meant to denote that the p_j argument is missing. Thus the Lagrange multiplier approach gives (n-1)st degree symmetric polynomials with a single missing argument all being equal at the extrema, as you say. Then e_{n-1}(¬j) = e_{n-1}(¬k), after canceling terms with no p_k or p_j shared on both sides, implies p_k e_{n-2}(¬j,¬k) = p_j e_{n-2}(¬j,¬k), which then gives p_k = p_j.
That's excellent. Exploiting symmetry made this simpler that I anticipated. The proof I've seen used another kind of convexity (although it's actually a monotonicity condition generalized to be more dimensions.)

[Kuroneko]

Any additional problems are quite welcome. So far, the only problem with no posted solution at all is the convex side-derivative one (although some of them may also have solutions quite different from the ones posted).

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4. There is an oft-repeated claim about the geometric mean, such as the following (taken from a random website): ... when evaluating investment returns and fluctuating interest rates, it is the geometric mean, not the arithmetic mean, that tells you what the average financial rate of return would have had to have been over the entire investment period to achieve the end result. This is perfectly (almost trivially) true, but potentially misleading due to missing information.
(a) Interpreting the above statement in the naive manner, with the usual rate of return 0<r_k = (interest earnings for period k)/(balance at period k), for compounded interest, show that replacing each rate of return by their geometric mean underestimates the actual return, while using the arithmetic mean over-estimates it.
(b) What quantities should one take the geometric mean of in order to make the statement true?
... alright, the latter may be too easy.

5. Evaluate I = Int[ log(sin x) dx ], where the integral is from 0 to π/2.

Some less formal texts state a different l'Hôspital's rule from what they prove. Is their oversight correctable or just plain wrong?
6. (Lazy l'Hôspital.) Let all limits be x→0, and let f,g be continuously differentiable functions on x≠0, with g,g' never 0 for 0<|x|<ε and lim[ f(x) ] = lim[ g(x) ] = 0. Then does lim[ f(x)/g(x) ] = L imply that lim[ f'(x)/g'(x) ] = L?

7. (Double periodicity.) Suppose f is a real continuous function with f(x) = f(x+α) = f(x+β), where α/β is not rational. Show that f is constant.

[Surlethe]

5. Evaluate I = Int[ log(sin x) dx ], where the integral is from 0 to π/2.

Wasn't this in a Testing thread a while ago? I seem to recall that there was some super-clever solution to it that I didn't see.

Some less formal texts state a different l'Hôspital's rule from what they prove. Is their oversight correctable or just plain wrong?
6. (Lazy l'Hôspital.) Let all limits be x→0, and let f,g be continuously differentiable functions on x≠0, with g,g' never 0 for 0<|x|<ε and lim[ f(x) ] = lim[ g(x) ] = 0. Then does lim[ f(x)/g(x) ] = L imply that lim[ f'(x)/g'(x) ] = L?

Isn't L'Hopital's rule that lim [f'/g'] = lim [f/g]? Nonetheless, put f(x) = x and g(x) = log(x+1). Both are C1 in a neighborhood of 0, with g and g' never 0 in that neighborhood sans 0, with both f, g tending to 0 as x tends to 0. But lim[ x/log(x+1) ] = 0 while lim[ x+1 ] = 1.

7. (Double periodicity.) Suppose f is a real continuous function with f(x) = f(x+α) = f(x+β), where α/β is not rational. Show that f is constant.

[I'm not putting this in a small font since it's more of a sketch than an actual proof.] Consider the set G={x : x=mα+nβ, m,n∈ℤ}. If G is dense in ℝ, then f will be constant. It is obvious that G is, in fact, dense in ℝ. Therefore, f is constant.

[Kuroneko]

Wasn't this in a Testing thread a while ago? I seem to recall that there was some super-clever solution to it that I didn't see.

There may have been. I'm sure I've posted it somewhere before; I just can't recall if it was on this board or not.

Isn't L'Hopital's rule that lim [f'/g'] = lim [f/g]?

No, actually, it's not (even under all the assumptions stated that problem), although it's certainly true that l'Hôspital's rule is closely related to that statement. I would say more, but that might give the problem away.

Nonetheless, put f(x) = x and g(x) = log(x+1). Both are C1 in a neighborhood of 0, with g and g' never 0 in that neighborhood sans 0, with both f, g tending to 0 as x tends to 0. But lim[ x/log(x+1) ] = 0 while lim[ x+1 ] = 1.

That's not true; lim[ f/g ] = 1 in your case, because g(x) = x - O(x²).

Consider the set G={x : x=mα+nβ, m,n∈ℤ}. If G is dense in ℝ, then f will be constant. It is obvious that G is, in fact, dense in ℝ. Therefore, f is constant.

That looks to be a good approach. It's not the only approach, though.

[Grog]

I'm sorry for the thread necromancy but this thread has so much potential I think and I'm in the mood for some math. I have some problems someone might be interested in:

Let a,b,c be positive real numbers such that abc=1 prove that (a+1/b-1)(b+1/c-1)(c+1/a-1)<=1. I often find it hard to find nice solutions to this kind of inequality problems; my solution did require a differentiation or quite a bit of annoying manipulations but I suspect there is a simple solution.

There is also a double integral I once calculated in two different ways and I'm not happy with either (one required a lot of calculations but was straightforward and the other used a trick that required me to check a lot of lemmas or theorems to be sure it was valid). Integrate 1/(1+(x+y)^4) over the first quadrant.

I can also give a solution to one of the old problems.

5. Evaluate I = Int[ log(sin x) dx ], where the integral is from 0 to π/2.

Int[ log(sin x) dx ]over the given interval is the same as Int[ log(cos x) dx ] over the same interval and so is Int[ log(sin x) dx/2 ] over the interval 0 to π. If we do the substitution x=2y in the last integral we get I=Int[ log(sin x) dx/2 ]=Int[ log(sin 2y) dy ]=Int[ log(2 sin y cos y) dy ]=2I+log(2)π/2 so I=-log(2)π/2.
Maybe not the most clever way to do it but it works.

[Grog]

7. (Double periodicity.) Suppose f is a real continuous function with f(x) = f(x+α) = f(x+β), where α/β is not rational. Show that f is constant.

Let A be the shortest possible periodicity and assume that at α/A is not an integer then there is an integer N such that 0<α-AN<A and it is now easy to see that f is α-AN periodic which is impossible!
Something like that should work.

I'm sorry for the double post.

[fnord]

I'm sorry for the thread necromancy but this thread has so much potential I think and I'm in the mood for some math. I have some problems someone might be interested in:

Let a,b,c be positive real numbers such that abc=1 prove that (a+1/b-1)(b+1/c-1)(c+1/a-1)<=1. I often find it hard to find nice solutions to this kind of inequality problems; my solution did require a differentiation or quite a bit of annoying manipulations but I suspect there is a simple solution.

Is each term of the form
x + 1 / (y-1)
or
(x + 1) / (y - 1) ?

[Kuroneko]

fnord: I don't think he means either of those, but rather in the form [(x)+(1/y)-1].

Let a,b,c be positive real numbers such that abc=1 prove that (a+1/b-1)(b+1/c-1)(c+1/a-1)<=1.

Interesting. If this was turned into a symmetric inequality (by summing over all permutations of a,b,c), it would probably be an easy one (or, at least, I have a good guess how to do it), but right now I don't see a simple way for this particular one.

Integrate 1/(1+(x+y)^4) over the first quadrant.

This is practically a one-liner. Spoiler: The integrand is constant over diagonals--make use of this.

Int[ log(sin x) dx ]over the given interval is the same as Int[ log(cos x) dx ] over the same interval and so is Int[ log(sin x) dx/2 ] over the interval 0 to π. If we do the substitution x=2y in the last integral we get I=Int[ log(sin x) dx/2 ]=Int[ log(sin 2y) dy ]=Int[ log(2 sin y cos y) dy ]=2I+log(2)π/2 so I=-log(2)π/2.

That's right, and that's pretty much the cleanest way I know to do that one.

Let A be the shortest possible periodicity and assume that at α/β is not an integer then there is an integer N such that 0<α-AN<A and it is now easy to see that f is α-AN periodic which is impossible!

A neat approach. To flesh it out, one would verify that A is well-defined in all cases (and it is, as an relevant infinitum of non-negatives will turn out to be itself a period) and that having two periods with irrational ratio means that α/A is irrational for some period α (which is also correct). The special case A = 0 would be trivial.

[Grog]

fnord: I don't think he means either of those, but rather in the form [(x)+(1/y)-1].

This is correct.

This is practically a one-liner. Spoiler: The integrand is constant over diagonals--make use of this.

that was embarrassingly simple. I did this integral when I first learned calculus and I don't know why I didn't use this, I just remembered it as much work and hadn't thought much about it since.

[Kuroneko]

Let a,b,c be positive real numbers such that abc=1 prove that (a+1/b-1)(b+1/c-1)(c+1/a-1)<=1. I often find it hard to find nice solutions to this kind of inequality problems; my solution did require a differentiation or quite a bit of annoying manipulations but I suspect there is a simple solution.

I went back to thinking about this over morning tea, and it seems that my earlier hunch does work after all, if we use the condition abc = 1 to transform the inequality into a more general one that happens to be completely symmetric. No calculus is required, although it involve fairly intense if short algebraic gymnastics. Massive Spoiler:
Define A = a+1/b-1, B = b+1/c-1, C = c+1/a-1, and some triple (x,y,z) satisfying a = x/y, b = y/z, c = z/x, and finally D = ABCxyz. Then the inequality is D ≤ xyz, which in turn is equivalent to x²y + x²z + yz² + y²z + xz² + xy² ≤ x³+y³+z³ + 3xyz. In symmetric-sum notation {...} (summation over all permutations of exponents),
LHS² = {x^2y^1z^0}² = {x^4y^2z^0} + {x^4y^1z^1} + {x^3y^3z^0} + 2{x^3y^2z^1} + {x^2y^2z^2} ≤ 6{x^4y^2z^0} ≤ [ (1/2){x^3y^0z^0} + (1/2){x^1y^1z^1} ]² = RHS²,
where the first inequality follows from Muirhead's inequality and the second from the AM-GM inequality.
There might be other solutions.

[Surlethe]

Here's a problem, which I'm sure has a simple solution, that I haven't quite yet figured out:

Let A, B, C be coplanar vectors. Prove, without using a coordinate system, that the cross product and the dot product are distributive.

[Grog]

There might be other solutions.

When I proved it I turned it into a symmetric inequality by seting ab=x and b=y then we get (x+y-1)(x-y+1)(-x+y+1)≤xy for all positive x and y.

[Kuroneko]

Let A, B, C be coplanar vectors. Prove, without using a coordinate system, that the cross product and the dot product are distributive.

Well, it would depend on which definition of the cross product one starts with. Taking as a starting point either the Levi-Civita pseudotensor or the Hodge star makes this immediate. However, since this nerfs the problem more than a little bit, I assume the intent is to start with something like the following:
-- Def. The cross product of two vectors in R³ is a pseudovector with magnitude equal to the area of the parallelogram formed by those two vectors and direction perpendicular to both compliant with the right-hand rule (or left-hand rule, as long as this is consistent).
I remember this problem from Griffiths' EM textbook, except that Griffiths also required to prove the non-coplanar case as well; you might want generalize the problem in this manner.
Hint: It might be helpful to first show that for a cross product with one vector held constant, the parallel part of the other vector does not contribute at all.

When I proved it I turned it into a symmetric inequality by seting ab=x and b=y then we get (x+y-1)(x-y+1)(-x+y+1)≤xy for all positive x and y.

I considered this as well, but I couldn't find a way to homogenize it, and disassembling the inequality by grouping terms of the same degree produced something that was false, so I couldn't find a way to do it without calculus. I'm not completely happy with my solution, although squaring that six-term sum isn't too bad because it's symmetry, making it unnecessary to explicitly write out all the terms.

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Since the double-periodicity problem already had several solutions or outlines of such, here's the solution I had:
7. Without loss of generality, we can assume β = 1, and so we can calculate the Fourier coefficients with an integral over any interval of size 1. Further letting y = x+α, where α is a period of f, F[n] = Int_0^1[ f(x) exp(-2nπix) dx] = Int_α^{1+α}[ f(y) exp(-2nπi(y-α)) dy ] = exp(2nπαi)F[n]. Hence, if α is irrational, F[n] = 0 for all nonzero n.

[Surlethe]

Let A, B, C be coplanar vectors. Prove, without using a coordinate system, that the cross product and the dot product are distributive.

Well, it would depend on which definition of the cross product one starts with. Taking as a starting point either the Levi-Civita pseudotensor or the Hodge star makes this immediate. However, since this nerfs the problem more than a little bit, I assume the intent is to start with something like the following:
-- Def. The cross product of two vectors in R³ is a pseudovector with magnitude equal to the area of the parallelogram formed by those two vectors and direction perpendicular to both compliant with the right-hand rule (or left-hand rule, as long as this is consistent).
I remember this problem from Griffiths' EM textbook, except that Griffiths also required to prove the non-coplanar case as well; you might want generalize the problem in this manner.

Well, we didn't have to do that part of the problem.

Hint: It might be helpful to first show that for a cross product with one vector held constant, the parallel part of the other vector does not contribute at all.

Thanks; I actually did solve it. It becomes intuitive when the vectors are drawn correctly.

[Kuroneko]

New and improved with actual spoiler tags. (I'm hoping the brief resurgence of math problems in OT might carry over here as well.)

As always, everyone is welcome to post any mathematical problems they find either interesting or challenging. Any level whatsoever.

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8. (Fractional derivative) Assuming that [(d/dx)^ν][exp(ax)] = a^ν exp(ax) holds even for noninteger ν, show that the derivates of sine and cosine are phase shifts by νπ/2, i.e., [(d/dt)^ν][sin(t)] = sin(t+νπ/2), etc.

9. Let f(ξ) be a second-differentiable function. Prove that f(r±ct)/r solves the wave equation u_tt = c²∇²u, where ∇² = ∇·∇ is the Laplacian and r is the distance from the origin.

Credits: #8 has been inspired by a comment by drachefly, while #9 has been refitted from Col. Olrik.

10. Kuroneko is forced to take an organic chemistry exam, part of which is involves matching questions (1,2,...) to a list of answers (A,B,...), each being right for exactly one question. Knowing nothing about the subject matter, he decides to mark them completely at random and use the extra time to doodle some math instead. (a) If n = 10, what is the probability that he got at least eight questions right? Assume all valid answer-sequences have the same probability. (b) His doodles led him to believe that no matter what n is, the expected number of correctly answered questions is always 1. Is he right?

11. (Periodicity redux) Let f be a continuous real periodic function. For what values of a does the equation f(x+a)= f(x) have a solution?

[Grog]

Yay, great thread!
Hopefully I can come up with some interesting problems this time around.

11. (Periodicity redux) Let f be a continuous real periodic function. For what values of a does the equation f(x+a)= f(x) have a solution?

Spoiler

f is continuous and periodic so it has a global maximum at x_1 and a global minimum at x_2.
f(x_1+a)-f(x_1)<0 and f(x_2+a)-f(x_2)>0
f(x+a)-f(x) is continous for all a so the there is a solution for all a.

[Surlethe]

8. (Fractional derivative) Assuming that [(d/dx)^ν][exp(ax)] = a^ν exp(ax) holds even for noninteger ν, show that the derivates of sine and cosine are phase shifts by νπ/2, i.e., [(d/dt)^ν][sin(t)] = sin(t+νπ/2), etc.

Spoiler

Use sin(x) = (1/2i)[exp(ix)-exp(-ix)] and i^v=exp[ivπ/2]. Cosine is similar.

9. Let f(ξ) be a second-differentiable function. Prove that f(r±ct)/r solves the wave equation u_tt = c²∇²u, where ∇² = ∇·∇ is the Laplacian and r is the distance from the origin.

Spoiler

In spherical coordinates, since f has no angular dependence, ∇²=(1/r²)(d/dr)[r²(d/dr)]. Using this, it is trivial to check that f does indeed satisfy the differential equation.

Grog, your answer is simple, but is it possible to phrase it in terms of Fourier series and make it simpler?