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SuicideFuel Math thread problem (official)

Let f be a function from the reals to the reals and let f^n denote the n-fold composition of f with itself. We say that x is a fixed point of order n if f^n(x) = x but f^k(x) ≠ x for all positive integers k < n. Show that if f has a fixed point of order 3 it must also have a fixed point of order 2.
 
A problem courtesy of @mindlessselfindulge
probability-problem.png
 
Let f be a function from the reals to the reals and let f^n denote the n-fold composition of f with itself. We say that x is a fixed point of order n if f^n(x) = x but f^k(x) ≠ x for all positive integers k < n. Show that if f has a fixed point of order 3 it must also have a fixed point of order 2.
I forgot to add that f has to be continuous like a dumbass. My bad.
 
Let f be a function from the reals to the reals and let f^n denote the n-fold composition of f with itself. We say that x is a fixed point of order n if f^n(x) = x but f^k(x) ≠ x for all positive integers k < n. Show that if f has a fixed point of order 3 it must also have a fixed point of order 2.
I forgot to add that f has to be continuous like a dumbass. My bad.

f(a) = b, f(b) = c, f(c) = a

Case 1: a < b < c

The points (a, b) and (b, c) are above the line y = x, and (c, a) is below. By the intermediate value theorem, there exists m with b < m < c and f(m) = m. Let C be the curve on f from (a, b) to (m, m). Reflect C across the line y = x to get the curve C'. C' contains the points (c, b) and (b, a), and is a function from y to x. Let D be the curve on f from (m, m) to (c, a). Since b < m < c, D and C' intersect by the intermediate value theorem. The intersection is a fixed point of order 2 since it can be mapped back onto the mirror image of the curve C.

Case 2: a > b > c

Same argument, except the direction of the curves and orientation is flipped.
 
For all functions f: R^2 -> R, prove that there exist vectors v1 and v2 with |f(v2) - f(v1)| < |v2 - v1|.

Bonus:

Replace the inequality above with |f(v2) - f(v1)| < |v2 - v1|^p. For what values of p is the statement still true?
 
C' contains the points (c, b) and (b, a), and is a function from y to x. Let D be the curve on f from (m, m) to (c, a). Since b < m < c, D and C' intersect by the intermediate value theorem. The intersection is a fixed point of order 2 since it can be mapped back onto the mirror image of the curve C.
  • I'm not convinced the IVT applies here. After all, the IVT applies to a function. What's your function in question?
  • How do you know that the intersection point ain't (m, m)?
  • Naively there should be 6 cases. How did you reduce down to 2?
 
For all functions f: R^2 -> R, prove that there exist vectors v1 and v2 with |f(v2) - f(v1)| < |v2 - v1|.

Bonus:

Replace the inequality above with |f(v2) - f(v1)| < |v2 - v1|^p. For what values of p is the statement still true?
I'll assume the image of f is an interval.

We want to find out for which p > 0 there are no f such that |f(x) − f(y)| ⩾ |x − y|^p for all x and y in R^2. If such an f were to exist, it would clearly be injective. Let g be the left inverse of f. Then g must satisfy |g(u) − g(v)| ⩽ |u − v|^(1/p) for all u and v in the image of f. In other words, g must be (1/p)-Hölder continuous. However, g is also surjective, meaning that g is a space filling curve. It is known that (1/p)-Hölder continuous space filling curves exist iff 0 < 1/p ⩽ 1/2 -- i.e., p ⩾ 2. Indeed, if p < 1, no (1/p)-Hölder continuous functions exist at all (which is a fun exercise).
 
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Prove that no Hölder continuous functions with exponent > 1 exist.
 
  • I'm not convinced the IVT applies here. After all, the IVT applies to a function. What's your function in question?
  • How do you know that the intersection point ain't (m, m)?
  • Naively there should be 6 cases. How did you reduce down to 2?

You are correct, I had wrong logic about the 2nd application of the IVT.

Fix:

C' contains a path going from (c, b) to (b, a). Since f is continuous everywhere and contains the points (c, a) and (b, c), the point (c, b) is above the curve of f and (b, a) is below the curve of f, which means that C' intersects f on that path.

The path from (c, b) to (b, a) on C' does not contain (m, m), since (m, m) is the other endpoint of C' and m =/= c and m =/= b.

Only the orientation of the triangle containing (a, b), (b, c), (c, a) matters. The case I did had a clockwise orientation and the other case has a counterclockwise orientation.
 
Prove that no Hölder continuous functions with exponent > 1 exist.

I think you mean no non-constant functions.

For this proof use the property that all Holder continuous functions with a positive exponent are differentiable.

Let a be a positive value with:

|f(y) - f(x)| <= C |y - x|^(1 + a)

=> |(f(y) - f(x)) / (y - x)| <= C |y - x|^a

Let y = x + dx and take the limit as dx approaches 0. Then:

|f'(x)| <= lim(dx -> 0) C |dx|^a.

Since a is positive, the limit on the right side evaluates to 0, which means that f is a constant function.
 
Integrate (1 - cos(x)) / (x * sqrt(x)) from x = 0 to infinity.
 
For a positive integer n, given n integers (not necessarily distinct), prove that there is at least 1 non empty subset sum that is divisible by n.
 
Integrate (1 - cos(x)) / (x * sqrt(x)) from x = 0 to infinity.
sqrt(2π). One way to see this is to apply Maz identity and do substitutions until you arrive at a multiple of the integral of ln(x + 1) / x^(5/4) from x = 0 to infinity. Now you can do IBP and more substitutions to arrive at the integral of x^2 / (x^4 + 1) from x = 0 to infinity. Since the integrand is even, this integral can be solved using Cauchy-Schlömilch.
 
Given a ball B of radius 1 in R^d, prove that there exists a set of 6^d balls of radius 1/2 with centers on the surface of B which cover the entire surface of B.
 
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Given a ball B of radius 1 in R^d, prove that there exists a set of 6^d balls of radius 1/2 with centers on the surface of B which cover the entire surface of B.
Too crude for my taste. For those who want more of a challenge
  • harder -- prove that S^(n − 1) can be covered by 5^n balls with radius ½
  • hardest -- prove that S^(n − 1) can be covered by
    1709296604773
    balls with radius ½
 
another problem from @mindlessselfindulge
1709457747412
 
prove that the expectation of the maximum of n iid Gaussian random variables is O(sqrt(ln(n)))
 
For a positive integer n, given n integers (not necessarily distinct), prove that there is at least 1 non empty subset sum that is divisible by n.
Idk, can this be solved using modular arith and dirichlet's?. Or is the solution in the realm of higher math
 
Anyways, comfort problem for the masses:
Let (S) be the sphere
(x-a)^2+(y-b)^2+(z-c)^2=1
With a,b,c satisfying
a^2+(b-5)^2+(c-200)^2=16
For all a,b,c (S) is tangent to two stationary spheres with radii R1, R2 (R2>R1)
Calculate P=sqrt(R2^2-R1^2)
 
Idk, can this be solved using modular arith and dirichlet's?. Or is the solution in the realm of higher math

There are 2 ways: a simple way, and a more complicated way (that I came up with).

Both are elementary.

Anyways, comfort problem for the masses:
Let (S) be the sphere
(x-a)^2+(y-b)^2+(z-c)^2=1
With a,b,c satisfying
a^2+(b-5)^2+(c-200)^2=16
For all a,b,c (S) is tangent to two stationary spheres with radii R1, R2 (R2>R1)
Calculate P=sqrt(R2^2-R1^2)

This doesnt make sense.
 
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Anyways, comfort problem for the masses:
Let (S) be the sphere
(x-a)^2+(y-b)^2+(z-c)^2=1
With a,b,c satisfying
a^2+(b-5)^2+(c-200)^2=16
For all a,b,c (S) is tangent to two stationary spheres with radii R1, R2 (R2>R1)
Calculate P=sqrt(R2^2-R1^2)

The tangency would imply that the distance from the centre of sphere to the centre of each stationary sphere equals the sum of their radii, drawing out a diagram would be proof for this.

Since S has radius 1, the distance from the centre of S to the centre of the stationary spheres is R1+1 and R2+1.

The sphere defined by the condition on a,b,c can be seen as one of the stationary spheres. This sphere has radius 4, so either R1=4 or R2=4. Since R2 > R1, R2 = 4 and R1 = 0, having radius 0 is permissible because a point touching the sphere means it is tangent to it. To visualise everything here what we have is a sphere with radius 4 and a point that is tangent to S.

Now we just substitute it, sqrt(4^2-0^2) = 4

P = 4
 
For a positive integer n, given n integers (not necessarily distinct), prove that there is at least 1 non empty subset sum that is divisible by n.
honestly no idea if this is right

n = 1 case is satisfied because any number can be divided by 1

cases where n > 1 would need pigeonhole principle

consider a set of n integers {a1, a2 .. ,an}

and the amount of sums it can produce:

a1 - 1 st
a1 + a2 - 2 nd
a1 + a2 + a3 .. + an = n th

the total amount of possible sums here is n. and the total amount of distinct non-zero remainders would be n-1 because that's the maximum possible number of remainders you can get out of a number n. the pigeonhole principle states that at least two of these sums would have the same remainder when divided by n. This is good because if you subtract the two sums you are left with a remainder of zero meaning that the new subset would be divisible by n.

therefore we can prove that there exists a subset within {a1, a2 .. ,an} that it has a sum divisible by n (which is equivalent to the initial statement)
 
When Chad and Stacy have a baby it will inherit 50% of Chad DNA.

Chad's grandkid will be only 25% his DNA, and so on...

At some point down the genetic road, his DNA will be like a single chromosome that can mutate after being hit by cosmic radiation, making Chad's offspring indistinguishable from a theorethical descendant of you and I.

How many generations it takes for Chad's offspring to accidentally become my genetic son?
 
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When Chad and Stacy have a baby it will inherit 50% of Chad DNA.

Chad's grandkid will be only 25% his DNA, and so on...

At some point down the genetic road, his DNA will be like a single chromosome that can mutate after being hit by cosmic radiation, making Chad's offspring indistinguishable from a theorethical descendant of you and I.

How many generations it takes for Chad's offspring to accidentally become my genetic son?
719746294729748173917391739273
 
honestly no idea if this is right

n = 1 case is satisfied because any number can be divided by 1

cases where n > 1 would need pigeonhole principle

consider a set of n integers {a1, a2 .. ,an}

and the amount of sums it can produce:

a1 - 1 st
a1 + a2 - 2 nd
a1 + a2 + a3 .. + an = n th

the total amount of possible sums here is n. and the total amount of distinct non-zero remainders would be n-1 because that's the maximum possible number of remainders you can get out of a number n. the pigeonhole principle states that at least two of these sums would have the same remainder when divided by n. This is good because if you subtract the two sums you are left with a remainder of zero meaning that the new subset would be divisible by n.

therefore we can prove that there exists a subset within {a1, a2 .. ,an} that it has a sum divisible by n (which is equivalent to the initial statement)
100% correct, very well done
 
nice, what are the other ways of doing it?
The only other way I know of is significantly more cumbersome and involves first proving the statement for primes and subsequently showing that it holds for a product if it holds for the factors of said product. I learned it from @mindlessselfindulge
 
Integrate sqrt(x) * e^(-k * (x + (1 / x))) from 0 to infinity given k > 0
 
Integrate sqrt(x) * e^(-k * (x + (1 / x))) from 0 to infinity given k > 0
sqrt(π/k) exp(-2k) (1+1/(2k))

one way to obtain this is to use example 2 at <https://sos440.blogspot.com/2017/01/glassers-master-theorem.html> and differentiate under the integral sign w.r.t. s twice

another way is to substitute u = x + 1/x - 2 for 0 < x < 1 and x > 1 separately and simplify until you end up with exp(-2k) times the Laplace transform of (u+1) / sqrt(u)
 
sqrt(π/k) exp(-2k) (1+1/(2k))

one way to obtain this is to use example 2 at <https://sos440.blogspot.com/2017/01/glassers-master-theorem.html> and differentiate under the integral sign w.r.t. s twice

another way is to substitute u = x + 1/x - 2 for 0 < x < 1 and x > 1 separately and simplify until you end up with exp(-2k) times the Laplace transform of (u+1) / sqrt(u)

So I wrote the problem wrong, it was a variant of Putnam 1985 B5, and the integrand should have sqrt(1 / x), not sqrt(x)

This makes it significantly easier to solve

For the new problem, my answer is wrong according to wolfram alpha:

 
So I wrote the problem wrong, it was a variant of Putnam 1985 B5, and the integrand should have sqrt(1 / x), not sqrt(x)

This makes it significantly easier to solve

For the new problem, my answer is wrong according to wolfram alpha:

my answer for the sqrt(x) integral is the same as that of Wolfram Alpha and both of the methods I outlined can easily be adapted to solve for the sqrt(1/x) integral, again concurring with Wolfram Alpha
 
prove that for every y > 1 there exist distinct x > 1 and z > 1 such that x^y^z = z^y^x
 
prove that for every y > 1 there exist distinct x > 1 and z > 1 such that x^y^z = z^y^x

x * ln(y) + ln(ln(z)) = z * ln(y) + ln(ln(x))

ln(y) = (ln(ln(z)) - ln(ln(x))) / (z - x)

The derivative of ln(ln(x)) takes all positive real values.

There exists a tangent line of ln(ln(x)) with slope ln(y). Shifting this line down and finding the points of intersection with ln(ln(x)) gives values of x and z.
 
x is uniformly distributed on (0, 1)

Find E[x * floor(1 / x)]
 
x * ln(y) + ln(ln(z)) = z * ln(y) + ln(ln(x))

ln(y) = (ln(ln(z)) - ln(ln(x))) / (z - x)

The derivative of ln(ln(x)) takes all positive real values.

There exists a tangent line of ln(ln(x)) with slope ln(y). Shifting this line down and finding the points of intersection with ln(ln(x)) gives values of x and z.
correct
 
x is uniformly distributed on (0, 1)

Find E[x * floor(1 / x)]
this is just the integral of x * floor(1/x) as x runs from 0 to 1, which can be written as the sum of n times the integral of x from x = 1/(n+1) to 1/n as n goes from 1 to infinity = the sum of (2n+1) / (2n(n+1)^2) as n goes from 1 to infinity = ½ * ( sum ( 1/n - 1/(n+1) ) + sum 1/(n+1)^2 ) = ½ * sum 1/n^2 = pi^2/12
 
more than 1000 posts of math :feelsLightsaber:
 

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