Solved by Shantanu Chakrabar

Question:

This question was posed by Professor Doyle during the Math Club meeting on 4/8/98.

Prove that along the equator of the earth there will be at least one point where the temperature is the same as at the point directly opposite of it on the equator. (One assumes that the temperature function of a point varies continuously with the location.)

Mathematically the problem is: consider a function, f, defined on a circle, prove that there exists some point, x, on the circle (designated by the angle x to that point) where f(x)=f(x+pi). The function f has two constraints on it: it is continuous and f(x+2*pi)=f(x) for all x.

Note: this problem is very accessible in that it only requires material from math 20A; this problem is challenging in that it requires an original use of that material. Good Luck!

Answer:

Integral Proof:

Let Int[f(x)] { 0 , pi } be a definite integral from 0 to pi. and g(x) = f(x+pi) - f(x) so Int[g(x)] { 0, pi } = Int[ f(x+pi) - f(x) ] { 0 , pi }

=> = Int[ f(x+pi) - f(x+2pi) + f(x+2pi) - f(x)] { 0, pi}

=> = Int[ f(x+pi) - f(x+2pi) ] { 0, pi}

=> = Int[ f(x) - f(x+pi) ] { pi , 2pi }

=> = -Int[ f(x+pi) - f(x) ] { pi, 2pi }

=> Int[g(x)] { 0, pi } + Int[g(x)] { pi, 2pi } = 0

 

Now because f(x) is well behaved function => g(x) is well behaved and hence has no discontinuities.

Therefore

 

Int [ g(x) ] { 0, 2pi } = 0

Here comes the result. That g(x) must have at least one zero crossing to achieve that which means that

 

There exists an x for which g(x) = 0 or f(x+pi) - f(x) = 0 or f(x) = f(x+pi).



A more simple proof:

 

Let g(x) be = f(x+pi) - f(x)

=> g(x+pi) = f(x+2pi) - f(x+pi)

 

Now adding both we get g(x) = -g(x+pi).

 

Because of the well behaving functionality of g(x) => the g(x) is

reflective at phase change of pi and hence must cross zero at least once.

and LO we get the result.

-- Shantanu Chakrabar