1. ## Intermediate-value theorem

can someone help me on this question.

2. ## Re: Intermediate-value theorem

Define $\displaystyle r(\theta)$ to be the radiation at angle $\displaystyle \theta$ from whatever starting point you want. Define $\displaystyle f(\theta)$ to be $\displaystyle r(\theta)- r(\theta+ \pi)$. Then $\displaystyle f(\theta+\pi)= r(\theta+ \pi)- r(\theta+ 2\pi)= r(\theta+ \pi)- r(\theta)= -f(\theta)$. So either f is identically 0, in which case $\displaystyle r(\theta)= r(\theta+ \pi)$ for all $\displaystyle \theta$ or there exist [tex]\theta[tex] such that $\displaystyle f(\theta)$ is positive and $\displaystyle f(\theta+ \pi)$ is negative. Since $\displaystyle r(\theta)$ is continuous, so is $\displaystyle f(\theta)$ and there must be some $\displaystyle \theta$ such that $\displaystyle f(\theta)= 0$

3. ## Re: Intermediate-value theorem

What if we considered the function $\displaystyle g(\theta) = f(\theta) - f(\theta + \pi)$? It follows from a fairly fundamental theorem about functions that $\displaystyle g$ is continuous on $\displaystyle [0,\pi]$. We must then be in one of three cases:
1. g(0) = 0
2. g(0) < 0
3. g(0) > 0

In the first case, we have $\displaystyle f(0) = f(\pi)$ and we are done.

Now, for the other two cases we must be more clever. We must first realise that $\displaystyle f(0) = f(2\pi)$ and then apply the intermediate value theorem to $\displaystyle g$. Do you think you can finish this off?

Edit: I was writing my answer at the same time as another user. I'll leave it here anyway, it might be useful.

4. ## Re: Intermediate-value theorem

thanks both of you