Dan's Segment 30

To Calculate

1. In the slides we were using a concave function, but $x^2$ is convex, so we reverse the inequality and have $function(interpolation) \le interpolation(function)$

In this case the function is $x^2$ and the interpolation is the mean, so the specific inequality here is:

$(\sum_{i=1}^{N} \frac{x_i}{N})^2 \le \sum_{i=1}^{N} \frac{{x_i}^2}{N}$

And the problem statement is satisfied.

2. This problem is best approached by looking at the case where g(xm) = f(xm) as the case where g > f is self-explanatory from the problem statement. If f(xm) is at a local maximum, its derivative is 0. If g = f, and its derivative is not 0, it would cross over the function f and thus not be greater than or equal to f at all values of x. Thus, the derivative of g must be zero and therefor it is at a local maximum.

I do have a question about this though. What is to stop the function g from being at a local minimum at that particular x value or even just at some sort of inflection point? If the function always had to be concave for all values of x that would not be a problem, but for the general case I'm not sure this holds.