## Category Archives: calculus

### Question on an infinite series

Hello Sir,

I was in your algebra 3 course last year and found this blog useful so I was hoping you could provide me with some assistance on the following problem from my Theory of Numbers Course.

How would you show that Sigma(1/p^2) is less that or equal to 1. Where p is a prime.

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First of all, it’s better to say that the $p$ in the sum *runs over* the set of primes. If you say $p$ is a prime, it sounds like we’re speaking just of one.

Anyways, I’m hoping you learned a bit about the Riemann zeta function

$\zeta (s)=\sum_{n=1}^{\infty} n^{-s}.$

It is easy to see that this sum converges for $Re(s)>1$ and, importantly, can be written also as an infinite product in this range:

$\zeta(s)=\prod_p\frac{1}{(1-p^{-s})},$

where again the $p$ runs over the primes. In particular,

$\zeta (2)=\sum_{n=1}^{\infty} n^{-2}=\prod_p\frac{1}{(1-p^{-2})}$

If you write the last quantity as

$\prod_p(1+p^{-2}+p^{-4}+\cdots),$

and expand the product, you will see that it’s greater than

$1+\sum_p p^{-2}.$

Thus, the sum you’re interested in has shown up. Hence,

$\sum_p p^{-2} < \zeta(2) -1.$

Actually, it’s possible to evaluate $\zeta(2)$ precisely, and get $\pi^2/6$. However, for your inequality, it’s not necessary. All you need to know is $\zeta(2) \leq 2.$ Try to show this by bounding the sum for $\zeta(2)$ by an integral. (Recall the idea in the integral test for convergence of a positive series.)

### Local and global maxima

Hi Professor!

Just very quickly, what’s the difference between the global maximum/minimum of a function and the local maximum/minimum of a function? I understand that if a function achieves a global maximum at a point then it automatically achieves a local maximum, but vice versa is not necessarily true.

Also, the definitions look identical besides the local maximum def. containing epsilon.

Cheers!

A function f has a local maximum at c if f(c) is $\geq$ all the values at nearby points. f has a global maximum at c if f(c) is $\geq$ the values at all points (in the domain). Therefore, a global maximum is a local maximum, but *not* vice versa. The best way to show the difference would be to draw a picture, but my graphic skills on the computer are very limited. So I’ll just have to write down a formula. Consider
$f(x)=x^3-x$
You should be able to see that f(x) has a local maximum at $c=-1/\sqrt{3}$. Near that point, the graph of f(x) is like a hill with apex above that point. What is the global maximum? Of course there is none, because f(x) keeps getting bigger as you move to the right along the x axis (at least starting from the point $x=1/\sqrt{3}$). One sometimes turns this into a problem with a global maximum by restricting the domain. That is, if we take the same function except with domain only on the closed interval [-100,100], you should be able to see that the global maximum is at x=100.