Category Archives: calculus

Question on an infinite series

January 8, 2009 – 8:03 pm

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.

I would really appreciate any help you could give me.

———————————————————————–
Reply:

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.)

By minhyong kim | Also posted in analysis, number theory | Comments (0)

Local and global maxima

March 21, 2008 – 9:36 pm

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!

Reply:

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.

By minhyong kim | Comments (0)
Follow

Get every new post delivered to your Inbox.