## Category Archives: analysis

### 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.

———————————————————————–

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.)

### Student reports, end of August

Because of a rather hectic travel schedule, I was slow in putting up the reports that were submitted assiduously by Zhe and Alex. I apologize.

Zhe’s reports on the book Riemann’s Zeta Function’ by Edwards:

Zhe’s report 4

Zhe’s report 5

Zhe’s report 6

Zhe’s report 7

Alex’s reports on elliptic curves:

Points of finite order (exercises)

Group of rational points

Mordell’s theorem

### Limits, one and two-sided

Hi Sir!

Sorry for this very elementary question. I want to ask about Analysis 1101
Topic: Limits of functions. What’s the difference between the these two
definitions?

1) Say lim f(x)= L as x→b‾. If given ε>0, we can find δ>0 such that when

b-δ< x <b =>│f(x)-L│< ε.

2) Assume f(x) is defined in an interval (a,b) with ξ ϵ (a,b) but possibly f is not defined at ξ. Say lim f(x)=L as x→ξ. If given ε >0, we can find δ>0 such that when 0 <│x-ξ│< δ =>│f(x)-L│< ε.

Thanks lots!
Smile always!
Vanessa =)

Oh! Is the difference in that (1) only accounts for the limit as x tends to a number from the negative side, so you will need to work out in a similar way to find the limit as it tends to the same number from the positive side?

Whereas (2) takes into account both limits as x tends to a number from both the negative and positive sides? But what happens when the graph experiences a drastic change when it goes from the negative side of a number to the positive side of the same number? Would the second definition still hold? If it doesn’t, would we have to split it into two parts like in (1)? And how would you know when the second definition holds or not?

An example of a graph with drastic change from lecture is:

f(x) = 4-x, x < 1
f(x)=4x, x ≥ 1.

Thanks and sorry to bother you on non-tutorial days…

Smile always!
Vanessa =)

It seems you’ve already figured out the main point. I strongly advise all students to take note of this process: When the effort was expended to formulate the question in precise terms, the question essentially resolved itself! This is one of the main reasons I am recommending that people submit their questions in writing.

In fact, to define the meaning of lim_{x->c}f(x)=L, f never needs to be actually defined at c. Also, as you’ve figured out, the condition

0<|x-c|< delta

can be broken up as

c< x < c+ delta (x is quite close to the right of c)
*or*
c- delta < x < c (x is quite close to the left of c).

So the meaning of

lim_{x->c}f(x)=L

which is

for all x close to c, f(x) is close to L,’

can be broken down into

for all x close to c on the right *or* on the left, f(x) is close to L,’

On the other hand, the meaning of

lim_{x->c-}f(x)=L

is

for all x close to the left of c, f(x) is close to L.’

So clearly, the first condition is stronger than the second.

Of course you can define the notion of right limits as well:

lim_{x->c+}f(x)=L means …’

In fact, a minor theorem is:

lim_{x->c)f(x) exists if and only if
lim_{x->c-)f(x) and lim_{x->c+)f(x)

both exist *and* are equal to each other. In that case, of course,

lim_{x->c)f(x) =lim_{x->c-)f(x)=lim_{x->c+)f(x)

In the example you give above,

lim_{x->1+}f(x)=4 while lim_{x->1-}f(x)=3.

So lim_{x->1}f(x) does *not* exist.

If you take a function like

f(x)= x for x< 0, f(x)=1/x for x> 0,

then lim_{x->0-}f(x) exists, but {x->0+}f(x) does not. A fortiori, {x->0}f(x) does not exist.

If you define f(x)=1/x for all non-zero x, then neither the left nor right limit exists at 0. (They both exist and are equal at all other points, however.)

Note that in the preceding paragraphs, I’m being quite informal in describing the definitions using the phrase close to.’ It’s important to acquire the ability to move freely between such casual descriptions that make meanings transparent, and the precise notions using epsilons and deltas.

MK

### A comment on the previous post

It occurred to me that the previous post might have given the wrong impression, for example, that you need to understand the article on the Chinese remainder theorem for the exam. This was far from my intention. I hope it’s obvious that they are actually recommended either for your general amusement, or if you wish to go farther than the constraints of the courses themselves allow.

MK

### Three pedagogical articles and one more

Three articles available on my website discuss topics that may be of interest to my current students. But they’re a bit hidden, so I’m adding links here.

Comments on the Chinese remainder theorem

Some matrix groups

Why everyone should know number theory

All three are a bit dated and many sentiments expressed there (especially in the third one) don’t look quite right anymore, but maybe someone will find them amusing, nevertheless.

I’ve received now several inquiries about my actual research. Perhaps I’ll attempt sometime to make the main ideas at least more accessible by writing something expository. In the meanwhile, here is the presentation I gave at the London-Paris Number Theory Seminar this fall. You can throw it a casual glance if it seems worth the bother.

MK

### Serge Lang

While conversing with Acyr Locatelli about the textbook on linear algebra, I ended up elaborating a bit on the author Serge Lang, who was mentioned earlier as the supervisor for my Ph.D. thesis. I thought therefore to provide links to two articles that appeared in the Notices of the American Mathematical Society a short while after his death. One concentrates on the personality, and the other, on mathematics. My contribution appears in the latter, which, unfortunately, is a bit involved. The former, on the other hand, presents a vivid portrait of an exceedingly colorful mathematician. So it may be of interest, especially to students thinking seriously about an academic career. In any case, it’s well-known that familiarity with the author can make books come alive, even when (especially when, I like to think) it’s about fundamental mathematics.

MK