# Weierstrass M test

1. Mar 21, 2004

### robousy

Hi,

I don't really grasp the M test for convergence and I have several Hmwk problems and figured that if someone could explain how to do one of them then I should be able to figure the rest out.

ok - the problem is this:

Use the M test to establish the uniform convergence of:

(Sum from n - infinite) exp(in^2x) / n! for |z| < r where r < infinite.

Any help would be really appreciated

rachel.

Last edited: Mar 22, 2004
2. Mar 21, 2004

### cookiemonster

I think you need to clear up your notation first.

What is "exp(in^2x / n!"?

cookiemonster

3. Mar 22, 2004

### matt grime

The test states that

$$\sum a_n(x)$$

converges absolutely and uniformly on some set of values of X if for all x in X |a_n(x)|<= M_n and

$$\sum M_n$$

converges.

So all you need to do is show the n'th term's absolute value satisfies this rule for some constant M_n, and the sum of these constants is convergent.

4. Mar 22, 2004

### robousy

Hi, the notation exp(in^2x) / n!
means e to the power of i times n squared times x, all dividewd by n factorial where e is the exponential function, i is the aquare root of minus 1, n runs from 0 to infinity.

I still dont understand how to do this problem. :(

Last edited: Mar 22, 2004
5. Mar 22, 2004

### matt grime

Reread the statement of teh Weierstrass M test. It takes no understnading to actually use the test:

a_n(x) = exp(ixn^2)/n!

So what can you say about the absolute value of the n'th term in the series? So what is the sum of these values? Hence...?

6. Mar 22, 2004

### robousy

well - i probably wouldnt have posted a message if I understood the M test now would I?

well - if the nth term is less than infinity, then exp(in^2x) is very big for some huge but not infinite value, and the denominator n! is also very big. Great, I still dont know if it converges or not!!

7. Mar 22, 2004

### matt grime

Firstly can you tell me if x is a real number or a complex number? your initial post half indicates complex by saying |z|<r some r, yet you use x inside the sum.

If x is real this question is easy, though it appears from your last post that it might be complex.

Understanding why the M test works and using it are two entirely different things.

8. Mar 22, 2004

### matt grime

Let us assume x is real because I think the question is wrong if we assume x is complex.

the n'th term in the series is

a(n) = exp{ixn^2}/n!

|a(n)| = 1/n!, let this be M_n

then sum M_n is convergent so the initial series is unifomrly convergent on the whole of the real line.

I don't think it works for complex x since x = -it, for t some real number requires us to sum:

e^{tn^2}/n!

and that series deoesn't even converge as the ratio of consecutive terms is:

e^{t(n+1)^2 -tn^2}/(n+1) = e^(2tn+t)/(n+1) which tends to infinity as n tends to infinity.

Last edited: Mar 22, 2004
9. Mar 22, 2004

### robousy

Hi,

Thank-you Matt. I appreciate the time you've spent looking at this.

Yes - it is a complex question but with x real.

So - essentially all you have to do is take the modulus of whatever you are summing over, and then see of the number gets closer to zero as n goes to infinity. Is it is straightforward as that?

10. Mar 22, 2004

### matt grime

Not quite - you must find for each n, a number M_n such that |a_n(x)|< M_n for all x in the domain, where a_n(x) is the n'th term in the series. It will be a different M_n for each n.

Then the sum from 1 to infinity of the M_n must be finite (which is not the same as the M_n tending to 0).

Example:

$$\sum x^n/n!$$

for x in the complex numbers and |x| < 10, say.

a_n(x) = (x^n)/n!

so |a_n(x)| < 10^n/n! := M_n

now sum 1 to infinity of M_n converges hence the function is unifomrly continuous on the disc |x|<10

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