I don't understand Dirichlet's Test

woodoo · 2007-02-11 19:47:42

I can't grasp the concept of Dirichlet's Test,

I've looked online, but I don't understand it. Can someone give me a simple example to explain it.

Thanks

edit: I think what is confusing me is this:

Am I supposed to prove that is true or do I just assume it is?

Last edited by woodoo (2007-02-11 19:56:17)

Zhylliolom · 2007-02-11 20:09:07

We have two sequences of real numbers {a[sub]n[/sub]} and {b[sub]n[/sub]} such that {a[sub]n[/sub]} is a decreasing positive sequence (a[sub]n[/sub] ≥ a[sub]n+1[/sub] > 0) which converges to 0 and the sum of countably many terms of {b[sub]n[/sub]} is bounded (|∑b[sub]n[/sub]| ≤ M for some constant M). Then

converges.

I will use the problem you posted in your other thread to give a simple example:

Let {a[sub]n[/sub]} = 1/n and {b[sub]n[/sub]} = sin n. Clearly, 1/n ≥ 1/(n + 1) > 0 and lim(1/n) = 0, so that {a[sub]n[/sub]} satisfies the conditions of Dirichlet's test. Since |sin x| ≤ 1 for any real x, we have that

for every positive integer N. Then {b[sub]n[/sub]} satisfies the conditions for Dirichlet's test. Thus

converges.

Last edited by Zhylliolom (2007-02-11 20:12:38)

woodoo · 2007-02-11 20:15:28

Oh I understand now, I thought that {bn} was going to have to equal (sin n)/n, I didn't realize you were supposed to break down (sin n)/n into {an} = 1/n and {bn} = sin n to represent

$1e5a14e85c161d005de3ce4877f65d1.gif$

Thank you Zhylliolom

Last edited by woodoo (2007-02-11 20:16:46)

MathsIsFun · 2007-02-11 22:55:00

Cool. I learned something, too!

mathsyperson · 2007-02-12 01:31:41

I'm not sure I get this either.

If we used the same reasoning as we did for sin n/n, but replaced sin n with 1, then according to Dirichlet's test it would converge, because all of the same reasoning still holds.

1≤1 for all x, so

And yet, it's been proven that 1/n diverges. What am I doing wrong?

George,Y · 2007-02-12 01:53:46

mathsyperson wrote:

I'm not sure I get this either.
If we used the same reasoning as we did for sin n/n, but replaced sin n with 1, then according to Dirichlet's test it would converge, because all of the same reasoning still holds.
1≤1 for all x, so
And yet, it's been proven that 1/n diverges. What am I doing wrong?

Great point!

And I don't think f(N) is bounded because |f(N)|≤N.

Dross · 2007-02-12 02:11:43

mathsyperson wrote:

I'm not sure I get this either.
If we used the same reasoning as we did for sin n/n, but replaced sin n with 1, then according to Dirichlet's test it would converge, because all of the same reasoning still holds.
1≤1 for all x, so
And yet, it's been proven that 1/n diverges. What am I doing wrong?

I think a minor mistake/typo has been made: for Dirichlet's test, you need

for every M, and for N a constant. (note the change in limit of the summation from N to M)

This is the case for b[sub]n[/sub] = sin(n), but not for b[sub]n[/sub]n = 1.

Last edited by Dross (2007-02-12 02:18:17)

mathsyperson · 2007-02-12 03:01:29

Ah, that makes a bit more sense. But then again, surely adding up lots of sin n's would eventually get you to a number greater than any M you choose? How does that bit work?

George,Y · 2007-02-12 03:30:21

That is hard

Dross · 2007-02-12 03:45:38

mathsyperson wrote:

Ah, that makes a bit more sense. But then again, surely adding up lots of sin n's would eventually get you to a number greater than any M you choose? How does that bit work?

Some of the sin(n) terms will be positive, and some will be negative (remember you only take the modulo after adding all the terms), so they will cancel each other out.

mathsyperson · 2007-02-12 04:22:52

Gah! Of course, I completely forgot about sin x being sometimes negative. Now it makes perfect sense.

woodoo · 2007-02-12 05:32:57

So if Zhylliolom right? Because you guys are confusing me

mathsyperson · 2007-02-12 09:01:57

Zhylliolom is nearly right. The only change you need to his explanation is to use Dross's formula instead of his, when talking about a[sub]n[/sub].

Sorry, I kind of hijacked your thread a bit there.

George,Y · 2007-02-12 13:04:34

Dross wrote:

mathsyperson wrote:
Ah, that makes a bit more sense. But then again, surely adding up lots of sin n's would eventually get you to a number greater than any M you choose? How does that bit work?
Some of the sin(n) terms will be positive, and some will be negative (remember you only take the modulo after adding all the terms), so they will cancel each other out.

That is not rigorous. Look at this sequence:
2, -1, 2, -2, 2, -0.5,...
Are you sure of convergence?

Dross · 2007-02-12 21:17:58

George,Y wrote:

That is not rigorous. Look at this sequence:
2, -1, 2, -2, 2, -0.5,...
Are you sure of convergence?

Oh, it's not rigorous at all - more a view to see that it's possible. I just can't remember how to prove it at the moment, is all.

Math Is Fun Forum

#1 2007-02-11 19:47:42

I don't understand Dirichlet's Test

#2 2007-02-11 20:09:07

Re: I don't understand Dirichlet's Test

#3 2007-02-11 20:15:28

Re: I don't understand Dirichlet's Test

#4 2007-02-11 22:55:00

Re: I don't understand Dirichlet's Test

#5 2007-02-12 01:31:41

Re: I don't understand Dirichlet's Test

#6 2007-02-12 01:53:46

Re: I don't understand Dirichlet's Test

#7 2007-02-12 02:11:43

Re: I don't understand Dirichlet's Test

#8 2007-02-12 03:01:29

Re: I don't understand Dirichlet's Test

#9 2007-02-12 03:30:21

Re: I don't understand Dirichlet's Test

#10 2007-02-12 03:45:38

Re: I don't understand Dirichlet's Test

#11 2007-02-12 04:22:52

Re: I don't understand Dirichlet's Test

#12 2007-02-12 05:32:57

Re: I don't understand Dirichlet's Test

#13 2007-02-12 09:01:57

Re: I don't understand Dirichlet's Test

#14 2007-02-12 13:04:34

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#15 2007-02-12 21:17:58

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