Range of Uniform convergence of dirchlet series

In summary: The textbook you're using is Arfken et al. (2013), and the person who wrote the answer you are asking for is Euge J. Weierstrass.
  • #1
ognik
643
2
Find the Range of Uniform convergence of $ \zeta\left(x\right) = \sum_{n=1}^{\infty}\frac{1}{{n}^{x}} $
Using the Weierstrass-M test, I get this converges for $ 1 \lt x \lt \infty $
But the book's answer is $ 1 \lt s \le x \lt \infty $? I have scoured the book but can't see why they say it this way (with the 's')?
 
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  • #2
Hi ognik,

Your answer is correct, but the book's answer doesn't make sense -- after all, the domain of convergence is a subset of $\Bbb R$, not a subset of $\Bbb R^2$. (Smile)
 
  • #3
ognik said:
Find the Range of Uniform convergence of $ \zeta\left(x\right) = \sum_{n=1}^{\infty}\frac{1}{{n}^{x}} $
Using the Weierstrass-M test, I get this converges for $ 1 \lt x \lt \infty $
But the book's answer is $ 1 \lt s \le x \lt \infty $? I have scoured the book but can't see why they say it this way (with the 's')?

What is required is the range of uniform convergence and not simply the range of convergence, that You have found... now the function $\displaystyle \zeta(x)= \sum_{n=1}^{\infty} \frac{1}{n^{x}}$ has a singularity in x=1 and that means that the range of uniform convergence is $s \le x < \infty$, where s is strictly greater than 1...

Kind regards

$\chi$ $\sigma$

P.S.: at first I wrote 's is stricly less that 1'... of course that is, as specified by Euge, a [trivial] mistake... very sorry!(Doh)...
 
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  • #4
I apologize for misreading -- the range $1 < x < \infty$ gives the domain of pointwise convergence, but $\sum\limits_{n = 1}^\infty n^{-x}$ is not uniformly convergent on $(1,\infty)$. However, by the Weierstrass M-test the series is uniformly convergent on every interval of the form $[1 + \delta, \infty)$, $\delta > 0$. The book's answer should have specified that $s$ is a parameter, not a variable of $\zeta$.
 
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  • #5
Thanks both, but I'm still not quite there. (Book is Arfken at al.)

I used the Weierstrass-M test, and as far as I can tell, that should only show Uniform (and incidentally Absolute) convergence? So I don't follow how "the range 1<x<∞ gives the domain of pointwise convergence"?

My main lack is - doesn't 1<x<∞ mean exactly the same as 1<s≤x<∞ ? Both say x cannot be 1, but can be any Real > 1. I suspect there is a mathematical subtlety there that I just don't know?

Finally - I can't see why x=1 is a singularity? It just makes $ \sum_{n=1}^{\infty}\frac{1}{{n}^{x}} $ 1n x the harmonic series doesn't it?
 
  • #6
ognik said:
I used the Weierstrass-M test, and as far as I can tell, that should only show Uniform (and incidentally Absolute) convergence? So I don't follow how "the range 1<x<∞ gives the domain of pointwise convergence"?
The series representing $\zeta(x)$ converges pointwise for $1 < x < \infty$ (by the $p$-test), but not uniformly on $1 < x < \infty$.

My main lack is - doesn't 1<x<∞ mean exactly the same as 1<s≤x<∞ ?
No. Take for example $s = 2$. This would give you the interval $[2,\infty)$, which is different from $(1,\infty)$. The domain of uniform convergence is more precisely the set $A = \bigcap\limits_{s > 1} [s,\infty)$.

Finally - I can't see why x=1 is a singularity? It just makes $ \sum_{n=1}^{\infty}\frac{1}{{n}^{x}} $ 1n x the harmonic series doesn't it?
The harmonic series does not converge.
 
  • #7
Ah, gotcha, singularity in terms of convergence (I was incorrectly thinking of the term itself).

I can see the sense in what you say w.r.t. ($ 1 \lt s \le x \lt ∞ $), but not why. If I explain what I am seeing, hopefully you can see what I am missing...

If someone wrote $ 1 \lt x \lt ∞ $, I would deduce x any real, > 1. Then if someone else wrote $ 1\lt s\le x \lt ∞ $, I would deduce the same thing. I don't know any reason for the parameter 's' ...
 
  • #8
Euge said:
...No. Take for example $s = 2$. This would give you the interval $[2,\infty)$, which is different from $(1,\infty)$. The domain of uniform convergence is more precisely the set $A = \bigcap\limits_{s > 1} [s,\infty)$.

I have really scoured the textbook & web and found nothing on why this parameter is used. I understand that with s=2 it would give the interval [2,∞), and s cannot = 1 etc., what I don't get is why we need to use s in the first place?
My only idea is - sometimes when using the integral test, we use a parameter t when the top bound is $ \infty $, replacing $ \infty $ with t and them finding the limit as t -> $ \infty $ ... is this parameter something similar?
 
  • #9
Hi ognik, sorry for the late response. What is the title of the textbook you're using, and who is the author? I'd like to take a look at the issue myself. Also, I made an error in writing the expression for $A$; certainly, there is no real number $x$ such that $x \ge s$ for all $s > 1$. Please ignore it.
 
  • #10
Thanks Euge, it's 'Essential Mathematical Methods for Physics', Weber & Arfken, International Students Edition (2003). This is exercise 5.5.1

BTW - hope I haven't mislead, your explanation is probably correct - I just don't understand ...
 
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  • #11
Hi, another way of looking at what I am not understanding is - why would an answer of $ 1 \lt x \lt \infty $ be wrong or incomplete?

My book says for the Weierstrass M test: " If we can construct a series of numbers $ \sum_{1}^{\infty} {M}_{i} $ in which $ {M}_{i} \ge \left| {U}_{i}\left(x\right) \right| $ for all x in the interval [a, b] and $ \sum_{1}^{\infty} {M}_{i} $ is convergent, our series $ {U}_{i}(x) $ will be uniformly convergent in [a, b]. I can't therefore see why the series is NOT uniformly convergent in $ 1 \lt x \lt \infty $ , which is what the W_M test gave me?

I found a line later in the textbook that seems to cover this - "Suppose our power series [Eq. (5.84)] has been found convergent for −R <
x < R; then it will be uniformly and absolutely convergent in any interior interval, −S ≤ x ≤ S, where 0 < S < R." I can see how this would explain it, but again - I found the series already uniformly convergent for $ 1 \lt x \lt \infty $ - confused, me.
 
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  • #12
The thing is, the series for $\zeta(x)$ does not converge uniformly on $(1,\infty)$, but it converges uniformly on every subinterval of the form $[t, \infty)$ where $t > 1$.
 
  • #13
I still can't figure out what I'm missing, why is [1+δ,∞), δ>0 not the same as saying (1,∞)?
 
  • #14
Given $\delta > 0$, the interval $[1 + \delta, \infty)$ is strictly contained in $(1,\infty)$.
 
  • #15
I must look rather dense here, but I'm still not getting this. Maybe its the W-M test itself I have not correctly understood.
The function is $ \zeta\left(x\right) = \sum_{n=1}^{\infty}\frac{1}{{n}^{x}} $ So that you can check my steps:

I chose to compare with the series $ \frac{1}{{n}^{p}} $, which converges for p > 1

So I reason, $ |\frac{1}{{n}^{x}}| \le \frac{1}{{n}^{p}} $ for all x, where $ x \gt p, and \: p \gt 1$
And from here I can't see the need for the parameter s, or the strictly inclusive condition; it seem to converge for x > 1
Am I missing something?
 
  • #16
Let me ask you this first: do you think the series for $\zeta$ converges uniformly on $(1,\infty)$?
 
  • #17
ognik said:
I still can't figure out what I'm missing, why is [1+δ,∞), δ>0 not the same as saying (1,∞)?
Suppose you have a collection of intervals $\{I_\alpha\}$ whose union is $I$. Suppose also that there is a function $f(x)$ that is uniformly continuous on each of the intervals $I_\alpha$. It does not necessarily follow that $f$ is uniformly continuous on $I$.

The function $\zeta(x)$ illustrates that fact. It is uniformly continuous on each interval $[1+\delta,\infty)$, $\delta>0$. The union of all those intervals is $[1,\infty)$, but $\zeta$ is not uniformly continuous on $[1,\infty)$.
 
  • #18
Thanks both, Euge first.
I believe that the W-M test shows if the series is uniformly convergent so, yes. I don't know of a reason it shouldn't - although obviously there is ...

Opalg:
Is the collection of intervals not continuous?
I know the argument here is that the series is only point wise continuous in (1, inf), yet my book says the W-M test shows uniform convergence?

I also thought - What is special about the zeta function? It has not been covered in my course, just used as an exercise. In searching I find that it "... converges when the real part of s is greater than 1'. Here we have Zeta(x) which is the real part anyway, so that again suggests x > 1 to me.

Incidentally I came across this as an intuition example of the use of the complex Zeta(s) function:
"As s gets closer to 1 , the temperature of this system increases until it would require infinite energy to make s equal to 1 . But this limit is extremely important to understand: it is the limit in which the probability distribution above gets closer and closer to uniform." So I think zeta in practice can approach 1 but that becomes less and less probable...
 
  • #19
ognik said:
Thanks both, Euge first.
I believe that the W-M test shows if the series is uniformly convergent so, yes. I don't know of a reason it shouldn't - although obviously there is ...

If you reread your argument, what you have shown is that for fixed $p > 1$, the series of $\zeta$ converges uniformly on $(p,\infty)$. In your M-test argument, you supposed $x > p$ for some $p > 1$ -- this is a restriction that does omits all values of $x$ in the interval $(1,p]$. Be vary careful when applying the M-test, because it does not work for every series of functions on every domain.
 
  • #20
I don't know why I am struggling to wrap my head around this.
Did I (accidentally) say that if x > p and p> 1, then there is some infinitely small 'gap' between 1 and p? Is that it?

So, if the condition was $ x \ge p $ and $ p \gt 1 $, only then would the interval be $ (1, \infty) $?
 
  • #21
You can't include $x = 1$ because the harmonic series diverges. Also, if you were to prove $\sum\limits_{n=1}^\infty n^{-x} $ converges uniformly on $(1,\infty)$ using the M- test, you would need to find a positive sequence $a_n$ (depending only on $n$) such that $|n^{-x}| \le a_n$ such that $\sum\limits_{n=1}^\infty a_n$ converges. You chose $a_n = n^{-p}$, which is a sequence which depends on $p$. This is fundamentally the flaw in your assertion that your proof by the M-test justifies uniform convergence of $\sum\limits_{n=1}^\infty n^{-x}$ on all of $(1,\infty)$.

Notice also that this is a different issue than saying $(1, p)\cup [p,\infty ) = (1,\infty)$, which is simply a fact about intervals.
 
  • #22
Sorry, now even more confused (BTW, I've never thought x=1 was included). So with the W-M test, I must always find a sequence that only depends on n?
 
  • #23
No, not always, but in this case, yes. Think of it this way. By choosing $a_n = n^{-p}$ (where $p$ is some real number greater than one) you have $|n^{-x}| \le a_n$ for which values of $x$? If you solve the inequality, you get $x \ge p$. But what if $x = p/2$? In that case you get $|n^{-x}| > a_n$. However, you're supposed to have $|n^{-x}| \le a_n$ for all $x\in (1,\infty)$. See now where the problem is? Try not to overthink the issue. :-)
 
  • #24
Thanks Euge;
Euge said:
No, not always, but in this case, yes. Think of it this way. By choosing $a_n = n^{-p}$ (where $p$ is some real number greater than one) you have $|n^{-x}| \le a_n$ for which values of $x$?
Is there a difference if I choose $a_n = n^{-x}$ ? Then I know this converges for x>1 by comparison with the p-series?

I don't think I am over thinking this, it is more that I feel strongly that I have some basic gap in my knowledge that everyone would assume I know, so please keep that in mind - I will happily accept 'do you know's at a basic level...for example what you said on limits I will have to look up, that is more detail on limits than I remember ever seeing.

May I ask a big favour - that you show me in detail, and assuming I am seeing it for the first time, exactly how you would work through the original exercise? That should let me see what I am missing (I could already write down the solution believably, but I really want to understand this)
 
  • #25
You cannot choose $a_n = n^{-x}$ because $x$ is the domain variable, which would mean that $a_n$ is not a numerical sequence but a sequence of functions. Also, you would gain nothing from the inequality $|a_n| \le a_n$.

Recall that a sequence $f_n : A \subset \Bbb R \to \Bbb R$ converges uniformly to a function $f : A \to \Bbb R$ if given $\epsilon > 0$, there exists a positive integer $N = N(\epsilon)$ such that $|f_n(x) - f(x)| < \epsilon$ for all $x\in A$ and $n\ge N$. Notice in this definition that $N$ depends only on $\epsilon$ and not on $x$ (in pointwise convergence, the $N$ may depend on $x$).

Now a series $\sum f_n$ converge uniformly to a function $f$ if its sequence of partial sums converge uniformly to $f$. The Weierstrass M-test gives a useful criterion for uniform convergence of series: If $\{f_n(x)\}$ is a sequence of functions on a set $A\subset \Bbb R$ and $\{M_n\}$ is a sequence of positive constants such that $|f_n(x)| \le M_n$ for all $x\in A$ and $\sum M_n$ converges, then the series $\sum_{n = 1}^\infty f_n(x)$ converges uniformly on $A$.

You're given the series $\sum_{n=1}^\infty n^{-x}$. Fix $p> 1$ and let $A = [p,\infty)$. Set $M_n = n^{-p}$. Then $M_n$ is a numerical sequence such that $|n^{-x}| \le M_n$ for all $x\in A$ and $\sum_{n=1}^\infty M_n$ converges. Hence by the M-test, the series $\sum_{n=1}^\infty n^{-x}$ converges uniformly on $[p\infty)$. Since $p$ was arbitrarily chosen, your series converges uniformly on every closed interval $[p,\infty)$ where $p> 1$.
 
  • #26
Euge said:
You're given the series $\sum_{n=1}^\infty n^{-x}$. Fix $p> 1$ and let $A = [p,\infty)$. Set $M_n = n^{-p}$. Then $M_n$ is a numerical sequence such that $|n^{-x}| \le M_n$ for all $x\in A$ and $\sum_{n=1}^\infty M_n$ converges. Hence by the M-test, the series $\sum_{n=1}^\infty n^{-x}$ converges uniformly on $[p\infty)$. Since $p$ was arbitrarily chosen, your series converges uniformly on every closed interval $[p,\infty)$ where $p> 1$.
Hi Euge, thanks for your patience. I quoted the above para. from you, because it is similar to how I arrived at my own answer.

So what is the reason to use the parameter s?
Looking at the above, does it have to do with your choosing a closed interval [p,∞), where p > 1? I know that I am thinking of choosing an open interval (p, ∞) with p > 1, so I might just be missing some notational fact of intervals?
Thanks again
 
  • #27
I have used $p$ instead of $s$, and to avoid confusion, I simply noted that the series converges uniformly on $[p,\infty)$ for every $p > 1$. I did not write $1 < s \le x < \infty$ as written in your book because of its ambiguity.
 
  • #28
Sorry Euge, wasn't referring to s or p specifically - just trying to see what in the analysis I quoted makes it necessary to say 1 < p ≤ x < ∞, instead of 1 < x < ∞?

So - does it have to do with your choosing a closed interval [p,∞), where p > 1? I know that I am thinking of choosing an open interval (p, ∞) with p > 1, so I might just be missing some notational fact of intervals to make 1 < p ≤ x < ∞ correct and 1 < x < ∞ wrong?
 
  • #29
I see that you're trying to make sense of the answer in the book, but like I've said, the answer is ambiguous as written. So try not to make sense out of it. The solution I gave is the one you need to know.
 
  • #30
Thanks Euge, you must be pretty weary of this by now, really sorry to drag it out.

Its not so much about this particular question, I just feel sure that the exercise has highlighted a hole in my maths, and I'd like to fill it if I can. While I follow your solution, I can't quite explain to myself why you chose "converges uniformly on [p,∞), where p > 1", instead of (p,∞), where p > 1 ?
 
  • #31
ognik said:
While I follow your solution, I can't quite explain to myself why you chose "converges uniformly on [p,∞), where p > 1", instead of (p,∞), where p > 1 ?

Hmm, I'm not sure if this question has do with a weak knowledge about intervals. The statement would be true if you put $(p,\infty)$ instead of $[p,\infty)$, but what I proved is the stronger result, since $(p,\infty)$ is a proper subset of $[p,\infty)$ ($p\in [p,\infty)$ but $p\notin (p,\infty)$).
 
  • #32
Is it because you chose p earlier in your analysis, so it would be wrong to then use (p, inf) because p is not in (p, inf)?
 
  • #33
I didn't say it would be wrong to use $(p,\infty)$. The point is that my argument shows uniform convergence is valid on $[p,\infty)$ for all $p> 1$. In particular, uniform convergence is valid on $(p,\infty)$ for all $p > 1$.
 
  • #34
sorry, I should have said something like weaker ...

But I still can't grasp why one is different from the other, OR why one is better than the other.

My lecturer finally responded to my nagging and said "What this statement means is that you have two parameters s and x. However, the values of s are restricted by the condition indicated. When x is varied s should be fixed."

While that seems to make sense, I still am not getting it ...I keep hoping there is just a piece of math theory that I don't know, that will make this clear to me?
 
  • #35
When you say that you can't grasp why one is different from the other, do you mean you don't understand why $(p,\infty)$ is different from $[p,\infty)$?
 

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