# Ramanujan Summation and ways to sum ordinarily divergent series

• A
• bhobba
In summary, the inverse problem is to guess what problem an infinite series was trying to solve and solve it that way.
Swamp Thing said:
Although "D" is not a number, it does take a function f(x) and give you f'(x), so in a sense we can think of D as a sort of number that represents the local value of f'(x)/f(x). If we plug that local ratio into a power series, then ##e^{Dh}## sort of makes sense. How far can we get if we try to run with this ball?

That's fine intuitively. You hit on the exact difference between applied and pure (or is that puerile ) math. In applied math degrees you still do some pure math simply as background (you should anyway). For example in the degree I did we did linear algebra as pure math before we did another subject Applied Linear Algebra. They don't tend to do that in Engineering and Physics courses so it makes following advanced pure math papers hard. Its actually a problem in applied math type degrees nowadays as well. My old alma mater doesn't even do the basic background pure math subjects because students thought it was just mind games. IMHO it's a big issue.

If you want to understand it you need a course in analysis, (colloquially called doing your epsilonics) which my alama mata once did but no longer does. A good reference and well written path into it for those already mathematically advanced is:
http://matrixeditions.com/5thUnifiedApproach.html
If you just have been exposed to basic calculus, and not gone deeper into applied math, I would suggest the following first:
https://www.amazon.com/dp/0691125333/?tag=pfamazon01-20

Thanks
Bill

Swamp Thing said:
Perhaps you have seen this blog post by Terence Tao.

Always read Terry's Blog. It is probably my favorite blog on the whole internet

My favorite way into this stuff is Borel Summation which is so simple I can't resist detailing a link to it here:
https://www.nbi.dk/~polesen/borel/node7.html
It can't be used to directly sum the Zeta function, but there is another function called the Eta function defined by η(s) = 1 - 1/2^s + 1/3^s - 1/4^s ... that you can easily derive a simple relation to the zeta function:
https://proofwiki.org/wiki/Riemann_Zeta_Function_in_terms_of_Dirichlet_Eta_Function

The Eta function is Borel Summable so low and behold you have summed the Zeta function.

Its interesting to see where analytic continuation has been used. I will leave those silly enough to read my ramblings to think about it.

Thanks
Bill

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bhobba said:
I like Hardy and his 'chatty' style, but it's a bit dated.
I think that Hardy's '1729' story was not true. I think that Hardy was aware of Ramanujan's prior writings regarding that number when the 2 men met in Punjab. I firmly disbelieve Hardy's anecdote to the effect that he had remarked to Ramanujan that he (Hardy) on his way to the meeting had ridden in taxicab number 1729, and that he (Hardy) thought 1729 was a dull or boring number, and that Ramanujan had immediately replied that 1729 was the least number that could be expressed as the sum of 2 cubes in 2 different ways -- despite Hardy's self-effacement and praise of the great mathematician Ramanujan, I believe that he just plain made up that story.

A characteristic of 1729 that Hardy did not report Ramanujan to have stated is that in duodecimal it is 1001 which numeral sequence in binary logic may be used to represent the 'if and only if' relation.

I think that Hardy intentionally fabricated the 1729 Hardy-Ramanujan anecdote.

Let's take the series 1 + 2 + 3 + ...

Let's add to it another series, that looks like

which is based on a normal distribution centered around say 20, with a spread of around 4, and the sum of the terms is 1.

Firstly, is this sum of two functions a valid candidate for a summation attempt? If so, it should result in -1/12 + 1, right?

If we look at the Ramanujan sum, then the bump around 20 would not contribute much to the result, would it? We would have -1/12 from the first series, and then we would have contributions from various derivatives of the "bump" taken at zero. These derivatives are going to be pretty negligible so the final result would still be about -1/12.

So is this connected to why Hardy cautioned about using the Ramanujan sum?

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Swamp Thing said:
So is this connected to why Hardy cautioned about using the Ramanujan sum?

See the Wikipedia article on Divergent Series:
https://en.wikipedia.org/wiki/Divergent_series#Ramanujan_summation

Note what it says about Ramanujan Summation:
The Ramanujan sum of a series f(0) + f(1) + ... depends not only on the values of f at integers, but also on values of the function f at non-integral points, so it is not really a summation method in the sense of this article.

That's its problem, although I do not know an example, you can get a different answer for the same sum by using a different f, so requires caution. In your example, since linearity does not apply to Ramanujan summation, you can't add the results like you want to.

Thanks
Bill

bhobba said:
you can get a different answer for the same sum by using a different f
That's an interesting point that I've missed in my (unsystematic) reading and video-viewing so far.
One way to generate new functions passing through the same sequence would be to add in one or more sine waves that go through zero at every integer. I would imagine that the derivatives of the sine wave around zero would dominate that puny -1/12 and change everything. (Unless they canceled out in some weird way).
It seems that the Ramanujan sum is really just a property of f around zero, that happens to equal the series sum for certain well-behaved types of f.

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Swamp Thing said:
It seems that the Ramanujan sum is really just a property of f around zero, that happens to equal the series sum for certain well-behaved types of f.

Could be. I have a copy of Hardy's book but have not studied it as thoroughly as I would like amongst all the myriad of other things I want to study. He likely has a more detailed analysis. It so bad I set myself the goal 10 years ago of studying Weinberg's masterpiece on QFT - but am lagging well behind - at the moment I am still stuck on Banks - Modern QFT after QFT for the Gifted Amateur. The sojourn I did with Zee didn't help. At first I rather liked Zee, but as time went by it was too disjointed and not that well explained for my taste, Banks I found a lot better.

Thanks
Bill

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