Ramanujan and his e power equation

evene · 2016-11-02 09:43:22

Ramanujan found that How did he do that? Is there an underlying method that is both quick and efficient at calculating digits precisely?

Note that Ramanujan found this a long time before computers. Thus, he had to have a method for calculating these transcendental numbers!

zetafunc · 2016-11-02 09:44:29

My first guess would be the Taylor series for exp.

bobbym · 2016-11-02 09:49:50

Ramanujan came up with lots of identities and approximations just like that one. You will remember that both Hardy and Littlewood, premier mathematicians of that time were baffled by them, In many cases no one knows how he did it. When asked (according to legend) he said that a goddess would whisper them in his ear.

In modern times we might try a pslq but it was unknown when he was around.

evene · 2016-11-02 10:12:42

@bobbym What's plsq?

@zetafunc I understand the part where you said Taylor series expansion, brilliant that is! But how would you calculate the infinite sums?
Given

, how would you find the converging point when ?
We have
How would you calculate the infinite sum? I haven't learned taylor expansion yet, so sorry if that question seems a bit stupid.

bobbym · 2016-11-02 10:17:59

Take care with convergence. Although the series for e^x converges for all values of x, when computing actual numbers it may need many, many terms and questions of round off error also apply.

zetafunc · 2016-11-02 10:18:50

The best you can do is compute it to a finite number of terms, if you're doing it by hand. If Ramanujan used the Taylor series (which he may not have), then you'd need only the first 20 terms or so of that series to get the accuracy of his result.

bobbym · 2016-11-02 14:21:43

@bobbym What's plsq?

Called the greatest algorithm of the 20th century, the pslq was discovered by Ferguson and Bailey. It is an integer relation algorithm

http://www2.lbl.gov/Science-Articles/Ar … rithm.html

The algorithm I like to think was inspired by Ramanujan's incredible relationships. Anyway, it takes a decimal number and tries to come up with a relationship between that number and known constants. If you had the decimal 3.65028153987288474 the pslq would find this relationship.

This is a very simple example of its power. The ramifications of this process are enormous but still largely unknown.

evene · 2016-11-03 05:30:04

Actually, I now think the problem is a bit "redundant"..?

Wikipedia describes a simple procedure. And I know what to do, but I don't understand why that method works and why they got the steps. Unfortunately, it only works on Heegner numbers. (Implying that you can't evaluate with it)

The concept is simple.

For d is a Heegner number (19, 43, 67, 163) You can find the approximation using the q-expansion (). But I don't know how to evaluate it.

Last edited by evene (2016-11-03 05:31:18)

bobbym · 2016-11-03 06:42:23

Der Heegner numbers? Never heard of them. Got a link to that Wiki page?

evene · 2016-11-03 07:40:12

https://en.m.wikipedia.org/wiki/Heegner_number#Detail

bobbym · 2016-11-03 09:32:51

Looks complicated. What happens next?

evene · 2016-11-03 09:36:14

Wait, what do you mean by "what happens next"?

I'm just as stumped. I don't know why the formula works and how it works. I just know that it works.

bobbym · 2016-11-03 10:04:39

I guess they plug into that q series. I wished they would have shown how to get more terms of that q series but they say 2 terms is enough.

evene · 2016-11-03 14:41:49

And unfortunately, it only works with Heegner numbers.

Implying one cannot evaluate

At all...

bobbym · 2016-11-03 15:23:37

Not by that method.

zetafunc · 2016-11-04 02:18:31

evene wrote:

And unfortunately, it only works with Heegner numbers.
Implying one cannot evaluate

At all...

Depends what you mean by "evaluate". There's no reason why you can't just use the Taylor series for exp(x) to get as many decimal places as you like. The other series you mentioned might have a faster rate of convergence, but you can use either series to get the accuracy you want.

evene · 2016-11-05 09:32:05

If J(q) denotes Klein's absolute invariant, with and , then the fourier coefficients have the form

Now, let d be a positive square free integer in . And let

Thus

Respectively. Therefore, j(q) is an algebraic integer of degree h(d) where h(d) is the class number of Q(i*√d).

I need help breaking down this theorem.

Last edited by evene (2016-11-05 09:34:14)

zetafunc · 2016-11-06 03:15:56

Is there a specific question you have?

evene · 2016-11-06 11:52:31

Yes.

What is a Klein Absolute Invariate?
What is an algebraic integer?
What is a class number?
And what is the prerequisite knowledge to this.

zetafunc · 2016-11-06 20:58:53

An algebraic integer is any number which is the root of some monic polynomial with integer coefficients. is an algebraic integer, for example. The ring of algebraic integers is usually written or sometimes just

I'd only heard of the j-invariant before, but it seems that Klein's absolute invariant means the same thing. You can find the definition here. I've given a simplified version in the elliptic curves thread in Euler's Avenue.

Suppose is a field. Then the class group of is defined as where is the group of non-zero fractional ideals of and is the group of principal fractional ideals. Then the class number is just the size of that group. If the class group happens to be trivial, then that tells us that is a principal ideal domain and has unique factorisation of elements. So in a sense the class number tells us how far is from being a principal ideal domain (and all principal ideal domains are also unique factorisation domains). In practice you can compute the class group by finding all the ideals whose norm is not greater than the Minkowski bound.

A course in algebraic number theory would be helpful, and maybe a course in elliptic curves and modular forms.

Last edited by zetafunc (2016-11-07 01:43:34)

Math Is Fun Forum

#1 2016-11-02 09:43:22

Ramanujan and his e power equation

#2 2016-11-02 09:44:29

Re: Ramanujan and his e power equation

#3 2016-11-02 09:49:50

Re: Ramanujan and his e power equation

#4 2016-11-02 10:12:42

Re: Ramanujan and his e power equation

#5 2016-11-02 10:17:59

Re: Ramanujan and his e power equation

#6 2016-11-02 10:18:50

Re: Ramanujan and his e power equation

#7 2016-11-02 14:21:43

Re: Ramanujan and his e power equation

#8 2016-11-03 05:30:04

Re: Ramanujan and his e power equation

#9 2016-11-03 06:42:23

Re: Ramanujan and his e power equation

#10 2016-11-03 07:40:12

Re: Ramanujan and his e power equation

#11 2016-11-03 09:32:51

Re: Ramanujan and his e power equation

#12 2016-11-03 09:36:14

Re: Ramanujan and his e power equation

#13 2016-11-03 10:04:39

Re: Ramanujan and his e power equation

#14 2016-11-03 14:41:49

Re: Ramanujan and his e power equation

#15 2016-11-03 15:23:37

Re: Ramanujan and his e power equation

#16 2016-11-04 02:18:31

Re: Ramanujan and his e power equation

#17 2016-11-05 09:32:05

Re: Ramanujan and his e power equation

#18 2016-11-06 03:15:56

Re: Ramanujan and his e power equation

#19 2016-11-06 11:52:31

Re: Ramanujan and his e power equation

#20 2016-11-06 20:58:53

Re: Ramanujan and his e power equation

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