Talk:Leibniz formula for π

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The justification of the term-by-term integration here is not actually trivial. Charles Matthews 12:34, 9 Mar 2005 (UTC)

${\displaystyle 1-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}...}$

can be rewritten as:

${\displaystyle 1+{\frac {1}{3}}+{\frac {1}{5}}+{\frac {1}{7}}...-{\frac {2}{3}}-{\frac {2}{7}}-{\frac {2}{9}}...}$

ie: as the sum of two overlapping infinite summations, where one is all positive and the other is all negative.

This necessarily converges far worse than the original formulation, as the two sums step at different rates. It works only because the infinities are both of natural numbers and thus equal infinities, so it doesn't matter that the step rate is different. But continuing with it anyway, we get:

${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{2n-1}}-\sum _{n=1}^{\infty }{\frac {2}{4n-1}}}$

Reduce to a single summation:

${\displaystyle =\sum _{n=1}^{\infty }({\frac {1}{2n-1}}-{\frac {2}{4n-1}})}$

Get the denominators equal:

${\displaystyle =\sum _{n=1}^{\infty }({\frac {4n-1}{(2n-1)(4n-1)}}-{\frac {4n-2}{(2n-1)(4n-1)}})}$

Combine the two fractions:

${\displaystyle =\sum _{n=1}^{\infty }{\frac {(4n-1)-(4n-2)}{(2n-1)(4n-1)}}}$

Simplify:

${\displaystyle =\sum _{n=1}^{\infty }{\frac {1}{(2n-1)(4n-1)}}}$

Since Pi is equal to four times this, we get that Pi then equals:

${\displaystyle 4\sum _{n=1}^{\infty }{\frac {1}{(2n-1)(4n-1)}}}$

As best as I can tell, this summation is completely useless, but seems easier to scan with the eye than a lot I've seen. —Preceding unsigned comment added by 75.164.145.226 (talk) 04:07, 17 February 2008 (UTC)[reply]

Since the series is only conditionally and not absolutely convergent, this whole remark is rubbish from the start. Reordering a conditonally convergent series allows to converge to any real number or to diverge.--LutzL (talk) 19:00, 12 May 2011 (UTC)[reply]

using same start point and step

${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{4n-3}}-\sum _{n=1}^{\infty }{\frac {1}{4n-1}}}$
simplifies to
${\displaystyle \sum _{n=1}^{\infty }{\frac {2}{16n^{2}-16n+3}}}$ 2600:1003:B84B:D477:14A6:A3D9:895A:5D12 (talk) 18:23, 12 March 2021 (UTC)[reply]

I was trying to find π using a power series in one way or another in order to get the Leibniz formula, and I found this to be a better method of showing how:

We know that ${\displaystyle \arctan 1={\frac {\pi }{4}}}$

Take the first derivative of arctan(x) and put it in a geometric series:

${\displaystyle {\frac {d}{dx}}\arctan x={\frac {1}{1+x^{2}}}=\sum _{n=1}^{\infty }(-x^{2})^{n-1}=\sum _{n=1}^{\infty }(-1)^{n-1}x^{2n-2}}$

Integrate that:

${\displaystyle \int _{1}^{x}(-1)^{n-1}t^{2n-2}\,dt={\frac {(-1)^{n-1}}{2n-1}}x^{2n-1}}$

So, we have a power series for arctan(x):

${\displaystyle \arctan x=\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{2n-1}}x^{2n-1}}$

Plug in 1 for x and get π/4:

${\displaystyle \arctan 1=\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{2n-1}}={\frac {\pi }{4}}}$

For aesthetics (article uses n=0 to ∞):

${\displaystyle \arctan 1=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{2n+1}}}$

Q.E.D.

-Matt 20:11, 18 March 2006 (UTC)[reply]

"However, if the series is truncated at the right time, the decimal expansion of the approximation will agree with that of π for many more digits, except for isolated digits or digit groups."

I feel there should be more elaboration on this.

Hint: if N is a power of ten, each term in the right sum will be a finite decimal fraction. I agree that there is room for elaboration in the article. Feel free to edit it. Fredrik Johansson 11:29, 13 September 2006 (UTC)[reply]

Is there a reason that this article is not at

Seriously, if the claims made by the author were even half valid then Indian mathematics would have allowed us to have worm-hole technology by now. Xp fun (talk)

Can anyone comment on the validity of these claims? I can try to suggest a less controversial rewrite of the article, also is there a better forum than this area for such broad changes? Xp fun (talk) 21:44, 14 August 2009 (UTC)[reply]

"You can wake up a person who is sleeping, but never a person who pretends to do so!"

1. If a book, or a source is not "available in Google Books or Scholar" does it mean it does not exist?!

2. If you clicked through the (presently ALSO ill-named) "Gregory Series" you will see 4 references to books published in the 1500s in India. Gregory's floruit was 1668.

3. It is a fact that Madhava's floruit is 2 FURTHER centuries older than that.

4. STOP the blatant WHITEWASHING AND RACISM! Stop the BLATANT EUROCENTRISM.

Thanks! Peace Out!! Shivasundar (talk) 23:13, 14 April 2022 (UTC)[reply]

I found this paragraph on Google books, from Mathematics in India by Kim Plofker.

However, I should point out that "Madhava–Leibniz series" has only two hits on Google books and one hit on Google scholar, all of which are references specifically about Indian mathematics. I see no evidence that the term "Madhava–Leibniz series" is used by mathematicians, and Wikipedia is not an appropriate forum for advocating a more correct name for a mathematical concept. Since the material is of historical interest, I have moved it from the lead to a new section on history. Jim (talk) 16:44, 23 October 2009 (UTC)[reply]

It seems that "Gregory-Leibniz" is the more common ordering. In particular, "Gregory-Leibniz series" gets

• 27,700 Google hits,
• 41 hits on Google books, and
• 23 hits on Google scholar,

while "Leibniz-Gregory series" only gets

• 9,560 Google hits,
• 8 hits on Google books, and
• 4 hits on Google scholar.

Jim (talk) 16:28, 23 October 2009 (UTC)[reply]

"To give the rightful place to this great mathematician, the series should be named 'mAdhava srENi' or Madhava Series. But now Madhava's work is sometimes known as the Madhava–Leibniz series."

Great? Rightful place? Should? "But now"? I believe it would be better suited for this articule to discuss other names the series could be known by in the opening paragraph (with redirections as necessary). History should include all of the history of the series rather than just the history of one "great" mathematician over others.

-Ben0mega (talk) 00:08, 19 May 2010 (UTC)[reply]

I edited the article to make it shorter and less controversial. I removed the part of the proof that was not a proof. I removed wishful thinking about honoring Madhava. I changed the heading 'history' to 'names'. Bo Jacoby (talk) 06:22, 1 June 2010 (UTC).[reply]

The reference: George E. Andrews, Richard Askey, Ranjan Roy (1999), Special Functions, Cambridge University Press, p. 58,

The following discussion is an archived discussion of a

requested move

. Please do not modify it. Subsequent comments should be made in a new section on the talk page. No further edits should be made to this section.

The result of the move request was: no consensus for move, partially in light of Talk:Pi#Requested_move Skomorokh 14:43, 12 May 2011 (UTC)[reply]

---

talk) 08:12, 4 May 2011 (UTC)[reply

]

• Support. To prevent confusion with Leibniz' famous apple pie recipe. Hans Adler 08:51, 4 May 2011 (UTC)[reply]
• Oppose. The consistency argument is illogical when the main title is
  chi squared distribution, Omega constant, etc, etc, etc. Kauffner (talk) 15:25, 7 May 2011 (UTC)[reply
  ]

  You never give up, do you? All the articles that you just mentioned are named after terms, per

  WP:COMMON. The articles that you have disruptively moved from π to pi have descriptive names, and for such titles "pi" is plain weird and eccentric. And this is a matter of mathematical typography, not spelling. Dictionaries are totally unqualified for that. Hans Adler 16:37, 7 May 2011 (UTC)[reply

  ]

• Support. This issue has been widely discussed elsewhere and consensus among people dealing with maths is that the symbol is a better look for non-defining articles. Xanthoxyl ^< 16:33, 8 May 2011 (UTC)[reply]

The above discussion is preserved as an archive of a

requested move

. Please do not modify it. Subsequent comments should be made in a new section on this talk page. No further edits should be made to this section.

Could someone please clean this up? 108.2.128.112 (talk) 03:31, 14 July 2018 (UTC)[reply]

print "π≈";Numerator;"/"; Denominator

S=-S 

N=N+2

Numerator=N*Numerator+S*Denominator

Denominator=Denominator*N

Series is a limit of partial sums, IT IS NOT A SUM. Just by definition. There is no such concept as infinite sum, at least until summation rules are introduced, like Cesaro summation. 2A00:1FA0:48BD:B662:FD9A:5A04:FE1B:E20D (talk) 12:04, 4 April 2021 (UTC)[reply]