Talk:Leibniz formula for pi

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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)

[edit] Alternative summations

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

can be rewritten as:

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:

\sum_{n=1}^\infty \frac{1}{2n - 1} - \sum_{n=1}^\infty \frac{1}{4n - 1}

Reduce to a single summation:

= \sum_{n=1}^\infty (\frac{1}{2n - 1} - \frac{1}{4n - 1})

Get the denominators equal:

= \sum_{n=1}^\infty (\frac{4n - 1}{(2n - 1)(4n - 1)} - \frac{2n - 1}{(2n - 1)(4n - 1)})

Combine the two fractions:

= \sum_{n=1}^\infty \frac{(4n - 1) - (2n - 1)}{(2n - 1)(4n - 1)}

Simplify:

= \sum_{n=1}^\infty \frac{2n}{(2n - 1)(4n - 1)}

Move the constant outside the summation and the classical sequence becomes:

= 2\sum_{n=1}^\infty \frac{n}{(2n - 1)(4n - 1)}

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

8\sum_{n=1}^\infty \frac{n}{(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)

[edit] Better proof

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 \arctan 1 = \frac{\pi}{4}

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

\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:

\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):

\arctan x = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{2n-1} x^{2n-1}

Plug in 1 for x and get π/4:

\arctan 1 = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{2n-1} = \frac{\pi}{4}

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

\arctan 1 = \sum_{n=0}^\infty \frac{(-1)^n}{2n+1}

Q.E.D.

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


"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)