How to prove the Gregory-Leibniz series expansion to evaluate the value of pi?
4/1 – 4/3 + 4/5 – 4/7 + 4/9 – 4/11 + … = π
Thanks for Wikipedia links… I couldnt have done that by myself. Now, will a non-point-gamer give a real answer, please?
Its really not all that difficult of a proof.
First you find the Taylor Expansion for the arctan function
∞
∑ (−1)^n ⋅ x^(2n + 1) / (2n + 1)
n=0
= arctan(x)
It can be prove that this series holds for all |x| ≤ 1
If you evaluate the equation at x = 1 you get an arctan(1) = π / 4… and the series expands into 1/1 − 1/3 + 1/5 − 1/7 …
π / 4 = 1/1 − 1/3 + 1/5 − 1/7…
Then pi is easy enough to isolate.
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On a side note, you do not need to use Taylor expansions to arrive at the equation
∞
∑ (−1)^n ⋅ x^(2n + 1) / (2n + 1) = arctan(x)
n=0
You can also arrive at this via the simple expression 1/(1 + x²)
If you use synthetic division to divide 1 + x² into 1, and decompile the fraction into a sum of separate fractions…
… you will arrive at this…
1/(1 + x²) = 1 − x² + x⁴ − x⁶ + x⁸…
Now you can integrate both sides with respect to x. The integral comes out to
∫ 1/(1 + x²) dx = ∫ 1 − x² + x⁴ − x⁶ + x⁸… dx
arctan(x) = x − x³/3 + x⁵/5 − x⁷/7 + x⁹/9…
Evaluated over the integral 0 → 1:
arctan(1) = 1 − 1/3 + 1/5 − 1/7 + 1/9…
π/4 = 1 − 1/3 + 1/5 − 1/7 + 1/9…
And we are back where the other method left off. Again, pi is easy enough to isolate.
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