Pi Day, 3.14 approximation

[lookagain]

Using \(\displaystyle \ 4\bigg(1 - \dfrac{1}{3} + \dfrac{1}{5} - \ ... \ \bigg) \ = \ \pi, \ \) you can estimate it to 3.14 without too much effort on a scientific calculator by working out

\(\displaystyle 4\bigg[1 - \dfrac{1}{3} + \ ... \ + \dfrac{1}{33} - \dfrac{1}{35} + \bigg(\dfrac{1}{2}\bigg)\bigg(\dfrac{1}{37}\bigg) \bigg], \ \ \) especially if it has the inverse key, \(\displaystyle \ X^{-1}.\)


If you increase the denominators, and allow the last denominator to be 41, or 45, or 49, for example, then you will get
better results, but I think it will require even more terms on average than before to achieve correct additional digits of pi.

 

[fresh_42]

I find it amazing that Archimedes already had [imath]3.14, \text{ to be exact } 3+\dfrac{10}{71} < \pi < 3+\dfrac{10}{70} [/imath] in 250 B.C. without a calculator!

 

[topsquark]

I find it amazing that Archimedes already had [imath]3.14, \text{ to be exact } 3+\dfrac{10}{71} < \pi < 3+\dfrac{10}{70} [/imath] in 250 B.C. without a calculator!

He did have an abacus, I believe.

-Dan

 

[mario99]

I am a calculus student, so I attack pi in this way:

[imath]\displaystyle \int_{0}^{1} \frac{16x - 16}{x^4 -2x^3 + 4x - 4} \ dx = \pi[/imath]

 

[fresh_42]

I am a calculus student, so I attack pi in this way:

[imath]\displaystyle \int_{0}^{1} \frac{16x - 16}{x^4 -2x^3 + 4x - 4} \ dx = \pi[/imath]

Nice area.