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Archive for April, 2016

Ingredient: \frac{1}{1+x^2}

Divide gently in a long division sauce pan:

\frac{1}{1+x^2}=1-x^2+x^4-x^6+x^8-x^{10}...

Integrate briskly over a low flame:

\int (1-x^2+x^4-x^6+x^8-x^{10}...)dx=

=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\frac{x^9}{9}-\frac{x^{11}}{11}+...

Evaluate for x=1 and let stand at room temperature for 1000 terms for accuracy to three decimal places.

=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}...

Add lemon zest to taste.

But this doesn’t look like a pi! or taste like a pi!

or a quarter pi!

Ah, but if we return to the original integral

\int \frac{1}{1+x^2}dx

and make the trigonometric substitution x=tan \theta so that dx=sec^2 \theta d \theta then:

\int \frac{1}{1+x^2}dx=\int \frac{1}{1+tan^2 \theta}sec^2 \theta d \theta

and

\int \frac{1}{1+tan^2 \theta}sec^2 \theta d \theta=\int \frac{1}{sec^2 \theta}sec^2 \theta d \theta

=\int d \theta=\theta

which, if we return to the original substitution x=tan \theta, we see that \tan^{-1} x=\theta

So, \int \frac{1}{1+x^2}dx=\tan^{-1}x, which means that:

\tan^{-1}x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\frac{x^9}{9}-\frac{x^{11}}{11}+...

and since \tan^{-1}(1)=\frac{\pi}{4}, then

\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}...

Enjoy your pi!

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