Leibniz's Formula for Pi/Leibniz's Proof

Theorem

 * $$\frac \pi 4 = 1 - \frac 1 3 + \frac 1 5 - \frac 1 7 + \frac 1 9 - \cdots $$

That is:
 * $$\pi = 4 \sum_{k \ge 0} \left({-1}\right)^k \frac 1 {2 k + 1}$$

Proof

 * LeibnizFormula.png

The area $$OAT$$ is a quarter-circle whose area is $$\frac \pi 4$$ by Area of a Circle.

Now consider the area $$C$$ of the segment $$OPQT$$, bounded by the arc $$OT$$ and the chord $$OT$$.

Consider the area $$OPQ$$, bounded by the line segments $$OP$$ and $$OQ$$ and the arc $$PQ$$.

As $$P$$ and $$Q$$ approach each other, the arc $$PQ$$ tends towards the straight line segment $$\mathrm d s = PQ$$.

We can therefore consider the area $$OPQ$$ as a triangle.

We extend the line segment $$PQ$$ and drop a perpendicular $$OR$$ to $$O$$.

Using Area of a Triangle in Terms of Side and Altitude, we see that the area $$\mathrm d C$$ of $$\triangle OPQ$$ is given by:
 * $$\mathrm d C = \triangle OPQ = \frac {OR \cdot PQ} 2 = \frac {OR \cdot \mathrm d s} 2$$

We also note from elementary Euclidean geometry that $$\triangle ORS$$ is similar to the small triangle on $$PQ$$.

Thus:
 * $$\frac {\mathrm d s}{\mathrm d x} = \frac {OS} {OR} \iff OR \cdot \mathrm d s = OS \cdot \mathrm d x$$

Thus:
 * $$\mathrm d C = \frac {OS \cdot \mathrm d x} 2 = \frac {y \mathrm d x} 2$$

where $$y = OS$$.

We set the horizontal coordinate of $$P$$ as equal to $$x$$.

Thus the total area $$C$$ is equal to the total of all the areas of these small triangles as $$x$$ increases from $$0$$ to $$1$$.

So:
 * $$C = \int \mathrm d C = \frac 1 2 \int_0^1 y \mathrm d x$$

Now we use Integration by Parts to swap $$x$$ and $$y$$:
 * $$C = \left[{\frac 1 2 x y}\right]_0^1 - \frac 1 2 \int_0^1 x \mathrm d y = \frac 1 2 - \frac 1 2 \int_0^1 x \mathrm d y$$

It can be seen that the limits on this new integral have to be $$0$$ and $$1$$ from the geometry of the situation.

Now we note that:
 * $$y = \tan \frac \phi 2$$
 * $$x = 1 - \cos \phi = 2 \sin^2 \frac \phi 2$$ (using the double angle formulas)

Thus:

$$ $$ $$

and so:
 * $$\frac x 2 = \frac {y^2}{1 + y^2}$$

Using Sum of Geometric Progression:
 * $$\frac {y^2}{1 + y^2} = y^2 - y^4 + y^6 - y^8 + \cdots$$

This gives us:
 * $$C = \frac 1 2 - \int_0^1 \left({y^2 - y^4 + y^6 - y^8 + \cdots}\right) \mathrm d y$$

$$ $$ $$ $$

Remember, $$C$$ is the area of the segment segment $$OPQT$$.

Now we add to it the area of $$\triangle OTA$$, which trivially equals $$\frac 1 2$$, to get the area of the quarter circle which we know as equal to $$\frac \pi 4$$.

Putting it all together, this gives us:
 * $$\frac \pi 4 = 1 - \frac 1 3 + \frac 1 5 - \frac 1 7 + \frac 1 9 - \cdots $$