Leibniz's Formula for Pi/Leibniz's Proof

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Theorem

$\dfrac \pi 4 = 1 - \dfrac 1 3 + \dfrac 1 5 - \dfrac 1 7 + \dfrac 1 9 - \cdots \approx 0 \cdotp 78539 \, 81633 \, 9744 \ldots$

This sequence is A003881 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).


That is:

$\displaystyle \pi = 4 \sum_{k \mathop \ge 0} \left({-1}\right)^k \frac 1 {2 k + 1}$


Proof

LeibnizFormula.png

The area $OAT$ is a quarter-circle whose area is $\dfrac \pi 4$ by Area of 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 Triangle in Terms of Side and Altitude, we see that the area $\mathrm d C$ of $\triangle OPQ$ is given by:

$\mathrm d C = \triangle OPQ = \dfrac {OR \cdot PQ} 2 = \dfrac {OR \cdot \mathrm d s} 2$

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

Thus:

$\dfrac {\mathrm d s} {\mathrm d x} = \dfrac {OS} {OR} \iff OR \cdot \mathrm d s = OS \cdot \mathrm d x$

Thus:

$\mathrm d C = \dfrac {OS \cdot \mathrm d x} 2 = \dfrac {y \mathrm d x} 2$

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:

$\displaystyle C = \int \mathrm d C = \frac 1 2 \int_0^1 y \mathrm d x$

Now we use Integration by Parts to swap $x$ and $y$:

$\displaystyle C = \left[{\frac 1 2 x y}\right]_0^1 - \frac 1 2 \int_0^1 x \mathrm d y = \frac 1 2 - \frac 1 2 \int_0^1 x \mathrm d y$

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:

$y = \tan \dfrac \phi 2$
$x = 1 - \cos \phi = 2 \sin^2 \dfrac \phi 2$ (from Double Angle Formula for Cosine: Corollary 2)

Thus:

\(\displaystyle \tan^2 \frac \phi 2\) \(=\) \(\displaystyle \frac {\sin^2 \frac \phi 2}{\cos^2 \frac \phi 2}\) $\quad$ Tangent is Sine divided by Cosine $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sin^2 \frac \phi 2 \sec^2 \frac \phi 2\) $\quad$ Secant is Reciprocal of Cosine $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sin^2 \frac \phi 2 \left({1 + \tan^2 \frac \phi 2}\right)\) $\quad$ Difference of Squares of Secant and Tangent $\quad$

and so:

$\dfrac x 2 = \dfrac {y^2} {1 + y^2}$

Using Sum of Geometric Progression:

$\dfrac {y^2} {1 + y^2} = y^2 - y^4 + y^6 - y^8 + \cdots$

This gives us:

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


\(\displaystyle C\) \(=\) \(\displaystyle \frac 1 2 - \int_0^1 \left({y^2 - y^4 + y^6 - y^8 + \cdots}\right) \mathrm d y\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 2 - \left[{\frac {y^3} 3 - \frac {y^5} 5 + \frac {y^7} 7 - \frac {y^9} 9 + \cdots}\right]_0^1\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 2 - \left({\frac 1 3 - \frac 1 5 + \frac 1 7 - \frac 1 9 + \cdots}\right)\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 2 - \frac 1 3 + \frac 1 5 - \frac 1 7 + \frac 1 9 - \cdots\) $\quad$ $\quad$


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

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

Putting it all together, this gives us:

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

$\blacksquare$


Sources