# Leibniz's Formula for Pi

## Theorem

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

That is:

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

## Elementary Proof

First we note that:

$\displaystyle (1): \quad \frac 1 {1+t^2} = 1 - t^2 + t^4 - t^6 + \cdots + t^{4n} - \frac {t^{4n + 2}}{1+t^2}$

which is demonstrated here.

Now consider the real number $x \in \R: 0 \le x \le 1$.

We can integrate expression $(1)$ WRT $t$ from $0$ to $x$:

$\displaystyle (2) \qquad \int_{0}^{x}\frac {\mathrm{d}{t}}{1+t^2} = x - \frac {x^3} 3 + \frac {x^5} 5 - \frac {x^7} 7 + \cdots + \frac {x^{4n + 1}}{4n + 1} = R_n \left({x}\right)$

where:

$\displaystyle R_n \left({x}\right) = \int_{0}^{x} \frac {t^{4n + 2}}{1+t^2}\mathrm{d}{t}$

From Even Powers are Positive we have that $t^2 \ge 0$ and so $1 \le 1 + t^2$.

From Relative Sizes of Definite Integrals, we have:

$\displaystyle 0 \le R_n \left({x}\right) \le \int_{0}^{x} t^{4n + 2} \mathrm{d}{t}$

that is:

$\displaystyle 0 \le R_n \left({x}\right) \le \frac {x^{4n + 3}}{4n + 3}$

But as $0 \le x \le 1$ it is clear that $\dfrac {x^{4n + 3}}{4n + 3} \le \dfrac 1 {4n + 3}$.

So:

$0 \le R_n \left({x}\right) \le \dfrac 1 {4n + 3}$

From Basic Null Sequences and the Squeeze Theorem, $\dfrac 1 {4n + 3} \to 0$ as $n \to \infty$.

This leads us directly to:

$\displaystyle (2): \quad \int_{0}^{x}\frac {\mathrm{d}{t}}{1+t^2} = x - \frac {x^3} 3 + \frac {x^5} 5 - \frac {x^7} 7 + \frac {x^9} 9 \cdots$

But from Derivative of Arctangent Function, we also have that:

$\displaystyle \frac{\mathrm{d}{}}{\mathrm{d}{x}} \arctan t = \frac 1 {1+t^2}$

and thence from the Fundamental Theorem of Calculus we have:

$\displaystyle \arctan x = \int_{0}^{x}\frac {\mathrm{d}{t}}{1+t^2}$

From $(2)$ it follows immediately that:

$\displaystyle (3): \quad \arctan x = x - \frac {x^3} 3 + \frac {x^5} 5 - \frac {x^7} 7 + \frac {x^9} 9 \cdots$

Now all we need to do is plug $x = 1$ into $(3)$.

$\blacksquare$

## Leibniz's Proof

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:

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

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

Thus:

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

Thus:

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

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:

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

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:

Thus:

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

and so:

$\displaystyle \frac x 2 = \frac {y^2}{1 + y^2}$
$\displaystyle \frac {y^2}{1 + y^2} = y^2 - y^4 + y^6 - y^8 + \cdots$

This gives us:

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

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

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:

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

$\blacksquare$

## Proof by Taylor Expansion

From Taylor Series Expansion for Arctangent Function, we obtain:

$\displaystyle \arctan x = x - \frac {x^3} 3 + \frac {x^5} 5 - \frac {x^7} 7 + \frac {x^9} 9 - \cdots$

Substituting $x = 1$ gives the required result.

$\blacksquare$

## Also known as

Leibniz's Formula for Pi is also known as the Gregory Series for James Gregory.

## Source of Name

This entry was named for Gottfried Wilhelm von Leibniz.