# Leibniz's Formula for Pi

## Theorem

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

That is:

$\displaystyle \pi = 4 \sum_{k \mathop \ge 0} \paren {-1}^k \frac 1 {2 k + 1}$

## Elementary Proof

First we note that:

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

which is demonstrated here.

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

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

$\displaystyle (2): \quad \int_0^x \frac {\d t} {1 + t^2} = x - \frac {x^3} 3 + \frac {x^5} 5 - \frac {x^7} 7 + \cdots + \frac {x^{4 n + 1} } {4 n + 1} - \map {R_n} x$

where:

$\displaystyle \map {R_n} x = \int_0^x \frac {t^{4 n + 2} } {1 + t^2} \rd t$

From Square of Real Number is Non-Negative we have that:

$t^2 \ge 0$

and so:

$1 \le 1 + t^2$

From Relative Sizes of Definite Integrals, we have:

$\displaystyle 0 \le \map {R_n} x \le \int_0^x t^{4 n + 2} \rd t$

that is:

$\displaystyle 0 \le \map {R_n} x \le \frac {x^{4n + 3} } {4 n + 3}$

But as $0 \le x \le 1$ it is clear that:

$\dfrac {x^{4 n + 3} } {4 n + 3} \le \dfrac 1 {4 n + 3}$

So:

$0 \le \map {R_n} x \le \dfrac 1 {4 n + 3}$

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

$(3): \quad \displaystyle \int_0^x \frac {\d t} {1 + t^2} = x - \frac {x^3} 3 + \frac {x^5} 5 - \frac {x^7} 7 + \frac {x^9} 9 \cdots$

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

$\dfrac \d {\d x} \arctan t = \dfrac 1 {1 + t^2}$

and thence from the Fundamental Theorem of Calculus we have:

$\displaystyle \arctan x = \int_0^x \frac {\d t} {1 + t^2}$

From $(3)$ it follows immediately that:

$(4): \quad \arctan x = x - \dfrac {x^3} 3 + \dfrac {x^5} 5 - \dfrac {x^7} 7 + \dfrac {x^9} 9 \cdots$

Now all we need to do is plug $x = 1$ into $(4)$.

$\blacksquare$

## 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 $\d s = PQ$.

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 $\d C$ of $\triangle OPQ$ is given by:

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

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

Thus:

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

Thus:

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

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:

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

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

$\displaystyle C = \intlimits {\frac 1 2 x y} 0 1 - \frac 1 2 \int_0^1 x \rd y = \frac 1 2 - \frac 1 2 \int_0^1 x \rd y$

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:

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

and so:

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

This gives us:

 $\displaystyle C$ $=$ $\displaystyle \frac 1 2 - \int_0^1 \paren {y^2 - y^4 + y^6 - y^8 + \cdots} \d y$ $\displaystyle$ $=$ $\displaystyle \frac 1 2 - \intlimits {\frac {y^3} 3 - \frac {y^5} 5 + \frac {y^7} 7 - \frac {y^9} 9 + \cdots} 0 1$ $\displaystyle$ $=$ $\displaystyle \frac 1 2 - \paren {\frac 1 3 - \frac 1 5 + \frac 1 7 - \frac 1 9 + \cdots}$ $\displaystyle$ $=$ $\displaystyle \frac 1 2 - \frac 1 3 + \frac 1 5 - \frac 1 7 + \frac 1 9 - \cdots$

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:

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

$\blacksquare$

## Proof by Taylor Expansion

From Power Series Expansion for Real Arctangent Function, we obtain:

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

Substituting $x = 1$ gives the required result.

$\blacksquare$

## Also known as

Some sources refer to this as Gregory's series for $\pi$.

## Source of Name

This entry was named for Gottfried Wilhelm von Leibniz.

## Historical Note

Leibniz discovered his formula for $\pi$ in $1673$.

He took great pleasure and pride in this discovery.

It's as if, by this expansion, the veil which hung over that strange number had been drawn aside.