Leibniz's Formula for Pi/Elementary Proof

From ProofWiki
Jump to: navigation, search

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

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 $x$:

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

where:

$\displaystyle R_n \left({x}\right) = \int_0^x \frac {t^{4 n + 2} } {1 + t^2} \mathrm d t$

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:

$\displaystyle 0 \le R_n \left({x}\right) \le \int_0^x t^{4 n + 2} \mathrm d t$

that is:

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


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:

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

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


This leads us directly to:

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


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

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

and thence from the Fundamental Theorem of Calculus we have:

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


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 $(4)$.

$\blacksquare$


Comment

Note that we did not just take the Sum of Infinite Geometric Progression:

$\dfrac 1 {1 - \left({-t^2}\right)} = 1 + \left({-t^2}\right) + \left({-t^2}\right)^2 + \left({-t^2}\right)^3 + \cdots$

and integrate it term by term, as we have not at this stage proved that this is permissible.


Sources