📂Calculus

Series Expansion of the Arctangent Function

Theorem¹

$$\tan ^{ -1 } x = \sum _{ n=0 }^{ \infty }{ \frac { (-1) ^{ n } { x } ^ { 2n+1 } } { 2n+1 } }$$

Description

Whether it is written as $\arctan$ or as $\tan ^{-1}$ does not matter. Among the several inverse trigonometric functions, arctan is particularly interesting because it provides a series that converges to $\pi$. When $x=1$ is substituted,

$${ \pi \over 4 } = \tan ^{-1} 1 = 1 - {1 \over 3} + {1 \over 5} - {1 \over 7} + \cdots$$

Multiplying both sides by $4$ gives an infinite series that converges to $\pi$.

$$\pi = 4 \left( 1 - {1 \over 3} + {1 \over 5} - {1 \over 7} + \cdots \right)$$

Calculating Pi

Although the series expansion of arctan has no direct relationship, it shares aspects with calculating pi $\pi$ through series. Newton made significant contributions in this regard.

Proof

From $-1<t<1$,

$$\frac { 1 }{ 1-t }=\sum _{ n=0 }^{ \infty }{ { t ^ n } } \implies \frac { 1 }{ 1+{ t ^ 2 } }=\sum _{ n=0 }^{ \infty }{ { \left( -{ t ^ 2 } \right) } ^{ n } }=\sum _{ n=0 }^{ \infty }{ { (-1) ^ { n } }{ t ^ {2n} } }$$

Hence,

$$\begin{align*} \tan^{ -1 }x =& \int _{ 0 }^{ x }{ \frac { 1 }{ 1+{ t ^ 2 } } }dt \\ =& \int _{ 0 }^{ x }{ \sum _{ n=0 }^{ \infty }{ { (-1) ^ { n } }{ t ^ {2n} } } }dt \\ =& { \left[ \sum _{ n=0 }^{ \infty }{ \frac { { (-1) ^ { n } }{ t ^ {2n+1} } }{ 2n+1 } } \right] }_{ 0 }^{ x } \\ =& \sum _{ n=0 }^{ \infty }{ \frac { { (-1) ^ { n } }{ x ^ {2n+1} } }{ 2n+1 } } \end{align*}$$

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