# Power Series Expansion for Tangent Function/Sequence

## Theorem

$\tan x = x + \dfrac 1 3 x^3 + \dfrac 2 {15} x^5 + \dfrac {17} {315} x^7 + \dfrac {62} {2835} x^9 + \cdots$

## Proof

 $\displaystyle \tan x$ $=$ $\displaystyle \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} 2^{2 n} \paren {2^{2 n} - 1} B_{2 n} } {\paren {2 n}!} x^{2 n - 1}$ $\displaystyle$ $=$ $\displaystyle \frac {2^2 \paren {2^2 - 1} B_2} {2!} x - \frac {2^4 \paren {2^4 - 1} B_4} {4!} x^3 + \frac {2^6 \paren {2^5 - 1} B_6} {6!} x^5 - \frac {2^8 \paren {2^7 - 1} B_8} {8!} + \frac {2^{10} \paren {2^9 - 1} B_{10} } {10!} - \cdots$

Enumerating the Bernoulli numbers:

 $\displaystyle B_2$ $=$ $\displaystyle \dfrac 1 6$ $\displaystyle B_4$ $=$ $\displaystyle -\dfrac 1 {30}$ $\displaystyle B_6$ $=$ $\displaystyle \dfrac 1 {42}$ $\displaystyle B_8$ $=$ $\displaystyle -\dfrac 1 {30}$ $\displaystyle B_{10}$ $=$ $\displaystyle \dfrac 5 {66}$

Thus the appropriate arithmetic is performed on each coefficient:

 $\displaystyle \frac {2^2 \paren {2^2 - 1} B_2} {2!}$ $=$ $\displaystyle \frac {4 \times 3} {2} \times \dfrac 1 6$ $\displaystyle$ $=$ $\displaystyle 1$

 $\displaystyle -\frac {2^4 \paren {2^4 - 1} B_4} {4!}$ $=$ $\displaystyle -\dfrac {16 \times 15} {24} \times \dfrac {-1} {30}$ $\displaystyle$ $=$ $\displaystyle \dfrac {\paren {2^4} \times \paren {3 \times 5} } {\paren {2^3 \times 3} \times \paren {2 \times 3 \times 5} }$ $\displaystyle$ $=$ $\displaystyle \dfrac 1 3$

 $\displaystyle \frac {2^6 \paren {2^5 - 1} B_6} {6!}$ $=$ $\displaystyle \dfrac {64 \times 63} {720} \times \dfrac 1 {42}$ $\displaystyle$ $=$ $\displaystyle \dfrac {\paren {2^6} \times \paren {3^2 \times 7} } {\paren {2^4 \times 3^2 \times 5} \times \paren {2 \times 3 \times 7} }$ $\displaystyle$ $=$ $\displaystyle \dfrac 2 {15}$

 $\displaystyle -\frac {2^8 \paren {2^7 - 1} B_8} {8!}$ $=$ $\displaystyle -\dfrac {256 \times 255} {40 \, 320} \times \dfrac {-1} {30}$ $\displaystyle$ $=$ $\displaystyle \dfrac {\paren {2^8} \times \paren {3 \times 5 \times 17} } {\paren {2^7 \times 3^2 \times 5 \times 7} \times \paren {2 \times 3 \times 5} }$ $\displaystyle$ $=$ $\displaystyle \dfrac {17} {3^2 \times 5 \times 7}$ $\displaystyle$ $=$ $\displaystyle \dfrac {17} {315}$

 $\displaystyle \frac {2^{10} \paren {2^9 - 1} B_{10} } {10!}$ $=$ $\displaystyle \dfrac {1024 \times 1023} {3 \, 628 \, 800} \times \dfrac 5 {66}$ $\displaystyle$ $=$ $\displaystyle \dfrac {\paren {2^{10} } \times \paren {3 \times 11 \times 31} \times 5} {\paren {2^8 \times 3^4 \times 5^2 \times 7} \times \paren {2 \times 3 \times 11} }$ $\displaystyle$ $=$ $\displaystyle \dfrac {2 \times 31} {\paren {3^4 \times 5 \times 7} }$ $\displaystyle$ $=$ $\displaystyle \dfrac {62} {2835}$

Hence the result.

$\blacksquare$

## Sources

Beware: there is a mistake in the $5$th term as reported here.