How to find the Maclaurin series for $f(x)=\ln(1+\tan {x})$. I tried it by successively differentiating it up to four times but it is very time consuming.
[Math] Maclaurin series for $f(x)=\ln(1+\tan{x})$
calculuspower seriestaylor expansion
calculuspower seriestaylor expansion
How to find the Maclaurin series for $f(x)=\ln(1+\tan {x})$. I tried it by successively differentiating it up to four times but it is very time consuming.
Best Answer
According to OEIS, the series can be written as $$\log (1+\tan x) = \sum_{n=0}^\infty \frac{a_n}{n!} x^n,$$ where $$a_n = \sum_{m=0}^{(n-1)/2} \sum_{j=0}^{2m} \binom{j+n-2m-1}{n-2m-1} \frac{(j+n-2m)!}{n-2m} \, 2^{2m-j} (-1)^{n-m+j-1} \left\{{n} \atop {j+n-2m}\right\}$$ where $\{{a \atop b}\}$ is a Stirling number of the second kind. Numeric calculation with Mathematica supports this; e.g.,
Then
S[50]
for example gives a list of zeros, indicating equality.