Thread: Maclaurin/ Taylor Series for Cotangent

1. Maclaurin/ Taylor Series for Cotangent

I was wondering how the Maclaurin series expansion for cotangent is derived using the Bernoulli numbers.

It is given as:

$\cot(x)=\sum^{\infty}_{n=0}\frac{(-1)^n2^{2n}B_{2n}x^{2n-1}}{(2n)!}$

I cannot normally find the Maclaurin expansion since $\cot(0)$ is undefined.

Could I derive this cotangent representation from the series representation of sine and cosine? Or tangent? If not, how would I?

Thank you very much.

2. Originally Posted by Anthonny
I was wondering how the Maclaurin series expansion for cotangent is derived using the Bernoulli numbers.

It is given as:

$\cot(x)=\sum^{\infty}_{n=0}\frac{(-1)^n2^{2n}B_{2n}x^{2n-1}}{(2n)!}$

I cannot normally find the Maclaurin expansion since $\cot(0)$ is undefined.
Right, there is no Maclaurin expansion for $\cot(x)$ for just that reason: the above series expansion for $\cot(x)$ is a so called Laurent-series, not a Maclaurin series.

Could I derive this cotangent representation from the series representation of sine and cosine? Or tangent? If not, how would I?
For one, you could determine the Maclaurin expansion for $x\cdot\cot(x)$ the usual way.
For another: much depends on what definition of the Bernoully numbers you are using. The derivation that I know of uses the "definition" $\frac{z}{e^z-1}=\sum_{n=0}^\infty \frac{B_n}{n!}z^n$ of $B_n$, then first derives the Maclaurin expansion for $z\cdot \mathrm{coth}(z)$, and finally gets the Maclaurin expansion for $z\cdot\cot(h)$ via the identity $z\cdot\cot(z)=\mathrm{i}z\cdot\mathrm{coth}(\mathr m{i}z)$.

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maclaurim cotx expansion

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