How to write Taylor series of $\coth(x+\frac{\pi}{4})$?

Question

How to write the hyperbolic cotangent function around the point $x=\frac\pi4$?

I first thought about using the formula:

$$f(x+a)=f(a)+f'(a)x+f''(a)\frac {x^2} 2+f^{(3)}(a)\frac{x^3}{3!}+\cdots$$

Which leads to:

$$\coth(x+\frac\pi4)=\coth(\frac\pi4)+[1-\coth^2(\frac\pi4)]x+\cdots$$

That I think is not acceptable because of the term $\coth(\frac\pi4)$.

The other way is to consider:

$$\coth(t)=\frac{e^t+e^{-t}}{e^t-e^{-t}}$$

And use the expansion:

$$e^x=1+x+\frac{x^2}2+\frac{x^3}{3!}+\cdots$$

writing the expansions up to the fourth degree we get a fraction whose denominator is always of one degree less than that of the numerator. which means if we compute the result, it'll always be a polynomial of first degree.

Which isn't acceptable in my opinion.

The question is that: Is there any other way to compute the expansion using the Taylor series?

Am I wrong anywhere?

Answer 1

You can always perform the polynomial version of "Euclidean" division (or long division) of the expansion of $\cosh(x)$ by that of $\sinh(x)$ and extract as many terms as you want. This is much easier, since the $n$-th derivative of both $\sinh(x)$ and $\cosh(x)$ are extremely easy to find! I'll start you off around $x=\pi/4$:

$$\cosh(x)=\cosh(\frac \pi4)+(x-\frac \pi4)\sinh(\frac\pi4)+\frac 12(x-\frac \pi4)^2\cosh(\frac\pi4)+\cdots$$ $$\sinh(x)=\sinh(\frac \pi4)+(x-\frac \pi4)\cosh(\frac\pi4)+\frac 12(x-\frac \pi4)^2\sinh(\frac\pi4)+\cdots$$

First multiply the second expansion by $\coth(\pi/4)$ and then subtract it from the first, you have thus eliminated the $\cosh(\pi/4)$ and obtained a rest $R(x)$. You repeat the process as many times as you want. The first term as mentioned is $\coth(\pi/4)$, then $(1-\coth^2(\pi/4))$ etc... The coefficients of the Taylor series are not necessarily rational functions, or factorials... They can be any function...

Answer 2

There's nothing unacceptable about complicated coefficients. The way you're going about it is perfectly fine!

What the Taylor expansion does is it expresses a function as a power series in the argument with constant coefficients. Those coefficients needn't be pretty, just constant. And if your $x$ above is very small compared to $1$, then you can discard all but the lowest powers of $x$ and still have a good approximation.