Taylor series exp. & a couple of other questions

[DivGradCurl]

Problem

Find the sum of the series

s(x) = \sum _{n=1} ^{\infty} \frac{1}{2^{n}} \tan \frac{x}{2^n}

Solution

If

s(x) = \sum _{n=1} ^{\infty} \frac{1}{2^{n}} \tan \frac{x}{2^n} = \frac{x}{3} + \frac{x^3}{45} + \frac{2x^5}{945} + \dotsb

\cot x = \frac{1}{x} - \frac{x}{3} - \frac{x^3}{45} - \frac{2x^5}{945} - \dotsb

Then

s(x) = \frac{1}{x} - \cot x \quad (x \neq n\pi \quad n \in \mathbb{N}) \mbox{ and } s(0) = 0

Questions

I found the maclaurin series expansion for s(x) and \cot x with the aid of mathematica. I know how to obtain the latter through long division, but I'm not sure on how to find the former. The only concept that I have in mind right now is that of a generic Taylor series:

\sum _{n=0} ^{\infty} \frac{f^{(n)}(a)}{n!} (x-a) ^n

Any help is highly appreciated.

 

[wais]

Response

To find the Maclaurin series for s(x) , we can use the geometric series formula:

\frac{1}{1-r} = \sum _{n=0} ^{\infty} r^n

Applying this formula to the series \frac{1}{2^n} , we get:

\sum _{n=0} ^{\infty} \frac{1}{2^n} = \frac{1}{1-\frac{1}{2}} = 2

Now, we can substitute this into the original series for s(x) and we get:

s(x) = \sum _{n=1} ^{\infty} \frac{1}{2^{n}} \tan \frac{x}{2^n} = \sum _{n=1} ^{\infty} \frac{1}{2^{n}} \frac{\sin \frac{x}{2^n}}{\cos \frac{x}{2^n}} = \frac{1}{2} \sum _{n=1} ^{\infty} \frac{\sin \frac{x}{2^n}}{\cos \frac{x}{2^n}}

Using the substitution u = \frac{x}{2^n} , we can rewrite this as:

s(x) = \frac{1}{2} \sum _{n=1} ^{\infty} \frac{\sin u}{\cos u} = \frac{1}{2} \sum _{n=1} ^{\infty} \tan u

Now, we can use the Maclaurin series for \tan u :

\tan u = u + \frac{1}{3} u^3 + \frac{2}{15} u^5 + \dotsb

Substituting this back into the series for s(x) , we get:

s(x) = \frac{1}{2} \sum _{n=1} ^{\infty} \tan u = \frac{1}{2} \sum _{n=1} ^{\infty} \left(u + \frac{1}{3} u^3 + \frac{2}{15} u^5 + \dotsb \right)

Finally, we can substitute back in our original variable x and we get the Maclaurin series for