- #1
DivGradCurl
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Problem
Find the sum of the series
[tex] s(x) = \sum _{n=1} ^{\infty} \frac{1}{2^{n}} \tan \frac{x}{2^n} [/tex]
Solution
If
[tex] 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 [/tex]
[tex] \cot x = \frac{1}{x} - \frac{x}{3} - \frac{x^3}{45} - \frac{2x^5}{945} - \dotsb [/tex]
Then
[tex] s(x) = \frac{1}{x} - \cot x \quad (x \neq n\pi \quad n \in \mathbb{N}) \mbox{ and } s(0) = 0 [/tex]
Questions
I found the maclaurin series expansion for [tex] s(x) [/tex] and [tex] \cot x [/tex] 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:
[tex] \sum _{n=0} ^{\infty} \frac{f^{(n)}(a)}{n!} (x-a) ^n [/tex]
Any help is highly appreciated.
Find the sum of the series
[tex] s(x) = \sum _{n=1} ^{\infty} \frac{1}{2^{n}} \tan \frac{x}{2^n} [/tex]
Solution
If
[tex] 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 [/tex]
[tex] \cot x = \frac{1}{x} - \frac{x}{3} - \frac{x^3}{45} - \frac{2x^5}{945} - \dotsb [/tex]
Then
[tex] s(x) = \frac{1}{x} - \cot x \quad (x \neq n\pi \quad n \in \mathbb{N}) \mbox{ and } s(0) = 0 [/tex]
Questions
I found the maclaurin series expansion for [tex] s(x) [/tex] and [tex] \cot x [/tex] 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:
[tex] \sum _{n=0} ^{\infty} \frac{f^{(n)}(a)}{n!} (x-a) ^n [/tex]
Any help is highly appreciated.