Let-f-x-e-x-cos-x-Find-n-0-f-n-x-2-n-

Question Number 145339 by qaz last updated on 04/Jul/21

Let f (x) = e^{x} \cos x, Find \underset{n = 0}{\sum^{\infty}} \frac{f^{(n)} (x)}{2^{n}} = ?

Answered by mathmax by abdo last updated on 04/Jul/21

f (x) = e^{x} cosx \Rightarrow f^{(n)} (x) = \sum_{k = 0}^{n} C_{n}^{k} {(cosx)}^{(k)} {(e^{x})}^{(n - k)}

= \sum_{k = 0}^{n} C_{n}^{k} \cos (x + \frac{k π}{2}) e^{x} \Rightarrow

\sum_{n = 0}^{\infty} \frac{f^{(n)} (x)}{2^{n}} = e^{x} \sum_{n = 0}^{\infty} \frac{1}{2^{n}} (\sum_{k = 0}^{n} C_{n}^{k} \cos (x + \frac{k π}{2}))

\sum_{k = 0}^{n} C_{n}^{k} \cos (x + \frac{k π}{2}) = Re (\sum_{k = 0}^{n} C_{n}^{k} e^{i (x + \frac{k π}{2})})

\sum_{k = 0}^{n} (\dots) = e^{ix} \sum_{k = 0}^{n} C_{n}^{k} {(i)}^{k} = e^{ix} {(1 + i)}^{n}

= e^{ix} {(\sqrt{2})}^{n} {(e^{\frac{i π}{4}})}^{n} = {(\sqrt{2})}^{n} e^{ix} e^{\frac{in π}{4}} = {(\sqrt{2})}^{n} e^{i (x + \frac{n π}{4})}

\Rightarrow \sum_{k = 0}^{n} C_{n}^{k} \cos (x + \frac{k π}{2}) = {(\sqrt{2})}^{n} \cos (x + \frac{n π}{4}) \Rightarrow

\sum_{n = 0}^{\infty} \frac{f^{(n)} (x)}{2^{n}} = e^{x} \sum_{n = 0}^{\infty} {(\frac{\sqrt{2}}{2})}^{n} \cos (x + \frac{n π}{4})

\sum_{n = 0}^{\infty} {(\frac{1}{\sqrt{2}})}^{n} \cos (x + \frac{n π}{4}) = Re (\sum_{n = 0}^{\infty} {(\frac{1}{\sqrt{2}})}^{n} e^{i (x + \frac{n π}{4}}))

Σ (\dots) = e^{ix} \sum_{n = 0}^{\infty} {(\frac{1}{\sqrt{2}})}^{n} {(e^{\frac{i π}{4}})}^{n}

= e^{ix} \sum_{n = 0}^{\infty} {(\frac{e^{\frac{i π}{4}}}{\sqrt{2}})}^{n} = e^{ix} \times \frac{1}{1 - \frac{1}{\sqrt{2}} e^{\frac{i π}{4}}}

= \frac{\sqrt{2} e^{ix}}{\sqrt{2} - \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}} = \frac{{2 e}^{ix}}{1 - i} = \frac{{2 e}^{ix} (1 + i)}{2}

= e^{ix} \sqrt{2} e^{\frac{i π}{4}} = \sqrt{2} e^{i (x + \frac{π}{4})} \Rightarrow Re (\dots .) = \sqrt{2} \cos (x + \frac{π}{4}) \Rightarrow

\sum_{n = 0}^{\infty} \frac{f^{(n)} (x)}{2^{n}} = \sqrt{2} e^{x} \cos (x + \frac{π}{4})