Develop-the-function-f-x-e-x-sinx-Then-Deduce-that-k-0-n-1-2-1-k-C-n-2k-1-2-n-2-sin-npi-4-

Question Number 124483 by snipers237 last updated on 03/Dec/20

D e v e l o p t h e f u n c t i o n f (x) = e^{x} s i n x

T h e n D e d u c e t h a t

\underset{k = 0}{\sum^{[\frac{n - 1}{2}]}} {(- 1)}^{k} C_{n}^{2 k + 1} = 2^{\frac{n}{2}} s i n (\frac{n π}{4})

Answered by mindispower last updated on 03/Dec/20

f (x) = e^{x} s i n (x) = I m (e^{x (1 + i)})

f^{(n)} = I m {(1 + i)}^{n} e^{x (1 + i)}

= I m {2^{\frac{n}{2}} e^{i \frac{n π}{4}} e^{x (1 + i)}}_{x = 0}

= 2^{\frac{n}{2}} s i n (\frac{n π}{4})

{(s i n (x) e^{x})}^{(n)} = \underset{k = 0}{\sum^{n}} C_{n}^{k} {s i n}^{(k)} (x) {(e^{x})}^{(n - k)}

(a) i s t h e a t h d e r i v a t i o n

{(s i n (x) e^{x})}^{(n)} = f^{(n)} (x) = \sum_{k ⩽ n} C_{n}^{k} s i n (x + \frac{k π}{2}) e^{x} ∣_{x = 0}

= s i n (0 + \frac{k π}{2}) = 0, k = 2 m

= {(- 1)}^{m} ∣ k = 2 m + 1

= \underset{k = 0}{\sum^{[\frac{n - 1}{2}]}} s i n (\frac{(2 k + 1) π}{2}) C_{n}^{2 k + 1} = f^{(n)} (0) = 2^{\frac{n}{2}} s i n (\frac{n π}{4})

\Rightarrow \sum_{k ⩽ [\frac{n - 1}{2}]} {(- 1)}^{k} C_{n}^{2 k + 1} = 2^{\frac{n}{2}} s i n (\frac{n π}{4})

Answered by mathmax by abdo last updated on 03/Dec/20

f (x) = Im (e^{x + ix}) = Im (e^{(1 + i) x}) but

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

= \sum_{n = 0}^{\infty} \frac{{(\sqrt{2})}^{n}}{n!} {\cos (\frac{n π}{4}) + isin (\frac{n π}{4})} x^{n} \Rightarrow

e^{x} sinx = \sum_{n = 0}^{\infty} \frac{2^{\frac{n}{2}}}{n!} \sin (\frac{n π}{4}) x^{n} \Rightarrow f^{(n)} (0) = 2^{\frac{n}{2}} \sin (\frac{n π}{4})

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

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

= \sum_{k = 0}^{[\frac{n - 1}{2}]} C_{n}^{2 k + 1} \sin (\frac{(2 k + 1) π}{2}) = \sum_{k = 0}^{[\frac{n - 1}{2}]} {(- 1)}^{k} C_{n}^{2 k + 1} \Rightarrow

2^{\frac{n}{2}} \sin (\frac{n π}{4}) = \sum_{k = 0}^{[\frac{n - 1}{2}]} {(- 1)}^{k} C_{n}^{2 k + 1}