Develop the function f(x)=e^x sinx Then Deduce that Σ_(k=0) ^([((n−1)/2)]) (−1)^k C


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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
	$f(x)=e^x sin(x)=Im(e^(x(1+i)) ) f^((n)) =Im(1+i)^n e^(x(1+i)) =Im{2^(n/2) e^(i((nπ)/4)) e^(x(1+i)) }_(x=0) =2^(n/2) sin(((nπ)/4)) (sin(x)e^x )^((n)) =Σ_(k=0) ^n C_n ^k sin^((k)) (x)(e^x )^((n−k)) (a) is the a th derivation (sin(x)e^x )^((n)) =f^((n)) (x)=Σ_(k≤n) C_n ^k sin(x+((kπ)/2))e^x ∣_(x=0) =sin(0+((kπ)/2))=0,k=2m =(−1)^m ∣k=2m+1 =Σ_(k=0) ^([((n−1)/2)]) sin((((2k+1)π)/2))C_n ^(2k+1) =f^((n)) (0)=2^(n/2) sin(((nπ)/4)) ⇒Σ_(k≤[((n−1)/2)]) (−1)^k C_n ^(2k+1) =2^(n/2) sin(((nπ)/4))$
	$f (x) = e^{x} s i n (x) = I m (e^{x (1 + i)})$ $f^{(n)} = I m {(1 + i)}^{n} e^{x (1 + i)}$ $= I m {2^{\frac{n}{2}} e^{i \frac{n π}{4}} e^{x (1 + i)}}_{x = 0}$ $= 2^{\frac{n}{2}} s i n (\frac{n π}{4})$ ${(s i n (x) e^{x})}^{(n)} = \underset{k = 0}{\sum^{n}} C_{n}^{k} {s i n}^{(k)} (x) {(e^{x})}^{(n - k)}$ $(a) i s t h e a t h d e r i v a t i o n$ ${(s i n (x) e^{x})}^{(n)} = f^{(n)} (x) = \sum_{k ⩽ n} C_{n}^{k} s i n (x + \frac{k π}{2}) e^{x} ∣_{x = 0}$ $= s i n (0 + \frac{k π}{2}) = 0, k = 2 m$ $= {(- 1)}^{m} ∣ k = 2 m + 1$ $= \underset{k = 0}{\sum^{[\frac{n - 1}{2}]}} s i n (\frac{(2 k + 1) π}{2}) C_{n}^{2 k + 1} = f^{(n)} (0) = 2^{\frac{n}{2}} s i n (\frac{n π}{4})$ $\Rightarrow \sum_{k ⩽ [\frac{n - 1}{2}]} {(- 1)}^{k} C_{n}^{2 k + 1} = 2^{\frac{n}{2}} s i n (\frac{n π}{4})$
	Answered by mathmax by abdo last updated on 03/Dec/20
	$f(x)=Im(e^(x+ix) )=Im(e^((1+i)x) ) but e^((1+i)x) =Σ_(n=0) ^∞ (((1+i)^n x^n )/(n!)) =Σ_(n=0) ^∞ (1/(n!))((√2))^n e^((inπ)/4) x^n =Σ_(n=0) ^∞ ((((√2))^n )/(n!)){cos(((nπ)/4))+isin(((nπ)/4))}x^n ⇒ e^x sinx =Σ_(n=0) ^∞ (2^(n/2) /(n!))sin(((nπ)/4))x^n ⇒f^((n)) (0) =2^(n/2) sin(((nπ)/4)) f^((n)) (x)=Σ_(k=0) ^n C_n ^k (sinx)^((k)) (e^x )^((n−k)) =Σ_(k=0) ^n C_n ^k sin(x+((kπ)/2))e^x ⇒f^((n)) (0) =Σ_(k=0) ^n C_n ^k sin(((kπ)/2)) =Σ_(k=0) ^([((n−1)/2)]) C_n ^(2k+1) sin((((2k+1)π)/2)) =Σ_(k=0) ^([((n−1)/2)]) (−1)^k C_n ^(2k+1) ⇒ 2^(n/2) sin(((nπ)/4))=Σ_(k=0) ^([((n−1)/2)]) (−1)^k C_n ^(2k+1)$
	$f (x) = Im (e^{x + ix}) = Im (e^{(1 + i) x}) but$ $e^{(1 + i) x} = \sum_{n = 0}^{\infty} \frac{{(1 + i)}^{n} x^{n}}{n!} = \sum_{n = 0}^{\infty} \frac{1}{n!} {(\sqrt{2})}^{n} e^{\frac{in π}{4}} x^{n}$ $= \sum_{n = 0}^{\infty} \frac{{(\sqrt{2})}^{n}}{n!} {\cos (\frac{n π}{4}) + isin (\frac{n π}{4})} x^{n} \Rightarrow$ $e^{x} sinx = \sum_{n = 0}^{\infty} \frac{2^{\frac{n}{2}}}{n!} \sin (\frac{n π}{4}) x^{n} \Rightarrow f^{(n)} (0) = 2^{\frac{n}{2}} \sin (\frac{n π}{4})$ $f^{(n)} (x) = \sum_{k = 0}^{n} C_{n}^{k} {(sinx)}^{(k)} {(e^{x})}^{(n - k)}$ $= \sum_{k = 0}^{n} C_{n}^{k} \sin (x + \frac{k π}{2}) e^{x} \Rightarrow f^{(n)} (0) = \sum_{k = 0}^{n} C_{n}^{k} \sin (\frac{k π}{2})$ $= \sum_{k = 0}^{[\frac{n - 1}{2}]} C_{n}^{2 k + 1} \sin (\frac{(2 k + 1) π}{2}) = \sum_{k = 0}^{[\frac{n - 1}{2}]} {(- 1)}^{k} C_{n}^{2 k + 1} \Rightarrow$ $2^{\frac{n}{2}} \sin (\frac{n π}{4}) = \sum_{k = 0}^{[\frac{n - 1}{2}]} {(- 1)}^{k} C_{n}^{2 k + 1}$
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