Question Number 124483 by snipers237 last updated on 03/Dec/20
![Develop the function f(x)=e^x sinx Then Deduce that Σ_(k=0) ^([((n−1)/2)]) (−1)^k C_n ^(2k+1) = 2^(n/2) sin(((nπ)/4))](https://www.tinkutara.com/question/Q124483.png)
$$\:{Develop}\:{the}\:{function}\:{f}\left({x}\right)={e}^{{x}} {sinx} \\ $$$${Then}\:{Deduce}\:{that}\: \\ $$$$\underset{{k}=\mathrm{0}} {\overset{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} {\sum}}\left(−\mathrm{1}\right)^{{k}} {C}_{{n}} ^{\mathrm{2}{k}+\mathrm{1}} \:=\:\mathrm{2}^{\frac{{n}}{\mathrm{2}}} {sin}\left(\frac{{n}\pi}{\mathrm{4}}\right) \\ $$
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))](https://www.tinkutara.com/question/Q124490.png)
$${f}\left({x}\right)={e}^{{x}} {sin}\left({x}\right)={Im}\left({e}^{{x}\left(\mathrm{1}+{i}\right)} \right) \\ $$$${f}^{\left({n}\right)} ={Im}\left(\mathrm{1}+{i}\right)^{{n}} {e}^{{x}\left(\mathrm{1}+{i}\right)} \\ $$$$={Im}\left\{\mathrm{2}^{\frac{{n}}{\mathrm{2}}} {e}^{{i}\frac{{n}\pi}{\mathrm{4}}} {e}^{{x}\left(\mathrm{1}+{i}\right)} \right\}_{{x}=\mathrm{0}} \\ $$$$=\mathrm{2}^{\frac{{n}}{\mathrm{2}}} {sin}\left(\frac{{n}\pi}{\mathrm{4}}\right) \\ $$$$\left({sin}\left({x}\right){e}^{{x}} \right)^{\left({n}\right)} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{n}} ^{{k}} {sin}^{\left({k}\right)} \left({x}\right)\left({e}^{{x}} \right)^{\left({n}−{k}\right)} \\ $$$$\left({a}\right)\:{is}\:{the}\:{a}\:{th}\:{derivation} \\ $$$$\left({sin}\left({x}\right){e}^{{x}} \right)^{\left({n}\right)} ={f}^{\left({n}\right)} \left({x}\right)=\underset{{k}\leqslant{n}} {\sum}{C}_{{n}} ^{{k}} {sin}\left({x}+\frac{{k}\pi}{\mathrm{2}}\right){e}^{{x}} \mid_{{x}=\mathrm{0}} \\ $$$$={sin}\left(\mathrm{0}+\frac{{k}\pi}{\mathrm{2}}\right)=\mathrm{0},{k}=\mathrm{2}{m} \\ $$$$=\left(−\mathrm{1}\right)^{{m}} \mid{k}=\mathrm{2}{m}+\mathrm{1} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} {\sum}}{sin}\left(\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{2}}\right){C}_{{n}} ^{\mathrm{2}{k}+\mathrm{1}} ={f}^{\left({n}\right)} \left(\mathrm{0}\right)=\mathrm{2}^{\frac{{n}}{\mathrm{2}}} {sin}\left(\frac{{n}\pi}{\mathrm{4}}\right) \\ $$$$\Rightarrow\underset{{k}\leqslant\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} {\sum}\left(−\mathrm{1}\right)^{{k}} {C}_{{n}} ^{\mathrm{2}{k}+\mathrm{1}} =\mathrm{2}^{\frac{{n}}{\mathrm{2}}} {sin}\left(\frac{{n}\pi}{\mathrm{4}}\right) \\ $$$$ \\ $$
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)](https://www.tinkutara.com/question/Q124516.png)
$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{Im}\left(\mathrm{e}^{\mathrm{x}+\mathrm{ix}} \right)=\mathrm{Im}\left(\mathrm{e}^{\left(\mathrm{1}+\mathrm{i}\right)\mathrm{x}} \right)\:\:\mathrm{but} \\ $$$$\mathrm{e}^{\left(\mathrm{1}+\mathrm{i}\right)\mathrm{x}} \:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(\mathrm{1}+\mathrm{i}\right)^{\mathrm{n}} \mathrm{x}^{\mathrm{n}} }{\mathrm{n}!}\:\:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{n}!}\left(\sqrt{\mathrm{2}}\right)^{\mathrm{n}} \:\mathrm{e}^{\frac{\mathrm{in}\pi}{\mathrm{4}}} \:\mathrm{x}^{\mathrm{n}} \\ $$$$=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(\sqrt{\mathrm{2}}\right)^{\mathrm{n}} }{\mathrm{n}!}\left\{\mathrm{cos}\left(\frac{\mathrm{n}\pi}{\mathrm{4}}\right)+\mathrm{isin}\left(\frac{\mathrm{n}\pi}{\mathrm{4}}\right)\right\}\mathrm{x}^{\mathrm{n}} \:\Rightarrow \\ $$$$\mathrm{e}^{\mathrm{x}} \mathrm{sinx}\:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2}^{\frac{\mathrm{n}}{\mathrm{2}}} }{\mathrm{n}!}\mathrm{sin}\left(\frac{\mathrm{n}\pi}{\mathrm{4}}\right)\mathrm{x}^{\mathrm{n}} \:\Rightarrow\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{0}\right)\:=\mathrm{2}^{\frac{\mathrm{n}}{\mathrm{2}}} \:\mathrm{sin}\left(\frac{\mathrm{n}\pi}{\mathrm{4}}\right) \\ $$$$\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{x}\right)=\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \:\mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} \:\left(\mathrm{sinx}\right)^{\left(\mathrm{k}\right)} \:\left(\mathrm{e}^{\mathrm{x}} \right)^{\left(\mathrm{n}−\mathrm{k}\right)} \\ $$$$=\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \:\mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} \:\:\mathrm{sin}\left(\mathrm{x}+\frac{\mathrm{k}\pi}{\mathrm{2}}\right)\mathrm{e}^{\mathrm{x}} \:\Rightarrow\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{0}\right)\:=\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \:\mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} \:\mathrm{sin}\left(\frac{\mathrm{k}\pi}{\mathrm{2}}\right) \\ $$$$=\sum_{\mathrm{k}=\mathrm{0}} ^{\left[\frac{\mathrm{n}−\mathrm{1}}{\mathrm{2}}\right]} \:\mathrm{C}_{\mathrm{n}} ^{\mathrm{2k}+\mathrm{1}} \mathrm{sin}\left(\frac{\left(\mathrm{2k}+\mathrm{1}\right)\pi}{\mathrm{2}}\right)\:=\sum_{\mathrm{k}=\mathrm{0}} ^{\left[\frac{\mathrm{n}−\mathrm{1}}{\mathrm{2}}\right]} \left(−\mathrm{1}\right)^{\mathrm{k}} \:\mathrm{C}_{\mathrm{n}} ^{\mathrm{2k}+\mathrm{1}} \:\:\Rightarrow \\ $$$$\mathrm{2}^{\frac{\mathrm{n}}{\mathrm{2}}} \:\mathrm{sin}\left(\frac{\mathrm{n}\pi}{\mathrm{4}}\right)=\sum_{\mathrm{k}=\mathrm{0}} ^{\left[\frac{\mathrm{n}−\mathrm{1}}{\mathrm{2}}\right]} \:\left(−\mathrm{1}\right)^{\mathrm{k}} \:\mathrm{C}_{\mathrm{n}} ^{\mathrm{2k}+\mathrm{1}} \\ $$