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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))

Developthefunctionf(x)=exsinxThenDeducethat[n12]k=0(1)kCn2k+1=2n2sin(nπ4)

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)=exsin(x)=Im(ex(1+i))f(n)=Im(1+i)nex(1+i)=Im{2n2einπ4ex(1+i)}x=0=2n2sin(nπ4)(sin(x)ex)(n)=nk=0Cnksin(k)(x)(ex)(nk)(a)istheathderivation(sin(x)ex)(n)=f(n)(x)=knCnksin(x+kπ2)exx=0=sin(0+kπ2)=0,k=2m=(1)mk=2m+1=[n12]k=0sin((2k+1)π2)Cn2k+1=f(n)(0)=2n2sin(nπ4)k[n12](1)kCn2k+1=2n2sin(nπ4)

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(ex+ix)=Im(e(1+i)x)bute(1+i)x=n=0(1+i)nxnn!=n=01n!(2)neinπ4xn=n=0(2)nn!{cos(nπ4)+isin(nπ4)}xnexsinx=n=02n2n!sin(nπ4)xnf(n)(0)=2n2sin(nπ4)f(n)(x)=k=0nCnk(sinx)(k)(ex)(nk)=k=0nCnksin(x+kπ2)exf(n)(0)=k=0nCnksin(kπ2)=k=0[n12]Cn2k+1sin((2k+1)π2)=k=0[n12](1)kCn2k+12n2sin(nπ4)=k=0[n12](1)kCn2k+1

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