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Question Number 31748 by abdo imad last updated on 13/Mar/18

1)find the value of  Σ_(n=0) ^∞  (x^(3n) /((3n)!))  2) find Σ_(n=0) ^∞    (8^n /((3n)!))  .

1)findthevalueofn=0x3n(3n)!2)findn=08n(3n)!.

Commented by rahul 19 last updated on 14/Mar/18

solution for 1) part plz??  2) it can be done by putting x=2.

solutionfor1)partplz??2)itcanbedonebyputtingx=2.

Commented by abdo imad last updated on 16/Mar/18

let put f(x)= e^x  +e^(jx)  +e^(j^2 x)  we have  f(x)= Σ_(n=0) ^∞  (x^n /(n!)) +Σ_(n=0) ^∞  ((j^n  x^n )/(n!)) +Σ_(n=0) ^∞  ((j^(2n)  x^n )/(n!))   (j=e^(i((2π)/3)) )  =Σ_(n=0) ^∞ (1+j^n  +j^(2n) ) (x^n /(n!))  let study A_n =1+j^n  +j^(2n)   n=3k ⇒ A_n =1+j^(3k)  +j^(6k)  =3    n=3k+1 ⇒ A_n = 1+j^(3k+1)  +j^(6k+2)  =1+j +j^2 =0  n=3k+2 ⇒ A_n =1+j^(3k+2)  +j^(6k+4)  =1+j^2  +j=0?so  f(x) =3Σ_(k=0) ^∞  (x^(3k) /((3k)!))  ⇒ Σ_(k=0) ^∞  (x^(3k) /((3k)!)) =(1/3)( e^x  +e^(jx)  +e^(j^2 x) ) .  2) let take x=2 inside f(x) we get  Σ_(k=0) ^∞  (x^(3k) /((3k)!)) =(1/3)( e^2   +e^(2j)  +e^(2j^2 ) )  but  e^(2j) =e^(2(−(1/2)+i((√3)/2)))  = e^(−1) (cos((√3))+isin((√3)))  e^(2j^2 )  = e^(2(−(1/2)−i((√3)/2))) = e^(−1)  (cos((√3) )−isin((√3)))  Σ_(k=0) ^∞   (x^(3k) /((3k)!))  =(1/3) ( e^2   +2e^(−1) cos((√3))) .

letputf(x)=ex+ejx+ej2xwehavef(x)=n=0xnn!+n=0jnxnn!+n=0j2nxnn!(j=ei2π3)=n=0(1+jn+j2n)xnn!letstudyAn=1+jn+j2nn=3kAn=1+j3k+j6k=3n=3k+1An=1+j3k+1+j6k+2=1+j+j2=0n=3k+2An=1+j3k+2+j6k+4=1+j2+j=0?sof(x)=3k=0x3k(3k)!k=0x3k(3k)!=13(ex+ejx+ej2x).2)lettakex=2insidef(x)wegetk=0x3k(3k)!=13(e2+e2j+e2j2)bute2j=e2(12+i32)=e1(cos(3)+isin(3))e2j2=e2(12i32)=e1(cos(3)isin(3))k=0x3k(3k)!=13(e2+2e1cos(3)).

Commented by abdo imad last updated on 16/Mar/18

we have e^x  +e^(jx)  +e^(j^2 x)  =e^x  +e^(x(−(1/2)+i((√3)/2)))   +e^(x(−(1/2)−i((√3)/2)))   = e^x   + e^(−(x/2)) (cos(x((√3)/2)) +isin(x((√3)/2))) +e^(−(x/2)) (cos(x((√3)/2))−isin(x((√3)/2)))  ⇒Σ_(k=0) ^∞   (x^(3k) /((3k)!)) =(1/3)( e^x   +2 e^(−(x/2))  cos(x((√3)/2)))  .

wehaveex+ejx+ej2x=ex+ex(12+i32)+ex(12i32)=ex+ex2(cos(x32)+isin(x32))+ex2(cos(x32)isin(x32))k=0x3k(3k)!=13(ex+2ex2cos(x32)).

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