Saturday, January 2, 2016

power series - Calculate sumlimitsinftyn=0fracx3n(3n)!




n=0x3n(3n)! should be calculated using complex numbers I think, the Wolfram answer is :



13(ex+2ex/2cos(3x2))



How to approach this problem?


Answer



We have that by f(x)=n=0x3n(3n)!




f(x)=ddxn=0x3n(3n)!=n=1x3n1(3n1)!



f



f'''(x)=\frac{d}{dx}\sum\limits_{n=1}^{\infty} \frac{x^{3n-2}}{(3n-2)!}=\sum\limits_{n=1}^{\infty} \frac{x^{3n-3}}{(3n-3)!}=f(x)



and f'''(x)=f(x) has solution



f(x)=c_1e^x+c_2e^{-x/2}\cos\left(\frac{\sqrt 3 x}{2}\right)+c_3e^{-x/2}\sin\left(\frac{\sqrt 3 x}{2}\right)




with the initial conditions f(0)=1, f'(0)=0, f''(0)=0.


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