∞∑n=0x3n(3n)! should be calculated using complex numbers I think, the Wolfram answer is :
13(ex+2e−x/2cos(√3x2))
How to approach this problem?
Answer
We have that by f(x)=∞∑n=0x3n(3n)!
f′(x)=ddx∞∑n=0x3n(3n)!=∞∑n=1x3n−1(3n−1)!
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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