How can we prove that? ∑n=1∞x3n(3n−1)!=13e−x2x(e3x2−2sin⁡(π+33x6))

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How can we prove that?  ∑n=1∞x3n(3n−1)!=13e−x2x(e3x2−2sin⁡(π+33x6))

Lakisha Archer · Accepted Answer

Consider the third root of unity ρ=e2πi3=−1+i32. You haveeρz=∑k=0∞ρkzkk!=∑m=0∞z3m(3m)!+ρ∑m=0∞z3m+1(3m+1)!+ρ2∑m=0∞z3m+2(3m+2)!since ρ3m=1, ρ3m+1=ρ, ρ3m+2=ρ2. You have something similar for eρ2z. Also consider 1+ρ+ρ2=0. Then a suitable combination of eρkx gives you∑n=1∞x3n−1(3n−1)!Using Euler&#039;s formula eit=cos⁡t+isin⁡t then gives you the right hand side.

Suhadolahbb · Accepted Answer

Consider the functionf(x)=∑n=1∞x3n−1(3n−1)!Consider this a Maclaurin series for f(x). So we have f(0)=0, f′(0)=0 and f″(0)=1. Finally, take note that f‴(x)=f(x). Solve this initial value problem and multiply by x to get your answer.It looks like it may take some more work to get this into the desired form, however, so I&#039;ll continue. The characteristic equation for our problem is s3−1=0, whose roots are the third roots of unity 1 and −12±32i. This suggestsf(x)=k1ex+k2e−x2sin⁡(x32)+k3e−x2cos⁡(x32)=k1ex+e−x2[k2sin⁡(x32)+k3cos⁡(x32)]f′(x)=k1ex+e−x2[(−12k2−32k3)sin⁡(x32)+(−12k3+32k2)cos⁡(x32]f″(x)=k1ex+e−π2[(14k2+34k3−34k2)sin⁡

alenahelenash · Accepted Answer

This is actually calling for Laplace transform. Set f(x)=∑n=1∞x3n(3n−1)! then F(s)=∫0∞f(x)xe−sxdx =∑n=1∞1s3n∫0∞u3n−1e−u(3n−1)!du =1s3−1 Now f(x)x=12πi∫γ−i∞γ+i∞F(s)exsds where γ&amp;gt;1. The contour can be closed along the half-circle to the left and using the residue-theorem you will get f(x)x=e−x23(e2x2+et3x2+2πt3+e−t3x2−2πt3) =e−x23(e3x2+2cos⁡(3x2+2π3))

How can we prove that? \sum_{n=1}^\infty\frac{x^{3n}}{(3n-1)!}=\frac{1}{3}e^{\frac{-x}{2}}x(e^{\frac{3x}{2}}-2\sin(\frac{\pi+3\sqrt{3}x}{6}))

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