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

I think if we write the taylor expansion of $\sin(u)$ and $e^u$, we can arrive from RHS to LHS, but I am looking for a way to prove it from LHS.

user91500
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  • This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. – Did Dec 17 '13 at 15:28
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    Not my area of expertise. I could probably devise a solution myself through differential equations, but I imagine that's off the table. – Mike Dec 17 '13 at 15:50
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    This is a general trick to extract a (regular) sub-series from a series : the multiplication formula. For example to obtain the expansion of $\zeta$ around $0$ keeping only every third term you could do series(zeta(x)+zeta(xexp(-2PII/3))+zeta(xexp(2PII/3)),x)/3 (add three times the function taken at x multiplied by a root of unity). – Raymond Manzoni Dec 17 '13 at 15:50
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    (in your case the function zeta should be replaced by $;\displaystyle x\mapsto x,e^x$) – Raymond Manzoni Dec 17 '13 at 16:47
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    @Did: What thing is strange in my question that you want to close my question? – user91500 Dec 20 '13 at 17:46
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    Would you describe "How can we obtain following identity? [Identity]" and nothing else, as the ideal MSE question? – Did Dec 20 '13 at 18:03

4 Answers4

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Consider the third root of unity $\rho = e^{2\pi i/3} = \frac{-1+i\sqrt{3}}{2}$. You have

$$e^{\rho z} = \sum_{k=0}^\infty \frac{\rho^k z^k}{k!} = \sum_{m=0}^\infty \frac{z^{3m}}{(3m)!} + \rho\sum_{m=0}^\infty \frac{z^{3m+1}}{(3m+1)!} + \rho^2\sum_{m=0}^\infty \frac{z^{3m+2}}{(3m+2)!}$$

since $\rho^{3m} = 1,\; \rho^{3m+1} = \rho,\; \rho^{3m+2} = \rho^2$. You have something similar for $e^{\rho^2 z}$. Also consider $1 + \rho + \rho^2 = 0$. Then a suitable combination of $e^{\rho^k x}$ gives you

$$\sum_{n=1}^\infty \frac{x^{3n-1}}{(3n-1)!}.$$

Using Euler's formula $e^{it} = \cos t + i\sin t$ then gives you the right hand side.

Daniel Fischer
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    Thanks very much. I found suitable combination:$$\sum_{m=0}^\infty\frac{z^{3m+2}}{(3m+2)!}=\frac{\rho e^{\rho^2z}-\rho e^z+e^{\rho z}-e^z}{2\rho^2-\rho-1}.$$ – user91500 Dec 18 '13 at 07:53
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Too long for a comment: Just to put things into proper perspective:

$$\sum_{n=0}^\infty\frac{x^{an+b}}{(an+b)!}= \begin{cases} e^x,&(a,b)=(1,0)\\ \\ \cosh x,&(a,b)=(2,0)\\ \\ \sinh x,&(a,b)=(2,1)\\ \\ \dfrac13\bigg[e^x+2e^{^{-\tfrac x2}}\cos\bigg(x\dfrac{\sqrt3}2\bigg)\bigg],&(a,b)=(3,0)\\ \\ \dfrac13\bigg\{e^x-e^{^{-\tfrac x2}}\bigg[\cos\bigg(x\dfrac{\sqrt3}2\bigg)-\sqrt3\sin\bigg(x\dfrac{\sqrt3}2\bigg)\bigg]\bigg\},&(a,b)=(3,1)\\ \\ \dfrac13\bigg\{e^x-e^{^{-\tfrac x2}}\bigg[\cos\bigg(x\dfrac{\sqrt3}2\bigg)+\sqrt3\sin\bigg(x\dfrac{\sqrt3}2\bigg)\bigg]\bigg\},&(a,b)=(3,2)\\ \\ \tfrac12(\cosh x+\cos x),&(a,b)=(4,0)\\ \\ \tfrac12(\sinh x+\sin x),&(a,b)=(4,1)\\ \\ \tfrac12(\cosh x-\cos x),&(a,b)=(4,2)\\ \\ \tfrac12(\sinh x-\sin x),&(a,b)=(4,3) \end{cases}$$

Lucian
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  • Also, for $(a,b)=\bigg(1, \dfrac12\bigg),~$ we have $S=e^x~\text{erf} \big(\sqrt x\big).$ – Lucian May 15 '15 at 19:56
  • Also, for $(a,b)=\bigg(\dfrac12, 0\bigg),~$ starting from $n=0,$ as well as for $(a,b)=\bigg(\dfrac12, \dfrac12\bigg),~$ starting from $n=-1,$ we have $S=e^x~\Big[1+\text{erf} \big(\sqrt x\big)\Big].$ – Lucian Jan 25 '17 at 13:56
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Consider the function

$$f(x)=\sum_{n=1}^\infty\frac{x^{3n-1}}{(3n-1)!}$$

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'll continue. The characteristic equation for our problem is $s^3-1=0$, whose roots are the third roots of unity $1$ and $-\frac12\pm\frac{\sqrt3}2i$. This suggests

$$f(x)=k_1e^x+k_2e^{-\frac x2}\sin(\frac{x\sqrt3}2)+k_3e^{-\frac x2}\cos(\frac{x\sqrt3}2)=$$ $$k_1e^x+e^{-\frac x2}[k_2\sin(\frac{x\sqrt3}2)+k_3\cos(\frac{x\sqrt3}2)]$$ $$f'(x)=k_1e^x+e^{-\frac x2}[(-\frac12k_2-\frac{\sqrt3}2k_3)\sin(\frac{x\sqrt3}2)+(-\frac12k_3+\frac{\sqrt3}2k_2)\cos(\frac{x\sqrt3}2)]$$ $$f''(x)=k_1e^x+e^{-\frac x2}[(\frac14k_2+\frac{\sqrt3}4k_3+\frac{\sqrt3}4k_3-\frac34k_2)\sin(\frac{x\sqrt3}2)+(\frac14k_3-\frac{\sqrt3}4k_2-\frac{\sqrt3}4k_2-\frac34k_3)\cos(\frac{x\sqrt3}2)]=$$ $$k_1e^x+e^{-\frac x2}[(-\frac12k_2+\frac{\sqrt3}2k_3)\sin(\frac{x\sqrt3}2)+(-\frac12k_3-\frac{\sqrt3}2k_2)\cos(\frac{x\sqrt3}2)]$$

Plugging in initial conditions

$$k_1+k_3=0$$ $$k_1+\frac{\sqrt3}2k_2-\frac12k_3=0$$ $$k_1-\frac{\sqrt3}2k_2-\frac12k_3=1$$

From the first equation, we get $k_3=-k_1$. Adding the second and third gives $k_3=2k_1-1$. So $k_1=\frac13,k_3=-\frac13$ and

$$\frac13+\frac{\sqrt3}2k_2+\frac16=0$$ $$\frac{\sqrt3}2k_2=-\frac12$$ $$k_2=-\frac{\sqrt3}3$$

So the last step is to prove

$$\sqrt3\sin(\frac{x\sqrt3}2)+\cos(\frac{x\sqrt3}2)=2\sin(\frac{x\sqrt3}2+\frac\pi6)$$

This last step can be done by solving $a\cos b=\sqrt3$ and $a\sin b=1$. This gives $\tan b=\frac{\sqrt3}3$ and $a^2=3+1=4$.

Mike
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This is actually calling for Laplace transform. Set $f(x)=\sum_{n=1}^\infty \frac{x^{3n}}{(3n-1)!}$ then \begin{align} F(s) &= \int_0^\infty \frac{f(x)}{x} \, e^{-sx} \, {\rm d}x \\ &= \sum_{n=1}^\infty \frac{1}{s^{3n}} \int_0^\infty \frac{u^{3n-1} \, e^{-u}}{(3n-1)!} \, {\rm d}u \\ &= \frac{1}{s^3-1} \end{align} Now \begin{align} \frac{f(x)}{x} = \frac{1}{2\pi i} \int_{\gamma -i\infty}^{\gamma + i\infty} F(s) \, e^{xs} \, {\rm d}s \end{align} where $\gamma > 1$. The contour can be closed along the half-circle to the left and using the residue-theorem you will get \begin{align} \frac{f(x)}{x} &= \frac{e^{-\frac{x}{2}}}{3} \left( e^{\frac{3x}{2}} + e^{\frac{i\sqrt{3}x}{2}+\frac{2\pi i}{3}} + e^{-\frac{i\sqrt{3}x}{2}-\frac{2\pi i}{3}} \right) \\ &= \frac{e^{-\frac{x}{2}}}{3} \left( e^{\frac{3x}{2}} + 2\cos\left(\frac{\sqrt{3}x}{2}+\frac{2\pi}{3}\right) \right) \end{align}

Diger
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