# Find the complex (or real) roots of $e^{\frac{3 x}{2}}+2 \cos \left(\frac{\sqrt{3} x}{2}\right)$

Define for natural $n\geq 2$ $$G(x,n)= \sum _{k=0}^\infty \frac{x^{k n}}{(k n)!}= \frac{\sum _{k=0}^{n-1} e^{x e^{\frac{2 i \pi k}{n}}}}{n}= G(x e^{\frac{2 i \pi}{n}},n)= \prod_{m=1}^\infty \left(1+\left(\frac{x}{r(n,m)}\right)^n\right)$$ where $r(n,m)$ is the $m$th biggest absolute value of a root of $G(x,n)$.

The complex roots for $G(x,2)$ and $G(x,4)$ are trivial to find. $r(2,m)=\left(m-\frac{1}{2}\right)\pi$ and $r(4,m)=\left(m-\frac{1}{2}\right)\pi\sqrt{2}$.

Because of rotational symmetry $G(x,3)=G(x e^{\frac{2 i \pi}{3}},3)$, we only need to find the roots on the real line. Note that $$G(x,3)=\frac{1}{3} \left(e^x+2 e^{-\frac{x}{2}} \cos \left(\frac{\sqrt{3} x}{2}\right)\right)$$ From this we can see that as $x\rightarrow -\infty$, the roots of $G(x,3)$ become arbitrarily close to the roots of $\cos \left(\frac{\sqrt{3} x}{2}\right)$

Because of this, we can use Newton's Method to find the roots of $G(x,3)$ by starting with the roots of $\cos \left(\frac{\sqrt{3} x}{2}\right)$.

The Mathematica code for $r(3,m)$ is

-x /. FindRoot[E^(3 x/2) + 2 Cos[(Sqrt x)/2], {x, -2 Pi (-1/2 + m)/Sqrt}]

This is not good enough though. I desire a formula for $r(3,m)$ which does not not involve a FindRoot. Also, because you could define Newton's method exactly with a limit, limits are not allowed in the final formula.

Because of input from the comments, we now have a new formula for $r(3,m)$: $$\frac{\pi(2m-1)}{\sqrt{3}}-\sum _{n=1}^{\infty } \frac{\left(-e^{\frac{1}{2} \sqrt{3} (\pi -2 \pi m)}\right)^n \left(\lim_{x\to \frac{\pi -2 \pi m}{\sqrt{3}}} \, \frac{\partial ^{n-1}}{\partial x^{n-1}}\left(\frac{x-\frac{\pi -2 \pi m}{\sqrt{3}}}{-e^{\frac{1}{2} \sqrt{3} (\pi -2 \pi m)}+e^{\frac{3 x}{2}}+2 \cos \left(\frac{\sqrt{3} x}{2}\right)}\right)^n\right)}{n!}$$ Though Mathematica cannot calculate this, the corresponding Mathematica code is:

-(([Pi] - 2 m [Pi])/Sqrt) - Sum[(-E^(1/2 Sqrt ([Pi] - 2 m [Pi])))^n/n!* Assuming[{Element[m, Integers]}, Limit[D[((x - ([Pi] - 2 m [Pi])/Sqrt)/(E^(3 x/2) + 2 Cos[(Sqrt x)/2] - E^( 1/2 Sqrt ([Pi] - 2 m [Pi]))))^n, {x, n - 1}], x -> ([Pi] - 2 m [Pi])/Sqrt]], {n, 1, Infinity}]

The remaining work is to find the general $(n-1)$th derivative in the sum and to get rid of the limit in the sum.

Here is a contour plot showing the roots of $G(x,3)$. The orange curve is $\Im(G(x,3))=0$. The blue curve is $\Re(G(x,3))=0$. The plot shows $-10<\Re(x)<10,-10<\Im(x)<10$. Also, by comparing the sum and product forms of $G(x,n)$, we can see $$\frac{1}{n!}=\sum_{m=1}^\infty\frac{1}{r(n,m)^n}$$

• You don't want a FindRoot. What sort of non-algebraic operations will you permit for this transcendental root? – Eric Towers Nov 23 '15 at 2:31
• @Eric Pretty much anything goes. Could be infinite sums/products, any documented math functions, integrals. What I don't like about FindRoot is that it is an algorithm, rather than a function. – Nazgand Nov 23 '15 at 2:35
• This boils down to solving $\cos\bigg(\frac{x\sqrt{3}}{2}\bigg) = -\frac{1}{2} e^{\frac{3x}{2}}$. – Brevan Ellefsen Nov 25 '15 at 9:33
• @Nazgand WA is telling me that $cos(x)=-\frac{1}{2}e^{x\sqrt 3}$ has different solutions than you are looking for... how did you get to that conclusion? The equation I give in my comment above still yields the correct solutions. – Brevan Ellefsen Nov 25 '15 at 10:21
• Infinite sum representations for the roots, expressed in terms of the zeros of the cosine term, aren't too hard to obtain using the Lagrange inversion theorem but they aren't pretty. – Antonio Vargas Nov 25 '15 at 10:53