# Closed form for zeros of a function

I need a closed form for the zeros of $$f(x)=2\sin\left(\frac{\pi}{6}-\frac{\sqrt{3} x}{2} \right)-e^{-\frac{3x}{2}}$$

Putting $$x=0$$, we see that $$f(0)=0$$. For the closed form of the remaining zeros we use the series expansion of $$f(x)$$ about $$x=0$$ and finally "reversion of series with nth term" (see here) to get a solution $$x=x_i$$, $$i=1,2,3...$$

$$f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!} x^n$$ where $$f^{(0)}(0)=f(0)=0$$ . So we have $$f^{(n)}(x)=\left(-\frac{\sqrt{3}}{2}\right)^n\sin\left(-\frac{\sqrt{3}x}{2}+\frac{\pi}{6}+\frac{n\pi}{2}\right)-\left(-\frac{3}{2}\right)^n e^{-\frac{3x}{2}}$$

$$f^{(n)}(0)=\left(-\frac{\sqrt{3}}{2}\right)^n\left[\sin\left(\frac{\pi}{6}+\frac{n\pi}{2}\right)-(\sqrt{3})^n\right] \tag{1}$$ Now we discuss two cases:

Case $$1$$: $$n$$ is even or $$n=2m$$ where $$m\in \mathbb{N}\cup \{0\}$$

$$f^{(n)}(0)=\left(-\frac{\sqrt{3}}{2}\right)^{2m}\left[\sin\left(\frac{\pi}{6}+m\pi\right)-(\sqrt{3})^{2m}\right]$$

$$f^{(n)}(0)=\left(\frac{\sqrt{3}}{2}\right)^{2m}\left[\frac{(-1)^m}{2}-(\sqrt{3})^{2m}\right]$$ $$f^{(n)}(0)=\left(\frac{3}{4}\right)^{m}\left[\frac{(-1)^m}{2}-3^{m}\right] \tag{2}$$

Case $$2$$: $$n$$ is odd or $$n=2k+1$$ where $$k\in \mathbb{N}\cup\{0\}$$ $$f^{(n)}(0)=-\left(\frac{\sqrt{3}}{2}\right)^{2k+1}\left[\sin\left(\frac{\pi}{6}+\frac{(2k+1)\pi}{2}\right)-(\sqrt{3})^{2k+1}\right]$$ Now we have $$\sin\left(\frac{\pi}{6}+\frac{(2k+1)\pi}{2}\right)=\frac{\sqrt{3}}{2}(-1)^k$$ $$f^{(n)}(0)=-\left(\frac{\sqrt{3}}{2}\right)^{2k+1}\left[\frac{\sqrt{3}}{2}(-1)^k-(\sqrt{3})^{2k+1}\right]$$ $$f^{(n)}(0)=-\left(\frac{3}{2}\right)\left(\frac{3}{4}\right)^{k}\left[\frac{(-1)^k}{2}-3^{k}\right] \tag{3}$$ $$f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n=\sum_{m=0}^\infty\frac{f^{(2m)}(0)}{(2m)!}x^{2m}+\sum_{k=0}^\infty\frac{f^{(2k+1)}(0)}{(2k+1)!}x^{2k+1}$$ So by $$(2)$$ and $$(3)$$ $$f(x)=\sum_{m=0}^\infty\frac{\left(\frac{3}{4}\right)^{m}\left(\frac{(-1)^m}{2}-3^{m}\right)}{(2m)!}x^{2m}-\frac{3}{2}\sum_{k=0}^\infty\frac{\left(\frac{3}{4}\right)^{k}\left(\frac{(-1)^k}{2}-3^{k}\right)}{(2k+1)!}x^{2k+1}$$ Edit I tried using Langrange inversion theorem but was unable to simply it further. Any help would be appreciated. Thank you.

• @TymaGaidash You are right. The reversion formula may be too long. I looked at the links provided by you. Can you please explain how to use Fourier series to find an exact solution. If possible, please write as an answer.
– Max
Apr 2, 2023 at 16:56
• What makes you think that the "closed form" exists? I doubt that very much. If not a secret, what were you going to do with it anyway? Apr 5, 2023 at 0:39
• @TymaGaidash Thank you for the link. I could not find the solution for the question. I would really appreciate if you could write as an answer. I will with utmost respect accept it.
– Max
Apr 5, 2023 at 2:56
• You mean that's homework? Then I think you've got some hints already ;-) I'll just add that in all likelihood all your Professor is looking for is either a fast way to find the roots (Newton is good) or the first few terms in an asymptotic expansion, but you'd better ask him/her to figure out what is the task exactly. Nothing nice and clean seems to be within reach anyway... Apr 5, 2023 at 4:19
• A good approximation to the $n$th positive zero $x_n$ is $$x_n \approx \frac{{2\pi }}{{\sqrt 3 }}n + \frac{\pi }{{3\sqrt 3 }} + \frac{{( - 1)^n }}{{2\sqrt 3 }}\exp \left( { - \sqrt 3 \pi n - \frac{\pi }{{2\sqrt 3 }}} \right).$$
– Gary
Apr 6, 2023 at 2:24

If you use what @Gary proposed in comments $$x_n^{(0)} = \frac{{2\pi }}{{\sqrt 3 }}n + \frac{\pi }{{3\sqrt 3 }} + \frac{{( - 1)^n }}{{2\sqrt 3 }}\exp \left( { - \sqrt 3 \pi n - \frac{\pi }{{2\sqrt 3 }}} \right)$$ as the starting point of any Newton-like method (Newton, Halley, Householder,$$\cdots$$) method, the first iterate is fully explicit.

Using (as the simplest) Newton method $$\left( \begin{array}{ccc} n & x_n^{(1)} & \text{solution} \\ 1 & \color{red}{4.2332071}8615947279466285745754 & 4.23320719243895656091540892155 \\ 2 & \color{red}{7.859792867351540}66752184229490 & 7.85979286735154015594186277305 \\ 3 & \color{red}{11.4873959924535109930019}408091 & 11.4873959924535109930019824389 \\ 4 & \color{red}{15.1149947018696098633343674540} & 15.1149947018696098633343674540 \\ \end{array} \right)$$ and you can do much better with any method of higher order.

Considering the case of the first root, playing with the order $$p$$ of the method $$\left( \begin{array}{ccc} p & \text{estimate for order p} & \text{method} \\ 2 & \color{red}{ 4.2332071}8615947279466285746 &\text{Newton} \\ 3 & \color{red}{ 4.2332071}8983134900127366436 &\text{Halley} \\ 4 & \color{red}{ 4.2332071924389}3072546444189&\text{Householder} \\ 5 & \color{red}{ 4.23320719243895}485340805279&\text{no name} \\ 6 & \color{red}{4.233207192438956560}86892954 &\text{no name} \\ 7 & \color{red}{ 4.233207192438956560914}41334&\text{no name} \\ 8 & \color{red}{ 4.233207192438956560915408}87 &\text{no name} \\ \cdots & \cdots & & \\ \infty & 4.23320719243895656091540892 &\text{solution} \\ \end{array} \right)$$

Edit

It is possible to make better approximations. Let $$x=\frac{\pi -6 t}{3 \sqrt{3}}\qquad \text{and} \qquad k=2 e^{\frac{\pi }{2 \sqrt{3}}}\qquad \implies \qquad k \sin (t)-e^{\sqrt{3} t}=0$$

Expanded around $$t=n\pi$$ $$k \sin (t)-e^{\sqrt{3} t}=\sum_{m=0}^\infty \left( (-1)^n\, k\sin \left(\frac{\pi m}{2}\right)-3^{m/2} e^{\sqrt{3} \pi n}\right) (t-n\pi)^m$$ which allows to perform a complete power series reversion using Morse and Feshbach general formulation.

This would give for the $$n^{\text{th}}$$ root $$t=n\pi+\sum_{m=0}^\infty a_m\, Z^m \qquad \text{where} \qquad Z=\frac{e^{\sqrt{3} \pi n}}{\sqrt{3} e^{\sqrt{3} \pi n}-(-1)^n\,k}$$

Update (for an example of the reversion) As an example, around $$t=3\pi$$, we have as a series $$-e^{3 \sqrt{3} \pi }+\left(-k-\sqrt{3} e^{3 \sqrt{3} \pi }\right) (t-3 \pi )-\frac{3}{2} e^{3 \sqrt{3} \pi } (t-3 \pi )^2+\frac{1}{6} \left(k-3 \sqrt{3} e^{3 \sqrt{3} \pi }\right) (t-3 \pi )^3+O\left((t-3 \pi )^4\right)$$ Let $$y=(t-3\pi)$$ and $$a=e^{3 \sqrt{3} \pi }$$ to make $$f(y)=-a-\left(a\sqrt{3}+k\right)y-\frac{3 a }{2}y^2+\frac{k-3 \sqrt{3} a}{6} y^3+O\left(y^4\right)$$ Now, using the formulae given here $$y=-\frac{a+f(y)}{\sqrt{3} a+k}-\frac{3 a (a+f(y))^2}{2 \left(\sqrt{3} a+k\right)^3}-\frac{(a+f(y))^3 \left(18 a^2-2 \sqrt{3} a k+k^2\right)}{6 \left(\sqrt{3} a+k\right)^5}+O\left((f(y)+a)^4\right)$$ Since we want $$f(y)=0$$, then $$y=-\frac{a}{\sqrt{3} a+k}-\frac{3 a^3}{2 \left(\sqrt{3} a+k\right)^3}-\frac{a^3 \left(18 a^2-2 \sqrt{3} a k+k^2\right)}{6 \left(\sqrt{3} a+k\right)^5}+\cdots$$ To make it looking nicer, let $$z=\frac{a}{\sqrt{3} a+k}$$ $$y=-z-\frac{3 a }{2 \left(\sqrt{3} a+k\right)}z^2-\frac{\left(18 a^2-2 \sqrt{3} a k+k^2\right)}{6 \left(\sqrt{3} a+k\right)^2}z^3+O\left(z^4\right)$$

• Thank you so much for your beautiful answer. I would really appreciate if you please provide a form which is exact solution may be in the form of some infinite series.
– Max
Apr 6, 2023 at 13:31
• @Max. Have a look at paragraph $4.1$ in this paper old.sztaki.hu/~bozoki/oktatas/nemlinearis/… This is what I used for the second part. "Just" make $p\to \infty$. Apr 6, 2023 at 13:51
• +1 for your answer. But my question is to how get an exact solution/ closed form may be in the form of infinite sum or some special function or something else?
– Max
Apr 6, 2023 at 13:56
• @Max. I wrote the infinite summation Apr 7, 2023 at 3:03
• I am really thankful to you for your recent elegant edit. I have one last question: How did you get the last equality? What is your $a_m$? Can you please explain the intermediate steps involved. I mean how did you get this :$$t=n\pi+\sum_{m=0}^\infty a_m\, Z^m \qquad \text{where} \qquad Z=\frac{e^{\sqrt{3} \pi n}}{\sqrt{3} e^{\sqrt{3} \pi n}-(-1)^n\,k}$$
– Max
Apr 7, 2023 at 3:58

Set up: Later on is an explicit Fourier series solution, not needing the series reversion formula for:

$$2\sin\left(\frac\pi6-\frac{\sqrt3x}2\right)-e^{-\frac{3x}2}=0$$

Substitute $$x\to\frac2{\sqrt 3}x$$ and transform the function around the zeros:

$$g(x)=2\sin\left(x-\frac\pi6\right)+e^{-\sqrt3x}=0\implies g_j(x)=(-1)^j(g(x+\pi j)-g(\pi j))=(-1)^j\left(2\sin\left(x+\pi j-\frac\pi6\right)+e^{-\sqrt3(x+\pi j)}-e^{-\sqrt3\pi j}+(-1)^{j+1}\right)$$

The Fourier sine series expands $$\sin(3h_j(x))$$ on $$0\le x

where $$h_j(x)$$ inverts $$g_j(x)$$. We solve $$g_j(x)=(-1)^{j+1}g(\pi j)$$ to find the transformed zeros illustrated here.

$$j=1$$: case

$$g_1(x)=2\sin\left(x-\frac\pi6\right)-e^{-\sqrt3(x+\pi)}+e^{-\sqrt3\pi}+1=e^{-\sqrt3\pi}+1\implies \sin(3 h_1(x))=-\frac2{e^{-\sqrt3\pi}-e^{-\frac{4\pi}{\sqrt3}}+2}\sum_{n=1}^\infty a_n\sin\left(\frac{\pi n x}{L_1}\right)$$

$$t\to h_1(t)$$ and integration by parts gives:

\begin{align}a_n=\int_0^{L_1}\sin(3h_1(t))\sin\left(\frac{\pi n t}{L_1}\right)dt=\int_0^\frac\pi3 \sin\left(\frac{\pi n}{L_1}\left(e^{-\sqrt3(t+\pi)}-2\sin\left(t-\frac\pi6\right)-e^{-\sqrt3 \pi}-1\right)\right)\left(2\cos\left(t-\frac\pi6\right)+\sqrt3 e^{-\sqrt3(t+\pi)}\right)\sin(3t)dt\end{align}

to get:

\begin{align}\sin(3h_1(t))=-\sum_{n=1}^\infty \frac6{\pi n} \sin\left(\frac{\pi n x}{e^{-\sqrt3\pi}-e^{-\frac{4\pi}{\sqrt3}}+2}\right)\int_{-\frac\pi 6}^\frac\pi6\sin(3t) \cos\left(\frac{\pi n \left(e^{-\sqrt3\left(t+\frac{7\pi}6\right)}-2\sin(t)-e^{-\sqrt3\pi}-1\right)}{e^{-\sqrt3\pi}-e^{-\frac{4\pi}{\sqrt3}}+2}\right)dt\end{align}

$$\sin(3 h_1((g_1(x)))=\sin(3x)$$ plot where $$e^{-\sqrt3\pi}-e^{-\frac{4\pi}{\sqrt3}}+2\approx 2.00362$$ for $$n=10$$:

The series gives $$h_1\left(1+e^{-\sqrt3\pi}\right)\approx 0.524$$ and $$x_1=\frac2{\sqrt3}\left(h_1\left(1+e^{-\sqrt3\pi}\right)+\pi\right)\approx 4.233$$ correctly.

General case:

The roots become ever closer to those of $$\sin\left(\frac\pi6-\frac{\sqrt3x}2\right)$$, so the setup for $$h_j(x)$$ is similar like when $$j=1$$

$$g_j(x)=(-1)^j\left(2\sin\left(x+\pi j-\frac\pi6\right)+e^{-\sqrt3(x+\pi j)}-e^{-\sqrt3\pi j}\right)-1\implies \sin(3h_j(x))=\frac 2{L_j}\sum_{n=1}^\infty b_n\sin\left(\frac{\pi n x}{L_j}\right)$$

Using the same techniques:

\begin{align}b_n=\int_0^{L_j}\sin(3h_j(t))\sin\left(\frac{\pi n t}{L_j}\right)dt=\int_0^\frac\pi3\sin(3t)\sin\left(\frac{\pi n}{L_j}g_j(x)\right)dg_j(x)\end{align}

and use this branch of the inverse sine for the solution:

$$3 h_j(t)=\pi+\sin^{-1}\left(\frac6\pi\sum_{n=1}^\infty\frac1n\sin\left(\frac{\pi nx}{e^{-\sqrt 3\pi j}\left(e^{-\frac\pi{\sqrt3}}-1\right)+2(-1)^j}\right)\int_{-\frac\pi 6}^\frac\pi6\sin(3t)\cos\left(\frac{\pi n\left(2\sin(t)+(-1)^je^{-\sqrt3\pi j}\left(e^{-\sqrt3\left(t+\frac\pi6\right)}-1\right)+1\right)}{e^{-\sqrt 3\pi j}\left(e^{-\frac\pi{\sqrt3}}-1\right)+2(-1)^j}\right)\right)dt$$

therefore:

$$2\sin\left(\frac\pi6-\frac{\sqrt3x}2\right)-e^{-\frac{3x}2}=0\implies x_j=\frac2{\sqrt 3}(h_j((-1)^{j+1}g(\pi j))+\pi j)$$

Setting $$j=1$$ should match the result in the previous section verifying the above general solution. Fourier series like these require many terms for better accuracy. A simple way to evaluate the coefficients of the form $$\int_0^\frac\pi3\sin(3t)\sin(ae^{bt}+c\sin(t)+d)dt$$ expands $$\sin(y+d)=\sin(y)\cos(d)+\cos(y)\sin(d)$$ and then $$\cos(y),\sin(y)$$ as series. Afterwards, use binomial theorem and integrate, but there may be a simpler result, so the integrals are left alone.

• There is a closed form for $x$ in $\sin(x)-ae^{-\sqrt3x}=0$ with $a$ about between $1$ and $40$. We invert the beta function in this setup using inverse beta regularized: $I^{-1}_s\left(\frac23,\frac13\right)$ Apr 12, 2023 at 11:39

I prefer to write a separate answer for the most general case $$\alpha\,\sin(\beta+\gamma\,x)-e^{\delta\,x}=0\qquad \text{with}\qquad \delta <0$$

Let $$x=\frac{t-\beta }{\gamma }\qquad\qquad a=\frac \delta \gamma\qquad \qquad\epsilon=\frac {e^{-a\beta}} \alpha$$ to write the equation $$\epsilon \sin (t)-e^{a t}=0$$ The $$n^{\text{th}}$$ zero being close to $$n\pi$$, let $$t_{(n)}=n\pi +y_n\qquad\qquad b_n=\frac{e^{an\pi}} \epsilon$$ to write the equation $$(-1)^n\sin(y_n)-b_n\, e^{ay_n}=0$$ From here, I shall use $$(b,y)$$ in place of $$(b_n,y_n)$$.

Around $$y=0$$, the series expansion is $$(-1)^n\sin(y)-b\, e^{ay}=\sum_{k=0}^\infty c_k\, y^k\qquad\text{where}\qquad c_k=\frac{(-1)^n \sin \left(\frac{k\pi }{2}\right)-b \,a^k}{k!}$$ Then, power series reversion for the result as a infinite summation

$$y=\sum_{m=1}^\infty d_m\,b^m$$ coefficients $$d_m$$ being computed using Morse and Feshbach formulation.

So, $$y_n$$ is known and then $$t_{(n)}=n\pi =y_n$$ and$$x_{(n)}$$ from its definition.