Find the root of $f(t)=0.05-e^{-2t}(1+2t+2t^2+\tfrac{4}{3}t^3)$

Question

I am looking for the root of $f(t)=0.05-e^{-2t}\left(1+2t+2t^2+\tfrac{4}{3}t^3\right)$, with $t>0$.

I know it is possible to numerically find the root $t_0$ using Newton's method. However, is it possible to derive the closed-form solution?

I came up with the following equivalence: $$ f(t) = 0.05 - e^{-2t}\left(e^{2t}-\sum_{i=4}^\infty \frac{(2t)^i}{i!} \right) $$ but I fail to see how that may be useful to find the root.

Answer 1

The

$$e^{-2t}\sum_{n=0}^3 \frac{(2t)^n}{n!}-0.05$$

looked very similar to a gamma regularized $Q(a,x)$ series represenation:

$$Q(4,2x)=e^{-2x}\left(\frac{4x^3}3+2x^2+2x+1\right)=0.05$$

Now use inverse gamma regularized $Q^{-1}(a,x)$ to get:

$$x=\frac{Q^{-1}\left(4,\frac1{20}\right)}2$$

shown here. An interpretation is a $5$th percentile for a chi squared distribution of $8$ degrees of freedom. Maybe someone else can find an integral/sum representation.

Answer 2

$$0.05-e^{-2t}\left(1+2t+2t^2+\frac{4}{3}t^3\right)=0$$

We see, this is an algebraic equation of more than one algebraically independent monomials ($t,e^{-2t}$). The main theorem in [Ritt 1925] cannot help in these cases: We cannot read an elementary inverse function over a non-discrete domain directly from the equation because we don't know how to rearrange the equation for $t$ by applying only finite numbers of elementary functions (operations) we can read from the equation.

$$0.05\ \left(e^t\right)^2-1-2t-2t^2-\frac{4}{3}t^3=0$$

We see, this is an irreducible algebraic equation of both $t$ and $e^t$. According to the theorems in [Lin 1983] and [Chow 1999], such kind of equations cannot have solutions except $0$ that are elementary numbers or explicit elementary numbers respectively.

$$0.05\ (e^t)^2=1+2t+2t^2+\frac{4}{3}t^3$$ $t\to\frac{x}{2}$: $$0.05e^x=1+x+\frac{1}{2}x^2+\frac{1}{6}x^3$$ $$\frac{0.05}{1+x+\frac{1}{2}x^2+\frac{1}{6}x^3}e^x=1$$

We see, this equation cannot be solved in terms of Lambert W but in terms of Generalized Lambert W:

$$\frac{6\cdot 0.05}{6+6x+3x^2+x^3}e^x=1$$ $$\frac{0.3}{(x-x_1)(x-x_2)(x-x_2)}e^x=1$$

$x_{1,2,3}$ are the solutions of the polynomial equation $6+6x+3x^2+x^3$. You can get them in closed form.

$$x=W\left(^{\ \ \ \ 0.3}_{x_1,x_2,x_3};1\right)$$ $$t=\frac{1}{2}W\left(^{\ \ \ \ 0.3}_{x_1,x_2,x_3};1\right)$$

So we have a closed form for $t$, and the series representations of Generalized Lambert W give some hints for calculating $t$.
$\ $

[Ritt 1925] Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90

[Lin 1983] Ferng-Ching Lin: Schanuel's Conjecture Implies Ritt's Conjectures. Chin. J. Math. 11 (1983) (1) 41-50

[Chow 1999] Chow, T.: What is a closed-form number. Am. Math. Monthly 106 (1999) (5) 440-448

[Mező 2017] Mező, I.: On the structure of the solution set of a generalized Euler-Lambert equation. J. Math. Anal. Appl. 455 (2017) (1) 538-553

[Mező/Baricz 2017] Mező, I.; Baricz, Á.: On the generalization of the Lambert W function. Transact. Amer. Math. Soc. 369 (2017) (11) 7917–7934 (On the generalization of the Lambert W function with applications in theoretical physics. 2015)

[Castle 2018] Castle, P.: Taylor series for generalized Lambert W functions. 2018