Closed-form solution to the transcendental equation

Question

Could you give me advice on how to find a closed-form solution $t>0$ to the following transcendental equation:

$$(t+1)^a - t^a = g$$

where $a>1$ and $g>1$. An accurate closed-form approximation for $t>0$ to see how $t$ depends on $a$ and $g$ could be also fine. I guess the above equation defines for a fixed $a>1$ a special function of $g>1$ $$t=t_a(g).$$ Maybe, it is possible to make an appropriate substitution to express $t_a$ by other special functions.

Many Thanks! Piotr

Answer 1

Your equation is an equation of elementary functions. It's an irreducible algebraic equation in dependence of $(t+1)^a$ and $t^a$. Because the terms $(t+1)^a,t^a$ are algebraically independent, we don't know how to rearrange the equation for $t$ by only elementary operations (means elementary functions).
I don't know when the equation has solutions in the elementary numbers for $a\notin\mathbb{Q}$.

For rational $a$, the equation is related to an algebraic equation over the reals, and we can use the known solution methods in radicals, Bring radicals etc.

see also:
Can we list explicitly all types of polynomial equations of one unknown that are solvable by radicals?

For $|t|\le 1$, we can use $(t+1)^a=\sum_{n=0}^\infty\binom{a}{n}t^n$ and apply Lagrange inversion.

Answer 2

I think that it could be possible to have decent approximations letting $t=\sinh^2(x)$ and then to look for the zero of function $$f(x)=\log \left(\cosh ^{2 a}(x)-\sinh ^{2 a}(x)\right)-k \qquad \text{with} \quad k=\log(g)$$ Around $x=0$ $$f(x)+k=a x^2 + O(x^4)$$

With $x_0=\sqrt{\frac{k}{a}}$, the first iterate of Newton method should probably be good.

Trying

• $a=e$ and $k=\pi$ give $x_0=0.9302$, $x_1=1.3158$ while the solution is $x=1.3125$
• $a=\pi$ and $k=e$ give $x_0=1.0751$, $x_1=1.0494$ while the solution is $x=1.0475$

Answer 3

$\def\B{\operatorname B}\def\P{\operatorname P}$ Fourier Series: Maybe someone who knows Mellin transforms better can find a shorter solution than with a Fourier sine series solution:

$$(t+1)^a-t^a=g\implies t_a(g)=\frac2{2^a-1}\sum_{n=1}^\infty\sin\left(\frac{\pi n g}{2^a-1}\right)\int_0^1 at((t+1)^{a-1}-t^{a-1})\sin\left(\frac{\pi n((t+1)^a-t^a)}{2^a-1}\right)dt;1\le g<2^a-1$$

graph of $t_e((t+1)^e-t^e)=t$ to $100$ terms:

$t_e(x)$ plot:

Sine Taylor series:

$$\int_0^1 at((t+1)^{a-1}-t^{a-1})\sin\left(\frac{\pi n((t+1)^a-t^a)}{2^a-1}\right)dt= a\sum_{m=0}^\infty\frac{(-1)^m}{(2m+1)!}\left(\frac{\pi n}{2^a-1}\right)^{2m+1}\int_0^1 t((t+1)^{a-1}-t^{a-1})((t+1)^a-t^a)^{2m+1}dt$$

binomial theorem: $$a\sum_{m=0}^\infty\frac{(-1)^m}{(2m+1)!}\left(\frac{\pi n}{2^a-1}\right)^{2m+1}\sum_{k=0}^{2m+1}\binom{2m+1}k(-1)^k \int_0^1 t((t+1)^{a-1}-t^{a-1})(t+1)^{a(2m-k+1)} t^{ak}dt$$ Finally, use the incomplete beta $\B_x(a,b)$ function: $$\int_0^1 t((t+1)^{a-1}-t^{a-1})(t+1)^{a(2m-k+1)} t^{ak}dt=(-1)^{-ak}\B_{-1}(ak+2,a(2m-k+2))+(-1)^{-a(k+1)}\B_{-1}(a(k+1)+1,a(2m-k+1)+1)$$

When $n=3$, the series coefficient integral equals our series expansion of it. You can show the same for any other $n$ series coefficient. Therefore: $$\boxed{(t+1)^a-t^a=g\implies t=t_a(g)=\frac{2a}{2^a-1}\sum_{n=1}^\infty\sum_{m=0}^\infty\sum_{k=0}^{2m+1}\frac{(-1)^{m+k}}{(2m-k+1)!k!}\left(\frac{\pi n}{2^a-1}\right)^{2m+1} \sin\left(\frac{\pi n g}{2^a-1}\right)\left((-1)^{-ak}\B_{-1}(ak+1,a(2m-k+2))+(-1)^{-a(k+1)}\B_{-1}(a(k+1)+1,a(2m-k+1)+1)\right),1\le g<2^a-1}$$

If $k$’s upper bound is infinity, then possibly all three sums are interchangeable. Also, Fourier series is slow, but does eventually converge to the wanted value, as shown by the graphs. Expanding the incomplete beta function a sum and switching sums may give another series solution.

Legendre Fourier Series: Using Legendre $\P_v(x)$ polynomial transforms:

$$t_a(g)=\frac a{2^a-1}\sum_{n=0}^\infty (4n+3) \P_{2n+1}\left(\frac g{2^a-1}\right)\int_0^1 \P_{2n+1}\left(\frac{(t+1)^a-t^a}{2^a-1}\right) t ((t+1)^{a-1}-t^{a-1})dt,1\le g<2^a-1$$

Integrating from $-1$ to $0$ likely will not simplify the integral as much. Composing $t_e(x)$ with the inverse gives the identity:

$t_e(x)$:

Expanding $\P_{2n+1}(y)$ as a series and then as a binomial series will again give incomplete beta functions from the remaining integral. There should be a simpler solution than an infinite series and 2 finite series maybe through complex integrals or with Teixeira’s theorem etc, but I have little experience with contour integrals.