Closed-form solution to the transcendental equation 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
 A: Your equation is an equation of elementary functions. It's an 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.
A: 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 $x=1.0475$
