What is the solution of $x^a+(1+x)^b=0 $?

Question

I am not a mathematician, but a theoretical physicist. I am faced with this equation coming from some plasma phenomenon and I am unable to 'recognise' it. Mathematica software cannot solve it. I have searched the web and it might have relation with the Lambert function, but I still can't clearly see how. I apologise before hand if this is too trivial a question or perhaps too difficult, but would enormously appreciate any hint. Many thanks. Cheers. $($I am not even sure under what TAG to classify this question... :-( $)$

To make it clearer $($sorry for not having mentioned it$)$: $a$ and $b$ are real positive numbers and $x$ $($which is actually a function$)$ has real values.

Answer 1

Questions about expressions in which negative numbers are raised to non-integer exponents are highly sensitive to the conventions used in defining such an operation. I'll illustrate with an example for the OP's expression, $x^a+(1+x)^b$.

One common convention (the one used by Mathematica, I believe) is to define $x^a$ for (real) $x\lt0$ by thinking of such $x$ as $|x|e^{i\pi}$, so that $x^a=|x|^ae^{i\pi a}=|x|^a(\cos\pi a+i\sin\pi a)$. Note that in this convention $(-1)^{1/3}={1+\sqrt3i\over2}$, so that the real and imaginary parts of $x^{1/3}$ are both positive when $x$ is negative. It follows that the equation $x^{1/3}+(1+x)^{1/3}=0$ has no real solutions $x$ under this convention: $x^{1/3}$ and $(1+x)^{1/3}$ both have both a non-negative real part and a non-negative imaginary part, and they are not both $0$.

But there is another common convention for the expression $x^{1/3}$ for (real) $x\lt0$, namely $x^{1/3}=-|x|^{1/3}$, so that $(-1)^{1/3}=-1$. (I.e., this convention thinks of $x^{1/3}$ as $\sqrt[3]x$, and thinks of that as asking for a real number whose cube is $x$.) In this convention, we do have a real solution to $x^{1/3}+(1+x)^{1/3}$, namely $x=-1/2$, since

$$(-1/2)^{1/3}+(1-1/2)^{1/3}=(-1/2)^{1/3}+(1/2)^{1/3}=-(1/2)^{1/3}+(1/2)^{1/3}=0$$

In summary, it certainly makes sense to ask about solutions to equations such as $x^a+(1+x)^b=0$, but the first step is to ask, What does the notation mean?

Answer 2

Assuming (as per comments) that $a$ and $b$ are actually positive integers then we have four cases:

1) $a$ even, $b$ even - no real solution.

2) $a$ even, $b$ odd - one real solution in $(-\infty, -2)$ if $a<b$, otherwise no real solutions.

3) $a$ odd, $b$ even - one real solution in $(-1,0)$.

4) $a$ odd, $b$ odd - one real solution in $(-1,0)$ - special case if $a=b$ then solution is $x=-\frac12$.

Where a real solution exists it shouldn't be difficult to find a good numerical approximation using Newton's method.