Analytic Solution to ##x^{\alpha} +x =1##?

[bob012345]

TL;DR

Is there an analytic solution to the simple equation $x^{\alpha} +x =1$ where $\alpha$ is a constant ?

Is there an analytic solution to the simple equation $x^{\alpha} +x =1$? I can get to a solution by iteration or by graphical methods but I wish to find a closed form exact solution. $\alpha$ is a constant. I tried to put it into a form where I could use the quadratic formula but that didn't work i.e something like $c^{2\beta} + c^{\beta} =1$. Thanks.

 

[fresh_42]

Summary: Is there an analytic solution to the simple equation $x^{\alpha} +x =1$ where $\alpha$ is a constant ?

Is there an analytic solution to the simple equation $x^{\alpha} +x =1$? I can get to a solution by iteration or by graphical methods but I wish to find a closed form exact solution. $\alpha$ is a constant. I tried to put it into a form where I could use the quadratic formula but that didn't work i.e something like $c^{2\beta} + c^{\beta} =1$. Thanks.

There is generally no analytic solution. E.g. consider the case $\alpha=7$ then $x^7+x-1=0$ cannot be solved analytically.

 

[bob012345]

There is generally no analytic solution. E.g. consider the case $\alpha=7$ then $x^7+x-1=0$ cannot be solved analytically.

Thanks. I suppose it is the same if $\alpha$ is less than 2 with the obvious exceptions of 0, and 1? My interest in this question is non-integer exponents.

 

[fresh_42]

Thanks. I suppose it is the same if $\alpha$ is less than 2 with the obvious exceptions of 0, 1 and 2?

There is a theorem that says that there are no solutions with radicals ($\sqrt[n]{.}$) for arbitrary integer polynomials of degree $5$ or higher. We can solve it for $\alpha \in \{0,1,2,3,4\}$ by roots. So $\alpha \geq 5$ has good chances of not being solvable. And what about non-integer values?

 

[bob012345]

There is a theorem that says that there are no solutions with radicals ($\sqrt[n]{.}$) for arbitrary integer polynomials of degree $5$ or higher. We can solve it for $\alpha \in \{0,1,2,3,4\}$ by roots. So $\alpha \geq 5$ has good chances of not being solvable. And what about non-integer values?

Well, turning the problem around it is easy to solve for $\alpha$ given any $x$.

$$\large x^{\alpha} + x=1$$

$$ \alpha log(x) = log (1-x)$$

$$\alpha = \large \frac{log(1-x)}{log(x)}$$

I just thought one might be able to transform the equation around to solve for $x$ given an $\alpha$ for this specific equation form by some kind of substitution.

 

[fresh_42]

Well, turning the problem around it is easy to solve for $\alpha$ given any $x$.

$$\large x^{\alpha} + x=1$$

$$ \alpha log(x) = log (1-x)$$

$$\alpha = \large \frac{log(1-x)}{log(x)}$$

I just thought one might be able to transform the equation around to solve for $x$ given an $\alpha$ for this specific equation form.

If you put it that way, we are given an equation with only one occurrence of the variable, so that we only have $c_{1}^\alpha = c_{2}$ which is basically the definition of the logarithm.

The problem with $x^\alpha +x-1=0$ when we ask for $x$ is, that we have a mixture of some arbitrary multiplication and a few simple additions. Whenever multiplication and addition meet, things get complicated. There is actually only one single law that combines them:
$$
a\cdot (b+c) = a\cdot b + a \cdot c
$$
That's it. It is all we have. And every combination of the two that differs from the distributive law is a complex problem, in our case literally complex.

 

[mfb]

All alpha values where I see a solution:

$\alpha \in \{-3,-2,-1,0,1,2,3,4\}$ as we get a solvable polynomial, multiply by powers of x for negative $\alpha$.
$\alpha \in \{-\frac 1 3, -\frac 1 2, \frac 1 2, \frac 1 3, \frac 1 4\}$ as we get a solvable polynomial in the 2nd, 3rd, 4th root of x.
For $\alpha=5+6z$ for integer z, $e^{\pm \pi i/3}$ is a solution. There is no general closed solution for polynomials of that degree, but some special cases can still have closed solutions.

 

[bob012345]

Where this was all coming from is from a similar equation used to derive it
$$\large a^y + b^y =1$$ where $a,b$ are constants and I let $\large x= b^y$ then we can write $\large a^y = b^{\alpha y} = (b^y)^{\alpha} = x^{\alpha}$ giving $$\large x^{\alpha} + x=1$$

I suppose the first form is just as problematic.

 

[fresh_42]

I suppose the first form is just as problematic.

Yes. Can you at least say where $a,b,y$ are taken from?

 

[bob012345]

Yes. Can you at least say where $a,b,y$ are taken from?

Yes, there was a problem but in its original form used $x$ which goes as $$16^x + 20^x =25^x$$ solve for $x$. I divided both sides by $25^x$.



I solved that exactly analytically but then wanted to generalize the coefficients a bit.

 

[fresh_42]

\begin{align*}
16^x+20^x=25^x &\Longleftrightarrow (5^x-4^x)(5^x+4^x)=4^x\cdot 5^x \\
&\Longleftrightarrow \left(\left(\dfrac{1}{4}\right)^x-\left(\dfrac{1}{5}\right)^x\right)
\left(\left(\dfrac{1}{4}\right)^x+\left(\dfrac{1}{5}\right)^x\right) =1\\
&\Longleftrightarrow \left(\dfrac{1}{4^x}\right)^2+1^2=\left(\dfrac{1}{5^x}\right)^2
\end{align*}
is a very specific form: $(a^x)^2+1^2=(b^x)^2.$ Now we have a quadratic equation and we can e.g. look for Pythagorean triples that are all known.

As a rule of thumb, you can say that, whenever there are symmetries, then there is a good chance to solve a system.

 

[WWGD]

There is a theorem that says that there are no solutions with radicals ($\sqrt[n]{.}$) for arbitrary integer polynomials of degree $5$ or higher. We can solve it for $\alpha \in \{0,1,2,3,4\}$ by roots. So $\alpha \geq 5$ has good chances of not being solvable. And what about non-integer values?

Is that no Real solutions? Is $\alpha$ Real? How about Complex ones? Seems like something like the W function may work?

 

[fresh_42]

Is that no Real solutions? How about Complex ones? Seems like something like the W function may work?

This is irrelevant. $p(x)\in \mathbb{Z}[x]$ is the restriction.

 

[WWGD]

This is irrelevant. $p(x)\in \mathbb{Z}[x]$ is the restriction.

But $Z[x]$ refers to polynomials, doesn't it? So we assume $\alpha \in \mathbb Z$?

 

[WWGD]

Well, turning the problem around it is easy to solve for $\alpha$ given any $x$.

$$\large x^{\alpha} + x=1$$

$$ \alpha log(x) = log (1-x)$$

$$\alpha = \large \frac{log(1-x)}{log(x)}$$

I just thought one might be able to transform the equation around to solve for $x$ given an $\alpha$ for this specific equation form by some kind of substitution.

That will require $x>1$. And I assume you want to solve for $x$, not for $\alpha$.

 

[fresh_42]

You mean for the problem here? No, we didn't assume anything. Integers are simply a domain we know something about.

 

[bob012345]

\begin{align*}
16^x+20^x=25^x &\Longleftrightarrow (5^x-4^x)(5^x+4^x)=4^x\cdot 5^x \\
&\Longleftrightarrow \left(\left(\dfrac{1}{4}\right)^x-\left(\dfrac{1}{5}\right)^x\right)
\left(\left(\dfrac{1}{4}\right)^x+\left(\dfrac{1}{5}\right)^x\right) =1\\
&\Longleftrightarrow \left(\dfrac{1}{4^x}\right)^2+1^2=\left(\dfrac{1}{5^x}\right)^2
\end{align*}
is a very specific form: $(a^x)^2+1^2=(b^x)^2.$ Now we have a quadratic equation and we can e.g. look for Pythagorean triples that are all known.

As a rule of thumb, you can say that, whenever there are symmetries, then there is a good chance to solve a system.

Yes, this was a special case since it could be reduced to a quadratic because of the specific values of the coefficients. BTW, the way I did it was divide by the RHS, combine to get

$$(16/25)^x + (20/25)^x =1$$ which is

$$(0.64)^x + (0.8)^x =1$$

which is $$(0.8^x)^2 + (0.8)^x =1$$ Then substituted $z=0.8^x$

$$z^2 + z =1$$ which gives $z= \frac{-1±5^{1/2}}{2}$

and $x=\large \frac{log(z)}{log(0.8)}$Then, I wanted to see if I changed the 16 to say, 18, if it had an analytic solution. The answer is no in general.

$$18^x+20^x=25^x$$

 

[bob012345]

BTW, $x= -∞$ might work too for $a^x + b^x = c^x$.

 

[fresh_42]

BTW, $x= -∞$ might work too for $a^x + b^x = c^x$.

As does $x=+\infty$ but that should only be used if we explicitly state that hyperreal numbers are allowed. In any other, i.e. standard case, there is no number $\pm \infty$ and we cannot pretend there was.

 

[PeroK]

BTW, $x= -∞$ might work too for $a^x + b^x = c^x$.

In the sense that for $a, b, c > 1$:$$\lim_{x \to -\infty} (a^x + b^x) = \lim_{x \to -\infty} (c^x) = 0$$