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

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The equation x^{\alpha} + x = 1 generally lacks an analytic solution for most values of α, particularly when α is 5 or higher. While specific cases for integer values of α (0, 1, 2, 3, 4) can be solved, non-integer and higher-degree cases typically do not yield closed-form solutions. The discussion highlights that transforming the equation to solve for α given x is straightforward, but finding x from the original equation remains complex due to the mixture of operations. Special cases can sometimes be reduced to solvable forms, but these are exceptions rather than the rule. Overall, the equation presents significant challenges for finding analytic solutions across a broad range of parameters.
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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.
 
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bob012345 said:
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.
 
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fresh_42 said:
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.
 
bob012345 said:
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?
 
fresh_42 said:
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.
 
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bob012345 said:
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.
 
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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.
 
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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.
 
bob012345 said:
I suppose the first form is just as problematic.
Yes. Can you at least say where ##a,b,y## are taken from?
 
  • #10
fresh_42 said:
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.
 
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  • #11
\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.
 
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  • #12
fresh_42 said:
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?
 
  • #13
WWGD said:
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.
 
  • #14
fresh_42 said:
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##?
 
  • #15
bob012345 said:
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##.
 
  • #16
You mean for the problem here? No, we didn't assume anything. Integers are simply a domain we know something about.
 
  • #17
fresh_42 said:
\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.

$$18^x+20^x=25^x$$
 
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  • #18
BTW, ##x= -∞## might work too for ##a^x + b^x = c^x##.
 
  • #19
bob012345 said:
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.
 
  • #20
bob012345 said:
BTW, ##x= -∞## might work too for ##a^x + b^x = c^x##.
In the sense that for ##a, b, c > 1##:$$\lim_{x \to -\infty} (a^x + b^x) = \lim_{x \to -\infty} (c^x) = 0$$
 
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