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

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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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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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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.
 
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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?
 
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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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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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  • #8
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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.
 
  • #10
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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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\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
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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?
 
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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.
 
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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##?
 
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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##.
 
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You mean for the problem here? No, we didn't assume anything. Integers are simply a domain we know something about.
 
  • #17
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\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.

$$18^x+20^x=25^x$$
 
  • #18
bob012345
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BTW, ##x= -∞## might work too for ##a^x + b^x = c^x##.
 
  • #19
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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.
 
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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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