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Is it possible to solve for $a$ in the following equation: $a^\alpha=b^\alpha-a$? Currently, I have resorted to using Excel to approximate $a$ (I am given values for $b$ and $\alpha$), but am wondering if it is possible to pinpoint $a$ exactly.

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In general, no. It's impossible (depending on what you mean by solve). Let $C=b^\alpha$. You have to solve

$$a^\alpha + a = C$$.

This polynomial is, in general, not solvable by radicals (Abel-Ruffini Theorem). For example $C=1, \alpha = 5$. This means that you cannot find a finite expression involving sums, multiplication, division, exponentiation and roots which expresses the solution.

You can calculate the solution with arbitrary percussion using numerics. Try, for example WolframAlpha.

EDIT: Be careful using approximations. The equation might have multiple roots.

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  • $\begingroup$ I agree. Although it can be solved if $\alpha \in \{ -1,0, \frac 12,1,2 \}$ $\endgroup$ – tomi Apr 23 '16 at 21:53
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    $\begingroup$ Yes. It can be solved by radicals for many specific values of $\alpha$ and $C$. The point is that that's not possible for arbitrary $\alpha$ and $C$. $\endgroup$ – mlainz Apr 23 '16 at 22:00
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I would use the Newton-Raphson method to find an approximate answer.

Rewrite the equation in the form $f(a)=0$.

This is $a^\alpha +a -b^\alpha=0$.

Then start with an estimate $a_0$. I suggest $a_0=b$ since that's the solution to $a^\alpha -b^\alpha=0$.

Then proceed with the iteration $a_{n+1}=a_n-\displaystyle \frac {f(a_n)}{f'(a_n)}$

Which in this case is $a_{n+1}=a_n-\displaystyle \frac {a_n^\alpha +a_n-b^\alpha}{\alpha a_n^{\alpha-1} +1}$

Or $a_{n+1}=\displaystyle \frac {(\alpha -1)a_n^\alpha +b^\alpha}{\alpha a_n^{\alpha-1} +1}$

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Although it can not be solved generally, it can be solved if $\alpha \in \{ -1,0, \frac 12,1,2 \}$

Let's consider these in turn:


  1. $\alpha = -1$

$a^{-1} =b^{-1}-a$

$b =a-a^2b$

$a^2b-a+b=0$

$a= \displaystyle \frac {1 \pm \sqrt{(-1)^2-4(b)(b)}}{2b}$

$a= \displaystyle \frac {1 \pm \sqrt{1-4b^2}}{2b}$


  1. $\alpha = 0$

$a^0 =b^0-a$

$1 =1-a$

$a=0$


  1. $\alpha = \frac 12$

$a^{\frac 12} =b^{\frac 12}-a$

Let $x=a^{\frac 12} \Rightarrow x^2=a$

$x =b^{\frac 12}-x^2$

$x^2+x-b^{\frac 12}=0$

$x= \displaystyle \frac {-1 \pm \sqrt{1+4b^{\frac 12}}}{2}$

$a= \displaystyle \left(\frac {-1 \pm \sqrt{1+4b^{\frac 12}}}{2}\right)^2$


  1. $\alpha = 1$

$a^{1} =b^{1}-a$

$a =b-a$

$2a=b$

$a= \displaystyle \frac b2$


  1. $\alpha = 2$

$a^{2} =b^{2}+a$

$a^2-a-b^2=0$

$a= \displaystyle \frac {1 \pm \sqrt{(-1)^2-4(-b^2)}}{2}$

$a= \displaystyle \frac {1 \pm \sqrt{1+4b^2}}{2}$

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