Equations involving sigma sums of powers

Question

Is there a trick to solving equations involving sigma sums of powers like:

$\sum\limits_{i=0}^{5}g^i = 5$ ?

I make it to

$g + g^2 + g^3 + g^4 + g^5 = 4$

But I don't know where to go from there outside of guess and check

Answer 1

Remember that $\sum_{i=0}^m g^i =\dfrac{1-g^{m+1}}{1-g} $, so that, if $\sum_{i=0}^m g^i =a $ then $a=\dfrac{1-g^{m+1}}{1-g} $ so $a(1-g) =1-g^{m+1} $ or $g^{m+1}-ag+a-1=0 $.

In your case this can only be solved numerically. Note that $g=1$ is always a root.

Answer 2

To solve the equation you gave $$g + g^2 + g^3 + g^4 + g^5 = 4$$ consider that you look for the zero of function $$f(g)=g^5+g^4+g^3+g^2+g-4$$ and,even without inspection or plotting, there is a root just lower that $1$ since $g(1)=1$.

If you want to avoid numerical method, make $g=1-x$ to make $$h(x)=-x^5+6 x^4-15 x^3+20 x^2-15 x+1$$ and, being (myself) tricky, consider that $h(x)$ is a Taylor expansion to $O(x^6)$. Now, using series reversion $$x=t+\frac{4 t^2}{3}+\frac{23 t^3}{9}+\frac{754 t^4}{135}+\frac{5284 t^5}{405}+O\left(t^{6}\right)$$ where $t=\frac{1-h(x)}{15}$.

Making, as desired, $h(x)=0$ gives, as an approximation $$x=\frac{22597714}{307546875}=0.07347730001\implies g=0.926523$$ while the exact solution (given by Newton method) is $g=0.926519$.

But, I can be more tricky considering for example that $h(x)$ is a Taylor expansion to $O(x^k)$ with $k>6$. For example, using $k=7$, the series reversion will give $$x=t+\frac{4 t^2}{3}+\frac{23 t^3}{9}+\frac{754 t^4}{135}+\frac{5284 t^5}{405}+\frac{12838 t^6}{405}+O\left(t^{7}\right)$$ which now will lead to $g=0.926520$ which is still better. And so on.