Simplifying an equation with summation

Question

Could you help me simplify the following equation?

$\displaystyle 5500^{0.7}=\sum_{n=1}^\infty{(5500 + 2^n - x)^{0.7}\frac {1}{2^n}}$

What would be some good sources where I could get help regarding problems like this?

Answer 1

The only solution I can see is inspection (with the problem of the infinite summation).

For an approximation, consider that you look for the zero of function $$f(x)= \sum_{n=1}^\infty{\frac{(5500 + 2^n - x)^{ 7/{10}}}{2^n}}-5500^{ 7/{10}}$$ Assuming that $x$ is small, expand as a Taylor series around $x=0$ to get $$f(x)=-5500^{ 7/{10}}+\sum_{n=1}^\infty 2^{-n} \left(2^n+5500\right)^{7/10}-\frac 75x\sum_{n=1}^\infty\frac{ 2^{-(n+1)} }{ \left(2^n+5500\right)^{3/10}}+O\left(x^2\right)$$

Using a speadsheet, the partial sums converge quite fast $$\left( \begin{array}{ccc} p & \Sigma_1 & \Sigma_2 \\ 10 & 415.317 & 0.0376882 \\ 20 & 416.058 & 0.0377188 \\ 30 & 416.118 & 0.0377188 \\ 40 & 416.125 & 0.0377188 \\ 50 & 416.126 & 0.0377188 \end{array} \right)$$ which means that we are left with $$-415.197+416.126-1.4\times 0.0377188\,x=0 \implies x=17.5926$$

Working with exact arithmetic and summations to infinity, the solution is $x=17.5921$.

For $x=17.5926$, $f(x)=-2.50\times 10^{-5}$ and for $x=17.5921$, $f(x)=1.46\times 10^{-6}$.