How to solve nonlinear systems of equation with summation

Question

I'm stuck with this

$$ \sum _{n=0}^{80}\:0.1+b^{n+k}=100 $$ $$ 0.1+b^{80+k}=10 $$

Is it possible to solve?

Answer 1

The summation of the first equation contains $81$ terms, from $n=0$ to $n=80$. Starting to count the terms from the last one and calling them $a_0, a_1, a_2...a_{i}$ with $i=80-n$, we have $a_0=10$.

Now we have $$a_1=(a_0-0.1)/b+0.1$$ and in general

$$a_{i+1}=(a_i-0.1)/b+0.1$$

This recurrence has solution

$$a_i = \frac{99/b^i + 1}{10} $$

So we can write

$$\sum_{i=0}^{80} \frac{99/ b^{i} + 1}{10}=100 $$

$$\sum_{i=0}^{80} \left(\frac{99}{ b^{i}} + 1\right)=1000 $$

$$\sum_{i=0}^{80} \frac{99}{b^{i}}=1000-81=919 $$

$$\sum_{i=0}^{80} \frac{1}{ b^{i}}=\frac{919}{99} $$

$$\frac{b^{81}-1}{ b^{80}(b-1)}=\frac{919}{99} (\text{with } b\neq 1)$$

$$b^{81}-1 =\frac{919}{99} b^{80}(b-1)$$

$$\frac{820}{99} b^{81} - \frac{919}{99} b^{80}+1=0$$

$$820 b^{81} - 919 b^{80}+99=0$$

This last equation has three real solutions, that WA numerically determines here : $$b=1$$ $$b=-0.96501...$$ $$b=1.12071...$$

The first solution cannot be accepted because of the above condition $b\neq 1$ (also note that $b=1$ trivially could not satisfy the initial equations of the OP).

Because from the second equation of the OP we get $$k=\log(9.9)/\log(b)-80$$

only the third solution is acceptable, providing

$$k\approx \frac{\log(9.9)}{\log(1.12071)}-80 \approx -59.884...$$

Here WA computes the summation in the LHS of the first equation of the OP using the values of $b$ and $k$ reported above, giving confirmation of the expected result.