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I have the following equation that I need to solve:

$$1000.00116=\frac{1000}{\left(1+x\right)^{16}}+\frac{1-\left(1+x\right)^{-16}}{x}$$

However, software I use is refusing to do it. Which software/web is capable of solving it? Or could you please show me the answer?

Thanks

Blue
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Emanuel
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2 Answers2

1

Hint.

$$ 1000.00116-\frac{1-(x+1)^{16}}{x}-\frac{1000}{(x+1)^{16}}\approx 16.0012+16120 x-135440 x^2+O\left(x^3\right) $$

solving for $x$ we have

$$ x = \cases{ -0.000984485\\ 0.120004 } $$

The first value gives an error of $0.000783992$. Now we can proceed using an iterative procedure like Newton-Raphson to approximate arbitrarily those values.

Cesareo
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This looks like a finance problem. Consider

$$y=\frac{1000}{\left(1+x\right)^{16}}+\frac{1-\left(1+x\right)^{-16}}{x}$$ and expand as a series around $x=0$ to get

$$y=1016-16136 x+136816 x^2-819876 x^3+3891504 x^4-15558264 x^5+54434544 x^6+O\left(x^7\right)$$

Using series reversion $$x=t+\frac{17102 t^2}{2017}+\frac{756491143 t^3}{8136578}+\frac{9312920857507 t^4}{8205738913}+O\left(t^5\right)$$ where $t=\frac{1016-y}{16136}$

Working with whole numbers and the value of $y$, this gives $$t=\frac{399971}{403400000}$$ for which the truncated inverse series gives $$x=\frac{217284880812466746896433664885508741308467}{217300772456778118887476800000000 000000000000}\approx 0.000999926868$$ while the exact solution should be $0.000999926882$

Claude Leibovici
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