# How do you find 4-sd rounded approximation(s) to the solution(s) of e^x-1/x=pi?

Aug 12, 2017

$x \approx 1.356 \text{ }$ or $\text{ } x \approx - 0.4042$

#### Explanation:

Let:

$f \left(x\right) = {e}^{x} - \frac{1}{x} - \pi$

Then:

$f ' \left(x\right) = {e}^{x} + \frac{1}{x} ^ 2$

Using Newton's method, then if we have an approximate zero ${a}_{i}$ of $f \left(x\right)$, a better one is given by:

${a}_{i + 1} = {a}_{i} - \frac{f \left({a}_{i}\right)}{f ' \left({a}_{i}\right)}$

$\textcolor{w h i t e}{{a}_{i + 1}} = {a}_{i} - \frac{{e}^{{a}_{i}} - \frac{1}{a} _ i + \pi}{{e}^{{a}_{i}} + \frac{1}{a} _ {i}^{2}}$

What to use as an initial approximation?

$f \left(1\right) = e - 1 - \pi \approx - 1.42$

$f \left(2\right) = {e}^{2} - \frac{1}{2} - \pi \approx 3.75$

So roughly linearly interpolating, we can choose ${a}_{0} = 1.4$

Then:

${a}_{1} \approx 1.35634086$

${a}_{2} \approx 1.35564215$

${a}_{3} \approx 1.35564198$

${a}_{4} \approx 1.35564198$

This is not the only solution, putting ${a}_{0} = - 0.4$ we also find a negative solution $x \approx - 0.4042$