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

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Answer 1 · 2017-08-12T12:07:56

$x ~~ 1.356" "$ or $" "x ~~ -0.4042$

Let:

$f(x) = e^x-1/x-pi$

Then:

$f'(x) = e^x+1/x^2$

Using Newton's method, then if we have an approximate zero $a_i$ of $f(x)$ , a better one is given by:

$a_(i+1) = a_i-(f(a_i))/(f'(a_i))$

$color(white)(a_(i+1)) = a_i-(e^(a_i)-1/a_i+pi)/(e^(a_i)+1/a_i^2)$

What to use as an initial approximation?

$f(1) = e-1-pi ~~ -1.42$

$f(2) = e^2-1/2-pi ~~ 3.75$

So roughly linearly interpolating, we can choose $a_0 = 1.4$

Then:

$a_1 ~~ 1.35634086$

$a_2 ~~ 1.35564215$

$a_3 ~~ 1.35564198$

$a_4 ~~ 1.35564198$

This is not the only solution, putting $a_0 = -0.4$ we also find a negative solution $x ~~ -0.4042$

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