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#