How do you solve # e^(-x)-x+2=0#?

1 Answer
Jun 21, 2016

Answer:

2.1200, nearly.

Explanation:

If LHS is #f(x), f(2)=0.135... > 0 and f(3)=-0.950...<0#.

So, the root in (2, 3)

Use the method of successive approximation, from the difference

equation

#x_n=2+e^(-(x_(n-1))), n = 1, 2, 3, ..#,

with the starter guess-value #x_0=2#,

we obtain the sequence of approximations

2.21..., 2.118..., 2.1202.., 2.1200.., 2.1200..

#f(2.12)=O(10^(-5))#.

f(x) is a decreasing function and, therefore, the root is unique.

Important note:
Despite that we get a good approximation to the solution of the given equation, this sequence converges to the solution of the difference equation and not the solution of the given equation, in mathematical exactitude. .