# For what range of x is (e^x-1)/2x=0.5 correct to 15 decimal digits?

• ver_mathstats

#### ver_mathstats

Homework Statement
For what range of x is (e^x-1)/2x=0.5 correct to 15 decimal digits?
Relevant Equations
(e^x-1)/2x=0.5
We have ex=1 + x + x2/2 + x3/3! + ...

ex - 1 = x + x2/2 + x3/3! + ...

(ex - 1)/(2x) = 0.5 + x/4 + x2/(2⋅3!) + ...

((ex - 1)/(2x)) - 0.5 = + x/4 + x2/(2⋅3!) + ...

After this, I am unsure of how to proceed to find my error any help would be appreciated thank you. Would we just be trying to isolate x, but that seems incorrect?

Homework Statement:: For what range of x is (e^x-1)/2x=0.5 correct to 15 decimal digits?
Relevant Equations:: (e^x-1)/2x=0.5

We have ex=1 + x + x2/2 + x3/3! + ...

ex - 1 = x + x2/2 + x3/3! + ...

(ex - 1)/(2x) = 0.5 + x/4 + x2/(2⋅3!) + ...

((ex - 1)/(2x)) - 0.5 = + x/4 + x2/(2⋅3!) + ...

After this, I am unsure of how to proceed to find my error any help would be appreciated thank you. Would we just be trying to isolate x, but that seems incorrect?
Yes, that's incorrect. You're not going to be able to solve for x in any of those equations.
It's better to write your last equation as ##\frac{e^x - 1}{2x} = 0.5 + \frac x 4 + \frac {x^2}{12} + \dots##. The dominant variable term on the right side is the ##\frac x 4## term. For small values of x, the ##x^2## and higher-degree terms will be relatively insignificant. If you can make ##\frac x 4## small enough, the contributions of the higher-degree terms shouldn't make any difference.

So how small should ##\frac x 4## be so that its contribution won't affect the first 15 decimal digits of your approximation?

Last edited:
How precisely do you have to specify ##x##?

• SammyS