Linear Approx. of e^-x/1+x

E

epicjacob1123

New member
Joined
Aug 12, 2015
Messages
7
I have to find the linear approx of e^(-x)/(1+x). For my answer, I got 1 and I don't know where I went wrong.

I know that e^(-x)/(1+x) is the same as e^(-x)*(1+x)^-1. I can approximate both of the components and then multiply them.

e^(-x) is in the form e^x. The approximation of e^x is 1+x near x=0, so the approximation of e^-x is 1-x/

(1+x)^-1 is in the form (1+x)^n. The approximation of (1+x)^n is 1+nx near x=0, so the approximation of (1+x)^-1 is 1-x.

I know have (1-x)(1-x). Simplifying, I get 1+x^2, and I remove the x^2 because it is not linear.

I also used f(x) ≈ f(x_0) + f'(x_0)*(x_0-x), and I still got 1.


Can anyone show me where I made a mistake?
 
epicjacob1123 said:
I have to find the linear approx of e^(-x)/(1+x). For my answer, I got 1 and I don't know where I went wrong.

I know that e^(-x)/(1+x) is the same as e^(-x)*(1+x)^-1. I can approximate both of the components and then multiply them.

e^(-x) is in the form e^x. The approximation of e^x is 1+x near x=0, so the approximation of e^-x is 1-x/

(1+x)^-1 is in the form (1+x)^n. The approximation of (1+x)^n is 1+nx near x=0, so the approximation of (1+x)^-1 is 1-x.

I know have (1-x)(1-x). Simplifying, I get 1+x^2, and I remove the x^2 because it is not linear.

I also used f(x) ≈ f(x_0) + f'(x_0)*(x_0-x), and I still got 1.


Can anyone show me where I made a mistake?
Click to expand...


What you have used is the Taylor series expansion for the function f (assuming the f(x_o)* is the derivative of f). So let
\(\displaystyle f(x)\, =\, \frac{1}{e^{x}\, (1+x)}\)
then
\(\displaystyle f(x_0)\, =\, \frac{1}{e^{x_0}\, (1+x_0)}\)
If x0 were zero then we would have f(0)=1.

You will also need f'(x) [the derivative of f]. What did you get for that?
 
Ishuda said:
What you have used is the Taylor series expansion for the function f (assuming the f(x_o)* is the derivative of f). So let
\(\displaystyle f(x)\, =\, \frac{1}{e^{x}\, (1+x)}\)
then
\(\displaystyle f(x_0)\, =\, \frac{1}{e^{x_0}\, (1+x_0)}\)
If x0 were zero then we would have f(0)=1.

You will also need f'(x) [the derivative of f]. What did you get for that?
Click to expand...

For the derivative, I got f'(x) = -(x+2)*e^(-x)/(x+1)^2
 
Ishuda said:
O.K. So, again letting x0=0, f'(x0)=?. After you have that use your formula
f(x) = f(0) + f'(0) (x-0) = 1 + ? x
Click to expand...

SO I get f'(0) = -2. SO the linear approximation is 1-2x. Ok, thanks for helping me work out this problem
 
You must log in or register to reply here.
Top