Calculus/Linear approximation
Expert: Paul Klarreich - 3/25/2011
QuestionHi,
I'm trying to linear the following eq. about x=0 up to the 3rd term:
x / (L^2 - x^2)
I know the formula is:
f(x) = f(0) + f'(0)x
If I keep my original f(x) as it is, I wind up after the 3rd expansion at 1 / L^2, which I don't think is correct. I THINK the correct answer should be x/L
Then I thought that I should approximate the original function as x / (L^2 - x ), for small x, but that doesn't seem to work.
Can you help?
Answer
Questioner: Mike
Country: United States
Category: Calculus
Private: No
Subject: linearize a function
Question: Hi,
I'm trying to linear the following eq. about x=0 up to the 3rd term:
x / (L^2 - x^2)
I know the formula is:
f(x) = f(0) + f'(0)x
If I keep my original f(x) as it is, I wind up after the 3rd expansion at 1 / L^2, which I don't think is correct. I THINK the correct answer should be x/L
Then I thought that I should approximate the original function as x / (L^2 - x ), for small x, but that doesn't seem to work.
Can you help?
.........................................
f(0) = 0, obviously.
L^2 - x^2 - (-2x)
f'(x) = -------------------
(L^2 - x^2)^2
L^2 - x^2 + 2x
f'(x) = ----------------
(L^2 - x^2)^2
Now
L^2 1
f'(0) = ------ = -----
L^4 L^2
So your 'linearization' would be:
f(x) ~~ 0 + x/L^2
(I attached a graph of these functions, with L = 3.)
.......................
I am not sure what you mean by 'up to the third term'. But I can suggest that you might want to look into the Taylor(Maclaurin?) polynomial. You can get it by either:
1. Doing long division:
L^2 - x^2 INTO x. (Yes, this is allowed, even though your fourth grade teacher would throw up.)
2. Use a binomial expansion of:
(L^2 - x^2)^-1, then times x.
Let me know what you did.