Taylor Series Expansion - Don't understand how to use

[jumbogala]

Homework Statement

This is actually not a problem, it's something in my notes. The function I am supposed to be approximating is

V(x) = V₀(1 - e^x/a)² - V₀

V₀ and a are constants.

Homework Equations

The Attempt at a Solution

It says that the function given is not a parabola. But it can be approximated to a parabola by approximating to the 2nd order term.

Then it says, 2nd order V = V₀[(x/a)² - 1].

Maybe I copied it wrong, but I thought I needed to approximate about a certain value. If you don't know that value, how can you linearize the function?

Also, my understanding is that Taylor expansion uses the formula given here http://mathworld.wolfram.com/TaylorSeries.html

But when I differentiate V(x) I get something completely different.

Can someone either help me get to the result in my notes, or if that result is wrong, help me get the right answer?

 

[jumbogala]

It looks like they may have taken V₀ = 1 and a = 1, and approximated around 0, if that helps.

Even when I do that though, I get zeros as my coefficients.

 

[vela]

They're expanding about x=0. Instead of grinding out the series for V(x) using the definition of the Taylor series, try expanding the exponential in a series about x=0, keeping only terms that will in the end contribute to the x² term, and then doing some algebra.

 

[jumbogala]

Thanks for your reply. What do you mean by try expanding the exponential in a series about x=0?

You mean just take e^(x/a) and expand that, still using a Taylor series? If I do that I get
the second derivative is (e^(x/a))/a^3

So the term that will contribute to the x^2 term is (1/2a^3)x^2.

Then would I replace the e^(x/a) term in my original equation? If I do I still don't get the same thing. I still don't quite get it...

 

[vela]

What is the Taylor series for e^x about the point x=0? Plug in x/a for x. That's the series for e^x/a. Then plug that into your potential function.

 

[jumbogala]

Ok.

The taylor series for e^x about x = 0 is:
1 + x + (1/2)x² + (1/6)x³ + ...
Keeping only the x² term, we have (1/2)x².

Now, subbing in x/a for x, we get (1/2)(x/a)².
How come what I did first didn't work? Shouldn't it work out to be the same thing...?

So now plugging my answer into my potential function, I get:
V(x) = V₀(1 - (1/2a²)x²) - V₀

Still doesn't match =(

Actually with a bit more algebra I get V₀ /2 = [-(x/a)²]

 

[vela]

Ok.

The taylor series for e^x about x = 0 is:
1 + x + (1/2)x² + (1/6)x³ + ...
Keeping only the x² term, we have (1/2)x².

The idea is that when x<<1, only the first few terms of the series contribute significantly to the sum, so you can approximate the sum well enough by neglecting higher-order terms. That doesn't mean you can just toss everything except one term.

Try keeping just the first three terms of the series and see how it works out.

Now, subbing in x/a for x, we get (1/2)(x/a)².
How come what I did first didn't work? Shouldn't it work out to be the same thing...?

So now plugging my answer into my potential function, I get:
V(x) = V₀(1 - (1/2a²)x²) - V₀

Still doesn't match =(

Actually with a bit more algebra I get V₀ /2 = [-(x/a)²]