# Taylor expansion

1. Oct 23, 2006

### bemigh

Hey Everyone.
I'm ALMOST finished this problem...
To spare you the long story, I need to take the difference between an gravitational acceleration, and the same gravitational acceleration at a slightly larger height.
The two functions are a(r) and a(r+d), where d is very small
Now... VERY SMALL tells me one thing... Taylor expansion. And this is what i have been advised to do.
I have the function f(a) = constant/r^2 which is just Newtons inverse square law of gravitation.

Now, i know that the function for a Taylor Series is going to be something like this:
F(x) = f(a) + xf'(a) + x^2/2 * f''(a) + ....
But how do i go about differentiating f(a), and what is x???
Thanks for the help

2. Oct 23, 2006

### quasar987

You seem to have lost yourself in the sea of variables. Let's start by clearing that up. You know that the acceleration 'g' as a function of the distance r from the center of the earth is of the form

$$g(r) = -Kr^{-2}$$

for K>0 a constant.

You want to calculate $g(r+d)-g(r)$, for d<<r. Since g(r+d) can be write as

$$g(r+d)=-K(r+d)^{-2}=-\frac{K}{r^2}\left(1+\frac{d}{r}\right)^{-2}$$

You will want to develop the function $f(x)=(1+x)^{-2}[/tex] in a Taylor series and find its radius of convergence (hint: it's gonna be 1, so as soon as d<r, the series converges), and then substitute back x=r/d to get the Taylor expansion of g(r+d). 3. Oct 23, 2006 ### quasar987 Upon reflexion, since this is a physics HW, you will not want to find the series of f(x), but rather you will want to just look it up on wiki and thurst that it converges to f(x) for |x|<1: http://en.wikipedia.org/wiki/Binomial_series ;p 4. Oct 23, 2006 ### tim_lou i guess you just need the fist and second terms of talyor series. stare at $$1/r^2$$, it's derivative is $$-2/r^3$$ since you only want the difference... its quite easy, derivative=df/dx, so df=df/dx * dx, and the difference is approximately: $$-2d/r^3$$ 5. Oct 23, 2006 ### quasar987 What tim_lou is saying is that the easiest approach to this problem is simply to say, ok since the derivative of a function is $$\frac{df}{dx}(x)=\lim_{\Delta x\rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}$$, it should be a reasonable approximation to say that for [itex]\Delta x$<<1,

$$f(x+\Delta x)-f(x) \approx \frac{df}{dx}(x)\Delta x$$

Incidentally, this is just a Taylor approximation of first order to $f(x+\Delta x)$ and is also sometimes refered to as "Euler's approximation".

6. Oct 23, 2006

### bemigh

Thanks for the input,
but what good would the radius of convergence give me? Couldnt I just truncate the expansion, and solve for the difference between g(r)?

7. Oct 23, 2006

### quasar987

I guess you have a point!

8. Oct 24, 2006

### quasar987

But I think it's important to know the radius of convergence. Take for exemple the (geometric) series

$$\sum_{n=0}^{\infty} x^n$$

which we know without knowledge about Taylor series to converge to $f(x)=1/(1-x)$ for |x|<1 and diverge otherwise. Now say we want to get an approximation of $f(-2)$ (which is 1). The first term approximation is 1. the second is 1-2=-1, third is -1+4=3, etc. The first order approx is right on, while the other terms make the approximation worse.

Whereas when we want to approximate a value inside the interval of convergence, we know that the more term we take, the better the approximation. ...well, I think!*

So it matters to know the radius of convergence even though we will troncate the series. If you know the series to diverge, it might not be such a good idea to take many terms to your approximation.

*At least we know that no matter how small a number 'h' we take, we can always find an interger N such that for all n>N, the error of nth taylor approximation is lesser than h.

Last edited: Oct 24, 2006
9. Oct 24, 2006

### quasar987

But I'd be interested in a formal result concerning the behavior of the "error function"

$$E(x,n) = \left|\sum_{i=1}^n \frac{f^{(i)}(0)}{i!}x^i - f(x) \right|$$

Is the sequence {$E(x_0,n)$} stricly decreasing for x_0 in the radius of convergence?

I'll ponder on it tomorrow. In the meantime if someone wants to make an input, please do.