Taylor Expansion for Gravitational Acceleration Problem

In summary, the "Taylor approximation" of a function is a good approximation for values within the radius of convergence.
  • #1
bemigh
30
0
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.
My only problem is... how do i go about this?
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
 
Physics news on Phys.org
  • #2
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

[tex]g(r) = -Kr^{-2}[/tex]

for K>0 a constant.

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

[tex]g(r+d)=-K(r+d)^{-2}=-\frac{K}{r^2}\left(1+\frac{d}{r}\right)^{-2}[/tex]

You will want to develop the function [itex]f(x)=(1+x)^{-2}[/tex] in a Taylor series and find its radius of convergence (hint: it's going to 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
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
i guess you just need the fist and second terms of talyor series.
stare at [tex]1/r^2[/tex], it's derivative is [tex]-2/r^3[/tex]
since you only want the difference... its quite easy, derivative=df/dx, so df=df/dx * dx, and the difference is approximately:
[tex]-2d/r^3[/tex]
 
  • #5
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

[tex]\frac{df}{dx}(x)=\lim_{\Delta x\rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}[/tex],

it should be a reasonable approximation to say that for [itex]\Delta x[/itex]<<1,

[tex]f(x+\Delta x)-f(x) \approx \frac{df}{dx}(x)\Delta x[/tex]

Incidentally, this is just a Taylor approximation of first order to [itex]f(x+\Delta x)[/itex] and is also sometimes referred to as "Euler's approximation".
 
  • #6
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
I guess you have a point!
 
  • #8
But I think it's important to know the radius of convergence. Take for exemple the (geometric) series

[tex]\sum_{n=0}^{\infty} x^n[/tex]

which we know without knowledge about Taylor series to converge to [itex]f(x)=1/(1-x)[/itex] for |x|<1 and diverge otherwise. Now say we want to get an approximation of [itex]f(-2)[/itex] (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:
  • #9
But I'd be interested in a formal result concerning the behavior of the "error function"

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

Is the sequence {[itex]E(x_0,n)[/itex]} 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.
 

Related to Taylor Expansion for Gravitational Acceleration Problem

1. What is the Taylor Expansion for Gravitational Acceleration Problem?

The Taylor Expansion for Gravitational Acceleration Problem is a mathematical tool used to approximate the value of gravitational acceleration at a certain point based on its values at nearby points. It is derived from the Taylor series, which represents a function as an infinite sum of its derivatives at a given point.

2. How is the Taylor Expansion for Gravitational Acceleration Problem used in scientific research?

The Taylor Expansion for Gravitational Acceleration Problem is used in various scientific fields, such as physics and astronomy, to calculate the gravitational acceleration at a specific point in space. It allows researchers to make precise predictions and calculations related to the motion of objects under the influence of gravity.

3. What are the assumptions made in the Taylor Expansion for Gravitational Acceleration Problem?

The Taylor Expansion for Gravitational Acceleration Problem assumes that the gravitational acceleration is a continuous function and that its derivatives exist at the point of interest. It also assumes that the gravitational field is conservative, meaning that the work done by gravity on an object is independent of the path taken.

4. Can the Taylor Expansion for Gravitational Acceleration Problem be used for any point in space?

Yes, the Taylor Expansion for Gravitational Acceleration Problem can be used for any point in space as long as the assumptions mentioned above hold true. However, the accuracy of the approximation decreases as the distance from the known points increases, so it is typically used for points in close proximity.

5. Are there any limitations to using the Taylor Expansion for Gravitational Acceleration Problem?

One limitation of the Taylor Expansion for Gravitational Acceleration Problem is that it is only accurate for small changes in position. It also assumes that the gravitational field is static, meaning that the positions of the masses causing the gravitational field do not change over time. Additionally, it is a mathematical approximation and may not accurately represent the true values in certain scenarios.

Similar threads

  • Advanced Physics Homework Help
Replies
1
Views
678
  • Advanced Physics Homework Help
Replies
2
Views
1K
Replies
3
Views
715
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
Replies
3
Views
3K
Replies
3
Views
1K
Replies
2
Views
2K
  • Linear and Abstract Algebra
Replies
5
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Advanced Physics Homework Help
Replies
10
Views
3K
Back
Top