Understanding Taylor Series Approximation with Taylor's Theorem Explanation

rmc240
Messages
4
Reaction score
0
I'm reading a derivation and it says that the following approximation can be used:

attachment.php?attachmentid=69764&stc=1&d=1400020033.png


I do not under stand how Taylor's theorem allows for this approximation. Can anyone explain this a little?
 

Attachments

  • Taylor.png
    Taylor.png
    896 bytes · Views: 604
Physics news on Phys.org
rmc240 said:
I'm reading a derivation and it says that the following approximation can be used:

attachment.php?attachmentid=69764&stc=1&d=1400020033.png


I do not under stand how Taylor's theorem allows for this approximation. Can anyone explain this a little?

If you let ##f(r) = \frac{dv}{dr}## you have ##f(r+dr) = f(r) + f'(r)dr##. Do you recognize that?
 
Yea, my problem was realizing that I was supposed to approximate the function around r and evaluate the function at r + dr. Should have seen that. Thank you for your help.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top