| Topic: |
Science > Physics |
| User: |
"Zinc Potterman delete 123s to reply" |
| Date: |
20 Mar 2006 04:30:36 AM |
| Object: |
Taylor approximations |
In my physics book I can't follow the derivation ### below
#########
f(x + delta x) = f(x) exp[-a(x) delta x]
(1)
If we approximate f(x) by the first two terms of its Taylor expansion, and
likewise approximate the exponential on the right hand side, we have
f(x) + f '(x) delta x =approx= f(x) [1 - a(x) delta x] (2)
#########
I thought I understood Taylor approximations .....
f(x + a) =approx= f(a) + (x-a) f '(a) + 1/2! (x-a)^2 f ''(a) + ......
(3)
but I can't seem to connect equations 1 and 2, and they don't match up to
equation 3.
Could somebody fill in some steps for me.
Zinc
--
zincnews at tiscali.co.uk
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| User: "Henning Makholm" |
|
| Title: Re: Taylor approximations |
20 Mar 2006 06:59:28 AM |
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Scripsit "Zinc Potterman" <zincnews@tiscali.c123o.uk. (delete 123's to reply)>
f(x + delta x) = f(x) exp[-a(x) delta x] (1)
f(x) + f '(x) delta x =approx= f(x) [1 - a(x) delta x] (2)
I thought I understood Taylor approximations .....
f(x + a) =approx= f(a) + (x-a) f '(a) + 1/2! (x-a)^2 f ''(a) + ...... (3)
Um, no. Either you add a to x on the RHS or you subtract it on the
LHS. Doing both at once is wrong.
but I can't seem to connect equations 1 and 2, and they don't match up to
equation 3.
View either side of (1) as a function of delta-x (treating x as a
constant), and use the first-order Taylor expansion
f(x0+a) =approx= f(x0) + f'(x0)*a
on each side of (1) separately, setting x0=0 and a=delta-x.
The resulting equation is (2).
--
Henning Makholm "There is a danger that curious users may
occasionally unplug their fiber connector and look
directly into it to watch the bits go by at 100 Mbps."
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