Taylor expansion in physics


by quietrain
Tags: expansion, physics, taylor
quietrain
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#1
Apr1-12, 03:01 AM
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i am very confuse how my profs always use taylor expansion in physics which somehow doesn't follow the general equation of

f(x) = f(a) + f'(a)(x-a) + 1/2! f''(a)(x-a)2 and so on...

like for example, what is the taylor expansion of x - kx where k is small

it was given as something like

x - kx f'(x) + (1/2) k2 x2 f''(x) + ...

is this taylor expansion? but there is no 'about which point, i.e, a=? '

i don't even understand how the first term x is gotten. f(a) = x??

please help thank you!
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Apr1-12, 04:36 AM
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hi quietrain!
Quote Quote by quietrain View Post
something like
hmm but not exactly like!

the (x)s should be a, and the x at the start should be f(a)

(and k = x - a)
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Apr1-12, 05:03 AM
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Hi quietrain! :)

Suppose you expand f(a-kx).

You would get: f(a-kx) = f(a) - kx f'(a) + (1/2) k2 x2 f''(a) + ...

Now replace "a" with "x"...
Apparently you've got a typo in the first term.

EDIT: Oops, TM got here first!

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Apr1-12, 05:34 AM
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Taylor expansion in physics


Quote Quote by quietrain View Post
like for example, what is the taylor expansion of x - kx where k is small
TT and ILS have done a great job of guessing what you meant, but you should probably explain what function you really wanted to expand. (The Taylor expansion of x-kx is x-kx, so that's probably not it ).
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Apr2-12, 04:39 AM
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Quote Quote by quietrain View Post
i am very confuse how my profs always use taylor expansion in physics which somehow doesn't follow the general equation of

f(x) = f(a) + f'(a)(x-a) + 1/2! f''(a)(x-a)2 and so on...

like for example, what is the taylor expansion of x - kx where k is small

it was given as something like

x - kx f'(x) + (1/2) k2 x2 f''(x) + ...

is this taylor expansion? but there is no 'about which point, i.e, a=? '

i don't even understand how the first term x is gotten. f(a) = x??

please help thank you!
Fix x, and let g(k) = f(x-kx). Expand as g(k) = g(0) + k*g'(0) + (k^2/2)*g''(0) + ...
g(0) = f(x), g'(k) = -x*f'(x-kx) --> g'(0) = -x*f'(x), g''(k) = x^2*f''(x-kx) --> g''(0) = x^2*f''(x), etc.

RGV
quietrain
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#6
Apr2-12, 05:29 AM
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hi everyone thanks for helping

but the taylor expansion was for x-kx

or perhaps only -kx was expanded?

but anyway it was given as x - kx f'(x) + ...


Quote Quote by I like Serena View Post
Hi quietrain! :)

Suppose you expand f(a-kx).

You would get: f(a-kx) = f(a) - kx f'(a) + (1/2) k2 x2 f''(a) + ...

Now replace "a" with "x"...
Apparently you've got a typo in the first term.
with regards to the above, how did you get that formula?

because i only know this f(x) = f(a) + f'(a)(x-a) + 1/2! f''(a)(x-a)2 as the general formula?


in any case, does it mean if i want to expand say 1 + kx

then it would be f(1+kx) = f(1) + kx f'(1) + ...

so f(1) is 1 + k(1) = 1+k?
then f'(1) is k? then where do i put my 1 since i don't have x
also does it mean f'' onwards are all 0?




Quote Quote by Ray Vickson View Post
Fix x, and let g(k) = f(x-kx). Expand as g(k) = g(0) + k*g'(0) + (k^2/2)*g''(0) + ...
g(0) = f(x), g'(k) = -x*f'(x-kx) --> g'(0) = -x*f'(x), g''(k) = x^2*f''(x-kx) --> g''(0) = x^2*f''(x), etc.

RGV
YES! this is the one. so it was an expansion about the small value k ? and not x? i see.

but in particular, how did you get g'(k) and g'(k)?

is it through the chain rule differentiation?

thanks!
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Apr2-12, 05:54 AM
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Quote Quote by quietrain View Post
with regards to the above, how did you get that formula?

because i only know this f(x) = f(a) + f'(a)(x-a) + 1/2! f''(a)(x-a)2 as the general formula?


in any case, does it mean if i want to expand say 1 + kx

then it would be f(1+kx) = f(1) + kx f'(1) + ...

so f(1) is 1 + k(1) = 1+k?
then f'(1) is k? then where do i put my 1 since i don't have x
also does it mean f'' onwards are all 0?
Okay, let's start with the formula you have:
f(x) = f(a) + f'(a)(x-a) + 1/2! f''(a)(x-a)2 + ...
but let's rewrite it with y instead of x to eliminate the ambiguity between the x's.

So:
f(y) = f(a) + f'(a)(y-a) + 1/2! f''(a)(y-a)2 + ...

You want to expand f(x-kx).
To do this, first define y=x-kx, and define a=x.
Then replace all occurrences of y by (x-kx), and replace all occurrences of a by x.

What do you get?


Edit: Btw, an alternative formula for Taylor expansion is: f(x+h)=f(x) + h f'(x) + 1/2! h2 f''(x) + ...
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Quote Quote by quietrain View Post
in any case, does it mean if i want to expand say 1 + kx

then it would be f(1+kx) = f(1) + kx f'(1) + ...
Your terminology is inaccurate. What you're doing here is to expand ##f## around the point 1 in its domain. The expansion of 1+kx (i.e. the function ##x\mapsto 1+kx##) around 0 (the point I'd assume you have in mind unless you specify another one) is just 1+kx.
quietrain
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Apr2-12, 07:30 AM
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Quote Quote by I like Serena View Post
Okay, let's start with the formula you have:
f(x) = f(a) + f'(a)(x-a) + 1/2! f''(a)(x-a)2 + ...
but let's rewrite it with y instead of x to eliminate the ambiguity between the x's.

So:
f(y) = f(a) + f'(a)(y-a) + 1/2! f''(a)(y-a)2 + ...

You want to expand f(x-kx).
To do this, first define y=x-kx, and define a=x.
Then replace all occurrences of y by (x-kx), and replace all occurrences of a by x.

What do you get?


Edit: Btw, an alternative formula for Taylor expansion is: f(x+h)=f(x) + h f'(x) + 1/2! h2 f''(x) + ...
oh, we can just define a as x?

does it mean i can define a as -kx or -k also?

but i thought a is the point that we evaluate the taylor expansion on?

am i right to say if i taylor expand on a point say x =1 for a curve graph. then as my orders of taylor expansion become greater, the approximation is better too?
quietrain
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Apr2-12, 07:33 AM
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Quote Quote by Fredrik View Post
Your terminology is inaccurate. What you're doing here is to expand ##f## around the point 1 in its domain. The expansion of 1+kx (i.e. the function ##x\mapsto 1+kx##) around 0 (the point I'd assume you have in mind unless you specify another one) is just 1+kx.
oh... do you mean since 1+kx is a straight line, so at the point 0, it is just 1 + kx ?

so if i had a curve, would taylor expansion ( approximation? ) make more sense here?

that means the first term of the expansion gives me a straight line, the 2nd makes it more curve, the 3rd makes it even more like the original curve function?

does it then mean i cannot taylor expand things like 1+kx? i have to expand only curves?
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Apr2-12, 07:34 AM
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Quote Quote by quietrain View Post
oh, we can just define a as x?

does it mean i can define a as -kx or -k also?

but i thought a is the point that we evaluate the taylor expansion on?

am i right to say if i taylor expand on a point say x =1 for a curve graph. then as my orders of taylor expansion become greater, the approximation is better too?
"x" and "a" are just letters.
You can replace them by anything you want, as long as you do so consistently and do not mix letters up.
(Note that "x" has 2 different meanings in your problem statement. That's why I introduced "y" - to eliminate one of the two.)

In your original formula f(x)=f(a)+f'(a)(x-a)+...
the expansion is around "a".

In the formula f(x-kx)=f(x)+f'(x)((x-kx)-x)+...= f(x) - kx f'(x) + ...
the expansion is around "x".
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Apr2-12, 08:20 AM
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Quote Quote by quietrain View Post
does it then mean i cannot taylor expand things like 1+kx? i have to expand only curves?
If you expand 1+kx around x=0, you are effectively defining f(x)=1+kx, and a=0.
So using your formula to expand it, you get:

f(x)=1+kx
f'(x)=k
f''(x)=0
f'''(x)=0

So:
f(x)=f(a)+f'(a)(x-a)+1/2! f''(a)(x-a)2+...
f(x)=f(0)+f'(0)(x-0)+1/2! f''(0)(x-0)2+...
f(x)=1+k(x-0)+1/2! 0.(x-0)2+ 0 + ...
f(x)=1+kx

Hey! But you already had that! :)
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Apr2-12, 02:47 PM
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Quote Quote by quietrain View Post
i am very confuse how my profs always use taylor expansion in physics which somehow doesn't follow the general equation of

f(x) = f(a) + f'(a)(x-a) + 1/2! f''(a)(x-a)2 and so on...

like for example, what is the taylor expansion of x - kx where k is small

it was given as something like

x - kx f'(x) + (1/2) k2 x2 f''(x) + ...

is this taylor expansion? but there is no 'about which point, i.e, a=? '

i don't even understand how the first term x is gotten. f(a) = x??

please help thank you!
I think your expression was copied wrong, or had a misprint: you should have written
f(x) - kx f'(x) + (1/2) k2 x2 f''(x) + ... [with f(x) as the first term, not just x]. You did say the expression was "something like...", which says to me that you were not sure.

RGV
quietrain
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#14
Apr3-12, 08:21 PM
P: 651
alright thanks everyone!


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