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Homework Help: Taylor expansion in physics

  1. Apr 1, 2012 #1
    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!
  2. jcsd
  3. Apr 1, 2012 #2


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    hi quietrain! :smile:
    hmm :rolleyes: … but not exactly like!

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

    (and k = x - a)
  4. Apr 1, 2012 #3

    I like Serena

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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, :smile:TM got here first!
  5. Apr 1, 2012 #4


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    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 :smile:).
  6. Apr 2, 2012 #5

    Ray Vickson

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    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.

  7. Apr 2, 2012 #6
    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) + ...

    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?

    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?

  8. Apr 2, 2012 #7

    I like Serena

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    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.

    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) + ...
    Last edited: Apr 2, 2012
  9. Apr 2, 2012 #8


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    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.
  10. Apr 2, 2012 #9
    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?
  11. Apr 2, 2012 #10
    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?
  12. Apr 2, 2012 #11

    I like Serena

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    "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".
  13. Apr 2, 2012 #12

    I like Serena

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    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)=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 + ...

    Hey! But you already had that! :)
  14. Apr 2, 2012 #13

    Ray Vickson

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    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.

  15. Apr 3, 2012 #14
    alright thanks everyone!
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