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Taylor formula. Need help

  1. Apr 15, 2007 #1
    Taylor formula. Need help!!

    we have f(a+h), where a-is a point, and h is a very small term, h->0. And we have the formula to evaluate the function y=f(x), around the point a, which is

    f(a+h)=f(a)+f'(a)h+o(h) --------(*)

    however when we want to take in consideration o(h) this formula does not work, so we want to come up with sth more appropriate. So we want to write it in terms of the value of a polynom of the form

    P_n(h)=b_1+b_2h+b_3h^2+.....+b_n h^n,where b-s are coeficients, that are not depended on h.
    So it also says that when h->0, than
    When we have n=1, than the value of the polynom is f(a)+f'(a)h, according to (*). What i do not understand comes right here, when we want to generalize this for n.
    It says that if the n-th derivative of f(x) exists, in particular f^(n) (a), than the polynom can be writen like this, or actually the polynom is:

    P_n(h)=f(a)+[f'(a)h/1!]+[f"(a)h^2/2!]+.....+[(f^(n) (a)/n!)h^n],

    Now my question is how did we come to this? In particular wher did we, or how did we derive this coefficients like f'(a)/1! , f"(a)/2! etc???

    I would really appreciate anyones help!!
    For two days i am stuck with this, i cannot just fathom it.

    Thnx in return
  2. jcsd
  3. Apr 15, 2007 #2

    matt grime

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    We did it by presuming that such an expansion exists, then trying to work out how to get the terms in the expansion.


    [tex] f(x)=a_0 + a_1 x + a_2 x^2 + \ldots +a_n x^n + \ldots[/tex]

    so an infinte series of powers. How do you find a_0, now how aobut a_1?
  4. Apr 16, 2007 #3

    I guess it should be easy to find those terms,but i must be too stupid because i cannot find them. Moreover i have no idea on how to begin.

    And if you could suggest me where i could find some material about these things??

  5. Apr 16, 2007 #4
    or, i have an idea on how to find those terms. I think it has to do something with the binomial expansion, using Newton's formula right?

    so, let's say the a_0 term will be n!/n!*0! multiplied by something, or the a_1 will be n!/(n-n+1)!*(n-1)! multiplied by sth.

    I do not really know if i am on the right track,i doubt it though.

    Can you give me some more help!!
  6. Apr 16, 2007 #5
    You have

    [tex] f(x)=a_0 + a_1 x + a_2 x^2 + \ldots +a_n x^n + \ldots[/tex]

    What happens if you differentiate both sides of that equation?
  7. Apr 16, 2007 #6
    (I copied and pasted this from another reply I made in a different thread last week)

    This is the way I think about it.

    imagine that a function f(x) can be written as a power series in x about x=0

    [itex] f(x)=A_0 +A_1 x+A_2 x^2+A_3 x^3 +.....[/itex]

    Then you can find the constants [itex]A_0,A_1,A_2...[/itex] from repeated differentiation.

    Finding A0 is the easiest


    because, when x=0 all the other terms are zero.

    Next, find A1

    So, at x=0
    [itex]\frac{df(x)}{dx} (x=0)=A_1[/itex]

    Then find A2


    When x=0

    And so on.

    You now substitute in the values of A0,A1,A2 etc to get
    [itex]f(x)=f(0)+\frac{df}{dx} x +1/2 \frac{d^2f}{dx^2} x^2+1/6 \frac{d^2 f}{dx^3} x^3+...[/itex]

    where the derivatives are evaluated at (x=0)
  8. Apr 16, 2007 #7

    matt grime

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    Didn't you think about what you wrote in your own first post? You know that to find the n'th term you differentiate n times and put in 0. You said that. All I wanted to show you was why that 'worked'.
  9. Apr 24, 2007 #8


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    But...to develop a Tayor series..must f(x) be 'strictly' differentiable ??.. i mean perhaps what would happen if for example the n-th derivative of f(x) exist but only in the sense of a 'distribution' ?? let's say:

    [tex] \frac{d^{n} f(x)}{dx^{n}}= \delta (x) ^{n}=(-1)^{n}\phi (0) [/tex]
  10. Apr 24, 2007 #9

    matt grime

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    For pity's sake Jose, won't you stop polluting other people's threads?
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