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Proof of Taylor's formula for polynomials

  1. Apr 12, 2009 #1
    My book reads as follows:

    "This is an entirely elementary algebraic formula concerning a polynomial in x or order n, say
    [tex] f(x) = a_0 + a_1x + a_2x^2 + ... + a_nx^n [/tex].

    If we replace x by a + h = b and expand each term in powers of h, there results immediately a representation of the form [tex] f(a+h) = c_0 = c_1h + c_2h^2 + ... + c_nh^n [/tex].

    Taylor's formula is the relation: [tex] c_v = \frac{1}{v!}f^v(a) [/tex], for the coefficients c_v in terms of f and its derivatives at x = a. To prove this fact we consider the quantity h = b - a as the independent variable, and apply the chain rule which shows that differentiation with respect to h is the same as differentiation with respect to b = a + h."

    I don't get how f'(h) = f'(a + h)?
    Last edited: Apr 12, 2009
  2. jcsd
  3. Apr 12, 2009 #2
    I think it's saying that if you have f'(h) and want f'(h+a), just substitute in h+a where you see h to get f'(h+a). This works because [tex]\frac{d}{dh} h+a = 1[/tex]
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