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Nth Derivatives and Taylor Polynomials

  1. Sep 16, 2013 #1
    1. The problem statement, all variables and given/known data
    Show that if [itex]f^{(n)}(x_0)[/itex] and [itex]g^{(n)}(x_0) [/itex] exist and
    [itex] \lim_{x \rightarrow x_0} \frac{f(x)-g(x)}{(x-x_0)^n} = 0 [/itex] then
    [itex]f^{(r)}(x_0) = g^{(r)}(x_0), 0 \leq r \leq n [/itex].

    2. Relevant equations
    If f is differentiable then [itex] \lim_{x \rightarrow x_0}\frac{f(x)-T_n(x)}{(x-x_0)^n}=0 [/itex], where Tn is the nth Taylor polynomial.


    3. The attempt at a solution I'm stuck on how to start the proof at all. I tried induction on r, but didn't get very far with that since I had trouble showing that [itex]f(x_0) = g(x_0)[/itex]. Any ideas on what direction to go to get started?

    Thanks.
     
    Last edited by a moderator: Sep 16, 2013
  2. jcsd
  3. Sep 17, 2013 #2
    Let's get you started here. First, you can regard f(x) and g(x) as the 0th derivatives, so that [tex] lim_{x \rightarrow x_0} \frac{f(x) - g(x)}{(x-x_0)^0} = 0 \Rightarrow f(x_0) -g(x_0) = 0[/tex] so we have [tex]f(x_0) = g(x_0)[/tex]

    For the first derivatives we have [tex] \frac {f(x)-f(x_0) -(g(x) - g(x_0))}{x-x_0} = \frac {f(x) -(g(x) }{x-x_0} \rightarrow 0 \hspace{50px}(1)[/tex]

    But the limit in (1) also goes to [tex]f'(x_0) - g'(x_0)[/tex]

    We conclude that [tex]f'(x_0) = g'(x_0)[/tex]

    If you were going to use this method to prove the result for f'' and g'' you could repeat the above argument for f' and g' instead of f and g. Does this tell you how to do the induction step?
     
  4. Sep 17, 2013 #3
    I understand how this works for the 0th derivative and first derivative, but I am having trouble seeing how it would work for the second and higher derivatives. What confuses me is the exponent on [itex](x-x_0)^n[/itex].

    For the second derivative I am writing[itex]\frac{f'(x)-f'(x_0)-(g'(x)-g'(x_0))}{(x-x_0)}[/itex] and using [itex]f'(x)=g'(x)[/itex] and using the defintion of the second derivative, but then I don't have [itex](x-x_0)^2[/itex] in the denominator or and f(x) or g(x) on top. Do I have to multiply through by something to get those to appear?
     
  5. Sep 17, 2013 #4
    Ah, actually I think I see how those appear. I broke down the first derivative the one step further into limit form.
     
  6. Sep 17, 2013 #5
    Sorry, I wrote this carelessly. When you go back to the definition of f' and g' your original hypothesis should appear. If you can show that [tex]f''(x_0) = g''(x_0)[/tex] then you can see how to do the induction step.
     
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