Nth Derivatives and Taylor Polynomials

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SUMMARY

The discussion focuses on proving that if the nth derivatives of functions f and g exist at a point x0 and the limit of their difference normalized by (x - x0)^n approaches zero, then their derivatives up to order n are equal at that point. The proof begins by establishing that f(x0) = g(x0) using the limit definition for the 0th derivative. It continues with the first derivative, showing that f'(x0) = g'(x0) through a similar limit approach. The participants discuss extending this argument to higher derivatives, specifically addressing the challenges in applying induction for the second derivative and beyond.

PREREQUISITES
  • Understanding of limits and continuity in calculus
  • Familiarity with derivatives and their definitions
  • Knowledge of Taylor polynomials and their properties
  • Experience with mathematical induction techniques
NEXT STEPS
  • Study the formal definition of Taylor polynomials and their applications
  • Learn about mathematical induction and its use in proofs
  • Explore the concept of limits in calculus, particularly in relation to derivatives
  • Investigate higher-order derivatives and their significance in function analysis
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Students studying calculus, particularly those focusing on derivatives and Taylor series, as well as educators looking for effective proof strategies in mathematical analysis.

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Homework Statement


Show that if f^{(n)}(x_0) and g^{(n)}(x_0) exist and
\lim_{x \rightarrow x_0} \frac{f(x)-g(x)}{(x-x_0)^n} = 0 then
f^{(r)}(x_0) = g^{(r)}(x_0), 0 \leq r \leq n.

Homework Equations


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

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 f(x_0) = g(x_0). Any ideas on what direction to go to get started?

Thanks.
 
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Let's get you started here. First, you can regard f(x) and g(x) as the 0th derivatives, so that lim_{x \rightarrow x_0} \frac{f(x) - g(x)}{(x-x_0)^0} = 0 \Rightarrow f(x_0) -g(x_0) = 0 so we have f(x_0) = g(x_0)

For the first derivatives we have \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)

But the limit in (1) also goes to f'(x_0) - g'(x_0)

We conclude that f'(x_0) = g'(x_0)

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?
 
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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 (x-x_0)^n.

For the second derivative I am writing\frac{f'(x)-f'(x_0)-(g'(x)-g'(x_0))}{(x-x_0)} and using f'(x)=g'(x) and using the definition of the second derivative, but then I don't have (x-x_0)^2 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?
 
Ah, actually I think I see how those appear. I broke down the first derivative the one step further into limit form.
 
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 f''(x_0) = g''(x_0) then you can see how to do the induction step.
 

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