# Nth Derivatives and Taylor Polynomials

## 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 defintion 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.