Nth Derivatives and Taylor Polynomials

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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 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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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