Confused about taylor approximation

[LostInSpace]

I am a bit confused about taylor approximation. Taylor around x_0 yields
<br /> f(x) = f(x_0) + f&#039;(x_0)(x-x_0) + O(x^2)<br />

which is the tangent of f in x_0, where
<br /> f&#039;(x) = f&#039;(x_0) + f&#039;&#039;(x_0)(x-x_0) + O(x^2)<br />

which adds up to
<br /> f(x) &amp;=&amp; f(x_0) + (f&#039;(x_0) + f&#039;&#039;(x_0)(x-x_0) + O(x^2))(x-x_0)+O(x^2) \\ &amp;=&amp; f(x_0) + f&#039;(x_0)(x-x_0) + f&#039;&#039;(x_0)(x-x_0)^2 + O(x^3)<br />
But it should be
<br /> f(x) = f(x_0) + f&#039;(x_0)(x-x_0) + \frac{f&#039;&#039;(x_0)}{2!}(x-x_0)^2 + O(x^3)<br />

Where does the 2! come from? Is this approach completely incorrect?

 

[arildno]

You have ignored in line 3 the O(x^(2))-term from the expansion of f(x).
Hence line 3 is not accurate to O(x^(3)), it's only accurate to O(x^(2)).

 

[Hurkyl]

Do you remember how to derive, from the limit definition of the derivative, the differential approximation formula:

<br /> f(x+\epsilon ) = f(x) + \epsilon f&#039;(x) + \epsilon \delta(x, \epsilon)<br />

Where \lim_{\epsilon \rightarrow 0} \delta(x, \epsilon) = 0?

Try writing the second derivative with limits, and see if any approach suggests itself.