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Alright, so I was messing around with the linear approximation for a function; I wanted to see if I could get the formula for the 2nd degree approximation from it. I went about it like this:

$$ f(x + \Delta x) \approx f(x) + f'(x) \Delta x $$ $$ \Rightarrow f'(x + \Delta x) \approx f'(x) + f''(x) \Delta x $$ $$ \frac{f(x + 2\Delta x)-f(x + \Delta x)}{\Delta x} \approx f'(x) + f''(x)\Delta x $$ $$ f(x + 2\Delta x) \approx f(x + \Delta x) + f'(x)\Delta x + f''(x)\Delta x^2 $$ $$ f(x + 2\Delta x) \approx \left[f(x) + f'(x)\Delta x \right] + f'(x)\Delta x + f''(x)\Delta x^2 $$ $$ f(x + 2\Delta x) \approx f(x) + 2f'(x)\Delta x + f''(x)\Delta x^2 $$

Then I do the replacement: $ h = 2\Delta x $:

$$ f(x + h) \approx f(x) + 2f'(x)\frac{h}{2} + f''(x)\left(\frac{h}{2}\right)^2 $$ $$ f(x + h) \approx f(x) + f'(x)\cdot h + f''(x) \cdot \frac{h^2}{4} $$

Which just seems outright wrong. Shouldn't I be getting the 2nd degree Taylor polynomial? I guess one of the assumptions I made was incorrect?

Anyway, thanks for reading through if you did. Any helpful pointer would be very much appreciated.

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  • $\begingroup$ Why do you say $\frac{f(x + 2\Delta x)-f(x + \Delta x)}{\Delta x} \approx f'(x) + f''(x)\Delta x?$ $\endgroup$ – mfl Mar 14 '15 at 22:54
  • $\begingroup$ There I did $ f'(x + \Delta x) \approx \frac{f(x + \Delta x + \Delta x) - f(x + \Delta x)}{\Delta x} $; I treated $ f(x + \Delta x ) $ as an arbitrary function and approximated its derivative $\endgroup$ – shost71 Mar 14 '15 at 23:01
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In the approximation

$$f(x + 2\Delta x) \approx f(x + \Delta x) + f'(x)\Delta x + f''(x)\Delta x^2$$

you are considering terms up to the second order. Thus, the problem appears when you approach

$$f(x + \Delta x) \approx f(x) + f'(x) \Delta x.$$ This is correct, but it is an approximation up to first order. So, when you put together both approximations there is a missing term. If you approach

$$f(x + \Delta x) \approx f(x) + f'(x) \Delta x+\frac{f''(x)}{2}(\Delta x)^2$$ the problem dissapears. This expression is exactly what you were looking for. In conclusion, the problem is the use of approximations up to different order.

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  • $\begingroup$ I see! That's it, exactly what I was looking for. Thank you very much :D $\endgroup$ – shost71 Mar 15 '15 at 0:45
  • $\begingroup$ You're welcome. $\endgroup$ – mfl Mar 15 '15 at 0:46

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