# Quadratic approximation of a function

Here en.wikipedia.org/wiki/Taylor_theorem i have found that linear approximation of f at the point a is

$$P_1(x)=f(a)+f'(a)(x-a)$$

For the quadratic approximation the quadratic polynomial is

$$P_2(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2}(x-a)^2$$

Please explain me how did we get the 1/2 multiplier near the second derivative.

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It comes from the fact that $f''(x)$ at $a$ should come to $f''(a)$ even if we go by the expansion instead of just by the definition. If we put

$f(x) = a_0 + a_1(x-a) + a_2(x-a)^2 + a_3(x-a)^3 + ...$

then

$f''(x) = 2a_2 + 6a_3(x-a) + ...$

Evaluating at $x = a$, $f''(a) = 2a_2$. So $a_2 = f''(a)/2$. So basically the $2$ here comes from the multiplier of $2$ for $xdx$ when differentiating $x^2$.

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