2nd order Taylor expansion of 2nd symmetric derivative

I'm trying to derive "intuitively" the approximation $$ f(x)\approx f(a)+ f'(a)(x-a) + \frac{1}{2}f''(a)(x-a)^2 $$ from the second symmetric derivative formula $$ f''(a)=\lim_{x\to a} \frac{f(a+h)+f(a-h)-2f(a)}{h^2}. $$ My goal is to obtain the same "derivation" as $$ f(x)\approx f(a)+f'(a)(x-a) $$ from the definition of the derivative.

Using $h=x-a$, I can rewrite this as $$ f''(a)=\lim_{x\to a} \frac{f(x)+f(2a-x)-2f(a)}{(x-a)^2} $$ Which yields

$$ \begin{aligned} f''(a)&\approx \frac{f(x)+f(2a-x)-2f(a)}{(x-a)^2} \\ f''(a)(x-a)^2&\approx f(x)+f(2a-x)-2f(a) \\ f(x)&\approx 2f(a)-f(2a-x)+f''(a)(x-a)^2 \end{aligned} $$ Using a first-order Taylor expansion, we have

$$ \begin{aligned} f(2a-x)&\approx f(a)+f'(a)(a-x) \\ &\approx f(a)-f'(a)(x-a) \end{aligned} $$ We thus have

$$ \begin{aligned} f(x)\approx f(a) + f'(a)(x-a)+f''(a)(x-a)^2 \end{aligned} $$ Which is almost right, except for the missing $1/2$ factor. I'm not sure what I'm missing here and why I can't recover the proper formula. Could you help me?

Thanks!

You can't use a first order Taylor expansion in the middle if you want a second order expansion. That is what's completely messing you up. You would get the correct answer if you did \begin{align} f(2a-x)&\approx f(a)+f'(a)(a-x) +\frac{1}{2}f''(a)(a-x)^2\\ &=f(a)-f'(a)(x-a)+\frac{1}{2}f''(a)(x-a)^2. \end{align} I strongly suggest you read the answer I wrote regarding small-angle approximations. There, the OP also gets an incorrect answer, because they neglected higher order contributions.

To be precise, let's not use the $\approx$ symbol. Rather let us write out explicitly the order to which the equation is true. The equation \begin{align} f''(a)&=\lim_{x\to a}\frac{f(x)+f(2a-x)-2f(a)}{(x-a)^2} \end{align} can be written as \begin{align} f''(a)(x-a)^2 + o((x-a)^2)&= f(x)+f(2a-x)-2f(a) \end{align} Here, $o((x-a)^2)$ refers to any function $\rho$ such that $\lim\limits_{x\to a}\frac{\rho(x)}{(x-a)^2}=0$. Rearranging this, we see that \begin{align} f(x)&=2f(a)-f(2a-x)+f''(a)(x-a)^2+o((x-a)^2) \end{align} If you now decide to use only a first order expansion for $f(2a-x)$, then \begin{align} f(x)&=2f(a)-[f(a)+f'(a)(a-x)+o((x-a))] + f''(a)(x-a)^2+o((x-a)^2)\\ &=f(a)+f'(a)(x-a)+o((x-a)). \end{align} The fact that you used a first order expansion for one of the terms means you no longer have any knowledge regarding second order contributions. So your error was in claiming that the second order term is $f''(a)(x-a)^2$. I drive this point home in my linked answer above, in a much simpler context, so I strongly suggest you read it carefully.

It's kind of like rounding off a decimal too early, and then using that imprecise rounded off value and doing further calculations. You'll then end up with a not-so correct answer. For example, suppose I give you the number $y=1.23456789$, and I ask you to calculate $y^2$ up to two decimal places. Well, the exact answer is $y^2=1.52415787502$, so the answer correct to two decimal places is $1.52$. However, if you approximate $y$ too crudely, say $y\approx 1.2$ and then you square it you'll say $y^2\approx 1.44$, but obviously $1.44$ is not the first two decimal approximation.

Also, the symmetric second derivative formula is derived using the second order Taylor polynomial, which is why there is no reason to expect the reasoning to go the opposite direction.