Help making sense of Taylor series using MVT only

Question

I know the coefficients in Taylor series are cooked up to be equal to the derivatives of the original function. But it's still a bit vague to me how information at one single point give the entire function... so I'm trying to derive it using MVT. Kindly have a look at the attached picture.

original image

From my attempt so far, I don't see any way of getting factorial terms in the denominators. Although it seems $\theta_i = \frac{1}{i+1}$ should fix the problem, but this proof looks beyond my depth. Any help?

If I apply MVT again on $f''$ : $$f(a+h) = f(a) + hf'(a) + h^2\theta_1 f''(a) + h^3\theta_1\theta_2f'''(a+h\theta_1\theta_2\theta_3) $$ for some $\theta_i \in (0,1)$

My question is how to get rid of these $\theta_i$...

Answer 1

Consider the defect in how good you can guess $f(a)$ from $x$ using function value and the first $k$ derivatives in the Taylor polynomial (that you get by considering coordinate shifts in polynomials): $$ \phi_a(x)=f(x)-f(a)+f'(x)(a-x)+\frac12f''(x)(a-x)^2+...+\frac1{k!}f^{(k)}(x)(a-x)^k. $$ Obviously, $\phi_a(a)=0$.

This function has a derivative where almost everything cancels \begin{align} \phi_a'(x)&=f'(x)+[f''(x)(a-x)-f'(x)]+\left[\frac12f'''(x)(a-x)^2-f''(x)(a-x)\right]+...+\left[\frac1{k!}f^{(k+1)}(x)(a-x)^k-\frac1{(k-1)!}f^{(k)}(x)(a-x)^{k-1}\right] \\ &=\frac1{k!}f^{(k+1)}(x)(a-x)^k \end{align}

Now apply the extended mean value theorem $$ \frac{\phi_a(x)-\phi_a(a)}{(a-x)^m-(a-a)^m}=\frac{\phi_a'(\tilde x)}{-m(a-\tilde x)^{m-1}}=-\frac1{m\cdot k!}f^{(k+1)}(\tilde x)(a-\tilde x)^{k+1-m}. $$

Now rename $x=x_0$, then $a=x$ to get $$ f(x)=f(x_0)+f'(x_0)(x-x_0)+\frac12f''(x_0)(x-x_0)^2+...+\frac1{k!}f^{(k)}(x_0)(x-x_0)^k + \frac1{m\cdot k!}f^{(k+1)}(\tilde x)(x-\tilde x)^{k+1-m}(x-x_0)^m $$

Answer 2

"I know the coefficients in Taylor series are cooked up to be equal to the derivatives of the original function. But it's still a bit vague to me how information at one single point give the entire function..."

It doesn't! You also need the information that the entire function is "analytic". That is, that it can be written as a power series and that the power series is equal to the function. For example, the function equal to $e^{-1/x^2}$ if $x\ne 0$, f(0)= 0 is infinitely differentiable and the value or every derivative is 0 at x= 0. So the "Taylor series" at x= 0 exists- and is identically 0. But f(x) is not equal to 0 anywhere except at x= 0!