# Help making sense of Taylor series using MVT only

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.

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 $$\theta_i$$...

• Rewrite the MVT in terms of an integral: $$\int_a^b f(t) dt = (b-a)f(\xi)$$ for $a\leq\xi\leq b$ and use repeated partial integration on the fundamental theorem of calculus. – Count Iblis Jul 13 at 17:39
• [quote]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...[/quote] – user247327 Jul 13 at 19:26

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 $$f(x)=f(x_0)+f'(x_0)(x-x_0)+\frac12f''(x_0)(x-x_0)^2+...+\frac1{k!}f^{(k)}(x_0)(x-x_0)^k + \frac1{m\cdot k!}f^{(k+1)}(\tilde x)(x-\tilde x)^{k+1-m}(x-x_0)^m$$

• Exactly what I have been looking for. I get till $${\phi_a}'(x) = \frac1{k!}f^{(k+1)}(x)(a-x)^k$$. Pretty clever way to use telescoping! I'll look up extended mean value theorem and get back. Thank you so much :) – rsadhvika Jul 14 at 18:52
• I've seen it first in an "calculus for engineers" book that my father used in the 1960ies, the idea is certainly older. – LutzL Jul 14 at 18:59
• If I understand correctly you're choosing these two functions for the extended MVT : $$f(t) = \phi_a(t)$$ $$g(t) = (a-t)^m$$ then letting $m = k+1$ should fix the remainder term. Looks I get this. Your father sounds like a great man and you're definitely lucky to still have access to his books! Old is indeed gold! Thanks again for sharing this cool telescoping trick :) It made my day XD – rsadhvika Jul 14 at 19:59

"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!