# Is this form of Taylor's theorem correct? Güler's Foundations of Optimization theorem 1.3

Let $$f: I = (c, d) \xrightarrow{} \mathbb{R}$$ be an $$n-$$times differentiable function. Let $$a, b$$ be distinct points in $$I$$. Then

$$f(b) = f(a) + f'(a)(b-a) + ... + \frac{f^{(n-1)}(a)}{(n-1)!}(b-a)^{n-1} + \int_{a}^{b} \int_{a}^{s_{1}} ... \int_{a}^{s_{n-1}} f^{(n)}(s_{n}) ds_{n}...ds_{1}.$$

Proof. By the Fundamental Thm of Calculus, we have $$f(b) = f(a) + \int_{a}^{b}f'(s_{1})ds_{1}$$. Similarly, $$f'(s_{1}) = f'(a) + \int_{a}^{s_{1}}f''(s_{2})ds_{2}$$, which gives

$$f(b) = f(a) + \int_{a}^{b}f'(s_{1})ds_{1} = f(a) + \int_{a}^{b}(f'(a) + \int_{a}^{s_{1}}f''(s_{2})ds_{2})ds_{1} = f(a) + f'(a)(b-a) + \int_{a}^{b} \int_{a}^{s_{1}} f''(s_{2}) ds_{2} ds_{1}.$$

Continuing this fashion, we obtain $$f(b) = f(a) + f'(a)(b-a) + f''(a)\int_{a}^{b} \int_{a}^{s_{1}} \int_{a}^{s_{2}} ds_{3}ds_{2}ds_{1} +...+ f^{(n-1)}(a)\int_{a}^{b}\int_{a}^{s_{1}} ... \int_{a}^{s_{n-1}}ds_{n}...ds_{1} + \int_{a}^{b}\int_{a}^{s_{1}}...\int_{a}^{s_{n-1}}f^{(n)}(s_{n})ds_{n}...ds_{1}.$$

...

My question is, in the last step the author obviously used $$f^{(n-1)}(s_{n-1}) - f^{(n-1)}(a) = \int_{a}^{s_{n-1}}f^{(n)}(s_{n})ds_{n}$$, but why is this true? The Fundamental theorem says if $$f$$ is integrable on the interval $$[a, b]$$ and $$F' = f$$ there, then $$\int_{a}^{b} f dx = F(b) - F(a)$$. But here, $$f^{(n)}$$ may not be integrable, so how do we get $$f^{(n-1)}(s_{n-1}) - f^{(n-1)}(a) = \int_{a}^{s_{n-1}}f^{(n)}(s_{n})ds_{n}$$?

Should we assume $$f^{(n)}$$ is continuous or at least Riemann integrable to apply the Fundamental Thm of Calculus?

I suspect the author of this book may be sloppy with hypotheses.

Yes, it is standard to state such results with the assumption that $$f\in C^n([a,b])$$.

You can certainly give counterexamples. It's easiest to give the counterexample for $$n=1$$. If you take $$f(x) = \begin{cases} x^2\sin(1/x^2), & x\ne 0 \\ 0, & x=0\end{cases},$$ then $$f$$ is differentiable but $$f'$$ is unbounded on $$[0,1]$$, and so the Riemann integral of $$f'$$ is not defined.

• I am convinced now that the author is being sloppy. In Lemma 1.17 the author lets $M >$ the supremum of something without showing the supremum is finite...
– Tom
Jan 15 at 17:29
• Maybe the author is more of an engineer :D Jan 15 at 17:38
• This is shocking. But this textbook is a GTM, how come...
– Tom
Jan 15 at 19:33
• There are plenty of mathematicians who write horribly, not just research papers, but also books. Jan 15 at 19:34
• Glad that I have you to help with my questions. It's only the introductory chapter. I just hope the next chapters are not like this. And thanks a ton for your help!
– Tom
Jan 15 at 19:49