Application of the mean value theorem to a Taylor series

I am working on a practice problem and there is step in the solution that deals with the application of the mean value theorem (MVT) in a Taylor series. The problem is asking for a condition on $f''(x)$ s.t. $\{(x,y)\in\mathbb{R}:y\ge f(x)\}$ is convex if $f:\mathbb{R}\rightarrow\mathbb{R}$ and $f$ is twice differentiable.

Taking the Taylor series up to the second term and applying $y\ge f(x)$ as is given in the problem statement gives $$y\ge f(x)=f(a)+f'(a)(x-a)+f''(a)\frac{(x-a)^2}{2!}$$ but the solution instead proceeds to the following step instructing the reader to use the MVT $$y\ge f(x)=f(a)+f'(a)(x-a)+f''(a^*)\frac{(x-a)^2}{2!},a^*\in(a,x).$$ This seems to be indicating to me that $f''(a)=f''(a^*)$ but since $a^*\in(a,x)$ not $\in[a,x]$ I don't see how this can be true. My understanding of the MVT simply says that if $f$ is differentiable over $(a,b)$ and cts. over $[a,b]$ then there exists a $f'(c)=\frac{f(b)-f(a)}{b-a},c\in(a,b)$.

Am I incorrect in my understanding of this application of the MVT?

• First you should note that your equation $f(x)=f(a)+f'(a)(x-a)+f''(a)\frac{(x-a)^2}{2!}$ is simply not right - it implies that $f$ is a polynomial. – David C. Ullrich Dec 30 '17 at 18:29
• @DavidC.Ullrich is it true if $x$ is a specific point on $\mathbb{R}$? If it's not true then the solution manual is incorrect because that's how they write it. – WnGatRC456 Dec 30 '17 at 18:33
• No. And that's not how they write it, at least not according to what you said: They wrote $f''(a^*)$, not $f''(a)$. – David C. Ullrich Dec 30 '17 at 18:36
• Because the value of $a^*$ depends on $x$. For every $x$ there exists $a^*$ such that right version is true... – David C. Ullrich Dec 30 '17 at 18:59
• Perhaps a different letter could have been used instead of $a^{\text{*}}$ otherwise it looks so similar to $a$ and creates confusion. Some textbooks even use subscript like $a_{x}$ in place of $a^{\text{*}}$ to indicate the dependence on $x$. – Paramanand Singh Dec 31 '17 at 5:44

It may be relevant to note that there exists a version of Taylor's Theorem that requires the existence of $f^{(k)}$ at only one point. Should be better known than it seems to be, in any case:

Theorem If $f:\mathbb R\to\mathbb C$, $k\ge1$, and $f^{(k)}(a)$ exists then $$f(x)=P_{a,k}(x)+o((x-a)^k)\quad(x\to a).$$

Thanks to Paramanand Singh: It appears that this is known as Taylor's Theorem with Peano's form of the remainder.

Here of course $P_{a,k}(x)=\sum_{j=0}^k\frac{f^{(j)}(a)}{j!}(x-a)^j$.

(Note that the hypothesis implies that $f^{(j)}$ exists in some neighborhood of $a$ for $j<k$.)

The conclusion means by definition that $f(x)=P_{a,k}(x)+E(x)$ where $$\lim_{x\to a}\frac{E(x)}{(x-a)^k}=0.$$

It's enough to prove this:

If $f:\mathbb R\to\mathbb R$, $k\ge1$, and $f(0)=f'(0)=\dots=f^{(k)}(0)=0$ then $f(x)=o(x^k)$ as $x\to0$.

The proof is by induction on $k$. For $k=1$ this is just the definition of the derivative: $$0=f'(0)=\lim_{x\to0}\frac{f(x)}x,$$which says exactly that $f(x)=o(x)$.

Now suppose that $k>1$. Note that $g=f'$ satisfies the hypothesis with $k-1$ in place of $k$, so by induction we have $$f'(x)=o(x^{k-1}).$$

Now MVT shows that if $x\ne0$ (and $x$ is small enough that we're inside the interval about $a$ where $f$ is differentiable) there exists $\xi$ with $|\xi|<|x|$ such that $$\frac{f(x)}x=f'(\xi).$$ So $$f(x)=xf'(\xi)=o(x^k)$$(since $|\xi|\le|x|$).

• That's another novel proof +1. I had asked a question in past to get various proofs for this theorem. See math.stackexchange.com/q/1809293/72031 – Paramanand Singh Dec 31 '17 at 16:55
• Can you explain what $E(x)$ represents here? – WnGatRC456 Jan 1 '18 at 1:48
• @NWernerC456 $E(x)$ is the error in the Taylor approximation. It's defined by the equation $f(x) = P(x) + E(x)$. – David C. Ullrich Jan 1 '18 at 2:37
• Ok so it's just the remainder then, or the higher order terms. – WnGatRC456 Jan 1 '18 at 2:38