Conditions of the Taylor Theorem

Question

I'm confused on the assumptions behind the Taylor Theorem because I found different versions of them across several books.

Consider the function $f:\mathbb{R}\rightarrow \mathbb{R}$

(1) If and only if $f$ is infinitely many times differentiable at $a$ I can write $$f(x)=f(a)+\sum_{k=1}^{\infty}\frac{f^{(k)}(a)(x-a)^k}{k!}$$ Correct?

(2) If and only if $f$ is $n$ times continuously differentiable at $a$ (which implies that $f$ is $n$ times differentiable in a neighbourhood of $a$) I can write $$ f(x)=f(a)+\sum_{k=1}^{n}\frac{f^{(k)}(a)(x-a)^k}{k!}+ o(||x-a||^n) $$ Correct?

(3) If and only if $f$ is $n$ times continuously differentiable at each point between $x$ and $a$ I can write $$ f(x)=f(a)+\sum_{k=1}^{n}\frac{f^{(k)}(a)(x-a)^k}{k!}+ \frac{f^{(n+1)}(c)(x-a)^{n+1}}{(n+1)!} $$ for $c$ between $x$ ans $a$. Correct?

My confusion is related in particular to the necessity of conditions.

Thanks, yes, I agree but I don't know how to formalise it. Do you have any source explaining that? I have never found a version of the Taylor Theorem putting restrictions on $x$. — Star, Dec 10 '15 at 14:51
Each statement starts with "If and only if...", what are you trying to say that the statements are equivalent to? — charlestoncrabb, Jan 15 '16 at 03:40
One reason you don't see necessary conditions (i.e., properties that "a polynomially-approximable function must satisfy") is that a function can vanish to infinite order at a point (i.e., be approximated to infinite order at some point by the zero function) without even being measurable: Multiply Chappers' example by the characteristic function of a dense, "nowhere-measurable" set. — Andrew D. Hwang, Jan 17 '16 at 17:08

score 5 · Accepted Answer · edited Apr 13 '17 at 12:21

I suspect you need to think more about how much $C^\infty$ functions can be wilder than analytic functions.

(1) False. A $C^\infty$ function need not be faithfully represented by its power series at a point. As @Chappers observes in a comment, $\mathrm{e}^{-1/x^2}$ is not faithfully represented by its power series at $0$. For a little more discussion on this example, see Why doesn't the identity theorem for holomorphic functions work for real-differentiable functions?.

(2) False. Let $n$ be given, set $N > n$, and let $H(x)$ be the (Heaviside) step function. Let $g(x) =(1+H(x))^N - 1$. Integrate $g$ $n$ times. The result is $n$ times continuously differentiable, but your error estimate is hopeless. (Want to violate it more? Increase $N$.) The error term is still $h(x)(x-a)^k$ with $\lim_{x\rightarrow a} h(x) = 0$, so the Peano form of the remainder still works.

(3) Probably false. This is the Lagrange form of the remainder term. This is usually stated with an additional hypothesis. The hypotheses are "$f$ is $k+1$ times differentiable on $(x,a)$ and $f^{(k)}$ is continuous on $[a,x]$. I suspect the example I used in (2) can be adapted to satisfy the hypotheses you give, but fail to satisfy the hypotheses I give. Probably need to set the step at the end of the interval, but arrange for that point to be the $c$ needed in the error estimate.

Conditions of the Taylor Theorem

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