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What does Taylor's theorem say? How do we use Taylor's theorem to get to Taylor's series? I need the basic idea behind these two. What's going on between them?

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Rougly, the Taylor theorem says: if $f$ has enough derivatives in a nhood of the point $a$, then $f$ is the sum of a polynomial and a remainder: $$f(x) = \sum_{k=0}^n\frac{f^{(k)}(a)}{k!}(x - a)^k + R_k(x).$$ If you can calculate the derivatives (see What do Taylor series accomplish?) and the remainder is "small", $$f(x) \approx\hbox{ a polynomial.}$$ Why this is important? Because polynomials are calculable.

If $f$ is $C^\infty$, you can write the Taylor series $$ \sum_{k=0}^\infty\frac{f^{(k)}(a)}{k!}(x - a)^k, $$ but $$ f(x) = \sum_{k=0}^\infty\frac{f^{(k)}(a)}{k!}(x - a)^k $$ can be false. When is true (this depends obviously of the behavior of the remainder) we say that the function is analytic. But being $C^\infty$ isn't enough. The function $f(x) = e^{-1/x^2}$ for $x\ne 0$, $f(0) = 0$ is $C^\infty$ but its Taylor series at $x = 0$ is zero. Even worse: there are examples of functions with divergent Taylor series (Real Mathematical Analysis by C. Pugh).

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    $\begingroup$ I like the contrast you make between the Taylor approximation by a Taylor polynomial and the series actually converging to the function. This remainder going to 0 condition is often neglected; it should be mention even if it is not needed to state Taylor's theorem. I think it would be really helpful to mention them together within the same theorem (at least I know that baby Rudin doesn't do so). $\endgroup$
    – tzy
    Dec 4, 2019 at 13:57
  • $\begingroup$ @Martin-Blas Why we want that $f$ is differentiable in a neighborhood of $a$? I think it is enough to say that $f$ is differentiable at $a$. Please reply I am confused at this point. $\endgroup$
    – MANI
    Jun 4, 2022 at 12:28
  • $\begingroup$ @MANI, differentiability in a neighborhood of $a$ isn't strictly necessary, but it is useful. See en.wikipedia.org/wiki/…. $\endgroup$ Jun 4, 2022 at 16:34
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Although the exact syntax of Taylor theorem is not agreed upon generally the basic of the theorem is:

For every $k\in\Bbb N$ differentiable function at some point $a$, $f$, exists $h_k$ such that $$\lim_{x\to a}h_k(x)=0,\;f(x)=\sum_{n=0}^k\frac{f^{(n)}(a)}{n!}(x-a)^n+h_k(x)(x-a)^k$$

Taylor series is:

  • if $f$ is of class $C^\omega$, then $h_k(x)=\sum_{n=k+1}^\infty\frac{f^{(n)}(a)}{n!}(x-a)^{n-k-1}$

The difference is that not all functions are class $C^\omega$


Few more things things that the theorem says is that if there exists polynomial, $p(x)$, such that $f(x)=p(x)+h_k(x)(x-a)^k$, then $p(x)=\sum_{n=0}^k\frac{f^{(n)}(a)}{n!}(x-a)^{n}$(uniqueness)

If so we can calculate the approximation error: $R_k(x)=f(x)-p(x)$ and $\lim_{x\to a}R_k(x)=o(|x-a|^k)$.

Using this we can get $|R_k(x)|\le M\frac{|x-a|^{k+1}}{(k+1)!}$

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