I would like to get some help with the next problem:
I'm trying to learn how to develop the real function into the power series. After reading my book, I want to check if I understood correctly what is the process of the developing of the given real function. So, this is how I understand what I need to do in order to develop the given real function $f$:
1) Check if the given function $f$ is infinitely diferentiable and where.
2) Chose the point $x_0$ in which we are going to develop the function.
3) Because now the first condition is fulfilled we have that the given function $f$ is continuous with all of its derivatives, up to the $n$-th order, in some neighborhood of the point $x_0$. With this, we have that all conditions from the following theorem are met:
Theorem:
Let the function $f(x)$, continuous with all of its derivatives up to the $n$ – th order inclusive in some neighborhood $U$ of the point $a$, has the derivative of the $(n + 1)$ – st order in that neighborhood. If $x \in U$ and $p \in \mathbb{N}$, than the following formula is true:
$$f(x) = f(a) + \frac{f'(a)}{1!}(x – a) + \frac{f^{(2)}(a)}{2!}(x – a)^2 + \cdot \cdot \cdot + \frac{f^{(n)}(a)}{n!}(x – a)^n + R_n(x),$$
where $R_n(x) = \left( \frac{x – a}{x – \xi} \right)^p \frac{(x – \xi)^{(n + 1)}}{p \cdot n!} f^{(n + 1)}(\xi)$, for some $\xi$ between $a$ and $x$.
4) Check if $\lim\limits_{n \to \infty} R_n(x) = 0$. In my book there is a following lemma that is given for help:
Lemma:
If real function $f$ is infinitelly diferentiable on the segment $[x_0 – h, x_0 + h]$ and there exist the constant $M$, such that for every $n \in \mathbb{N}$ and for every $x \in [x_0 – h, x_0 + h]$ it is $|f^{(n)}(x)| \le M$, than it is $\lim\limits_{n \to \infty} R_n(x) = 0$ for all $x \in [x_0 – h, x_0 + h]$.
5) Check if $f(x_0) = P_n(x, x_0) = \sum_{n = 0}^{\infty} \frac{f^{(n)}(x_0)}{n!}(x – x_0)^n$. This means that we have to check the convergence of the Taylor series that we got and to calculate the sum of the series if the series is convergent.
6) If all conditions are fulfilled, than we can say that function $f$ can be developed into power series $\sum_{n = 0}^{\infty} \frac{f^{(n)}(x_0)}{n!}(x – x_0)^n$ and we can call that function analytic.
Please, could you tell me if I understood this process correctly and if not where did I make a mistake?
Best Answer
This looks fine! Just a bit of minor nitpicking:
1) Looks good!
2) Looks good!
3) Your formula for the remainder term looks off. It should be $R_n(x) = \frac{(x - x_0)^{n + 1}}{(n+1)!} f^{(n + 1)}(\xi)$ [wikipedia]
Edit: as OP mentioned, their version reduces to this version when $p=n+1$. I couldn't find any references to validate your version though, but as long as it comes from a credible source it should be fine.
4) Remember, $\lim\limits_{n\to\infty} R_n(x)=0$ needs to hold for all $x$
5) Really, this is the same as step 4 -- if the error term tends to $0$ then the functions are the same. So step $5$ isn't even necessary.
Edit: This is only the same as step 4 if for step 4 you proved that the error converges to $0$ for all $x\in\mathbb{R}$, not just for a single interval. Of course, if you can prove that it converges to $0$ for any interval then that'd work too. Otherwise, if you only did step 4 for a specific interval, it tells you nothing about what happens outside that interval.
6) If we followed the steps correctly, and all conditions were met,
$$f(x)=\sum_{n = 0}^{\infty} \frac{f^{(n)}(x_0)}{n!}(x - x_0)^n$$
is indeed the Taylor (power) series around $x_0$, it converges for all $x$ to $f(x)$, and $f$ is indeed analytic.
But, there's an important point I want to make -- some functions (e.g $\log$) are analytic without a corresponding Taylor series equal to the function. Your argument can only find the Taylor series for a function, and prove that it's analytic, but there are analytic functions which you might miss if you use this method.
Your understanding on this topic is very good :) the best thing to do now is to do some examples. I would recommend trying $x^2$ as a warm up, or $e^x$ if you're feeling more confident.