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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?

MathsLearner
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1 Answers1

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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.

auscrypt
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  • I don't think the formula looks off, because it is a general look of the remainder. If we have $a = x_0$ and $p = n + 1$, than we will get what you wrote.
    • – MathsLearner May 25 '19 at 07:34
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    ah, you're right. sorry about that, i've not seen the more general form before and i'd have to think more about it to see why it works. Thanks for teaching me something new :) i'll edit the answer to reflect this – auscrypt May 25 '19 at 07:43
  • I can see that it is implied through the lemma, becuse $h$ is arbitrarily chosen so segment $[x_0 - h, x_0 + h]$ can expand into the $\mathbb{R}$. Am I correct about this? Also, I wouldn't see easily what you said, without your remark. Also, thank you for compliments about my understanding of power series and analisys.
    • – MathsLearner May 25 '19 at 07:47
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    I've edited my answer to address step 4 -- if you prove that it holds for a specific $h$, then you're right, 5 is different from 4. But if you prove that it's true for arbitrary $h$, then step 4 is equivalent to 5 – auscrypt May 25 '19 at 07:54
  • In my book I have that I need to prove that $\lim\limits_{n \to \infty}R_n(x) = 0$ and doesn't say anything about all $x \in \mathbb{R}$. Because we develop function in only one point, I guess lemma is only given to help to prove that remainder fulfilles the conditions in the arbitrarily neighborhood of the chosen point $x_0$. On the other side in lemma is said 'If real function $f$ is infinitelly diferentiable on the segment $[x_0 − h, x_0 + h]$', so the segment can be even $\mathbb{R}$. – MathsLearner May 25 '19 at 08:00
  • mmm, I kinda get what you're saying. If you prove it only in a neighbourhood, then your function is only guaranteed to work in a neighbourhood of $x_0$ as well. For instance, the taylor series for log(1+x) is x-x^2/2+x^3/3-x^4/4+... which converges if |x|<1 or x=1 and diverges otherwise. If you proved it in the neighbourhood of x, you'd be fine but you wouldn't be able to apply this outside the nieghbourhood. – auscrypt May 25 '19 at 08:06
  • So, if segment $[x_0 - h, x_0 + h] \ne \mathbb{R}$ than I always have to do step 5) from my question? – MathsLearner May 25 '19 at 08:08
  • auscrypt, thank you very much for your time, but now I have to go and will not be able to respond to you in the next few hours. – MathsLearner May 25 '19 at 08:20
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    No worries, also I'm getting system messages to move to a chatroom if you want to discuss much further. You have to do step 5 if you want to prove that $f$ and it's taylor series are the same for all $x$, but in order to do step $5$ the method is essentially the same as proving $R_n(x)$ converges for all $x$ which is just step $4$. If you'd like I can walk you through an example in chat. – auscrypt May 25 '19 at 08:28
  • It would be great if you could do that. I am going to move discussion to chat and present how I did one example, $f(x) = e^x$. – MathsLearner May 25 '19 at 10:05
  • Let us continue this discussion in chat. – MathsLearner May 25 '19 at 10:05