# Remainder asymptotic in Taylor's expansion. Big O form.

I've studied Taylor series about half a year ago and know I realized, I didn't understand it properly. I'll mark my questions with numbers so you can understand what I'm actually asking:)

So, let's take a look at the Taylor's Expansion at zero:

$$f(x)=f(0)+\frac{f'(0)}{1!}\cdot x + \frac{f''(0)}{2!} \cdot x^2 + ... + \frac{f^{(n)}(0)}{1!} \cdot x ^ n + R_n(x)$$

Where the remainder $$R_n(x)$$ may be expressed in Lagrange Form:

$$R_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!} \cdot x^{n+1}$$

But people often like to write $$O(x^{n+1})$$. (Big $$O$$ here, not Peano form).

And it for me was a little bit "scary". Well, firstly, Big $$O$$ notation is used to express this kind of relation: $$\text{In neighbourhood of some point a we can say that } \varphi(x)=O(\psi(x)) \text{ if and only if } \\ \text{there exist a constant } C>0 \text{ s.t. } |\varphi(x)| \le C|(g(x)| \text{ for all } x \text{ in small enough neighbourhood of that point.}$$

The defintion of Big O implies that Big O notation is binded with the point where we approximate our function. But nobody writes anything about the point when they say $$R_n(x)=O(x^{n+1})$$!

1. Does it mean that for every point in some set (function domain or whatever else) this relation remains right?

Ok, let's forget about the first issue. Why does such a constant $$C$$ exist? I mean, let's fix some point $$a$$ where we approximate our function $$R_n(x)$$ with Big O.

Why we can say that $$|R_n(x)| \le C|x^{n+1}|$$ for all $$x \in (a-\delta, a+\delta)$$? I have doubts on this because $$\xi$$ varies between $$0$$ and $$x$$. And the only thing we know about the location of $$\xi$$ is that it is between $$0$$ and $$x$$! Doesn't that mean that $$f(\xi)$$ can be unpredictably big?

Well, I thought a little bit on this and realized that we can take $$C=\sup_\limits{\text{domain}} f^{(n+1)}(x)$$. And then this comes to true for every bounded set(since we require (n+1)-derivative be continuous).

1. But what if such a supremum doesn't exist? Nevertheless $$f^{(n+1)}(x)$$ is continuous on it's domain, what happens at infinity? It may be not bounded there.

So, for example, why does people use an asymptotic in improper integral $$\int_1^{\infty} f(x)dx=\int_1^{\infty} \text{ first n terms in Taylor's expansion } + O(x^{n+1})dx$$

1. Or we use it only for functions $$f(x)$$, which has bounded at infinity (n+1)-derivative?
• For your first issue, I believe the big $O$ relation holds as $x \rightarrow 0$. I suppose they don't write it because the context suggests that.
– fwd
Apr 11, 2021 at 10:11
• Can you give a specific example for your last point?
– fwd
Apr 11, 2021 at 10:15
• @fwd sorry, I can't. It's hard for me to write all this latex stuff. Apr 11, 2021 at 12:03
• @fwd do you really think that we need $x \to 0$? It doesn't look like a true since $R_n(x)$ is an "error term" and it would be nice if it shows an error at every point. And not only in neighbourhood of 0. I believe it should represent something like that: "The more we are from point of expansion, the greater is error of approximation" Apr 11, 2021 at 12:09

1. I agree the notation is sloppy because strictly when using big-O or little-o one should specify the neighbourhood in question. In this case it means in the neighbourhood of $$x=0$$. Then it is only true if $$f^{(n+1)}(x)$$ is bounded for small enough $$x$$.
2. If continuous, $$f^{(n+1)}(x)$$ will be bounded on the interval $$[-\delta,\delta]$$, and that is all you need for strict application of the big-O notation. The bound of course depends on $$\delta$$. For example, is you take the first term of the Taylor's series for $$e^x$$ say there is no $$C$$ such that $$|e^x-1| \leqslant C|x|$$ for all $$x$$. Only for bounded $$x$$.
3. The sloppy use of big-O can lead to trouble when you try to interchange the Taylor's series expansion and an improper integration. If the intent is to allow the sum to become infinite, care is needed when handling the remainder. It is hard to comment further without a specific example. Trouble is likely to arise because $$x^{n+1}$$ is not integrable over $$[1,\infty)$$ and since the point $$\xi$$ is not precisely positioned, you will struggle to show the remainder term converges when integrated. A different form of the remainder will probably be more useful.
• Well, nice answer. Do you really think that if we expand function at x=0 using Taylor's formula, then it means that big O assumes that $x$ is around zero? Apr 11, 2021 at 12:01
• I mean, $\sin(x)=x-x^3/6+O(x)$ - isn't that true for every $x$ in $\mathbb{R}$ (I suppose it is)? Or it is correct only for small enough x? Apr 11, 2021 at 12:02
• I think the context does make it reasonably clear that we are talking about $O(x^{n+1})$ near zero (or often written as $x \to 0$). But near may include far! In the example of $\sin(x)=x-x^3/6$ the remainder is $O(x^5)$ around $x = 0$ but here the interval can be as large as you like, only because the $n$th derivative is always bounded for real arguments. The same cannot be said for $e^x$ unless $x$ is limited to some range. Back to $\sin x$ - the $O(x^5)$ bound is not that useful though for large $x$. Hope that makes sense. Apr 11, 2021 at 15:08