# Maclaurin approximations

1. Jun 13, 2010

### estro

Suppose f(x) is differentiable 2 times, can I still use Maclaurin polynomial approximations and write:

$$f(x)=f(0)+f'(0)x+\frac {f''(0)x^2} {2!}+R_2(x)$$

If yes why? (Lagrange and Cauchy reminder theorem are using in this case the 3'rd order derivative...)

Last edited: Jun 13, 2010
2. Jun 13, 2010

### rock.freak667

Last edited by a moderator: Apr 25, 2017
3. Jun 13, 2010

### estro

Are you sure about that?
I remember reading proof for a theoretical problem using this concept. (Can't remember where it was).

I hope I'll find it tomorrow and post it here.
Thanks for the quick reply!

Last edited: Jun 13, 2010
4. Jun 13, 2010

### rock.freak667

I think if you do it on a polynomial, you will just get back the polynomial.

5. Jun 13, 2010

### estro

Yes, I understand and know it.

I asked this question from theoretical point of view.
f(x) is unknown function we only have the numerical information about f(0), f'(0) and f''(0).

6. Jun 13, 2010

### rock.freak667

Yes I don't see why you can't use the series if you have those values.

7. Jun 13, 2010

### estro

But $$R_2(x)$$ is defined using $$f^{(3)} (x)$$ And we don't know if this function exists. [Lagrange Reminder Theorem]

Last edited: Jun 13, 2010
8. Jun 13, 2010

### DrRocket

You can always do that.
The difficulty comes the determination of a useful expression for $$R_2(x)$$. Probably the easiest way to determine such an expression is the integral form of the remainder which comes from integrating by parts repeatedly.The result is an expression in whch the remainder involves the integral of the third derivative. So you need at least three times differentiability to get a McClaurin polynomial of degree two with the usual integral form of the remainder.

$$f(x)-f(0)=\int_{0}^{x}f'(t)dt$$

= $$x f(0) - \int_{0}^{x}f''(t)(x-t) dt$$
.
.
.
= $$f(0) + f'(0)x + 1/2f''(0)x^2 + 1/2 \int_{0}^{x} f'''(t)(t-x)^2dt$$

Please excuse the sloppy Tex. I seem to be getting no correlation between the code, the preview and the post. The first expressions should be the remainder, not f and the expressions for the integrals are pretty much a mess. But I hope you can see through this disaster and recreate the proof.

Last edited: Jun 13, 2010
9. Jun 13, 2010

### estro

I'm not sure I understood your idea.
Basically if I knew that f(x) is differentiable 3 times, I would use Lagrange Reminder Theorem to determine $$R_2(x)$$. The problem arise what to do when I don't know such thing.

10. Jun 13, 2010

### DrRocket

The idea is that first, you can write any function as a quadratic polynomial plus a remainder. That is trivial, since the remainder is just the difference between the function and the polynomial. A Maclaurin expansion is useful only because there is a useful expression for the remainder -- which is "small" and therefore makes the approximation by the polynomial "close".

The problem is really one of finding a useful expression for the remainder. There are various forms for that remainder. What I showed (as best I could with problem that I was having with the Tex software) is a simple way to get the remainder by integrating by parts a couple of times. The process works iteratively and will produce a Maclaurin expansion of any degree that you like, so long as you can keep integrating by parts. That requires some level of differentiability -- it requires n+1 derivatives for an nth degree polynomial.

If you know nothiing whatever about the differentiability of the function then you are basically out of luck. That is because you have no good way to estimate the remainder.

Now, if you know something else about the function then depending on what that is one might be able to do something else -- I am not sure what at the moment.

Someone noted that a full blown Taylor series requires infinite differentiability. That is also true. In fact it requires more, It requires the function be not only infinitely differentiable but actually analytic, a much more restrictive condition. The point being that to approximate a function by a polynomial requires that you know something about the function.

What is the fundamental problem that you are trying to solve ?

11. Jun 13, 2010

### estro

Oh, I think I get it now, and you answered the next question I was about to ask.

I don't have any fundamental problem I'm trying to solve, just want to get idea and intuition with Taylor approximations and the Reminder Theorem of Lagrange.

I think because of this "analytic" thing I have so much trouble feeling this idea with intuition. (I don't have problems solving exercises in this topic in my book, But I don't "feel" this topic to the extend I want)

Thank you very much for your detailed answer.
I will find myself over the moon if you can refer me to a good source about this topic.
(I'm self learner student with Calc 1-2 knowlage)

Last edited: Jun 13, 2010
12. Jun 13, 2010

### DrRocket

The first source is your calculus book, assuming that I understand what Calc 1-2 are.

I am not a big fan of most calculus books, but Spivak's book is pretty good.

The next level would be an introductory text on real analysis. There a couple that I like. Those are Elements of Real Analysis by Bartle and Principles of Mathematical Analysis by Rudin. These are generally considered junior-senior level books for mathematics majors.

Analytic functions are usually encountered in a class on complex analysis. It turns out that if a complex valued function is differentiable even one time, then it is differentiable infinitely often and in fact is analytic -- locally representable by a power series. So to see this you would need to read a book on complex analysis. There are several good texts on this subject, but a classic is Complex Analysis by Ahlfors. But to read it you may need a bit more than just Calc 1-2. It is considered at advanced undergraduate or beginnin graduate level book.

As far as the proof that I tried to sketch, I am sure that I have seen it in a book somewhere, but most texts take a slightly different tack and I am not sure of a ready reference off the top of my head. But in any case it is not difficult to work out for yourself -- just fill in the small blanks in what I posted (It is just integration by parts).

Basically what you get is a function minus some polynomial as the remainder, and at that point you have to know something to limit the size of the remainder. Knowing something about its differentiabillity gives you some knowledge of that sort. Without any knowledge at all, anything can happen. So the important part of such approximation theorems revolves around showing how some piece of information regarding the function limits the size of the remainder.

13. Jun 14, 2010

### estro

I'm from Israel, Calculus here is some sort of mix between "Real Analysis and "Calculus 2" tending heavily towards "Calculus 2 [In US]"

I'm too, they tend to concentrate more on solving very similar problems rather understanding theory.

I'll try to get my hands on one of these.

I have very long way to Complex Analysis.

I was never exposed to this approach, but I think now I get the idea and this is quite trivial.

About this I have to think more.

Thank you!

Last edited: Jun 14, 2010
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