Question about Taylor Series?

I'm getting a little confused about Taylor series.

For a function

\displaystyle f

to have a Taylor series then

\displaystyle f

must have a power series representation and must be infinitely differentiable at some point that is where the Taylor series is centered on. But how do we know that such a power series exists for a given function.

Take sin x for example, obviously it has a power series/Taylor series representation but where does that come from? How do we know

\displaystyle sin(x)

can be expressed as

\displaystyle sin(x)=a_0+a_1(x-a)+a_2(x-a)^2+...+a_r(x-a)^r+...

thus leaving it open to a Taylor expansion?

Also

\displaystyle e^x

can be defined in terms of a power series so is the taylor series of

\displaystyle e^x

the same as

\displaystyle e^x

in this context (about zero)?

I'm guessing there is something fundamental I'm not understanding here if anyone could give me some reasoning and try to clear this up for me I would be grateful.

I tried looking online but I couldn't really find an answer.

Thanks.

(edited 9 years ago)

Reply 1

brianeverit

9

Original post by alex2100x

I'm getting a little confused about Taylor series.

For a function

\displaystyle f

to have a Taylor series then

\displaystyle f

must have a power series representation and must be infinitely differentiable at some point that is where the Taylor series is centered on. But how do we know that such a power series exists for a given function.

Take sin x for example, obviously it has a power series/Taylor series representation but where does that come from? How do we know

\displaystyle sin(x)

can be expressed as

\displaystyle sin(x)=a_0+a_1(x-a)+a_2(x-a)^2+...+a_r(x-a)^r+...

thus leaving it open to a Taylor expansion?

Also

\displaystyle e^x

can be defined in terms of a power series so is the taylor series of

\displaystyle e^x

the same as

\displaystyle e^x

in this context (about zero)?

I'm guessing there is something fundamental I'm not understanding here if anyone could give me some reasoning and try to clear this up for me I would be grateful.

I tried looking online but I couldn't really find an answer.

Thanks.

We assume that a function has a power series expansion and then justify that assumption.

Reply 2

alex2100x

OP

2

Original post by brianeverit

We assume that a function has a power series expansion and then justify that assumption.

So you assume it can be represented as a power series and then show that it can. Could you tell me an example where this fails and a power series representation of a function doesn't exist? I have only seen ones that work like

\displaystyle e^x

,

\displaystyle sin(x)

etc. I would be interested to see that.

Thanks.

Reply 3

brianeverit

9

Original post by alex2100x

So you assume it can be represented as a power series and then show that it can. Could you tell me an example where this fails and a power series representation of a function doesn't exist? I have only seen ones that work like

\displaystyle e^x

,

\displaystyle sin(x)

etc. I would be interested to see that.

Thanks.

Try y=ln x.

Reply 4

atsruser

11

Original post by alex2100x

I'm getting a little confused about Taylor series.

For a function

\displaystyle f

to have a Taylor series then

\displaystyle f

must have a power series representation and must be infinitely differentiable at some point that is where the Taylor series is centered on. But how do we know that such a power series exists for a given function.

1. It's easy to show that *if* a function can be represented by a power series centered at

a

then that series must be a Taylor series i.e. the coefficients in the power series must be of the form

\frac{f^{(n)(a)}}{n!}

.

You can show this by assuming that a power series exists and evaluating the coefficients by repeated differentiation. Note that a function must be infinitely differentiable for this to work (though infinitely many of the derivatives may be 0, as in the case of the Taylor series of a polynomial).

2. If you calculate a finite Taylor series for

f(x)

at

a

up to derivative

n

, then you get a polynomial of order

n

called the Taylor polynomial,

T_n(x,a)

3. Then the question arises: if I write down a Taylor series for a function, then does that series converge at any set of points to the value of the function at those set of points. You decide this by appealing to Taylor's theorem which says, more-or-less, that a function is equal to a Taylor polynomial + a remainder term:

f(x) = T_n(x,a) + R_n(x,a)

There are 3(? I think) ways to write down the remainder term: Cauchy form, Lagrange form, integral form.

4. If

\lim_{n \rightarrow \infty} R_n(x,a) = 0

then

\displaystyle \lim_{n \rightarrow \infty} (f(x)-T_n(x,a)) = \lim_{n \rightarrow \infty} R_n(x,a) = 0

\displaystyle \Rightarrow f(x) = \lim_{n \rightarrow \infty} T_n(x,a) = \sum_{n=0}^{\infty} \frac{f^{(n)(a)}}{n!} (x-a)^n

so if the remainder term disappears in the limit, a function is equal to its Taylor series over some range in which the series converges.

5. You find the region of convergence of the Taylor series by applying the usual tests (ratio, nth-root, etc). Since a Taylor series is a power series, it must converge at most in an interval centered on

a

, or on all of

\mathbb{R}

.

6. The famous function for which a Taylor series exists, but which does not converge to the value of the function anywhere is

f(x) = e^{-{1/x^2}}

. You can look up the details on the internet.

7. Since to find a Taylor series for a function we need to have an infinitely differentiable function, we must show that a function is infinitely differentiable by some other means than pointing to its Taylor series as a justification, else we have a circular argument.

I think that all makes sense. I suspect that others will correct me if I omitted anything important, or screwed up some of the details. In general, however, I think it pretty much summarises what you need to remember when dealing with Taylor series.