What Does It Mean to Expand a Function About a Point?

[fa2209]

I'm a 2nd year Physics undergrad and am ashamed to admit I don't fully understand the taylor expansion.

I have seen the Maclaurin expansion derived which I understand but the Taylor expansion is a little weird to me. I don't really understand what it means to expand a function "about a point". I also have seen people write the Taylor expansion in a few different ways which is also a little confusing.

Could someone explain:

1). What it means to expand a function "about a point".
2). How to find the taylor expansion of a function f(x) at a nearby point f(x+dx).

 

[HallsofIvy]

The MacLaurin series for f(x) is
$$\displaytype \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n$$

The Taylor series "about x= a" is
[tex]\displaytype \sum_{n=0}^\infty \frac{f^{(n)}(a)}{a!}(x- a)^n[/itex]

In other words, the MacLauren series of any function is just the Taylor series "about x= a= 0".

 

[Fredrik]

Note that if f is differentiable infinitely many times, then for any x, f(x) is equal to both of the series in HallsofIvy's posts. If you only keep terms with n≤N, what you're doing is to approximate f by a polynomial of degree n. The approximation is only guaranteed to be good near the point around which you expand (i.e. 0 if you use the first series in HallsofIvy's post, and a if you use the second). So if you're only going to keep a finite number of terms, you should choose a to be the point where you need the approximation to be good.

Edit: A few comments about the terminology:

2). How to find the taylor expansion of a function f(x) at a nearby point f(x+dx).

f is a function. x is a point in its domain. f(x) is a point in its range, called the value of f at x. The expansion is always around a point in the domain, not around a point in the range. If you e.g. write f(x+dx)=f(x)+f'(x)dx+(1/2)f''(x)dx²+..., you're expanding the function f around the point x in its domain.

 

[lurflurf]

f(x+h)~exp(h D)f(x)

The taylor expansion is a local approximation. That means we chose a point of expansion x and fit a polynomial to f by requiring the the function and polynomials agree to as many orders of derivative as the degree of the polynomial allows. This (assuming some conditions are met) makes for a good approximation near x that gets better and stays good longer the higher the order, but a poor approximation far from x.

 

[HallsofIvy]

Note that if f is differentiable infinitely many times, then for any x, f(x) is equal to both of the series in HallsofIvy's posts.

No, this is not true. A function that is equal to its Taylor's series (about a) for all x is a (real) "analytic" function. But being infinitely differentiable is not sufficient to be analytic. An example is
$$f(x)= e^{-1/x^2}$$
if $x\ne 0$, f(0)= 0. That function is infinitely differentiable for all x, all of its dertivatives at x= 0 are 0 so its Taylor expansion is just 0. But the function is only equal to 0 at x= 0.

If you only keep terms with n≤N, what you're doing is to approximate f by a polynomial of degree n. The approximation is only guaranteed to be good near the point around which you expand (i.e. 0 if you use the first series in HallsofIvy's post, and a if you use the second). So if you're only going to keep a finite number of terms, you should choose a to be the point where you need the approximation to be good.

Edit: A few comments about the terminology:

f is a function. x is a point in its domain. f(x) is a point in its range, called the value of f at x. The expansion is always around a point in the domain, not around a point in the range. If you e.g. write f(x+dx)=f(x)+f'(x)dx+(1/2)f''(x)dx²+..., you're expanding the function f around the point x in its domain.

 

[y.moghadamnia]

I know this might be a little not-so-mathematical, but u know whenever I need to expand a function, taylor or whatever there is, I imagine that so many functions can have values there, like strings floating in the plane :-)
now in a point and a little bit farther(of a known function), our function has a value. and if we search carefully we can catch some of those floating functions and add them all up to get the value our one function gives us. so now we know adding so many other functions gives us the exact or good enough approximate result.so there would be the expansion.
its nothing scientific, its just a seeable example.hope it helps.

 

[fa2209]

Hey thanks for all the help guys. Frederick, I ment to say "a nearby point x+dx". Sorry for that.

So, anyway I know how to derive the Maclaurin expansion. How do I then generalise this to a Taylor expansion.

So if i have:

f(x) = f(0) + xf'(0) + ...

How does the Taylor expansion follow? Or is it just that when we derive Maclaurin by assuming any function can be written as an infinite power series of the form a0 +a1x+..., our assumption is not general and corresponds to an expansion about 0, whereas we know that if the x's were (x-a)'s, our series would be more general as it refers to a more generally defined polynomial which can be centered around any point a which we can choose?

 

[Fredrik]

Define a new function g by g(x)=f(x+a) for all x. Now McLarurin's formula implies that for all x,

$$f(x)=f(x-a+a)=g(x-a)=g(0)+(x-a)g'(0)+\dots=f(a)+(x-a)f'(a)+\dots$$