1. Sep 17, 2010

### danik_ejik

Hello,
if I understand correctly the Taylor approximation for a=0 gives me the possibility to approximate a function, say sin(x), at any x.
But, what gives me Taylor polynomial at some point [PLAIN]http://latex.codecogs.com/gif.latex?a\neq0 [Broken] [Broken],[/URL] what's the difference? what does it mean centred about [URL]http://latex.codecogs.com/gif.latex?a\neq0[/URL] ?

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2. Sep 17, 2010

### Char. Limit

While approximating sin(x) with a Taylor series gives an approximation for any x, this doesn't generally hold true for any function. for example, the function 1/(1-x) has a Taylor series that's only valid if |x|<1.

Well, if the formula for a Taylor series centered at x=0 can be written as this...

$$\sum_{n=0}^\infty \frac{f^{(n)}(0) x^n}{n!}$$

Then the formula for a taylor series centered at x=a can be written as this...

$$\sum_{n=0}^\infty \frac{f^{(n)}(a) (x-a)^n}{n!}$$

I think there's a proof of this, but I'm not sure.

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3. Sep 17, 2010

### Staff: Mentor

A Taylor series is a representation of a function in powers of x - a, (or centered at a) as in c0 + c1(x - a) + c2(x - a)2 + ... + cn(x - a)n + ...

A Taylor series like the above is centered at a.

If the Taylor series is in powers of x (centered at 0), it's called a Maclaurin series. A famous one is the series for ex = 1 + x + (1/2!)x2 + (1/3!)x3 + ...

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4. Sep 18, 2010

### danik_ejik

i understand the mathematical formulas, but i wasn't sure about the meaning.
so, correct me if i'm wrong:
for functions like
where [URL]http://latex.codecogs.com/gif.latex?|x|%3C1,[/URL] in order to approximate the function at some point greater than one, i.e. x=3.5, i need to develop taylor series for
this function around the point a=3.
is this correct?

Last edited by a moderator: Apr 25, 2017
5. Sep 18, 2010

### HallsofIvy

Staff Emeritus
Basically, yes. However, I should point out that the Taylor series does not approximate a function- it is exactly equal to it. It is the Taylor polynomials, that you get by truncating the series at a finite number of terms, that approximate the function.

By the way, another way of getting the Taylor's series for
$$\frac{1}{1+ x}$$
around x= 3.5 is to rewrite it as
$$\frac{1}{1+ x- 3.5+ 3.5}= \frac{1}{4.5+ (x- 3.5)}= \frac{1}{4.5+ y}$$
where, of course, y= x- 3.5.

That is the same as
$$\frac{\frac{1}{4.5}}{1+ \frac{y}{4.5}}$$

Of course, the geometric series
$$\sum_{n=0}^\infty ar^n$$
has sum
$$\frac{a}{1- r}$$
and converges for -1< r< 1.

Here we have a= 1/4.5 and r= -y/4.5 so the geometric series has sum
$$\sum_{n=0}^\infty \frac{1}{4.5}\left(-\frac{y}{4.5}\right)^n= \sum_{n=0}^\infty \frac{(-1)^n}{4.5^{n+1}}y^{n}$$
$$= \sum_{n=0}^{\infty}\frac{(-1)^n}{4.5^{n+1}}(x- 3.5)^n$$
the Taylor's series for 1/(1+ x) around x= 3.5.

Since the geometric series converged for -1< r< 1, it converges for $-1< -y/4.5< 1$ or $-4.5< y< 4.5$ and so for $-4.5< x- 3.5< 4.5$ or $-1< x< 8,$ which is exactly what we would expect- 1/(1+ x) is not defined at x= -1. And the "radius of convergence" really is a radius- since it cannot converge at x= -1= 3.5- 4.5, it also cannot converge at or beyond x= 3.5+ 4.5= 8.

Last edited: Sep 18, 2010
6. Sep 21, 2010

### danik_ejik

I understand now, thank you all

7. Sep 21, 2010

Hey I see you got your answer but if you think of the proof of the Taylor series it makes
perfect sense. The whole proof involves integration by parts mixed with the fundamental
theorem of calculus.

Btw: I wrote this out months ago in an old post, I wouldn't have spent so long rewriting
this when you seem to be satisfied :tongue: Thought it couldn't hurt to copy & paste the
old post anyway!

Using the Fundamental Theorem of Calculus:

$\int_{a}^{b} f ' (t)\,dt \ = \ f(b) \ - \ f(a)$

We'll rewrite this in integration by parts form:

$u \ = \ f'(t) \ , \ du \ = \ f''(t) dt \ ,\ dv \ = \ dt \ , \ v \ = \ - (b - t)$

(The "v" term includes the minus due to a chain rule differentiation removing it anyway! Very sneaky thing that confused me for a few minutes but I like it! ).

$\int_{a}^{b} u dv \ = \ u v |^b_a \ - \ \int_{a}^{b} v du$

$\int_{a}^{b} f ' (t)\,dt \ = \ - f'(t)(b \ - \ t) |^b_a \ - \ \int_{a}^{b} - f''(t)(b \ - \ a)dt$

$\int_{a}^{b} f ' (t)\,dt \ = \ f'(a)(b \ - \ a) \ + \ \int_{a}^{b} f''(t)(b \ - \ a)dt$

$\int_{a}^{b} f ' (t)\,dt \ = \ f'(a)(b \ - \ a) \ - \ f''(t)\frac{(b \ - \ a)^2}{2} |^b_a \ - \ \int_{a}^{b} - f'''(t)\frac{(b \ - \ a)^2}{2}dt$

$\int_{a}^{b} f ' (t)\,dt \ = \ f'(a)(b \ - \ a) \ + \ f''(a)\frac{(b \ - \ a)^2}{2} \ + \ \int_{a}^{b} f'''(t)\frac{(b \ - \ a)^2}{2}dt$

$\int_{a}^{b} f ' (t)\,dt \ = \ f'(a)(b \ - \ a) \ + \ f''(a)\frac{(b \ - \ a)^2}{2} \ + \ \int_{a}^{b} f'''(t)\frac{(b \ - \ a)^2}{2}dt$

Okay, before I go on I'd like to qualify that the left hand side can be
rewritten as

$\int_{a}^{b} f ' (t)\,dt \ = \ f(b) \ - \ f(a)$

so that the above becomes

$\ f(b) \ - \ f(a) \ = \ \ f'(a)(b \ - \ a) \ + \ f''(a)\frac{(b \ - \ a)^2}{2} \ + \ \int_{a}^{b} f'''(t)\frac{(b \ - \ a)^2}{2}dt$

or better yet:

$\ f(b) \ = \ f(a) \ + \ \ f'(a)(b \ - \ a) \ + \ f''(a)\frac{(b \ - \ a)^2}{2} \ + \ \int_{a}^{b} f'''(t)\frac{(b \ - \ a)^2}{2}dt$

Okay, well you'd keep going with the right hand term until you've
calculated the (n - 1) term & then you want the "n"th term

$\ f(b) \ = \ f(a) \ + \ \ f'(a)(b \ - \ a) \ + \ f''(a)\frac{(b \ - \ a)^2}{2} \ + ... \ + \ \int_{a}^{b} \frac{(b \ - \ t)}{(n \ - \ 1)!}^{(n - 1)} \ f^n (t) dt$

$\ f(b) \ = \ f(a) \ + \ \ f'(a)(b \ - \ a) \ + \ f''(a)\frac{(b \ - \ a)^2}{2} \ + ... \ - \ \ f^n (t) \frac{(b \ - \ t)}{(n )!}^{(n)} |^b_a \ - \ \ f^{(n + 1)}(t) dt \frac{(b \ - \ t)}{(n )!}^{(n)}$

$\ f(b) \ = \ f(a) \ + \ \ f'(a)(b \ - \ a) \ + \ f''(a)\frac{(b \ - \ a)^2}{2} \ + ... \ + \ f^n (a) \frac{(b \ - \ a)}{(n )!}^{(n)} \ + \ \int_{a}^{b} \ f^{(n + 1)}(t) \frac{(b \ - \ t)}{(n )!}^{(n)} dt$

Do you see how we can choose any origin f(a), i.e. a = 0 makes (b - a)/2 = b/2 and
all of the rest of them get rid of the a, explaining what Char. Limit meant.
f(a) becomes whatever it's equal to, in the case of ℯˣ we get ℯ⁰ = 1