ATAR Notes: Forum

VCE Stuff => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematics => Topic started by: xZero on April 12, 2011, 03:14:25 pm

Title: Power series
Post by: xZero on April 12, 2011, 03:14:25 pm: I'm really confused about power series, if someone can explain it to me it will be greatly appreciated :D

Taylor series

on my lecture notes it said, f(x) at the point x = 0 can be expressed as
$f(x) = \sum_{m=0}^\infty x^m\frac{f^{(m)}(0)}{m!}$

this confused the hell out of me... am I correct to say that f(0) = the summation of $x^m\frac{f^{(m)}(0)}{m!}$ where m started from 0 to infinity? If so then whats the $f^{(m)}(0)$ suppose to be? Also is $f^{(m)}(0)$ somehow related to B(m) (Bernoulli Numbers)?

Thanks in advance
Title: Re: Power series
Post by: moekamo on April 12, 2011, 03:59:51 pm: $f^{(m)}(x)$ is the $m$ th derivative of $f$ at $x$

so the value at $x$ is $\sum_{m=0}^{\infty} \frac{f^{(m)}(x)}{m!} x^m$

also note that $0^0=1$ , otherwise your series for x=0 will be 0 as each term is multiplied by $0^m$ which is 0 except for the case when m=0

so the value at x=0 is simply $\frac{f^{0}(x)}{0!} 0^0=f(0)$ which is pretty useless really...
Title: Re: Power series
Post by: TrueTears on April 12, 2011, 04:29:08 pm: Also just notice when you take the taylor series around the point a = 0, then it's usually called the Maclaurin series.
Title: Re: Power series
Post by: xZero on April 12, 2011, 04:37:42 pm: Ohhhh I see :)

but then after the Taylor series the lecture notes have $f(x)=\frac{x}{e^x-1}$ which is indefinitely differentiable at x = 0

hence $\frac{x}{e^x-1}=\sum_{m=0}^\infty x^m\frac{B(m)}{m!}$ where $B(m)=\frac{1}{2}-\sum_{j=o}^{m-1}\frac{m!\times2^{m-j}\times B(j)}{j!\times(m+1-j)!}$

I'm confused (again), does $\frac{x}{e^x-1}=\sum_{m=0}^\infty x^m\frac{B(m)}{m!}$ work for all x? not just for x=0.

And for $B(m)=\frac{1}{2}-\sum_{j=o}^{m-1}\frac{m!\times2^{m-j}\times B(j)}{j!\times(m+1-j)!}$ , is this only for $\frac{x}{e^x-1}$ or for all functions?

and as for Taylor series, is it just a really complicated way to approx a value of f(x)?

Sorry for all these questions, I don't have any knowledge of power series until now and the lecturer didn't really explain what it is :-[
Title: Re: Power series
Post by: TrueTears on April 12, 2011, 04:51:00 pm: f(x) may be indefinitely differentiable at x = 0, but it is a real valued function for all x. Infact if we limit $x \to 0$ of f(x) we get a value of 1 (try prove this yourself). One way to define the Bernoulli numbers by a Taylor series expansion about 0 for $f(x)$ and looking at the coefficient $B_m$ , thus $\frac{x}{e^x-1}=\sum_{m=0}^\infty x^m\frac{B(m)}{m!}$ . Don't treat it as if you have to put a real value input for $x$ . You will see that even if you try to compute the Bernoulli numbers using the Taylor series, it gets really damn hard because when you differentiate each f(x), you will have to limit each to 0 (as you can not compute it by just substituting in x = 0, see why this is the case by expanding out the Taylor series yourself :P)

There are many definitions (and thus computations) for the Bernoulli numbers, the ones you have listed is the recursive definition, others can be found out here http://en.wikipedia.org/wiki/Bernoulli_number#Definitions, although the proofs are not given, you should try search them out as they are quite interesting :)

There are so many other interesting facts about the Bernoulli numbers and in fact I just did a huge topic on them not so long ago, they're DAMN COOL! haha
Title: Re: Power series
Post by: xZero on April 12, 2011, 05:24:31 pm: haha we computed the second Bernoulli number by differentiating f(x) and its ... really messy :P Thanks guys for the help, needed that clarification after being bombarded by recursion, Fibonacci sequences, Bernoulli and Catalan numbers within 2 lectures
Title: Re: Power series
Post by: TrueTears on April 12, 2011, 05:35:36 pm: Is this mth1112 you're doing? If so, dw about these fibonnaci, bernoulli and catalan numbers, heck most of the stuff you do in mth1112 ain't even on the final exam lol but it's really interesting area for pure mathematicians :D
Title: Re: Power series
Post by: xZero on April 12, 2011, 05:45:20 pm: Yup haha, I chose it so I experience a bit of pure maths stuff but at this rate I don't think I'm suited for pure maths >.< but its good to know that the exam won't be as hard as I imagined :D
Title: Re: Power series
Post by: TrueTears on April 12, 2011, 06:45:17 pm: haha yeah, the exam is really simple, take a look at the past exam papers, it's just same questions different numbers.
Title: Re: Power series
Post by: Mao on April 12, 2011, 08:08:51 pm: Quote from: TrueTears on April 12, 2011, 04:51:00 pm
f(x) may be indefinitely differentiable at x = 0, but it is a real valued function for all x. Infact if we limit $x \to \infty$ of f(x) we get a value of 1 (try prove this yourself).

isn't it zero?
Title: Re: Power series
Post by: TrueTears on April 12, 2011, 08:17:04 pm: fk opps, i meant limit x to 0 you get 1 lol, dont even know why i limited x to infinity, it's not even useful!