Talk:Power series

From Wikipedia, the free encyclopedia

You have new messages (last change).

Contents

[edit] Formulas converging to f(x+1) and f(x-1)

If x, f(x), and f(x+1) are all real numbers, and f can be differentiated infinitely many times, then the following series converges to f(x+1)
f(x) + f '(x) + f(double prime)(x)/2 + f(triple prime)(x)/6 + f(quadruple prime)(x)/24... where the numerator of each term is the function differentiated n times and the denominator is n!
A similar series, the only difference is that the terms alternate in signs, converges to f(x-1)

Yes, that's Taylor's theorem, in special cases. However you'd need more than a smooth function for f.

Do you have a similar formula converging to f(x+c) where x, c, f(x), and f(x+c) are real numbers?

This is what the page on Taylor's theorem goes into, just with some small changes of notation. You can't say that convergence occurs for just any function; but it does for many of the common functions like exponential or sine.

Charles Matthews 19:07, 7 Mar 2004 (UTC)

Can you name some functions it doesn't work for?

There are smooth functions (all derivatives exist) that are not analytic functions (power series). See the smooth function page for constructions; also discussed at An infinitely differentiable function that is not analytic. The power series for a function such as log (1+x) has a radius of convergence that is only 1; so it can't be used if |x| > 1.

Charles Matthews 08:14, 8 Mar 2004 (UTC)

[edit] Formula of a Power Series

I am reading Penrose's "Road To Reality" where he states (but doesn't demonstrate) that

1 + x2+x4+x6+x8+... = (1-x2)-1.

I understand how both of these are different ways of looking at the same function, but how is it possible to get from one to the other ? (Penrose has a site set up for the solutions to the problems in the book, but is 'too busy' to actually post the solutions there...)

That's a geometric series on the left, with common ratio x2. Charles Matthews 16:23, 20 Nov 2004 (UTC)

[edit] Power series with x values to positive powers

Is, for example, f(x) = \sum_{n=0}^\infty a_n \left( x^2 \right)^n a power series? If so, why is this function a power series but not a series with x to a fractional power? Isn't this function like a power series with a cn value that depends on x? --Ott0 00:28, 13 December 2005 (UTC)

Ah, got the answer. This series satisfies the form if a_n is 1 for even powers of n and 0 for odd powers on n. --Ott0 00:52, 13 December 2005 (UTC)
Yeah, you can do that. It also converges: |x| < 1 \Rightarrow \sum_{n=0}^\infty a_n \left( x^2 \right)^n = \frac{a_n}{1 - x^2}

[edit] List of Known Series'

I think it would be extremely useful to link to a list of known series' on this page. For example, the list could give series-form definitions of e^x, sin(x), cos(x), etc. If it exists on wikipedia, I couldn't find this list. Anyone else think this is a good idea? Fresheneesz 23:17, 12 January 2006 (UTC)

Some of them are listed on the Taylor Series page.

[edit] A Question in Modern Physics by Power Series

Why does \frac{1}{e^{-x}} \approx 1+x  ? Condition:when \mathcal ,x<<1,

The \mathcal X can be as \frac{h\nu}{kT}(in Modern Physics).

Myself proof: (not official but helps me in studying)

Take both \mathcal ,ln \Longrightarrow ln1-lne^{-x}=x, which just is quite as the right side:\mathcal ,lnx \approx x, when \mathcal ,x<<1,

Can anyone tell me that might be right or not?--HydrogenSu 11:43, 4 February 2006 (UTC)

Simplify to \frac{1}{e^{-x}} = e^x. Then take the first two terms of the Taylor series around 0 (hence x << 1) of the exponential function ex. Fredrik Johansson - talk - contribs 11:55, 4 February 2006 (UTC)
Thanks. Doing it immediately: (by Taylor)
e^{x}=\frac{x^0}{0!}+\frac{x}{1!}+\frac{x^2}{2!}+...
With \mathcal, x<<1, \Longrightarrow e^{x} \approx 1+x That's right!!! (Higher terms vanished!!)--HydrogenSu 15:10, 4 February 2006 (UTC)

[edit] Include information for usage of power series'?

For instance, scientific calculators can only exist due to the usage of power series (i.e. the lowest-level math concept that can only be simplified into simple mathematics). Also, what about their relation to transcendental functions? -Matt 19:46, 5 March 2006 (UTC)

[edit] Merge with radius of convergence

Since it looks as if a radius of convergence is a term only used for power series, and this page already has something on radius of convergence, it seems only natural to merge the two. Any comments? Fresheneesz 10:30, 29 March 2006 (UTC)

Discussed at talk:radius of convergence. Oleg Alexandrov (talk) 19:35, 29 March 2006 (UTC)
Retrieved from "http://en.wikipedia.org../../../p/o/w/Talk%7EPower_series_908d.html"