Talk:Power series

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

Removed this:

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.

Charles Matthews 09:32, 7 Mar 2004 (UTC)

Do you have a similar formula converging to f(x+c) where x, c, f(x), and f(x+c) are real numbers?
—Preceding unsigned comment added by 66.245.127.196 (talk • contribs) 7 March 2004

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? —Preceding unsigned comment added by 66.32.150.241 (talk • contribs) 7 March 2004

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...) —Preceding unsigned comment added by 217.137.214.85 (talk • contribs) 20 November 2004

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}
—Preceding unsigned comment added by JVz (talk • contribs) 5 March 2006

[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. —Preceding unsigned comment added by 153.90.114.135 (talk • contribs) 6 March 2006

[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)

Actually, calculators don't use power series. I used to think that, too. See CORDIC. - dcljr (talk) 22:44, 3 May 2007 (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)

[edit] Removed false statement

The article claimed that there were power series which are not the Taylor series of any function, and purported to give an example. This is false: a famous (though evidently not famous enough) theorem of Borel asserts that if {a_n} is any real sequence, there exists a C-infinity function f on the real line whose Taylor series at x = 0 is sum_n a_n x^n. If the a_n's grow large enough (like a_n = n!), this gives an example of a Taylor series with 0 radius of convergence. It would be nice to add this information to the article... —Preceding unsigned comment added by Plclark (talk • contribs) 10:44, 20 November 2007 (UTC)