# Talk:Power series

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### 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 (talkcontribs) 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 (talkcontribs) 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)

## 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 (talkcontribs) 20 November 2004

That's a geometric series on the left, with common ratio x2. Charles Matthews 16:23, 20 Nov 2004 (UTC)
Aren't these infinite "polynomial" series containing all integer powers? For example, ex =
${\displaystyle e^{x}=\sum {_{n}=0}{^{\infty }}{\frac {x^{n}}{n!}}=1+{\frac {x}{1!}}+{\frac {x^{2}}{2!}}+\cdots }$
Generally, the formula of a power series is
${\displaystyle \sum {_{n}=0}{^{\infty }}a_{n}x^{n}}$
Cheers, The Doctahedron, 00:06, 18 December 2011 (UTC)

## Power series with x values to positive powers

Is, for example, ${\displaystyle f(x)=\sum _{n=0}^{\infty }a_{n}\left(x^{2}\right)^{n}}$ 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 ${\displaystyle c_{n}}$ 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: ${\displaystyle |x|<1\Rightarrow \sum _{n=0}^{\infty }a_{n}\left(x^{2}\right)^{n}={\frac {a_{n}}{1-x^{2}}}}$
—Preceding unsigned comment added by JVz (talkcontribs) 5 March 2006
It's an even function. Cheers, The Doctahedron, 00:07, 18 December 2011 (UTC)

## 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 (talkcontribs) 6 March 2006

## A Question in Modern Physics by Power Series

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

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

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

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

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

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

## 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)

## 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)

## 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 (talkcontribs) 10:44, 20 November 2007 (UTC)

## Summation over Integers

.. and with the summation ranging over the integers instead of the rationals. Imo, summation over Z should also be treated in this article, or there should be a link to a page which covers that case (I haven't looked for it yet) Illegal604 (talk) 17:42, 5 September 2008 (UTC)

## Continuity or not

In case someone is curious concerning my removal of the claim of continuity on the boundary, here is a simple counterexample:

${\displaystyle G(z)=\sum _{n=1}^{\infty }{\frac {1}{n}}(-z^{3^{n}})^{n}}$

The series converges at ${\displaystyle z=1}$, but is unbounded along each ray given by ${\displaystyle te^{i\pi /3^{m}}}$ for ${\displaystyle t\in [0,1)}$, and so it is unbounded in any neighbourhood of ${\displaystyle z=1}$. [Argh, that came out ugly as inline math. Also, how do I write “less than” in math mode?] Hanche (talk) 09:57, 5 November 2008 (UTC)

## Explanation of Symbols Needed.

There is an uneven distribution of levels of understanding on this (and it seems, any mathematics related) page.

The problem is that many of the symbols are not explained. There is a lot that is easy to understand on this page, and the hypothetical you, the reader, will be going along fine "yeah yeah, uhuh, I get that," until all of a sudden, "What the hell is that??!"

And if you don't know, then you just don't know.

I would go through and fix this problem, But I somehow managed to get to Multivariable Calculus without seeing some of these symbols.

DoomedToBeTeaching (talk) 05:46, 8 March 2009 (UTC)

Which ones? 67.198.37.16 (talk) 17:37, 5 July 2016 (UTC)