Talk:Power series
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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)
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)