| Marcos |
If
then are the coefficients an unique for convergent f? I mean can we express a function as a power series in more than one way? My intuition immediately tells me "yes", thinking twice though leads to serious doubts (eg. in considering the case where for integers m > some integer, am = 0 - ie. a polynomial)... I haven't been able to prove/disprove anything though... Thanks, Marcos |
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| Andre
Rzym |
In order to create a power series expansion about a, our starting point is to require that all derivatives of f exist at x=a. In your case, a=0. Suppose there are two power series expansions of f(x) about x=0:
Firstly, substitute x=0. This forces a0=b0. Now differentiate both series:
Substitute x=0 and you get a1=b1. Repeating, you find that all terms are equal. Andre |
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| Marcos |
Thanks Andre... Does this show that you can't have an everywhere- convergent power series expansion for (1+x)a where a is any real, other than the one for |x| < 1? Marcos |
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| Marcos |
Or, to stress my point more, in general, if there's a power series representation of a function that is convergent for only some values then there can't exist another function which is convergent for all values, eg. the ln(1 + x) series, etc... Is that right? Marcos |
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| Andre
Rzym |
If we expand a function as a power series about x=a, then the coefficients of the (Taylor) series are unique, and therefore there can only be one radius of convergenge [i.e. set of values for x for which the series converges]. However, there are a couple of of points to note: 1) If we expand about a different a, the new series may converge for different values of x to the original series. 2) Suppose we start with ln(1+z)=z -z2 /2 - ... which diverges for |z| > 1. We can transform this series (the transformation being due to Euler) to get ln(1+z) = (z/(1+z)) + (z/(1+z))2 /2 + (z/(1+z))3 /3 + ... convergent for all positive z. 3) If you fancy something more radical, there are continued fractions: ln(1+z) = z / (1+12 z/(2+12 z/(3+22 z/(... convergent for z> -1 The point I am trying to make is that if you stick to Taylor series, then for a given a, the radius of convergence is fixed, however changing 'a' or changing the 'type' of series may change things. Andre |
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| Marcos |
Thanks, I was basically referring specifically to forms I've mentioned at the beginning of the thread, ie. power series in x (so they're particular Taylor expansions)... The thing is with Taylor expansions around x=a, isn't it really the same thing? If we get a power series in (x-a) then can't we expand the (x-a)n and group together like terms to get the same thing (unless anything is undefined/doesn't converge)? Haven't we (well, you) just shown this? So even by expanding around x=a isn't it the same thing? Ie. same radius of convergence? Thanks, Marcos P.S. I realise that if you express it using some other way other than a power series there can be an everywhere-convergent series |
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| Andre
Rzym |
OK so your proposal is to create a series, S1 which is a power series in (x-a), then expand the powers of (x-a), collect terms to get a second series, S2 which looks like a power series in x. Does S2 have the same radius of convergence as S1 ? The answer is not necessarily. The problem is that you are rearranging infinite series, and these may be conditionally convergent. Depending on how you rearrange the series, it may converge to different things (or not at all). The reason you are doomed to failure is the one I gave initially -the power series expansion is unique. So once you have found it (Taylor expansion about x=0) and computed its radius of convergence, you are done -you can do no better (and no worse!). Andre |
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| Marcos |
Thanks Andre, I guess that clears things up now. Marcos |