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[MUG] Re: Radius of Convergence and Power Series
| [MUG] Re: Radius of Convergence and Power Series |
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Author: Maple Group
Posted: 15/11/2000 21:00:17 GMT
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>> From: Maple Group "maple_gr"
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Date: Mon, 13 Nov 2000 15:52:40 +0100
From: Adri van der Meer "A.W.J.vanderMeer"
To: "maple-list"
Subject: Radius of Convergence and Power Series
For linear equations you can use formal power series.
Maple V.5:
> restart; with(powseries):
> DE := diff(y(x),x,x) - x*diff(y(x),x) - y(x) = 0:
> a := powsolve( DE );
a := proc(powparm) ... end
recurrence relation:
> a(_k);
a(_k - 2)
---------
_k
Truncated power series:
> tpsform(a,x,10);
2 3 4 5
C0 + C1 x + 1/2 C0 x + 1/3 C1 x + 1/8 C0 x + 1/15 C1 x +
6 7 8 9 10
1/48 C0 x + 1/105 C1 x + 1/384 C0 x + 1/945 C1 x + O(x )
Your second equation is not linear, so powsolve doesn't work:
> DE2 := (1-y(x))*diff(y(x),x,x) + y(x) = 0:
> b := powsolve( DE2 );
Error, (in spcldif) final value in for loop must be numeric or character
A possibility is (without rewriting the equation):
> dsolve( {DE2,y(0)=A,D(y)(0)=B}, {y(x)}, 'type=series' );
2
A 2 B 3 A - 2 B
y(x) = A + B x + 1/2 ------ x - 1/6 --------- x - 1/24 ---------
-1 + A 2 3
(-1 + A) (-1 + A)
2
4 B (1 + 6 A - 6 B ) 5 6
x + 1/120 ------------------ x + O(x )
4
(-1 + A)
--
A. van der Meer
Dept. Applied Mathematics
University of Twente Phone +31 (53) 4893427
P.O. Box 217 Fax +31 (53) 4894824
7500 AE Enschede
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Date: Wed, 15 Nov 2000 11:53:07 +0100 (MET)
From: Gottfried Barthel "Gottfried.Barthel"
To: "''" "maple-list"
Subject: Radius of Convergence and Power Series
Hello Chuck,
here is how to answer your "Radius of Convergence" question using Maple:
By the Cauchy-Hadamard formula, the radius of convergence R of the power
series sum(a_n z^n, n=0..infty) satisfies 1/R = lim sup |a_n|^(1/n). In
your example, the sequence actually converges, and Maple can compute the
limit:
> limit((n/2^n)^(1/n), n=infinity);
1/2
Similarly for the other example: The linear substitution u := x - 1/2 does
not change the radius of convergence, so you have to consider the series
sum((2*u)^n/(n^2), n=1..infty) (most likely, you would not really wish to
start the series with n=0 for good reasons). Then
> limit(( 2^n / (n^2))^(1/n), n=infinity);
2
Alternatively, you may use a formula of Euler: If a_n <> 0 for all n (at
least for all that are sufficiently large) and if lim |a_n| / |a_{n+1}|
exists, then this limit is the radius. In our case:
> limit((n/2^n)/((n+1)/2^(n+1)), n=infinity);
2
and
> limit(( 2^n / (n^2)) / (2^(n+1)/(n+1)^2), n=infinity);
1/2
Hope that helps! Regards,
Gottfried Barthel
Fachbereich Mathematik, Universitaet Konstanz
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