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[MUG] Radius of Convergence and Power Series
| [MUG] Radius of Convergence and Power Series |
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Author: Chuck Baker
Posted: 09/11/2000 19:49:53 GMT
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>> From: Chuck Baker "geogra4"
Hello all,
I have computed the radius of convergence of these two problems, but
I'm having trouble trying to arrive at the same answer using Maple.
Can anyone describe how to solve using Maple?
inf
Sigma n/2^n * x^n
n=0
After computing by hand, I have a radius of convergence = 2
inf
Sigma (2x + 1)^n/n^2
n=0
After computing by hand, I have a radius of convergence = 1/2
Regarding Power series:
If I have:
y'' - xy' - y = 0, Xo = 0
After computing by hand using the Taylor series, I get
Recurrence relation: a sub n+2 = a sub n / n+2
y1(x) = 1+x^2/2 + x^4/2*4 + x^6/2*4*6 + .....
y2(x) = x+x^3/3 + x^5/3*5 + x^7/3*5*7 + .....
How can this be done using Maple so I get the same answer.
Finally, if I have:
(1 - y)y'' + y = 0, Xo = 0
After computing by hand using the Taylor series, I get:
Recurrence relation: (n+2)(n+1)a sub n+2 - n(n+1)a sub n+1 + a sub n = 0
Y1(x) = 1 - 1/2 x^2 - 1/6 x^3 - 1/24 x^4 - .....
Y2(x) = x - 1/6 x^3 - 1/12 x^4 - ....
Again, how can this be done using Maple so I get the same answer.
Thanks,
Chuck
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| [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"
-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-
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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