NetNews Usenet Archive 1993 #3/NN_1993_3.iso/spool/sci/math/18719

Text File | 1993-01-25 | 1.9 KB | 49 lines

Newsgroups: sci.math
Path: sparky!uunet!pipex!pavo.csi.cam.ac.uk!emu.pmms.cam.ac.uk!rgep
From: [e-mail] (Richard Pinch)
Subject: Re: Can Anyone Solve this?????
Message-ID: <[e-mail]>
Sender: [e-mail] (USENET news)
Nntp-Posting-Host: emu.pmms.cam.ac.uk
Organization: Department of Pure Mathematics, University of Cambridge
References: <[e-mail]>
Date: Mon, 25 Jan 1993 10:44:44 GMT
Lines: 36

In article <[e-mail]> [e-mail] (Mark R Freiman) writes:
>
>             If W={f such that f"-2f'+f=0}
>               What is the dimension? and what is a Basis?
>
>I know the answer for the dimension is 2, and a basis is
>               x          x
>              e  and    xe
>
>             but how would one go about obtaining these answers?

Assume that the solution exists and can be represented by a power series
in x, and that differentiation of the power series term-by-term is valid
(these are not trivial assumptions).  Suppose that f = \sum_n a_n x^n;
then equating coefficients of power of x in the equation gives

          2.1.a_2 + 1.a_1 + a_0 = 0    ... coefficient of x^0

          3.2.a_3 + 2.a_2 + a_1 = 0    ... coefficient of x^1

etc

          (n+2)(n+1)a_n+2 + (n+1)a_n+1 + a_n = 0 ...      x^n

and it is clear that the a_n for n >= 2 depend only on the values
of a_0 and a_1, which can be chosen freely.  Choosing a_0 = a_1 = 1
gives a_n = 1/n! and choosing a+0 = 1, a_1 = 1/2 gives a_n = 1/(n+1)!
and so these two solutions span the two-dimensional vector space of
possible sequences of coefficients a_n.  Since the first corresponds
to the function e^x, and the second to the function xe^x, and since
these two functions do indeed satisfy the original equation (this _has_
to be verified) then they forms a basis for the two-dimensional space
of solutions.

Richard Pinch