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- Newsgroups: sci.math
- Subject: Re: Functional Equation
- Message-ID: <a_rubin.716591769@dn66>
- From: a_rubin@dsg4.dse.beckman.com (Arthur Rubin)
- Date: 15 Sep 92 21:16:09 GMT
- References: <424@nttcvg.ntt.JP>
- Distribution: sci.math
- Nntp-Posting-Host: dn66.dse.beckman.com
- Lines: 35
-
- In <424@nttcvg.ntt.JP> herve@nttcvg.NTT.JP (Herve Delingette) writes:
-
-
- >I would like to find all the functions f C1 continuous such that :
-
- >f'(x)= [f(x+ x0)-f(x-x0)]/2x0 where f'(x) is the derivative of f
- >and where x0 is a real number ..
-
- >I am pretty sure that the only solutions are the linear functions
- >f(x)=ax+b but I am not sure.
-
- >Please answer at herve@nttcvg.ntt.jp
-
- Still looks like homework, but the answer is:
-
- If x0 can vary, then differentiating the equation
-
- 2x0 f'(x) = f(x+x0) + f(x-x0) by x0, we get
-
- 2 f'(x) = f'(x+x0) + f'(x-x0),
-
- so f' must be linear, and f must be quadradic. Evaluating, we determine
- that f must be linear.
-
- If x0 is fixed, then let c=a+ib be a solution of sinh(c) = c. (They DO
- exist.) Then
-
- f(x) = (E^(a x/x0)) cos(b x/x0)
-
- is a solution.
- --
- Arthur L. Rubin: a_rubin@dsg4.dse.beckman.com (work) Beckman Instruments/Brea
- 216-5888@mcimail.com 70707.453@compuserve.com arthur@pnet01.cts.com (personal)
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