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- Newsgroups: sci.math
- Path: sparky!uunet!sun-barr!ames!elroy.jpl.nasa.gov!aero.org!Aero.org!campbell
- From: campbell@Aero.org (Andrew Campbell)
- Subject: Re: Need help!
- Message-ID: <1992Jul30.180317.8306@aero.org>
- Sender: news@aero.org
- Organization: The Aerospace Corporation
- References: <chepan.712505966@gsusgi1.gsu.edu>
- Date: Thu, 30 Jul 1992 18:03:17 GMT
- Lines: 46
-
- -----------------------
- In article <chepan.712505966@gsusgi1.gsu.edu>, chepan@gsusgi1.gsu.edu (Parthasarathy Nambi) writes:
- |>
- |>
- |> Hi,
- |>
- |> I need help in solving the equation:
- |>
- |>
- |>
- |>
- |> e^(-x) - x = e^(-y) + y
- |>
- |>
- |> For what values of x and y this equation is true?
- |>
- |> P.Nambi
-
- f(x) = e^(-x) -x has derivative -e^(-x) - 1 so it is
- monotone decreasing. Since the limits as x -> +/- oo
- of f(x) are -/+ oo, f is a one-to-one map of R onto all of R.
- Let g be the inverse function. Then the set of solutions
- can be described by x = g(e^(-y)+y). Unfortunately, g has
- no common name or elementary expression that I am aware of.
- Presumably, this is the sort of answer you were after (with
- the set of (x,y) satisfying the equation described in terms of
- a function of one the variables). From the form of the answer
- given and the fact that the function e^(-y)+y has a single
- global minimum at y = 0, where its value is 1, and assumes
- every value greater than 1 twice, one can deduce that:
-
- for every y, there is a unique x satisfying the equation
-
- every x for which there is a solution satisfies f(x) => 1
-
- if f(x) = 1 then x = 0 and (0,0) is the only solution for this x
-
- if f(x) > 1, there are exactly two values of y for which
- there is a solution (x,y)
-
- Hope this is what you were after.
-
- zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz
- |L. Andrew Campbell internet: campbell@aerospace.aero.org |
- |M1-102 PO Box 92957 organization: The Aerospace Corp. |
- |Los Angeles CA 90009 telephone: (310) 336-8642 |
-