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Re: Functions whose sum and product are equalMore
Posted:
Jan 13, 1998 9:31 AM


Here's a solution which should be accessible to all precalc students.
By abstracting the relationship between Tan[x] and Cot[x], you can see that if you take any old function h[x] and put f[x] = 1 + h[x] and g[x] = 1 + 1/h[x], then
f[x] g[x] = f[x] + g[x] for all x at which h[x] is not 0..
It turns out that if f[x] g[x] = f[x] + g[x] for all x, then f[x] = 1 + h[x] and g[x] = 1 + 1/h[x] for all x's for which f[x] is not 1. for some other function h[x] ( = f[x]  1).
To see why start with
f[x] g[x] = f[x] + g[x]. Replace f[x] with 1 + h[x].
This gives (1 + h[x]) g[x] = 1 + h[x] + g[x].
This is the same as g[x] + h[x] g[x] = 1 + h[x] + g[x].
This is the same as h[x] g[x] = 1 + h[x] ,
And this gives g[x] = 1+ 1/h[x] for all x's for which h[x] is not 0 (i.e. f[x] is not 1).
 Jerry Uhl juhl@ncsa.uiuc.edu Professor of Mathematics 1409 West Green Street University of Illinois Urbana,Illinois 61801 Calculus&Mathematica Development Team
http://wwwcm.math.uiuc.edu http://wwwcm.math.uiuc.edu/dep
[If] logic is the hygiene of the mathematician, it is not his source of food. Andre Weil
Only professional mathematicians learn anything from proofs. Other people learn from explanations. R. P. Boas



