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Topic: Re: Functions whose sum and product are equal-More
Replies: 2   Last Post: Jan 21, 1998 6:24 PM

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Jerry Uhl

Posts: 1,267
Registered: 12/3/04
Re: Functions whose sum and product are equal-More
Posted: Jan 13, 1998 9:31 AM
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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://www-cm.math.uiuc.edu
http://www-cm.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






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