Guest Posted June 4, 2009 Report Share Posted June 4, 2009 There are two functions f(x) and g(x,y), that work in the real domain. f(x) returns the absolute value of its input. ( e.x. f(-1) = 1 ). g(x,y) returns numerically bigger of the two input numbers ( e.x. g(1, −2) = 1 ). 1. f(x) in terms of g(x,y) 2. g(x,y) in terms of f(x) Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 4, 2009 Report Share Posted June 4, 2009 There are two functions f(x) and g(x,y), that work in the real domain. f(x) returns the absolute value of its input. ( e.x. f(-1) = 1 ). g(x,y) returns numerically bigger of the two input numbers ( e.x. g(1, −2) = 1 ). 1. f(x) in terms of g(x,y) 2. g(x,y) in terms of f(x) f(x) = g(x, -x) G(x, y) = y + f(x) + f(y) Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 4, 2009 Report Share Posted June 4, 2009 f(x) can b written as f(x) = x + 2*g(0, x) working on the second one Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 4, 2009 Report Share Posted June 4, 2009 [spoiler='for g(x, y) ']g(x,y) = f(x) - f(y) + y Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 4, 2009 Report Share Posted June 4, 2009 looks suspiciously like you are trying to get us to do your homework... Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 4, 2009 Report Share Posted June 4, 2009 (edited) [spoiler='for g(x, y) ']g(x,y) = f(x) - f(y) + y i think i got it wrong, don't know about second Edited June 4, 2009 by tarunark Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 4, 2009 Report Share Posted June 4, 2009 g(x,y)= (x+y+f(x-y))/2 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 4, 2009 Report Share Posted June 4, 2009 Be sure it works when both x and y are negative. g(x,y)=x*[f(x-y)+(x-y)]/[2*(x-y)]+y*[f(y-x)+(y-x)]/[2*(y-x)] Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 4, 2009 Report Share Posted June 4, 2009 (edited) Mine reduces down to mghoffman78's, so ignore mine and use the simplified one. Edited June 4, 2009 by eel_electric Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 4, 2009 Report Share Posted June 4, 2009 (edited) Be sure it works when both x and y are negative. g(x,y)=x*[f(x-y)+(x-y)]/[2*(x-y)]+y*[f(y-x)+(y-x)]/[2*(y-x)] If you go through the math, you will find that your solution simlifies exactly to my solution above. EDIT: Ah, you beat me to it! Edited June 4, 2009 by mghoffman78 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 4, 2009 Report Share Posted June 4, 2009 Kudos to tkn (1) AND mghoffman78(2).. that was very fast Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted June 5, 2009 Report Share Posted June 5, 2009 See also. Quote Link to comment Share on other sites More sharing options...
Question
Guest
There are two functions f(x) and g(x,y), that work in the real domain.
f(x) returns the absolute value of its input. ( e.x. f(-1) = 1 ).
g(x,y) returns numerically bigger of the two input numbers ( e.x. g(1, −2) = 1 ).
1. f(x) in terms of g(x,y)
2. g(x,y) in terms of f(x)
Link to comment
Share on other sites
11 answers to this question
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.