|


Finding the FunctionDate: 12/17/96 at 11:03:39 From: Bolibar Elourdi, Adolfo Subject: Functions First of all, many thanks for your interest in our problem. We know the dependence of Y with each one of these five variables (a,b,c,d,e) as follows: Y= f(a) = 0.5* a^2 (with b,c,d,e as constants) Y= f(b) = 2*b^1.5 (with a,c,d,e as constants) Y= f(c) = 3+2*c+c^2 (with a,b,d,e as constants) Y= f(d) = 2+d+d^2 (with a,b,c,e as constants) Y= f(e) = 4+3*e+1.5*e^2 (with a,b,c,d as constants) We would like to find a global expression which gives: Y = f(a,b,c,d,e) Thank you in advance again. Date: 12/17/96 at 17:35:32 From: Doctor Jerry Subject: Re: Functions Hi Bolibar Elourdi, Adolfo, I believe there would be many possible answers. I'll try to describe my thoughts and assumptions, beginning with the two-variable case. I'll use the notation p_q to mean p with a subscript q and p^q to mean p to the power q. We consider a fixed point (a_0,b_0) of two-space and a function f(a,b). We want: (1) f(a,b_0) = a^2/2 (2) f(a_0,b) = 2b^(3/2) Now, what about the value of f at (a_0,b_0)? If we calculate f(a_0,b_0) from (1), we get: (3) f(a_0,b_0) = (a_0)^2/2 If we calculate f(a_0,b_0) from (2), we get: (4) f(a_0,b_0) = 2(b_0)^(3/2) Presumably, we would want these values to be equal. This will place some constraints on the form of the functions given. It happens that for a_0=2 and b_0=1, (3) and (4) are satisfied. We find f(2,1) = 2. We can define a function f that satisfies (1), (2), (3), and (4) as follows: Define f(2,1) = 2; for all (a,b) other than (2,1). Define f(a,b) = (a^2/2)*(cos(t))^2 + (2b^(3/2))*(sin(t))^2, where t is the angle defined as follows: Letting L_1 be the line from (a_0,b_0) to (a_0+1,b_0) and L_2 the line from (a_0,b_0) to (a,b), t is the angle from L_1 to L_2. For all points (a,b_0), where a is not equal to a_0, t is 0 or pi. So, sin(t) = 0 and (cos(t))^2 = 1. So, f(a,b_0) = a^2/2. Similarly, for points (a_0,b), where b is not equal to b_0. For the higher-dimensional cases, I don't know how you might choose (a_0,b_0,c_0,d_0,e_0) so that the five equations analogous to (3) and (4) have a common value at (a_0,b_0,c_0,d_0,e_0). However, if this can be done, then a definition of f analogous to the above can be given, using the components of the unit vector from (a_0,b_0,c_0,d_0,e_0) to (a,b,c,d,e). These components are the cosines of the direction angles from the line from (a_0,b_0,c_0,d_0,e_0) to (a,b,c,d,e) to the five coordinate axes. I have made several interpretations in your statement: Y= f(a) = 0.5* a^2 (with b,c,d,e as constants) Y= f(b) = 2*b^1.5 (with a,c,d,e as constants) Y= f(c) = 3+2*c+c^2 (with a,b,d,e as constants) Y= f(d) = 2+d+d^2 (with a,b,c,e as constants) Y= f(e) = 4+3*e+1.5*e^2 (with a,b,c,d as constants) For example, I assumed you meant f(a,b_0,c_0,d_0,e_0), not f(a), etc. I also could not understand the constants unless they were based on a single point (a_0,b_0,c_0,d_0,e_0). I hope this is useful. -Doctor Jerry, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]


Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/