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Finding the Function


Date: 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/   
    
Associated Topics:
High School Functions

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