Thoughts on an identity.

1. Jun 16, 2008

ssd

Given that
1/ f(cx) = k - g(x) and
2/ the above is an identity,
where f(.) and g(.) are two functions
and c, k are real valued constants.

The problem is to infer upon the types of f(.) and g(.).
I have a hunch that f(.) and g(.) are logarithimic functions. Can any one provide any analytical proof or counter example?

2. Jun 16, 2008

arildno

What do you mean by "2/above is an identity"????

3. Jun 16, 2008

matt grime

Surely there should be some mention of c on the RHS? Otherwise only constant functions work.

For f(c)=f(c.1)=k-g(1) for all c, i.e. f is identically equal to k-g(1), whatever g(1) is. And then g(x)=k-f(cx)=g(1), for all x.

4. Jun 16, 2008

dodo

Well, making c part of the definition of g(x) would do.

The obvious choice is to take g(x) = k - 1/f(cx), for whatever definition of f(x) you wish; in other words, I don't think the condition limits the actual form of the functions in any way (except by relating one to the other). For example, if f(x) = x^2, then g(x) = k - 1/(cx)^2. And so on.

Edit: Oh, sorry, I read the original condition as 1/f(cx) = k - g(x). But the argument stays the same. Just take g(x) = k - f(cx). If f(x) = x^2, then g(x) = k - (cx)^2.

(Yet another edit:)
Or... (psychic powers for the win)... by his logarithmic hunch, the OP meant something like f(cx) = h(c) - g(x), with f(x), g(x), h(x) functions and c a constant.

Last edited: Jun 16, 2008
5. Jun 17, 2008

ssd

We are given with two conditions. 1st condition is the expression given in point 1. The 2nd condition is given in point 2, which states that the expression given point 1 is an identity.

6. Jun 17, 2008

ssd

'k' in RHS can involve 'c' in some suitable form. Similarly 'c' can involve 'k' in some form. Or, g(.) may contain c or f(.) may contain k..... I dont really know. I know that 'k' and 'c' are two numerical constants obtained from an experiment, and they donot change with change of the study parameters of the experiment.

Last edited: Jun 17, 2008
7. Jun 17, 2008

matt grime

Oh, it can, can it? Are there any other things you've not mentioned?

8. Jun 17, 2008

ssd

No, nothing is appearing in my mind at this moment except that f(.) and g(.) are real valued.

Last edited: Jun 17, 2008
9. Jun 17, 2008

dodo

Is there a table of numeric results for f() and g(), coming out of experiments, that you can post?

10. Jun 18, 2008

ssd

Yes it is there. Infact there are a number of such tables obtained from different experiment stations by different scientiests under similar set up. The values are small in number (like 10 to 12) in one set. Practically may functions fit (numerically) the data (of one set) well. There are moderate variations in data from different stations. But the problem is to find a theoritical form. One has to infer only on the basis of information I gave already. I (we) have theoritically simplified the work upto this. The usual forms of the expressions and the method (using which c & k are found in current practice) are really complicated.

Last edited: Jun 18, 2008