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        riddles >> putnam exam (pure math) >> Finding the subject of the equation
        
(Message started by: mattian on Jul 19^th, 2004, 8:09am)

Title: Finding the subject of the equation
Post by mattian on Jul 19^th, 2004, 8:09am

Given an equation

f(a) = 1 - f(a, b)

How would one make a the subject of the equation such that a is given as a function of b?

a = f(b)

For example:

Express y in terms of x such that y = f(x)

log y = 1 - 2xy

The inverse is easy - i.e.

x = f(y) = (1 - (log y)) / 2y

But simplifying and making y the subject of the equation - not so easy.

Title: Re: Finding the subject of the equation
Post by mattian on Jul 19^th, 2004, 8:17am

Alternatively, what techniques could be used to find y, given x?

Title: Re: Finding the subject of the equation
Post by towr on Jul 19^th, 2004, 2:14pm

I don't think there is a general way to find either a as a function of b, nor b as a function of a, since it might not be a function in the first place..

And for instance you could have
log(y) = 1 - (2 xy + log(x) )
Which is equal to log(x) = 1 - (2 xy + log(y) ), so you have the same problem for x and y

Title: Re: Finding the subject of the equation
Post by mattian on Jul 19^th, 2004, 5:03pm

So what does one do in such a situation? Successive approximations? Newton's Method, perhaps? How would one find a, given b - in a way that can be expressed mathematically?

Title: Re: Finding the subject of the equation
Post by Jack Huizenga on Jul 19^th, 2004, 6:52pm

Another good example of the question you are asking is provided by the gamma function, which is an extension of the factorial function to the real line. If you want to invert the expression x=gamma(y), you don't :) . The gamma function is not 1-1, and therefore there is no "inverse gamma" function. Often times in this type of situation a numerical approximation is the best you can do. Also, the correct way of dealing with such a problem is often determined by the context of the question. Recently I was proving a theorem where the conclusion was that "if r>m!, then f(r)>m". The most desirable format for the final theorem would have been something more like "f(r)>g(r)" for some function g. However, as there is no nice way to invert the factorial, stating the theorem as such would have required a loss of information. Hence, in this case, the best thing to do is just to leave the solution in the slightly less desirable but much more exact formulation. There are other situations where a particular numerical answer is more desired. However, keep in mind that something such as Newton's method will only provide a single value for a upon fixing b; it will never yield an actual function for a in terms of b.

Sadly, the vast majority of mathematical expressions cannot be inverted. We just have to deal with it :).

Title: Re: Finding the subject of the equation
Post by mattian on Jul 19^th, 2004, 7:14pm

Quote:

However, keep in mind that something such as Newton's method will only provide a single value for a upon fixing b;

Yes - I know - I thought I might have to resort to the synthetic solution rather than rigidly hang on to the analytical one.

Well how about I post the actual problem and see how I could best represent (or solve for) y in terms of x.

Here is the equation:

sin (y/4) = 1 - x(2[pi] - y/2)

What is the best way to find y?

Title: Re: Finding the subject of the equation
Post by Leonid Broukhis on Jul 19^th, 2004, 8:15pm

According to gnuplot, y is not a function of x:

splot [-0.3:0.3][-10:10] 1-x*(2*pi-y/2)-sin(y/4),0

X axis is the closer one. Observe the contour line corresponding to 0.

Title: Re: Finding the subject of the equation
Post by mattian on Jul 20^th, 2004, 6:00am

Hi Leonid -

Welcome to my maths question :-).

Regarding your graph - that's pretty much what I've done so far - and it works well. I guess I'm just one of those people who likes things (like maths expressions) to look pretty.

Thanks.