Re: How Do I Invert These Two Functions

On Wed, 28 Nov 2007 18:47:41 -0500, quasi <quasi@xxxxxxxx> wrote:

On Wed, 28 Nov 2007 18:35:14 -0500, quasi <quasi@xxxxxxxx> wrote:

On Wed, 28 Nov 2007 23:23:02 GMT, John Schutkeker
<jschutkeker@xxxxxxxxxxxxxxxxxxxx> wrote:


If I define u1(r1,r2)=(1/r1)+(1/r2) and u2(r1,r2)=1/(r2-r1). How would I
go about solving these two equations for r1(u1,u2) and r2(u1,u2), which
would constitute "inverting the functions"? Thanks In Advance!

Elementary algebra! Try it before giving up.

Write down 2 equations:

u1 = (1/r1)+(1/r2)

u2 = 1/(r2-r1)

Viewing r1,r2 as the unknowns, you have 2 equations in 2 unknowns.

How hard can it be?

Of course, there are always tricks, but before looking for tricks, how
about trying the most basic method (from elementary algebra):

Choose one equation and solve that equation for one of the unknowns in
terms of the other. Then use the result as a replacement for that
unknown in the other equation, thus yielding one equation in one
unknown. You should be able to figure out the rest.

You try it.

Also, for a function to be invertible, it has to be one-to-one. But as
you'll see when you solve algebraically, your function is mostly
two-to-one (on R^2).

Looking more closely at your original question, I should point out
that the title of your thread is not accurate.

You are not inverting _two_ functions.

Instead you are (locally) inverting a single function with 2 component
functions.

You can regard the map

(r1,r2) --> (u1,u2)

defined by the equations

u1 = (1/r1)+(1/r2)

u2 = 1/(r2-r1)

as a map from R^2 to R^2.

Call it u. Thus u has 2 components functions, u = (u1,u2).

So you can talk about inverting u, but you are _not_ inverting the
functions u1,u2. Of course, as mentioned, u is not one-to-one, so u
doesn't really have an inverse (except locally). But you can still
solve for algebraically for (r1,r2) in terms of (u1,u2), and as you'll
see, for most (but not all) pairs (u1,u2), there are 2 inverse images.

quasi
.

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  • Re: How Do I Invert These Two Functions
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    (sci.math)