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Equation with x,y as exponential functions

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Mar17-12, 04:00 AM
P: 44
Quote Quote by epenguin View Post
Can you show us what the family looks like, so we see what happens as you reduce this something to 1?
I can't.
Mar17-12, 08:43 AM
HW Helper
P: 3,561
I've just been playing with things for a bit, and just thought I'd dumb things down for simplicity, which could also mean I dumbed things down a little too far.
If we look at the exponents only, and take the addition of them we have a function

[tex]=\left( x+\frac{1}{2} \right)^2+\left( y+\frac{1}{2} \right)^2-\frac{1}{2}[/tex]

So we clearly have a symmetry about y=x (as we already knew) and the minimum value f can take is -1/2.

At this point [itex](x,y)=(-1/2,-1/2)[/itex] which is where the minimum of f occurs, we found that


And so if we take some other value, say, [itex]x=\frac{-1}{2}+k[/itex], [itex]y=\frac{-1}{2}-k[/itex], [itex]k\neq0[/itex] ***
then with these values we have



Therefore, g becomes



And for all k


So we only have the minimum at (-1/2,-1/2) which means this is the only combination (x,y) that satisfies [itex]g(x,y)=1[/itex]

*** I'm uneasy about making this assumption that if we take some small linear increase of k in one variable, we should decrease the other variable with the same magnitude k.
Mar17-12, 11:33 AM
P: 44
Holy Menials! Mentallic, you're a genius!!!
I love you!

I just thought of something new (credits go to you of course)

We know that,
Arithmetic mean >= geometric mean.


[16^(x*x + y) + 16^(y*y + x)]/2 >= { [16^(x*x + y)]*[16^(y*y + x)] } ^1/2

g(x,y) >= 2*[16^( x*x + y*y + x + y)]^1/2

Now (x*x + y*y + x + y) >= -1/2

Thus g(x,y) >= 2 *[16^(-1/4)]

And g(x,y)>= 1

But at (x,y) = (-1/2,-1/2)


Mar17-12, 06:37 PM
HW Helper
P: 3,561
Oh awesome! You found the answer without using any crude assumptions like I did

And I'm hardly a genius, I just gave you the kick start you needed. Using the geometric mean <= arithmetic mean was the real genius here

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