f(x,y) = X^2 for instance, For the sake of a concrete proof. Would it be sufficient to say that if f(x) is unbounded and f(x,y) >f(x) then f(x,y) is also unbounded. Seems incomplete, or have we just skimmed the surface in terms of reasoning?
I'm assuming by f(x,y) you're referring to the original equation and no it isn't bounded above. I'm sorry, i have a feeling I'm supposed to make some intellectual leap with that example,but I am still somewhat lost.
Homework Statement
Show that this function has no absolute max by showing that it is unbounded
Homework Equations
f(x,y) = (x-1)^2 + (y+2)^2 -4
The Attempt at a Solution
my initial idea is to construct a sequence of points {(xk, yk)} so that the sequence {f(xk, yk)} becomes unbounded.
to...
Taking what you said into consideration;
given f(x,y): x^2 -2x +1 +y^2 +4y
The completed square should look like this -> f(x,y)= (x-1)^2 +(y+2)^2 - 4
Assuming this holds, intuitively it seems the smallest possible value would occur at (1,-2). which results in -4 being the absolute minimum.?
Here are the steps i went through to arrive at my new version of f(x,y).
Given :f(x, y) = x^2 + y^2 − 2x + 4y + 1
By taking the half the coefficients of (-2x) and (4y) then squaring them to find the missing values for (__) within the brackets.
(x^2 - 2x + __ )+ (y^2 + 4y + __) = -1
which...
Perhaps it was a typo on your part,because my solution was (x-1)^2 + (y+2)^2, I have done it several times.However if it is still incorrect I'm more than willing to attempt to complete the square again.
The issue of using this new version of the function to find the absolute minimum, The first...
Homework Statement
Use technique of completing squares to Show that this function has an absolute minimum.
f(x, y) = x^2 + y^2 − 2x + 4y + 1
Homework Equations
Not entirely sure how completing the squares will indicate an absolute minimum.Is there some additional reasoning required?
The...