# Converting function to Quadratic form

• Feb 13th 2010, 01:59 PM
scg4d
Hi everyone. I'm looking for help with converting this function to quadratic form.

The function is f(x1,x2)=(x2-x1)^4 + (12*x1*x2) - x1 + x2 - 3.

The quadratic form I need to convert to is: f(x)=(1/2)x'Qx - x'b + h

where x is a vector=[x1 x2]', '=transpose, Q and b are vectors, and h is the constant. Also, Q is symmetric and positive definite (Q=Q' > 0).

The trouble I'm running into is that f(x1,x2) is 4th order, and the examples I have convert only 2nd order functions to the f(x) quadratic objective format above.

In case your interested, I'm looking for this info to solve a steepest descent problem, where the varying ak value is ak=g(k)'g((k))/( g(k)'Qg(k) ) where g(k)=Qx(k) - b.

• Feb 14th 2010, 03:09 AM
HallsofIvy
Quote:

Originally Posted by scg4d
Hi everyone. I'm looking for help with converting this function to quadratic form.

The function is f(x1,x2)=(x2-x1)^4 + (12*x1*x2) - x1 + x2 - 3.

The quadratic form I need to convert to is: f(x)=(1/2)x'Qx - x'b + h

where x is a vector=[x1 x2]', '=transpose, Q and b are vectors, and h is the constant. Also, Q is symmetric and positive definite (Q=Q' > 0).

The trouble I'm running into is that f(x1,x2) is 4th order, and the examples I have convert only 2nd order functions to the f(x) quadratic objective format above.

Well, yes. That's because 2nd order functions are quadratic! You cannot write a function that is NOT quadratic in a quadratic format.

Quote:

In case your interested, I'm looking for this info to solve a steepest descent problem, where the varying ak value is ak=g(k)'g((k))/( g(k)'Qg(k) ) where g(k)=Qx(k) - b.