Download A < B by Kiran Kedlaya PDF

By Kiran Kedlaya

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Xn , we call the directional derivative along the i-th standard basis vector the partial derivative of f with respect to i and denote it by ∂f . In other words, the partial derivative is the derivative of f as a function of xi along, ∂xi regarding the other variables as constants. TOTAL DERIVATIVE Caveat! Since the derivative is not a “function” in our restricted sense (it has takes values in a vector space, not R) we cannot take a “second derivative”—yet. ASSUMING the derivative exists, it can be computed by taking partial derivatives along a basis.

It is even more critical here than in the one-variable case that the Lagrange multiplier condition is a necessary one only for an interior extremum. Unless one can prove that the given function is convex, and thus that an interior extremum must be a global one, one must also check all boundary situations, which is far from easy to do when (as often happens) these extend to infinity in some direction. For a simple example, let f (x, y, z) = ax + by + cz with a, b, c constants, not all zero, and consider the constraint g(x, y, z) = 1, where g(x, y, z) = x2 + y 2 + z 2 .

BLAH. 5 1. (IMO 1968/2) Prove that for all real numbers x1 , x2 , y1 , y2 , z1 , z2 with x1 , x2 > 0 and x1 y1 > z12 , x2 y2 > z2 , the inequality 8 1 1 ≤ + 2 2 (x1 + x2 )(y1 + y2 ) − (z1 + z2 ) x1 y1 − z1 x2 y2 − z22 is satisfied, and determine when equality holds. (Yes, you really can apply the material of this section to the IMO! 6 Constrained extrema and Lagrange multipliers In the multivariable realm, a new phenomenon emerges that we did not have to consider in the one-dimensional case: sometimes we are asked to prove an inequality in the case where the variables satisfy some constraint.

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