Advanced Calculus by Phil Dyke (auth.)

By Phil Dyke (auth.)

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Xn, A1, A-z, ... , Am) are those of f subject to the constraints g, = 0, r = I, 2, ... , m. The A/s are called undetermined multipliers (or Lagrange multipliers). 1 Consider the function f(x, y) = 1n(l + xy). Find the Taylor Series expansion to quadratic order for f(x, y) about the following points: (a) (0, 0), (b) (0, 1), (c) (0, -1) and deduce the expansions about the points (1, 0) and (-1, 0). ls there an expansion about the point (1, -1)? Solution When finding Taylor Series expansions, especially severa1 about different points as required here, it is wise to find all first-order and second-order partial derivatives.

Written as an algorithm, it Iooks rather involved. 9 if you want actually to learn how the method works. If one starts with the iterative scheme: where then this satisfies the iteration by setting X(r+l) = X(r) and + aU(UTdg(r)) ß

Q_+l_ and a -a= a~ - -a+ a11 -- dX d1J d~ Clt at a~ at a11 {. 17 d2 C/J Thus 1/>(x, t) satlsfies ;--z ax ~ c at a2 ~ Jo~ e-=:inbx 0 b2 where a and b are constants, a > 0, evaluate the integrals dx by using Leibniz's Rule and its extensiono Hence deduce the sinx Io -X- dxo Starting with the result r e-ax sinbxdx = a 2 : b 2 , we differentiate both sides with respect to a to o ~~ -ax o b dx _ d ( b ) _ -2ab g1ve 0 -xe Sill x . - da a 2 + b 2 . - (a 2 + b2) 2 Wh 0 ence J~ 0 xe -ax o b dx _ Sill x - (a 2 2ab + b 2) 2 0 For the second integral, we use integration under the integral sign, an extension of Leibniz's Ruleo 31 J: Again starting with e -ax sinbxdx limits 0 and a to obtain [- ~ !

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