By Phil Dyke (auth.)
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Extra resources for Advanced Calculus
Example text
Terms o r r an d u: ll . First of all, we need to fin d the first denvatlves -arp an d -arp m ax ay arp arp ar arp ae ax ar arp ax arp cose-ar arp ar ae ax sin8 arp --r ae arp ae ay ar ae -== -+-- -=--+-= 22 ay ay sin8 arp + cose arp ar r ae ~~ + ~; = 0. Taking these as operator identities gives d2t/J ax2 = (cose ~ - sin8 _i_)(cose dt/J - sine dt/J) ar r ae ar r ae cose~ (cose at/J _ sin8 at/J) _ sin8 ~ (cose at/J _ sin8 at/J) ar = cos ar r ae r ae ar r ae 2 d2t/J cos8sin8 dt/J cos8sin8 d2t/J e+ - -2 ar2 r ae r arae sin8 ( .
1 at the top of page 37. 2 Find the general Taylor Series expansion about an arbitrary point (u 1, u 2 , u 3 , u4 ) for the function of four variables f(x 1, x 2, x 3, x 4) = x 1x 2x 3x 4. Hence deduce the Taylor Series expansion about the arbitrary point (u 1, U2, ... , Un) for the n-variable function f(x 1, X 2, ... , Xn) = X 1X 2 ... Xn. Solution This question is a little theoretical, but gives some experience in using Taylor's Theorem in many variables. Differentiating f gives the following expressions: fx 1 = x 2x 3x 4,fx2 = x 1x 3x 4,fx3 = x 1x 2x 4 and fx 4 = x 1x 2x 3.
Since the point ( t, +) corresponds to an extreme value of the function f for any t, the O(e) term must vanish and we must be prepared to expand to O(tr) at least. ) be employed, but for those without access to this, the following gives the details. Note that (t + ecos9) ( so that + e -(r+Ecose)( f +Esin9) + esin() ) = e + e ( I - e ( tsin () = l + e (rsin() + cos()) + ;! coso) E t Slflu t e -E 2cos8sin9 Hence f + ! ; < 0 for all t and for all 8. Thus the hyperbola xy = 1( t, 1) = ~ thus I is entirely composed of maxima.