Question

5. Find the 2nd-order Taylor approximation of f(x, y) = el+22 –y? around the critical point (which you have to find) of f(x,y). Us- ing this approximation explain why the critical point is a saddle point. Hint: f(xo + h) = f(x0) + Df(xo)h + ihBh, where B is the matrix with elements on your ; i, j = 1, 2, ..., n.

Answer 1

Gradient off is of 12 1 2xetary . (ex i fa 8 At critical point, x=o=y zxeltainy = 0 a yelta- (.el+x stor, 4) ETR²) X=(,,)=(0,0) . =(,, , , ) . for f(xth) = f (x0) +Df (%) h + ² 5 T / ar a m 2 22f 4 => erthithese *o*o* + ( Cros , (x) ) + < Chi na celom this) where Of&o) =($. (X) + (x))+ (0 o) -4an eltary & B /2eltatay at 42² e 1+x-% 4ny - 2 Clash +44²2 423 o e thitheae tot hi hy) = ethitha (Regd. Approximation) f (Xo th)-f(x) =hr-histo when h, > h24 (high2) >h 6, orto, but <o when h, che & hohishe) C •)=x. © Thus f(Xoth ) >fcxo) when his un 37 Xo= 0) is smalle f(xath) < f(xo) when hiche 5 point

Answer 2

f(x,y) = .eltu²nyz od 2x eltal-y? y = -2y e'tu?ng? At critical point at a 20 = 24 >) VODY A OVVERT Lou) in the critical point. f(0,0)= e. 2212= 20 l tulny taule tulnya a ; . 199= 2e auf 2-2etal-yltay' eltulny' ; 377 (0) = -2e. a = 2x1-24) p'tuany! andy - lou) lou) dyan By, Taylor and order approximation Ø f (m, y)= f(0,0) + De [ ] [ sta ] 1,111 . ayin Jyi low) =e+0+ / En y] ze onrur et en?-ey? z excitn?-y?) Here the matran Bore i 7 hun eigenvalue. 2e and -2e . ce so So, the motor the matrix i i neither positive definite nor negative de finite. Hence it is saddle point. for n2(0,1), Buso, n= (0,0), ut Bn >o]