高维微分学—无限小增量公式 谢锡麟 13多元函数多项式逼近的实际获得方法 131乘积函数的多项式逼近 如有 f(x,y)=Ao+(A10x+A1y)+(A20x2+A1y+A2y2)+o(x2+y2) 9(x,y)=B00+(B0x+Boy)+(B20x2+B1xy+B2y2)+o(x2+y2) 则有 (f9)(x,y)=[4+(A10x+Ao1y)+(420x2+A1y+Amy2)+o(x2+y2) [Bo0+(B10x+Bo1y)+(B20x2+B1y+Bm2y2)+o(x2+y2) A000+[(40030+Bo0410)x+(B01+AoBo1)y +[(A0020+40B10+B0420)x2+(A0Bn+Boon)xy +(Aoo Bo2+Ao1 Bo1+ BooA02)y2+o(a2+y2) 1关分析中,利用关系式 P+g-1 Vp,q∈N 分析:考虑到 g(x2+y2)≤(x2+y2) (x2+y2) (x2+y2)5(x2+y2) 132除法函数的多项式逼近 )(x,y)=4+(4ax+Auy)+(4x0x2+Anxy+Ay2)+o(x2+y2) Boo +(B1oc+ Bo1g)+(B20 2+Bllry+B02y2)+o(r2+y2 40+(A10x+Ay)+(20x2+A1xy+A02y2)+o(x2+y2 Boo1_(B1oz+Bo1y)+(B202+Buay+ Bo2y2)+o(22+y2) A0+(10x+A1y)+(A20x2+A1y+A02y2)+o(x2+y2) Boo[1+e(, y) 而 1+(x0)1=1+∑())+((,) 133复合函数的多项式逼近 如有 f(x,y)=(410x+A01y)+(A20x2+A1xy+A2y2)+o(x2+y o(zP)微积分讲稿 谢锡麟 高维微分学—— 无限小增量公式 谢锡麟 1.3 多元函数多项式逼近的实际获得方法 1.3.1 乘积函数的多项式逼近 如有    f(x, y) = A00 + (A10x + A01y) + (A20x 2 + A11xy + A02y 2 ) + o(x 2 + y 2 ) g(x, y) = B00 + (B10x + B01y) + (B20x 2 + B11xy + B02y 2 ) + o(x 2 + y 2 ) 则有 (fg)(x, y) = [ A00 + (A10x + A01y) + (A20x 2 + A11xy + A02y 2 ) + o(x 2 + y 2 ) ] · [ B00 + (B10x + B01y) + (B20x 2 + B11xy + B02y 2 ) + o(x 2 + y 2 ) ] = A00B00 + [ (A00B10 + B00A10)x + (B00A01 + A00B01)y ] + [ (A00B20 + A10B10 + B00A20)x 2 + (A10B01 + B10A01)xy + (A00B02 + A01B01 + B00A02)y 2 ] + o(x 2 + y 2 ) 相关分析中，利用关系式 x p y q = o ( ( x 2 + y 2 ) p+q−1 2 ) , ∀ p, q ∈ N 分析：考虑到 |x py q | (x 2 + y 2) p+q−1 2 = |x p | (x 2 + y 2) p 2 |y q | (x 2 + y 2) q 2 (x 2 + y 2 ) 1 2 6 (x 2 + y 2 ) 1 2 1.3.2 除法函数的多项式逼近 ( f g ) (x, y) = A00 + (A10x + A01y) + (A20x 2 + A11xy + A02y 2 ) + o(x 2 + y 2 ) B00 + (B10x + B01y) + (B20x 2 + B11xy + B02y 2) + o(x 2 + y 2) = A00 + (A10x + A01y) + (A20x 2 + A11xy + A02y 2 ) + o(x 2 + y 2 ) B00 [ 1 + (B10x + B01y) + (B20x 2 + B11xy + B02y 2 ) + o(x 2 + y 2 ) B0 ] = A00 + (A10x + A01y) + (A20x 2 + A11xy + A02y 2 ) + o(x 2 + y 2 ) B00 [1 + θ(x, y)] 而 [1 + θ(x, y)]−1 = 1 +∑ p k=1 1 k! (p k ) θ k (x, y) + o(θ p (x, y)) 1.3.3 复合函数的多项式逼近 如有    f(x, y) = (A10x + A01y) + (A20x 2 + A11xy + A02y 2 ) + o(x 2 + y 2 ) Θ(z) = C0 + ∑p k=1 Ckz k + o(z p ) 6