3. In class we have seen the linear approximation of a function f : R2 + R. This

Question

3. In class we have seen the linear approximation of a function f : R2 + R. This can be seen as the Taylor polynomial of degree 1 of a function of two variables. The goal of this exercise is to study Taylor polynomials of degree two. The Taylor polynomial of a function f around (xo, yo) is given by the formula P2 [f: (20, Yo)] (x, y) = f(xo, Y0) + f(x0, yo)(x - x0) ++fy(xo, Yo)(y - y0) + fax (20, 90) (2 - x0) + fy(10, 90)(y - y0)² + fxy(x0, 90) (2 - x0)(y - y0). (a) Check that the formula makes sense by showing that if f(x, y) = x2 + y² and g(x, y) = x+y+1 then P: [f: (x0, y0)] = f(x, y) and P2 [g: (xo, y0)] = g(, y). (b) We are going to justify the formula using the one variable Taylor expansion. Define g(t) = f(x(t), y(t)), where r(t) = xo + (x – xo)t, y(t) = Yo + (y - Yo)t. i. Write down the second order Taylor polynomial of a one variable function h(t) around a generic point to. Assume that the Taylor remainder is 0. ii. Apply the formula you just wrote to the function g(t) around to = 0. Hint: Use the chain rule to compute the derivatives of g. iii. Evaluate the approximation of g at t = 1 and see what you get.

Explanation / Answer

a) P2[f;(x0,y0)](x,y) = f(x0,y0) + fx(x0,y0)(x-x0) + fy(x0,y0)(y-y0) + 1/2fxx(x0,y0)(x-x0)2 + 1/2fyy(x0,y0)(y-y0)2 + fxy(x0,y0)(x-x0)(y-y0)

f(x,y) = x2 + y2

fx(x,y) = d(x2 + y2)/dx = 2x ; fy(x,y) = d(x2 + y2)/dy = 2y

fxx(x,y)=d2(x2 + y2)/dx2 = 2 ; fyy(x,y)=d2(x2 + y2)/dy2 = 2

fxy(x,y)=d2(x2 + y2)/dxy= 0

SUBSTITUTE the above values in this equation

P2[f;(x0,y0)](x,y) = f(x0,y0) + fx(x0,y0)(x-x0) + fy(x0,y0)(y-y0) + 1/2fxx(x0,y0)(x-x0)2 + 1/2fyy(x0,y0)(y-y0)2 + fxy(x0,y0)(x-x0)(y-y0)

P2[f;(x0,y0)](x,y) = x02+y02 + 2x0(x-x0) + 2y0(y-y0) + 1/2*2(x-x0)2 + 1/2*2(y-y0)2 + 0*(x-x0)(y-y0)

P2[f;(x0,y0)](x,y) =  x02+y02+2xx0-2x02+2yy0-2y02+x2+x02-2xx0+y2+y02-2yy0

By simplifying we get

P2[f;(x0,y0)](x,y) =  x2+y2 (x2+y2 = f(x,y))

So, P2[f;(x0,y0)](x,y) = f(x,y) Hence proved

P2[g;(x0,y0)](x,y) = g(x0,y0) + gx(x0,y0)(x-x0) + gy(x0,y0)(y-y0) + 1/2gxx(x0,y0)(x-x0)2 + 1/2gyy(x0,y0)(y-y0)2 + gxy(x0,y0)(x-x0)(y-y0)

g(x,y) = x+y+1

gx(x,y) = d(x+y+1)/dx = 1 ; gy(x,y) = d(x+y+1)/dy = 1

gxx(x,y)=d2(x+y+1)/dx2 = 0 ; gyy(x,y)=d2(x+y+1)/dy2 = 0

gxy(x,y)=d2(x+y+1)/dxy= 0

Substitute the above values in P2[g;(x0,y0)](x,y) = g(x0,y0) + gx(x0,y0)(x-x0) + gy(x0,y0)(y-y0) + 1/2gxx(x0,y0)(x-x0)2 + 1/2gyy(x0,y0)(y-y0)2 + gxy(x0,y0)(x-x0)(y-y0)

P2[g;(x0,y0)](x,y) = x0+y0+1 + 1(x-x0) + 1(y-y0) + 1/2*0(x-x0)2 + 1/2*0(y-y0)2 + 0*(x-x0)(y-y0)

P2[g;(x0,y0)](x,y) =x+y+1 = f(x,y) Hence Proved

b) g(t)=f(x(t),y(t)). ; x(t)=x0+(x-x0)t , y(t)=y0+(y-y0)t.

i) P2(t) = h(t0)+h'(t0)(t-t0)/1!+h''(t0)(t-t0)2/2!

Taylor remainder = R2(t)=h(t)-P2(t) ; given R2(t)=0

so h(t)=P2(t) ;

ii) At t0 = 0 g(t0=0) = f(x(t0),y(t0)) = f(x0,y0)

g(t)=P2(t) ; (as remainder is 0) ; f(x(t),y(t)) = g(t0)+g'(t0)(t-t0)/1!+g''(t0)(t-t0)2/2!

g'(t) = df/dx*dx/dt + df/dy*dy/dy = df/dx*(x-x0)+df/dy*(y-y0)  (since dx/dt=d(x0+(x-x0)t)/dt=x-x0 : dy/dt=d(y0+(y-y0)t)/dt=y-y0)

g''(t) = dg'(t)/dt = 0.

P2(t) = g(t0)+g'(t0)(t-t0)/1!+g''(t0)(t-t0)2/2! = f(x0,y0) + df/dx*(x-x0) + df/dy*(y-y0) = f(x0,y0) + fx(x-x0) + fy(y-y0)

iii) g(1) = f(x(1),y(1)) = f(x,y)

P2(t) =

P2[f;(x0,y0)](x,y) = f(x0,y0) + fx(x0,y0)(x-x0) + fy(x0,y0)(y-y0) + 1/2fxx(x0,y0)(x-x0)2 + 1/2fyy(x0,y0)(y-y0)2 + fxy(x0,y0)(x-x0)(y-y0)

P2[f;(x0,y0)](x,y) = g(1)+0+0+0+0=g(1) since f(x,y)=constant for any value of x and y , the constant value is g(1) , so fx, f y , fxy is zero since derivative of constant is zero

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