Question

Use Taylor’s formula for f x, y = ln 2x + y + 1 at the origin to find quadratic and cubic approximations of f near the origin.

Answer 1

Answer:

Step-by-step explanation:

Use Taylor’s formula for f (x, y) = ln(2x + y +1) to find quadratic and cubic approximations of

f near the origin.

Answer 2

Answer:

Step-by-step explanation:

Concept:

The Taylor polynomial, also known as the kth-order Taylor polynomial, is a polynomial of degree k that approximates a k-times differentiable function around a given point in calculus according to Taylor's theorem. The Taylor polynomial for a smooth function is the truncation of the function's Taylor series at order k. The function's linear approximation is represented by the first-order Taylor polynomial, and the quadratic approximation is frequently represented by the second-order Taylor polynomial.

Given: $f x, y = ln 2x + y + 1$

Find: To find quadratic and cubic approximations of f near to origin

Solution:

Given that

$f(x,y) = ln (2x+y+1)$  let it be equation 1

we know that the quadratic approximation of f near the origin is given by

$f(x,y) = f(0,0) + x fx+ y fy + 1/2!(x^{2} fxx + 2xyfxy+y^{2}fyy)$ let it be equation 2

and the cubic approximation is

$f(x,y) = [ Quadratic approximation]$

$+ 1/3!(x^{3}fxxx+3x^{2} fxxy+3xy^{2} fxyy+y^{3} fyyy)$ let it be equation 3

Now, $fx = 2/2x+y+1$ ⇒ $fx(0,0) = 2$

$fy = 1/2x+y+1$ ⇒ $fy(0,0) = 1$

$fxy = -2*1/(2x+y+1)^{2}$⇒ $fxy(0,0) = -2$ $= fyx(0,0)$

$fxx = -2*2/(2x+y+1)^{2}$ ⇒  $fxx(0,0) = -4$

$fyy = -1*1/(2x+y+1)^{2}$ ⇒ $fyy(0,0) = -1$

$fxxy = -4*-2/(2x+y+1)^{3}$ ⇒ $fxxy(0,0) = 8$

$fyyx = (-1)*(-2)*(2) / (2x+y+1)^{3}$ ⇒ $fyyx(0,0) = -4$

$fxxx = -4*(-2)*2 / (2x+y+1)^{3}$ ⇒ $fxxx(0,0) = 16$

$fyyy = -1*(-2)*1/(2x+y+1)^{3}$ ⇒ $fyyy(0,0) = 2$

$fxyy(0,0) = 4$

∴ The quadratic approximation is

$f(x,y) = 0+(x*2)+(y*1)+1/2![x^{2}*(-4)+2xy*(-2)+y^{2} *(-1)]$

$= 2x+y+1/2(-4x^{2} -4xy-y^{2} )$ using equation 1

∴$f(x,y) = 2x+y-2x^{2} -2xy-1/2y^{2}$

The cubic approximation is

$f(x,y) = [2x+y-2x^{2} -2xy-1/2y^{2} ]+1/3![x^{3} *16+3x^{2} *8+3xy^{2} *4+y^{3}*2]$

$f(x,y) = 2x+y-2x^{2} -2xy-1/2y^{2} +1/6[16x^{3} +24x^{2} +12xy^{2} +2y^{3} ]$

#SPJ3