Firstly, to compute both the quadratic and cubic approximations we need to calculate the function’s derivatives up to the 3rd order, at the origin (0,0).
Let us start with evaluating f(x, y) = 5x * e^(3y). Here are the steps you need to follow:
The partial derivatives of the function f with respect to x and y at the point (0,0):
- f_x = 5 * e^(3y)
- f_y = 15x * e^(3y)
Let's calculate the quadratic approximation:
The Hessian matrix, which is the square matrix of second-order mixed partial derivatives, is used to achieve the quadratic approximation.
The Hessian matrix is:
[ [f_xx, f_xy], [f_yx, f_yy] ]
Under the condition f_xy = f_yx, we get:
- f_xx = 0
- f_xy = 5 * e^(3y)
- f_yx = 5 * e^(3y)
- f_yy = 45 * x * e^(3y)
Quadratic approximation for the function f(x,y) is given by:
Taylor(2, f) = f(0,0) + ((f_x(0,0)*x) + (f_y(0,0)*y)) / 2! + ((f_xx * x^2 + 2*f_xy * x*y + f_yy * y^2 ) / 2)
This simplifies to 5x + 22.5 * x * y * e(3y) + 15.0 * x * y * e(3y).
Now, let's calculate the cubic approximation:
Calculate the third-order derivatives:
- f_xxx = 0
- f_xxy = 5 * e^(3y)
- f_xyy = 15 * e^(3y)
- f_yyy = 135 * x * e^(3y)
Substitute these values into the Taylor series approximation for the function f up to the 3rd order:
Taylor(3, f) = Taylor(2,f) + (f_xxx * x^3 + 3 * f_xxy * x^2 * y + 3 * f_xyy * x * y^2 + f_yyy * y^3) / 3!
The cubic approximation now simplifies to 5x + 22.5 * x * y * e(3y) + 15.0 * x * y * e(3y) + 22.5 * x * y^2 + 22.5 * x * y^2 * e(3y)
So, the quadratic approximation of the function 5x * e^(3y) near the origin (0,0) is "22.5*x*y^2*exp(3*y) + 15.0*x*y*exp(3*y) + 5x", and the cubic approximation is "22.5*x*y^2*exp(3*y) + 22.5*x*y^2 + 15.0*x*y*exp(3*y) + 5x".