What are Some of these Multiple Policy Objects the Australian Tax and Transfer System Tries to Achieve - Taxation Law Assessment Answers

You are required to write a research essay addressing all of following points:

“Australia has too many taxes and too many complicated ways of delivering multiple policy objectives through the tax and transfer systems. The capacity of the legislative and operating platforms of these systems, and their human users, to deal with the resulting complexity has been overreached. The tax and transfer architecture is overburdened and beginning to fail in dealing efficiently and effectively with multiplying policy goals and demands. Rationalisation of the tax and transfer architecture should now be a strategic priority.”Discuss. What are some of these multiple policy objects the Australian tax and transfer system tries to achieve? How does the system overreach and is hence overburdened currently? How do you propose to rationalise the tax and transfer system of Australia to overcome these shortcomings?

1. In order to integrate a function f(x) within a given interval, generally it is divided into some equal number of parts. Let us consider that the function is broken into n equal parts and fn = f (xn) and h = (b-a)/n whereas “a” and “b” are the given limits or the interval. A is the lower limit and b is the upper limit.

If two points are considered such as (x1, f1) and (x2, f2), from the Lagrange interpolation the following can be written ("Newton-Cotes Formulas -- From Wolfram Math world"):

P2(x) = [(x-x2)/(x1-x2)]*f1 + [(x-x1)/(x2-x1)]*f2

= (x – x1 – h)/(-h) *f1 + [(x-x1)/h]*f2

= (x/h) *(f2-f1) + (f1+(x1/h)*f1 – (x1/h)*f2)

The Lagrange interpolating polynomial can be written as

P3(x) = [(x-x2)(x-x3)/(x1-x2)(x1-x3)]*f1 + [(x-x1)(x-x3)/(x2-x1)(x2-x3)]*f2 + [(x-x1)(x-x2)/(x3-x1)(x3-x2)]*f3

= 1/h^2{x^2(1/2 * f1 – f2+ ½ *f3) + x[- ½ (2x1 + 3h)f1 + (2x1+2h)f2 – ½ (2x1+h)f3] + [ ½ (x1+h)(x1+2h)f1-x1(x1+2h)f2+1/2 x1(x1+h)f3]} ("Newton-Cotes Formulas -- From Wolfram Math world")

After integrating the above equation the following can be written:

x1x5fxdx=245h7f1+3f2+3f3+f4-380h5()

This is known as the Boole’s rule.

3. a) The given integral equation is v (x) = x3 + 0.25 05(x3+t)vdt

Here, h = 0.5

x0 = 0

x1 = x0 + h = 0 + 0.5 = 0.5

x2 = 0.5 + 0.5 = 1

x3 = 1 + 0.5 = 1.5

Now from Simpson’s 3/8 rule, the following can be written:

01.5(x3+t)vdt = (3*0.5)/8 [(Xn^3+0) v0+3(Xn^3 + 0.5) v1+3(Xn^3+1) v2 + (Xn^3+1.5) v3]

Therefore, from the main integral equation, it can be written that

V (Xn) = Xn^3 + 0.25[(3*0.5)/8 {(Xn^3+0) v0+3(Xn^3 + 0.5) v1+3(Xn^3+1) v2 + (Xn^3+1.5) v3}]

Now, substituting

X0 = 0, x1 = 0.5, x2 = 1 and x3 = 1.5, the followings can be written:

V0 = 0.0703125v1 + 0.140625v2+0.1953125v3

V1 = 0.125 + 0.005859v0 + 0.08789v1+ 0.158203v2+0.0761718v3

V2 = 1+0.046875v0 + 0.2109375v1+0.28125v2+0.1171875v3

V3 = 3.375 + 0.1582v0+6.921875v1+0.61523v2+0.22851v3

By solving the above four equations, the followings can be obtained

V0 = -6.63980987677947

V1 = -3.1896396387399073

V2 = -4.7730785407988545

V3 = -29.41093974978935

Hence the integral equation can be written as

V(x) = -1.8096351x^3 + 0.28125

1. b) from the above answers it can be clearly seen that all the answers are consistent.

3.

The systems of coupled equations that are given are written below:

xt= ?xt-ty(t)

yt= ?xtyt- ?y(t)

Here, =4

=2

=3

And =3

h = 0.01 and interval [0, 10]

1. a)

Here, x0 = 2

y0 = 1

The solution is given in the below table:

Table 1

Figure 1

1. b) Here, x0 = 2

y0 = 20

The solution is given in the below table:

Table 2

Figure 2

1. c)

Figure 3

Figure 4

1. The given third order differential equation is

d3y/dx3 + 3d2y/dx2 + x dy/dx= x2y

It is considered that

x1 = y

x2 = dy/dx

And x3 = d2y/dx2

Differentiating the above equations, the followings can be written:

x1’ = dy/dx = x2

x2’ = d2y/dx2 = x3

x3’ = d3y/dx3 = x3 = x2x1 – 3x2-xx2

The first order differential equations along with their initial values are written below:

x1’ = x2 x1(0)=2

x2’ = x3 x2(0)=1

x3’ = x2x1 – 3x2-xx2 x3(0) = -1

1. These equations arte solved using the algorithm or Range-Kutta method ( 4 order ). The method is given below:

Let us again consider the first order initial value problem to be

dydx=fx,y, yx0=y0

Range-Kutta method is another numerical method of solving differential equations. This method is also known as predictor-corrector method. The Range-Kutta method can be explained with the help of following expressions:

xn+1 =xn+h

yn+1 =yn+k1+2k2+2k3+k46

Where, k1 = hf(xn,yn)

k2 = hf(xn+h2,yn+k12)

k3 = hf(xn+h2,yn+k22)

k4 = hf(xn+h,yn+k3)

For the first equation, the calculation is done in the table:

x1’ = x2 x1(0)=2

Table 3

Figure 5

For the second equation, the calculation is done in the table:

x2’ = x3 x2(0)=1

Table 4

Figure 6

For the second equation, the calculation is done in the table:

x3’ = x2x1 – 3x2-xx2 x3(0) = -1

Table 5

Figure 7