Question

Represent the function f(x) = x^4 - 12x^3 + 24x^2 - 30x +9 and all its successive differences into factorial notation. Hence show that ∆^5f(x) = 0.​

Answer 1

Given :

Function f(x) :

$f(x) = {x}^{4} - 12 {x}^{3} + 24 {x}^{2} + 30x + 9$

To find :

Prove :

${ \Delta}^{5} f(x) = 0$

Solution :

Given function is :

$f(x) = {x}^{4} - 12 {x}^{3} + 24 {x}^{2} + 30x + 9$

(equation 1)

To show successive differences of this function into factorial notation :

we will take that

$f(x) = a{x}^{(4)} + b{x}^{(3)} + c{x}^{(2)} + dx^{(1)} + e$

So by successive differences this function becomes:

$f(x) = a(x)( x - 1)(x - 2)(x - 3) + b(x)(x - 1)(x - 2) + c(x)(x - 1) + dx + e$

(equation 2)

Equation 1 is equal to equation 2 then

${x}^{4} - 12 {x}^{3} + 24 {x}^{2} + 30x + 9 = a(x)( x - 1)(x - 2)(x - 3) + b(x)(x - 1)(x - 2) + c(x)(x - 1) + dx + e \:$

(equation 3)

Putting x = 0,

we get,

e = 9

Then putting e = 9 in equation 3 and x = 1,

we get

d = 1

Then putting e = 9 and d = 1 in equation 3 and x = 2,

we get

c = 13

Then putting e = 9, d = 1 , c = 13 and x = 3,

we get

b = -6

Then comparing both equations, we get

a = 1

so,

by successive difference method

successive difference methodfunction f(x) becomes,

$f(x) = {x}^{(4)} - 6{x}^{(3)} + 13{x}^{(2)} + 1x^{(1)} + 9$

Now, on differentiating f(x)

$\: f(x) = {x}^{(4)} - 6{x}^{(3)} + 13{x}^{(2)} + 1x^{(1)} + 9 \\ \Delta \: f(x) = 4{x}^{(3)} - 18{x}^{(2)} + 26x^{(1)}+ 1$

now differentiating function upto 5th differencial,

we get :

$\Delta ^{2} \: f(x) = 12{x}^{(2)} - 36{x}^{(1)} + 26$

$\Delta ^{3} \: f(x) = 24{x}^{(1)} - 36$

$\Delta ^{4} \: f(x) = 24$

So fifth diffrentiation of f (x) is :

$\bf \: \Delta ^{5} \: f(x) = 0$

proved.