- #1

relinquished™

- 79

- 0

[tex]

\Delta^n p(x) = a_o n!

[/tex]

My proof: By mathematical induction

[tex]

\Delta^1 p(x) = [a_0(x+1) + a_1] - [a_0x + a_1]

[/tex]

[tex]

\Delta^1 p(x) = [a_0x + a_0 + a_1] - [a_0x + a_1]

[/tex]

[tex]

\Delta^1 p(x) = a_0

[/tex]

[tex]

\Delta^1 p(x) = a_0 \cdot 1!

[/tex]

hence, S(1) is true

This is where I have a problem. I assume that [tex]\Delta^n p(x) = a_o n! [/tex] is true... how do i show that S(n+1) is also true? The degree of the polynomial becomes n+1 and my S(n) becomes inapplicable already...