- #1
relinquished™
- 79
- 0
I am to prove by mathematical induction that for a polynomial of degree n p(x) with leading coefiicient a_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...
[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...