Can Mathematical Induction Prove the Finite Difference Formula for Polynomials?

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The discussion centers on proving the finite difference formula for polynomials using mathematical induction. The formula states that for a polynomial of degree n, \( \Delta^n p(x) = a_0 n! \), where \( a_0 \) is the leading coefficient. The initial proof demonstrates that \( \Delta^1 p(x) = a_0 \cdot 1! \) holds true. The challenge arises in extending the proof to \( S(n+1) \), where the polynomial's degree increases, necessitating a demonstration that the first difference of a polynomial of degree \( n+1 \) is indeed a polynomial of degree \( n \).

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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...
 
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What is the first difference of a polynomial of degree n+1? Isn't it a polynomial of degree n? What is its leading coefficient?
 
First show that n=1,2,...
[tex]\Delta x^{n}=n x^{n-1} +P_{n-2}[/tex]
where P_{n} means some polynomial of degree at most n
and thus P_0=0
and
[tex]\Delta P_{0}=0[/tex]
then show by induction that for k=0,...,n
[tex]\Delta^k x^n=\frac{n!}{(n-k)!}x^{n-k}+P_{n-1-k}[/tex]
Take your result as a special case
 

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