I am to prove by mathematical induction that for a polynomial of degree n p(x) with leading coefiicient a_0,(adsbygoogle = window.adsbygoogle || []).push({});

[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...

**Physics Forums - The Fusion of Science and Community**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Finite Differences proving

Loading...

Similar Threads for Finite Differences proving | Date |
---|---|

Showing ADM angular momentum to be well-defined and finite | Apr 25, 2013 |

Example of cover (of a set) having finite sub-covers in collection. | Dec 30, 2012 |

Product of a finite complex and a point | Apr 16, 2011 |

Why can finite elements handle complex geometries, but finite differences can't? | Apr 3, 2011 |

Finite rank operators | Jan 22, 2011 |

**Physics Forums - The Fusion of Science and Community**