Difference between revisions of "User talk:PaoloMartini"

Revision as of 15:43, 14 September 2006

Show that if $p(x)$ is a polynomial of degree $n$ , then $p(x - 1)$ is a polynomial of the same degree.

Definition of polynomial.

p(x) = \sum_{i=0}^n a_i x^i

Binomial theorem.

(a + b)^n = \sum_{i=0}^n {n \choose i} a^{n-1} b^i

Special case.

(x - 1)^n = \sum_{i=0}^n {n \choose i} x^{n-i} (-1)^i

Binomial coefficient simmetry.

{n \choose k} = {n \choose n-k}

Hence:

(x - 1)^n = \sum_{i=0}^n {n \choose i} x^i (-1)^i

p(x-1) = \sum_{i=0}^n a_i (x - 1)^i = \sum_{i=0}^n (a_i \sum_{k=0}^n {n \choose k} x^k (-1)^k).

zZzZ...

@@ Line 1: / Line 1: @@
+Show that if <math>p(x)</math> is a polynomial of degree <math>n</math>, then <math>p(x - 1)</math> is a polynomial of the same degree.
+Definition of polynomial.
 :<math>p(x) = \sum_{i=0}^n a_i x^i </math>
-:<math>(x - 1)^n = \sum_{i=0}^n {n \choose i } x^{n-i} (-1)^i </math>
+Binomial theorem.
+:<math>(a + b)^n = \sum_{i=0}^n {n \choose i} a^{n-1} b^i </math>
+Special case.
+:<math>(x - 1)^n = \sum_{i=0}^n {n \choose i} x^{n-i} (-1)^i </math>
+Binomial coefficient simmetry.
 :<math>{n \choose k} = {n \choose n-k} </math>
+Hence:
 :<math>(x - 1)^n = \sum_{i=0}^n {n \choose i} x^i (-1)^i </math>
@@ Line 10: / Line 24: @@
 = \sum_{i=0}^n (a_i \sum_{k=0}^n {n \choose k} x^k (-1)^k). </math>
-:<math>p(x-1) = \sum_{i=0}^n a_i (x-1)^i </math>
+zZzZ...
-:<math>t = x-1</math>
-:<math>p(t) = \sum_{i=0}^n a_i t^i </math>
+----