Question

Someone explains me, please
\(f(x) =x^{p-1}+x^{p-2}+.....+x^2+x+1\)
\(f(x+1) = \sum_{s=0}^{p-1}\left(\begin{matrix}p\\s+1\end{matrix}\right) x^s\)
How to get f(x+1) from f(x) as above?

Answer 1

@tkhunny

Answer 2

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1

p = 2 -- f(x+1) = x+2 = [(x+1)^2 - 1]/x
p = 3 -- f(x+1) = x^2 + 3x + 3 = [(x+1)^3 - 1]/x
p = 4 -- f(x+1) = x^3 + 4x^2 + 6x + 4 = [(x+1)^4 - 1]/x
p = 5 -- f(x+1) = x^4 + 5x^3 + 10x^2 + 10x + 5 = [(x+1)^5 - 1]/x

Starting to seem like you can just collect the like terms. It might seem tedious, at first.

Answer 3

I know pascal, and I know how to expand the sum. But I don't see the link between f(x) and f(x+1)
For example: if p =3, then f(x) = x^2+x+1
f(x+1) =(x+1)^2 +(x+1)+1= x^2+3x+3

Answer 4

\[f(x)=\sum_{k=0}^{p-1}x^k\]

\(\displaystyle f(x+1)=\sum_{k=0}^{p-1}(x+1)^k=\sum_{k=0}^{p-1}\sum_{s=0}^{k}{k\choose s}x^k\) [binomial expansion of \((x+1)^k\)]

\[0\le s\le k\le p-1\]

\[=\sum_{s=0}^{p-1}\sum_{k=s}^{p-1}{k\choose s}x^s\]

\[=\sum_{s=0}^{p-1}x^s\sum_{k=s}^{p-1}{k\choose s}\]

by classifying the s+1 subsets of {1,2,3,...,p} according to the last element s+1 we have...

\[\sum_{k=s}^{p-1}{k\choose s}={p\choose s+1}\]

and thus

\[=\sum_{s=0}^{p-1}x^s\sum_{k=s}^{p-1}{k\choose s}=\sum_{s=0}^{p-1}x^s{p\choose s+1}=\sum_{s=0}^{p-1}{p\choose s+1}x^s\]

Answer 5

Thank you.

Answer 6

Just noticed...the second line has a typo...should be

\[\Large f(x+1)=\sum_{k=0}^{p-1}(x+1)^k=\sum_{k=0}^{p-1}\sum_{s=0}^{k}{k\choose s}x^{\color{red}{s}}\]

Answer 7

Rewriting like that is the only step I was seeing.
Glad you knew where to go from there.