Problem: If $f$ is a polynomial of degree $4$ such that $$f(0) = f(1) = f(2) = f(3) = 1$$ and $$f(4) = 0,$$ then determine $f(5)$.

f(0)  = f(1) = f(2) = f(3)   = 1

And f(4)  = 0

In the form

ax^4 + bx^3 + cx^2 + dx + e        

If f(0)  = 1, then  e  = 1

And we have this system

a + b + c + d  =  0

16a + 8b + 4c + 2d  = 0

81a + 27b  + 9c + 3d  =  0

256a + 64b + 16c + 4d  = -1 

The solution to this system is

a = -1/24     b  =  1/4      c  =  -11/24   d = 1/4

So...the polynomial  is

(-1/24)x^4  + (1/4)x^3 - (11/24)x^2 +  (1/4)x  + 1

And   f(5)  =  - 4

 If $f$ is a polynomial of degree $4$ such that $$f(0) = f(1) = f(2) = f(3) = 1$$ and $$f(4) = 0,$$ then determine $f(5)$.

\(\small{ \begin{array}{lrrrrrrrrrrrrr} & f(0) && f(1) && f(2) && f(3) && f(4) && {\color{red}f(5)} && \cdots \\ & d_0=1 && 1 && 1 && 1 && 0 && {\color{red}-4} && \cdots \\ \text{1. Difference } && d_1=0 && 0 && 0 && -1 && {\color{red}-4} && \cdots \\ \text{2. Difference } &&& d_2=0 && 0 && -1 && {\color{red}-3} && \cdots \\ \text{3. Difference } &&&& d_3=0 && -1 && {\color{red}-2} && \cdots \\ \text{4. Difference } &&&&& {\color{red}d_4=-1 } && {\color{red}-1} && \cdots \\ \end{array} }\)

\(\text{ ${\color{red}d_4=-1}$ is constant, if a polynomial is of degree 4.} \)

\(f(5) = {\color{red}-1}+(-1)+(-1)+(-1)+0 = -4\)

polynomial:

\(\small{ \begin{array}{rcll} \hline a_n &=& \binom{n-1}{0}\cdot {\color{red}d_0 } + \binom{n-1}{1}\cdot {\color{red}d_1 } + \binom{n-1}{2}\cdot {\color{red}d_2 } + \binom{n-1}{3}\cdot {\color{red}d_3 } + \binom{n-1}{4}\cdot {\color{red}d_4 } \\\\ && \quad {\color{red}d_0 } = 1 \quad {\color{red}d_1}={\color{red}d_2}={\color{red}d_3} = 0 \quad {\color{red}d_4 } = -1 \\\\ a_n &=& \binom{n-1}{0}\cdot 1 + \binom{n-1}{1}\cdot 0 + \binom{n-1}{2}\cdot 0 + \binom{n-1}{3}\cdot 0 + \binom{n-1}{4}\cdot (-1) \\ a_n &=& \binom{n-1}{0}\cdot 1 - \binom{n-1}{4} \quad & | \quad \binom{n-1}{0} = 1 \\ a_n &=& 1 - \dbinom{n-1}{4} \quad & | \quad n = x+1 \\ \mathbf{f(x)} &\mathbf{=}& \mathbf{ 1 - \binom{x}{4} } \\\\ && \quad \dbinom{x}{4} = \left(\dfrac{x}{4}\right)\left(\dfrac{x-1}{3}\right)\left(\dfrac{x-2}{2}\right)\left(\dfrac{x-3}{1}\right) \\\\ \mathbf{f(x)} &\mathbf{=}& \mathbf{ 1 - \dfrac{x(x-1)(x-2)(x-3)}{24} } \\ \end{array} } \)

Hi Heureka and Chris,

You have each given a really good answer here.  

I would have done it similar to Chris but Heureka's answer is really incitful.

I am talking about the first half of Hereka's answer.  I never would have thought to do that and it is so simple!

I will admit that I do not understand the second half of what you have done though Heureka. 

How did that combinatory maths sneak in there??

Thanks very much Heureka :)

This link has both the original Latin and English translation for Part 2 of Ars Conjectandi:  The Doctrine of Permutations and Combinations: Being an Essential and Fundamental Doctrine of Changes.  (1795). (Navigate to page 217 for the English translation of the page Heureka references.) Despite the archaic spelling, syntax, and mathematical notation, this text is very readable. This book includes several related essays and theories from notable mathematicians, contemporary to, and proceeding, Jakob Bernoulli.

Complete English translations of Ars Conjectandi are rare.  This is required reading for Lancelot Link’s school of Mathematics, Physics, Chemistry, and Trolling. I had to brush-up on my Latin.

For anyone interested, here’s Part 4 of  Ars Conjectandi (English) Link

JB, will you sign my printed copies?

GA

correction and the sum:

page 211, 212, 213: https://books.google.de/books?id=B05LAAAAMAAJ&dq=doctrine+of+permutations&pg=PP9&redir_esc=y#v=onepage&q&f=false