Question

If `p(x)` is a polynomial of degree `3` such that `p(i)=1/(i+1)` for all values of `i in {1,2,3,4}` then find `p(5)`?

Answer 1

`p(5) = 2/15`

Explanation:

Since the sample points are evenly spaced, we can analyse this using a method of differences:

Write down the initial sequence of values:

`1/2color(white)(00000)1/3color(white)(00000)1/4color(white)(00000)1/5`

Write the sequence of differences between consecutive terms underneath:

`1/2color(white)(00000)1/3color(white)(00000)1/4color(white)(00000)1/5`

`color(white)(0)-1/6color(white)(00)-1/12color(white)(00)-1/20`

Write the sequence of differences of differences underneath that:

`1/2color(white)(00000)1/3color(white)(00000)1/4color(white)(00000)1/5`

`color(white)(0)-1/6color(white)(00)-1/12color(white)(00)-1/20`

`color(white)(000000)1/12color(white)(0000)1/30`

Write the sequence of differences of those differences underneath:

`1/2color(white)(00000)1/3color(white)(00000)1/4color(white)(00000)1/5`

`color(white)(0)-1/6color(white)(00)-1/12color(white)(00)-1/20`

`color(white)(000000)1/12color(white)(0000)1/30`

`color(white)(0000000)-1/20`

Since the polynomial `p(x)` is cubic, the last row should be constant. So to find another term, add an extra `-1/20` on the end, then reconstruct the terms in the rows above:

`1/2color(white)(00000)1/3color(white)(00000)1/4color(white)(00000)1/5color(white)(00000)color(red)(2/15)`

`color(white)(0)-1/6color(white)(00)-1/12color(white)(00)-1/20color(white)(00)color(red)(-1/15)`

`color(white)(000000)1/12color(white)(0000)1/30color(white)(000)color(red)(-1/60)`

`color(white)(0000000)-1/20color(white)(000)color(red)(-1/20)`

Bonus

If we would like a formula for `p(i)` then we can write one down using the initial term of each of the sequences above as coefficients to find:

`p(i) = color(blue)(1/2)/(0!)+color(blue)(-1/6)/(1!)(i-1)+color(blue)(1/12)/(2!)(i-1)(i-2)+color(blue)(-1/20)/(3!)(i-1)(i-2)(i-3)`

`color(white)(p(i)) = 1/2-1/6i+1/6+1/24i^2-1/8i+1/12-1/120i^3+1/20i^2-11/120i+1/20`

`color(white)(p(i)) = 1/120 (-i^3 + 11i^2 - 46i + 96)`

Answer 2

First find the 4 first "p"s:
`p_(1)=1/(1+1)=1/2`
`p_(2)=1/(2+1)=1/3`
`p_(3)=1/(3+1)=1/4`
`p_(4)=1/(4+1)=1/5`

Given the function `p(i)`:

`p_((i))=a*i^3+b*i^2+c*i+d`
`=>`
`p_((1))=a+b+c+d=1/2`
`p_((2))=8a+4b+2c+d=1/3`
`p_((3))=27a+9b+3c+d=1/4`
`p_((4))=64a+16b+4c+d=1/5`

Therefore:
`a=1/2-b-c-d`
`=> 8(1/2-b-c-d)+4b+2c+d=1/3 => b=...`
`=> ... => c= ...`
`=> ... => d= ...`

you get a number for d, then for c, then for b, then for a

then you put them (a,b,c,d you got) back in the function:

`p_((i))=a*i^3+b*i^2+c*i+d`