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)#?
2 Answers
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
#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) = 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)#
First find the 4 first "p"s:
Given the function
Therefore:
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: