This problem explores how to scale the data to help improve the computability of the interpolation polynomial. We consider the direct approach to determine
n
(
), and as usual the data points are (
1,
1), (
2,
2), ··· , (
n+1,
n+1), where
1
2
<>
n+1. The
values are going to be scaled by letting
where
and
are given numbers with
> 0. The data point (
i
,
i
) in this case changes to (
i
,
i
), where
i
= (
i
−
)/. Also, the interpolation polynomial also changes to
(a) The original data interval is
1
≤
≤
n+1. What is the data interval when using
? What matrix equation must be solved to find the ai’s in the above formula for
n
()?
(b) The values for
and
are going to be selected so the
data interval is −1 ≤
≤ 1. What are
and
in this case?
(c) Using the population data from Exercise 5.14, plot the interpolation function using the direct approach on the original
i
data set. Also compute the condition number for
V.
(d) Using the population data from Exercise 5.14, scale the data based on the result from part (b), and then find the coefficients for
n
(). What is the condition number of the matrix in this case? Once the
i
’s are computed then in terms of the original
variable,
Plot this function and compare the result with what you found in part (c).
Exercise 5.14
The data considered here are the population of a country for the years
1
= 1900,
2
= 1910,
3
= 1920,
4
= 1930, ··· ,
12
= 2010. The
i
’s are the corresponding population values, and they should be given per million. For example, the population of the USA is given in Table 5.7, and this is from the Wikipedia page Demographics of the United States.
(a) Fit this data with: i) a global polynomial using Lagrange interpolation, and ii) a natural cubic spline. Plot these two curves and the data on the same axis. Make sure to include a legend in your plot.
(b) What do each of the two interpolation functions give as the population in 2005?
(c) What do each of the two interpolation functions predict the population will be in 2015?