Answered: These two functions make way more sense…

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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These two functions make way more sense now but I'm still confused about how I can write a function "newtinterp" out of these two. (Input arguments (x,y,z) where (xi, yi), i = 1,..., n+1, (interpolation data points) and z=(z1, ..., zm)(vectors containing m points on which we want to evaluate the interpolating polynomial)).

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I tried to make an estimate of a population with the newtinterp function above but it somehow does not work. Can somebody please tell me what is wrong with my code?

In [248] import pandas as pd
import matplotlib.pyplot as plt
from scipy. interpolate import CubicSpline
x= [1991, 1996, 2001, 2006, 2011, 2016]
y = [3516000,3762300, 3916200, 4209100, 4399400,4747200]
Z = 0
data = pd.DataFrame({ 'Years': x, 'Population': y })
f = newtinterp(x, y, z)
for x in range (1991, 2017):
print("Population in ",x," is: ", f(x))
[ 3.51600000E+06 4.92600000e+04 -1.84800000e+03
-3.15333333e+01 2.59920000e+00]
TypeError
Input In [248], in <cell line: 11>()
10 f = newtinterp(x, y, z)
11 for x in range (1991, 2017):
---> 12
print("Population in ",x," is: ", f(x))
TypeError: 'numpy. float64' object is not callable
3.08533333e+02
Traceback (most recent call last)

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Follow-up Question

How can I implement this part of the question in the newtinterp function?

Let
f(x)=
1
1+25x²
−1≤ x ≤ 1.
=
Using the newtinterp compute the interpolating polynomial p € Pn, which in-
terpolates the points (xi, f(xi)), i 1,2,..., n + 1 of a uniform partition of [-1,1] for
n = : 5, 10, 20, 100. Let zi, i = 1,2,..., 201, equi-distributed points of [-1,1]. Plot your
interpolants and comment on the results.

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