#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Mon Apr 19 15:40:37 2021
Plot intermediate points in a linear 'fhat' approximation to 'f'
@author: neil
"""

import numpy as np
import matplotlib.pyplot as plt

def points(x,x0,x1,a0,a1):     # function to return the line or a point within an element
    # x is a point or list of points, x0,x1, are the edges of the element, a0,a1 are the
    # values of the 'f' function at the element edges
    temp = (x - x0)/(x1 - x0)    # note that x can be a list of numbers
    return a0*(1-temp) + a1*temp # can return a list, if x is a list

xi = [0.5,1.5,4,5.5,7]                   # list of element boundaries
xpoints = np.linspace(xi[0],xi[-1],100)  # for plotting, make a list of points
f = np.sin(xpoints) + 0.1*xpoints + 2    # this is the 'f' function we are approximating

fig = plt.figure()           # plot the 'f' function in red, leave the axis to plot 'fhat'
ax  = fig.add_subplot(111)  
#ax.plot(xpoints,f,"r-*")  #
ax.set_ylabel("f(x) or fhat(x)")
ax.set_title("element by element fhat or f")

ao = np.sin(xi[0])+0.1*xi[0]+2
n=10
for i in range(len(xi)-1):              # do each element in sequence
    x = np.linspace(xi[i],xi[i+1],n)    # make 10 x points inside the lement
    f = np.sin(x) + 0.1*x + 2           # produce the f curve inside the element
    s = 0
    t = 0
    for j in range(0,n):
        s += (f[j]*(xi[i+1]-xi[i]) - ao*(xi[i+1]-x[j]))
        t += x[j]-xi[i]
    a1 = s/(t)
    fhat = points(x,xi[i],xi[i+1],ao,a1)
    ax.plot(x,fhat,"g-+")               # plot fhat that was returned by points
    ax.plot(x,f,"r-*")  #
    ax.plot([xi[i],xi[i]],[1,3.5],"k:") # add the element edges
    ao = a1
        
ax.plot([xi[-1],xi[-1]],[1,3.5],"k:")