#!/usr/bin/env python
# coding: utf-8

# # Accuracy of Newton-Cotes
# 
# Copyright (C) 2020 Andreas Kloeckner
# 
# <details>
# <summary>MIT License</summary>
# Permission is hereby granted, free of charge, to any person obtaining a copy
# of this software and associated documentation files (the "Software"), to deal
# in the Software without restriction, including without limitation the rights
# to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
# copies of the Software, and to permit persons to whom the Software is
# furnished to do so, subject to the following conditions:
# 
# The above copyright notice and this permission notice shall be included in
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# 
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# IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
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# OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN
# THE SOFTWARE.
# </details>

# In[1]:


import numpy as np
import numpy.linalg as la
import matplotlib.pyplot as pt


# A function to make Vandermonde matrices:
#     
# (Note that the ordering of this matrix matches the convention in our class but *disagrees* with `np.vander`.)

# In[2]:


def vander(nodes, ncolumns=None):
    if ncolumns is None:
        ncolumns = len(nodes)
    result = np.empty((len(nodes), ncolumns))
    for i in range(ncolumns):
        result[:, i] = nodes**i
    return result


# Fix a set of nodes:

# In[3]:


# nodes = [0.5] # Midpoint
# nodes = [0]
#nodes = [0, 1] # Trapezoidal
nodes = [0, 0.5, 1] # Simpson's
#nodes = [0, 1/3, 1]


# Find the weights for the Newton-Cotes rule for the given nodes on $[0,1]$:

# In[4]:


(a, b) = (0, 1)
nodes = np.array(nodes)
n = len(nodes)

degs = np.arange(n)
rhs = 1/(degs+1)*(b**(degs+1 - a**(degs+1)))
weights = la.solve(vander(nodes).T, rhs)
print(weights)


# Here is a function and its definite integral from $0$ to $x$:
# 
# $$\text{int_f}(x)=\int_0^x f(\xi)d\xi$$

# In[5]:


fdeg = 9
def f(x):
    return sum((-x)**i for i in range(fdeg + 1))
def int_f(x):
    return sum(
        (-1)**i*1/(i+1)*(
            (x)**(i+1)-0**(i+1)
            )
        for i in range(fdeg + 1))


# Plotted:

# In[6]:


plot_x = np.linspace(0, 1, 200)

pt.plot(plot_x, f(plot_x), label="f")
pt.fill_between(plot_x, 0*plot_x, f(plot_x),alpha=0.3)
pt.plot(plot_x, int_f(plot_x), label="$\int f$")
pt.grid()
pt.legend(loc="best")


# This here plots the function, the interpolant, and the area under the interpolant:

# In[7]:


# fix nodes
h = 1
x = nodes * h

# find interpolant
coeffs = la.solve(vander(x), f(x))

# evaluate interpolant
plot_x = np.linspace(0, h, 200)
interpolant = vander(plot_x, len(coeffs)) @ coeffs

# plot
pt.plot(plot_x, f(plot_x), label="f")
pt.plot(plot_x, interpolant, label="Interpolant")
pt.fill_between(plot_x, 0*plot_x, interpolant, alpha=0.3, color="green")
pt.plot(x, f(x), "og")
pt.grid()
pt.legend(loc="best")


# Compute the following:
# 
# *   The true integral as `true_val` (from `int_f`)
# *   The quadrature result as `quad` (using `x` and `weights` and `h`)
# *   The error as `err` (the difference of the two)
# 
#     (Do not be tempted to compute a relative error--that has one order lower.)
# 
# Compare the error for $h=1,0.5,0.25$. What order of accuracy do you observe?

# In[8]:



errors = []

for h in [1, 0.5, 0.25, 0.125, 0.125*0.5]:
    true_val = int_f(h)
    quad = h * weights @ f(h * nodes)
    error = abs(quad - true_val)
    print(h, true_val, quad, error)
    
    errors.append(error)


# Estimate the order of accuracy:
# 
# We assume that the error depends on the mesh spacings $h$ as
# $E(h)\approx C h^p$ for some unknown power $p$.  Taking the $\log$
# of this approximate equality reveals a linear function in $p$:
# $$
# E(h) \approx C h^p \quad \iff \quad \log E(h) \approx \log(C) +
# p\log(h).
# $$
# You can now either do a least-squares fit for $\log C$ and $p$ from
# a few data points $(h,E(h))$ (more accurate, more robust), or you
# can use just two grid sizes $h_1$ and $h_2$, and estimate the slope:
# (less accurate, less robust)
# $$
#    p \approx \frac{ \log(\frac{E(h_2)}{E(h_1)}) } {\log(\frac{h_2}{h_1})}.
# $$
# This is called the *empirical order of convergence* or EOC.
# 
# 

# In[9]:


for i in range(len(errors)-1):
    print(np.log(errors[i+1]/errors[i])/np.log(1/2))


# In[ ]: