from __future__ import division
import numpy as np
from scipy import interpolate
import sympy as sym
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D

%matplotlib notebook
sym.init_printing()

"""
Explanation: One-dimensional Lagrange Interpolation
The problem of interpolation or finding the value of a function at an arbitrary point $X$ inside a given domain, provided we have discrete known values of the function inside the same domain is at the heart of the finite element method. In this notebooke we use Lagrange interpolation where the approximation $\hat f(x)$ to the function $f(x)$ is built like:
\begin{equation}
\hat f(x)={L^I}(x)f^I
\end{equation}
In the expression above $L^I$ represents the $I$ Lagrange Polynomial of order $n-1$ and $f^1, f^2,,...,f^n$ are the $n$ known values of the function. Here we are using the summation convention over the repeated superscripts.
The $I$ Lagrange polynomial is given by the recursive expression:
\begin{equation}
{L^I}(x)=\prod_{J=1, J \ne I}^{n}{\frac{{\left( {x - {x^J}} \right)}}{{\left( {{x^I} - {x^J}} \right)}}} 
\end{equation}
in the domain $x\in[-1.0,1.0]$.
We wish to interpolate the function $ f(x)=x^3+4x^2-10 $ assuming we know its value at points $x=-1.0$, $x=1.0$ and $x=0.0$.
End of explanation
"""


def basis_lagrange(x_data, var, cont):
    """Find the basis for the Lagrange interpolant"""
    prod = sym.prod((var - x_data[i])/(x_data[cont] - x_data[i])
                    for i in range(len(x_data)) if i != cont)
    return sym.simplify(prod)

"""
Explanation: First we use a function to generate the Lagrage polynomial of order $order$ at point $i$
End of explanation
"""


fun = lambda x: x**3 + 4*x**2 - 10

x = sym.symbols('x')
x_data = np.array([-1, 1, 0])
f_data = fun(x_data)

"""
Explanation: we now define the function $ f(x)=x^3+4x^2-10 $:
End of explanation
"""


basis = []
for cont in range(len(x_data)):
    basis.append(basis_lagrange(x_data, x, cont))
    sym.pprint(basis[cont])


"""
Explanation: And obtain the Lagrange polynomials using:
End of explanation
"""


npts = 101
x_eval = np.linspace(-1, 1, npts)
basis_num = sym.lambdify((x), basis, "numpy")  # Create a lambda function for the polynomials

plt.figure(figsize=(6, 4))
for k in range(3):    
    y_eval = basis_num(x_eval)[k]
    plt.plot(x_eval, y_eval)

y_interp = sym.simplify(sum(f_data[k]*basis[k] for k in range(3)))
y_interp

"""
Explanation: which are shown in the following plots/
End of explanation
"""


y_interp = sum(f_data[k]*basis_num(x_eval)[k] for k in range(3))
y_original = fun(x_eval)

plt.figure(figsize=(6, 4))
plt.plot(x_eval, y_original)
plt.plot(x_eval, y_interp)
plt.plot([-1, 1, 0], f_data, 'ko')
plt.show()

"""
Explanation: Now we plot the complete approximating polynomial, the actual function and the points where the function was known.
End of explanation
"""


from IPython.core.display import HTML
def css_styling():
    styles = open('../styles/custom_barba.css', 'r').read()
    return HTML(styles)
css_styling()

"""
Explanation: The next cell change the format of the Notebook.
End of explanation
"""


