# Week 43: Solving Differential Equations with Deep Learning

**Morten Hjorth-Jensen**, Department of Physics, University of Oslo and Department of Physics and Astronomy and Facility for Rare ion Beams, Michigan State University

Date: **Apr 23, 2021**

Copyright 1999-2021, Morten Hjorth-Jensen. Released under CC Attribution-NonCommercial 4.0 license

## Ordinary Differential Equations

An ordinary differential equation (ODE) is an equation involving functions having one variable.

In general, an ordinary differential equation looks like

$$
\begin{equation} \label{ode} \tag{1}
f\left(x, \, g(x), \, g'(x), \, g''(x), \, \dots \, , \, g^{(n)}(x)\right) = 0
\end{equation}
$$

where $g(x)$ is the function to find, and $g^{(n)}(x)$ is the $n$-th derivative of $g(x)$.

The $f\left(x, g(x), g'(x), g''(x), \, \dots \, , g^{(n)}(x)\right)$ is just a way to write that there is an expression involving $x$ and $g(x), \ g'(x), \ g''(x), \, \dots \, , \text{ and } g^{(n)}(x)$ on the left side of the equality sign in ([1](#ode)).
The highest order of derivative, that is the value of $n$, determines to the order of the equation.
The equation is referred to as a $n$-th order ODE.
Along with ([1](#ode)), some additional conditions of the function $g(x)$ are typically given
for the solution to be unique.

## The trial solution

Let the trial solution $g_t(x)$ be

$$
\begin{equation}
	g_t(x) = h_1(x) + h_2(x,N(x,P))
\label{_auto1} \tag{2}
\end{equation}
$$

where $h_1(x)$ is a function that makes $g_t(x)$ satisfy a given set
of conditions, $N(x,P)$ a neural network with weights and biases
described by $P$ and $h_2(x, N(x,P))$ some expression involving the
neural network.  The role of the function $h_2(x, N(x,P))$, is to
ensure that the output from $N(x,P)$ is zero when $g_t(x)$ is
evaluated at the values of $x$ where the given conditions must be
satisfied.  The function $h_1(x)$ should alone make $g_t(x)$ satisfy
the conditions.

But what about the network $N(x,P)$?

As described previously, an optimization method could be used to minimize the parameters of a neural network, that being its weights and biases, through backward propagation.

## Minimization process

For the minimization to be defined, we need to have a cost function at hand to minimize.

It is given that $f\left(x, \, g(x), \, g'(x), \, g''(x), \, \dots \, , \, g^{(n)}(x)\right)$ should be equal to zero in ([1](#ode)).
We can choose to consider the mean squared error as the cost function for an input $x$.
Since we are looking at one input, the cost function is just $f$ squared.
The cost function $c\left(x, P \right)$ can therefore be expressed as

$$
C\left(x, P\right) = \big(f\left(x, \, g(x), \, g'(x), \, g''(x), \, \dots \, , \, g^{(n)}(x)\right)\big)^2
$$

If $N$ inputs are given as a vector $\boldsymbol{x}$ with elements $x_i$ for $i = 1,\dots,N$,
the cost function becomes

$$
\begin{equation} \label{cost} \tag{3}
	C\left(\boldsymbol{x}, P\right) = \frac{1}{N} \sum_{i=1}^N \big(f\left(x_i, \, g(x_i), \, g'(x_i), \, g''(x_i), \, \dots \, , \, g^{(n)}(x_i)\right)\big)^2
\end{equation}
$$

The neural net should then find the parameters $P$ that minimizes the cost function in
([3](#cost)) for a set of $N$ training samples $x_i$.

## Minimizing the cost function using gradient descent and automatic differentiation

To perform the minimization using gradient descent, the gradient of $C\left(\boldsymbol{x}, P\right)$ is needed.
It might happen so that finding an analytical expression of the gradient of $C(\boldsymbol{x}, P)$ from ([3](#cost)) gets too messy, depending on which cost function one desires to use.

Luckily, there exists libraries that makes the job for us through automatic differentiation.
Automatic differentiation is a method of finding the derivatives numerically with very high precision.

## Example: Exponential decay

An exponential decay of a quantity $g(x)$ is described by the equation

$$
\begin{equation} \label{solve_expdec} \tag{4}
  g'(x) = -\gamma g(x)
\end{equation}
$$

with $g(0) = g_0$ for some chosen initial value $g_0$.

The analytical solution of ([4](#solve_expdec)) is

$$
\begin{equation}
  g(x) = g_0 \exp\left(-\gamma x\right)
\label{_auto2} \tag{5}
\end{equation}
$$

Having an analytical solution at hand, it is possible to use it to compare how well a neural network finds a solution of ([4](#solve_expdec)).

## The function to solve for

The program will use a neural network to solve

$$
\begin{equation} \label{solveode} \tag{6}
g'(x) = -\gamma g(x)
\end{equation}
$$

where $g(0) = g_0$ with $\gamma$ and $g_0$ being some chosen values.

In this example, $\gamma = 2$ and $g_0 = 10$.

## The trial solution
To begin with, a trial solution $g_t(t)$ must be chosen. A general trial solution for ordinary differential equations could be

$$
g_t(x, P) = h_1(x) + h_2(x, N(x, P))
$$

with $h_1(x)$ ensuring that $g_t(x)$ satisfies some conditions and $h_2(x,N(x, P))$ an expression involving $x$ and the output from the neural network $N(x,P)$ with $P $ being the collection of the weights and biases for each layer. For now, it is assumed that the network consists of one input layer, one hidden layer, and one output layer.

## Setup of Network

In this network, there are no weights and bias at the input layer, so $P = \{ P_{\text{hidden}},  P_{\text{output}} \}$.
If there are $N_{\text{hidden} }$ neurons in the hidden layer, then $P_{\text{hidden}}$ is a $N_{\text{hidden} } \times (1 + N_{\text{input}})$ matrix, given that there are $N_{\text{input}}$ neurons in the input layer.

The first column in $P_{\text{hidden} }$ represents the bias for each neuron in the hidden layer and the second column represents the weights for each neuron in the hidden layer from the input layer.
If there are $N_{\text{output} }$ neurons in the output layer, then $P_{\text{output}} $ is a $N_{\text{output} } \times (1 + N_{\text{hidden} })$ matrix.

Its first column represents the bias of each neuron and the remaining columns represents the weights to each neuron.

It is given that $g(0) = g_0$. The trial solution must fulfill this condition to be a proper solution of ([6](#solveode)). A possible way to ensure that $g_t(0, P) = g_0$, is to let $F(N(x,P)) = x \cdot N(x,P)$ and $A(x) = g_0$. This gives the following trial solution:

$$
\begin{equation} \label{trial} \tag{7}
g_t(x, P) = g_0 + x \cdot N(x, P)
\end{equation}
$$

## Reformulating the problem

We wish that our neural network manages to minimize a given cost function.

A reformulation of out equation, ([6](#solveode)), must therefore be done,
such that it describes the problem a neural network can solve for.

The neural network must find the set of weights and biases $P$ such that the trial solution in ([7](#trial)) satisfies ([6](#solveode)).

The trial solution

$$
g_t(x, P) = g_0 + x \cdot N(x, P)
$$

has been chosen such that it already solves the condition $g(0) = g_0$. What remains, is to find $P$ such that

$$
\begin{equation} \label{nnmin} \tag{8}
g_t'(x, P) = - \gamma g_t(x, P)
\end{equation}
$$

is fulfilled as *best as possible*.

## More technicalities

The left hand side and right hand side of ([8](#nnmin)) must be computed separately, and then the neural network must choose weights and biases, contained in $P$, such that the sides are equal as best as possible.
This means that the absolute or squared difference between the sides must be as close to zero, ideally equal to zero.
In this case, the difference squared shows to be an appropriate measurement of how erroneous the trial solution is with respect to $P$ of the neural network.

This gives the following cost function our neural network must solve for:

$$
\min_{P}\Big\{ \big(g_t'(x, P) - ( -\gamma g_t(x, P) \big)^2 \Big\}
$$

(the notation $\min_{P}\{ f(x, P) \}$ means that we desire to find $P$ that yields the minimum of $f(x, P)$)

or, in terms of weights and biases for the hidden and output layer in our network:

$$
\min_{P_{\text{hidden} }, \ P_{\text{output} }}\Big\{ \big(g_t'(x, \{ P_{\text{hidden} }, P_{\text{output} }\}) - ( -\gamma g_t(x, \{ P_{\text{hidden} }, P_{\text{output} }\}) \big)^2 \Big\}
$$

for an input value $x$.

## More details

If the neural network evaluates $g_t(x, P)$ at more values for $x$, say $N$ values $x_i$ for $i = 1, \dots, N$, then the *total* error to minimize becomes

$$
\begin{equation} \label{min} \tag{9}
\min_{P}\Big\{\frac{1}{N} \sum_{i=1}^N  \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2 \Big\}
\end{equation}
$$

Letting $\boldsymbol{x}$ be a vector with elements $x_i$ and $C(\boldsymbol{x}, P) = \frac{1}{N} \sum_i  \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2$ denote the cost function, the minimization problem that our network must solve, becomes

$$
\min_{P} C(\boldsymbol{x}, P)
$$

In terms of $P_{\text{hidden} }$ and $P_{\text{output} }$, this could also be expressed as

$$
\min_{P_{\text{hidden} }, \ P_{\text{output} }} C(\boldsymbol{x}, \{P_{\text{hidden} }, P_{\text{output} }\})
$$

## A possible implementation of a neural network

For simplicity, it is assumed that the input is an array $\boldsymbol{x} = (x_1, \dots, x_N)$ with $N$ elements. It is at these points the neural network should find $P$ such that it fulfills ([9](#min)).

First, the neural network must feed forward the inputs.
This means that $\boldsymbol{x}s$ must be passed through an input layer, a hidden layer and a output layer. The input layer in this case, does not need to process the data any further.
The input layer will consist of $N_{\text{input} }$ neurons, passing its element to each neuron in the hidden layer.  The number of neurons in the hidden layer will be $N_{\text{hidden} }$.

## Technicalities

For the $i$-th in the hidden layer with weight $w_i^{\text{hidden} }$ and bias $b_i^{\text{hidden} }$, the weighting from the $j$-th neuron at the input layer is:

$$
\begin{aligned}
z_{i,j}^{\text{hidden}} &= b_i^{\text{hidden}} + w_i^{\text{hidden}}x_j \\
&=
\begin{pmatrix}
b_i^{\text{hidden}} & w_i^{\text{hidden}}
\end{pmatrix}
\begin{pmatrix}
1 \\
x_j
\end{pmatrix}
\end{aligned}
$$

## Final technicalities I

The result after weighting the inputs at the $i$-th hidden neuron can be written as a vector:

$$
\begin{aligned}
\boldsymbol{z}_{i}^{\text{hidden}} &= \Big( b_i^{\text{hidden}} + w_i^{\text{hidden}}x_1 , \ b_i^{\text{hidden}} + w_i^{\text{hidden}} x_2, \ \dots \, , \ b_i^{\text{hidden}} + w_i^{\text{hidden}} x_N\Big)  \\
&=
\begin{pmatrix}
 b_i^{\text{hidden}}  & w_i^{\text{hidden}}
\end{pmatrix}
\begin{pmatrix}
1  & 1 & \dots & 1 \\
x_1 & x_2 & \dots & x_N
\end{pmatrix} \\
&= \boldsymbol{p}_{i, \text{hidden}}^T X
\end{aligned}
$$

## Final technicalities II

The vector $\boldsymbol{p}_{i, \text{hidden}}^T$ constitutes each row in $P_{\text{hidden} }$, which contains the weights for the neural network to minimize according to ([9](#min)).

After having found $\boldsymbol{z}_{i}^{\text{hidden}} $ for every $i$-th neuron within the hidden layer, the vector will be sent to an activation function $a_i(\boldsymbol{z})$.

In this example, the sigmoid function has been chosen to be the activation function for each hidden neuron:

$$
f(z) = \frac{1}{1 + \exp{(-z)}}
$$

It is possible to use other activations functions for the hidden layer also.

The output $\boldsymbol{x}_i^{\text{hidden}}$ from each $i$-th hidden neuron is:

$$
\boldsymbol{x}_i^{\text{hidden} } = f\big(  \boldsymbol{z}_{i}^{\text{hidden}} \big)
$$

The outputs $\boldsymbol{x}_i^{\text{hidden} } $ are then sent to the output layer.

The output layer consists of one neuron in this case, and combines the
output from each of the neurons in the hidden layers. The output layer
combines the results from the hidden layer using some weights $w_i^{\text{output}}$
and biases $b_i^{\text{output}}$. In this case,
it is assumes that the number of neurons in the output layer is one.

## Final technicalities III

The procedure of weighting the output neuron $j$ in the hidden layer to the $i$-th neuron in the output layer is similar as for the hidden layer described previously.

$$
\begin{aligned}
z_{1,j}^{\text{output}} & =
\begin{pmatrix}
b_1^{\text{output}} & \boldsymbol{w}_1^{\text{output}}
\end{pmatrix}
\begin{pmatrix}
1 \\
\boldsymbol{x}_j^{\text{hidden}}
\end{pmatrix}
\end{aligned}
$$

## Final technicalities IV

Expressing $z_{1,j}^{\text{output}}$ as a vector gives the following way of weighting the inputs from the hidden layer:

$$
\boldsymbol{z}_{1}^{\text{output}} =
\begin{pmatrix}
b_1^{\text{output}} & \boldsymbol{w}_1^{\text{output}}
\end{pmatrix}
\begin{pmatrix}
1  & 1 & \dots & 1 \\
\boldsymbol{x}_1^{\text{hidden}} & \boldsymbol{x}_2^{\text{hidden}} & \dots & \boldsymbol{x}_N^{\text{hidden}}
\end{pmatrix}
$$

In this case we seek a continuous range of values since we are approximating a function. This means that after computing $\boldsymbol{z}_{1}^{\text{output}}$ the neural network has finished its feed forward step, and $\boldsymbol{z}_{1}^{\text{output}}$ is the final output of the network.

## Back propagation

The next step is to decide how the parameters should be changed such that they minimize the cost function.

The chosen cost function for this problem is

$$
C(\boldsymbol{x}, P) = \frac{1}{N} \sum_i  \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2
$$

In order to minimize the cost function, an optimization method must be chosen.

Here, gradient descent with a constant step size has been chosen.

## Gradient descent

The idea of the gradient descent algorithm is to update parameters in
a direction where the cost function decreases goes to a minimum.

In general, the update of some parameters $\boldsymbol{\omega}$ given a cost
function defined by some weights $\boldsymbol{\omega}$, $C(\boldsymbol{x},
\boldsymbol{\omega})$, goes as follows:

$$
\boldsymbol{\omega}_{\text{new} } = \boldsymbol{\omega} - \lambda \nabla_{\boldsymbol{\omega}} C(\boldsymbol{x}, \boldsymbol{\omega})
$$

for a number of iterations or until $ \big|\big| \boldsymbol{\omega}_{\text{new} } - \boldsymbol{\omega} \big|\big|$ becomes smaller than some given tolerance.

The value of $\lambda$ decides how large steps the algorithm must take
in the direction of $ \nabla_{\boldsymbol{\omega}} C(\boldsymbol{x}, \boldsymbol{\omega})$.
The notation $\nabla_{\boldsymbol{\omega}}$ express the gradient with respect
to the elements in $\boldsymbol{\omega}$.

In our case, we have to minimize the cost function $C(\boldsymbol{x}, P)$ with
respect to the two sets of weights and biases, that is for the hidden
layer $P_{\text{hidden} }$ and for the output layer $P_{\text{output}
}$ .

This means that $P_{\text{hidden} }$ and $P_{\text{output} }$ is updated by

$$
\begin{aligned}
P_{\text{hidden},\text{new}} &= P_{\text{hidden}} - \lambda \nabla_{P_{\text{hidden}}} C(\boldsymbol{x}, P)  \\
P_{\text{output},\text{new}} &= P_{\text{output}} - \lambda \nabla_{P_{\text{output}}} C(\boldsymbol{x}, P)
\end{aligned}
$$

## The code for solving the ODE

```python
%matplotlib inline

import autograd.numpy as np
from autograd import grad, elementwise_grad
import autograd.numpy.random as npr
from matplotlib import pyplot as plt

def sigmoid(z):
    return 1/(1 + np.exp(-z))

# Assuming one input, hidden, and output layer
def neural_network(params, x):

    # Find the weights (including and biases) for the hidden and output layer.
    # Assume that params is a list of parameters for each layer.
    # The biases are the first element for each array in params,
    # and the weights are the remaning elements in each array in params.

    w_hidden = params[0]
    w_output = params[1]

    # Assumes input x being an one-dimensional array
    num_values = np.size(x)
    x = x.reshape(-1, num_values)

    # Assume that the input layer does nothing to the input x
    x_input = x

    ## Hidden layer:

    # Add a row of ones to include bias
    x_input = np.concatenate((np.ones((1,num_values)), x_input ), axis = 0)

    z_hidden = np.matmul(w_hidden, x_input)
    x_hidden = sigmoid(z_hidden)

    ## Output layer:

    # Include bias:
    x_hidden = np.concatenate((np.ones((1,num_values)), x_hidden ), axis = 0)

    z_output = np.matmul(w_output, x_hidden)
    x_output = z_output

    return x_output

# The trial solution using the deep neural network:
def g_trial(x,params, g0 = 10):
    return g0 + x*neural_network(params,x)

# The right side of the ODE:
def g(x, g_trial, gamma = 2):
    return -gamma*g_trial

# The cost function:
def cost_function(P, x):

    # Evaluate the trial function with the current parameters P
    g_t = g_trial(x,P)

    # Find the derivative w.r.t x of the neural network
    d_net_out = elementwise_grad(neural_network,1)(P,x)

    # Find the derivative w.r.t x of the trial function
    d_g_t = elementwise_grad(g_trial,0)(x,P)

    # The right side of the ODE
    func = g(x, g_t)

    err_sqr = (d_g_t - func)**2
    cost_sum = np.sum(err_sqr)

    return cost_sum / np.size(err_sqr)

# Solve the exponential decay ODE using neural network with one input, hidden, and output layer
def solve_ode_neural_network(x, num_neurons_hidden, num_iter, lmb):
    ## Set up initial weights and biases

    # For the hidden layer
    p0 = npr.randn(num_neurons_hidden, 2 )

    # For the output layer
    p1 = npr.randn(1, num_neurons_hidden + 1 ) # +1 since bias is included

    P = [p0, p1]

    print('Initial cost: %g'%cost_function(P, x))

    ## Start finding the optimal weights using gradient descent

    # Find the Python function that represents the gradient of the cost function
    # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer
    cost_function_grad = grad(cost_function,0)

    # Let the update be done num_iter times
    for i in range(num_iter):
        # Evaluate the gradient at the current weights and biases in P.
        # The cost_grad consist now of two arrays;
        # one for the gradient w.r.t P_hidden and
        # one for the gradient w.r.t P_output
        cost_grad =  cost_function_grad(P, x)

        P[0] = P[0] - lmb * cost_grad[0]
        P[1] = P[1] - lmb * cost_grad[1]

    print('Final cost: %g'%cost_function(P, x))

    return P

def g_analytic(x, gamma = 2, g0 = 10):
    return g0*np.exp(-gamma*x)

# Solve the given problem
if __name__ == '__main__':
    # Set seed such that the weight are initialized
    # with same weights and biases for every run.
    npr.seed(15)

    ## Decide the vales of arguments to the function to solve
    N = 10
    x = np.linspace(0, 1, N)

    ## Set up the initial parameters
    num_hidden_neurons = 10
    num_iter = 10000
    lmb = 0.001

    # Use the network
    P = solve_ode_neural_network(x, num_hidden_neurons, num_iter, lmb)

    # Print the deviation from the trial solution and true solution
    res = g_trial(x,P)
    res_analytical = g_analytic(x)

    print('Max absolute difference: %g'%np.max(np.abs(res - res_analytical)))

    # Plot the results
    plt.figure(figsize=(10,10))

    plt.title('Performance of neural network solving an ODE compared to the analytical solution')
    plt.plot(x, res_analytical)
    plt.plot(x, res[0,:])
    plt.legend(['analytical','nn'])
    plt.xlabel('x')
    plt.ylabel('g(x)')
    plt.show()
```

## The network with one input layer, specified number of hidden layers, and one output layer

It is also possible to extend the construction of our network into a more general one, allowing the network to contain more than one hidden layers.

The number of neurons within each hidden layer are given as a list of integers in the program below.

```python
import autograd.numpy as np
from autograd import grad, elementwise_grad
import autograd.numpy.random as npr
from matplotlib import pyplot as plt

def sigmoid(z):
    return 1/(1 + np.exp(-z))

# The neural network with one input layer and one output layer,
# but with number of hidden layers specified by the user.
def deep_neural_network(deep_params, x):
    # N_hidden is the number of hidden layers

    N_hidden = np.size(deep_params) - 1 # -1 since params consists of
                                        # parameters to all the hidden
                                        # layers AND the output layer.

    # Assumes input x being an one-dimensional array
    num_values = np.size(x)
    x = x.reshape(-1, num_values)

    # Assume that the input layer does nothing to the input x
    x_input = x

    # Due to multiple hidden layers, define a variable referencing to the
    # output of the previous layer:
    x_prev = x_input

    ## Hidden layers:

    for l in range(N_hidden):
        # From the list of parameters P; find the correct weigths and bias for this layer
        w_hidden = deep_params[l]

        # Add a row of ones to include bias
        x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0)

        z_hidden = np.matmul(w_hidden, x_prev)
        x_hidden = sigmoid(z_hidden)

        # Update x_prev such that next layer can use the output from this layer
        x_prev = x_hidden

    ## Output layer:

    # Get the weights and bias for this layer
    w_output = deep_params[-1]

    # Include bias:
    x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0)

    z_output = np.matmul(w_output, x_prev)
    x_output = z_output

    return x_output

# The trial solution using the deep neural network:
def g_trial_deep(x,params, g0 = 10):
    return g0 + x*deep_neural_network(params, x)

# The right side of the ODE:
def g(x, g_trial, gamma = 2):
    return -gamma*g_trial

# The same cost function as before, but calls deep_neural_network instead.
def cost_function_deep(P, x):

    # Evaluate the trial function with the current parameters P
    g_t = g_trial_deep(x,P)

    # Find the derivative w.r.t x of the neural network
    d_net_out = elementwise_grad(deep_neural_network,1)(P,x)

    # Find the derivative w.r.t x of the trial function
    d_g_t = elementwise_grad(g_trial_deep,0)(x,P)

    # The right side of the ODE
    func = g(x, g_t)

    err_sqr = (d_g_t - func)**2
    cost_sum = np.sum(err_sqr)

    return cost_sum / np.size(err_sqr)

# Solve the exponential decay ODE using neural network with one input and one output layer,
# but with specified number of hidden layers from the user.
def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb):
    # num_hidden_neurons is now a list of number of neurons within each hidden layer

    # The number of elements in the list num_hidden_neurons thus represents
    # the number of hidden layers.

    # Find the number of hidden layers:
    N_hidden = np.size(num_neurons)

    ## Set up initial weights and biases

    # Initialize the list of parameters:
    P = [None]*(N_hidden + 1) # + 1 to include the output layer

    P[0] = npr.randn(num_neurons[0], 2 )
    for l in range(1,N_hidden):
        P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias

    # For the output layer
    P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included

    print('Initial cost: %g'%cost_function_deep(P, x))

    ## Start finding the optimal weights using gradient descent

    # Find the Python function that represents the gradient of the cost function
    # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer
    cost_function_deep_grad = grad(cost_function_deep,0)

    # Let the update be done num_iter times
    for i in range(num_iter):
        # Evaluate the gradient at the current weights and biases in P.
        # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases
        # in the hidden layers and output layers evaluated at x.
        cost_deep_grad =  cost_function_deep_grad(P, x)

        for l in range(N_hidden+1):
            P[l] = P[l] - lmb * cost_deep_grad[l]

    print('Final cost: %g'%cost_function_deep(P, x))

    return P

def g_analytic(x, gamma = 2, g0 = 10):
    return g0*np.exp(-gamma*x)

# Solve the given problem
if __name__ == '__main__':
    npr.seed(15)

    ## Decide the vales of arguments to the function to solve
    N = 10
    x = np.linspace(0, 1, N)

    ## Set up the initial parameters
    num_hidden_neurons = np.array([10,10])
    num_iter = 10000
    lmb = 0.001

    P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb)

    res = g_trial_deep(x,P)
    res_analytical = g_analytic(x)

    plt.figure(figsize=(10,10))

    plt.title('Performance of a deep neural network solving an ODE compared to the analytical solution')
    plt.plot(x, res_analytical)
    plt.plot(x, res[0,:])
    plt.legend(['analytical','dnn'])
    plt.ylabel('g(x)')
    plt.show()
```

## Example: Population growth, comparing Autograd, TensorFlow, and Euler's scheme

A logistic model of population growth assumes that a population converges toward an equilibrium.
The population growth can be modeled by

$$
\begin{equation} \label{log} \tag{10}
	g'(t) = \alpha g(t)(A - g(t))
\end{equation}
$$

where $g(t)$ is the population density at time $t$, $\alpha > 0$ the growth rate and $A > 0$ is the maximum population number in the environment.
Also, at $t = 0$ the population has the size $g(0) = g_0$, where $g_0$ is some chosen constant.

In this example, similar network as for the exponential decay using Autograd has been used to solve the equation. However, as the implementation might suffer from e.g numerical instability
and high execution time (this might be more apparent in the examples solving PDEs),
a network has been constructed using TensorFlow also.
For comparison, the forward Euler method has been implemented in order to see how the networks performs compared to a numerical scheme.

## Setting up the problem

Here, we will model a population $g(t)$ in an environment having carrying capacity $A$.
The population follows the model

$$
\begin{equation} \label{solveode_population} \tag{11}
g'(t) = \alpha g(t)(A - g(t))
\end{equation}
$$

where $g(0) = g_0$.

In this example, we let $\alpha = 2$, $A = 1$, and $g_0 = 1.2$.

## The trial solution

We will get a slightly different trial solution, as the boundary conditions are different
compared to the case for exponential decay.

A possible trial solution satisfying the condition $g(0) = g_0$ could be

$$
h_1(t) = g_0 + t \cdot N(t,P)
$$

with $N(t,P)$ being the output from the neural network with weights and biases for each layer collected in the set $P$.

The analytical solution is

$$
g(t) = \frac{Ag_0}{g_0 + (A - g_0)\exp(-\alpha A t)}
$$

## The program using Autograd

The network will be the similar as for the exponential decay example, but with some small modifications for our problem.

```python
import autograd.numpy as np
from autograd import grad, elementwise_grad
import autograd.numpy.random as npr
from matplotlib import pyplot as plt

def sigmoid(z):
    return 1/(1 + np.exp(-z))

# Function to get the parameters.
# Done such that one can easily change the paramaters after one's liking.
def get_parameters():
    alpha = 2
    A = 1
    g0 = 1.2
    return alpha, A, g0

def deep_neural_network(P, x):
    # N_hidden is the number of hidden layers
    N_hidden = np.size(P) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer

    # Assumes input x being an one-dimensional array
    num_values = np.size(x)
    x = x.reshape(-1, num_values)

    # Assume that the input layer does nothing to the input x
    x_input = x

    # Due to multiple hidden layers, define a variable referencing to the
    # output of the previous layer:
    x_prev = x_input

    ## Hidden layers:

    for l in range(N_hidden):
        # From the list of parameters P; find the correct weigths and bias for this layer
        w_hidden = P[l]

        # Add a row of ones to include bias
        x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0)

        z_hidden = np.matmul(w_hidden, x_prev)
        x_hidden = sigmoid(z_hidden)

        # Update x_prev such that next layer can use the output from this layer
        x_prev = x_hidden

    ## Output layer:

    # Get the weights and bias for this layer
    w_output = P[-1]

    # Include bias:
    x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0)

    z_output = np.matmul(w_output, x_prev)
    x_output = z_output

    return x_output

def cost_function_deep(P, x):

    # Evaluate the trial function with the current parameters P
    g_t = g_trial_deep(x,P)

    # Find the derivative w.r.t x of the trial function
    d_g_t = elementwise_grad(g_trial_deep,0)(x,P)

    # The right side of the ODE
    func = f(x, g_t)

    err_sqr = (d_g_t - func)**2
    cost_sum = np.sum(err_sqr)

    return cost_sum / np.size(err_sqr)

# The right side of the ODE:
def f(x, g_trial):
    alpha,A, g0 = get_parameters()
    return alpha*g_trial*(A - g_trial)

# The trial solution using the deep neural network:
def g_trial_deep(x, params):
    alpha,A, g0 = get_parameters()
    return g0 + x*deep_neural_network(params,x)

# The analytical solution:
def g_analytic(t):
    alpha,A, g0 = get_parameters()
    return A*g0/(g0 + (A - g0)*np.exp(-alpha*A*t))

def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb):
    # num_hidden_neurons is now a list of number of neurons within each hidden layer

    # Find the number of hidden layers:
    N_hidden = np.size(num_neurons)

    ## Set up initial weigths and biases

    # Initialize the list of parameters:
    P = [None]*(N_hidden + 1) # + 1 to include the output layer

    P[0] = npr.randn(num_neurons[0], 2 )
    for l in range(1,N_hidden):
        P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias

    # For the output layer
    P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included

    print('Initial cost: %g'%cost_function_deep(P, x))

    ## Start finding the optimal weigths using gradient descent

    # Find the Python function that represents the gradient of the cost function
    # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer
    cost_function_deep_grad = grad(cost_function_deep,0)

    # Let the update be done num_iter times
    for i in range(num_iter):
        # Evaluate the gradient at the current weights and biases in P.
        # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases
        # in the hidden layers and output layers evaluated at x.
        cost_deep_grad =  cost_function_deep_grad(P, x)

        for l in range(N_hidden+1):
            P[l] = P[l] - lmb * cost_deep_grad[l]

    print('Final cost: %g'%cost_function_deep(P, x))

    return P

if __name__ == '__main__':
    npr.seed(4155)

    ## Decide the vales of arguments to the function to solve
    Nt = 10
    T = 1
    t = np.linspace(0,T, Nt)

    ## Set up the initial parameters
    num_hidden_neurons = [100, 50, 25]
    num_iter = 1000
    lmb = 1e-3

    P = solve_ode_deep_neural_network(t, num_hidden_neurons, num_iter, lmb)

    g_dnn_ag = g_trial_deep(t,P)
    g_analytical = g_analytic(t)

    # Find the maximum absolute difference between the solutons:
    diff_ag = np.max(np.abs(g_dnn_ag - g_analytical))
    print("The max absolute difference between the solutions is: %g"%diff_ag)

    plt.figure(figsize=(10,10))

    plt.title('Performance of neural network solving an ODE compared to the analytical solution')
    plt.plot(t, g_analytical)
    plt.plot(t, g_dnn_ag[0,:])
    plt.legend(['analytical','nn'])
    plt.xlabel('t')
    plt.ylabel('g(t)')

    plt.show()
```

## Using forward Euler to solve the ODE

A straight-forward way of solving an ODE numerically, is to use Euler's method.

Euler's method uses Taylor series to approximate the value at a function $f$ at a step $\Delta x$ from $x$:

$$
f(x + \Delta x) \approx f(x) + \Delta x f'(x)
$$

In our case, using Euler's method to approximate the value of $g$ at a step $\Delta t$ from $t$ yields

$$
\begin{aligned}
  g(t + \Delta t) &\approx g(t) + \Delta t g'(t) \\
  &= g(t) + \Delta t \big(\alpha g(t)(A - g(t))\big)
\end{aligned}
$$

along with the condition that $g(0) = g_0$.

Let $t_i = i \cdot \Delta t$ where $\Delta t = \frac{T}{N_t-1}$ where $T$ is the final time our solver must solve for and $N_t$ the number of values for $t \in [0, T]$ for $i = 0, \dots, N_t-1$.

For $i \geq 1$, we have that

$$
\begin{aligned}
t_i &= i\Delta t \\
&= (i - 1)\Delta t + \Delta t \\
&= t_{i-1} + \Delta t
\end{aligned}
$$

Now, if $g_i = g(t_i)$ then

$$
\begin{equation}
  \begin{aligned}
  g_i &= g(t_i) \\
  &= g(t_{i-1} + \Delta t) \\
  &\approx g(t_{i-1}) + \Delta t \big(\alpha g(t_{i-1})(A - g(t_{i-1}))\big) \\
  &= g_{i-1} + \Delta t \big(\alpha g_{i-1}(A - g_{i-1})\big)
  \end{aligned}
\end{equation} \label{odenum} \tag{12}
$$

for $i \geq 1$ and $g_0 = g(t_0) = g(0) = g_0$.

Equation ([12](#odenum)) could be implemented in the following way,
extending the program that uses the network using Autograd:

```python
# Assume that all function definitions from the example program using Autograd
# are located here.

if __name__ == '__main__':
    npr.seed(4155)

    ## Decide the vales of arguments to the function to solve
    Nt = 10
    T = 1
    t = np.linspace(0,T, Nt)

    ## Set up the initial parameters
    num_hidden_neurons = [100,50,25]
    num_iter = 1000
    lmb = 1e-3

    P = solve_ode_deep_neural_network(t, num_hidden_neurons, num_iter, lmb)

    g_dnn_ag = g_trial_deep(t,P)
    g_analytical = g_analytic(t)

    # Find the maximum absolute difference between the solutons:
    diff_ag = np.max(np.abs(g_dnn_ag - g_analytical))
    print("The max absolute difference between the solutions is: %g"%diff_ag)

    plt.figure(figsize=(10,10))

    plt.title('Performance of neural network solving an ODE compared to the analytical solution')
    plt.plot(t, g_analytical)
    plt.plot(t, g_dnn_ag[0,:])
    plt.legend(['analytical','nn'])
    plt.xlabel('t')
    plt.ylabel('g(t)')

    ## Find an approximation to the funtion using forward Euler

    alpha, A, g0 = get_parameters()
    dt = T/(Nt - 1)

    # Perform forward Euler to solve the ODE
    g_euler = np.zeros(Nt)
    g_euler[0] = g0

    for i in range(1,Nt):
        g_euler[i] = g_euler[i-1] + dt*(alpha*g_euler[i-1]*(A - g_euler[i-1]))

    # Print the errors done by each method
    diff1 = np.max(np.abs(g_euler - g_analytical))
    diff2 = np.max(np.abs(g_dnn_ag[0,:] - g_analytical))

    print('Max absolute difference between Euler method and analytical: %g'%diff1)
    print('Max absolute difference between deep neural network and analytical: %g'%diff2)

    # Plot results
    plt.figure(figsize=(10,10))

    plt.plot(t,g_euler)
    plt.plot(t,g_analytical)
    plt.plot(t,g_dnn_ag[0,:])

    plt.legend(['euler','analytical','dnn'])
    plt.xlabel('Time t')
    plt.ylabel('g(t)')

    plt.show()
```

## Using TensorFlow to model logistic population growth

TensorFlow is a library widely used in the machine learning community.

Tensorflow is an open source library machine learning library
developed by the Google Brain team for internal use. It was released
under the Apache 2.0 open source license in November 9, 2015.

Tensorflow is a computational framework that allows you to construct
machine learning models at different levels of abstraction, from
high-level, object-oriented APIs like Keras, down to the C++ kernels
that Tensorflow is built upon. The higher levels of abstraction are
simpler to use, but less flexible, and our choice of implementation
should reflect the problems we are trying to solve.

[Tensorflow uses](https://www.tensorflow.org/guide/graphs) so-called graphs to represent your computation
in terms of the dependencies between individual operations, such that you first build a Tensorflow *graph*
to represent your model, and then create a Tensorflow *session* to run the graph.

In this guide we will analyze the same data as we did in our NumPy and
scikit-learn tutorial, gathered from the MNIST database of images. We
will give an introduction to the lower level Python Application
Program Interfaces (APIs), and see how we use them to build our graph.
Then we will build (effectively) the same graph in Keras, to see just
how simple solving a machine learning problem can be.

To install tensorflow on Unix/Linux systems, use pip as

```python
pip3 install tensorflow
```

and/or if you use **anaconda**, just write (or install from the graphical user interface)
(current release of CPU-only TensorFlow)

```python
conda create -n tf tensorflow
conda activate tf
```

To install the current release of GPU TensorFlow

```python
conda create -n tf-gpu tensorflow-gpu
conda activate tf-gpu
```

## Using Keras

Keras is a high level [neural network](https://en.wikipedia.org/wiki/Application_programming_interface)
that supports Tensorflow, CTNK and Theano as backends.
If you have Anaconda installed you may run the following command

```python
conda install keras
```

You can look up the [instructions here](https://keras.io/) for more information.

## The general program flow in TensorFlow

Usually, a program in TensorFlow is divided into two parts; the
*construction phase* and the *execution phase*.  In the construction
phase, the computational graph that TensorFlow uses to perform its
calculations are set up.  In the execution phase, TensorFlow evaluates
any procedure that was defined in the construction phase.

## Program flow in TensorFlow - Construction phase

Here, the architecture for the neural network will be set up, along
with the cost function and an optimizer class used during training of
the network.  Note that TensorFlow uses a different convention for the
weighting done in each neuron in each layer within the network than in
the implementation using Autograd.  The matrix-vector multiplication
between the input from the previous layer and the weighting at the
neuron at current layer in the program using Autograd, is the
transpose of the convention used in TensorFlow. But it will not affect
that much our construction, as TensorFlow takes care of most of the
computations. The only thing we have to be aware of, is how the
dimensions are for our inputs.

## Program flow in TensorFlow - Execution phase

The computation graph has been defined, and is ready to be evaluated.
In order to get access to the graph, it has to be initialized and be runned within a Session.

## The full program modeling logistic population growth using TensorFlow

```python
# import necessary packages
import numpy as np
import matplotlib.pyplot as plt
import tensorflow as tf

# Just to reset the graph such that it is possible to rerun this in a
# Jupyter cell without resetting the whole kernel.
tf.reset_default_graph()

# Set a seed to ensure getting the same results from every run
tf.set_random_seed(4155)

Nt = 10
T = 1
t = np.linspace(0,T, Nt)

## The construction phase

# Convert the values the trial solution is evaluated at to a tensor.
t_tf = tf.convert_to_tensor(t.reshape(-1,1),dtype=tf.float64)
zeros = tf.reshape(tf.convert_to_tensor(np.zeros(t.shape)),shape=(-1,1))

# Define the parameters of the equation
alpha = tf.constant(2.,dtype=tf.float64)
A = tf.constant(1.,dtype=tf.float64)
g0 = tf.constant(1.2,dtype=tf.float64)

num_iter = 100000

# Define the number of neurons at each hidden layer
num_hidden_neurons = [100,50,25]
num_hidden_layers = np.size(num_hidden_neurons)

# Construct the network.
# tf.name_scope is used to group each step in the construction,
# just for a more organized visualization in TensorBoard
with tf.name_scope('dnn'):

    # Input layer
    previous_layer = t_tf

    # Hidden layers
    for l in range(num_hidden_layers):
        current_layer = tf.layers.dense(previous_layer, num_hidden_neurons[l], name='hidden%d'%(l+1), activation=tf.nn.sigmoid)
        previous_layer = current_layer

    # Output layer
    dnn_output = tf.layers.dense(previous_layer, 1, name='output')

# Define the cost function
with tf.name_scope('cost'):
    g_trial = g0 + t_tf*dnn_output
    d_g_trial = tf.gradients(g_trial,t_tf)

    func = alpha*g_trial*(A - g_trial)
    cost = tf.losses.mean_squared_error(zeros, d_g_trial[0] - func)

# Choose the method to minimize the cost function, along with a learning rate
learning_rate = 1e-2
with tf.name_scope('train'):
    optimizer = tf.train.GradientDescentOptimizer(learning_rate)
    traning_op = optimizer.minimize(cost)

# Set up a referance to the result from the neural network:
g_dnn_tf = None

# Define a node that initializes all of the other nodes in the computational graph
# used by TensorFlow:
init = tf.global_variables_initializer()

## Execution phase

# Start a session where the graph defined from the construction phase can be evaluated at:
with tf.Session() as sess:
    # Initialize the whole graph
    init.run()

    # Evaluate the initial cost:
    print('Initial cost: %g'%cost.eval())

    # The training of the network:
    for i in range(num_iter):
        sess.run(traning_op)

        # If one desires to see how the cost function behaves for each iteration:
        #if i % 1000 == 0:
        #    print(cost.eval())

    # Training is done, and we have an approximate solution to the ODE
    print('Final cost: %g'%cost.eval())

    # Store the result
    g_dnn_tf = g_trial.eval()

# Compare with analytical solution
def get_parameters():
    alpha = 2
    A = 1
    g0 = 1.2
    return alpha, A, g0

def g_analytic(t):
    alpha,A, g0 = get_parameters()
    return A*g0/(g0 + (A - g0)*np.exp(-alpha*A*t))

g_analytical = g_analytic(t)
diff_tf = g_dnn_tf - g_analytical.reshape(-1,1)

print('\nMax absolute difference between the analytical solution and solution from TensorFlow DNN: %g'%np.max(np.abs(diff_tf)))

# Plot the result
plt.figure(figsize=(10,10))

plt.title('Numerical solutions of the ODE')

plt.plot(t, g_dnn_tf)
plt.plot(t, g_analytical)

plt.legend(['dnn, tensorflow', 'exact'])
plt.xlabel('Time t')
plt.ylabel('g(t)')

plt.show()
```

```python

# ensure the same random numbers appear every time
np.random.seed(0)

# display images in notebook
%matplotlib inline
plt.rcParams['figure.figsize'] = (12,12)

# download MNIST dataset
digits = datasets.load_digits()

# define inputs and labels
inputs = digits.images
labels = digits.target

print("inputs = (n_inputs, pixel_width, pixel_height) = " + str(inputs.shape))
print("labels = (n_inputs) = " + str(labels.shape))

# flatten the image
# the value -1 means dimension is inferred from the remaining dimensions: 8x8 = 64
n_inputs = len(inputs)
inputs = inputs.reshape(n_inputs, -1)
print("X = (n_inputs, n_features) = " + str(inputs.shape))

# choose some random images to display
indices = np.arange(n_inputs)
random_indices = np.random.choice(indices, size=5)

for i, image in enumerate(digits.images[random_indices]):
    plt.subplot(1, 5, i+1)
    plt.axis('off')
    plt.imshow(image, cmap=plt.cm.gray_r, interpolation='nearest')
    plt.title("Label: %d" % digits.target[random_indices[i]])
plt.show()
```

```python
from tensorflow.keras.layers import Input
from tensorflow.keras.models import Sequential      #This allows appending layers to existing models
from tensorflow.keras.layers import Dense           #This allows defining the characteristics of a particular layer
from tensorflow.keras import optimizers             #This allows using whichever optimiser we want (sgd,adam,RMSprop)
from tensorflow.keras import regularizers           #This allows using whichever regularizer we want (l1,l2,l1_l2)
from tensorflow.keras.utils import to_categorical   #This allows using categorical cross entropy as the cost function

from sklearn.model_selection import train_test_split

# one-hot representation of labels
labels = to_categorical(labels)

# split into train and test data
train_size = 0.8
test_size = 1 - train_size
X_train, X_test, Y_train, Y_test = train_test_split(inputs, labels, train_size=train_size,
                                                    test_size=test_size)
```

```python

epochs = 100
batch_size = 100
n_neurons_layer1 = 100
n_neurons_layer2 = 50
n_categories = 10
eta_vals = np.logspace(-5, 1, 7)
lmbd_vals = np.logspace(-5, 1, 7)
def create_neural_network_keras(n_neurons_layer1, n_neurons_layer2, n_categories, eta, lmbd):
    model = Sequential()
    model.add(Dense(n_neurons_layer1, activation='sigmoid', kernel_regularizer=regularizers.l2(lmbd)))
    model.add(Dense(n_neurons_layer2, activation='sigmoid', kernel_regularizer=regularizers.l2(lmbd)))
    model.add(Dense(n_categories, activation='softmax'))

    sgd = optimizers.SGD(lr=eta)
    model.compile(loss='categorical_crossentropy', optimizer=sgd, metrics=['accuracy'])

    return model
```

```python
DNN_keras = np.zeros((len(eta_vals), len(lmbd_vals)), dtype=object)

for i, eta in enumerate(eta_vals):
    for j, lmbd in enumerate(lmbd_vals):
        DNN = create_neural_network_keras(n_neurons_layer1, n_neurons_layer2, n_categories,
                                         eta=eta, lmbd=lmbd)
        DNN.fit(X_train, Y_train, epochs=epochs, batch_size=batch_size, verbose=0)
        scores = DNN.evaluate(X_test, Y_test)

        DNN_keras[i][j] = DNN

        print("Learning rate = ", eta)
        print("Lambda = ", lmbd)
        print("Test accuracy: %.3f" % scores[1])
        print()
```

```python
# optional
# visual representation of grid search
# uses seaborn heatmap, could probably do this in matplotlib
import seaborn as sns

sns.set()

train_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))
test_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))

for i in range(len(eta_vals)):
    for j in range(len(lmbd_vals)):
        DNN = DNN_keras[i][j]

        train_accuracy[i][j] = DNN.evaluate(X_train, Y_train)[1]
        test_accuracy[i][j] = DNN.evaluate(X_test, Y_test)[1]

fig, ax = plt.subplots(figsize = (10, 10))
sns.heatmap(train_accuracy, annot=True, ax=ax, cmap="viridis")
ax.set_title("Training Accuracy")
ax.set_ylabel("$\eta$")
ax.set_xlabel("$\lambda$")
plt.show()

fig, ax = plt.subplots(figsize = (10, 10))
sns.heatmap(test_accuracy, annot=True, ax=ax, cmap="viridis")
ax.set_title("Test Accuracy")
ax.set_ylabel("$\eta$")
ax.set_xlabel("$\lambda$")
plt.show()
```


Computational Seismology

Numerical derivatives based on the Fourier Transform

---

This notebook is part of the supplementary material
to [Computational Seismology: A Practical Introduction](https://global.oup.com/academic/product/computational-seismology-9780198717416?cc=de&lang=en&#),
Oxford University Press, 2016.

##### Authors:
* Fabian Linder ([@fablindner](https://github.com/fablindner))
* Heiner Igel ([@heinerigel](https://github.com/heinerigel))
* David Vargas ([@dvargas](https://github.com/davofis))
---

## Basic Equations
The derivative of function $f(x)$ with respect to the spatial coordinate $x$ is calculated using the differentiation theorem of the Fourier transform:

\begin{equation}
\frac{d}{dx} f(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} ik F(k) e^{ikx} dk
\end{equation}

In general, this formulation can be extended to compute the n−th derivative of $f(x)$ by considering that $F^{(n)}(k) = D(k)^{n}F(k) = (ik)^{n}F(k)$. Next, the inverse Fourier transform is taken to return to physical space.

\begin{equation}
f^{(n)}(x) = \mathscr{F}^{-1}[(ik)^{n}F(k)] = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} (ik)^{n} F(k) e^{ikx} dk
\end{equation}

```
# Import all necessary libraries, this is a configuration step for the exercise.
# Please run it before the simulation code!
import numpy as np
import matplotlib
# Show Plot in The Notebook
matplotlib.use("nbagg")
import matplotlib.pyplot as plt
```

#### Exercise 1

Define a python function call "fourier_derivative(f, dx)" that compute the first derivative of a function $f$ using the Fourier transform properties.

```

```

```
def fourier_derivative(f, dx):
    # Length of vector f
    nx = np.size(f)
    # Initialize k vector up to Nyquist wavenumber
    kmax = np.pi/dx
    dk = kmax/(nx/2)
    k = np.arange(float(nx))
    k[: int(nx/2)] = k[: int(nx/2)] * dk
    k[int(nx/2) :] = k[: int(nx/2)] - kmax
    # Fourier derivative
    ff = np.fft.fft(f); ff = 1j*k*ff
    df_num = np.real(np.fft.ifft(ff))
    return df_num
```

#### Exercise 2

Calculate the numerical derivative based on the Fourier transform to show that the derivative is exact. Define an arbitrary function (e.g. a Gaussian) and initialize its analytical derivative on the same spatial grid. Calculate the numerical derivative and the difference to the analytical solution. Vary the wavenumber content of the analytical function. Does it make a difference? Why is the numerical result not entirely exact?

```

```

```
# Basic parameters
# ---------------------------------------------------------------
nx = 128
x, dx = np.linspace(2*np.pi/nx, 2*np.pi, nx, retstep=True)
sigma = 0.5
xo = np.pi

# Initialize Gauss function
f = np.exp(-1/sigma**2 * (x - xo)**2)

# Numerical derivative
df_num = fourier_derivative(f, dx)

# Analytical derivative
df_ana = -2*(x-xo)/sigma**2 * np.exp(-1/sigma**2 * (x-xo)**2)

# To make the error visible, it is multiply by 10^13
df_err = 1e13*(df_ana - df_num)

# Error between analytical and numerical solution
err = np.sum((df_num - df_ana)**2) / np.sum(df_ana**2) * 100
print('Error: %s' %err)
```

#### Exercise 3

Now that the numerical derivative is available, we can visually inspect our results. Make a plot of both, the analytical and numerical derivatives together with the difference error.

```

```

```
# Plot analytical and numerical derivatives
# ---------------------------------------------------------------

plt.subplot(2,1,1)
plt.plot(x, f, "g", lw = 1.5, label='Gaussian')
plt.legend(loc='upper right', shadow=True)
plt.xlabel('$x$')
plt.ylabel('$f(x)$')
plt.axis([2*np.pi/nx, 2*np.pi, 0, 1])

plt.subplot(2,1,2)
plt.plot(x, df_ana, "b", lw = 1.5, label='Analytical')
plt.plot(x, df_num, 'k--', lw = 1.5, label='Numerical')
plt.plot(x, df_err, "r", lw = 1.5, label='Difference')
plt.legend(loc='upper right', shadow=True)
plt.xlabel('$x$')
plt.ylabel('$\partial_x f(x)$')
plt.axis([2*np.pi/nx, 2*np.pi, -2, 2])

plt.show()
#plt.savefig('Fig_5.9.png')
```

notebooks/Computational Seismology/The Pseudospectral Method/ps_fourier_derivative_exercise.ipynb


# Estimating the Effect of Schooling on Wages: Intrumental Variables Application

### Summary of Contents:
1. [Introduction](#intro)
2. [NLSYM Dataset](#data)
3. [A Gentle Start: The Naive Approach](#naive)
4. [Using Instrumental Variables: 2SLS](#2sls)
5. [Bonus: Deep Instrumental Variables](#deepiv)

**Important:** This notebook provides some room for experimentation in solving this problem. If you are looking for the end-to-end solution, look for the same file name with the "Complete" suffix added.

# 1. Introduction

To measure true causal effects of a treatment $T$ on an outcome $Y$ from observational data, we need to record all features $X$ that might influence both $T$ and $Y$. These $X$'s are called confounders.

When some confounders are not recorded in the data, we might get biased estimates of the treatment effect. Here is an example:
* Children of high-income parents might attain higher levels of education (e.g. college) since they can afford it
* Children of high-income parents might also obtain better paying jobs due to parents' connections and knowledge
* At first sight, it might appear as if education has an effect on income, when in fact this could be fully explained by family background

There are several reasons for not recording all possible confounders, such as incomplete data or a confounder that is difficult to quantify (e.g. parental involvement). However, not all is lost! In cases such as these, we can use instrumental variables $Z$, features that affect the outcome only through their effect on the treatment.

In this notebook, we use a real-world problem to show how treatment effects can be extracted with the help of instrumental variables.

# 2. NLSYM Dataset

The **causal impact of schooling on wages** had been studied at length. Although it is generally agreed that there is a positive impact, it is difficult to measure this effect precisely. The core problem is that education levels are not assigned at random in the population and we cannot record all possible confounders. (Think about the value parents assign to education. How would you quantify how valuable parents think their children's education is?).

To get around this issue, we can use **proximity to a 4-year college** as an instrumental variable. Having a college nearby can allow individuals (especially low-income ones) to complete more years of education. Hence, if there was a positive treatment effect, we would expect these individuals to have higher wages on average. Note that college proximity is a valid IV since it does not affect wages directly.

We use data from the National Longitudinal Survey of Young Men (NLSYM, 1966) to estimate the average treatment effect (ATE) of education on wages (see also [Card, 1999](https://www.nber.org/papers/w4483)). The NLSYM data contains entries from men ages 14-24 that were interviewed in 1966 and again in 1976.

The dataset contains the following variables:
* $Y$ (outcome): wages (log)
* $T$ (treatment): years of schooling
* $Z$ (IV): proximity to a 4-year college (binary)
* $X$ (heterogeneity): e.g. parental education
* $W$ (controls): e.g. family composition, location, etc.

The world can then be modelled as:
$$
\begin{align}
Y & = \theta(X) \cdot T + f(W) + \epsilon\\
T & = g(Z, W) + \eta
\end{align}
$$
where $\epsilon, \eta$ are uncorrelated error terms.

```python
# Python imports
import keras
import matplotlib.pyplot as plt
import numpy as np
import pandas as pd
from scipy.stats import pearsonr
from sklearn.ensemble import RandomForestRegressor
from sklearn.linear_model import LinearRegression
from sklearn.model_selection import KFold
from sklearn.preprocessing import PolynomialFeatures, StandardScaler

# EconML imports
from econml.dml import DMLCateEstimator
from econml.two_stage_least_squares import NonparametricTwoStageLeastSquares
from econml.deepiv import DeepIVEstimator

%matplotlib inline
```

    Using TensorFlow backend.

```python
# Data processing
df = pd.read_csv("data/card.csv", dtype=float)
# Filter out individuals with low education levels (outliers)
data_filter = df['educ'].values >= 6
# Define some variables
T = df['educ'].values[data_filter]
Z = df['nearc4'].values[data_filter]
Y = df['lwage'].values[data_filter]

# Impute missing values with mean, add dummy columns
# Filter outliers (interviewees with less than 6 years of education)
X_df = df[['exper', 'expersq']].copy()
X_df['fatheduc'] = df['fatheduc'].fillna(value=df['fatheduc'].mean())
X_df['fatheduc_nan'] = df['fatheduc'].isnull() * 1
X_df['motheduc'] = df['motheduc'].fillna(value=df['motheduc'].mean())
X_df['motheduc_nan'] = df['motheduc'].isnull() * 1
X_df[['momdad14', 'sinmom14', 'reg661', 'reg662',
        'reg663', 'reg664', 'reg665', 'reg666', 'reg667', 'reg668', 'reg669', 'south66']] = df[['momdad14', 'sinmom14',
        'reg661', 'reg662','reg663', 'reg664', 'reg665', 'reg666', 'reg667', 'reg668', 'reg669', 'south66']]
X_df[['black', 'smsa', 'south', 'smsa66']] = df[['black', 'smsa', 'south', 'smsa66']]
columns_to_scale = ['fatheduc', 'motheduc', 'exper', 'expersq']
# Scale continuous variables
scaler = StandardScaler()
X_df[columns_to_scale] = scaler.fit_transform(X_df[columns_to_scale])
X = X_df.values[data_filter]
```

```python
# Explore data
X_df.head()
```

      exper
      expersq
      fatheduc
      fatheduc_nan
      motheduc
      motheduc_nan
      momdad14
      sinmom14
      reg661
      reg662
      ...
      reg665
      reg666
      reg667
      reg668
      reg669
      south66
      black
      smsa
      south
      smsa66

      0
      1.725159
      1.896133
      5.439188e-16
      1
      0.000000
      1
      1.0
      0.0
      1.0
      0.0
      ...
      0.0
      0.0
      0.0
      0.0
      0.0
      0.0
      1.0
      1.0
      0.0
      1.0

      1
      0.034739
      -0.172321
      -6.134540e-01
      0
      -0.786159
      0
      1.0
      0.0
      1.0
      0.0
      ...
      0.0
      0.0
      0.0
      0.0
      0.0
      0.0
      0.0
      1.0
      0.0
      1.0

      2
      1.725159
      1.896133
      1.223740e+00
      0
      0.553046
      0
      1.0
      0.0
      1.0
      0.0
      ...
      0.0
      0.0
      0.0
      0.0
      0.0
      0.0
      0.0
      1.0
      0.0
      1.0

      3
      0.276228
      0.052254
      3.051432e-01
      0
      0.553046
      0
      1.0
      0.0
      0.0
      1.0
      ...
      0.0
      0.0
      0.0
      0.0
      0.0
      0.0
      0.0
      1.0
      0.0
      1.0

      4
      1.725159
      1.896133
      -6.134540e-01
      0
      -1.120960
      0
      1.0
      0.0
      0.0
      1.0
      ...
      0.0
      0.0
      0.0
      0.0
      0.0
      0.0
      0.0
      1.0
      0.0
      1.0

5 rows × 22 columns

# 3. A Gentle Start: The Naive Approach

Let's assume we know nothing about instrumental variables and we want to measure the treatment effect of schooling on wages. We can apply an IV-free method like Double Machine Learning (DML) to do this and extract a treatment effect.

```python
### DIY Section!
# Define a DML Estimator
dml_est = None
```

```python
# Fit the DML estimator and calculate the heterogeneous treatment effect
dml_te = None
# Uncoment after you calculated dml_te
"""
dml_ate = dml_te.mean()
"""
```

    '\ndml_ate = dml_te.mean()\n'

```python
# Uncomment after you calculated the dml_ate
"""
print("Average treatment effect: {0:.3f}".format(dml_ate))
"""
```

    '\nprint("Average treatment effect: {0:.3f}".format(dml_ate))\n'

This treatment effect is smaller than other values obtained in literature via IV. Why could that be?

Because DML (like all IV-free methods) assumes that the residual errors are uncorrelated (i.e. $Y - \hat{Y}$ is uncorrelated with $T - \hat{T}$). Let's test this assumption:

```python
### DIY Section
# Calculate the residuals Y_res = Y - Y_pred and T_res = T - T_pred
# Bonus: Split data in 2 parts, train on one part and predict on the other to avoid over-fitting
T_res, Y_res = np.zeros(T.shape[0]), np.zeros(Y.shape[0])
```

```python
# Uncomment this after you calculated T_res and Y_res
"""
plt.scatter(T_res, Y_res)
plt.show()
"""
```

    '\nplt.scatter(T_res, Y_res)\nplt.show()\n'

```python
# Uncomment this after you calculated T_res and Y_res
"""
corr_coeficient = pearsonr(T_res, Y_res)[0]
print("Correlation coefficient between T and Y errors: {0:.2f}".format(corr_coeficient))
"""
```

    '\ncorr_coeficient = pearsonr(T_res, Y_res)[0]\nprint("Correlation coefficient between T and Y errors: {0:.2f}".format(corr_coeficient))\n'

The correlation coefficient between the residuals is quite large, which means that there is some unobserved variables that affect both $T$ and $Y$. To get an accurate estimate in this case, we need to use IVs.

# 4. Using Intrumental Variables: 2SLS

Two stage least square regression procedure (2SLS):
1. Fit a model $T \sim W, Z$
2. Fit a linear model $Y \sim \hat{T}$ where $\hat{T}$ is the prediction of the model in step 1.
The coefficient from 2. above is the average treatment effect.

If interested in heterogeneous treatment effects, fit a model $Y \sim \hat{T}\otimes h(X)$, where $h(X)$ is a chosen featurization of the treatment effect.

For more information, see the `econml` [documentation](https://econml.azurewebsites.net).

```python
# For average treatment effects, X is a column of 1s
W = X
Z = Z.reshape(-1, 1)
T = T.reshape(-1, 1)
X_ate = np.ones_like(Z)
```

```python
# We apply 2SLS from the EconML library
two_sls_est = NonparametricTwoStageLeastSquares(
    t_featurizer=PolynomialFeatures(degree=1, include_bias=False),
    x_featurizer=PolynomialFeatures(degree=1, include_bias=False),
    z_featurizer=PolynomialFeatures(degree=1, include_bias=False),
    dt_featurizer=None) # dt_featurizer only matters for marginal_effect
```

```python
two_sls_est.fit(Y, T, X_ate, W, Z)
two_sls_ate = two_sls_est.effect(np.ones((1,1)))[0]
```

```python
print("Average treatment effect: {0:.3f}".format(two_sls_ate))
```

    Average treatment effect: 0.134

# 5. Bonus: Deep Instrumental Variables

For very flexible, but fully non-parametric IV methods, you can use neural networks for the two models in 2SLS and a mixture of gaussians for the featurizer $h(X)$. In `econml`, this method is called DeepIV.

The NLSYM dataset is small (on neural net scale) so applying DeepIV is a bit of a stretch. Nevertheless, we apply DeepIV the NLSYM data as an example. You should not read too much into the results.

```python
# Define treatment model, T ~ X, Z
treatment_model = keras.Sequential([keras.layers.Dense(64, activation='relu', input_shape=(X.shape[1] + 1,)),
                                    keras.layers.Dropout(rate=0.17),
                                    keras.layers.Dense(32, activation='relu'),
                                    keras.layers.Dropout(rate=0.17),
                                    keras.layers.Dense(1)])
# Define outcome model, Y ~ T_hat, X
response_model = keras.Sequential([keras.layers.Dense(64, activation='relu', input_shape=(X.shape[1] + 1,)),
                                   keras.layers.Dropout(rate=0.17),
                                   keras.layers.Dense(32, activation='relu'),
                                   keras.layers.Dropout(rate=0.17),
                                   keras.layers.Dense(1)])
```

```python
keras_fit_options = { "epochs": 30,
                      "validation_split": 0.3,
                      "callbacks": [keras.callbacks.EarlyStopping(patience=2, restore_best_weights=True)]}

deepIvEst = DeepIVEstimator(n_components = 10, # number of gaussians in our mixture density network
                            m = lambda z, x : treatment_model(keras.layers.concatenate([z,x])), # treatment model
                            h = lambda t, x : response_model(keras.layers.concatenate([t,x])),  # response model
                            n_samples = 1, # number of samples to use to estimate the response
                            use_upper_bound_loss = False, # whether to use an approximation to the true loss
                            n_gradient_samples = 1, # number of samples to use in second estimate of the response (to make loss estimate unbiased)
                            optimizer='adam', # Keras optimizer to use for training - see https://keras.io/optimizers/
                            first_stage_options = keras_fit_options, # options for training treatment model
                            second_stage_options = keras_fit_options) # options for training response model
```

```python
deepIvEst.fit(Y, T, X, Z)
deepIv_effect = deepIvEst.effect(X)
```

    WARNING:tensorflow:From C:\Users\moprescu\AppData\Local\Continuum\anaconda3\lib\site-packages\tensorflow\python\ops\math_ops.py:3066: to_int32 (from tensorflow.python.ops.math_ops) is deprecated and will be removed in a future version.
    Instructions for updating:
    Use tf.cast instead.
    Train on 2093 samples, validate on 898 samples
    Epoch 1/30
    2093/2093 [==============================] - 1s 654us/step - loss: 23.8329 - val_loss: 5.9042
    Epoch 2/30
    2093/2093 [==============================] - 0s 95us/step - loss: 5.0409 - val_loss: 3.9612
    Epoch 3/30
    2093/2093 [==============================] - 0s 118us/step - loss: 3.8898 - val_loss: 3.4648
    Epoch 4/30
    2093/2093 [==============================] - 0s 116us/step - loss: 3.5212 - val_loss: 3.2288
    Epoch 5/30
    2093/2093 [==============================] - 0s 118us/step - loss: 3.3116 - val_loss: 3.0616
    Epoch 6/30
    2093/2093 [==============================] - 0s 77us/step - loss: 3.1305 - val_loss: 2.9131
    Epoch 7/30
    2093/2093 [==============================] - 0s 118us/step - loss: 2.9753 - val_loss: 2.7573
    Epoch 8/30
    2093/2093 [==============================] - 0s 116us/step - loss: 2.8158 - val_loss: 2.5689
    Epoch 9/30
    2093/2093 [==============================] - 0s 111us/step - loss: 2.5800 - val_loss: 2.2856
    Epoch 10/30
    2093/2093 [==============================] - 0s 112us/step - loss: 2.2167 - val_loss: 1.7570
    Epoch 11/30
    2093/2093 [==============================] - 0s 114us/step - loss: 1.7293 - val_loss: 1.4546
    Epoch 12/30
    2093/2093 [==============================] - 0s 110us/step - loss: 1.6448 - val_loss: 1.4081
    Epoch 13/30
    2093/2093 [==============================] - 0s 113us/step - loss: 1.6269 - val_loss: 1.4076
    Epoch 14/30
    2093/2093 [==============================] - 0s 117us/step - loss: 1.5911 - val_loss: 1.3996
    Epoch 15/30
    2093/2093 [==============================] - 0s 87us/step - loss: 1.5620 - val_loss: 1.3765
    Epoch 16/30
    2093/2093 [==============================] - 0s 119us/step - loss: 1.5518 - val_loss: 1.3627
    Epoch 17/30
    2093/2093 [==============================] - 0s 110us/step - loss: 1.5245 - val_loss: 1.3463
    Epoch 18/30
    2093/2093 [==============================] - 0s 110us/step - loss: 1.4822 - val_loss: 1.3022
    Epoch 19/30
    2093/2093 [==============================] - 0s 113us/step - loss: 1.5112 - val_loss: 1.3304
    Epoch 20/30
    2093/2093 [==============================] - 0s 108us/step - loss: 1.5124 - val_loss: 1.3415
    Train on 2093 samples, validate on 898 samples
    Epoch 1/30
    2093/2093 [==============================] - 3s 1ms/step - loss: 9.3704 - val_loss: 1.8564
    Epoch 2/30
    2093/2093 [==============================] - 0s 166us/step - loss: 5.7598 - val_loss: 8.5731
    Epoch 3/30
    2093/2093 [==============================] - 0s 164us/step - loss: 5.2696 - val_loss: 1.6304
    Epoch 4/30
    2093/2093 [==============================] - 0s 167us/step - loss: 34.8642 - val_loss: 1.8287
    Epoch 5/30
    2093/2093 [==============================] - 0s 168us/step - loss: 5.5516 - val_loss: 564.5439

```python
print("Average treatment effect: {0:.3f}".format(deepIv_effect.mean()))
```

    Average treatment effect: 0.191

```python
# Heterogeneity of treatment effects
plt.hist(deepIv_effect)
plt.show()
```

```python

```

notebooks/Hands-on Part II -- Estimating the Effect of Schooling on Wages.ipynb