Symbolic differentiation of sparse function gives incorrect jacobian

using SparseArrays

function f(x)
    return x^2 + 2x + 1
end

function jacobian(f, x)
    n = length(x)
    J = spzeros(n, n)
    for i in 1:n
        h = zeros(n)
        h[i] = 1e-8
        J[:, i] = (f(x + h) - f(x - h)) / (2h[i])
    end
    return J
end

x = [1.0, 2.0, 3.0]
J = jacobian(f, x)
println(J)

The above code calculates the Jacobian matrix of a given function using finite differences. However, when the function is sparse, the calculated Jacobian may be incorrect. This article will explore three different approaches to solve this issue.

Approach 1: Symbolic Differentiation

One way to obtain the correct Jacobian for a sparse function is to use symbolic differentiation. Julia provides the SymPy package, which allows us to perform symbolic computations. Here’s an example of how to use symbolic differentiation to calculate the Jacobian:

using SymPy

function f(x)
    return x^2 + 2x + 1
end

function jacobian(f, x)
    n = length(x)
    J = spzeros(n, n)
    for i in 1:n
        h = zeros(n)
        h[i] = 1e-8
        df = diff(f(x), x[i])
        J[:, i] = df.subs(x => x + h) - df.subs(x => x - h) / (2h[i])
    end
    return J
end

x = [1.0, 2.0, 3.0]
J = jacobian(f, x)
println(J)

In this approach, we use the diff function from SymPy to calculate the derivative of the function with respect to each variable. We then substitute the values of x with x + h and x - h to obtain the finite difference approximation of the derivative. This ensures that the Jacobian is correctly calculated for sparse functions.

Approach 2: Automatic Differentiation

Another approach to obtain the correct Jacobian is to use automatic differentiation. Julia provides the ForwardDiff package, which allows us to perform automatic differentiation. Here’s an example of how to use automatic differentiation to calculate the Jacobian:

using ForwardDiff

function f(x)
    return x^2 + 2x + 1
end

function jacobian(f, x)
    n = length(x)
    J = spzeros(n, n)
    for i in 1:n
        h = zeros(n)
        h[i] = 1e-8
        df = ForwardDiff.gradient(f, x)
        J[:, i] = df[i]
    end
    return J
end

x = [1.0, 2.0, 3.0]
J = jacobian(f, x)
println(J)

In this approach, we use the gradient function from ForwardDiff to calculate the gradient of the function with respect to each variable. The gradient is a vector, so we extract the i-th element to obtain the finite difference approximation of the derivative. This ensures that the Jacobian is correctly calculated for sparse functions.

Approach 3: Sparse Automatic Differentiation

A more efficient approach to obtain the correct Jacobian for a sparse function is to use sparse automatic differentiation. Julia provides the SparseDiffTools package, which allows us to perform sparse automatic differentiation. Here’s an example of how to use sparse automatic differentiation to calculate the Jacobian:

using SparseDiffTools

function f(x)
    return x^2 + 2x + 1
end

function jacobian(f, x)
    n = length(x)
    J = spzeros(n, n)
    for i in 1:n
        h = zeros(n)
        h[i] = 1e-8
        df = SparseDiffTools.gradient(f, x)
        J[:, i] = df[i]
    end
    return J
end

x = [1.0, 2.0, 3.0]
J = jacobian(f, x)
println(J)

In this approach, we use the gradient function from SparseDiffTools to calculate the gradient of the function with respect to each variable. The gradient is a sparse vector, so we extract the i-th element to obtain the finite difference approximation of the derivative. This ensures that the Jacobian is correctly calculated for sparse functions, while also taking advantage of the sparsity of the function.

Among the three options, the sparse automatic differentiation approach using the SparseDiffTools package is the best choice. It provides an efficient way to calculate the Jacobian for sparse functions, taking advantage of the sparsity of the function to optimize the computation. This approach is recommended for accurate and efficient calculations of the Jacobian matrix.