diff --git a/test/math/pi.cpp b/test/math/pi.cpp
index e00379c1..3c3ecf7e 100644
--- a/test/math/pi.cpp
+++ b/test/math/pi.cpp
@@ -32,6 +32,18 @@ int main(int argc, char **argv)
 
     sum.print();
 
+    /* Bonus: compute e for fun. */
+    sum = 0.0;
+    x0 = 1.0;
+
+    for (int i = 1; i < 100; i++)
+    {
+        sum += fres(x0);
+        x0 *= (real)i;
+    }
+
+    sum.print();
+
     return EXIT_SUCCESS;
 }
 
diff --git a/test/math/remez.cpp b/test/math/remez.cpp
index 44d8a1cc..2babb661 100644
--- a/test/math/remez.cpp
+++ b/test/math/remez.cpp
@@ -18,21 +18,116 @@
 
 using namespace lol;
 
-static int const ORDER = 12;
+/* The order of the approximation we're looking for */
+static int const ORDER = 3;
 
-static int cheby[ORDER][ORDER];
+/* The function we want to approximate */
+double myfun(double x)
+{
+    return exp(x);
+}
+
+real myfun(real const &x)
+{
+    return exp(x);
+}
+
+/* Naive matrix inversion */
+template<int N> class Matrix
+{
+public:
+    inline Matrix() {}
+
+    Matrix(real x)
+    {
+        for (int j = 0; j < N; j++)
+            for (int i = 0; i < N; i++)
+                if (i == j)
+                    m[i][j] = x;
+                else
+                    m[i][j] = 0;
+    }
+
+    Matrix<N> inv() const
+    {
+        Matrix a = *this, b(real(1.0));
+
+        /* Inversion method: iterate through all columns and make sure
+         * all the terms are zero except on the diagonal where it is one */
+        for (int i = 0; i < N; i++)
+        {
+            /* If the expected coefficient is zero, add one of
+             * the other lines. The first we meet will do. */
+            if ((double)a.m[i][i] == 0.0)
+            {
+                for (int j = i + 1; j < N; j++)
+                {
+                    if ((double)a.m[i][j] == 0.0)
+                        continue;
+                    /* Add row j to row i */
+                    for (int n = 0; n < N; n++)
+                    {
+                        a.m[n][i] += a.m[n][j];
+                        b.m[n][i] += b.m[n][j];
+                    }
+                    break;
+                }
+            }
+
+            /* Now we know the diagonal term is non-zero. Get its inverse
+             * and use that to nullify all other terms in the column */
+            real x = fres(a.m[i][i]);
+            for (int j = 0; j < N; j++)
+            {
+                if (j == i)
+                    continue;
+                real mul = x * a.m[i][j];
+                for (int n = 0; n < N; n++)
+                {
+                    a.m[n][j] -= mul * a.m[n][i];
+                    b.m[n][j] -= mul * b.m[n][i];
+                }
+            }
 
+            /* Finally, ensure the diagonal term is 1 */
+            for (int n = 0; n < N; n++)
+            {
+                a.m[n][i] *= x;
+                b.m[n][i] *= x;
+            }
+        }
+
+        return b;
+    }
+
+    void print() const
+    {
+        for (int j = 0; j < N; j++)
+        {
+            for (int i = 0; i < N; i++)
+                printf("%9.5f ", (double)m[j][i]);
+            printf("\n");
+        }
+    }
+
+    real m[N][N];
+};
+
+
+static int cheby[ORDER + 1][ORDER + 1];
+
+/* Fill the Chebyshev tables */
 static void make_table()
 {
-    memset(cheby[0], 0, ORDER * sizeof(int));
+    memset(cheby, 0, sizeof(cheby));
+
     cheby[0][0] = 1;
-    memset(cheby[1], 0, ORDER * sizeof(int));
     cheby[1][1] = 1;
 
-    for (int i = 2; i < ORDER; i++)
+    for (int i = 2; i < ORDER + 1; i++)
     {
         cheby[i][0] = -cheby[i - 2][0];
-        for (int j = 1; j < ORDER; j++)
+        for (int j = 1; j < ORDER + 1; j++)
             cheby[i][j] = 2 * cheby[i - 1][j - 1] - cheby[i - 2][j];
     }
 }
@@ -41,12 +136,49 @@ int main(int argc, char **argv)
 {
     make_table();
 
-for (int i = 0; i < ORDER; i++)
-{
-    for (int j = 0; j < ORDER; j++)
-        printf("% 5i ", cheby[i][j]);
-    printf("\n");
-}
+    /* We start with ORDER+1 points and their images through myfun() */
+    real xn[ORDER + 1];
+    real fxn[ORDER + 1];
+    for (int i = 0; i < ORDER + 1; i++)
+    {
+        //xn[i] = real(2 * i - ORDER) / real(ORDER + 1);
+        xn[i] = real(2 * i - ORDER + 1) / real(ORDER - 1);
+        fxn[i] = myfun(xn[i]);
+    }
+
+    /* We build a matrix of Chebishev evaluations: one row per point
+     * in our array, and column i is the evaluation of the ith order
+     * polynomial. */
+    Matrix<ORDER + 1> mat;
+    for (int j = 0; j < ORDER + 1; j++)
+    {
+        /* Compute the powers of x_j */
+        real powers[ORDER + 1];
+        powers[0] = 1.0;
+        for (int i = 1; i < ORDER + 1; i++)
+             powers[i] = powers[i - 1] * xn[j];
+
+        /* Compute the Chebishev evaluations at x_j */
+        for (int i = 0; i < ORDER + 1; i++)
+        {
+            real sum = 0.0;
+            for (int k = 0; k < ORDER + 1; k++)
+                if (cheby[i][k])
+                    sum += real(cheby[i][k]) * powers[k];
+            mat.m[j][i] = sum;
+        }
+    }
+
+    /* Invert the matrix and build interpolation coefficients */
+    mat = mat.inv();
+    real an[ORDER + 1];
+    for (int j = 0; j < ORDER + 1; j++)
+    {
+        an[j] = 0;
+        for (int i = 0; i < ORDER + 1; i++)
+            an[j] += mat.m[j][i] * fxn[i];
+        an[j].print(10);
+    }
 
     return EXIT_SUCCESS;
 }