Convert all %timeit instances to qe.timeit() with runs=3

Co-authored-by: mmcky <8263752+mmcky@users.noreply.github.com>

Author: Copilot

lectures/parallelization.md

@@ -166,7 +166,7 @@ x, y = np.meshgrid(grid, grid)
 ```
 
 ```{code-cell} ipython3
-%timeit np.max(f(x, y))
+qe.timeit(lambda: np.max(f(x, y)), runs=3)
 ```
 
 If you have a system monitor such as htop (Linux/Mac) or perfmon
@@ -198,7 +198,7 @@ np.max(f_vec(x, y))  # Run once to compile
 ```
 
 ```{code-cell} ipython3
-%timeit np.max(f_vec(x, y))
+qe.timeit(lambda: np.max(f_vec(x, y)), runs=3)
 ```
 
 At least on our machine, the difference in the speed between the
@@ -248,7 +248,7 @@ np.max(f_vec(x, y))  # Run once to compile
 ```
 
 ```{code-cell} ipython3
-%timeit np.max(f_vec(x, y))
+qe.timeit(lambda: np.max(f_vec(x, y)), runs=3)
 ```
 
 Now our code runs significantly faster than the NumPy version.

lectures/scipy.md

@@ -64,6 +64,7 @@ However, it's more common and better practice to use NumPy functionality explici
 
 ```{code-cell} python3
 import numpy as np
+import quantecon as qe
 
 a = np.identity(3)
 ```
@@ -320,11 +321,11 @@ brentq(f, 0, 1)
 Here the correct solution is found and the speed is better than bisection:
 
 ```{code-cell} ipython
-%timeit brentq(f, 0, 1)
+qe.timeit(lambda: brentq(f, 0, 1), runs=3)
 ```
 
 ```{code-cell} ipython
-%timeit bisect(f, 0, 1)
+qe.timeit(lambda: bisect(f, 0, 1), runs=3)
 ```
 
 ### Multivariate Root-Finding