Minor changes to scripts in 4th module.

2 files changed, 17 insertions, 17 deletions

diff --git a/accessing_parts_of_arrays/script.rst b/accessing_parts_of_arrays/script.rst
index aeb1938..48ed2e9 100644
--- a/accessing_parts_of_arrays/script.rst
+++ b/accessing_parts_of_arrays/script.rst
@@ -60,14 +60,14 @@ At the end of this tutorial, you will be able to,
 
 Before beginning this tutorial,we would suggest you to complete the 
 tutorial on "Getting started with arrays".
- 
-.. L4
-
-{{{ Open the terminal }}}
 
 .. R4
 
 As usual, we start IPython, using 
+
+.. L4
+
+{{{ Open the terminal }}}
 ::
 
     ipython -pylab 
@@ -211,7 +211,7 @@ just one particular element. We access the third column by saying,
 .. L16
 ::
   
-  C[:, 2]
+    C[:, 2]
 
 .. R17
 
@@ -225,12 +225,12 @@ Pause the video here, try out the following exercise and resume the video.
 
 .. R18
 
- Change the last column of C to zeroes. 
+ Change the last column of C to zeros. 
 
 .. R19
 
 Switch to the terminal for solution.To change the entire last column of 
-C to zeroes, we simply say,
+C to zeros, we simply say,
 
 .. L19
 
diff --git a/least_square_fit/script.rst b/least_square_fit/script.rst
index 429b640..acf6432 100644
--- a/least_square_fit/script.rst
+++ b/least_square_fit/script.rst
@@ -115,6 +115,10 @@ values of m and c.
 .. L8
 
 {{{ Switch to the terminal }}}
+::
+
+    inter_mat = array((l, ones_like(l)))
+    inter_mat
 
 .. R8
 
@@ -122,18 +126,12 @@ let us now generate the A matrix with l values.
 We shall first generate a 2 x 90 matrix with the first row as l values 
 and the second row as ones. Then take the transpose of it. Type
 
-.. L9
-::
-
-    inter_mat = array((l, ones_like(l)))
-    inter_mat
-
 .. R9
 
 We see that we have intermediate matrix. Now we need the transpose. 
 Type
 
-.. L10
+.. L9
 ::
 
     A = inter_mat.T
@@ -145,7 +143,7 @@ Now we have both the matrices A and tsq. We only need to use
 the ``lstsq``
 Type
 
-.. L11
+.. L10
 ::
 
     result = lstsq(A, tsq)
@@ -155,7 +153,7 @@ Type
 The result is a sequence of values. The first item in this sequence,
 is the matrix p i.e., the values of m and c. 
 
-.. L12
+.. L11
 ::
 
     m, c = result[0]
@@ -167,7 +165,7 @@ is the matrix p i.e., the values of m and c.
 Now that we have m and c, we need to generate the fitted values of t^2. 
 Type
 
-.. L13
+.. L12
 ::
 
     tsq_fit = m * l + c
@@ -178,6 +176,8 @@ Type
 
 We get the least square fit of l vs t^2.
 
+.. L13
+
 .. L14
 
 {{{ Show summary slide }}}