diff -r 9bc78792904b -r 0ff3f1a97068 lstsq.rst
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-.. Author              : Nishanth
-   Internal Reviewer 1 : Puneeth
-   Internal Reviewer 2 : 
-   External Reviewer   :
-
-Hello friends and welcome to the tutorial on Least Square Fit
-
-{{{ Show the slide containing title }}}
-
-{{{ Show the slide containing the outline slide }}}
-
-In this tutorial, we shall look at generating the least square fit line for a
-given set of points.
-
-First let us have a look at the problem.
-
-{{{ Show the slide containing problem statement. }}}
-
-We have an input file generated from a simple pendulum experiment.
-
-It contains two columns of data. The first column is the length of the
-pendulum and the second is the corresponding time period of the pendulum.
-
-As we know, the square of time period of a pendulum is directly proportional to
-its length, we shall plot l vs t^2 and verify this. 
-
-#[Puneeth:] removed the explanation about loadtxt and unpack
- option. It's been done in another LO already. simple dependency 
- should work?
-
-To read the input file and parse the data, we are going to use the
-loadtxt function.  Type 
-::
-
-    l, t = loadtxt("/home/fossee/pendulum.txt", unpack=True)
-    l
-    t
-
-We can see that l and t are two sequences containing length and time values
-correspondingly.
-
-Let us first plot l vs t^2. Type
-::
-
-    tsq = t * t
-    plot(l, tsq, 'bo')
-
-{{{ switch to the plot window }}}
-
-#[Puneeth:] Moved explanation of least square fit here. seems more
-apt. 
-
-We can see that there is a visible linear trend, but we do not get a
-straight line connecting them. We shall, therefore, generate a least
-square fit line.
-
-{{{ show the slide containing explanation on least square fit }}}
-
-As shown in the slide, we are first going to generate the two matrices
-tsq and A. Then we are going to use the ``lstsq`` function to find the
-values of m and c.
-
-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
-::
-
-    inter_mat = array((l, ones_like(l)))
-    inter_mat
-
-We see that we have intermediate matrix. Now we need the transpose. Type
-::
-
-    A = inter_mat.T
-    A
-
-Now we have both the matrices A and tsq. We only need to use the ``lstsq``
-Type
-::
-
-    result = lstsq(A, tsq)
-
-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. Hence, 
-::
-
-    m, c = result[0]
-    m
-    c
-
-Now that we have m and c, we need to generate the fitted values of t^2. Type
-::
-
-    tsq_fit = m * l + c
-    plot(l, tsq, 'bo')
-    plot(l, tsq_fit, 'r')
-
-We get the least square fit of l vs t^2
-
-{{{ Pause here and try out the following exercises }}}
-
-%% 2 %% change the label on y-axis to "y" and save the lines of code
-        accordingly
-
-{{{ continue from paused state }}}
-
-{{{ Show summary slide }}}
-
-This brings us to the end of the tutorial.
-we have learnt
-
- * how to generate a least square fit
-
-{{{ Show the "sponsored by FOSSEE" slide }}}
-
-#[Nishanth]: Will add this line after all of us fix on one.
-This tutorial was created as a part of FOSSEE project, NME ICT, MHRD India
-
-Hope you have enjoyed and found it useful.
-Thank you
-
-Questions
-=========
-
- 1. What does ones_like([1, 2, 3]) produce
-
-   a. array([1, 1, 1])
-   #. [1, 1, 1]
-   #. [1.0, 1.0, 1.0]
-   #. Error
-   
- 2. What does ones_like([1.2, 3, 4, 5]) produce
-
-   a. [1.2, 3, 4, 5]
-   #. array([1.0, 1.0, 1.0, 1.0])
-   #. array([1, 1, 1, 1])
-   #. array([1.2, 3, 4, 5])
-
- 3. What is the shape of the