OBJECTIVE OF THIS LECTURE

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• OBJECTIVE OF THIS LECTURE
• What is Curve Fitting? Why?
• Methods of Curve Fitting
• by Least Square Principle
• How to fit a Straight Line
• How to fit a parabola?

Dr. S. R. Jayaram, SBMJCE, Bangalore
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• Introduction
• Given n points ( xi, yi)
• It may be values observed/measured from
  experiments
• x and y may be linearly related
• How to fit a straight line?

Dr. S. R. Jayaram, SBMJCE, Bangalore
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• 5. Does the line pass through all points?
• 6. Let yi f(xi), i 0,1, 2, ..n
• 7. No value between xi and xi1 is known
• 8. If true value is yi and tabulated value is yi
  then error is ei yi yi.
• In method of least squares sum of all
• the errors is minimised.

Dr. S. R. Jayaram, SBMJCE, Bangalore
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The values a and b of the linear equation
. Is obtained by solving the
normal equations.
S.R.Jayaram, Sbmjce, Jakkasandra post, Bangalore
rural
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Solution We fit a straight line of the form
according to Least Square Principle,
i.e. the sum of squares of difference between
actual values and the observed values is least.
Here, are the actual values and are the observed
values are respectively.
S.R.Jayaram, Sbmjce, Jakkasandra post, Bangalore
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According to Least Square Principle,
least. Note that
here, S is dependent on two parameters a and b.
From Differential Calculus, it is known that a
function of two variables attains an extreme
value only at points where the first order
partial differential derivates vanishes.

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Now, differentiating S with respect a and b
partially and equating these to zero, we obtain
the following, called as normal equations.

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• By solving these two equations, we obtain values
  of the parameters a and b, hence the best
  straight line .
• Steps to set a line for tabulated points y f(x)
• Find n
• Find sum of x, sum of y, sum of x2 , sum of xy.
  Substitute these in normal equation.

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.

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By solving these two equations, we obtain values
of the parameters a and b, hence the best
straight line . An Illustrative
example Fit a straight line of the form
to the data given below

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Solution As we have to fit a line of the form y
ax b, the two normal equations are
and


S.R.Jayaram, Sbmjce, Jakkasandra post, Bangalore
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S.R.Jayaram, Sbmjce, Jakkasandra post, Bangalore
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The two normal equations are
By Solving these two equations, yields, a
-1.3 and b 15. 8. Therefore, required
straight line matching the given data is y
-1.3x 15.8
S.R. Jayaram, Sbmjce, Jakkasandra post, Bangalore
rural
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Fit a straight line of the form y a bx to
the data given below For this problem, the
two normal equations are
x 1 2 3 4 5
6 y 9 8 10 12 11 13
S.R.Jayaram, Sbmjce, Jakkasandra post, Bangalore
rural
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As to fit a line of the form y a b x,

Here, n 6. First, we
shall prepare these two normal equations. For
this consider the table,
S.R.Jayaram, Sbmjce, Jakkasandra post, Bangalore
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D S.R.Jayaram, Sbmjce, Jakkasandra post,
Bangalore rural
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The
normal equations become
and . After solving
these two equations for a and b, we obtain a
7 and b 1. Thus, the required straight line fit
is y 7 x.



S.R.Jayaram, Sbmjce, Jakkasandra post, Bangalore
rural
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• Fit a straight line of the form y ax b to the
  data given below.
• x 1 2 3 4 5
  6
• y 8 7 6 10 12
  15
• Solution Here, the normal equations are found
  to be
• 21a 6b 60, 91a 21b 239,
• a 1.6571, b 4.2.
• The answer is 1.6571x 4.2

S.R.Jayaram, Sbmjce, Jakkasandra post, Bangalore
rural
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Discussion on fitting a Parabola to a given
data Consider a data obtained from an
experiment, say, To fit a mathematical
equation of the form
. Applying least square principle, we
obtain normal equations as
S.R.Jayaram, Sbmjce, Jakkasandra post, Bangalore
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A solution of this system of linear equations
yields a, b, c. Hence, the required parabolic
fit.
S.R. Jayaram, Sbmjce, Jakkasandra post, Bangalore
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An illustrative example Fit a parabola of the
form, to the following
data to the following data. Solution
S.R.Jayaram, Sbmjce, Jakkasandra post, Bangalore
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Solution As to fit a parabola, The required
normal equations are
S.R.Jayaram, Sbmjce, Jakkasandra post, Bangalore
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S.R.Jayaram, Sbmjce, Jakkasandra post, Bangalore
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Thus, normal equations take the form Solving
these equations, we obtain a 2.250, b 2.850
and c -0.75. Hence,
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Fit a parabola of the form
to the data given below x 1 2
3 4 5 y 2
6 7 8 10 Solution As in
the previous problem, the normal equations are
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S.R.Jayaram, Sbmjce, Jakkasandra post, Bangalore
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S.R.Jayaram, Sbmjce, Jakkasandra post, Bangalore
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A solution of these equations yields, a
-66.80, b 60.8057 and c -9.7143. Hence, the
required parabola is

S.R.Jayaram, Sbmjce, Jakkasandra post, Bangalore
rural