Fitting a Parabola or Second Degree Polynomial - Business Mathematics and Statistics Video Lecture | Business Mathematics and Statistics - B Com

today we are going to discuss fitting a
parabola or second-degree polynomial by
the matter of least squares fitting a
parabola okay so he has given the some
data so the value of the value of x is 0
the value of y is 1 with the value of x
t is one failure y is 1 point 8 4 2 it
is 1 point 3 3 it is 2 point 5 4 4 it is
6 point 3 now we have a fit a parabola
using that matter of least squares the
first point what we have to do the first
point what we have to do is let the
parabola equation be let the parabola be
what you have to consider it as y is
equal to a plus BX plus C X square this
is the parabola equation okay next what
you have to do you have to consider the
normal equations Y you have to consider
the normal equations means here if you
know the a.b.c values we can frame a
parable to get the a B values see how we
require the three normal equations we
require so what are the three normal
equations means Sigma Y ug equal to ne a
plus B Sigma X plus C Sigma X square C
Sigma X square similarly Sigma X Y Z
equal to a sigma X plus B Sigma X square
plus C Sigma X cube ok C Sigma X cube
similarly Sigma X square Y Z equal to e
Sigma X square plus B Sigma X cube plus
C Sigma X power 4 C Sigma X power 4
these are the three normal equation to
solve the values of a B C we require the
three equation these are the three
equations one two
and the three if you come mock or
compare this one with the straight-line
there you require the only a be valise
that is why they have there you have the
two normal equations here you come
across with the three normal equations
okay now the third step is you have to
prepare a table so what the table that
we have to prepare means X so you read
the values of X here okay similarly you
have to read the values of Y here the
values of X are 0 1 2 3 & 4 the values
of Y are 1 1 point 8 and 1 point 3 and
two point five and six point three okay
if you nest what you have to do either
if you sum the values here you are going
to get the Sigma X values okay here you
are going to get the Sigma Y values next
what we require let us consider Sigma Y
already over Sigma X over Sigma X square
therefore you come across with the X
square okay now you frame these values X
square mister similarly you come across
with the X cube also therefore you read
the X cube ok next what you have to do
the you frame the X power 4 X power 4
because here you require the X power 4
values okay therefore its X power 4 next
one what you require is the Sigma XY
that is why you frame the X into y
column
similarly you frame the X square Y
column X square is this one by easy some
so you frame the X square into y column
also okay if you frame and if you add
these values you are going to get the
Sigma X Sigma Y so on so values so x
square values what are the different x
square values you are going to
zero one four nine and sixteen
so similarly XQ means 0 cube means 0 1
cube is 1 2 key VJ 2 3 cube is 27 4
cubed is 64
similarly X power 4 X performance 0
power 4 0 1 power 4 1 2 power 4
16 so 3 power 481 say for power 4 means
256 256
similarly X into y X addition X into y 0
into 1 0 1 into 1 point 8 is 1 point 8
is 1 2 into 1 point 3 is a 2 point 6 s 1
3 into 2 point 5 is 7 and half that is 7
point 5 6 point 3 into 4 that is nothing
but the 24 you pi into 2 ok
similarly X square into y X square X
square Y this one and the ye the Sun 0
into 1 is 0 s 1 1 into 1 point 8 is 1
point 8 4 into 1 point 2 3 4 into 1
point 3 is nothing but 5 pi into 2
similarly 22 point 2 5
similarly 100 point 800 point 80 will
get up so the value of x square by E is
nothing but 1 30 point 2 3 if you sum
these column similarly if you X Y this
is thirty seven point one next one X
power 4 it is nothing but 3 5 4 okay
similarly X cube column that is nothing
but 100
similarly X square column it is nothing
but 30 similarly Y column that is
nothing but at all open to 9 similarly X
column so it is 10 okay now we have
prepared the table next to fourth step
what you have to do that the value of n
is equal
Hammacher so one two three four five
five sets of data is given therefore n
is equal to five
first we prepare the first normal
equation Sigma by Sigma I is equal to
how much follow point nine is equal to
twelve point nine is equal to n the N
stands for five evenly Sigma X Sigma X
stands for ten ten B plus Sigma X square
stands for 30 thirty
C this is a first normal equation next
what you have to do is Sigma X into y
that is nothing but the thirty seven
point one thirty seven point one z equal
to e ay Sigma X that is nothing but a
ten times of a plus B Sigma X square
that is nothing but thirty times of B
plus C Sigma X cube hundred times of C
okay one hundred times of C this is the
second equation you are going to get
them similarly the third equation you
are going to get Sigma X square Y that
is nothing but 130 PI into three is
equal to a sigma x squared that is
nothing but 30 a plus Sigma X QB b sigma
x cube one hundred be hundred B plus C
Sigma X power 4 C Sigma X / 354 C three
fifty four times of C this is the third
equation you are going to get so if you
solve these three equations using the
Cal C the values of a you are going to
get it as a is equal to 1.4 to V is
equal to minus one point zero seven and
C is equal to zero point two double five
we are going to get there further
equation of the straight line he is
nothing but the Y is equal to a plus BX
plus C X square that is nothing but Y is
equal to H stands for 1.42 plus B stands
for minus one
five to 0-7 times of X plus C C really
is nothing but zero point double 5 into
X square this is the equation of the
parabola using the method of least
squares okay so this includes the only
five steps first one is a parabolic
equation next one is the normal
equations
Esther setting of the table fourth one
is a substituting the table values in
the normal equation and solving and the
finally you substitute in the parabola
equation okay thank you