interpolation

Dear Friends,

This is just an introduction letter.My name is J. Sudhindra Simha. My qualification is M.Sc. in Statistics from the University of Mysore, Karnataka, India. I have also obtained Certificate in Population Studies, IIPS, Mubai. I am retired now.
I have about 9-10 published papers in family planning methods while I was working for the Christian Medical Association of India, Bangalore.
I have also worked as Lecturer for BBM course in Bangalore. I have developed some simple modules in the application of statistics and demography in different topics which may be helpful to under graduate students and professional courses also. Hence, I would like to share this with interested groups who would benefit from this. As I have opened a blog, I want to display it.

The topics covered are :-

1. Interpolation
2. Central Tendency
3. Dispersion
4. Regression analysis
5. Probability
6. Testing of Hypothesis
7. Life expectancy
8. Age at marriage.

As a first step I am attaching herewith “Interpolation”.

Regards,
J. Sudhindra Simha

9/25/2009

Find a suitable method of interpolation to find the value of y when x=7?


















x 2 4 6 8







y 7 21 43 73







x y ∆ ∆2 ∆3







2 7












14

use newton's backward





4 21
8
formula








22
0

y n 73



6 43
8


vΔy n-1 -15





30

+{ v(v+1)/1.2} Δ2y n-2 -1



8 73





57



x=7 xn=8 v=(x-xn)/h v=(7-8)/2 v=-0.5







h=2


h
















Applying Newton’s Backward Interpolation formula we get
































y7 = 73 -15 -1 57








y7 = 57






















































































x y ∆ ∆2 ∆3






















0

use newton's backward


-1




0
formula








0
0

y n= 0






0


vΔy n-1= #DIV/0!





0

+{ v(v+1)/1.2} Δ2y n-2= #DIV/0!










= #DIV/0!



x=
xn=









h=
v= (x-xn)/v










v= #DIV/0!








Applying Newton’s Backward Interpolation formula we get
































y7 = #DIV/0!
























Using Newton's method of interpolation find from the data given below, the number of




the number of persons in the income group between Rs.20 and Rs.25?

Wages No. of










(inRs.) persons ∆ ∆2 ∆3 ∆4






Below10 20












25









Below20 45
45










70
-20







Below30 115
25
15








95
-5







Below40 210
20










115









Below 50 325























h= 10 x= 20 x0= 10 u=(x-x0)/h= 1





10 x= 25 x0= 10 u=(x-x0)/h= 1.5



u= 1.5 (u/1)∆Y0= 37.5










(u)(u-1)/(1*2)*(∆2y0)= 16.88









(u)(u-1)(u-2)/(1*2*3)*(∆3y0)= 1.25








(u)(u-1)(u-2)(u-3)/(1*2*3*4)*(∆4y0)= 0.35



















u= 1 (u/1)∆Y0= 25
y0 = 20







(u)(u-1)/(1*2)*(∆2y0)= 0









Applying Newton’s forward interpolation formula, we get






Y = y0 + (u/1) Dy0 + {u(u-1)/1.2}D2y0 + {u(u-1)(u-2)/1.2.3}D3y0 + -------------



--------------+ {u(u-1)u-2)…………..(u-n+1)/1.2.3…..n}Dny0







y20 = 20 + 25 = 45



y25 = 20 + 37.5 + 16.88 + 1.25 + 0.35 = 75.98
















y25 - y20 = 30.98 = 31



Hence the number of persons in the income group between Rs.25 and Rs.20 are 31.






























DATA: First two columns are given






















To estimate the candidates getting more than 48 marks and less than 50 marks. rks but not more than 50 marks.



























Marks No. of










upto candidates ∆ ∆2 ∆3 ∆4






45 447












37









50 484
-16










21
1







55 505
-15
11








6
12







60 511
-3










3









65 514


















z



h= 5 x= 50 x0= 45 u=(x-x0)/h= 1




h= 5 x= 48 x0= 45 u=(x-x0)/h= 0.6




0.6 (u/1)∆Y0= 22.2










(u)(u-1)/(1*2)*(∆2y0)= 1.92









(u)(u-1)(u-2)/(1*2*3)*(∆3y0)= 0.056








(u)(u-1)(u-2)(u-3)/(1*2*3*4)*(∆4y0)= -0.37




















1 (u/1)∆Y0= 37
y0 = 447







(u)(u-1)/(1*2)*(∆2y0)= 0









(u)(u-1)(u-2)/(1*2*3)*(∆3y0)= 0








(u)(u-1)(u-2)(u-3)/(1*2*3*4)*(∆4y0)=
0







Applying Newton’s forward interpolation formula, we get






Y = y0 + (u/1) Dy0 + {u(u-1)/1.2}D2y0 + {u(u-1)(u-2)/1.2.3}D3y0 + -------------



--------------+ {u(u-1)u-2)…………..(u-n+1)/1.2.3…..n}Dny0







y48= 447 + 22.2 = 469



y50= 447 + 37 + 0.00 + 0.00 + 0.00 = 484.00
















y50 - y48 = 14.80 = 13




The number of candidates who get more than 48 marks but not more than 50 marks are 13.


The number of candidates who get more than
marks but not more than



marks are













































































∆ ∆2 ∆3 ∆4





















0












0










0
0










0
0








0
0










0










0



































h=
x= 0 x0= 10 u=(x-x0)/h= #DIV/0!






x=
x0= 10 u=(x-x0)/h= #DIV/0!




#DIV/0! (u/1)∆Y0= #DIV/0!










(u)(u-1)/(1*2)*(∆2y0)= #DIV/0!









(u)(u-1)(u-2)/(1*2*3)*(∆3y0)= #DIV/0!








(u)(u-1)(u-2)(u-3)/(1*2*3*4)*(∆4y0)= #DIV/0!




















#DIV/0! (u/1)∆Y0= #DIV/0!
y0 = 20







(u)(u-1)/(1*2)*(∆2y0)= 0









Applying Newton’s forward interpolation formula, we get






Y = y0 + (u/1) Dy0 + {u(u-1)/1.2}D2y0 + {u(u-1)(u-2)/1.2.3}D3y0 + -------------



--------------+ {u(u-1)u-2)…………..(u-n+1)/1.2.3…..n}Dny0







y20 = 20 + #### = ####




20 + #DIV/0! + #DIV/0! + #DIV/0! + #DIV/0! = #DIV/0!
















y25 - y20 = #DIV/0! =

















Hence the number of persons in the income group between Rs.
and Rs.


are











DATA: First two columns are given






















To estimate the candidates getting more than 48 marks and less than 50 marks. rks but not more than 50 marks.




























Marks No. of










upto candidates ∆ ∆2 ∆3 ∆4





















0












0










0
0










0
0








0
0










0










0






























z



h=
x=
x0= 0 u=(x-x0)/h= #DIV/0!




h=
x=
x0= 0 u=(x-x0)/h= #DIV/0!




#DIV/0! (u/1)∆Y0= #DIV/0!










(u)(u-1)/(1*2)*(∆2y0)= #DIV/0!









(u)(u-1)(u-2)/(1*2*3)*(∆3y0)= #DIV/0!








(u)(u-1)(u-2)(u-3)/(1*2*3*4)*(∆4y0)= #DIV/0!




















#DIV/0! (u/1)∆Y0= #DIV/0!
y0 = 0







(u)(u-1)/(1*2)*(∆2y0)= #DIV/0!









(u)(u-1)(u-2)/(1*2*3)*(∆3y0)= #DIV/0!








(u)(u-1)(u-2)(u-3)/(1*2*3*4)*(∆4y0)=
#DIV/0!







Applying Newton’s forward interpolation formula, we get






Y = y0 + (u/1) Dy0 + {u(u-1)/1.2}D2y0 + {u(u-1)(u-2)/1.2.3}D3y0 + -------------



--------------+ {u(u-1)u-2)…………..(u-n+1)/1.2.3…..n}Dny0




y50= 0 + #DIV/0! + #DIV/0! + #DIV/0! + #DIV/0! = #DIV/0!
y48= 0 + #DIV/0! + #DIV/0! + #DIV/0! + #DIV/0! = #DIV/0!
















y50 - y48 = #DIV/0! = 13

















The number of candidates who get more than .
marks but not more than




marks are



































.

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