# Interpolation of adjusted square roots...emergency

• Aug 6th 2010, 06:07 PM
yashsr
We've got a project where we're required to interpolate. We've got 10 values, square roots of 10 to 1. These roots are then adjusted by adding 50% to root of 10, 40% to root of 9, 30 % to root of 8, 20% to root of 7 and 10% to root of 6. Then again, 10% to root of 5, 20% to root of 4, 30% to root of 3, 40% to root of 2 and 50% to root of 1.

It looks like this:
3.16 + 50%=4.74
3 + 40%=4.2
2.83 + 30%=3.68
2.65 + 20%=3.17
2.45 + 10%=2.69
2.24 + 10%=2.46
2 + 20%=2.4
1.73 + 30%=2.25
1.41 + 40%=1.98
1 + 50%=1.5

Now we take cumulative totals and the final total is 29.08.
Now, if 10 is 4.74, 20 is 8.94(sum of 4.74 and 4.2)....90 is 27.58(sum of 1st 9 items or say 29.08-1.5)...

Can I interpolate and find values of say 3, 27, 67, 83 etc. now that we can values of all multiples of 10?

I actually tried using Newton's Approximation Formula but it either didn't work all the time or it seems it is giving approximate values which is of little use.
• Aug 7th 2010, 04:37 AM
yashsr
Well its basically this:

x sqrt(x) %inc y sum z
--- ------- ---- ---- ----- ----
10 3.16 50% 4.74 4.74 10
9 3.00 40% 4.20 8.94 20
8 2.83 30% 3.68 12.62 30
7 2.65 20% 3.17 15.79 40
6 2.45 10% 2.69 18.48 50
5 2.24 10% 2.46 20.94 60
4 2.00 20% 2.40 23.34 70
3 1.73 30% 2.25 25.59 80
2 1.41 40% 1.98 27.57 90
1 1.00 50% 1.50 29.07 100

So now I want to find values of say when z is equal to 3, 25, 47,94 etc. Is Interpolation possible? How? And note you cannot do something like the following which is linear Interpolation which is just not accurate.

z sum
--- -----
20 8.94
27 ?
30 12.62

The approximation would be

8.94 + (27-20)/(30-20)*(12.62-8.94) = 8.94 + 0.7*3.68 = 11.516

This is simply 7/10 of the way from 8.95 to 12.62.

Obviously this is not correct. The value for z=27 would be something more than 11.516.