Ordinary least squares?
When you say "Order=4", you mean to the 4 power?
Hi all,
This is my first post on this forum.
Lets say I have,
x vectors (-0.0028, 5.0028, 10.0085, 15.0085, 20.0028)
y vectors (0, 5, 10, 15, 20)
Order = 4
y = b0 + b1x + b2x² + b3x^3 + b4 x^4
Is there any equation to calculate b0, b1, b2, b3, b4 values?
Thanks and Regards,
Hemnath.D
Ordinary least squares?
When you say "Order=4", you mean to the 4 power?
I want to calculate the co-efficients of fourth order equation. since i have 5 data points.
You could just use any software that fits linear models to get the coefficients. But if you want to do this by hand you might want to read this: https://en.wikipedia.org/wiki/Polynomial_interpolation
I don't have emotions and sometimes that makes me very sad.
hemnath (09-25-2015)
Yes I have tried with software and manually calculated by putting matrices to calculate the co-efficients. Both the results are same.
But the thing is, Is anyone evaluated the equation to find the co-efficients. Because I'm developing an application that doesn't support matrices. Only arithmetic calculations. If there is equation to find the co-efficients, I can easily put into my application. Can you please take me in the right path.
Thanks in Advance.
Hi,
depending on what you want, this couild be useful: https://mat.iitm.ac.in/home/sryedida.../lagrange.html
regards
hemnath (09-30-2015)
I don't think you realize what it is you are actually really doing. Specifically, you are using curvilinear regression analysis using the highest-degree polynomial possible, which will make your R^2 equal to eta^2 - your accounting for both linear and non-linear effects - since both analyses permit as many bends in the curve as there are degrees of freedom minus one for the between-sums of squares.
In short, you are doing nothing more than conducting a regression analysis using "dummy coding". Thus, the coefficients associated with your model are straight forward to compute: b0=20 (value of y asssigned 0's), b1=-20 (0-20), b2=-15 (5-20), b3=-10 (10-20), and b4=-5 (15-20).
As such, the coefficients are nothing more that the differences between the value assigned "zeroes" and the other values of y.
Note: One of the goals of scientific research, however, is parsimony. The interest is not so much in the highest-degree polynomial equation possible - but rather - in the highest-degree polynomial equation necessary to describe a set of data.
I hope this helps.
Thanks guys. Used the Lagrange's Interpolation formula.
I have Y vectors = (0, 5, 10, 15, 20)
At Condition1, x Vectors = (-0.0056, 5.0000, 10.0057, 15.0057, 20.0000)
At Condition2, x vectors = (-0.0028, 5.0028, 10.0085, 15.0085, 20.0028)
At Condition3, x vectors = ( 0.0000, 5.0056, 10.0113, 15.0113, 20.0056)
There can happen many conditions between condition1 and condition2 or between condition2 and condition3.
I calculated the co-efficients for conditon2.
When I apply x = 10.0085 for condition1, I get y = 10.0000
When I apply x = 10.0057 for condition2, I get y = 9.99720
When I apply x = 10.0113 for condition3, I get y = 10.002798
To get y = 10.0000 for any condition, Do I have to calculate co-efficients individually? Or is there any mathematical model or which regression technique should i use?
Please guide me.
Thanks in advance.
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