Create three fitting functions: const - tries to fit with f(2)= lin - tries to fit...

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Create three fitting functions: const - tries to fit with f(2)= lin - tries to fit with f(2)=2+Co quad - tries to fit with f(2)=C₂ 2² +₂·1+% Quantifying a Fit We may ask ourselves "how good is that fit?" To answer that, we need to define the error of our fit relative to the provided data. There are many ways to do this, but we will use the "root mean square" method - we will calculate the square root of the mean of the squares of the error at all points. The sum of the squares of the errors (se) is: QQQQQQQ se= 1-0 data-u)² The mean (average) square error (mse) is: mse= - sq_err And the square root of the mean square error (RMS) is: RMS = √mse Putting it all together: 5550000000 (datal-ful¹])² RMS N For the data above, if our fit line was y=1-2 +0, then the RMS can be calculated as follows: sq_err = (0-0)²+(1-2)²+(2-2)²+(3-3)²+...+(8-7)²+(99)²) = 2 mean_sq_err = 0·2=0.2 10 RMS = √0.2 = 0.4472 The actual fit line calculated by curve fit is y=0.915-r+0.382. Does this give a lower RMS? Create three fitting functions: const - tries to fit with f(2)= lin - tries to fit with f(2)=2+Co quad - tries to fit with f(2)=C₂ 2² +₂·1+% Quantifying a Fit We may ask ourselves "how good is that fit?" To answer that, we need to define the error of our fit relative to the provided data. There are many ways to do this, but we will use the "root mean square" method - we will calculate the square root of the mean of the squares of the error at all points. The sum of the squares of the errors (se) is: QQQQQQQ se= 1-0 data-u)² The mean (average) square error (mse) is: mse= - sq_err And the square root of the mean square error (RMS) is: RMS = √mse Putting it all together: 5550000000 (datal-ful¹])² RMS N For the data above, if our fit line was y=1-2 +0, then the RMS can be calculated as follows: sq_err = (0-0)²+(1-2)²+(2-2)²+(3-3)²+...+(8-7)²+(99)²) = 2 mean_sq_err = 0·2=0.2 10 RMS = √0.2 = 0.4472 The actual fit line calculated by curve fit is y=0.915-r+0.382. Does this give a lower RMS?

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Replace the fitparams tuple with the coefficients from your actual curve fit in this case the coeffi... View the full answer