Regression v. interpolation

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Chris Jeong
Chris Jeong 2017-12-13
回答： kaitlyne diaz 2018-4-24
Fit a line using linear regression to the following data set. Then linearly interpolate.
x = [ 0.13653 0.14551 0.28696 0.47473 0.48740 0.75441 0.88754 0.91808 0.94291 0.97388 ];
y = [ 0.29250 0.79586 0.78081 0.12643 0.14409 0.50125 0.40181 0.83697 0.34130 0.27120 ];
Evaluate both at point x = 0.5. Store the absolute value of difference of these two in a variable `lin_diff`.

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kaitlyne diaz
kaitlyne diaz 2018-4-24
x =1

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