Recent content by zollen

Ans(b): Given ##\hat{x}## = bavg (from Ans(a)) Therefore e = b - a \hat{x} = \begin{bmatrix} b_1\\ b_2\\ ..\\ b_m \end{bmatrix} ~-~ \begin{bmatrix} 1\\ 1\\ ..\\ 1 \end{bmatrix} ~b_{avg} = \begin{bmatrix} e_1\\ e_2\\ ..\\ e_m \end{bmatrix} ||e||^2~=~\frac{{e_1}^2 + {e_2}^2 + ... + {e_m}^2}{m}...

This problem projects b = (b1,b2...,bm) onto the line through a = (1, 1, 1, ...1). We solve m equations ax = b in 1 unknown (by least squares). (a) Solve aT a ##\hat{x}## = aT b to show that ##\hat{x}## is the mean (the average) of the b’s. (b) Find e = b - a ##\hat{x}## and the variance ||e||2...

Hi All, This question is about vector calculus, gradient, directional derivative and normal line. If the gradient is the direction of the steepest ascent: >> gradient(x, y) = [ derivative_f_x(x, y), derivative_f_y(x, y) ] Then it really confuse me as when calculating the normal line...