GPY

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9 years, 169 days

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These are questions asked by GPY

I need to find the local maxima and minima of f(x,y)=x(x+y)*e^(y-x). I have tried to look for an appropriate method that I could use to achieve this, but got stuck. I also don't quite understand the math behind tying to obtain the local maxima and  minima for a function of this type.

I'm trying to evaluate the multidimensional limit: 

(1+y)^(x-1)-1/(1-cos((x-1)^2+y^2)^(1/4)) as (x,y)->(1,0) using the limit command :

limit((1+y)^(x-1)/(1-cos((x-1)^2+y^2)^(1/4)),{x=1,y=0});

 

but don't seem to get any output. Also, I think the limit for this function doesn't exist or is indeterminate on R2. Where am I wrong?

I need to show that the least square solution x that I obtained as x=pseudoinverse(A).b is the solution of Transpose(A).A.x=Transpose(A).b with the smallest norm. I've obtained the norm for the RHS of this expression as well as the norm of x but I'm unsure of how to conclude that this is the lowest possible norm using these values

 

I've got a vector x=[0.36,1.3279,1.6882] (1*3) obtained as x:=pseudoInverse(A)*b where A=[<2,4,1>|<1,-1,1>|<3,3,2>] where <2,4,1>,<1,-1,1>,<3,3,2> are the columns of A and b is [8,5,4](1*3). Now, when I find the rowspace of A using RowSpace(A) I get the row vectors <1,0,1> and <0,1,1>, neither of which are equivalent to x. How do I arrive at my result that x is in the rowspace?

where A^T is the transpose of A and it's given that (A^T).A is not invertible.

I am stuck as to how to arrive at the solution for x in this case. I initially thought I could multiply both sides by the inverse of A^T reducing it to Ax=b but that was obviously wrong since A^T is itself not invertible(it is singular).

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