I kinda liked that substitute TA as she was kind of funny. BTW for the 64bit long machine number to decimal conversion. The formula is as follows: (-1)^1*2^(c-1023)*(1+f) 1st bit for sign, 11 bits for exponent and the rest for the decimal mantissa. where c is the exponent total and the f is the total on the mantissa. Found this to be more succinct then the explanation from the tutorial.
Also did anybody find that the homework did nothing to prepare you for the quiz1? That is how I felt at least.
An eigenvector of an n x n (square) A matrix is a non-zero vector v that, when multiplied by A, yields the original vector multiplied by a number Lambda (multiplier)
Comments
Find the roots of the following non-linear equation
| x^2 -3 | = 4
One can use
http://www.wolframalpha.com/
plot ( abs(x^2-3), 4 ) , x=-3 to 3
to visualize the solution
BTW for the 64bit long machine number to decimal conversion. The formula is as follows:
(-1)^1*2^(c-1023)*(1+f)
1st bit for sign, 11 bits for exponent and the rest for the decimal mantissa.
where c is the exponent total and the f is the total on the mantissa. Found this to be more succinct then the explanation from the tutorial.
Also did anybody find that the homework did nothing to prepare you for the quiz1? That is how I felt at least.
The understanding of this following question:
It takes 60s to compute LU factorization of a 6000x6000 matrix. How long for a 1500x1500 one?
What I understand so far:
1) O(n^3) for LU factorization
2) 1500x1500 is 1/16 the area of a 6000x6000
3) ratio of 6000/1500 is 4.
The answer is 1 second but I am trying to figure out how that is obtained.
about the number of floating point operations to compute the LU for
a matrix of size n x n.
The number of operations is proportional to ~ n^3.
LU order is O(n^3). This means that if we reduce n by 4, we obtain
(1/4)^3 = 1/64 reduction
in the number of floating point operations needed to obtain the LU
decomposition. Indeed, we
need 1/64 of the initial time. i.e. ~1sec