Section 11.3 Details of Broyden's method
Recall from Section 11.1 that we seek to improve our guess for the inverse of Jacobian by finding a matrix B1 such that
B1(F(x1)−F(x0))=x1−x0
To simplify notation, let h=x1−x0 and w=F(x1)−F(x0); both these vectors are known. We need
B1w=h
Question 11.3.1. Why would not an outer product work here?
We already know how to find a matrix that satisfies (11.3.1), by taking an outer product multiplied by a scalar (11.2.2). But this would be a bad, illogical choice for \(B_1\text{.}\) Why?
B0w+MB0w=h
so
MB0w=h−B0w
According to (11.2.2) we can achieve this by letting
M=1hTB0w(h−B0w)hT
where the chose the term b to be h. To summarize, the formula for updating our guess for inverse Jacobian is
B1=B0+1hTB0w(h−B0w)hTB0
It looks messy, but at least the derivation was logical.
The matrix B1 is still a guess, it may be very different from the actual inverse of Jacobian matrix. But it is a more educated guess, and as this process repeats, these repeated corrections will produce more accurate guesses.