We want to find A satisfying the following expression with the symmetric and positive-definite condition
Jn=Jn−1+A⇒Δfn=(Jn−1+A)Δxn=Jn−1Δxn+AΔxn⇒AΔxn=Δfn−Jn−1Δxn by the definition
So, solve the minimization problem with the symmetric and positive-definite conditions using method of Lagrange multipliers
min∣∣A∣∣w2=∣∣MAM⊺∣∣F2 subject to A=A⊺ and AΔxn=Δfn−Jn−1Δxn
where W=M⊺M:=Hk
Then, the solution is
A=Δxn⊺(W−1Δxn)(W−1Δxn)(Δfn−Jn−1Δxn)⊺+(Δfn−Jn−1Δxn)(W−1Δxn)⊺−(Δxn⊺(W−1Δxn))2Δxn⊺(Δfn−Jn−1Δxn)(W−1Δxn)(W−1Δxn)⊺
where W−1Δxn=Δfn by the definition
=Δxn⊺ΔfnΔfn(Δfn−Jn−1Δxn)⊺+(Δfn−Jn−1Δxn)Δfn⊺−(Δxn⊺Δfn)2Δxn⊺(Δfn−Jn−1Δxn)ΔfnΔfn⊺=(I−Δfn⊺ΔxnΔfnΔxn⊺)Jn−1(I−Δfn⊺ΔxnΔxnΔfn⊺)+Δfn⊺ΔxnΔfnΔfn⊺
Therefore, updating formula is
Jn=Jn−1+(I−Δfn⊺ΔxnΔfnΔxn⊺)Jn−1(I−Δfn⊺ΔxnΔxnΔfn⊺)+Δfn⊺ΔxnΔfnΔfn⊺
And by the Sherman–Morrison Formula,
Hn=Hn−1+Δxn⊺ΔfnΔxnΔxn⊺−Δfn⊺Hn−1ΔfnHn−1Δfn(Hn−1Δfn)⊺
and
xn+1=xn−Hnf(xn)