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I think it is strange that 𝛽 is also a variable. Obviously the minimum will happen at 𝛽=0, and you lose constraint on the sum of 𝑥𝑖. Although 𝛽 will stop at 𝛽=𝜖, that is still not right, it means very weak constraint on the sum of 𝑥𝑖.

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(Too long for a comment.) The solution $\theta$ (which is presumably real) may not exist. This occurs, e.g., when $A=0$ and $B=-I$. I don't know any sufficient existential condition that isn't too demanding. (It is easy to cook up a sufficient condition, such as $B\succeq2\|A\|_2I$, but that would make every $\theta$ a solution.) Here is a reformulation of ...

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There are histogram estimators in Matlab's "histogram" function, which do some automatic selection of bins (this is the recommended replacement for hist/histc): https://www.mathworks.com/help/matlab/creating_plots/replace-discouraged-instances-of-hist-and-histc.html You should probably use these rather than rolling out your own histogram estimator. ...

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Yes, this has been studied before. The replacement of a (computationally) heavy cost function with a cheaper, but less accurate version, is known as surrogate cost/fitness function. You might look for Surrogate-Based Modeling and Optimization techniques (in this reference are quite some available).

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Having a solver for equality constrained QP only will not help you in solving the general inequality constrained problem (unless you develop a QP solver, which can have an equality-constrained QP solver as a required sub-routine). In other words, your problem cannot be reformulated to a problem with only equalities.

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One can read the Braatz, Morari's paper here (cf. Jiaqi's post) https://www.researchgate.net/publication/238877232_Minimizing_the_Euclidean_Condition_Number They consider results about the $2$ norm condition number. The minimization about the $\infty$ norm condition number is standard and due to L. Bauer (cf. reference in the above one). Essentially, it ...

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This answer is a bit late. I think the following paper may help you Braatz, Richard D., and Manfred Morari. "Minimizing the Euclidean condition number." SIAM Journal on Control and Optimization 32.6 (1994): 1763-1768.

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In the last displayed inequality, notice that the RHS is squared. Rewrite the inequality: $$\|\nabla f(x^{(k+1)})\|_2\le\frac L{2m^2}\|\nabla f(x^{(k)})\|_2^2.\tag1$$ If $\|\nabla f(x^{(k)})\|_2$ is less than $\eta$, then its square is less than $\eta^2$. But $\eta<m^2/L$ so the RHS of (1) is less than  \frac L{2m^2}\eta\eta<\frac L{2m^2}\eta\frac{...

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A non-zero polynomial $\sum_{k=0}^n a_k z^k$ has at most $n$ roots on any subset of $z\in \mathbb C$ including the circle $z=e^{ix}, x\in(0,2\pi]$ corresponding to at most $n$ distinct values of $x$ that produce such roots. That is, the polynomial $\sum_{k=0}^n a_k e^{ixk}$ has at most $n$ roots on the interval $x\in(0,2\pi]$.

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