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It's a bit confusing to take ##b < a##; usually we have ##a < b##. My approach would be to define $$ g : [0,1]\to\mathbb{R} : x \mapsto...

A Hessian matrix must be symmetric; this matrix is not. This is a consequence of the problem I noted in my earlier post: the vector...

It may help to note that $$\frac{n^2}{n + \frac15} = n - \frac{1}{5 + \frac 1n}.$$

Yes. Expanding ##(z - a)^{-k}## in binomial series can be done in two ways; one converges for ##|z| < |a|## and the other for ##|z| > |a|##.

A lot o numerical analysis involves dealing with spaces of polynomials of degree at most $$N$$ on $$[-1,1]$$. There are various families...