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Calculus is a tough discipline. No wonder our students don't get it. On the other hand, analysis is an easy subject (so easy that we find it unnecessary to teach it to our students). So, let's do ana …

Sorry for making you wait 14 hours unnecessarily but you are partially guilty yourself: if you posted a correct and full version of the question from the beginning, you would get the answer in 5 minut …

I'll change the variable $y=\frac 12-x$ to make typing easier. Since, as Peter already observed, the condition $Q(0)=0$ is worthless and since $\frac 1{12}$ is just an additive constant, we are just t …

Here is just a recipe. It has its advantages and disadvantages, so try to implement it and see if it gives what you want or there are some undesirable features. It minimizes the integral of the gradie …

The first maximum of $f(s)=\sin s+\sin\sqrt 2 s$ is attained at some $s_0\in [1.2,1.3]$ and exceeds $1.9$. The derivative of $f$ is at most $1+\sqrt 2\le\frac 52$. One obvious strategy is to go at the …

Do you have an easy argument... Sorry for the late reply, but it is, actually, a rather simple story. Let $R=1$ and suppose that we have declared some $M$ and $\delta$. Then the adversary starts at s …

I have no doubt that someone will come with some brighter idea but here are my 2 cents anyway. If you don't aim at something very fast, I would just use the inequality $(a+b)^2\le (ta^2+(1-t)b^2)(t^{- …

Yes, of course (provided that you mean that the maximum is attained on a diagonal matrix, not that every matrix on which it is attained is diagonal). First notice that it is a bit more convenient to …

$x^2-y^2$ is positive on $[2,3]\times [-1,1]$ but not convex there. This creates problems for any convex sets not containing the origin. You are, probably, after something else not so obviously false. …

If $k>0$, it becomes elementary algebra. As Arthur pointed out, the equation is $P(z)=z^3-kz^2+(bk)z-ak=0$. On the one hand, assume that all roots have positive real part. Then we either have 3 positi …

This is just about the asymptotic behavior for $a\to+\infty$. I claim that the minimal length $\ell$ is about $\sqrt{2a}$ for large $a$. The upper bound Consider $2n$ horizontal lines splitting the …

Just use Cauchy-Schwarz and the identity $$ \sum_{i,j}(u_iu_j-v_iv_j)^2=\left[\sum_i u_i^2\right]^2+\left[\sum_i v_i^2\right]^2-2\left[\sum_i u_iv_i\right]^2=2 $$ to get $f(u,v)\le \sqrt2 n$ for all $ …

If $m=k+\ell$, then the question reduces to when the intervals $(a_i-m,a_i+m)\mod p$ do not cover the full interval $[0,p]$. The answer is, of course, a union of finitely many intervals and the intere …

It turns out that the algorithm I was trying works for the minimization of the sum $\sum_{j=1}^m F(r_{k_{j-1}+1}+\dots+r_{k_j})$ (we put $k_0=0$) for an arbitrary strictly convex function $F$, so I'll …

Since the functional is convex, I would try modified gradient descent. The main problem with this approach is that you'll really have to compute the inverse matrix a few times, which may be out of que …