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 Mar 17 awarded Good Answer Dec 23 awarded Yearling Sep 27 answered Property of a spanning list in a finite-dimensional vector space Sep 20 revised Solving a generic second order differential equation added 5 characters in body Sep 19 answered Solving a generic second order differential equation Sep 15 answered First class example and i have no clue what to do Sep 15 revised First class example and i have no clue what to do added 67 characters in body Sep 15 comment A simple question about ring theory @JohnLee $$x+y=(x+y)^2\\x+y=x^2+y^2+xy+yx\\x+y=x+y+xy+yx\\xy+yx=0$$ Sep 9 comment What is a moving average system? @Guesswhoitis. the OP is a causal filter, yours is not. the way to see how this works is to consider a sliding window of 3 samples and at each point, the output $y[n]$ is the mean of the previous 3 input samples $x[n],x[n-1],x[n-2]$ in the window Sep 9 revised Definition of characteristic polynomial edited body Sep 1 answered Change of variable using dirac delta function Sep 1 comment Proving $[(P\lor Q)\land(P\to R)\land(Q\to R)]\to R$ is a tautology without using a truth table? what inference rules do you have? surely implication elimination, i.e. $(a\rightarrow b)\land a\implies b$ Sep 1 comment Find $\sum\limits_{n=1}^{\infty}\frac{n^4}{4^n}$ Fourier series instead of generating functions -- nice Sep 1 comment Understanding Little Oh Notation Proof - Prove the function$f(n) = 12n^2 + 6n\ \ is\ \ o(n^3)$ if for any $c$ we can find $n_0$ such that $0\le f(n)\le cg(n)$ holds for $n\ge n_0$ then it follows that $n\ge n_0$ is sufficient to give a strict bound $0\le f(n)c$. So if we want to show for $d$ that there is $n_0$ so that $0\le f(n)< dg(n)$ for $n\ge n_0$, just pick a $c$ such that $0a$ how are you defining $\lfloor a\rfloor$? once you specifically have that it should be obvious Aug 30 comment Generalizing limits of sums, products, and quotients of sequences to abstract topological spaces? @EthanAlvaree en.wikipedia.org/wiki/Algebra_over_a_field Aug 22 comment Polynomial equations of degree larger than 4 finding cube roots is trivial by de Moivre's theorem: $z^3=re^{it}$ gives $z\in\{\sqrt[3]{r}e^{it/3},\sqrt[3]{r}e^{i(t+2\pi)/3},\sqrt[3]{r}e^{i(t+4\pi)/3}‌​\}$ Jul 31 comment Asymptotic direction just an fyi: the second derivative doesn't live in the tangent space or bundle