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 Nov 23 comment Consider $f : \mathbb{N} ā \mathbb{Z}$ defined as $f (n) = \frac{(ā1)^n (2nā1)+1}{4}$. Find its inverse. Looks to me that bot $0$ and $1$ are mapped to $0$, so that $f$ is not injective. Perhaps you are not including $0$ in the natural numbers? Nov 23 comment Convergence of numerical method without stability? This is pretty common in numerical analysis. When the stability constant is independent on the discretization parameters, it's great! You only need stability to converge (for linear methods). Otherwise, you need to prove that the degeneration of the stability constant (as the discretization parameter goes to 0) is not too bad, meaning that the "good" properties of the consistency can make up for it. Nov 23 comment Convergence of numerical method without stability? Sure, you don't need "uniform" stability. What you need, as you already guessed, is a proper balance between stability constant and consistency. In the case $\sigma=1+k$, then the sum in parentheses adds up to $((1+k)^n-1)/k$. So, multiplying by $k\tau$, and assuming $\tau\to 0$, then you still achieve convergence. This is because $k\tau\to 0$ "faster" than $k$ (but you don't have any hint on how much faster), so you can accept a stability constant growing at most linearly with $k$. Nov 23 answered Convergence of numerical method without stability? Oct 28 comment Theoretical Approach to iteration matrix multiplication on a vector It may be worth to point out that the iteration does not "always" converge to the eigenvector corresponding to the largest eigenvalue. In particular, if $a=cw_2$, you converge to the other eigenvector. However, almost surely, you converge to the first one, and, in any case, you end up converging towards one of the two. Oct 28 comment How do you classify the singularities of a complex function by just looking at the equation? The only thing you have to be careful is whether or not the "candidate poles" are indeed poles, since they may cancel with something in the numerator. For instance, $\sin(z)/z$ does not have a pole at $z=0$. Other than that, your eyeballing gave you the right idea. Oct 13 reviewed Edit Limit of a Limit Oct 13 revised Limit of a Limit formatting Oct 13 reviewed Approve Differentiability question wouldn't $f(0,0)$ be undefined because Oct 12 answered If $A_{n \times n}x=b$ has no solutions then $Ax=0$ has infinitely many solutions Oct 12 revised Counterexample to show that the interior of union may be larger than the union of interiors added 4 characters in body Oct 12 comment Are upper and lower Lebesgue integrals equal for every $f$? If $f$ is unbounded, then all the "dominant" simple functions would be infinite on a set of non-zero measure. All the "dominant" integrals would then be infinite. Oct 12 revised ODE Hyperbolic fixed point study grammar Oct 8 comment Use only Archimedean Property of $\mathbb{R}$ to give a direct $\epsilon$-$N$ verification for $\lim\limits_{n\rightarrow\infty}\frac{1}{\sqrt{n}}=0$ I think verification and proof are pretty similar expressions... Oct 8 comment How to find the first root of a cubic equation in matlab Ok, I see it now, thanks. Btw, I think you should have $-|c|$ on the right of the first inequality, no? Oct 8 comment How to find the first root of a cubic equation in matlab I don't get why that inequality implies the roots are in $[-R,R]$. Can you explain that? Oct 2 comment How do I find the $n^{th}$ derivative of the given function? So, I understand the last part (it basically is partial fractions, with complex roots). But I do not understand how the last line "follows" from the infinite series. I think it would be much clearer for the user without the series/residue arguments, simply put as a partial fraction with complex roots. But perhaps that's just me... Oct 2 comment How do I find the $n^{th}$ derivative of the given function? Given the tags to the question, I "assumed" the user was only interested in computing "a few" derivatives of the function. For small $n$, I think this expression is easier, since the deriatives of $g(x)$ are much easier than those of the original function. Then one only has to glue the pieces together.. Oct 2 comment How do I find the $n^{th}$ derivative of the given function? Yeah, that was my point. The question is, since $f(x)$ is still defined for $|x|>1$, does your final expression still hold? Oct 2 comment How do I find the $n^{th}$ derivative of the given function? Provided $|x|<1$...