Hamid Enki
• Member for 6 years, 6 months
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Yes, it is always a field. Because you are just adjoining an algebraic (in your case) or transcendental element to F. $\mathbf{Z}[\sqrt{-2}]$ Is not a field because $\mathbf{Z}$ is not a field.

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Pick a large circle around the origin containing all the zero's. Then your sum equals to $\frac{1}{2\pi i} \int_{|z|=R} \frac{1}{f}$ by residue theorem. Now apply ML inequality to see that the ...

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Hint: Consider a base of $R^n$ like $\{e_i\}$ and apply $A$ to each $e_i$, assuming such $y$ exists. What should be this $Ae_i$?

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Hint: Based on your proof of convergence for $a_1\in(1-\sqrt a, 1+\sqrt a)$ break the problem into few steps: $a_1 \geq 1+\sqrt a$; show that for all $n$, $a_n\geq1+\sqrt a$ as well. Then show that $... View answer 5 answers votes 187 views 1 votes Use polar form for$1-i$which is$\sqrt2e^{-i \pi/4}$. View answer 1 answers votes 51 views 0 votes How about$A=\begin{pmatrix} 1 &amp; 0\\ 1 &amp; 1 \end{pmatrix}$?$A^n=\begin{pmatrix} 1 &amp; 0\\ n &amp; 1 \end{pmatrix}$. View answer 2 answers votes 65 views 0 votes As you pointed out we cannot. I would put$n_0=\lfloor\frac{1}{\epsilon}\rfloor+1$. View answer 3 answers votes 42 views 0 votes$\log(n) $stands as counter example to both. View answer 2 answers votes 261 views 0 votes If they represent$\gamma, \alpha$and$\beta$correspondingly and if$\gamma \gt \alpha+\beta$, then there should be m such that$\alpha+\beta\lt a_m +b_m \le\gamma$which means$(\alpha-a_m)+(\beta-...
I think the $\lambda_1 \lambda_2 n$ should be equal to $(-1)^n (-1)^n \det(M)$. Now putting them in a quadratic equation may help a little bit.