Spook
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If the Taylor series about 0 does not terminate, $f(1/z)$ has an essential singularity at $0$ (why?) Then from the Casorati–Weierstrass theorem (have you learnt this?) you know $f(1/z)$ cannot be ...

Suppose not. Then $x^2=y^3-2$, hence $x$, is even. Therefore, $2=y^3-x^2$ is divisible by $2^2$, which is a contradiction.

Picture a sinusoidal curve, but stretched in the horizontal direction more and more as $x\rightarrow\infty$. The amplitude is fixed but the velocity decreases. Something like $\sin (\sqrt{x})$.

$(n+1)n!<(n^n)(n+1)<((n+1)^n)(n+1)=(n+1)^{(n+1)}$. Try to direct your algebraic manipulations so that the expressions gradually look like the desired result.

No. Imagine strectching the plane in the $x$-direction. Formally $(x,y)\mapsto (2x,y)$. The angle $(1,1)$ makes with $(1,0)$, which, originally is 45 degrees, is reduced.

No, you have not understood the notation. It just means adding up all possible evaluations of ij under the restriction that $1\le i<j\le 3$.

In the context of computational complexity, we want to describe the behaviour of $f(n)$ as $n\rightarrow \infty$. $\frac{\log 4n }{n}\rightarrow 0$ as $n\rightarrow \infty$ so $f(n)/n\rightarrow 1/3$ ...

All the $x_i/x_j$'s in the sum is 1 or -1. Therefore you need an even number of them to make the sum 0, since every time you have a 1 you must have another -1 to cancel it out and vice versa. ...

No. A subring of $\mathbb{Z}$ must contain $1$. Then it must contain all elements $1+1+\cdots+1$ and all their additive inverses. And also zero. So any subring of $\mathbb{Z}$ must be $\mathbb{Z}$ and ...

A set $S$ being linearly independent by definition means no non-trivial linear combination of elements in $S$ is zero. And a linear combination is a FINITE sum.Therefore if all finite subsets are LI, $... View answer 1 votes The key is that$A_n$are all finite. Can you see that if an ant approaches$a$without touching$a$, for any$n$, eventually it will leave$A_1, A_2, \cdots, A_n$? If you see that then you are one ... View answer 1 votes Generally the one-step subgroup test is faster but in this case you can just check the group axioms: the only non-trivial one is closure. If$a^2=b^2=e$, can you see that$ab$is its own inverse, ... View answer 0 votes Assume$x\ge y$by symmetry. We want $$x^x-x^y\ge y^x-y^y.$$ i.e. $$x^y(x^{x-y}-1)\ge y^y(y^{x-y}-1),$$ which is then easy to show by listing all possible cases according to how$x,y$compare to 1. View answer Accepted answer 0 votes Hint: Express$|x|$as a square root of scalar product of vectors. Then try to make the scalar product look like$f(x)+A(x)+o(x)$where$A$is a linear transformation. Then use the fact that$x\mapsto ...