Understandably there are a lot of answers, but if you still have any further questions maybe this will help. An embedding of a topological space $X$ into a topological space $Y$ is a continuous map $... View answer 8 votes In$\mathbb{R}^2$, Schonflies theorem ensures that any two embeddings of Cantor sets are equivalently embedded, or equivalently that any homeomorphism$h$between the two Cantor sets (it can be proven ... View answer 4 votes$\frac{4^{2x-2}}{2^{x-2}} = \frac{(2^2)^{2x-2}}{2^{x-2}} = 2^{(4x-4)-(x-2)} = 2^{3x-2}$View answer Accepted answer 3 votes Either would be a valid equation but you'd get different values of C; I.e. if you take C to be the value that satisfies:$a=Cb$Then we could also write this as:$aC^{-1}=b$So you can view it ... View answer 2 votes I think that part of this question answers itself when you get "into" a field. When I was writing my MSc. dissertation I really had no idea what I was interested in, but my supervisor pointed me in a ... View answer Accepted answer 2 votes Its like compound interest! The general formula would be$10 \times 1.25^n$where n is the amount of times you are "compounding the interest". View answer 2 votes So if you consider::$$\frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} \times \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}} = \frac{(\sqrt{3} + \sqrt{2})^2}{(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2})... View answer 1 votes Simply put isn't it 10^(3.8) ? View answer 0 votes In$\mathbb{R}^2$, for any two Cantor sets$A$and$B$we can find a homeomorphism$h \colon \mathbb{R}^2 \to \mathbb{R}^2$which carries$h(A)$onto$B$. This essentially says they're equivalently ... View answer 0 votes We initially use the tan function:$\tan \left( \frac{5}{3} \right) = \frac{b}{a} = \frac{b}{16}$, so$b = 16 \tan(\frac{5}{3})$. You can then directly apply Pythagoras' Theorem, so:$c^2 = 16^2 + (...

This may be a different approach but you could consider your formula as: $f(x) =\frac{(x^3)^{\frac{4}{3}}} {(2-x)^{\frac{4}{3}}} = (x^3)^{\frac{4}{3}} (2-x)^{-\frac{4}{3}} = x^4 (2-x)^{-\frac{4}{3}}$ ...

I'd say it very much depends on the context; if its a field you are comfortable in and you have a good grasp of the basic concepts then it definitely is. If, however on the other hand, you're ...

Any point on the boundary will have a neighbourhood homeomorphic to a neighbourhood in $\mathbb{R}^2_+$; think of this pictorally - its neighbourhood will be a collection of interior points of the ...

Youre meant to be evaluating the modulus of your answer, so $|\frac{3i+1}{5}| = \sqrt{\frac{9}{25}+\frac{1}{25}} = \sqrt{\frac{10}{25}} = \frac{\sqrt{10}}{5} = \frac{\sqrt{2}}{\sqrt{5}} = B$

Try evaluating the gradient at points very close to $x = -2$, and see what happens as you approach $x$, to get some hints!
Maybe a totally different approach would be to consider your elements as rotations of $\frac{2 \pi}{3}$ acting on a regular 3-gon i.e. an equilateral triangle?