Playing around with exponents on $x^{n}-y^{n}=1$ suggests a trend toward cleaner and cleaner corners as n increases on one graph for even numbers and a second graph for odd numbers. Therefore, is it possible to graph the case of $n=\infty$ for odd and/or even numbers (separately)?
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$x^\infty$ is meaningless – mrf Sep 26 '13 at 07:35
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4@mrf But the limits of the curves $x^{2n} - y^{2n} = 1$ and $x^{2n+1} - y^{2n + 1} = 1$ aren't, and I believe it's pretty obvious that that's what he meant. – Arthur Sep 26 '13 at 07:36
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$\infty$ is not a number so you wont be able to find one, you can probably find the limit as it APPROACHES $\infty$ – user93089 Sep 26 '13 at 07:37
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The question is perfectly clear,but we shouldn't encourage people writing things like $x^\infty-y^\infty$. – mrf Sep 26 '13 at 14:25
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Odd case:
For every $k$, $y_{2k+1}(x)=(x^{2k+1}-1)^{1/(2k+1)}$ hence, when $k\to\infty$, $y_{2k+1}(x)\to x$ if $x\leqslant-1$, $y_{2k+1}(x)\to-1$ if $-1\leqslant x\lt1$, $y_{2k+1}(1)=0$, and $y_{2k+1}(x)\to x$ if $x\gt1$.
Even case:
For every $k$, $y_{2k}(x)=\pm(x^{2k}-1)^{1/(2k)}$ hence, when $k\to\infty$, $y_{2k}(x)$ does not exist if $|x|\lt1$, $y_{2k}(x)=0$ if $|x|=1$ and $y_{2k}(x)\to\pm x$ if $|x|\gt1$.
Proofs:
Use only the fact that $|x|^n\to\infty$ if $|x|\gt1$ and $|x|^n\to0$ if $|x|\lt1$.
Did
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When you say that $y_{2k}(x)$ doesn't exist if $|x|<1$, do you mean that it is imaginary? Would taking the limit of $x^{2n}+y^{2n}=1$ as $n\to \infty$ work like this: $y_{2k}(x)=\pm (x^{2k}+1)^{\frac{1}{2k}}$ so as $k\to \infty$, $y_{2k}(x)=\infty$ if $x\not = 0$. – User3910 Sep 30 '13 at 17:52
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Yes $y_{2k}(x)$ does not exist in the reals when $|x|\lt1$. The mention of "graphs" in the question strongly suggests real functions defined on the real line. – Did Sep 30 '13 at 18:36
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It does not make sense to use $\infty$ in a function. You may however have a limiting curve for either case.
mau
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