How do I draw $x^{2n-1}+y^{2n-1}=r^{2n-1}$? Drawing the $x^{2n}+y^{2n}=r^{2n}$ is possible by proving that each side is a straight line.
but I thought that it would be slightly different because $2n-1$ is an odd number.
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The sides of $x^{2n}+y^{2n}=r^{2n}$ are not perfectly straight. – Akiva Weinberger Jul 14 '17 at 03:50
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by using limit n goes to infinity we can say that it nearly is almost a straight line, right? – RINY Jul 14 '17 at 05:01
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In the limit, yes. – Akiva Weinberger Jul 14 '17 at 11:03
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For the first one, $$ x^3 + y^3 = (x+y)\left( x^2 - xy + y^2 \right). $$ Not only is $ x^2 - xy + y^2 \geq 0, $ we have $$ x^2 - xy + y^2 \geq \frac{3}{4} \; x^2, $$ $$ x^2 - xy + y^2 \geq \frac{3}{4} \; y^2. $$ We conclude that (when $x^3 + y^3 = 1$) $x+y$ stays small and positive, while $x+y$ gets close to $0$ when either $|x|$ or $|y|$ is large.
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Will Jagy
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How can I prove the abstract shape of x^(2n-1)+y^(2n-1)=r^(2n-1) algebraically? – RINY Jul 14 '17 at 00:31
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How can you even express it algebraically, other than as the equation $x^{2n - 1} + y^{2n - 1} = r^{2n - 1}$? If you can pick out a geometric feature from the graphs that has some concrete definition (e.g. has a slant asymptote of $y = x$, is symmetric about $y = -x$, etc), then you might be able to prove it. – Theo Bendit Jul 14 '17 at 03:38