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I ran the function $(x^2+y^2)^z=z$, where $z$ is a constant. The function produced a circle, and the $z$ value where the radius of the circle turned out to be the largest was $e$. At that point, the radius of said circle was $1.201~943\dots$ Since it correlates to $e$ in that way, I wonder of $1.201~943\dots$ has any significance, or a special name. If it does, I could potentially use it somehow. Is this number important?

mrtaurho
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Nip Dip
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    This is the constant $e^{1/(2e)}$ which is just the square root of the maximum of the function $f(z)=z^{1/z}$ at $(e,e^{1/e})$. – Peter Foreman Mar 15 '20 at 00:19

2 Answers2

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This number is just $e^{\frac1{2e}}$. Nothing especially noteworthy about it as far as I know.

Note your equation can be written as $$r=z^{\frac1{2z}}$$ and you have maximized this.

Addendum: Note that $$\frac{dr}{dz}=\tfrac12 z^{(2z)^{-1}-2}(1-\ln z)$$ The sign of this expression is determined by $1-\ln z$, which shows the function $r(z)$ is increasing to the left of $z=e$ and decreasing to the right of $z=e$. The function is therefore maximized at $z=e$.

MPW
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  • Okay, how did you find that number so quickly? Is it memory or something else? I couldn't find what that number means. – Nip Dip Mar 15 '20 at 00:50
  • The function $r(z)=z^{\frac1{2z}}$ attains its maximum at $z=e$ (this is where the derivative $dr/dz$ vanishes). So the maximum value there is $r(e)=e^{\frac1{2e}}$. Do you want to see how to compute that derivative? – MPW Mar 15 '20 at 01:42
  • Im only in algebra 2 so yes, yes I would. – Nip Dip Mar 15 '20 at 20:39
  • Added some comments on maximizing this function. See my answer here for details on computing this derivative. But note if you don’t know calculus, you won’t understand it. I can’t explain calculus here in the comments. – MPW Mar 16 '20 at 08:47
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Your equation is $$x^2+y^2= \left(z^{\frac{1}{2z}}\right)^2$$

The function $z^{\frac{1}{2z}}$ attains its maximum when $z=e$ and the value is $e^{\frac{1}{2e}}$, your number must be equal to $e^{\frac{1}{2e}}$.

mrtaurho
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N. S.
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  • @mrtaurho Thank you, I wanted to write explicitely the radius and forgot to square it :) – N. S. Mar 15 '20 at 00:24