2

I graphed this family of equations with range $0<y<1$ and domain $0<x<1$. Each of the curves has has a point for which its derivative is one; for perspective, I also graphed the line with a slope of one which is tangent to these points. How would one go about finding the equation for the curve that connects every such point? (Namely, all the points for which the derivative is 1)

I could always have wolfram|alpha fit a polynomial equation by feeding it the points, but I would like to know how one would one calculate the equation, if possible.

Thanks in advance. enter image description here

Misha Lavrov
  • 142,276
Alex D
  • 1,214

2 Answers2

1

The slope of $x^n$ is, at any point, equal to $nx^{n-1}$ (this is true even for non-integer $n$). So if the slope is equal to $1$, then (for $n\neq 1$) we have $$ nx^{n-1}=1\\ x=\sqrt[n-1]{\frac1n} $$ As for the $y$-coordinate of this point, we just insert this $x$-value into the original expression for the curve: $$ y=\sqrt[n-1]{\frac1n}^n $$ So now we have a parametric expression for the curve going through all your tangent points (again, this is not valid for $n=1$) $$ \cases{x(n)=\sqrt[n-1]{\frac1n}\\ y(n)=\sqrt[n-1]{\frac1n}^n} $$ In principle, you could probably solve the first equation for $n$, and insert that into the second equation to get something of the form $y=f(x)$, but I don't know how possible it is in practice.

Arthur
  • 199,419
  • Funny explicit equation, isn't it ? If I may ask a favor, could you put a plot of it ? Being almost blind, I cannot do it (at least properly). Thanks and cheers. – Claude Leibovici Oct 04 '17 at 08:16
  • I saw it. I don't really like $W$ personally, it smells to much of being not really a solution but rather just a hidden rewriting. I know that the same can really be said about $\sqrt[n]{{}\cdot{}}$, so it's hypocritical of me. – Arthur Oct 04 '17 at 08:18
  • If I may ask : why don't you like $W(x)$ ? To me, it is one of the most beautiful function after $\log(x)$ and $e^x$. Moreover, it has so many practical applications in so many areas of engineering (and more). – Claude Leibovici Oct 04 '17 at 08:21
  • @ClaudeLeibovici As I said, to me it smells too much of a rewriting, saying that "the solution of this equation is that $x$ is the number that solves this equation". It's not a logical thing. I acknowledge that it is very useful, and there isn't really a substitute, and I did upvote your answer. I just don't like it. Perhaps it's because I haven't used it much myself. – Arthur Oct 04 '17 at 08:29
1

Starting from Arthur's answer, you could make things explicit since $$x=\sqrt[n-1]{\frac1n}\implies n=\frac{W(x \log (x))}{\log (x)}$$ where appears Lambert function. This makes $$\Large{y=x^{\frac{W(x \log (x))}{\log (x)}}}$$

Claude Leibovici
  • 260,315