Valentine’s Reflections: Mathematical Matters of the Heart

[neilparker62]

Table of Contents
ToggleIntroductionThe Function (and its Inverse)Reflectionsy = -x(x-1) and Derivativesy=-x(x-3) and DerivativesHeart...

[url="https://www.physicsforums.com/insights/valentines-reflections-mathematical-matters-of-the-heart/"]Continue reading...

 

[Useful nucleus]

Please suggest a citation for anyone who wish to use these expressions (e.g. for homework problems...).

 

[neilparker62]

Please suggest a citation for anyone who wish to use these expressions (e.g. for homework problems...).

Bibtex Code follows - is that ok ?

@misc{Valentine Reflections online,
author = {Neil Parker},
title = {Valentine's Reflections: Mathematical Matters of the Heart},
howpublished = {\url{https://www.physicsforums.com/insights/valentines-reflections-mathematical-matters-of-the-heart/}},
month = {March},
year = {2021},
note = {(Accessed on 03/02/2021)}
}

 

[Useful nucleus]

Thank you!

 

[epenguin]

"needing to solve a polynomial of order 4: y = − x ( x − 4 ) = y ( y − 4 ) ( − y ( y − 4 ) − 4 ) , or − y 4 + 8 y 3 − 20 y 2 + 15 y = 0 . Since such a task is well out of my mathematical reach, "It is not beyond your reach when you already know, as you do, two of the solutions!

 

[neilparker62]

"needing to solve a polynomial of order 4: y = − x ( x − 4 ) = y ( y − 4 ) ( − y ( y − 4 ) − 4 ) , or − y 4 + 8 y 3 − 20 y 2 + 15 y = 0 . Since such a task is well out of my mathematical reach, "It is not beyond your reach when you already know, as you do, two of the solutions!

True enough - should have seen that one! Perhaps a little bit 'knee jerk' to send it off to WA!

 

[neilparker62]

It is not beyond your reach when you already know, as you do, two of the solutions!

Ok - better give it a go then!

$$y=-x(x-k) ; x=-y(y-k) \implies
y=y(y-k)(-y(y-k)-k)$$ $$ \implies
-y=y(y-k)(y(y-k)+k)=(y^2-ky)(y^2-ky+k)$$ $$\implies -y=y^4-ky^3+ky^2-ky^3+ky^2-k^2y$$ $$\implies 0=y^4-2ky^3+2ky^2-(k^2-1)y=y(y-(k-1))(y^2-(k+1)y+(k+1))$$ From which we obtain the two solutions at (0;0) and (0;k-1). The discriminant of the third factor quadratic is $$(k+1)^2-4(k+1)=k^2-2k-3=(k-3)(k+1)$$ in accordance with the WA solution.