To deduce *geometrically*

[Hill]

Homework Statement

The roots of a general cubic equation in X may be viewed (in the XY-plane) as the intersections of the X-axis with the graph of a cubic of the form, Y = X^3 + AX^2 + BX + C.
(i) Show that the point of inflection of the graph occurs at X = −A/3 .
(ii) Deduce (geometrically) that the substitution X = x − A/3 will reduce the above equation to the form Y = x^3 + bx + c.
(iii) Verify this by calculation.

Relevant Equations

Y = X^3 + AX^2 + BX + C
Y = x^3 + bx + c

The (i) is straightforward: take the second derivative to 0.
The (iii) is obvious: after the substitution, $x^2$ comes from the $X^3$ with the coefficient $-3A/3$ and from the $AX^2$ with the coefficient $A$, and they cancel.

Here is my attempt for the (ii).
The substitution translates the graph so that the inflection point sits at $x=0$. Thus, the curvature of the graph is zero at $x=0$.

The curvature of $x^3$ is zero at $x=0$. The terms $bx$ and $c$ do not affect the curvature. But $ax^2$ with $a \neq 0$ has a non-zero curvature. So, if there were such a term, the curvature of the graph would be zero when the curvature of the $x^3$ cancels the curvature of $ax^2$, i.e., when the curvature of $x^3$ is not zero, i.e., not at $x=0$. Done.

Do you think this derivation is geometric? Any ideas for a different geometric derivation?

 

[Mark44]

The substitution translates the graph so that the inflection point sits at
$x=0$. Thus, the curvature of the graph is zero at $x=0$.

The curvature of $x^3$ is zero at $x=0$. The terms $bx$ and $c$ do not affect the curvature. But $ax^2$ with $a \neq 0$ has a non-zero curvature. So, if there were such a term, the curvature of the graph would be zero when the curvature of the $x^3$ cancels the curvature of $ax^2$, i.e., when the curvature of $x^3$ is not zero, i.e., not at $x=0$.

I don't see anything wrong with this, but I would add that the terms bx and c do not affect the curvature, because the curvature involves the second derivative, for which the second derivative of bx + c vanishes. I don't see another way of making the point that there cannot be an $ax^2$ term, but I would try for a cleaner explanation of why this can't happen.

 

[Hill]

the terms bx and c do not affect the curvature, because the curvature involves the second derivative, for which the second derivative of bx + c vanishes

This is of course true. However, trying to keep it geometrical, I'd rather point to the fact that $bx+c$ is a straight line, and that does not have a (non-zero) curvature.

 

[pasmith]

Homework Statement: The roots of a general cubic equation in X may be viewed (in the XY-plane) as the intersections of the X-axis with the graph of a cubic of the form, Y = X^3 + AX^2 + BX + C.
(i) Show that the point of inflection of the graph occurs at X = −A/3 .
(ii) Deduce (geometrically) that the substitution X = x − A/3 will reduce the above equation to the form Y = x^3 + bx + c.
(iii) Verify this by calculation.
Relevant Equations: Y = X^3 + AX^2 + BX + C
Y = x^3 + bx + c

The (i) is straightforward: take the second derivative to 0.
The (iii) is obvious: after the substitution, $x^2$ comes from the $X^3$ with the coefficient $-3A/3$ and from the $AX^2$ with the coefficient $A$, and they cancel.

Here is my attempt for the (ii).
The substitution translates the graph so that the inflection point sits at $x=0$.

And you already know from (i) that the unique point of inflection of $x^3 + ax^2 + bx + c$ is at $-a/3$, so you must have $-a/3 = 0$.

The "geometric" derivation comes from translation of the graph by $A/3$ units to the left, as opposed to the algebraic derivation of (iii). This does of course assume that a translated cubic remains a cubic, but I don't think it is possible to show that without doing some algebra.

 

[Hill]

This does of course assume that a translated cubic remains a cubic, but I don't think it is possible to show that without doing some algebra.

This is a challenge.

In case all three roots are real we don't need algebra to show that the translated cubic is at least cubic: the horizontal translation cannot change the number of intersections with the x-axis.

 

[Hill]

This does of course assume that a translated cubic remains a cubic, but I don't think it is possible to show that without doing some algebra.

We don't need algebra to show that the translated cubic is at least cubic: translation cannot remove inflection point.