Consider the function f(x,y)....?

[Trenters4325]

Consider the function f(x,y) = 4y^3 - 3y - x = 0.

The Implicit Function Theorem guarantees that y can be locally defined as a function of x. The "trouble" points (which in this case have vertical tangents) are (1,-1) and (-1,1). Thus, we have three sets of equations: one for (-infinity,-1), one for (-1,1), and one for (1,infinity).

The function between negative infinity and -1 was given. I found the function between 1 and infinity by reflecting this over the origin.

There are three equations we will need for the portion of the curve between -1 and 1. One of them was given as f(x)=cos(1/3arccos(x)) and the second I derived by reflecting this over the origin. I need to find the final equation with domain (-1,1) which will connect f(x) and its inverse and will travel through the origin. Can you help me?

 

[JakeD]

Consider the function f(x,y) = 4y^3 - 3y - x = 0.

The Implicit Function Theorem guarantees that y can be locally defined as a function of x. The "trouble" points (which in this case have vertical tangents) are (1,-1) and (-1,1). Thus, we have three sets of equations: one for (-infinity,-1), one for (-1,1), and one for (1,infinity).

The function between negative infinity and -1 was given. I found the function between 1 and infinity by reflecting this over the origin.

There are three equations we will need for the portion of the curve between -1 and 1. One of them was given as f(x)=cos(1/3arccos(x)) and the second I derived by reflecting this over the origin. I need to find the final equation with domain (-1,1) which will connect f(x) and its inverse and will travel through the origin. Can you help me?

The three equations you mention will give you the three roots of \(\displaystyle \L 4y^3 - 3y - x = 0\) for given \(\displaystyle \L x.\) You know two equations, so you know two of those roots. Using the fundamental theorem of algebra, you can then calculate the third root: write

\(\displaystyle \L 4y^3 - 3y - x = 4(y-r_1 )(y-r_2 )(y-r_3 )\)

where the \(\displaystyle \L r_i\) are the roots. Expand the right hand side enough so that you can equate coefficients and thus get an equation determining the third root from the other two and \(\displaystyle \L x.\)

 

[Trenters4325]

So you want me to set \(\displaystyle $$
4(y - r_1 )(y - r_2 )
$$\) equal to \(\displaystyle $$
\cos (1/3\arccos (x))* - \cos (1/3\arccos ( - x))
$$\)?

How will I be able to equation coefficients when one side of the equation is a polynomial and the other is not.

 

[JakeD]

So you want me to set \(\displaystyle $$
4(y - r_1 )(y - r_2 )
$$\) equal to \(\displaystyle $$
\cos (1/3\arccos (x))* - \cos (1/3\arccos ( - x))
$$\)?

I'm confused by this. Please explain how you derived it.

How will I be able to equation coefficients when one side of the equation is a polynomial and the other is not.

I agree that equating coefficients only applies to equations where both sides are polynomials. So do that to the equation in polynomials I gave.

 

[Trenters4325]

If I expanded correctly, the equation becomes

\(\displaystyle 4y^3 - 3y - x = 4y^3 - 4y^2 (r_1 + r_2 + r_3 ) + 4y(r_1 r_2 + r_2 r_3 + r_3 r_1 ) - 4r_1 r_2 r_3\)

Setting coefficients of like terms equal, we get

\(\displaystyle - 3 = 4(r_1 r_2 + r_2 r_3 + r_3 r_2 )\)

and

\(\displaystyle - x = - 4y^2 (r_1 + r_2 + r_3 ) -4r_1 r_2 r_3\)

Now, I could solve the second equation for y, but it would still contain three undefined variables. :?:

 

[JakeD]

If I expanded correctly, the equation becomes

\(\displaystyle 4y^3 - 3y - x = 4y^3 - 4y^2 (r_1 + r_2 + r_3 ) + 4y(r_1 r_2 + r_2 r_3 + r_3 r_1 ) - 4r_1 r_2 r_3\)

Setting coefficients of like terms equal, we get

\(\displaystyle - 3 = 4(r_1 r_2 + r_2 r_3 + r_3 r_2 )\)

and

\(\displaystyle - x = - 4y^2 (r_1 + r_2 + r_3 ) -4r_1 r_2 r_3\)

Now, I could solve the second equation for y, but it would still contain three undefined variables. :?:

You're getting there, but why include the term in \(\displaystyle \L y^2\) in with the constant terms?

This is what I had in mind:

\(\displaystyle \L
- x = -4r_1 r_2 r_3 .\)

Solve it for the third root \(\displaystyle \L r_3 .\) Go back a reread my first post. In effect you said you know what the other two roots are.

BTW, equating the coefficients for the \(\displaystyle \L y^2\) terms gives another simple way to solve for the third root:

\(\displaystyle \L 0 = r_1 + r_2 + r_3.\)

 

[JakeD]

Another BTW. You don't really need to fully expand the RHS of

\(\displaystyle \L 4y^3 - 3y - x = 4(y-r_1 )(y-r_2 )(y-r_3 ).\)

Can you see by inspection that the constant term of the RHS must be \(\displaystyle \L 4 (-r_1 )(- r_2 )(- r_3) ,\) which in turn must equal \(\displaystyle \L -x\)?

 

[Trenters4325]

How can you just ignore the \(\displaystyle $$
y^2
$$\) term? Since, the we are dealing with polynomials, we should be able to set the coefficients of each term equal on the RHS to the coefficients of each term on LHS. How can \(\displaystyle $$
y^2
$$\) only be on one side?

But let's continue with your suggestion:

Solve it for the third root...

\(\displaystyle $$
r_3 = \frac{x}
{{4r_1 r_2 }}
$$\)

So, I think you are claiming that

\(\displaystyle \L
r_1 = \cos \left( {1/3\arccos (x)} \right)\)

and

\(\displaystyle \L
r_2 = -\cos \left( {1/3\arccos (-x)} \right)\)

from which the answer would be immediate.

However, I still have some qualms. First, as I stated above, how can we just get ride of the y^2 term. Second, how do you know that each of the roots defines a function for this curve over the (-1,1)? Why shouldn't the roots define the function over the entire domain (recall that there are five functions total)? Does this mean that each root is only "in effect" when the other two are equal to y? That I could understand, but where are the other two functions that are "in effect" beyond 1 and -1?

 

[JakeD]

How can you just ignore the \(\displaystyle $$
y^2
$$\) term? Since, the we are dealing with polynomials, we should be able to set the coefficients of each term equal on the RHS to the coefficients of each term on LHS. How can \(\displaystyle $$
y^2
$$\) only be on one side?

It's on the other side. It just has a zero coefficient on that side. That implies an equation that the sum of the roots must be zero that I gave you.

But let's continue with your suggestion:

Solve it for the third root...

\(\displaystyle $$
r_3 = \frac{x}
{{4r_1 r_2 }}
$$\)

So, I think you are claiming that

\(\displaystyle \L
r_1 = \cos \left( {1/3\arccos (x)} \right)\)

and

\(\displaystyle \L
r_2 = -\cos \left( {1/3\arccos (-x)} \right)\)

from which the answer would be immediate.

However, I still have some qualms. First, as I stated above, how can we just get ride of the y^2 term.

See above.

Second, how do you know that each of the roots defines a function for this curve over the (-1,1)? Why shouldn't the roots define the function over the entire domain (recall that there are five functions total)? Does this mean that each root is only "in effect" when the other two are equal to y? That I could understand, but where are the other two functions that are "in effect" beyond 1 and -1?

Your original post said that some of the functions that solved \(\displaystyle \L 4y^3 - 3y - x = 0\) were given and you solved for some more. You just asked for how to solve for that last function. I only know that it and the other functions must be roots of the original equation and that if you know two roots of a cubic equation you can solve for the third. That's all I showed you how to do.

I don't know how those cosine-related functions were derived or even whether they are correct. Do you? You also did not give formulas for the other functions beyond 1 and -1. Presumably when \(\displaystyle \L |x| > 1,\) there is only one root of multiplicity 3 for the cubic equation. I don't know enough about roots of polynomials to be able to prove that.

You have derived 3 equations for the 3 roots:

\(\displaystyle \L \begin{array}{rll}
- 3 &=& 4(r_1 r_2 + r_2 r_3 + r_3 r_2 ), \\
x &=& 4r_1 r_2 r_3 , \\
0 &=& r_1 + r_2 + r_3.
\end{array}\)

Since you know the formulas for the roots, perhaps you can prove using these equations that when \(\displaystyle \L |x| > 1,\) the 3 roots must all be equal.

 

[Trenters4325]

I don't know how those cosine-related functions were derived or even whether they are correct. Do you?

I don't know how they were derived either but they are correct. I assume that they were derived by solving that system of equations you gave at the end of your last post.

I now understand what happens to the y^2, but I still do not see why the roots of an equation can be used to define. Does it make sense to you that each root is only "in effect" on the interval (-1,1) when the other two are equal to y, thus making their factors equal to zero?

The roots of the equation f(x)=x^2-y^2 =(y-x)(y+x) are x and -x, but x and -x will never define the function on any interval. I guess I do not understand why we took the roots in the first place.