Inverse of polynomial in dy^2 + ey + f form

In summary, the conversation discusses finding the inverse polynomial of y = ax^2 + bx + c in the form of x = dy^2 + ey + f. Suggestions are given to use the quadratic formula and Chebychev Polynomials for approximation. There is also a clarification on the definition of inverse and possible misunderstandings of the original question.
  • #1
stevec
2
0
Hello,

I'm trying to find the inverse polynomial of y = ax^2 + bx + c in the form of x = dy^2 + ey + f.

I'm able to approximate this using Excel, but would prefer a more elegant solution. Any suggestions?

Steve
 
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  • #2
Isolate x in terms of y using the quadratic formula should give you the inverse.
 
  • #3
I don't think I'm understanding your problem properly because the inverse of [tex]y=ax^2+bx+c[/tex] IS [tex]x=ay^2+by+c[/tex] and so d=a, e=b, f=c
 
  • #5
Thanks for the replies, I apologize for not being more descriptive in my question.

Gerenuk's reply is close to what I am looking for, although I think I may need to increase the terms - a quick set of data against the equation was off.
 
  • #6
Mentallic is correct: the "inverse function" to [itex]y= ax^2+ bx+ c[/itex] is just [itex]x= ay^2+ by+ c[/itex]. Now use the quadratic formula to solve for y:
[tex]y= \frac{-b\pm\sqrt{b^2- 4ac}}{2a}[/tex]

But notice the "[itex]\pm[/itex]". The quadratic function is not one-to-one and so does not have a true "inverse". You could restrict x to one side or the other of the vertex of the parabolic graph, thus using either the "+" or the "-".
 
  • #7
HallsofIvy said:
Mentallic is correct
Neither of you is correct, because you don't understand the question. He might not have used the proper wording, but it's not hard to guess what he is really looking for.

@SteveC: Maybe you want to look at Chebychev Polynomials and their Approximation theory. They provide a method to vaguely minimize the maximum total error. Whereas the series expansion I wrote down only aims to be best a y=0.
 
  • #8
Gerenuk said:
Neither of you is correct, because you don't understand the question.

Yep, I already acknowledged that I might not be understanding it properly.

Mentallic said:
I don't think I'm understanding your problem properly because the inverse of [tex]y=ax^2+bx+c[/tex] IS [tex]x=ay^2+by+c[/tex] and so d=a, e=b, f=c

The way he asked it, my answer is correct. I was only sceptical about my answer because it was too simple.
 

1. What is the inverse of a polynomial in dy^2 + ey + f form?

The inverse of a polynomial in dy^2 + ey + f form is another polynomial that, when multiplied by the original polynomial, results in a constant term. In other words, the inverse polynomial "undoes" the effects of the original polynomial.

2. How do you find the inverse of a polynomial in dy^2 + ey + f form?

To find the inverse of a polynomial in dy^2 + ey + f form, you can use the substitution method. Start by setting the polynomial equal to a variable, and then solve for that variable. The resulting expression will be the inverse polynomial.

3. Why is finding the inverse of a polynomial important?

Finding the inverse of a polynomial can be useful in solving equations and understanding the behavior of functions. It can also help in simplifying complicated expressions and making calculations easier.

4. Are there any restrictions when finding the inverse of a polynomial in dy^2 + ey + f form?

Yes, there are certain restrictions when finding the inverse of a polynomial in dy^2 + ey + f form. The polynomial must be a one-to-one function, meaning that for every value of y, there is exactly one corresponding value of x. Additionally, the leading coefficient of the polynomial cannot be equal to zero.

5. Can the inverse of a polynomial in dy^2 + ey + f form ever be undefined?

Yes, the inverse of a polynomial in dy^2 + ey + f form can be undefined. This can occur when the polynomial does not have an inverse, either because it is not a one-to-one function or because the leading coefficient is equal to zero.

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