Finding the inverse of y = x^7 + x^5

[wtf]

It's just that if some horizontal line intersects the graph at more than one point, then the function can't be invertible; simply because there are two x-values that produce a single y.

Edited February 13, 2016 by wtf

[Lightmeow]

It's just that if some horizontal line intersects the graph at more than one point, then the function can't be invertible; simply because there are two x-values that produce a single y.

I understand what you mean, that the inverse won't be a function. My question was, how would you solve y = x⁷ + x⁵ for x. I used the word inverse improperly, as this function does not have an inverse(unless you were to restrict the domain). But it should still have a relation that, when graphed, is a reflection of the function y = x⁷ + x⁵ over the line y = x. Just like the graph x²=y has the graph y²=x.

Edited February 14, 2016 by Lightmeow

[wtf]

I understand what you mean, that the inverse won't be a function. My question was, how would you solve y = x⁷ + x⁵ for x. I used the word inverse improperly, as this function does not have an inverse(unless you were to restrict the domain). But it should still have a relation that, when graphed, is a reflection of the function y = x⁷ + x⁵ over the line y = x. Just like the graph x²=y has the graph y²=x.

John Cuthber was talking about [math]x^7 - x^5[/math]. Note the minus sign. That function is not injective. That means that it takes the same value at more than one point, so that it does not have an inverse unless you restrict the domain.

 

Your function [math]x^7 + x^5[/math] does have an inverse. It never takes the same value twice. However, we can't solve for the inverse in terms of elementary functions or algebraic operations. I don't have a proof for that fact, only a meta-proof; namely, that if there were some closed-form or elementary solution, Wolfram Alpha would know it. Of course that's not really a proof of anything.

Edited February 14, 2016 by wtf

[Xerxes]

That function is not injective. That means that it takes the same value at more than one point,

It is not clear what you mean here. If by "taking the same value" you mean that for each [math]x[/math] in the domain, there are more than one [math]f(x)[/math] in the codomain, then we do not have a function at all!

 

If you mean that more that one [math]x[/math] in the domain has the same image in the codomain, then you are talking about a surjection, not an injection

 

Your function [math]x^7 + x^5[/math] does have an inverse. It never takes the same value twice.

But then, neither does an injection, but it may not be a bijection. But yes, we may assume that a bijection admits of an inverse. (Even though the expression as you wrote it here is not a function - it is just that, an expression.)

 

Why is [math]f(x) = x^7+x^5[/math] a bijection?

 

Well, note that for [math]f(x)=x^n[/math] with [math]n[/math] an odd integer, then the sign in the domain is preserved in the codomain.

But when [math]n[/math] is even, all values, both positive a negative in the codomain are positively signed. (easy proof by induction on [math]n[/math])

 

Hence only in the former case can there be a bijection - this will apply equally to the function in question here.

 

What the image of the inverse is, I cannot say - the usual methods seem to give rubbish answers!

[wtf]

It is not clear what you mean here. If by "taking the same value" you mean that for each [math]x[/math] in the domain, there are more than one [math]f(x)[/math] in the codomain, then we do not have a function at all!

That's a very weird comment relative to what I wrote, which is perfectly standard terminology and couldn't be more clear. Are you making a joke? I really don't get you.

 

If you mean that more that one [math]x[/math] in the domain has the same image in the codomain, then you are talking about a surjection, not an injection

Is that your definition of a surjection? Care to revise your statement?

Edited February 15, 2016 by wtf

[Xerxes]

OK. To simplify notation, consider a function [math]f:X \to Y[/math], where [math]X[/math]is called the domain, and [math]Y[/math] is called the codomain.

 

If for any [math]y \in Y[/math] there is at most one [math]x \in X[/math] such that [math]y =f(x) [/math] one calls this an injection.

 

If for any [math]y \in Y[/math] there is at least one [math]x \in X[/math] such that [math]f(x) =y[/math] one calls this a surjection.

 

Roll these together, and you have the bijection: for any [math]y \in Y[/math] there is at least one and at most one [math]x \in X[/math] such that [math]f(x) =y[/math].

 

Now, no function, on sets or otherwise, is allowed to have multiple images in [math]Y[/math] (these being the result of applying our function to its domain, here [math]X[/math]). But, by our surjection above, the preimage [math]f^{-1}(y)[/math] MUST be a non-empty set, say [math]f^{-1}(y) =\{x_1,x_2,...x_n\}\subseteq X[/math].

 

The preimage of our injection is a set also. So, by my above, for the injection [math]f:X \to Y[/math], then either [math] f^{-1}(y) = \O[/math] or [math]f^{-1}(y) = \{x\}[/math]. This called a "singleton set"

 

For the bijection we will therefore have that [math]f^{-1}(y) \ne \O,\,\,\, f^{-1} = \{x\}[/math].

 

In this particular circumstance is customary to ever so slightly abuse the notation, and to equate [math]\{x\} = x[/math] and by a further slight abuse to call [math]f^{-1}[/math] the inverse of [math]f[/math]

Edited February 15, 2016 by Xerxes

[deesuwalka]

You have to switch the [latex]x[/latex] and [latex]y[/latex], and then solve for [latex]y[/latex].

 

[latex]y=x^7+x^5 [/latex]

 

[latex]x=y^7+y^5 [/latex]

 

[latex]x=y^5(y^2+1) [/latex]

 

I think it can't be solved farther.

Edited November 8, 2016 by deesuwalka

[renerpho]

If one could calculate the Galois group of the septic equation [math]x^{7}+x^{5}+a{_0}=0[/math], there is a nice test:

The septic is solveable by radicals if and only if its Galois group is either the cyclic group of order 7, the dihedral group of order 14 or the metacyclic group of order 21 or 42.

Septics that have the Galois group [math]L(3,2)[/math] of order 168 can be solved using elliptic functions. All other septics (with Galois groups of higher order, 2520 or 5040) can not be solved with radicals or elliptic functions alone.

Unfortunately, to calculate the Galois group of a septic equation is difficult, and I don't know of any general algorithm.

 

To give a negative answer to the question if the function [math]y=x^{7}+x^{5}[/math] has an inverse that can be epressed by radicals is equivalent to finding a single number [math]a{_0}[/math] (a rational number is sufficient) such that the Galois group of the septic equation has order >42. I am pretty sure that this is the case for almost all choices of [math]a{_0}[/math], even though I don't have a single example.

 

EDIT: I would go even further, and conjecture that there does not exist a polynomial of degree >4 for which the inverse function can be expressed with radicals alone. (All known examples of such polynomials that can be solved that way require special values for [math]a{_0}[/math].)

Maybe somebody can provide a source to a proof of that "conjecture"?

 

EDIT2: As the OT only asks how the inverse can be found, and not if it can be found using radicals alone, here is an answer: The general septic equation can be solved using hyperelliptic curves. So you can make use of that to give an expression for the inverse function in terms of hyperelliptic functions. Maybe someone has a link that gives a formula for the general solution of the septic, then you could apply it to the particular example with [math]a{_7}=a{_5}=1[/math] and [math]a{_6}=a{_4}=a{_3}=a{_2}=a{_1}=0[/math] to get an expression for the inverse function.

Edited November 21, 2016 by renerpho