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

## Recommended Posts

Hey all. I am trying to figure something out that I can't. Basically, for some reason I wanted to find the inverse of y = x7+x5. How the heck would you do that. I know it is probably possible, but how?

##### Share on other sites

y = x7+x5

x = y7+y5

Then you can isolate y on the left hand side if you want.

##### Share on other sites

y = x7+x5

x = y7+y5

Then you can isolate y on the left hand side if you want.

I am aware of that. You would factor x = y5(y2+1)

But then how would you get y by itself?

##### Share on other sites

I'm not sure, but I think I'd start by taking logs of both sides and seeing if that got anywhere.

##### Share on other sites

I think it's likely impossible to do so using radicals (i.e. using only addition, multiplication, and taking nth roots); you'd be solving x^7 + x^5 - a = 0. I'd guess that the Galois group is probably some large subgroup of S7, and so likely isn't solvable.

##### Share on other sites

But then how would you get y by itself?

I don't know. I don't think it is possible.

so likely isn't solvable.

You can however solve it by brute force given x and testing loads of numbers.

Edited by fiveworlds
##### Share on other sites

I'm using "solvable" in the sense of group theory - the Galois group for this polynomial is likely not a solvable group. It turns out that this has implications for whether it's possible to find x using radicals.

Edited by uncool
##### Share on other sites

I can "brute force"it by simply swapping the axes. So, there is an inverse; there might not be an analytic one.

##### Share on other sites

I can "brute force"it by simply swapping the axes. So, there is an inverse; there might not be an analytic one.

Is there a way to prove that it is unsolvable by normal means? Why isn't it solvable from an analytic approach?

I think it's likely impossible to do so using radicals (i.e. using only addition, multiplication, and taking nth roots); you'd be solving x^7 + x^5 - a = 0. I'd guess that the Galois group is probably some large subgroup of S7, and so likely isn't solvable.

Thanks for saying that. I am trying to figure out what the Galois group is and S7 is. Could you shed some light or links, because Google isn't telling me that much(Or I am using wrong terms, which seems more likely)

##### Share on other sites

Is there a way to prove that it is unsolvable by normal means? Why isn't it solvable from an analytic approach?

I don't know if it's soluble or not. But I know that the inverse function exists.

##### Share on other sites

http://www.mathpages.com/home/kmath290/kmath290.htm

Math pages is often better for this than wikipedia

##### Share on other sites

Hey all. I am trying to figure something out that I can't. Basically, for some reason I wanted to find the inverse of y = x7+x5. How the heck would you do that. I know it is probably possible, but how?

You are trying to find the solution of a 7th degree polynomial equation. In general equations of degree > 4 do not have a closed form solution.

##### Share on other sites

In general, you can't.

##### Share on other sites

In general, you can't.

You might for specific values of y, but not in general as a function of y.

Edited by mathematic
##### Share on other sites

That's not quite what I meant.

There is a general solution to quadratic equations "minus b pluss or minus the square root of b squared minus four a c all divided by two a"

There is no general solution for 8th order polynomial.

But there is an inverse for the particular instance of y=x^8.

the solution is that y is the 8th root of x

and you can calculate it by pressing the "square root" button on your calculator a few times. (OK, it's messy for negative numbers, but that's not the point)

What I don't know is whether X^7+X^5 falls into the group of 7th order polynomials that dos have an analytical solution.

I can show, graphically, that x^7 - x^5 does not have an inverse.

Edited by John Cuthber
##### Share on other sites

What I don't know is whether X^7+X^5 falls into the group of 7th order polynomials that dI can show, graphically, that x^7 - x^5 does not have an inverse.

Can you not just flip it across the line y=x? There is only one point of inflexion (there is only one real solution to dy/dx - and the value of d2y/dx2 is zero) so that line would also be 1:1

##### Share on other sites According to wolfram alpha it looks like this.

I can show, graphically, that x^7 - x^5 does not have an inverse.

Edited by fiveworlds
##### Share on other sites

Thanks for drawing the graphs.

If you look at the blue line (which indicates what would be the "inverse function") at x= about -0.05, the blue line has 3 different values.

So, if I asked you what's the value of f^-1 of -0.05, which answer would you give?

The function isn't monotonic, so there's no reliable way to invert it.

If you do the same with y=x^5+ x^7 you will see what the difference is.

##### Share on other sites

Thanks for drawing the graphs.

If you look at the blue line (which indicates what would be the "inverse function") at x= about -0.05, the blue line has 3 different values.

So, if I asked you what's the value of f^-1 of -0.05, which answer would you give?

The function isn't monotonic, so there's no reliable way to invert it.

If you do the same with y=x^5+ x^7 you will see what the difference is.

"If you do the same with y=x^5+ x^7 you will see what the difference is." My problem was that I was doing it with y=x^7 + x^5 - brain fart. Your posts now make perfect sense once I have read it properly

##### Share on other sites

If you look at the blue line (which indicates what would be the "inverse function") at x= about -0.05, the blue line has 3 different values.

i would imagine they are generating the values via brute force though so they are not going to make the graph 100% accurate

##### Share on other sites

The graph only needs to be fairly accurate to illustrate the point.

There are three values of x where x^7- x^5 = -0.05

How do you invert that? Which one do you choose?

##### Share on other sites

The graph only needs to be fairly accurate to illustrate the point.

There are three values of x where x^7- x^5 = -0.05

How do you invert that? Which one do you choose?

I didn't understand your point about there being three values etc. Both x^7 + x^5 and its inverse are bijective from the real to the reals. Besides, you swapped the + for a -, that confused me too. In any event, there's no elementary inverse. Proof by Wolfram. If an elementary solution exists, Wolfram would know it.

ps -- It's clear that x^7 - x^5 is not injective.

Edited by wtf
##### Share on other sites

If I ask you to think of a number, double it and add ten then tell me the result I can work out what number you first thought of by applying the reverse process- subtract ten then halve it.

I can do that because the function that I asked you to apply ha a properly defined inverse.

If I tell you to think of a number and then subtract the fifth power from the seventh, then tell me the result I can't (reliably) say what number you thought of.

However if I asked you to add the fifth and seventh powers then tell me the answer I can work out which number you chose in the first place- I can look on the graph.
##### Share on other sites

If I ask you to think of a number, double it and add ten then tell me the result I can work out what number you first thought of by applying the reverse process- subtract ten then halve it.

I can do that because the function that I asked you to apply ha a properly defined inverse.

If I tell you to think of a number and then subtract the fifth power from the seventh, then tell me the result I can't (reliably) say what number you thought of.

However if I asked you to add the fifth and seventh powers then tell me the answer I can work out which number you chose in the first place- I can look on the graph.

Totally agreed. Some functions are invertible and some aren't. It's the horizontal line test from analytic geometry.

##### Share on other sites

I didn't know the test had a name.

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account