Lightmeow 45 Posted February 11, 2016 Share Posted February 11, 2016 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 = x^{7}+x^{5}. How the heck would you do that. I know it is probably possible, but how? Link to post Share on other sites

fiveworlds 91 Posted February 11, 2016 Share Posted February 11, 2016 y = x^{7}+x^{5} x = y^{7}+y^{5} ^{Then you can isolate y on the left hand side if you want.} Link to post Share on other sites

Lightmeow 45 Posted February 11, 2016 Author Share Posted February 11, 2016 y = x^{7}+x^{5} x = y^{7}+y^{5} ^{Then you can isolate y on the left hand side if you want.} I am aware of that. You would factor x = y^{5}(y^{2}+1) But then how would you get y by itself? Link to post Share on other sites

John Cuthber 3856 Posted February 11, 2016 Share Posted February 11, 2016 I'm not sure, but I think I'd start by taking logs of both sides and seeing if that got anywhere. Link to post Share on other sites

uncool 231 Posted February 11, 2016 Share Posted February 11, 2016 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. Link to post Share on other sites

fiveworlds 91 Posted February 11, 2016 Share Posted February 11, 2016 (edited) 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 February 11, 2016 by fiveworlds Link to post Share on other sites

uncool 231 Posted February 11, 2016 Share Posted February 11, 2016 (edited) 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 February 11, 2016 by uncool 1 Link to post Share on other sites

John Cuthber 3856 Posted February 11, 2016 Share Posted February 11, 2016 I can "brute force"it by simply swapping the axes. So, there is an inverse; there might not be an analytic one. 1 Link to post Share on other sites

Lightmeow 45 Posted February 11, 2016 Author Share Posted February 11, 2016 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) Link to post Share on other sites

John Cuthber 3856 Posted February 11, 2016 Share Posted February 11, 2016 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. Link to post Share on other sites

imatfaal 2481 Posted February 11, 2016 Share Posted February 11, 2016 http://www.mathpages.com/home/kmath290/kmath290.htm Math pages is often better for this than wikipedia Link to post Share on other sites

mathematic 104 Posted February 12, 2016 Share Posted February 12, 2016 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 = x^{7}+x^{5}. 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. Link to post Share on other sites

John Cuthber 3856 Posted February 12, 2016 Share Posted February 12, 2016 In general, you can't. But what about this specific case? Link to post Share on other sites

mathematic 104 Posted February 13, 2016 Share Posted February 13, 2016 (edited) In general, you can't. But what about this specific case? You might for specific values of y, but not in general as a function of y. Edited February 13, 2016 by mathematic Link to post Share on other sites

John Cuthber 3856 Posted February 13, 2016 Share Posted February 13, 2016 (edited) 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 February 13, 2016 by John Cuthber Link to post Share on other sites

imatfaal 2481 Posted February 13, 2016 Share Posted February 13, 2016 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 Link to post Share on other sites

fiveworlds 91 Posted February 13, 2016 Share Posted February 13, 2016 (edited) According to wolfram alpha it looks like this. I can show, graphically, that x^7 - x^5 does not have an inverse. Edited February 13, 2016 by fiveworlds Link to post Share on other sites

John Cuthber 3856 Posted February 13, 2016 Share Posted February 13, 2016 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. Link to post Share on other sites

imatfaal 2481 Posted February 13, 2016 Share Posted February 13, 2016 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 Link to post Share on other sites

fiveworlds 91 Posted February 13, 2016 Share Posted February 13, 2016 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 Link to post Share on other sites

John Cuthber 3856 Posted February 13, 2016 Share Posted February 13, 2016 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? Link to post Share on other sites

wtf 160 Posted February 13, 2016 Share Posted February 13, 2016 (edited) 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. https://www.wolframalpha.com/input/?i=inverse%5By+%3D+x%5E7+%2B+x%5E5%5D ps -- It's clear that x^7 - x^5 is not injective. https://www.wolframalpha.com/input/?i=x%5E7-+x%5E5 Edited February 13, 2016 by wtf Link to post Share on other sites

John Cuthber 3856 Posted February 13, 2016 Share Posted February 13, 2016 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. Link to post Share on other sites

wtf 160 Posted February 13, 2016 Share Posted February 13, 2016 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. Link to post Share on other sites

John Cuthber 3856 Posted February 13, 2016 Share Posted February 13, 2016 I didn't know the test had a name. Link to post Share on other sites

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