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  • one year ago

How do I find the inverse function of $$f(x)=x^5+x$$

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  1. jtvatsim
    • one year ago
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    The standard method is to replace the f(x) with a x and each x with a y. Then, solve for y. However, it seems like this would be a bit messy in this case... still thinking here.

  2. Astrophysics
    • one year ago
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    Replace f(x) with y Switch x's and y's, so put x where y is and x where y is. Solve for y Replace y with f^-1(x) Yeah I don't think this will work well

  3. jtvatsim
    • one year ago
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    And I'm assuming the question is to find the actual equation of the inverse correct?

  4. Empty
    • one year ago
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    Yeah this is my question haha.

  5. jtvatsim
    • one year ago
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    Obviously, I just said that because if we only need the graph, we can just reflect over y = x and obtain the visual solution. :)

  6. Astrophysics
    • one year ago
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    Hey quick question, so this function is 1 - 1 right

  7. jtvatsim
    • one year ago
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    agreed

  8. freckles
    • one year ago
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    We can invent a new function and call it the freckles function where the freckles function is the inverse function of f(x)=x^5+x. I guess that is what Lambert did.

  9. jtvatsim
    • one year ago
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    LOL

  10. jtvatsim
    • one year ago
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    Clearly, the inverse does exist, so we can call it whatever we want right? :)

  11. Empty
    • one year ago
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    hahaha :)

  12. dan815
    • one year ago
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    how about polar coords

  13. Empty
    • one year ago
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    I like what @freckles is saying but I'm not sure if I like this answer I want something more algebraic.

  14. dan815
    • one year ago
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    what we're looking for is a way to write out the inverse function with one equation without having to deal with the +/- sqrts right

  15. jtvatsim
    • one year ago
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    Is there a way to do it with +/- sqrts?

  16. dan815
    • one year ago
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    i dunno i didnt really do anything i just saw it had powers in it so lol

  17. freckles
    • one year ago
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    @Empty do you have an answer and are challenging us, or is this something you wanted to explore with us while challenging yourself and us?

  18. UsukiDoll
    • one year ago
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    @freckles probably the second option

  19. dan815
    • one year ago
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    |dw:1438056301096:dw|

  20. Empty
    • one year ago
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    Haha well I have found something that seems to be a lead just now, but I don't really know anything about this and it looks similar. It uses something called Lagrange Inversion Theorem. http://mathoverflow.net/a/32261 But I accidentally came to this just now since it is so simple and obviously has an inverse yet there does not seem to be a simple answer!

  21. dan815
    • one year ago
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    how about defining inversions in terms of a matrix transformation

  22. dan815
    • one year ago
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    whats stopping us from making a function that reflex across y=x line and saying that is the inversion of all y=f(x) functions

  23. dan815
    • one year ago
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    making a transformation*

  24. Empty
    • one year ago
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    Yeah there definitely needs to be a general thing because there are no doubt an infinitude of functions like this.

  25. dan815
    • one year ago
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    we did talk about how numbers are weak matrices

  26. dan815
    • one year ago
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    so why dont variables and functions

  27. jtvatsim
    • one year ago
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    It's a bit difficult since a 5th degree function like this one is technically in the 6th dimension. We would need to define an analogy to "reflecting over the y = x line" in this situation.

  28. freckles
    • one year ago
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    using what @Astrophysics said earlier is there anyway to find a function f such that \[x^4+1=\frac{1}{f'(x^5+x)}\]

  29. dan815
    • one year ago
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    hmm if i were to ask you what is the inverse of Y=f(x,z)

  30. dan815
    • one year ago
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    id think of more dimensions then, why consider more dimensions for 2 variables only

  31. freckles
    • one year ago
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    oops forgot the 5

  32. freckles
    • one year ago
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    \[5x^4+1=\frac{1}{f'(x^5+x)}\]

  33. Astrophysics
    • one year ago
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    Oh I'll put that up again haha, I thought I was totally wrong..If \[f'(a) \neq a ~~\text{and}~~f(a)=b\] then \[f^{-1}(b) = \frac{ 1 }{ f'(a) }= \frac{ 1 }{ f'(f^{-1}(b)) }\]

  34. freckles
    • one year ago
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    I don't know. I thought it was worth investigating @Astrophysics

  35. dan815
    • one year ago
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    so by the way for the inverse of * z=f(x,y) would we reflect across z+x+y=0 plane or somethingÉ

  36. freckles
    • one year ago
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    though you wind up with f(u)=x+C where u=x^5+x so we are kind of where we started

  37. dan815
    • one year ago
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    i think complex numbers and complex functions can be a nice way to deal with these reflections

  38. Astrophysics
    • one year ago
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    Oh to add to that, \[f(a) = b \implies f^{-1}f(a) = f^{-1}(b) \implies a = f^{-1}(b)\]

  39. jtvatsim
    • one year ago
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    @dan815 Possibly... I see your point from a geometric standpoint. I was considering the underlying algebraic structure, but that may have been unnecessary.

  40. dan815
    • one year ago
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    just some thoughts, im not sure how to consider complex numbers, but i feel like there is a way to deal with reflections in the algebra itself with it comes to complex numbers

  41. dan815
    • one year ago
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    like conjugating automatically reflects across the horizontal

  42. dan815
    • one year ago
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    we can deal with higher variable inverses too with higher degree complex numbers if we do figure this out

  43. jtvatsim
    • one year ago
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    So, we would need to take a + bi to b + ai to perform the equivalent reflection.

  44. dan815
    • one year ago
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    yeah

  45. Empty
    • one year ago
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    so you're saying @jtvatsim $$f(z)=i(z^*)$$

  46. jtvatsim
    • one year ago
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    Well, if you want to get fancy! ... lol exactly!

  47. jtvatsim
    • one year ago
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    That is actually a really neat representation. :)

  48. jtvatsim
    • one year ago
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    But, it seems we are still trapped in the obvious answer that the inverse should be associated with x + x^5 + (x)i, that is (x+x^5,x) provides the recipe for the coordinates. But that is the same as saying the inverse is just (y,x).

  49. jtvatsim
    • one year ago
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    Final (simple) ?answer?: The inverse is given by x = y^5 + y. This is a single equation. Plug in any y value you want, I will tell you the x value, and I have my graph. lol :)

  50. freckles
    • one year ago
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    http://www.wolframalpha.com/input/?i=inverse+of+y%3Dx%5E5%2Bx cany anyone tell me what those pound signs means or is that all gibberish

  51. Astrophysics
    • one year ago
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    Yeah I did the same thing didn't understand it haha

  52. jtvatsim
    • one year ago
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    whoa...

  53. UsukiDoll
    • one year ago
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    I have never seen that before. Maybe wolfram gave up.

  54. Empty
    • one year ago
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    Yeah same here hahaha

  55. Empty
    • one year ago
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    Wolfram is like "I'll put these fake plus signs on them and not explain them shhhhhhhhhhhhhh"

  56. UsukiDoll
    • one year ago
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    I did a problem last semester and I tried to compute the elasticity matrix of a 2 x 2 with all letters and wolfram was able to compute it, but wouldn't provide the details on how to do it >:/

  57. jtvatsim
    • one year ago
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    Sometimes the #1's stand for gigantic expressions that the computer can't fit on screen. They are a sort of place holder perhaps. MATLAB does this sometimes as well.

  58. jtvatsim
    • one year ago
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    The Root[stuff, 3] indicates that this is a third root however.

  59. jtvatsim
    • one year ago
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    I have nothing to say about the "&" though... :P

  60. UsukiDoll
    • one year ago
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    how did wolfram get the graph :/

  61. Astrophysics
    • one year ago
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    |dw:1438058243991:dw| I don't know if it's the same thing... https://en.wikipedia.org/wiki/List_of_mathematical_symbols

  62. anonymous
    • one year ago
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    you very wise

  63. UsukiDoll
    • one year ago
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    A#B is the connected sum of the manifolds A and B. If A and B are knots, then this denotes the knot sum, which has a slightly stronger condition. #X means the cardinality of the set X. ???

  64. jtvatsim
    • one year ago
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    Getting the graph is easy enough as long as it is smart enough to plug in y's first instead of x's.

  65. UsukiDoll
    • one year ago
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    ah ok :)

  66. jtvatsim
    • one year ago
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    I'd say no to the connected sum and not cardinality either... just doesn't make sense (if only math were that simple though). :)

  67. Empty
    • one year ago
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    Yeah It's not the knot sum, cardinality, or primorial. I've had experience with all these before but why isn't wolfram alpha's # in here?!

  68. jtvatsim
    • one year ago
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    @Empty Even though we haven't solved it, thanks for sharing this problem. It is so tantalizingly obvious, yet nontrivial to solve. It illustrates how the simplest mathematics can push our minds to the limits (pun intended). :)

  69. freckles
    • one year ago
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    http://mathematica.stackexchange.com/questions/19035/what-does-mean-in-mathematica I think this kind of explains the pound sign thing

  70. jtvatsim
    • one year ago
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    @freckles That makes sense, that's sort of what I was thinking.

  71. Empty
    • one year ago
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    Hmmm interesting, I think I might have found a method that will arbitrarily approximate the inverse by a Taylor series, Start by noting this is true of the inverse: \[y(0)=0\] Also since this function is odd its inverse will also be an odd function so we only have to look at the odd derivatives since there will be no even terms: Then we can start differentiating to build up the power series. But this is kinda gross lol

  72. Empty
    • one year ago
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    Is there anyway to show that the inverse function's even derivatives will definitely satisfy at x=0? Intuitively I am sure it's true, but I'd like to see it shown algebraically too. $$y^{(2n)}(0)=0$$

  73. Empty
    • one year ago
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    WARNING SKIP IF YOU DONT CARE ABOUT THE PROOF OF 2 CLAIMS I MADE LOL I just decided I should probably prove some things I've said since I've never really heard them before: proof that if \(f(x)\) is odd then \(f^{-1}(x)\) is odd: Def of odd function: \[f(x)=-f(-x)\] plug in \(f^{-1}(x)\) for x \[f(f^{-1}(x))=-f(-f^{-1}(x))\]\[x=-f(-f^{-1}(x))\]\[-x=f(-f^{-1}(x))\]\[f^{-1}(-x)=-f^{-1}(x)\] The inverse is odd, done! Ok next thing to prove is that the even derivatives of an odd function at x=0 are equal to zero. So to prove this I'll take an arbitrary odd function (it satisfies \(f(x)=-f(-x)\) ) \[f(x)=g(x)-g(-x)\] Now evaluating any odd function at x=0 is clear that it equals 0. So now taking the derivative twice gives us: \[f''(x)=g''(x)-g''(-x)\] Which is some new, but also arbitrary odd function. Since I already showed all odd functions evaluated at 0 are 0, so are all even derivatives of odd functions. (really this all comes from the simple fact that the derivative of an odd function is even and the derivative of an even function is odd)

  74. Empty
    • one year ago
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    Anyways the whole point of that was to make it simpler to try to construct the Taylor series of the inverse.

  75. jtvatsim
    • one year ago
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    seems legit to me

  76. Empty
    • one year ago
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    Yeah but this is ugly I want something just as nice as the original function we started with not some infinite garbage!

  77. Empty
    • one year ago
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    WOAH I found something! \[x=y+y^5\] I can go two routes with this: \[y=x-y^5\] \[y=x-(x-(x-\cdots)^5)^5)^5\] or I can go: \[y=\sqrt[5]{x-y}\] \[y=\sqrt[5]{x-\sqrt[5]{x-\sqrt[5]{x-\cdots}}}\] So here are two quick answers I think.

  78. freckles
    • one year ago
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    wow that is totally cute in a cute infinite way

  79. Empty
    • one year ago
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    ty! <3

  80. Empty
    • one year ago
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    I betcha that first of these two new ones will also end up generating the taylor series if you multiply out all those nested binomials too.

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