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

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

Mathematics
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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.
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
And I'm assuming the question is to find the actual equation of the inverse correct?

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Yeah this is my question haha.
Obviously, I just said that because if we only need the graph, we can just reflect over y = x and obtain the visual solution. :)
Hey quick question, so this function is 1 - 1 right
agreed
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.
LOL
Clearly, the inverse does exist, so we can call it whatever we want right? :)
hahaha :)
how about polar coords
I like what @freckles is saying but I'm not sure if I like this answer I want something more algebraic.
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
Is there a way to do it with +/- sqrts?
i dunno i didnt really do anything i just saw it had powers in it so lol
@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?
@freckles probably the second option
|dw:1438056301096:dw|
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!
how about defining inversions in terms of a matrix transformation
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
making a transformation*
Yeah there definitely needs to be a general thing because there are no doubt an infinitude of functions like this.
we did talk about how numbers are weak matrices
so why dont variables and functions
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.
using what @Astrophysics said earlier is there anyway to find a function f such that \[x^4+1=\frac{1}{f'(x^5+x)}\]
hmm if i were to ask you what is the inverse of Y=f(x,z)
id think of more dimensions then, why consider more dimensions for 2 variables only
oops forgot the 5
\[5x^4+1=\frac{1}{f'(x^5+x)}\]
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 \[f^{-1}(b) = \frac{ 1 }{ f'(a) }= \frac{ 1 }{ f'(f^{-1}(b)) }\]
I don't know. I thought it was worth investigating @Astrophysics
so by the way for the inverse of * z=f(x,y) would we reflect across z+x+y=0 plane or somethingÉ
though you wind up with f(u)=x+C where u=x^5+x so we are kind of where we started
i think complex numbers and complex functions can be a nice way to deal with these reflections
Oh to add to that, \[f(a) = b \implies f^{-1}f(a) = f^{-1}(b) \implies a = f^{-1}(b)\]
@dan815 Possibly... I see your point from a geometric standpoint. I was considering the underlying algebraic structure, but that may have been unnecessary.
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
like conjugating automatically reflects across the horizontal
we can deal with higher variable inverses too with higher degree complex numbers if we do figure this out
So, we would need to take a + bi to b + ai to perform the equivalent reflection.
yeah
so you're saying @jtvatsim $$f(z)=i(z^*)$$
Well, if you want to get fancy! ... lol exactly!
That is actually a really neat representation. :)
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).
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 :)
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
Yeah I did the same thing didn't understand it haha
whoa...
I have never seen that before. Maybe wolfram gave up.
Yeah same here hahaha
Wolfram is like "I'll put these fake plus signs on them and not explain them shhhhhhhhhhhhhh"
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 >:/
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.
The Root[stuff, 3] indicates that this is a third root however.
I have nothing to say about the "&" though... :P
how did wolfram get the graph :/
|dw:1438058243991:dw| I don't know if it's the same thing...https://en.wikipedia.org/wiki/List_of_mathematical_symbols
you very wise
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. ???
Getting the graph is easy enough as long as it is smart enough to plug in y's first instead of x's.
ah ok :)
I'd say no to the connected sum and not cardinality either... just doesn't make sense (if only math were that simple though). :)
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?!
@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). :)
http://mathematica.stackexchange.com/questions/19035/what-does-mean-in-mathematica I think this kind of explains the pound sign thing
@freckles That makes sense, that's sort of what I was thinking.
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
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. $$y^{(2n)}(0)=0$$
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)
Anyways the whole point of that was to make it simpler to try to construct the Taylor series of the inverse.
seems legit to me
Yeah but this is ugly I want something just as nice as the original function we started with not some infinite garbage!
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.
wow that is totally cute in a cute infinite way
ty! <3
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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