In Need of some brilliant ideas regarding ODD FUNCTIONS!!

Question1. Let Fn(x)=(f1f2fn)(x)F_n(x)= \left( f_1 \circ f_2 \circ \ldots \circ f_n \right) (x) where n2n \geq 2. Find all (kinds of) functions f1,f2,f_1,f_2, \ldots such that for all xx (in some nice interval), Fn(x)=Fn(x)F_n(-x)=-F_n(x)(we will now call this odd function) where fi(x)(i=1,2,,n)f_i(x) (i=1,2, \ldots, n) is continuous and not an odd function?

(DISCLAIMER: This discussion is NOT about finding all possible solutions but brainstorming as much examples as possbile. Can you find other interesting examples? Maybe some mathematical ingenuity and creativity comes into play?)

Here are some well-known examples for n=2n=2:

(Note: Let us exclude the case where we define f2f_2 as the inverse function of f1f_1 or vice versa, ie f1(x)=ln(x),f2(x)=exf_1(x)=\ln(x), f_2(x)=e^x Also, f1(x)=g(x+c),f2(x)=xcf_1(x)=g(x+c), f_2(x)=x-c where gg is some odd function and cc is constant.)

Case1: f1f_1 being a logarithmic function and f2(x)f2(x)=1f_2(-x)f_2(x)=1 so that we can use the identity ln1x=ln(x)\ln{\frac1x}=-\ln(x) f1(x)=ln(x),f2(x)=1+x1x(1<x<1)f_1(x)=\ln(x), f_2(x)=\frac{1+x}{1-x} (-1<x<1)

f1(x)=ln(x),f2(x)=x2+1+xf_1(x)=\ln(x), f_2(x)=\sqrt{x^2+1}+x

f1(x)=ln(x),f2(x)=1+tan(x+π8)2(3π8<x<3π8)f_1(x)=\ln(x), f_2(x)= \frac{1+ \tan(x+\frac{\pi}8)}{\sqrt2} (-\frac{3π}8<x< \frac{3π}8) (in fact, the third example is the only example I can find which involves two transcendal functions.)

Case2: (Peculiar stuff) f1(x)=x1x+1,f2(x)=exf_1(x)=\frac{x-1}{x+1}, f_2(x)=e^x f1(x)=cosx,f2(x)=π2xf_1(x)=\cos{x}, f_2(x)=\fracπ2-x (NOT INTERESTING)

Question2. T/F

(1) Define Gn(x)G_n(x) as the n-th iterate of gg. x,Gn(x)=Gn(x)    x,g(x)=g(x)\forall x, G_n(-x)=-G_n(x) \iff \forall x, g(-x)=-g(x) (-> FALSE STATEMENT)

(2) There always exists some gg for any odd function GnG_n, where gg is not an odd function.

Question3. Using the function Gn(x)G_n(x) from Q2, find a solution gg such that for all n, Gn(x)G_n(x) is not odd BUT limnGn(x)\lim \limits_{n\to \infty} G_n(x) is odd, where gg is not odd.

LEAVE A NOTE FOR THE FOLLOWING: Errors, better notations, more examples, anything helpful, etc

Note by Inquisitor Math
5 months, 3 weeks ago

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f(x)=2+xf(x)=\sqrt{2+x} is non-odd and non-even. When composed with itself infinitely, it gives F(x)=2F(x)=2 which is even. Can a trick like this be used to find an odd function?

Justin Travers - 5 months ago

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Seems that you are interested in this discussion!(Particularly question2, yes?) Well honestly, I haven’t thought about infinite compositions that approach some odd function. But here is a function that works out (kinda disappointing that it doesn’t approach something.. if you seek of a general solution I dunno.), assuming that we are focusing on Q2: g(x)={0(0x1)1(x<0or1<x) g(x) = \begin{cases} 0 & \quad (0 \leq x \leq1)\\ 1 & \quad (x<0 \text{or} 1<x) \end{cases}

(in fact, Gn(x)=0G_n(x)=0 for n2n \geq 2)

Inquisitor Math - 5 months ago

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For question 1, I'm not sure how you might define "kinds of composition". You can get a boring (but infinite!) set of examples by using functions that are translations of odd functions, then applying a translation as the final function of the composition. For example, f1(x)=(x1)3f_1(x)=(x-1)^3, f2(x)=x+1f_2(x)=x+1. Neither of these is odd, but their composition is. Similarly you can combine arbitrarily many linear functions.

For q2, I think induction might work: F1(x)=f(x)F_1(x)=f(x) is odd; then assuming Fn(x)=Fn(x)F_n(-x)=-F_n(x), Fn+1(x)=f(Fn(x))=f(Fn(x))=Fn+1(x)F_{n+1} (-x) = f(F_n (-x)) = f(-F_n(x))=-F_{n+1}(x)

A question back to you, though: is it possible to compose a non-odd function ff with itself infinitely many times and get an odd function as a result? (I've only just thought of this so it may be trivial!)

Chris Lewis - 5 months, 2 weeks ago

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Can you think of any solutions for n=2n=2 other than the example I posted? Thinking of a solution for this is challenging, let alone the general case.

Inquisitor Math - 5 months, 2 weeks ago

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Ah, sorry, I think I misread your q2. Well, the proof above works one way, at least.

Here's another n=2n=2 example; again it's part of a larger family, but, again, not a very interesting one... f1(x)=11x,      f2(x)=11+xf_1(x)=1-\frac{1}{x},\;\;\;f_2(x)=\frac{1}{1+x}

Actually, you also get f1(f1(f1(x)))=xf_1(f_1(f_1(x)))= x

which answers q2!

Chris Lewis - 5 months, 2 weeks ago

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Just in case you visit here... the updated questions are q2(2) and q3. Also q1 remains unsolved...

As for q1, since it is impalpable for now, it would suffice to find more interesting examples. For me, those are all the interesting examples I can come up with.

As for q2(2), I wonder if it is always possible to find a solution no matter what odd function GnG_n is.

As for q3, the person below has come up with this question and it seems kinda interesting too!

Inquisitor Math - 2 months, 3 weeks ago

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Yes, I'm interested in whether there's a non-odd function that approaches an odd function after infinite compositions with itself. I'm not any expert and have just scratched the surface of this so far. Your function gg above actually fits the bill. It feels wrong though, perhaps because a function that is zero everywhere (so is technically odd) seems trivial somehow. I have no idea how to approach Q2(2). I feel like there should be some trick to allow it to happen but it depends what is allowed. It is possible if the function can vary with nn e.g. (x+1n(mod3))(x+1-n (\mod 3))

Justin Travers - 4 months, 4 weeks ago

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I think that the question I am asking is this: Is there a function (real to real) that does not vary with n, is not odd, is continuous over the real numbers and when composed with itself infinitely many times approaches a non-zero odd function as a limit?

Justin Travers - 4 months, 3 weeks ago

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