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Find $\lim_{x \to 0} \dfrac{\tan^{-1}(\sin^{-1}(x))-\sin^{-1}(\tan^{-1}(x))}{\tan(\sin(x))-\sin(\tan(x))}$

I came across this limit a long time ago and could easily obtain a straightforward solution by finding the asymptotic expansion. But since the limit turns out to be nice despite the messy coefficients, I'm just curious if there is some reason other than just coincidental coefficients.

J. M. ain't a mathematician
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user21820
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    if $f'(0)=g'(0)=1$ and $f(x)-g(x)=a,x^n+O(x^{n+1})$ for some $a\neq0$ ($n>1$) then $\lim_{x\to0}(f^{-1}(x)-g(x))/(g^{-1}(x)-f(x))=\lim_{x\to0}(x-f(g(x)))/(x-g(f(x)))=1$. The reason is that the Taylor expansions of $f\circ g$ and of $g\circ f$ start with $x$ plus the same term (or if you wish, $f\circ g-g\circ f$ is smaller than everybody else involved). – user8268 May 26 '14 at 07:42
  • @user8268: I get the first and the last part but what's the reason for the equality between your two limits? – user21820 May 26 '14 at 07:49
  • $f'(0)=1$ gives $\lim(f^{-1}(x)-g(x))/(f^{-1}(f(x))-g(f(x)))=1$, similar for $g'(0)=1$ - so I actually flipped the numerator and the denominator in the 2nd limit in the comment above (though it doesn't change the result :). Anyway - drawing graphs is always helpful. – user8268 May 26 '14 at 08:01
  • @user8268: Unless I made a mistake your limit in the last comment can't exist, because $f(x) \in o(x)$. – user21820 May 26 '14 at 10:04
  • $f(x)=x+O(x^2)$ ($f'(0)=1$). $f(x)=o(x)$ is certainly not true. – user8268 May 26 '14 at 13:00
  • @user8268: Oh yea, but still I don't see why the limit should exist. If you can prove it, do add an answer! – user21820 May 26 '14 at 13:11
  • This is already a very famous question. See good amount of discussion and solution at http://mathoverflow.net/questions/20696/a-question-regarding-a-claim-of-v-i-arnold Also see another question based on the expression $\sin(\tan x) - \tan(\sin x)$ at http://math.stackexchange.com/q/741446/72031 – Paramanand Singh May 29 '14 at 11:23

3 Answers3

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Proposition. Consider two odd functions $f$ and $g$ defined in a neighborhood $(-\epsilon,\epsilon)$, of $0$. Suppose that $f$ and $g$ have the following asymptotic expansions $$ \eqalign{f(x)&=x+\alpha x^3+\beta x^5+\gamma x^7+{\cal O}(x^9)\cr g(x)&=x+\alpha' x^3+\beta' x^5+\gamma' x^7+{\cal O}(x^9) } $$ Then, if$~~ $ $\color{red}{3\alpha\alpha'(\alpha'-\alpha)\ne 2(\beta'\alpha-\beta\alpha')}$,$~$ we have $$\color{blue}{ \lim_{x\to 0}\frac{f\circ g(x)-g\circ f(x)}{f^{-1}\circ g^{-1}(x)-g^{-1}\circ f^{-1}(x)}=1.} $$

Proof.

$\qquad$ Note that the assumptions on $f$ and $g$ insure that they define invertible functions in a neighborhood of $0$, (may be smaller than a $(-\epsilon,\epsilon)$). Moreover, $f^{-1}$ and $g^{-1}$ have asymptotic expansion in the neighborhood of $0$ to the same order as $f$ and $g$ respectively. This expansion, can be calculated using the the method of undetermined coefficients. We find $$ \eqalign{f^{-1}(x)&=x-\alpha x^3+(3\alpha^2-\beta) x^5+(-12\alpha^3+8\alpha\beta-\gamma) x^7+{\cal O}(x^9)\cr g^{-1}(x)&=x-\alpha' x^3+(3\alpha'^2-\beta') x^5+(-12\alpha'^3+8\alpha'\beta'-\gamma') x^7+{\cal O}(x^9) }$$

Now, it is easy to find that $$ f\circ g(x) =x+(\alpha+\alpha') x^3+(3\alpha\alpha'+\beta+\beta') x^5+(5\alpha'\beta+3\alpha(\alpha'^2+\beta')+\gamma+\gamma') x^7+{\cal O}(x^9) $$ So subtracting the similar formula for $g\circ f$ we obtain $$ f\circ g(x)-g\circ f(x)=(3\alpha\alpha'(\alpha'-\alpha)- 2(\beta'\alpha-\beta\alpha'))x^7 +{\cal O}(x^9)\tag{1} $$ Similarly, (but harder), we find $$ f^{-1}\circ g^{-1}(x) =x-(\alpha+\alpha') x^3+(3(\alpha^2+\alpha\alpha'+\alpha'^2)-\beta-\beta') x^5+(-12\alpha^3-12\alpha'^3-15\alpha^2\alpha'-12\alpha\alpha'^2+5\alpha'\beta+3\alpha\beta'+8\alpha\beta+8\alpha'\beta'-\gamma-\gamma') x^7+{\cal O}(x^9) $$ So subtracting the similar formula for $g^{-1}\circ f^{-1}$ we obtain $$ f^{-1}\circ g^{-1}(x)-g^{-1}\circ f^{-1}(x)=(3\alpha\alpha'(\alpha'-\alpha)- 2(\beta'\alpha-\beta\alpha'))x^7 +{\cal O}(x^9)\tag{2} $$ Now the result follows by comparing $(1)$ and $(2)$.$\qquad\square$

Omran Kouba
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  • This is just a generalization of asymptotic expansion and to me doesn't really explain why the coefficients are the same. Moreover I believe the theorem works with any two twice differentiable functions that have gradient 1 at 0 and do not intersect on a punctured neighbourhood of 0. – user21820 May 26 '14 at 12:02
  • Sorry no intersection is not enough, but certainly it seems like the condition necessary for the theorem to hold is much weaker than yours. – user21820 May 26 '14 at 12:12
  • @user21820 May be you are right, at least for analytic functions near $0$. – Omran Kouba May 26 '14 at 13:34
  • As user8268 showed, analyticity and distinctness would suffice, but I believe much weaker conditions are enough, and will post my solution when I've checked it. – user21820 May 26 '14 at 14:31
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(this is the answer I gave in the comments below the question. I guess a nicer solution exists.)

If $f,g$ are say analytic functions of one variable such that $f(0)=g(0)=0$ and $f'(0)=g'(0)=1$, and $f\neq g^{-1}$ (the Taylor series of $f^{-1}$ and $g$ are not equal) then $$\lim_{x\to 0}\frac{f^{-1}(x)-g(x)}{g^{-1}(x)-f(x)}=1,$$ hence the answer is $1$.

To see it (the best way is with pictures, but formulas will do):

Since $f(x)=x+O(x^2)$ we have $$\lim_{x\to 0}\frac{f^{-1}(x)-g(x)}{f^{-1}(f(x))-g(f(x))}=1$$ as the Taylor series of $f^{-1}(x)-g(x)$ and $f^{-1}(f(x))-g(f(x))$ have the same leading term $ax^n$ (whatever $a$ and $n$ are).

On the other hand $$\lim_{x\to 0}\frac{g(g^{-1}(x))-g(f(x))}{g^{-1}(x)-f(x)}=1,$$ because $g'(0)=1$ and $g^{-1}(x),f(x)\to 0$ as $x\to 0$.

Now just use $f^{-1}(f(x))=g(g^{-1}(x))=x$.

user8268
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  • Oh so you use infinite differentiability, in which case it's easier to just attack the original problem directly. Your observation that there are inverses certainly makes things simpler, but I believe I have a much more general solution now, requiring only $f,g$ to be $C^1$ and on opposite sides of the identity function except at $0$ where all three are tangent. It'll take me a while to write it up though! – user21820 May 26 '14 at 14:33
  • @user21820: what is "on the opposite sides of the identity function" and why do your $f,g$ satisfy it? As for differentiability - I only used it because it's true in your problem, so why not use it; the arguments why the two limits are 1 don't use it so much. "It's easier just to attack ... directly" - you mean by computing explicitly the Taylor series? I guess that's what you wanted to avoid and see an argument without calculation. I probably don't understand your original question. – user8268 May 26 '14 at 14:57
  • By "directly" I meant just observing that the inverse of an analytic function $f$ such that $f'(0) = 1$ has exactly the same two initial terms in the series expansion except for a sign flip. This immediately gives the limit we are looking for. – user21820 May 26 '14 at 15:37
  • @user21820 The method you suggest would fail if (in my notation) $f(x)=x-ax^n+O(x^{n+1})$, $g(x)=x+ax^n+O(x^{n+1})$ ($a\neq0$). I think that this happens in your limit - and if not, you'd have to verify it. – user8268 May 26 '14 at 15:56
  • You're right. Anyway my method at http://math.stackexchange.com/q/810079/21820 that doesn't require analyticity doesn't work on this example because they are both 'on the same side'. – user21820 May 26 '14 at 16:08
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I suppose and think that this is really related to the coincidental coefficients. We can show it building for each piece a Taylor expansion built at $x=0$. As results, we have $$\tan ^{-1}\left(\sin ^{-1}(x)\right)=x-\frac{x^3}{6}+\frac{13 x^5}{120}-\frac{173 x^7}{5040}+O\left(x^{9}\right)$$ $$\sin ^{-1}\left(\tan ^{-1}(x)\right)=x-\frac{x^3}{6}+\frac{13 x^5}{120}-\frac{341 x^7}{5040}+O\left(x^{9}\right)$$ $$\tan (\sin (x))=x+\frac{x^3}{6}-\frac{x^5}{40}-\frac{107 x^7}{5040}+O\left(x^{9}\right)$$ $$\sin (\tan (x))=x+\frac{x^3}{6}-\frac{x^5}{40}-\frac{55 x^7}{1008}+O\left(x^{9}\right)$$ As a result $$ \dfrac{\tan^{-1}(\sin^{-1}(x))-\sin^{-1}(\tan^{-1}(x))}{\tan(\sin(x))-\sin(\tan(x))} \approx 1$$ In my opinion, it is quite important to notice for the terms appearing in numerator and denominator the identity of the first three terms (this is why a fourth term had to be introduced in order to avoid a $\frac{0}{0}$ situation).

Pushing the expansion to much higher orders would give $$ \dfrac{\tan^{-1}(\sin^{-1}(x))-\sin^{-1}(\tan^{-1}(x))}{\tan(\sin(x))-\sin(\tan(x))}=1-\frac{5 x^2}{3}+\frac{3937 x^4}{1890}-\frac{24779 x^6}{11907}+O\left(x^8\right)$$ If we modify the problem to $$ \dfrac{\tan^{-1}(\sin^{-1}(a x))-\sin^{-1}(\tan^{-1}(a x))}{\tan(\sin(b x))-\sin(\tan(b x))}$$ the result of the expansion would be $$\frac{a^7}{b^7}-\frac{5 x^2 \left(13 a^9+29 a^7 b^2\right)}{126 b^7}+O\left(x^4\right)$$ For a completely general formulation such as $$ \dfrac{\tan^{-1}(\sin^{-1}(a x))-\sin^{-1}(\tan^{-1}(b x))}{\tan(\sin(c x))-\sin(\tan(d x))}$$ ($a,b,c,d$ being different) the result of the expansion would be $$\frac{a-b}{c-d}-\frac{x^2 \left((a-b) \left(a^2+a b+b^2+c^2+c d+d^2\right)\right)}{6 (c-d)}+O\left(x^4\right)$$

Claude Leibovici
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