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I was wondering how one could derive the

$$\tan(x)\simeq \frac{x}{1-x^2/3}$$

valid for small $x$ values.

This is similar to the ratio of the small $x$ expansions of $\sin(x)$ an $\cos(x)$, however that would yield

$$\tan(x)\simeq \frac{x}{1-x^2/2}$$

so I have been left slightly confused. Many thanks in advance.

EDIT: In my notes this seems to be some sort of recursive fraction approximation. A second version I have written is:

$$ \tan(x)\simeq %%% \frac{\lambda} {1-\frac{\lambda^2}{3-\frac{\lambda^2}{5-\lambda^2/2}}} $$

EDIT: Thanks for your great answers! You can also find this derived in the references to Equation 33 here: http://mathworld.wolfram.com/Tangent.html

Wall, H. S. (1948). Analytic theory of continued fractions. pg. 349

C.D., O. (1963). Continued fractions. pg. 138

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  • $\begingroup$ If you use the Taylor expansion of sin(x) and cos(x) I believe you would get the one you suggested too. $\endgroup$
    – Henry Lee
    May 23 '18 at 12:28
  • $\begingroup$ @HenryLee: Yes, it's quite frustrating to see where this other one came from. $\endgroup$
    – Freeman
    May 23 '18 at 12:31
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    $\begingroup$ Somehow, $3$ gives a smaller error, even though the Taylor expansions would suggest $2$. i.imgur.com/tczqY6G.png $\endgroup$
    – Jam
    May 23 '18 at 12:31
  • $\begingroup$ Yes, I think that's why I used it. I'm digging through thesis notes currently trying to write up. Past me apparently didn't think it was useful to write where this came from. $\endgroup$
    – Freeman
    May 23 '18 at 12:33
  • $\begingroup$ what is this for? $\endgroup$
    – Henry Lee
    May 23 '18 at 12:34
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This relation con be derived using a Pade approximant. Define a function

$$ R(x) = \frac{a_0 + a_1 x + \cdots + a_m x^m}{1 + b_1 x + \cdots + b_n x^n} $$

That agrees with $f$ up to some order

$$ \left.\frac{{\rm d}^kf}{{\rm d}x^k}\right|_{x=0} = \left.\frac{{\rm d}^kR}{{\rm d}x^k}\right|_{x=0} ~~~\mbox{for}~~ k = 0, 1, \cdots $$

So in your example, take $m=1$ and $n=2$ and $f(x) = \tan(x)$, this would lead to the equations

\begin{eqnarray} a_0 &=&0 \\ a_1 - a_0 b_1 &=& 1 \\ -2a_1 b_1 + a_0(2b_1^2 - 2b_2) &=&0 \\ 3a_1(2 b_1^2 - 2b_2) + a_0(-6b_1^3 + 12 b_1 b_2) &=& 2 \end{eqnarray}

whose solution is

$$ a_0 = 0, a_1 =1, b_1 = 0~~\mbox{and}~~ b_2=-1/3 $$

That is

$$ \tan(x) \approx \frac{x}{1 - x^2/3} $$

which aggress up to the fourth derivative!

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  • $\begingroup$ Ah awesome! That's really helpful of you, and I learnt something new! Thank you very much for your help. $\endgroup$
    – Freeman
    May 23 '18 at 12:43
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    $\begingroup$ @Freeman Happy to help $\endgroup$
    – caverac
    May 23 '18 at 12:45
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    $\begingroup$ Today I learned! :) $\endgroup$
    – Macrophage
    May 23 '18 at 12:47
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Recall that

$$\tan(x)= x+\frac13x^3+o(x^3)$$

and by binomial expansion as $x\to 0$

$$\frac{x}{1-x^2/3}=x(1-x^2/3)^{-1}\sim x+\frac13x^3$$

thus

$$\tan(x)\sim x+\frac13x^3\sim \frac{x}{1-x^2/3}$$

Note that from here

$$\tan(x)=\frac{\sin x}{\cos x}$$

to obtain the same result we need to expand to the 3rd order that is

$$\tan(x)=\frac{\sin x}{\cos x}=\frac{x-\frac16x^3+o(x^3)}{1-\frac12x^2+o(x^3)}=(x-\frac16x^3+o(x^3))(1-\frac12x^2+o(x^3))^{-1}=(x-\frac16x^3+o(x^3))(1+\frac12x^2+o(x^3))=x-\frac16x^3+\frac12x^3+o(x^3)=x+\frac13x^3+o(x^3)$$

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  • $\begingroup$ That's pretty neat! Thank you! caverac also posted a nice proof about the same time as you, ideally I would like to accept both answers, but as caverac has less reputation than you I will accept theirs. Thanks so much! $\endgroup$
    – Freeman
    May 23 '18 at 12:42
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    $\begingroup$ @Freeman You are welcome! Bye $\endgroup$
    – user
    May 23 '18 at 12:44
  • $\begingroup$ @Freeman I add something for the derivation from $\sin x/\cos x$ $\endgroup$
    – user
    May 23 '18 at 12:50
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I think your derivation is correct. Please refer to this former post for more information! :)

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  • $\begingroup$ Thanks, yes, it is very clear how to derive that one. See my edit, I think it's from a recursive fraction approximation. $\endgroup$
    – Freeman
    May 23 '18 at 12:32
  • $\begingroup$ I don't think this answer quite addresses the question, or where the $1/3$ term comes from. $\endgroup$
    – Jam
    May 23 '18 at 13:20
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For an alternative derivation, you may use the Shafer-Fink inequality and compute the inverse function of an algebraic function. This gives $$ \tan(x) \approx \frac{3x+2x\sqrt{9-3x^2}}{9-4x^2}\quad\text{for }x\approx0 $$ which is even more accurate than $\tan x\approx \frac{x}{1-x^2/3}$. As already shown, the last approximation can be derived from Padé approximants or generalized continued fractions.

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  • $\begingroup$ I'm not sure whether this is an answer to the question or just a note that there are other possible approximations. $\endgroup$
    – Jam
    May 23 '18 at 13:22
  • $\begingroup$ @Jam: This proves a sharper approximation, from which $\tan(x)\approx \frac{x}{1-x^2/3}$ can be easily derived in a algebraic fashion. $\endgroup$ May 23 '18 at 13:24
  • $\begingroup$ A better approximation could be $\tan(x)=\frac{ x-\frac{x^3}{15}}{1-\frac{2 x^2}{5} }$ $\endgroup$ May 24 '18 at 8:36

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