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I wanted to solve $\tan(x) = x$ without Newton's method. Along I have found various questions here on Mathematics like :

Solution of tanx = x? (Answer of JJacquelin)

Derivation of asymptotic solution of $\tan(x) = x$. (Answer of Antonio Vargas, Lagrange inversion formula) - This way I understand, how they get the result.

and

Location of the extrema of the sinc function.

In the last one the K. B. Dave named some sources like :

https://soundandscience.de/text/theory-sound-volume-1 (Page 278 or PDF-Page 293) (Rayleigh 1877)

and

https://www.digi-hub.de/viewer/image/BV044241472/325/ (Page 319 or PDF-Page 325) (Euler 1748)

Euler for example does not explain how he solved this. (or is it me who does not get it?) How did they do this?

$$s = x - \frac{1}{3}x^3 + \frac{1}{5}x^5 - \frac{1}{7}x^7 ... (1)$$ $$(2n+1)q= \frac{1}{x} + x - \frac{1}{3}x^3 + \frac{1}{5}x^5 - \frac{1}{7}x^7 ... (2) $$

$$ \frac{1}{x} = (2n+1)q - s = (2n+1)q - \frac{1}{(2n+1)q} - \frac{2}{3(2n+1)^3q^3} - \frac{13}{15(2n+1)^5q^5} - \frac{146}{105(2n+1)^7q^7} etc. $$

How to get from (1) and (2) to the last equation?

Further helpful links:

https://groups.google.com/g/sci.math/c/1KOaZXdFAG4/m/Ehqn-G6pJY8J?pli=1

and

https://mathworld.wolfram.com/TancFunction.html

OcK
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    I have a manuscript that I worked on a bit in 2006, which is based on the talk I gave (the one K. B. Dave gave links to, along with my sci.math post that you found on the topic), that has a number of references that might be of use to you. I don't really have time now to look into your actual question, but I can send you a .pdf file of the manuscript if you send me an email (so that I'll know where to send the file). You can find my email address at my profile. – Dave L. Renfro Dec 31 '20 at 21:34
  • The explanation by Euler is in Latin. What part don't you understand? – Somos Dec 31 '20 at 22:49
  • @DaveL.Renfro Hi, very kind of you. I found a link to your presentation of April 2006 but it was not valid anymore. – OcK Jan 01 '21 at 09:00
  • @Somos It's not the language actually but the math. Around the middle. Where do the coefficients 2/3, 13/15, 156/105 etc. come from? The way he got s and (2n+1)q is clearly understandable. But then 1/x = (2n+1)q-s= ... not anymore. – OcK Jan 01 '21 at 09:10

1 Answers1

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Here is an explanation. Suppose we let $\,y := \frac1{(2n+1)q}.\,$ Euler's equation is $$ \frac1y = \frac1x + x - \frac13 x^3 + \frac15 x^5 + \cdots. \tag{1}$$ He wants $\,\frac1x\,$ as a Laurent series in $\,y.\,$ First take the reciprocal of equation $(1)$ to get $$ y = x - x^3 + \frac43 x^5 - \frac{28}{15}x^7 + \cdots. \tag{2}$$ Use series reversion to express $\,x\,$ as a power series in $\,X\,$ to get $$ x = y + y^3 + \frac53 y^5 + \frac{16}5 y^7 + \cdots. \tag{3}$$ Take the reciprocal of $(3)$ to get $$ \frac1x = \frac1y - y - \frac23 y^3 - \frac{13}{15} y^5 + \cdots. \tag{4}$$ Euler had long experience with power series and series reversion so he did not include the intermediate steps from $(1)$ to $(4)$.

Note that OEIS sequence A079330 "Numerators of coefficients of odd powers of 1/q in the solution series for Tan[x]/x=1" is relevant.

Somos
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  • So to obtain (2) you take the reciprocal of (1) and then the Taylor-Series at x=0. Then via this https://mathworld.wolfram.com/SeriesReversion.html you obtain (3) and then the reciprocal of (3) and Taylor-Series at x=0 for (4) I guess? Numerators https://oeis.org/A079330 , Denominators https://oeis.org/A079330 is just a look up table, so you don't have to calculate them by yourself via series reversion right?

    Thanks for clarifying!

     – OcK Jan 01 '21 at 17:06
  • Laurent-Series at x=0 for (4), like you said – OcK Jan 01 '21 at 17:21
  • Okay I tested it. Thank you very much for the details. How one decides to do these steps to solve it? – OcK Jan 01 '21 at 17:26
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    @Ock As with many other tasks, if you once see how others do it, or you figure it out, you can do it much easier the next time you need to do it. – Somos Jan 01 '21 at 18:09