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We know that transcendent equations $$\tan{x}=bx$$ and $$x\tan{x}=b$$ can not be solved exactly. But what I concerned most is the relationship between their non-trival roots $x_{n}^{(1)}$ and $x_{n}^{(2)}$, where $x_{n}^{(1)}$ and $x_{n}^{(2)}$ refer to the $n$-th positive solution to the first and second of these equations respectively. For example $x_{n}^{(1)}=x_{n}^{(2)}+\pi/2$( which is incorrect). Any suggestions?

Similar question is the roots of $W(x)\exp(W(x))=x$ and $M(x)\ln(M(x))=x$.

Roger209
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  • Are there more conditions for the first equation? There are trivial solutions for $b=0$ and $b=1$. – Jam Aug 04 '14 at 06:07
  • By "be solved exactly" do you mean "be expressed using elementary functions"? What would it even mean "to solve something exactly"? What's an exact solution to $x^2-2=0$? We just made up special notation for them: $\pm2^{1/2}$ or $\pm\sqrt{2}$. So if that's OK, then just make up notation for whatever the solutions to $x\tan x=b$ are. We can find decimal approximations to arbitrarily high degree, and even express solutions definitively as the limit of a certain sequence. – 2'5 9'2 Aug 04 '14 at 06:07
  • Regarding $\tan x=bx$, as $x\to\infty$, the graph of $y=\tan x$ becomes closer and closer to an array of vertical lines at $x=\pi/2+\pi n$. So the $n$th solution gets closer and closer to $x=\pi/2+\pi n$. – 2'5 9'2 Aug 04 '14 at 06:11
  • Sorry for the barrage of comments, but does $x_n^{(1)}$ and $x_n^{(2)}$ refer to the $n$th positive solution to the first and second of these equations respectively? – 2'5 9'2 Aug 04 '14 at 06:13
  • @2'59'2 What is $M(x)$ in this case? – Тyma Gaidash Apr 11 '23 at 17:17

3 Answers3

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In this article you can find the exact solutions of both your equations using an extension of Riemann method http://www4.ncsu.edu/~ces/pdfversions/52.pdf

AlienRem
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When the roots are close to $P_n=(n+\frac12)\pi$, we can approximate the tangent as $$\tan(P_n+h)\approx-\frac1h,$$ and the equations become $$-\frac1h=b(P_n+h),$$ $$-\frac1{h'}(P_{n'}+h')=b.$$ Eliminating $b$, this gives the equation of a cubic $$-\frac1h=-\frac1{h'}(P_{n'}+h')(P_n+h),$$ $$h'=h(P_{n'}+h')(P_n+h).$$ For large $n$ and $n'$, it simplifies to $$h'=hP_{n'}P_n,$$ so that $$\frac{x_n^{(2)}-P_n}{x_n^{(1)}-P_n}\approx P_n^2.$$ Nothing really exciting as this is just an approximation. But my guess is that if a simple relation between the roots had existed, the linearized analysis would have revealed it.

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In the particular case $b=1$, it is known that the roots of $\tan(x_n)=x_n$ are close to : $x_n=(n-\frac{1}{2})\pi-\frac{1}{(n-\frac{1}{2})\pi}-\frac{2}{3}(\frac{1}{(n-\frac{1}{2})\pi})^3-\frac{13}{15}(\frac{1}{(n-\frac{1}{2})\pi})^5-\frac{146}{105}(\frac{1}{(n-\frac{1}{2})\pi})^7-\frac{781}{315}(\frac{1}{(n-\frac{1}{2})\pi})^9-...$

The trivial root $x=0$ is not included in the list. The first positive root $x_1$ is not accurate. The accuracy increasses quickly for larger $\vert n \vert$. The first negative root is denoted $x_0$ ( not accurate) and the next negative roots correspond to $n<0$ so that $x_{-n}=-x_{n+1}$

With the same method of series expansion, I got the roots of $\tan(x_n)=b x_n$ :

$x_n=(n-\frac{1}{2})\pi-\frac{1}{b}\frac{1}{(n-\frac{1}{2})\pi}-\frac{3b-1}{3b^3}(\frac{1}{(n-\frac{1}{2})\pi})^3-\frac{30b^2-20b+3}{15b^5}(\frac{1}{(n-\frac{1}{2})\pi})^5$ $-\frac{525b^3-252b^2+161b-15}{105b^7}(\frac{1}{(n-\frac{1}{2})\pi})^7-\frac{4410b^4-5880b^3+2744b^2-528b+35}{315b^9}(\frac{1}{(n-\frac{1}{2})\pi})^9-...$

and the roots of $x_n \tan(x_n)=b$ with $n>1$ :

$x_n=(n-1)\pi+\frac{b}{(n-1)\pi}-\frac{b^2(b+3)}{3}(\frac{1}{(n-1)\pi})^3+\frac{b^3(3b^2+20b+30)}{15}(\frac{1}{(n-1)\pi})^5$ $-\frac{b^4(15b^3+161b^2+525b+525)}{105}(\frac{1}{(n-1)\pi})^7+\frac{b^5(35b^4+528b^3+2744b^2+5880b+4410)}{315}(\frac{1}{(n-1)\pi})^9-...$

Case $n=1$ : The serie is not convergent for the first root $(n=1)$ because $x_1$ is equivalent to $\sqrt{b}$ when $x$ tends to $0$. So, the serie cannot be expressed in terms of integer powers of $b$ as it was above. An approximate for the first root is :

$x_1=\sqrt b \big( 1-\frac{1}{6}b+ \frac{11}{360}b^2- \frac{17}{5040}b^3- \frac{281}{604800}b^4+ \frac{44029}{119750400}b^5- \frac{12147139}{130767436800}b^6$ $+ \frac{2030761}{784604620800}b^7+ \frac{1381516351}{313841848320000}b^8+…\big)$

This serie is slowly convergent. So, a large number of terms are required if $b$ is not small.

JJacquelin
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  • Thanks. Have you used the Langrange inverse theorem? – Roger209 Aug 06 '14 at 04:49
  • It could be done with the Lagrange inversion theorem. In fact, I didn't used this method. I proceed by identification of coefficients of limited series. For example, in case of $\tan(x)=bx$, let $\epsilon=\frac{1}{(n-\frac{1}{2})\pi}$ and $x=\sigma f_n(b)*\epsilon^n$. Pluging this $x$ into $\tan(x)=bx$ and expending it in terms of negative and positive powers of $\epsilon$ leads to the equations to be solved for $f_n(b)$ – JJacquelin Aug 06 '14 at 05:05
  • Besides, in case of the last formula, can we give the form of $x_0$ with arbitrary $b>0$?. Saying $x_0=0.8603$ and $x_0=1.0769$ if $b=1$ and $b=2$, respectively. – Roger209 Aug 06 '14 at 05:14
  • I cannot correct a typing mistake in this equation because the 5 mn delay for edition are exceeded : $x=\sum_n f_n(b)*\epsilon^n$ – JJacquelin Aug 06 '14 at 05:14
  • I cannot understand your preceeding question. Put $n=1$ in the formula and $b=1$ : the result is the root $x_1=0.8603$ . With $b=2$ , the result is the root $x_1=1.0769$ . For this formula $x_0$ doesn't exists. The first root is $x_1$ , the second $x_2$ and so on – JJacquelin Aug 06 '14 at 05:41
  • Why? if$b=1,n=1$, $x_1=\pi+1/\pi-4/(3\pi^3)=3.4169$ according to your formula. How come $x_1=0.8603$ as you said? – Roger209 Aug 06 '14 at 05:42
  • Sorry for my late answer : I have few time available today. You are right. Your remark is justified. In fact, in order to solve $x\tan(x)=b$, I was using the result of the solving of $\cot(x)=Bx$ with $B=1/b$. But I made some confusion in the formulas. In fact, $x_1$ is not the first root but the second. The formula doesn't works for the first root due to non-convergence. I will look more carefully at this point as soon as I'll have more time. – JJacquelin Aug 06 '14 at 08:04
  • The anwer for $x_n\tan (x_n)=b$ has been corrected : change of numbering of the roots and an additional formula for the first root. – JJacquelin Aug 07 '14 at 09:12
  • Thanks for your enthusiastic help. But would you love to share some details on the first root of $x\tan(x)=b$ in order to promote me? you can post those details as a new reply to my question to end the interminable comments. – Roger209 Aug 07 '14 at 11:12
  • Nothing mysterious, just boring calculus. First, prove that $x_1$ is equivalent to $\sqrt{b}$ for $x$ tending to $0$. So, the patern for the series is $x= \sqrt{b}( a_0+a_1 x+a_2 x^2+… )$. Here, the series is limited to few terms in interest of space. $x\tan(x)=b= \sqrt{b}(a_0+a_1 x+a_2 x^2+…) \tan \left( \sqrt(b)(a_0+a_1 x+a_2 x^2+…)\right)$. Expend it to series : $b=a_0^2 b+(2a_0a_1+\frac{1}{3} a_0^4)b^2+…$ So,\ $a_0=1$ and $(2a_0a_1+\frac{1}{3} a_0^4)=0$ giving $a_1=-\frac{1}{6}$. With more terms in the series, you obtain the successive coefficients. – JJacquelin Aug 08 '14 at 06:22
  • Hi Jean ! If you have a look at my most recent question, you will notice that I reinvented the wheel ! I am sorry to have missed this answer. Cheers. – Claude Leibovici Sep 02 '19 at 09:11
  • Hi Claude ! In fact a formal solution of $\tan(x)=bx$ is : $$x=\text{tanc}^{-1}(b)$$ with special function tanc : http://mathworld.wolfram.com/TancFunction.html . A magic trick which gives nothing more. Cheers. – JJacquelin Sep 02 '19 at 11:19