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I have been researching the series solution to Johann Lambert's trinomial equation, $x^m+px=q$.

The series solution that I have discovered does not converge. Can someone tell me the exact series and explain it? I have read his original analysis in Latin but the analysis for the stated equation is not there. $x=\frac qp-....$ I am trying to solve a simple equation, $x^4+2x=3$. The obvious solution is $x=1$ and I cannot calculate it with series I discovered. Then I would use polynomial long division and get a cubic, which I would hopefully depress with Cardano's method. This should be a real solution and then I can use the quadratic formula. This is my first attempt at this.

David G. Stork
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Elliott
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  • This Mathoverflow post may help. Please post the series you found. – Тyma Gaidash Nov 25 '23 at 17:12
  • see Szabó and Belkić in https://math.stackexchange.com/questions/4657862/how-to-solve-a-trinomial-with-a-rational-exponent-xq-x-c-0/4659046#4659046 – IV_ Nov 25 '23 at 18:26
  • That series $$x={p}^{-1}q-{p}^{-5}{q}^{4}+4,{p}^{-9}{q}^{7} -22,{p}^{-13}{q}^{10}+140,{p}^{-17}{q}^{13} +\dots $$is only for the case $|q^3/p^4| <1$. You will need a different series to get your answer $1$. – GEdgar Nov 25 '23 at 18:46
  • Use the rational root theorem? – Тyma Gaidash Nov 25 '23 at 19:47
  • GEdgar Thank you very much. I reread the appropriate section of Observationes Variae In Mathesin Puram, (m-1)^m p^m is greater than m^m q^m-1. My Latin is rusty. Has this been written in English? – Elliott Nov 25 '23 at 22:27

2 Answers2

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Apply Lagrange reversion. Taking the multivalued root on both sides can give all complex roots, but for one $x\in\Bbb R$ solution: $$x^m+ax=b\implies x=\sqrt[m]{b-ax}=\sum_{n=1}^\infty\frac1{n!}\left.\frac{d^{n-1}}{dx^{n-1}}(b-ax)^\frac nm\right|_0$$ and factorial power $n^{(m)}$. The radius of convergence was found using the ratio test and software: $$\boxed{x^m+ax=b\implies x=\sum_{n=0}^\infty\frac{b^\frac nm}{n!} \left(-\frac ab\right)^{n-1}\left(\frac nm\right)^{(n-1)},\left|-\frac am\left(\frac b{1-m}\right)^{\frac1m-1}\right|<1}$$ shown here. As for the asker’s equation:

$$\boxed{x^4+2x=3\implies x=\sum_{n=0}^\infty\frac{(-2)^{n-1}}{3^{\frac{3n}4-1}n!}\left(\frac n4\right)^{(n-1)}=\sqrt[4]3-\frac1{2\sqrt3}-\frac1{24\sqrt[4]3}+\frac7{1152\cdot3^\frac34}+\frac1{432\sqrt3}+\dots=1}$$

which also converges since $\left|-\frac 24\left(\frac 3{1-4}\right)^{\frac14-1}\right|=\frac12<1$

Тyma Gaidash
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As @Tyma Gaidash has mentioned, you can look at some of my other answers. Your generic problem can be written as: $$x= \frac{\frac{q}{p}}{1 + \frac{W_r(z)}{r}}$$ with $r=m-1$ and $z=r\frac{a_n}{a_m} (-\frac{a_0}{a_m})^r$.

To your numeric equation: $$2x(1 + \frac{x^3}{2})=3$$ $$\implies x \cdot \exp_1(\frac{x^3}{2})=\frac{3}{2}$$ which raising to $r=3$, you have $$x^3 \cdot \exp_3(\frac{3x^3}{2})=(\frac{3}{2})^3$$ $$\implies x = \frac{3/2}{1 + \frac{W_3\left( \frac{3}{2}\left (\frac{3}{2} \right)^3 \right)}{3}}$$ resulting in $W_3(5.0625)=\{ 1.5000 + 0.0000i; -2.3212 + 3.1888i; -2.3212 - 3.1888i; -5.8576 + 0.0000i\}$ and $x=\{ 1.0000 + 0.0000i; 0.2874 - 1.3500i; 0.2874 + 1.3500i; -1.5747 + 0.0000i \}$

Gary
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ZKZ
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