1

Doing some exercises I have encountered an equation I could not solve. I know it has a numerical solution of about 1.835 but I am missing the algebraic methods to get to that solution analyticaly.

$$\ln({x})=\frac{1+x}{1+2x} \ \implies \ x=e^{\frac{1+x}{1+2x}}$$

I would like to know what algebraic methods I am missing and which should I learn next to solve equations like this one.

Harish Chandra Rajpoot
  • 37,450
mathsandscience
  • 82
  • 1
    Do you have reason to believe there is a nice formula for the solution? – aschepler Jul 04 '23 at 13:29
  • You can compare with similar equations, e.g. here, and the methods used. – Dietrich Burde Jul 04 '23 at 13:30
  • You probably can only prove that there is a solution and exhibit a “nice” convergent sequence to this solution, but I don’t think you will get a closes form – julio_es_sui_glace Jul 04 '23 at 13:31
  • 2
    Few root solving problems involving transcendental functions can be solved algebraically. Those that can are typically "wired" to do so. That is, they are carefully designed so that a solution drops out easily and then one just has to argue that other solutions do not exist. – lulu Jul 04 '23 at 13:31
  • I never said I believed there was a formula, I just want to know what should I learn to solve this type of problems – mathsandscience Jul 04 '23 at 13:38
  • 2
    So, learn numerical methods. Newton's method, for example, or simple binary search to start. In this particular case, the left hand is strictly increasing while the right hand decreases to $\frac 12$. Easy to see that the solution is unique and simple methods get you in the ball park. – lulu Jul 04 '23 at 13:42
  • In terms of some specific transformations, you might find the RHS more or less interesting by changing it into $1-\frac x{1+2x}$ or $\frac 12+\frac 1{2+4x}$. This does not lead immediately to any interesting solutions as far as I know, but might be worth further investigation. – abiessu Jul 04 '23 at 13:45

3 Answers3

2

One way is by applying Lagrange reversion and letting $x=\frac1w$:

$$\ln(x)=\frac{x+1}{2x+1}\iff ew=\sqrt[w+2]e\implies w=\sum_{n=1}^\infty\frac{e^{-n}}{n!}\left.\frac{d^{n-1}}{dw^{n-1}}e^\frac nw\right|_2$$

If we wanted to expand $\frac1x$ as a series only, it would diverge. Maclaurin series uses factorial power $n^{(m)}$.

$$\frac{d^{n-1}}{dw^{n-1}}e^\frac nw =\sum_{m=0}^\infty\frac{n^m (-m)^{(n-1)}}{y^{m+n-1}m!}$$

which is just a confluent hypergeometric $\,_1\text F_1(a;b;x)$ or Laguerre $\operatorname L_n^r(x)$ function. Expanding $\operatorname L_n^1(x)$’s integral representation gives us an integral solution:

$$\bbox[4px,border:4px #90EE90 double]{x=\frac1{\frac1e-\sum_\limits{n=1}^\infty(-2e)^{-n}\,_1\text F_1\left(n;2;\frac n2\right)}=\frac1{\frac1{\sqrt e}-\frac1{4\pi}\int_0^{2\pi} e^{\frac{ e^{it}}2+it-\frac12}\ln\left(\frac12(e^{it}+1)e^{\frac{e^{it}}2-it-\frac12}+1\right)dt}}$$

Typing in RealDigits[1/(1/E-NSum[(-2E)^(-n) Hypergeometric1F1[n,2,n/2],{n,1,7000}])] into Mathematica matches the many first true digits:

enter image description here

while the, rounded, integral is shown here

Another way uses the Ramanujan master theorem, which would give the same series expansion or Mellin inversion which would possibly give another integral representation with a similar integrand.

Тyma Gaidash
  • 12,081
2

By inspection, you know that the zero of function $$f(x)=\log(x)-\frac{x+1}{2x+1}$$ is rather close to $x=2$.

By Taylor around this point $$f(x)= \left(\log (2)-\frac{3}{5}\right)+\sum_{n=1}^\infty (-1)^n \frac{4^n\, n+2\ 5^{n+1} }{10^{n+1}\,n}\, (x-2)^n$$ Truncating to some order $(p+1)$ and using power series reversion $$x=2+\sum_{n=1}^p a_n \,t^n+O(t^{p+1}) \qquad \text{where} \qquad t=f(x)-\left(\log (2)-\frac{3}{5}\right)$$ the first coefficient being $$\left\{\frac{50}{27},\frac{5875}{6561},\frac{479375}{1594323},\frac {4371875}{57395628},\frac{5668796875}{376572715308},\frac{450991484375}{183014339639688}\right\}$$

Since we want $f(x)=0$, for this level of expansion, converted to decimals, $$x_{(6)}=\color{red}{1.8350370674}51$$

Claude Leibovici
  • 260,315
2

$$\log(x)=\frac{x+1}{2x+1}$$ $$\frac{x+1}{2x+1}=t \implies x=\frac{1-t}{2t-1}\implies \log \left(\frac{1-t}{2 t-1}\right)=t$$ Exponentiate and reverse $$e^{-t}=\frac{2 t-1}{1-t}$$ The only explicit solution is given in terms of the generalized Lambert function (have a look at equation $(4)$ in the linked paper).

Claude Leibovici
  • 260,315