Can anyone help me solve this equation to isolate $d_f$?
$$ U=\frac{2\cdot λ_g}{\pi\cdot B+d_f}\cdot \ln\left(\frac{\pi\cdot B}{d_f}+1\right) $$
(original equation from ISO13370)
I can get the values from the natural log with inverse log, but then I'm still left with the $d_f$ term in the other part of the equation.
Help please!!!
Changing notations : $a=2\lambda_g$, $b=\pi B$, $d_f=x$, $y=U$, you want to solve for $x$ the equation $$y=\frac a {b+x}\log \left(\frac{b}{x}+1\right)$$ where (having read the $ISO 13370$ norm) I suppose that all $(a,b,c,x,y)$ are positive.
Now, let $x=\frac b z$ to make $$\frac b a y=\frac{z }{z+1}\log (z+1) < \log(z+1)$$ and the rhs is a nice, smooth, increasing function. So we have an upper bound of the solution $$z_0=e^{\frac{b }{a}y}-1$$
Now, starting with this guess $z_0=(e^k-1)$, start Newton iteration for the function $$f(z)=\frac{z }{z+1}\log (z+1)-k\qquad \text{where} \quad k=\frac b a y$$ $$f'(z)=\frac{z+\log (z+1)}{(z+1)^2}$$ that is to say $$z_{n+1}=z_n-\frac {f(z_n)}{f'(z_n)}$$ It would converge very fast.
Let me try for $k=12.34$, the iterates would be $$\left( \begin{array}{cc} n & z_n \\ 0 & 228660.9521 \\ 1 & 228673.2914 \\ 2 & 228673.2918 \end{array} \right)$$ For the case where $z$ is small compared to $1$ (so $k$), $z_0$ could be too far from the solution. So, using series expansion close to $z=0$, $$k=z^2-\frac{3 z^3}{2}+\frac{11 z^4}{6}-\frac{25 z^5}{12}+\frac{137 z^6}{60}-\frac{49 z^7}{20}+\frac{363 z^8}{140}+O\left(z^9\right)$$ and using series reversion $$z=\sqrt{k}+\frac{3 k}{4}+\frac{47 k^{3/2}}{96}+\frac{7 k^2}{24}+\frac{14939 k^{5/2}}{92160}+\frac{49 k^3}{576}+\frac{876719 k^{7/2}}{20643840}+O\left(k^4\right)$$ Trying for $k=0.1234$, the above truncated series would give $z=0.470552$ while the exact solution is $z=0.470558$.
