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 Mar 12 asked Scaling a function with two 'asymptotes' of which one is non-constant Mar 12 asked Relation of common divisors leading to integer results Jan 29 accepted Approximating $\tanh(B\sqrt{A} )$ for small $A$ and arbitrary $B$ by correcting for asymptotic behaviour of $\tanh$ Dec 21 awarded Yearling Dec 4 comment Approximating $\tanh(B\sqrt{A} )$ for small $A$ and arbitrary $B$ by correcting for asymptotic behaviour of $\tanh$ I just found this question, I think that could be helpful for this particular case: math.stackexchange.com/questions/107292/… Dec 4 comment Approximating $\tanh(B\sqrt{A} )$ for small $A$ and arbitrary $B$ by correcting for asymptotic behaviour of $\tanh$ How did you come up with $\tanh x \approx 1 - e^{-1.48x}$? If there is some strategy to that which I can also apply to the complete function which contains the $\tanh$ that would be perfect. Using the $\tanh$ itself is not an option because I need to solve an equation of the type $A + \tanh(B \sqrt{A})=1$ explicitly for $A$. Dec 4 comment Approximating $\tanh(B\sqrt{A} )$ for small $A$ and arbitrary $B$ by correcting for asymptotic behaviour of $\tanh$ Strictly it can be anywhere between 0 and $\infty$, but the range I am interested in is roughly 0 to 50. Dec 4 revised Approximating $\tanh(B\sqrt{A} )$ for small $A$ and arbitrary $B$ by correcting for asymptotic behaviour of $\tanh$ edited title Dec 4 asked Approximating $\tanh(B\sqrt{A} )$ for small $A$ and arbitrary $B$ by correcting for asymptotic behaviour of $\tanh$ Nov 1 comment Why does the moment of approximation matter for the end result? @AntonioVargas - I have tried the Newton's method approach for the full equation I am working on (the one here is just a minimal example to reproduce the behaviour) and it does indeed give a somewhat better approximation, although at the expense of having a significantly bigger equation. Thanks! Oct 30 comment Why does the moment of approximation matter for the end result? So would it be appropriate then to take again a Taylor series of (4) because this function is no longer linear in $a$? Because doing that would indeed result in the same answer as (3) Oct 30 revised Why does the moment of approximation matter for the end result? choose more appropriate tags Oct 30 asked Why does the moment of approximation matter for the end result? Oct 15 comment Approximations other than taylor series and pade approximation Yes, I would like to have an approximate solution to get a feel for the way the result scales with $K_i$ and $Q_i$. Without having to scan a whole range of these parameters. Oct 15 asked Approximations other than taylor series and pade approximation Sep 16 answered Are there any surfaces that contain both positive and negative Gaussian curvature? Jun 4 awarded Enlightened Jun 3 awarded Nice Answer May 26 comment A Math function that draws water droplet shape? @Henry, indeed, a falling liquid droplet will be close to spherical with a slightly flattened bottom part. The shape with the tail is something you will only see for a droplet running down a surface. In that case the tail forms due to a balance of capillary and viscous forces. May 21 revised Determine if the Series Converges and Explain approach removed mathematically slopiness