# Thread: solving y=tanh(y-x) in terms of x

1. ## solving y=tanh(y-x) in terms of x

Is it possible to obtain analytic solution for y=tanh(y-x) in terms of x?

Tanh function can be used to model input-output voltage characteristics of op-amp. The above equation describes the case, in which x represents input voltage to negative input, & y is the output & the positive input.

Thank you very much for your help!

2. Probably not. $tanh(x)= \frac{e^x- e^{-x}}{e^{x}+ e^{-x}$

$tanh(x- y)= \frac{e^{x- y}- e^{y- x}}{e^{x-y}+ e^{y- x}}$[tex]= \frac{e^xe^{-y}- e^{-x}e^y}{e^xe^{-y}+ e^{-x}e^y}[tex]
I don't see any good way to separate the "x" and "y" terms and if you could you would still be left with y both in the exponential and linear so you would probably Lambert's W function to solve that.

3. Why not solve for x first: $x=y-\text{arctanh}(y)$ and then state the inverse series for that function in its radius of convergence is the analytic expression for the inverse? I do have a slight problem when I attempt to extract the inverse series in Mathematica using the code:

myInverse = Normal[InverseSeries[Series[y - ArcTanh[y], {y, 0, 10}],x]]

which gives me:

$(-3)^{1/3} x^{1/3}+\frac{3 x}{5}-\frac{9}{175} (-3)^{2/3} x^{5/3}+\frac{2}{175} (-3)^{1/3} x^{7/3}$

but the exponents on -3 result in complex numbers. Something ain't right.

4. Thanks guys for the help!

I asked Dr. Math, & the reply was it's impossible to analytically solve y in terms of x.