# Need help with rearranging

Hey I need help with rearranging the following equation for y in terms of x.

$$ax+b=\frac{\sqrt c(e^{2xyc}-1)}{\sqrt y(e^{2xyc}+1)}$$

a, b and c are constants by the way

• What is your work on the subject ? Have you noticed for example that a $\tanh$ (hyperbolic tangent) is hidden there ? Commented Jan 30, 2020 at 8:12

As said in comments $$\frac{e^{2cxy}-1}{e^{2cxy}+1}=\tanh(c x y)$$ So, you want to solve for $$y$$ the equation $$a x+b =\frac{\sqrt c}{\sqrt y}\tanh(c x y)$$ Let $$c xy=t$$ to make $$\frac{a x+b}{c \sqrt{x}}=\frac{\tanh(t)}{\sqrt t}$$ which will not show explicit solution even using special functions.

Considering, for a given $$x$$, the equation $$k=\frac{\tanh(t)}{\sqrt t}$$ if $$k>0.7632712$$, there will nor be any solution. In the other case, there are two solutions, one between $$0$$ and $$1.088659$$ (which is the maximum) and another one between $$1.088659$$ and $$\infty$$.

If you are concerned by the samllest root, you could use Taylor series to get $$k=\sqrt t\left(1-\frac{1}{3}t^2+\frac{2 }{15}t^4-\frac{17 }{315}t^6+O\left(t^8\right) \right)$$ and use series reversion to get $$t=k^2+\frac{2 }{3}k^6+\frac{43 }{45}k^{10}+\frac{1642}{945} k^{14}+O\left(k^{18}\right)$$

Trying for $$k=0.5$$, the above truncated expansion would give $$t=\frac{404809}{1548288}\approx 0.261456$$ while the exact solution, obtained using Newton method, would be $$0.261472$$.

However, for this case, the second solution is close to $$t=4$$.

Edit

When $$x$$ starts to be large, we could use $$k\sim\frac{1}{\sqrt t}$$ which would give $$t \sim \frac 1 {k^2}$$ which, for the worked example would exactly give $$x=4$$ while the exact solution is $$3.99458$$.

Now, remains the intermediate range that is to say around $$t=1$$. In such a case, wa can again use Taylor expansion $$\frac{\tanh(t)}{\sqrt t}=\alpha+(t-1) \left(-\alpha^2-\frac{\alpha}{2}+1\right)+(t-1)^2 \left(\alpha^3+\frac{\alpha^2}{2}-\frac{5 \alpha}{8}-\frac{1}{2}\right)+O\left((t-1)^3\right)$$ where $$\alpha=\tanh(1)$$. Now, we just face a quadratic equation in $$(t-1)$$. For example, using $$k=0.7$$ would give the roots $$0.571656$$ and $$1.58875$$ while the exact solutions are $$0.620209$$ and $$1.84781$$. The estimates are suffciently good to make Newton method converging quite fast.