Problem solving nonlinear equation of magnetic saturation Hi everyone I've found a model of magnetic saturation in a reference like this $$\ B=\mu_0 *\mu_r*H $$ $$\ \mu_r = \frac{a}{b*B^4-c*B^2+d}$$ in which $a,b,c,d,\mu_0$ are constant integers with known values $a  = 10750; b  = 2.4; c  = 0.65; d  = 2.5;\mu_0=4e-7*\pi $
but I'm looking for a model based on $H$ as an independent variable
so I can say that $$\ \mu_r = \frac{a}{b*(\mu_0\mu_rH)^4-c*(\mu_0\mu_rH)^2+d} $$and I need to know $\mu_r(H)$
it means that I have to obtain $\mu_r$ in the form of $H$ but without any $\mu_r$ at the left side.lets change the variables for better insight of the problem
($\mu_r \rightarrow y , H \rightarrow x$)
$$\ y = \frac{a}{b*(\mu_0*y*x)^4-c*(\mu_0*y*x)^2+d} $$ change this in the form of $y=f(x)$
B is changing from-2.5 to 2.5 and H is changing from -1e5 to 1e5 .
thanks for your answer
B-H curveand Mur-B curves
 A: $$\ y = \frac{a}{b*(\mu_0*y*x)^4-c*(\mu_0*y*x)^2+d} $$
Rearrange:
$$ y[  \frac{b*(\mu_0*y*x)^4-c*(\mu_0*y*x)^2+d}{a}] = 1$$
Now, if you have a put $ x=a$, then you can write $y$ is a polynomial in $x$ around some domain locally as:
$$ y(x) = \sum_{k=0}^{\infty} \left( \frac{d^ky}{dx^k} \right)|_{a} \frac{(x-a)^k}{k!}$$
For generating the coefficients, you can take implicit derivatives of original expression and evaluate at zero. However note that this approximation only holds for a smol domain.

Example:
Circle centered at origin: $$ x^2 +y^2 =r^2$$
So, say I want the approximation/ linearization around $x=a$,
$$ a^2 + y^2 = r^2$$
Now an immediate complexity which pops up is that there is a positive and negative $y$ value corresponding to this. So you have to choose both $ (x,y)$ around which you want to approximate. Let's take the top semi circle i.e $y= \sqrt{r^2 -a^2}$
Now the derivative:
$$ (\frac{dy}{dx}) = - \frac{x}{\sqrt{r^2 -x^2}}  $$
Now one nice simplification is that to be on the semi circle, $ x<r$ and hence we can use binomial series:
$$ \frac{1}{\sqrt{r^2 -x^2 } } = \frac{1}{x} [  1 + \frac{1}{2} \frac{x^2}{r^2} +O(\frac{x^4}{r^4}) ]$$
So, this makes further derivatives an easier task since you will be differentiating a polynomial with the above, also depending on the order of the $x$ value , you can choose where you want to chop off the series.
Anyhow,
$$y(x) = y(a) - \frac{a}{\sqrt{r^2 - a^2}} (x-a) + O((x-a)^2)$$
Is an approximation which holds true on the upper semi circle around some interval about a point $ x=a$

For more about Taylor Series,  Check out this blog which I wrote (Here)
