Generating an inverse function from the given one

[clumps tim]

Hi, I have a relationship

$$P \cong \Bigg[\Big(K_1\rho^{\frac{5}{3}}\Big)^{-2}+ \Big(K_2\rho^{\frac{4}{3}}\Big)^{-2}\Bigg]^{-\frac{1}{2}}$$I need to find the inverse as $$\rho= \rho(P)$$.

I made a detailed calculation and came up to this

$$y^5+\Big(\frac{P}{K_2}\Big)^2 y+ \Big(\frac{P}{K_1}\Big)^2=0$$
here $$ y= \rho^{\frac{2}{3}}$$
Now I am stuck, I need a general solution for 5th order polynomail or is there aany other suggested method to solve, plaes help

 

[Ackbach]

Hmm. Using your substitution $y=\rho^{2/3}$, I get
$$y^5-\left(\frac{P}{K_1}\right)^{\! 2}-\left(\frac{P}{K_2}\right)^{\!2}y=0.$$
There is no general solution to the quintic, as Abel and Ruffini proved.

If you have numerical values for the constants, you could solve this numerically. If the coefficients are all real, then you're guaranteed at least one real solution.

 

[clumps tim]

Hmm. Using your substitution $y=\rho^{2/3}$, I get
$$y^5-\left(\frac{P}{K_1}\right)^{\! 2}-\left(\frac{P}{K_2}\right)^{\!2}y=0.$$
There is no general solution to the quintic, as Abel and Ruffini proved.

If you have numerical values for the constants, you could solve this numerically. If the coefficients are all real, then you're guaranteed at least one real solution.

Thanks a lot. Now I have never done any numerical solution. if I assume k1 and k2 to be 1 and then take different values for P to get different values for y, and plot a graph, will it be a reasonable solution.

can you please guide how to get a numerical solution for this quintic equation?

 

[mathbalarka]

A standard way to get a numeric solution is to use the Taylor expansion. Let

$$x^5 + ax + b = 0$$

Be your quintic. Applying the affine transformation $x = \sqrt[4]{a} \cdot z$ gives

$$\sqrt[4]{a^5} \cdot z^5 + \sqrt[4]{a^5} \cdot z + b = 0$$

Dividing out by $\sqrt[4]{a^5}$ and setting $t = b \cdot a^{-5/4}$ result the one-parameter Bring-Jerrard form

$$z^5 + z = -t \tag{+}$$

Write $z = z(-t)$ for one of the 5 solutions of the quintic dependent on parameter $t$. Clearly, if $t = 0$ then $z(t)$ must satisfy $z(z^4 + 1) = 0$ the real solution of which is $z = 0$. Thus, $z(0) = 0$. For calculating $z'(0)$, differentiate $(+)$ with respect to $t$ to get

$$5z(t)^4 \cdot z'(t) + z'(t) + 1 = 0 \Rightarrow z'(t) = \frac1{1 + 5z(t)^4}$$

Substituting $t = 0$ above and noting that $z(0) = 0$, we get $z'(0) = 1$. One can in this way derive the higher derivatives at $t = 0$. First few values of the derivatives are $z(0) = 0, z'(0) = 1, z''(0) = 0, z'''(0) = 0, z^\text{IV}(0) = 0, z^\text{V}(0) = -1$ etc. The nonzero derivatives are $z'(0) = 1, z^\text{V}(0) = -1, z^\text{IX}(0) = 5, z^\text{XIII}(0) = -35$ etc.

Finding a closed form for this is a pain in the neck. Of course, we can cheat. The first entry is precisely the sequence involved (see Sloan's comment on the formula section). Thus, our desired Maclaurin series is

$$z(t) = \sum_{k = 0}^\infty \binom{5k}{k} \frac{(-1)^{k+1} t^{1+4k}}{1+4k}$$

This is not entirely convergent outside the radius $|t| = 4 \cdot 5^{-5/4}$, so you have to be careful. Let's try out formula for $t = 1/2$ :

Using the PARI/GP snippet $\colorbox{LightGray}{$\texttt{solve(x=0,-1,x^5+x+1)}$}$ one gets the approximate solution of $-0.47565274353960478548185664$. Whereas using the snippet $\colorbox{LightGray}{$\texttt{sum(k=0,10,binomial(5*k,k)*(-1)^(k+1)*(1/2)^(1+4*k)/(1+4*k)}$}$, one gets $-0.47569803996611881302669644$ which is correct upto 4 decimal digits. This is not really THAT efficient, however one can use some series accelerations to get an rapid convergence of the series.

[PS : We can actually prove the above formula rigorously without any cheating whatsoever, but that'd require a fair bit of complex analysis. Google "Lagrange-Burmann inversion"]

 

[Ackbach]

Another method is the following: differentiate the equation w.r.t. $P$, treating $y=y(P)$ thus:
\begin{align*}
y^5-\frac{P^2}{K_2^2}y-\frac{P^2}{K_1^2}&=0 \\
5y^4 y'-\frac{1}{K_2^2}\left(2Py+P^2 y'\right)-\frac{2P}{K_1^2}&=0 \\
5y^4 y'-\frac{P^2}{K_2^2} y'&=\frac{2P}{K_1^2}+\frac{2Py}{K_2^2}=2P\left[ \frac{1}{K_1^2}+\frac{y}{K_2^2}\right] \\
y'&=\frac{2P\left[ \frac{1}{K_1^2}+\frac{y}{K_2^2}\right]}{5y^4-\frac{P^2}{K_2^2}}.
\end{align*}
Now solve this differential equation numerically for $y(P)$, and plot the solution. I know that Mathematica will do this. I would guess MATLAB could do it as well.

 

[mathbalarka]

That reminded me of the Cockle-Harley type methods of the mid 19s! [this and this]