# Differential equation for function involving hyperbolic Sine

Let $y(z) = 2 \frac{sinh^{-1}(z/(2a)^{1/2})}{(z^2 +2a)^{1/2}}$

where $sinh(z)$ is the sine hyperbolic function.

$$K(z):=\frac{1}{z(y(z) - y(-z))} \rightarrow 1/2\,{\frac {a}{{z}^{2}}}+1/6-{\frac {{z}^{2}}{90\,a}}+{\frac {{z}^{4} }{378\,{a}^{2}}}-{\frac {23\,{z}^{6}}{28350\,{a}^{3}}}+{\frac {263\,{z }^{8}}{935550\,{a}^{4}}}-{\frac {133787\,{z}^{10}}{1277025750\,{a}^{5} }}+{\frac {157009\,{z}^{12}}{3831077250\,{a}^{6}}}-{\frac {16215071\,{ z}^{14}}{976924698750\,{a}^{7}}}+{\frac {2689453969\,{z}^{16}}{ 389792954801250\,{a}^{8}}}+O \left( {z}^{18} \right)$$

How can I find the linear differential equation in $\frac{\partial}{\partial z }$ with coefficent in the polynomial ring $\mathbb{C}[z]$ that annihilates $K(z)$. I am unable to do it by hand I think some software in sagemath might help.

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$$K(z) := \frac{1}{z(y(z)-y(-z))}$$

I can't make sense of the denominator, where $z$ is simultaneously a function and a dependent variable... unleess you have meant :

$$K(z) := \frac{1}{(y(z)-y(-z))\cdot z}$$

( 2022-05-15 17:18:07 +0100 )edit

BTW...

Is that homework ?

( 2022-05-15 17:33:33 +0100 )edit

Isn't $y(z)$ an odd function with $y(z)-y(-z)=2y(z)$?

( 2022-05-16 02:33:36 +0100 )edit

Yes I have meant what you wrote. No it's homework question, it's part of a question I ask in last post, which I cannot solve. So I trying to break it in parts and analyse.

( 2022-05-16 08:48:31 +0100 )edit

The last post https://ask.sagemath.org/question/623... I realise that if there is a recursion in the coefficient of $K(z)$ that might help me to solve my question.

( 2022-05-16 08:49:54 +0100 )edit

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Are you sure $K(z)$ satisfies a linear ODE? Below I derive an ODE but it's not linear.

Let us first rewrite and simplify $K(z)$: $$K(z) = \frac{\sqrt{z^2+2a}}{2z(2\log(z+\sqrt{z^2+2a})-\log(2a))}.$$ It is important to notice that expressions for $K(z)$ and its derivatives are rational functions in $z$, $u:=\log(2a)-2\log(z+\sqrt{z^2+2a})$ and $v:=\sqrt{z^2+2a}$. We will need to eliminate the latter two. To so, we will compute $K^{(n)}(z)$ for $n=0,1,2$ and turn them into polynomial equations in $z$, $K^{(n)}(z)$, and $u,v$ and then eliminate $u,v$ from these polynomials.

R = PolynomialRing(QQ,6,names='K0,K1,K2,z,u,v')
k = R.gens()[:3]
Z,u,v = R.gens()[3:]

z,a = var('z a')
K = sqrt(z^2+2*a)/(z*(4*log(z+sqrt(z^2+2*a))-2*log(2*a)))

pol = []
for i in (0..2):
S = str(K.diff(z,i)).replace('log(2*a) - 2*log(z + sqrt(z^2 + 2*a))','u').replace('sqrt(z^2 + 2*a)','v').replace('(z^2 + 2*a)^(3/2)','v^3')
#print(i,S)
pol.append( numerator( sage_eval(S,locals={'z':Z,'u':u,'v':v}) - k[i] ) )

p2 = [pol[0].subresultants(e,u)[0] for e in pol[1:]]     # eliminating u
r = p2[0].subresultants(p2[1],v)[0]                          # eliminating v
print( factor(r) )


As a result it prints:

(-1/65536) * v * K0^3 * (4*K0 - 1)^3 * z^17 * (-8*K0*K1^2*z^2 + 4*K0^2*K2*z^2 - 8*K0^2*K1*z - K1^2*z^2 - K0*K2*z^2 - 8*K0^3 - 3*K0*K1*z)^2


Ignoring linear factors (which correspond to extraneous solutions), we get that $K(z)$ satisfies the following ODE: $$-8z^2KK'^2 + 4z^2K^2K'' - 8zK^2K' - z^2K'^2 - z^2KK'' - 8K^3 - 3zKK'=0,$$ which is not linear however.

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Thanks a lot, I guess it should ODE not linear ODE.

( 2022-05-21 04:12:07 +0100 )edit

I am bit confused about your claim that derivative of $K(z)$ is rational? I am having the log at the denominator.

( 2022-05-24 04:52:21 +0100 )edit

I said it is rational in $z$, $u$, and $v$ that I defined. After replacing logs and roots with $u$'s and $v$'s as explained (see also how it's done in the code), you'll get a rational function.

( 2022-05-24 04:56:28 +0100 )edit

I see thank. you.

( 2022-05-24 05:15:13 +0100 )edit

Also one thing to notice is that it never involves any $a$ in the coefficient of the ode.

( 2022-05-24 05:17:19 +0100 )edit