Level 40

$$\begin{align}
j_{40A}(\tau) &= \sqrt{j_{20B}(\tau)} = \sqrt[4]{j_{10A}(2\tau)}\\
j_{40B}(\tau) &= \sqrt{j_{20A}(2\tau)}= \left(\frac{d_4^2\, d_{20}^2}{d_2\,  d_8\, d_{10}\, d_{40}}\right)^2\\
&= j_{40C}(\tau)+\frac1{j_{40C}(\tau)}\\
j_{40C}(\tau) &= \sqrt{j_{20F}(2\tau)}= \left(\frac{d_8\, d_{10}}{d_2\,  d_{40}}\right)
\end{align}$$
There is no linear relation between these functions. \(j_{40A}\) is one of the few that involves a 4th root.