The $2n$ th derivative of $\frac{1}{1+x^2y^2}$ with respect to $x$ in a way that does not require the imaginary unit.

Question

I wanted to know what is the $\frac{d^{2n}}{dx^{2n}}$ of \begin{align} \frac{1}{1+x^2y^2} \end{align} But the answer I got from another post was a function relaying on the imaginary unit $i^2=-1$, and I wonder if we can find the $2n$ th derivative in a different way.

Answer 1

I haven't double checked the simplifications between steps. So please verify as you read. If there is a mistake, I can correct and I believe the approach is still going to work out.

When you have a composite function $f\circ g$, its $n$th derivative is given by Faà di Bruno's formula. That formula looks bad, but here I would take $f(x)=\frac{1}{1+x}$ and $g(x)=y^2x^2$, and that means that third derivatives and higher of $g$ are zero. So the formula becomes

$$\frac{d^{2n}}{dx^{2n}}\frac{1}{1+y^2x^2}=\sum\frac{(2n)!}{m_1!m_2!2^{m_2}}f^{(m_1+m_2)}\mathopen{}\left(y^2x^2\right)\mathclose{}\cdot \left(2y^2x\right)^{m_1}\left(2y^2\right)^{m_2}$$

where the sum is over all nonnegative $(m_1,m_2)$ with $m_1+2m_2=2n$. Some simplifications:

$$\frac{d^{2n}}{dx^{2n}}\frac{1}{1+y^2x^2}=\sum\frac{(2n)!2^{m_1}x^{m_1}y^{2(m_1+m_2)}}{m_1!m_2!}f^{(m_1+m_2)}\mathopen{}\left(y^2x^2\right)\mathclose{}$$

Since $f(x)=(1+x)^{-1}$, then $f^{(N)}(x)=(-1)^NN!(1+x)^{-1-N}$. So the above is:

$$\frac{d^{2n}}{dx^{2n}}\frac{1}{1+y^2x^2}=\sum\frac{(2n)!2^{m_1}x^{m_1}y^{2(m_1+m_2)}}{m_1!m_2!}(-1)^{m_1+m_2}(m_1+m_2)!\frac{1}{\left(1+x^2y^2\right)^{1+m_1+m_2}}$$

which is simpler as:

$$\frac{d^{2n}}{dx^{2n}}\frac{1}{1+y^2x^2}=(2n)!\sum(2x)^{m_1}\left(-y^2\right)^{(m_1+m_2)}\binom{m_1+m_2}{m_1}\frac{1}{\left(1+x^2y^2\right)^{1+m_1+m_2}}$$

Now the sum is still over all nonnegative $(m_1,m_2)$ with $m_1+2m_2=2n$. These pairs are easy to index. $m_2$ can range from $0$ to $n$, and then $m_1=2n-2m_2$. So we have:

$$\frac{d^{2n}}{dx^{2n}}\frac{1}{1+y^2x^2}=(2n)!\sum_{m_2=0}^{n}(2x)^{2n-2m_2}\left(-y^2\right)^{(2n-m_2)}\binom{2n-m_2}{m_2}\frac{1}{\left(1+x^2y^2\right)^{1+2n-m_2}}$$

Simplified a little as:

$$\frac{d^{2n}}{dx^{2n}}\frac{1}{1+y^2x^2}=\frac{(2n)!(2xy^2)^{2n}}{\left(1+x^2y^2\right)^{1+2n}}\sum_{m_2=0}^{n}\binom{2n-m_2}{m_2}\left(\frac{1}{4x^2y^2}+\frac{1}{4}\right)^{m_2}$$