Expressing the $n$–th derivative of $y=\frac{1}{(1+x^2)^2}$

Question

In an attempt to find a way to express the n-th derivative of $y=\frac{1}{(1+x^2)^2}$ using the binomial theorem I got stuck in the last computation.

I'll explain what I did:

$$f(x)=\frac{1}{(1+x^2)^2}=\frac{1}{(x+i)^2(x-i)^2}=\\ \frac{i}{4}\left( \frac{1}{x+i} \right)-\frac{1}{4}\left( \frac{1}{(x+i)^2} \right)-\frac{i}{4}\left( \frac{1}{x-i} \right)-\frac{1}{4}\left( \frac{1}{(x-i)^2} \right)$$

Then I used $y=\frac{1}{ax+b} \Rightarrow y_{n}=\frac{(-1)^nn!a^n}{(ax+b)^{n+1}}$ and $y=\frac{1}{(ax+b)^2} \Rightarrow y_{n}=\frac{(-1)^n(n+1)!a^n}{(ax+b)^{n+2}}$ to get:

$$\frac{(-1)^n}{4}\left( in!\left( \frac{(x-i)^{n+1}-(x+i)^{n+1}}{(x^2+1)^{n+1}} \right)-(n+1)!\left( \frac{(x-i)^{n+2}+(x+i)^{n+2}}{(x^2+1)^{n+2}} \right) \right)$$

Then I stopped for a second and I analized $(x-i)^{n+1}-(x+i)^{n+1}$ and $(x-i)^{n+2}+(x+i)^{n+2}$ using the binomial theorem:

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$(x-i)^{n+1}-(x+i)^{n+1}$ = $$\sum_{k=0}^{n+1}\binom{n+1}{k}x^{n+1-k}\cdot i^k((-1)^k-1)$$

$i^k((-1)^k-1) =0$ for $k=2n$ ,whereas for $k=2n+1$ $\longrightarrow$$-2i(-1)^{\frac{k-1}{2}}$

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$(x-i)^{n+2}+(x+i)^{n+2}$ = $$\sum_{k=0}^{n+2}\binom{n+2}{k}x^{n+2-k}\cdot i^k((-1)^k+1)$$

$i^k((-1)^k+1) =0$ for $k=2n+1$ , whereas for $k=2n$ $\longrightarrow$$2(-1)^{\frac{k}{2}}$.

And now I got stuck sorting out the indeces $n,k$ of summations; what I obtain is:

$$\frac{d^n}{dx^n}(f(x))=\\ \frac{(-1)^nn!}{2(x^2+1)^{n+1}}\cdot \sum_{k=1}^{n+1}\binom{n+1}{k}x^{n+1-k}(-1)^{\frac{k-1}{2}} - \frac{(-1)^n(n+1)!}{2(x^2+1)^{n+2}}\cdot \sum_{k=2}^{n+2}\binom{n+2}{k}x^{n+2-k}(-1)^{\frac{k}{2}}$$

But testing it I can see it’s wrong. Can anyone help me please sorting the $n,k$ indices and perhaps have a look at my computations also? Thank you

Answer 1

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Update (28/08/2021): I eventually managed to solve my problem. This is the (tested) result obtained: $$\frac{d^n}{dx^n}\left(\frac{1}{(1+x^2)^2} \right)=$$ $$ \fbox{$\begin{array}{rcl}\displaystyle \frac{(-1)^{n+1}n!}{2(1+x^2)^{n+2}}\left\{ (n+1)\sum_{k=0/even}^{n+2}\binom{n+2}{k}x^{n+2-k}(-1)^{\frac{k}{2}}-(1+x^2)\sum_{k=0/odd}^{n+1}\binom{n+1}{k}x^{n+1-k}(-1)^{\frac{k-1}{2}} \right\}\end{array}$} $$

Thanks a lot to everyone who invested time and effort into answering me.

Answer 2

I will check later about your indexes and the calculations.

I know you asked explicitly for a Binomial Series solution, yet I just wanted to point out that the $n$-th derivative of the function above can be easily found with

$$\dfrac{\text{d}^n}{\text{d}x^n} \dfrac{1}{(1+x^2)^2} = \dfrac{n!}{4}\left[-\dfrac{2 + n + ix}{(-x + i)^{n+2}} - \dfrac{2+n - ix}{(-x-i)^{n+2}}\right]$$

Answer 3

The analysis and the result stated in OPs answer are sound (+1). Since I also did already the calculation, here it is together with a few hints.

We obtain \begin{align*} &\color{blue}{D^n}\color{blue}{f(x)}\\ &=\frac{i}{4}(-1)^nn!\frac{(x-i)^{n+1}-(x+i)^{n+1}}{\left(x^2+1\right)^{n+1}}\\ &\qquad-\frac{1}{4}(-1)^n(n+1)!\frac{(x-i)^{n+2}+(x+i)^{n+2}}{\left(x^2+1\right)^{n+2}}\tag{1}\\ &=\frac{i(-1)^nn!}{4\left(x^2+1\right)^{n+1}}\sum_{k=0}^{n+1}\binom{n+1}{k}x^{n+1-k}i^k\left((-1)^k-1\right)\\ &\qquad-\frac{i(-1)^n(n+1)!}{4\left(x^2+1\right)^{n+2}}\sum_{k=0}^{n+2}\binom{n+2}{k}x^{n+2-k}i^k\left((-1)^k+1\right)\tag{2}\\ &=-\frac{(-1)^{n}n!}{4\left(x^2+1\right)^{n+1}}\sum_{k\geq 0}\binom{n+1}{2k+1}x^{n-2k}(-1)^k(-2)\\ &\qquad-\frac{(-1)^n(n+1)!}{4\left(x^2+1\right)^{n+2}}\sum_{k\geq 0}\binom{n+2}{2k}x^{n+2-2k}(-1)^k\cdot 2\tag{3}\\ &\,\,\color{blue}{=\frac{(-1)^nn!}{2\left(x^2+1\right)^{n+2}}\left(\left(x^2+1\right)\sum_{k\geq 0}\binom{n+1}{2k+1}(-1)^kx^{n-2k}\right.}\\ &\qquad\qquad\qquad\qquad\color{blue}{-\left.(n+1)\sum_{k\geq 0}\binom{n+2}{2k}(-1)^kx^{n+2-2k}\right)}\tag{4} \end{align*}

Comment:

• In (2) we apply the binomial theorem and collect terms with equal power.

• In (3) we observe that in the left-hand sum terms with even powers vanish. We take odd indices $k\to 2k+1$ only. Note we take the index range $k\geq 0$ which is admissible, since $\binom{n}{k}=0$ if $k>n$. In the right-hand sum terms with odd powers vanish and we take even indices $k\to 2k$ only.

• In (4) we simplify further by factoring out the common denominator $\left(x^2+1\right)^{n+2}$.

Note: The answer stated by @Turing is nice and correct. We observe there is just one common denominator $(x-i)^{n+2}(x+i)^{n+2}=\left(x^2+1\right)^{n+2}$. We could think to also transform (1) this way to get another binomial expansion which is somewhat different to (4).

Answer 4

I am sorry, i haven't comprehend everything in detail, but you have to be very careful, when you would like to use $(a\cdot b)^n=a^n\cdot b^n$ for complex numbers $a,b$ instead of positive real numbers. There is a popular example here Why $\sqrt{-1 \cdot {-1}} \neq \sqrt{-1}^2$?

However, you can find the power series. We know $\frac{1}{1-t}=\sum_{k=0}^\infty t^k$ for $|t|<1$. If we calculate the derivative, we get $$ \frac{1}{(1-t)^2}=\sum_{k=0}^\infty kt^{k-1}=\sum_{k=0}^\infty (k+1)t^{k}. $$ Now, let $t=-x^2$, then \begin{align} \frac{1}{(1+x^2)^2} &=\sum_{k=0}^\infty (k+1)(-x^2)^{k}\\ &=\sum_{k=0}^\infty (-1)^k(k+1)x^{2k}\\ &=\sum_{2\mid n}^\infty (-1)^{n/2}\left(\frac{n}{2}+1\right)x^{n}\\ &=\sum_{2\mid n}^\infty \frac{n!(-1)^{n/2}(\frac{n}{2}+1)}{n!}x^{n}. \end{align} Hence by taylor's formula the $n$-th derivative of $f$ is $$f^{(n)}(0)=\begin{cases}0&2\nmid n\\ n!\cdot (-1)^{n/2}\cdot \left(\frac{n}{2}+1\right)&2\mid n \end{cases}.$$