Wolfram Math (http://mathworld.wolfram.com/InverseCosecant.html) had a breakdown of Inverse Cosecant (arcsc), a series that looks almost identical to the inverse Sine (arcsin). It (the series) seemed straight forward enough, but I'm trying to understand the generating Sigma/Sum below for that series:


I'm tripped up here 3 times in a row:

1st: the "i^n+1" What is the "i"? Is it imaginary "i"? And if so, what do I do when the sums include positive and negative i on n=2,4,6, etc.? I assume if this is imaginary "i" there's a rule I don't know that makes this work in a series.

2nd & 3rd: P_n-1 and (0) here, according to wolfram:

"P_n(x) is a Legendre Polynomial and (x)_n is a Pochhammer symbol."

At this point, I'm not sure where the Legendre Polynomial ends and where the Pochhammer symbol begins. My textbooks on Elementary Differential Equations and Advanced Engineering Mathematics cover Legendre Polynomials but I couldn't find any reference explaining Pochhammer symbols. Arcsine is similar, and again, lack of understanding Pochhammer symbols is a road block to my understanding:


I believe that part that looks like a fraction (1/2) is pochhammer notation, and not a real fraction. Wolfram again says "where (x)_n is a pochhammer symbol"

EDIT (for inverse Cosecant): this formula below (for each term) makes more sense. I only need to know what the P and (0) stand for and how they convert into the following:


With it though, the explanation of fractions multiplying [for interpreting the (1/2)_n] works.

  • $\begingroup$ mathworld.wolfram.com/PochhammerSymbol.html $\endgroup$ – James Arathoon Sep 5 '18 at 7:52
  • $\begingroup$ Another way of writing these series is $$sin^{-1}x=\sum_{n=0}^{\infty}\frac{\binom{2n}{n}}{(2n+1)\,2^{2n}}x^{2n+1}=\sum_{n=0}^{\infty}\frac{(2n)!}{(2n+1)\,(n!)^2\,2^{2n}}x^{2n+1}$$ $\endgroup$ – James Arathoon Sep 5 '18 at 8:10
  • $\begingroup$ Your second (Pochhammer) formula should read $$\sin^{-1}x=\sum_{n=0}^{\infty}\frac{(\frac{1}{2})_n}{(2n+1)n!}x^{2n+1}$$ $\endgroup$ – James Arathoon Sep 5 '18 at 8:16
  • $\begingroup$ Your first formula should read $$\csc^{-1}x=-\sum_{n=1}^{\infty}\frac{i^{n+1}P_{n-1}(0)}{n}x^{-n}$$ $\endgroup$ – James Arathoon Sep 5 '18 at 8:40
  • $\begingroup$ thanks for finding those typos. i missed them typing this late at night. $\endgroup$ – Tristian Sep 6 '18 at 2:22

Note that this is an expansion at $x=\infty!\;$ $P_{n-1}(0)$ is the value at $x=0$ not a mixture of Legendre polynomial and Pochhammer symbol. $(\frac{1}{2})_{n-1}$ is a Pochhammer symbol and means $(\frac{1}{2})_{n-1} = \frac{1}{2} \times \frac{3}{2}\times \frac{5}{2} \cdots$ or generally for $a\in \mathbb{R}:$ $$(a)_n=a(a+1)(a+2)\cdots(a+n-1),$$ by convention $(a)_0=1.$ For odd $n$ the term $i^{n+1}$ in $i^{n+1}P_{n-1}(0)$ is real, and for even $n=2m$ the Legendre polynomial $P_{2m-1}(x)$ is zero at $x=0$

  • $\begingroup$ The question of how to prove the equivalence of the two series forms remains. $\endgroup$ – James Arathoon Sep 5 '18 at 8:46
  • $\begingroup$ @JamesArathoon: What two series shall be equivalent? There is only one for $\mathrm{csc}.$ I answered the three questions. A connection to the $\arcsin$ series is the change of variable $x \rightarrow x^{-1}$ and the value of the Legendre polynomials $P_{2n}(0)=(-1)^n(\tfrac{1}{2})_n / n!$ (see dlmf.nist.gov/18.6) and $P_{2n+1}(0)=0.$ $\endgroup$ – gammatester Sep 5 '18 at 10:05
  • $\begingroup$ Sorry in hindsight my comment is far from clear. I did in fact mean proof of $P_{2n}(0)=(-1)^n(\tfrac{1}{2})_n / n!$ and $P_{2n+1}(0)=0$ would complete your answer. Don't worry I should have let the OP clarify their question (and correct the typo's) first, before making my, perhaps inaccurate, interpretation of their question. $\endgroup$ – James Arathoon Sep 5 '18 at 11:10
  • $\begingroup$ i have the formula, and i have what it produces. Your reply alludes to the idea that imaginary plane solutions are converted to x1, which allows us to skip the even numbers, producing 1,3,5,7, etc. But I do not understand why the numerator changes by odd values, but the denominator 2, remains a 2. What does the P mean and where does it go? and why do you have a factorial/! next to your lemniscate/infinity? $\endgroup$ – Tristian Sep 8 '18 at 3:45
  • $\begingroup$ As already said, the $P_n$ are the Legendre polynomials (see en.wikipedia.org/wiki/Legendre_polynomials if you do not like the DLMF link) and for odd $n$ you have $P_n(0)=0,$ so every second term is zero. $\endgroup$ – gammatester Sep 8 '18 at 7:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.