Alternative method for expanding $\dfrac{1}{1+\sin(x)}$

Question

I was browsing on topics about Taylor series and come across this thread: Maclaurin series of $\frac{1}{1+\sin x}$

In it,

I come across this answer by Marc van Leeuwen. I have two questions regarding his answer.

The first is what is $1$ mod $x^5$?

The second is how does he develop the equation $a+b$, $b+c$, $-\dfrac{1}{6}a+c+d$ and so on?

The third is why he sets all these equations equal zero?

I have seen this method of obtaining indeterminate coefficient of power series but I still don't have an exact understanding of it.

Answer 1

The notation $1 \bmod x^5$ doesn't have anything in particular to do with the answer.

The notation $$F(x) \equiv 1 \pmod {x^5},$$ where $F$ is a polynomial function, means there is some polynomial $Q(x)$ such that $$ F(x) = 1 + x^5 Q(x). $$ A little less formally, it means $F(x)$ is a polynomial with a constant term $1,$ and all other non-zero terms of $F(x)$ are degree $5$ or higher. That is, $$ F(x) = 1 + 0x + 0x^2 + 0x^3 + 0x^4 + f_5x^5 + f_6x^6 + \cdots .\tag1$$

The reason we're interested in this in this answer is that we're really looking for the Taylor series of a function $p$ such that $$ p(x) \times (1 + \sin x) = 1. $$ But since we're only looking for the terms up through $x^4$, the $x^5$ and higher terms don't matter. So by saying we don't care about errors the product of our functions for terms in $x^5$ or higher, we are able to ignore the terms in $x^5$ or higher in the Taylor expansions of $p(x)$ and $1 + \sin x.$ We can truncate the Taylor series of $p(x)$ to just the polynomial $P(x) = a+bx+cx^2+dx^3+ex^4$, and we can truncate the Taylor series of $1 + \sin x$ to just $1 + x - \frac16 x^3,$ because everything else only affects the $x^5$ and higher terms.

So we end up with $$(a+bx+cx^2+dx^3+ex^4) \times \left(1 + x - \frac16 x^3\right) \equiv 1 \pmod {x^5}.$$

Now if you take the trouble to multiply out the thing on the left to write it as a single polynomial in standard form, you'll find that the constant term and the coefficients for $x,$ $x^2,$ $x^3,$ and $x^4$ are the left-hand sides of the equations in the red box. And Equation $(1)$ in this answer says that the constant term is $1$ and the next four coefficients are all zero, so those are the numbers on the right-hand side of the equations in the red box. (Two polynomials are equal if their coefficients are equal term by term.)

Answer 2

$$\dfrac{1}{1+\sin x} = P$$
Cross multiplying
$$P(1+\sin x) = 1$$
Plugin Maclaurin series for $\sin x$
$$P(1+x - \frac{x^3}{6} + \frac{x^5}{120}-\cdots)=1$$
Compare coefficients both sides upto $x^4$ term.
($\mod x^5$ is just a fancy of way saying abive line. You may forget about it if it confuses.)

Answer 3

A general answer to this question is the series expansion \begin{equation}\label{sin-recip-ser-expan}\tag{SQE} \frac{1}{1+\sin x}=1+\sum_{k=1}^{\infty}(-1)^{k}\Biggl[\sum_{\ell=0}^{\lfloor{(k-1)/2}\rfloor} (-1)^{\ell} (k-2\ell)! 2^{2\ell} R\biggl(k,k-2\ell,-\frac{k-2\ell}{2}\biggr)\Biggr]\frac{x^k}{k!} \end{equation} for $|x|<\frac{\pi}{2}$, where \begin{equation}\label{S(n,k,x)-satisfy-eq} R(n,k,r)=\frac1{k!}\sum_{j=0}^k(-1)^{k-j}\binom{k}{j}(r+j)^n \end{equation} for $r\in\mathbb{R}$ and $n\ge k\ge0$ is called weighted Stirling numbers of the second kind, see Carlitz's paper [1] below, and $\bigl\lfloor{\frac{k-1}{2}}\bigr\rfloor$ denotes the floor function whose value is equal to the largest integer less than or equal to $\frac{k-1}{2}$.

For proving the series expansion \eqref{sin-recip-ser-expan}, we need the Faa di Bruno formula and the following theorem.

Theorem. For $k,m\in\mathbb{N}$, partial Bell polynomials $\textrm{B}_{m,k}$ satisfy \begin{equation}\label{bell-polyn-2m-1}\tag{BQF1} \textrm{B}_{2m+k-1,k}(1,0,-1,0,\dotsc, \sin(m\pi))=0 \end{equation} and \begin{equation}\label{bell-polyn-2m}\tag{BQF2} \textrm{B}_{2m+k,k}\biggl(1,0,-1,0,\dotsc, \sin\frac{(2m+1)\pi}{2}\biggr) =(-1)^{m} 2^{2m} R\biggl(2m+k,k,-\frac{k}{2}\biggr). \end{equation}

About the formulas \eqref{bell-polyn-2m-1} and \eqref{bell-polyn-2m}, please see Theorem 1.2 in [2] and Section 1.6 in [3] below.

References

1. L. Carlitz, Weighted Stirling numbers of the first and second kind, I, Fibonacci Quart. 18 (1980), no. 2, 147--162.
2. F. Qi, Derivatives of tangent function and tangent numbers, Appl. Math. Comput. 268 (2015), 844--858; available online at http://dx.doi.org/10.1016/j.amc.2015.06.123.
3. F. Qi, D.-W. Niu, D. Lim, and Y.-H. Yao, Special values of the Bell polynomials of the second kind for some sequences and functions, J. Math. Anal. Appl. 491 (2020), no. 2, Article 124382, 31 pages; available online at https://doi.org/10.1016/j.jmaa.2020.124382.