Maclaurin series for $\sqrt{\frac{1}{1 - (x + \frac{1}{2})^2}}$

Question

So, the problem was to find Maclaurin series for $\arctan\sqrt{\frac{1-2x}{1+2x}}$ using series of its derivative. The most obvious way to do it is to take derivative of $\arctan$, then find Maclaurin series for it and integrate it. But I wanted to go another way, so I derived the following formula: $\arctan\sqrt{\frac{1-2x}{1+2x}} = \arccos\left(\frac{1+2x}{2}\right)$...which turned out to be wrong (Thanks to Mycroft for noticing it). The right formula is: $$\arctan\sqrt{\frac{1-2x}{1+2x}} = \arccos\sqrt{\frac{1+2x}{2}}$$. Its derivative is $$- \frac{1}{\sqrt{1-\left(\frac{1+2x}{2}\right)}}\cdot \frac{1}{2\sqrt{x+\frac{1}{2}}}$$ and to find MacLaurin series for this function is not so hard. The question remains the same: how to find MacLaurin series for the derivative of first formula with $\arccos$ which is $$-\frac{2}{\sqrt{3-4x-4x^2}}$$. I can find its representation only for $x = -\frac{1}{2}$ that I don't need.

Answer 1

Taylor series at $x = x_0$:

$$f(x) \approx f(x_0) + \sum_{k = 1}^{+\infty} f^{(k)}(x = x_0) \frac{(x-x_0)^k}{k!}$$

MacLaurin series is a special case where $x_0 = 0$.

For what concerns your function, it's all about derivatives:

$$f(x) = \arctan\left(\sqrt{\dfrac{1-2x}{1+2x}}\right)$$

$$f'(x) = -\frac{1}{\sqrt{1-4 x^2}}$$

$$f''(x) = -\frac{4 x}{\left(1-4 x^2\right)^{3/2}}$$

$$f'''(x) = -\frac{48 x^2}{\left(1-4 x^2\right)^{5/2}}-\frac{4}{\left(1-4 x^2\right)^{3/2}}$$

And so on.

At $x = 0$ you can easily see that

$$f(0) = \dfrac{\pi}{4}$$

$$f'(0) = -1$$

$$f''(0) = 0$$

$$f'''(0) = -4$$

Go on with derivatives and you will get

$$f(x) \approx \frac{\pi }{4}-x-\frac{2 x^3}{3}-\frac{6 x^5}{5}-\frac{20 x^7}{7}-\frac{70 x^9}{9}-\frac{252 x^{11}}{11}+O\left(x^{12}\right)$$

The other function

Let's instead search for MacLaurin series of

$$g(x) = \sqrt{\dfrac{1}{1 - (x + 1/2)^2}}$$

The same speech holds.

$$g'(x) = \left(x+\frac{1}{2}\right) \left(\frac{1}{\frac{3}{4}-x (x+1)}\right)^{3/2}$$

$$g''(x) = 16 \left(\frac{1}{3-4 x (x+1)}\right)^{5/2} (4 x (x+1)+3)$$

$$g'''(x) = -\frac{96 (2 x+1) \sqrt{\frac{1}{3-4 x (x+1)}} (4 x (x+1)+7)}{(4 x (x+1)-3)^3}$$

And so on, if you want to have fun.

Again

$$g(0) = \frac{2}{\sqrt{3}}$$

$$g'(0) = \frac{4}{3 \sqrt{3}}$$

$$g''(0) = \frac{16}{3 \sqrt{3}}$$

$$g'''(0) = \frac{224}{9 \sqrt{3}}$$

And so on.

MacLaurin Series for this function is

$$g(x) \approx \frac{2}{\sqrt{3}}+\frac{4 x}{3 \sqrt{3}}+\frac{8 x^2}{3 \sqrt{3}}+\frac{112 x^3}{27 \sqrt{3}}+\frac{608 x^4}{81 \sqrt{3}}+\frac{1088 x^5}{81 \sqrt{3}}+\frac{6016 x^6}{243 \sqrt{3}}+\frac{33536 x^7}{729 \sqrt{3}}+\frac{20992 x^8}{243 \sqrt{3}}+\frac{3214336 x^9}{19683 \sqrt{3}}+\frac{18335744 x^{10}}{59049 \sqrt{3}}+\frac{35024896 x^{11}}{59049 \sqrt{3}}+O\left(x^{12}\right)$$

Answer 2

Best bet is to make use of the fact that whatever way you get the series, it is the same.

$\begin{align*} \sqrt{\frac{1}{1 - (x + 1/2)^2}} &= (1 - (x + 1/2)^2)^{-1/2} \\ &= \sum_{n \ge 0} (-1)^n \binom{-1/2}{n} (x + 1/2)^{2 n} \\ &= \sum_{n \ge 0} (-1)^n \binom{-1/2}{n} \sum_k 2^{- (2 n - k)} \binom{2 n}{k} x^k \\ &= \sum_{n \ge 0} \sum_k 2^{2 n} \binom{2 n}{n} \binom{2 n}{k}2^{- (2 n - k)} x^k \\ &= \sum_k \left( 2^{-k} \sum_{n \ge 0} \binom{2 n}{n} \binom{2 n}{k} \right) x^k \end{align*}$

Will need some serious machinery to get a closed form for the coefficients, though. I'd take a peek at Petkovsek, Wilf, Zeilberger's A = B (A. K. Peters, 1996), the underlying algorithms to get closed forms for many such sums (or proving they don't exist) are available in most computer algebra systems.

Answer 3

Considering $$y=\tan^{-1}\left(\sqrt{\frac{1-2x}{1+2x}}\right)$$ you could compose the series. Start with $$\tan^{-1}(t)=\frac{\pi }{4}+\frac{t-1}{2}-\frac{1}{4} (t-1)^2+\frac{1}{12} (t-1)^3-\frac{1}{40} (t-1)^5+\frac{1}{48} (t-1)^6+O\left((t-1)^7\right)$$ Now, long division $$\frac{1-2x}{1+2x}=1-4 x+8 x^2-16 x^3+32 x^4-64 x^5+128 x^6+O\left(x^7\right)$$ Binomial theorem $$\sqrt{\frac{1-2x}{1+2x}}=1-2 x+2 x^2-4 x^3+6 x^4-12 x^5+20 x^6+O\left(x^7\right)$$ makes $$t-1=-2 x+2 x^2-4 x^3+6 x^4-12 x^5+20 x^6+O\left(x^7\right)$$ Computing the powers of $(t-1)$ to $O\left(x^7\right)$ gives finally $$y=\frac{\pi }{4}-x-\frac{2 x^3}{3}-\frac{6 x^5}{5}+O\left(x^7\right)$$ which is in fact $$y=\frac \pi 4-\sum_{n=0}^\infty \frac{\binom{2 n}{n}}{2 n+1} x^{2 n+1}$$

Answer 4

By the Faa di Bruno formula and some properties of the partial Bell polynomials, we have \begin{align*} \biggl(\frac{1}{\sqrt{3-4x-4x^2}\,}\biggr)^{(n)} &=\sum_{k=0}^n \biggl\langle-\frac12\biggr\rangle_k\frac1{(3-4x-4x^2)^{1/2+k}} B_{n,k}(-4-8x,-8,0,\dotsc,0)\\ &=\sum_{k=0}^n \biggl\langle-\frac12\biggr\rangle_k\frac{(-1)^k 8^k}{(3-4x-4x^2)^{1/2+k}} B_{n,k}\biggl(x+\frac12,1,0,\dotsc,0\biggr)\\ &=\sum_{k=0}^n \biggl\langle-\frac12\biggr\rangle_k\frac{(-1)^k 8^k}{(3-4x-4x^2)^{1/2+k}} \frac{(n-k)!}{2^{n-k}}\binom{n}{k}\binom{k}{n-k}\biggl(x+\frac12\biggr)^{2k-n}\\ &\to \sum_{k=0}^n (-1)^k\frac{(2k-1)!!}{2^k}\frac{(-1)^k 8^k}{3^{1/2+k}} \frac{(n-k)!}{2^{n-k}}\binom{n}{k}\binom{k}{n-k}\frac1{2^{2k-n}}, \quad x\to0\\ &=\frac{n!}{\sqrt{3}\,}\sum_{k=0}^n \frac{(2k-1)!!}{k!} \biggl(\frac{2}{3}\biggr)^k \binom{k}{n-k} \end{align*} for $n\ge0$, where we used the formula \begin{equation}\label{Bell-x-1-0-eq}\tag{BQ} B_{n,k}(x,1,0,\dotsc,0) =\frac{(n-k)!}{2^{n-k}}\binom{n}{k}\binom{k}{n-k}x^{2k-n}. \end{equation} Consequently, we obtain \begin{equation*} \frac{1}{\sqrt{3-4x-4x^2}\,} =\frac{1}{\sqrt{3}\,} \sum_{n=0}^\infty\Biggl[\sum_{k=0}^n \frac{(2k-1)!!}{k!} \biggl(\frac{2}{3}\biggr)^k \binom{k}{n-k}\Biggr]x^n, \quad |x|<\frac12. \end{equation*} The formula \eqref{Bell-x-1-0-eq} was first established in the paper [1] and then reviewed in the article [2] below.

References

1. F. Qi and B.-N. Guo, Explicit formulas for special values of the Bell polynomials of the second kind and for the Euler numbers and polynomials, Mediterr. J. Math. 14 (2017), no. 3, Article 140, 14 pages; available online at https://doi.org/10.1007/s00009-017-0939-1.
2. 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.

Answer 5

Since $\arctan\sqrt{\frac{1-2x}{1+2x}}\,\to\frac\pi4$ as $x\to0$ and \begin{align*} \biggl(\frac1{\sqrt{1-4x^2}\,}\biggr)^{(n)} &=\sum_{k=0}^n\biggl\langle-\frac12\biggr\rangle_k\frac1{(1-4x^2)^{1/2+k}} B_{n,k}(-8x,-8,0,\dotsc,0)\\ &=\sum_{k=0}^n\biggl\langle-\frac12\biggr\rangle_k\frac{(-8)^k}{(1-4x^2)^{1/2+k}} B_{n,k}(x,1,0,\dotsc,0)\\ &=\sum_{k=0}^n(-1)^k\frac{(2k-1)!!}{2^k}\frac{(-8)^k}{(1-4x^2)^{1/2+k}} \frac{(n-k)!}{2^{n-k}}\binom{n}{k}\binom{k}{n-k}x^{2k-n}\\ &=\frac{n!}{2^n}\sum_{k=0}^n\frac{1}{(1-4x^2)^{1/2+k}} \frac{2^{3k}(2k-1)!!}{k!}\binom{k}{n-k}x^{2k-n}\\ &\to\begin{cases} 0, & n=2m+1\\ 2^{2m}[(2m-1)!!]^2, & n=2m \end{cases} \end{align*} as $x\to0$ for $m,n\ge0$, we acquire \begin{equation*} \arctan\sqrt{\frac{1-2x}{1+2x}}\, =\arccos\sqrt{\frac{1+2x}{2}}\, =\frac\pi4+\sum_{m=1}^\infty\frac{(2m-1)!!}{(2m)!!}(2x)^{2m}, \quad |x|<\frac12. \end{equation*} Replacing $2x$ by $x$ gives \begin{equation*} \arctan\sqrt{\frac{1-x}{1+x}}\, =\arccos\sqrt{\frac{1+x}{2}}\, =\frac\pi4+\sum_{m=1}^\infty\frac{(2m-1)!!}{(2m)!!}x^{2m}, \quad |x|<1. \end{equation*} The references for looking up concepts and notions used above are the same ones as at https://math.stackexchange.com/a/4672287.