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.
Maclaurin series for $\sqrt{\frac{1}{1 – (x + \frac{1}{2})^2}}$
taylor expansion
Best Answer
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
$$g(x) \approx \frac{2}{\sqrt{3}}+\frac{4 x}{3 \sqrt{3}}+\frac{8 x^2}{3 \sqrt{3}}+\frac{112 x^3}{27 \sqrt{3}}+\frac{608 x^4}{81 \sqrt{3}}+\frac{1088 x^5}{81 \sqrt{3}}+\frac{6016 x^6}{243 \sqrt{3}}+\frac{33536 x^7}{729 \sqrt{3}}+\frac{20992 x^8}{243 \sqrt{3}}+\frac{3214336 x^9}{19683 \sqrt{3}}+\frac{18335744 x^{10}}{59049 \sqrt{3}}+\frac{35024896 x^{11}}{59049 \sqrt{3}}+O\left(x^{12}\right)$$