How to derive the gregory series for inverse tangent function?

Question

How to derive the Gregory series for the inverse tangent function? Why is Gregory series applicable only to the set $ [-\pi/4,\pi/4] $ ?

Answer 1

A fast way is by exploiting $\arctan'(x)=\dfrac1{1+x^2}$.

The Taylor series of that derivative is easily established to be the sum of a geometric series of common ratio $-x^2$,

$$\sum_{k=0}^\infty(-x^2)^k=\frac1{1-(-x^2)},$$ which only converges for $x^2<1$.

Then integrating term-wise,

$$\arctan(x)=\int_0^x\frac{dx}{1+x^2}=\sum_{k=0}^\infty\frac{(-1)^kx^{2k+1}}{2k+1}.$$

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Alternatively, assume that you know

$$\ln(1+z)=-\sum_{k=1}^\infty\frac{(-z)^k}{k}.$$

Then with $z=ix$,

$$\ln(1+ix)=\ln\left(\sqrt{1+x^2}\right)+i\arctan(x)=-\sum_{k=1}^\infty\frac{(-ix)^k}{k}.$$

The imaginary part of this identity gives

$$\arctan(x)=\sum_{odd\ k}(-1)^{(k-1)/2}\frac{x^k}k$$

while the real part gives the extra

$$\ln\left(\sqrt{1+x^2}\right)=\sum_{even\ k>0}(-1)^{k/2-1}\frac{x^k}k.$$

Answer 2

You need to know that $$\arctan x = \int_{0}^{x}\frac{dt}{1 + t^{2}}\tag{1}$$ In fact this equation can be used as a starting point to develop the theory of circular functions. But let's not worry about that here.

We know (via the formula for sum of a geometric progression) that $$1 - t^{2} + t^{4} - t^{6} + \cdots + (-1)^{n - 1}t^{2n - 2} = \frac{1 - (-1)^{n}t^{2n}}{1 + t^{2}}\tag{2}$$ and therefore $$\frac{1}{1 + t^{2}} = 1 - t^{2} + t^{4} - t^{6} + \cdots + (-1)^{n - 1}t^{2n - 2} + (-1)^{n}\cdot\frac{t^{2n}}{1 + t^{2}}\tag{3}$$ and hence on integrating the above equation between $0$ and $x$ and using $(1)$ we get $$\arctan x = x - \frac{x^{3}}{3} + \cdots + (-1)^{n - 1}\cdot\frac{x^{2n - 1}}{2n - 1} + (-1)^{n}\int_{0}^{x}\frac{t^{2n}}{1 + t^{2}}\,dt$$ or $$\arctan x = x - \frac{x^{3}}{3} + \cdots + (-1)^{n - 1}\cdot\frac{x^{2n - 1}}{2n - 1} + (-1)^{n}R_{n}(x)\tag{4}$$ where $$R_{n}(x) = \int_{0}^{x}\frac{t^{2n}}{1 + t^{2}}\,dt\tag{5}$$ Now from the above equation it is easy to check that $R_{n}(-x) = -R_{n}(x)$. We will prove that if $0 \leq x \leq 1$ then $R_{n}(x) \to 0$ as $n \to \infty$. Since $R_{n}(-x) = -R_{n}(x)$ it will imply that the $\lim_{n \to \infty}R_{n}(x) = 0$ holds for all $x$ with $|x|\leq 1$.

Now let's assume that $0 \leq x \leq 1$. Then we know that $$\frac{t^{2n}}{1 + t^{2}} \leq t^{2n}$$ for all $t$ with $0 \leq t \leq x$. Hence on integrating the above between $0$ and $x$ we get $$0 \leq R_{n}(x) \leq \int_{0}^{x}t^{2n}\,dt = \frac{x^{2n + 1}}{2n + 1} \leq \frac{1}{2n + 1}$$ and taking limits as $n \to \infty$ and using Squeeze theorem we get $$\lim_{n \to \infty}R_{n}(x) = 0$$ It follows that $$\lim_{n \to \infty}(-1)^{n}R_{n}(x) = 0$$ and as noted earlier this equation holds for all values of $x$ with $|x|\leq 1$. Now taking limits as $n \to \infty$ on both sides of equation $(4)$ we get $$\arctan x = x - \frac{x^{3}}{3} + \cdots = \sum_{n = 1}^{\infty}(-1)^{n - 1}\frac{x^{2n - 1}}{2n - 1}$$ for all $x$ with $|x| \leq 1$.

Answer 3

I think the most straightforward way is to yes the expansions of log(1+x) and log(1-x) log(1+x)=x - x^2/2 + x^3/3....and so on.

log(1-x)=-x-x^2/2-x^3/3....and so on.

Now subtract the two series to get log[(1+x)/(1-x)]= 2[x+x^3/3+x^5/5+x^7/7......inf ] Now replace x with [¡.arctanx] After a little bit of manipulation [using Euler's form] You will get the famous Madhav-Gregory Series .