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

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  • $\begingroup$ Is this an assignment? What tools are you expected to use? What did you try already? $\endgroup$
    – Yuriy S
    Commented Jun 10, 2016 at 11:52
  • $\begingroup$ I tried to use eulers formula of complex to derive the series value for pi/4. But I don't have any idea how to derive the series. No,this is not an assignment. $\endgroup$ Commented Jun 10, 2016 at 11:57
  • $\begingroup$ Eh? The radius of convergence of $\sum_{n\geq 0}\frac{(-1)^n}{2n+1}x^{2n+1}$ is $1$, not $\frac{\pi}{4}$. That comes from the simple poles at $x=\pm i$ of $\frac{d}{dx}\arctan(x)=\frac{1}{x^2+1}$. $\endgroup$ Commented Jun 10, 2016 at 11:57
  • $\begingroup$ Since the power series has a convergence radius of 1 it applicable for arguments with magnitude between $\pi/4$ and 1. $\endgroup$ Commented Jun 10, 2016 at 11:58

3 Answers 3

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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,

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


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

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

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  • $\begingroup$ This approach has a minor problem at $x = \pm 1$ which needs to be fixed by further analysis. It is better to analyze the error term during integration of geometric series for $1/(1 + x^{2})$ as done in my answer. $\endgroup$
    – Paramanand Singh
    Commented Jun 10, 2016 at 13:07
  • $\begingroup$ @ParamanandSingh: you are right. $\endgroup$
    – user65203
    Commented Jun 10, 2016 at 13:23
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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 $|x| \leq 1$.

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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 .

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  • $\begingroup$ Any thoughts about the radius of convergence? $\endgroup$
    – Moritz
    Commented Sep 11, 2016 at 15:14

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