Does the Taylor series for arctan converge at x = 1?

[Euge]

Show that $$\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots$$

 

[julian]

Spoiler

We have

\begin{align*}
\arctan (z) = z - \frac{z^3}{3} + \frac{z^5}{5} - \frac{z^7}{7} + \cdots = \sum_{n=0}^\infty \dfrac{(-1)^n z^{2n+1}}{2n+1} ; \quad |z| \leq 1 \quad z \not= \pm i
\end{align*}

So that

\begin{align*}
\frac{\pi}{4} = \arctan (1) = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots = \sum_{n=0}^\infty \dfrac{(-1)^n}{2n+1}
\end{align*}

 

[wrobel]

Spoiler: Just a remark

The expansion $\arctan z=z-\frac{z^3}{3}+\ldots$ follows from general theory for $|z|<1.$ The assertion $\arctan(1)=1-\frac{1}{3}+\ldots$ needs some additional work. Namely, one must show that the series in the right side converges to $\arctan(1)$ indeed.

 

[julian]

Spoiler: Just a remark

The expansion $\arctan z=z-\frac{z^3}{3}+\ldots$ follows from general theory for $|z|<1.$ The assertion $\arctan(1)=1-\frac{1}{3}+\ldots$ needs some additional work. Namely, one must show that the series in the right side converges to $\arctan(1)$ indeed.

Because $\dfrac{1}{1+z^2}$ only converges for $|z| < 1$?

 

[wrobel]

Because $\dfrac{1}{1+z^2}$ only converges for $|z| < 1$?

Because it is insufficient to prove that 1-1/3+... converges, one must prove that it converges to $\arctan 1$

 

[WWGD]

Because it is insufficient to prove that 1-1/3+... converges, one must prove that it converges to $\arctan 1$

Well, since arctanz is Complex-analytic; entire actually, that should be enough.

 

[Svein]

On another note: Create the Fourier series transform of f(x)=x between -π and π. The series converges at π/2, since f(x) is differentiable in the open interval <-π, π>. Evaluate the series at π/2.
For a more complete discussion, see: https://www.physicsforums.com/insights/sums-found-fourier-series/

 

[wrobel]

Well, since arctanz is Complex-analytic; entire actually, that should be enough.

By which theorem is it enough? Consider me as a wet blanket but I think that if we are solving a math problem, we must refer to mathematical theorems and apply them. For example, the result follows from
Theorem (Abel). Assume we have an analytic function
$$f(z)=\sum_{k=0}^\infty a_kz^k,\quad a_k\in\mathbb{R}$$ and
the Taylor series is convergent in $\{|z|<1\}$. If a series
$a=\sum_{k=0}^\infty a_k$ converges then
$$\lim_{\mathbb{R}\ni x\to 1-}f(x)=a.$$

 

[PeroK]

There's a proof here:

https://proofwiki.org/wiki/Power_Series_Expansion_for_Real_Arctangent_Function

 

[WWGD]

By which theorem is it enough? Consider me as a wet blanket but I think that if we are solving a math problem, we must refer to mathematical theorems and apply them. For example, the result follows from
Theorem (Abel). Assume we have an analytic function
$$f(z)=\sum_{k=0}^\infty a_kz^k,\quad a_k\in\mathbb{R}$$ and
the Taylor series is convergent in $\{|z|<1\}$. If a series
$a=\sum_{k=0}^\infty a_k$ converges then
$$\lim_{\mathbb{R}\ni x\to 1-}f(x)=a.$$

I thought we were working over the Complexes, where analiticity is guaranteed by differentiability.

 

[wrobel]

I thought we were working over the Complexes, where analiticity is guaranteed by differentiability.

We are working over complexes. Please present your argument as a formal proof.

 

[WWGD]

We are working over complexes. Please present your argument as a formal proof.

You mean that a differentiable function is infinitely-differentiable and analytic?

 

[WWGD]

Isn't this one of the mainstream results in Complex Analysis? Differentiability implies $C^{\infty}$, implies Analytic?

 

[wrobel]

You mean that a differentiable function is infinitely-differentiable and analytic?

I mean posts #5, 6
How does your answer from #6 solve the problem formulated in #5?

 

[PeroK]

Isn't this one of the mainstream results in Complex Analysis? Differentiability implies $C^{\infty}$, implies Analytic?

The issue is whether the Taylor series for $\arctan$ converges to the function at the radius of convergence (that is to say at $x = 1$). We can take that the Taylor series converges to the function value within the radius of convergence, but not at the radius itself. That's a standard result. In this case we have:
$$\arctan(x) = \sum_{n = 0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1} \ \ \ (|x| < 1)$$The question is what happens at $x = 1$? The series for $x = 1$ converges, by the alternating series test, but does it converge to $\arctan(1)$?

Note that I posted a proof from Wiki above (post #9), that uses integration over the interval $(0, 1)$ to prove the equality at the endpoint.

I'd also be interested to see an alternative proof using complex numbers, if you could supply one.

 

[julian]

Spoiler

We will evaluate

\begin{align*}
\sum_{k=1}^\infty \frac{(-1)^k \sin \dfrac{\pi k}{2}}{k} = - \sum_{k=1}^\infty \dfrac{(-1)^{k-1}}{2k-1}
\end{align*}

I will express the sum via a complex contour integration. It employs the fact that the function

\begin{align*}
\dfrac{\sin \dfrac{\pi z}{2}}{\sin \pi z}
\end{align*}

has simple poles at all integer values except when $\sin \dfrac{\pi z}{2} = 0$.

This allows us to write

\begin{align*}
\sum_{k=1}^\infty \frac{(-1)^k \sin \dfrac{\pi k}{2}}{k} = \frac{1}{2i} \oint_C \dfrac{\sin \dfrac{\pi z}{2}}{z \sin \pi z} dz
\end{align*}

where the contour $C$ is defined in fig (a).

We have

\begin{align*}
2 \sum_{k=1}^\infty (-1)^k \frac{\sin \dfrac{\pi k}{2}}{k} & = \sum_{k=1}^\infty (-1)^k \frac{\sin \dfrac{\pi k}{2}}{k} + \sum_{k=-1}^{-\infty} (-1)^k\frac{\sin \dfrac{\pi k}{2}}{k}
\nonumber \\
& = \frac{1}{2i} \oint_{C+C'} \dfrac{\sin \dfrac{\pi z}{2}}{z \sin \pi z} dz
\end{align*}

where the contour $C'$ is defined in fig (a). We complete the path of integration along semicircles at infinity (see fig (b)) since the integrand vanishes there. Since the resulting enclosed area contains no singularities except at $z=0$, we can shrink this contour down to an infinitesimal circle $C_0$ around the origin (see fig (c)). So that

\begin{align*}
\sum_{k=1}^\infty (-1)^k \frac{\sin \dfrac{\pi k}{2}}{k} = \frac{1}{4 i} \oint_{C_0} \dfrac{\sin \dfrac{\pi z}{2}}{z \sin \pi z} dz
\end{align*}

We expand the integrand in powers of $z$ about $z=0$ and isolate the $z^{-1}$ term. We get

\begin{align*}
\dfrac{\sin \dfrac{\pi z}{2}}{z \sin \pi z} & = \dfrac{\frac{\pi z}{2} - \cdots}{z [\pi z - \frac{1}{3!} \pi^3 z^3 + \cdots]}
\nonumber \\
& = \dfrac{\frac{\pi z}{2}- \cdots}{z^2 \pi [1 - \frac{1}{6} \pi^2 z^2 + \cdots]}
\nonumber \\
& = \dfrac{\frac{\pi z}{2} - \cdots}{z^2 \pi} (1 + \frac{1}{6} \pi^2 z^2 + \cdots)
\nonumber \\
& = \frac{1}{2z} + \cdots
\end{align*}

So that,

\begin{align*}
\sum_{k=1}^\infty (-1)^k \frac{\sin \dfrac{\pi k}{2}}{k} & = \frac{1}{4 i} \oint_{C_0} \dfrac{\sin \dfrac{\pi z}{2}}{z \sin \pi z} dz
\nonumber \\
& = \frac{1}{4 i} (-2 \pi i) \frac{1}{2}
\nonumber \\
& = - \frac{\pi}{4} .
\end{align*}

So that

\begin{align*}
\frac{\pi}{4} = \sum_{k=1}^\infty \dfrac{(-1)^{k-1}}{2k-1} .
\end{align*}

 

[wrobel]

I think that the shortest proof is via the Taylor series for arctan and Abel's theorem.

 

[bob012345]

Given a circle and a square which circumscribes that circle, the ratio of areas of that circle to that square is $\frac{\pi}{4}$.
I wonder if there is a purely geometric construction based series approach to this problem?

 

[WWGD]

The issue is whether the Taylor series for $\arctan$ converges to the function at the radius of convergence (that is to say at $x = 1$). We can take that the Taylor series converges to the function value within the radius of convergence, but not at the radius itself. That's a standard result. In this case we have:
$$\arctan(x) = \sum_{n = 0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1} \ \ \ (|x| < 1)$$The question is what happens at $x = 1$? The series for $x = 1$ converges, by the alternating series test, but does it converge to $\arctan(1)$?

Note that I posted a proof from Wiki above (post #9), that uses integration over the interval $(0, 1)$ to prove the equality at the endpoint.

I'd also be interested to see an alternative proof using complex numbers, if you could supply one.

Thanks for clarifying. I missed that obvious point( not being sarcastic) . Embarrassed.

 

[julian]

Same method I used here:

POTW

can be used here as well:

Spoiler

We have

\begin{align*}
\sum_{n=0}^\infty \dfrac{(-1)^n}{2n+1} & = \sum_{n=1}^\infty \dfrac{\sin \dfrac{\pi n}{2}}{n}
\nonumber \\
& = \sum_{n=1}^\infty \dfrac{\sin \dfrac{\pi n}{2}}{n} \int_0^\infty e^{-y} dy
\nonumber \\
& = \sum_{n=1}^\infty \sin \dfrac{\pi n}{2} \int_0^\infty e^{-nx} dx
\nonumber \\
& = \frac{1}{2i} \sum_{n=1}^\infty \int_0^\infty (e^{-nx + \frac{i\pi n}{2}} - e^{-nx + \frac{-i\pi n}{2}}) dx
\nonumber \\
& = \frac{1}{2i} \int_0^\infty \left( \dfrac{1}{1-e^{-x + \frac{i\pi}{2}}} - \dfrac{1}{1-e^{-x + \frac{-i\pi}{2}}} \right) dx
\nonumber \\
& = \int_0^\infty \dfrac{1} {e^x + e^{-x}} dx
\nonumber \\
& = \frac{1}{2} \int_{-\infty}^\infty \dfrac{1} {e^x + e^{-x}} dx
\end{align*}

EDIT: Instead of using complex analysis to evaluate the integral, as I originally did, you can use the fact that $\frac{d}{dx} \tan^{-1} (\sinh x) = \frac{1}{\cosh x}$ to obtain:

\begin{align*}
\frac{1}{4} \int_{-\infty}^\infty \dfrac{2}{e^x + e^{-x}} dx = \frac{1}{4} [\tan^{-1} (\sinh x)]_{-\infty}^\infty = \frac{\pi}{4} .
\end{align*}

----

We will evaluate

\begin{align*}
\int_{-\infty}^\infty \dfrac{1} {e^x + e^{-x}} dx
\end{align*}

by considering the rectangular contour integral (see figure) of

\begin{align*}
\oint_C \dfrac{1} {e^z + e^{-z}} dz
\end{align*}

The integral along the vertical edges vanishes as:

\begin{align*}
f(z) = \dfrac{1}{e^{x+iy} + e^{-x-iy}} =
\begin{cases}
e^{- (x+iy)} & x \rightarrow \infty \\
e^{(x+iy)} & x \rightarrow - \infty \\
\end{cases}
\end{align*}

So that

\begin{align*}
\oint_C \dfrac{1}{e^z + e^{-z}} dz & = \int_{-\infty}^\infty \dfrac{1}{e^x + e^{-x}} dx + \int_{-\infty+ i \pi}^{\infty + i \pi} \dfrac{1}{e^x + e^{-x}} dx
\nonumber \\
& = 2 \int_{-\infty}^\infty \dfrac{1} {e^x + e^{-x}} dx
\end{align*}

and so

\begin{align*}
\frac{1}{2} \int_{-\infty}^\infty \dfrac{1}{e^x + e^{-x}} dx & = \frac{i \pi}{2} \frac{1}{2 \pi i} \oint_C \dfrac{1}{e^z + e^{-z}} dz
\nonumber \\
& = \frac{i \pi}{2} \lim_{z \rightarrow \frac{i \pi}{2}} (z - \frac{i \pi}{2}) \dfrac{1}{e^z + e^{-z}}
\nonumber \\
& = \frac{i \pi}{2} \lim_{z \rightarrow \frac{i \pi}{2}} \dfrac{1}{e^z - e^{-z}}
\nonumber \\
& = \frac{\pi}{4} .
\end{align*}


So finally,

\begin{align*}
\sum_{n=0}^\infty \dfrac{(-1)^n}{2n+1} = \frac{\pi}{4} .
\end{align*}