My solution of KAIST POW 2012-1

Posted on February 19, 2012 by hunminpark

Problem(KAIST POW 2012-1)

Compute \displaystyle \tan^{-1}{\frac{1}{1}}-\tan^{-1}{\frac{1}{3}}+\tan^{-1}{\frac{1}{5}}-\tan^{-1}{\frac{1}{7}}+\cdots

(See KAIST POW 2012-1)

My solution

Let

\displaystyle S:=\tan^{-1}{\frac{1}{1}}-\tan^{-1}{\frac{1}{3}}+\tan^{-1}{\frac{1}{5}}-\tan^{-1}{\frac{1}{7}}+\cdots

\displaystyle F(x):=\tan^{-1}{\frac{x}{1}}-\tan^{-1}{\frac{x}{3}}+\tan^{-1}{\frac{x}{5}}-\tan^{-1}{\frac{x}{7}}+\cdots \quad (0 \leq x \leq 1)

\displaystyle f(x):=\frac{1}{1^2+x^2}-\frac{3}{3^2+x^2}+\frac{5}{5^2+x^2}-\frac{7}{7^2+x^2}+\cdots \quad (0 \leq x \leq 1)

By alternating series test(see Mathworld), F(x) and f(x) converge when 0 \leq x \leq 1.

Now let’s observe that

\displaystyle S=F(1),\quad \frac{d}{dx}F(x)=f(x)

This gives us

\displaystyle \int_{0}^{1} f(x)dx=S

Now we use the following well-known formula about series expansion of hyperbolic secant function(see Wolframfunction);

\displaystyle \sec h {z}=\pi \sum_{k=0}^{\infty}\frac{(-1)^{k} (1+2k)}{(1/2+k)^2 \pi^2+z^2} \left(-\frac{1}{2}+\frac{iz}{\pi} \not \in \mathbb{Z}\right)

(it is sech(z), not sec(hz)…why LaTeX doesn’t have \sech operator command?? -_-)

Substitute z=\pi x/2, then we can get a closed expression of f(x);

\displaystyle f(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}(2n-1)}{x^2+(2n-1)^2}=\frac{\pi}{2} \cdot \frac{1}{e^{\pi x/2}+e^{-\pi x/2}} \quad (0 \leq x \leq 1)

Hence

\displaystyle S=\int_{0}^{1}f(x)dx=\int_{0}^{1}{\frac{\pi}{2} \cdot \frac{1}{e^{\pi x/2}+e^{-\pi x/2}}}dx

Substitute u=e^{\pi x/2}, then we finally get

\displaystyle S=\frac{\pi}{2}\int_{1}^{e^{\pi/2}}{\frac{1}{1+u^2}}du=\tan^{-1}\left(\tanh{\frac{\pi}{4}}\right)

Hence, the answer is \displaystyle \tan^{-1}\left(\tanh{\frac{\pi}{4}}\right)  \square

Remark

We can see some of others’ solutions at the KAIST POW homepage (KAIST POW 2012-1). Most of the solutions I can see contain some algebraical identities about trigonometric functions. I think that there may exist a proof which uses telescoping; find a_n such that

\displaystyle a_n-a_{n+1}=(-1)^{n-1} \tan^{-1}{\frac{1}{2n-1}}

If \lim_{n \to \infty}a_{n} exists, then the summation may become

\displaystyle S=a_1-\lim_{n \to \infty}a_{n}

I tried to find some properties of \left\{a_n \right\}…But it was hard for me;;;

I can’t know there is a proof with telescoping method as I can’t see all of others’ solutions… but I believe that telescoping method may works if we find \displaystyle \lim_{n \to \infty}a_{n}.

This entry was posted in KAIST POW and tagged , . Bookmark the permalink.

4 Responses to My solution of KAIST POW 2012-1

  1. tetration says:
    September 9, 2012 at 2:14 pm

    \sech가 안 먹히면 \operatorname{sech}를 써 주면 됩니다.

    Reply
  2. hunminpark says:
    September 9, 2012 at 2:53 pm

    답변 감사합니다 ^^ {\rm sech}도 먹히는 듯 하군요. 근데 \rm과 \operatorname의 차이가 무엇인가요? \rm을 자주 사용했지만 커맨드가 정확히 어떤 짓을(?) 하는 건지 잘 모르겠네요.

    Reply
    • tetration says:
      September 9, 2012 at 5:26 pm

      \rm (혹은 \mathrm)은 로만 체를 입력할 때 씁니다. \operatorname{}은 함수를 쓸 때 주로 쓰는 걸로 보입니다. 적어도 위키피디아 상에선 표시상 큰 차이가 나진 않는 걸로 압니다.

      Reply

Leave a comment