I was born and raised in Daegu, S. Korea. I majored in electronics and math in Seoul from 2007 to 2012. I've had a great interest in math since freshman year, and I studied PDE in Osaka, Japan from 2012-2014. I worked at a science museum and HUFS from 2014 in Seoul. Now I'm studying PDE in Tokyo, Japan. I also developed an interest in music, as I met a great piano teacher Oh in 2001, and joined an indie metal band in 2008. In my spare time, I enjoy various things, such as listening music, blogging, traveling, taking photos, and playing Go and Holdem. Please do not hesitate to contact me with comments, email, guestbook, and social medias.

한큐 이시바시역의 늦은 오후 (阪急石橋駅, Hankyu Ishibashi Stn.)

오사카대학 이시바시문에서 5분 거리에 있는 이시바시역입니다.
비가 와서 흑마를 탈 수 없었기에, 오랜만에 이시바시역을 이용했네요.

이시바시역 바로 앞에 있는 한큐 소바(阪急そば).
역 안에 하나 더 있는데, 저도 몇 번 이용해 봤습니다.
뭐 가격 대비 맛은 그럭저럭.

이시바시역 앞 상점가 (石橋駅前).

왼쪽은 몇번 이용했던 반찬 가게.

한큐 이시바시역 플랫폼의 모습.

왼쪽에 한큐 소바(阪急そば)가 또 보이네요.

한큐 미노센 플랫폼의 모습 (箕面線).

늦은 오후, 마침 중고딩들의 하교 시간이네요.

드디어 미노행 전차가 도착했네요.

[Calculation 15] Introduction to the Ramanujan Theta Functions

In this post, we introduce the Ramanujan theta functions $f(a,b)$, which generalize the form of the Jacobi theta functions. Here we define the Ramanujan theta function, and introduce some elementary properties.

First, we define the Ramanujan theta function as
$$f(a,b) = 1+\sum_{k=1}^\infty (ab)^{k(k-1)/2} (a^k + b^k) = \sum_{k=-\infty}^\infty a^{k(k+1)/2} b^{k(k-1)/2}$$
where $|ab|<1$. Then the following holds.

 Theorem. We have \begin{align*} &\text{(i) } f(a,b) = f(b,a),\\ &\text{(ii) } f(1,a) = 2f(a,a^3),\\ &\text{(iii) } f(-1,a) = 0,\\ &\text{(iv) } \forall n \in \mathbb Z, \; f(a,b) = a^{n(n+1)/2} b^{n(n-1)/2} f(a(ab)^n, b(ab)^{-n}). \end{align*}

Proof. (i) is trivial. For (ii), we have
\begin{eqnarray*}
f(1,a)
&=&
2+\sum_{k=1}^\infty a^{k(k+1)/2} + \sum_{k=2}^\infty a^{k(k-1)/2}\\
&=&
2 \left( 1+ \sum_{k=1}^\infty a^{k(k+1)/2}\right)\\
&=&
2 \left(1+ \sum_{k=1}^\infty a^{k(2k+1)}+\sum_{k=1}^\infty a^{k(2k-1)} \right)\\
&=&
2f(a,a^3).
\end{eqnarray*}
For (iii), we have
\begin{eqnarray*}
f(-1,a)
&=&
\sum_{k=2}^\infty (-1)^{k(k+1)/2} a^{k(k-1)/2}
+
\sum_{k=1}^\infty (-1)^{k(k-1)/2} a^{k(k+1)/2}\\
&=&
\sum_{k=1}^\infty (-1)^{(k+1)(k+2)/2} a^{k(k+1)/2}
+
\sum_{k=1}^\infty (-1)^{k(k-1)/2} a^{k(k+1)/2}\;\;=\;\;0.
\end{eqnarray*}
And for (iv), we have
\begin{eqnarray*}
f(a,b)
&=&
\sum_{k=-\infty}^\infty a^{k(k+1)/2} b^{k(k-1)/2}\\
&=&
\sum_{k=-\infty}^\infty a^{(k+n)(k+n+1)/2} b^{(k+n)(k+n-1)/2}\\
&=&
a^{n(n+1)/2} b^{n(n-1)/2} \sum_{k=-\infty}^\infty a^{k(k+2n+1)/2} b^{k(k+2n-1)/2}\\
&=&
a^{n(n+1)/2} b^{n(n-1)/2} \sum_{k=-\infty}^\infty (a(ab)^n)^{k(k+1)/2} (b(ab)^{-n})^{k(k-1)/2}\\
&=&
a^{n(n+1)/2} b^{n(n-1)/2} f(a(ab)^n, b(ab)^{-n}).
\end{eqnarray*}

References.
[1] Bruce C. Berndt, Ramanujan’s Notebooks, Part III, Springer-Verlag, pp. 34-35.

[Calculation 14] q-Series and the q-Binomial Theorem

In this post, we introduce q-Series and the q-Binomial theorem. For any complex number $a$, we write
$$(a;q)_k = (1-a)(1-aq)(1-aq^2) \cdots (1-aq^{k-1})$$
where $|q|<1$. Also we write
$$(a;q)_\infty = \prod_{k=0}^\infty (1-aq^k).$$
With these notations, we will prove the following theorem, which is called q-Binomial Theorem discovered by Rothe.

 Theorem. (q-Binomial Theorem) If $|x|, |q|<1$ then $$\sum_{k=0}^\infty \frac{(a;q)_k}{(q;q)_k} x^k = \frac{(ax;q)_\infty}{(x;q)_\infty}.$$

Proof. First, we set the function $f_a$ as follows:
$$f_a(x) = \sum_{k=0}^\infty \frac{(a;q)_k}{(q;q)_k} x^k.$$
Then we have
\begin{eqnarray*}
\frac{f_a(x) – f_a (qx)}{x}
&=&
\sum_{k=0}^\infty \frac{(a;q)_k}{(q;q)_k} (1-q^k) x^{k-1}\\
&=&
(1-a) \sum_{k=1}^\infty \frac{(aq;q)_{k-1}}{(q;q)_{k-1}} x^{k-1}\\
&=&
(1-a) \sum_{k=0}^\infty \frac{(aq;q)_k}{(q;q)_k} x^k\\
&=&
(1-a) f_{aq} (x)
\end{eqnarray*}
so that
\tag{1}
f_a(x) – f_a (qx) = (1-a)xf_{aq} (x).

Also we have
\begin{eqnarray*}
f_a (x) – f_{aq} (x)
&=&
\sum_{k=0}^\infty \frac{(aq;q)_{k-1}}{(q;q)_k} (1-a – 1+aq^k)x^k\\
&=&
-a \sum_{k=0}^\infty \frac{(aq;q)_{k-1}}{(q;q)_k} (1-q^k) x^k\\
&=&
-a \sum_{k=1}^\infty \frac{(aq;q)_{k-1}}{(q;q)_{k-1}} x^k\\
&=&
-a \sum_{k=0}^\infty \frac{(aq;q)_{k}}{(q;q)_{k}} x^{k+1}\\
&=&
-axf_{aq} (x)
\end{eqnarray*}
so that
\tag{2}
f_a (x) = (1-ax) f_{aq} (x).

Solving (1) and (2) to eliminate $f_{aq}(x)$, we get
$$f_a (x) = \frac{1-ax}{1-x} f_a (qx).$$
Iterating this relation $n$ times, we arrive at
$$f_a (x) = \frac{(ax;q)_n}{(x;q)_n} f_a (q^n x).$$
Taking the limit $n\to \infty$, we obtain
$$f_a (x) = \lim_{n \to \infty} \frac{(ax;q)_n}{(x;q)_n} f_a (q^n x) = \frac{(ax;q)_\infty}{(x;q)_\infty},$$
which proves the theorem.$\square$

References.
[1] G. E. Andrews, Special Functions, Cambridge University Press, pp. 487-489.
[2] Bruce C. Berndt, Ramanujan’s Notebooks, Part III, Springer-Verlag, pp. 11-12.

Elsevier Journal Finder

논문 제목과 abstract를 입력하면 Elsevier의 어느 저널에 어울리는지 찾아주는 웹사이트.
우연히 페이스북을 통해 알게되었습니다.

http://journalfinder.elsevier.com

덤으로 각 저널의 impact factor와 acceptance rate 도 나오는군요.
저도 한 번 테스트(?) 해 보았습니다.

상당히 똑똑한 놈이군요?!
음 이번에 투고한 Nonlinear Analysis는 2순위로 뜨는군요.

[Calculation 13] A Simple Formula Related to Digamma Functions

 Theorem. The following holds: $$\pi \;{}_2 F_1 \left( \frac{1}{2}, \frac{1}{2}; 1; 1-x\right) = \log \left( \frac{16}{x}\right) {}_2 F_1 \left( \frac{1}{2}, \frac{1}{2};1;x\right) – 4 \sum_{k=1}^\infty \frac{(\frac{1}{2})_k^2}{(k!)^2} \sum_{j=1}^k \frac{x^k}{(2j-1)(2j)}.$$

Proof. First, we recall the Corollary in [2] with $a=b=-\frac{1}{2}$ and $n=0$:

{}_2 F_1 \left( \frac{1}{2}, \frac{1}{2}; 1; 1-x\right) =- \frac{1}{\pi} \sum_{k=0}^\infty \frac{(\frac{1}{2})_k^2}{(k!)^2} \left(2\psi\left(\frac{1}{2}+k\right) – 2\psi(k+1) + \log x \right) x^k.

Thus we obtain
\begin{align*}
&\pi \; {}_2 F_1 \left( \frac{1}{2}, \frac{1}{2}; 1; 1-x\right)\\
&=
– \log x \; {}_2 F_1 \left( \frac{1}{2}, \frac{1}{2}; 1; x\right)
– 2 \sum_{k=0}^\infty \frac{(\frac{1}{2})_k^2}{(k!)^2} \left( \psi \left(k+ \frac{1}{2} \right) – \psi (k+1) \right) x^k\\
&=- \log x \; {}_2 F_1 \left( \frac{1}{2}, \frac{1}{2}; 1; x\right)
– 2 \sum_{k=0}^\infty \frac{(\frac{1}{2})_k^2}{(k!)^2} \left(
\sum_{j=1}^k \frac{2}{2j-1} – 2\log 2 – \sum_{j=1}^k \frac{1}{j} \right) x^k\\
&= \log \left( \frac{16}{x}\right) {}_2 F_1 \left( \frac{1}{2}, \frac{1}{2}; 1; x\right) – 2 \sum_{k=1}^\infty \frac{(\frac{1}{2})_k^2}{(k!)^2} \sum_{j=1}^k \frac{x^k}{j(2j-1)},
\end{align*}
where we used the expressions
$$\psi (k+1) = \sum_{j=1}^k \frac{1}{j} – \gamma$$ and $$\psi \left( k + \frac{1}{2} \right) = -\gamma – 2\log 2 + \sum_{j=1}^k \frac{2}{2j-1}$$
for the digamma functions at the second equality.$\square$

Taking exponential to the both sides of the equation in Theorem, we obtain the following corollary.

 Corollary. $$\exp \left( -\pi \frac{{}_2 F_1 (\frac{1}{2}, \frac{1}{2}; 1; 1-x)}{{}_2 F_1 (\frac{1}{2}, \frac{1}{2}; 1; x)} \right) = \frac{x}{16} \exp \left( \frac{4 \sum_{k=1}^\infty \frac{(\frac{1}{2})_k^2}{(k!)^2} \sum_{j=1}^k \frac{x^k}{2j(2j-1)}}{{}_2 F_1 (\frac{1}{2},\frac{1}{2};1;x)}\right)$$

References.
[1] Bruce C. Berndt, Ramanujan’s Notebooks, Part II, Springer-Verlag, pp. 78-79.
[2] Leun Kim, [Calculation 12] A Basic Formula for Hypergeometric Functions.
[3] Bruce C. Berndt, Ramanujan’s Notebooks, Part III, Springer-Verlag, p. 91.

[오사카 미노오] 미노산 아래 우리마을 (大阪箕面, Minoh)

잠깐 짬을 내어 흑마를 타고 마을 한 바퀴 돌아 봤습니다.
저도 이 마을의 주민이 된지 벌써 2년이 다 되어 가네요..

미노역 앞 버스 정류장의 모습입니다.
대충 짐작하셨겠지만 물론 이 곳이 종점입니다.

미노산을 따라서 줄지어 자리잡고 있는 집들.

흑마를 잠시 주차해 놓고, 경치를 감상합니다.

아름다운 하늘, 아름다운 마을.

ロボットの米屋さん이 보이네요.
마을에 쌀을 공급해주고 있는 방앗간입니다.

해가 지고 있네요.
이제 슬슬 내려가야지..

뒤 돌아서도 한 번 찍어주고..

한큐 버스의 모습.

이제 거의 다 내려 왔습니다.
산이라 그런지 언덕길이 상당히 가파르죠?

다시 한큐 미노역에 도착했습니다.
우리 마을의 상징은 “단풍”이라 그런지 곳곳에 은행나무들이 보이네요.

한큐 미노역의 모습입니다.
여기서 저희집까지는 자전거로 5분 정도 걸리지요.

미노역 앞 늘어선 상점가들.
저기 보이는 もみぢや에서는 아마 단풍과자를 팔고 있을 겁니다.
이 마을에서는 단풍나무 잎을 튀겨서 과자로 만들어 먹기도 합니다!

우리 마을 또 하나의 명물 미노맥주.
미노 맥주 종류는 8가지 정도 되는데, 저는 아직도 맛을 못봤네요 흑흑.

오사카대학 스이타 캠퍼스, 힘들게 갔더니 휴강이라니!

이번 학기 수업을 들으러 매주 한 번씩 스이타 캠퍼스를 들리고 있습니다. 오사카대학의 경우 캠퍼스가 3개(토요나카, 스이타, 미노오)로 나뉘어져 있어서, 다른 캠퍼스로 이동해서 수업을 들으려면, 학내연락버스(学内連絡バス)를 이용해야 하지요.

아침에는 이처럼 늘 만원이지요.
물론 무료고, 30분마다 캠퍼스간 버스가 있기 때문에,
설령 학생이 아니더라도 잘만 이용하면 교통비 절약까지도..?

30분 정도 걸려서 도착했는데 아무도 없군요? OTL.

쓸쓸히 다시 스이타 캠퍼스 버스 정류장으로.

토요나카 캠퍼스행 버스에 탔습니다.
출발하기를 기다리는 중..

그렇게 30분을 달려서..

제 본거지(?) 토요나카 캠에 다시 돌아왔습니다.

아래는 버스에서 심심해서 허접하게 찍어본 그 날(10월 중순) 여정.

[Calculation 12] A Basic Formula for Hypergeometric Functions

 Theorem. Let $n \notin \mathbb Z$. Then we have \begin{align}\tag{1} &{}_2 F_1 \left( a+n+1, b+n+1; a+b+n+2; 1-z\right)\\ &\, \qquad\qquad\qquad=\frac{\Gamma(a+b+n+2)\Gamma(-n)}{\Gamma(a+1)\Gamma(b+1)}\; {}_2 F_1 (a+n+1,b+n+1;n+1;z)\\ &\, \qquad\qquad\qquad\qquad+\frac{\Gamma(a+b+n+2)\Gamma(n)z^{-n}}{\Gamma(a+n+1)\Gamma(b+n+1)} \;{}_2 F_1 (a+1,b+1;-n+1;z). \end{align}

Proof. We consider the following ODE, which is called hypergemoetric differential equation:
\tag{2}
z(1-z)y”(z) + (c-(a+b+1)z) y'(z) – ab y(z)=0.

Then we see that the general solution for (2) is the form of
\tag{3}
y= A_1\; {}_2 F_1 (a,b;c;z) + A_2 z^{1-c} {}_2 F_1 (a+1-c,b+1-c;2-c;z)

valid for $|z|<1$, where $A_1, A_2$ are constants. Changing $z$ into $1-\zeta$ in (2), we obtain
\tag{4}
\zeta (1-\zeta) y”(\zeta) + ((a+b+1-c) – (a+b+1)\zeta)y'(\zeta)-aby(\zeta)=0

which has the solutions
\tag{5}

Thus comparing the coefficients of (3) and (5), we obtain the relation
\begin{align*}
&{}_2 F_1 (a,b;c;z) \\
&= C_1 \; {}_2 F_1 (a,b;a+b+1-c;1-z) + C_2 (1-z)^{c-a-b} \, {}_2 F_1 (c-a,c-b;1+c-a-b;1-z)
\end{align*}
valid for $|z|<1$ and $|1-z|<1$, where $C_1, C_2$ are constants. The constant $C_1, C_2$ can be found by limiting $z \to 0$ and $z \to 1$ with Gauss Summation Formula [2]. Then we obtain
$$\frac{\Gamma(a+b+1-c)\Gamma(1-c)}{\Gamma(a+1-c)\Gamma(b+1-c)} C_1 + \frac{\Gamma(1+c-a-b)\Gamma(1-c)}{\Gamma(1-a)\Gamma(1-b)}C_2 =1,$$
$$C_1 = \frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}.$$
Therefore we find that
\begin{align*}
&{}_2 F_1 (a,b;c;z) =
\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}\; {}_2 F_1 (a,b;a+b+1-c;1-z)\\
&+
\frac{\Gamma(c)\Gamma(a+b-c)}{\Gamma(a)\Gamma(b)} (1-z)^{c-a-b} {}_2 F_1 (c-a,c-b;1+c-a-b;1-z).
\end{align*}
Replacing $a \to a+n+1$, $b \to b+n+1$ and $c \to a+b+n+2$ yields (1).$\square$

 Corollary. If $n$ is a nonnegative integer, then \begin{align*} &{}_2 F_1 \left( a+n+1, b+n+1; a+b+n+2; 1-z\right)\\ &\, \qquad = \frac{\Gamma(a+b+n+2)\Gamma(n)z^{-n}}{\Gamma(a+n+1)\Gamma(b+n+1)} \sum_{k=0}^{n-1} \frac{(a+1)_k (b+1)_k}{(-n+1)_k k!} z^k \\ &\, \qquad – \frac{(-1)^n \Gamma(1+b+n+2)}{\Gamma(a+1)\Gamma(b+1)\Gamma(n+1)} \sum_{k=0}^\infty \frac{(a+n+1)_k (b+n+1)_k}{(n+1)_k k!}\\ &\, \qquad \times \left( \psi(a+n+k+1) + \psi (b+n+k+1) – \psi (n+k+1) – \psi (k+1) + \log x \right) x^k \end{align*} where $\psi$ is Digamma function. If $n=0$ then the first expression on the right side above is understood to be equal to $0$.

For the proof of this corollary, see for example Andrews [4]. The method in [4] can be applied to this case also.

References.
[1] W. N. Bailey, Generalized Hypergeometric Series, Cambridge University Press, 1964, pp.1-4.
[2] Leun Kim, [Calculation 5] Gauss’s Summation Formula.
[3] Bruce C. Berndt, Ramanujan’s Notebooks, Part II, Springer-Verlag, pp. 77-78.
[4] G. E. Andrews, Special Functions, Cambridge University Press, pp. 82-84.

[Calculation 11] Simple Examples from the Dixon Theorem

Here we introduce some examples from the Dixon Theorem. By setting suitable coefficients, we can obtain simple formulas of infinite series, which are related to the Gamma functions.

 Example. From the Dixon Theorem, we have \begin{align*} &\text{(i) } 1 + \frac{1}{5} \left( \frac{1}{2}\right)^2 + \frac{1}{9} \left( \frac{1\cdot 3}{2\cdot 4}\right)^2 + \frac{1}{13} \left(\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\right)^2+ \cdots =\frac{\pi^2}{4\Gamma^4(\frac{3}{4})}, \\ &\text{(ii) } 1 + \frac{1}{5^2} \left( \frac{1}{2}\right) + \frac{1}{9^2} \left( \frac{1\cdot 3}{2\cdot 4}\right) + \frac{1}{13^2} \left( \frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\right) +\cdots =\frac{\pi^{5/2}}{8\sqrt{2}\Gamma^2(\frac{3}{4})}, \\ &\text{(iii) } 1 + \left( \frac{1}{2}\right)^3 + \left( \frac{1\cdot 3}{2\cdot 4}\right)^3 + \left( \frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\right)^3 +\cdots =\frac{\pi}{\Gamma^4(\frac{3}{4})}. \end{align*}

Proof. First, we recall the Dixon Theorem [2]:
\begin{align}\tag{1}
&{}_3 F_2 (a,b,c;1+a-b,1+a-c;1) \\
&=
\frac{\Gamma(1+\frac{a}{2})\Gamma(1+a-b)\Gamma(1+a-c)\Gamma(1+\frac{a}{2}-b-c)}{\Gamma(1+a)\Gamma(1+\frac{a}{2}-b)\Gamma(1+\frac{a}{2}-c)\Gamma(1+a-b-c)}.
\end{align}
(i) Setting $a=b=\frac{1}{2}$ and $c=\frac{1}{4}$ in the equation (1), we obtain
\begin{eqnarray*}
{}_3 F_2 \left(\frac{1}{2},\frac{1}{2},\frac{1}{4};1,\frac{5}{4};1\right)
&=&
\frac{\Gamma(\frac{5}{4})\Gamma(1)\Gamma(\frac{5}{4})\Gamma(\frac{1}{2})}{\Gamma(\frac{3}{2})\Gamma(\frac{3}{4})\Gamma(1)\Gamma(\frac{3}{4})}\\
&=&
\frac{2\Gamma^2(\frac{5}{4})}{\Gamma^2 (\frac{3}{4})}\\
&=&
\frac{\pi^2}{4\Gamma^4(\frac{3}{4})},
\end{eqnarray*}
where we used the Legendre Duplication Formula:
$$\Gamma \left( \frac{3}{4}\right) \Gamma \left( \frac{5}{4}\right) = 2^{1-2\cdot \frac{3}{4}}\sqrt{\pi}\Gamma\left(\frac{3}{2}\right) = 2^{-\frac{3}{2}}\pi$$
for the last equality.

(ii) Setting $a=\frac{1}{2}$ and $b=c=\frac{1}{4}$ in the equation (1) yields the result by the similar steps.

(iii) Setting $a=b=c=\frac{1}{2}$ in the equation (1) yields the result.$\square$

References.
[1] Bruce C. Berndt, Ramanujan’s Notebooks, Part II, Springer-Verlag, p.24.
[2] Leun Kim, [Calculation 10] Dixon Theorem.

[오늘의 야식] 우마이봉 (うまい棒) ~

오늘의 야식 (아니 조식인가?) 은 우마이봉(うまい棒)~
1979년부터 꾸준한 인기를 누려 아직도 생존해 있는 과자라는군요.
근처 패밀리마트에서 우마이봉 개당 10엔에 구입했습니다.
사진은 타코야끼 맛(たこ焼)과 명란젓(めんたい) 맛!

이것도 상당히 중독성이 있군요..

가을비에 젖은 토요나카 캠퍼스 (大阪大学豊中キャンパス)

이제 이 곳도 가을이 왔습니다.
토요나카 캠퍼스에도 단풍이 물들었길래, 한 번 이곳저곳 찍어 보았습니다.

기초공학부 앞에서 (基礎工学部前)

공통교육센터 앞 (共通教育センター前)

중앙도서관 앞 공원 1 (中央図書館前)

중앙도서관 앞 공원 2 (中央図書館前)

기초공학부에서 이학부로 가는 길 (基礎工学部から理学部へ)

이학부 맞은 편에 있는 언덕 올라가는 길 (理学部前坂道)

기초공학부 앞 호수에서 (基礎工学部前の湖)

시그마홀 (シグマホール)

토요나카 캠퍼스 이시바시 문 쪽으로 하교하는 길 (石橋門の方へ)

[M2 Seminar II] Accepted !

2013년 11월 26일 (월)

투고한 저널로부터 최종 acceptance 회신이 도착했다! accept된 저널은

이다. 내심 1지망으로 투고한 곳이라 별 기대를 하지 않고 있었건만, 행운이 따라준 듯 싶다. 지도 교수에게 큰 절이라도 올려야 하나?!호호

올해는 여러가지 일로 꽤나 정신이 없었는데, 어느새 학교도 가을로 물들어 있다.. 혼자서 뒷산으로 단풍 놀이나 갈까나..OTL

[Calculation 10] Dixon Theorem

Here we note Dixon’s theorem, which gives some special values of ${}_3 F_2$, since the proof is almost automatic by using Gauss and Kummer’s formulas which we’ve shown before.

 Theorem. (Dixon’s Theorem) $${}_3 F_2 (a,b,c;1+a-b,1+a-c;1) = \frac{\Gamma(1+\frac{a}{2})\Gamma(1+a-b)\Gamma(1+a-c)\Gamma(1+\frac{a}{2}-b-c)}{\Gamma(1+a)\Gamma(1+\frac{a}{2}-b)\Gamma(1+\frac{a}{2}-c)\Gamma(1+a-b-c)}$$

Proof. First, we recall the Gauss Summation Formula ([2]) and Kummer’s Theorem ([3]) as below:
\begin{align*}
&{}_2 F_1 (\alpha,\beta;\gamma;1) = \frac{\Gamma(\gamma) \Gamma(\gamma-\alpha-\beta)}{\Gamma(\gamma-\alpha)\Gamma(\gamma-\beta)}\quad\text{(Gauss)}\\
\end{align*}
Using these formulas, we obtain
\begin{align*}
&{}_3 F_2 (a,b,c;1+a-b,1+a-c;1)\\
&=
\sum_{k=0}^\infty \frac{\Gamma(a+k)\Gamma(b+k)\Gamma(c+k)\Gamma(1+a-b)\Gamma(1+a-c)}{\Gamma(a)\Gamma(b)\Gamma(c)\Gamma(1+a-b+k)\Gamma(1+a-c+k)k!}\\
&=
\frac{\Gamma(1+a-b)\Gamma(1+a-c)}{\Gamma(a)\Gamma(b)\Gamma(c)}\sum_{k=0}^\infty \frac{\Gamma(a+k)\Gamma(b+k)\Gamma(c+k){}_2 F_1 (b+k, c+k;1+a+2k;1)}{\Gamma(1+a+2k)\Gamma(1+a-b-c)k!}\\
&=
\frac{\Gamma(1+a-b)\Gamma(1+a-c)}{\Gamma(a)\Gamma(b)\Gamma(c)}\sum_{k=0}^\infty \sum_{m=0}^\infty \frac{\Gamma(a+k)\Gamma(b+k+m)\Gamma(c+k+m)}{k!m!\Gamma(1+a+2k+m)\Gamma(1+a-b-c)}\\
&=
\frac{\Gamma(1+a-b)\Gamma(1+a-c)}{\Gamma(a)\Gamma(b)\Gamma(c)}\sum_{p=0}^\infty \sum_{k=0}^p \frac{\Gamma(a+k)\Gamma(b+p)\Gamma(c+p)}{k!(p-k)!\Gamma(1+a+k+p)\Gamma(1+a-b-c)}\\
&=
\frac{\Gamma(1+a-b)\Gamma(1+a-c)}{\Gamma(a)\Gamma(b)\Gamma(c)}
\sum_{p=0}^\infty \frac{\Gamma(a)\Gamma(b+p)\Gamma(c+p)}{p!\Gamma(1+a-b-c)\Gamma(1+a+p)} {}_2 F_1(a,-p;1+a+p;-1)\\
&=
\frac{\Gamma(1+a-b)\Gamma(1+a-c)}{\Gamma(a)\Gamma(b)\Gamma(c)}
\sum_{p=0}^\infty \frac{\Gamma(a)\Gamma(b+p)\Gamma(c+p)\Gamma(1+\frac{a}{2})}{p!\Gamma(1+a-b-c)\Gamma(1+a)\Gamma(1+\frac{a}{2}+p)}\\
&=
\frac{\Gamma(1+a-b)\Gamma(1+a-c)}{\Gamma(1+a)\Gamma(1+a-b-c)} \sum_{p=0}^\infty \frac{(b)_p (c)_p}{(1+\frac{a}{2})_p p!}\\
&=
\frac{\Gamma(1+a-b)\Gamma(1+a-c)}{\Gamma(1+a)\Gamma(1+a-b-c)} {}_2 F_1 (b,c; 1+ \frac{a}{2};1)\\
&=
\frac{\Gamma(1+a-b)\Gamma(1+a-c)}{\Gamma(1+a)\Gamma(1+a-b-c)}
\cdot
\frac{\Gamma(1+\frac{a}{2}) \Gamma(1+\frac{a}{2}-b-c)}{\Gamma(1+\frac{a}{2}-b)\Gamma(1+\frac{a}{2}-c)}
\end{align*}
where we applied Gauss’s Summation Formula ([2]) for the second equality and the last equality, Kummer’s Theorem ([3]) for the 6th equality.$\square$

References.
[1] W. N. Bailey, Generalized Hypergeometric Series, Cambridge University Press, 1964, pp. 13-14.
[2] Leun Kim, [Calculation 5] Gauss’s Summation Formula.
[3] Leun Kim, [Calculation 6] Kummer’s Theorem.

[Calculation 9] Simple Corollaries from Gauss and Bailey Formula

 Corollary 1. For $\frac{1}{2} < z < 2$, $${}_2 F_1 \left(\frac{1}{2}, \frac{1}{2};1;1-\frac{1}{z}\right) = \sqrt{z} {}_2 F_1 \left(\frac{1}{2}, \frac{1}{2};1;1-z\right).$$

Proof. We recall Bailey’s Formula ((i) in [2]) for $w\in\mathbb R$:
\tag{1}
(1-w)^{-a} {}_2 F_1 \left( a,b;c; – \frac{w}{1-w}\right) = {}_2 F_1 (a,c-b;c;w), \qquad -1< w <\frac{1}{2} Setting $a=b=\frac{1}{2}$, $c=1$ and $z=1-w$ at (1), we obtain $$z^{-1/2} {}_2 F_1 \left(\frac{1}{2}, \frac{1}{2};1;1-\frac{1}{z}\right) = {}_2 F_1 \left(\frac{1}{2}, \frac{1}{2};1;1-z\right), \qquad \frac{1}{2} < z < 2.$$ $\square$

 Corollary 2. $${}_2 F_1 \left( \frac{1}{2}, \frac{1}{2}; 1; 1 – \left( \frac{1-z}{1+z}\right)^2\right) = (1+z) {}_2 F_1 \left(\frac{1}{2}, \frac{1}{2}; 1; z^2\right)$$

Proof. Replacing
$$z \to \left(\frac{1-z}{1+z}\right)^2$$
in Corollary 1 above, we have

{}_2 F_1 \left( \frac{1}{2}, \frac{1}{2}; 1; 1 – \left( \frac{1-z}{1+z}\right)^2\right)
=
\frac{1+z}{1-z} {}_2 F_1\left( \frac{1}{2}, \frac{1}{2};1; – \frac{4z}{(1-z)^2}\right).

We remember that Gauss’s Quadratic Transformation ([3]) with $a=b=\frac{1}{2}$ and $z \to -z$ is

{}_2 F_1 \left( \frac{1}{2}, \frac{1}{2}; 1; – \frac{4z}{(1-z)^2}\right)
=
(1-z) {}_2 F_1 \left( \frac{1}{2}, \frac{1}{2}; 1; z^2 \right).

Piecing (2) and (3) together, we obtain Corollary 2.$\square$

References.
[1] Bruce C. Berndt, Ramanujan’s Notebooks, Part II, Springer-Verlag, pp.92-93.
[2] Leun Kim, [Calculation 7] Bailey’s Formulas for Hypergeometric Functions.
[3] Leun Kim, [Calculation 8] Gauss’s Quadratic Trasformation.

오사카 미노 심야 마을 산책 (Osaka Minoh, 大阪箕面)

밤 늦게 바람도 쐴 겸 동네 산책을 갔다 왔습니다.
그런데 오후 8시 쯤인데도 불구하고 대부분의 상점이 문을 닫았군요 OTL.

우리 마을에서 그나마(?) 가장 번화한 상점가..

한산한 동네의 모습.

골목 구석에 위치한 음식점도 찍어주고..

오른쪽에 보이는 건 반찬가게인데, 예전에 자주 이용했습죠 ㅎㅎ.
밤 날씨가 꽤 쌀쌀해진 듯 싶습니다.

[M2 Seminar II] 논문 / Beamer 최종본 완성

2013년 11월 22일 (금)

저번 주에 투고한 저널의 referee들로부터 받은 리포트들을 참고하여, 거기서 요구하는 수정내용들을 수정했다. 추가로 remainder $Q_j$의 형태를 구체화시키고, 사소한 수정도 가미했다. 이 날 저녁 저널에 수정안을 보냈으니, 지도교수 말로는 아마 1주일 정도면 회신이 오지 않을까라고 한다.

오전 10시 반에는 야스에군의 세미나가 있었다. 30분 정도 결과 보고회를 가졌는데, 역시 동(同)학부 출신답게 근사한 결과를 낸 듯 하다. 파라미터 $\gamma_1$과 $\gamma_2$의 변화에 따른 슈뢰딩거 방정식 해의 안정성에 대한 연구였는데, 저번 주 곤란했던 부분을 말끔히 해결했다고 한다!

131122-Beamer

오후 1시부터는 세미나 최종 연습이 있었다. 원래 계획한 것은 지도교수와 나만의 간단한 연습 정도의 느낌이었는데, 의외로 항상 몇 명씩 꼭 참석해 준다. 오늘은 박사과정 3년차 와카스기 선배와, 늘 참석해 주는 오오무카이군이 참석했다. 지도교수는 항상 똑같은 발표를 듣고 있으니 꽤 지루하기도 할 것이다 -_-;

토요나카 캠퍼스 흡연장소들 중 한 곳, 이제 이 곳도 슬슬 가을이 온다..

 Theorem. (Gauss’s Quadratic Transformation) \tag{1} (1+z)^{-2a} {}_2 F_1 \left(a,b; 2b; \frac{4z}{(1+z)^2}\right) = {}_2 F_1 \left(a, 1+\frac{1}{2}-b;b+\frac{1}{2};z^2\right).

Proof. The proof is almost similar to that of [2]. Note that the left hand side of (1) can be expanded in powers of $z$, since it is analytic in a certain neighborhood of $z=0$. Thus by the binomial theorem, we have
\begin{eqnarray*}
(1+z)^{-2a} {}_2 F_1 \left(a,b; 2b; \frac{4z}{(1+z)^2}\right)
&=&
\sum_{k=0}^\infty \frac{(a)_k (b)_k 4^k}{(2b)_k k!} z^k (1+z)^{-2a-2k}\\
&=&
\sum_{k=0}^\infty \sum_{r=0}^\infty \frac{(a)_k (b)_k (-4)^k}{(2b)_k k!} \frac{(2a+2k)_r}{r!} z^{k+r}.
\end{eqnarray*}
Calculating the coefficient of $z^n$, we get
\tag{2}
\sum_{k=0}^n \frac{(a)_k (b)_k 4^k (-1)^k}{(2b)_k k!} \frac{(2a+2k)_{n-k}}{(n-k)!} = \frac{(2a)_n}{n!}\sum_{k=0}^n \frac{(b)_k (2a+n)_k (-n)_k}{(2b)_k (\frac{1}{2} + a)_k k!}

where we used the identities
$$(2a+2k)_{n-k} = \frac{(2a)_n (2a+n)_k}{4^k (a)_k (\frac{1}{2} +a)_k}, \qquad \frac{(-1)^k}{(n-k)!} = \frac{(-n)_k}{n!}.$$
Note that the right hand side of (2) vanishes when $n$ is odd. Also we can easily verify that
\tag{3}
\frac{(2a)_n}{n!}\sum_{k=0}^n \frac{(b)_k (2a+n)_k (-n)_k}{(2b)_k (\frac{1}{2} + a)_k k!}
=
\frac{(a)_{n/2} (a+\frac{1}{2}-b)_{n/2}}{(b+\frac{1}{2})_{n/2} n!}

for all even $n$. But the right hand side of (3) is exactly the coefficients of $z^n$, in the right hand side of (1). This proves the theorem. For the another proof, I recommend Erdelyi [1].$\square$

References.
[1] A. Erdelyi, Higher Transcendental Functions, Vol. 1, McGraw-Hill, New York, 1953, pp.64-68.
[2] Leun Kim, [Calculation 6] Kummer’s Theorem.
[3] Bruce C. Berndt, Ramanujan’s Notebooks, Part II, Springer-Verlag, p.63.

[Calculation 7] Bailey’s Formulas for Hypergeometric Functions

 Theorem. (Bailey) The followings are valid: \begin{align*} &\text{(i) } (1-z)^{-a} {}_2 F_1 \left( a,b;c; – \frac{z}{1-z}\right) = {}_2 F_1 (a,c-b;c;z),\quad|z|<1,\;\text{Re}z< \frac{1}{2},\\ &\text{(ii) } {}_2 F_1 \left( a,b; \frac{a+b+1}{2}; \frac{1}{2}\right) = \frac{\Gamma(\frac{1}{2}) \Gamma(\frac{1+a+b}{2})}{\Gamma(\frac{1+a}{2})\Gamma(\frac{1+b}{2})},\\ &\text{(iii) } {}_2 F_1 \left(a,1-a;c;\frac{1}{2}\right) = \frac{\Gamma(\frac{1}{2}c)\Gamma(\frac{c+1}{2})}{\Gamma(\frac{c+a}{2})\Gamma(\frac{1+c-a}{2})}. \end{align*}

Proof. (i) By the direct calculations, the left hand side of (i) becomes
\begin{eqnarray*}
(1-z)^{-a} {}_2 F_1 \left( a,b;c; – \frac{z}{1-z}\right)
&=&
\sum_{k=0}^\infty \frac{(a)_k (b)_k (-1)^k}{(c)_k k!} z^k (1-z)^{-a-k}\\
&=&
\sum_{k=0}^\infty \sum_{r=0}^\infty \frac{(a)_k (b)_k (-1)^k}{(c)_k k!} \frac{(a+k)_r}{r!} z^{k+r}
\end{eqnarray*}
for $|z|< \frac{1}{2}$. Calculating the coefficient of $z^n$, we get
\begin{eqnarray*}
\sum_{k=0}^n \frac{(a)_k (b)_k (-1)^k}{(c)_k k!} \frac{(a+k)_{n-k}}{(n-k)!}
&=&
\frac{(a)_n}{n!} \sum_{k=0}^n \frac{(b)_k (-n)_k}{k!(c)_k}\\
&=&
\frac{(a)_n}{n!} \frac{(c-b)_n}{(c)_n}
\end{eqnarray*}
where we applied the identities
$$(a)_k (a+k)_{n-k} = (a)_n, \qquad \frac{(-1)^k}{(n-k)!} = \frac{(-n)_k}{n!}$$
for the first equality, and the Chu-Vandermonde identity (in [2]) for the second equality. Finally we obtain
\begin{eqnarray*}
(1-z)^{-a} {}_2 F_1 \left( a,b;c; – \frac{z}{1-z}\right)
&=&
\sum_{n=0}^\infty \frac{(a)_n}{n!}\frac{(c-b)_n}{(c)_n} z^n\\
&=&
{}_2 F_1 (a,c-b;c;z),
\end{eqnarray*}
which proves (i) for $|z|< \frac{1}{2}$. Complete result follows by analytic continuation.$\square$

(ii) Letting $z \to -1$ in (i), we find that
\tag{1}
{}_2 F_1 \left(a,b;c;\frac{1}{2}\right) =
2^a {}_2 F_1 (a,c-b;c;-1).

Setting $c=\frac{1}{2}(a+b+1)$, we obtain
\begin{eqnarray*}
{}_2 F_1 \left( a,b; \frac{a+b+1}{2}; \frac{1}{2}\right)
&=&
2^a {}_2 F_1 \left( a, \frac{1}{2}(1+a-b); \frac{1}{2}(1+a+b);-1\right)\\
&=&
2^a \frac{\Gamma(\frac{1+a+b}{2})\Gamma(1+\frac{1}{2}a)}{\Gamma(1+a)\Gamma(\frac{1+b}{2})}\\
&=&
\frac{\Gamma(\frac{1}{2}) \Gamma(\frac{1+a+b}{2})}{\Gamma(\frac{1+a}{2})\Gamma(\frac{1+b}{2})}
\end{eqnarray*}
where we used Kummer’s Theorem for the second equality, and the Duplication Formula of Gamma functions for the last equality.$\square$

(iii) Setting $b=1-a$ in the equation (1), we have
\begin{eqnarray*}
{}_2 F_1 \left(a,1-a;c;\frac{1}{2}\right)
&=&
2^a {}_2 F_1 (a, c+a-1; c; -1)\\
&=&
2^a {}_2 F_1 (c+a-1,a; c; -1)\\
&=&
2^a \frac{\Gamma(c)\Gamma(\frac{c+a+1}{2})}{\Gamma(c+a)\Gamma(\frac{c-a+1}{2})}\\
&=&
\frac{\Gamma(\frac{1}{2}c)\Gamma(\frac{c+1}{2})}{\Gamma(\frac{c+a}{2})\Gamma(\frac{1+c-a}{2})}
\end{eqnarray*}
where we used Kummer’s Theorem for the third equality, and the Duplication Formula of Gamma functions twice for the last equality.$\square$

References.
[1] W. N. Bailey, Generalized Hypergeometric Series, Cambridge University Press, 1964, pp.10-11.
[2] Leun Kim, [Calculation 5] Gauss’s Summation Formula.
[3] Leun Kim, [Calculation 6] Kummer’s Theorem.

[Calculation 6] Kummer’s Theorem

 Theorem. (Kummer’s Theorem) $${}_2 F_1 (a,b;1+a-b;-1) = \frac{\Gamma(1+a-b)\Gamma\left(1+\frac{1}{2}a\right)}{\Gamma(1+a)\Gamma\left(1+\frac{1}{2}a-b\right)}$$

To prove Kummer’s theorem, we introduce the following lemma, which is called Kummer’s quadratic transformation:

 Lemma. (Kummer’s Quadratic Transformation) \tag{1} {}_2 F_1 (a,b;1+a-b;z) = (1-z)^{-a} {}_2 F_1 \left( \frac{1}{2}a, \frac{1}{2}+\frac{1}{2}a-b; 1+a-b; – \frac{4z}{(1-z)^2} \right) holds for $|z|<1$.

Proof of Lemma. Note that the right hand side of (1) is analytic inside the loop of the curve $|4z|=|1-z|^2$ which surrounds the origin as below.

Therefore the right hand side of (1) can be expanded in powers of $z$ when $|z|<3-2\sqrt 2$. Thus
\begin{eqnarray*}
\text{(rhs)}
&=&
(1-z)^{-a} \sum_{k=0}^\infty \frac{(\frac{1}{2}a)_k (\frac{1}{2} + \frac{1}{2}a-b)_k}{(1+a-b)_k k!}\left( -\frac{4z}{(1-z)^2}\right)^k\\
&=&
\sum_{k=0}^\infty \frac{(\frac{1}{2}a)_k (\frac{1}{2} + \frac{1}{2}a-b)_k (-4)^k}{(1+a-b)_k k!}z^k (1-z)^{-a-2k}\\
&=&
\sum_{k=0}^\infty \sum_{r=0}^\infty \frac{(\frac{1}{2}a)_k (\frac{1}{2} + \frac{1}{2}a-b)_k (-4)^k}{(1+a-b)_k k!} \frac{(a+2k)_r}{r!}z^{k+r}.
\end{eqnarray*}
Calculating the coefficient of $z^n$, we get
\tag{2}
\sum_{k=0}^n \frac{(\frac{1}{2}a)_k (\frac{1}{2} + \frac{1}{2}a-b)_k (-4)^k}{(1+a-b)_k k!} \frac{(a+2k)_{n-k}}{(n-k)!}
=
\frac{(a)_n}{n!}\sum_{k=0}^n \frac{(\frac{1}{2}+\frac{1}{2}a-b)_k (a+n)_k (-n)_k}{(1+a-b)_k (\frac{1}{2} + \frac{1}{2}a)_k k!},

where we applied the relations
$$\frac{1}{(n-k)!} = \frac{(-1)^k (-n)_k}{n!}$$
and
$$(a+2k)_{n-k} = \frac{(a)_n(a+n)_k}{4^k (\frac{1}{2}a)_k (\frac{1}{2} + \frac{1}{2}a)_k}.$$
Therefore using Saalschütz’s Theorem to the right hand side of (2), we obtain
\begin{eqnarray*}
\frac{(a)_n}{n!}\sum_{k=0}^n \frac{(\frac{1}{2}+\frac{a}{2}-b)_k (a+n)_k (-n)_k}{(1+a-b)_k (\frac{1}{2} + \frac{a}{2})_k k!}
&=&
\frac{(a)_n}{n!} {}_3 F_2 \left( \frac{1}{2} + \frac{a}{2} -b, a+n, -n; 1+a-b, \frac{1}{2} + \frac{a}{2};1 \right)\\
&=&
\frac{(a)_n}{n!} \frac{(\frac{1}{2}+\frac{1}{2}a)_n (a-b-n)_n}{(1+a-b)_n(\frac{1}{2}-\frac{1}{2}a-n)_n}\\
&=&
\frac{(a)_n (b)_n}{n!(1+a-b)_n}
\end{eqnarray*}
so that
$$\text{(rhs)} = \sum_{n=0}^\infty \frac{(a)_n (b)_n}{n!(1+a-b)_n} z^n = {}_2 F_1 (a,b;1+a-b;z),$$
which proves Lemma for $|z|<3-2\sqrt 2$, and the complete result follows by analytic continuation.$\square$

Proof of Theorem. Limiting $z\to -1$ in Lemma, we obtain
\begin{eqnarray*}
{}_2 F_1 (a,b;1+a-b;-1)
&=&
2^{-a} {}_2 F_1 \left( \frac{1}{2}a, \frac{1}{2}+\frac{1}{2}a-b; 1+a-b; 1\right)\\
&=&
2^{-a} \frac{\Gamma(1+a-b)\Gamma(\frac{1}{2})}{\Gamma(1+\frac{1}{2}a-b)\Gamma(\frac{1}{2}+\frac{1}{2}a)}\\
&=&
\frac{\Gamma(1+a-b)\Gamma\left(1+\frac{1}{2}a\right)}{\Gamma(1+a)\Gamma\left(1+\frac{1}{2}a-b\right)},
\end{eqnarray*}
where we used the Gauss Summation Formula for the second equality, and the Duplication Formula for the Gamma function:
$$\Gamma\left(\frac{1}{2} + \frac{1}{2}a\right) = \frac{2^{-a}\Gamma(\frac{1}{2})\Gamma(1+a)}{\Gamma(1+\frac{1}{2}a)}$$
for the last equality.$\square$

Errata : $(0,3-2\sqrt 2)$ should be replaced by $(3-2\sqrt 2,0)$.

[1] W. N. Bailey, Generalized Hypergeometric Series, Cambridge University Press, 1964, pp.9-10.
[2] Bruno Gauthier, A Proof of Kummer’s Theorem, 2008. (arXiv:math/9904061v2)
[3] Leun Kim, Saalschütz’s Theorem.
[4] Leun Kim, Gauss’s Summation Formula.

[Calculation 5] Gauss’s Summation Formula

 Theorem. (Gauss’s Summation Formula) For $\text{Re}c>\text{Re}b>0$, $${}_2 F_1 (a,b;c;1) = \frac{\Gamma(c) \Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}$$ holds.

Proof. We remember that the Euler Integral Representation for the hypergeometric function is
$$_2 F_1 (a,b;c;z) = \frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)} \int_0^1 \frac{t^{b-1} (1-t)^{c-b-1}}{(1-tz)^a}\,dt.$$
Taking the limit $z\to 1$ both sides, we obtain
\begin{eqnarray*}
_2 F_1 (a,b;c;1)
&=&
\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}
\int_0^1 t^{b-1} (1-t)^{c-a-b-1}\,dt\\
&=&
\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}
\frac{\Gamma(b)\Gamma(c-b-a)}{\Gamma(c-a)}\\
&=&
\frac{\Gamma(c) \Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)},
\end{eqnarray*}
which proves the theorem.$\square$

 Corollary. (Chu–Vandermonde Identity) $${}_2 F_1 (-n,b;c;1) = \frac{(c-b)_n}{(c)_n}$$

Proof. Taking $a = -n$ in Gauss’s Summation Formula, we have
$${}_2 F_1 (-n,b;c;1) = \frac{\Gamma(c)\Gamma(c-b+n)}{\Gamma(c+n)\Gamma(c-b)} = \frac{(c-b)_n}{(c)_n}.$$

References.
[1] en.wikipedia.org/wiki/Hypergeometric_function
[2] Leun Kim, Fundamentals of Hypergeometric Functions.