Introduction

Before we find out more about our topic, let's learn a little bit about the Gamma function and the Beta function.

The gamma function is defined as such:

$$\Gamma(z)=\int_{0}^{\infty}t^{z-1}e^{-t}dt$$

with integration by parts, we can see

$$\int_{0}^{\infty}t^{n-1}e^{-t}dt=n! (n\in\mathbb{N})$$

It is a natural extension of the factorial function's domain from natural numbers to the reals. For more, visit jongwhi's blogpost here

Gamma Function

Introduction 이 글에서는 감마 함수에 대해서 다루기는 하지만, 나중에 쓸 글에 선행되어야 할 지식만 언급을 할 것이기 때문에, 감마 함수의 성질이나 파생된 정리에 대해서는 일절 다루지 않는다. 필자는 감마 함수의 정의만 소개할 것이다. 감마 함수는 나중에 n 차원 구의 표면적을 구할 때 유용하게 사용되기 때문에 잘 알아두도록 하자. Definition 감마

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The Beta function is defined as such:
$$B(z_1,z_2)=\int_{0}^{1}t^{z_1-1}(1-t)^{z_2-1}dt$$

It has a very close connection with the Gamma function because of the following equality.
$$B(z_1,z_2)=\frac{\Gamma(z_1)\Gamma(z_2)}{\Gamma(z_1+z_2)}$$

Proof

$$\Gamma(z_1)\Gamma(z_2)= \int_0^{\infty} e^{-u} u^{z_1 - 1} du\cdot \int_0^{\infty} e^{-v} v^{z_2 - 1}dv $$

$$= \int_0^{\infty} \int_0^{\infty} e^{-u - v} u^{z_1 - 1} v^{z_2 - 1} du dv$$

let's make some change of variables

$$ u = st\quad v = s(1 - t) \Rightarrow u + v = s\quad \frac{u}{u+v} = t$$

$$= \int_0^{\infty} \int_0^1 e^{-s} (st)^{z_1 - 1} (s(1 - t))^{z_2 - 1} s dt ds$$

$$= \int_0^{\infty} e^{-s} s^{z_1 + z_2 - 1} ds \cdot \int_0^1 t^{z_1 - 1} (1 - t)^{z_2 - 1} dt$$

$$= \Gamma(z_1 + z_2) \cdot B(z_1, z_2)$$

$$\therefore \quad B(z_1, z_2) = \frac{\Gamma(z_1)\Gamma(z_2)}{\Gamma(z_1+z_2)}$$

To start the proof, let's look back at how the 3-dimensional ball volume formula is obtained.

$$V_3(R)=\int_{-R}^{R}\pi(\sqrt{1-r^2})^2dr=\int_{-R}^{R}V_2(\sqrt{1-r^2})dr$$

Believing that this kind of extension works for all dimensions,

$$V_n(R)=\int_{-R}^{R} V_{n-1}(\sqrt{R^2-r^2})dr $$
With this in hand, let's truly begin the proof.

Let's use mathematical induction to prove it
when n=1

$$\rightarrow V_n(Rt)=V_1(Rt)=2Rt=V_1(t)R $$
let us hypothesize when n=k
$$V_k(Rt)=R^kV_{k}(t)$$
Then, when n=k+1
$$ \Rightarrow V_{k+1}(Rt)=\int_{-Rt}^{Rt}V_{k}(\sqrt{(Rt)^2-x^2})dx $$

let us make the substitution x=Rs

$$=\int_{-t}^{t}V_{k}(R\sqrt{t^2-s^2})Rds$$

$$=\int_{-t}^{t}R^kV_{k}(\sqrt{t^2-s^2})Rds$$

$$=R^{k+1}\int_{-t}^{t}V_{k}(\sqrt{t^2-s^2})ds$$
$$=R^{k+1}V_{k+1}(t)$$
therfore, by induction

$$V_n(Rt)=R^nV_n(t)$$

so when t=1, the lemma holds.

$$V_n(1)=\int_{-1}^{1}V_{n-1} (\sqrt{1^2-t^2})dt$$

using Lemma 1

$$=V_{n-1}(1)\int_{-1}^{1} (\sqrt{1^2-t^2})^{n-1}dt $$

$$=2V_{n-1}(1)\int_{0}^{1} (\sqrt{1^2-t^2})^{n-1}dt $$

making the substitution t=u^1/2

$$=V_{n-1}(1)\int_{0}^{1} (1-u)^\frac{n-1}{2}u^{-\frac{1}{2}} du $$

$$=V_{n-1}(1)B(\frac{1}{2},\frac{n+1}{2}) $$

$$=V_{n-1}(1) \frac{\Gamma(\frac{1}{2}) \Gamma(\frac{n+1}{2})}{\Gamma(\frac{n+2}{2})}$$

therefore, using Lemma 1 and Lemma 2

$$ V_n(R)=R^nV_n(1)$$

$$=R^nV_{n-1}(1)\frac{\Gamma(\frac{1}{2}) \Gamma(\frac{n+1}{2})}{\Gamma(\frac{n+2}{2})}$$

$$=R^nV_{n-2}(1)\frac{\Gamma(\frac{1}{2}) \Gamma(\frac{n+1}{2})}{\Gamma(\frac{n+2}{2})}\frac{\Gamma(\frac{1}{2}) \Gamma(\frac{n}{2})}{\Gamma(\frac{n+1}{2})}$$

$$\cdots$$

$$ =R^n(\Gamma(1/2))^n \prod_{k=0}^n\frac{\Gamma(\frac{k+1}{2})}{\Gamma(\frac{k+2}{2})}$$

using the telecscoping method and since Γ(1/2)=π^1/2

$$= {\frac {\pi ^{n/2}}{\Gamma {\bigl (}{\tfrac {n}{2}}+1{\bigr )}}}R^{n}$$

$$\blacksquare$$

lastly to end with a fun quiz,

Conclusion

To sum stuff up, Jongwhi did the proof for the surface area of an n-dimensional sphere which by integrating, you can get the volume of an n-dimensional sphere. The proof that he showed is quite interesting, so I recommend you check it out if you enjoyed this blogpost.

Surface Area of n-sphere

Introduction 이 글에서는 n 차원 초구(hypersphere)의 겉부피를 구해볼 것이다. 나중에 열 및 통계 역학에서 나오는 식을 유도할 때 사용되기 때문에 잘 알아두도록 하자. 일단 n 차원 구의 겉부피는 다음과 같다. 💡 \[S_{n-1} = \frac{2\pi^{\frac{n}{2}}}{\Gamma\left(\frac{n}{2}\right)} R^{n-1} \] 이때, \(S_

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Thank you for reading!