Proof of euler reflection formula

Here's another proof of Euler reflection formula :

Lemma1:Lemma1\quad :

P(n):i=1n1(xai)=i=1n1(xai)j=1,jin1(aiaj) P(n) : \displaystyle \prod _{ i=1 }^{ n }{ \dfrac { 1 }{ (x-{ a }_{ i }) } } =\sum _{ i=1 }^{ n }{ \dfrac { 1 }{ (x-{ a }_{ i }) } \prod _{ j=1,j\neq i }^{ n }{ \dfrac { 1 }{ ({ a }_{ i }-{ a }_{ j }) } } } for all nn belonging to NN

Proof :

I will prove it by induction :

It can be easily verified that P(1)P(1) is true :

Assume P(n)P(n) is true :

than i=1n1(xai)=i=1n1(xai)j=1,jin1(aiaj)\displaystyle \prod _{ i=1 }^{ n }{ \dfrac { 1 }{ (x-{ a }_{ i }) } } =\sum _{ i=1 }^{ n }{ \dfrac { 1 }{ (x-{ a }_{ i }) } \prod _{ j=1,j\neq i }^{ n }{ \dfrac { 1 }{ ({ a }_{ i }-{ a }_{ j }) } } }

Multiplying both sides with 1(xan+1)\dfrac{1}{(x-{a}_{n+1})} we have :

i=1n+11(xai)=i=1n1(xai)(1(xan+1)j=1,jin1(aiaj)) \displaystyle \prod _{ i=1 }^{ n+1 }{ \dfrac { 1 }{ (x-{ a }_{ i }) } } =\sum _{ i=1 }^{ n }{ \dfrac { 1 }{ (x-{ a }_{ i }) } (\dfrac { 1 }{ (x-{ a }_{ n+1 }) } \prod _{ j=1,j\neq i }^{ n }{ \dfrac { 1 }{ ({ a }_{ i }-{ a }_{ j }) } ) } }

It is easy to check that :

1(xai)1(xan+1)=1(aian+1)1(xai)+1(an+1ai)1(xan+1) \dfrac { 1 }{ (x-{ a }_{ i }) } \dfrac { 1 }{ (x-{ a }_{ n+1 }) } =\dfrac { 1 }{ ({ a }_{ i }-{ a }_{ n+1 }) } \dfrac { 1 }{ (x-{ a }_{ i }) } +\dfrac { 1 }{ ({ a }_{ n+1 }-{ a }_{ i }) } \dfrac { 1 }{ (x-{ a }_{ n+1 }) }

Using this we have :

i=1n+11(xai)=i=1n1(xai)j=1,jin+11(aiaj)+1(xan+1)i=1n1(an+1ai)j=1,jin1aiaj \displaystyle \prod _{ i=1 }^{ n+1 }{ \dfrac { 1 }{ (x-{ a }_{ i }) } } =\sum _{ i=1 }^{ n }{ \dfrac { 1 }{ (x-{ a }_{ i }) } \prod _{ j=1,j\neq i }^{ n+1 }{ \dfrac { 1 }{ ({ a }_{ i }-{ a }_{ j }) } } } +\dfrac { 1 }{ (x-{ a }_{ n+1 }) } \sum _{ i=1 }^{ n }{ \dfrac { 1 }{ { (a }_{ n+1 }-{ a }_{ i }) } } \prod _{ j=1, j \neq i }^{ n }{ \dfrac { 1 }{ { a }_{ i }-{ a }_{ j } } }

Now using P(n) we can say that :

i=1n1(an+1ai)j=1,jin1aiaj=i=1n1(an+1ai)=i=1,in+1n+11(an+1ai) \displaystyle \sum _{ i=1 }^{ n }{ \dfrac { 1 }{ { (a }_{ n+1 }-{ a }_{ i }) } } \prod _{ j=1,j\neq i}^{ n }{ \dfrac { 1 }{ { a }_{ i }-{ a }_{ j } } } =\prod _{ i=1 }^{ n }{ \dfrac { 1 }{ ({ a }_{ n+1 }-{ a }_{ i }) } } =\prod _{ i=1,i\neq n+1\quad }^{ n+1 }{ \dfrac { 1 }{ ({ a }_{ n+1 }-{ a }_{ i }) } }

Using this in our upper result we have :

i=1n+11(xai)=i=1n1(xai)j=1,jin+11(aiaj)+1(xan+1)j=1,jn+1n+11an+1aj=i=1n+11(xai)j=1,jin+11(aiaj) \displaystyle \prod _{ i=1 }^{ n+1 }{ \dfrac { 1 }{ (x-{ a }_{ i }) } } =\sum _{ i=1 }^{ n }{ \dfrac { 1 }{ (x-{ a }_{ i }) } \prod _{ j=1,j\neq i }^{ n+1 }{ \dfrac { 1 }{ ({ a }_{ i }-{ a }_{ j }) } } } +\dfrac { 1 }{ (x-{ a }_{ n+1 }) } \prod _{ j=1,j\neq n+1 }^{ n+1 }{ \dfrac { 1 }{ { a }_{ n+1 }-{ a }_{ j } } } =\sum _{ i=1 }^{ n+1 }{ \dfrac { 1 }{ (x-{ a }_{ i }) } \prod _{ j=1,j\neq i }^{ n+1 }{ \dfrac { 1 }{ ({ a }_{ i }-{ a }_{ j }) } } }

Hence P(n)P(n) is true implies P(n+1)P(n+1) is true implies P(n)P(n) is true for all nn belongs to NN

Lemma2:Lemma2 \quad :

j=1,jin1(aiaj)=1P(ai) \displaystyle \prod _{ j=1,j\neq i }^{ n }{ \dfrac { 1 }{ ({ a }_{ i }-{ a }_{ j }) } } =\dfrac { 1 }{ P'({ a }_{ i }) } , where P(x)P(x) is a monic polynomial having roots {ak{a}_{k}}, for 1kn 1 \leq k \leq n

Proof : P(x)=j=1n(xaj) \displaystyle P(x) = \prod _{j=1}^{ n }{ (x-{ a }_{ j }) }

Dividing both sides by (xai)(x-{a}_{i}) and taking limit xai x \rightarrow {a}_{i} we have :

limxaiP(x)xai=j=1,jin(aiaj) \displaystyle \lim _{ x\rightarrow { a }_{ i } }{ \dfrac { P(x) }{ x-{ a }_{ i } } } =\prod _{ j=1,j\neq i }^{ n }{ ({ a }_{ i }-{ a }_{ j }) }

Applying L-Hopitals rule we have our result :

Now we start with the integral :

I(n)=0dx1+xn \displaystyle I(n) = \int _{ 0 }^{ \infty }{ \dfrac { dx }{ 1+{ x }^{ n } } }

Let it's roots be {ak{a}_{k}}, for 1kn 1 \leq k \leq n

Then we have :

I(n)=0j=1n1(xaj)dx \displaystyle I(n) = \int _{ 0 }^{ \infty }{ \prod _{ j=1 }^{ n }{ \dfrac { 1 }{ (x-{ a }_{ j }) } }dx }

Applying lemma 1 we have :

I(n)=0i=1n1(xai)j=1,jin1(aiaj)dx \displaystyle I(n) = \int _{ 0 }^{ \infty }{ \sum _{ i=1 }^{ n }{ \dfrac { 1 }{ (x-{ a }_{ i }) } \prod _{ j=1,j\neq i }^{ n }{ \dfrac { 1 }{ ({ a }_{ i }-{ a }_{ j }) } } } dx }

Applying lemma 2 we have :

I(n)=0i=1n1(xai)1nain1dx\displaystyle I(n) = \int _{ 0 }^{ \infty }{ \sum _{ i=1 }^{ n }{ \dfrac { 1 }{ (x-{ a }_{ i }) } \dfrac { 1 }{ n{ a }_{ i }^{ n-1 } } } dx }

Using the fact that ain=1{a}_{i}^{n}=-1 we have :

1n0i=1nai(xai)dx\displaystyle -\dfrac { 1 }{ n } \int _{ 0 }^{ \infty }{ \sum _{ i=1 }^{ n }{ \dfrac { { a }_{ i } }{ (x-{ a }_{ i }) } } dx }

Changing the order of integration and summation and integrating it we have :

I(n)=1nlimpi=1nailn(xai)p0 \displaystyle I(n) = \dfrac { 1 }{ n } \lim _{ p\rightarrow \infty }{ \sum _{ i=1 }^{ n }{ { a }_{ i }ln(x-{ a }_{ i }) } { | }_{ p }^{ 0 } }

I(n)=1n(i=1nailn(ai)limpi=1nln(pai)) \displaystyle I(n) = \dfrac { 1 }{ n } (\sum _{ i=1 }^{ n }{ { a }_{ i }ln(-{ a }_{ i }) } -\lim _{ p\rightarrow \infty }{ \sum _{ i=1 }^{ n }{ ln(p-{ a }_{ i }) } } )

Now limpi=1nailn(pai)=limpln(p)i=1naii=1nailn(1aip)=0 \displaystyle \lim _{ p\rightarrow \infty }{ \sum _{ i=1 }^{ n }{ { a }_{ i }ln(p-{ a }_{ i }) } } =\lim _{ p\rightarrow \infty }{ ln(p)\sum _{ i=1 }^{ n }{ { a }_{ i } } -\sum _{ i=1 }^{ n }{ { a }_{ i }ln(1-\dfrac { { a }_{ i } }{ p } ) } } =0

Here we have used the fact that sum of roots is 0 and applied the limit.

Now the value of our integral becomes :

I(n)=1n(i=1nailn(ai))=1n(i=1nailn(ai)+ln(1)i=1nai)=1n(i=1nailn(ai))\displaystyle I(n) = \dfrac { 1 }{ n } (\sum _{ i=1 }^{ n }{ { a }_{ i }ln(-{ a }_{ i }) } )=\dfrac { 1 }{ n } (\sum _{ i=1 }^{ n }{ { a }_{ i }ln({ a }_{ i }) } +ln(-1)\sum _{ i=1 }^{ n }{ { a }_{ i } } )=\dfrac { 1 }{ n } (\sum _{ i=1 }^{ n }{ { a }_{ i }ln({ a }_{ i }) } )

Using the general form of root ak=e((2k1)πi2) \large {a}_{k} = { e }^{ \left({ \frac { (2k-1)\pi i }{ 2 } }\right) }

Our integral becomes :

I(n)=1n(k=1n(2k1)πine(2k1)πin)=2πin2eπink=1nke2πikn \displaystyle I(n) = \dfrac { 1 }{ n } \left(\sum _{ k=1 }^{ n }{ \frac { (2k-1)\pi i }{ n } { e }^{ \frac { (2k-1)\pi i }{ n } } } \right)= \large \dfrac { 2\pi i }{ { n }^{ 2 }{ e }^{ \frac { \pi i }{ n } } } \sum _{ k=1 }^{ n }{ k{ e }^{ \frac { 2\pi ik }{ n } } }

Here I have again used the fact that sum of roots is 0.

Using the concept of Arithmetic-Geometric series it can be easily shown that :

k=1nkxk=x(xn1)(1x)2nxn+11x \displaystyle \sum _{ k=1 }^{ n }{ k{ x }^{ k } } =\dfrac { x({ x }^{ n }-1) }{ { (1-x) }^{ 2 } } -\dfrac { n{ x }^{ n+1 } }{ 1-x }

Using this we have :

I(n)=2πineπin(e2πine2πin1) I(n) = \large \dfrac { 2\pi i }{ n{ e }^{ \frac { \pi i }{ n } } } \left(\dfrac { { e }^{ \frac { 2\pi i }{ n } } }{ { e }^{ \frac { 2\pi i }{ n } }-1 } \right)

Simplifying it we get :

I(n)=πncscπn \displaystyle I(n) = \dfrac { \pi }{ n } \csc { \dfrac { \pi }{ n } }

Again back to our integral :

I(n)=0dx1+xn \displaystyle I(n) = \int _{ 0 }^{ \infty }{ \dfrac { dx }{ 1+{ x }^{ n } } }

Use the substitution y=dx1+xn y = \dfrac { dx }{ 1+{ x }^{ n } } , we have :

I(n)=1n01y1n(1y)1n1dy \displaystyle I(n) = \dfrac { 1 }{ n } \int _{ 0 }^{ 1 }{ { y }^{ \frac { -1 }{ n } }{ (1-y) }^{ \frac { 1 }{ n } -1 }dy }

By defintion of beta function this evaluates to :

I(n)=1nΓ(1n)Γ(11n)Γ(1) \displaystyle I(n) = \dfrac { 1 }{ n } \dfrac { \Gamma \left(\dfrac { 1 }{ n } \right)\Gamma \left(1-\dfrac { 1 }{ n } \right) }{ \Gamma (1) }

Comparing our two results we have :

1nΓ(1n)Γ(11n)=πncscπn \dfrac { 1 }{ n } \Gamma \left(\dfrac { 1 }{ n } \right)\Gamma \left(1-\dfrac { 1 }{ n } \right)=\dfrac { \pi }{ n } \csc { \dfrac { \pi }{ n } }

I have used Γ(1)=1 \Gamma(1)=1

Finally we have :

Γ(1n)Γ(11n)=πcscπn \Gamma \left(\dfrac { 1 }{ n } \right)\Gamma \left(1-\dfrac { 1 }{ n } \right)=\pi \csc { \dfrac { \pi }{ n } }

Take 1n=x \dfrac{1}{n} = x to get :

Γ(x)Γ(1x)=πcscπx \Gamma (x)\Gamma (1-x)=\pi \csc { \pi x }

HENCEPROVED \LARGE HENCE \quad PROVED

Sorry for the fact that the proof looks too messy, and that I have jumped many steps. It's largely based on manipulations on complex numbers.

Note by Ronak Agarwal
3 years, 10 months ago

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Comments

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Great! Can you put that inside here?

Pi Han Goh - 3 years, 10 months ago

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Nice. And I really appreciate your patience while writing LATEXLATEX

Pranjal Prashant - 3 years, 10 months ago

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Thanks, how's the proof.

Ronak Agarwal - 3 years, 10 months ago

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It's really nice . And since I am talking to jee topper and I am aspirant for JEE 2016 , I would like to ask the same question, which books do you referred to in Inorganic , Is it all about just remembering things up or analysing the concept of reaction mechanism ?

Pranjal Prashant - 3 years, 10 months ago

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Isn't there a proof for lemma 1 which doesn't use induction?

Aditya Kumar - 3 years, 10 months ago

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Note that in Lemma 1 after multiplying both sides by i=1n(xai)\displaystyle \prod _{ i=1 }^{ n }{ (x-{ a }_{ i }) } and subtracting 11 from both sides, the polynomial generated on R.H.S. is of n1n-1 degrees, but has nn roots. Thus it is an identity. See Lagrange's Interpolation Polynomial.

Ishan Singh - 3 years, 10 months ago

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Thanks!

Aditya Kumar - 3 years, 10 months ago

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You can try that for yourself.

Ronak Agarwal - 3 years, 10 months ago

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It is obvious if you use covering lemma

Praneeth Kacham - 3 years, 9 months ago

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Did you mean Γ(1)=1\Gamma(1)=1?

Also, your prove only shows that the Euler reflection formula is valid for x=1nx=\frac{1}{n}, where n is an integer more than 0.

Great prove though.

Julian Poon - 3 years, 10 months ago

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I know, I am trying to generalize this.

Ronak Agarwal - 3 years, 10 months ago

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Can you explain how you got to that step after you wrote 'using this we have' in the first lemma?

donkey kong - 3 years, 9 months ago

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