Gamma reflection formula proof
$begingroup$
For $xinmathbb{C}setminus0,-1,...$ If we know that $$Gamma(x)=lim_{nto +infty} frac{n^x.n!}{prod_{k=0}^n(k+x)}$$
and $$sin(pi x)=pi xprod_{k=1}^{infty}left(1- frac{x^2}{k^2}right)$$
Then we have
$$ Gamma(x)Gamma(1-x)=lim_{nto +infty} frac{n^x.n!.n^{1-x}.n!}{prod_{k=0}^n(k+x)(k+1-x)} =frac1xlim_{nto +infty} frac{n}{n+1-x}prod_{k=1}^nfrac{k^2}{k^2-x^2}=$$$$=frac1xleft(prod_{k=1}^{infty}left(1- frac{x^2}{k^2}right)right)^{-1}=frac{pi}{sin(pi x)} $$
Is this a correct way to prove the famous Euler's reflection formula ?
proof-verification special-functions gamma-function
edited Dec 19 '18 at 1:46
Yadati Kiran
2,1261622
asked Dec 16 '17 at 4:38
KamouloxKamoulox
185
$endgroup$
add a comment |
$begingroup$
For $xinmathbb{C}setminus0,-1,...$ If we know that $$Gamma(x)=lim_{nto +infty} frac{n^x.n!}{prod_{k=0}^n(k+x)}$$
and $$sin(pi x)=pi xprod_{k=1}^{infty}left(1- frac{x^2}{k^2}right)$$
Then we have
$$ Gamma(x)Gamma(1-x)=lim_{nto +infty} frac{n^x.n!.n^{1-x}.n!}{prod_{k=0}^n(k+x)(k+1-x)} =frac1xlim_{nto +infty} frac{n}{n+1-x}prod_{k=1}^nfrac{k^2}{k^2-x^2}=$$$$=frac1xleft(prod_{k=1}^{infty}left(1- frac{x^2}{k^2}right)right)^{-1}=frac{pi}{sin(pi x)} $$
Is this a correct way to prove the famous Euler's reflection formula ?
proof-verification special-functions gamma-function
edited Dec 19 '18 at 1:46
Yadati Kiran
2,1261622
asked Dec 16 '17 at 4:38
KamouloxKamoulox
185
$endgroup$
add a comment |
$begingroup$
For $xinmathbb{C}setminus0,-1,...$ If we know that $$Gamma(x)=lim_{nto +infty} frac{n^x.n!}{prod_{k=0}^n(k+x)}$$
and $$sin(pi x)=pi xprod_{k=1}^{infty}left(1- frac{x^2}{k^2}right)$$
Then we have
$$ Gamma(x)Gamma(1-x)=lim_{nto +infty} frac{n^x.n!.n^{1-x}.n!}{prod_{k=0}^n(k+x)(k+1-x)} =frac1xlim_{nto +infty} frac{n}{n+1-x}prod_{k=1}^nfrac{k^2}{k^2-x^2}=$$$$=frac1xleft(prod_{k=1}^{infty}left(1- frac{x^2}{k^2}right)right)^{-1}=frac{pi}{sin(pi x)} $$
Is this a correct way to prove the famous Euler's reflection formula ?
proof-verification special-functions gamma-function
edited Dec 19 '18 at 1:46
Yadati Kiran
2,1261622
asked Dec 16 '17 at 4:38
KamouloxKamoulox
185
$endgroup$
For $xinmathbb{C}setminus0,-1,...$ If we know that $$Gamma(x)=lim_{nto +infty} frac{n^x.n!}{prod_{k=0}^n(k+x)}$$
and $$sin(pi x)=pi xprod_{k=1}^{infty}left(1- frac{x^2}{k^2}right)$$
Then we have
$$ Gamma(x)Gamma(1-x)=lim_{nto +infty} frac{n^x.n!.n^{1-x}.n!}{prod_{k=0}^n(k+x)(k+1-x)} =frac1xlim_{nto +infty} frac{n}{n+1-x}prod_{k=1}^nfrac{k^2}{k^2-x^2}=$$$$=frac1xleft(prod_{k=1}^{infty}left(1- frac{x^2}{k^2}right)right)^{-1}=frac{pi}{sin(pi x)} $$
Is this a correct way to prove the famous Euler's reflection formula ?
proof-verification special-functions gamma-function
proof-verification special-functions gamma-function
edited Dec 19 '18 at 1:46
Yadati Kiran
2,1261622
asked Dec 16 '17 at 4:38
KamouloxKamoulox
185
edited Dec 19 '18 at 1:46
Yadati Kiran
2,1261622
asked Dec 16 '17 at 4:38
KamouloxKamoulox
185
edited Dec 19 '18 at 1:46
Yadati Kiran
2,1261622
edited Dec 19 '18 at 1:46
Yadati Kiran
2,1261622
edited Dec 19 '18 at 1:46
Yadati Kiran
2,1261622
2,1261622
asked Dec 16 '17 at 4:38
KamouloxKamoulox
185
asked Dec 16 '17 at 4:38
KamouloxKamoulox
185
asked Dec 16 '17 at 4:38
KamouloxKamoulox
185
185
add a comment |
add a comment |
1 Answer
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votes
$begingroup$
Other than the fact that the formula for $$sin pi x = pi x , , prod_{k=1}^{infty} left(1-frac{x^2}{k^2}right)$$ should be corrected, I think your proof is fine!
answered Dec 16 '17 at 4:43
RohanRohan
27.8k42444
$endgroup$
$begingroup$
My bad.. Thanks!
$endgroup$
– Kamoulox
Dec 16 '17 at 4:49
$begingroup$
would the downvoter care to comment on what is wrong with this answer?
$endgroup$
– robjohn♦
Dec 16 '17 at 4:58
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Other than the fact that the formula for $$sin pi x = pi x , , prod_{k=1}^{infty} left(1-frac{x^2}{k^2}right)$$ should be corrected, I think your proof is fine!
answered Dec 16 '17 at 4:43
RohanRohan
27.8k42444
$endgroup$
$begingroup$
My bad.. Thanks!
$endgroup$
– Kamoulox
Dec 16 '17 at 4:49
$begingroup$
would the downvoter care to comment on what is wrong with this answer?
$endgroup$
– robjohn♦
Dec 16 '17 at 4:58
add a comment |
$begingroup$
Other than the fact that the formula for $$sin pi x = pi x , , prod_{k=1}^{infty} left(1-frac{x^2}{k^2}right)$$ should be corrected, I think your proof is fine!
answered Dec 16 '17 at 4:43
RohanRohan
27.8k42444
$endgroup$
$begingroup$
My bad.. Thanks!
$endgroup$
– Kamoulox
Dec 16 '17 at 4:49
$begingroup$
would the downvoter care to comment on what is wrong with this answer?
$endgroup$
– robjohn♦
Dec 16 '17 at 4:58
add a comment |
$begingroup$
Other than the fact that the formula for $$sin pi x = pi x , , prod_{k=1}^{infty} left(1-frac{x^2}{k^2}right)$$ should be corrected, I think your proof is fine!
answered Dec 16 '17 at 4:43
RohanRohan
27.8k42444
$endgroup$
Other than the fact that the formula for $$sin pi x = pi x , , prod_{k=1}^{infty} left(1-frac{x^2}{k^2}right)$$ should be corrected, I think your proof is fine!
answered Dec 16 '17 at 4:43
RohanRohan
27.8k42444
answered Dec 16 '17 at 4:43
RohanRohan
27.8k42444
answered Dec 16 '17 at 4:43
RohanRohan
27.8k42444
answered Dec 16 '17 at 4:43
RohanRohan
27.8k42444
27.8k42444
$begingroup$
My bad.. Thanks!
$endgroup$
– Kamoulox
Dec 16 '17 at 4:49
$begingroup$
would the downvoter care to comment on what is wrong with this answer?
$endgroup$
– robjohn♦
Dec 16 '17 at 4:58
add a comment |
$begingroup$
My bad.. Thanks!
$endgroup$
– Kamoulox
Dec 16 '17 at 4:49
$begingroup$
would the downvoter care to comment on what is wrong with this answer?
$endgroup$
– robjohn♦
Dec 16 '17 at 4:58
$begingroup$
My bad.. Thanks!
$endgroup$
– Kamoulox
Dec 16 '17 at 4:49
$begingroup$
My bad.. Thanks!
$endgroup$
– Kamoulox
Dec 16 '17 at 4:49
$begingroup$
would the downvoter care to comment on what is wrong with this answer?
$endgroup$
– robjohn♦
Dec 16 '17 at 4:58
$begingroup$
would the downvoter care to comment on what is wrong with this answer?
$endgroup$
– robjohn♦
Dec 16 '17 at 4:58
add a comment |
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