Question

State and prove Legendre's Duplication formula.
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Answer 1

The duplication formula can be written as

Γ(x)Γ(x+12)Γ(2x)=Γ(12)22x−1=π−−√22x−1.

We want to derive this formula using the Weierstrass definition for the gamma function,

1Γ(x)=xeγx∏k=1∞(1+xk)e−x/k.

We have

Γ(x)Γ(x+12)Γ(2x)=2xe2γxxeγx(x+12)eγxeγ/2∏∞k=1(1+2xk)e−2x/k∏∞k=1(1+xk)e−x/k∏∞k=1(1+xk+12k)e−x/ke−1/2k=1eγ/2limn→∞2x∏2nk=1(1+2xk)x(x+12)∏nk=1(1+xk)∏nk=1(1+xk+12k)∏2nk=1e−2x/k(∏nk=1e−x/k)2∏nk=1e−1/2k=1eγ/2limn→∞Pn(x)Qn(x).

First simplify Pn(x) as follows:

Pn(x)=2x∏2nk=1(1+2xk)x(x+12)∏nk=1(1+xk)∏nk=1(1+xk+12k)=(n!)2(2n)!(x+n+12)∏nk=0(2x+2k)∏n−1k=0(2x+2k+1)∏nk=0(x+k)∏n−1k=0(x+k+12)=(n!)222n+1(2n)!(x+n+12)

Next consider Qn(x):

Qn(x)=∏2nk=1e−2x/k(∏nk=1e−x/k)2∏nk=1e−1/2k=n1/222x(2n)2x∏2nk=1e−2x/k(nx∏nk=1e−x/k)2n1/2∏nk=1e−1/2k

Reassembling we get

Γ(x)Γ(x+12)Γ(2x)=1eγ/2limn→∞(n!)222n+1(2n)!(x+n+12)n1/222x(2n)2x∏2nk=1e−2x/k(nx∏nk=1e−x/k)2n1/2∏nk=1e−1/2k=122x−1limn→∞n(x+n+12)(n!)222n(2n)!n1/2(2n)2x∏2nk=1e−2x/keγ/2(nx∏nk=1e−x/k)2n1/2∏nk=1e−1/2k.

We can evaluate the limit in three parts

First,

limn→∞n(x+n+12)=1.

Second, using a well-known identity for the Euler-Mascheroni constant,

limn→∞(2n)2x∏2nk=1e−2x/keγ/2(nx∏nk=1e−x/k)2n1/2∏nk=1e−1/2k=e−2γx(e−γx)2e−γ/2eγ/2=1.

Third using Stirlings's asymptotic formula n!∼2π−−√nn+1/2e−n,

limn→∞(n!)222n(2n)!n1/2=π−−√,

and finally we get

Γ(x)Γ(x+12)Γ(2x)=π−−√22x−1.

HENCE PROVED ......MARK AS BRAINLIEST

Answer 2

Answer:

prove of duplication formula