Show that In (A) = In ( (N!)² (2N-n)! (2N)! [(N-n)!]² ) = In (2-2) using Sterlings approx N! ~ Nin N-N where Ука (A) ~ N² (2N-2n)² N-2n (2N) N(N)2N-Zn = 22N-24 =2"?" 2N ↑ Prove this
Show that In (A) = In ( (N!)² (2N-n)! (2N)! [(N-n)!]² ) = In (2-2) using Sterlings approx N! ~ Nin N-N where Ука (A) ~ N² (2N-2n)² N-2n (2N) N(N)2N-Zn = 22N-24 =2"?" 2N ↑ Prove this
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
Prove. Show all steps to get to ln (2^(-2n))
![Show that
In (A) = In ( (N!)² (2N-n)!
(2N)! [(N-n)!]²
)
= In (2-2)
using Sterlings approx N! ~ Nin N-N
where
Ука
(A)
~ N² (2N-2n)² N-2n
(2N) N(N)2N-Zn
=
22N-24
=2"?"
2N
↑
Prove
this](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd085bffa-52c2-484f-adda-21395b3d213f%2F5f398709-0fca-4826-a764-ed50b64e0a18%2Ffaqpizk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Show that
In (A) = In ( (N!)² (2N-n)!
(2N)! [(N-n)!]²
)
= In (2-2)
using Sterlings approx N! ~ Nin N-N
where
Ука
(A)
~ N² (2N-2n)² N-2n
(2N) N(N)2N-Zn
=
22N-24
=2"?"
2N
↑
Prove
this
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