Looking for help understanding the asymptotic expansion of

Overath097p

Overath097p

Answered question

2022-03-21

Looking for help understanding the asymptotic expansion of the digamma function
I was recently given an example using this asymptotic expansion of the digamma function where:
ddx(lnΓ(x))=ψ(x)ln(x)12x112x2
Here's the example:
ψ(x4)4ψ(x5+12)5ψ(x20+12)20ln(4)4+ln(5)5+ln(20)2012x118x2
I'm unclear on the following points:
What happened to each x term?
Why is the first term negative and the rest of the terms positive? Why isn't the signs of the original terms would be preserved?
I would have expected something like this:
ln(x4)4ln(x4)4ln(x4)4
How is 118x2 being determined? Why does 112x2 change but 12x stays the same?
Sorry for the elementary questions. The explanation will really help! :-)
Thanks,

Answer & Explanation

pautmndu

pautmndu

Beginner2022-03-22Added 10 answers

The log(x) terms cancel because 1415120=0 In somewhat more detail,
ψ(x/4)4=-12ln(2)+14ln(x)-12x-13x2+O(x-4)
ψ(x/5+1/2)5=-15ln(5)+15ln(x)+524x2+O(x-4)
ψ(x20+12)20=120ln(20)+120ln(x)+56x2+O({x}4)

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