I need to find the principal part of the Laurent series of the following function:
1/(e^z - 1)
I found the power series for (e^z - 1) = z + z²/2! + z³/3! + ...
Can I manipulate 1 / (z + z²/2! + z³/3! + ...) to figure out the principal part (negative values) of the Laurent series? I'm clearly failing at algebra tonight. Any help is appreciated!
1/(e^z - 1)
I found the power series for (e^z - 1) = z + z²/2! + z³/3! + ...
Can I manipulate 1 / (z + z²/2! + z³/3! + ...) to figure out the principal part (negative values) of the Laurent series? I'm clearly failing at algebra tonight. Any help is appreciated!
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Due to z being the term of smallest positive degree, we see that the principal part of 1/(e^z - 1) is 1/z.
Why?
If 1/(e^z - 1) = 1/z + A + Bx + ..., then
(e^z - 1) (1/z + A + Bx + ...) = 1
==> (1/z + A + Bx + ...)(z + z²/2! + z³/3! + ...) = 1
==> 1 + (constant term and terms of higher degree) = 1, via distributive law.
(If needed, we see that the constant term and beyond must equal 0; so we could solve for A, B, etc. as seen fit.)
I hope this helps!
Why?
If 1/(e^z - 1) = 1/z + A + Bx + ..., then
(e^z - 1) (1/z + A + Bx + ...) = 1
==> (1/z + A + Bx + ...)(z + z²/2! + z³/3! + ...) = 1
==> 1 + (constant term and terms of higher degree) = 1, via distributive law.
(If needed, we see that the constant term and beyond must equal 0; so we could solve for A, B, etc. as seen fit.)
I hope this helps!