Question

Hi, I am trying to learn how a power series can be inverted. I was following the example from this website:
http://planetmath.org/encyclopedia/SeriesInversion.html

However I cannot understand how the author got from Equation (3) to Equation (4) and from Equation (4) to Equation (5).

Can someone please explain it to me or maybe suggest other techniques and resources?

Answer 1

i think from 3 to 4 you expand the cube

Answer 2

you get \(w^3+O(z^2)(w^2+\text{lower terms}\))

Answer 3

well actually you get some coefficients which are irrelevant because it is asymptotic

Answer 4

then \(w^2+\text{lower terms} \) is \(O(w^2)\)

Answer 5

but doesn't expanding the cube yield to \[w^3 + O(z^3)w^2 + O(z^3)^2w + O(z^3)^3\] Why is the power in O(z^3) ignored?

Answer 6

i am lost after that, because i am not sure what all this stuff is
need to read it carefully
but there is is an explanation to go from 4 to 5

Answer 7

Ok, I think I get the step from 3 to 4, because the coefficients which are \[1, O(z^3), O(z^3)^2\] can be ignored since we are looking at 'w' as a variable

Answer 8

oh good point. i guess we need to look more carefully at the big O notation and find out

Answer 9

Yeah the step from 4 to 5 looses me completely..

Answer 10

ok no i don't really see it and we are expanding power series in any case, so i guess it is a bit more complicated

Answer 11

oh now i think i do see it. we are expanding about 0 so the higher powers are not involved

Answer 12

from 4 to 5 it looks like they have simply replaced \(z\) by \(w\)

Answer 13

the reason they give is that the constant is 0, so the first term in the expansion of \(z\) must begin with a \(w\) and so \(O(z^3)\) must be \(O(w^3)\) and also \(z^5=O(w^3)\)
why i do not know

Answer 14

oops i meant \(z^5=O(w^5)\) of course

Answer 15

i should really be quiet because i do not know this and have nor read carefully, but i am pretty sure about the last reason

Answer 16

oh ok I see!
Thanks, I can see it a bit clearer now. I will continue reading about it tomorrow cos I think my mind needs a bit of rest.
Thanks for helping!
If I still find it confusing I will let you know.

Answer 17

good luck
a quick google search shows that this gets complicated rather fast, but it is a computation