Using term-by-term differentiation, expand in a Laurent series valid for . Ok, so, suppose I get a few derivatives: ... There's infinitely many. What next?
Follow Math Help Forum on Facebook and Google+
Originally Posted by scorpion007 Using term-by-term differentiation, expand in a Laurent series valid for . Ok, so, suppose I get a few derivatives: ... There's infinitely many. What next? Pattern. Do you have to do it this way? Note that
I guess I have to, since that's what the question asks. I know the pattern is: but I don't know what to do with it.
Doing it the other way, I suppose what happens is: for . But the whole series is squared. How am I suppose to write out the first few terms?
Originally Posted by scorpion007 Doing it the other way, I suppose what happens is: for . But the whole series is squared. How am I suppose to write out the first few terms? Hint: ...
Ok, so do I subs that in one of the factors and write it in terms of d/dz ? I'm not sure what do with that...
Originally Posted by scorpion007 Ok, so do I subs that in one of the factors and write it in terms of d/dz ? I'm not sure what do with that... Adapt that.
I guess you want me to take the deriv of both sides? is that right? I could simplify this and all, but I don't quite see where this is going...
Originally Posted by Drexel28 Hint: ... Maybe it's easier to use Newton's binomial theorem to expand .
Ah, I figured it out guys. The example I needed was on this page: Pauls Online Notes : Calculus II - Power Series and Functions
View Tag Cloud