Hey mates, another Series question. Will ask on the next post
\[f(x) =(1-3x)^{-5}\]\[f(x)=\ln{(1-x)}\]Find the MacLaurin representation of both functions and their respective radius of convergence. The first one I am pretty much stuck. The second one I tried to use e^x expansion, but it became way to messy. Any ideas, mates? Thanks in advance
the second one has a neat trick; take the derivaitve and long divide it; then integrate the long dividion to get back to ln(1-x)
If I may ask a personal question you don't have to reply but where are you from, I have not met anyone that says mates besides someone from Australia.
-1/(1-x) -1-x-x^2-x^3-x^4.... -------------- 1-x) -1 (-1+x) ------ -x (-x+x^2) --------- -x^2 \[\int -1-x-x^2-x^3-x^4....dx=-x-x^2/2-x^3/3-x^4/4-x^5/5....\]
\[ln(1-x)=\sum_{n=0}^{inf} \frac{-x^{n+1}}{n+1}\]
@ParisWinter Actually, I am from Brazil, so English is not my native language. I tend to use a bit of old of style words because I generally don't like slangs. :-)
Oh that's awesome!!!! I tell people all the time if i could go anywhere i would want to go to Brazil!!!!!
@amistre64 Thank you very much. Gonna try it myself. I am having issues conceptualizing these series expasions, but, well, have to deal with it. @ParisWinter It's a nice country, but very hot, even in the Southern part of it, haha.
\[\lim_{n\to inf} \frac{-x^{n+1}}{n+1}\frac{n}{-x^{n}}\] \[\lim_{n\to inf} \frac{nx}{n+1}\] \[|x|\ \lim_{n\to inf} \frac{n}{n+1}\] \[|x|*1\]
|x|<1 defines the radius of convergence i believe
or the 1 is the radius :)
True!!! I am from Florida though it gets pretty hot but probably not as hot as Brazil!!!!
@amistre64 Indeed, I got 1 for the radius, but because I cheated and looked up the series expansion. At least, I got part of it right. Anyway, I can't work it out from e^x expansion? It seemed more natural, yet very messy. @ParisWinter Ah, Florida. I think that there are a lot of Brazilian in Miami nowadays.
I wouldn't know I haven't been to Miami, I live in Ocala which is in central Florida. I would love to go though very soon!
@ParisWinter Isn't Ocala near Gainesville and McIntosh?
Yes it is!!!
I see. :-) Well, at least it's more or less near Tampa and Daytona. Anyway, gotta study for my exams. See you around, @ParisWinter And once again, thank you very much @amistre64 Will try to fiddle around the series a bit more.
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