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anonymous
 4 years ago
how to take the derivative of series? for example ((1)^n+1)(x2)^n)/n
anonymous
 4 years ago
how to take the derivative of series? for example ((1)^n+1)(x2)^n)/n

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but you are a higher level than zarkon haha.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0is (1)^n+1(x2)^n1 right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes i believe so , let me write it so i don't do anything silly are we starting at n = 0 or n = 1?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes that is right. what you have

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.01. okay and then i got the interval of that to be 1<x<3, but i am having trouble with the convergence of course.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh but not at the endpoints of course

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0next question...can you explain how to find the power series of a function?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i can show you a trick for that one if you like

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes please! i have an exam tomorrow and i need as much help as possible.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well then i don't want to waste your time with this. but if you know that \[\frac{1}{1x}=\sum x^n\] then \[\frac{1}{1+x}=\sum (1)^nx^n\] and so \[\frac{1}{x1}=\frac{1}{1+(x2)}=\sum(1)^n(x2)^n\] so that is your derivative, and therefore your original series was the log

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but if you have an exam forget that mess

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and also your derivative will not convege at the endpoints, because there is not way that \[\sum 1^n\] or \[\sum (1)^n\] converges

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this is not wasting my time at alll. umm soo what about htis: use the binomial series to find the maclaurin series of the function f(x) = \[\sqrt[4]{1+x}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i guess you are supposed to write this as \[(1+x)^{\frac{1}{4}}\] first and then use "general binomial series

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0er \[1+\frac{1}{4}x+\frac{\frac{1}{4}1}{2!}x^2\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[+\frac{(\frac{1}{4}1)(\frac{1}{4}2)}{3!}x^3 +...\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0probably some algebra will turn up a nice pattern

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but i dont understand the maclaurin part?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that is the maclaurin series

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0expand about zero, get \[a_0+a_1x+a_2x^2+a_3x^3 + ...\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you can do this using the usual derivative method, but the generalized binomial formula will work in this case

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wait what about the thing where you take the derivative multiple times and plug it in?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes you can do that, but it is a pain. problem said "binomial" so i used it. take a look here http://www.proofwiki.org/wiki/Binomial_Theorem/General_Binomial_Theorem

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the hint was "binomial" formula

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ohh oops. i am terrible at math! and context clues i guess haha. well, do you have any testtaking tips ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0get some sleep and don't study right before the test, it will only freak you out when you see what you don't know ok easier said than done, i know relax though, it helps

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but i do not believe you are terrible at math because this is fairly advanced stuff, and in any case you found the intervals of convergence yourself also look at assigned homework problmes, and especialy quizzes because professors tend to repeat themselves, or at least ask the same types of questions

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that first part of the tip is so true! anyway thank you so much. you are a lifesaver. seriously.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0your quite welcome, and really good luck and relax and don't stay up all night studying
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