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anonymous

  • 4 years ago

how to take the derivative of series? for example ((-1)^n+1)(x-2)^n)/n

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  1. anonymous
    • 4 years ago
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    but you are a higher level than zarkon haha.

  2. anonymous
    • 4 years ago
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    hell no term by term

  3. anonymous
    • 4 years ago
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    is (-1)^n+1(x-2)^n-1 right?

  4. anonymous
    • 4 years ago
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    yes i believe so , let me write it so i don't do anything silly are we starting at n = 0 or n = 1?

  5. anonymous
    • 4 years ago
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    yes that is right. what you have

  6. anonymous
    • 4 years ago
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    1. okay and then i got the interval of that to be 1<x<3, but i am having trouble with the convergence of course.

  7. anonymous
    • 4 years ago
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    same

  8. anonymous
    • 4 years ago
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    oh but not at the endpoints of course

  9. anonymous
    • 4 years ago
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    next question...can you explain how to find the power series of a function?

  10. anonymous
    • 4 years ago
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    i can show you a trick for that one if you like

  11. anonymous
    • 4 years ago
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    well maybe not

  12. anonymous
    • 4 years ago
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    yes please! i have an exam tomorrow and i need as much help as possible.

  13. anonymous
    • 4 years ago
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    well then i don't want to waste your time with this. but if you know that \[\frac{1}{1-x}=\sum x^n\] then \[\frac{1}{1+x}=\sum (-1)^nx^n\] and so \[\frac{1}{x-1}=\frac{1}{1+(x-2)}=\sum(-1)^n(x-2)^n\] so that is your derivative, and therefore your original series was the log

  14. anonymous
    • 4 years ago
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    but if you have an exam forget that mess

  15. anonymous
    • 4 years ago
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    and also your derivative will not convege at the endpoints, because there is not way that \[\sum 1^n\] or \[\sum (-1)^n\] converges

  16. anonymous
    • 4 years ago
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    this 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) = \[\sqrt[4]{1+x}\]

  17. anonymous
    • 4 years ago
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    i guess you are supposed to write this as \[(1+x)^{\frac{1}{4}}\] first and then use "general binomial series

  18. anonymous
    • 4 years ago
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    er \[1+\frac{1}{4}x+\frac{\frac{1}{4}-1}{2!}x^2\]

  19. anonymous
    • 4 years ago
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    \[+\frac{(\frac{1}{4}-1)(\frac{1}{4}-2)}{3!}x^3 +...\]

  20. anonymous
    • 4 years ago
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    probably some algebra will turn up a nice pattern

  21. anonymous
    • 4 years ago
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    but i dont understand the maclaurin part?

  22. anonymous
    • 4 years ago
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    that is the maclaurin series

  23. anonymous
    • 4 years ago
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    expand about zero, get \[a_0+a_1x+a_2x^2+a_3x^3 + ...\]

  24. anonymous
    • 4 years ago
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    you can do this using the usual derivative method, but the generalized binomial formula will work in this case

  25. anonymous
    • 4 years ago
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    wait what about the thing where you take the derivative multiple times and plug it in?

  26. anonymous
    • 4 years ago
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    yes 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

  27. anonymous
    • 4 years ago
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    the hint was "binomial" formula

  28. anonymous
    • 4 years ago
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    ohh oops. i am terrible at math! and context clues i guess haha. well, do you have any test-taking tips ?

  29. anonymous
    • 4 years ago
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    get 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

  30. anonymous
    • 4 years ago
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    but 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

  31. anonymous
    • 4 years ago
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    good luck

  32. anonymous
    • 4 years ago
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    that first part of the tip is so true! anyway thank you so much. you are a lifesaver. seriously.

  33. anonymous
    • 4 years ago
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    your quite welcome, and really good luck and relax and don't stay up all night studying

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