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anonymous

  • 5 years ago

use (1+x)^p= 1+px+p(p-1)/2!x^2+ p(p-1)(p-2)/3!x^3+... to find the first four nonzero terms of the taylor series about 0 for arcsinx

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  1. anonymous
    • 5 years ago
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    hey u found me lol

  2. anonymous
    • 5 years ago
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    Haha yeah.

  3. anonymous
    • 5 years ago
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    Can you find the derivative of arcsinx for me?

  4. anonymous
    • 5 years ago
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    I'll help you with the idea of this question, and then you try and let me know if you face any problem.

  5. anonymous
    • 5 years ago
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    \[{d \over dx}\sin^{-1}x={1 \over \sqrt{1-x^2}}=(1+(-x^2))^{1/2}\] Which is in the given form, after substituting with p=1/2

  6. anonymous
    • 5 years ago
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    Now find the first 4 terms of the expansion of the derivative of arcsinx using the given formula you wrote above. After that, just integrate these four terms, and that would be your first five term of the taylor series of arcsinx.

  7. anonymous
    • 5 years ago
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    You following?!

  8. anonymous
    • 5 years ago
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    Hello?! :)

  9. anonymous
    • 5 years ago
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    so do i plug in (-x)^2 in for x in the formula

  10. anonymous
    • 5 years ago
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    Exactly, and 1/2 for p.

  11. anonymous
    • 5 years ago
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    simple. right?!

  12. anonymous
    • 5 years ago
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    then i should find the antiderivative to get the orginal

  13. anonymous
    • 5 years ago
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    Yep!

  14. anonymous
    • 5 years ago
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    Does that make sense?

  15. anonymous
    • 5 years ago
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    yah just a general question when do first use the derivative to find the taylor series

  16. anonymous
    • 5 years ago
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    Well, that depends on the question. In this case, the question tells us what method to use to find the Taylor expansion of arcsinx. Here we should look for anything related to arcsin, that's similar to the given formula. Its derivative is the best choice, I believe.

  17. anonymous
    • 5 years ago
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    okay fantastic ty you for your help.

  18. anonymous
    • 5 years ago
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    You're welcome!

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