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Nora

  • 5 years ago

f(x)=x sin x(x^2) Find f^(2011) (0).

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  1. amistre64
    • 5 years ago
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    2011/360=?

  2. amistre64
    • 5 years ago
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    5.58611 degrees; but is 2011 in radians or degrees for that?

  3. nora
    • 5 years ago
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    I believe you have to work this using Taylor series

  4. amistre64
    • 5 years ago
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    oh.... then you prolly want someone else to help you ; like the cow :)

  5. nora
    • 5 years ago
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    rude.

  6. amistre64
    • 5 years ago
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    no..... rude would be me doing something absurd and uncalled for; this is just me not being good with taylor series and suggesting you find someone smarter than me...

  7. amistre64
    • 5 years ago
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    like...the cow here :)

  8. nora
    • 5 years ago
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    oh ok ok i mis understood u then . thanks for attempting

  9. amistre64
    • 5 years ago
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    sorry i couldnt help out :) good luck with it tho ;)

  10. anonymous
    • 5 years ago
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    Just for clarification, is it \[x \sin (x*x ^{2})\]

  11. nora
    • 5 years ago
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    no its (x)(sin(x^2)

  12. nora
    • 5 years ago
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    )

  13. anonymous
    • 5 years ago
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    so if you taylor expand this, you will get\[x(x^2 - (x^2)^3/3! - (x^2)^5/5! + ...)\]

  14. anonymous
    • 5 years ago
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    and if you plug in 0 for this you get 0 .-.

  15. nora
    • 5 years ago
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    did u use a formula

  16. anonymous
    • 5 years ago
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    well, I just used the taylor expansion of sin x and then I plugged in x^2 for that x

  17. nora
    • 5 years ago
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    then multiplied the series by x right??

  18. anonymous
    • 5 years ago
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    yup!

  19. anonymous
    • 5 years ago
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    and because you can always factor out an x term, when you plug in x, you'll get 0, or at least this is my idea of how it works

  20. anonymous
    • 5 years ago
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    "no its (x)(sin(x^2)" f(x)= x*sin(x^2) f(0) = 0. At the origin both x x^2 are zero. Thus the product of x and Sin[x^2] is zero. Refer to the attached plot, Nora_1.pdf .

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  21. anonymous
    • 5 years ago
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    Robtobey, I think the problem isn't taking the function to the 2011 power, but rather taking the 2011 derivative

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