Nora
  • Nora
f(x)=x sin x(x^2) Find f^(2011) (0).
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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amistre64
  • amistre64
2011/360=?
amistre64
  • amistre64
5.58611 degrees; but is 2011 in radians or degrees for that?
Nora
  • Nora
I believe you have to work this using Taylor series

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amistre64
  • amistre64
oh.... then you prolly want someone else to help you ; like the cow :)
Nora
  • Nora
rude.
amistre64
  • amistre64
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...
amistre64
  • amistre64
like...the cow here :)
Nora
  • Nora
oh ok ok i mis understood u then . thanks for attempting
amistre64
  • amistre64
sorry i couldnt help out :) good luck with it tho ;)
anonymous
  • anonymous
Just for clarification, is it \[x \sin (x*x ^{2})\]
Nora
  • Nora
no its (x)(sin(x^2)
Nora
  • Nora
)
anonymous
  • anonymous
so if you taylor expand this, you will get\[x(x^2 - (x^2)^3/3! - (x^2)^5/5! + ...)\]
anonymous
  • anonymous
and if you plug in 0 for this you get 0 .-.
Nora
  • Nora
did u use a formula
anonymous
  • anonymous
well, I just used the taylor expansion of sin x and then I plugged in x^2 for that x
Nora
  • Nora
then multiplied the series by x right??
anonymous
  • anonymous
yup!
anonymous
  • anonymous
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
anonymous
  • anonymous
"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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anonymous
  • anonymous
Robtobey, I think the problem isn't taking the function to the 2011 power, but rather taking the 2011 derivative

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