## Nora 5 years ago f(x)=x sin x(x^2) Find f^(2011) (0).

1. amistre64

2011/360=?

2. amistre64

5.58611 degrees; but is 2011 in radians or degrees for that?

3. nora

I believe you have to work this using Taylor series

4. amistre64

oh.... then you prolly want someone else to help you ; like the cow :)

5. nora

rude.

6. 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...

7. amistre64

like...the cow here :)

8. nora

oh ok ok i mis understood u then . thanks for attempting

9. amistre64

sorry i couldnt help out :) good luck with it tho ;)

10. anonymous

Just for clarification, is it $x \sin (x*x ^{2})$

11. nora

no its (x)(sin(x^2)

12. nora

)

13. anonymous

so if you taylor expand this, you will get$x(x^2 - (x^2)^3/3! - (x^2)^5/5! + ...)$

14. anonymous

and if you plug in 0 for this you get 0 .-.

15. nora

did u use a formula

16. anonymous

well, I just used the taylor expansion of sin x and then I plugged in x^2 for that x

17. nora

then multiplied the series by x right??

18. anonymous

yup!

19. 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

20. 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 .

21. anonymous

Robtobey, I think the problem isn't taking the function to the 2011 power, but rather taking the 2011 derivative