## anonymous 5 years ago calc II: find the taylor series (at x=0) of cos(x^2) can anyone help me i need reassurance to make sure my answers is correct ?

1. amistre64

whatcha got so far?

2. anonymous

didn't do it yet im doing it now i just did the first problem which was cos(x)

3. amistre64

wouldnt x=0 be a Maclaurin series?

4. anonymous

can you explain in your way of understanding when i would use Maclaurin series and taylor and which would be easier because my professor said to find it using taylor series

5. anonymous

6. anonymous

Maclaurin series is just s special case of taylor series when it's centered at 0.

7. amistre64

its been awhile, but i think when f(a), that taylor series is used; but when a=0; then f(0); and the Maclaurin is used...

8. anonymous

because ive got a final tomorrow and im freaking out been studying and idk everything that im supposed to yet also yes i think thats what he said

9. amistre64

cos(0) -sin(0)-cos(0)+sin(0)+cos(0)....like that rght?

10. amistre64

forgot the factorials....

11. amistre64

cos(x^2) = 1- sin(0)x -cos(0)x^2+sin(0)x^3+cos(0)x^4 -------- -------- --------... 2! 3! 4!

12. amistre64

right?

13. amistre64

every other one goes to zero since sin(0)=0... which leaves us with: 1 -1 +1 -1 +1 -1 +1 -- -- -- -- -- -- ... right? 2! 4! 6! 8! 10! 12!

14. amistre64

x^2...x^4....x^6....

15. anonymous

i believe so

16. anonymous

i havent finished yet

17. anonymous

I think the point here is the derivatives. I have does up to 8th derivatives, all are zeros at x=0 except the fourth and the eighth.

18. amistre64

i got no idea what to do after that :)

19. anonymous

Oh I got an idea. We can derive the Taylor series of cos(x^2) from that of cosx. Since tge expansion of cosx is well known

20. anonymous

So the Taylor series of cos x centered at 0 is: $\cos x=\sum_{n=0}^{\infty}(-1)^n{x^{2n} \over (2n)!}$ Then:$\cos(x^2)=\sum_{n=0}^{\infty}(-1)^n{(x^2)^{2n} \over (2n)!}=\sum_{n=0}^{\infty}(-1)^n{x^{4n} \over (2n)!}$

21. anonymous

Do you know how to find the Taylor series of cos x at x=0?

22. anonymous

23. anonymous

here

24. anonymous

Good. So, I just used that expansion and replaced x by x^2.

25. anonymous

F^(1) (x^2)=-sin x^2?

26. anonymous

f2 = -cosx^2?

27. anonymous

What's that?

28. anonymous

derivatives of the function

29. anonymous

of cos(x^2)?

30. anonymous

then your evaluate them then move on to the generate a formul

31. anonymous

part

32. anonymous

Yeah. But the first derivative is not -sin(x^2). You should use chain rule here. It should be: -2xsin(x^2)

33. anonymous

kk i didnt know that

34. anonymous

And for the second derivative, you should apply both the product rule and chain rule.

35. anonymous

I did it and found that all derivatives are zeros except those who are multiple of 4. I mean the 4th, 8th, 12th.. derivatives.

36. anonymous

kk sec....... im going to try to do it now and thanks for this your a life saver

37. anonymous

what did you get for the second derivative

38. anonymous

-4x^2cos(x^2)-2sin(x^2)

39. anonymous

the third derivative

40. anonymous

and you said the fourth derivative should be the one that isnt zero when evaluating it right?

41. anonymous

Yep. You should not bother yourself with finding these derivatives. I am pretty sure the question wants you to use the method I used.

42. anonymous

hey buddy also as i try to comprehend and understand this can you take a look at this Let f(x) = the cube root of x. Find the area of the surface generated by revolving the curve y = f(x) around the x-axis, for x ranging from 0 to 1.

43. anonymous

ok explain to me why it really doesnt matter to find the derivatives

44. anonymous

that is if you can

45. anonymous

You told me you know the Taylor series of cosx, then you just have to substitute for each x in that expansion of cosx by x^2. That's all.

46. anonymous

yes so the ans should be 1-1/2x^4+1/24x^8 then whats the nxt one

47. anonymous

-x^12/720

48. anonymous

ok now explain to me why not to worry about the derivative part because i must know this for tomorrow

49. anonymous

is there any tricks to getting the correct answere fast4r then going through each an evaluating it

50. anonymous

because after you find this i must know if it converges or diverges

51. anonymous

I mean you shouldn't worry about it in this particular problem, since you can find the Taylor series without doing any derivatives.

52. anonymous

oh ok

53. anonymous

can you help me figure out if it now converges or diverges?

54. anonymous

You know how to use the ratio test?

55. anonymous

is there an easy way to know it so i know? or no

56. anonymous

because i dont understand exactly bc my professor did a poor job explaining it

57. anonymous

hmm I am not sure if there is a simpler way. All I know that you need to know the ratio test and the root test to find the radius and interval of convergence for power series such as Taylor series.

58. anonymous

is that*

59. anonymous

like ive got the definitions i front of me but explain why it converges or diverges

60. anonymous

ill talk to you on the other page this one i need for this page