Question

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 ?

Answer 1

whatcha got so far?

Answer 2

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

Answer 3

wouldnt x=0 be a Maclaurin series?

Answer 4

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

Answer 5

as well as the answer

Answer 6

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

Answer 7

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...

Answer 8

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

Answer 9

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

Answer 10

forgot the factorials....

Answer 11

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

Answer 12

right?

Answer 13

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!

Answer 14

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

Answer 15

i believe so

Answer 16

i havent finished yet

Answer 17

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.

Answer 18

i got no idea what to do after that :)

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

yes i did alread

Answer 23

here

Answer 24

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

Answer 25

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

Answer 26

f2 = -cosx^2?

Answer 27

What's that?

Answer 28

derivatives of the function

Answer 29

of cos(x^2)?

Answer 30

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

Answer 31

part

Answer 32

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

Answer 33

kk i didnt know that

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

what did you get for the second derivative

Answer 38

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

Answer 39

the third derivative

Answer 40

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

Answer 41

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

Answer 42

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.

Answer 43

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

Answer 44

that is if you can

Answer 45

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.

Answer 46

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

Answer 47

-x^12/720

Answer 48

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

Answer 49

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

Answer 50

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

Answer 51

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

Answer 52

oh ok

Answer 53

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

Answer 54

You know how to use the ratio test?

Answer 55

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

Answer 56

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

Answer 57

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.

Answer 58

is that*

Answer 59

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

Answer 60

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