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 ?

- anonymous

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- amistre64

whatcha got so far?

- anonymous

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

- amistre64

wouldnt x=0 be a Maclaurin series?

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## More answers

- 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

- anonymous

as well as the answer

- anonymous

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

- 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...

- 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

- amistre64

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

- amistre64

forgot the factorials....

- amistre64

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

- amistre64

right?

- 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!

- amistre64

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

- anonymous

i believe so

- anonymous

i havent finished yet

- 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.

- amistre64

i got no idea what to do after that :)

- 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

- 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:\[\cos(x^2)=\sum_{n=0}^{\infty}(-1)^n{(x^2)^{2n} \over (2n)!}=\sum_{n=0}^{\infty}(-1)^n{x^{4n} \over (2n)!}\]

- anonymous

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

- anonymous

yes i did alread

- anonymous

here

- anonymous

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

- anonymous

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

- anonymous

f2 = -cosx^2?

- anonymous

What's that?

- anonymous

derivatives of the function

- anonymous

of cos(x^2)?

- anonymous

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

- anonymous

part

- anonymous

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

- anonymous

kk i didnt know that

- anonymous

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

- 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.

- anonymous

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

- anonymous

what did you get for the second derivative

- anonymous

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

- anonymous

the third derivative

- anonymous

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

- 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.

- 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.

- anonymous

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

- anonymous

that is if you can

- 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.

- anonymous

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

- anonymous

-x^12/720

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

oh ok

- anonymous

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

- anonymous

You know how to use the ratio test?

- anonymous

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

- anonymous

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

- 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.

- anonymous

is that*

- anonymous

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

- anonymous

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

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