anonymous
  • anonymous
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 ?
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
whatcha got so far?
anonymous
  • anonymous
didn't do it yet im doing it now i just did the first problem which was cos(x)
amistre64
  • amistre64
wouldnt x=0 be a Maclaurin series?

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anonymous
  • 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
  • anonymous
as well as the answer
anonymous
  • anonymous
Maclaurin series is just s special case of taylor series when it's centered at 0.
amistre64
  • 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
  • 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
  • amistre64
cos(0) -sin(0)-cos(0)+sin(0)+cos(0)....like that rght?
amistre64
  • amistre64
forgot the factorials....
amistre64
  • amistre64
cos(x^2) = 1- sin(0)x -cos(0)x^2+sin(0)x^3+cos(0)x^4 -------- -------- --------... 2! 3! 4!
amistre64
  • amistre64
right?
amistre64
  • 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
  • amistre64
x^2...x^4....x^6....
anonymous
  • anonymous
i believe so
anonymous
  • anonymous
i havent finished yet
anonymous
  • 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
  • amistre64
i got no idea what to do after that :)
anonymous
  • 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
  • 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
  • anonymous
Do you know how to find the Taylor series of cos x at x=0?
anonymous
  • anonymous
yes i did alread
anonymous
  • anonymous
here
anonymous
  • anonymous
Good. So, I just used that expansion and replaced x by x^2.
anonymous
  • anonymous
F^(1) (x^2)=-sin x^2?
anonymous
  • anonymous
f2 = -cosx^2?
anonymous
  • anonymous
What's that?
anonymous
  • anonymous
derivatives of the function
anonymous
  • anonymous
of cos(x^2)?
anonymous
  • anonymous
then your evaluate them then move on to the generate a formul
anonymous
  • anonymous
part
anonymous
  • anonymous
Yeah. But the first derivative is not -sin(x^2). You should use chain rule here. It should be: -2xsin(x^2)
anonymous
  • anonymous
kk i didnt know that
anonymous
  • anonymous
And for the second derivative, you should apply both the product rule and chain rule.
anonymous
  • 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
  • anonymous
kk sec....... im going to try to do it now and thanks for this your a life saver
anonymous
  • anonymous
what did you get for the second derivative
anonymous
  • anonymous
-4x^2cos(x^2)-2sin(x^2)
anonymous
  • anonymous
the third derivative
anonymous
  • anonymous
and you said the fourth derivative should be the one that isnt zero when evaluating it right?
anonymous
  • 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
  • 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
  • anonymous
ok explain to me why it really doesnt matter to find the derivatives
anonymous
  • anonymous
that is if you can
anonymous
  • 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
  • anonymous
yes so the ans should be 1-1/2x^4+1/24x^8 then whats the nxt one
anonymous
  • anonymous
-x^12/720
anonymous
  • anonymous
ok now explain to me why not to worry about the derivative part because i must know this for tomorrow
anonymous
  • anonymous
is there any tricks to getting the correct answere fast4r then going through each an evaluating it
anonymous
  • anonymous
because after you find this i must know if it converges or diverges
anonymous
  • 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
  • anonymous
oh ok
anonymous
  • anonymous
can you help me figure out if it now converges or diverges?
anonymous
  • anonymous
You know how to use the ratio test?
anonymous
  • anonymous
is there an easy way to know it so i know? or no
anonymous
  • anonymous
because i dont understand exactly bc my professor did a poor job explaining it
anonymous
  • 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
  • anonymous
is that*
anonymous
  • anonymous
like ive got the definitions i front of me but explain why it converges or diverges
anonymous
  • anonymous
ill talk to you on the other page this one i need for this page

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