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 ?

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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
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whatcha got so far?
didn't do it yet im doing it now i just did the first problem which was cos(x)
wouldnt x=0 be a Maclaurin series?

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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
as well as the answer
Maclaurin series is just s special case of taylor series when it's centered at 0.
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...
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
cos(0) -sin(0)-cos(0)+sin(0)+cos(0)....like that rght?
forgot the factorials....
cos(x^2) = 1- sin(0)x -cos(0)x^2+sin(0)x^3+cos(0)x^4 -------- -------- --------... 2! 3! 4!
right?
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!
x^2...x^4....x^6....
i believe so
i havent finished yet
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.
i got no idea what to do after that :)
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
So the Taylor series of cos x centered at 0 is: \[\cos x=\sum_{n=0}^{\infty}(-1)^n{x^{2n} \over (2n)!}\] Then:\[\cos(x^2)=\sum_{n=0}^{\infty}(-1)^n{(x^2)^{2n} \over (2n)!}=\sum_{n=0}^{\infty}(-1)^n{x^{4n} \over (2n)!}\]
Do you know how to find the Taylor series of cos x at x=0?
yes i did alread
here
Good. So, I just used that expansion and replaced x by x^2.
F^(1) (x^2)=-sin x^2?
f2 = -cosx^2?
What's that?
derivatives of the function
of cos(x^2)?
then your evaluate them then move on to the generate a formul
part
Yeah. But the first derivative is not -sin(x^2). You should use chain rule here. It should be: -2xsin(x^2)
kk i didnt know that
And for the second derivative, you should apply both the product rule and chain rule.
I did it and found that all derivatives are zeros except those who are multiple of 4. I mean the 4th, 8th, 12th.. derivatives.
kk sec....... im going to try to do it now and thanks for this your a life saver
what did you get for the second derivative
-4x^2cos(x^2)-2sin(x^2)
the third derivative
and you said the fourth derivative should be the one that isnt zero when evaluating it right?
Yep. You should not bother yourself with finding these derivatives. I am pretty sure the question wants you to use the method I used.
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.
ok explain to me why it really doesnt matter to find the derivatives
that is if you can
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.
yes so the ans should be 1-1/2x^4+1/24x^8 then whats the nxt one
-x^12/720
ok now explain to me why not to worry about the derivative part because i must know this for tomorrow
is there any tricks to getting the correct answere fast4r then going through each an evaluating it
because after you find this i must know if it converges or diverges
I mean you shouldn't worry about it in this particular problem, since you can find the Taylor series without doing any derivatives.
oh ok
can you help me figure out if it now converges or diverges?
You know how to use the ratio test?
is there an easy way to know it so i know? or no
because i dont understand exactly bc my professor did a poor job explaining it
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.
is that*
like ive got the definitions i front of me but explain why it converges or diverges
ill talk to you on the other page this one i need for this page

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