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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 ?
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 ?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0didn't do it yet im doing it now i just did the first problem which was cos(x)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0wouldnt x=0 be a Maclaurin series?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can 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
 5 years ago
Best ResponseYou've already chosen the best response.0as well as the answer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Maclaurin series is just s special case of taylor series when it's centered at 0.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0its 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
 5 years ago
Best ResponseYou've already chosen the best response.0because 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
 5 years ago
Best ResponseYou've already chosen the best response.0cos(0) sin(0)cos(0)+sin(0)+cos(0)....like that rght?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0forgot the factorials....

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0cos(x^2) = 1 sin(0)x cos(0)x^2+sin(0)x^3+cos(0)x^4   ... 2! 3! 4!

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0every 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!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i havent finished yet

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
 5 years ago
Best ResponseYou've already chosen the best response.0i got no idea what to do after that :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh 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
 5 years ago
Best ResponseYou've already chosen the best response.0So 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
 5 years ago
Best ResponseYou've already chosen the best response.0Do you know how to find the Taylor series of cos x at x=0?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Good. So, I just used that expansion and replaced x by x^2.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0F^(1) (x^2)=sin x^2?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0derivatives of the function

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then your evaluate them then move on to the generate a formul

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah. But the first derivative is not sin(x^2). You should use chain rule here. It should be: 2xsin(x^2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0And for the second derivative, you should apply both the product rule and chain rule.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I did it and found that all derivatives are zeros except those who are multiple of 4. I mean the 4th, 8th, 12th.. derivatives.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0kk sec....... im going to try to do it now and thanks for this your a life saver

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what did you get for the second derivative

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.04x^2cos(x^2)2sin(x^2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and you said the fourth derivative should be the one that isnt zero when evaluating it right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yep. You should not bother yourself with finding these derivatives. I am pretty sure the question wants you to use the method I used.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hey 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 xaxis, for x ranging from 0 to 1.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok explain to me why it really doesnt matter to find the derivatives

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You 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
 5 years ago
Best ResponseYou've already chosen the best response.0yes so the ans should be 11/2x^4+1/24x^8 then whats the nxt one

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok now explain to me why not to worry about the derivative part because i must know this for tomorrow

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is there any tricks to getting the correct answere fast4r then going through each an evaluating it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because after you find this i must know if it converges or diverges

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I mean you shouldn't worry about it in this particular problem, since you can find the Taylor series without doing any derivatives.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can you help me figure out if it now converges or diverges?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You know how to use the ratio test?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is there an easy way to know it so i know? or no

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because i dont understand exactly bc my professor did a poor job explaining it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmm 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
 5 years ago
Best ResponseYou've already chosen the best response.0like ive got the definitions i front of me but explain why it converges or diverges

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ill talk to you on the other page this one i need for this page
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