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Prometheus
 Post subject: MacLaurin expansion  |  Posted: Sat Apr 28, 2012 4:21 pm

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Joined: Sun Aug 07, 2011 8:58 am
Posts: 314

 I found the MacLaurin expansion of $\displaystyle y=cos^2+4$ up to the $\displaystyle x^4$ term to be $\displaystyle 5-x^2-2x^4+...$ Assuming no arithmetical errors, i think this is right.However, now i've been asked to find the expansion of $\displaystyle y=3sin^2$up to the $\displaystyle x^4$ term using the answer i got above. I've not been shown how to do this. The closest i could get was to try to find some function that sends $\displaystyle y=cos^2+4$ to $\displaystyle y=3sinx$, and then apply the same function to the expansion.In this case i found the third derivative (how do you write that with latex?) to be $\displaystyle 8sin^2x$. Then times by $\displaystyle \frac{3}{8}$ to get $\displaystyle 3sin^2$. Taking the third derivative of the expansion and times by the same i got $\displaystyle 3sin^2x\equiv-9+...$Is this even remotely correct - the process if not the actual answer? Hope again this is a difficult one, its taken me most of the day.
DrRocket
 Post subject: Re: MacLaurin expansion  |  Posted: Sat Apr 28, 2012 11:31 pm
Original Member

Joined: Fri Aug 05, 2011 2:22 am
Posts: 477

 Prometheus wrote:I found the MacLaurin expansion of $\displaystyle y=cos^2+4$ up to the $\displaystyle x^4$ term to be $\displaystyle 5-x^2-2x^4+...$ Assuming no arithmetical errors, i think this is right.However, now i've been asked to find the expansion of $\displaystyle y=3sin^2$up to the $\displaystyle x^4$ term using the answer i got above. I've not been shown how to do this. The closest i could get was to try to find some function that sends $\displaystyle y=cos^2+4$ to $\displaystyle y=3sinx$, and then apply the same function to the expansion.In this case i found the third derivative (how do you write that with latex?) to be $\displaystyle 8sin^2x$. Then times by $\displaystyle \frac{3}{8}$ to get $\displaystyle 3sin^2$. Taking the third derivative of the expansion and times by the same i got $\displaystyle 3sin^2x\equiv-9+...$Is this even remotely correct - the process if not the actual answer? Hope again this is a difficult one, its taken me most of the day.$\displaystyle cos^2x + sin^2x = 1$ _________________gone
Prometheus
 Post subject: Re: MacLaurin expansion  |  Posted: Tue May 01, 2012 8:47 am

Original Member

Joined: Sun Aug 07, 2011 8:58 am
Posts: 314