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


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DrRocket
Post  Post subject: Re: MacLaurin expansion  |  Posted: Sat Apr 28, 2012 11:31 pm
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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$

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Prometheus
Post  Post subject: Re: MacLaurin expansion  |  Posted: Tue May 01, 2012 8:47 am
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Hmm... I'll have to think about this one.


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