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What is the taylor series in sigma notation of: #sin^2(x)#?

Answer 1

Samantha Adkison

You can start by using the trig identity of #sin^2x = (1 - cos2x)/2#

we know the Maclurin series of #cosx# is #sum_(n=0)^oo (-1)^n (x^(2n))/((2n)!)#

Keep in mind here that 0!=1, so the case of n=0 is still valid.

So we can then sub in #2x# in place of #x# to solve for #cos2x#

#cos2x = sum_(n=0)^oo (-1)^n ((2x)^(2n))/((2n)!)#

thus we get:

#sin^2x = (1-cos2x)/2 =1/2- 1/2sum_(n=0)^oo (-1)^n ((2x)^(2n))/((2n)!)#

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Answer 2

Emma Aldridge

The Taylor series in sigma notation for $\sin^2(x)$ is:

$\sin^2(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}x^{2n}$

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.