# How do you find the Maclaurin series for #cos^2 (x)#?

There are two approaches.

According to the Product Rule and/or Chain Rule, we have

etc...

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To find the Maclaurin series for ( \cos^2(x) ), you can start with the Maclaurin series for ( \cos(x) ), which is ( \cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}x^{2n} ).

Now, to find ( \cos^2(x) ), you simply square each term of the series for ( \cos(x) ). Thus:

[ \begin{split} \cos^2(x) & = \left( \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}x^{2n} \right) \cdot \left( \sum_{m=0}^{\infty} \frac{(-1)^m}{(2m)!}x^{2m} \right) \ & = \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \frac{(-1)^n}{(2n)!}x^{2n} \cdot \frac{(-1)^m}{(2m)!}x^{2m} \ & = \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \frac{(-1)^{n+m}}{(2n)! (2m)!}x^{2(n+m)} \ & = \sum_{k=0}^{\infty} \left( \sum_{n=0}^{k} \frac{(-1)^{n+k-n}}{(2n)! (2(k-n))!} \right) x^{2k} \end{split} ]

Therefore, the Maclaurin series for ( \cos^2(x) ) is:

[ \sum_{k=0}^{\infty} \left( \sum_{n=0}^{k} \frac{(-1)^{n+k-n}}{(2n)! (2(k-n))!} \right) x^{2k} ]

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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.

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.

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.

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