How do you find the maclaurin series expansion of #cos (x)^2#?

Answer 1

# cos^2x = 1 -x^2+1/3x^4-2/45x^6 + ...#

We seek a Maclaurin series expansion of #cos^2x#

Using the well-known sequence:

# cosx = 1 - (x^2)/(2!) + (x^4)/(4!) - (x^6)/(6!) + ... #

The product can be written as:

# cos^2x = {1 - (x^2)/(2!) + (x^4)/(4!) - ...}{1 - (x^2)/(2!) + (x^4)/(4!) - ...} #
# \ \ \ \ \ \ \ \ \ = {1 - (x^2)/(2) + (x^4)/(24) - (x^6)/(720) +...} # # \ \ \ \ \ \ \ \ \ \ \ \ \ \ - (x^2)/(2){1 - (x^2)/(2) + (x^4)/(24) - (x^6)/(720) +...} # # \ \ \ \ \ \ \ \ \ \ \ \ \ \ + (x^4)/(24){1 - (x^2)/(2) + (x^4)/(24) - (x^6)/(720) +...} # # \ \ \ \ \ \ \ \ \ \ \ \ \ \ - (x^6)/(720){1 - (x^2)/(2) + (x^4)/(24) - (x^6)/(720) +...} - ... #
# \ \ \ \ \ \ \ \ \ = {1 - (x^2)/(2) + (x^4)/(24) - (x^6)/(720) +...} # # \ \ \ \ \ \ \ \ \ \ \ \ \ \ - {(x^2)/(2) - (x^4)/(4) + (x^6)/(48) - ...} # # \ \ \ \ \ \ \ \ \ \ \ \ \ \ + {(x^4)/(24) - (x^6)/(48) + ...} # # \ \ \ \ \ \ \ \ \ \ \ \ \ \ - {(x^6)/(720) - ...} - ... #
# \ \ \ \ \ \ \ \ \ = (1) + (-1/2-1/12)x^2+(1/24+1/4+1/24)x^4# # \ \ \ \ \ \ \ \ \ \ \ \ \ \ + (-1/720-1/48-1/48-1/720)x^6 + ...#
# \ \ \ \ \ \ \ \ \ = 1 -x^2+1/3x^4-2/45x^6 + ...#
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Answer 2

To find the Maclaurin series expansion of cos(x)^2, we can start with the Maclaurin series expansion of cos(x), which is:

cos(x) = 1 - x^2/2! + x^4/4! - x^6/6! + ...

Then, we square this series term by term:

(cos(x))^2 = (1 - x^2/2! + x^4/4! - x^6/6! + ...) * (1 - x^2/2! + x^4/4! - x^6/6! + ...)

Multiplying and collecting like terms, we get:

(cos(x))^2 = 1 - x^2 + 2x^4/4! - x^6 + ...

So, the Maclaurin series expansion of cos(x)^2 is:

1 - x^2 + x^4/3 - x^6/3! + ...

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Answer 3

To find the Maclaurin series expansion of ( \cos(x)^2 ), we can use known Maclaurin series expansions and algebraic manipulations.

Starting with the Maclaurin series expansion of ( \cos(x) ), which is:

[ \cos(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!} ]

Then, to find the series expansion of ( \cos(x)^2 ), we square each term of the expansion for ( \cos(x) ) and sum them up:

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

Expanding this expression using the distributive property, we get:

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

Simplifying the expression, we have:

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

Now, we can combine the indices of the double sum into a single index ( k = n + m ):

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

This gives us the Maclaurin series expansion of ( \cos(x)^2 ).

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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.

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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