How do you find the Maclaurin Series for #f(x)=cos(5x^2)#?

Answer 1
The Maclaurin series for #cos(x)# is well known:
#cos(x)=sum_(n=0)^oo(-1)^n/((2n)!)x^(2n)=1-x^2/(2!)+x^4/(4!)-x^6/(6!)+...#

So:

#cos(5x^2)=sum_(n=0)^oo(-1)^n/((2n)!)(5x^2)^(2n)=1-(5x^2)^2/(2!)+(5x^2)^4/(4!)-(5x^2)^6/(6!)+...#
#color(white)(cos(5x^2))=sum_(n=0)^oo((-1)^n5^(2n))/((2n)!)x^(4n)#
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Answer 2

To find the Maclaurin series for (f(x) = \cos(5x^2)), we first need to find the derivatives of (f(x)) at (x = 0) to determine the coefficients of the series. Then, we express the function as an infinite series using those coefficients. The Maclaurin series for (f(x) = \cos(5x^2)) is:

[ \cos(5x^2) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n ]

To find the derivatives of (f(x)), we use the chain rule. The (n)th derivative of (\cos(5x^2)) is a combination of (\sin(5x^2)) and (\cos(5x^2)) terms. After evaluating these derivatives at (x = 0), we obtain the coefficients for the Maclaurin series.

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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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