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Use a Maclaurin series in this table to obtain the Maclaurin series for the given function. f(x) = x cos(4x)

User JorenHeit
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Final Answer:

The Maclaurin series for the given function f(x) = x cos(4x) is f(x) = 1 - 16x².

Step-by-step explanation:

The Maclaurin series expansion for a function involves expressing the function as an infinite sum of its derivatives evaluated at a specific point (usually 0). In this case, we can use the Maclaurin series for cos(x) and multiply it by x to obtain the series for x cos(4x).

The Maclaurin series for cos(x) is
\( \cos(x) = 1 - (x^2)/(2!) + (x^4)/(4!) - (x^6)/(6!) + \ldots \).

Multiplying this by x gives
\( x \cos(x) = x - (x^3)/(2!) + (x^5)/(4!) - (x^7)/(6!) + \ldots \).

Now, substitute 4x for x to obtain the series for xcos(4x):


\[ f(x) = 4x - ((4x)^3)/(2!) + ((4x)^5)/(4!) - ((4x)^7)/(6!) + \ldots \]

Simplify each term and combine like terms:


\[ f(x) = 4x - (64x^3)/(2) + (1024x^5)/(24) - (16384x^7)/(720) + \ldots \]

Further simplification gives the final answer f(x) = 1 - 16x². This is obtained by keeping only the terms up to x² and combining them. Therefore, the Maclaurin series for f(x) = xcos(4x) is f(x) = 1 - 16x².

User Turophile
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The Maclaurin series for the function
\( f(x) = x \cos(4x) \) is given by
\( \sum_(n=0)^(\infty) (-1)^n (2^(2n)x^(2n+1))/((2n+1)!) \).

The Maclaurin series is obtained by expressing the given function as an infinite sum of its derivatives evaluated at the origin. For
\( f(x) = x \cos(4x)\), we use the Maclaurin series expansion for
\( \cos(4x) \), which is
\( \sum_(n=0)^(\infty) (-1)^n ((4x)^(2n))/((2n)!) \). Multiplying this by
\( x \), we get
\( \sum_(n=0)^(\infty) (-1)^n (2^(2n)x^(2n+1))/((2n+1)!) \).

In this series, each term corresponds to the nth derivative of
\( f(x) \)evaluated at
\( x = 0 \). The general term follows the pattern of the nth derivative of
\( \cos(4x) \)multiplied by
\( x \). The alternating signs come from the expansion of
\( \cos(4x) \). This series converges for all real numbers, providing an approximation of the original function
\( f(x) \) in the vicinity of the origin. The accuracy of the approximation increases as more terms are included in the series.

User Rcrogers
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