Maclaurin series for cos 2x
Understand the Problem
The question is asking for the Maclaurin series expansion of the function cos(2x). The Maclaurin series is a special case of the Taylor series, which expresses a function as an infinite sum of terms calculated from the derivatives of that function at a single point (in this case, at x=0). We will derive the series using the formula for the Maclaurin series.
Answer
The series is given by: $$ \cos(2x) = \sum_{n=0}^{\infty} \frac{(-1)^n (2x)^{2n}}{(2n)!} $$
Answer for screen readers
The Maclaurin series expansion of $\cos(2x)$ is given by: $$ \cos(2x) = \sum_{n=0}^{\infty} \frac{(-1)^n (2x)^{2n}}{(2n)!} $$
Steps to Solve
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Identify the function and derivatives
The function we want to expand is $f(x) = \cos(2x)$. We will need to find the derivatives at $x = 0$.
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Calculate the function and its derivatives at zero
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First derivative: $$ f'(x) = -2\sin(2x) $$ $$ f'(0) = -2\sin(0) = 0 $$
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Second derivative: $$ f''(x) = -4\cos(2x) $$ $$ f''(0) = -4\cos(0) = -4 $$
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Third derivative: $$ f'''(x) = 8\sin(2x) $$ $$ f'''(0) = 8\sin(0) = 0 $$
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Fourth derivative: $$ f^{(4)}(x) = 16\cos(2x) $$ $$ f^{(4)}(0) = 16\cos(0) = 16 $$
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Apply the Maclaurin series formula
The Maclaurin series for $f(x)$ is given by: $$ f(x) = f(0) + \frac{f'(0)}{1!} x + \frac{f''(0)}{2!} x^2 + \frac{f'''(0)}{3!} x^3 + \frac{f^{(4)}(0)}{4!} x^4 + \ldots $$
Substituting the values we found:
- $f(0) = \cos(0) = 1$
- $f'(0) = 0$
- $f''(0) = -4$
- $f'''(0) = 0$
- $f^{(4)}(0) = 16$
This results in: $$ f(x) = 1 + 0 \cdot x + \frac{-4}{2!} x^2 + 0 \cdot x^3 + \frac{16}{4!} x^4 + \ldots $$
Simplifying gives: $$ f(x) = 1 - 2x^2 + \frac{2}{3} x^4 + \ldots $$
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Write the general term of the series
Notice a pattern; the coefficients for even powers of $x$ can be expressed: $$ f(x) = \sum_{n=0}^{\infty} \frac{(-1)^n (2x)^{2n}}{(2n)!} $$
or alternatively, $$ f(x) = \sum_{n=0}^{\infty} \frac{(-1)^n 2^{2n} x^{2n}}{(2n)!} $$
The Maclaurin series expansion of $\cos(2x)$ is given by: $$ \cos(2x) = \sum_{n=0}^{\infty} \frac{(-1)^n (2x)^{2n}}{(2n)!} $$
More Information
The Maclaurin series is particularly useful in approximating functions around $x=0$. The cosine function exhibits periodic properties, and this expansion can provide accurate approximations for small values of $x$. The factor of 2 in the $2x$ modifies the frequency of the oscillation, affecting the coefficients in the series.
Tips
- Forgetting to evaluate derivatives at $x=0$ can lead to incorrect coefficients in the series.
- Confusing the signs of the coefficients derived from derivatives, especially when dealing with sine and cosine functions.
