Math Problem Statement

maclaurin series of cos(3x)

Solution

To find the Maclaurin series of cos(3x)\cos(3x), we need to expand cos(3x)\cos(3x) around x=0x = 0. The Maclaurin series for a function f(x)f(x) is:

f(x)=f(0)+f(0)x+f(0)2!x2+f(0)3!x3+f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots

Step-by-Step Solution

  1. Identify f(x)=cos(3x)f(x) = \cos(3x) and compute derivatives at x=0x = 0:

    We know: f(x)=cos(3x)f(x) = \cos(3x)

  2. Compute the derivatives and evaluate them at x=0x = 0:

    • f(x)=cos(3x)f(x) = \cos(3x), so f(0)=cos(0)=1f(0) = \cos(0) = 1.

    • First derivative: f(x)=3sin(3x)f'(x) = -3 \sin(3x).
      At x=0x = 0: f(0)=3sin(0)=0f'(0) = -3 \sin(0) = 0.

    • Second derivative: f(x)=9cos(3x)f''(x) = -9 \cos(3x). At x=0x = 0: f(0)=9cos(0)=9f''(0) = -9 \cos(0) = -9.

    • Third derivative: f(x)=27sin(3x)f'''(x) = 27 \sin(3x). At x=0x = 0: f(0)=27sin(0)=0f'''(0) = 27 \sin(0) = 0.

    • Fourth derivative: f(4)(x)=81cos(3x)f^{(4)}(x) = -81 \cos(3x). At x=0x = 0: f(4)(0)=81cos(0)=81f^{(4)}(0) = -81 \cos(0) = -81.

  3. Construct the series:

    Now substitute these values into the Maclaurin series formula:

    f(x)=f(0)+f(0)x+f(0)2!x2+f(0)3!x3+f(4)(0)4!x4+f(x) = f(0) + f'(0) x + \frac{f''(0)}{2!} x^2 + \frac{f'''(0)}{3!} x^3 + \frac{f^{(4)}(0)}{4!} x^4 + \cdots

    Substitute the values we calculated:

    cos(3x)=1+0x+92x2+0x3+8124x4+\cos(3x) = 1 + 0 \cdot x + \frac{-9}{2} x^2 + 0 \cdot x^3 + \frac{-81}{24} x^4 + \cdots

  4. Simplify terms:

    This simplifies to:

    cos(3x)=192x28124x4+\cos(3x) = 1 - \frac{9}{2} x^2 - \frac{81}{24} x^4 + \cdots

    Further simplifying 8124=278\frac{81}{24} = \frac{27}{8}, we get:

    cos(3x)=192x2+278x4+\cos(3x) = 1 - \frac{9}{2} x^2 + \frac{27}{8} x^4 + \cdots

Final Answer

The Maclaurin series for cos(3x)\cos(3x) is:

cos(3x)=192x2+278x4+\cos(3x) = 1 - \frac{9}{2} x^2 + \frac{27}{8} x^4 + \cdots

Would you like more details on each step or have any questions?


Related Questions

  1. What is the Maclaurin series of sin(3x)\sin(3x)?
  2. How does the Maclaurin series for cos(x)\cos(x) compare to cos(3x)\cos(3x)?
  3. What are the convergence properties of the Maclaurin series for cosine functions?
  4. How would you find the Maclaurin series of e3xe^{3x}?
  5. How can the Taylor series expansion of cos(x)\cos(x) help in calculating higher-order derivatives?

Tip:

The Maclaurin series for trigonometric functions like sin(x)\sin(x) and cos(x)\cos(x) always have alternating signs and factorial denominators, which is a helpful pattern to remember.

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Math Problem Analysis

Mathematical Concepts

Calculus
Maclaurin Series
Power Series Expansion
Derivatives

Formulas

Maclaurin series formula: f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots
Trigonometric identities, such as \cos(3x)

Theorems

Maclaurin Series Expansion
Derivative Rules

Suitable Grade Level

Undergraduate Calculus