Math Problem Statement

khai triển maclaurin của xsinx-cosx đến bậc 6

Solution

Để khai triển Maclaurin của xsinxcosxx\sin x - \cos x đến bậc 6, ta thực hiện như sau:

  1. Maclaurin của sinx\sin x: sinx=xx33!+x55!+O(x7)\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} + O(x^7)

  2. Maclaurin của cosx\cos x: cosx=1x22!+x44!x66!+O(x7)\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + O(x^7)

  3. Biểu thức xsinxx\sin x: xsinx=x(xx33!+x55!)=x2x46+x6120+O(x7)x\sin x = x \cdot \left(x - \frac{x^3}{3!} + \frac{x^5}{5!}\right) = x^2 - \frac{x^4}{6} + \frac{x^6}{120} + O(x^7)

  4. Biểu thức xsinxcosxx\sin x - \cos x: xsinxcosx=(x2x46+x6120)(1x22+x424x6720)x\sin x - \cos x = \left(x^2 - \frac{x^4}{6} + \frac{x^6}{120}\right) - \left(1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720}\right)

  5. Rút gọn đến bậc 6: xsinxcosx=x2x46+x61201+x22x424+x6720x\sin x - \cos x = x^2 - \frac{x^4}{6} + \frac{x^6}{120} - 1 + \frac{x^2}{2} - \frac{x^4}{24} + \frac{x^6}{720}

    Gom các bậc tương ứng:

    • Hệ số x2x^2: x2+x22=3x22x^2 + \frac{x^2}{2} = \frac{3x^2}{2}
    • Hệ số x4x^4: x46x424=5x424-\frac{x^4}{6} - \frac{x^4}{24} = -\frac{5x^4}{24}
    • Hệ số x6x^6: x6120+x6720=7x6720=x6103\frac{x^6}{120} + \frac{x^6}{720} = \frac{7x^6}{720} = \frac{x^6}{103}.

    Do đó, khai triển Maclaurin của xsinxcosxx\sin x - \cos x đến bậc 6 là:

  6. Recall the expansions: [ \sin x = x - \frac{x^3}{6} + \frac{x^5}{120} + O(x^7) \cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720} + O(x^7)$$

  7. Compute xsinxx \sin x: xsinx=x(xx36+x5120)x \sin x = x \cdot \left(x - \frac{x^3}{6} + \frac{x^5}{120}\right) xsinx=x2x46+x6120+O(x7)x \sin x = x^2 - \frac{x^4}{6} + \frac{x^6}{120} + O(x^7)

  8. Compute xsinxcosxx \sin x - \cos x: xsinxcosx=(x2x46+x6120)(1x22+x424x6720)x \sin x - \cos x = \left(x^2 - \frac{x^4}{6} + \frac{x^6}{120}\right) - \left(1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720}\right)

  9. Combine terms:

    • Constant term: 1-1
    • x2x^2 term: x2(x22)=x2+x22=3x22x^2 - \left(-\frac{x^2}{2}\right) = x^2 + \frac{x^2}{2} = \frac{3x^2}{2}
    • x4x^4 term: x46x424=4x424x424=5x424-\frac{x^4}{6} - \frac{x^4}{24} = -\frac{4x^4}{24} - \frac{x^4}{24} = -\frac{5x^4}{24}
    • x6x^6 term: x6120x6720=6x6720x6720=5x6720=x6144\frac{x^6}{120} - \frac{x^6}{720} = \frac{6x^6}{720} - \frac{x^6}{720} = \frac{5x^6}{720} = \frac{x^6}{144}

    So: xsinxcosx=1+3x225x424+x6144+O(x7)x \sin x - \cos x = -1 + \frac{3x^2}{2} - \frac{5x^4}{24} + \frac{x^6}{144} + O(x^7)

Final Answer:

xsinxcosx=1+3x225x424+x6144x \sin x - \cos x = -1 + \frac{3x^2}{2} - \frac{5x^4}{24} + \frac{x^6}{144}

Questions for Deeper Understanding:

  1. Can you confirm the general formula for the Maclaurin series expansion?
  2. How is the factorial function involved in the derivation of Maclaurin series?
  3. What role does truncation play in approximation?
  4. What is the difference between Maclaurin and Taylor series?
  5. How does the error term O(x7)O(x^7) affect practical applications?

Tip:

When performing Maclaurin expansions, align terms carefully to avoid sign and coefficient errors during subtraction.

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Math Problem Analysis

Mathematical Concepts

Calculus
Maclaurin Series
Taylor Series Expansion

Formulas

sin(x) ≈ x - x^3/3! + x^5/5! + O(x^7)
cos(x) ≈ 1 - x^2/2! + x^4/4! - x^6/6! + O(x^7)
Maclaurin expansion of a function f(x): f(x) = f(0) + f'(0)x + f''(0)x^2/2! + ...

Theorems

Maclaurin Series Theorem: A special case of the Taylor series centered at x = 0.

Suitable Grade Level

College Level (Calculus I/II)