Solution
Để khai triển Maclaurin của xsinx−cosx đến bậc 6, ta thực hiện như sau:
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Maclaurin của sinx:
sinx=x−3!x3+5!x5+O(x7)
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Maclaurin của cosx:
cosx=1−2!x2+4!x4−6!x6+O(x7)
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Biểu thức xsinx:
xsinx=x⋅(x−3!x3+5!x5)=x2−6x4+120x6+O(x7)
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Biểu thức xsinx−cosx:
xsinx−cosx=(x2−6x4+120x6)−(1−2x2+24x4−720x6)
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Rút gọn đến bậc 6:
xsinx−cosx=x2−6x4+120x6−1+2x2−24x4+720x6
Gom các bậc tương ứng:
- Hệ số x2: x2+2x2=23x2
- Hệ số x4: −6x4−24x4=−245x4
- Hệ số x6: 120x6+720x6=7207x6=103x6.
Do đó, khai triển Maclaurin của xsinx−cosx đến bậc 6 là:
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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)$$
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Compute xsinx:
xsinx=x⋅(x−6x3+120x5)
xsinx=x2−6x4+120x6+O(x7)
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Compute xsinx−cosx:
xsinx−cosx=(x2−6x4+120x6)−(1−2x2+24x4−720x6)
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Combine terms:
- Constant term: −1
- x2 term: x2−(−2x2)=x2+2x2=23x2
- x4 term: −6x4−24x4=−244x4−24x4=−245x4
- x6 term: 120x6−720x6=7206x6−720x6=7205x6=144x6
So:
xsinx−cosx=−1+23x2−245x4+144x6+O(x7)
Final Answer:
xsinx−cosx=−1+23x2−245x4+144x6
Questions for Deeper Understanding:
- Can you confirm the general formula for the Maclaurin series expansion?
- How is the factorial function involved in the derivation of Maclaurin series?
- What role does truncation play in approximation?
- What is the difference between Maclaurin and Taylor series?
- How does the error term O(x7) affect practical applications?
Tip:
When performing Maclaurin expansions, align terms carefully to avoid sign and coefficient errors during subtraction.