Math Problem Statement
Find the Maclaurin series for sin2(x) using the series for cos(2x). Hence, find limx→0 sin2(x) x2−x4
Solution
Step 1: Find the Maclaurin Series for
We will begin by using the known series expansion for , which is:
This simplifies to:
We want to express . We can use the identity:
Substitute the Maclaurin series for :
Simplify this:
Step 2: Find the Limit
Now, we compute the limit:
Using the Maclaurin series expansion for :
So, the limit becomes:
Factor out from both the numerator and denominator:
Canceling the common term:
As , both the numerator and denominator approach 1:
Final Answer:
Do you need more details on any part of this process or have any questions?
Here are some related questions for further exploration:
- How do you find the Maclaurin series for other trigonometric functions, like ?
- Can you derive the Maclaurin series for and compare it with the series for ?
- What happens if you apply a higher-order approximation for in the limit expression?
- How do you expand and into series and use them for approximations in limits?
- How would you derive a series for using a similar method?
Tip: Always check the first few terms of your series expansion before using them for approximations in limits!
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Math Problem Analysis
Mathematical Concepts
Maclaurin series
Limits
Trigonometric identities
Series expansion
Formulas
cos(2x) = 1 - 2x² + (2x⁴)/3 - (2x⁶)/45 + ...
sin²(x) = (1 - cos(2x)) / 2
limₓ→0 f(x) / g(x) = f(0) / g(0) for well-defined functions
Theorems
Maclaurin Series Expansion
Trigonometric Identities
Limit Laws
Suitable Grade Level
Undergraduate Mathematics or Advanced High School
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