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 sin2(x)\sin^2(x)

We will begin by using the known series expansion for cos(2x)\cos(2x), which is:

cos(2x)=1(2x)22!+(2x)44!(2x)66!+\cos(2x) = 1 - \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} - \frac{(2x)^6}{6!} + \cdots

This simplifies to:

cos(2x)=12x2+2x432x645+\cos(2x) = 1 - 2x^2 + \frac{2x^4}{3} - \frac{2x^6}{45} + \cdots

We want to express sin2(x)\sin^2(x). We can use the identity:

sin2(x)=1cos(2x)2\sin^2(x) = \frac{1 - \cos(2x)}{2}

Substitute the Maclaurin series for cos(2x)\cos(2x):

sin2(x)=1(12x2+2x43)2\sin^2(x) = \frac{1 - \left( 1 - 2x^2 + \frac{2x^4}{3} - \cdots \right)}{2}

Simplify this:

sin2(x)=2x22x43+2\sin^2(x) = \frac{2x^2 - \frac{2x^4}{3} + \cdots}{2}

sin2(x)=x2x43+\sin^2(x) = x^2 - \frac{x^4}{3} + \cdots

Step 2: Find the Limit limx0sin2(x)x2x4\lim_{x \to 0} \frac{\sin^2(x)}{x^2 - x^4}

Now, we compute the limit:

limx0sin2(x)x2x4\lim_{x \to 0} \frac{\sin^2(x)}{x^2 - x^4}

Using the Maclaurin series expansion for sin2(x)\sin^2(x):

sin2(x)=x2x43+\sin^2(x) = x^2 - \frac{x^4}{3} + \cdots

So, the limit becomes:

limx0x2x43+x2x4\lim_{x \to 0} \frac{x^2 - \frac{x^4}{3} + \cdots}{x^2 - x^4}

Factor out x2x^2 from both the numerator and denominator:

=limx0x2(1x23+)x2(1x2)= \lim_{x \to 0} \frac{x^2 \left(1 - \frac{x^2}{3} + \cdots \right)}{x^2 (1 - x^2)}

Canceling the common x2x^2 term:

=limx01x23+1x2= \lim_{x \to 0} \frac{1 - \frac{x^2}{3} + \cdots}{1 - x^2}

As x0x \to 0, both the numerator and denominator approach 1:

limx01x231x2=11=1\lim_{x \to 0} \frac{1 - \frac{x^2}{3}}{1 - x^2} = \frac{1}{1} = 1

Final Answer:

limx0sin2(x)x2x4=1\lim_{x \to 0} \frac{\sin^2(x)}{x^2 - x^4} = 1


Do you need more details on any part of this process or have any questions?

Here are some related questions for further exploration:

  1. How do you find the Maclaurin series for other trigonometric functions, like cos(x)\cos(x)?
  2. Can you derive the Maclaurin series for sin(x)\sin(x) and compare it with the series for sin2(x)\sin^2(x)?
  3. What happens if you apply a higher-order approximation for sin2(x)\sin^2(x) in the limit expression?
  4. How do you expand sin(x)\sin(x) and cos(x)\cos(x) into series and use them for approximations in limits?
  5. How would you derive a series for sin3(x)\sin^3(x) 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