Math Problem Statement

given sin(x^2)dx [0,1] give an upper bound for the absolute error that results from approximations and round to 6 decimal places. also, how large do we need to choose n so that each of the approximations is accurate to within 0.001?

Solution

Step 1: Understanding the Problem

We need to approximate the integral:

I=01sin(x2)dxI = \int_0^1 \sin(x^2)\, dx

We aim to:

  1. Estimate an upper bound for the absolute error that results from using numerical approximations.
  2. Determine how large to choose nn (number of intervals) so that the approximation is accurate to within 0.0010.001.

Step 2: Use of Trapezoidal Rule for Error Bound

For a smooth function like sin(x2)\sin(x^2), we can use the Trapezoidal Rule to approximate the integral. The error in the Trapezoidal Rule for nn intervals is given by:

ET=(ba)312n2f(ξ)E_T = -\frac{(b-a)^3}{12n^2} \cdot f''(\xi)

where ξ[a,b]\xi \in [a, b], and f(x)=sin(x2)f(x) = \sin(x^2). The second derivative of f(x)f(x), f(x)f''(x), will help in estimating the error bound.

Step 3: Deriving f(x)f'(x) and f(x)f''(x)

  1. First derivative f(x)f'(x): f(x)=ddxsin(x2)=2xcos(x2)f'(x) = \frac{d}{dx} \sin(x^2) = 2x \cos(x^2)

  2. Second derivative f(x)f''(x): f(x)=ddx(2xcos(x2))=2cos(x2)4x2sin(x2)f''(x) = \frac{d}{dx} \left( 2x \cos(x^2) \right) = 2\cos(x^2) - 4x^2\sin(x^2)

Step 4: Estimating Maximum of f(x)f''(x)

To estimate the error, we need to determine the maximum value of f(x)|f''(x)| over the interval [0,1][0, 1]. Since sin(x2)\sin(x^2) and cos(x2)\cos(x^2) are bounded between -1 and 1, we can calculate:

Maximum of f(x) on [0,1]2\text{Maximum of } |f''(x)| \text{ on } [0, 1] \approx 2

This is a rough upper estimate, based on the fact that both terms 2cos(x2)2\cos(x^2) and 4x2sin(x2)-4x^2\sin(x^2) are bounded in magnitude by 2.

Step 5: Error Bound Formula

The error bound for the Trapezoidal Rule is now approximately:

ET(10)312n22=16n2E_T \approx \frac{(1-0)^3}{12n^2} \cdot 2 = \frac{1}{6n^2}

Step 6: Solving for nn

We want the error to be less than 0.0010.001:

16n20.001\frac{1}{6n^2} \leq 0.001

Solving for nn:

n216×0.001=10.006=166.67n^2 \geq \frac{1}{6 \times 0.001} = \frac{1}{0.006} = 166.67

n166.6712.91n \geq \sqrt{166.67} \approx 12.91

Thus, n13n \geq 13 will ensure that the error is less than 0.001.

Step 7: Conclusion

  • Upper bound for absolute error: 16n2\frac{1}{6n^2}
  • Minimum value of nn for an error less than 0.0010.001: n13n \geq 13

Final Answer:

  • The absolute error can be bounded by 16n2\frac{1}{6n^2}, where nn is the number of intervals.
  • To ensure an error of less than 0.0010.001, n13n \geq 13.

Would you like more details on how to refine these bounds, or perhaps a demonstration of this approximation? Here are some related questions:

  1. How does the Simpson’s rule compare with the Trapezoidal rule for this problem?
  2. What is the geometric intuition behind why the Trapezoidal Rule error decreases with increasing nn?
  3. How does the behavior of f(x)=sin(x2)f(x) = \sin(x^2) affect the error in different numerical methods?
  4. Can we use adaptive quadrature methods for better accuracy without increasing nn?
  5. What are some common pitfalls in estimating error bounds for oscillating functions like sin(x2)\sin(x^2)?

Tip: When approximating integrals, increasing the number of intervals nn or switching to a more accurate rule (e.g., Simpson’s Rule) can drastically reduce error.

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Math Problem Analysis

Mathematical Concepts

Numerical Integration
Error Estimation
Trapezoidal Rule
Calculus

Formulas

Integral: I = ∫_0^1 sin(x^2) dx
Error Bound (Trapezoidal Rule): E_T = -((b - a)^3 / (12n^2)) ⋅ f''(ξ)
Second Derivative: f''(x) = 2cos(x^2) - 4x^2sin(x^2)

Theorems

Error Bound Theorem for Trapezoidal Rule

Suitable Grade Level

Undergraduate Level