Chapter 9: Problem 30
The trapezium rule estimate for \(\int_{0}^{1} \mathrm{e}^{x^{2}} \mathrm{~d} x\) with \(h=0.25\) is \(1.49068\) to \(5 \mathrm{dp}\). Estimate the size of the global truncation error in this approximation and show that $$ 1.40 \leqslant \int_{0}^{1} \mathrm{e}^{x^{2}} \mathrm{~d} x<1.48 $$ What value of \(h\) will give an answer correct to \(4 \mathrm{dp}\) ?
Short Answer
Step by step solution
Global Truncation Error Formula
Find the Second Derivative
Evaluate the Maximum of \(|f''(x)|\)
Calculate the Global Truncation Error
Utilize Trapezium Rule Estimate
Determine New \(h\) for Desired Precision
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Trapezium Rule
You would start by choosing a step size \( h \), and over the interval \([0, 1]\), calculate the function values at each point \( x_i \), with \( x_i = a + i \, h \) for \( i = 0, 1, \, ..., \, n \) where \( n \) is the number of segments. The trapezium rule formula for this integral would be:
\[ \int_a^b f(x) \, dx \approx \frac{h}{2} \left( f(a) + 2 \sum_{i=1}^{n-1} f(x_i) + f(b) \right) \]
Using this approach, we create an estimate for \( \int_{0}^{1} e^{x^2} \) which yields the value of \( 1.49068 \) with \( h = 0.25 \). The smaller \( h \) is, the more precise the estimate tends to be.
Global Truncation Error
To compute the global truncation error for the trapezium rule, we use:
\[ -\frac{(b-a)^3}{12n^2}f''(\xi) \]
where \( f''(\xi) \) is an evaluation of the second derivative at some point within the interval \([a, b]\). This value is crucial as it determines the size of the error.
After evaluating, we found that the second derivative's maximum on \([0, 1]\) occurred at \( x = 1 \), leading to \( 6e \) as the maximum value. This resulted in an estimated global truncation error of approximately \(-0.08541\), which helped verify the range of the integral from \(1.40527\) to \(1.57609\).
Derivative Evaluation
First, we found the first derivative:
\[ f'(x) = 2x \cdot e^{x^2} \]
This was followed by the second derivative:
\[ f''(x) = 2e^{x^2} + 4x^2 \cdot e^{x^2} = 2e^{x^2} (1+2x^2) \]
With these derivatives, we can evaluate their values across the interval to identify maximum values, using them for calculating the global truncation error. Evaluating at \( x = 1 \), the second derivative peaked at \( 6e \), providing a key component in error analysis.
Precision in Calculations
In our example, we wanted to find \( h \) that results in an error less than \( 0.00005 \) to ensure precision to four decimal places. After rearranging and solving the error formula:
\[ \frac{1}{12n^2} \cdot 6e = 0.00005 \]
We calculated \( n \approx 45.71 \), needing some approximation to the next whole number. Hence, \( n = 46 \). Consequently, the new step size \( h \approx 0.02174 \) satisfies the required precision, assuring our results are accurate within four decimal places.