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Magazine Chapter 5: Problem 47
[T] Use a computer algebra system to compute the Riemann sum, \(L_{N}, \quad\) for \(\quad N=10,30,50\) for \(f(x)=\frac{1}{\sqrt{1+x^{2}}}\) on \([-1,1]\)
Short Answer
Expert verified
Use a computer algebra system to evaluate the left Riemann sum, \(L_N\), for each \(N = 10, 30, 50\).
Step by step solution
01
Understand the Problem
We need to calculate the left Riemann sum, denoted as \(L_N\), for the function \(f(x)=\frac{1}{\sqrt{1+x^{2}}}\) on the interval \([-1, 1]\), for three different values of \(N\): 10, 30, and 50.
02
Find the Interval Width
The interval from \(-1\) to \(1\) is \(2\) units long. The width \(\Delta x\) of each sub-interval with \(N\) divisions is given by: \[\Delta x = \frac{b-a}{N} = \frac{2}{N}.\]
03
Set Up the Summation for \(L_N\)
Since this is a left Riemann sum, the sample points \(x_i\) will be the left endpoints of each sub-interval. Therefore, \(x_i = a + i\cdot \Delta x\), where \(i = 0, 1, 2, \ldots, N-1\). The left Riemann sum is calculated as: \[L_N = \sum_{i=0}^{N-1} f(x_i) \cdot \Delta x.\]
04
Apply the Formula for Each \(N\)
For each \(N\), calculate \(\Delta x\) and plug into the Riemann sum formula. Use a computer algebra system to perform these computations. For instance, the function \(f(x) = \frac{1}{\sqrt{1+x^2}}\) is substituted for each \(x_i\).
05
Compute \(L_{10}\)
Using \(N=10\), \(\Delta x = 0.2\):\[L_{10} = 0.2\left(\frac{1}{\sqrt{1+(-1)^2}} + \frac{1}{\sqrt{1+(-0.8)^2}} + \cdots + \frac{1}{\sqrt{1+(-0.2)^2}}\right).\] Calculate this using your computer algebra system.
06
Compute \(L_{30}\)
Using \(N=30\), \(\Delta x = 0.0667\):\[L_{30} = 0.0667\left(\frac{1}{\sqrt{1+(-1)^2}} + \frac{1}{\sqrt{1+(-0.9333)^2}} + \cdots + \frac{1}{\sqrt{1+(-0.1333)^2}}\right).\] Calculate this using your computer algebra system.
07
Compute \(L_{50}\)
Using \(N=50\), \(\Delta x = 0.04\):\[L_{50} = 0.04\left(\frac{1}{\sqrt{1+(-1)^2}} + \frac{1}{\sqrt{1+(-0.96)^2}} + \cdots + \frac{1}{\sqrt{1+(-0.04)^2}}\right).\] Calculate this using your computer algebra system.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Computer Algebra System
A computer algebra system (CAS) is a powerful tool that aids in symbolic mathematics. It can perform algebraic calculations, including simplifications, expansions, and exact solutions of equations. For Riemann sums, a CAS can handle complex summations, especially useful when the number of sub-intervals is large.
When computing the left Riemann sum, sometimes manually calculating the sum for each sub-interval can be time-consuming. This is where a CAS comes in handy.
When computing the left Riemann sum, sometimes manually calculating the sum for each sub-interval can be time-consuming. This is where a CAS comes in handy.
- With a CAS, you can input a function and specify the intervals and summation limits.
- The system then automates the calculation, providing quick and precise results.
Interval Width
The interval width, often represented by \( \Delta x \), is a critical part of calculating Riemann sums. It defines the thickness of each slice of the interval being evaluated. The greater the number of sub-intervals (divisions within the main interval), the smaller the interval width, leading to a more precise approximation of the integral.
To find the interval width in the context of the given Riemann sum, use the formula: \[ \Delta x = \frac{b-a}{N} \] where \( a \) and \( b \) are the endpoints of the interval, and \( N \) is the number of sub-intervals.
Here's why interval width matters:
To find the interval width in the context of the given Riemann sum, use the formula: \[ \Delta x = \frac{b-a}{N} \] where \( a \) and \( b \) are the endpoints of the interval, and \( N \) is the number of sub-intervals.
Here's why interval width matters:
- Smaller widths mean more computations but yield results closer to the true value of the integral.
- In our example, with \( N = 50 \), the interval width is 0.04; with \( N = 10 \), it's 0.2. Notice how a smaller \( \Delta x \) allows for a finer approximation.
Sub-Intervals
Sub-intervals subdivide the total interval into smaller segments. In a Riemann sum, the approximate integral's precision depends significantly on these sub-divisions. Each sub-interval hosts a data point that contributes to the sum. The number of sub-intervals is determined by \( N \), the preselected quantity, which dictates how many slices you cut the entire interval into.
In the exercise, sub-intervals help express the function \( f(x)=\frac{1}{\sqrt{1+x^{2}}} \), like a series of rectangles whose areas are added. With an ideal number of sub-intervals,
In the exercise, sub-intervals help express the function \( f(x)=\frac{1}{\sqrt{1+x^{2}}} \), like a series of rectangles whose areas are added. With an ideal number of sub-intervals,
- The approximation becomes closer to the integral's true value.
- Sub-intervals define each rectangle, and more sub-intervals result in a finer approximation.
- For \( N = 10, 30, 50 \), their respective sub-interval counts allow for different precision levels, with \( N = 50 \) being the most precise.
Left Endpoints
In Riemann sums, particularly the left Riemann sum, computations begin at each sub-interval's leftmost side. These are known as left endpoints. For each sub-interval, the left endpoint is calculated, at which the function is evaluated to estimate its contribution to the total sum.
In mathematical terms, the left endpoint \( x_i \) for the \( i \)-th sub-interval is given by: \[ x_i = a + i \cdot \Delta x \] where \( a \) is the start of the interval and \( i \) ranges from 0 to \( N-1 \. \)
Why left endpoints are important:
In mathematical terms, the left endpoint \( x_i \) for the \( i \)-th sub-interval is given by: \[ x_i = a + i \cdot \Delta x \] where \( a \) is the start of the interval and \( i \) ranges from 0 to \( N-1 \. \)
Why left endpoints are important:
- They define where the slices begin for calculation purposes in a left Riemann sum.
- The choice of endpoint (left, right, or midpoint) affects the terminus of where the function value is calculated.
- In our exercise, using left endpoints means starting evaluations from the segment's beginning, leading to slightly different sums than if using right or midpoints.
