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The fraction in a single precision word has 23 bits (alas, less than half the length of the double precision word). Show that the corresponding rounding unit is approximately \(6 \times 10^{-8}\).

Short Answer

Expert verified
Answer: The rounding unit for a single-precision word with 23 bits is approximately \(6 \times 10^{-8}\).

Step by step solution

01

Identify the number of bits in the fraction

In this exercise, the fraction in a single-precision word has 23 bits.
02

Use the formula for the rounding unit

The formula for the rounding unit is given as \(0.5 \times 2^{-p}\), where p is the number of bits in the fraction part of the single-precision number. In this case, we have p = 23.
03

Calculate the rounding unit

Plugging the value of p (23) into the formula \(0.5 \times 2^{-p}\), we get: \(0.5 \times 2^{-23} = \frac{1}{2} \times 2^{-23} = 2^{-24} = \frac{1}{2^{24}} \approx 5.96 \times 10^{-8}\)
04

Compare the result to the given approximation

The calculated rounding unit is approximately \(5.96 \times 10^{-8}\). Comparing this to the given approximation, \(6 \times 10^{-8}\), we can see that they are approximately equal, confirming that the rounding unit for a single-precision word with 23 bits is indeed approximately \(6 \times 10^{-8}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Floating Point Representation
In computing, numbers are often represented in a format known as floating point representation. This is particularly important when working with very large or very small numbers, which might not fit into an integer format. The floating point representation approximates real numbers by using a fixed number of significant digits and a floating radix point (or decimal point).

A single precision floating point number typically uses 32 bits, where:
  • 1 bit represents the sign of the number (positive or negative).
  • 8 bits make up the exponent, determining the scale or magnitude.
  • 23 bits form the mantissa or fractional part, capturing the significant digits of the number.
The mantissa's 23 bits imply that the precision is limited, leading to possible approximations when representing certain decimals. This means some numbers can only be approximately stored, especially those requiring more than 23 bits of precision.
Rounding Error
Rounding errors occur when calculations involve numbers with a degree of precision exceeding what can be represented with limited bits. In single precision, the 23-bit fraction means that computations need to "round" numbers to fit this representation.

The rounding unit can help quantify the error introduced through rounding. It's the smallest difference between numbers that can be represented. Using the formula \(0.5 \times 2^{-p}\), where \(p\) is the number of bits in the mantissa (23), we get:
  • \(0.5 \times 2^{-23} = 2^{-24}\)
  • This equals approximately \(5.96 \times 10^{-8}\).
Such small errors, while tiny, can accumulate in iterative computations, necessitating careful consideration during numerical analysis.
Numerical Analysis
Numerical analysis focuses on creating algorithms that can solve mathematical problems with floating point numbers. Given the limitations of floating point representation, especially in single precision, errors can appear and affect the result's accuracy.

To manage and minimize errors, numerical analysts:
  • Use algorithms that are stable, meaning small changes in initial data result in small errors in the results.
  • Employ methods that reduce the impact of rounding; for example, compensating for errors in iterative processes.
  • Conduct error analysis to understand and mitigate the effects of rounding errors, ensuring the outcome remains as accurate as needed for the application.
Numerical analysis provides the tools to make accurate scientific and engineering calculations despite the innate limitations of floating point arithmetic.

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Most popular questions from this chapter

Write a MATLAB program that receives as input a number \(x\) and a parameter \(n\) and returns \(x\) rounded to \(n\) decimal digits. Write your program so that it can handle an array as input, returning an array of the same size in this case. Use your program to generate numbers for Example \(2.2\), demonstrating the phenomenon depicted there without use of single precision.

The function \(f_{1}(x, \delta)=\cos (x+\delta)-\cos (x)\) can be transformed into another form, \(f_{2}(x, \delta)\), using the trigonometric formula $$ \cos (\phi)-\cos (\psi)=-2 \sin \left(\frac{\phi+\psi}{2}\right) \sin \left(\frac{\phi-\psi}{2}\right) . $$ Thus, \(f_{1}\) and \(f_{2}\) have the same values, in exact arithmetic, for any given argument values \(x\) and \(\delta\). (a) Show that, analytically, \(f_{1}(x, \delta) / \delta\) or \(f_{2}(x, \delta) / \delta\) are effective approximations of the function \(-\sin (x)\) for \(\delta\) sufficiently small. (b) Derive \(f_{2}(x, \delta)\). (c) Write a MATLAB script which will calculate \(g_{1}(x, \delta)=f_{1}(x, \delta) / \delta+\sin (x)\) and \(g_{2}(x, \delta)=\) \(f_{2}(x, \delta) / \delta+\sin (x)\) for \(x=3\) and \(\delta=1 . \mathrm{e}-11 .\) (d) Explain the difference in the results of the two calculations.

(a) Explain in detail how to avoid overflow when computing the \(\ell_{2}\) -norm of a (possibly large in size) vector. (b) Write a MATLAB script for computing the norm of a vector in a numerically stable fashion. Demonstrate the performance of your code on a few examples.

Write a MATLAB program that (a) sums up \(1 / n\) for \(n=1,2, \ldots, 10,000\); (b) rounds each number \(1 / n\) to 5 decimal digits and then sums them up in 5 -digit decimal arithmetic for \(n=1,2, \ldots, 10,000 ;\) (c) sums up the same rounded numbers (in 5 -digit decimal arithmetic) in reverse order, i.e., for \(n=10,000, \ldots, 2,1\). Compare the three results and explain your observations. For generating numbers with the requested precision, you may want to do Exercise 6 first.

For the following expressions, state the numerical difficulties that may occur, and rewrite the formulas in a way that is more suitable for numerical computation: (a) \(\sqrt{x+\frac{1}{x}}-\sqrt{x-\frac{1}{x}}\), where \(x \gg 1\). (b) \(\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}}\), where \(a \approx 0\) and \(b \approx 1\).

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