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Chapter 7: Problem 6

If \(\psi:[a, b] \rightarrow \mathbb{R}\) takes on only a finite number of distinct values, is \(\psi\) a step function?

Short Answer

Expert verified
No, the function \(\psi:[a, b] \rightarrow \mathbb{R}\) that takes only a finite number of distinct values is not necessarily a step function as it doesn't necessarily meet all the properties of a step function.

Step by step solution

01

Understanding the Description of the Function

Let's first understand the description of the function \(\psi\). The function \(\psi:[a, b] \rightarrow \mathbb{R}\) is defined over the interval [a, b] (inclusive of a and b) and the function takes real values. Due to the assumption that the number of distinct values it takes is finite, we have a limited and enumerable set of outputs for the function. This does not directly tell us if it's a step function, but it gives us a characteristic quite common in step functions.
02

Comparing with the Definition of a Step Function

Next, we will compare our understanding of the function \(\psi\) with the definition of a step function. A step function is defined such that, on certain intervals, the function is constant. Now, note that our function \(\psi\) doesn't necessarily conform to this description. While \(\psi\) takes a finite number of values, there is not necessarily a consistent pattern or 'steps', as in a step function.
03

Formulating the Conclusion

Having compared the properties of the function \(\psi\) with that of a step function, we can reach a conclusion. While \(\psi\) shares some properties with a step function, it does not meet all of the needed properties to be considered a step function itself. A step function is only constant on certain finite subintervals, with abrupt 'steps' or changes at the end of each subinterval. The function \(\psi\) can potentially have changes at any point, not necessarily with the segmentation of intervals like a step function.

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Most popular questions from this chapter

If \(g \in \mathcal{R}[a, b]\) and if \(f(x)=g(x)\) except for a finite number of points in \([a, b]\), prove that \(f \in \mathcal{R}[a, b]\) and that \(\int_{a}^{b} f=\int_{a}^{b} g\)

Show that \(g(x):=\sin (1 / x)\) for \(x \in(0.1]\) and \(g(0):=0\) belongs to \(\mathcal{R}[0,1]\).

Let \(B(x):=-\frac{1}{2} x^{2}\) for \(x<0\) and \(B(x):=\frac{1}{2} x^{2}\) for \(x \geq 0 .\) Show that \(\int_{a}^{b}|x| d x=B(b)-B(a)\).

If \(f \in \mathcal{R}[a, b]\) and \(c \in \mathbb{R}\), we define \(g\) on \([a+c, b+c]\) by \(g(y):=f(y-c)\). Prove that \(g \in \mathcal{R}[a+c, b+c]\) and that \(\int_{a+c}^{b+c} g=\int_{a}^{b} f .\) The function \(g\) is called the \(c\) -translate of \(f\).

Use the Simpson Approximation with \(n=4\) to evaluatc \(\ln 2=\int_{1}^{2}(1 / x) d x\). Show that \(0.6927 \leq\) \(\ln 2 \leq 0.6933\) and that $$ 0.000016<\frac{1}{2^{5}} \cdot \frac{1}{1920} \leq S_{4}-\ln 2 \leq \frac{1}{1920}<0.000521 $$
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