/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 Let \(S=\\{0,1\\}\) and \(F=R\).... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 7

Let \(S=\\{0,1\\}\) and \(F=R\). In \(\mathcal{F}(S, R)\), show that \(f=g\) and \(f+g=h\), where \(f(t)=2 t+1, g(t)=1+4 t-2 t^{2}\), and \(h(t)=5^{t}+1 .\)

Short Answer

Expert verified
In the vector space \(\mathcal{F}(S, R)\), after evaluating \(f(t)=2t+1\), \(g(t)=1+4t-2t^2\), and \(h(t)=5^t+1\) for the set \(S=\{0,1\}\), we find the function values: - \(f(0) = 1\), \(f(1) = 3\) - \(g(0) = 1\), \(g(1) = 3\) - \(h(0) = 2\), \(h(1) = 6\) Since \(f(t)\) and \(g(t)\) have equal function values for all elements in the set \(S\), we conclude that \(f = g\). Furthermore, as \(f(t) + g(t)\) equals \(h(t)\) for all elements in the set \(S\), we conclude that \(f + g = h\).

Step by step solution

01

Evaluate functions on S

Given \(f(t) = 2t + 1\), \(g(t) = 1 + 4t - 2t^2\), and \(h(t) = 5^t + 1\), we first need to evaluate these functions for each element of the set \(S = \{0, 1\}\). For \(f(t)\), when \(t = 0\): \(f(0) = 2(0) + 1 = 1\) When \(t = 1\): \(f(1) = 2(1) + 1 = 3\) For \(g(t)\), when \(t = 0\): \(g(0) = 1 + 4(0) - 2(0)^2 = 1\) When \(t = 1\): \(g(1) = 1 + 4(1) - 2(1)^2 = 3\) For \(h(t)\), when \(t = 0\): \(h(0) = 5^0 + 1 = 2\) When \(t = 1\): \(h(1) = 5^1 + 1 = 6\) Now we have the following function values for the elements of the set \(S\): - \(f(0) = 1\), \(f(1) = 3\) - \(g(0) = 1\), \(g(1) = 3\) - \(h(0) = 2\), \(h(1) = 6\)
02

Compare functions f and g

Now that we've computed the function values for each element of S, we should compare the results of \(f(t)\) and \(g(t)\) for each element in S. For \(t = 0\), we found that \(f(0) = 1\) and \(g(0) = 1\). As they are equal, this part is true. For \(t = 1\), we found that \(f(1) = 3\) and \(g(1) = 3\). As they are equal, this part is true as well. Since the function values of \(f(t)\) and \(g(t)\) are equal for all elements in the set \(S\), we can conclude that \(f = g\).
03

Verify if f + g = h

Next, we should verify if the sum \(f(t) + g(t)\) equals \(h(t)\) for each element of the set \(S\). For \(t = 0\): \(f(0) + g(0) = 1 + 1 = 2\) \(h(0) = 2\) For \(t = 1\): \(f(1) + g(1) = 3 + 3 = 6\) \(h(1) = 6\) Upon comparing, we can see that \(f(0) + g(0) = h(0)\) and \(f(1) + g(1) = h(1)\). Thus, we can conclude that \(f + g = h\).

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