I gave ChatGPT the midterm. It scored around 60/100. (Update: I just graded the midterm and was more generous, so 65/100.) It failed miserably on the graphing question, which is maybe expected since it doesn’t have the ability scan in a graph. After some prodding, it produced the following:
which is a somewhat heroic effort in the circumstances.
Let me give ChatGPT’s answers as generated entirely by ChatGPT, making some comments as we go along.
Question 1:
Suppose that \(\alpha\) and \(\beta\) are numbers such that \(\alpha \beta\) and \(\alpha + \beta\) are both integers. Prove that \(\alpha^n + \beta^n\) is an integer for each \(n \in \mathbb{N}\).
Solution:
We are given that \(\alpha \beta\) and \(\alpha + \beta\) are integers. We want to show that \(\alpha^n + \beta^n\) is an integer for each \(n \in \mathbb{N}\).
Let \(s_n = \alpha^n + \beta^n\). We will use a recurrence relation to prove that \(s_n\) is an integer for all \(n \geq 0\).
From the quadratic equation with roots \(\alpha\) and \(\beta\):
\[
x^2 – (\alpha + \beta)x + \alpha \beta = 0,
\]
we have the recurrence relation:
\[
s_{n+2} = (\alpha + \beta) s_{n+1} – \alpha \beta s_n.
\]
Initial conditions:
\[
s_0 = \alpha^0 + \beta^0 = 2, \quad s_1 = \alpha + \beta.
\]
Both are integers by assumption. By induction, assume \(s_n\) and \(s_{n+1}\) are integers for some \(n \geq 0\). Then, by the recurrence relation:
\[
s_{n+2} = (\alpha + \beta) s_{n+1} – \alpha \beta s_n,
\]
which is an integer because \(\alpha + \beta\), \(\alpha \beta\), \(s_n\), and \(s_{n+1}\) are all integers.
Thus, by the principle of induction, \(s_n = \alpha^n + \beta^n\) is an integer for all \(n \geq 0\).
Frank’s Comments: I would prefer not being so cavalier with writing \(\alpha^0 = 1\) when \(\alpha\) could be zero, but this answer is more or less correct. I would also have preferred that instead of talking about recurrence relations one could just prove that formula directly. That said, with generous grading \(25/25\).
Question 2:
Determine if the following claims are true or false. Justify your answer with a proof.
(a) There exists a function with domain \(\mathbb{R}\) and range the integers.
Solution:
This statement is true. A simple example is the floor function \(\lfloor x \rfloor\), which takes a real number \(x\) and returns the largest integer less than or equal to \(x\). The floor function has domain \(\mathbb{R}\) and range \(\mathbb{Z}\).
(b) The zero function is defined as the function which returns zero for all real numbers — it has domain \(\mathbb{R}\) and range \(0\). Suppose that \(f\) and \(g\) are two functions with domain \(\mathbb{R}\) whose product is equal to the zero function. Then one of \(f\) or \(g\) is the zero function.
Solution:
This statement is false. Consider the functions \(f(x) = x\) and \(g(x) = 0\). Their product is zero for all \(x\), but neither function is identically zero over \(\mathbb{R}\).
Frank’s Comments The function \(g(x)\) is the zero function here! One can easily modify this with \(g(x)=0\) for \(x \ne 0\) and \(g(0) = 1\). With generous grading \(5/10\).
(c) If the domain of \(f\) is all of \(\mathbb{R}\), then the domain of the function \(g(x) = \frac{1}{1 + f(x)^2}\) is also all of \(\mathbb{R}\).
Solution:
This statement is true. Since \(f(x)^2 \geq 0\) for all \(x \in \mathbb{R}\), we have \(1 + f(x)^2 \geq 1\), which means the denominator of \(g(x)\) is never zero. Therefore, the function \(g(x)\) is well-defined for all \(x \in \mathbb{R}\).
Question 3:
Let \(\Gamma(x)\) be a function with domain \((-1, 0) \cup (0, 5]\), such that \(\Gamma(1) = 1\) and \(\Gamma(x+1) = x \Gamma(x)\) for all \(x\) in the domain of \(\Gamma(x)\). Draw the graph of \(\Gamma(x)\), assuming that for \(x \in [1, 5]\), the graph is provided.
Solution:
We can compute the values of \(\Gamma(x)\) for \(x \in [1, 5]\) using the recurrence relation. We know that \(\Gamma(1) = 1\). From the relation \(\Gamma(x+1) = x \Gamma(x)\), we get:
\[
\Gamma(2) = 1 \cdot \Gamma(1) = 1,
\]
\[
\Gamma(3) = 2 \cdot \Gamma(2) = 2,
\]
\[
\Gamma(4) = 3 \cdot \Gamma(3) = 6,
\]
\[
\Gamma(5) = 4 \cdot \Gamma(4) = 24.
\]
Thus, we have the points \(\Gamma(1) = 1\), \(\Gamma(2) = 1\), \(\Gamma(3) = 2\), \(\Gamma(4) = 6\), and \(\Gamma(5) = 24\).
Frank’s Comments: Since ChatGPT couldn’t read the graph, there is not much content here; \(0/20\).
Question 4:
Let \(a_n\) and \(b_n\) be the sequences defined by \(a_0 = b_0 = 1\) and the recurrence relations:
\[
a_{n+1} = 3^{a_n}, \quad b_{n+1} = 2^{b_n}.
\]
Prove the following:
(a) Prove that \(a_n > b_{n+1}\) for \(n \geq 2\).
Solution:
We will use induction to prove that \(a_n > b_{n+1}\) for \(n \geq 2\).
Base case:
For \(n = 2\), we have:
\[
a_2 = 3^{a_1} = 3^1 = 3, \quad b_3 = 2^{b_2} = 2^2 = 4.
\]
Thus, \(a_2 = 3 < b_3 = 4\), which satisfies the inequality.
Inductive step:
Assume \(a_n > b_{n+1}\) for some \(n \geq 2\). We want to show that \(a_{n+1} > b_{n+2}\). We have:
\[
a_{n+1} = 3^{a_n}, \quad b_{n+2} = 2^{b_{n+1}}.
\]
By the inductive hypothesis, \(a_n > b_{n+1}\), so:
\[
a_{n+1} = 3^{a_n} > 3^{b_{n+1}} > 2^{b_{n+1}} = b_{n+2}.
\]
Thus, the inequality holds for \(n+1\), completing the induction.
(b) Prove that \(b_{n+2} > a_n\) for all \(n \geq 0\).
Solution:
We proceed similarly by induction.
Base case:
For \(n = 0\), we have:
\[
b_2 = 2^{b_1} = 2^1 = 2 > a_0 = 1.
\]
Thus, \(b_2 > a_0\) holds.
Inductive step:
Assume \(b_{n+2} > a_n\) for some \(n \geq 0\). We want to show that \(b_{n+3} > a_{n+1}\). We have:
\[
b_{n+3} = 2^{b_{n+2}}, \quad a_{n+1} = 3^{a_n}.
\]
By the inductive hypothesis, \(b_{n+2} > a_n\), so:
\[
b_{n+3} = 2^{b_{n+2}} > 3^{a_n} = a_{n+1}.
\]
Thus, the inequality holds for \(n+1\), completing the induction.
Frank’s Comments: Of course if \(b > a\) then it is not necessarily the case that \(2^b > 3^a\), so the second part is wrong, \(15/25\), for a total of
\[25+25+15=65\]