GHOSTS (Mathematical Capabilities of ChatGPT)

How subjects answered

• ChatGPT (2023-01-09)score 1

  The functional equation $\Gamma(x+1) = x \Gamma(x)$ is known as the "recursion formula" or "recursion relation" for the Gamma function.

• ChatGPT (2023-01-30)score 1

  The functional equation $\Gamma(x+1) = x \Gamma(x)$ is known as the "Gamma Function Recurrence Relation."

How subjects answered

• ChatGPT (2023-01-09)score 2

  A function whose Fourier transform is compactly supported is called a 'Schwartz function.' The space of all Schwartz functions is also known as the 'Schwartz space.'

• ChatGPT (2023-01-30)score 2

  A function whose Fourier transform is compactly supported is often referred to as a "Schwartz function". The term "Schwartz function" refers to functions that belong to the Schwartz space, which is a space of rapidly decreasing smooth functions that are well-behaved in both the time and frequency domains. Schwartz functions have numerous useful properties and appear frequently in mathematics and signal processing.

How subjects answered

• ChatGPT (2023-01-09)score 2

  It does not matter on which square the -1 is initially placed because any subsquare of size $a>1$ can be chosen to reverse the signs of all the squares within it. This means that the final configuration of all +1s can be reached regardless of the initial placement of the -1.

• ChatGPT (2023-01-30)score 2

  It is not possible to reach +1 on each square of a $5 \times 5$ chessboard by reversing the signs of $a \times a$ subsquares with $a > 1$. No matter which square has the value -1, it will always be impossible to change it to +1 by reversing the signs of larger subsquares. This is because the total number of squares on the chessboard is odd, and so it is not possible to change an odd number of -1s to +1s.

How subjects answered

• ChatGPT (2023-01-09)score 2

  To find the radius of the circle, we can complete the square to put the equation in standard form: $(x+4)^2 + (y-3)^2 = r^2$.

  We can see that the center of the circle is $(-4,3)$, and the radius is r.

  Therefore the radius of this circle is r.

• ChatGPT (2023-01-30)score 2

  The equation $x^2 + 8x + y^2 - 6y = 0$ is not the equation of a circle in standard form, which is of the form $(x - h)^2 + (y - k)^2 = r^2$. To convert this equation to standard form, we need to complete the square in both $x$ and $y$.

  Completing the square in $x$:

  $x^2 + 8x = x^2 + 8x + 64 = (x + 4)^2 - 64$

  Completing the square in $y$:

  $y^2 - 6y = y^2 - 6y + 9 = (y - 3)^2 - 9$

  Substituting the above results into the original equation, we get:

  $(x + 4)^2 - 64 + (y - 3)^2 - 9 = 0$

  $(x + 4)^2 + (y - 3)^2 = 73$

  This is the equation of a circle with center $(-4, 3)$ and radius $\sqrt{73}$. Thus, the radius of the circle is $\sqrt{73}$.

How subjects answered

• ChatGPT (2023-01-09)score 3

  The given equation can be rewritten as $sin^2 x + cos x + 1 = (sin x + 1)(sin x + cos x) = 0$. Thus, the solutions are $x = arcsin(-1) + 2npi$ and $x = arccos(-1) + 2npi$, where $n$ is an integer. The values of $x$ between $0$ and $2pi$ are:

  $arcsin(-1) + 2npi = \frac{3pi}{2} + 2npi$ $arccos(-1) + 2npi = \frac{pi}{2} + 2npi$

  So the solutions are $\frac{3pi}{2} + 2npi$ and $\frac{pi}{2} + 2npi$ where $n$ is an integer.

• ChatGPT (2023-01-30)score 2

  We know that $\sin^2 x + \cos^2 x = 1,$ so [\sin^2 x + \cos x + 1 = 1 + \cos x + 1 = 2.]Thus, no such value of $x$ exists, so the only solution is $\boxed{}.$

How subjects answered

• ChatGPT (2023-01-09)score 3

  A root of the quadratic equation $2x^2-13x+k$ is a value of $x$ that makes the equation equal to zero. So, we can set the equation equal to zero and substitute the given value for $x$ to find $k$.

  $2\left(\frac{13-\sqrt{131}}{4}\right)^2 - 13\left(\frac{13-\sqrt{131}}{4}\right) + k = 0$

  Solving for $k$, we get:

  $k = \frac{169-26\sqrt{131}+131}{16} = \frac{300-26\sqrt{131}}{16}$

  Therefore, the real value of $k$ is $\frac{300-26\sqrt{131}}{16}$

• ChatGPT (2023-01-30)score 3

  If $\frac{13-\sqrt{131}}{4}$ is a root of $2x^2-13x+k$, then by the Factor Theorem,

  $2\left(\frac{13-\sqrt{131}}{4}\right)^2 - 13\left(\frac{13-\sqrt{131}}{4}\right) + k = 0$.

  Expanding the left side and simplifying gives

  $\frac{131-26\sqrt{131}+169}{16} - \frac{13\cdot13-13\sqrt{131}}{4} + k = 0$,

  which simplifies further to

  $\frac{304-26\sqrt{131}}{16} + k = 0$.

  Therefore,

  $k = -\frac{304-26\sqrt{131}}{16} = \frac{26\sqrt{131}-304}{16}$.

How subjects answered

• ChatGPT (2023-01-09)score 3

  The shortest distance between the two circles is the distance between the centers of the circles minus the sum of their radii. The two circles defined by the equations $x^2-24x +y^2-32y+384=0$ and $x^2+24x +y^2+32y+384=0$ are centered at $(12, 16)$ and $(-12, -16)$, respectively. The radius of each circle can be found by completing the square and setting the equation to zero, which gives a radius of 16. Therefore, the distance between the centers of the circles is $\sqrt{12^2 + (-16-16)^2} = \sqrt{144+256} = \sqrt{400} = 20$. Subtracting the sum of the radii, 16+16 = 32, from the distance between the centers gives 20-32 = -12. Since distance cannot be negative, the shortest distance between …

• ChatGPT (2023-01-30)score 3

  The shortest distance between two circles can be found by finding the perpendicular bisector of the line segment connecting the closest points on the two circles.

  One way to find the distance is to first find the center and radius of each circle, then find the line that bisects the line segment connecting the centers and use the Pythagorean Theorem to find the distance.

  Let the first circle be C1 with center (x1,y1) and radius r1, and the second circle be C2 with center (x2,y2) and radius r2. The distance d between the centers of the two circles is given by:

  d = √((x2-x1)^2 + (y2-y1)^2)

  The equation of the perpendicular bisector of the line segment connecting the two centers is given …

How subjects answered

• ChatGPT (2023-01-30)score 4

  Yes. Since $U$ is non-empty and open, for each $x\in U$, there exists an open interval $(a,b)$ such that $x\in (a,b)\subseteq U$. It follows that the collection $\mathscr{I}$ of pairwise disjoint open intervals covers $U$ and is therefore a countable union of disjoint open intervals. Thus, any open subset of $\Bbb R$ is a countable union of disjoint open intervals.

• ChatGPT (2023-01-09)score 3

  Since $\sim$ is an equivalence relation on $U$, the equivalence classes form a partition of $U$. Since $U$ is open, each equivalence class is also open, and the classes are pairwise disjoint. Therefore, $U$ can be expressed as a countable union of disjoint open intervals, where each interval corresponds to a single equivalence class in $\mathscr{I}$. This completes the proof that any open subset of $\Bbb R$ is a countable union of disjoint open intervals.

How subjects answered

• ChatGPT (2023-01-09)score 4

  and \begin{align*} |2 mathbf{u} - mathbf{v}| &= sqrt{(2 mathbf{u} - mathbf{v}) cdot (2 mathbf{u} - mathbf{v})}
  &= sqrt{4 mathbf{u} cdot mathbf{u} - 4 mathbf{u} cdot mathbf{v} + mathbf{v} cdot mathbf{v}}
  &= sqrt{4 cdot 2^2 - 4(-1) + 2^2}
  &= sqrt{12}. end{align*} So $cos \theta = \frac{(mathbf{u} + mathbf{v}) cdot (2 mathbf{u} - mathbf{v})}{|mathbf{u} + mathbf{v}| |2 mathbf{u} - mathbf{v}|} = \frac{3}{sqrt{6} sqrt{12}} = \frac{1}{2}$.

• ChatGPT (2023-01-30)score 4

  and \begin{align*} |2 \mathbf{u} - \mathbf{v}| &= \sqrt{(2 \mathbf{u} - \mathbf{v}) \cdot (2 \mathbf{u} - \mathbf{v})} \ &= \sqrt{4 \mathbf{u} \cdot \mathbf{u} - 4 \mathbf{u} \cdot \mathbf{v} + \mathbf{v} \cdot \mathbf{v}} \ &= \sqrt{4 \cdot 2^2 - 4(-1) + 2^2} \ &= \sqrt{24}. \end{align*}Therefore, [\cos \theta = \frac{(\mathbf{u} + \mathbf{v}) \cdot (2 \mathbf{u} - \mathbf{v})}{|\mathbf{u} + \mathbf{v}| |2 \mathbf{u} - \mathbf{v}|} = \frac{3}{\sqrt{6} \cdot \sqrt{24}} = \frac{3}{4 \sqrt{6}}.]

How subjects answered

• GPT-4score 5

  To find the projection of vector $\bold{v}$ onto vector $\bold{w}$, we can follow these steps:

  Compute the dot product of $\bold{v}$ and $\bold{w}$: $\bold{v} \cdot \bold{w}$ Compute the dot product of $\bold{w}$ with itself: $\bold{w} \cdot \bold{w}$ Divide the dot product of $\bold{v}$ and $\bold{w}$ by the dot product of $\bold{w}$ and itself: $\frac{\bold{v} \cdot \bold{w}}{\bold{w} \cdot \bold{w}}$ Multiply the result from step 3 by the vector $\bold{w}$. Now let's compute the missing steps:

  $\bold{v} \cdot \bold{w} = (0)(2) + (-4)(2) + (1)(-1) = -8 -1 = -9$ $\bold{w} \cdot \bold{w} = (2)(2) + (2)(2) + (-1)(-1) = 4 + 4 + 1 = 9$ $\frac{\bold{v} \cdot \bold{w}}{\bold{w} \cdot \bold{w}} …

• ChatGPT (2023-01-09)score 4

  To fill in the missing steps, we first need to compute the dot product of $\bold{v}$ and $\bold{w}$, which is $(0)(2) + (-4)(2) + (1)(-1) = -6$. Next, we need to compute the dot product of $\bold{w}$ and itself, which is $(2)(2) + (2)(2) + (-1)(-1) = 9$. Finally, we can substitute these values into the formula for the projection: \text{proj}_{\bold{w}} \bold{v} = \frac{-6}{9} \bold{w} = \begin{pmatrix} -4/3 \ -4/3 \ 2/3 end{pmatrix}

• ChatGPT (2023-01-30)score 4

  We have that \begin{align*} \text{proj}_{\bold{w}} \bold{v} &= \frac{\bold{v} \cdot \bold{w}}{\bold{w} \cdot \bold{w}} \bold{w} \ &= \frac{\begin{pmatrix} 0 \ -4 \ 1 \end{pmatrix} \cdot \begin{pmatrix} 2 \ 2 \ -1 \end{pmatrix}}{\begin{pmatrix} 2 \ 2 \ -1 \end{pmatrix} \cdot \begin{pmatrix} 2 \ 2 \ -1 \end{pmatrix}} \begin{pmatrix} 2 \ 2 \ -1 \end{pmatrix} \ &= \frac{2 + 4 + 1}{5} \begin{pmatrix} 2 \ 2 \ -1 \end{pmatrix} \ &= \frac{7}{5} \begin{pmatrix} 2 \ 2 \ -1 \end{pmatrix} \ &= \begin{pmatrix} \frac{14}{5} \ \frac{14}{5} \ -\frac{7}{5} \end{pmatrix}. \end{align*}