Challenge Math Challenge - November 2020

[member 587159]

What is a Euclidean normal form, and what do we mean by $(x,y)^{\tau}$?

I believe it means the vector $(x,y)$ transposed, so written as column vector $\begin{pmatrix}x \\ y\end{pmatrix}$.

 

[fresh_42]

What is a Euclidean normal form, and what do we mean by $(x,y)^{\tau}$?

Yes, it is only a habit in this case to write vectors or coordinates as column. The Euclidean normal form is, when the axis of the conic sections are the coordinate axis, and its center the origin of the Cartesian coordinate system: rotations, translations; basically such that all interesting quantities can directly be read.

 

[etotheipi]

So if I understand correctly,

Spoiler: P5

$$A_Q = \begin{pmatrix} 34 & 12 & 10 \\ 12 & 41 & 55 \\ 10 & 55 & 50\end{pmatrix}$$and $f(x,y) = \mathbf{x}^{\tau}A_Q \mathbf{x}$, $\text{det}(A_Q) = -31250$, and $\text{det}(A_{33}) = 1250$, and $\frac{\text{det}(A_Q)}{\text{det}(A_{33})} = -25$, so I guess when you say you want it in normal form, since $A_{33}$ has eigenvalues $\lambda = 25, 50$, it will be$$50u^2 + 25v^2 = 25 \iff 2u^2 + v^2 = 1 \iff \left(\frac{u}{1/\sqrt{2}}\right)^2 + \left(\frac{v}{1}\right)^2 = 1$$so vertices at $(u,v) = \left( \pm 1/\sqrt{2}, 0\right), \left( 0, \pm 1\right)$ and foci at $(0, \pm \frac{1}{2})$? Is that what you had in mind, @Fred Wright?

 

[fresh_42]

So if I understand correctly,

Spoiler: P5

$$A_Q = \begin{pmatrix} 34 & 12 & 10 \\ 12 & 41 & 55 \\ 10 & 55 & 50\end{pmatrix}$$and $f(x,y) = \mathbf{x}^{\tau}A_Q \mathbf{x}$, $\text{det}(A_Q) = -31250$, and $\text{det}(A_{33}) = 1250$, and $\frac{\text{det}(A_Q)}{\text{det}(A_{33})} = -25$, so I guess when you say you want it in normal form, since $A_{33}$ has eigenvalues $\lambda = 25, 50$, it will be$$50u^2 + 25v^2 = 25 \iff 2u^2 + v^2 = 1 \iff \left(\frac{u}{1/\sqrt{2}}\right)^2 + \left(\frac{v}{1}\right)^2 = 1$$so vertices at $(u,v) = \left( \pm 1/\sqrt{2}, 0\right), \left( 0, \pm 1\right)$ and foci at $(0, \pm \frac{1}{2})$? Is that what you had in mind, @Fred Wright?

Well, you could have shown how you did it. It's a bit difficult to learn anything from your solution. And the foci are $(0,\pm 1/\sqrt{2})$ if I wasn't mistaken.

 

[Office_Shredder]

As far as making questions more accessible, you could have just asked something like construct a sequence of functions that are differentiable whose limit is not differentiable, and not scared off people that haven't heard of a Banach space.

 

[fresh_42]

As far as making questions more accessible, you could have just asked something like construct a sequence of functions that are differentiable whose limit is not differentiable, and not scared off people that haven't heard of a Banach space.

That isn't what the problem asks for.

 

[etotheipi]

Well, you could have shown how you did it. It's a bit difficult to learn anything from your solution. And the foci are $(0,\pm 1/\sqrt{2})$ if I wasn't mistaken.

Yes, sorry, I forgot to square root. Anyway, I'll give a more complete method here for anyone that wants it. First we need to find the centre of $$f(x,y) = 34x^2 + 24xy +41y^2 + 20x + 110y + 50=0$$Set $f_y = 0$ and $f_x = 0$ to get$$\begin{align*}68x + 24y + 20 &= 0 \\
82y + 24x + 110&=0
\end{align*}$$this is solved by $\mathbf{x}_c = (x_c,y_c)^{\tau} = (\frac{1}{5}, \frac{-7}{5})^{\tau}$, i.e. this is the centre of the ellipse. Now we perform a coordinate transformation $\mathbf{x}' = \mathbf{x} -\mathbf{x}_c$ so that the centre is at our new origin $\mathcal{O}'$, i.e.$$34(x' + \frac{1}{5}) + 24(x' + \frac{1}{5})(y' - \frac{7}{5}) + 41(y' - \frac{7}{5})^2 + 20(x' + \frac{1}{5}) + 110(y' - \frac{7}{5}) + 50 = 0$$ $$34x'^2 + 24x'y' + 41y'^2 = 25$$Now we perform a further coordinate transformation, this time a rotation (preserving the origin, $\mathcal{O}'' = \mathcal{O}'$), so that the new axis intersect the ellipse at the vertices. This will be $\mathbf{x}'' = R\mathbf{x}'$, where $R = \begin{pmatrix} c & s \\ -s & c\end{pmatrix}$, and $c=\cos{\theta}$, $s = \sin{\theta}$ for some angle $\theta$ to be determined. Running through the algebra with $x' = cx'' - sy''$ and $y' = sx'' + cy''$ yields$$34(cx'' -sy'')^2 + 24(cx'' - sy'')(sx'' + cy'') + 41(sx'' + cy'') = 25$$ $$x''^2 [ 34 c^2 + 41s^2 + 24cs] + y''^2 [41c^2 + 34s^2 - 24cs] + x''y''[24c^2 - 24s^2 + 14cs]$$We require that the coefficient of $x''y''$ is brought to be zero, i.e. that $s = 4/5$ and $c=3/5$. Now it's just a case of substituting those back into the equation,$$50x''^2 + 25y''^2 = 25 \iff 2x''^2 + y''^2 = 1$$and for brevity I'll re-label $x'' \equiv u$ and $y'' \equiv v$. And that completes the transformation!

 

[Mayhem]

Yes, sorry, I forgot to square root. Anyway, I'll give a more complete method here for anyone that wants it. First we need to find the centre of $$f(x,y) = 34x^2 + 24xy +41y^2 + 20x + 110y + 50=0$$Set $f_y = 0$ and $f_x = 0$ to get$$\begin{align*}68x + 24y + 20 &= 0 \\
82y + 24x + 110&=0
\end{align*}$$this is solved by $\mathbf{x}_c = (x_c,y_c)^{\tau} = (\frac{1}{5}, \frac{-7}{5})^{\tau}$, i.e. this is the centre of the ellipse. Now we perform a coordinate transformation $\mathbf{x}' = \mathbf{x} -\mathbf{x}_c$ so that the centre is at our new origin $\mathcal{O}'$, i.e.$$34(x' + \frac{1}{5}) + 24(x' + \frac{1}{5})(y' - \frac{7}{5}) + 41(y' - \frac{7}{5})^2 + 20(x' + \frac{1}{5}) + 110(y' - \frac{7}{5}) + 50 = 0$$ $$34x'^2 + 24x'y' + 41y'^2 = 25$$Now we perform a further coordinate transformation, this time a rotation (preserving the origin, $\mathcal{O}'' = \mathcal{O}'$), so that the new axis intersect the ellipse at the vertices. This will be $\mathbf{x}'' = R\mathbf{x}'$, where $R = \begin{pmatrix} c & s \\ -s & c\end{pmatrix}$, and $c=\cos{\theta}$, $s = \sin{\theta}$ for some angle $\theta$ to be determined. Running through the algebra with $x' = cx'' - sy''$ and $y' = sx'' + cy''$ yields$$34(cx'' -sy'')^2 + 24(cx'' - sy'')(sx'' + cy'') + 41(sx'' + cy'') = 25$$ $$x''^2 [ 34 c^2 + 41s^2 + 24cs] + y''^2 [41c^2 + 34s^2 - 24cs] + x''y''[24c^2 - 24s^2 + 14cs]$$We require that the coefficient of $x''y''$ is brought to be zero, i.e. that $s = 4/5$ and $c=3/5$. Now it's just a case of substituting those back into the equation,$$50x''^2 + 25y''^2 = 25 \iff 2x''^2 + y''^2 = 1$$and for brevity I'll re-label $x'' \equiv u$ and $y'' \equiv v$. And that completes the transformation!

How did you learn this stuff when you're a first year uni student? I'm at the top of my first year math course for chemistry, but I haven't even seen half of this stuff before.

 

[etotheipi]

How did you learn this stuff when you're a first year uni student? I'm at the top of my first year math course for chemistry, but I haven't even seen half of this stuff before.

Coordinate transformations I learned from Douglas Gregory, Classical Mechanics! It's not so bad, you just need to be careful about when you're leaving the vector alone but changing its representation i.e. w.r.t. a new coordinate system (passive), or if you're actually transforming the vector (active), and not getting the signs mixed up!

 

[fresh_42]

How did you learn this stuff when you're a first year uni student? I'm at the top of my first year math course for chemistry, but I haven't even seen half of this stuff before.

It is not too difficult:

Given the red one, asked for the blue one. You have to rotate the curve, stretch it, and shift the center into the origin. You can either handle this with linear algebra (hint in post #29, or wait for the long version in my solution manual), or simply perform coordinate transformations (post #44) until it fits. We want to have an ellipse at the end: $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$.

Imagine the ellipse is a crystal instead, and the red one is what you see under the microscope, whereas you need the blue version to identify it by the pictures in the book.

 

[LCSphysicist]

How did you learn this stuff when you're a first year uni student? I'm at the top of my first year math course for chemistry, but I haven't even seen half of this stuff before.

In physics i risk to say there is change of coordinates in all place, mainly in mechanic, since it is important to all the laws be the same in all coordinates, i think the college starts to talk about it very early when the course is physics. For example, in my first semester of physics i saw it, probably the same with ethoteipi

 

[fresh_42]

In physics i risk to say there is change of coordinates in all place, mainly in mechanic, since it is important to all the laws be the same in all coordinates, i think the college starts to talk about it very early when the course is physics. For example, in my first semester i saw it.

I would even go as far as to say: physics is all about coordinates. Without coordinates, how do you measure? And it is the reason why general relativity is hard to understand: all of a sudden, coordinates become dependent on the observer! Mathematically you just want to know whether it is an ellipse, a parabola or an hyperbola.

 

[etotheipi]

Problem 5, I can do the first two parts today, too tired to attempt the harder final parts,$$\begin{align*}

\text{det}\begin{bmatrix}5 - \lambda &0&1&6\\3&3-\lambda&5&2\\0&0&3-\lambda&0\\6&0&3&-\lambda\end{bmatrix} &= \text{det}\begin{bmatrix}0&0&3-\lambda&0\\6&0&3&-\lambda\\5 - \lambda &0&1&6\\3&3-\lambda&5&2\end{bmatrix} \\ \\

&= (3-\lambda) \text{det} \begin{bmatrix} 6 & 0 & -\lambda \\ 3-\lambda & 0 & 6 \\ 3 & 3-\lambda & 2 \end{bmatrix} \\ \\

&= (3-\lambda)\left( 36(\lambda -3) -\lambda(\lambda-3)^2 \right) \\

&= (\lambda - 3)^2 (\lambda + 4)(\lambda -9) = 0
\end{align*}$$so $\lambda_1 = 3$, $\lambda_2 = -4$, $\lambda_3 = 9$ are the eigenvalues of $A$. Now for the eigenvectors, $\vec{u}, \vec{v}$ and $\vec{w}$,$$\begin{align*}\begin{bmatrix} -4 & 0 & 1 & 6 \\ 3 & -6 & 5 &2 \\ 0 & 0 & -6 & 0 \\ 6 & 0 & 3 & 9\end{bmatrix} \begin{bmatrix} u_1 \\ u_2 \\ u_3 \\ u_4 \end{bmatrix} &= \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} \iff u_1 = \frac{3}{2} u_4, u_2 = \frac{13}{12}u_4, u_3 = 0 \\ \\\begin{bmatrix} 9 & 0 & 1 & 6 \\ 3 & 7 & 5 &2 \\ 0 & 0 & 7 & 0 \\ 6 & 0 & 3 & 4\end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \\ v_4 \end{bmatrix} &= \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} \iff v_1 = -\frac{2}{3}v_4, v_2 =0, v_3 = 0 \\ \\ \begin{bmatrix} 2 & 0 & 1 & 6 \\ 3 & 0 & 5 &2 \\ 0 & 0 & 0 & 0 \\ 6 & 0 & 3 & -3\end{bmatrix} \begin{bmatrix} w_1 \\ w_2 \\ w_3 \\ w_4 \end{bmatrix} &= \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} \iff w_1 = w_3 = w_4 = 0\end{align*}$$

So $\vec{u} = \begin{bmatrix} 18 \\ 13 \\ 0 \\ 12 \end{bmatrix}$, $\vec{v} = \begin{bmatrix} -2 \\ 0 \\ 0 \\ 3 \end{bmatrix}$, $\vec{w} = \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}$

 

[fresh_42]

Problem 5, I can do the first two parts today, too tired to attempt the harder final parts,$$\begin{align*}

\text{det}\begin{bmatrix}5 - \lambda &0&1&6\\3&3-\lambda&5&2\\0&0&3-\lambda&0\\6&0&3&-\lambda\end{bmatrix} &= \text{det}\begin{bmatrix}0&0&3-\lambda&0\\6&0&3&-\lambda\\5 - \lambda &0&1&6\\3&3-\lambda&5&2\end{bmatrix} \\ \\

&= (3-\lambda) \text{det} \begin{bmatrix} 6 & 0 & -\lambda \\ 3-\lambda & 0 & 6 \\ 3 & 3-\lambda & 2 \end{bmatrix} \\ \\

&= (3-\lambda)\left( 36(\lambda -3) -\lambda(\lambda-3)^2 \right) \\

&= (\lambda - 3)^2 (\lambda + 4)(\lambda -9) = 0
\end{align*}$$so $\lambda_1 = 3$, $\lambda_2 = -4$, $\lambda_3 = 9$ are the eigenvalues of $A$. Now for the eigenvectors, $\vec{u}, \vec{v}$ and $\vec{w}$,$$\begin{align*}\begin{bmatrix} -4 & 0 & 1 & 6 \\ 3 & -6 & 5 &2 \\ 0 & 0 & -6 & 0 \\ 6 & 0 & 3 & 9\end{bmatrix} \begin{bmatrix} u_1 \\ u_2 \\ u_3 \\ u_4 \end{bmatrix} &= \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} \iff v_1 = -\frac{2}{3} v_4, v_2 = \frac{13}{12}u_4, v_3 = 0 \\ \\\begin{bmatrix} 9 & 0 & 1 & 6 \\ 3 & 7 & 5 &2 \\ 0 & 0 & 7 & 0 \\ 6 & 0 & 3 & 4\end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \\ v_4 \end{bmatrix} &= \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} \iff u_1 = \frac{3}{2}u_4, u_2 =0, u_3 = 0 \\ \\ \begin{bmatrix} 2 & 0 & 1 & 6 \\ 0 & 7 & 5 &2 \\ 0 & 0 & 0 & 0 \\ 6 & 0 & 3 & -3\end{bmatrix} \begin{bmatrix} w_1 \\ w_2 \\ w_3 \\ w_4 \end{bmatrix} &= \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} \iff w_1 = w_3 = w_4 = 0\end{align*}$$So $\vec{u} = \begin{bmatrix} 18 \\ 13 \\ 0 \\ 12 \end{bmatrix}$, $\vec{v} = \begin{bmatrix} -2 \\ 0 \\ 0 \\ 3 \end{bmatrix}$, $\vec{w} = \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}$

Sleep over it. It is problem 4. And what is the difference between 3 and -4? And what are 9,12,13,18?

 

[etotheipi]

Sleep over it. It is problem 4. And what is the difference between 3 and -4? And what are 9,12,13,18?

Think I got the numbers mixed up, they should be ok now!

 

[fresh_42]

Think I got the numbers mixed up, they should be ok now!

Let's agree on a single set of representatives, not on many at the same time.

 

[mathwonk]

historical hints: 3-weierstrass, inverse mapping theorem, 7-sylow, 9-nakayama.

 

[songoku]

Spoiler: Question 11

Let $f(x)=x(x+1)(x+2)(x+3)=x^4 +6x^3 11x^2 + 6x$

1. Finding the stationary points of $f(x) \rightarrow f'(x)=0$
$4x^3 + 18x^2 + 22x + 6 = 0$
$(2x+3)(4x^2+12x+4)=0$
$x=-\frac{3}{2} , x=\frac{-3 \pm \sqrt{5}}{2}$

Checking the sign diagram of first derivative:
a. For $x<\frac{-3 - \sqrt{5}}{2} \rightarrow f'(x) = (-)$

b. For $\frac{-3 - \sqrt{5}}{2}<x<-\frac{3}{2} \rightarrow f'(x) = (+)$

c. For $-\frac{3}{2} <x<\frac{-3 + \sqrt{5}}{2} \rightarrow f'(x) = (-)$

d. For $x>\frac{-3 + \sqrt{5}}{2} \rightarrow f'(x) = (+)$

$x=\frac{-3 - \sqrt{5}}{2} \rightarrow y=-1$ so $(\frac{-3 - \sqrt{5}}{2} , -1)$ is local minimum

$x=-\frac{3}{2} \rightarrow y=\frac{9}{16}$ so $(-\frac{3}{2} , \frac{9}{16})$ is local maximum

$x=\frac{-3 + \sqrt{5}}{2} \rightarrow y=-1$ so $(\frac{-3 + \sqrt{5}}{2} , -1)$ is local minimum2. Checking the end behaviour of $f(x)$
$\lim_{x \to \pm \infty} {f(x)=\infty}$

So the local minimum found above is also global minimum

3.
$f(x)=a$ will have no solution if $a$ is below global minimum $\rightarrow a < -1$

No $a$ will result in $f(x)$ has one solution

$f(x)=a$ will have two solutions if $a$ is global minimum or $a$ is above global maximum $\rightarrow a=-1$ or $a>\frac{9}{16}$

$f(x)=a$ will have three solutions if $a$ is global maximum $\rightarrow a=\frac{9}{16}$

$f(x)=a$ will have four solutions if $a$ is between local minimum and local maximum $\rightarrow -1<a<\frac{9}{16}$

I am not sure how to write good conclusion for final answer

 

[songoku]

Spoiler: Question13

Let $a=65536$ and $b=32768$ so $z=\frac{(a-3)^3 +(a-2)^3 + (a-1)^3 + a^3 + (a+1)^3 (a+2)^3 + (a+3)^3}{(b-3)(b-2)+(b-1).b+b.(b+1)+(b+2)b+3)}$

After expanding everything: $z=\frac{7a^3+84a}{4b^2 +12}$

Since $a=2b$ :
$z=\frac{56b^3 +168b}{4b^2 +12}$

$=\frac{56b(b^2+3)}{4(b^2+3)}$

$=14b$

 

[fishturtle1]

Spoiler: question 7

Note $351 = 3^3\cdot 13$. Let $n_3$ denote the number of Sylow $3$ subgroups of $G$ and $n_{13}$ denote the number of Sylow $13$ subgroups of $G$. For a given prime $p$ that divides $\vert G \vert$, the Sylow $p$ subgroups are conjugates. So, it's enough to show $n_3 = 1$ or $n_{13} = 1$. By Sylow's theorem, $n_{13} \vert 27$ and $n_{13} = 1 \text{mod 13}$. So, $n_{13} = 1$ or $27$. If $n_{13} = 1$, we are done. Suppose $n_{13} = 27$. This gives $27\cdot 12 = 324$ distinct elements of order $13$. This leaves $351 - 324 = 27$ elements unaccounted for. Each Sylow $3$ subgroup has order $27$, and we know $n_3 \vert 13$. (And no element of a Sylow $3$ subgroup contains an element of order $13$). Combining the last two lines, we have $n_3 = 1$. []
[\spoiler]

 

[PeroK]

Spoiler: Question 11

Let $f(x)=x(x+1)(x+2)(x+3)=x^4 +6x^3 11x^2 + 6x$

1. Finding the stationary points of $f(x) \rightarrow f'(x)=0$
$4x^3 + 18x^2 + 22x + 6 = 0$
$(2x+3)(4x^2+12x+4)=0$
$x=-\frac{3}{2} , x=\frac{-3 \pm \sqrt{5}}{2}$

Checking the sign diagram of first derivative:
a. For $x<\frac{-3 - \sqrt{5}}{2} \rightarrow f'(x) = (-)$

b. For $\frac{-3 - \sqrt{5}}{2}<x<-\frac{3}{2} \rightarrow f'(x) = (+)$

c. For $-\frac{3}{2} <x<\frac{-3 + \sqrt{5}}{2} \rightarrow f'(x) = (-)$

d. For $x>\frac{-3 + \sqrt{5}}{2} \rightarrow f'(x) = (+)$

$x=\frac{-3 - \sqrt{5}}{2} \rightarrow y=-1$ so $(\frac{-3 - \sqrt{5}}{2} , -1)$ is local minimum

$x=-\frac{3}{2} \rightarrow y=\frac{9}{16}$ so $(-\frac{3}{2} , \frac{9}{16})$ is local maximum

$x=\frac{-3 + \sqrt{5}}{2} \rightarrow y=-1$ so $(\frac{-3 + \sqrt{5}}{2} , -1)$ is local minimum2. Checking the end behaviour of $f(x)$
$\lim_{x \to \pm \infty} {f(x)=\infty}$

So the local minimum found above is also global minimum

3.
$f(x)=a$ will have no solution if $a$ is below global minimum $\rightarrow a < -1$

No $a$ will result in $f(x)$ has one solution

$f(x)=a$ will have two solutions if $a$ is global minimum or $a$ is above global maximum $\rightarrow a=-1$ or $a>\frac{9}{16}$

$f(x)=a$ will have three solutions if $a$ is global maximum $\rightarrow a=\frac{9}{16}$

$f(x)=a$ will have four solutions if $a$ is between local minimum and local maximum $\rightarrow -1<a<\frac{9}{16}$

I am not sure how to write good conclusion for final answer

You should have drawn a graph of the function first. Then, is there a symmetry in the function? What happens if you let $z = x + \frac 3 2$?

This should give you more insight and save at least some of the algebra.

 

[fresh_42]

Spoiler: Question 11

Let $f(x)=x(x+1)(x+2)(x+3)=x^4 +6x^3 11x^2 + 6x$

1. Finding the stationary points of $f(x) \rightarrow f'(x)=0$
$4x^3 + 18x^2 + 22x + 6 = 0$
$(2x+3)(4x^2+12x+4)=0$
$x=-\frac{3}{2} , x=\frac{-3 \pm \sqrt{5}}{2}$

Checking the sign diagram of first derivative:
a. For $x<\frac{-3 - \sqrt{5}}{2} \rightarrow f'(x) = (-)$

b. For $\frac{-3 - \sqrt{5}}{2}<x<-\frac{3}{2} \rightarrow f'(x) = (+)$

c. For $-\frac{3}{2} <x<\frac{-3 + \sqrt{5}}{2} \rightarrow f'(x) = (-)$

d. For $x>\frac{-3 + \sqrt{5}}{2} \rightarrow f'(x) = (+)$

$x=\frac{-3 - \sqrt{5}}{2} \rightarrow y=-1$ so $(\frac{-3 - \sqrt{5}}{2} , -1)$ is local minimum

$x=-\frac{3}{2} \rightarrow y=\frac{9}{16}$ so $(-\frac{3}{2} , \frac{9}{16})$ is local maximum

$x=\frac{-3 + \sqrt{5}}{2} \rightarrow y=-1$ so $(\frac{-3 + \sqrt{5}}{2} , -1)$ is local minimum2. Checking the end behaviour of $f(x)$
$\lim_{x \to \pm \infty} {f(x)=\infty}$

So the local minimum found above is also global minimum

3.
$f(x)=a$ will have no solution if $a$ is below global minimum $\rightarrow a < -1$

No $a$ will result in $f(x)$ has one solution

$f(x)=a$ will have two solutions if $a$ is global minimum or $a$ is above global maximum $\rightarrow a=-1$ or $a>\frac{9}{16}$

$f(x)=a$ will have three solutions if $a$ is global maximum $\rightarrow a=\frac{9}{16}$

$f(x)=a$ will have four solutions if $a$ is between local minimum and local maximum $\rightarrow -1<a<\frac{9}{16}$

I am not sure how to write good conclusion for final answer

Can you summarize this to check the answers which are somehow hidden in your text, which is hard to follow, since you didn't say what you are doing there?
No solution: $a = ...$ or $a\in ...$
1 solution: $a = ...$ or $a\in ...$
2 solutions: $a = ...$ or $a\in ...$
etc.

I know I haven't explicitly asked for, but what are the solutions? At which values of $x$ does the graph intersect the $x-$axis in the five cases?

 

[fresh_42]

Spoiler: Question13

Let $a=65536$ and $b=32768$ so $z=\frac{(a-3)^3 +(a-2)^3 + (a-1)^3 + a^3 + (a+1)^3 (a+2)^3 + (a+3)^3}{(b-3)(b-2)+(b-1).b+b.(b+1)+(b+2)b+3)}$

After expanding everything: $z=\frac{7a^3+84a}{4b^2 +12}$

Since $a=2b$ :
$z=\frac{56b^3 +168b}{4b^2 +12}$

$=\frac{56b(b^2+3)}{4(b^2+3)}$

$=14b$

... and $14b = 458752$. This would have actually answered the question!

 

[Mayhem]

For problem 4, what does $\mathbb{M_4}(\mathbb{Z_4})$ mean?

 

[fresh_42]

$\mathbb{M}_4(\mathbb{Z}_7)$ is the vector space of all $4\times 4$ matrices, where the numbers are taken from the field $\mathbb{Z}_7=\mathbb{Z}/7\mathbb{Z}=\{0,1,2,3,4,5,6\}$.

 

[Mayhem]

I solved part a of 4. I have no idea what "basis of an eigenspace is" so I'm going to look into that.

Spoiler: Problem 4, a

Let
$$A = \begin{pmatrix}
5 & 0 & 1 & 6\\
3 & 3 & 5 & 2\\
0 & 0 & 3 & 0 \\
6 & 0 & 3 & 0
\end{pmatrix}$$
And $I$ denote a 4x4 identity matrix. Then
$$
t I - A = \begin{pmatrix}
t- 5 & 0 & 1 & 6\\
3 & t- 3 & 5 & 2\\
0 & 0 & t- 3 & 0 \\
6 & 0 & 3 & t
\end{pmatrix}
$$
The determinant can be determined through a series of calculations:
$$
\det(t I - A) = (t-5)\begin{vmatrix}
t-3 & 5 & 2\\
0 & t-3 & 0 \\
0 & 6 & t
\end{vmatrix} + \begin{vmatrix}
3 & t-3 & 2 \\
0 & 0 & 0 \\
6 & 0 & t
\end{vmatrix} - 6\begin{vmatrix}
3 & t-3 & 5 \\
0 & 0 & t-3 \\
6 & 0 & 3
\end{vmatrix}
$$
which is then reduced to
$$
\det(t I - A) =
(t-5)\left((t-3)\begin{vmatrix}
t-3 & 0 \\
3 & t
\end{vmatrix}
-5\begin{vmatrix}
0 & 0\\
0 & t
\end{vmatrix} +
2\begin{vmatrix}
0 & t-3 \\
0 & 3
\end{vmatrix}
\right ) + 3\begin{vmatrix}
0 & 0 \\
0 & t \end{vmatrix} - (t-3)\begin{vmatrix}
0 & 0 \\
6 & t
\end{vmatrix} + 2\begin{vmatrix}
0 & 0\\
6 & 0
\end{vmatrix} -

6\left(3\begin{vmatrix}
0 & t-3 \\
0 & 3
\end{vmatrix} - (t-3)\begin{vmatrix}
0 & t-3 \\
6 & 3
\end{vmatrix} + 5 \begin{vmatrix}
0 & 0\\
6 & 0
\end{vmatrix}\right)
$$
It should be reasonably easy to see that all 2x2 determinants with zeroes in both diagnols equal 0, and are therefore canceled out. This yields
$$
\chi_A (x) = (t-5)(t-3)^2+6(t-3)(-6(t-3))
$$
which can be expanded to
$$
\chi_A (x)= t^4 -11t^3 + 3t^2 + 171t-324
$$
which is the characteristic polynomial of $A$

 

[PeroK]

I solved part a of 4. I have no idea what "basis of an eigenspace is" so I'm going to look into that.

Spoiler: Problem 4, a

Let
$$A = \begin{pmatrix}
5 & 0 & 1 & 6\\
3 & 3 & 5 & 2\\
0 & 0 & 3 & 0 \\
6 & 0 & 3 & 0
\end{pmatrix}$$
And I denote a 4x4 identity matrix. Then
$$
t I - A = \begin{pmatrix}
t- 5 & 0 & 1 & 6\\
3 & t- 3 & 5 & 2\\
0 & 0 & t- 3 & 0 \\
6 & 0 & 3 & t
\end{pmatrix}
$$
The determinant can be determined through a series of calculations:
$$
\det(t I - A) = (t-5)\begin{vmatrix}
t-3 & 5 & 2\\
0 & t-3 & 0 \\
0 & 6 & t
\end{vmatrix} + \begin{vmatrix}
3 & t-3 & 2 \\
0 & 0 & 0 \\
6 & 0 & t
\end{vmatrix} - 6\begin{vmatrix}
3 & t-3 & 5 \\
0 & 0 & t-3 \\
6 & 0 & 3
\end{vmatrix}
$$
which is then reduced to
$$
\det(t I - A) =
(t-5)\left((t-3)\begin{vmatrix}
t-3 & 0 \\
3 & t
\end{vmatrix}
-5\begin{vmatrix}
0 & 0\\
0 & t
\end{vmatrix} +
2\begin{vmatrix}
0 & t-3 \\
0 & 3
\end{vmatrix}
\right ) + 3\begin{vmatrix}
0 & 0 \\
0 & t \end{vmatrix} - (t-3)\begin{vmatrix}
0 & 0 \\
6 & t
\end{vmatrix} + 2\begin{vmatrix}
0 & 0\\
6 & 0
\end{vmatrix} -

6\left(3\begin{vmatrix}
0 & t-3 \\
0 & 3
\end{vmatrix} - (t-3)\begin{vmatrix}
0 & t-3 \\
6 & 3
\end{vmatrix} + 5 \begin{vmatrix}
0 & 0\\
6 & 0
\end{vmatrix}\right)
$$
It should be reasonably easy to see that all 2x2 determinants with zeroes in both diagnols equal 0, and are therefore canceled out. This yields
$$
\chi_A (x) = (t-5)(t-3)^2+6(t-3)(-6(t-3))
$$
which can be expanded to
$$
\chi_A (x)= t^4 -11t^3 + 3t^2 + 171t-324
$$
which is the characteristic polynomial of $A$

What are these numbers $11, 171$ and $324$? There are no such numbers in $\mathbb Z_7$.

 

[Mayhem]

What are these numbers $11, 171$ and $324$? There are no such numbers in $\mathbb Z_7$.

How do I circumvent that issue?

 

[PeroK]

How do I circumvent that issue?

You learn the mathematics of modular arithmetic.

 

[Mayhem]

You learn the mathematics of modular arithmetic.

So if I rewrite those numbers in the correct mod, it's otherwise correct?