# Challenge Math Challenge - December 2018

#### lpetrich

I will solve problem 9.
The integral
$$I = \int_1^{\infty} \frac{dx}{p(x)}$$
can be done in indefinite form for n = 0 or 1.

For n = 0, p(x) = 1, and the indefinite integral is x. It diverges for $x \to \infty$.

For n = 1, p(x) = x + a1, and the indefinite integral is $\log(x + a_1)$. It also diverges for $x \to \infty$.

Thus, the integral $I$ is always divergent for $n \leq 1$.

This leaves the case where $n \geq 2$. This problem will be solved by bounding the integrand, and showing that this gives a finite integral. That will then make $I$ finite.

If p(x) has all-negative roots, then it must be uniformly positive or negative over x in the domain of integration of the problem's integral, $[1,\infty)$. Let us choose a form for p(x) that will make it possible to show that $I$ is bounded from above. This form is p(x) = xn q(1/x), where
$$q(u) = 1 + a_{n-1} u + a_{n-2} u^2 + \cdots + a_0 u^n$$
As $x \to \infty$, $q(1/x) \to 1$, which is positive, thus, p(x) is uniformly positive over x in $[1,\infty)$. That also makes q(1/x) uniformly positive over that interval. This also means that 1/p(x) is uniformly positive, thus making $I$ also positive, bounded from below by 0.

We must now show that 1/p(x) is bounded from above, and bounded from above in a way that makes $I$ finite. This is equivalent to showing that p(x) is bounded from below in this sort of fashion, and this can be done by bounding q(u) for u in $(0,1]$. By the extreme value theorem, q(u) must have a minimum value for some value of u in that interval. Since q(u) is always positive in that interval, that means that its minimum value in that integral must also be positive. I will call that value Q. Thus, $q(1/x) \geq Q$ for all x in $[1,\infty)$, giving $p(x) = x^n q(1/x) \geq Q x^n$ for all x in that interval. This makes $1/p(x) \leq 1/(Q x^n)$ over that interval, and thus,
$$I = \int_1^{\infty} \frac{dx}{p(x)} \leq \int_1^{\infty} \frac{dx}{Q x^n} = \frac{1}{(n-1)Q}$$
Thus, $I$ is bounded from both above and below, and the integral always converges for $n \geq 2$.

In summary, this integral always diverges for $n \leq 1$ and always converges for $n \geq 2$.

#### fresh_42

Mentor
2018 Award
I will solve problem 13.
Consider a group Gpq with order pq, where both p and q are prime numbers and p > q.

If Gpq is abelian, then according to a well-known theorem...
Fundamental theorem of finitely generated Abelian groups.
https://en.wikipedia.org/wiki/Finitely_generated_abelian_group#Classification
... it is a product of cyclic groups Z(xm), where each x is a prime factor of |G|, the order of G, and m is between 1 and the power where x appears in |G|, inclusive. This means that if Gpq is abelian, then it must be composed of a product of powers of Z(p) and Z(q). By Cauchy's theorem, Gpq has elements of order p and q, thus subgroups Z(p) and Z(q). This means that Gpq must be at least Z(p)*Z(q). But the order of Z(p)*Z(q) is pq, the order of Gpq. So abelian Gpq is Z(p)*Z(q).

By Bézout's identity, since p and q are relatively prime, there is some u and v such that up + vq = 1. So if a generates a Z(p) and b generates a Z(q), and if c = avbu, then cq = avqbuq = a1-up = a and cp = avpbup = b1-vq = b.

So Z(p)Z(q) = Z(pq), and thus abelian Gpq is cyclic.

For possible nonabelian Gpq, we use Sylow's theorems. One of these theorems states that for every prime x dividing |G|, the group has a subgroup with order xm, where x appears in |G| with power m.
1st of Sylow's theorems.
https://en.wikipedia.org/wiki/Sylow_theorems
Another of these theorems states that these subgroups need not be unique, and that all such subgroups in a group are related by conjugation.
The 2nd of Sylow's theorems says, that all maximal p-groups contain a conjugate of any other p-subgroup. In our case, this means that all p-subgroups, resp. q-subgroups are conjugate. The general statement is a corollary to Sylow's theorems.
Still another states ...
3rd of Sylow's theorem.
... that there are kx+1 of these subgroups ...
Sylow x-groups, i.e. a maximal x-subgroup
... where kx+1 evenly divides |G|/xm. If k = 0, then the subgroup is normal.
A corollary of Sylow's theorems.
Since p and q are the only prime factors, the Sylow subgroups must be the Z(p)'s and Z(q)'s in Gpq. Since p > q, that means that there is no k that makes (kp +1) = q, therefore, the Z(p) is normal and there is only one of it in Gpq. But it may happen that (kq+1) = p, and in that case, there are (kq+1) = p copies of the Z(q) subgroup of Gpq, copies that are all conjugates of each other. That makes the group nonabelian. But that is only possible if p mod q = 1. If not, then there is only one of the Z(q) in Gpq, and it is normal.

If Gpq has only one Z(p) subgroup and only one Z(q) subgroup
Haven't you already shown, that both have to be normal? In this case $\mathbb{Z}_p \times \mathbb{Z}_q$ is a direct product and isomorph to a subgroup of $G$. Since both have $pq$ elements, they have to be equal.
... then it has the identity, (p-1) elements with order p, (q-1) elements with order q, and (p-1)(q-1) elements with a different order. Since pq is only evenly divisible by itself, 1, p, and q, that additional order must be pq, and an element with that order will generate the entire group. Thus arriving at the cyclic Gpq group.

Thus, if p mod q is not 1, then the only possible group Gpq is the cyclic one.

#### fresh_42

Mentor
2018 Award
I will solve problem 9.

The integral
$$I = \int_1^{\infty} \frac{dx}{p(x)}$$
can be done in indefinite form for n = 0 or 1.

For n = 0, p(x) = 1, and the indefinite integral is x. It diverges for $x \to \infty$.

For n = 1, p(x) = x + a1, and the indefinite integral is $\log(x + a_1)$. It also diverges for $x \to \infty$.

Thus, the integral $I$ is always divergent for $n \leq 1$.

This leaves the case where $n \geq 2$. This problem will be solved by bounding the integrand, and showing that this gives a finite integral. That will then make $I$ finite.

If p(x) has all-negative roots, then it must be uniformly positive or negative over x in the domain of integration of the problem's integral, $[1,\infty)$. Let us choose a form for p(x) that will make it possible to show that $I$ is bounded from above. This form is p(x) = xn q(1/x), where
$$q(u) = 1 + a_{n-1} u + a_{n-2} u^2 + \cdots + a_0 u^n$$
As $x \to \infty$, $q(1/x) \to 1$, which is positive, thus, p(x) is uniformly positive over x in $[1,\infty)$. That also makes q(1/x) uniformly positive over that interval. This also means that 1/p(x) is uniformly positive, thus making $I$ also positive, bounded from below by 0.

We must now show that 1/p(x) is bounded from above, and bounded from above in a way that makes $I$ finite. This is equivalent
Proof?
to showing that p(x) is bounded from below in this sort of fashion, and this can be done by bounding q(u) for u in $(0,1]$.
Proof?
By the extreme value theorem, q(u) must have a minimum value for some value of u in that interval.
What about $x=0$? I think you must first get rid of the open end of this interval, at least formally.
Since q(u) is always positive in that interval, that means that its minimum value in that integral must also be positive. I will call that value Q. Thus, $q(1/x) \geq Q$ for all x in $[1,\infty)$, giving $p(x) = x^n q(1/x) \geq Q x^n$ for all x in that interval. This makes $1/p(x) \leq 1/(Q x^n)$ over that interval, and thus,
$$I = \int_1^{\infty} \frac{dx}{p(x)} \leq \int_1^{\infty} \frac{dx}{Q x^n} = \frac{1}{(n-1)Q}$$
Thus, $I$ is bounded from both above and below, and the integral always converges for $n \geq 2$.

In summary, this integral always diverges for $n \leq 1$ and always converges for $n \geq 2$.
I think you should work out the three points I mentioned. You claim equivalences which can probably easily be shown, and you must deal with the boundary $0$.

#### Infrared

Gold Member
I will attempt to solve problem 12.
Since $x_{n+1} = n (x_n + x_{n-1})$, it is evident that every solution is a linear combination of two solutions, a combination that can be specified with the values of $x_0$ and $x_1$.

It is easy to show that $x_n = n!$ is a solution.
$n(n! + (n-1)!) = n(n + 1)\cdot(n-1)! = (n+1)!$.

With Mathematica and a lot of trial and error with series, I came up with a trial solution that seems to be linearly independent of the first one:
$$x_n = \sum_{k=0}^\infty \frac{(-1)^{k+n} (k+1)}{n(n+1) \cdots (n+k)}$$
Since k = (n+k) - n, this solution can be split in two: $x_n = y_n + (1-1/n)y_{n+1}$ where
$$y_n = \sum_{k=0}^\infty \frac{(-1)^{k+n}}{n(n+1) \cdots (n+k-1)} = - (n-1)! \sum_{k=(n-1)}^\infty \frac{(-1)^k}{k!} = - (n-1)! \left( \frac{1}{e} - \sum_{k=0}^{n-2} \frac{(-1)^k}{k!} \right)$$
One can verify the x-y equation with the help of
$$y_{n+1} = - \sum_{k=0}^\infty \frac{(-1)^{k+n}}{(n+1)(n+2) \cdots (n+k)}$$
The y expression with 1/e in it gives us
$$x_n = - n! \left( \frac{1}{e} - \sum_{k=0}^n \frac{(-1)^k}{k!} \right)$$
It is now time to verify this solution, to show that it satisfies the problem's recurrence relation. For this purpose, we find
$$x_{n-1} = - \frac{1}{n} n! \left( \frac{1}{e} - \sum_{k=0}^n \frac{(-1)^k}{k!} \right) - \frac{(-1)^n}{n}$$
$$x_n = - (n+1) n! \left( \frac{1}{e} - \sum_{k=0}^n \frac{(-1)^k}{k!} \right) - (-1)^n$$
It can easily be shown that this solution satisfies the recurrence. The general solution for x can now be expressed as
$$x_n = n! \left(A + B\sum_{k=0}^n \frac{(-1)^k}{k!}\right)$$
for some A and B. Solving $x_0 = 0 = A + B$ and $x_1 = 1 = A$, we get $A = 1$ and $B = -1$. This gives us the solution for the problem:
$$x_n = n! \left(1 - \sum_{k=0}^n \frac{(-1)^k}{k!}\right)$$
and in the limit of large n,
$$\frac{x_n}{n!} = 1 - \sum_{k=0}^n \frac{(-1)^k}{k!} \to 1 - \frac{1}{e} \approx 0.632121$$
Yes, this is right! Can you find a way to solve the problem without guessing the second linearly independent solution?

#### YoungPhysicist

Problem 14:
First, the number must start with 1or 2, since everything bigger than that will make it a six digit number after multiplying by 4.

But 1 isn’t possible, since no number from 0 to 9 can end with 1 after multiplying by 4.

2xxxx

Then,it must end with a 8 or 9,since that’s the only two possibility of 2xxxx multiplying by 4.

But 9 isn’t correct,because 9x4 = 36, which doesn’t end with 2.

2xxx8

The second digit can be 0,1, or 2.

After checking all multiples of 4 by 8,18,28.....98, I found 0 and 2 impossible, since the second two digits of the previous multiple only consists 3,7,1 and 5.

That leads to only two possibilities:

21x28
21x78

Then I found the first one impossible since 21xxx x 4 always end up with 84xxx or bigger, but 21x28 requires the first two digits to be 82, which can’t happen.

21x78

After trying all ten possibilities, I get 21978.

#### Math_QED

Homework Helper
15:

In what follows, I will assume that already is established that Boolean rings are commutative.

Let $P$ be a prime ideal. Our goal here is to prove that $P$ is maximal. I.e., we want to show that $R/P$ is a field.

Let's examine the quotient ring a little further.

Take an element $[x] \in R/P$.

Observe that $(x-1)^2 = x-1$, and thus by distributivity: $x-1 = (x-1)(x-1) = x^2 -x -x +1$

Hence, $(x-(1+1))(x-1) = x^2 - x - x - x +1 +1 = 0 \in P$, and because $P$ is prime it follows that $(x - (1 +1)) \in P$ or $x - 1 \in P$. But now observe that $1+1 = 0$, because $1 +1 = (1+1)^2 = 1 + 1 +1 +1$. Thus we either have $x \in P$ or $x-1 \in P$, which means that $[x] = [0] = 0$ or $[x] = [1] = 1$.

We thus have shown that the quotient ring contains 2 elements, that is $R/P \cong \mathbb{F}_2$ and we are done.

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#### fresh_42

Mentor
2018 Award
15:

In what follows, I will assume that already is established that Boolean rings are commutative.

Let $P$ be a prime ideal. Our goal here is to prove that $P$ is maximal. I.e., we want to show that $R/P$ is a field.

Let's examine the quotient ring a little further.

Take an element $[x] \in R/P$.

Observe that $(x-1)^2 = x-1$, and thus by distributivity: $x-1 = (x-1)(x-1) = x^2 -x -x +1$

Hence, $(x-(1+1))(x-1) = x^2 - x - x - x +1 +1 = 0 \in P$, and because $P$ is prime it follows that $(x - (1 +1)) \in P$ or $x - 1 \in P$. But now observe that $1+1 = 0$, because $1 +1 = (1+1)^2 = 1 + 1 +1 +1$. Thus we either have $x \in P$ or $x-1 \in P$, which means that $[x] = [0] = 0$ or $[x] = [1] = 1$.

We thus have shown that the quotient ring contains 2 elements, that is $R/P \cong \mathbb{F}_2$ and we are done.
Just in case you're interested in a shorter version:

Let $x,y \in \mathcal{B}/P - \{\,0\,\}$ and $z:=x\cdot y\,.$ Then $xz=x(xy)=(xx)y=xy$ and $0=x(z-y)\,.$ As $\mathcal{B}/P$ is an integral domain and $x\neq 0$, we have $y=z$ and similarly $x=z$. So all elements different from $0$ are identical. Because $P \neq\mathcal{B}$ we get $\mathcal{B}/P \cong \mathbb{Z}_2\,.$

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#### lpetrich

I will attempt to solve problem 16.
I first establish coordinate conventions. The center of the square is at (x=0, y=0) for coordinates x and y, and the four edges are at x = 1, y = 1, x = -1, and y = -1. This makes the area of the square $(1 - (-1))^2 = 4$.

The solution comes in two parts:
1. Find which region is closer to an edge than to the center
2. Find the area of that region

The first part requires finding for each edge, the region that is closer to the center than to the edge. Once that is done, then the region that is closer to the center than to any edge is the intersection of all these four regions.

For finding the region closer to the center than to some edge, I will do the x = 1 case, and then generalize with the help of symmetries. For x = 1, the inequality that gives the region is
$$\sqrt{x^2 + y^2} < 1 - x$$
Squaring both sides gives
$$x^2 + y^2 < 1 - 2x + x^2$$
Solving for x gives
$$x < \frac12 (1 - y^2)$$
A parabola. Using reflection symmetries, the solution region is
• (x = 1): $x < (1/2)(1 - y^2)$
• (y = 1): $y < (1/2)(1 - x^2)$
• (x = -1): $x > - (1/2)(1 - y^2)$
• (y = -1): $y > - (1/2)(1 - x^2)$
Since the region's boundary is given in piecewise fashion, we must find where the boundary pieces intersect. For (x=1) and (x=-1), it is easy: $y = \pm 1$, and similarly for (y=1) and (y=-1). That is well outside the region. For (x=1) and (y=1), I find two real solutions:
• $x = y = \sqrt{2} - 1$
• $x = y = - (\sqrt{2} + 1)$
The second solution is clearly outside of the region, as one can tell from the inequalities derived from the third and fourth edges, but the first one is not. Thus, the four intersection points are
$$x = \pm (\sqrt{2} - 1) ,\ y = \pm (\sqrt{2} - 1)$$

Having found the solution region, we now find its area. This can be done with the help of the square's symmetries. The square can be divided into eight triangles that have vertices (center), (edge center), (nearby vertex). All eight triangles have equal area, and for convenience, I will choose the one with vertices (0,0), (0,1), and (1,1). The part of the solution region in this triangle is given by lines (x = 0) and (y = x), and the (y = 1) region boundary, $y = (1/2)(1-x^2)$. The second two boundaries intersect at $x = y = \sqrt{2} - 1$. So the integral for area A is
$$A = \int_0^{\sqrt{2}-1} \left( \int_x^{(1/2)(1-x^2)} 1 \, dy \right) \, dx = \int_0^{\sqrt{2}-1} \frac12 (1 - 2x - x^2) \, dx = \frac{4\sqrt{2} - 5}{6}$$
The area of the entire region is thus $8A = (4/3)(4\sqrt{2} - 5)$, and dividing by the square's area gives us
$$\frac{\text{(area of solution region)}}{\text{(area of square)}} = \frac{4\sqrt{2} - 5}{3} \approx 0.218951$$
This is the fraction of success of winning an extra point. I have approximately verified it by doing some simple numerical integration.

#### lpetrich

I will solve problem 18.
The problem may be restated as: find all nonnegative-integer x less than 1000 such that the power identity $x^n = 1000 m + x$ for positive integers n and nonnegative integers m. It can be reduced to a much simpler problem, because the power identity is true for n = 2, and because its truth for n = 2 implies its truth for all n. This can easily be shown by mathematical induction. Let $x^2 = 1000 m_0 + x$, and use the power identity's truth for some value of n. Then
$x^{n+1} = x^n x = 1000 m x + x^2 = 1000 (m x + m_0) + x$
Since m and x are both nonnegative integers, their product will also be a nonnegative integer, and thus $m x + m_0$ will be a nonnegative integer. That proves the power identity for n+1, and thus for all values of n.

So we only need to find some x that satisfies the power identity for n = 2: $x^2 = 1000 m + x$.

We first consider the possible ones digits. There are four possibilities: 0, 1, 5, and 6.

02 = 0, 12 = 1, 52 = 5, 62 = 6.

For the tens digit, 0 has only one possibility: 0. Likewise, 1 has 0, 5 has 2, and 6 has 7.

002 = 00, 012 = 01, 252 = 625, 762 = 5776

For the hundreds digit, 00 has only one possibility, 0. Likewise, 01 has 0, 25 has 6, and 76 has 3.

0002 = 000, 0012 = 001, 6252 = 390625, 3762 = 141376

So this problem's solutions are 000, 001, 625, and 376.

#### mfb

Mentor
There is an interesting pattern in problem 18:
The remainder from a division by 23=8 depends on the last three digits only. To stay the same it has to be either 0 or 1.
The remainder from a division by 53=125 depends on the last three digits only. To stay the same it has to be either 0 or 1.

=> We get the trivial 000 and 001 and the option to have multiples of 125 or 8.
625 = 5*125, the only one in range with a remainder of 1 in a division by 8.
376 = 47*8, the only one in range with a remainder of 1 in a division by 125.

This is a general pattern for bases that are the product of two primes independent of the number of trailing digits we look at. In other bases things are different. If the base is a prime or prime power then 0 and 1 are the only options, for example.

#### lpetrich

I will solve Problem 19.
The equation $x^4 − 18 x^3 + k x^2 + 200 x − 1984 = 0$ is satisfied by four values of x, and for some k, two of those values have the product -32.

Let one of these roots be r and another one be s. One now has three variables to solve for: k, r, and s. However, it is easy to solve for s in terms of k and r. The product condition, $r s = -32$, gives us $s = -32/r$.

Since both r and s must satisfy the problem's equation, I substitute them in. For r:
$$r^4 − 18 r^3 + k r^2 + 200 r − 1984 = 0$$
For s:
$$s^4 − 18 s^3 + k s^2 + 200 s − 1984 = 0 \to \\ (-32/r)^4 − 18 (-32/r)^3 + k (-32/r)^2 + 200 (-32/r) − 1984 = 0 \to \\ - \frac{64}{r^4}(31 r^4 + 100 r^3 - 16 k r^2 - 9216 r - 16384) = 0$$
Extracting its polynomial part gives us
$$31 r^4 + 100 r^3 - 16 k r^2 - 9216 r - 16384 = 0$$
Both r and k must solve both this equation and the earlier r equation. It is easier to eliminate k from these two equations than it is to eliminate r. So to find k, I first eliminate it, find the possible r values, and then find the corresponding k values.

Adding 16 times the r equation gives us
$$47 r^4 - 188 r^3 - 6016 r - 48128 = 0 \to \\ 47 (r - 8) (r + 4) (r^2 + 32) = 0$$
This gives us these solutions: $r = 8,\ r = -4 ,\ r = 4i\sqrt{2} ,\ r=-4i\sqrt{2}$. Plugging these numbers into the product condition gives the corresponding values of s: $s = -4,\ s = 8 ,\ s = 4i\sqrt{2} ,\ s=-4i\sqrt{2}$. Plugging these numbers into the problem's equation gives us what we seek: values of k. Here are the solutions:
• The two related roots are 8 and -4, and k = 86
• The two related roots are both $4i\sqrt{2}$, and $k = -30 + 97i\sqrt{2}$
• Like the above, but with complex conjugates: $i \to -i$.

#### fresh_42

Mentor
2018 Award
How did you solved this step:
I will solve Problem 19.
$$47 r^4 - 188 r^3 - 6016 r - 48128 = 0 \to \\ 47 (r - 8) (r + 4) (r^2 + 32) = 0$$

Gold Member

#### lpetrich

I did that step with Mathematica's Factor[] function, but I will do that more explicitly here.
I wish to factor the polynomial $47 r^4 - 188 r^3 - 6016 r - 48128$.

The first step is to look for a constant factor in its coefficients. The smallest one, 47, is a prime, so I test dividing all the coefficients by 47. I find 188/47 = 4, 6016/47 = 128, and 48128/47 = 1024. This gives us
$$47 (r^4 - 4 r^3 - 128 r - 1024)$$
After dividing out the 47, it is evident that the coefficients other than 0 and 1 are powers of 2. 4 = 22, 128 = 27, 1024 = 210. These powers are 2, 7, and 10, while when one divides the polynomial by r4, their r powers becomes -1, -3, and -4. Multiplying by 2 gives 2, 6, and 8, all less than or equal to the coefficients' powers of 2. Thus, we substitute r = 4t, turning
$$r^4 - 4 r^3 - 128 r - 1024$$
into
$$256 t^4 - 256 t^3 - 512 t - 1024 = \\ 256 (t^4 - t^3 - 2t - 4)$$
We now look for a factorization of the t polynomial over the integers. We first look for linear factors, factors with the form $a t + b$. Of coefficients a and b, a must divide the highest-degree coefficient, 1, and b must divide the lowest-degree coefficient, -4. We can set a = 1 without loss of generality, while b is less specified. From the divisors of 4 and the possible signs, its possible values are $\pm 1 ,\ \pm 2 ,\ \pm 4$. We can thus test for such factors by checking on whether t = - (any of those b values) is a root. Trying
$$t \to 1 ,\ -1 ,\ 2 ,\ -2 ,\ 4 ,\ -4$$
gives polynomial values
$$-6 ,\ 0 ,\ 0 ,\ 24 ,\ 180 ,\ 324$$
This means that t = -1 and t = 2 are roots, giving factors $t + 1$ and $t -2$. Dividing $t^4 - t^3 - 2t - 4$ by both of those polynomials gives $t^2 + 2$.
Thus,
$$t^4 - t^3 - 2t - 4 = (t + 1) (t - 2) (t^2 + 2) \\ r^4 - 4 r^3 - 128 r - 1024 = (r + 4) (r - 8) (r^2 + 32) \\ 47 r^4 - 188 r^3 - 6016 r - 48128 = 47(r + 4) (r - 8) (r^2 + 32)$$
Which is what I'd found earlier with Mathematica.

#### jbstemp

Problem 2(b)

Let $A$ be an element of the tangent space of $SU(n)$ at the identity, $I$, and let $M: \mathbb{R} \rightarrow SU(n)$ be a smooth path such that $M(0) = I$ and $M'(0) = A$. Now let $M^{\dagger}$ be a path in $SU(n)$ such that $M^{\dagger}(t) = M(t)^{\dagger} = M(t)^{-1}$. Given that inversion in a Lie group is smooth, it follows that $M^{\dagger}$ is a smooth path in $SU(n)$. Taylor expanding $M$ and $M^{\dagger}$ about $t = 0$ gives: $M(t) = I + tA + O(t^2)$ and $M^{\dagger}(t) =I + tA^{\dagger} + O(t^2)$. Thus, $M(t) M^{\dagger}(t) = I + t(A + A^{\dagger} ) + O(t^2) = I$ and so $A = - A^{\dagger}$. Hence, the tangent space of $SU(n)$ at the identity element consists of $n \times n$ skew-symmetric matrices.

#### fresh_42

Mentor
2018 Award
Problem 2(b)

Let $A$ be an element of the tangent space of $SU(n)$ at the identity, $I$, and let $M: \mathbb{R} \rightarrow SU(n)$ be a smooth path such that $M(0) = I$ and $M'(0) = A$. Now let $M^{\dagger}$ be a path in $SU(n)$ such that $M^{\dagger}(t) = M(t)^{\dagger} = M(t)^{-1}$. Given that inversion in a Lie group is smooth, it follows that $M^{\dagger}$ is a smooth path in $SU(n)$. Taylor expanding $M$ and $M^{\dagger}$ about $t = 0$ gives: $M(t) = I + tA + O(t^2)$ and $M^{\dagger}(t) =I + tA^{\dagger} + O(t^2)$. Thus, $M(t) M^{\dagger}(t) = I + t(A + A^{\dagger} ) + O(t^2) = I$ and so $A = - A^{\dagger}$. Hence, the tangent space of $SU(n)$ at the identity element consists of $n \times n$ skew-symmetric matrices.
Well, I expected some messy differentiations of coordinate functions, but that was easier to check, thanks.

#### lpetrich

I will solve problem 21.
We are given a curve C in the form of a space position $\mathbf{x}$ that is a function of a parameter $t$:
$$\mathbf{x} = \left( t, \ t^2, \ \frac23 t^3 \right)$$
Here is an outline of my solution:
1. Find the tangent vector from the derivative of the position. Doing so gives us the distance along the curve, something necessary for later steps.
2. Find the normal vector from the tangent vector. Doing so gives us the curvature.
3. Find the binormal vector from the tangent and normal vectors.
4. Find the torsion from the binormal vector.

The first step is to find the tangent vector $\mathbf{t}$. It is defined in terms of the distance $s$ along the curve as
$$\mathbf{t} = \frac{d\mathbf{x}}{ds} = \frac{d\mathbf{x}/dt}{ds/dt}$$
Since the tangent vector has norm 1, it is easy to show that
$$\frac{ds}{dt} = \left| \frac{d\mathbf{x}}{dt} \right|$$
For C,
$$\frac{d\mathbf{x}}{dt} = (1,\ 2t,\ 2t^2)$$
yielding $ds/dt = \sqrt{1 + 4t^2 + 4t^4} = 1 + 2t^2$. This gives us $s = t + \frac23 t^3$. This solution is difficult to invert, so I will continue using t as the independent variable. This also gives us the tangent vector:
$$\mathbf{t} = \frac{(1,\ 2t,\ 2t^2)}{1+2t^2}$$
The second step is to find the normal vector from the tangent vector. The normal vector $\mathbf{n}$ is given by
$$\frac{d\mathbf{t}}{ds} = \kappa \mathbf{n}$$
where $\kappa$ is the curvature. Since the normal vector has norm 1, we can find the curvature by taking the norm of both sides:
$$\kappa = \left| \frac{d\mathbf{t}}{ds} \right|$$
For C,
$$\frac{d\mathbf{t}}{ds} = \frac{d\mathbf{t}/dt}{ds/dt} = \frac{ (-4t ,\ 2(1-2t^2) ,\ 4t)}{(1 + 2t^2)^3}$$
From its norm, we find the curvature:
$$\kappa = \frac{2}{(1 + 2t^2)^2}$$
and the normal vector:
$$\mathbf{n} = \frac{ (-2t ,\ 1-2t^2 ,\ 2t) }{1 + 2t^2}$$
The third step is to find the binormal vector $\mathbf{b}$. It is given by
$$\mathbf{b} = \mathbf{t} \times \mathbf{n} = \frac{ (2t^2 ,\ -2t ,\ 1) }{1 + 2t^2}$$
The fourth step is to find the torsion $\tau$ from the binormal vector. It is found from
$$\frac{d\mathbf{t}}{ds} = - \tau \mathbf{n}$$
For C,
$$\frac{d\mathbf{b}}{ds} = \frac{d\mathbf{b}/dt}{ds/dt} = \frac{ (4t, -2(1 - 2t^2), -4t) }{(1 + 2t^2)^3} = - \frac{2}{(1 + 2t^2)^2} \mathbf{n}$$
Thus giving its value of torsion,
$$\tau = \frac{2}{(1 + 2t^2)^2}$$
Collecting the results requested in the problem statement, the curvature and torsion are
$$\kappa = \tau = \frac{2}{(1 + 2t^2)^2}$$

#### lpetrich

I will solve problem 22.
Our major clue is which of the divisibility statements are clue and which false. There are only two falsehoods, and they are contiguous in the last. Let us use this condition to narrow down the possibilities.

I will use this shorthand: "n is divisible by m" is (n div m) and "n is not divisible by m" is (n not div m). If the m is a list, then the appropriate statement is true for all members of that list: (n div m1, m2, m3) is (n div m1) and (n div m2) and (n div m3).

If (n not div 2), then (n not div 4, 6, 8), and since 2 and 4 are not contiguous, that is contrary to the problem condition. Thus, (n div 2). Likewise, if (n not div 3), then (n not div 6, 9, 12), and from non-contiguity, (n div 3). Because 2 and 3 are relatively prime, (n div 2) and (n div 3) imply (n div 6),. Combining these results, (n div 2, 3, 6).

Trying (n not div 4), it implies (n not div 8, 12), and from non-contiguity, (n div 4). Since 3 is relatively prime with 4, combining with (n div 3) gives (n div 12). Thus, (n div 2, 3, 4, 6, 12).

Since divisibility by 13 is contiguous with a true divisibility, (n div 12), thus (n div 13). Likewise, divisibility by 5 is contiguous with (n div 4) and (n div 6), and thus, (n div 5). That gives us (n div 2, 3, 4, 5, 6, 12, 13).

Since 2 and 5 are relatively prime, (n div 2) and (n div 5) combine to give (n div 10), for a result of (n div 2, 3, 4, 5, 6, 10, 12, 13). In this list, 11 is contiguous with 10 and 12, and since (n div 10) and (n div 12), thus, (n div 11): (n div 2, 3, 4, 5, 6, 10, 11, 12, 13).

This gives undecided divisibility only for 7, 8, and 9, and of these, the possible false sets are (n not div 7, 8) and (n not div 8, 9). This gives us these two solutions:

(n div 2, 3, 4, 5, 6, 9, 10, 11, 12, 13) and (n div 2, 3, 4, 5, 6, 7, 10, 11, 12, 13)

The smallest possible n that will satisfy one of these condition is the least common multiple (LCM) of the numbers. Any other solutions will be multiples of that LCM. Thus, these two solutions are 25,740 and 60,060 and their multiples.

Since we want a number less than 50,000, that rules out 60,060, leaving 25,740. Its first multiple is 2 times that value, or 51,480, and that is also ruled out.

Thus, the solution is 25,740.