Challenge Math Challenge - December 2018

[lpetrich]

I will solve Problem 19.

Spoiler

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]

How did you solved this step:

I will solve Problem 19.
$$ 47 r^4 - 188 r^3 - 6016 r - 48128 = 0 \to \\ 47 (r - 8) (r + 4) (r^2 + 32) = 0 $$

 

[QuantumQuest]

Well done @lpetrich

 

[lpetrich]

I did that step with Mathematica's Factor[] function, but I will do that more explicitly here.

Spoiler

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)

Spoiler

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]

Problem 2(b)

Spoiler

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.

Spoiler

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.

Spoiler

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.