Shankar，基礎物理，PHYS 200，Problem Set I, #4

題目原文 

[Difficult] Ball A is dropped from rest from a building of height H exactly when ball B is thrown up vertically. When they collide A has double the speed of B. If the collision occurs at height h what is h/H? 

Hint: Write equations for heights $y_A, y_B$ and velocities $v_A$ and $v_B$. What can you say about these at the time of the collision?

題目翻譯

[難題] 球A從高度為H的建築物自由落下之際，地面上的球B正好同時垂直上拋。 兩球在高度h的位置相撞，撞擊時A的速率是B的兩倍，則h/H是多少？

提示：分別對高度$y_A, y_B$以及速度$v_A$和$v_B$寫出方程式。在發生碰撞時，你有什麼結論？

我的解答

首先寫出$y_A$與$y_B$：

$$y_A(t) = H + 0t - \frac{1}{2}gt^2,$$

$$y_B(t) = 0 + v_B(0)t - \frac{1}{2}gt^2.$$

假定在時刻$t^*$時發生碰撞，利用A球的位置函數，得

$$h = H - \frac{1}{2}g{t^*}^2.$$

解出$t^* = \sqrt{\frac{2(H - h)}{g}}$。代入B球的位置函數，可以解出B球上拋的初始速度：

$$h = v_B(0) \cdot \sqrt{\frac{2(H - h)}{g}} - \frac{1}{2}g\cdot \left( \sqrt{\frac{2(H - h)}{g}} \right)^2,$$

$$v_B(0) = \sqrt{\frac{g}{2(H - h)}}H.$$

到目前為止都還沒用到題目的另個條件「撞擊時A的速率是B的兩倍」，現在來使用它。寫出$v_A$與$v_B$：

$$v_A(t) = 0 - gt,$$

$$v_B(t) = \sqrt{\frac{g}{2(H - h)}}H - gt.$$

代入撞擊時刻$t^* = \sqrt{\frac{2(H - h)}{g}}$，得

$$\left| 0 - g \cdot \sqrt{\frac{2(H - h)}{g}} \right| = 2 \times \left| \sqrt{\frac{g}{2(H - h)}}H - g \cdot \sqrt{\frac{2(H - h)}{g}} \right|.$$

（注意左式絕對值內的式子必為負，右式絕對值內的式子必為正）化簡可得

$$\frac{h}{H} = \frac{2}{3}.$$

Shankar的解答

11. n次單位根的本原根(Primitive n-th Roots of Unity)

重點提要

定義：若某個n次單位根所有正整數乘冪中，可化為1的最小次數為n，則稱該單位根為本原單位根(primitive)。

規約：$R = \cos \frac{2\pi}{n} + i \sin \frac{2\pi}{n}$。（式(7)，第10節）

事實：1的所有n次單位根有：$R, R^2, \cdots, R^n$。（第10節）

定理：1的n次單位根$R, R^2, \cdots, R^n$中，$R^k$為本原單位根若且唯若指數k與n互質。

習題4

1. Show that the primitive cube roots of unity are $\omega$ and $\omega^2$.

2. For $R$ given by (7), prove that the primitive nth roots of unity are

    (i) for $n = 6$, $R, R^5$;

    (ii) for $n = 8$, $R, R^3, R^5, R^7$;

    (iii) for $n = 12$, $R, R^5, R^7, R^{11}$.

3. When n is a prime, prove that any nth root of unity, other than 1, is primitive.

4. Let $R$ be a primitive nth root (7) of unity, where n is a product of two different primes p and q. Show that $R, \cdots, R^n$ are primitive with the exception of $R^p, R^{2p}, \cdots, R^{qp}$, whose qth power are unity, and $R^q, R^{2q}, \cdots, R^{pq}$, whose pth power are unity. These two sets of exceptions have only $R^{qp}$ in common. Hence there are exactly $pq - p - q + 1$ primitive nth roots of unity.

5. Find the number of primitive nth roots of unity if n is a square of a prime p.

6. Extend Ex. 4 to the case in which n is a product of three distinct primes.

7. If $R$ is a primitive 15th root (7) of unity, verify that $R^3, R^5, R^9, R^{12}$ are the primitive fifth roots of unity, and $R^5$ and $R^{10}$ are the primitive cube roots of unity. Show that their eight products by pairs give all the primitive 15th roots of unity.

8. If $\rho$ is any primitive nth root of unity, prove that $\rho, \rho^2, \cdots, \rho^n$ are distinct and give all the nth roots of unity. Of these show that $\rho^k$ is a primitive nth root of unity if and only if k is relatively prime to n.

9. Show that the six primitive 18th roots of unity are the negative of the primitive ninth roots of unity.

習題解答

1. $n = 3, R = \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3}, \omega = R = \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3}, \omega = R$。由定理可知，本原單位根有$R^1, R^2$，亦即$\omega, \omega^2$。

2. (i) 不小於6而與6互質的數有1, 5。

    (ii) 不小於8而與8互質的數有1, 3, 5, 7。

    (iii) 不小於12而與12互質的數有1, 5, 7, 11。

3. 不小於質數p而與質數p互質的數有$p - 1$個。

4. #(1的$n = pq$次本原單位根) 

= #(不小於pq且與pq互質的數) 

= n $-$ #(不小於pq且與pq不互質的數) 

= n $-$ #(不小於pq，p的倍數或q的倍數) 

= n $- \left( \lfloor {\frac{pq}{p}} \rfloor + \lfloor {\frac{pq}{q}} \rfloor - 1 \right)$ 

= $n - (q + p - 1)$ 

= $pq - p - q + 1$ 

= $(p - 1)(q - 1)$ 。

5. #(1的$n = p^2$次本原單位根)

= #(不小於$p^2$且與$p^2$互質的數)

= $p^2 -$ #(不小於$p^2$且與$p^2$不互質的數)

= $p^2 -$ #(不小於$p^2$，p的倍數或$p^2$的倍數)

= $p^2 - \left( \lfloor \frac{p^2}{p} \rfloor + \lfloor \frac{p^2}{p^2} \rfloor - 1 \right)$

= $p^2 - p - 1 + 1$

= $p(p - 1)$

6. 命$n = pqr$。

#(1的$n = pqr$次本原單位根) 

= #(不小於pqr且與pqr互質的數) 

= pqr $-$ #(不小於pqr且與pqr不互質的數) 

= pqr $-$ #(不小於pq，p的倍數或q的倍數或r的倍數) 

= pqr $- \left( \lfloor {\frac{pqr}{p}} \rfloor + \lfloor {\frac{pqr}{q}} \rfloor  + \lfloor {\frac{pqr}{r}} \rfloor - \lfloor {\frac{pqr}{pq}} \rfloor - \lfloor {\frac{pqr}{pr}} \rfloor - \lfloor {\frac{pqr}{qr}} \rfloor +  \lfloor {\frac{pqr}{pqr}} \rfloor \right)$ 

= $pqr - (qr + pr + pq - r - q - p + 1)$ 

= $(p - 1)(q - 1)(r - 1)$ 。

7. $R = \cos \frac{2\pi}{15} + i \sin \frac{2\pi}{15}$。

$R^3 = \cos \frac{2\pi}{5} + i \sin \frac{2\pi}{5} = \left( \cos \frac{2\pi}{5} + i \sin \frac{2\pi}{5} \right)^1$,

$R^6 = \cos \frac{4\pi}{5} + i \sin \frac{4\pi}{5} = \left( \cos \frac{2\pi}{5} + i \sin \frac{2\pi}{5} \right)^2$,

$R^9 = \cos \frac{6\pi}{5} + i \sin \frac{6\pi}{5} = \left( \cos \frac{2\pi}{5} + i \sin \frac{2\pi}{5} \right)^3$,

$R^{12} = \cos \frac{8\pi}{5} + i \sin \frac{8\pi}{5} = \left( \cos \frac{2\pi}{5} + i \sin \frac{2\pi}{5} \right)^4$.

與5互質的數：1, 2, 3 ,4。

$R^5 = \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} = \left( \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} \right)^1$,

$R^{10} = \cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3} = \left( \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} \right)^2$.

與3互質的數：1, 2。

$R^3 \times R^5 = R^8$,

$R^3 \times R^{10} = R^{13}$,

$R^6 \times R^5 = R^{11}$,

$R^6 \times R^{10} = R^{16} = R^1$,

$R^9 \times R^5 = R^{14}$,

$R^9 \times R^{10} = R^{19} = R^4$,

$R^{12} \times R^5 = R^{17} = R^2$,

$R^{12} \times R^{10} = R^{22} = R^7$,

與15互質的數：1, 2, 4, 7, 8, 11, 13, 14。

8. 因為$\rho$是本原單位根，所以存在正整數a滿足$1 \le a \le n$且$\gcd (a, n) = 1$使得$\rho = R^a$，其中$R = \cos \frac{2\pi}{n} + i \sin \frac{2\pi}{n}$。

於是$\{ \rho, \rho^2, \cdots, \rho^n \} = \{ R^a, R^{2a}, \cdots, R^{na} \}$。任取兩相異正整數$k, j$滿足$1 \le k < j \le n$。若$\rho^j = \rho^k$，即$R^{ja} = R^{ka}$，則有$\cos \frac{2ja\pi}{n} + i \sin \frac{2ja\pi}{n} = \cos \frac{2ka\pi}{n} + i \sin \frac{2ka\pi}{n}$。因此$\frac{2ja\pi}{n}$與$\frac{2ka\pi}{n}$為同界角，得$j - k$為n的倍數。然而$1 \le j-k \le n$，所以$j - k$不可能是n的倍數。這意味著$R^a, R^{2a}, \cdots, R^{na}$完全相異。

因$\left( \rho^k \right)^n = \left( \rho^n \right)^k = 1^k = 1$，所以$\{ \rho, \rho^2, \cdots, \rho^n \}$是1的n次單位根。

假定$\rho^k$是本原根，由前文$\rho$與$R$的關係，得$R^{ak}$是本原根。再由定理可知$\gcd (ak, n) = 1$。因為$\gcd (a, n) = 1$，所以$\gcd (k, n) = 1$。

假定k與n互質，則$\gcd (ak, n) = 1$，故$R^{ak}$是本原根，即$\rho^k$是本原根。

9. $n = 18$，與18互質的數有1, 5, 7, 11, 13, 17。取$R_{18} = \cos \frac{2\pi}{18} + i \sin \frac{2\pi}{18}$。

$R_{18}^1 = \cos \frac{2\pi}{18} + i \sin \frac{2\pi}{18} = \cos \frac{\pi}{9} + i \sin \frac{\pi}{9}$,

$R_{18}^5 = \cos \frac{5\pi}{9} + i \sin \frac{5\pi}{9}$,

$R_{18}^7 = \cos \frac{7\pi}{9} + i \sin \frac{7\pi}{9}$,

$R_{18}^{11} = \cos \frac{11\pi}{9} + i \sin \frac{11\pi}{9}$,

$R_{18}^{13} = \cos \frac{13\pi}{9} + i \sin \frac{13\pi}{9}$,

$R_{18}^{17} = \cos \frac{17\pi}{9} + i \sin \frac{17\pi}{9}$.

$n = 9$，與9互質的數有1, 2, 4, 5, 7, 8。取$R_{9} = \cos \frac{2\pi}{9} + i \sin \frac{2\pi}{9}$。

$R_{9}^1 = \cos \frac{2\pi}{9} + i \sin \frac{2\pi}{9}$,

$R_{9}^2 = \cos \frac{4\pi}{9} + i \sin \frac{4\pi}{9}$,

$R_{9}^4 = \cos \frac{8\pi}{9} + i \sin \frac{8\pi}{9}$,

$R_{9}^5 = \cos \frac{10\pi}{9} + i \sin \frac{10\pi}{9}$,

$R_{9}^7 = \cos \frac{14\pi}{9} + i \sin \frac{14\pi}{9}$,

$R_{9}^8 = \cos \frac{16\pi}{9} + i \sin \frac{16pi}{9}$.

觀察幅角，有

$\frac{10\pi}{9} - \frac{\pi}{9} = \pi$,

$\frac{14\pi}{9} - \frac{5\pi}{9} = \pi$,

$\frac{16\pi}{9} - \frac{7\pi}{9} = \pi$,

$\frac{11\pi}{9} - \frac{2\pi}{9} = \pi$,

$\frac{13\pi}{9} - \frac{4\pi}{9} = \pi$,

$\frac{17\pi}{9} - \frac{8\pi}{9} = \pi$.

所以互為相反數。