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2020年9月28日 星期一

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 yA,yB and velocities vA and vB. 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yA,yB以及速度vAvB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yAyB
yA(t)=H+0t12gt2,
yB(t)=0+vB(0)t12gt2.
假定在時刻t時發生碰撞,利用A球的位置函數,得
h=H12gt2.
解出t=2(Hh)g。代入B球的位置函數,可以解出B球上拋的初始速度:
h=vB(0)2(Hh)g12g(2(Hh)g)2,
vB(0)=g2(Hh)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vAvB
vA(t)=0gt,
vB(t)=g2(Hh)Hgt.
代入撞擊時刻t=2(Hh)g,得
|0g2(Hh)g|=2×|g2(Hh)Hg2(Hh)g|.
(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
hH=23.

Shankar的解答

2020年9月11日 星期五

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=cos2πn+isin2πn。(式(7),第10節)

事實:1的所有n次單位根有:R,R2,,Rn。(第10節)

定理1的n次單位根R,R2,,Rn中,Rk為本原單位根若且唯若指數kn互質

習題4

1. Show that the primitive cube roots of unity are ω and ω2.

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

    (i) for n=6, R,R5;
    (ii) for n=8, R,R3,R5,R7;
    (iii) for n=12, R,R5,R7,R11.

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,,Rn are primitive with the exception of Rp,R2p,,Rqp, whose qth power are unity, and Rq,R2q,,Rpq, whose pth power are unity. These two sets of exceptions have only Rqp in common. Hence there are exactly pqpq+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 R3,R5,R9,R12 are the primitive fifth roots of unity, and R5 and R10 are the primitive cube roots of unity. Show that their eight products by pairs give all the primitive 15th roots of unity.

8. If ρ is any primitive nth root of unity, prove that ρ,ρ2,,ρn are distinct and give all the nth roots of unity. Of these show that ρk 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=cos2π3+isin2π3,ω=R=cos2π3+isin2π3,ω=R。由定理可知,本原單位根有R1,R2,亦即ω,ω2

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p1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pqp的倍數或q的倍數) 
= n (pqp+pqq1) 
= n(q+p1) 
= pqpq+1 
= (p1)(q1) 。

5. #(1的n=p2次本原單位根)
= #(不小於p2且與p2互質的數)
= p2 #(不小於p2且與p2不互質的數)
= p2 #(不小於p2p的倍數或p2的倍數)
= p2(p2p+p2p21)
= p2p1+1
= p(p1)

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pqp的倍數或q的倍數或r的倍數) 
pqr (pqrp+pqrq+pqrrpqrpqpqrprpqrqr+pqrpqr) 
= pqr(qr+pr+pqrqp+1) 
= (p1)(q1)(r1) 。

7. R=cos2π15+isin2π15
R3=cos2π5+isin2π5=(cos2π5+isin2π5)1,
R6=cos4π5+isin4π5=(cos2π5+isin2π5)2,
R9=cos6π5+isin6π5=(cos2π5+isin2π5)3,
R12=cos8π5+isin8π5=(cos2π5+isin2π5)4.
與5互質的數:1, 2, 3 ,4。

R5=cos2π3+isin2π3=(cos2π3+isin2π3)1,
R10=cos4π3+isin4π3=(cos2π3+isin2π3)2.
與3互質的數:1, 2。

R3×R5=R8,
R3×R10=R13,
R6×R5=R11,
R6×R10=R16=R1,
R9×R5=R14,
R9×R10=R19=R4,
R12×R5=R17=R2,
R12×R10=R22=R7,
與15互質的數:1, 2, 4, 7, 8, 11, 13, 14。

8. 因為ρ是本原單位根,所以存在正整數a滿足1angcd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}\frac{2ka\pi}{n}為同界角,得j - kn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。

\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\rhoR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kn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。