Homework #6: Due Thursday November 3
Chapter 7:
. 2. Solve the following problems of al-Khwārizmī by applying the appropriate formula:
a) ((1/3)x+1)((1/4)x+1) = 20
x2/12 + 7x/12 + 12/12 = 20
x2 + 7x + 12 = 240
x2 + 7x = 228.
This equation is of the fourth kind, so halve the number of roots – 3.5 – multiply it by itself – 12.25 – add it to 228 – 240.25. Take the root of this – 15.5 – and subtract from it half the number of roots – 12. (x = Ã((7/2)2 + 228) – 7/2 = Ã(49/4 + 912/4) – 7/2 = Ã(961/4) – 7/2 = 31/2 – 7/2 = 24/2 = 12).
b) x2 + (10-x)2 = 58
x2 + 100 – 20x + x2 = 58
2x2 – 20x + 100 = 58
x2 – 10x + 50 = 29
x2 + 21 = 10x
This equation is of the fifth kind, so
x = b/2 +/- Ã((b/2)2 – c) = 5 +/- Ã(25-21) = 5 +/- 2. x = 7 or 3.
4. Solve this problem of al-Khwārizmī: I have divided 10 into two parts, and have divided the first by the second, and the second by the first and the sum of the quotients is 2 and 1/6. Find the parts.
Call the first part x and the second part 10-x, then this becomes
x/(10-x) + (10-x)/x = 13/6
x2/(10x-x2) + (10-x)2/(10x-x2) = (2x2 + 100 – 20x)/(10x-x2) = 13/6
12x2 + 600 – 120x = 130x – 13x2
25x2 + 600 = 250x
x2 + 24 = 10x
This equation is of the fifth kind, so
x = b/2 +/- Ã((b/2)2 – c) = 5 +/- Ã(25 – 24) = 5 +/- 1, so the parts are 4 and 6.
17. Give a modern proof by induction on n of ibn al-Bannā's result that Cnk = ((n-(k-1))/(k))Cnk-1.
Basis: n = 1, k = 1 LHS: C(1,1) = 1. RHS (1-0)/1C(1,0) = 1*1 = 1.
Induction: Assume C(n,k) = (n – (k -1)/k * C(n, k -1) for k = 1 to n.
Then prove C(n+1, k) = (n+1 – (k -1))/k * C(n+1, k -1)/
LHS: C(n+1, k) = C(n, k) + C(n, k – 1) = (n – (k -1)/k * C(n, k -1) + C(n, k -1) =
(n – k + 1 –k)//k C(n, k -1) = (n+1)/k C(n, k -1 ) = (n+1)/k *(n+2 -k)/(k -1) C(n, k-2).
RHS: (n+1 – (k -1))/k * C(n+1, k -1) = (n+2-k)/k * (C(n, k-1) + C(n, k- 2))
= (n+2-k)/k (n- (k+2) +(k-1)/(k-1) C(n, k-2) = (n+2-k)/k* (n+1)/(k-1) C(n, k -2).
So LHS = RHS and the induction carries through.
Chapter 8
. 2. A man must ferry a wolf, a goat, and a head of cabbage across a river. The available boat, however, can only carry the man and one other thing. The goat cannot be left alone with the cabbage, nor the wolf with the goat. How should the man ferry his three items across the river?
The man should ferry the goat across the river, then return alone. He should then take either the cabbage or the wolf across the river, and ferry the goat back to the original side. Leaving the goat, he should take the remaining wolf or cabbage across, then return empty to ferry the goat across:
wolf, cabbage ------> goat
wolf, cabbage <-------- goat
cabbage ------------> wolf, goat
cabbage, goat <------------ wolf
goat -------------> wolf, cabbage
goat <-------------- wolf, cabbage
nothing -----------> wolf, cabbage, goat
15. Leonardo has the following problem in the Liber Abaci: There is a lion at the bottom of a pit 50 feet deep. The lion climbs up 1/7 of a foot each day and then falls back 1/9 of a foot each night. How long will it take him to climb out of the pit? (not 1575 days)
On the last day, the lion will get free and not slide down, so there will be one more day of climbing than of falling, so x is the number of the days it will take to get free, and
x/7 – (x-1)/9 ³ 50
(9x – 7x + 7)/63 ³ 50
2x + 7 ³ 3150
2x ³ 3143
x ³ 1571.5
x = 1572
On the 1571st day, the lion reaches 1571/7 – 1570/9 = 49.984 feet, and during the night slides back down to 49.984 – 1/9 = 49.873 feet. On the 1572nd day, the 1/7 of a foot would put him at 49.873 + 1/7 = 50.016 feet, so he has reached the edge and gotten free.
16. The Fibonacci sequence (the sequence of rabbit pairs) is determined by the recursive rule
F0 = F1 = 1 and Fn = Fn-1+Fn-2.
Show that
a) Fn+1*Fn-1 = F2n =(-1)n
Rewrite this as Fn+1*Fn-1 - F2n = (-1)n+1 and prove by induction.
Basis: n = 1 F2*F0 - F12 = 2* 1 – 1 = 1 = (-1)2.
Induction: Assume that Fn+1*Fn-1 - F2n = (-1)n+1.
Fn+2*Fn
- F2n+1 = (Fn
+ F n+1)*Fn –
Fn+12 = FnFn
+ Fn+1Fn - Fn+1* Fn+1
= Fn+1(Fn – Fn+1) + FnFn
=
Fn+1*
(- Fn-1) + FnFn = - (Fn+1*Fn-1 - F2n) = -1* (-1)n+1 = (-1)n+2
b) that lim(n->°) Fn/Fn-1 = (1+Ã5)/2.
First use a) to show that this limit exists.
Fn+1/Fn
- Fn/Fn-1 = ( Fn+1*Fn+1 - Fn*Fn) /Fn*Fn-1
= (-1)n+1/Fn*Fn-1. This clearly ->0 as n -> °.
So let x = lim(n->°)Fn/Fn-1 = lim(n->°)(Fn-1+Fn-2)/Fn-1 = lim(n->°)1 + Fn-2/Fn-1
= 1 + lim(n->°)Fn-2/Fn-1 = 1 + lim(n->°) 1/(Fn-1/Fn-2),
x = 1 + lim(n->°)1/(Fn/Fn-1) = 1 + 1/(lim(n->°)Fn/Fn-1) = 1 + 1/x
x – 1 = 1/x
x2 – x = 1
x2 – x – 1 = 0
x = (1 ± Ã(1 + 4))/2 = (1 ± Ã5)/2, and as this number must be positive (since the ratio of positive numbers must be positive), x = (1 + Ã5)/2
Chapter 9:
. 16. Prove that the equation x3 + cx = d always has one positive solution and no negative ones.
Using calculus: Let f(x) = x3 + cx – d; then fÕ(x) = 3 x2 + c , which is always >0 (since c is).
So f(x) is always increasing. f(0) = -d < 0 and f(x) is negative for x negative.
Since f increases without limit, it will cross the x axis once, for a positive value of x.
21. It is obvious that 3 is a root of x3 + 3x = 36. Show that Cardano's formula gives 3Ã(Ã(325)+18) - 3Ã(Ã(325)-18). Using Bombelli's methods, show that this number is in fact equal to 3.
x = 3Ã(n/2 + Ã(n2/4 + m3/27)) - 3Ã(-n/2 + Ã(n2/4 + m3/27))
= 3Ã(36/2 + Ã(362/4 + 33/27)) - 3Ã(-36/2 + Ã(362/4 + 33/27))
= 3Ã(18 + Ã(1296/4 + 27/27)) - 3Ã(-18 + Ã(1296/4 + 27/27))
= 3Ã(18 + Ã(324 + 1)) - 3Ã(-18 + Ã(324 + 1))
= 3Ã(18 + Ã(325)) - 3Ã(-18 + Ã(325)) = 3Ã(Ã325 + 18) - 3Ã(Ã325 – 18)
Set 3Ã(Ã325 + 18) = Ãb + a and 3Ã(Ã325 – 18) = Ãb – a
Multiplying these together gives b – a2 = 3Ã(325-324) = 3Ã1 = 1
b – a2 = 1, b = 1 + a2
Cubing Ãb + a gives a3 + 3ab + (3a2+b)Ãb
Setting the parts without the square root equal gives a3 + 3ab = 18.
Integer a.b wonÕt work. If a = ½, then b = 5/4 and a3 + 3ab = 1/8 + 15/8 = 2, which is too small.
Try a = 3/2. Then b = 13/4 and a3 + 3ab = 27/8 + 117/8 = 144/8 = 18. This works.
Then (Ãb + a) – (Ãb – a) = 2a = 3.