~jacks/dal-classwork unlisted

533c1a65bd6d1ea0cbe6b7e879cf790e9cf07c08 — jacksarick 10 months ago 73f0e31
ass8
A CSCI 2112/assignment8/a8_W19.pdf => CSCI 2112/assignment8/a8_W19.pdf +0 -0

A CSCI 2112/assignment8/assignment8.md => CSCI 2112/assignment8/assignment8.md +42 -0
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1. Prove $\forall n > n_0; n! > 2^n$
	1. $\{1,2,3,4,5\}! = \{1,2,6,24,120\}$
	2. $\{1,2,3,4,5\}^2 = \{1,4,9,16,25\}$
	3. $n_0 = 4$
	4. $n! = n * (n-1) * (n-2) * (n-3)!$
	5. $(n - 1) * (n - 2) \geq n; n > 2 = n_0 - 2$
	6. $n! > n * n * (n-3)! > n^2$

2. Elbonia's currency (@)
	1. 1x 2x 3x 4x 5 6x 7x 8x 9 10 11x 12x 13x 14 15 16x 17x 18 19x 20 21x

3. Induction:
	1. Floors:
		1. $D_1 = 2$ `[y, b]`
		2. $D_2 = 3$ `[yy, yb, by]`
		3. $D_3 = 5$ `[yyb, yyy, yby, byy, byb]`
		4. $D_4 = 8$ `[yyby, yyyb, yyyy, ybyy, ybyb, byyy, byyb byby]`
		5. $D_n = D_{n-1} + D_{n-2}$
	
	2.  Proofs:
		1. $F_0 - F_1 + F_2 - F_{2n-1} + F_{2n} = F_{2n-1} -1$
			1. $F_0 = 0$
			2. $F_1 = F_2$
		2. $\sum^n_{j=1}F_{2j-1}=F_{2n}$
			1. $\sum^n_{j=1}F_{2j-1} = F_0 + F_1 + F_3 + ... + F_{2n-1}$
		3. $F_{2n+1} = 3F_{2n-1}-F_{2n-3}$
		4. ${F_{n+1}}^2 = {F_n}^2 + F_{n-1} * F_{n+2}, n \geq 2$
	
	3. $\{1200, 1800, 2700, 4050\}$ 
		1. $C_{n+1} = C_n + \frac{C_n}{2}, C_0 = 1200$ 
	
	4. Recursion
		1. $a^{2^n}$
			1. $\{a^2, a^4, a^8, ...,  a^{2(n-1)} \}$
			2. $b_n(a) = a^{2^n}$
			3. $c_0(a) = a^2; c_n(a)=c_{n-1}(a)^2$
			4. $c_n(a) = b_n(a) \because \{a^2, a^4, a^8, ...,  a^{2(n-1)} \}$
		2. Efficient powers:
			1. $\text{pow}(a, 0) = 1$
			3. $\text{pow}(a, n) = a * \text{pow}(a, n - 1)$ 
			4. It's efficient because it always runs in n+1 steps
			5. It's correct because $\text{pow}(a, 3) = a * \text{pow}(a, 2) = a * (a * \text{pow}(a, 1)) = a * (a * (a * 1)) = a^3$
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A CSCI 2112/assignment8/assignment8.pdf => CSCI 2112/assignment8/assignment8.pdf +0 -0