1.8 #2,3,7,8,16,18,20,23,32 
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#2. a,b,c,f, but not d,e.
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#8.
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<item>a. 4 -- tricky! There is no least _real_ n (anything >3 works)
<item>b. 5
<item>c. 0
<item>d. -1
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#16. In short, since O(x) = O(x^2), f(x) = O(x) implies f(x) = O(x^2).
Proceeding from the definition, though, if we have an E and a B such that
|f(x)/x| < B for all x>E, then we can use the same B (and replace E by 1,
if E<1) to say |f(x)/x^2| < B.
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#18. 1^k + 2^k + ... + n^k < n^k + n^k + ... + n^k = n * n^k = n^{k+1}.
So we can even use B=1 in this case.
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#20a. First simplify to O(n^3 * log n) + O(log n * n^3), then add them together
to just get O(n^3 * log n). <br>
b. First get to O(2^n) O(3^n), then go to O(6^n). <br>
c. Already n^2 = o(2^n), so 2^{n^2} = o(2^{2^n}). So toss out that tiny term
and get O(2^{2^n}).
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#32. 1*3*...*(2n-1) < 2*4*...*(2n) = 2^n n!. So O(2^n n!) is a good answer.<br>
A lamer answer is 1*3*...*(2n-1) < (2n-1)^n, giving O((2n)^n).<br>
A _really_ lame answer is 1*3*...*(2n-1) < 1*2*3*4*...*2n = (2n)!.<br>
A downright feeble answer is O((2n)^(2n)). I hope nobody resorted to that.