Quiz #9 for Math 55, Dec. 4 2002
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1. Let f(n) = 1 * 3 * 5 * ... * (2n-1). 
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<li> Is f(n) = O(2^n)?
<li> Is f(n) = O(n^n)?
<li> Is f(n) = O(n!)?
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In each case, either give a B and E (such that
for all x greater than or equal to  E, 
|f(x)/g(x)| is less than B) or explain why none exist.

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A. No, yes, no. Let's recall that f(n) = (2n)! / (2^n n!).
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f/2^n = 1/2 * 3/2 * 5/2 * ... * (2n-1)/2. Each time we increase n by
one, we multiply the ratio by about n, so it keeps going up -- it isn't
bounded.
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f/n^n = (2n)! / (2^n n! n^n) = (2n!)/n! / (2n)^n. This numerator is
the product (n+1)*...*2n, and the denominator is 2n*2n*...*2n (n terms),
so the ratio is always less than 1. Take B=E=1.
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f/n! = 1/1 * 3/2 * 5/3 * ... * (2n-1)/n. Each time we increase n by one,
we multiply the ratio by about 2, so it keeps going up -- it isn't bounded.


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2. For each of the following functions f, find the smallest <i>integer</i> n
such that f = O(x^n).
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<li> f(x) = (2 x^3 + 3 x log x) (x+2)^{10}
<li> f(x) = (x^2 + x^2 log x) (x - log x)^3
<li> f(x) = sqrt{x+1} - sqrt{x}
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A. <menu>
<li> n = 13. The x^3 and x^10 terms dominate.
<li> n = 6. The x^2 log x "rounds up" to x^3.
<li> There are two ways to do this. One is to factor out and use the
binomial theorem:
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f(x) = sqrt(x) ( sqrt(1+1/x) - 1 ) = sqrt(x) O(1/x) = O(x^(-1/2))
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The other is to multiple "top and bottom" by sqrt(x+1) + sqrt(x):
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f(x) = (sqrt{x+1} - sqrt{x}) (sqrt(x+1) + sqrt(x)) / (sqrt(x+1) + sqrt(x))
     = 1 / (sqrt(x+1) + sqrt(x)) = O(x^{-1/2})
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Either way, the smallest integer n is n=0.
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3. What is the set of all <i> real</i> numbers n such that 
 x + sqrt{x} = O(x^n)? 
Make sure you find all of them, and that you don't include any that
aren't there.

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A. Any real strictly more than 1 will work, whereas 1 will not work.