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# quiz9ans - integer n such that f = O(x^n • f(x =(2 x^3 3...

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Quiz #9 for Math 55, Dec. 4 2002 1. Let f(n) = 1 * 3 * 5 * . .. * (2n-1). Is f(n) = O(2^n)? Is f(n) = O(n^n)? Is f(n) = O(n!)? 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. A. No, yes, no. Let's recall that f(n) = (2n)! / (2^n n!). 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. 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. 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. 2. For each of the following functions f, find the smallest
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Unformatted text preview: integer n such that f = O(x^n). • f(x) = (2 x^3 + 3 x log x) (x+2)^{10} • f(x) = (x^2 + x^2 log x) (x - log x)^3 • f(x) = sqrt{x+1} - sqrt{x} A. • n = 13. The x^3 and x^10 terms dominate. • n = 6. The x^2 log x "rounds up" to x^3. • There are two ways to do this. One is to factor out and use the binomial theorem: f(x) = sqrt(x) ( sqrt(1+1/x) - 1 ) = sqrt(x) O(1/x) = O(x^(-1/2)) The other is to multiple "top and bottom" by sqrt(x+1) + sqrt(x): 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}) Either way, the smallest integer n is n=0. 3. What is the set of all real 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. A. Any real strictly more than 1 will work, whereas 1 will not work....
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