ndix A: Summations vation: Evaluating and/or bounding sums are frequen ed in the solution of recurrences types of evaluation problems: Prove by induction that formula is correct Find the function that the sum equals or is bounde untered both types in analysis of insertion sort
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Appendix A: Summations Motivation: Evaluating and/or bounding sums are frequently needed in the solution of recurrences Two types of evaluation problems:
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Appendix A: Summations
Motivation: Evaluating and/or bounding sums are frequently needed in the solution of recurrences
Two types of evaluation problems:
Prove by induction that formula is correct
Find the function that the sum equals or is bounded by
Encountered both types in analysis of insertion sort
Prove by induction that j=1 to n j = n(n+1)/2 Called the arithmetic sumText p 1059
Use the arithmetic sum to evaluate the sums in the analysis of insertion sort runtime
Important sums to remember
Arithmetic k=1 to n k = n(n+1)/2 = (n2)
Geometric k=0 to n xk = (xn+1 – 1)/(x – 1) when x 1
Harmonic k=1 to n (1/k) = ln(n) + (1)
Alternate forms of geometric sum useful in tree analysis
k=0 to n-1 xk = (xn – 1)/(x – 1) when x 1
How do we show this is true?
k=0 to ∞ xk = 1/(1 – x)
when |x| < 1
Integration and differentiation can be used to evaluate sums
derivative: d{ f(x)}/dx = df/dx
integral: dx {f(x)} = dx f(x)
Example: eq. A.8 p1148
Show k=0 to ∞ k xk = x/(1 – x)2
when 0< |x| < 1
See bottom p1147 for simpler approach
Bounding sums
Prove a bound by induction
Bound ever term in sum
Bound by integrationmonotone increasing and decreasing summands
Prove by induction on integers that k=0 to n 3k = O(3n)
there exist c=4/3 such that 0<4<3c Similar argument applies n=2, etc.
Property of sums independent of what we are trying to prove
and (1/3 +1/c) < 1, which is true if c > 3/2; therefore, c=3/2 or larger will work in the definition of big OHence k=0 to n 3k = O(3n) by definition
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Base case n=0 is true
< c3n which implies
Example of bound sum by bounding every term
Show that (n/2)2 < k=1 to n k < n2
Bound by integration: monotone increasing summand
Shaded area is integral ofcontinuous function f(x)
Sum equals area of “upper sum”rectangles
Same f(x) different limits on integration
Sum equals area of “lower sum”rectangles
Note the difference for monotone increasing and decreasing summand
Method not applicable if summand is not monotone increasing or decreasing
Use bounding by integrals for informal proof that k=1 to n k-1 = (ln(n))
CptS 450 Spring 2015[All problems are from Cormen et al, 3nd Edition]Homework Assignment 3: due 2/4/151. ex A.1-3 p 11492. ex A.1-6 p 11493. ex A.2-1 p 1156 (hint: use integration)4. part a of prop A-1 p 1156 using bound each term