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Section 5.2 Summations and Closed FormsA closed form is an expression that can be computed by applying a fixed number of familiar operations to the arguments. For example, the expression 2 + 4 + … + 2n is not a closed form, but the expression n(n+1) is a closed form.

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akk=1

n

∑ = a1 +L + an .Summation Notation:

Some Useful Closed Forms

Summation Facts

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ak+ik=m

n

∑ = akk=m+i

n+i

∑ .(4)

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(ak + bk )∑ = ak +∑ bk∑ .(2)

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ak x i+k∑ = x i ak x k∑ .(3)

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cak∑ = c ak∑ .(1)

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ck=m

n

∑ = (n − m +1)c.(1)

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kk=1

n

∑ =n(n +1)

2.(2)

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k 2

k=1

n

∑ =n(n +1)(2n +1)

6.(3) (4)

(5)

(5) Collapsing Sums)

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(ak − ak−1)k=1

n

∑ = an − a0 and (ak−1 − ak )k=1

n

∑ = a0 − an .

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a k

k=0

n

∑ =a n+1 −1a −1

(where a ≠1).

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ka k

k=1

n

∑ =a − (n +1)a n+1 + nan+2

(a −1)2(where a ≠1).

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(k −1)2k+1

k=2

n

∑ = k2k+2

k=1

n−1

∑

= 22 k2k

k=1

n−1

∑

= 22(2 − n2n +(n −1)2n+1)

= 23 − (2 − n)2n+2.

(Fact 4)

(Fact 3)

(Form 5)

Example. Find a closed form for the expression

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(k −1)2k+1

k=2

n

∑ .

Solution:

Solution: The sum has the form

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2 + 2k (k −1) ⋅7k=2

n

∑

= 2 + 7 (k −1)2k

k=2

n

∑

= 2 + 7 k2k+1

k=1

n−1

∑

= 2 +14 k2k

k=1

n−1

∑

= 2 +14(2 − n2n +(n −1)2n+1).

Example. Find a closed form for

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2+22 ⋅7 +23 ⋅14 +L +2n(n−1)⋅7.

(Fact 1)

(Fact 4)

(Fact 3)

(Form 5)

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Solution.

Quiz. Use summation facts and forms to prove that

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2+3+...+n=(n−1)(n+2)2

.

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2 + 3+...+ n = kk=2

n

∑ = (k +1)k=1

n−1

∑

= kk=1

n−1

∑ + 1k=1

n−1

∑

=(n −1)(n)

2+(n −1)

=(n −1)(n + 2)

2.

Solution.

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(3+ 4k)k=0

n

∑ = 3k=0

n

∑ + 4kk=0

n

∑

= 3k=0

n

∑ + 4 kk=0

n

∑

= 3(n +1)+4n(n +1)

2= (3+ 2n)(n +1).

Quiz. Use summation facts and forms to find a closed form for

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3+7+L +(3+4n).

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Example. Let count(n) be the number of := statements executed by the following algorithm as a function of n, where n N. Find a closed form for count(n).

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count(n) =1+ (n −1) + (2+ 3+L + n)

= (n −1) + (1+ 2+ 3+L + n)

= (n −1) +n(n +1)

2.

Quiz. Let count(n) be the number of executions of S in the preceding algorithm as a function of n. Find a closed form for count(n).

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count(n) = (2+ 3+L + n)

= (1+ 2+ 3+L + n) −1

=n(n +1)

2−1.

Solution.

i := 1;while i < n do

i := i + 1;for j := 1 to i do S od

od

The expressions in parentheses indicate the number of times that := is executed.

(1)

(n – 1)(2 + 3 + … + n)

Therefore, count(n) is the sum:

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Example. Let count(n) be the number of times S is executed by the following algorithm as a function of n, where n N. Find a closed form for count(n).

i := 1;while i < n do

i := i + 2;for j := 1 to i do S od

odSolution: Each time through the while-loop i is incremented by 2. So the values of i at the start of each for-loop are 3, 5, …, (2k + 1), where i = 2k + 1 ≥ n represents the stopping point for the while-loop. So we have

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count(n) = 3+5+L + (2k +1)

= (2i +1)i=1

k

∑ = 2 i + 1i=1

k

∑i=1

k

∑

=2k(k +1)

2+ k = k(k +1) + k = k(k + 2).

But we need to write count(n) in terms of n. Since 2k + 1 ≥ n is the stopping point for the while-loop it follows that 2k –1< n is the last time the while-condition is true. In other words, we have the inequality 2k – 1 < n ≤ 2k + 1. Solving for k, we have 2k – 2 < n – 1 ≤ 2k, which gives k – 1 < (n – 1)/2 ≤ k. Therefore k = (n – 1)/2. Now we can write count(n) in terms of n as

count(n) = k(k + 2) = (n – 1)/2((n – 1)/2 + 2).

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Approximating Sums. Suppose we have a sum that doesn’t have a closed form or wecan’t find a closed form. Then we might find an approximation to suit our needs. Forexample, consider the following sum:

Hn = 1 + 1/2 + 1/3 + ... + 1/n.

This sum is called the nth harmonic number and it has no closed form. It is closelyapproximated by ln n because the definite integral of 1/x from 1 to n is ln n. A constant,called Euler’s constant, with a value close to 0.58, approximates the difference between Hn

and ln n for large n. In other words, we have Hn > ln n and Hn – ln n is close to Euler’s constant for large n. For example,

H10 – ln 10 2.93 – 2.31 = 0.62 H20 – ln 20 3.00 – 2.60 = 0.60H40 – ln 40 4.28 – 3.69 = 0.59

Example (Overlapping windows). Suppose we have set of files in the form of windows on acomputer screen that are to be displayed as a stack of overlapped windows. How muchtotal space is needed for the display? For example, if A is the area of the top window andeach overlapped window has area A/2, then a stack of n files will have area

A + A(n – 1)/2.This might take too much space for even small values of n, which would force A to bequite small. In this case we could decrease the size of each overlapped window to a smallerfraction of A. But it might be useful to make the overlapped files progressively smaller withsizes such as A/2, A/3, A/4, ... . In this case the n files will have total area

A + A/2 + A/3 + ... + A/(n – 1) = AHn–1,

which is approximately equal to A(ln(n – 1) + 0.58).