105/05/23

MATH 224 – Discrete Mathematics

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i

i1

n

j

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Sequences and Sums

A sequence of the form ar0, ar1, ar2, ar3, ar4, … , arn, is called a geometric sequence and occurs quite often in computer science applications. Another common sequence is of the form a, 2a, 3a, 4a, … , na is called an arithmetic sequence. We have seen this when examining the number of steps executed by selection sort. In our example, the variable a was equal to 1, so we just had 1, 2, 3, … . N. Often we are interested in the sum of one of these sequences as in the arithmetic sum and the geometric sum.

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ra n

1)1( 1

=

Arithmetic Sum Geometric Sum (as long as r ≠ 1)

What will be the formula for the geometric sum when a = 1. Compare this to the formula on Page 155 from our text. Are they equivalent?

205/05/23

MATH 224 – Discrete Mathematics

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Basic Properties of Sums

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305/05/23

MATH 224 – Discrete MathematicsMathematical Induction

Mathematical induction is a property of the integers and any well ordered set. It may be used to prove properties hold for an infinite set of values. For example it may be used to prove something such as

Proofs by induction require two steps: the Basis Step and the Inductive Step. In addition, a proof of the inductive step, makes use of the an Inductive Hypothesis. All three of these elements will be required in your proofs.

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jr0 r

r n

1)1( 1= when r ≠ 1

405/05/23

MATH 224 – Discrete MathematicsMathematical Induction Basis Step

The Basis Step requires the proof of some property P(x0) for a starting value x0. P(x) might involve an equation as in the summation above or it might be a statements such as a C++ function sorts all input arrays from smallest to largest. So for example the basis step might require one to prove that

Or it might require one to prove that a C++ function such as binary_search works correctly for a sorted array with just one element.

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1 10

rr=

505/05/23

MATH 224 – Discrete MathematicsMathematical Induction Inductive Step

The Inductive Step requires a proof that P(x) → P(x+1) for any x ≥ x0. To do this you assume that the Inductive Hypothesis P(x) is true for some arbitrary value x ≥ x0. You are not assuming this for all x, but just for some x with no restrictions. Note that x is greater than or equal to the starting value.

In your proofs you must state the inductive hypothesis explicitly. So for example, in a proof of the geometric sum, the inductive hypothesis would be something like:

Assume that for some x ≥ 0 the following holds as long as r ≠ 1.

Note the word some. We are only assuming this for one value of x. Also be sure to note the x is set greater than or equal to the starting value. Both of these are critical for a correct proof.

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r x

1)1( 1

=

605/05/23

MATH 224 – Discrete Mathematics

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Proof without Using Induction

The following is a proof of the formula for the Geometric Sum without using induction.

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1)1( 1=

Consider

11322 1)...(...1 nnn rrrrrrrr

How do we get the expression on the right?

Then dividing the left and right sides by 1 – r we get the final result.

What is wrong with dividing by 1 – r if r = 1?

when r ≠ 1

705/05/23

MATH 224 – Discrete Mathematics

k

j

jr0

n

j

jr0

Proof Using InductionThe following is a proof of the formula for the Geometric Sum using induction.

rr n

1)1( 1

=

Basis: (Sometimes called the base case.) Consider the sum when n = 0. Then the sum has only one term r0 =1. And the fraction on the right is just

(1–r)/(1–r) since n+1 = 1. Thus the equation is correct for the base case.

Inductive Hypothesis: Assume that for some k ≥ 0 the equation holds. In other words,

when r ≠ 1

rr k

1)1( 1

=

805/05/23

MATH 224 – Discrete Mathematics

1

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Proofs Using InductionIt is critical that the inductive hypothesis does not restrict the value of k. In other words it must be possible for k to take on the base value. Why would k > 0 be wrong for the Inductive Hypothesis in this proof?

Inductive Step: Consider the following sum : for some k ≥ 0. By algebra this sum may be written as

The right side of the equality follows from the inductive hypothesis. Why?

Then adding the two fractions and simplifying yields the fraction:

And from this we may conclude that the original equation is correct by induction. If we substitute k+1 for n, what does the original equation give?

rrk

11 2