Introduction to Algorithms

Appendix N: Derivation of the Logarithm Change of Base Formula

We set out to prove the logarithm change of base formula: logb x = loga x loga b. To do so we let y = logb x and apply these as exponents on the base.

MATHEMATICS 0110A CHANGE OF BASE Suppose that we have

Let y = logb a. Then we know that this means that by = a. We can take logarithms to base c
Change of Base

6.2 Properties of Logarithms

out the inverse relationship between these two change of base formulas. To change the base of Prove the Quotient Rule and Power Rule for Logarithms.
S&Z . & .

Lesson 5-2 - Using Properties and the Change of Base Formula

You can prove the Change of Base. Formula blog X x because exponents and logarithms are inverses. Take the log base a of both sides: log

Elementary Functions The logarithm as an inverse function

The positive constant b is called the base (of the logarithm.) Smith (SHSU) Let's call this the “change of base” equation or “change of base” property.
. Logarithms (slides to )

Introduction to Algorithms

I can prove this using the definition of big-Omega: This tells us that every positive power of the logarithm of n to the base b where b ¿ 1
cs lect fall notes

3.3 The logarithm as an inverse function

Let's call this the “change of base” equation. Example. Suppose we want to compute log2(17) but our calculator only allows us to use the natural logarithm ln.
Lecture Notes . Logarithms

CS 161 Lecture 3 - 1 Solving recurrences

n ≥ 2 and the base cases of the induction proof (which is not the same as Plugging the numbers into the recurrence formula
lecture

Logarithms Math 121 Calculus II

Proof. By the inverse of the Fundamental Theorem of Calculus since lnx is defined as an In particular
logs

Section 4.3 Logarithmic Functions

expression. Properties of Logs: Change of Base. Proof: Let . Rewriting as an exponential gives . Taking the log base c of both sides of this equation gives.
logarithms

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Outils d`évaluation clinique L`anxiété généralisée. Cours 3. 23/03/03

Outils d`expérimentation - Cégep de Sainte-Foy

Outils d`extraction d`informations à partir de documents numérisés

Outils d`intervention Compétences psychosociales Centre de

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OUTILS ÉLECTRIQUES

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CS 5002: Discrete Structures Fall 2018

Lecture 9: November 8, 2018

Instructors: Adrienne Slaughter, Tamara BonaciDisclaimer:These notes have not been subjected to the usual scrutiny reserved for formal publications.

They may be distributed outside this class only with the permission of the Instructor.

Introduction to Algorithms

Readings for this week:

Rosen, Chapter 2.1, 2.2, 2.5, 2.6

Sets, Set Operations, Cardinality of Sets, Matrices9.1 Overview

Objective: Introduce algorithms.

Review logarithms

Asymptotic analysis

Dene Algorithm

Ho wto express or describ ean algorithm

Run time, space (Resource usage )

Determining Correctness

In troducerepresen tativeproblems

1. fo o

9.2 Asymptotic Analysis

The goal with asymptotic analysis is to try to nd a bound, or an asymptote, of a function. This allows

us to come up with an \ordering" of functions, such that one function is denitely bigger than another,

in order to compare two functions. We do this by considering the value of the functions asngoes to innity, so for very large values ofn, as opposed to small values ofnthat may be easy to calculate. Once we have this ordering, we can introduce some terminology to describe the relationship of two functions. 9-1

9-2Lecture 9: November 8, 201823456781248163264128256512102420484096n!n

n2 nn

2nlog(n)log(n)n

1Growth of Functions

From this chart, we see:

1lognpnnnlog(n)n22nn!nn(9.1)

ComplexityTerminology

(1)Constant (logn)Logarithmic (n)Linear (nlogn)Linearithmic nbPolynomial (bn) (whereb >1)Exponential (n!)Factorial

9.2.1 Big-O: Upper BoundDenition 9.1 (Big-O: Upper Bound)f(n) =O(g(n))means there exists some constantcsuch

thatf(n)cg(n), for large enoughn(that is, asn! 1).

We sayf(n) =O(g(n))

Example:I claim 3n2100n+ 6 =O(n2). I can prove this using the denition of big-O:

Lecture 9: November 8, 20189-3

f(n) = 3n2100n+ 6 (9.2) g(n) =n2(9.3) )3n2100n+ 6cn2for somec(9.4)

Ifc= 3 : 3n2100n+ 63n2(9.5)

To prove using Big-O:

Determinef(n) andg(n)

Write the equation based on the denition

Choose acsuch that the equation is true.

{If you can nd ad, thenf(n) =O(g(n)). If not, thenf(n)6=O(g(n)).

These statements are all true:

3n2100n+ 6 =O(n2) (9.6)

3n2100n+ 6 =O(n3) (9.7)

3n2100n+ 66=O(n) (9.8)

Proving

9.7 f(n) = 3n2100n+ 6 (9.9) g(n) =n3(9.10) )3n2100n+ 6 =cn3(for somec) (9.11)

Ifc= 1 : 3n2100n+ 6n3(whenn >3) (9.12)

We also know this to be true because order istransitive: iff(n) =O(g(n)), andg(n) =O(h(n)), then f(n) =O(h(n)). Sincen2=O(n3), then anyf(n) =O(n2) is alsoO(n3).

Proving

9.8 f(n) = 3n2100n+ 6 (9.13) g(n) =n(9.14)

For anyc:cn <3n2(whenn > c) (9.15)

9.2.2 Big-Omega: Lower BoundDenition 9.2 (Big-Omega: Lower Bound)f(n) =

(g(n))means there exists some constantc such thatf(n)cg(n), for large enoughn(that is, asn! 1).

We sayf(n) =

(g(n))or \fofnis Big Omega ofgofn"

9-4Lecture 9: November 8, 2018

Example:I claim 3n2100n+ 6 =

(n2). I can prove this using the denition of big-Omega: f(n) = 3n2100n+ 6 (9.16) g(n) =n2(9.17) )3n2100n+ 6cn2for somec(9.18)

Ifc= 2 : 3n2100n+ 62n2(9.19)

We show Big-Omega the same way we show Big-O.

These statements are all true:

3n2100n+ 6 =

(n2) (9.20)

3n2100n+ 66=

(n3) (9.21)

3n2100n+ 6 =

(n) (9.22)

Proving

9.21 f(n) = 3n2100n+ 6 (9.23) g(n) =n3(9.24) )3n2100n+ 6cn3(for somec) (9.25)

Ifc= 1 : 3n2100n+ 6n3(whenn >3) (9.26)

Proving

9.22 f(n) = 3n2100n+ 6 (9.27) g(n) =n(9.28)

For anyc:cn <3n2(whenn >100c) (9.29)

9.2.3 Big-Theta: \Tight" BoundDenition 9.3 (Big-Theta: \Tight" Bound)f(n) = (g(n))means there exists some constantsc1

andc2such thatf(n)c1g(n)andf(n)c2g(n).

We sayf(n) = (g(n))or \fofnis Big-Theta ofgofn".

Denition 9.4 (Theta and \order of")Whenf(x) = (g(x)), it is the same as sayingf(x)is the order ofg(x), or thatf(x)andg(x)are the same order.

3n2100n+ 6 = (n2)Both Oand

apply

Lecture 9: November 8, 20189-5

3n2100n+ 66= (n3)Only Oapplies

3n2100n+ 66= (n)Only

app lies Interesting AsideDonald Knuth popularized the use of Big-O notation. It was originally inspired by the use of \ell" numbers, written asL(5), which indicates a number that we don't know the exact

value of, but is less than 5. That allows us to reason about the value without knowing the exact value:

we knowL(5)<100, for example. Theorem 9.5Iff(x) =anxn+an1xn1++a1x+a0, thenf(x) =O(n) a)f(x) = 17x+ 11 b)f(x) =x2+ 1000 c)f(x) =xlogxd)f(x) =x4=2 e)f(x) = 2xf)f(x) =bxc dxe

9.2.4 Logs, Powers, Exponents

We've seenf(n) =O(nd). Ifd > c >1, thennc=O(nc).ncisOnd;butndis notO(nc). log bnisO(n) wheneverb >1. Wheneverb >1, c and d are positive: (log bn)cisOnd;butndis not (O(logbn)c) (9.30)

This tells us that every positive power of the logarithm of n to the base b, where b > 1, is big-O of every

positive power of n, but the reverse relationship never holds. In Example 7, we also showed that n is

O(2n). More generally, whenever d is positive and b > 1, we have n disO(bn);butbnis notOnd(9.31)

This tells us that every power of n is big-O of every exponential function of n with a base that is greater

than one, but the reverse relationship never holds. Furthermore, we have when c > b > 1, b nisO(cn) butcnis notO(bn) (9.32)

This tells us that if we have two exponential functions with dierent bases greater than one, one of these

functions is big-O of the other if and only if its base is smaller or equal.

9.2.5 Adding Functions

There are a set of rules that govern combining functions together.

O(f(n)) +O(g(n))!O(max(f(n);g(n))) (9.33)

(f(n)) +