MATHEMATICS 0110A CHANGE OF BASE Suppose that we have
So we get the following rule: Change of Base Formula: logb a = logc a logc b. Example 1. Express log3 10 using natural logarithms. log3 10 =.
Change of Base
6.2 Properties of Logarithms
(x − 1). Page 10. 446. Exponential and Logarithmic Functions. In Exercises 30 - 33 use the appropriate change of base formula to convert the given expression
S&Z . & .
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.
Logarithms - changing the base
This leaflet gives this formula and shows how to use it. A formula for change of base. Suppose we want to calculate a logarithm to base 2. The formula states.
mc logs
Change-of-Base Formula. For any logarithmic bases a and b and
Problem #1. Use your calculator to find the following logarithms. Show your work with Change-of-Base Formula. a) b). 2 log 10. 1. 3 log 9 c). 7 log 11.
Lecture
Properties of Exponents and Logarithms
Most calculators can directly compute logs base 10 and the natural log. For any other base it is necessary to use the change of base formula: logb a =.
Exponents and Logarithms
What is a logarithm? Log base 10
Now we have a new set of rules to add to the others: Table 4. Functions of log base 10 and base e. Exponents. Log base 10. Natural Logs sr.
logarithms
Precalculus: 4.3 Rules of Loagrithms Concepts: rules of logarithms
Concepts: rules of logarithms change of base
. RulesofLogarithms
Introduction to Algorithms
The Power Rule loga(xy) = y loga(x). When the term of a logarithm has an exponent it can be pulled out in front of the log. Change of Base Rule.
cs lect fall notes
Change of Base
Problem 1 – Relating log functions with different bases. Execute the DIFFBASE program. You have found a formula for changing the base of a logarithm.
CS 5002: Discrete Structures Fall 2018
Lecture 9: November 8, 2018
1Instructors: 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 OverviewObjective: Introduce algorithms.
1.Review logarithms
2.Asymptotic analysis
3.Dene Algorithm
4.Ho wto express or describ ean algorithm
5.Run time, space (Resource usage )
6.Determining Correctness
7.In troducerepresen tativeproblems
1. fo o9.2 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. 9-19-2Lecture 9: November 8, 201823456781248163264128256512102420484096n!n
n2 nn2nlog(n)log(n)n
pn1Growth 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!)Factorial9.2.1 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
applyLecture 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 exactvalue 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 dxe9.2.4 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)) + (g(n))! (max(f(n);g(n))) (9.34) (f(n)) + (g(n))!(max(f(n);g(n)) (9.35) These statements express the notion that the largest term of the statement is the dominant one. For example,n3+ 2n2+ 3 =O(n3).Example:Prove thatn2=O(2n).
9-6Lecture 9: November 8, 2018Example:Prove that iff1(n) =O(g1(n)) andf2(n) =O(g2(n)), thenf1(n)+f(n) =O(g1(n)+g2(n)).Example:
f(n) =n+ logn(9.36) g(n) =p(n) (9.37)Isf(n) =O(g(n)),g(n) =O(f(n)), or both?Example:Iff(n) =n+ logn+pn, nd a simple functiongsuch thatf(n) = (g(n)).
Lecture 9: November 8, 20189-7Summary
f(n) =O(g(n)) meanscg(n) is an upper bound onf(n). Thus there exists some constantcsuch thatf(n) is alwayscg(n), for large enoughn(i.e.,nn0for some constantn0). f(n) = (g(n)) meanscg(n) is a lower bound onf(n). Thus there exists some constantcsuchCS 5002: Discrete Structures Fall 2018
Lecture 9: November 8, 2018
1Instructors: 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 OverviewObjective: Introduce algorithms.
1.Review logarithms
2.Asymptotic analysis
3.Dene Algorithm
4.Ho wto express or describ ean algorithm
5.Run time, space (Resource usage )
6.Determining Correctness
7.In troducerepresen tativeproblems
1. fo o9.2 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. 9-19-2Lecture 9: November 8, 201823456781248163264128256512102420484096n!n
n2 nn2nlog(n)log(n)n
pn1Growth 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!)Factorial9.2.1 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
applyLecture 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 exactvalue 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 dxe9.2.4 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)) + (g(n))! (max(f(n);g(n))) (9.34) (f(n)) + (g(n))!(max(f(n);g(n)) (9.35) These statements express the notion that the largest term of the statement is the dominant one. For example,n3+ 2n2+ 3 =O(n3).Example:Prove thatn2=O(2n).
9-6Lecture 9: November 8, 2018Example:Prove that iff1(n) =O(g1(n)) andf2(n) =O(g2(n)), thenf1(n)+f(n) =O(g1(n)+g2(n)).Example:
f(n) =n+ logn(9.36) g(n) =p(n) (9.37)Isf(n) =O(g(n)),g(n) =O(f(n)), or both?Example:Iff(n) =n+ logn+pn, nd a simple functiongsuch thatf(n) = (g(n)).
Lecture 9: November 8, 20189-7Summary
f(n) =O(g(n)) meanscg(n) is an upper bound onf(n). Thus there exists some constantcsuch thatf(n) is alwayscg(n), for large enoughn(i.e.,nn0for some constantn0). f(n) = (g(n)) meanscg(n) is a lower bound onf(n). Thus there exists some constantcsuch- logarithm base change rule proof
- log base change rule
- logarithm base change formula
- log base change formula
- log base change formula proof
- log change base law
- log base change rule proof
- logarithm base change formula proof