Chapter3 Growth of Function

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The following is from "Introduction to the algorithm" LZ Novice, there are various errors and a variety of unreasonable places to hope that everyone points out

3.1 Asymptotic notation

Θ notation θ (g (n)) ={f (n): There is a normal number c1,c2,n0, so that for all n≥n0, there is 0≤c1g (n) ≤f (n) ≤c2g (n)}.

o Mark O (g (n)) ={f (n): There is a normal number c,n0, so that for all n≥n0, there is 0≤f (n) ≤CG (n)}.

Ω Mark Ω (g (n)) ={f (n): There is a normal number c,n0, so that for all n≥n0, there is 0≤CG (n)}≤f (n)}.

o Mark O (g (n)) ={f (n): For any normal number c>0, there is n0, so that for all n≥n0, there is 0≤f (n) ≤CG (n)}.

Ω Mark Ω (g (n)) ={f (n): For any normal number c>0, there is n0, so that for all n≥n0, there is 0≤CG (n)}≤f (n)}.

Several properties of existence:

Transitivity

Reflexivity: F (n) =k (f (n))---K stands for θ,o,ω three cases

Symmetry: F (n) =θ (g (n))---when and only if G (n) =θ (f (n))

Transpose symmetry: F (n) =o (g (n))---when and only if G (n) =ω (f (n))

Exercise

3.1-1
Prove:
Step1: By definition, the upper formula is equivalent to
Step2: And according to know, C1=1/2,c2=1 can meet the requirements, so the proof

3.1-2
Prove to any real constant A and B, where b>0, there is
Step1: Convert the up-to-

Step2: Further change, thus requiring only

Step3: When N0=⌈a+1⌉ n≥n0,k1→0,k2=3 satisfies the condition, the proof.

3.1-3
Because o this symbol contains the meaning of the upper limit.

3.1-4
Established, not established.

3.1-5
Conditions f (n) =o (g (n)) and f (n) =ω (g (n))
Verification f (N) =θ (g (n))
Proof: According to the definition 0<f (n) ≤c2g (n) and 0<c1g (n) ≤f (n), so that the existence of C1 and C2 makes C1G (n) ≤f (n) ≤c2g (n), thereby proving.

3.1-6
According to Theorem 3.1.

3.1-7
Prove it from the definition.

3.1-8
Slightly

3.2 Standard notation and common function index

Correlation between polynomial and exponential growth rate: (where a>1) is thus available, an exponential function with an arbitrary bottom greater than 1 is faster than any polynomial function.

Natural logarithm E: (to all real numbers x)

Thus there are inequalities

For all x, we have

Logarithmic

The polynomial is correlated with the growth of multiple logarithms:, thus

Factorial

The Stirling approximate formula for factorial:

N! The upper and lower bounds:

For all n≥1 there are:, which

Fibonacci number

Defined:

The golden ratio, and its conjugate number, the value of two root and

The relationship between Fibonacci number and Golden ratio:

And because, therefore, there is, thus, a Fibonacci number that grows exponentially .

Exercise

3.2-1

Conditions: F (n) and g (n) are monotonically increasing functions
Required: F (n) +g (n) and F (g (n)) are also monotonically increasing
It's a very obvious feeling! Slightly
If the increase condition F (n) and g (n) are non-negative, the F (n) ∗g (n) is also monotonically increasing
Slightly

3.2-2
Prove:

Both sides of the equation take ln at the same time: left =, right =. Thus to the Left = right, proof

3.2-3
According to the sterling formula, you can get:

Thus, for example, the requirements can be met and thus proven. The following two proofs are easier, slightly.

3.2-4

Didn't understand.?

3.2-5

By definition

3.2-6

The general solution of the two-second equation can be.

3.2-7
Prove

Step1: When I=0 is satisfied, when I=1 is satisfied.

Step2: Assumptions and N are satisfied before

Step3: proof (Note: The use of and)

3.2-8
The use of nature. Thus, there is only evidence
According to thereby the proof.

Study Questions

3-1
Slightly

3-2
A. Based on the relationship between the polynomial and the logarithm

B. Based on the relationship of the polynomial to the exponent (where a>1)

C. There is no upper and lower bounds due to cyclical fluctuations (e.g. every half cycle value returns to 1)
D.

E. According to the known, thus

F. According to Available

3-3

Slightly

3-4
A. errors, such as but
B. Errors, if not satisfied
C. Correct, and thereby simply satisfy the existence of C1,C2 and n0 make it tenable. Obviously exists.
D. errors, such as

E. Correct,

F. Correct, transpose symmetry

G. Errors, if not satisfied when taking

H. Correct, can meet

3-5

3-6

Chapter3 Growth of Function