Based on Theorem 4.4.3 P.169, when n > 2, which of the following is biggest? 1. n! 2. n^n 3. (n!)^2 4. n x lg(n) 5. lg(n!) Answer 1, 2, 3, 4, or 5 based on your choice.
Based on Theorem 4.4.3 P.169, when n > 2, which of the following is biggest? 1. n! 2. n^n 3. (n!)^2 4. n x lg(n) 5. lg(n!) Answer 1, 2, 3, 4, or 5 based on your choice.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
100%
Based on Theorem 4.4.3 P.169, when n > 2, which of the following is biggest?
1. n!
2. n^n
3. (n!)^2
4. n x lg(n)
5. lg(n!)
Answer 1, 2, 3, 4, or 5 based on your choice.
Transcribed Image Text:Theorem 4.4.3: For all positive integers
(1) n! <= n"
but
n! < n"
when n> 1;
(n!)²
(3) Ig(n!)
and (4) (½)n x lg(n) < lg(n!)
n" < (n!)² when n> 2;
2 x lg(n!)
n x lg(n)
(2) п" <-
but
< nx lg(n)
when n> 2;
when n> 2.
Proof. |/ We'll prove these results in order.
(using algebra, not MI)
п! 3D пх (п —
1) x (п — 2) х.x (2) x (1)
x (п) x (п) — п".
<nx (п) x (п) х... х
Equality holds only when n =
1.
// We've proven (1).
// What about (2)?
// If n = 5, then
// and
// so
5! = (5)(4)(3)(2)(1)
5!
= (1)(2)(3)(4)(5),
(5!) = (5!)(5!) = (5 × 1)(4 × 2)(3 × 3)(2 × 4)(1 × 5)
= (5)
> (5)
//
(8)
(9)
(8)
(5)
(5)
(5)
(5) = 55.
//
//
(5)
// Does this generalize?
![In general,
in
n!
— пх (п— 1) x ( - 2) х... x (2)х (1) —D П (п+1—r)
r=1
and
n! 3D (1) x (2) х (3) х... x (п—1)x (п) %3D I7,
r=1
where II (the Greek capital letter n) is used in the way the Greek capital letter sigma
is used in sigma-notation for sums, but it applies to products. Then,
n
n
(n!)' = 1I (n+1– r) x
r=1
r=1
n
=II {(n+1– r) × r}.
r=1
But (n+1– r)r
(n – r+1)r = (n – r)r +r
(n – r)([r – 1] +1)
(n – r)(r – 1) + (n – r) +r
= (n – r)(r – 1) +n
+r
// Both (n – r) >= 0 and (r – 1) >= 0
// when 1 <= r = n
>= n.
Thus,
(n!)“ >=
n = n".
r=1
1 or 2. If n >= 3, there is at least one r-value in the
Equality holds only when n =
product where 1 < r < n and then n < (n + 1 – r)r.
-
// Now we've proven (2). What about the logarithms in (3) and (4)?
Suppose that n >= 3. Since n! < n" < (n!)°, taking logarithms, we get (3),
lg(n!) <пx1g(n) <2х1g(n!).
// Recall log, (x) = y x log, (x)
Dividing the last two terms by 2, we get
(½)n x lg(n) < lg(n!)
and so we’ve proven result (4) as well.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa4792360-22b4-40c0-a48a-02ce03c9f6f7%2F0c98b0f0-ae51-492b-b9e3-0af603e4e072%2F7ftbzjd_processed.png&w=3840&q=75)
Transcribed Image Text:In general,
in
n!
— пх (п— 1) x ( - 2) х... x (2)х (1) —D П (п+1—r)
r=1
and
n! 3D (1) x (2) х (3) х... x (п—1)x (п) %3D I7,
r=1
where II (the Greek capital letter n) is used in the way the Greek capital letter sigma
is used in sigma-notation for sums, but it applies to products. Then,
n
n
(n!)' = 1I (n+1– r) x
r=1
r=1
n
=II {(n+1– r) × r}.
r=1
But (n+1– r)r
(n – r+1)r = (n – r)r +r
(n – r)([r – 1] +1)
(n – r)(r – 1) + (n – r) +r
= (n – r)(r – 1) +n
+r
// Both (n – r) >= 0 and (r – 1) >= 0
// when 1 <= r = n
>= n.
Thus,
(n!)“ >=
n = n".
r=1
1 or 2. If n >= 3, there is at least one r-value in the
Equality holds only when n =
product where 1 < r < n and then n < (n + 1 – r)r.
-
// Now we've proven (2). What about the logarithms in (3) and (4)?
Suppose that n >= 3. Since n! < n" < (n!)°, taking logarithms, we get (3),
lg(n!) <пx1g(n) <2х1g(n!).
// Recall log, (x) = y x log, (x)
Dividing the last two terms by 2, we get
(½)n x lg(n) < lg(n!)
and so we’ve proven result (4) as well.
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 3 steps with 2 images
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
