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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
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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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.

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?
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.
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.
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