The value of ln ( 5000 ! ) is to be evaluated. The corresponding value is to be compared with the given value. The closeness of the corresponding different values is to be predicted. Concept introduction: Stirling’s approximation is used to estimate the natural logarithm of factorial of a number. The Stirling’s approximation is represented as, ln N ! ≈ N ln N − N It is assumed in the Stirling’s approximation that N is a large number.

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Answer 1

Question

Chapter 17, Problem 17.8E

Interpretation Introduction

Interpretation:

The value of ln(5000!) is to be evaluated. The corresponding value is to be compared with the given value. The closeness of the corresponding different values is to be predicted.

Concept introduction:

Stirling’s approximation is used to estimate the natural logarithm of factorial of a number. The Stirling’s approximation is represented as,

lnN!≈NlnN−N

It is assumed in the Stirling’s approximation that N is a large number.

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Answer 2

Question

Chapter 17, Problem 17.8E

Interpretation Introduction

Interpretation:

The value of ln(5000!) is to be evaluated. The corresponding value is to be compared with the given value. The closeness of the corresponding different values is to be predicted.

Concept introduction:

Stirling’s approximation is used to estimate the natural logarithm of factorial of a number. The Stirling’s approximation is represented as,

lnN!≈NlnN−N

It is assumed in the Stirling’s approximation that N is a large number.

Expert Solution & Answer

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Check out a sample textbook solution

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Answer 3

Question

Chapter 17, Problem 17.8E

Interpretation Introduction

Interpretation:

The value of ln(5000!) is to be evaluated. The corresponding value is to be compared with the given value. The closeness of the corresponding different values is to be predicted.

Concept introduction:

Stirling’s approximation is used to estimate the natural logarithm of factorial of a number. The Stirling’s approximation is represented as,

lnN!≈NlnN−N

It is assumed in the Stirling’s approximation that N is a large number.

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Answer 4

Question

Chapter 17, Problem 17.8E

Interpretation Introduction

Interpretation:

The value of ln(5000!) is to be evaluated. The corresponding value is to be compared with the given value. The closeness of the corresponding different values is to be predicted.

Concept introduction:

Stirling’s approximation is used to estimate the natural logarithm of factorial of a number. The Stirling’s approximation is represented as,

lnN!≈NlnN−N

It is assumed in the Stirling’s approximation that N is a large number.

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Answer 5

Chapter 17, Problem 17.8E

Interpretation Introduction

Interpretation:

The value of ln(5000!) is to be evaluated. The corresponding value is to be compared with the given value. The closeness of the corresponding different values is to be predicted.

Concept introduction:

Stirling’s approximation is used to estimate the natural logarithm of factorial of a number. The Stirling’s approximation is represented as,

lnN!≈NlnN−N

It is assumed in the Stirling’s approximation that N is a large number.

Answer 6

Chapter 17, Problem 17.8E

Interpretation Introduction

Interpretation:

The value of ln(5000!) is to be evaluated. The corresponding value is to be compared with the given value. The closeness of the corresponding different values is to be predicted.

Concept introduction:

Stirling’s approximation is used to estimate the natural logarithm of factorial of a number. The Stirling’s approximation is represented as,

lnN!≈NlnN−N

It is assumed in the Stirling’s approximation that N is a large number.

Answer 7

Chapter 17, Problem 17.8E

Interpretation Introduction

Interpretation:

The value of ln(5000!) is to be evaluated. The corresponding value is to be compared with the given value. The closeness of the corresponding different values is to be predicted.

Concept introduction:

Stirling’s approximation is used to estimate the natural logarithm of factorial of a number. The Stirling’s approximation is represented as,

lnN!≈NlnN−N

It is assumed in the Stirling’s approximation that N is a large number.

Answer 8

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Answer 9

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Answer 10

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Answer 11

Ch. 17 - Prob. 17.1E Ch. 17 - Prob. 17.2E Ch. 17 - Prob. 17.3E Ch. 17 - Prob. 17.4E Ch. 17 - Prob. 17.5E Ch. 17 - Prob. 17.6E Ch. 17 - Prob. 17.7E Ch. 17 - Prob. 17.8E Ch. 17 - Prob. 17.9E Ch. 17 - Prob. 17.10E

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