Math Problem Statement

is this true?

Is n! asymptotically equivalent to (n+1)/2 raised to the power n?

Solution

The question in the image seems to be asking whether the factorial n!n! is asymptotically equivalent to the expression (n+12)n\left( \frac{n+1}{2} \right)^n. The symbol \sim indicates an asymptotic approximation, meaning we are interested in whether n!n! behaves similarly to (n+12)n\left( \frac{n+1}{2} \right)^n for large values of nn.

Comparing n!n! with (n+12)n\left( \frac{n+1}{2} \right)^n:

To analyze this, let's compare these two expressions asymptotically.

Stirling's Approximation:

A well-known approximation for large nn is Stirling's approximation for factorials: n!2πn(ne)nn! \sim \sqrt{2\pi n} \left( \frac{n}{e} \right)^n This provides a good approximation of the growth rate of n!n!.

Expression (n+12)n\left( \frac{n+1}{2} \right)^n:

The expression in question is: (n+12)n\left( \frac{n+1}{2} \right)^n

For large nn, this behaves similarly to: (n2)n\left( \frac{n}{2} \right)^n

Comparing Growth Rates:

  • n!2πn(ne)nn! \sim \sqrt{2\pi n} \left( \frac{n}{e} \right)^n
  • (n+12)n(n2)n\left( \frac{n+1}{2} \right)^n \sim \left( \frac{n}{2} \right)^n

We see that the growth rates are not exactly the same because Stirling's approximation has an additional factor of ne\frac{n}{e} in the exponent, while the expression (n+12)n\left( \frac{n+1}{2} \right)^n grows more slowly due to the factor of n+12\frac{n+1}{2} instead of ne\frac{n}{e}.

Conclusion:

No, the approximation n!(n+12)nn! \sim \left( \frac{n+1}{2} \right)^n is not true. The factorial n!n! grows faster than the expression (n+12)n\left( \frac{n+1}{2} \right)^n because of the difference in the factors inside the exponent.

Would you like further details or clarification on any part of this?


Here are 5 related questions that expand on this:

  1. What is Stirling's approximation, and how is it derived?
  2. How does the growth of factorials compare with exponential functions?
  3. Can the approximation n!(n+12)nn! \sim \left( \frac{n+1}{2} \right)^n be useful in any specific cases?
  4. How do other asymptotic approximations compare with Stirling's approximation?
  5. What are the applications of asymptotic approximations like this in real-world problems?

Tip: When comparing two functions asymptotically, always examine the dominant term in their growth rates for large values of nn.

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Math Problem Analysis

Mathematical Concepts

Factorial Growth
Asymptotic Approximation
Stirling's Approximation

Formulas

Stirling's approximation: n! ≈ √(2πn) (n/e)^n
(n+1)/2^n

Theorems

Stirling's Approximation

Suitable Grade Level

Undergraduate Mathematics or Advanced High School