# Factorials approximation problem

• karnten07
In summary, a factorial approximation problem involves finding an approximate value for a factorial function, which calculates the product of all positive integers up to a given number. Approximating factorials is important because calculating large factorials exactly can be difficult. Methods such as Stirling's approximation, the gamma function, and the logarithmic approximation can be used to estimate factorials. Stirling's approximation, proposed by James Stirling, uses a formula involving the natural logarithm of the factorial number. Factorial approximation has practical applications in fields like statistics, physics, and computer science.
karnten07

## Homework Statement

How is,

[(N+Q)!Q!]/[(Q+1)!(N+Q-1)!] equal to (N+Q)/(Q+1) when N,Q>>1 ??

It looks like the Q!/(N+Q-1)! cancels but i don't see how, I am going from my lecturers notes here.

## The Attempt at a Solution

Last edited:
Break it up - look at the fraction as

$$\frac{Q!}{(Q+1)!} \cdot \frac{(N+Q)!}{(N+Q-1)!}$$

For the first fraction, ask yourself "what number you would multiply $Q!$ by to get $(Q+1)!$"?

Do a similar analysis for the second fraction.

As N and Q get larger, the terms Q! and (N+Q-1)! will also become very large. In this case, we can use Stirling's approximation to approximate these factorials. Stirling's approximation states that for large values of n, the factorial n! can be approximated by n^(n+1/2)e^(-n), where e is the mathematical constant approximately equal to 2.71828.

Using this approximation, we can rewrite Q! as Q^(Q+1/2)e^(-Q) and (N+Q-1)! as (N+Q-1)^(N+Q-1+1/2)e^(-(N+Q-1)).

Plugging these approximations into the original expression, we get:

[(N+Q)!Q!]/[(Q+1)!(N+Q-1)!] = [(N+Q)(N+Q-1)^(N+Q-1+1/2)e^(-(N+Q-1))] / [(Q+1)Q^(Q+1/2)e^(-Q)]

= [(N+Q)(N+Q-1)^(N+Q-1)e^(-(N+Q-1)) / (Q+1)Q^Qe^-Q] * [e^(1/2)]

= [(N+Q)(N+Q-1)^(N+Q-1)/ (Q+1)Q^Q] * [e^(-(N+Q-1+1/2)+Q+1/2)]

= [(N+Q)(N+Q-1)^(N+Q-1)/ (Q+1)Q^Q] * [e^(-(N+Q)+Q+1/2)]

= [(N+Q)(N+Q-1)^(N+Q-1)/ (Q+1)Q^Q] * [e^(1/2)]

Since N and Q are very large, we can assume that (N+Q)(N+Q-1)^(N+Q-1) is much larger than (Q+1)Q^Q, making the expression approximately equal to (N+Q)(N+Q-1)^(N+Q-1) / (Q+1)Q^Q.

Now, using the approximation (N+Q-1

## What is a factorial approximation problem?

A factorial approximation problem is a mathematical problem that involves finding an approximate value for a factorial function. This function calculates the product of all positive integers from 1 up to a given number.

## Why is it important to approximate factorials?

Approximating factorials is important because the exact value of large factorials can be difficult to calculate and may require advanced mathematical techniques. By approximating factorials, we can obtain a close enough value for practical purposes.

## What methods can be used to approximate factorials?

There are several methods that can be used to approximate factorials, including Stirling's approximation, the gamma function, and the logarithmic approximation. These methods use different mathematical formulas to estimate the value of a factorial.

## What is Stirling's approximation and how does it work?

Stirling's approximation is a method for approximating factorials proposed by Scottish mathematician James Stirling. It uses a formula that involves the natural logarithm of the factorial number to estimate its value. The formula is n! ≈ √(2πn)(n/e)^n.

## What are some applications of factorial approximation?

Factorial approximation has many practical applications in fields such as statistics, physics, and computer science. It can be used to estimate the number of possible outcomes in a probability experiment, calculate the number of ways to arrange a set of objects, and analyze the complexity of algorithms.

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