/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 57 The Gamma Function The gamma fun... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The Gamma Function The gamma function is defined by the formula $$ \Gamma(x)=\int_{0}^{+\infty} t^{x-1} e^{-t} d t $$ a. Find \(\Gamma(1)\) and \(\Gamma(2)\). b. Use integration by parts to show that for every positive integer \(n, \Gamma(n+1)=n \Gamma(n)\). c. Deduce that \(\Gamma(n)=(n-1) ![=(n-1)(n-2) \cdots 2 \cdot 1]\) for every positive integer \(n\).

Short Answer

Expert verified
In short, we compute \( \Gamma(1) = 1 \), \( \Gamma(2) = 1 \), and show \( \Gamma(n+1) = n \Gamma(n) \) for every positive integer \( n \) using integration by parts. From this, we deduce that \( \Gamma(n) = (n-1)! \) for every positive integer \( n \).

Step by step solution

01

Computing \( \Gamma(1) \)

We plug \( x = 1 \) into the formula. So, \[ \Gamma(1) = \int_{0}^{+\infty} t^{1-1} e^{-t} dt = \int_{0}^{+\infty} e^{-t} dt \] For this integral, we can use the formula for Gaussian shape integral where \( \int_{0}^{+\infty} e^{-t} dt = 1 \). So, \( \Gamma(1) = 1 \).
02

Computing \( \Gamma(2) \)

Next, we plug \( x = 2 \) into the formula: \[ \Gamma(2) = \int_{0}^{+\infty} t^{2-1} e^{-t} dt \] This integral can be solved by using the integration by parts, which states \[ \int u dv = uv - \int v du \] Here \( u=t \), \( dv=e^{-t} dt \), with \( du=dt \) and \( v=-e^{-t} \). Now proceeding with the integration by parts: \[ \Gamma(2) = [-t e^{-t}]_{0}^{+\infty} - \int_{0}^{+\infty} -e^{-t} dt = 0 - \Gamma(1) = 0 - 1 = 1 \]
03

Showing \( \Gamma(n+1) = n \Gamma(n) \)

We plug \( x = n+1 \) into the formula: \[ \Gamma(n+1) = \int_{0}^{+\infty} t^{n+1-1} e^{-t} dt \] Now, using integration by parts with \( u= t^n \), \( dv= t e^{-t} dt \), \( du= n t^{n-1} dt \), and \( v= -e^{-t} \): \[ \Gamma(n+1) = \left[-t^n e^{-t}\right]_{0}^{+\infty} - \int_{0}^{+\infty} -n t^{n-1} e^{-t} dt = 0 + n \Gamma(n) \]
04

Deducing \( \Gamma(n) = (n-1)! \)

Based on the above step, we already know that \( \Gamma(n+1) = n \Gamma(n) \). Using the definitions of \( \Gamma(1) = 1 \) and \( \Gamma(2) = 1 \), we can iterate through the identities created by the equation \( \Gamma(n+1) = n \Gamma(n) \) to deduce that \[ \Gamma(n) = (n-1) \Gamma(n-1) = (n-1)(n-2) \Gamma(n-2) = \ldots = (n-1)(n-2)\ldots*2*1 = (n-1)! \] which is exactly the definition of the factorial of \( n-1 \). Thus, we have shown that the Gamma function at any positive integer \( n \) is equal to the factorial of \( n-1 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration by Parts
When we deal with integrals involving products of functions, integration by parts can be a very useful tool. This technique is an essential part of calculus and helps us solve complex integrals by transforming them into simpler forms. Here's how it works:
  • Choose two parts of the integral: one part to differentiate (\( u \)) and one part to integrate (\( dv \)).
  • The formula for integration by parts is: \( \int u\, dv = uv - \int v\, du \).
To apply integration by parts effectively:1. Identify which part of the integral will simplify when differentiated. This is your \( u \).2. Choose the \( dv \) such that its integral \( v \) is simpler.3. Substitute back each part into the integration by parts formula.In the context of the Gamma function, integration by parts helps illustrate the recursive property of \( \Gamma(n+1) = n \Gamma(n) \) by selecting appropriate \( u \) and \( dv \). This approach transforms the original integral into a form involving \( \Gamma(n) \), highlighting the connection between different values of the Gamma function.
Factorial Function
The factorial function, denoted as \( n! \), is a fundamental concept in mathematics, especially in combinatorics and algebra. It represents the product of all positive integers up to \( n \):\[ n! = n \times (n-1) \times (n-2) \times \ldots \times 2 \times 1 \]The factorial function is defined for non-negative integers, with the special case of \( 0! = 1 \).
  • Factorial is used in permutations and combinations to calculate possible arrangements.
  • Its properties make it suitable for applications in series expansion and probability.
The relationship between the Gamma function and the factorial function is significant because it extends the concept of factorials to non-integer values. Whereas the factorial function only applies to integers, the Gamma function bridges the gap by generalizing the concept:\[ \Gamma(n) = (n-1)! \]for any positive integer \( n \). Understanding this relationship allows mathematicians to solve more complex problems involving continuous data.
Definite Integral
A definite integral calculates the net area under a curve between two points. It is an essential tool in calculus, enabling the computation of quantities like area, volume, and accumulation.The notation for a definite integral is:\[ \int_{a}^{b} f(t)\, dt \]Where \( a \) and \( b \) are the bounds of integration. The integral provides:
  • The total "accumulation" of the function \( f(t) \) from \( t = a \) to \( t = b \).
  • A geometric interpretation, as the signed area between the curve of \( f(t) \) and the horizontal axis.
When applied to the Gamma function, the definite integral from 0 to infinity of \( t^{x-1} e^{-t} \) gives profound insights into both pure and applied mathematics. Understanding how to manipulate definite integrals is key to evaluating the Gamma function, especially in showing recursive relationships like \( \Gamma(n+1) = n \Gamma(n) \). These integrals offer insights into how abstract mathematical concepts translate into practical applications.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Cooling A bowl of clam chowder at \(190^{\circ} \mathrm{F}\) is placed in a room whose air temperature is \(75^{\circ} \mathrm{F}\). After 10 minutes, the soup has cooled to \(150^{\circ} \mathrm{F}\). Find the value of \(k\) in Newton's Law of Cooling, and hence find the temperature of the chowder as a function of time.

Linear Differential Equationsare based on first order linear differential equations with constant coefficients. These have the form $$ \frac{d y}{d t}+p y=f(t) \quad(p \text { constant }) $$ and the general solution is $$ y=e^{-p t} \int f(t) e^{p t} d t . \quad \text { (Check this by substituting!) } $$ Solve the linear differential equation $$ 2 \frac{d y}{d t}+y=-t . \quad y=1 \text { when } t=0 $$

Determining Demand Nancy's Chocolates estimates that the elasticity of demand for its chocolate strawberries is \(E=0.02 p-0.5\) where \(p\) is the price per pound. It sells 30 pounds of chocolate strawberries per week when the price is \(\$ 30\) per pound. Find the formula expressing the demand \(q\) as a function of \(p\). Recall that the elasticity of demand is given by $$ E=-\frac{d q}{d p} \times \frac{p}{q} $$

Decide whether or not the given integral converges. If the integral converges, compute its value. $$\int_{-1}^{2} \frac{2 x}{x^{2}-1} d x$$

Use technology to approximate the given integrals with \(M=10,100,1,000, \ldots\) and hence decide whether the associated improper integral converges and estimate its value to four significant digits if it does. $$\int_{1}^{M} \frac{1}{x} d x$$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.