/*! 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} Q16P Prove that the functions Ln(x) ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Prove that the functions Ln(x)are orthogonal on(0,∞)with respect to the weight functione−x

Hint: Write the differential equation(22.21)asexddxxe−xy'+ny=0, and see Sections7and19 .

Short Answer

Expert verified

The required equation is∫0∞e−xLm(x)Ln(n)dx=0.

Step by step solution

01

Concept of Differential equation and Orthogonal functions 

Differential equations are equations that connect one or more derivatives of a function. This implies that their answer is a function.

Two mathematical functions whose definite integral of their product is 0 given proper bounds are known as orthogonal functions.

02

Step 2:Determining Orthogonal function

Consider the differential equation xy''+(1−x)y'+ny=0

This can be written as exddxxe−xy'+ny=0.

Write 1for Ln(x)and Lm(x)

exddxxe−xLn'(x)+nLn(x)=0…exddxxe−xLm'(x)+mLm(x)=0.

Multiply 2by Ln(x)and 3by Lm(x)and subtract

Lm(x)exddxxe−xLn'(x)−Ln(x)exddxxe−xLm'(x)+(n−m)Lm(x)Ln(n)=0

Step 2: Integrating the limits

Integrate in the limits (0,∞)

∫0∞ddxxe−xLm(x)Ln'(x)−Ln(x)Lm'(x)dx+(n−m)∫0∞e−xLm(x)Ln(n)dx=0xe−xLm(x)Ln'(x)−Ln(x)Lm'(x)0∞+(n−m)∫0∞e−xLm(x)Ln(n)dx=01+∫0∞e−xLm(x)Ln(n)dx=0

The integration term above will be equal to zero because for lower limit,x=0 and for higher e−x=0.

Hence,∫0∞e−xLm(x)Ln(n)dx=0.

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!

One App. One Place for Learning.

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

Get started for free

Study anywhere. Anytime. Across all devices.