Find study content
Learning Materials

Discover learning materials by subject, university or textbook.

Explanations

Textbooks

Exams
All Subjects

Anthropology

Archaeology

Architecture

Art and Design

Bengali

Biology

Business Studies

Greek

Chemistry

Chinese

Combined Science

Computer Science

Economics

Engineering

English

English Literature

Environmental Science

French

Geography

German

History

Hospitality and Tourism

Human Geography

Italian

Japanese

Law

Macroeconomics

Marketing

Math

Media Studies

Medicine

Microeconomics

Music

Nursing

Nutrition and Food Science

Physics

Polish

Politics

Psychology

Religious Studies

Sociology

Spanish

Sports Sciences

Translation

All Subjects

Biology

Business Studies

Chemistry

Combined Science

Computer Science

Economics

English

Environmental Science

Geography

History

Math

Physics

Psychology

Sociology

EXAM TYPES

GCSE

IGCSE

AS

A Level

International A Level

University Admissions Tests

GCSE SUBJECTS

GCSE 1

GCSE 2

GCSE 3

GCSE 4

GCSE 5

SOME-TEXT

IGCSE SUBJECTS

IGCSE 1

AS SUBJECTS

AS 1

A Level SUBJECTS

A Level 1

International A Level SUBJECTS

International A Level 1

University Admissions Tests SUBJECTS

University Admissions Tests 1
Features
Features

Discover all of these amazing features with a free account.

Flashcards

Vaia AI

Notes

Study Plans

Study Sets
What’s new?

Flashcards
Study your flashcards with three learning modes.

Study Sets
All of your learning materials stored in one place.

Notes
Create and edit notes or documents.

Study Plans
Organise your studies and prepare for exams.
Resources
Discover

All the hacks around your studies and career - in one place.

Find a job

Find a degree

Student Deals

Magazine

Mobile App
Featured

Magazine
Trusted advice for anyone who wants to ace their studies & career.

Job Board
The largest student job board with the most exciting opportunities.

Vaia Deals
Verified student deals from top brands.

Our App
Discover our mobile app to take your studies anywhere.

Go to App

Learning Materials

Features

Discover

Chapter 12: Q9P (page 590)

To show that, $\sqrt{ττx /} 2 J_{1 / 2} (x) = \sin x$ .

Short Answer

Expert verified

Hence, $\sqrt{\frac{πx}{2}} J_{\frac{1}{2}} (x) = \sin x$ is proved.

Step by step solution

01

Concept of the Infinite series:

Infinite series:

$J_{p} (x) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{⌜ (n + 1) ⌜ (n + 1 + p)} {(\frac{x}{2})}^{2 n + p}$

02

Calculation of the function ττx/2J1/2(x) :

Considering the provided statement to show that the provided infinite series for $J_{p} (x)$ converges for all the values of x by the ratio test.

The infinite series is,

$\sqrt{\frac{π x}{2}} J_{\frac{1}{2}} (x) = \sin x$

Substituting the series for $\frac{1}{2}$ for p in the equation (1) as follows:

$\sqrt{\frac{πx}{2}} J_{\frac{1}{2}} (x) = \sqrt{\frac{πx}{2}} \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{⌜ (n + 1) ⌜ (n + 1 + \frac{1}{2})} {(\frac{x}{2})}^{2 π + \frac{1}{2}} = \sqrt{\frac{πx}{2}} \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n! (n + \frac{1}{2}) (n + \frac{1}{2} - 1) . . . . . . . . . . \frac{1}{2} ⌜ (\frac{1}{2}) 2^{π} \sqrt{2}} x^{2 n} = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n + 1)!} x^{2 n + 1} = \sin x$

$\sqrt{\frac{πx}{2}} J_{\frac{1}{2}} (x) = \sqrt{\frac{πx}{2}} \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{⌜ (n + 1) ⌜ (n + 1 + \frac{1}{2})} {(\frac{x}{2})}^{2 π + \frac{1}{2}} = \sqrt{\frac{πx}{2}} \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n! (n + \frac{1}{2}) (n + \frac{1}{2} - 1) . . . . . . . . . . \frac{1}{2} ⌜ (\frac{1}{2}) 2^{π} \sqrt{2}} x^{2 n} = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n + 1)!} x^{2 n + 1} = \sin x$

Hence, $\sqrt{\frac{π x}{2}} J_{\frac{1}{2}} (x) = \sin x is proved$

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

For Problems 1 to 4, find one (simple) solution of each differential equation by series, and then find the second solution by the "reduction of order" method, Chapter 8, Section 7 (e).

$(x^{2} + 1) y " - x y' + y = 0$

Verify the recursion relations $(22.24)$ as follows:

a) Differentiate $(22 .23)$ With respect toto get $h Φ = (h - 1) (\partial Φ / \partial x)$ equate coefficients ofrole="math" localid="1654857725406" $h^{n + 1}$

b) Differentiate $(22 .23)$ with respect to to get ${(1 - h)}^{2} (\partial Φ / \partial h) = (1 - h - x) Φ$ equate coefficients of $h$

c) Combine (a) and (b) to get $x (\partial Φ / \partial x) + h Φ - h (1 - h) \partial Φ / \partial h = 0$ . Substitute the series for $Φ$ and equate coefficients of $h^{n}$

Use the Section 15 recursion relations and (17.4) to obtain the following recursion relations for spherical Bessel functions. We have written them for $j_{n}$ , but they are valid for $y_{n}$ and for the $h_{n,} (\frac{d}{dx}) j_{n} (x) = {nj}_{n} (x) / x - j_{n + 1} (x)$ ,.

Solve the differential equations in Problems 5 to 10 by the Frobenius method; observe that you get only one solution. (Note, also, that the two values of are equal or differ by an integer, and in the latter case the larger gives the one solution.) Show that the conditions of Fuchs's theorem are satisfied. Knowing that the second solution is X times the solution you have, plus another Frobenius series, find the second solution.

$x (x + 1) y^{''} - (x - 1) y^{'} + y = 0$

Prove $J_{p} (x) J'_{- P} (x) - J_{- p} (x) J'_{p} (x) = - \frac{2}{π x} s i n p π$ as follows:

Write Bessel's equation (12.1) with $y = J_{p}$ and with $y = J_{- p}$ ; multiply the $J_{p}$ equation by $J_{- p}$ and the $J_{- p}$ equation by $J_{p}$ and subtract to get $\frac{d}{d x} ['' x (J_{p} {J_{- p}}^{'} - J_{- p} J_{p}^{'})^{''}]^{''} = 0$ . Then $J_{p} {J_{- p}}^{'} - J_{- p}^{'} J_{p}^{'} = \frac{c}{x}$ . To find , use equation for each of the four functions and pick out the $1 / x$ terms in the products.

See all solutions

Recommended explanations on Physics Textbooks

Space Physics

Read Explanation

Medical Physics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Classical Mechanics

Read Explanation

Geometrical and Physical Optics

Read Explanation

Force

Read Explanation

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.

Sign-up for free