Q8P From 聽聽equation 聽聽 (15.4) ... [FREE SOLUTION]

Chapter 12: Q8P (page 593)

From equation (15.4) show that ${鈭�}_{0}^{鈭�} J_{1} (x) dx = {鈭�}_{0}^{鈭�} J_{3} (x) dx = . . . = {鈭�}_{0}^{鈭�} J_{2 n + 1} (x) dx$ and
${鈭�}_{0}^{鈭�} J_{0} (x) dx = {鈭�}_{0}^{鈭�} J_{2} (x) dx = . . . = {鈭�}_{0}^{鈭�} J_{2 x} (x) dx$ . Then, by Problem 7, show that
${鈭�}_{0}^{鈭�} J_{n} (x) dx = 1$ for all integral. $n . (15.4) J_{p - 1} (x) - J_{P + 1} (x) = 2 J_{p}^{'} (x)$ .

Short Answer

Expert verified

This equation has been proved.

Step by step solution

Concept of Differential equations:

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

Determine equations and integrate them:

The equations are

${鈭�}_{0}^{鈭�} J_{1} (x) dx = {鈭�}_{0}^{鈭�} J_{3} (x) dx = . . . = {鈭�}_{0}^{鈭�} J_{2 n + 1} (x) dx$

And

${鈭�}_{0}^{鈭�} J_{0} (x) dx = {鈭�}_{0}^{鈭�} J_{2} (x) dx = . . . = {鈭�}_{0}^{鈭�} J_{2 n} (x) dx$

Calculate from each equation.

$2 J_{P} (x) = J_{P - 1} (x) - J_{P - 1} (x) 2 J_{P} (x) = J_{P - 1} (x) - J_{P - 1} (x) 2 J_{1} (x) = J_{0} (x) - J_{2} (x)$

...... (1)

Put P = 1 in equation (1) as follows:

${鈭�}_{0}^{蟺} 2 J_{1} (x) dx = {鈭�}_{0}^{蟺} J_{0} (x) dx - {鈭�}_{0}^{蟺} J_{2} (x) dx$ 鈥�.. (2)

Integrating from limits 0 to 0 as follows:

role="math" localid="1659327415234" ${[2 J_{1} (x)]}_{0} = {鈭�}_{0}^{蟺} J_{0} (x) dx - {鈭�}_{0}^{m} J_{2} (x) dx 2 {[{limJ}_{1} (x) - J_{1} (0)]}_{0} = {鈭�}_{0 鈫� 鈭�}^{蟺} J_{0} (x) dx - {鈭�}_{0}^{m} J_{2} (x) dx$

Since. $J_{1} (x) 鈫� 0$ as $J_{1} (x) 鈫� 0$ .

Put 3 for p in the equation (1) as follows:

role="math" localid="1659328288884" $2 J_{3} (x) = J_{2} (x) - J_{4} (x) {鈭�}_{0}^{鈭�} 2 J_{3} (x) dx = {鈭�}_{0}^{鈭�} J_{2} (x) dx - {鈭�}_{0}^{鈭�} J_{4} (x) dx$

鈥�.. (3)

Integrating with limits 0 to $鈭�$ as follows:

From equation (2) and (3) obtain:

role="math" localid="1659328451089" ${鈭�}_{0}^{鈭�} J_{0} (x) dx = {鈭�}_{0}^{鈭�} J_{2} (x) dx = . . . = {鈭�}_{0}^{鈭�} J_{4} (x) dx$

Similarly obtain:

${鈭�}_{0}^{蟺} J_{4} (x) dx = {鈭�}_{0}^{蟺} J_{6} (x) dx {鈭�}_{0}^{蟺} J_{6} (x) dx = {鈭�}_{0}^{蟺} J_{8} (x) dx$

Hence,

${鈭�}_{0}^{鈭�} J_{0} (x) dx = {鈭�}_{0}^{鈭�} J_{2} (x) dx = . . . = {鈭�}_{0}^{鈭�} J_{2 n} (x) dx$ ...... (4)

Identify the equation:

Indentify the equation to show as follows:

${鈭�}_{0}^{鈭�} J_{1} (x) dx = {鈭�}_{0}^{鈭�} J_{3} (x) dx = . . . {鈭�}_{0}^{鈭�} J_{2 n + 1} (x) dx 2 J_{2} (x) = J_{1} (x) - J_{3} (x) {鈭�}_{0}^{鈭�} 2 J_{2} (x) dx = {鈭�}_{0}^{鈭�} J_{1} (x) dx = . . . {鈭�}_{0}^{鈭�} J_{3} (x) dx 12 J_{4} (x) = J_{3} (x) - J_{5} (x)$

Simplify further as follows:

${[2 J_{4} (x)]}_{0}^{鈭�} = {鈭�}_{0}^{鈭�} J_{3} (x) dx - {鈭�}_{0}^{鈭�} J_{5} (x) dx 2 [l i m J_{4} (x) - J_{4} (0)] = {鈭�}_{0}^{鈭�} J_{3} (x) - {鈭�}_{0}^{鈭�} J_{5} (x) dx {鈭�}_{0}^{鈭�} J_{3} (x) dx = {鈭�}_{0}^{鈭�} J_{5} (x) dx = 0$

Put 2 for $尾$ in equation (1).

$2 J_{2} (x) = J_{1} (x) - J_{3} (x) {鈭�}_{0}^{鈭�} 2 J_{2} (x) dx = {鈭�}_{0}^{鈭�} J_{1} (x) dx - {鈭�}_{0}^{鈭�} J_{3} (x) dx$

...... (5)

Integrating with limits 0 to $鈭�$ as follows:

Put for in equation (5) as follows:

$2 J_{4} (x) = J_{3} (x) - J_{5} (x)$

Integrating with limits 0 to $鈭�$ as follows:

${[2 J_{4} (x)]}_{0}^{x} = {鈭�}_{0}^{鈭�} J_{3} (x) dx - {鈭�}_{0}^{鈭�} J_{5} (x) dx 2 [{limJ}_{4} (x) - J_{4} (0)] = {鈭�}_{0}^{鈭�} J_{3} (x) - {鈭�}_{0}^{鈭�} J_{5} (x) dx {鈭�}_{0}^{鈭�} J_{3} (x) dx = {鈭�}_{0}^{鈭�} J_{5} (x) dx = 0$

Since, $J_{4} (x) 鈫� 0$ .

$x 鈫� 鈭� J_{4} (0) 鈫� {鈭�}_{0}^{鈭�} J_{3} (x) dx = {鈭�}_{0}^{鈭�} J_{5} (x) dx$ ...... (6)

From equation (5) and (6), obtain:

${鈭�}_{0}^{鈭�} J_{1} (x) dx = {鈭�}_{0}^{鈭�} J_{3} (x) dx = . . . {鈭�}_{0}^{鈭�} J_{5} (x) dx$

Similarly,

${鈭�}_{0}^{鈭�} J_{1} (x) dx = {鈭�}_{0}^{鈭�} J_{7} (x) dx {鈭�}_{0}^{鈭�} J_{7} (x) dx = {鈭�}_{0}^{鈭�} J_{9} (x) dx$

Obtain the equation:

Hence,

${鈭�}_{0}^{鈭�} J_{1} (x) dx = {鈭�}_{0}^{鈭�} J_{3} (x) dx = . . . {鈭�}_{0}^{鈭�} J_{2 n + 1} (x) dx {鈭�}_{0}^{鈭�} J_{n} (x) dx = 1$ ...... (7)

To show that, ${鈭�}_{0}^{鈭�} J_{1} (x) dx = 1$ , from equation (7).

Since,

${鈭�}_{0}^{鈭�} J_{2 n + 1} (x) dx = 1 {鈭�}_{0}^{鈭�} J_{0} (x) dx = 1$

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 91影视!

One App. One Place for Learning.

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

Get started for free

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

91影视

Short Answer

Step by step solution

Concept of Differential equations:

Determine equations and integrate them:

Identify the equation:

Obtain the equation:

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Thermodynamics

Rotational Dynamics

Physical Quantities And Units

Work Energy and Power

Mechanics and Materials

Astrophysics

Study anywhere. Anytime. Across all devices.

91影视

From equation (15.4) show that 鈭�0鈭�J1(x)dx=鈭�0鈭�J3(x)dx=...=鈭�0鈭�J2n+1(x)dx and 鈭�0鈭�J0(x)dx=鈭�0鈭�J2(x)dx=...=鈭�0鈭�J2x(x)dx. Then, by Problem 7, show that鈭�0鈭�Jn(x)dx=1for all integral.n.(15.4)Jp-1(x)-JP+1(x)=2Jp'(x) .

Short Answer

Step by step solution

Concept of Differential equations:

Determine equations and integrate them:

Identify the equation:

Obtain the equation:

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Thermodynamics

Rotational Dynamics

Physical Quantities And Units

Work Energy and Power

Mechanics and Materials

Astrophysics

Study anywhere. Anytime. Across all devices.