Book edition 9th

Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages 616 pages

ISBN 9780321977069

Answers without the blur.

Just sign up for free and you're in.

Short Answer

Question: Show that, $28 . \sum_{n = 0}^{\infty} a_{n} x^{n + 1} + \sum_{n = 1}^{\infty} n b_{n} x^{n - 1} = b_{1} + \sum_{n - 1}^{\infty} [2 a_{n - 1} + (n + 1) b_{n + 1}] x^{n}$

We showed that 2 . $\sum_{n = 0}^{\infty} a_{n} x^{n + 1} + \sum_{n = 1}^{\infty} n b_{n} x^{n - 1} = b_{1} + \sum_{n - 1}^{\infty} [2 a_{n - 1} + (n + 1) b_{n + 1}] x^{n}$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Power series

A power series is an infinite series of the form, $\sum_{n = 0}^{\infty} a_{n} {(x - c)}^{n} = a_{0} + a_{1} (x - c) + a_{2} (x - c)^{2} + . . . . .$ Where,a_n represents the coefficient term of the nth term and c is a constant.

Step 2: Changing the index of the first term

We have to show that,

2. $\sum_{n = 0}^{\infty} a_{n} x^{n + 1} + \sum_{n = 1}^{\infty} n b_{n} x^{n - 1} = b_{1} + \sum_{n - 1}^{\infty} [2 a_{n - 1} + (n + 1) b_{n + 1}] x^{n}$

Simplifying the L.H.S expression,

L.H.S =2. $\sum_{n = 0}^{\infty} a_{n} x^{n + 1} + \sum_{n - 1}^{\infty} n b_{n} x^{n - 1}$

Now changing the index of the first term, let,

n + 1 = k

n = k -1

Then,

$2 \sum_{n = 0}^{\infty} a_{n} x^{n + 1} = 2 \sum_{k - 1 = 0}^{\infty} a_{k - 1} x^{k} = 2 \sum_{k = 0}^{\infty} a_{k - 1} x^{k}$

The index is a dummy variable, so we can replace ^kwithⁿ , the first term of the L.H.S becomes $, 2 \sum_{k = 0}^{\infty} a_{k - 1} x^{k}$

Step 3: Changing the index of the second term

Now changing the index of the second term, let,

n - 1 = k

n = k + 1

Then,

$\sum_{n - 1}^{\infty} n b_{n} x^{n - 1} = \sum_{k + 1 = 1}^{\infty} (k + 1) b_{k + 1} x^{k} = \sum_{k = 0}^{\infty} (k + 1) b_{k + 1} x^{k}$

The index is a dummy variable, so we can replace ^kwithⁿ , the first term of the L.H.S becomes $= \sum_{k = 0}^{\infty} (k + 1) b_{k + 1} x^{k}$

The second expression in L.H.S, starts with index 0 , in order to combine both the expressions, we will take the second expression from n=1

, $2 . \sum_{n = 0}^{\infty} a_{n} x^{n + 1} + \sum_{n = 1}^{\infty} n b_{n} x^{n - 1} = 2 . \sum_{n = 1}^{\infty} a_{n - 1} x^{n} + b_{1} + \sum_{n = 0}^{\infty} (n + 1) b_{n + 1} x^{n} = b_{1} + \sum_{n = 1}^{\infty} [2 a_{n - 1} + (n + 1) b_{n + 1}] x^{n}$

which is equal to the R.H.S of the given statement.

Therefore, by simplifying the L.H.S of the expression, we can prove that both the expressions are the same.

Find Study Materials for

Create Study Materials

Select your language

Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

Question: Show that, $28 . \sum_{n = 0}^{\infty} a_{n} x^{n + 1} + \sum_{n = 1}^{\infty} n b_{n} x^{n - 1} = b_{1} + \sum_{n - 1}^{\infty} [2 a_{n - 1} + (n + 1) b_{n + 1}] x^{n}$

Step by Step Solution

Step 1: Power series

Step 2: Changing the index of the first term

Step 3: Changing the index of the second term

Most popular questions for Math Textbooks

Want to see more solutions like these?

Recommended explanations on Math Textbooks

Select your language

Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

Question: Show that,28. ∑n=0∞anxn+1+∑n=1∞nbnxn-1=b1+∑n-1∞[2an-1+(n+1)bn+1]xn

Step by Step Solution

Step 1: Power series

Step 2: Changing the index of the first term

Step 3: Changing the index of the second term

Most popular questions for Math Textbooks

Want to see more solutions like these?

Recommended explanations on Math Textbooks

Decision Maths

Calculus

Probability and Statistics

Statistics

Mechanics Maths

Geometry

Pure Maths

Question: Show that, $28 . \sum_{n = 0}^{\infty} a_{n} x^{n + 1} + \sum_{n = 1}^{\infty} n b_{n} x^{n - 1} = b_{1} + \sum_{n - 1}^{\infty} [2 a_{n - 1} + (n + 1) b_{n + 1}] x^{n}$