C₁ + ₁[(n + 1)cn+1 +2cn_1]x" = 0 2Cn-1 relation: Cn+1 = n+1 From: I have the recurrence From this, I have: n = 1: C₂ = n = 2: c3 = 0 Co n = 3: C4 2! n = 4: c5 = 0 C5 n = 5: c6 = n = 6: c = 0 n = 7: Cg = n = 8: c9 = II. Co 4! III = 0 Co 1! Co n = 9: C₁0 - 5 5! Co 3!
C₁ + ₁[(n + 1)cn+1 +2cn_1]x" = 0 2Cn-1 relation: Cn+1 = n+1 From: I have the recurrence From this, I have: n = 1: C₂ = n = 2: c3 = 0 Co n = 3: C4 2! n = 4: c5 = 0 C5 n = 5: c6 = n = 6: c = 0 n = 7: Cg = n = 8: c9 = II. Co 4! III = 0 Co 1! Co n = 9: C₁0 - 5 5! Co 3!
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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I have gotten as far as the attached. From there I need to:
- come up with a sum formula that works for any n (induction maybe???)
- identify the elementary power series function that matches this formula
- find its radius of convergence
Thank you.
![**Recurrence Relation in Series Expansion**
Given the expression:
\[ c_1 + \sum_{n=1}^{\infty} [(n + 1)c_{n+1} + 2c_{n-1}] x^n = 0 \]
We derive the recurrence relation:
\[ c_{n+1} = -\frac{2c_{n-1}}{n+1} \]
From this recurrence relation, the coefficients are determined as follows:
For \( n = 1 \):
\[ c_2 = -\frac{c_0}{1!} \]
For \( n = 2 \):
\[ c_3 = 0 \]
For \( n = 3 \):
\[ c_4 = \frac{c_0}{2!} \]
For \( n = 4 \):
\[ c_5 = 0 \]
For \( n = 5 \):
\[ c_6 = -\frac{c_0}{3!} \]
For \( n = 6 \):
\[ c_7 = 0 \]
For \( n = 7 \):
\[ c_8 = \frac{c_0}{4!} \]
For \( n = 8 \):
\[ c_9 = 0 \]
For \( n = 9 \):
\[ c_{10} = -\frac{c_0}{5!} \]
This pattern continues, alternating between zero and the fraction of \( c_0 \) over factorial terms depending on whether \( n \) is even or odd. The non-zero terms are defined by the factorial series involving \( c_0 \).
This derivation is fundamental in understanding series solutions to differential equations or other mathematical contexts where recurrence relations play a key role.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3b9e8ebd-ceba-4fff-be05-4ffc155e08e3%2F5802339e-4c2f-4006-9448-df2033adfb87%2Fbzkqx6_processed.png&w=3840&q=75)
Transcribed Image Text:**Recurrence Relation in Series Expansion**
Given the expression:
\[ c_1 + \sum_{n=1}^{\infty} [(n + 1)c_{n+1} + 2c_{n-1}] x^n = 0 \]
We derive the recurrence relation:
\[ c_{n+1} = -\frac{2c_{n-1}}{n+1} \]
From this recurrence relation, the coefficients are determined as follows:
For \( n = 1 \):
\[ c_2 = -\frac{c_0}{1!} \]
For \( n = 2 \):
\[ c_3 = 0 \]
For \( n = 3 \):
\[ c_4 = \frac{c_0}{2!} \]
For \( n = 4 \):
\[ c_5 = 0 \]
For \( n = 5 \):
\[ c_6 = -\frac{c_0}{3!} \]
For \( n = 6 \):
\[ c_7 = 0 \]
For \( n = 7 \):
\[ c_8 = \frac{c_0}{4!} \]
For \( n = 8 \):
\[ c_9 = 0 \]
For \( n = 9 \):
\[ c_{10} = -\frac{c_0}{5!} \]
This pattern continues, alternating between zero and the fraction of \( c_0 \) over factorial terms depending on whether \( n \) is even or odd. The non-zero terms are defined by the factorial series involving \( c_0 \).
This derivation is fundamental in understanding series solutions to differential equations or other mathematical contexts where recurrence relations play a key role.
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