claurin series formula is (0) + F(0)x+0+0³+¹(0)¹ +.. 2(1-x)2 has the following derivatives. F"(x)= F(x)=4- A²(x)-[240✔ the general formula is E✓ 323 (1-x) 1: 12 (1-x) (0)-2✔ 2 (0) 44(0)- 1212 (0) 48 48 and (0) 240 240 - - +10 10+ lim 40 48 (1-x) ore, the Maclaurin Series begins as follows. - lim 240 240 (1-x) 2(n+1)x" the Ratio Test to find the radius of convergence, we have 2(+2)+1 2(n+1)x) 2(n+1) 2(n+1)

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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Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that has a power series expansion. Do not show that R(x) → 0.] Find the associated radius of convergence R.
f(x) = 2(1-x)-2
Step 1
The Maclaurin series formula is f(0) + F'(0)x+
f(x) = 2(1-x)-2 has the following derivatives.
f'(x) = 4✔
f"(x) = 12✔✔
F"(x) = 48✔
Step 2
Now, f(0) = 2✔
Step 4
f(4)(x) = 240✔✔
Step 6
Step 3
Therefore, the Maclaurin Series begins as follows.
+6✔
+4✔
tot
+10✔
Thus, the general formula is
lim
an +1
an
= lim
F'(0) = 4✔
Simplifying, we get |x|lim
F"(0) 2
2!
-4✔
Step 5
Using the Ratio Test to find the radius of convergence, we have
2(n + 2)x+1
0+ 0 +1
2(n+1)x²
+2 = |x| · |
2(n+1)
+
n+1
F(0)3 + F(4)(0) 4
F"(0) = 12✔
2(n+1)x"
2(n + 1)
X
13
8x²
f(0) = 48
48 and f(4)(0) = 240
240
Transcribed Image Text:Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that has a power series expansion. Do not show that R(x) → 0.] Find the associated radius of convergence R. f(x) = 2(1-x)-2 Step 1 The Maclaurin series formula is f(0) + F'(0)x+ f(x) = 2(1-x)-2 has the following derivatives. f'(x) = 4✔ f"(x) = 12✔✔ F"(x) = 48✔ Step 2 Now, f(0) = 2✔ Step 4 f(4)(x) = 240✔✔ Step 6 Step 3 Therefore, the Maclaurin Series begins as follows. +6✔ +4✔ tot +10✔ Thus, the general formula is lim an +1 an = lim F'(0) = 4✔ Simplifying, we get |x|lim F"(0) 2 2! -4✔ Step 5 Using the Ratio Test to find the radius of convergence, we have 2(n + 2)x+1 0+ 0 +1 2(n+1)x² +2 = |x| · | 2(n+1) + n+1 F(0)3 + F(4)(0) 4 F"(0) = 12✔ 2(n+1)x" 2(n + 1) X 13 8x² f(0) = 48 48 and f(4)(0) = 240 240
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