Use the Maclaurin series representation of sinh(x) from the table to create a Maclaurin series of sinh(x^2). Next, differentiate it to create a Maclaurin series for 2xcosh(x^2)
Use the Maclaurin series representation of sinh(x) from the table to create a Maclaurin series of sinh(x^2). Next, differentiate it to create a Maclaurin series for 2xcosh(x^2)
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter2: Equations And Inequalities
Section2.6: Inequalities
Problem 80E
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Question
Use the Maclaurin series representation of sinh(x) from the table to create a Maclaurin series of sinh(x^2). Next,
![Table 11.5
1
= 1 + x + x² + . . . + x* + . .
8.
+*, for |x < 1
=1 x + x² -
+(-1)'r* + • . .
Σ(-1), for l지 < 1
1 + x
x²
e = 1 + x +
2!
.2.
·+
k!
for x < ∞
k!
k=0
(-1)*x²k+1
(2k + 1)!
- -1)*x*+1
(2k + 1)!
sin x x -
3!
for x < 0
5!
k=0
(-1)' x²*
(2k)!
2k
(-1)*x*
cos x = 1 -
2!
for x< ∞
4!
k=0
(2k)!
(-1)**'x*
(-1)k+l,k
In (1 + x) = x -
2
3.
for -1 < x < 1
k.
k=1
ナイ
+· • · +
.3.
–In (1 – x) = x +
for -1 < x < 1
k=1
(-1)*x²+1
+. .
(-1) x
3.
(-1)*x²*+1
nx= .
3.
-1
tan
for |x| < 1
2k + 1
2k + 1
5.
.3
2k+1
2k+1
sinh x = x +
3!
+· · ·+
5!
for x < ∞
%3D
(2k+ 1)!
k=0(2k + 1)!'
.2
2k
00
cosh x = 1+
2!
for x<∞
4!
(2k)!
(2k)!'
p(p - 1)(p – 2)- - (p – k + 1)
(1 + x) = E(")
, for |x| < 1 and
()
%3D
1
%3D
%3D
k!
8.
8.
||
:
+.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7a6e6fa2-624a-47a3-bf18-e2313eb69562%2F1bcd492f-68ba-44b4-b798-80682924a3c6%2Fkvag3ji.jpeg&w=3840&q=75)
Transcribed Image Text:Table 11.5
1
= 1 + x + x² + . . . + x* + . .
8.
+*, for |x < 1
=1 x + x² -
+(-1)'r* + • . .
Σ(-1), for l지 < 1
1 + x
x²
e = 1 + x +
2!
.2.
·+
k!
for x < ∞
k!
k=0
(-1)*x²k+1
(2k + 1)!
- -1)*x*+1
(2k + 1)!
sin x x -
3!
for x < 0
5!
k=0
(-1)' x²*
(2k)!
2k
(-1)*x*
cos x = 1 -
2!
for x< ∞
4!
k=0
(2k)!
(-1)**'x*
(-1)k+l,k
In (1 + x) = x -
2
3.
for -1 < x < 1
k.
k=1
ナイ
+· • · +
.3.
–In (1 – x) = x +
for -1 < x < 1
k=1
(-1)*x²+1
+. .
(-1) x
3.
(-1)*x²*+1
nx= .
3.
-1
tan
for |x| < 1
2k + 1
2k + 1
5.
.3
2k+1
2k+1
sinh x = x +
3!
+· · ·+
5!
for x < ∞
%3D
(2k+ 1)!
k=0(2k + 1)!'
.2
2k
00
cosh x = 1+
2!
for x<∞
4!
(2k)!
(2k)!'
p(p - 1)(p – 2)- - (p – k + 1)
(1 + x) = E(")
, for |x| < 1 and
()
%3D
1
%3D
%3D
k!
8.
8.
||
:
+.
![2xcosh(x²).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7a6e6fa2-624a-47a3-bf18-e2313eb69562%2F1bcd492f-68ba-44b4-b798-80682924a3c6%2Fndypssh.jpeg&w=3840&q=75)
Transcribed Image Text:2xcosh(x²).
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