x² x3 log(1 + x) = x – - 2 3 4 for -1 < x < 1, by first calculating by long division 7 the series 1 =1 – x + x² – x³+ .., 1+x odi and then integrating termwise between 0 and x. 5. Prove that 1+x log x7 +.. = 2 x + %3D 5 laimonylog for – 1 < x < 1, and hence :). 3 1 +.. 535 hg?-2+ ) 1 1 1 log 2 = 2 3. 1 8. %3D 3 33 . Supply the details of the following derivation, due to Euler, of the infinite series expansion for log(1+x): (a) Show that log(1 + x) can be given by the limit = lim n[(1+x)/" – 1]. log(1 + x) = lim n[(1 -

x² x3 log(1 + x) = x – - 2 3 4 for -1 < x < 1, by first calculating by long division 7 the series 1 =1 – x + x² – x³+ .., 1+x odi and then integrating termwise between 0 and x. 5. Prove that 1+x log x7 +.. = 2 x + %3D 5 laimonylog for – 1 < x < 1, and hence :). 3 1 +.. 535 hg?-2+ ) 1 1 1 log 2 = 2 3. 1 8. %3D 3 33 . Supply the details of the following derivation, due to Euler, of the infinite series expansion for log(1+x): (a) Show that log(1 + x) can be given by the limit = lim n[(1+x)/" – 1]. log(1 + x) = lim n[(1 -

Name: Number 5 x²x3log(1 + x) = x – -23 4for -1 < x < 1, by first calculatin
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Description: Number 5 x²x3log(1 + x) = x – -23 4for -1 < x < 1, by first calculating by long division7the series1=1 – x + x² – x³+ ..,1+xodiand then integrating termwise between 0 and x.5. Prove that1+xlogx7+..= 2 x +%3D5laimonylogfor – 1 < x < 1, and hence:).3 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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Question

Number 5

x²
x3
log(1 + x) = x – -
2
3 4
for -1 < x < 1, by first calculating by long division
7
the series
1
=1 – x + x² – x³+ ..,
1+x
odi
and then integrating termwise between 0 and x.
5. Prove that
1+x
log
x7
+..
= 2 x +
%3D
5
laimonylog
for – 1 < x < 1, and hence
:).
3 1
+..
535
hg?-2+ )
1 1
1
log 2 = 2
3.
1
8.
%3D
3 33
. Supply the details of the following derivation, due to
Euler, of the infinite series expansion for log(1+x):
(a) Show that log(1 + x) can be given by the limit
= lim n[(1+x)/" – 1].
log(1 + x) = lim n[(1 -

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