By MACLAURIN'S THEOREM PROVE THAT :x/sinx=1+x^2/6+x^4/40
Answers
Answer:
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Step-by-step explanation:
We have:
f
(
x
)
=
x
sin
x
First let us take a look at the graph and see how "well defined" the function is:
graph{x/sinx [-40, 40, -20, 20]}
The Maclaurin series can be expressed in the following way:
f
(
x
)
=
f
(
0
)
+
f
'
(
0
)
1
!
x
+
f
'
'
(
0
)
2
!
x
2
+
f
'
'
'
(
0
)
3
!
x
3
+
(
f
(
4
)
)
0
4
!
x
4
+
...
=
∞
∑
n
=
0
f
(
n
)
(
0
)
n
!
x
n
We also note from the graph that
f
is even, so we expect all odd powers of
x
in the series to vanish. So, Let us find the derivatives, and compute the values at
x
=
0
.
First Term
f
(
x
)
=
x
sin
x
We cannot evaluate
f
(
x
)
when
x
=
0
as it is of an indeterminate form
0
0
, so must examine the limit of
f
(
x
)
as
x
→
0
, which we can evaluate using L'Hôpital's rule , we have:
f
(
0
)
=
lim
x
→
0
f
(
x
)
=
lim
x
→
0
x
sin
x
=
lim
x
→
0
1
cos
x
=
1
Second Term:
f
(
x
)
=
x
sin
x
Using the quotient rule, we have:
f
'
(
x
)
=
(
sin
x
)
(
1
)
−
(
x
)
(
cos
x
)
(
sin
x
)
2
=
sin
x
−
x
cos
x
sin
2
x
Again, we cannot evaluate
f
'
(
x
)
at
x
=
0
, and again we can use L'Hôpital's rule, thus:
f
'
(
0
)
=
lim
x
→
0
f
'
(
x
)
=
lim
x
→
0
sin
x
−
x
cos
x
sin
2
x
=
lim
x
→
0
cos
x
−
(
x
)
(
−
sin
x
)
−
(
1
)
(
cos
x
)
2
sin
x
cos
x
=
lim
x
→
0
x
sin
x
2
sin
x
cos
x
=
lim
x
→
0
x
2
cos
x
=
0
Third Term:
f
'
(
x
)
=
sin
x
−
x
cos
x
sin
2
x
Using the quotient rule, and the product rule we have:
f
'
'
(
x
)
=
(
sin
2
x
)
(
x
sin
x
)
−
(
sin
x
−
x
cos
x
)
(
2
sin
x
cos
x
)
(
sin
2
x
)
2
=
(
x
sin
2
x
−
2
cos
x
sin
x
+
2
x
cos
2
x
)
sin
3
x
=
(
x
−
2
cos
x
sin
x
+
x
cos
2
x
)
sin
3
x
Again, we cannot evaluate
f
'
'
(
x
)
at
x
=
0
, and again we can use L'Hôpital's rule, thus:
f
'
'
(
0
)
=
lim
x
→
0
f
'
'
(
x
)
=
lim
x
→
0
(
x
−
sin
2
x
+
x
cos
2
x
)
sin
3
x
=
lim
x
→
0
(
1
−
2
cos
2
x
+
x
(
2
cos
x
)
(
−
sin
x
)
+
cos
2
x
)
(
3
sin
2
x
)
(
cos
x
)
=
lim
x
→
0
(
1
−
2
cos
2
x
−
x
sin
2
x
+
cos
2
x
)
(
3
sin
2
x
)
(
cos
x
)
=
2
3
lim
x
→
0
(
1
−
2
cos
2
x
−
x
sin
2
x
+
cos
2
x
)
(
sin
x
)
(
sin
2
x
)
And a repeated application of L'Hôpital's rule gives:
f
'
'
(
0
)
=
2
3
lim
x
→
0
(
4
sin
2
x
−
x
(
2
cos
2
x
)
−
sin
2
x
+
2
cos
x
(
−
sin
x
)
)
(
sin
x
)
(
2
cos
2
x
)
+
(
cos
x
)
(
sin
2
x
)
=
2
3
lim
x
→
0
(
4
sin
2
x
−
2
x
cos
2
x
−
sin
2
x
−
sin
2
x
)
(
sin
x
)
(
2
cos
2
x
)
+
(
cos
x
)
(
sin
2
x
)
=
4
3
lim
x
→
0
sin
2
x
−
x
cos
2
x
2
sin
x
cos
2
x
+
cos
x
sin
2
x
And a repeated application of L'Hôpital's rule gives:
f
'
'
(
0
)
=
4
3
lim
x
→
0
2
cos
2
x
−
x
(
−
2
sin
2
x
)
−
cos
2
x
2
sin
x
(
−
2
sin
x
)
+
(
2
cos
x
)
(
cos
x
)
+
cos
x
(
2
cos
2
x
)
+
(
−
sin
x
)
sin
2
x
=
4
3
2
+
0
−
1
2
+
2
−
0
=
1
3