Find maclaurin series of this question
F(x) = cos x
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Step-by-step explanation:
The Maclaurin series of
f
(
x
)
=
cos
x
is
f
(
x
)
=
∞
∑
n
=
0
(
−
1
)
n
x
2
n
(
2
n
)
!
.
Let us look at some details.
The Maclaurin series for
f
(
x
)
in general can be found by
f
(
x
)
=
∞
∑
n
=
0
f
(
n
)
(
0
)
n
!
x
n
Let us find the Maclaurin series for
f
(
x
)
=
cos
x
.
By taking the derivatives,
f
(
x
)
=
cos
x
⇒
f
(
0
)
=
cos
(
0
)
=
1
f
'
(
x
)
=
−
sin
x
⇒
f
'
(
0
)
=
−
sin
(
0
)
=
0
f
'
'
(
x
)
=
−
cos
x
⇒
f
'
'
(
0
)
=
−
cos
(
0
)
=
−
1
f
'
'
'
(
x
)
=
sin
x
⇒
f
'
'
'
(
0
)
=
sin
(
0
)
=
0
f
(
4
)
(
x
)
=
cos
x
⇒
f
(
4
)
(
0
)
=
cos
(
0
)
=
1
Since
f
(
x
)
=
f
(
4
)
(
x
)
, the cycle of
{
1
,
0
,
−
1
,
0
}
repeats itself.
So, we have the series
f
(
x
)
=
1
−
x
2
2
!
+
x
4
4
!
−
⋯
=
∞
∑
n
=
0
(
−
1
)
n
x
2
n
(
2
n
)
!
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