What is the
Taylor's series
of f(x)=x^2 - x + 1
about the
point x=-1
Answers
Answered by
0
Answer:
The Taylor series expansion, in general, is written as:
∞
∑
n
=
0
f
n
(
a
)
n
!
(
x
−
a
)
n
So, we will have to take
n
derivatives of
1
x
2
.
n
=
3
is the bare minimum in my opinion if you want to see a significant chunk of a pattern, but let's just stop at
n
=
4
; this derivative isn't too bad, I guess. You just get pretty large numbers past
n
=
4
.
f
0
(
x
)
=
f
(
x
)
=
x
−
2
=
1
x
2
f
'
(
x
)
=
−
2
x
−
3
=
−
2
x
3
f
'
'
(
x
)
=
6
x
−
4
=
6
x
4
f
'
'
'
(
x
)
=
−
24
x
−
5
=
−
24
x
5
f
'
'
'
'
(
x
)
=
120
x
−
6
=
120
x
6
Now, let's plug them in. Just remember:
x
→
a
, except for the term
(
x
−
a
)
n
, which remains as
(
x
−
a
)
n
.
n
does vary from
0
to your choice of
n
. We chose
n
e
n
d
=
4
.
a
is a constant we choose. We chose
a
=
1
for simplicity, and because it's close to
x
=
0
(if we let
x
→
a
as
a
→
0
, then
x
→
0
, and our Taylor series becomes more accurate, usually).
a
does not vary.
Generally, the sum is written out to be:
=
f
0
(
1
)
0
!
(
x
−
1
)
0
+
f
'
(
1
)
1
!
(
x
−
1
)
1
+
f
'
'
(
1
)
2
!
(
x
−
1
)
2
+
f
'
'
'
(
1
)
3
!
(
x
−
1
)
3
+
f
'
'
'
'
(
1
)
4
!
(
x
−
1
)
4
+
...
Now plug in your newly-acquired derivatives:
=
1
1
2
1
!
(
1
)
+
−
2
1
3
1
!
(
x
−
1
)
+
6
1
4
2
!
(
x
−
1
)
2
+
−
24
1
5
3
!
(
x
−
1
)
3
+
120
1
6
4
!
(
x
−
1
)
4
+
...
Simplify:
=
1
+
(
−
2
)
(
x
−
1
)
+
6
2
(
x
−
1
)
2
+
−
24
6
(
x
−
1
)
3
+
120
24
(
x
−
1
)
4
+
...
And simplify some more:
=
1
−
2
(
x
−
1
)
+
3
(
x
−
1
)
2
−
4
(
x
−
1
)
3
+
5
(
x
−
1
)
4
+
...
Similar questions
English,
2 months ago
English,
2 months ago
Hindi,
5 months ago
CBSE BOARD X,
11 months ago
India Languages,
11 months ago