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Taylor series in
C
\mathbb C
C
f
(
z
)
=
∑
n
=
0
∞
a
n
(
z
−
z
0
)
n
,
(
∣
z
−
z
0
∣
<
R
)
f(z)=\displaystyle\sum_{n=0}^{\infty} a_n(z-z_0)^n,\;\; (|z-z_0|<R)
f
(
z
)
=
n
=
0
∑
∞
a
n
(
z
−
z
0
)
n
,
(
∣
z
−
z
0
∣
<
R
)
f(z)=\displaystyle\sum_{n=0}^{\infty} a_n(z-z_0)^n,\;\; (|z-z_0|<R)
a
n
=
f
(
n
)
(
z
0
)
n
!
,
(
n
=
0
,
1
,
2
,
…
)
a_n=\frac{f^{(n)}(z_0)}{n!},\;\; (n=0,1,2,\ldots)
a
n
=
n
!
f
(
n
)
(
z
0
)
,
(
n
=
0
,
1
,
2
,
…
)
a_n=\frac{f^{(n)}(z_0)}{n!},\;\; (n=0,1,2,\ldots)
Taylor series in
R
\mathbb R
R
f
(
x
)
=
∑
n
=
0
∞
f
(
n
)
(
a
)
n
!
(
x
−
a
)
n
f(x)=\displaystyle\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n
f
(
x
)
=
n
=
0
∑
∞
n
!
f
(
n
)
(
a
)
(
x
−
a
)
n
f(x)=\displaystyle\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n
Taylor series in
R
\mathbb R
R
f
(
x
)
=
∑
n
=
0
∞
f
(
n
)
(
a
)
n
!
(
x
−
a
)
n
f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n
f
(
x
)
=
∑
n
=
0
∞
n
!
f
(
n
)
(
a
)
(
x
−
a
)
n
f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n
Geometric series
1
1
−
z
=
1
+
z
+
z
2
+
z
3
+
⋯
\dfrac{1}{1-z} = 1+ z + z^2 + z^3+ \cdots
1
−
z
1
=
1
+
z
+
z
2
+
z
3
+
⋯
\dfrac{1}{1-z} = 1+ z + z^2 + z^3+ \cdots
Geometric series
1
1
−
z
=
1
+
z
+
z
2
+
z
3
+
⋯
\dfrac{1}{1-z} = 1+ z + z^2 + z^3+ \cdots
1
−
z
1
=
1
+
z
+
z
2
+
z
3
+
⋯
\dfrac{1}{1-z} = 1+ z + z^2 + z^3+ \cdots
1
1
−
z
=
1
+
z
+
z
2
+
z
3
+
⋯
\dfrac{1}{1-z} = 1+ z + z^2 + z^3+ \cdots
1
−
z
1
=
1
+
z
+
z
2
+
z
3
+
⋯
\dfrac{1}{1-z} = 1+ z + z^2 + z^3+ \cdots
Resume presentation
Taylor series in C f ( z ) = ∑ n = 0 ∞ a n ( z − z 0 ) n , ( ∣ z − z 0 ∣ < R ) f(z)=\displaystyle\sum_{n=0}^{\infty} a_n(z-z_0)^n,\;\; (|z-z_0|<R) a n = f ( n ) ( z 0 ) n ! , ( n = 0 , 1 , 2 , … ) a_n=\frac{f^{(n)}(z_0)}{n!},\;\; (n=0,1,2,\ldots)
Lecture MATH3401
By Juan Carlos Ponce Campuzano
Lecture MATH3401
Geometric representation of series
5 years ago
1,103
Juan Carlos Ponce Campuzano
Independent Mathematics Educator
jcponce.com
jcponcemath
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