Created by Turbolearn AI
Taylor and Maclaurin Series
In the context of power series of the form:
∞
n
∑ Cn × X
n=0
Let's expand our understanding of these series by looking at the power expansion:
∞
n
∑ C n × (X − A)
n=0
Solving for Coefficients
To solve for the coefficients, find
F (A)
. All binomials become zero, except for the first coefficient,
C0
.
F (A) = C 0
Differentiation
Taking the derivative of the function:
Page 1
, Created by Turbolearn AI
The derivative of any constant is zero, so
C0
goes away.
The term
(X − A)
will go away, leaving
C1
.
Using the chain rule on the other terms:
d 2
[C 2 × (X − A) ] = 2 × C 2 × (X − A)
dx
d 3 2
[C 3 × (X − A) ] = 3 × C 3 × (X − A)
dx
d 4 3
[C 4 × (X − A) ] = 4 × C 4 × (X − A)
dx
Plugging in
A
will make all terms disappear, leaving
C1
.
′
F (A) = C 1
Taking the derivative again:
′′
F (A) = 2 × C 2
Therefore:
′′
F (A)
C2 =
2
Page 2
Taylor and Maclaurin Series
In the context of power series of the form:
∞
n
∑ Cn × X
n=0
Let's expand our understanding of these series by looking at the power expansion:
∞
n
∑ C n × (X − A)
n=0
Solving for Coefficients
To solve for the coefficients, find
F (A)
. All binomials become zero, except for the first coefficient,
C0
.
F (A) = C 0
Differentiation
Taking the derivative of the function:
Page 1
, Created by Turbolearn AI
The derivative of any constant is zero, so
C0
goes away.
The term
(X − A)
will go away, leaving
C1
.
Using the chain rule on the other terms:
d 2
[C 2 × (X − A) ] = 2 × C 2 × (X − A)
dx
d 3 2
[C 3 × (X − A) ] = 3 × C 3 × (X − A)
dx
d 4 3
[C 4 × (X − A) ] = 4 × C 4 × (X − A)
dx
Plugging in
A
will make all terms disappear, leaving
C1
.
′
F (A) = C 1
Taking the derivative again:
′′
F (A) = 2 × C 2
Therefore:
′′
F (A)
C2 =
2
Page 2
