Step 1:
Finding the first three derivatives of
1
f(x) =
x + 1
You can see that this is a composite of
g(x) = x + 1 with
h(g) = 1/g = g-1.
Following the chain rule
you find that
g'(x) = 1,
and
h'(g) = -1/g2 = -g-2. So
f'(x) = h'(g(x))g'(x) =
|
-1
(x + 1)2
|
You should be able to continue to apply the chain rule to arrive at
and
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Step 2:
As you multiply successive negative numbers, you get
(-1),
(-1)×(-2),
(-1)×(-2)×(-3),
(-1)×(-2)×(-3)×(-4), and so on.
If you factor out all the (-1)'s, you get, for the
nth number in this series,
(-1)n×n!.
Hence to find the nth derivative of f(x),
you will need this as a coefficient.
And since the power in the denominator increases by 1
with each derivative,
you will be taking the nth power in the denominator.
So
f(n)(x) =
|
(-1)n n!
(x + 1)n+1
|
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Step 3:
You have
Put in zero for
x and the denominator is
1.
So
f(0) = 1 = A0.
And you have
Put in zero for
x and the denominator is still
1, and
1 raised to any power is still
1.
So
f'(0) = -1 = A1.
And you have
so
f"(0) = 2 = A2.
And you have
so
f(3)(0) = -6 = A3.
And in general you have
f(n)(x) =
|
(-1)n n!
(x + 1)n+1
|
so
f(n)(0) = (-1)n n! = An.
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Step 4:
From the previous step you have that
Ak = (-1)k k!
So plug that into
∞
f(x) = ∑
k=0
|
1
Ak xk
k!
|
which is the Maclaurin formula.
Substituting gives you
∞
f(x) = ∑
k=0
|
(-1)k
k! xk
k!
|
Now cancel the
k!'s.
∞
f(x) = ∑ (-1)k xk
k=0
Which is the Maclaurin series for
If you wanted to write the series without the
sigma notation it would be
f(x) = lim
n → ∞
|
1 - x + x2 - x3 + x4 - ... ± xn
|
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