Gerald has taught engineering, math and science and has a doctorate in electrical engineering.
The Taylor Series Expression
The Taylor series represents a function of x where the function is evaluated about a value, a. As an infinite sum, we have this:
The Σ symbol means ''summation.'' We evaluate the expression to the right of Σ for n = 0, and then this result is added to the expression evaluated for n = 1, n = 2, and so on all the way to n = ∞. Let's expand this sum into the first 5 terms:
Taylor Series Steps
Here are the steps for finding the Taylor series of ln(1 + x).
Step 1: Calculate the first few derivatives of f(x).
We see in the formula, f(a). This is f(x) evaluated at x = a. Then, we see f '(a). This is the first derivative of f(x) evaluated at x = a.
Thus, the first step is to calculate derivatives of the function, f(x) = ln(1 + x).
The derivative of ln(1 + x) = 1/(1 + x) times the derivative of 1 + x. The derivative of 1 + x = 0 + 1 = 1. So far we have this:
To find the next derivative, note 1/(1 + x) is the same as (1 + x)-1. The derivative of (1 + x)-1 is (-1)(1 + x)-2, which equals -1/(1 + x)2. Thus:
Differentiating again, we get this:
Let's do one more.
We write 6 as 3!. The exclamation point is the factorial, meaning 3(2)(1). Note we could have written 2 as 2! because 2! = 2(1). By the way, 1! is 1 and 0! is also equal to 1.
Step 2: Evaluate the function and its derivatives at x = a.
Take each of the results from the previous step and substitute a for x. For f(x) = ln(1 + x) we get f(a) = ln(1 + a). For the first derivative:
Likewise, for each of the derivatives we evaluated earlier:
Step 3: Fill in the right-hand side of the Taylor series expression.
We now build the Taylor series from this:
This gives us this:
In the third line, we see 2! over 3!. This can be simplified because 2!/3! = 2(1)/3(2)(1) = 1/3. Likewise, in the next line, 3!/4! = 1/4. Our expression becomes this:
Step 4: Write the result using a summation.
Having a summation of a general term will be useful when determining the interval of convergence, or the set of x-values where a series converges. After the ln(1 + a) term, we see the numerator with an (x - a) term raised to increasingly higher powers. Likewise, the denominator has a (1 + a) term that grows with increasing powers. There is also an integer in the denominator that increases linearly. Lastly, the sign of the terms alternates.
You might try expanding this sum to convince yourself if it really does represent the terms of our expression.
So the Taylor series of ln(1 + x) about the point a is this:
Interval of Convergence
Let's find the interval of convergence. The general term, An, is this:
Replacing n with n + 1 we get An+1:
The ratio test says a series will converge provided:
Without showing all the details, we take the ratio of An+1 to An and simplify to get:
Take the absolute value and factor out n over n + 1:
Take the limit as n goes to ∞ and set this result < 1:
We have to consider both x and a. The first term of the series is ln(1 + a). Since the logarithm is valid only for positive arguments, we require 1 + a > 0; i.e. a > -1.
The absolute value of a quotient is the quotient of the absolute values.
This is the same as:
When we expand the absolute value, we get:
And then we add to each part of the inequality:
Now, for a > -1, 1 + a is never negative. Thus, the | 1 + a | is the same as (1 + a). This means:
We now check each of the endpoints. Without showing the details, when we substitute x = -1 into the series, we get the negative of the harmonic series, which does not converge. Thus, x = -1 is not a part of the interval. If we substitute 1 + 2a into the series, we get an alternative harmonic series which does converge. Thus, 1 + 2a is included. The interval of convergence is -1 < x ≤ 1 + 2a for a > -1.
Let's say we wanted a Taylor series approximation for ln(1 + x) about a = 2. Then, the series will converge for the values of x within the interval of convergence. The left-hand point is -1, and the right-hand point is 1 + 2a = 1 + 2(2) = 5. Thus, we expect a good match with the function when -1 < x ≤ 5.
The ln function in blue is being approximated with the first 6 terms of the Taylor series about a = 2 (in green). As predicted by the interval of convergence, these two curves are close between -1 and 5. Increasing the number of terms in the summation will improve the convergence at x = 5, but there will never be convergence at x ≤ -1. Thus, our choice of a lets us focus on an interval of interest provided a > -1.
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