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Taylor Series for ln(1+x): How-to & Steps

Lesson Transcript
Instructor: Gerald Lemay

Gerald has taught engineering, math and science and has a doctorate in electrical engineering.

In this lesson, we show how to find the Taylor series for ln(1+x). This is a series evaluated about a specific point so we also look at the interval of convergence about this point.

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:


infinite_sum_expression


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:


infinite_sum_first_five_terms


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  • 0:02 The Taylor Series Expression
  • 0:36 Taylor Series Steps
  • 5:31 Solution
  • 5:57 Interval of Convergence
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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:


f(x)


And this:


f_prime(x)


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:


f_double_prime(x)


Differentiating again, we get this:


f_triple_prime(x)


Let's do one more.


f_four(x)


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:


f_prime(a)


Likewise, for each of the derivatives we evaluated earlier:


f_double_prime(a)



f_triple_prime(a)



f_four_prime(a)


Step 3: Fill in the right-hand side of the Taylor series expression.

We now build the Taylor series from this:


infinite_sum_first_five_terms


This gives us this:


infinite_sum_first_five_terms


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:


simplified_infinite_sum_first_five_terms


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.


infinite_sum


You might try expanding this sum to convince yourself if it really does represent the terms of our expression.

Solution

So the Taylor series of ln(1 + x) about the point a is this:


infinite_sum


Interval of Convergence

Let's find the interval of convergence. The general term, An, is this:


A_n


Replacing n with n + 1 we get An+1:


A_n+1


The ratio test says a series will converge provided:


ratio_test


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