# Representing the ln(1-x) Power Series: How-to & Steps

Instructor: Gerald Lemay

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

In this lesson, we find the power series for ln(1-x) by deriving a simpler series and then integrating it. This lesson includes how to find the interval of convergence.

## Finding the Power Series for ln(1 - x)

A power series is the sum of an infinite number of terms. Each term is a power of x multiplied by a coefficient. To find the power series for ln(1 - x) we will

• derive the power series for a related function: 1/(1 - x)
• integrate to find the power series for ln(1 - x)
• determine the interval of convergence

#### Step 1: Use long division to find the power series for 1/(1 - x)

The derivative of ln(x) is 1/x. This ''1 over the argument'' suggests looking at how ln(1 - x) relates to 1/(1 - x).

1/(1 - x) means ''1 divided by 1 - x.'' Let's do a long division to find the power series for 1/(1 - x). The divisor, 1 - x, is written to the left of the box. The 1 goes in the box, and the quotient will appear above the box.

1 - x goes into 1, 1 times. We write a 1 above the division box.

Multiplying the divisor, 1 - x, by 1 gives 1 - x which we write under the 1.

Draw a line. Subtract 1 - x from 1. To subtract, change the sign and add. 1 - x becomes -1 + x. Adding -1 + x to 1 gives 0 + x or just x.

What goes into x? The divisor multiplied by x gives x - x2. The quotient becomes 1 + x and x - x2 goes below the x.

Draw a line, change the sign and add. The next term in the quotient is + x2.

See the pattern?

#### Step 2: Integrate.

Let's integrate the function 1/(1 - x):

First, let 1 - x = u. Differentiating, we get d(1 - x) = du. But the differential, d(1 - x), is 0 - dx. Thus, dx = - du.

dx over 1 - x is - du over u with the minus sign brought out in front of the intergral.

The integral of du/u is ln u.

The vertical bar reminds to replace u with 1 - x.

The constant of integration, C, is added because the integral does not have evaluation limits;i.e., we have an indefinite integral. No worry, we will get a value for C later.

Next, we make ln(1 - x) the subject of the equation.

Just to be clear, the integral of dx over 1 - x is the same as the integral of 1 over 1 - x. But 1 over 1 - x is the power series we derived earlier. Thus,

In general, xn integrates to xn+1 over n + 1. This is the power rule for integration.

Substitute this integration into our expression for ln(1 - x):

To find C, let x = 0. On the right-hand side, all the terms with an x will be 0. This leaves C on the right-hand side. On the left-hand side ln(1 - x) is ln(1 - 0) which is ln(1) which equals 0. Thus, C = 0.

Our result so far:

Note the Σ in the second line. This is summation notation meaning we substitute n = 1 into the xn / n to get x1 / 1 = x. We add the result of substituting n = 2 into the xn / n to get x + x2 / 2. This substituting for n and adding continues towards n = ∞. Don't forget the minus sign in front of the Σ.

#### Step 3: Determine the interval of convergence.

To find values of x which allow us to use the power series, we do a ratio test. The ratio test is based on:

An is the general term in our summation.

An+1 is n replaced with n + 1:

Next, divide An+1 by An:

In line 2, the exponents of x are grouped and then simplified to what we see in line 3.

Next, take the absolute value.

In our sum, n is always positive so it comes out of the absolute value.

Next, we take the limit as n goes to ∞:

The absolute value of x comes out of the limit because it does not depend on n.

As n gets larger and larger, n + 1 gets closer and closer to n. Thus, n divided by n + 1 becomes 1 in the limit as n goes to ∞.

The ratio test now says | x |, must be less than 1. This means we have convergence for values of x whose absolute value is less than 1. These are the values -1 < x < 1. Converge means the sum of the terms in the power series evaluated for an x in the interval of convergence, gives a number identical to what we get when substituting this x into ln(1 - x).

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