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_into_1


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


gives_1


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


multiplying_gives_1-x


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.


change_the_sign_and_add


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


x_times_1-x_gives_x-x^2


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


change_the_sign_add_goes_in_x^2


See the pattern?


power_series_for_1/(1-x)


Step 2: Integrate.

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


integral_dx_over_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.


minus_integral_du_over_u


The integral of du/u is ln u.


-ln_u_u=1-x


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


=ln(1-x)+C


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.


make_ln(1-x)_the_subject


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,


substitute_the_series


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


integrate_term_by_term


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


result+C


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:


result_up_to_interval_of_convergence


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:


ratio_test_equation


An is the general term in our summation.


A_n


An+1 is n replaced with n + 1:


A_(n+1)


Next, divide An+1 by An:


A_(n+1)/A_n


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

Next, take the absolute value.


abs


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

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


limit


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


abs_out_of_limit


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 ∞.


abs(x)


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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