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

An error occurred trying to load this video.

Try refreshing the page, or contact customer support.

Coming up next: Taylor Series for sin(x): How-to & Steps

You're on a roll. Keep up the good work!

Take Quiz Watch Next Lesson
 Replay
Your next lesson will play in 10 seconds
  • 0:00 Maclaurin Series: f(x)…
  • 7:57 The Summation Expression
  • 8:29 Lesson Summary
Add to Add to Add to

Want to watch this again later?

Log in or sign up to add this lesson to a Custom Course.

Login or Sign up

Timeline
Autoplay
Autoplay

Recommended Lessons and Courses for You

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 Maclaurin series for a particular function: ln(1 + x). In addition to the steps for finding this series, you will also learn how to determine its region of convergence.

The Maclaurin Series for f(x) = ln(1 + x)

The general expression for the Maclaurin series is given by the formula:


The_McLaurin_series


The formulation for the Maclaurin series is complete when we specify the region of convergence. Let's carefully detail five steps for determining the Maclaurin series of f(x) = ln(1 + x).

Step 1: Find Derivatives for f(x)

The derivative of ln(x) is 1/x. Thus, the derivative of ln(1 + x) is 1/(1 + x):


derivative_of_ln(1+x)


The second derivative of f(x) is the derivative of f '(x):


derivative_of_f_


We write 1/(1 + x) as (1 + x)-1:


(1+x)^-1


The derivative of (1 + x)-1 is (-1)(1 + x)-2 where the exponent, -1, goes in front and the exponent has been reduced by 1 to become -2:


derivative_of_(1+x)^-1


Moving the (1 + x)-2 to the denominator, we get:


(1+x)^-2_moved_to_denominator


Differentiating again will give 2/(1 + x)3:


Third_derivative_of_f(x)


The next derivative will be -6/(1 + x)4. The next derivative will be 24/(1 + x)5. Looking ahead, it will be definitely useful to have an general expression for the nth derivative. Let's walk through this.

  • Note the signs alternate between positive and negative. We express this as (-1)n+1 for n = 1, 2, 3, etc.
  • The numerator of the derivative is 1, 1, 2, 6, 24, . . . for n = 1, 2, 3, 4, 5, . . . which agrees with (n - 1)! Note 0! = 1, 1! = 1 and 2! = 2(1) = 2. The exclamation point symbol is called the factorial.
  • The denominator is (1 + x) raised to the n.

Thus, the nth derivative is as follows:


General_expression_for_the_derivative


Step 2: Evaluate These Derivatives and f(x) at x = 0

For f(x) = ln(1 + x), let x = 0. Thus, f(0) = ln(1 + 0) = ln(1) = 0 meaning the first term in this series is 0. The first 4 derivatives evaluated at x = 0.


First_four_derivatives


By the way, the expression for the nth derivative evaluated at x = 0.


General_expression_for_the_derivative_at_x=0


We'll use this later when we determine the region of convergence.

Step 3: Assemble the Sum of Products

Remember the first line of the general expression for the Maclaurin series?


The_McLaurin_series_line1


Let's substitute what we know into this expression.

As already shown, f(0) = 0.

The second term is:


ln(1+x)_second_term


The third term is a little more complicated:


ln(1+x)_third_term


Using the first line as a model, we can deduce the terms which follow. In the fourth term we will see a 3! in the denominator, which evaluates to 3(2)(1) = 6. With the 2 in the numerator gives 2/6, which reduces to 1/3. As you can see, this ultimately equals:


ln(1+x)_fourth_term


The fifth term will have 4! = 4(3)(2)(1) = 24. The numerator is 6 giving 6/24 = 1/4. As you can see, it ultimately turns into:


ln(1+x)_fifth_term


Thus, the Maclaurin series for ln(1 + x) is this:


The_McLaurin_series_for_ln(1+x)


Step 4: Find the General Term

We'll need the general term when we explore the region of convergence. From line two of the general expression, we see:


line_2


Recall the nth derivative evaluated at x = 0, which is:


General_expression_for_the_derivative_at_x=0


Thus, the general term ends up being:


the_general_form


Which is valid for n = 1, 2, 3, etc. Note that n! is n(n-1)(n-2). . . (1) or just n! = n(n-1)!. The (n-1)! terms cancel leaving 1/n.


the_(n-1)!_cancel


To unlock this lesson you must be a Study.com Member.
Create your account

Register for a free trial

Are you a student or a teacher?

Unlock Your Education

See for yourself why 30 million people use Study.com

Become a Study.com member and start learning now.
Become a Member  Back
What teachers are saying about Study.com
Free 5-day trial

Earning College Credit

Did you know… We have over 160 college courses that prepare you to earn credit by exam that is accepted by over 1,500 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level.

To learn more, visit our Earning Credit Page

Create an account to start this course today
Try it free for 5 days!
Create an account
Support