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College Algebra: Help and Review27 chapters | 228 lessons

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

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*).

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

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

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

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

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

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 *n*th 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 *n*th derivative is as follows:

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.

By the way, the expression for the *n*th derivative evaluated at *x* = 0.

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

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

Let's substitute what we know into this expression.

As already shown, f(0) = 0.

The second term is:

The third term is a little more complicated:

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:

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:

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

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

Recall the *n*th derivative evaluated at *x* = 0, which is:

Thus, the general term ends up being:

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 region of convergence tells us the domain values of *x* for which the Maclaurin series is valid. Using the ratio test, a series converges if the equation you're looking at on screen holds.

The A*n* is the general term. Thus,

A*n*+1 is A*n* with *n* = *n* + 1:

The ratio A*n*+1/ A*n*:

*x**n*+1 divided by *x**n* is *x* raised to the (n + 1) - *n* which is *x*1 = *x*. The (-1) in front comes from dividing (-1)*n*+2 by (-1)n+1 giving (-1)(*n*+2)-(*n*+1) = (-1)1 = (-1).

Taking the absolute value, like we do here, we get:

*n*/(*n* + 1) is always positive so it comes out of the absolute value.

Taking the limit as *n* goes to infinity:

The absolute value of *x* does not depend on *n* so it was moved outside of the limit.

The limit of *n*/(*n* + 1) as *n* goes to infinity is 1. Thus, the Maclaurin series we found will converge to ln(1 + *x*) provided |*x*| < 1. In terms of an region, |*x*| < 1 means -1 < *x* < 1. We now check the endpoints: 1 and -1.

At *x* = 1, the series, 1 - 1/2 + 1/3 - 1/4 + so on, is an alternating series which converges; *x* = 1 is included.

At *x*= -1, the series, -1 - 1/2 -1/3 - 1/4 -. . . , is a harmonic series, which does not converge; *x* = -1 is not included. The region of convergence is (-1, 1] also written as -1 < *x* â‰¤ 1.

So let's take a look at our final result. The Maclaurin series for ln(1 + x) is what you're looking at on your screen and the region of convergence is (-1, 1].

The third line in the general expression is a compact way to write the Maclaurin series, which you might recall from earlier. If not, you can see it laid out.

The general term is on the right and the Î£ means we sum from *n* = 0 to *n* = âˆž.

For ln(1 + *x*) the summation starts at *n* = 1 (the *n* = 0 term is 0):

This compact statement of ln(1 + *x*) when expanded gives the same Maclaurin series result we obtained earlier.

Let's take a quick moment to briefly recap what we've learned about solving the Maclaurin Series for ln(1+*x*). The five steps for determining the Maclaurin Series of f(*x*) = ln(1+*x*) are as follows.

- Find derivatives for f(
*x*) - Evaluate these derivatives and f(
*x*) at*x*= 0 - Assemble the sum of products
- Find the general term
- Find the region of convergence

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College Algebra: Help and Review27 chapters | 228 lessons

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