The series x-(x^2)/(2)+(x^3)/(3)-(x^4)/(4)+⋯can be used to approximate the value of ln(1+x) for

The series x-(x^2)/(2)+(x^3)/(3)-(x^4)/(4)+⋯can be used to...

Key Concepts

Error Estimation in Series Approximations

This involves techniques to estimate the difference between the true value of a function and its approximation by a finite number of series terms. In alternating series, the error is often bounded by the magnitude of the first omitted term, allowing practitioners to assess the accuracy of their approximations.

Convergence and Radius of Convergence

This concept addresses the set of x-values for which a power series converges to the function it represents. For the ln(1+x) series, the series converges for |x| < 1 (and conditionally when x = 1), ensuring the validity of approximations within this interval.

Natural Logarithm Series Expansion

This is the specific Taylor series for ln(1+x), expressed as x - x²/2 + x³/3 - x?/4 + ... for values of x in the interval (-1,1]. It provides a method to approximate logarithmic values using polynomial terms, which is especially useful when x is small.

Alternating Series

An alternating series is one in which the signs of the terms alternate between positive and negative. Its significance lies in the fact that, under certain conditions, the error introduced by truncating the series is less than the absolute value of the first omitted term, providing a simple error bound.

Taylor Series Expansion

A Taylor series represents a function as an infinite sum of terms calculated from the values of its derivatives at a single point. It is used to approximate functions by polynomials whose accuracy increases with the number of terms included, making it a fundamental tool in mathematical analysis and numerical methods.

Transcript

00:04 Here we have a series that can be used to approximate the value of the natural log of 1 plus x.

00:15 Obviously, the more terms of the series that you add together, the better the approximation is going to be.

00:22 But in this particular problem, they've asked us to use the first six terms to approximate the value of a couple different expressions.

00:32 The first one, they want us to approximate what the natural log of 1 .02 would be.

00:43 And they want us to use the first six terms.

00:46 So only the first four terms are listed, but you can see the pattern here.

00:52 So if we wanted to expand it out to six terms, the fifth term, let me write it here a little bit.

00:59 I'll just clean this up and move it over.

01:01 The fifth term, we'll do it in a different color here to show that we're adding it, would be x to the fifth over five, and then the sixth term, that should be a minus because the signs are alternating, x to the sixth, sixth over six.

01:22 And then, of course, that's going to be approximately the natural log of one plus x.

01:30 So if we want to approximate the natural log of 1 .02, this means we're going to put 0 .02 in for x, right? because we see here it's 1 plus x.

01:48 So if it's 1 .02 and that comes from 1 plus x, then x, of course, must be 0 .02.

01:55 And then we'll throw that expression in the calculator real quick and see how close we are.

02:00 So the approximation would be 0 .02 minus 0 .02 squared over 2 plus 0 .02 cubed over 3 minus 0 .02 to the 4 over 4 plus 0 .02 to the 5th over 5, minus 0 .02 to the 6th over 6 over 6.

02:31 Maybe it look a little better if we put those in parentheses, but not necessary.

02:39 Okay.

02:41 So we're going to throw that in the calculator and see what it comes out as, and then we'll compute the actual natural log of 1 .02 and see how close we are.

02:50 So i'm going to switch over to desmos here real quick for that.

02:55 All right.

02:57 So it was 0 .02, and then the first sign was a plus, right? no minus and then it's gonna be minus point zero two over squared right let's do that and over two plus point zero two to the third over three plus point zero two to the third over three plus point zero two to the fourth over four sorry plus minus plus that should get minus.

03:59 Sorry about that.

04:02 And then plus, that's going to be 0 .02 to the fifth over five.

04:15 And lastly, minus 0 .02 to the 6th over 6.

04:29 And that's the approximation.

04:32 Now, the actual natural log of 1 .02 would be 0 .01980.

04:49 That's an exact approximation.

04:52 Six terms nailed it.

04:54 So it depends.

04:55 Obviously, if you carried it out more decimal places, you know, you would see a difference between the two answers.

05:03 But that just shows you what a great approximation of that expression, that particular series is...