Find the Taylor series for the functions defined as follows. Give the interval of convergence fo

Find the Taylor series for the functions defined as follows....

Key Concepts

Taylor Series

The Taylor series is a representation of a function as an infinite sum of terms calculated from the function’s derivatives at a single point, often used to approximate the function near that point. It is a central tool in calculus that allows complex functions to be expressed as power series, facilitating analysis, differentiation, and integration.

Geometric Series Expansion

Geometric series expansion involves expressing functions in the form 1/(1 - u) as the infinite series sum of u^n for n = 0 to infinity, provided that |u| < 1. This method is particularly useful for deriving Taylor series quickly when a function can be manipulated into a form resembling a geometric series.

Interval of Convergence

The interval of convergence refers to the set of input values for which a power series converges to the function it represents. Determining this interval is essential because it defines the range where the Taylor series accurately approximates the function. Convergence tests, such as the ratio test, are often employed to determine this interval.

Transcript

00:01 And hello, we are going to do problem from chapter 12, section 5, and the problem number 9.

00:08 We are asked to find the taylor series expansion for the given function, in this case f of x equals x squared over 4 minus x, and find the interval of convergence as well.

00:20 So we start by using the elementary taylor series that most resembles our function that we're trying to expand.

00:30 So that would be 1 over 1 minus x, which expands into 1 plus x plus x squared plus and we keep going until we get to our nth term, x to the end, x to the end plus dot dot dot.

00:50 And this has a interval of convergence of negative 1 is less than x is less than 1.

00:57 So what we need to do is rewrite this so to match our function that we're trying to find.

01:08 So we're going to do a couple steps to do that.

01:11 First, i'm going to replace every x with x over 4.

01:25 And i'm choosing that because later i can multiply through the denominator by 4.

01:31 And that should wind up giving us 4 minus x, which is what we ultimately want.

01:35 So there's our first step.

01:38 So we're going to write this out as 1 over 1 minus x to the 4, x over 4, sorry, not x2 the 4.

01:49 And so we get that by replacing every x with x over 4.

01:55 So the 1 is going to stay 1.

01:58 Then we have x over 4 in place of x.

02:04 Then our next term would be x over 4 quantity squared.

02:10 And this is going to keep going to our nth term x over 4 to the n.

02:25 So we're not right to what we want, right? but now if i multiply the denominator by 4, i'm going to get what i want when i distribute, and i multiply the numerator by x squared.

02:37 So i'm going to ultimately multiply every term by x squared over 4.

02:50 So that's going to look like...

02:55 All right, x squared over 4 times here we had 1 over 1 minus x over 4.

03:11 So does that give us the function that we're looking for? so our numerator is x squared times 1 or x squared.

03:19 And our denominator, we're going to distribute the 4 and get 4 minus x, which looks good.

03:25 That's what we're trying to find.

03:26 So that means i need to take my green expansion here and multiply term by term by x squared over 4.

03:35 So i'm going to write 1 times x squared over 4.

03:42 I can write that out like so, x squared over 4 times 1...