Compute the Taylor series for f(x)=ln(1+x^2) about x0=0 . [Hint: Multiply the series for (1+x^2)

Compute the Taylor series for f(x)=ln(1+x^2) about x0=0 ....

Key Concepts

Power Series Manipulation

Manipulating power series includes operations such as multiplying the series by an algebraic factor and changing variables. This manipulation makes it possible to derive new series from known ones. For example, multiplying a known series by a polynomial term and then integrating term-by-term can lead to the Taylor series expansion of a related function, illustrating the flexibility of power series in solving problems.

Term-by-Term Integration of Power Series

Term-by-term integration involves integrating each term of a power series individually. This property is particularly useful when a function is expressed as a series and an antiderivative is needed. The procedure is valid within the radius of convergence, allowing one to safely integrate the series to obtain a new series representing the integral of the original function.

Taylor Series Expansion

A Taylor series is a representation of a function as an infinite sum of terms calculated from the function's derivatives at a single point. When this point is zero, the series is called a Maclaurin series. This method is useful for approximating functions with polynomials around a specific point, especially when dealing with analytic functions.

Geometric Series

A geometric series is a series of the form 1 + r + r^2 + r^3 + ... which converges when the absolute value of the ratio r is less than one. In the context of power series, geometric series principles are often used to represent functions like 1/(1 + x^2) by substituting r with an appropriate expression, facilitating further operations such as integration or term-by-term manipulation.

Transcript

00:02 Okay, so today we're finding the power series for the following series.

00:10 We've got f of x equals the natural log of 1 plus x squared.

00:23 So the first observation that we're going to make is that this function is actually the derivative of something we know, right? or rather, it's going to be the integral of something we know.

00:35 So consider this, right? say we want to integrate this.

00:46 So normally if you integrate this, you get like some kind of r -tan, right? which is great if our function was an r -10, but we don't want that.

00:54 Notice that the way to make this work is that if we replace the top of this with a 2x, then via u substitution, we actually get the following right.

01:08 So you can let u equals 1 plus x squared, and then d -u will be 2x -e -x.

01:28 Now, if you do that substitution and then finish integration, notice that you will get the function that we want.

01:38 So that's great.

01:40 From here, what we're going to do is we're going to substitute the series for 1 over 1 plus x squared.

01:47 Because we know that 1 over 1 plus x squared, well, that's the same thing as 1 over 1 minus negative x squared.

02:03 And we can know what the series for that.

02:05 That's the sum n from 0 to infinity of negative x squared to the n.

02:21 Now what we're going to do is we're going to take this expression here.

02:27 We're actually going to plug it into this equation over here.

02:32 So let me make some space and then we'll get that going.

02:44 So we've got the integral of 2x and that's going to be multiplied by this sum.

02:56 I'm going to go ahead and rewrite this integrate a little bit, right? so put this as a kind of a more familiar format.

03:02 So this would be negative 1 to the n...

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