6.2.4

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Instructor: Number 12, for f of x equals ln of 1 plus x find the power series representation using f prime of x equals 1 over 1 plus x. OK, so let's begin with f prime of x. f prime of x is 1 over 1 minus negative x squared. Oh. Whoops. That's the wrong one. Negative x, OK, so this is 1 over 1 minus negative x. 

So the representation of that would be negative x to the n. So term by term, OK, so f prime of x, term by term, that would be 1 minus x plus x squared minus x cubed plus, et cetera. So let's go ahead and take the integral of both sides. OK, with respect to x, so this will give us f of x is x minus x squared over 2 plus x cubed over 3 minus x to the 4th over 4 plus, et cetera. Now we want to have C our constant there. 

OK. Well, because this is natural log of 1 plus x, if we evaluated x equals 0, x equals 0 would tell us that we have 0 equals C plus 0. So therefore, C equals 0. All right, so that means natural log of 1 plus x is. Then we have alternating signs there. So this is the sum of n equals 1 to n equals 0. 

Now leave it at 1. OK, take care of the alternating. So negative 1 to the n plus 1, x to the n over n. Stare at that pattern for a little while and make some sense of that if it's not obvious at the moment. 

All right, last question in this section, for f of x equals tangent inverse, find the power series representation using f prime of x equals 1 over 1 plus x squared. So we've already dealt with this one. So let's start with f prime of x. 

All right, f prime of x equals 1 over 1 minus negative x squared, which is the sum n equals 0 to infinity of negative 1. Ah, OK, so we'll have negative x squared to the negative 1 to the n, x to the 2n. So in our closed form, that's what we have. 

So let's go ahead and write it out term by term. That is 1 minus x squared, OK, plus x to the 4th minus x the 6th plus, et cetera. Now we will take the integral with respect to x. So term by term, we get x minus x cubed over 3 plus x to the 5th over 5 minus x to the 7 over 7, et cetera. Now we need to have our plus C. 

And the integral of that is f of x, which we're saying is tangent inverse of x. Tangent inverse of x equals that. Now tangent inverse of 0 is 0, which means our value of C is 0. 

So that tangent inverse x is just x minus x cubed over 3 plus x to the 5th over 5. OK, now those are alternating signs. Alternating signs. And we have all odd exponents there. 

It's negative 1 to the n. Got x to the 2n plus 1. And it's divided by each of those same odd coefficients, 2n plus 1. OK, and there we are. 

And I should probably know this is valid for absolute value of x less than 1. Because going back here, this was for absolute value of x squared, negative x squared being less than 1, which is equivalent to the absolute value of x being less than 1. So since they are valid on the same interval, there we go.