Find a power series representation centered at 0 for the...

Find a power series representation centered at 0 for the following function using known power series. Give the interval of convergence for the resulting series. f(x) = 1 / (1 + x^2) Which of the following is the power series representation for f(x)? A. sum_{k=0}^infinity 2x^k B. sum_{k=0}^infinity x^(2k) C. sum_{k=0}^infinity (-2x)^k D. sum_{k=0}^infinity (-x^2)^k The interval of convergence is [ ]. (Simplify your answer. Type your answer in interval notation.)

Transcript

00:01 So we have a function 1 over 1 plus x squared, and that is our f of x.

00:07 We need to find a series representation for this, and so we'll use a maclaurin series, which means we're going to take the derivative of this, and we're going to do all the mechanics, and we're going to have the point a at x at a equals 0.

00:22 So the way that we'll do this, we'll say that f of x is approximately equal to, technically equal to, we'll start k equals 0 and go to infinity of this eighth derivative of f at the point 0 over k factorial x raised to the k.

00:45 That is our series.

00:49 Now the first derivative, f1, will be equal to negative 2x over x squared plus 1, all squared.

01:03 Second derivative, i'm only going to probably do a few of these.

01:07 The derivatives will get complicated after, well, really now, but then they'll definitely get complicated after the second derivative.

01:14 So the second derivative will be equal to 2 times 3x squared minus 1 over x squared plus 1 squared.

01:24 The third derivative will be equal to 24x 1 minus x squared over x squared plus 1 to the fourth power.

01:45 Top one should be 3 to the third power.

01:49 And we can start to see a pattern.

01:51 So f of 4, of the fourth derivative, we should already be able to guess that this is going to be x squared plus 1 to the fifth at the bottom.

02:06 That top is going to be profoundly more nasty.

02:10 I think i might even, i'll just keep it, i guess.

02:14 So f, the fourth derivative, will be 24 times 8x squared times x squared minus 1 plus 1 minus 3x squared times x squared minus 1.

02:32 Now all of this is pretty crazy, but really it doesn't matter, but it doesn't matter because a is equal to 0 for the maclaren series.

02:43 So we're going to plug in, quickly, x is equal to 0.

02:48 So we're going to plug in x equals 0.

02:49 I guess i'll put that.

02:50 I'm going to plug in x is equal to 0 for all of these.

02:57 And so when we do that, the first derivative, it just becomes 2 over 1.

03:10 And let's look at that.

03:11 We plug in 0.

03:15 Well, technically, i'm sorry, 2 over 1.

03:17 No, that gives us 0.

03:18 2 times 0 is equal to 0.

03:21 Our first one's done.

03:22 Didn't mean in my head or myself.

03:24 F, second derivative at 0, evaluated at 2, 0 will be equal to negative 2.

03:31 Third derivative evaluated at 0 will be equal to, let's see if we can see this.

03:39 Well, that will be equal to 0 since there's an x factor there.

03:43 So it looks like we have a pattern of evens that will show up that won't vanish.

03:51 Derivative of fourth derivative at x equal to 0 is going to give me positive 24.

03:57 So it looks like we're also alternating as well.

04:01 The fifth derivative evaluated at 0 is equal to 0.

04:08 And so from this, we can tell that only the even powers matter.

04:14 What we get is f of x is approximately equal to 1 over 0 factorial, 0, excuse me, x to the 0th power.

04:28 No derivative yet...

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