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# Power series of arctan(2x)

AP.CALC:
LIM‑8 (EU)
,
LIM‑8.E (LO)
,
LIM‑8.E.1 (EK)
,
LIM‑8.F (LO)
,
LIM‑8.F.1 (EK)

## Video transcript

so what I would like us to do in this video is find the power series representation of or find the power series approximate ation the power series approximation of arctangent of two x centered at zero and let's just say we want the first four nonzero terms of the power series approximation of arctangent of two x centered at zero so essentially the Maclaurin series of arctangent of two x the first four nonzero terms of it and if you feel good about it I encourage you to pause the video and try to work through it yourself so you might have tried to work through it and you probably took the first derivative of it you probably saw that hey you know the derivative with respect to X of arctangent of two x is equal to and this is a refresher if you didn't realize it the first time it's going to be the derivative of arctangent of X is one over X 1 over 1 plus x squared so this is going to be this is going to be the derivative of this which is 2 over 1 plus this whole thing squared so 1 plus I would say 1 plus 4 x squared and then as you try to find more of the terms of the Maclaurin series you would have taken more derivatives of this and it would have gotten very hairy very fast especially if you're looking for the first four nonzero terms so you probably realize hey there must be some type of an insight that I hadn't fully appreciated yet when I just try to just power through or no pun intended power through finding the power series the first four nonzero terms of the power series centered at zero of our tangent of 2x and you are right there is a key inside here the key inside here is well instead of doing it directly let's see if we could find the power series representation the first four terms of this thing right over here and then we can take the antiderivative of that to find the power series of our tangent of 2x making sure that we get the constant right that it satisfies the fact that we are being centered at zero so but I know what you're thinking now well that seems like what's getting us to the exact same issue if I want to find the power series representation of this the first four terms of it I'm still having to take the derivative of this multiple times which seems just as hard but this is the key inside I guess you could say the key inside is let's say say f of X which of course is the derivative of arctangent of two x is two over one plus four x squared now if we had another function if we had another function that kind of cleans it up a little bit so that we don't get all this hairiness when we take the anti when we take the derivatives so let's say we had another function G of X so I'm just in a color that I have not used so let's say that I have G of X is equal to 1 over 1 plus X so this is an interesting thing because it's really easy then this is the same thing as 1 plus X 1 plus X to the negative 1 power G of X is interesting because it's really easy to take its to take its derivatives for example G prime of X is going to be equal to chain rule derivative of 1 plus X is just 1 so it's going to be equal to negative 1 plus X to the negative 2 power if I want to take the second derivative of that G prime prime of X that's going to be negative 2 times negative 1 it's 2 times 1 plus X to the negative 3 power if I want to take the third derivative of that if I want to take the third derivative of that that's going to be let's see negative 3 times 2 it's negative 6 times 1 plus X to the negative 4 power and are we saying here salad isn't that aren't we worried about this why are you doing this well just just bear with me for a second so just that quickly I was able to find the first three derivatives of G of X and it's very easy then to find out the first four terms of its power series representation or especially its Maclaurin it's a Maclaurin series so for the power series centered at 0 we just have to evaluate each of these at 0 so G of 0 is equal to 1 G prime of 0 is equal to negative 1 G prime prime of 0 so 1 plus 0 to the negative 3 that's just 1 times 2 is equal to 2 and then the third derivative evaluated at 0 is equal to negative six so I could write the G of X G of X is approximately equal to I'm just going to do the first four terms here well it's going to be G of zero which is one minus G prime of zero times X so that's minus one times X so that's negative x plus plus G prime prime of zero two over two factorial times x squared well this is just one times x squared so let me just write that so that's just going to be plus x squared and then we have plus G prime prime prime of zero which is negative six over three factorial times X to the third well three factorial is just six so negative six divided by six is just negative one so that's going to be negative x to the third negative X to the third and I know what you're thinking all right Sal you just started with a hard problem and you gave yourself a much easier problem to find the power series representation how is this useful well this is the key insight that I've been promising throughout this video so far the key insight the product long-promised key insight is that and I'm finding the suitable color for key insight is that we can write f of X notice f of X is just two times f of X is two times G of 4x squared notice you take you take replace your X's with a 4x squared you're gonna have 1 over 1 plus 4x squared and then you multiply that whole thing times 2 you get this thing right over here and so if f of X is equal to that then f of X is power series representation it's just going to be it's just going to be taking this power series or at least the first four terms of it and replacing the X's with 4x squares and then multiplying the whole thing times 2 so let's do that so we can write we can write that f of X so we could write that f of X is going to be approximately equal to 2 times this thing evaluated when x is equal to 4x squared so it is one - instead of an X I'm going to write a four x squared plus x squared but instead of an X I have a four x squared squared so this is plus four x squared squared well that's going to be 16 X to the fourth so let me just write that that's going to be plus 16 X to the fourth and finally minus X to the third but now an X is four x squared so it's minus four x squared to the third power well that's going to be 64 X to the sixth so let me write that minus 64 X to the sixth power and then we could say that this is going to be so f of X is approximately equal to distribute the 2 to minus eight x squared plus 32 X to the fourth minus 128 X to the sixth so just like that with a little bit of a substitution I was able to reasonably in a non hairy fashion find the first four nonzero terms of the power series of two over one plus four x squared which is the derivative which is going to be the derivative of the power series of arctangent of two x so let's write this down so I'm going to write I'm going to write down so arc tangent of 2x arc tangent of 2x which is equal to an antiderivative of f of X DX which is going to be equal to the antiderivative of this whole all of this business equal to the antiderivative of two minus eight minus eight x squared plus 32 X to the fourth minus 128 X to the sixth actually let me make this approximate because now of course we're doing a approximation with the power series DX and what is this going to be equal to so we get approximately approximately equal to well I'm going to have a constant there let me write the constant fur because when we write our power series or Maclaurin series the first term is the constant term it's our function evaluated at zero we're going to have a constant if I take the antiderivative of two that's going to be plus 2x the antiderivative of this is C X to the third divided eight by three so it's negative 8/3 X to the third power then plus 32x to the fifth over five minus 128 X to the seventh over seven and we're in the homestretch we this is we've got at least four nonzero terms if this is nonzero there's going to be five nonzero terms but now let's just make sure that our constant is appropriate for arctangent of two x this should essentially evaluate to what arctangent of when arctic what when this function is when X is equal to zero and what is our tangent of zero well I remember this is centered at zero so we better get it right there that's you know that's just the the most basic thing if we're doing the Maclaurin series representation we're centering at zero so the so our approximate our approximation evaluated at zero better be the same thing as a function evaluated zero well if you well arctangent of 2 times zero is just going to be zero and so this thing when you evaluate it at zero this gets to C so this gets to C must be equal to zero C must be equal to zero if we want this thing to be zero when x is equal to zero so just like that we're done we're able to find we've been able to figure out that arctangent of two x is approximately equal to two X minus 8/3 X to the third power plus 32 over five x to the fifth minus 128 over seven x to the seventh and if we wanted more terms we could have gotten more terms by just doing what we just did but doing it four more terms so hopefully you enjoyed enjoyed that that fairly hairy problem but as you saw it's not as hairy as we thought it was going to be
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