Find the interval of convergence for the Maclaurin series for f(x)=tan−1xf\left(x\right)=\tan^{-1}xf(x)=tan−1x

( − 1 , 1 ) \left(-1,1\right)

Find the interval of convergence for the Maclaurin series for f ( x ) = tan − 1 x f\left(x\right)=\tan^{-1}x f ( x ) = tan − 1 x

we're asked to find the interval of convergence for the Maclaurin series for the function f of x equals the inverse tangent of x. Now before we can find that interval of convergence, we first just need to find the Taylor series representation of this function. We know that this is going to be the sum from n equals zero to infinity of the nth derivative of my function evaluated at the center. So here, because this is a Maclaurin series, that center is zero. Then dividing that by n factorial and multiplying by x raised to that power of n, again, because our center is zero. That's just x minus zero to the power of n. So in order to find this here, we need to determine those nth derivatives of our functions and evaluate them at the center. I'm gonna go ahead and set up a chart to do that over here off to the side. So derivative of our function and then specifically at x equals zero. Now starting with just the function itself, that's the inverse tangent of x. We can evaluate that at zero, the inverse tangent of of zero. So this is just going to be zero as we think about our unit circle. Now we wanna take that first derivative. The first derivative of the inverse tangent of x is going to be one over one plus x squared. We can go ahead and evaluate that at zero, and that's one over one plus zero squared, which just gives us here one. Now to find that second derivative. So finding our second derivative here, we can think of this as being one plus x squared raised to the power of negative one and just apply the chain rule here. So this is going to be negative two x over one plus x squared squared. We can go ahead and evaluate that quickly at zero as well. Negative two times zero over one plus zero squared squared. This still just gives us zero. Then finding our third derivative here is going to be a bit trickier. It's going to be easiest to just apply the quotient rule here thinking of our function negative two x over one plus x squared squared. So it might have been a while since we've applied the quotient rule. I like to remember a memory tool in order to do this. Low d high minus high d low over the square of what's below, where d just refers to taking the derivative and high and low refer to our numerator and denominator. So low d high, that's gonna be one plus x squared squared times the derivative of our numerator. So that's just going to be negative two minus high d low, so that's minus two x, times the derivative of one plus x squared squared. So again, having to apply our chain rule here, two times one plus x squared and then subtracting one from that power, and then multiplying by the derivative of what's inside. That's just going to be positive two x. Then we put this over the square of what's below, so that's a one plus x squared squared squared. So this just becomes to the power of four. Now we can do a lot of simplification here. So let's go ahead and do that. This first term, I'm gonna go ahead and move that negative two to the front. So that's negative two times one plus x squared squared. And then multiplying everything going on here, negative two x, positive two, and positive two x is going to give us eight x squared here. So we're adding eight x squared as that negative turns to a positive or the double negative turns to a positive, times one plus x squared, then dividing this by one plus x squared to the power of four. Now we can see some stuff start to cancel. We're going to get rid of one of these one plus x squareds everywhere that we look in our numerator and denominator. So now we just have a minus two times one plus x squared plus eight x squared over one plus x squared cubed. Now I'm gonna go ahead and distribute that negative two. So that's gonna become negative two minus two x squared. And now we can combine like terms, and this gives us here six x squared minus two over one plus x squared cubed. So finally, this gives us our third derivative here. I'm gonna go ahead and fill that into my table. This is six x squared minus two over one plus x squared cubed. Now we can evaluate this at zero, our center, that's six times zero squared minus two over one plus zero squared cubed. One plus zero squared is just one, one cubed is also one. In the top, that first term goes to zero, so I have negative two over one which is negative two. So I'm gonna go ahead and stop there having found that third derivative, and I'm also gonna go ahead and get rid of this work that I did on just finding the derivative using the quotient rule so I have a bit more space to work with my Taylor series. So now that we have these first couple of derivatives, let's go ahead and look at the first couple terms of our Taylor series. So starting with n equals zero, this is going to be our numerator here. It's just zero, so zero over over zero factorial times x to the power of zero. Then for my next term here, plugging in n equals one, that numerator is going to be that one that we found over here, one over one factorial times x to the power of one. Then for n equals two, we're plugging in a zero for our numerator, zero over two factorial times x squared. And this should be another addition happening. Then for our next term here n equals three, plugging in that negative two to our numerator, so plus negative two over three factorial times x cubed. And this will continue on, but we'll stop there. Now we want to do some simplification. I can see that these two terms are just going to completely cancel out because they're just full of zeros. Then in this next term, one over one factorial is just one. One times x to the power of one is x, so that's just x. Then for that next term, minus two over three factorial, factorial, that gives me minus one over or minus two over six, which is really minus one third x cubed. And that's really all I have to work with here. Now if you can't see a pattern start to emerge, you can do the next couple of terms. But here I can already sort of see what's happening. We want to go ahead and now write this in summation notation so that we can then find that interval of convergence. So this is going to be the sum from n equals zero to infinity, and I can see that this is going to be an alternating series. It starts off as positive, and then that second term is negative. So I need to take negative one and raise it to a particular power. Now should that power be n or n plus one? Well, because it starts out as positive and we're starting our series from n equals zero, that means we wanna start this just raised to the power of n so that first term is positive. Now how else do we account for what's going on in our coefficients here or in our constants? So this has a constant of one and this has a one third. So what exactly is going on here? Well, it seems that I'm just getting odd terms in my denominator here. So I wanna go ahead and divide this by two n plus one. Starting from n equals zero, that's my first term there, that would make my denominator just one. And for n equals one, if I plugged one into this, that would make my denominator three. So it looks like this is a good choice. Now one more thing to account for our x term here, we actually do start out with x raised to the power of one as opposed to x raised to the power of zero. So that means we don't just want to take x and raise it to the power of n. Now just like what's happening in our denominators here, I can see that I'm getting odd powers. So I have the power of one, then the power of three, and my next term would be to the power of five, seven, and so on. So that means I also want to go ahead and raise this to the power of two and plus one just like I put in my denominator here. And this gives me my Taylor series for the function the inverse tangent of x. So now that we have our Taylor series we wanna determine our convergence, and we can do that by applying a convergence test. I'm gonna go ahead and set up the ratio test here. So this is going to be the limit as n approaches infinity of my series with n plus one plugged in divided by my original series with just n. Now as I do this, because we're taking the absolute value here, this alternating piece of my series isn't going to be necessary to plug it. So I can go ahead and just focus on everything else going on. So in my numerator, plugging in n plus one, this is x times two times n plus one plus one. Then dividing this by two times n plus one plus one. Now we're dividing by our original series with just n. When we divide by a fraction, that's the same thing as multiplying by the reciprocal of that fraction. So this just becomes two n plus one over x to the power of two n plus one. We're checking that this is less than one in order to determine that convergence. So now we need to do some simplification based on exponential properties in order to get stuff to cancel. Now looking at this two times n plus one plus one, I can think of this as being two n plus two plus one. And I can go ahead and rearrange that further to stick that plus one in front of the plus two so that I can think of this as two n plus one plus two. Now why would I want to do that? Well, looking at my original exponent and what I have here, it's two n plus one. So in order to get this to cancel, I want another x raised to the power of two n plus one. So this allows me to rewrite this as the limit as n approaches infinity of x to the power of two n plus one times x to the power of just that remaining two. Then in my denominator, I can just go ahead and simplify this. Two n plus one plus one is the same thing as two n plus three. Then this is multiplying by two n plus one over x to the power of two n plus one. Now I can see what's going to cancel x to the power of two n plus one cancels in the top and the bottom here, and I'm left with the limit as n approaches infinity of grouping my n terms together. So two n plus one over two n plus three and then times x squared. So now looking at this limit as n approaches infinity, as n goes off to infinity, because I have these n's to the same degree, they're just to the power of one, that means that as I take my limit, I'm just going to get the ratio of the coefficients. So that's going to be two over two. So coming up with this inequality here, two over two, that's what I get from my limit, times x squared. And this is less than one. And we wanna do some simplification here. Two over two is just one, so really this just leaves me with x the absolute value of x squared being less than one. And I can go ahead and take the square root on both sides here to ultimately get that the absolute value of x is less than one. So this one gives me my radius of convergence here. So knowing that my radius of convergence is one, I can determine my interval of convergence. Because the center of this series is zero, if I mark that zero on my number line here and go one out in the positive direction, one out in the negative direction, this gives me my interval of convergence that goes from negative one to one, and my series will diverge everywhere else. So I can say that this series converges on the interval from negative one to positive one. Now depending on what calculus course you are in, you may be able to stop at this point, and this is a sufficient final answer. However, if you're in a more advanced calculus course, you may be required to take this one step further and test the convergence of those endpoints. So I'm gonna go ahead and continue on here and show you that process. We're going to be plugging these endpoints into our original series. So plugging in x equals negative one and x equals positive one. So for x equals negative one, plugging this into our original series, this will be the sum from n equals zero to infinity of negative one to the nth power times negative one to the power of two n plus one over two n plus one. So looking at this here, I have negative one to the power of n, negative one to the power of two n plus one. Let's see how this shakes out here. The sum from n equals zero to infinity, I'm going to go ahead and separate this out using exponential properties. So this is negative one to the power of n times negative one to the power of two n times negative one to the power of one, then dividing this by two n plus one. Now negative one to the power of two n is just always going to be one because I'm always raising it to an even exponent. So this doesn't really affect this here, this is always just going to be one. So now I'm left with the sum from n equals zero to infinity of negative one to the power of n plus one, just combining those remaining exponents using exponential properties and dividing this by two n plus one. Now here I have an alternating series and I can determine the convergence of this using the alternating series test, so taking the limit as n approaches infinity of the non alternating piece of my series. I'm going to go ahead and get out of the way while we do that here. So here by the alternating series test, we wanna take the limit as n approaches infinity of one over two n plus one. Again, that's the non alternating piece of my series. Now we can see here that this limit goes to zero. Now the other thing that I need to check here for the alternating series test is that a sub n plus one is less than a sub n. Now this just means that the next term up is going to be less than the current term. Now that is true here because the only time we see n is in that denominator. So because this limit went to zero and a sub n plus one is less than a sub n, that tells us that this series converges. And therefore, this our original series converges at this endpoint, x equals negative one. So that's just one of our endpoints that we now know that's included in that interval of convergence, but we need to check one more endpoint here. So checking our other endpoint, x equals positive one, plugging that in for x to our original series, this is the sum from n equals zero to infinity of negative one to the power of n times one to the power of two n plus one. And this is over two n plus one. Now one raised to any power is just going to be one. So this ends up giving me the series from n equals zero to infinity of negative one to the power of n over two n plus one. Now this is almost an identical series to what we saw with our last endpoint, except here negative one is raised to the power of n as opposed to n plus one. But the alternating series test would work the same exact way. And because both of these things would still be true, that means that this point also converges. So that tells us that for our original series, it converges not only at x equals negative one, but also x equals positive one. So that means we need to come back up to our interval of convergence and make sure those two endpoints are included by using our square brackets here. So we can now say that this converges on the interval from negative one to one with both of those endpoints included. Now if you have any questions, feel free to let us know in the comments, and I'll see you in the next one.

Find the interval of convergence for the Maclaurin series for f ( x ) = tan ⁡ − 1 x f\left(x\right)=\tan^{-1}x f ( x ) = tan − 1 x

( − 1 , 1 ) \left(-1,1\right)

( − 1 , 1 ] \left(-1,1\right\rbrack ( − 1 , 1 ]

( − 3 , 3 ) \left(-\sqrt3,\sqrt3\right) ( − 3 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> , 3 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> )

[ − 3 , 3 ) \left\lbrack-\sqrt3,\sqrt3\right) [ − 3 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> , 3 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> )

15. Power Series

Power Series & Taylor Series

Business Calculus

15. Power Series

Power Series & Taylor Series

Struggling with Business Calculus?

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Multiple Choice

Find the interval of convergence for the Maclaurin series for $f\left(x\right)=\tan^{-1}x$

$open paren negative 1 comma 1 close paren$

$\left(-1,1\right\rbrack$

$\left(-\sqrt3,\sqrt3\right)$

$\left\lbrack-\sqrt3,\sqrt3\right)$

Verified step by step guidance

Step 1: Recall that the Maclaurin series is a special case of the Taylor series centered at x = 0. The interval of convergence for a series is determined by testing the values of x for which the series converges.

Step 2: Write the Maclaurin series for f(x) = tan⁡−1(x). The series expansion is given by: f(x) = x - x³/3 + x⁵/5 - x⁷/7 + ... . This is an alternating series with terms decreasing in magnitude.

Step 3: Apply the ratio test to determine the interval of convergence. The ratio test states that a series Σaₙ converges absolutely if lim (n → ∞) |aₙ₊₁ / aₙ| < 1. Compute the ratio of successive terms for the series.

Step 4: Solve the inequality obtained from the ratio test to find the range of x values for which the series converges. This will give the interval of convergence.

Step 5: Check the endpoints of the interval separately to determine whether the series converges at x = -1 and x = 1. Use the alternating series test or direct substitution to verify convergence at these points.

5:58m

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