Question Video: Finding the Radius of Convergence for the Taylor Series of a Trigonometric Function Mathematics • Higher Education

Find the radius of convergence for the Taylor series of 𝑓(π‘₯) = sin 2π‘₯ about π‘₯ = πœ‹.

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Video Transcript

Find the radius of convergence for the Taylor series of 𝑓 of π‘₯ is equal to the sin of two π‘₯ about π‘₯ is equal to πœ‹.

The question is asking us to find the radius of convergence of the Taylor series of the function 𝑓 of π‘₯ is equal to the sin of two π‘₯ about π‘₯ is equal to πœ‹. And to find the radius of convergence of this series, we’re first going to need to find an expression for this series. So, we’ll start by recalling what we mean by the Taylor series of a function.

The Taylor series of a function about π‘₯ is equal to π‘Ž is given by the sum from 𝑛 equals zero to ∞ of the 𝑛th derivative of 𝑓 evaluated at π‘Ž divided by 𝑛 factorial all multiplied by π‘₯ minus π‘Ž all raised to the 𝑛th power. In our case, we want the Taylor series of the sin of two π‘₯ about π‘₯ is equal to πœ‹. So, we’ll set our value of π‘Ž equal to πœ‹. And we can update our Taylor series to have the value of π‘Ž set to be πœ‹. And we see to find our Taylor series, we need to find an expression for the 𝑛th derivative of 𝑓 of π‘₯.

To do this, we’ll start by finding an expression for 𝑓 prime of π‘₯. That’s the derivative of the sin of two π‘₯ with respect to π‘₯. And we know this will be equal to two times the cos of two π‘₯. We can differentiate this again. First, we need to bring down our coefficient of two and then differentiate the cos of two π‘₯. Well, the derivative of the cos of two π‘₯ with respect to π‘₯ is equal to negative two times the sin of two π‘₯. And of course, we can simplify this to two multiplied by negative two is equal to negative four.

And we can keep going in this manner to find that 𝑓 triple prime of π‘₯ will bring down our coefficient of negative four. And then, we need to differentiate the sin of two π‘₯. We know the derivative of the sin of two π‘₯ with respect to π‘₯ is equal to two times the cos of two π‘₯. And just as we did before, we can simplify. This time, our coefficient will be negative eight.

We’ll do this process one more time to find the fourth derivative of 𝑓 of π‘₯ with respect to π‘₯. First, we’ll bring down our coefficient of negative eight. Then, we differentiate the cos of two π‘₯ with respect to π‘₯. This is equal to negative two times the sin of two π‘₯. And we simplify this to get 16 times the sin of two π‘₯. And the reason we stop here is we can notice something interesting. The only part of this expression which varies with respect to π‘₯ is the sin of two π‘₯. And we can see this is equal to 𝑓 of π‘₯.

So, the fifth derivative of 𝑓 of π‘₯ with respect to π‘₯ will just be equal to 16 times 𝑓 prime of π‘₯. And of course, that pattern continues. The sixth derivative of 𝑓 of π‘₯ with respect to π‘₯ will be 16 times 𝑓 double prime of π‘₯. We can use this to find an expression for the 𝑛th of derivative of 𝑓 of π‘₯ evaluated at πœ‹. We’ll start by finding 𝑓 evaluated at πœ‹. Remember, we could write this as the zeroth derivative of 𝑓 evaluated at πœ‹; however, we’ll just leave this as 𝑓 evaluated at πœ‹.

We substitute π‘₯ is equal to πœ‹ into this expression, giving us the sin of two πœ‹, which we know is equal to zero. And because this pattern continues every four derivatives, we know the fourth derivative of 𝑓 evaluated at πœ‹ will be zero and the eighth derivative of 𝑓 evaluated at πœ‹ will be zero, and so on. Let’s now find the first derivative of 𝑓 evaluated at πœ‹. We substitute this into our expression. We get two cos of two πœ‹, which is equal to two.

And we’ll do the same to find 𝑓 double prime of πœ‹ and 𝑓 triple prime of πœ‹. We get negative four sin of two πœ‹, which is equal to zero, and negative eight times the cos of two πœ‹, which is equal to negative eight. And we know this pattern should continue. For example, the fourth derivative evaluated at πœ‹ will be equal to 16 times zero, which is, of course, equal to zero. But then the fifth derivative of 𝑓 evaluated at πœ‹ will be equal to 16 times two, which is equal to 32.

We can now use this information to find the 𝑛th derivative of 𝑓 evaluated at πœ‹. First, we need to notice if we don’t take any derivatives or if we take an even number of derivatives, then our derivative of 𝑓 evaluated at πœ‹ is equal to zero. Next, we need to notice the size of our coefficient is being multiplied by two at each step. For example, in the first derivative of 𝑓, we can write our coefficient as two to the first power. And in our second derivative, we can write negative four as negative two squared. And we could do the same with all of our derivatives of 𝑓.

And this gives us a hint about what this value should be. For example, 𝑓 prime of πœ‹ is two to the first power and 𝑓 triple prime of πœ‹ is negative two cubed. For the odd derivatives, we’re raising two to the power of the number of derivatives we’re taking. We just need to worry about whether we have a positive or a negative value. And to make this easier, we’ll start by writing it in our two 𝑛 plus one derivative of 𝑓 evaluated at πœ‹.

Remember, we’re only interested in the odd derivatives, so we’ll just write this as two 𝑛 plus one. As discussed, we want two raised to the power of the number of derivatives we’ve taken. So, when 𝑛 is equal to zero, we substitute zero in and we get two which is 𝑓 prime evaluated at πœ‹. But remember, we want the third derivative of 𝑓 evaluated at πœ‹ to be negative two cubed. So, when 𝑛 is equal to one, we want this to give us a negative value.

And to do this, all we need to do is multiply by negative one to the 𝑛th of power. Then, for our third derivative, when 𝑛 is equal to one, we get a coefficient of negative one. We can use this to find the Taylor series of the sin of two π‘₯ centered around π‘₯ is equal to πœ‹. First, we know all the even derivatives of 𝑓 evaluated at πœ‹ will be equal to zero. So, we can remove these from our series and we’re only interested in the odd terms. So, we’ll write two 𝑛 plus one.

Now, all we need to do is write in our expression for the two 𝑛 plus one for the derivative of 𝑓 evaluated at πœ‹. But before we do this, we’ll clear some space. So, all we have to do is substitute this expression into our power series. This gives us this representation for the Taylor series of 𝑓 of π‘₯ is equal to the sin of two π‘₯ about π‘₯ is equal to πœ‹. But the question isn’t just asking us to find the Taylor series of this function; it’s asking us to find its radius of convergence. And usually, the easiest way to find the radius of convergence of a power series is to use the ratio test.

So, let’s start by recalling the ratio test. For a series, the sum from 𝑛 equals zero to ∞ of π‘Ž 𝑛. We know if the limit as 𝑛 approaches ∞ of the absolute value of the ratio of successive terms π‘Ž 𝑛 plus one over π‘Ž 𝑛 is less than one, then our series must be convergent. However, if the limit as 𝑛 approaches ∞ of the absolute value of π‘Ž 𝑛 plus one over π‘Ž 𝑛 is greater than one, then our series is divergent. We also know if this limit is equal to one, then the ratio test is inconclusive.

We want to apply this to our Taylor series. So we need to set the value of π‘Ž 𝑛 to be the summand of our Taylor series. So, let’s clear some space and start evaluating this limit. First, instead of finding π‘Ž 𝑛 plus one divided by π‘Ž 𝑛, we’ll instead multiply π‘Ž 𝑛 plus one by the reciprocal of π‘Ž 𝑛. We need to start by finding an expression for π‘Ž 𝑛 plus one. We need to write 𝑛 plus one for every instance of 𝑛 in the summand of our Taylor series. Doing this gives us the following expression.

We then need to multiply this by the reciprocal of π‘Ž 𝑛, and remember, this means we invert our fraction. So, by doing all of this, this gives us the following limit we need to evaluate. And this is a very complicated looking limit. However, as we’ll see, this actually cancels very nicely. First, in our numerator and our denominator, we have 𝑛 shared factors of negative one we can cancel. This just leaves us with one factor of negative one in our numerator.

Next, we have three instances of two times 𝑛 plus one plus one. We need to evaluate all three of these expressions. If we distribute two over our parentheses and simplify, we see all three of these expressions are two 𝑛 plus three. So, we’ll replace all of these with two 𝑛 plus three to help us simplify. Next, we can see in our numerator and our denominator, we have two 𝑛 plus one shared factors of two. So, we can simplify this to give us a factor of two squared in our numerator.

In fact, we can do exactly the same with factors of π‘₯ minus πœ‹. We have two 𝑛 plus one shared factors in our numerator and our denominator. So, this leaves us with π‘₯ minus πœ‹ all squared in our numerator. The last thing we need to do is simplify our factorials. To do this, we need to recall that two 𝑛 plus three factorial can be rewritten as two 𝑛 plus three times two 𝑛 plus two times two 𝑛 plus one factorial. And then, we see our numerator and our denominator share a factor of two 𝑛 plus one factorial, and we can cancel this out.

So, after all of this cancellation, our limit simplifies to give us the limit as 𝑛 approaches ∞ of the absolute value of two squared times π‘₯ minus πœ‹ all squared all over two 𝑛 plus three times two 𝑛 plus two. And we can actually just evaluate this limit directly. Our limit is as 𝑛 is approaching ∞. And remember, the value of π‘₯ does not change as the value of 𝑛 changes. So, in fact, the entire numerator of this expression remains constant. It doesn’t change as the value of 𝑛 changes.

However, if we were to distribute over the parentheses in our denominator, we would see the highest power of 𝑛 we have is four 𝑛 squared. So, as our value of 𝑛 is growing, our denominator is growing without bound. This means this limit must evaluate to give us zero. And the important thing to notice here is this limit evaluates to give us zero regardless of which value of π‘₯ we use. So, by using the ratio test, we know if this limit is less than one, then our series must be convergent. But this limit is less than one for any value of π‘₯; in fact, it’s always equal to zero. Therefore, our Taylor series converges for any real value of π‘₯. And when a series converges for all real values of π‘₯, we say that its radius of convergence is ∞, or sometimes we write positive ∞.

Therefore, we were able to directly show from the definition of the radius of convergence and the definition of a Taylor series that the Taylor series for 𝑓 of π‘₯ is equal to the sin of two π‘₯ about π‘₯ is equal to πœ‹ will have an infinite radius of convergence.

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