Taylor Series and Maclaurin Series – Calculus 2
Taylor Series and Maclaurin Series – Calculus 2

Video Transcript

Find the radius of convergence for the Maclaurin series for 𝑓 of π‘₯ is equal to the cos of two π‘₯.

We’re given the function 𝑓 of π‘₯ is equal to the cos of two π‘₯. We need to find the radius of convergence for the Maclaurin series of this function. Let’s start by recalling what we mean by the Maclaurin series of a function 𝑓 of π‘₯. The Maclaurin series of a function 𝑓 of π‘₯ is the following power series. It’s the sum from 𝑛 equals zero to ∞ of the 𝑛th derivative of 𝑓 of π‘₯ with respect to π‘₯ evaluated at zero divided by 𝑛 factorial multiplied by π‘₯ to the 𝑛th power.

But this power series will only converge for certain values of π‘₯. In fact, for the Maclaurin series of a function the values of π‘₯ which our power series will converge always follows a special form. It will always follow one of the following three forms. Either it will converge for all real values of π‘₯ or there will exist some positive real constant 𝑅 such that our power series converges when the absolute value of π‘₯ is less than 𝑅 and it will diverge when the absolute value of π‘₯ is greater than 𝑅. And the third possibility is our Maclaurin series only converges when π‘₯ is equal to zero.

The radius of convergence for a Maclaurin series can be found by checking which of the three situations we’re in. If our Maclaurin series converges for all real values of π‘₯, we say that our radius of convergence 𝑅 is equal to ∞. If the power series converges when the absolute value of π‘₯ is less than 𝑅 and diverges when the absolute value of π‘₯ is greater than 𝑅, we just say the radius of convergence is 𝑅. Otherwise, if our power series only converges when π‘₯ is equal to zero, then we call the radius of convergence 𝑅 zero.

So, the first thing we’re going to want to do is find the Maclaurin series expansion for our function the cos of two π‘₯. We’ll do this directly from the definition. First, we see that 𝑓 evaluated at zero is equal to the cos of two times zero, which is the cos of zero, which we know is equal to one. Now to find the rest of the terms of our Maclaurin series, we’re going to need to start differentiating the cos of two π‘₯.

To start, we need to find 𝑓 prime of π‘₯. That’s the derivative of the cos of two π‘₯ with respect to π‘₯. And we know how to do this. For any constant π‘˜, the derivative of the cos of π‘˜π‘₯ with respect to π‘₯ is equal to negative π‘˜ times the sin of π‘˜π‘₯. In this case, our value of π‘˜ is two. So, we’ve shown 𝑓 prime of π‘₯ is equal to negative two sin of two π‘₯. Now, we need to find 𝑓 prime of zero. That’s negative two times the sin of two times zero, which is negative two times the sin of zero, which we know is equal to zero.

To find the next term in our Maclaurin series expansion for the cos of two π‘₯, we’re going to need to find 𝑓 double prime of π‘₯. That’s the derivative of negative two sin of two π‘₯ with respect to π‘₯. And we can do this because we know for any real constants π‘Ž and π‘˜, the derivative of negative π‘Ž times the sin of π‘˜π‘₯ with respect to π‘₯ is equal to negative π‘Ž times π‘˜ times the cos of π‘˜π‘₯. In our case, π‘Ž is two, and π‘˜ is also two. So, we get 𝑓 double prime of π‘₯ is equal to negative four cos of two π‘₯.

Now, we need to evaluate this at π‘₯ is equal to zero. We get negative four times the cos of two times zero. But the cos of two times zero is just equal to one, so this evaluates to give us negative four. We can find 𝑓 triple prime of π‘₯ in exactly the same way. This is the derivative of negative four cos of two π‘₯ with respect to π‘₯. But this time, we have the derivative of negative the cos of two π‘₯ is two times the sin of π‘₯.

But we know for real constants π‘Ž and π‘˜, the derivative of negative π‘Ž cos of π‘˜π‘₯ with respect to π‘₯ is equal to π‘Žπ‘˜ times the sin of π‘˜π‘₯. So, using this, we get 𝑓 triple prime of π‘₯ is eight times the sin of two π‘₯. We now need to evaluate this at π‘₯ is equal to zero. We get eight times the sin of two times zero, which we know is equal to zero.

Let’s find one more derivative of our function. That’s the derivative of eight sin of two π‘₯ with respect to π‘₯. This time, we know for real constants π‘Ž and π‘˜, the derivative of π‘Ž sin of π‘˜π‘₯ with respect to π‘₯ is equal to π‘Žπ‘˜ times the cos of π‘˜π‘₯. So, if π‘Ž equal to eight and π‘˜ equal to two, we get that our fourth derivative of 𝑓 of π‘₯ with respect to π‘₯ is equal to 16 times the cos of two π‘₯.

And now, we can see something interesting. This is a constant multiple of our function 𝑓 of π‘₯. So, if we were to start finding more derivatives of this function, we would do this in exactly the same way we did before. We would just get 16 times the answer. Keeping this in mind, we should be able to now find an expression for our Maclaurin series. Let’s first find the fourth derivative of 𝑓 of π‘₯ evaluated at zero. This is equal to 16 times the cos of zero, which we know is just equal to 16.

Let’s now write term by term what we found for our Maclaurin series for the cos of two π‘₯. Substituting in the values we found for our various derivatives of 𝑓 of π‘₯ evaluated at π‘₯ is equal to zero, we get that the first five terms of our Maclaurin expansion for the cos of two π‘₯ is equal to one plus zero divided by one factorial π‘₯ plus negative four divided by two factorial π‘₯ squared plus zero divided by three factorial π‘₯ cubed plus 16 divided by four factorial π‘₯ to the fourth power. And remember, we know this pattern will continue because we ended up with 16 times the cos of two π‘₯ which was just a multiple of our function 𝑓 of π‘₯.

So, let’s take a look at our Maclaurin series and see if we can find an expression for the 𝑛th term. We can see all of our odd terms will have the coefficient of zero. But what about the coefficients of our even terms? Well, each time we differentiate, we multiply by two. So, we know the absolute value of these coefficients is just going to be two to the 𝑛th power. But we also need to remember the derivative of the cos of π‘₯ is equal to negative the sin of π‘₯. In other words, alternate even terms are going to be negative. And we know this pattern continues, so we need to write this in 𝛴 notation. Remember, the first thing we showed is our odd terms will all have coefficient of zero.

Next, we can see that the exponent of our power of two is just equal to the exponent of π‘₯. So, we can just write these coefficients as two to the power of two 𝑛. Finally, we need to divide by 𝑛 factorial. This gives us the sum from 𝑛 equals zero to ∞ of two to the power of two 𝑛 divided by 𝑛 factorial multiplied by π‘₯ to the power of two 𝑛. But remember, we need to find the radius of convergence of this Maclaurin series. To do this, let’s clear some space and work out how we’ll find the radius of convergence of our Maclaurin series.

To do this, it’s often useful to try using the ratio test first. We recall the ratio test tells us if we let 𝐿 be equal to the limit as 𝑛 approaches ∞ of the absolute value of π‘Ž 𝑛 plus one divided by π‘Ž 𝑛 for the series the sum from 𝑛 equals zero to ∞ of π‘Ž 𝑛, then if the value of 𝐿 is less than one, we can conclude that our series must converge absolutely. However, if the value of 𝐿 is greater than one or equal to ∞, then we can say that our series is divergent.

To use the ratio test in our series, we’ll want to set π‘Ž 𝑛 to be equal to our summand. However, because both two and π‘₯ are raised to the power of two 𝑛, we’ll write π‘Ž 𝑛 as two π‘₯ all raised to the power of two 𝑛 divided by 𝑛 factorial. Let’s now find the value of 𝐿 for our series. That’s the limit as 𝑛 approaches ∞ of the absolute value of π‘Ž 𝑛 plus one divided by π‘Ž 𝑛. Instead of finding π‘Ž 𝑛 plus one divided by π‘Ž 𝑛, we’ll instead find π‘Ž 𝑛 plus one multiplied by the reciprocal of π‘Ž 𝑛.

Now, by finding expressions for π‘Ž 𝑛 plus one and π‘Ž 𝑛 and evaluating the reciprocal, we get the limit as 𝑛 approaches ∞ the absolute value of two π‘₯ raised to the power two times 𝑛 plus one divided by 𝑛 plus one factorial multiplied by 𝑛 factorial divided by two π‘₯ raised to the power of two 𝑛. And we can simplify this. First, we’ll distribute two over the parentheses in our first exponent. This gives us two 𝑛 plus two. Now, we can start simplifying. We see we have shared factors of two π‘₯ in our numerator and our denominator. In fact, we can cancel two 𝑛 of these shared factors. This just leaves us with two π‘₯ all squared in our numerator.

Next, remember, we can rewrite 𝑛 plus one factorial as 𝑛 plus one multiplied by 𝑛 factorial. Doing this means we can just cancel a shared factor of 𝑛 factorial in our numerator and our denominator. This leaves us with the limit as 𝑛 approaches ∞ of the absolute value of two π‘₯ all squared divided by 𝑛 plus one. And we can just evaluate this limit directly. Our limit is as 𝑛 is approaching ∞. π‘₯ does not vary as 𝑛 varies. So, our numerator is constant; however, our denominator is approaching ∞. So, the numerator is bounded; however, the denominator is approaching ∞. Therefore, this limit approaches zero.

So for any value of π‘₯, the limit of the absolute value of the ratio of successive terms is equal to zero. Therefore, by the ratio test, our power series must converge for any value of π‘₯. And if our power series converges for all real values of π‘₯, this means its radius of convergence is ∞. Therefore, we were able to show directly the radius of convergence for the Maclaurin series of the cos of two π‘₯ must be equal to ∞.

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