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Consider the function 𝑓(π‘₯) = cos π‘₯. What are the first four derivatives of 𝑓 with respect to π‘₯? Write the general form for the 𝑛th derivative of 𝑓 with respect to π‘₯. Hence, derive the Maclaurin series for cos of π‘₯.

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Video Transcript

Consider the function 𝑓 of π‘₯ equals cos of π‘₯. What are the first four derivatives of 𝑓 with respect to π‘₯? Write the general form for the 𝑛th derivative of 𝑓 with respect to π‘₯. And hence, derive the Maclaurin series for cos of π‘₯.

There is a fourth part to this question, which asks us to find the radius 𝑅 of convergence of the Maclaurin series for cos of π‘₯. And we’ll consider that at the very end. So we begin by finding the first four derivatives of 𝑓 with respect to π‘₯. We’re told that 𝑓 of π‘₯ is equal to cos of π‘₯. And so we quote the general result for the first derivative of cos of π‘₯ is negative sin π‘₯. And therefore, 𝑓 prime of π‘₯, the first derivative of 𝑓 with respect to π‘₯, is negative sin π‘₯.

Next, we recall the general result for the derivative of sin π‘₯. It’s cos π‘₯. And this means the derivative of negative sin π‘₯ must be negative cos of π‘₯. By returning to the first general result we quoted, we’ll be able to differentiate negative cos of π‘₯. It’s negative, negative sin of π‘₯, which is, of course, simply sin of π‘₯. So the third derivative is sin of π‘₯. And finally, we obtain the fourth derivative to be equal to cos of π‘₯. And so we have the first four derivatives. They are negative sin π‘₯, negative cos π‘₯, sin π‘₯, and cos π‘₯.

The second part of this question asks us to find a general form for the 𝑛th derivative of 𝑓 with respect to π‘₯. And so we need to spot a pattern with our derivatives. Firstly, we should notice that if we differentiate cos of π‘₯, we go back to negative sin π‘₯. And this cycle will continue. Next, we recall that sine and cosine are horizontal translations of one another. And so if we sketch the curve of 𝑦 equals negative sin of π‘₯, that, of course, is a reflection on the π‘₯-axis for the graph of 𝑦 equals sin π‘₯. We see that we can write 𝑦 equals negative sin π‘₯ as 𝑦 equals cos of π‘₯ plus πœ‹ by two. It’s a horizontal translation of the graph of 𝑦 equals cos of π‘₯, left by πœ‹ by two radians. So 𝑓 prime of π‘₯ is equal to cos of π‘₯ plus πœ‹ by two.

We’ll repeat this for the graph of 𝑦 equals negative cos of π‘₯. It’s a reflection again in the π‘₯-axis of the graph 𝑦 equals cos of π‘₯. And it, therefore, also can be represented by a horizontal translation of the graph of 𝑦 equals cos of π‘₯ left by πœ‹ radians. So we can write 𝑓 double prime of π‘₯, which was negative cos of π‘₯, as cos of π‘₯ plus πœ‹. By applying a similar thought process, we find that the third derivative, sin of π‘₯, can be written as cos of π‘₯ plus three πœ‹ by two. And the fourth derivative can be written as cos of π‘₯ plus two πœ‹.

Remember, cos itself is periodic with a period of two πœ‹ radians. So cos of π‘₯ plus two πœ‹ is exactly the same as cos of π‘₯. And we might be starting to spot a pattern. In fact, if we write the second derivative, cos of π‘₯ plus πœ‹, as cos of π‘₯ plus two πœ‹ over two and the fourth derivative as cos of π‘₯ plus four πœ‹ over two, we see we can write the 𝑛th derivative as cos of π‘₯ plus π‘›πœ‹ over two. And so the general form for the 𝑛th derivative of 𝑓 with respect to π‘₯ is cos of π‘₯ plus π‘›πœ‹ over two.

The third part of this question asks us to derive the Maclaurin series for cos of π‘₯. And so we recall this is the sum from 𝑛 equals zero to ∞ of the 𝑛th derivative of 𝑓 evaluated at zero over 𝑛 factorial times π‘₯ to the 𝑛th power. Well, in this case, the 𝑛th derivative of 𝑓 evaluated at zero is cos of zero plus π‘›πœ‹ over two. When 𝑛 is zero, we have 𝑓 of zero, which is cos of zero which is one. When 𝑛 is one, we have 𝑓 prime of zero, which is cos of πœ‹ by two, which is zero. When 𝑛 is two, we have cos of πœ‹, which is negative one. And when 𝑛 is equal to three, we have cos of three by two, which is once again zero. And of course, as we saw in the second part of this question, this cycle continues.

And this means the first few terms will be one minus π‘₯ squared over two factorial plus π‘₯ to fourth power over four factorial minus π‘₯ to the six over six factorial, which we can write as the sum from 𝑛 equals zero to ∞ of negative one to the 𝑛th power. Remember, this part will just give us the alternating signs. We go from positive to negative and back to positive again. This is all over two 𝑛 factorial. This bit gives us the even factorials and the denominator times π‘₯ to the power of two 𝑛. And so we’ve derived the Maclaurin series for cos of π‘₯. It’s the sum from 𝑛 equals zero to ∞ of negative one to the 𝑛th power over two 𝑛 factorial times π‘₯ to the power of two 𝑛. And, of course, if we chose to alternatively write this as shown, that would be absolutely fine too.

We’re now going to clear some space and consider the fourth and final part of this question. We’ll keep the Maclaurin series for cos of π‘₯ on screen as we’re going to use that in a moment. And this question says, what is the radius 𝑅 of convergence of the Maclaurin series for cos of π‘₯? We recall that, in general, there’s an open interval from negative 𝑅 to 𝑅, in which a power series converges. And that number 𝑅 is called the radius of convergence. And we can use the ratio test to find this value. The part of the test we’re interested in says that, given a series of sum of π‘Ž 𝑛, if the limit as 𝑛 approaches ∞ of the absolute value of π‘Ž 𝑛 plus one over π‘Ž 𝑛 is less than one, then the series is absolutely convergent and hence convergent.

In our case, we let π‘Ž 𝑛 be equal to negative one to the 𝑛th power times π‘₯ to the power of two 𝑛 over two 𝑛 factorial. Then, π‘Ž 𝑛 plus one is negative one to the power of 𝑛 plus one times π‘₯ to the power of two times 𝑛 plus one over two times 𝑛 plus one factorial. We’ll distribute the parentheses to make the next step easier. Two times 𝑛 plus one is two 𝑛 plus two. According to the ratio test, we want to find where the limit as 𝑛 approaches ∞ of the absolute value of the quotient of these is less than one.

We know that when dividing by a fraction, we simply multiply by the reciprocal of that fraction. So we can say that we want the limit as 𝑛 approaches ∞ of the absolute value of negative one to the power of 𝑛 plus one times π‘₯ to the power of two 𝑛 plus two over two 𝑛 plus two factorial times two 𝑛 factorial over negative one to the 𝑛th power times π‘₯ to the power of two 𝑛. And then, we recall that when dividing two numbers with equal basis, we simply subtract their exponents so that π‘₯ to the power of π‘Ž divided by π‘₯ to the power of 𝑏 is π‘₯ to the power of π‘Ž minus 𝑏. We’ll use this to simplify negative one to the power of 𝑛 plus one divided by negative one to the power of 𝑛. It’s simply negative one. Similarly, we’ll be able to simplify π‘₯ to the power of two 𝑛 plus two divided by π‘₯ to the power of two 𝑛. It’s π‘₯ squared.

And what about these factorials? Well, we know that two 𝑛 plus two factorial is the same as two 𝑛 plus two times two 𝑛 plus one times two 𝑛 times two 𝑛 minus one and so on. Alternatively, that’s the same as two 𝑛 plus two times two 𝑛 plus one times two 𝑛 factorial. And that allows us to divide through by two 𝑛 factorial. When we do, we’re left with two 𝑛 plus two times two 𝑛 plus one on the denominator of our fraction. So we want the limit as 𝑛 approaches ∞ of the absolute value of this fraction to be less than one. We spot that negative one times π‘₯ squared is independent event. And so we can take the absolute value of negative π‘₯ squared outside of the limit. And we have the absolute value of negative π‘₯ squared times the limit as 𝑛 approaches ∞ of one over two 𝑛 plus two times two 𝑛 plus one.

We’re now ready to evaluate this limit. As 𝑛 grows larger, one over two 𝑛 plus two times two 𝑛 plus one approaches zero. And so, in fact, we’re looking to find the values of π‘₯ such that the absolute value of negative π‘₯ squared times zero is less than one. Well, when we multiply the absolute value of negative π‘₯ squared by zero, we’ll always get zero. And this means our Maclaurin series converges for all values of π‘₯. And we can therefore say that 𝑅 equals ∞ or positive ∞.

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