### Video Transcript

Find the Taylor polynomials of the fourth degree approximating the function π of π₯ equals cos of three π₯ at the point π equals π over three.

Letβs start this question by writing out the formula which approximates the function with Taylor polynomials. And for this formula, π is the point where we center the approximation. So for this question, π is going to be π over three. Weβve been asked to go up to the fourth degree. So, this is the general expression for Taylor polynomials up to the fourth degree. As weβre centering the approximation around π over three, letβs replace π with π over three. Now from here, thereβs quite a few things weβre going to need to find. Weβre going to need to find the function evaluated at π over three, the first derivative of the function evaluated at π over three, and also the second, third, and fourth derivatives all evaluated at π over three.

We know our function π of π₯ is cos of three π₯. But we also need to find the first, second, third, and fourth derivatives of π of π₯. To do this, we recall two general rules that weβre going to use. If π¦ equals sin of ππ₯, then dπ¦ by dπ₯ equals π cos of ππ₯. And if π¦ equals cos of ππ₯, then dπ¦ by dπ₯ equals negative π sin of ππ₯. So using the second rule, we have that the first derivative of cos of three π₯ equals negative three sin of three π₯. To obtain the second derivative of π of π₯, we differentiate the first derivative. So, we need to differentiate negative three sin of three π₯.

Using the first rule here, we find that this is negative three multiplied by three cos of three π₯, which is just negative nine cos of three π₯. We differentiate the second derivative to find the third derivative. This gives us negative nine multiplied by negative three sin of three π₯, which is just 27 sin of three π₯. And then, we differentiate once more to find the fourth derivative. This is 27 multiplied by three cos of three π₯, which is 81 cos of three π₯. But what we actually need for our formula is π and its derivatives evaluated at π over three.

So, letβs start with π of π over three, and this is going to be cos of three multiplied by π over three. But the threes cancel, so this is just cos of π, which we know to be negative one. And we do the same for the first derivative. We substitute π₯ for π over three, and then we can cancel the threes. And because sin of π is zero, this is just zero. And we do the same for the rest of the derivatives of π substituting π over three. And in fact, we can cancel the threes in all of these. Remember that cos of π is negative one. So, negative nine cos of π is going to be negative nine multiplied by negative one, which is just nine. Sin of π is zero, so 27 sin of π is going to give us zero. And finally, 81 cos of π is going to give us 81 multiplied by negative one, which is negative 81.

So, now Iβm going to clear some space so we can put what we just found into our formula. When we do this, we actually find that two of the terms disappear. And thatβs because we found the coefficient to be zero. And now, we can simplify our answer by working out these factorials. Remember that a factorial of a number is the product of that number and all the integers below it down to one. So, we work out two factorial to be two and four factorial to be 24. And weβre almost done with this question, but we can actually simplify 81 over 24 to its reduced form of 27 over eight. And that gives us our final answer.

So, to recap, we started by writing out the Taylor polynomial for a general function up to the degree we wanted for our question and we replaced π with the point we were given in the question. We then found the derivatives of π up to the fourth derivative and evaluated each one at our point. We then simplified to obtain our final answer.