We want to find the first few coefficients of the Taylor series for f of X equals natural log cosine, X centered at zero. And then also find the air in approximating natural log cosine 0.2 by its fourth degree, Taylor polynomial, well, first because the Taylor series and the Taylor polynomial are centered at zero. We could also call this MacLaurin series and a MacLaurin polynomial let's begin with a brief review, a Taylor polynomial and Taylor series are defined as we see here. Both of these are.

Centered at C, the only difference here is a Taylor series is an infinite series. And a Taylor polynomial is formed by using a finite number of terms of the Taylor series. And now if we take the Taylor polynomial in Taylor series and Center them as zero, we have a MacLaurin polynomial and MacLaurin series, which is our case for this problem. So because we're trying to find C sub 0 through C sub, 4 we'll have to find the first four derivatives of our function. And then use this formula here to find the.

Series and in the degree four polynomial again, this is the formula for the polynomial. The formula for this series would be an infinite series so beginning with our function f of X equals natural log cosine. X we'll begin by determining the first four derivatives. So f, prime of X is equal to the derivative of natural log cosine X, which would be 1 over cosine X times the derivative of cosine X, which would be negative sine X. Notice how this would be negative sine x over cosine X, which is negative. Tangent X.

And now the second derivative would be the derivative of negative tangent X, which would be negative secant squared X in the third derivative would be the derivative of negative secant squared X. So we'd have negative two times secant to the first x times, the derivative of secant X, which is secant X, tangent X, which we could write as negative. Two secant squared X times, tangent, X, let's, perform a substitution here for secant squared X. So we have the third derivative in terms of one trig. Function, we could write this as negative two times secant squared X equals tan squared X, plus one, and we still have times tangent X. So the third derivative can be written in the form of we'd have negative.

Two tangent squared X times, tangent, X, that's negative, two tangent, cubed x and then minus two times tangent X. So minus two tangent X. And now for the fourth derivative will be the derivative of negative. Two tangent, cubed x, minus two, tangent X. So here, we'll, multiply by three that's negative, 6, Tangent squared X times the derivative of tangent X, which is secant squared X. Then we have minus the derivative of two tangent X, which would be two secant squared X. And now we'll evaluate each of these functions at zero since our Taylor polynomial is centered at zero.

So f of zero would be equal to natural log cosine zero or natural. Log 1, which is zero f. Prime of zero would be equal to negative tangent zero, which is zero f, double prime of zero is equal to negative secant squared zero, which would. Be a negative one and the fourth derivative F, triple prime of zero would be equal to negative tangent, cubes, 0, which is 0 minus 2, tangent 0, which is also and finally, the fourth derivative at zero is equal to well since tangent zero is zero. This product would be 0, but secant 0 is 1.

So the fourth derivative at 0 is equal to negative 2 times 1 or negative 2. So now we'll take the value of these derivatives and form the Taylor series in Taylor polynomial and let's do this on the next slide, natural log. Cosine X is equal to f of 0, which is 0 plus F prime of 0 times X that would be 0 times X, plus F double prime of 0, which is negative 1. So let's write this as minus 1 divided by 2 factorial times x squared. And we'd have plus the third derivative evaluated at 0, which is 0 divided by 3 factorial times X to the third plus the fourth derivative evaluated at 0, which is negative 2. So let's write -2, divided by 4 factorial times X to the fourth and so on.

So this would be the Taylor polynomial centered at.0, or if we want them to flora in polynomial, let's, go ahead and simplify this. We have natural log of cosine x equals this would be 0. This would be 0 2. Factorial is 2. So we have negative 1/2 x squared.

This would be 0 2 over 4 factorial. It will be 2 twenty-fourths or 112. So we have minus 1 12 X to the fourth and so on.

So this would be the Taylor series centered at 0 or the MacLaurin series going back to the question let's, go ahead and find C sub 0 through C sub, 4 c. Sub 0 would be the coefficient of. The degree 0 terms of the constant term, notice that we don't have a constant term. So C sub 0 is 0 C. Sub 1 is also 0 because we don't have a linear term or degree 1 term, C sub 2, it negatives 1/2, there's, no degree three-term.

So C sub 3 is 0 and C sub. 4 is the coefficient of the degree 4 terms. So C sub, 4 is negative 1 12. And now we'll find the exact error in approximating natural log cosine 0.2 by forming. The fourth degree Taylor polynomial using these terms of our Taylor series centered at zero. So. The fourth degree Taylor polynomial is formed by using just these first two terms of the Taylor series, because notice how the second term is degree four.

So the fourth degree Taylor polynomial is equal to negative one half x squared minus one, twelve X to the fourth. So the true function value. F of X is equal to the Taylor polynomial approximation, plus the heir. So if we solve this equation for R sub, N, or in our case, R sub, four, we'd have R sub. Four of 0.2 equals the true function value, f of zero. Point two minus degree, 4 Taylor, polynomial approximation. So R sub, four of zero point two is equal to natural log cosine 0.2, minus the Taylor polynomial approximation, which would be negative one half times 0.2 squared minus one.

Twelve zero point two to the fourth let's. First verify, we are in Radian mode. We'll press, the mode key Radian is highlighted. So now we'll go back to the home screen, and we'll enter a natural log of cosine. Zero point two and then minus the quantity negative 1/2 or negative.0.5 times 0.2 squared, and then we'll have minus 1/12 times 0.2 raised to the fourth and enter notice how the calculator is giving a scientific notation. This would be negative 1 point, 4 3, 9 7 2 times 10 raised to the power of negative 6.

Of course, this was rounded those changes to an approximation symbol. But do when I convert this to a decimal we have to move the decimal point to the left 6 times. So it could also at the approximate value as 0 point. You know, five zeros 1 2, 3, 4 5, 1, 4 3, 9 7 2 and. This is negative. So the Taylor polynomial approximation is a very good approximation for the exact value of natural log cosine 0.2.

We take a look at the graph of the original function, it's graphed here in blue. And our Taylor polynomial is graphed here in red. And notice how around x equals 0, the red function is a very good representation of the original function again, graphed in blue, I hope you found this helpful.