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 error 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 a Maclaurin series,
and a Maclaurin polynomial.
Let's begin with a brief review.
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 and Taylor series,
and center them at zero, we have a Maclaurin polynomial,
and Maclaurin series, which is our case for this problem.
So, because we are trying to find
c sub zero through c sub four, we'll have to find
the first four derivatives of our function, and then,
use this formula here to find the series,
and then, the degree four polynomial.
Again, this is the formula for the polynomial.
The formula for the 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 one 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.
And 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 our 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 could 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 six tangent squared x,
times the derivative of tangent x,
which is secant squared x,
and then, we have minus, the derivative of two tangent x,
which would be two secant squared x.
And now, we 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 one, 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 negative one,
and the fourth derivative, f-triple prime of zero
would be equal to negative tangent cubed zero,
which is zero, minus two tangent zero, which is also zero,
and then, finally,
the fourth derivative at zero is equal to,
well since tangent zero is zero, this product would be zero,
but secant zero is one, so the fourth derivative at zero
is equal to negative two times one, or negative two.
So now, we'll take the value of these derivatives,
and form the Taylor series and Taylor polynomial,
and let's do this on the next slide.
Natural log, cosine x
is equal to f of zero, which is zero,
plus f-prime of zero times x, that would be zero times x,
plus f-double prime of zero, which is negative one,
so let's write this as minus
one, divided by two factorial,
times x squared, then we'd have plus,
the third derivative evaluated at zero, which is zero,
divided by three factorial times x to the third,
plus the fourth derivative evaluated at zero,
which is negative two, so let's write minus two,
divided by four factorial times x to the fourth, and so on.
So, this would be the Taylor polynomial centered at zero,
or if we want, the Maclaurin polynomial.
Let's go ahead and simplify this.
We have natural log
of cosine x
equals, well, this should be zero, this should be zero.
Two factorial is two, so have negative 1/2 x squared.
This would be zero.
Two over four factorial would be 2/24, or 1/12,
so we have minus 1/12, x to the fourth, and so on.
So, this would be the Taylor series, centered at zero,
or the Maclaurin series.
Going back to the question,
let's go ahead and find c sub zero through c sub four.
C sub zero would be the coefficient
of the degree zero term, or the constant term.
Notice how we don't have a constant term,
so c sub zero is zero.
C sub one is also zero because we don't
have a linear term, or a degree one term.
C sub two would be negative 1/2.
There's no degree three term, so c sub three is zero,
and c sub four is the coefficient of the degree four term,
so c sub four 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 it'll be fourth degree Taylor polynomial
is equal to negative 1/2, x squared,
minus 1/12, x to the fourth.
So, the true function value f of x,
is equal to the Taylor polynomial approximation,
plus the error, 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 0.2,
minus degree four Taylor polynomial approximation.
So, R sub four of 0.2 is equal to
natural log, cosine 0.2,
minus your Taylor polynomial approximation, which would be
negative 1/2 times 0.2 squared,
minus 1/12, 0.2 to the fourth.
Let's first verify we are in radian mode,
so we'll press the mode key.
Radian is highlighted, so I'll go back to the home screen,
and we'll enter a natural log
of cosine 0.2
and then, minus
the quantity, negative 1/2, or negative .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 us scientific notation.
This would be negative 1.43972
times 10, raised to the power of negative six.
Of course, this was rounded.
Let's change this to an approximation symbol.
If we do have to convert this to a decimal,
we'd have to move the decimal point to the left six times.
So, we could also write the approximate value as
zero, point, we have five zeroes,
one, two, three, four, five,
one, four, three, nine, seven, two,
and this is negative.
So, the Taylor polynomial approximation
is a very good approximation for the exact value
of natural log, cosine 0.2.
If 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 zero,
the red function is a very good representation
of the original function, again, graphed in blue.
I hope you found this helpful.