Video Transcript
Find the error bound when using the
third Taylor polynomial for the function 𝑓 of 𝑥 is equal to the natural logarithm
of three 𝑥 at 𝑥 is equal to one-third to approximate the value of 𝑓 evaluated at
one-half. Give your answer in scientific form
to three significant figures.
The question gives us a function 𝑓
of 𝑥, and it wants us to determine the error bound if we were to approximate this
function by using the third Taylor polynomial. We want to center our Taylor
polynomial at one-third. And we’re going to use this to
approximate the value of 𝑓 evaluated at one-half. Finally, we need to give our answer
in scientific form to three significant figures.
To start, recall if we can
approximate 𝑓 of 𝑥 by using an 𝑛-term Taylor polynomial, we can also add on a
remainder term to help us see the difference between our function 𝑓 of 𝑥 and our
approximation 𝑇 𝑛 of 𝑥. And we can actually find a bound on
this remainder term. We need to recall the
following. For the 𝑛th Taylor polynomial
centered at a value of 𝑎 where we want to approximate at a value of 𝑥, we have the
absolute value of 𝑅 𝑛 of 𝑥 is less than or equal to the absolute value of 𝑀
times 𝑥 minus 𝑎 all raised to the power of 𝑛 plus one divided by 𝑛 plus one
factorial. And this value of 𝑀 will be an
upper bound on the absolute value of the 𝑛 plus oneth derivative of 𝑓 of 𝑥 on an
interval containing both 𝑎 and 𝑥.
And it’s worth pointing out there
are a few pieces of information we’re missing about this, for example, how many
times can we differentiate 𝑓 of 𝑥 and are these derivatives continuous on the
interval containing 𝑥 and 𝑎. These are definitely worth thinking
about. However, usually problems of this
will arise when we try and find our value of 𝑀 or we try and construct our Taylor
polynomial. So we’ll ignore this and move on to
find our bound. This is a very complicated-looking
expression. However, it can be made easier by
substituting in our values of 𝑎 and 𝑥.
First, we’re told to center our
Taylor polynomial at 𝑥 is equal to one-third. So we’ll set our value of 𝑎 equal
to one-third and update all of our bounds. Next, we see we’re using this to
estimate the value of 𝑓 at one-half. So we can set our value of 𝑥 equal
to one-half. And once again we can update our
bound to include the value of 𝑥 at one-half. We can also update our equation for
𝑓 of 𝑥 involving the Taylor polynomial. However, this is not necessary. Finally, the question tells us to
use the third Taylor polynomial. So we’ll set our value of 𝑛 equal
to three. And once again, we can update all
of our bounds with the value of 𝑛 set to be three.
The question is asking us to find a
bound on our error, which is the following inequality. In other words, all we need to do
is calculate the absolute value of 𝑀 times one-half minus one-third all raised to
the power of three plus one divided by three plus one factorial. And the only part of this we don’t
know is the value of 𝑀. So all we need to do is find our
value of 𝑀. And we know that 𝑀 is an upper
bound on the 𝑛 plus oneth derivative of 𝑥 on an interval.
In this case, since 𝑛 is three,
this will be the fourth derivative of 𝑓 of 𝑥 with respect to 𝑥. We know this needs to be an
interval containing both 𝑎 and 𝑥. So we need an interval containing
one-third and one-half. When we’re only using this on
individual values — for example, in this case, we’re only using this to approximate
𝑓 evaluated at one-half — we can just choose the closed interval between 𝑎 and
𝑥. In other words, we can just choose
the closed interval from one-third to one-half.
We’re now almost ready to find our
value of 𝑀. We just need to find an expression
for the fourth derivative of 𝑓 of 𝑥 with respect to 𝑥. To do this, we just need to
differentiate 𝑓 of 𝑥 four times. First, we need to find 𝑓 prime of
𝑥. That’s the derivative of the
natural logarithm of three 𝑥 with respect to 𝑥.
There’s a few different ways of
doing this. One way is to use the product rule
for logarithms to rewrite 𝑓 of 𝑥 as the natural algorithm of three plus the
natural logarithm of 𝑥. Then, the natural logarithm of
three is a constant. So its derivative is just going to
be equal to zero. And we know the derivative of the
natural algorithm of 𝑥 with respect to 𝑥 is the reciprocal function, one over
𝑥. So 𝑓 prime of 𝑥 is one over
𝑥. Because we’re going to
differentiate this again, we’ll write this as 𝑥 to the power of negative one.
We can now find all of our
remaining derivatives by using the power rule for differentiation. We want to multiply by our exponent
of 𝑥 and then reduce this exponent by one. We get 𝑓 double prime of 𝑥 will
be equal to negative one times 𝑥 to the power of negative two. Doing the same again, we get 𝑓
triple prime of 𝑥 will be equal to two times 𝑥 to the power of negative three. And doing this one more time, we
get the fourth derivative of 𝑓 of 𝑥 with respect to 𝑥 is equal to negative six
times 𝑥 to the power of negative four. And we’ll use our laws of exponents
to rewrite this as negative six divided by 𝑥 to the fourth power. The reason we do this is because
we’re going to want to find an upper bound for this expression. And it’s easier to do this in this
form.
First, it’s important to realize we
don’t want an upper bound on the fourth derivative of 𝑓 of 𝑥. We want an upper bound on the sides
of 𝑓 of 𝑥. So we need to take the absolute
value of this expression, so we’ll just take the absolute value of this. This gives us the absolute value of
the fourth derivative of 𝑓 of 𝑥 with respect to 𝑥 is equal to the absolute value
of negative six divided by 𝑥 to the fourth power.
And we can simplify this. First, in our denominator, 𝑥 to
the fourth power is equal to 𝑥 squared all squared. And we know this is greater than or
equal to zero for all values of 𝑥. So taking the absolute value of our
denominator is not going to change its value. This means all we need to do is
take the absolute value of our numerator. And of course, we know the absolute
value of negative six is just equal to six. So we can just simplify this
expression to get six divided by 𝑥 to the fourth power.
Now we’re ready to find our value
of 𝑀. We want to find an upper bound for
this expression on the closed interval from one-third to one-half. We can see on this interval, we
have a positive number divided by another positive number. We want to make this as big as
possible. So to make this number as big as
possible, we want to divide by the smallest number we can. The smallest number on our interval
is when 𝑥 is equal to one-third. Therefore, on the closed interval
from one-third to one-half, the absolute value of the fourth derivative of 𝑓 of 𝑥
with respect to 𝑥 will be less than or equal to six divided by one-third raised to
the fourth power.
All we’re really saying here is
we’re dividing by the smallest positive number we can on our interval, and this will
be our value of 𝑀. And we can just calculate this
expression. Six divided by one-third raised to
the fourth power is equal to 486. Now, all we need to do is
substitute this into our error bound. Substituting 𝑀 is equal to 486, 𝑎
is equal to one-third, 𝑥 is equal to one-half, and 𝑛 is equal to three into our
error bound formula, we get the absolute value of 𝑅 three will be less than or
equal to the absolute value of 486 times one-half minus one-third all raised to the
power of three plus one divided by three plus one factorial. And it’s worth pointing out since
we’re only interested in the error at our point, we will simplify our remainder
polynomial to just be 𝑅 three.
Now, all that’s left to do is
evaluate this expression. First, one-half minus one-third is
one-sixth, three plus one is equal to four, and we know that four factorial is equal
to 24. Then, if we just calculate this
expression, we get one divided by 64. But remember, the question doesn’t
want us to give this as an exact number. It wants us to give this in
scientific form to three significant figures. To do this, we’ll start by writing
this out as a decimal expansion. It’s equal to 0.015625. Next, we want this to three
significant figures. We need to remember what this
means. First, the initial zeros will never
count. So in this case, our three
significant figures will be the one, five, and six.
Next, we need to check the first
emitted digit to see if we need to round up. In this case, it’s two, so we don’t
need to round up. It’s less than five. So to three significant figures,
the absolute value of 𝑅 three is approximately 0.0156. Finally, we need to write this in
scientific notation. We start with a number between
negative 10 and 10 and multiply this by 10 to the power of some integer. In this case, by looking at our
three significant figures, we can see we need to start with 1.56. Then, we can see that 0.0156 is
just equal to 1.56 times 10 to the power of negative two, which is our final
answer.
Therefore, we were able to find the
error bound when using the third Taylor polynomial for the function 𝑓 of 𝑥 is
equal to the natural logarithm of three 𝑥 at 𝑥 is equal to one-third to
approximate the value of 𝑓 of one-half. In scientific form to three
significant figures, we got 1.56 times 10 to the power of negative two.
