Evaluate the limit.
Since we use the expansion (page 287) for ,
Therefore,
Evaluate the limit.
Since we use the expansion (page 287) for ,
Therefore,
Evaluate the limit.
We know (p. 287) the following expansions as ,
(Note that these are the same expansion when we use only the first order terms. This tells us that and behave similarly near 0. We would need to take higher order terms to differentiate between the two. For instance, if we wanted to include cubic terms we would have , but .) From here we compute the limit,
Evaluate the limit.
To evaluate this we use the trig identity to simplify
We have proved the limit earlier (at least once), but let’s do it again using the techniques of this section and -notation.
when .
Therefore, when we have
where
using the cubic Taylor polynomial approximation to .
prove that the number from part (a) satisfies
Determine if is positive or negative and prove the result.
So, to approximate the nonzero root of we have
Using the values for and given we have
is approximated by .
given that . Determine whether is positive or negative and prove the result.
This implies
Therefore, we can approximate the nonzero root by
So, for , and using the given inequality , we have
Furthermore, since
with the absolute value of each term in the sum strictly less than the absolute value of the previous term (since and ). Thus, each pair is positive, so the whole series is positive
Prove that the error term in the Taylor expansion of satisfies the following inequality.
Proof. To prove this we will work directly from the definition of the error as an integral,
We know for we have, (we need for the expansion of to be valid),
Therefore we have
So, we can bound the error term by bounding the integral,
Prove that the error of the Taylor expansion of satisfies the following inequality.
Proof. Since the derivatives of are always , or we know that for we have . (In other words, the st derivative is bounded above by 1 and below by .) Therefore, we can apply Theorem 7.7 (p. 280 of Apostol) to estimate the error in Taylor’s formula at with and . For this gives us
Next, (from the second part of Theorem 7.7) if we have
Show that
Using the trig identity suggested in the hint, we have
We know from Example 3 (page 275 of Apostol) that
Therefore, by Theorem 7.3 (the substitution property, page 276) we know
Then, using the linearity of the Taylor operator (Theorem 7.2) we have
Given constants such that , prove that
for some constants .
Proof. Define constants and by
(Since we know , so these definitions make sense.) Then
Therefore, we may write,
So, to evaluate the integral we have
For the integral on the right, we make the substitution , so . Therefore,