Consider the function . Determine the first five coefficients () in the Taylor series expansion of at .
First, we take some derivatives,
Then, since the coefficients in the Taylor series at are of the form we can compute as follows:
Consider the function . Determine the first five coefficients () in the Taylor series expansion of at .
First, we take some derivatives,
Then, since the coefficients in the Taylor series at are of the form we can compute as follows:
Consider the function
and .
where is a polynomial in .
by Theorem 7.11 (page 301 of Apostol) since implies as well
So, indeed the formula is valid in the case . Assume then that the formula holds for some positive integer . We want to show this implies the formula holds for the case .
But then the leading term
is still a polynomial in since the derivative of a polynomial in is still a polynomial in , and so is the sum of two polynomials in . Therefore, we have that the formula holds for the case ; hence, it holds for all positive integers
So, indeed and the statement is true for the case . Assume then that for some positive integer . Then, we use the limit definition of the derivative again to compute the derivative ,
This follow since is still a polynomial in , and by the definition of for . But then, by part (a) we know
Therefore,
Thus, the formula holds for the case , and hence, for all positive integers
Consider a function that satisfies the limit relation
and has a continuous third derivative everywhere. Determine the values
We do some simplification to the expression first.
Now, we apply the hint (that if then ),
So as we have
But then since has three derivatives at 0 we know its Taylor expansion at 0 is unique and is given by (Theorem 7.1, page 274 of Apostol)
Hence, equating the coefficients of like powers of we have
Next, to compute the limit
we write
Then, using the Taylor expansion of (page 287 of Apostol) we know as we have
Therefore we have
Evaluate the limit.
We have
Evaluate the limit.
First, we rewrite the expression using the definition of the exponential:
Next, we need to get a series expansion for as . The most straightforward way is to take the first few derivatives (we’ll only need the term).
Therefore, we have as
Hence, as
Since this is going to 1 as we may apply this exercise (Section 7.11, Exercise #4) to conclude
Now, we have the expansion as
Therefore,
Evaluate the limit.
First, we write the expression using the definition of the exponential,
Now, considering the expression in the exponent and using the expansion of as (page 287 of Apostol) we have as ,
Therefore, we have
Evalue the limit.
First, we have
As we use the expansion for (page 287 of Apostol) to write,
From this we see that as ; and so,
Hence, using the expansion for as (page 287 of Apostol) we have
Therefore, we have
Evaluate the limit.
From the definition of the exponential we have
So, first we use the expansion of as (page 287 of Apostol) to write
Therefore, as we have
Now, since as we can use the expansion (again, page 287) of as to write
Therefore, as we have
So, getting back to the expression we started with,
But, as in the previous exercise (Section 7.11, Exercise #23) we know . Hence,
Evaluate the limit.
First, we write
From this exercise (Section 7.11, Exercise #4) we know that as we have
Therefore, as ,
So, we then have
(Here we could say that since the exponential is a continuous function we can bring the limit inside and so this becomes . I’m not sure we know we can pass limits through continuous functions like that, so we continue on with expanding the exponential as in previous exercises.)
Since as we take the expansion of as ,
Therefore,