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# Determine the coefficients in the Taylor series of (2 + x2)5/2

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:

# Prove some properties of the function e-1/x2

Consider the function

and .

1. Prove that for every positive number we have

2. Prove that if then

where is a polynomial in .

3. Prove that

1. Proof. (A specific case of this general theorem is actually the first problem of this section, here. Maybe it’s worth taking a look since this proof is just generalizing that particular case.) We make the substitution , so that as and we have

by Theorem 7.11 (page 301 of Apostol) since implies as well

2. Proof. The proof is by induction on . In the case we have

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

3. Proof. The proof is by induction on . If then we use the limit definition of the derivative to compute the derivative at 0,

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

# Compute derivatives of a function given a limit equation that it satisfies

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

# Find the limit as x goes to 1 of (1 / log x – 1 / (x-1))

Evaluate the limit.

We use this exercise (Section 7.11, Exercise #4) to evaluate,

# Find the limit as x goes to 0 of ((1/x) – (1/(ex – 1)))

Evaluate the limit.

We have

# Find the limit as x goes to 0 of (arcsin x / x)1/x2

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,

# Find the limit as x to 0 of the given expression

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

# Find the limit as x goes to 0 of ((1+x)1/x – e) / x

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

# Find the limit as x goes to 0 of (x + e2x)1/x

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,

# Find the limit as x goes to 1 of x1 / (1-x)

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,