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# Prove properties of polynomials of a complex variable with real coefficients

Consider the polynomial of a complex variable with real coefficients.

1. Prove that

for all .

2. Using part (a) show that if has any nonreal zeros, they must occur in pairs of a complex number and its conjugate.

1. Proof. Let

Then, using the properties of conjugation we have,

(The final line follows since and then induction for to get for all .) This completes the proof

2. Proof. If is a non-real zero of then

and (since is not real by assumption). Hence, is also a zero of , so the non-real zeros come in pairs

# Derive some properties of the product of ex with a polynomial

Let

1. Prove that

where denotes the th derivative of .

2. Do part (a) in the case that is a cubic polynomial.
3. Find a similar formula and prove it in the case that is a polynomial of degree .

For all of these we recall from a previous exercise (Section 5.11, Exercise #4) that by Leibniz’s formula if then the th derivative is given by

So, in the case at hand we have and so

(Since the th derivative of is still for all and .)

1. Proof. From the formula above we have

But, since is a quadratic polynomial we have

Hence, we have

2. If is a cubic polynomial we may write,

Claim: If then

Proof. We follow the exact same procedure as part (a) except now we have the derivatives of given by

Therefore, we now have

3. Claim: Let be a polynomial of degree ,

Let . Then,

Proof. Using Leibniz’s formula again, we have

But for the degree polynomial , we know if and for all . Hence, we have

# Prove properties of the Bernoulli polynomials

The Bernoulli polynomials are defined by

1. Find explicit formulas for the first Bernoulli polynomials in the cases .
2. Use mathematical induction to prove that is a degree polynomial in , where the degree term is .
3. For prove that .
4. For prove that

5. Prove that

for .

6. Prove that for ,

7. Prove that for ,

Now, using the integral condition to find ,

Thus,

Next, using this expression for we have

Using the integral condition to find ,

Thus,

Next, using this expression for we have

Using the integral condition to find ,

Thus,

Next, using this expression for we have

Using the integral condition to find ,

Thus,

Finally, using this expression for we have

Using the integral condition to find ,

Thus,

2. Proof. We have shown in part (a) that this statement is true for . Assume then that the statement is true for some positive integer , i.e.,

Then, by the definition of the Bernoulli polynomials we have,

where for . Then, taking the integral of this expression

Hence, the statement is true for the case ; hence, for all positive integers

3. Proof. From the integral property in the definition of the Bernoulli polynomials we know for ,

Then, using the first part of the definition we have ; therefore,

Thus, we indeed have

4. Proof. The proof is by induction. For the case we have

Therefore,

Since , the stated difference equation holds for . Assume then that the statement holds for some positive integer . Then by the fundamental theorem of calculus, we have

Therefore,

Hence, the statement is true for the case , and so it is true for all positive integers

5. Proof. (Let’s assume Apostol means for to be some positive integer.) First, we use the definition of the Bernoulli polynomials to compute the integral,

Now, we want to express the numerator as a telescoping sum and use part (d),

Thus, we indeed have

6. Proof.

Incomplete. I’ll try to fix parts (f) and (g) soon(ish).

# Find a polynomial satisfying given conditions

1. Find a polynomial satisfying

Prove that there is only one such polynomial.

2. Given a polynomial , prove there is exactly one polynomial such that

1. Proof. (Finding the polynomial will prove that it is unique since we will not have any choices to make while deriving the polynomial .) First, we write

Thus, we have

Setting this equal to we have

But, this implies and since is the only term on the left (so if , then we couldn’t have the largest power of on the right). Therefore, is a degree polynomial and with , so we have

Hence we have

Thus we have the equations

These uniquely determine and ,

Hence, there is a unique satisfying this equation,

2. Proof. Let be a given polynomial and suppose there exist two polynomials and such that

This implies

Now, if then it is of degree for some . We know its derivative has degree (Apostol, Page 166). But then, this would imply

has degree (since the coefficient of in is zero since it is degree , and the coefficient of is nonzero since it has degree ). But we know this difference is 0, which means it cannot have degree for any . Thus, we must have or

# Find a quintic polynomial meeting given conditions

Find a polynomial of degree satisfying the following conditions:

Since must be a polynomial of degree we may write

where any of the may be 0 (since we could have a polynomial of degree strictly less than 5). First, let’s apply the condition to obtain

Now, let’s take the first two derivatives since we have conditions on and .

We can then apply the conditions and to obtain

So now we have and and so

Now we need to use the other three conditions

(If you know some linear algebra feel free to solve this in a more efficient way.) From the first equation we have

Plugging this into the second equation we have

Now plugging in our expressions for and into the third equation we have

Then using our expressions for and we have

Now, we have computed all of the constants so we can write down the formula for the polynomial

# Compute the derivatives of g(x) = xn f(x)

Assume is a polynomial with . Define

Compute the values of .

Assume is a non-negative integer (otherwise is undefined at ). Then, we make the following claim:

Claim: The polynomial has derivatives at 0 given by the following

Proof. Since is a polynomial we may write,

Furthermore, since is given we know . Now, multiplying by we have

Next, we will use induction to prove that the th derivative of for is given by

for constants . Since the derivative is given by

for constants , we see that the formula holds for . Assume then that it holds for some ,

Then, taking the derivative of this we have,

Hence, the formula holds for all . But then, if we have

If then for all ; hence,

# Prove there is no polynomial with derivative 1/x

Prove that there is no polynomial such that

Proof. We know from Example 1 of Section 4.5 in Apostol (p. 166) that every polynomial is differentiable everywhere on . (In that example we show that the derivative of a polynomial is a polynomial, and we know that polynomials are defined everywhere on .) However, the function is not defined for . Hence, this function cannot be the derivative of a polynomial

# Prove properties about the zeros of a polynomial and its derivatives

Consider a polynomial . We say a number is a zero of multiplicity if

where .

1. Prove that if the polynomial has zeros in , then its derivative has at least zeros in . More generally, prove that the th derivative, has at least zeros in the interval.
2. Assume the th derivative has exactly zeros in the interval . What can we say about the number of zeros of in the interval?

1. Proof. Let denote the distinct zeros of in and their multiplicities, respectively. Thus, the total number of zeros is given by,

By the definition given in the problem, if is a zero of of multiplicity then

Taking the derivative (using the product rule), we have

Thus, again using the definition given in the problem, is a zero of of multiplicity .
Next, we know from the mean-value theorem for derivatives, that for distinct zeros and of there exists a number (assuming, without loss of generality, that ) such that . Hence, if has distinct zeros, then the mean value theorem guarantees numbers such that . Thus, has at least:

By induction then, the th derivative has at least zeros.

2. If the th derivative has exactly zeros in , then we can conclude that has at most zeros in .

# Determine a polynomial based on values of its derivatives

Let such that

Find the values of the constants .

First, let’s compute the first and second derivatives,

Now, we start determining values based on the conditions given,

Next, using this value we have , so,

Then, plugging these values into the expression for we have . So,

and,

Therefore,

# Prove there is exactly one negative solution to an equation

Show there is exactly one such that for and an odd, positive integer.

Proof. Let , and let with . Then, (for odd ) so . Since and , by the Intermediate Value Theorem, we know takes every value between and 0 for some . Thus, we know there exists such that (since ). This implies for some .

We know this solution is unique since is strictly increasing on the whole real line for odd