Tag: Polynomials

Prove properties of polynomials of a complex variable with real coefficients

by RoRi

February 25, 2016

Consider the polynomial $f(z)$ of a complex variable with real coefficients.

Prove that
$\overline{f(z)} = f (\overline{z})$

for all $z \in \mathbb{C}$ .
Using part (a) show that if $f$ has any nonreal zeros, they must occur in pairs of a complex number and its conjugate.

Proof. Let
$f(z) = a_0 + a_1 z + a_2 z^2 + \cdots + a_n z^n \qquad a_i \in \mathbb{R}.$

Then, using the properties of conjugation we have,

$\begin{align*} \overline{f(z)} &= \overline{a_0 + a_1 z + \cdots + a_n z^n} \\ &= \overline{a_0} + \overline{a_1 z} + \cdots + \overline{a_n z^n} &(\text{since } \overline{z_1 + z_2} = \overline{z_1} + \overline{z_2}) \\ &= \overline{a_0} + \overline{a_1} \, \overline{z} + \cdots + \overline{a_n} \, \overline{z^n} &(\text{since } \overline{z_1 z_2} = \overline{z_1} \overline{z_2})\\ &= a_0 + a_1 \overline{z} + a_2 \overline{z^2} + \cdots + a_n \overline{z_n} &(\text{since } \overline{a} = a \text{ for } a \in \mathbb{R}) \\ &= a_0 + a_1 \overline{z} + a_2 (\overline{z})^2 + \cdots + a_n (\overline{z})^n \\ &= f(\overline{z}). \end{align*}$

(The final line follows since $\overline{z^2} = \overline{z \cdot z} = \overline{z} \cdot \overline{z} = (\overline{z})^2$ and then induction for to get $\overline{z^n} = (\overline{z})^n$ for all $n$ .) This completes the proof $. \qquad \blacksquare$
Proof. If $z$ is a non-real zero of $f$ then
$f(z) = 0 \quad \implies \quad \overline{f(z)} = 0 \quad \implies \quad f(\overline{z}) = 0,$

and $z \neq \overline{z}$ (since $z$ is not real by assumption). Hence, $\overline{z}$ is also a zero of $f$ , so the non-real zeros come in pairs $. \qquad \blacksquare$

Derive some properties of the product of e^x with a polynomial

by RoRi

December 19, 2015

Let

$f(x) = e^x p(x) \qquad \text{where} \qquad p(x) = c_0 + c_1 x + c_2x^2.$

Prove that
$f^{(n)} (0) = c_0 + nc_1 + n(n-1)c_2$

where $f^{(n)}$ denotes the $n$ th derivative of $f$ .
Do part (a) in the case that $p(x)$ is a cubic polynomial.
Find a similar formula and prove it in the case that $p(x)$ is a polynomial of degree $m$ .

For all of these we recall from a previous exercise (Section 5.11, Exercise #4) that by Leibniz’s formula if $h(x) = f(x)g(x)$ then the $n$ th derivative $h^{(n)}(x)$ is given by

$h^{(n)} (x) = \sum_{k=0}^n \binom{n}{k} f^{(k)} (x) g^{(n-k)}(x).$

So, in the case at hand we have $f(x) = e^x p(x)$ and so

$f^{(n)} (x) = \sum_{k=0}^n \binom{n}{k} p^{(k)}(x) e^x.$

(Since the $(n-k)$ th derivative of $e^x$ is still $e^x$ for all $n$ and $k$ .)

Proof. From the formula above we have
$f^{(n)}(0) = \sum_{k=0}^n \binom{n}{k} p^{(k)}(0) e^0 = \sum_{k=0}^n \binom{n}{k} p^{(k)}(0).$

But, since $p(x)$ is a quadratic polynomial we have

$\begin{align*} p(x) &= c_0 + c_1 x + c_2 x^2 \\[9pt] p'(x) &= c_1 + 2c_2 x \\[9pt] p''(x) &= 2c_2 \\[9pt] p^{(k)} (x) &= 0 & \text{for all } k \geq 3. \end{align*}$

Hence, we have

$\begin{align*} f^{(n)}(0) &= \binom{n}{0} p(0) + \binom{n}{1} p'(0) + \binom{n}{2} p''(0) \\[9pt] &= c_0 + n c_1 + \frac{n(n-1)}{2} 2c_2 \\[9pt] &= c_0 + n c_1 + n(n-1) c_2. \qquad \blacksquare \end{align*}$
If $p(x)$ is a cubic polynomial we may write,
$p(x) = c_0 + c_1 x + c_2 x^2 + c_3 x^3.$

Claim: If $f(x) = e^x p(x)$ then

$f^{(n)}(0) = c_0 + nc_1 + n(n-1)c_2 + n(n-1)(n-2)c_3.$

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

$\begin{align*} p(x) &= c_0 + c_1 x + c_2x^2 + c_3x^3 \\[9pt] p'(x) &= c_1 + 2c_2 x + 3c_3 x^2 \\[9pt] p''(x) &= 2c_2 + 6c_3 x \\[9pt] p^{(3)}(x) &= 6c_3 \\[9pt] p^{(k)}(x) &= 0 & \text{for all } k \geq 4. \end{align*}$

Therefore, we now have

$\begin{align*} f^{(n)}(0) &= \sum_{k=0}^n \binom{n}{k} p^{(k)}(0) e^0 \\[9pt] &= \sum_{k=0}^n \binom{n}{k} p^{(k)}(0) \\[9pt] &= \binom{n}{0} p(0) + \binom{n}{1} p'(0) + \binom{n}{2} p''(0) + \binom{n}{3} p'''(0) \\[9pt] &= c_0 + nc_1 + \frac{n(n-1)}{2} 2c_2 + \frac{n(n-1)(n-2)}{6} 6c_3 \\[9pt] &= c_0 + nc_1 + n(n-1)c_2 + n(n-1)(n-2)c_3. \qquad \blacksquare \end{align*}$
Claim: Let $p(x)$ be a polynomial of degree $m$ ,
$p(x) = \sum_{k=0}^m c_k x^k.$

Let $f(x) = e^x p(x)$ . Then,

$f^{(n)}(0) = \sum_{k=0}^m k! \binom{n}{k} c_k.$

Proof. Using Leibniz’s formula again, we have

$f^{(n)}(0) = \sum_{k=0}^m \binom{n}{k} p^{(k)}(0).$

But for the degree $m$ polynomial $p(x)$ , we know $p^{(k)}(0) = k!c_k$ if $0 \leq k \leq m$ and $p^{(k)} (0) = 0$ for all $k > m$ . Hence, we have

$f^{(n)}(0) = \sum_{k=0}^m k! \binom{n}{k} c_k. \qquad \blacksquare$

Prove properties of the Bernoulli polynomials

by RoRi

October 31, 2015

The Bernoulli polynomials are defined by

$P_0(x) = 1; \qquad P'_n(x) = n P_{n-1} (x) \qquad \text{and} \qquad \int_0^1 P_n (x) \, dx = 0 \qquad \text{if } n \geq 1.$

Find explicit formulas for the first Bernoulli polynomials in the cases $n = 1,2,3,4,5$ .
Use mathematical induction to prove that $P_n(x)$ is a degree $n$ polynomial in $x$ , where the degree $n$ term is $x^n$ .
For $n \geq 2$ prove that $P_n (0) = P_n (1)$ .
For $n \geq 1$ prove that
$P_n (x+1) - P_n (x) = nx^{n-1}.$
Prove that
$\sum_{r=1}^{k-1} r^n = \int_0^k P_n (x) \, dx = \frac{P_{n+1}(k) - P_{n+1}(0)}{n+1}$

for $n \geq 2$ .
Prove that for $n \geq 1$ ,
$P_n (1-x) = (-1)^n P_n (x).$
Prove that for $n \geq 1$ ,
$P_{2n+1} (0) = 0 \qquad \text{and} \qquad P_{2n-1}\left( \frac{1}{2} \right) = 0.$

We start with the initial condition $P_0(x) = 1$ . This gives us
$P'_1 (x) = 1 \cdot P_0(x) = 1 \quad \implies \quad P_1 (x) = \int dx = x + C_1.$

Now, using the integral condition to find $C_1$ ,

$\int_0^1 (x+C_1) \, dx = 0 \quad \implies \quad \left( \frac{1}{2}x^2 + C_1 x \right) \Bigr \rvert_0^1 = 0 \quad \implies \quad C_1 &= -\frac{1}{2}.$

Thus,

$P_1 (x) = x -\frac{1}{2}.$

Next, using this expression for $P_1(x)$ we have

$P'_2 (x) = 2 \cdot \left( x- \frac{1}{2} \right) = 2x - 1 \quad \implies \quad P_2 (x) = x^2 - x + C_2.$

Using the integral condition to find $C_2$ ,

$\int_0^1 (x^2 - x + C_2) \, dx = 0 \quad \implies \quad \frac{1}{3} - \frac{1}{2} + C_2 = 0 \quad \implies \quad C_2 = \frac{1}{6}.$

Thus,

$P_2 (x) = x^2 - x + \frac{1}{6}.$

Next, using this expression for $P_2 (x)$ we have

$P'_3 (x) = 3 \cdot \left( x^2 - x + \frac{1}{6} \right) = 3x^2 - 3x + \frac{1}{2} \quad \implies \quad P_3(x) = x^3 - \frac{3}{2}x^2 + \frac{1}{2}x + C_3.$

Using the integral condition to find $C_3$ ,

$\int_0^1 \left( x^3 - \frac{3}{2}x^2 + \frac{1}{2}x + C_3\right) \, dx = 0 \quad \implies \quad \frac{1}{4} - \frac{1}{2} + \frac{1}{4} + C_3 = 0 \quad \implies \quad C_3 = 0.$

Thus,

$P_3 (x) = x^3 - \frac{3}{2}x^2 + \frac{1}{2} x.$

Next, using this expression for $P_3 (x)$ we have

$P'_4(x) = 4 \cdot \left( x^3 - \frac{3}{2}x^2 + \frac{1}{2}x \right) = 4x^3 - 6x^2 + 2x \quad \implies \quad P_4 (x) = x^4 - 2x^3 + x^2 + C_4.$

Using the integral condition to find $C_4$ ,

$\int_0^1 \left( x^4 - 2x^3 + x^2 + C_4 \right) \, dx = 0 \quad \implies \quad \frac{1}{5} - \frac{1}{2} + \frac{1}{3} + C_4 = 0 \quad \implies \quad C_4 = \frac{1}{30}.$

Thus,

$P_4 (x) = x^4 - 2x^3 + x^2 - \frac{1}{30}.$

Finally, using this expression for $P_4 (x)$ we have

$P'_5(x) = 5 \cdot \left( x^4 - 2x^3 + x^2 - \frac{1}{30} \right) = 5x^4 - 10x^3 + 5x^2 - \frac{1}{6} \quad \implies \quad P_5 (x) = x^5 - \frac{5}{2} x^4 + \frac{5}{3}x^3 - \frac{1}{6}x + C_5.$

Using the integral condition to find $C_5$ ,

$\int_0^1 \left( x^5 - \frac{5}{2} x^4 + \frac{5}[3} x^3 - \frac{1}{6} x + C_5 \right) \, dx = \frac{1}{6} - \frac{1}{2} + \frac{5}{12} - \frac{1}{12} + C_5 = 0 \quad \implies \quad C_5 = 0.$

Thus,

$P_5 (x) = x^5 - \frac{5}{2}x^4 + \frac{5}{3}x^3 - \frac{1}{6}x.$
Proof. We have shown in part (a) that this statement is true for $n = 0, 1, \ldots, 5$ . Assume then that the statement is true for some positive integer $m$ , i.e.,
$P_m (x) = x^m + \sum_{k=0}^{m-1} c_k x^k.$

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

$P_{m+1}' (x) = (m+1) \cdot \left( x^m + \sum_{k=0}^{m-1} a_k x^k \right) = (m+1)x^m + \sum_{k=0}^{m-1} b_k x^k,$

where $b_k = (m+1)a_k$ for $k = 1, \ldots, m-1$ . Then, taking the integral of this expression

$P_{m+1} (x) = \int \left( (m+1)x^m + \sum_{k=0}^{m-1} b_k x^k \right) \, dx = x^{m+1} + \sum_{k=0}^m \frac{b_k}{k+1} x^{k+1}.$

Hence, the statement is true for the case $m+1$ ; hence, for all positive integers $n. \qquad \blacksquare$
Proof. From the integral property in the definition of the Bernoulli polynomials we know for $n \geq 1$ ,
$\int_0^1 P_n (x) \, dx = 0 \quad \implies \quad \int_0^1 (n+1) P_n (x) \, dx = 0.$

Then, using the first part of the definition we have $P'_{n+1} (x) = (n+1) P_n (x)$ ; therefore,

$0 = \int_0^1 (n+1) P_n (x) \, dx = \int_0^1 P'_{n+1} (x) \, dx = P_{n+1}(1) - P_{n+1}(0).$

Thus, we indeed have

$P_{n+1} (1) = P_{n+1}(0). \qquad \blacksquare$
Proof. The proof is by induction. For the case $n = 1$ we have
$P_1 (x) = x - \frac{1}{2}, \qquad \text{and} \qquad P_1 (x+1) = x + \frac{1}{2}.$

Therefore,

$P_1 (x+1) - P_1 (x) = 1.$

Since $nx^{n-1} = 1 \cdot x^0 = 1$ , the stated difference equation holds for $n =1$ . Assume then that the statement holds for some positive integer $m$ . Then by the fundamental theorem of calculus, we have

$P_{m+1} (x) = \int_0^x P'_{m+1}(t) \, dt = (m+1) \int_0^x P_m (t) \, dt.$

Therefore,

$\begin{align*} P_{m+1}(x+1) - P_{m+1}(x) &= (m+1) \left( \int_0^{x+1} P_m (t) \, dt - \int_0^x P_m (t) \, dt \right) \\[9pt] &= (m+1) \left( \int_0^1 P_m (t) \, dt + \int_1^{x+1} P_m (t) \, dt - \int_0^x P_m (t) \, dt \right) \\[9pt] &= (m+1) \left( 0 + \int_0^x P_m (t+1) \, dt - \int_0^x P_m (t) \, dt \right) &( \text{Integral condition})\\[9pt] &= (m+1) \left( \int_0^x (P_m (t+1) - P_m (t)) \, dt\right) \\[9pt] &= (m+1) \left( \int_0^x mt^{m-1} \, dt \right)&(\text{Induction Hypothesis})\\[9pt] &= (m+1) x^m. \end{align*}$

Hence, the statement is true for the case $m+1$ , and so it is true for all positive integers $n. \qquad \blacksquare$
Proof. (Let’s assume Apostol means for $k$ to be some positive integer.) First, we use the definition of the Bernoulli polynomials to compute the integral,
$\begin{align*} \int_0^k P_n (x) \, dx &= \int_0^k \frac{1}{n+1} P'_{n+1} (x) \, dx \\[9pt] &= \frac{P_{n+1}(k) - P_{n+1}(0)}{n+1}. \end{align*}$

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

$\begin{align*} P_{n+1}(k) - P_{n+1}(0) &= \sum_{r = 1}^{k-1} \left( P_{n+1}(r + 1) - P_{n+1}(r) \right) \\[9pt] &= \sum_{r=0}^{k-1} \left( (n+1)r^n \right) \\ &= (n+1) \sum_{r=1}^{k-1} r^n. \end{align*}$

Thus, we indeed have

$\sum_{r=1}^{k-1} r^n = \int_0^k P_n (x) \, dx = \frac{P_{n+1}(k) - P_{n+1}(0)}{n+1}. \qquad \blacksquare$
Proof.

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

Find a polynomial satisfying given conditions

by RoRi

October 31, 2015

Find a polynomial satisfying
$P'(x) - 3P(x) = 4 - 5x + 3x^2.$

Prove that there is only one such polynomial.
Given a polynomial $Q(x)$ , prove there is exactly one polynomial $P(x)$ such that
$P'(x) - 3P(x) = Q(x).$

Proof. (Finding the polynomial will prove that it is unique since we will not have any choices to make while deriving the polynomial $P(x)$ .) First, we write
$\begin{align*} P(x) = \sum_{k=0}^n c_k x^k && \implies && P'(x) &= \sum_{k=1}^n c_k (k) x^{k-1} \\ &&&&& = \sum_{k=0}^{n-1} (k+1)c_{k+1} x^k. \end{align*}$

Thus, we have

$P'(x) - 3P(x) = \sum_{k=0}^{n-1} (k+1)c_{k+1} x^k - \sum_{k=0}^n 3c_k x^k = \sum_{k=0}^{n-1} ((k+1)c_{k+1} - 3c_k)x^k - 3c_n x^n.$

Setting this equal to $4 - 5x + 3x^2$ we have

$\sum_{k=0}^{n-1} ((k+1)c_{k+1} - 3c_k)x^k - 3c_n x^n = 4 - 5x + 3x^2.$

But, this implies $n = 2$ and $c_2 = -1$ since $-3c_n x^n$ is the only $x^n$ term on the left (so if $n > 2$ , then we couldn’t have $x^2$ the largest power of $x$ on the right). Therefore, $P(x)$ is a degree polynomial and with $c_2 = -1$ , so we have

$P(x) = c_0 + c_1 x - x^2 \quad \implies \quad P'(x) = c_1 - 2x.$

Hence we have

$\begin{align*} P'(x) - 3P(x) = 4 - 5x + 3x^2 && \implies && c_1 - 2x - 3c_0 - 3c_1 x + 3x^2 &= 4 - 5x + 3x^2 \\ && \implies && (c_1 - 3c_0) - (2+3c_1)x &= 4 - 5x. \end{align*}$

Thus we have the equations

$c_1 - 3c_0 = 4 \quad \text{and} \quad 2+3c_1 = 5.$

These uniquely determine $c_0$ and $c_1$ ,

$c_1 = 1,\quad c_0 = -1.$

Hence, there is a unique $P(x)$ satisfying this equation,

$P(x) = -1 + x - x^2. \qquad \blacksquare$
Proof. Let $Q(x)$ be a given polynomial and suppose there exist two polynomials $P_1 (x)$ and $P_2(x)$ such that
$P'(x) - 3P(x) = Q(x) \quad \text{and} \quad R'(x) - 3R(x) = Q(x).$

This implies

$(P'(x) - R'(x)) - 3(P(x) - R(x)) = 0 \quad \implies \quad (P(x) - R(x))' - 3(P(x) - R(x)) = 0.$

Now, if $P(x) - R(x) \neq 0$ then it is of degree $n$ for some $n \geq 1$ . We know its derivative has degree $n-1$ (Apostol, Page 166). But then, this would imply

$(P(x) - R(x))' - 3(P(x) - R(x))$

has degree $n$ (since the coefficient of $x^n$ in $(P(x) - R(x))'$ is zero since it is degree $n-1$ , and the coefficient of $3(P(x) - R(x))$ is nonzero since it has degree $n$ ). But we know this difference is 0, which means it cannot have degree $n$ for any $n \geq 1$ . Thus, we must have $P(x) - R(x) = 0$ or $P(x) = R(x). \qquad \blacksquare$

Find a quintic polynomial meeting given conditions

by RoRi

October 24, 2015

Find a polynomial $P$ of degree $\leq 5$ satisfying the following conditions:

$\begin{align*} P(0) &= 1, \\ P(1) &= 2, \\ P'(0) &= P''(0) = P'(1) = P''(1) = 0. \end{align*}$

Since $P$ must be a polynomial of degree $\leq 5$ we may write

$P(x) = a_5 x^5 + a_4 x^4 + a_3 x^3 + a_2 x^2 + a_1 x + a_0$

where any of the $a_i$ may be 0 (since we could have $P$ a polynomial of degree strictly less than 5). First, let’s apply the condition $P(0) = 1$ to obtain

$P(0) = a_0 = 1.$

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

$\begin{align*} P(x) &= a_5 x^5 + a_4 x^4 + a_3 x^3 + a_2 x^2 + a_1 x + 1 \\ P'(x) &= 5a_5 x^4 + 4a_4 x^3 + 3a_3 x^2 + 2a_2 x + a_1 \\ P''(x) &= 20a_5 x^3 + 12a_4 x^2 + 6a_3 x + 2a_2. \end{align*}$

We can then apply the conditions $P'(0) = 0$ and $P''(0) = 0$ to obtain

$\begin{align*} P'(0) &= a_1 = 0 \\ P''(0) &= 2a_2 = 0. \end{align*}$

So now we have $a_0 = 1$ and $a_1 = a_2 = 0$ and so

$P(x) = a_5 x^5 + a_4 x^4 + a_3 x^3 + 1.$

Now we need to use the other three conditions

$\begin{align*} P(1) &= 2 & \implies && a_5 + a_4 + a_3 &= 1 \\ P'(1) &= 0 & \implies && 5a_5 + 4a_4 + 3a_3 &= 0 \\ P''(1) &= 0 & \implies && 20a_5 + 12a_4 + 6a_3 &= 0. \end{align*}$

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

$a_3 = 1 - a_4 - a_5.$

Plugging this into the second equation we have

$5a_5 + 4a_4 + 3(1 - a_4 - a_5) = 0 \quad \implies \quad 2a_5 + a_4 + 3 = 0 \quad \implies \quad a_4 = -3-2a_5.$

Now plugging in our expressions for $a_3$ and $a_4$ into the third equation we have

$20a_5 + 12(-3-2a_5) + 6(1-(-3-2a_5)-a_5) = 0 \quad \implies \quad a_5 = 6.$

Then using our expressions for $a_3$ and $a_4$ we have

$\begin{align*} a_4 &= -15 \\ a_3 &= 10. \end{align*}$

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

$P(x) = 6x^5 - 15x^4 + 10x^3 + 1.$

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

by RoRi

October 24, 2015

Assume $f$ is a polynomial with $f(0) = 1$ . Define

$g(x) = x^n f(x).$

Compute the values of $g(0), g'(0), \ldots, g^{(n)}(0)$ .

Assume $n$ is a non-negative integer (otherwise $g$ is undefined at $x = 0$ ). Then, we make the following claim:

Claim: The polynomial $g(x)$ has derivatives at 0 given by the following

$g^{(k)}(0) = \begin{cases} 0 & \text{if } 0 \leq k < n \\ n! & \text{if } k =n. \end{cases}$

Proof. Since $f$ is a polynomial we may write,

$f(x) = a_m x^m + a_{m-1} x^{m-1} + \cdots + a_1 x + a_0.$

Furthermore, since $f(0) = 1$ is given we know $a_0 = 1$ . Now, multiplying by $x^n$ we have

$\begin{align*} g(x) &= x^n f(x) = x^n (a_m x^m + \cdots + a_1 x + 1) \\ &= a_m x^{m+n} + \cdots + a_1 x^{n+1} + x^n. \end{align*}$

Next, we will use induction to prove that the $k$ th derivative of $g(x)$ for $0 \leq k < n$ is given by

$g^{(k)}(x) = c_m x^{m+n-k} + c_{m-1} x^{m+n-k-1} + \cdots + c_1 x^{n+1-k} + \frac{n!}{\cdot (n-k)!} x^{n-k},$

for constants $c_1, \ldots, c_m$ . Since the derivative $g'(x)$ is given by

$\begin{align*} g'(x) &= (m+n) a_m x^{m+n-1} + (m+n-1) a_{m-1} x^{m+n-2} + \cdots + (n+1) a_1 x^n + n x^{n-1} \\ &= b_m x^{m+n-1} + b_{m-1} x^{m+n-2} + \cdots + b_1 x^n + \frac{n!}{(n-1)!}x^{n-1}, \end{align*}$

for constants $b_1, \ldots, b_m$ , we see that the formula holds for $k = 1$ . Assume then that it holds for some $k$ ,

$g^{(k)}(x) = b_m x^{m+n-k} + b_{m-1} x^{m+n-k-1} + \cdots + b_1 x^{n+1-k} + \frac{n!}{(n-k)!}x^{n-k}.$

Then, taking the derivative of this we have,

$\begin{align*} g^{(k+1)}(x) &= b_m (m+n-k) x^{m+n-k-1} + b_{m-1} (m+n-k-1) x^{m+n-k-2} + \cdots + b_1 (n+1-k) x^{n+1-k-1} + \frac{n!}{(n-k)!} (n-k) x^{n-k-1)} \\ &= c_m x^{m+n-(k+1)} + c_{m-1} x^{m+n-(k+1)-1} + \cdots + c_1 x^{n+1-(k+1)} + \frac{n!}{(n-(k+1))!} x^{n-(k+1)}. \end{align*}$

Hence, the formula holds for all $0 \leq k \leq n$ . But then, if $0 \leq k < n$ we have

$\begin{align*} &&g^{(k)}(x) &= b_m x^{m+n-k} + b_{m-1} x^{m+n-k-1} + \cdots + b_1 x^{n+1-k} + \frac{n!}{(n-k)!}x^{n-k} \\ \implies g^{(k)}(0) &= 0. \end{align*}$

If $k =n$ then $x^{n-k} = x^0 = 1$ for all $x$ ; hence,

$g^{(n)}(0) &= \frac{n!}{0!} x^0 = n!. \qquad \blacksquare$

Prove there is no polynomial with derivative 1/x

by RoRi

October 14, 2015

Prove that there is no polynomial $f$ such that

$f'(x) = \frac{1}{x}.$

Proof. We know from Example 1 of Section 4.5 in Apostol (p. 166) that every polynomial is differentiable everywhere on $\mathbb{R}$ . (In that example we show that the derivative of a polynomial is a polynomial, and we know that polynomials are defined everywhere on $\mathbb{R}$ .) However, the function $\frac{1}{x}$ is not defined for $x = 0$ . Hence, this function cannot be the derivative of a polynomial $. \qquad \blacksquare$

Prove properties about the zeros of a polynomial and its derivatives

by RoRi

September 28, 2015

Consider a polynomial $f$ . We say a number $\alpha$ is a zero of multiplicity $m$ if

$f(x) = (x - \alpha)^m g(x),$

where $g(\alpha) \neq 0$ .

Prove that if the polynomial $f$ has $r$ zeros in $[a,b]$ , then its derivative $f'$ has at least $r-1$ zeros in $[a,b]$ . More generally, prove that the $k$ th derivative, $f^{(k)}$ has at least $r-k$ zeros in the interval.
Assume the $k$ th derivative $f^{(k)}$ has exactly $r$ zeros in the interval $[a,b]$ . What can we say about the number of zeros of $f$ in the interval?

Proof. Let $\alpha_1, \ldots, \alpha_k$ denote the $k$ distinct zeros of $f$ in $[a,b]$ and $m_1, \ldots, m_k$ their multiplicities, respectively. Thus, the total number of zeros is given by,
$r = \sum_{i=1}^k m_i.$

By the definition given in the problem, if $\alpha_i$ is a zero of $f$ of multiplicity $m_i$ then

$f(x) = (x - \alpha_i)^{m_i} g(x) \qquad \text{where } g(x) \neq 0.$

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

$\begin{align*} f'(x) &= m_i (x - \alpha_i)^{m_i - 1} g(x) + (x- \alpha_i)^{m_i} g'(x)\\ &= (x - \alpha_i)^{m_i - 1} (m_i g(x) + (x - \alpha_i) g'(x)) \end{align*}$

Thus, again using the definition given in the problem, $\alpha_i$ is a zero of $f'(x)$ of multiplicity $m_i -1$ .
Next, we know from the mean-value theorem for derivatives, that for distinct zeros $\alpha_i$ and $\alpha_j$ of $f$ there exists a number $c \in [\alpha_i, \alpha_j]$ (assuming, without loss of generality, that $\alpha_i < \alpha_j$ ) such that $f'(c) = 0$ . Hence, if $f$ has $k$ distinct zeros, then the mean value theorem guarantees $k-1$ numbers $c$ such that $f'(c) = 0$ . Thus, $f'$ has at least:

$\left(\sum_{i=1}^k (m_i -1)\right) + k - 1 = \left( \sum_{i=1}^k m_i \right) - 1 = r-1 \ \ \text{zeros}.$

By induction then, the $k$ th derivative $f^{(k)}(x)$ has at least $r-k$ zeros.
If the $k$ th derivative $f^{(k)}$ has exactly $r$ zeros in $[a,b]$ , then we can conclude that $f$ has at most $r+k$ zeros in $[a,b]$ .

Determine a polynomial based on values of its derivatives

by RoRi

September 18, 2015

Let $P(x) = ax^3 + bx^2 + cx + d$ such that

$P(0) = P(1) = -2, \qquad P'(0) = -1, \qquad P''(0) = 10.$

Find the values of the constants $a,b,c,d$ .

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

$\begin{align*} && P(x) &= ax^3 + bx^2 + cx + d \\ \implies && P'(x) &= 3ax^2 + 2bx + c \\ \implies && P''(x) &= 6ax + 2b. \end{align*}$

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

$P''(0) = 10 \quad \implies \quad 2b = 10 \quad \implies \quad b = 5.$

Next, using this value we have $P'(x) = 3ax^2 + 10x + c$ , so,

$P'(0) = -1 \quad \implies \quad c = -1.$

Then, plugging these values into the expression for $P$ we have $P(x) = ax^3 +5x^2 -x + d$ . So,

$P(0) = -2 \quad \implies \quad d = -2,$

and,

$P(1) = -2 \quad \implies \quad a + 5 - 1 -2 = -2 \quad \implies \quad a = -4.$

Therefore,

$a = -4, \quad b = 5, \quad c = -1, \quad d = -2.$

Prove there is exactly one negative solution to an equation

by RoRi

September 5, 2015

Show there is exactly one $b < 0$ such that $b^n = a$ for $a < 0$ and $n$ an odd, positive integer.

Proof. Let $f(x) = x^n$ , and let $c < -1$ with $c < a < 0$ . Then, $c^n < c$ (for odd $n$ ) so $c^n < c < a < 0$ . Since $f(0) = 0$ and $f(c) = c^n$ , by the Intermediate Value Theorem, we know $f(x)$ takes every value between $c^n$ and 0 for some $x \in [c,0]$ . Thus, we know there exists $b \in [c,0]$ such that $f(b) = a$ (since $c^n < a < 0$ ). This implies $b^n = a$ for some $b \in [c,0]$ .

We know this solution is unique since $f$ is strictly increasing on the whole real line for odd $n. \qquad \blacksquare$