# Applications of Calculus

The derivative of a polynomial
\( P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0 \) is given by
\[ P^{\prime}(x)=na_nx^{n-1}+(n-1)a_{n-1}x^{n-2}+\cdots+a_1.\] The inverse
operation, the indefinite integral, is given by
\[ \int P(x)dx=\frac{a_n}{n+1}x^{n+1}+\frac{a_{n-1}}nx^n+\cdots+
a_0x+C.\] If the polynomial \( P \) is not given by its coefficients but
rather by its canonical factorization, as
\( P(x)=(x-x_1)^{k_1}\cdots(x-x_n)^{k_n} \), a more suitable expression
for the derivative is obtained by using the logarithmic derivative
rule or product rule: \[ P^{\prime}(x)=P(x)\left(\frac{k_1}{x-x_1}+\cdots+
\frac{k_n}{x-x_n}\right).\] A similar formula can be obtained for
the second derivative.

** Problem 17 **

Suppose that real numbers \( 0=x_0< x_1< \cdots< x_n< x_{n+1}=1 \) satisfy
\[ \sum_{j=0,\,j\neq i}^{n+1}\frac{1}{x_i-x_j}=0\quad\mbox{for }
i=1,2,\dots,n.\quad\quad\quad\quad\quad(1)\] Prove that \( x_{n+1-i}=1-x_i \) for
\( i=1,2,\dots,n \).

Let \( P(x)=(x-x_0)(x-x_1)\cdots(x-x_n)(x-x_{n+1}) \). We have
\[ P^{\prime}(x)=\sum_{j=0}^{n+1}\frac{P(x)}{x-x_j}\quad\mbox{i}\quad
P^{\prime\prime}(x)=\sum_{j=0}^{n+1}\sum_{k\neq j}\frac{P(x)}
{(x-x_j)(x-x_k)}\;.\] Therefore \[ P^{\prime\prime}(x_i)=2P^{\prime}(x_i) \sum_{j\neq
i}\frac1{(x_i-x_j)}\] for \( i=0,1,\dots,n+1 \). Thus the condition of
the problem is equivalent to \( P^{\prime\prime}(x_i)=0 \) for \( i=1,2,\dots,n \).
Therefore \[ x(x-1)P^{\prime\prime}(x)=(n+2)(n+1)P(x).\] It is easy to see that
there is a unique monic polynomial of degree \( n+2 \) satisfying the
above differential equation. On the other hand, the monic polynomial
\( Q(x)=(-1)^nP(1-x) \) satisfies the same equation and has degree
\( n+2 \), so we must have \( (-1)^nP(1-x)=P(x) \), which implies the
statement.

What makes derivatives of polynomials especially suitable is their
property of preserving multiple zeros.

** Theorem 6.1 **

If \( (x-\alpha)^k\mid P(x) \), then \( (x-\alpha)^{k-1}\mid
P^{\prime}(x) \).

If \( P(x)=(x-\alpha)^kQ(x) \), then
\( P^{\prime}(x)=(x-\alpha)^kQ^{\prime}(x)+k(x-\alpha)^{k-1}Q(x) \)

** Problem 18 **

Determine a real polynomial \( P(x) \) of degree at most 5 which leaves
remainders \( -1 \) and 1 upon division by \( (x-1)^3 \) and \( (x+1)^3 \),
respectively.

If \( P(x)+1 \) has a triple zero at point 1, then its derivative
\( P^{\prime}(x) \) has a double zero at that point. Similarly, \( P^{\prime}(x) \) has a
double zero at point \( -1 \) too. It follows that \( P^{\prime}(x) \) is divisible
by the polynomial \( (x-1)^2(x+1)^2 \). Since \( P^{\prime}(x) \) is of degree at
most 4, it follows that \[ P^{\prime}(x)=c(x-1)^2(x+1)^2=c(x^4-2x^2+1)\] for
some constant \( c \). Now \( P(x)=c(\frac15x^5-\frac23x^3+x)+d \) for some
real numbers \( c \) and \( d \). The conditions \( P(-1)=1 \) and \( P(1)=-1 \) now
give us \( c=-15/8 \), \( d=0 \) and \[ P(x)=-\frac38x^5+\frac54x^3-
\frac{15}8x. \]

** Problem 19 **

For polynomials \( P(x) \) and \( Q(x) \) and an arbitrary \( k\in\mathbb{C} \),
denote \[ P_k=\{z\in\mathbb{C}\mid P(z)=k\}\quad\mbox{and}\quad
Q_k=\{z\in\mathbb{C}\mid Q(z)=k\}.\] Prove that \( P_0=Q_0 \) and
\( P_1=Q_1 \) imply that \( P(x)=Q(x) \).

Let us assume w.l.o.g. that \( n=\mathrm{deg}P\geq\mathrm{deg}Q \). Let
\( P_0=\{z_1,z_2,\dots,z_k\} \) and \( P_1 \) \( =\{z_{k+1} \), \( z_{k+2},\dots,
z_{k+m}\} \). Polynomials \( P \) and \( Q \) coincide at \( k+m \) points
\( z_1,z_2,\dots,z_{k+m} \). The result will follow if we show that
\( k+m> n \).

We have \[ P(x)=(x-z_1)^{\alpha_1}\cdots
(x-z_k)^{\alpha_k}=(x-z_{k+1})^{\alpha_{k+1}}\cdots
(x-z_{k+m})^{\alpha_{k+m}}+1\] for some natural numbers
\( \alpha_1,\dots,\alpha_{k+m} \). Let us consider \( P^{\prime}(x) \). We know that
it is divisible by \( (x-z_i)^{\alpha_i-1} \) for \( i=1,2,\dots,k+m \);
hence, \[ \prod_{i=1}^{k+m}(x-z_i)^{\alpha_i-1}\mid P^{\prime}(x).\]
Therefore, \( 2n-k-m=\deg\prod_{i=1}^{k+m}(x-z_i)^{\alpha_i-1}\leq
\mathrm{deg}P^{\prime}=n-1 \), i.e. \( k+m\geq n+1 \), as desired.

Even if \( P \) has no multiple zeros, certain relations between zeros
of \( P \) and \( P^{\prime} \) still hold. For example, the following statement
holds for all differentiable functions.

** Theorem 6.2 (Rolle’s Theorem) **

Between every two zeros of a
polynomial \( P(x) \) there is a zero of \( P^{\prime}(x) \).

Let \( a< b \) be two zeros of polynomial \( P \). Assume
w.l.o.g. that \( P^{\prime}(a)> 0 \) and consider the point \( c \) in the interval
\( [a,b] \) in which \( P \) attains a local maximum (such a point exists
since the interval \( [a,b] \) is compact). We know that
\( P(x)=P(c)+(x-c)[P^{\prime}(c)+o(1)] \). If for example \( P^{\prime}(c)> 0 \) (the case
\( P^{\prime}(c)< 0 \) leads to a similar contradiction), then \( P(x)> P(c) \) would
hold in a small neighborhood of \( c \), a contradiction. It is only
possible that \( P^{\prime}(c)=0 \), so \( c \) is a root of \( P^{\prime}(x) \) between \( a \) and
\( b \).

** Corollary **

If all zeros of \( P(x) \) are real, then so are all
zeros of \( P^{\prime}(x) \).