Tag: Polynomials

Use Bolzano’s theorem to isolate real roots of given polynomials

by RoRi

September 5, 2015

A real value $x$ is a root of a function $f$ if $f(x) = 0$ . We say we have isolated a real root if we find an interval $[a,b]$ such that $x \in [a,b]$ and no other real roots are in the interval. Use Bolzano’s Theorem to isolate the real roots of the following:

$3x^4 - 2x^3 - 36x^2+ 36x - 8 = 0$ .
$2x^4 - 14x^2 + 14x -1 = 0$ .
$x^4 + 4x^3 + x^2 - 6x + 2 = 0$ .

We have
$\begin{alignat*}{4} f(-4) &= 168, \quad &f(-3) &=& -143 &\qquad \implies \qquad & f(c_1) &=& 0 \quad \text{for} \quad c_1 \in [-4,-3].\\ f(0) &= -8, \quad &f(0.5) &=& \frac{15}{16} &\qquad \implies \qquad & f(c_2) &=& 0 \quad \text{for} \quad c_2 \in [0,0.5]\\ f(0.5) &= \frac{15}{16}, \quad &f(1) &=& -7 &\qquad \implies \qquad & f(c_3) &=& 0 \quad \text{for} \quad c_3 \in [0.5,1] \\ f(1) &= -7, \quad &f(4) &=& 200 &\qquad \implies \qquad & f(c_4) &=& 0 \quad \text{for} \quad c_4 \in [1,4] \end{alignat*}$
We have
$\begin{alignat*}{4} f(-4) &= 231, \quad &f(-3) &=& -7 &\qquad \implies \qquad & f(c_1) &=& 0 \quad \text{for} \quad c_1 \in [-4,-3].\\ f(0) &= -1, \quad &f(1) &=& 1 &\qquad \implies \qquad & f(c_2) &=& 0 \quad \text{for} \quad c_2 \in [0,1]\\ f(1) &= 1, \quad &f(1.5) &=& -\frac{11}{8} &\qquad \implies \qquad & f(c_3) &=& 0 \quad \text{for} \quad c_3 \in [1,1.5] \\ f(1.5) &= -\frac{11}{8}, \quad &f(2) &=& 3 &\qquad \implies \qquad & f(c_4) &=& 0 \quad \text{for} \quad c_4 \in [1.5,2] \end{alignat*}$
We have
$\begin{alignat*}{4} f(-3) &= 2, \quad &f(-2.5) &=& -\frac{3}{16} &\qquad \implies \qquad & f(c_1) &=& 0 \quad \text{for} \quad c_1 \in [-3,-2.5].\\ f(-2.5) &= -\frac{3}{16}, \quad &f(-2) &=& 2 &\qquad \implies \qquad & f(c_2) &=& 0 \quad \text{for} \quad c_2 \in [-2.5,-2]\\ f(0) &= 2, \quad &f(0.5) &=& -\frac{3}{16} &\qquad \implies \qquad & f(c_3) &=& 0 \quad \text{for} \quad c_3 \in [0,0.5] \\ f(0.5) &= -\frac{3}{16}, \quad &f(1) &=& 2 &\qquad \implies \qquad & f(c_4) &=& 0 \quad \text{for} \quad c_4 \in [0.5,1] \end{alignat*}$

Prove a polynomial with opposite signed first and last coefficients has at least one positive zero

by RoRi

September 4, 2015

Define a polynomial

$f(x) = \sum_{k=0}^n c_k x^k,$

such that $c_0$ and $c_n$ have opposite signs. Prove there is some $x > 0$ such that $f(x) = 0$ .

Proof. Since

$f(0) = \sum_{k=0}^n c_k 0^k = c_0$

we see that $f(0)$ has the same sign as $c_0$ .
Next,

$f(x) = c_n x^n + c_{n-1}x^{n-1} + \cdots + c_1 x + c_0 = c_n x^n \left( 1+ \frac{c_{n-1}}{c_n} \cdot \frac{1}{x} + \cdots + \frac{c_0}{c_n} \cdot \frac{1}{x^n} \right).$

Then we claim that for sufficiently large $x$ ,

$1 + \frac{c_{n-1}}{c_n} \frac{1}{x} + \cdots + \frac{c_0}{c_n} \frac{1}{x^n} > 0;$

hence, $f(x)$ will have the same sign as $c_n$ for sufficiently large $x$ , since $x > 0$ and so

$x^n \left( 1 + \frac{c_{n-1}}{c_n} \frac{1}{x} + \cdots + \frac{c_0}{c_n} \frac{1}{x^n} \right) > 0.$

(So, when we multiply a positive number by $c_n$ the result will have the same sign as $c_n$ .)

So, we need to show that the claimed term is indeed positive. First,

$\left( 1 + \frac{c_{n-1}}{c_n} \cdot \frac{1}{x} + \cdots + \frac{c_0}{c_n} \cdot \frac{1}{x^n} \right) \geq 1 - \left( \left| \frac{c_{n-1}}{c_n} \cdot \frac{1}{x} \right| + \cdots + \left| \frac{c_0}{c_n} \cdot \frac{1}{x^n} \right| \right).$

This is true since for each term in the sum

$\left| \frac{c_i}{c_n} \cdot \frac{1}{x^{n-i}} \right| \geq \frac{c_i}{c_n} \cdot \frac{1}{x^{n-i}}, \qquad \text{for all } 1 \leq i \leq n-1.$

Now, since we are showing there is sufficiently large $x$ such that our claim is true, we let

$x > \max \left\{ 1, \left| \frac{c_n}{n \cdot c_a} \right| \right\}, \qquad \text{where} \qquad |c_a| = \max\{ |c_0|, |c_1|, \ldots, |c_n|\}.$

Since $x > 1$ , we know $\frac{1}{x} > \frac{1}{x^n}$ for all $n \in \mathbb{Z}_{>0}$ , so

$\begin{align*} 1 - \left( \left| \frac{c_{n-1}}{c_n} \frac{1}{x} \right| + \cdots + \left| \frac{c_0}{c_n} \frac{1}{x^n} \right| \right) &> 1 - \left( \left| \frac{c_{n-1}}{c_n} \frac{1}{x} \right| + \cdots + \left| \frac{c_0}{c_n} \frac{1}{x} \right| \right) \\ &= 1 - \left( \left| \frac{c_{n-1}}{c_n} \frac{c_n}{n c_a} \right| + \cdots + \left| \frac{c_0}{c_n} \frac{c_n}{n c_a}\right| \right) \\ &= 1 - \left( \left| \frac{c_{n-1}}{n c_a} \right| + \cdots + \left| \frac{c_0}{nc_a} \right| \right) \\ &> 1 - \left( \left| \frac{c_a}{nc_a} \right| + \cdots + \left| \frac{c_a}{nc_a} \right| \right)\\ &= 1 - \left( \underbrace{\frac{1}{n} + \cdots + \frac{1}{n}}_{n-1\text{ terms}} \right) \\ &= 1 - \left( \frac{n-1}{n} \right) \\ &= \frac{1}{n} > 0. \end{align*}$

This proves our claim, and so $f(x)$ has the same sign as $c_n$ for sufficiently large $x$ . Hence, $f(0)$ and $f(x)$ have different signs (since $c_0$ and $c_n$ had different signs by assumption); thus, there is some $c>0$ such that $f(c) = 0. \qquad \blacksquare$

Compute the integral

by RoRi

August 21, 2015

Compute the following integral:

$\int_x^{x^2} (t^{1/2} + t^{1/4}) \, dt.$

We have

$\begin{align*} \int_x^{x^2} (t^{1/2} + t^{1/4} ) \, dt &= \left. \left( \frac{2t^{3/2}}{3} + \frac{4 t^{5/4}}{5} \right) \right|_x^{x^2} \\ &= \left( \frac{2x^3}{3} - \frac{4x^{5/2}}{5} \right) - \left( \frac{2x^{3/2}}{3} + \frac{4x^{5/4}}{5} \right) \\ &= \frac{2}{3} (x^3 - x^{3/2}) + \frac{4}{5} (x^{5/2} - x^{5/4}). \end{align*}$

Compute the integral

by RoRi

August 21, 2015

Compute the following integral:

$\int_1^x (t^{1/2} + 1)^2 \, dt.$

We have

$\begin{align*} \int_1^x (t^{1/2} + 1)^2 \, dt &= \left. \left( \frac{2t^{2/3}}{3} + t \right) \right|_1^x \\ &= \left( \frac{2x^{2/3}}{3} + x \right) - \left(1 + \frac{2}{3} \right) \\ &= \frac{2x^{2/3}}{3} + x - \frac{5}{3}. \end{align*}$

Compute the integral

by RoRi

August 20, 2015

Compute the following integral:

$\int_x^{x^2} (t^2+1)^2 \, dt.$

We have

$\begin{align*} \int_x^{x^2} (t^2+1)^2 \, dt &= \int_x^{x^2} (t^4 + 2t^2 + 1) \, dt \\ &= \left. \left( \frac{t^5}{5} + \frac{2t^3}{3} + t \right) \right|_x^{x^2} \\ &= \left( \frac{x^{10}}{5} + \frac{2x^6}{3} + x^2 \right) - \left( \frac{x^5}{5} + \frac{2x^3}{5} + x \right) \\ &= \frac{x^{10}}{5} + \frac{2x^6}{3} - \frac{x^5}{5} - \frac{x^3}{3} + x^2 - x. \end{align*}$

Compute the integral

by RoRi

August 20, 2015

Compute the following integral:

$\int_{-2}^x t^2(t^2+1)\, dt.$

We have

$\begin{align*} \int_{-2}^x t^2 (t^2+1) \, dt &= \int_{-2}^x (t^4 + t^2) \, dt \\ &= \left. \left( \frac{t^5}{5} + \frac{t^3}{3} \right) \right|_{-2}^x \\ &= \left( \frac{x^5}{5} + \frac{x^3}{3} \right) - \left( -\frac{32}{5} - \frac{8}{3} \right) \\ &= \frac{1}{5} x^5 + \frac{1}{3} x^3 + \frac{136}{15}. \end{align*}$

Compute the integral

by RoRi

August 20, 2015

Compute the following integral:

$\int_1^{1-x} (1-2t+3t^2) \, dt.$

We have

$\begin{align*} \int_1^{1-x} (1-2t+3t^2) \, dt &= (t - t^2 + t^3)\biggr \rvert_1^{1-x} \\ &= ((1-x) - (1-x)^2 + (1-x)^3) - 1 \\ &= (1-x - 1 + 2x - x^2 + 1 - 3x + 3x^2 - x^3) - 1 \\ &= -x^3 + 2x^2 -2x. \end{align*}$

Compute the integral

by RoRi

August 20, 2015

Compute the following integral:

$\int_{-1}^{2x} (1+t+t^2) \, dt.$

We have

$\begin{align*} \int_{-1}^{2x} (1+t+t^2) \, dt &= \left. \left( t + \frac{t^2}{2} + \frac{t^3}{3} \right) \right|_{-1}^{2x} \\ &= \left( 2x + 2x^2 + \frac{8}{3} x^3 \right) - \left(-1 + \frac{1}{2} -\frac{1}{3} \right) \\ &= 2x + 2x^2 + \frac{8}{3}x^3 + \frac{5}{6}. \end{align*}$