Tag Info

Hot answers tagged

16

For convenience let me set $k = 2n$. We can represent the binomial coefficient by $$ \binom{x}{2n} = \frac{1}{(2n)!}\prod_{j=0}^{2n-1} (x-j), $$ so to make this an even polynomial we will replace $x$ by $y+n-\tfrac{1}{2}$. This will make the roots of the equation $$ \binom{y+n-\tfrac{1}{2}}{2n} = c \tag{0} $$ symmetric about the real and imaginary axes ...


7

Not quite, I recommend going about this problem a different way. $$z^6=-i\\ \implies z^6 =e^{i\frac{(4k+3)\pi}{2}}\\ \implies z=\left(e^{i\frac{(4k+3)\pi}{2}}\right)^{\frac{1}{6}} \\ \implies z= e^{i\frac{(4k+3)\pi}{12}}$$ where we have six different solutions for $k \in \{0,1,2,3,4,5 \}$ which allows you to conclude that $$z \in ...


7

$p\left(\frac{1}{x}\right) = \frac{1}{x^4}p(x)$. So unless there are roots on the unit circle (which is not ruled out in the problem as stated), there are two inside and two outside the unit circle.


7

If a polynomial has complex roots, then in the real domain it will have less roots than would otherwise be "expected". When I say expected, we normally would think that a polynomial of degree $n$ would have $n$ roots, counting multiplicities, but this is only the case over an algebraically closed field (like $\mathbb C$, not $\mathbb R$). If you have $m$ ...


6

As rogerl said, by symmetry the number of roots inside the unit circle is equal to the number outside. Now, how many are on the unit circle? Let's suppose $1$ is not a root, i.e. $2 + 2 b + c \ne 0$. The Möbius transformation $ w = i(1+z)/(1-z)$ ($z = (w-i)/(w+i)$) takes the unit circle (except for the point $1$) to the real line, and $p(z) = 0$ becomes ...


5

First Part This is true for $n=1$. Suppose this is true for $n-1$. Since $\lambda_1e^{a_1x}\ne0$ we can divide by that to get $$ 1+\frac{\lambda_2}{\lambda_1}e^{(a_2-a_1)x}+\frac{\lambda_3}{\lambda_1}e^{(a_3-a_1)x}+\dots+\frac{\lambda_n}{\lambda_1}e^{(a_n-a_1)x} $$ taking the derivative gives $$ ...


5

A complex root of a polynomial can have some significance itself when the roots of the polynomial have significance in general. One example that comes to mind where the roots of polynomials have a meaningful interpretation is in the field of dynamical systems. Consider a matrix differential equation $$ X' = AX, $$ where $A$ is a constant real $2\times2$ ...


5

Hint: $f'(x)=3x^2\ge 0$ for all $x$. Note that the function is increasing and $f(0)<0$ and $f(2)>0$, what can you conclude using continuity of $f$? Side-note the roots of the equation are $2^{1/3}\cdot \omega ^i$ where $i=0,1,2$ and $\omega$ is the cube root of unity. i.e. there is only one real root.


4

$$A=\sqrt[3]3+\sqrt[3]2\implies A^3=3+2+3\sqrt[3]6(A)\implies A^3-5=3A\sqrt[3]6$$ Now cube both sides


3

Depends on the value of $a$. For instance, if $a \geq 1 $, then there is no solution. Why? well, $e^{ax} - x = 0 \iff e^{ax} = x $. Hence, your equation is asking where this two functions intersect. IF $a \geq 1 $, then $$ e^{ax} \geq ax + 1 > ax > x $$ Hence, there is no solution. Now, there is indeed a solution if $a \leq \frac{1}{e} $. To see ...


3

In this answer I showed that the polynomial $$ p(x) = x^4 + ax^3 + bx^2 + ax + 1 $$ has no real zeros if $|a| \leq 4$ and $b > 2 |a| - 2$. We have $$ 60x^4-44x^3-25x^2-44x+60 = 60 \left(x^4 - \frac{44}{60}x^3 - \frac{25}{60}x^2 - \frac{44}{60}x+1\right), $$ so in this case $a = -44/60$ and $b = -25/60$. Indeed, $|a| \leq 4$ and $$ b = ...


3

Use polynomial division by $(x-r)^2$ to write $$P(x)=(x-r)^2\cdot Q(x)+R(x)$$ where $\deg R<2$, i.e. $R(x)=ax+b$ for some $a,b$. Then $$P'(x)=2(x-r)Q(x)+(x-r)^2Q'(x)+R'(x) $$ From $P(r)=P'(r)=0$, we conclude $R(r)=R'(r)=0$. From $R'(r)=0$ we get $a=0$ and then also $b=0$.


3

You have that $$\begin{align*}α^3+β^3&=(α+β)(α^2-αβ+β^2)=(α+β)(α^2+2αβ+β^2-3αβ)=\\&=(α+β)((α+β)^2-3αβ)\end{align*}$$


3

Here's a way to fill in the gap in @Yiorgos' solution. The problem is local, so we may as well work on a simply connected domain. Let $v$ be a harmonic conjugate of $u$, so that $f = u+iv$ is holomorphic. It follows from Cauchy-Riemann's equations that $f' = u'_x -iu'_y$. In particular, the gradient of $u$ vanishes exactly at the points where $f' = 0$. But ...


3

If $$x^3+bx^2+cx+d=(x-x_1)(x-x_2)(x-x_3),$$ then $$x_1+x_2+x_3=-b$$ $$x_1x_2x_3=-d$$ instead of $$x_1+x_2+x_3=b$$ $$x_1x_2x_3=d$$ because $$(x-x_1)(x-x_2)(x-x_3)$$ $$=x^3-(x_1+x_2+x_3)x^2+(x_1x_2+x_2x_3+x_3x_1)x-x_1x_2x_3.$$


2

Hint: Set $u = x^{2}$ if you make this change you will have: $$u^{2} - 28 u +49$$ from here you can use the quadratic formula to factorize it and solve it, when you solve it for $u$ remember to go back to $x$


2

First, represent $-\rm i$ in exponential form $$ z^6 = -{\rm i} = {\rm e}^{- \frac{{\rm i}\pi}2} $$ Now you can multiply it by $1 = {\rm e}^{2\pi \rm i}$, actually you can do it $k$ times: $$ z^6 = {\rm e}^{2{\rm i} \pi k - \frac{{\rm i}\pi}2}, k =0,1,\ldots $$ now let us consider the $1/6$-th power of the left an of the right sides of the equation $$ z = ...


2

You're close: $$ z^6 =-i= e^{i\big( \frac{3\pi}{2} + 2\pi k\big)} $$ for $k \in \Bbb{Z}$. With $z = re^{i \theta}$ you find that $r^6 = 1$ implying that $r=1$ and $\theta = \frac{3\pi}{12} + \frac{\pi k}{3}$ for $k \in \Bbb{Z}$. Can you go from here..?


2

Put $t=\tan(x)$, so that your equation become: $t^3-3t^2+t+1=0$. Since $t^3-3t^2+t+1=(t-1)(t^2-2t-1)$, it follows that the solutions are $t=1$, $t=1\pm\sqrt 2$, from which $x\equiv\arctan(1)\equiv\frac\pi 4\pmod \pi$, $x\equiv\arctan(1+\sqrt 2)\equiv\frac{3\pi}8\pmod\pi$, $x\equiv\arctan(1-\sqrt 2)\equiv-\frac\pi8\pmod\pi$.


2

Choose an initial seed $x_0 \in [0, 1]$. You want to show that $$[x_{n+1}]^{n+1} \geq [x_n]^n,$$ for $n \geq 3$. Since $$\frac{{x_n}^n[1 - {x_n}^n]}{(1 - {x_n})n} = a$$ for $n \in \mathbb{N}$, we have $$\frac{{x_{n+1}}^{n+1}[1 - {x_{n+1}}^{n+1}]}{(1 - {x_{n+1}})(n + 1)} = a.$$ Dividing through both equations in $a$: $$\frac{{x_n}^n[1 - {x_n}^n]}{(1 ...


2

If $r$ is a root of $P(x)$, then $x - r$ is a factor of $P(x)$, that is, $$P(x) = (x - r) Q(x)$$ for some polynomial $Q$. Then, the product rule gives $$P'(x) = (x - r) Q'(x) + Q(x).$$ Now, since $r$ is a root of $P'(x)$, $$0 = P'(r) = (r - r) Q'(x) + Q(r) = Q(r),$$ that is, $r$ is also root of $Q$. So, $$Q(x) = (x - r) R(x)$$ for some polynomial $R(x)$, and ...


2

The lesson you have to learn here is the following: Newton's method is of no help in finding the global configuration of the zeros; but if you start sufficiently near a zero you're going to find it. Let $$f(x):=x^4-8x^2-x+16\ .$$ We have to undertake a global study of $f$ in the $x$-interval $[1,3]$. First note that $f(1)=9>0$, $\>f(2)=-2<0$, ...


2

The first two observations can be explained by the fact that if $r$ is the root of $a_nz^n+\cdots +a_0$ then $r^{-1}$ is the root of $a_0z^n+\cdots +a_n$. Since the joint probability density of the coefficients is symmetric under $a_k\mapsto a_{n-k}$ transformation the density of the roots should be symmetric under $r\mapsto r^{-1}$. It should be fairly ...


1

As the comments say, these are called Vieta's formulas, and the plus or minus sign is not difficult to remember, because it always alternates. In general, if you have an $n$th-degree polynomial $$p(x) = a_0x^n + a_{1}x^{n-1} + a_2x^{n-2} + \cdots + a_{n-1}x + a_n$$ whose roots are $r_1, r_2, \ldots r_n$ then Vieta's formulas tell you that you can find ...


1

I'll give examples, then you'll understand the general pattern better. Take a degree four polynomial. I'll denote a general fourth-degree polynomial by $P_4$. A general $5^{\text{th}}$ degree polynomial as $P_5$ and so on... Your $P_4$ is: $ a_0x^4 + a_1x^3 + a_2x^2 + a_3x^1 + a_4$ Then, I'll get $4$ formulas: $r_0 + r_1 + r_2 + r_3 = -\dfrac{a_1}{a_0}$ ...


1

You can get 20 exact digits with only 3 iterations, but it require a better initial guess ( do not use X0=1 ) This guess is easy to find with a simple table of square roots between 1 and 100. Then you start with 3,16 x 10^4 for X0... Apply Newton/Hero 3 times, and you get: 31426,968052931864079... that's 20 correct digits! Hello again, With my " 101 ...


1

Following the hint in the comments, we divide by $x^2$ and rearrange to get $$60\left(x^2+\frac{1}{x^2}\right)-44\left(x+\frac{1}{x}\right)-25=0$$ Set $y=x+\frac{1}{x}$ and note that $y^2=x^2+\frac{1}{x^2}+2$ to get $$60(y^2-2)-44y-25=0$$ which rearranges to the quadratic $$60y^2-44y-145=0$$ This has roots $y\approx -1.23, 1.96$. [Alternate answer: ...


1

Bairstow's method is of this nature, using Newton iterations to approximate two roots of a real polynomial simultaneously. However it is usually employed to find a pair of complex conjugate roots. Obviously if we find one quadratic real factor of the real quartic polynomial, the other factor will also be quadratic, and the quadratic formula can be used on ...


1

For polynomial of degree $3$ you can use the following procedure. Assume that you guessed the solution $x_1=4$ (indeed $4^3-6\cdot 4^2-2\cdot 4+40 = 64 -96-8+40 =0)$. You can use Horner's method to get the polynomial $p(x)=p_2x^2+p_1x+p_0$ such that $(x-4)\cdot p(x) = x^3-6x^2-2x+40$. You want to do that because $p(x)$ will be a polynomial of degree $2$ ...


1

Hint: This can give some information about the possible location of roots, to help eliminate what you actually have to test. (Note: everything here refers to real roots and real zeroes.) Write your polynomial as the function $$p(x) = x^3-6x^2-2x+40$$ and note that its derivative $$p'(x) = 3x^2 -12x-2$$ has exactly two distinct real zeroes $d_0<d_1$ given ...



Only top voted, non community-wiki answers of a minimum length are eligible