# Tag Info

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

$p\left(\frac{1}{x}\right) = \frac{1}{x^4}p(x)$. 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 ...

6

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 $$z \in ... 5 First Part This is true for n=1. Suppose this is true for n-1. Since \lambda_1e^{a_1x}\ne0 we can divide by that to get$$ 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

Your identity shows that the product of terms, each of which is a sum of two squares, is a sum of two squares. So we just have to write your polynomial as a product of terms, each of which is a sum of two squares of real polynomials. If $P(x) \ge 0$ for all real $x$, and $\alpha$ is a real root of $P$, then it must have even multiplicity (otherwise $P(x)$ ...

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 $2\times2$ ...

4

Clearly, each of the $n$ DISTINCT complex numbers $$\lambda_k=\exp\left(\frac{2k\pi i}{n}\right), \quad k=1,\ldots,n,$$ is a root of the polynomial $$p(z)=z^n-1.$$ Hence, as we have found $n$ DISTINCT roots of $p$, which is a polynomial of degree $n$, then we can analyze $p$ as $$p(z)=(z-\lambda_1)\cdots ... 4$$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

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

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

Your title says it all. Some operations work backwards, but others don't! Here's the reason not every polynomial has a root at $0$: to go from $a_1x+a_2x^2+...$ to $a_1+a_2x+...$ you'd divide by $x$, right? But you can't always divide by $x$-why?-because $x$ could be zero, and you can't divide by zero. The right way to solve $a_1x+a_2x^2+...$ is instead to ...

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

set $u = x^2$ and use the quadratic formula to factorize it. if you do the change, you will have $u^2 +2u +1$ and from here its easy to solve it, ones you factorize it remember to go back to $x$

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)} $$for k \in \Bbb{Z}. With z = re^{i \theta} you find that r^6 = 1 implying that r=1 and \theta = \frac{3\pi}{12} + \frac{\pi k}{3} for k \in \Bbb{Z}. Can you go from here..? 2 Choose an initial seed x_0 \in [0, 1]. You want to show that$$[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

Writing the function with its roots seems helpful. The derivative can be written with terms like $x_1-x_2$ which its product is related to the discriminant. Let the roots of $f_2(x)=ax^2+bx+c=0$ be $x_1$ and $x_2$. Then $f_2(x)=a(x-x_1)(x-x_2)$ and $\Delta=(a(x_1-x_2))^2$. Differentiation gives $f'_2(x_1)=a(x_1-x_2)=\sqrt{\Delta}$. Looking at the cubic ...

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

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

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 ...

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

I will elaborate the hint of @Liu Gang and the solution of @Yiorgos S. Smyrlis : First, we need a standard result due to Bezout: Let $P(z)$ a polynomial. The remainder of the division of $P(z)$ by $(z-\lambda)$ equals $P(\lambda)$. Indeed, write $$P(z) = (z-\lambda)Q(z) + R$$ where $R$ is the remainder, $\deg R < \deg (z-\lambda) = 1$ so $R$ is a ...

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: $r_0 + r_1 + r_2 + r_3 = -\dfrac{a_1}{a_0}$ ...

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