Hot answers tagged

62

$\exp(x)$ has no real roots, and no complex roots either. It is not difficult to find functions that have no roots: for any function $f(x)$, $g(x) = |f(x)|+1$ has no roots.


45

The answer to the question as asked is simply "yes", as others have said. I'd like to give a little more context and explain why (for one interpretation of the question) the answer comes close to being no. So, first of all, "function" is a very broad term. The usual definition in mathematics is that a function is any way of assigning output values to input ...


15

The Rational Root Test shows that the only possible rational solutions are $\pm 1$. Substituting gives that $x = -1$ is one (but $x = 1$ is not), so polynomial long division gives $p(x) = -(x + 1) q(x)$ for some quintic $q$. Substituting $x = -1$ gives that $-1$ is not a root of $q$, so if $q$ factors over $\Bbb Q$, it does so into an irreducible quadratic ...


13

These are properties of exponentiation. In particular, $$a^{-b} = \frac{1}{a^b}$$ combined with $$a^{\frac{m}{n}} = \sqrt[\leftroot{-2}\uproot{2}n]{a^m}.$$ In your case, $$\frac{1}{4}x^\frac{-3}{4} = \frac{1}{4x^\frac{3}{4}} = \frac{1}{4\sqrt[\leftroot{-2}\uproot{2}4]{x^3}}.$$


12

If $g$ is an entire function, then $f(z)=e^{g(z)}$ is entire and has no zeros in the complex plane. This is essentially the only example for $f$ entire because of the Weierstrass factorization theorem.


12

The gamma function $ z \mapsto \Gamma(z)$ has no roots on the complex plane. Observe that $$\cos \left(i\log(2+\sqrt{3})\right)=2$$ thus $\cos x -2=0$ does admit a complex root. Edit. From $\displaystyle \cos x =\frac12\left(e^{ix}+e^{-ix}\right)$, one has $$ \begin{align} \cos ...


9

$\text{Let } x = z + \frac{2}{z}\\\begin{align*}\left(z+\frac{2}{z}\right)^3 - 6\left(z+\frac{2}{z}\right) - 6 &= 0\\z^6 - 6z^3 + 8 &= 0\\z^3 &= 3\pm1\, (\text{using the quadratic formula})\\z^3 &= 4, 2,\\z &= \sqrt[3]{2},\sqrt[3]{4}\end{align*}$ Now just substitute back one value for $z$ to get one root: $x = \sqrt[3]{2} + 2\sqrt[3]{4}$ ...


8

Consider this system of polynomial equations: $$\begin{align} a+b+c &= 2 \\ a^3 &= 7x + 1 \\ b^3 &= 8 + x - x^2 \\ c^3 &= x^2 - 8 x - 1 \end{align}$$ Using the method of resultants (via the oh-so-convenient Resultant[] function in Mathematica), we can eliminate $a$, $b$, $c$ to get $$x^3\;(x-1)^6\;(x+1)^3\;(x - 9 )^3 \;=\; 0$$ Therefore, the ...


7

Why does this equation has $4$ roots? The short answer is it is an accident. For similar problem like $$(7x-6)^{1/3}+(8+x-x^2)^{1/3}+(x^2-8x-1)^{1/3}=1 $$ You can get $5$ instead of $4$ solutions. Define $a,b,c$ like Blue's answer. $$\begin{cases} a^3 &= 7x + 1 \\ b^3 &= 8 + x - x^2 \\ c^3 &= x^2 - 8 x - 1 \end{cases} $$ The key of the whole ...


7

Quoting from Locating the zeros of partial sums of $\exp(z)$ with Riemann-Hilbert methods: We denote by $p_{n}(z) := 1 + z + \cdots + \frac{z^{n}}{n!}$ the partial sums of the exponential series. The problem to describe the asymptotic distribution of the zeros of $p_n$ was posed and solved in the classical paper of Szegő [11]. He proved that the ...


6

Any nontrivial root of $x^5-1$ will be an element of order $5$ in $\mathbb{Z}_7^\times$, which has order $6$. This cannot happen because of Lagrange's theorem. Alternatively, if $x^5=1$ then $x\ne0$ and so $x^6=1$ by Fermat's theorem. But then $1=x^6=xx^5=x$.


6

hint: Use AM-GM inequality $4$ times: $$x^3+y^3+z^3 \geq 3xyz$$ with $(x,y,z) = (a,b,c), (a,c,d), (a,b,d), (b,c,d)$, and add up.


6

Suppose the roots are $\alpha,\beta,\gamma,\delta$. Then we have $\frac{\alpha+\beta+\gamma+\delta}{4}=\alpha\beta\gamma\delta=1$. But the arithmetic and geometric means of $\alpha,\beta,\gamma,\delta$ can only be equal if $\alpha=\beta=\gamma=\delta$.


6

The following trick springs to mind. We know that the minimal polynomials of seventh roots of unity are the cubic irreducibles $$ p_1(x)=x^3+x+1\qquad\text{and}\qquad p_2(x)=x^3+x^2+1. $$ Now, because $2$ is of order $21$ modulo $49$ (leaving it to you to check that), we can deduce (Galois theory is all we need here) that: The 49th roots of unity ...


5

You have actually (not) proved that if $0=0^0$, then $0^0\ne 1$ which needs no argument at all. Moreover, your argument is wrong: you use $\ln(t)$, which is not defined for $t=0$. So you can draw no conclusion about $t=0$. Under fairly standard conventions, the function $x\mapsto x^x$ is only defined for $x>0$; more generally, one sets $$ ...


5

You may show that the map $\psi:x\to x^5$ is bijective over $\mathbb{F}_7^*$, since it is an involution: $(x^5)^5 = x^{25}\equiv x^1 = x\pmod{7}$ by the Fermat's little theorem. That gives $\left|\psi^{-1}(1)\right|=1$.


4

It is a property of polynomials that a polynomial of degree $n$ must have exactly $n$ complex roots. For example, if you took a function like: $$ f(x) = 3x^2 + 2x -4$$ Its highest power is 2, so it will have two complex roots, and this can be said for any polynomial with real or complex coefficients. (However, some roots may have "multiplicity," so that a ...


4

If you are working in $\mathbb{R}$, then you can easily find that $x = 1$ is a solution, as you have found. But what about if we wish to find all soutions, including those in $\mathbb{C}$? How do we know that we haven't lost solutions, or that we have all of them? By the Fundamental Theorem of Algebra, you know that the equation $x^3 = 1$ (or equivalently, ...


4

Using the Berlekamp algorithm it follows that $$ x^{21}+x+1=(x^{14} + x^{12} + x^7 + x^6 + x^4 + x^3 + 1)(x^7 + x^5 + x^3 + x + 1) $$ over $\mathbb{F}_2$, hence it is reducible. However, $$ g(x)=x^{21}+x^2+1 $$ is irreducible over $\mathbb{F}_2$.


4

Let: $(7x+1)^{1/3} = a,(x^2-8x-1)^{1/3} = b, (8+x-x^2)^{1/3} = c.$ You have: $$\begin{cases} a+b+c = 2 \\ a^3+b^3+c^3 = 8 \end{cases} \Rightarrow a^3+b^3+c^3 = (a+b+c)^3$$ $$ \Leftrightarrow a^3+b^3 = (a+b+c)^3-c^3 = (a+b)[(a+b+c)^2+c(a+b+c)+c^2)]$$ $$ \Leftrightarrow a+b = 0 \text{ or }a^2-ab+b^2 = (a+b+c)^2+c(a+b+c)+c^2 $$ The first case is easy: $a = -b ...


4

General approach, you use Rouche theorem to count the number ofroots inside the circles of radius $1$ and $2$ and by substracting this quantity, you obtain the number of roots in the annulus. How to apply Rouches theorem to find the number of roots in the circle of radius $1$. You can chose $f(z)=6z^2$ and $g(z)=z^8+3z^7+1$. On the considerd contour we have ...


3

Hint: $x^3-6x-6=x^3-3bcx+b^3+c^3$ $$bc=-2, b^3+c^3=-6$$ $$b=-\sqrt[3]{2}, c=-\sqrt[3]{4}$$ So $(b+c)^3=b^3+c^3+3bc(b+c)$, then number $b+c$ - solution of $x^3-3bcx+b^3+c^3=0$


3

You can simply use that $x^3$ and $x^2$ are both strictly increasing for positive $x$. (What does this tell you about $f(x)$?)


3

Let $f(x) = a_{2n}x^{2n} + \cdots + a_1x + a_0$ be a polynomial with odd integer coefficients. By the rational root test, any rational root of $f(x)$ is of the form $c/d$ where $c$ divides $a_0$ and $d$ divides $a_{2n}$. Since $a_0$ and $a_{2n}$ are odd, so are $c$ and $d$. We have \begin{equation*} \begin{aligned} &\mathrel{\phantom{=}} ...


3

Edit : I divide this answer into 2 parts. The first one originates as a quick answer to the question; the second one has my preference because it provides an intuitive understanding of the situation. Part 1 : With your notations ($D$ being the discriminant): $$\dfrac{D}{4} = a^2 + b^2 + c^2 -ab - bc - ca=\dfrac{1}{4}(2a-b-c)^2+\dfrac{3}{4}(b-c)^2$$ ...


3

$$f(x)=(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)$$ Then, $$f(a)=(a-b)(a-c)$$ $$f(b)=(b-c)(b-a)$$ $$f(c)=(c-a)(c-b)$$ Clearly, the coefficient of $x^2$ in $f(x)$ is positive. Suppose there are no real roots. Then, $f(x)\gt0 \forall x\in\mathbb R$. $$(a-b)(a-c)\gt0$$ Without loss of generality, let $a\gt b$ and $a\gt c$. $$(b-c)(b-a)\gt 0\Rightarrow c\gt b$$ ...


3

On the unit circle we have $|f(z)-z+z| =|f(z)|< 1 = |z|$, hence $z \mapsto z$ and $z \mapsto f(z)-z$ have the same number of zeroes in the unit disk. Hence there is exactly one point in the unit disk for which $f(z) = z$.


3

Using the Extended Euclidean Algorithm as implemented in this answer, we get $$ \begin{array}{r} &&1&x+6&(x-8)/53\\\hline 1&0&1&-x-6&(x^2-2x+5)/53\\ 0&1&-1&x+7&(-x^2+x+3)/53\\ x^2-x-3&x^2-2x+5&x-8&53&0\\ \end{array} $$ That is, $$ (x+7)(x^2-2x+5)-(x+6)(x^2-x-3)=53\tag{1} $$ We can now use the ...


3

Using a complicated computer algebra system (Mathematica 10.4), we can get all the roots to this equation as radicals. Two roots are complex and the rest are all real (surprisingly). Module[{roots}, roots = Solve[-x^6 + x^5 + 2 x^4 - 2 x^3 + x^2 + 2 x - 1 == 0, x]; Transpose[{ N[x /. roots], FullSimplify[Element[x, Reals] /. roots], x /. ...


3

For $z^{3}+w z^{2}+\bar{w} z+1=0$, there's a paper discussed. Still need to check for your case.



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