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Questions tagged [roots]

Questions about the set of values at which a given function evaluates to zero. For questions about "square roots", "cube roots" and such, consider using the (radicals) and the (arithmetic) tag. For questions about roots of Lie algebras, use the (lie-algebra) tag instead.

23
votes
3answers
3k views

Square roots — positive and negative

It is perhaps a bit embarrassing that while doing higher-level math, I have forgot some more fundamental concepts. I would like to ask whether the square root of a number includes both the positive ...
9
votes
2answers
1k views

Why Rational Root Theorem requires integer coefficient polynomials? [closed]

Why does the rational root theorem only work when the polynomial has integer coefficients?
86
votes
6answers
11k views

Continuity of the roots of a polynomial in terms of its coefficients

It's commonly stated that the roots of a polynomial are a continuous function of the coefficients. How is this statement formalized? I would assume it's by restricting to polynomials of a fixed ...
75
votes
4answers
7k views

Are these solutions of $2 = x^{x^{x^{\:\cdot^{\:\cdot^{\:\cdot}}}}}$ correct?

Find $x$ in $$ \Large 2 = x^{x^{x^{\:\cdot^{\:\cdot^{\:\cdot}}}}}$$ A trick to solve this is to see that $$\large 2 = x^{x^{x^{\:\cdot^{\:\cdot^{\:\cdot}}}}} \quad\implies\quad 2 = x^{\Big(x^{x^{x^{...
101
votes
5answers
6k views

Why are the solutions of polynomial equations so unconstrained over the quaternions?

An $n$th-degree polynomial has at most $n$ distinct zeroes in the complex numbers. But it may have an uncountable set of zeroes in the quaternions. For example, $x^2+1$ has two zeroes in $\mathbb C$,...
25
votes
8answers
33k views

How to prove that a polynomial of degree $n$ has at most $n$ roots?

How can I prove, that a polynomial function $$f(x) = \sum_{0\le k \le n}a_k x^k\qquad n\in\mathbb N,\ a_k\in\mathbb C$$ is zero for at most $n$ different values of $x$? (Except $n=0$ where $f(x)$ is $...
17
votes
1answer
5k views

Derivation of asymptotic solution of $\tan(x) = x$.

An equation that seems to come up everywhere is the transcendental $\tan(x) = x$. Normally when it comes up you content yourself with a numerical solution usually using Newton's method. However, ...
7
votes
3answers
2k views

Complex zeros of the polynomials $\sum_{k=0}^{n} z^k/k!$, inside balls

this is a question from a Temple prelim exam, and i'm trapped in it! We have $p_n(z)=\sum_{k=0}^n\frac{z^k}{k!}$ and we have to prove that $\forall r>0 \quad \exists N\in\mathbb{N}$ s.t. $p_n(z)$ ...
27
votes
1answer
4k views

Showing that the roots of a polynomial with descending positive coefficients lie in the unit disc.

Let $P(z)=a_0+a_1z+\cdots+a_nz^n$ be a polynomial whose coefficients satisfy $$0<a_0<a_1<\cdots<a_n.$$ I want to show that the roots of $P$ live in unit disc. The obvious idea is to use ...
10
votes
2answers
1k views

Roots of the incomplete gamma function

Is there any way that one can describe all the roots of the incomplete gamma function $\Gamma(n,z)$, for $n\in \mathbb{N}$, analytically?
3
votes
2answers
289 views

On the trigonometric roots of a cubic

We have, $$x^3+x^2-2x-1=0,\quad\quad x =\sum_{k=1}^{2}\,\exp\Bigl(\tfrac{2\pi\, i\, (6^k)}{7}\Bigr)\\ x^3+x^2-4x+1=0,\quad\quad x =\sum_{k=1}^{4}\,\exp\Bigl(\tfrac{2\pi\, i\, (5^k)}{13}\Bigr)\\ x^3+x^...
33
votes
2answers
1k views

Something strange about $\sqrt{ 4+ \sqrt{ 4 + \sqrt{ 4-x}}}=x$ and its friends

This post can be generalized to, $$\begin{align} \sqrt{ 2+ \sqrt{ 2 + \sqrt{ 2-x}}}=x&,\quad\quad x = -2\cos\left(\frac{8\pi}{9}\right)=1.8793\dots\quad\quad\quad \\ \\ \sqrt{ 4+ \sqrt{ 4 + \sqrt{...
6
votes
7answers
59k views

How to find the roots of $x^4 +1$

I'm trying to find the roots of $x^4+1$. I've already found in this site solutions for polynomials like this $x^n+a$, where $a$ is a negative term. I don't remember how to solve an equation when $a$ ...
4
votes
2answers
1k views

Polynomial $p(a) = 1$, why does it have at most 2 integer roots?

The question that I am trying to answer is : Suppose is $p(x)$ is a polynomial with integer coefficients. Show that if $p(a) = 1$ for some integer a then $p(x)$ has at most two integer roots. I have ...
15
votes
8answers
23k views

Fastest Square Root Algorithm

What is the fastest algorithm for finding the square root of a number? I created one that can find the square root of "987654321" to 16 decimal places in just 20 iterations I've now tried Newton's ...
14
votes
3answers
4k views

Irreducibility of a polynomial if it has no root (Capelli) [duplicate]

Let $F$ be a field of arbitrary characteristic, $a\in F$, and $p$ a prime number. Show that $$f(X)=X^p-a$$ is irreducible in $F[X]$ if it has no root in $F$. This answer to a related question ...
15
votes
1answer
2k views

Approximating roots of the truncated Taylor series of $\exp$ by values of the Lambert W function

If you map the nth roots of unity $z$ with the function $-W(-z/e)$ you get decent starting points for some root finding algorithm to the roots of the scaled truncated taylor series of $\exp$. Here W ...
4
votes
2answers
983 views

Is there anything like “cubic formula”?

Just like if we have any quadratic equation which has complex roots, then we are not able to factorize it easily. So we apply quadratic formula and get the roots. Similarly if we have a cubic ...
37
votes
11answers
8k views

Why can a quadratic equation have only 2 roots?

It is commonly known that $ax^2+bx+c=0$ have two solutions $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ but how to prove that another root couldn't exist? I think derivation of quadratic formula is not enough......
6
votes
3answers
4k views

Proof the Legendre polynomial $P_n$ has $n$ distinct real zeros

I need a proof to show that the inequality $m < n$ leads to a contradiction and $P_n$ has $n$ distinct real roots, all of which lie in the open interval $(-1, 1)$.
6
votes
2answers
3k views

Create polynomial coefficients from its roots

Given some roots : $r_1,r_2,\ldots,r_n$, how can we reconstruct polynomial coefficients? I know the Horner scheme and that we can just go backwards receiving those coefficients. But I'm curious if ...
19
votes
1answer
12k views

Using Vieta's theorem for cubic equations to derive the cubic discriminant

Background: Vieta's Theorem for cubic equations says that if a cubic equation $x^3 + px^2 + qx + r = 0$ has three different roots $x_1, x_2, x_3$, then $$\begin{eqnarray*} -p &=& x_1 + x_2 +...
2
votes
2answers
794 views

Find Number Of Roots of Equation $11^x + 13^x + 17^x =19^x $

The Equation $11^x + 13^x + 17^x =19^x $ Has No Real Roots Only One Real Roots Exactly Two Real Roots More than Two Real Roots What I have done is The function $f(x)=11^x + 13^x + 17^x -19^x ...
5
votes
1answer
693 views

Proof that $\sqrt[m]{a} + \sqrt[n]{b}$ is irrational

Is there a way to prove that $\sqrt[m]{a} + \sqrt[n]{b}$ ($\sqrt[m]{a}$ and $\sqrt[n]{b}$ are irrational); $a, b, m, n \in \mathbb{N}$; $m, n \neq 2$; is irrational without using the theorem mentioned ...
4
votes
4answers
259 views

What is the solution to the equation $9^x - 6^x - 2\cdot 4^x = 0 $?

I want to solve: $$9^x - 6^x - 2\cdot 4^x = 0 $$ I was able to get to the equation below by substituting $a$ for $3^x$ and $b$ for $2^x$: $$ a^2 - ab - 2b^2 = 0 $$ And then I tried \begin{align}x ...
58
votes
2answers
3k views

Denesting radicals like $\sqrt[3]{\sqrt[3]{2} - 1}$

The following result discussed by Ramanujan is very famous: $$\sqrt[3]{\sqrt[3]{2} - 1} = \sqrt[3]{\frac{1}{9}} - \sqrt[3]{\frac{2}{9}} + \sqrt[3]{\frac{4}{9}}\tag {1}$$ and can be easily proved by ...
40
votes
3answers
7k views

Why is it so hard to find the roots of polynomial equations?

The question that follows was inspired by this question: When trying to solve for the roots of a polynomial equation, the quadratic formula is much more simple than the cubic formula and the cubic ...
15
votes
1answer
3k views

$\lambda-z-e^{-z}=0$ has one solution in the right half plane

Let $\lambda > 1$ , want to show that the equation $$\lambda-z-e^{-z}=0$$ has exactly one solution in the right half plane $\{z:Re(z)>0\}$. Moreover, the solution must be real.I tried to use ...
8
votes
2answers
11k views

How do you solve 5th degree polynomials?

I looked on Wikipedia for a formula for roots of a 5th degree polynomial, but it said that by Abel's theorem it isn't possible. The Abel's theorem states that you can't solve specific polynomials of ...
14
votes
2answers
904 views

What are the properties of the roots of the incomplete/finite exponential series?

Playing around with the incomplete/finite exponential series $$f_N(x) := \sum_{k=0}^N \frac{z^k}{k!} \stackrel{N\to\infty}\longrightarrow e^z$$ for some values on alpha (e.g. ...
5
votes
1answer
1k views

Solving a cubic polynomial equation.

Overview I have tried finding a solution to this problem myself and I have flailed. Its just a challenge for me. could you please tell me how far am I in solving this question? My approach for ...
5
votes
2answers
850 views

The trigonometric solution to the solvable DeMoivre quintic?

Using the relations for the Rogers-Ramanujan cfrac described in this post, $$\frac{1}{r}-r = x$$ $$\frac{1}{r^5}-r^5 = y$$ and eliminating $r$ yields, $$x^5+5x^3+5x = y$$ This is the case $a=1$ ...
8
votes
1answer
2k views

Why not write the solutions of a cubic this way?

For the solution of the cubic equation $x^3 + px + q = 0$ Cardano wrote it as: $$\sqrt[3]{-\frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}+\sqrt[3]{-\frac{q}{2} - \sqrt{\frac{q^2}{4} + \frac{p^3}...
17
votes
3answers
5k views

How to solve polynomial equations in a field and/or in a ring?

I'm studying for my exam, and I stuck on solving polynomials in a field and/or in a ring. Let me give you some examples: (1) Solve equation $x^2+4x+3=0$ in field $\mathbb{Z}_5$, $\mathbb{Z}_8$ and in ...
16
votes
5answers
70k views

Prove using Rolle's Theorem that an equation has exactly one real solution.

Prove that the equation $x^7+x^5+x^3+1=0$ has exactly one real solution. You should use Rolle’s Theorem at some point in the proof. Since $f(x) = x^7+x^5+x^3+1$ is a polynomial then it is continuous ...
5
votes
2answers
201 views

Asymptotic to a sequence of algebraic numbers.

Let $f(n)$ be the largest real solution of $$x^n - x^{n-1} = 1 $$ As $n$ grows to positive infinity we get the asymptotic : $$ f(n) = 1 + \frac{\ln(n)}{n} + \frac{\exp(2)}{n^2} + ...$$ Where the ...
2
votes
1answer
621 views

A positive polynomial is the sum of two squares in $\mathbb{R}[X]$ [duplicate]

Let $P\in\mathbb{R}[X]$ be a positive polynomial. I want to show that there exists $A,B\in\mathbb{R}[X]$ so that $P=A^2+B^2$ $\displaystyle P=a\prod_{k=0}^q(x-a_k)\prod_{k=0}^{2p}(x-b_k)=a\prod_{k=0}...
8
votes
3answers
462 views

$\sqrt[31]{12} +\sqrt[12]{31}$ is irrational

Prove that $\sqrt[31]{12} +\sqrt[12]{31}$ is irrational. I would assume that $\sqrt[31]{12} +\sqrt[12]{31}$ is rational and try to find a contradiction. However, I don't know where to start. Can ...
9
votes
5answers
966 views

Prove $\sqrt[3]{3} \notin \mathbb{Q}(\sqrt[3]{2})$ [duplicate]

I've tried solving $\sqrt[3]{3} =a + b* \sqrt[3]{2}+c* \sqrt[3]{4}$, but there is no obvious contradiction, even when taking the norms/traces of both sides. I can't think of another approach. This is ...
12
votes
5answers
579 views

Solve this tough fifth degree equation.

$$x^5+x^4-12x^3-21x^2+x+5=0$$ I think it can be solved by trigonometric ways but how?
11
votes
6answers
8k views

Complex roots of $z^3 + \bar{z} = 0$

I'm trying to find the complex roots of $z^3 + \bar{z} = 0$ using De Moivre. Some suggested multiplying both sides by z first, but that seems wrong to me as it would add a root ( and I wouldn't know ...
6
votes
3answers
2k views

Prove that $a^2 + b^2 \geq 8$ if $ x^4 + ax^3 + 2x^2 + bx + 1 = 0 $ has at least one real root.

If it is known that the equation $$ x^4 + ax^3 + 2x^2 + bx + 1 = 0 $$ has a (real) root, prove the inequality $$ a^2 + b^2 \geq 8. $$ I am stuck on this problem, though, it is a very easy problem for ...
5
votes
4answers
2k views

Finding all complex zeros of a high-degree polynomial

Given a large univariate polynomial, say of degree 200 or more, is there a procedural way of finding all the complex roots? By "roots", I mean complex decimal approximations to the roots, though the ...
0
votes
1answer
208 views

Find the root of the polynomial?

Consider the root of the polynomial $p(x) = x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\dots+a_1x -1$. Suppose that $p(x)$ has no roots in the open unit disc in a complex plane and $p(-1)=0$. Show that $p(1)...
0
votes
0answers
62 views

How to prove $a_{1}\sqrt[b_{1}]{c_{1}}+a_{2}\sqrt[b_{2}]{c_{2}}+…+a_{n}\sqrt[b_{n}]{c_{n}}$ is irrational?

Let's define the number $$A=a_{1}\sqrt[b_{1}]{c_{1}}+a_{2}\sqrt[b_{2}]{c_{2}}+.....+a_{n}\sqrt[b_{n}]{c_{n}}$$ where $a_{1}, a_{2}, ..., a_{n}$ are positive integers and $b_{1}, b_{2}, ..., b_{n}, ...
3
votes
2answers
148 views

Number of real roots of $2 \cos\left(\frac{x^2+x}{6}\right)=2^x+2^{-x}$

Find the number of real roots of $ \cos \,\left(\dfrac{x^2+x}{6}\right)= \dfrac{2^x+2^{-x}}{2}$ 1) 0 2) 1 3) 2 4) None of these My guess is to approach it in graphical way. But equation seems ...
3
votes
1answer
842 views

How do iterative methods applied to the companion matrix of a polynomial $p(\lambda)$ relate to $p$ itself?

A few days ago, I had a vague question in my mind about "matrix methods" for finding the roots of a polynomial. Now, I can ask at least a semi-precise question, thanks to the post How to calculate ...
26
votes
3answers
1k views

Prove that there exists $n\in\mathbb{N}$ such that $f^{(n)}$ has at least n+1 zeros on $(-1,1)$

Let $f\in C^{\infty}(\mathbb{R},\mathbb{R})$ such that $f(x)=0$ on $\mathbb{R}$\ $(-1,1)$. Prove that there exists $n\in\mathbb{N}$ such that $f^{(n)}$ has at least n+1 zeros on $(-1,1)$ My attempt : ...
16
votes
3answers
646 views

How can I prove $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}…}}}}=2$ [duplicate]

How can I prove $$\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}...}}}}=2$$ I don't know which method can be used for this?
22
votes
5answers
11k views

Roots of Legendre Polynomial

I was wondering if the following properties of the Legendre polynomials are true in general. They hold for the first ten or fifteen polynomials. Are the roots always simple (i.e., multiplicity $1$)? ...