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11 votes
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May a trinomial with degree exceeding four have only real roots?

Yes, there are trinomials of arbitrarily large degree whose roots are all real, e.g., $$x^{n + 4} - (r^2 + s^2) x^{n + 2} + r^2 s^2 x^n = (x + s) (x + r) x^n (x - r) (x - s) , \quad r, s \in \Bbb R .$$...
Travis Willse's user avatar
7 votes
Accepted

Proving that the set of $a$ for which the roots of $P(X)-a$ are all real is an interval

$\def\R{\mathbb{R}}\def\ge{\geqslant}\def\emptyset{\varnothing}\def\le{\leqslant}\def\paren#1{\left(#1\right)}\def\={\mathrel{\hphantom{=}}}$Lemma: For any $Q(x) \in \R[x] \setminus \{0\}$ and $a \in \...
Ѕᴀᴀᴅ's user avatar
  • 34.1k
6 votes

Solutions to equation $x^{\frac2 7}=16$

By the usual conventions used in College Algebra, $x^{a/b} = (x^a)^{1/b}$ if $x$ is real and $a/b$ is a rational number in lowest terms with $a$ even.
Robert Israel's user avatar
4 votes

May a trinomial with degree exceeding four have only real roots?

$$ax^{2n} + bx^n + c = 0 \quad \Longleftrightarrow \quad at^2 + bt + c = 0 \;\text{and}\; t = x^n$$ The cases are as follows (all solutions are computed in $\mathbb R$): If $b^2 - 4ac < 0$ no ...
Kroki's user avatar
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3 votes
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Roots and extrema of the polynomial $P_n(x)=\sum_{k=0}^n\binom{n+k}{2k}(-x)^k$.

Let $S_n(x) = \sum_{k=0}^n P_k(x)$, and $U_n(y) = S_n(2 - 2y)$, it is easy to prove that $U_0(y) = 1$, $U_1(y) = 1 + 1 - (2-2y) = 2y$ and $$U_{n+1}(y) = 2y U_n(y) - U_{n-1}(y)$$ $U_n$ is then the ...
Kroki's user avatar
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3 votes
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Tricky Application of Rouche's Theorem

For “small” $z$ is $e^z \approx 1 + z $ or $e^z - z \approx 1 \ne 0$. That suggests to apply Rouché's theorem to the functions $f(z) = e^z-z$ and $g(z) =1$: For $|z| = 1$ is $$ |f(z)-g(z)| = \left| \...
Martin R's user avatar
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3 votes
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How to check the convergence of $\sqrt[3]{5\sqrt[3]{3\sqrt[3]{5\sqrt[3]{3}}}}\cdots$

In order to show something, we need that something clearly defined in our hands. We need to show that an "expression converges". Well, it is a sequence. Then which are its terms? So as the ...
dan_fulea's user avatar
  • 32.3k
3 votes

Number of real roots of $f(x) = (x^6) + 2(x^4) + (x^2) - 2(x) + 1$

Note $x^6 \ge 0$ with equality only for $x=0$; $2x^4 \ge 0$ with equality only for $x=0$; $x^2-2x+1 = (x-1)^2 \ge 0$ with equality only for $x=1$. Adding, we conclude $f(x) \ge 0$ with no equality.
GEdgar's user avatar
  • 111k
3 votes
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How to find the solution for this integral equation?

$$I=\int_0^\infty \frac{t^{zi}+t^{-zi}}{e^t}\,dt=\Gamma(1+iz)+\Gamma(1-iz)$$ So, considering that we need to find the first zero of function $$f(z)=\Gamma(1+iz)+\Gamma(1-iz)$$ consider as a very ...
Claude Leibovici's user avatar
2 votes

Solution for $(4+2r) x^{(1+r)}−x−1=0$

If you look for approximate but accurate approximations, consider instead the function $$f(x)=\log \left(2( r+2)\, x^{r+1}\right)-\log (x+1)$$which is much closer to linearity than the original one. ...
Claude Leibovici's user avatar
2 votes
Accepted

Solution for $(4+2r) x^{(1+r)}−x−1=0$

$$(4+2r)x^{1+r}-x-1=0\ \ \ \ \ \ (r\in\mathbb{N})$$ Your equation is a polynomial equation and an algebraic equation, and you can use the known solution formulas and methods for algebraic equations. ...
IV_'s user avatar
  • 6,854
2 votes
Accepted

Critical value for the separation of two normal distributions.

From @OmnipotentEntity’s answer, let $x_0^2=e^w\implies z=\frac1{x_0}+x_0=2\cosh(\frac w2)$ $$y=x_0^2\exp\left(\frac{1}{2}\left(\frac{1}{x_0^2}-x_0^2\right)\right)=e^{\sinh(w)-w}$$ This post inverts $\...
Тyma Gaidash's user avatar
2 votes

Proving that the set of $a$ for which the roots of $P(X)-a$ are all real is an interval

Let $n$ be the degree of $P$. The case $n\le 1$ is trivial, so suppose that $n\ge 2$. Let $I'$ consists of values $a$ which $P$ attains exactly $n$ times. Clearly, $I'\subset I$. Let $y_1\le \dots\le ...
Alex Ravsky's user avatar
  • 88.7k
2 votes

General formulae for Roots of 5 degree polynomial

I've no idea what Wolfram Alpha is trying to say here. There is no formula for the roots of the general fifth degree polynomial. Solving quintic equations in terms of radicals (nth roots) was a major ...
Ethan Bolker's user avatar
  • 93.8k
2 votes

Solutions to equation $x^{\frac2 7}=16$

R returns NaN for complex numbers unless you specify as.complex() or use functions designed for complex numbers. You can use ...
Al Veolus's user avatar
2 votes

Solutions to equation $x^{\frac2 7}=16$

In analysis and calculus $b^x:= e^{x\ln b}$ and if $b \le 0$ you will get $NaN$ and that is probably what is going on with R. But in other context if $r\in \mathbb Q$ then $b^r=b^{\frac nm}:= (b^n)^{\...
fleablood's user avatar
  • 124k
2 votes

How to check the convergence of $\sqrt[3]{5\sqrt[3]{3\sqrt[3]{5\sqrt[3]{3}}}}\cdots$

From $x_0 = 5 \sqrt[3]{3}$, we repeatedly do $y_{i + 1} = 3\sqrt[3]{y_i}$ and $x_{i+1} = 5 \sqrt[3]{y_{i+1}}$, i.e., $x_{i+1} = 5 \sqrt[3]{3} \sqrt[9]{x_i}$. We can see that for positive $x$, (1) $x_{...
Vezen BU's user avatar
  • 1,812
2 votes
Accepted

Show that $a_n \underset{(+\infty)}{\sim} n$ with $a_n$ solution of $e^{-x}\sum_{k=0}^{n}\frac{x^k}{k!}=\frac{1}{2}$

Let $N_x$ denote a random variable having the Poisson distribution with rate $x$. Then \begin{align*} f_n(x) = \mathbf{P}(N_x \leq n). \end{align*} Moreover, by realizing the family $(N_x)_{x\geq 0}$ ...
Sangchul Lee's user avatar
2 votes

Tricky Application of Rouche's Theorem

You can try the symmetric version of Rouché': If $f$ and $g$ are analytic in a neighbourhood of $K$ and $|f(z) - g(z)| < |f(z)| + |g(z)|$ for $z \in \partial K$, then $f$ and $g$ have the same ...
Robert Israel's user avatar
2 votes

Solving $2mx^{2m+1}-(2m+1)x^{2m}+1=0$

As a hint$$2mx^{2m+1}-(2m+1)x^{2m}+1=0 \\2mx^{2m+1}-2mx^{2m}-x^{2m}+1=0 \\2m(x^{2m+1}-x^{2m})-(x^{2m}-1)=0\\2mx^{2m}(x-1)-(x^{2m}-1)=0\\$$now factor $(x-1)$ or take $f(x)=2mx^{2m+1}-(2m+1)x^{2m}+1$ $$...
Khosrotash's user avatar
  • 24.8k
2 votes
Accepted

how to find the roots for this equation?

When I have to solve a quartic equation $x^4+\alpha x^3+\beta x^2+\gamma x+\delta=0$, usually the first thing that I do is to apply the substitution $x=y-\frac\alpha4$, because then I get a quartic ...
José Carlos Santos's user avatar
2 votes

Why is it that: "If polynomials have integer coefficients, the roots of those polynomials will be a divisor (factor) of the constant term"?

The argument presented in the body of your question is perfectly sound [with a missing minus sign: a typo?] and is worth digesting. If you find that the notation is what's holding you back, then ...
Randall's user avatar
  • 18.5k
2 votes
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Understanding and Visualizing Complex Roots According to the Fundamental Theorem of Algebra

While FTOA tells us that every polynomial has a certain number of roots, it doesn't give us a method for finding them. In general, while the roots of any linear or quadratic polynomial can be found ...
ConMan's user avatar
  • 24k
2 votes
Accepted

Show that if $x=-1$ is a solution of $x^{3}-2bx^{2}-a^{2}x+b^{2}=0$, then $1-\sqrt{2}\le b\le1+\sqrt{2}$

You seem to have made a mistake during simplification and/or completing the square. Substituting $x = -1$ into the given equation, we get $$\begin{align*} -1 - 2b + a^2 + b^2 &= 0 \\[0.3cm] \...
Haris's user avatar
  • 2,910
1 vote

Roots and extrema of the polynomial $P_n(x)=\sum_{k=0}^n\binom{n+k}{2k}(-x)^k$.

We have the identity: $$\sin\left(\frac{x}{2}\right)P_n(2\cos(x) +2) = (-1)^{n+1}\sin\left(\frac{2n+1}{2}x\right)$$ This holds by induction on the recursion you provided. Now this should answer your ...
Lee Fisher's user avatar
  • 1,874
1 vote

Rouché's Theorem: How many roots does $\lambda-z=\frac{1}{3}e^{z^2}$ have in the strip Re$(z)\in [-1,1]$?

On the circle $|\lambda-z|=\frac{1}{3}e\;$we have \begin{align*} & |\lambda-z| < 1 \\[4pt] \implies\;\;& |\text{Re}(\lambda-z)| < 1 \\[4pt] \implies\;\;& |\text{Re}(\lambda)-\text{Re}...
quasi's user avatar
  • 58.6k
1 vote
Accepted

Some pattern in the gap of the extrema/root of a product using Stirling approximation

$x$ is a root of $f$ iff it makes one of the factors zero; that is, iff for some $n$ we have $g(x/n)=2$ where $g(t)=\frac{\Gamma(1+t)}{\sqrt{2\pi t}e^{-t}t^t}$. It happens that the function $g$ is ...
Gareth McCaughan's user avatar
1 vote

Number of real roots of $f(x) = (x^6) + 2(x^4) + (x^2) - 2(x) + 1$

I'm not sure but the last three terms we can factor them \begin{equation*} x^2-2x+1=(x-1)^2 \end{equation*} Hence, $f(x)=x^6+2x^4+(x-1)^2\geq 0$. Since each term is quadratic, we have $f(x)=0$ if and ...
Eduardo Maza's user avatar
1 vote

Deriving asymptotic for the roots of Digamma Function

After looking at a similar post, it seems to be a much better idea if we write $\psi(1+x_n)=\psi(-x_n)-\pi\cot(\pi x_n)$ and use the recurrence relation of $\psi$ to get $\frac1{x_n}=\psi(-x_n)-\pi\...
Kamal Saleh's user avatar
  • 6,403
1 vote

On the fractional parts of the roots of the Alternating Harmonic Numbers

I finally solved the conjecture. Notice that as $x$ approaches negative infinity, the reflection formula becomes $\ln2+\pi\cot(\pi x)$ and so $\ln2+\pi\cot(\pi x_n)\sim0$. Solving for $x_n$ we get ...
Kamal Saleh's user avatar
  • 6,403

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