# Tag Info

7

$f(0)=1$ and $f(1)=0$ for all $n$, so it suffices to show, for example, that $f(2/3)\lt0$. But if $n\gt1$ then $$f(2/3)=(2/3)^{n+1}-(4/3)+1\lt(2/3)^3-(1/3)=-1/27$$

7

Are you familiar with the rational roots theorem? It says that any rational roots of a polynomial with integer coefficients must have the form $\frac{p}{q}$ where $p$ is a factor of the polynomial's constant term and $q$ is a factor of its leading term. The constant term of your polynomial is $12$ and its leading term is $1$, so the possibilities for ...

5

Use the graph to find points of opposite sign for the intermediate value theorem. Use the derivative to prove strict monotonicity.

5

There are lots of powers of $3$ in there. Try $x=3y$, then $$27y^3-324y^2+1215y-1458=0\\y^3-12y^2+45y-54=0$$ Can you find a solution to that, or change it again?

5

OK, as zhoraster wrote, the first method is simply recognizing the Lagrange interpolation polynomial $$\prod_{j\ne i}^n \frac{z-r_j}{r_i-r_j} = \frac{P(x)}{P'(x)\cdot(x-r_j)}.$$ So, we are interpolating the points $(r_i,r_i^k)$ by a polynomial and and wonder why do we get $x^k$ as result. :-) A different approach, using complex analysis (so it will work ...

5

Look at $4x(x+1)$, in $\mathbb{Z}_8$. Oops, a quadratic! Multiply by something. For another example, this time monic, use $x(x+1)(x+2)(x+3)$, again in $\mathbb{Z}_8$.

5

Rouches theorem: let $f,g$ be homomorphic in open set $U$ and $C$ boundary in $U$, of a disc inside $U$. If $|f|>|g|$ for all $z$ on the circle $C$ then $f$ and $f+g$ have same number of zeros inside $C$ (counted with multiplicity). Let $g(z)=z^7+z+1$, and $f(z)=-4z^3$. Both are clearly holomorphic. Then for $|z|=1$ i.e. $z$ on the unit circle ...

5

\begin{align} \left(x-\frac{3}{\sqrt{2}}\right)^2 - \frac{1}{2} &= \left(x-\frac{3}{\sqrt{2}}\right)\left(x-\frac{3}{\sqrt{2}}\right) - \frac{1}{2} \\ \\ &= x^2 - \frac{3}{\sqrt{2}}x - \frac{3}{\sqrt{2}}x + \frac{9}{2} - \frac{1}{2} \\ \\ &= x^2 - \frac{2\cdot 3}{\sqrt{2}}x + \frac{8}{2} \\ \\ &= x^2 - 3\sqrt{2}x + 4 \end{align}

4

Along the unit circle $|x| = 1$, the third degree term dominates the rest. Therefore, there are three zeroes inside the circle (counting multiplicity). The idea is that compared to the polynomial $-4x^3$ (which has $3$ zeroes), the rest (namely $x^7 + x + 1$) is not large enough to "push" any of the zeroes out of the unit circle, and it's not large enough ...

4

Note that the polynomial factors as $$(x^2+x-a)(x^2-x+1-a)$$ The discriminants are both positive if and only if $a\gt 3/4$. And if $a\gt 3/4$, subtraction shows the two quadratics cannot have a common root. Well within high school range.

4

You've made a computational error. $$\frac{1}{x}=-1$$ multiply both sides by $x$ gives $$1=-x$$ Math is still safe to use.

3

Hint: $x = 2$ is a root. Can you take it from here? in general, there are some tricks out there, for example if you have the type I: $x^4+ ax^3+ bx^2+ ax + 1 = 0$, then you divide both sides by the middle $x^2$ and put $y = x+\dfrac{1}{x}$, reducing it to a quadratic equation immediately.

3

The result follows immediately if $a=0$, so suppose $a \neq 0$. The product of the zeros of the polynomial $$P(z) = az^n + z + 1 = a \left(z^n + \frac{1}{a} z + \frac{1}{a} \right)$$ is $1/a$. If all $n$ zeros of $P$ satisfy $|z| > 2$ then this product must be greater than $2^n$ in absolute value. In other words, this would imply that $$... 3 Naturally, the left hand-side is the Lagrange interpolation polynomial for the points r_1,\dots,r_n and the values r_1^k,\dots,r_n^k, hence the claim. But this is kind of cheating, because it uses the same idea you want to avoid: two polynomials of degree at most n-1 agree at n points, they must be equal. Thus, I give two alternative approaches. ... 3 The roots go off to infinity. One can show the roots of \displaystyle f(z) = \sum_{k=0}^{N-1} \frac{z^k}{k!} tend to a curve |z e^{1-z}| = 1 if you rescale by a factor of N. Please see MathOverflow: Roots of truncations of e^x - 1 Math.SE: Approximating roots of the truncated Taylor series of \exp by values of the Lambert W function arXiv: ... 3 You know, a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2+ab+bc+ca) a^3+b^3+c^3=(a+b+c)[(a+b+c)^2-3(ab+bc+ca)]+3abc Now I hope you know the relation between roots. 3 if x^{n+1} = 2x-1, then x^n = 2 - \frac{1}{x} Does this help? 3 The Omno's definition of \log(.) has two drawbacks: its domain is very small and practically, we cannot derive the series which defines it. Let \log(.) be the principal logarithm over \mathbb{C}. In the sequel, A\in M_n(\mathbb{C}) denotes any matrix with no eigenvalues in (-\infty,0]. STEP 1. If A is diagonalizable: A=Pdiag(\lambda_i)P^{-1}, ... 2 Hint: For x \rightarrow 1^- write the function as f(x)=x\left(x^n-2+\dfrac{1}{x}\right) and you can see that its value approximate 0 by negative numbers. 2 Let's take x = 1-\frac{1}{n+1} and prove that$$ x^{n+1}-2x+1<0 \Leftrightarrow \left(1-\frac{1}{n+1}\right)^{n+1}<2 \left(1-\frac{1}{n+1}\right)-1\Leftrightarrow \left(\frac{n}{n+1}\right)^n<1-\frac{1}{n}.$$The last inequality can be proved if we notice that for n=1 it is true and the left side is decreasing function and the right side is ... 2 Hint: Put t=x+\frac{1}{x} then: x^4+ax^3+bx^2+ax+1=0 has a real root if and only if t^2+at+(b-2)=0 has a real root t such that |t| \ge 2, which is equivalent to $$\tag{*} \left[\begin{matrix} \frac{-a-\sqrt{a^2-4(b-2)}}{2} \le -2 \\ \frac{-a+\sqrt{a^2-4(b-2)}}{2} \ge 2 \end{matrix}\right.$$ Using this:$$A \le ...

2

This is not intend to be a canonical answer, but rather a guide of discusion I still think there are many more roots. The reason is as follows: Let's imagine $q$ is purely imaginary, let $q=i(t-\pi/2), t\in \mathcal{R}$, $z=\sinh q=i\sin(t-\pi/2)=-i\cos t$. Then we get the following equation for $t$: $$\frac{1}{2}(\frac{t}{a}-\frac{a}{t})=\cot t$$ Note ...

2

It's clear that $x=0$ is one of the roots. Hence, if we prove there are atleast 2 zeros to $f(x) := x^4-1102x^3-2015$, we are done. Observe, $f(0) < 0$ and $f(-2) > 0$, so from Intermediate Value Theorem there exists at least one root between $-2$ and $0$. Now, lets say there is exactly one real root to $f$ which means that there are 3 non real ...

2

It's about the real zeros of $x q(x)$ with $q(x):=x^4-1102 x^3-2015$. There is the obvious zero $x=0$. Furthermore $q(0)<0$ and $\lim_{x\to\pm\infty} q(x)=+\infty$ guarantee two more real zeros.

2

You can use Descartes' rule of signs to tell you the number of real roots as long as you are not interested in the value of each. First observe that $x=0$ is a root for: $f(x)=x^5 − 1102x^4 − 2015x$ Second, count positive real roots by counting sign changes in $f(x)$, we have: (+-)(--), that is 1 sign change indicating 1 positive root. third, count ...

2

Suppose $a,b,c\in\mathbb{C}$ are the roots to your equation. There is this algebraic identity you should know: $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$ Do some manipulations to get this form: $$a^3+b^3+c^3=3abc+(a+b+c)\left((a+b+c)^2-3(ab+bc+ca)\right)$$ Vieta's formulas are formulas that relate the coefficients of a polynomial to sums and ...

2

Induction on $n\ge 2$. The base case $n=2$ is left to the readers. By the induction hypothesis $\deg f\ge n-1$. If $\deg f\ge n$ then there is nothing to prove, so we may assume $\deg f=n-1$. Write $f(X)=b_{n-1}X^{n-1}+\cdots+b_1X+b_0$ with $b_{n-1}\ne0$. We have $f(a_i)=0$ for $1\le i\le n$, that is, $b_{n-1}a_i^{n-1}+\cdots+b_1a_i+b_0=0$ for $1\le i\le ... 2 For monic examples, try$f(x) = x^4 + 2 x^3 + 7 x^2 + 6 x$or$x^4 + 6 x^3 + 3 x^2 + 6 x$in$\mathbb Z_8$. EDIT: And at the next level, there are monic sextics over$\mathbb Z_{16}$with$16$different roots. For example,${x}^{6}+3\,{x}^{5}+9\,{x}^{4}+5\,{x}^{3}+6\,{x}^{2}+8\,x$. 2 Given$x=\alpha,\beta,\gamma$are the roots of the equation$x^3+2x^2-3x+4=0$So using factor theorem,$x-\alpha,x-\beta,x-\gamma$are the factors of given equation. So $$x^3+2x^2-3x+4 = (x-\alpha)\cdot (x-\beta)\cdot (x-\gamma)$$$$\displaystyle x^3+2x^2-3x+4=x^3-(\alpha+\beta+\gamma)x^2+(\alpha\cdot \beta+\beta\cdot \gamma+\gamma\cdot ... 2 Now Given$\displaystyle x^2-2mx+m^2-1 = 0\Rightarrow (x-m)^2 = 1\Rightarrow x=\pm 1+m$So we get$m=x\mp 1\;,$Given$-1<m<3\bullet \;$If$ m=x-1\;,$Then$-1<x-1<3\Rightarrow 0<x<4\bullet \;$If$ m=x+1\;,$Then$-1<x+1<3\Rightarrow -2<x<2$so we have seen above$x$lies between$-2$to$4\$

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