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Here's an outline of how I got to this, rather than a pretty, cleaned up proof. First I worked out that $$(-\log{f})'' = \frac{f'^2}{f^2}-\frac{f''}{f},$$ which I thought might be useful because if you divide by $f^2 \geqslant 0$, you find terms that look like $f'/f$. Therefore the inequality is equivalent to showing $$n (-\log{f})'' \geqslant ... 0 Note that for all n, the polynomial z_n\in\Bbb Z[c] always is of the form c^{2^n}+\ldots +c, i.e., we clearly have c=0 as a root and after dividing out this linear factor we have a monic polynomial ending in \ldots +1. By the rational root theorem, any rational root must in fact be integer and a divisor of 1, i.e., the only candidates are c=1 ... 1 Let  c = \sqrt{\sqrt{2} + 1} . Then your given equation is written as :  x^2 - 2cx + c = 0  Solving it gives you :  Δ = 4c^2 - 4c > 0   x_{1,2} = \frac {- 2c \pm \sqrt{Δ}}{2}  You can then proceed with the calculations by substituting c back in. Just be careful with the square roots and your calculations :). 0 You can factor it if you know its roots. In general if r is a root of a polynomial f(x), then (x-r) divides it. For example, if you see that f(x)=3 x^4−8 x^3+16 has the root 2, that is, 3(2)^4-8(2)^3+16=48-64+16=0, then you know that (x-2) divides it. You can then use long division to show that 3 x^4−8 x^3+16= (x-2)(3x^3-2x^2-4x-8) . Now ... 1 You can use the rational root theorem: if this polynomial has a rational root, say p/q in irreducible form, p is a divisor of the constant term and q a divisor of the leading coefficient (this is because it has integer coefficients). This makes a finite number of possibilities: p=\pm 1,\pm 2,\pm 4,\pm 8,\pm 16 and q=1 or 3. Furthermore, it can't ... 1 if you are searching for integer solutions you must look at the divisors of$$16$$1 Hint \  Evaluating \,1 = 2 f(x) + x g(x)\, at \,x=0\, implies 1 is even, contradiction. 0 NTT can be useful for polynomial multiplication in a polynomial ring. If you transform coefficients to the NTT domain, the multiplication between coefficients of two polynomials is component-wise. 0 \frac{1}{2}x^4-\frac{1}{2}x^2-\frac{1}{2}x+\frac{1}{2} has two roots in [0,1] as well, there are many examples. (Unless I don't understand what you mean by integer degree). 1 Hint For c \neq 0, the polynomial$$c \left(x - \tfrac{1}{3}\right) \left(x - \tfrac{2}{3}\right) = c \left(x^2 - x + \frac{2}{9}\right)$$has two roots in [0, 1]. Can you choose c so that the polynomial is a counterexample? 4 (Edit after a bit more research:) Another widespread notation for R[\Bbb Z] is R\Bbb Z. The latter may be preferable if one wants to avoid confusion with the polynomial ring in either a single variable \Bbb Z or the polynomial ring in countably many variables (that happen to be integers, but cannot be added or multiplied as such); then again, for some ... 1 I think you have a typo. The isomorphism is$$\phi (\sum_{k=0}^n a_k x^k) = \prod_{k=0}^n p_{k+1}^{a_k}$$Then, \frac 12 = p_1^{-1} and so the corresponding polynomial is -1x^0. The point is that (\mathbb Q_{>0}, .) \cong (\Pi_p \mathbb Z, +). That is exactly what (\mathbb Z[x], +) is. 0 Based on the answers in this question, I'm able to finish the general case. Let x^3+px+q be a cubic with rational coefficients and having real splitting field of degree 3. Let its roots be a,b,c. Then$$b+c=-a, \qquad bc=-\dfrac{q}{a}$$and b,c are the roots of the quadratic$$ h(x)=(x-b)(x-c)=x^2+ax-\dfrac{q}{a} $$The discriminant of h is b-c. ... 0 In general, this type of problem can be very difficult. To see some of the difficulties, check out Wilkinson's Polyonmial, specifically, the conditioning and stability parts. In the case you describe, if you have real coefficients, then starting with 4 distinct real roots, the roots will move continuously as the coefficients vary and remain real until ... 1 Suppose otherwise, then \;x^2-5\; splits (there is no other option for a quadratic to be reducible) over \;\Bbb Q(\gamma')\; , which means \;\sqrt5\in\Bbb Q(\gamma')\; . But we know there exists a \;\Bbb Q\,- isomorphism \;\phi:\Bbb Q(\gamma')\to\Bbb Q(\gamma)\; since \;\gamma,\,\gamma'\; roots of the same irreducible polynomial \;p(x)\in\Bbb ... 1 Note x^3=2 in the quotient, so x\cdot x^2=2 (in the quotient) so x\cdot {1\over 2}x^2=1. 1 By well-known trigonometric formulas and a few transformations$$\cos(5t)=\cos(4t)\cos(t)-\sin(4t)\sin(t)\\ =(2\cos^2(2t)-1)\cos(t)-2\sin(2t)\cos(2t)\sin(t)\\ =(2(2\cos^2(t)-1)^2-1)\cos(t)-4\sin(t)\cos(t)(2\cos^2(t)-1)\sin(t)\\ =16\cos^5(t)- 20\cos^3(t)+ 5\cos(t)$$Then, ... 0 x^2-x = (x-1)^2 + (x-1), so if you call y = x-1, \Bbb Q[x]/(x^2-x) = \Bbb Q[x]/(y^2+y) \cong \Bbb Q[y]/(y^2+y) 0 Hint   r is a root of f(x) = x^2+x  iff \,-r is a root of f(-x) = x^2-x 2 Hagen's answer is great, if you prefer an alternative note that the Chinese Remainder theorem gives$$\Bbb Q[x]/(x(x+1))\cong \Bbb Q\oplus\Bbb Q\cong \Bbb Q[x]/(x(x-1))$$as rings since (x), (x+1), (x-1) are prime ideals. Edit The generalized CRT says that if you have an ideal I which is the product of coprime prime-power ideals i.e. ... 5 Map x+(x^2+x)\mapsto -x+(x^2-x) 5 As advised by @πr8, two of the Viète formulas allow to replace c by \dfrac{1}{ab} and d by -2a-b-c=-2a-b-\dfrac{1}{ab}. Therefore, the other two Viète formulas can be expressed with variables a and b only, under the form$$f(a,b)=-1 - 2\,a^2\,b - a\,b^2 - 2\,a^4\,b^2 - a^2\,b^3 - 2\,a^3\,b^3 - a^2\,b^4=0 \ \ \ (1)$$and$$g(a,b)=-a - b + ...

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The monomials $x_1^{d_1}x_2^{d_2}x_3^{d_3}$ with $d_1+d_2+d_3=d$ form a basis of $V$. By the Stars and Bars Theorem there are ${d+2\choose2}$ such monomials. It follows that ${\rm dim}(V)={d+2\choose 2}$.

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A basis for $V$ will be the following set: $$\{x_1^ix_2^jx_3^k : i+j+k=d\}$$ So what you looking is essentially number of different partitions of $d$ using 3 natural numbers (called a restricted partition). I think the following wikipedia page includes the answer to that https://en.wikipedia.org/wiki/Partition_(number_theory)

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It is known that $$x^2+1+x^5=(x-x_1)(x-x_2)(x-x_3)(x-x_4)(x-x_5)\tag1$$ Comparing the constants of each side, we have $-1=x_1x_2x_3x_4x_5$. Now, puting in $-x$ in $\text{(1)}$, we have $$x^2+1-x^5=-(x+x_1)(x+x_2)(x+x_3)(x+x_4)(x+x_5)\tag 2$$Multipling $\text{(1)}$ and $\text{(2)}$, we have ...

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The simplest function I can find seems $f(x) = \dfrac{2x+2}{x+2}$. It is easily verified this satisfies the double inequality for $x \ge \sqrt2$. This was obtained by looking at linear approximants, and then setting the conditions $f(\sqrt2) = \sqrt2, f(x) \ge \sqrt2$ and finally the RHS inequality. Even among linear functions, there are of course many ...

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Suppose that $|z_0|$ is larger than $\max(1, |a_0|+\cdots +|a_{n-1}|)$. Estimate the size of $a_{n-1}z_0^{n-1}+ \cdots + a_0$ and compare that to $|z_0^n|$. You should be able to show that $z_0^n$ is greater in absolute value, so $p(z_0)$ can't be zero.

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First of all, let me say that your approach works. That said, you can save some effort when you take an orthogonal basis, because the coefficient of $p_k$ in the projection of $f$ is $\frac{\langle f,p_k \rangle}{\langle p_k,p_k \rangle}$ provided $p_k$ are orthogonal to each other. Thus with $f=x^4$ and the basis you chose you have the coefficients: 1: ...

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If you allow non-monic polynomials the question is not very arithmetic in nature. Any class of integer functions closed under the operations $f(n) \to Af(n)+B$ with integer $A,B$, is guaranteed to have this property for any set that contains long arithmetic progressions (such as the primes, by the Green/Tao theorem as people have mentioned). This ...

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I would start by substituting one $x^2$ into the other equation and multiplying out the squared brackets. Then subtract one equation from the other. You will end up with two new equations which may well be simpler to combine and solve for $x$ or $n$.

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Let $y=e^\lambda$, then we have $$py + (1-p)y^{-2} = 1.$$ Multiplying by $y^2$ and simplifying yields $$py^3 -y^2+1-p=0.$$ Factoring, we have $$(1-y)(1-p+(1-p)y + py^2)=0.$$ The solution $y=1$ has $\lambda =0$, so taking the other root, we have $$py^2 + (1-p)y + 1-p = 0.$$ Completing the square yields $$p\left(y -\frac12(1-p)\right)^2 - ... 3 Using a complicated computer algebra system (Mathematica 10.4), we can get all the roots to this equation as radicals. Two roots are complex and the rest are all real (surprisingly). Module[{roots}, roots = Solve[-x^6 + x^5 + 2 x^4 - 2 x^3 + x^2 + 2 x - 1 == 0, x]; Transpose[{ N[x /. roots], FullSimplify[Element[x, Reals] /. roots], x /. ... 15 This follows by the Green-Tao theorem: there exist arbitrarily long arithmetic sequences of primes. If$$ak+b,a(k+1)+b,\ldots,a(k+m^2)+b$$are prime, we can take P(n)=a(k+n^2)+b, being prime for n=0,\ldots,m. Note that the same works if we require \deg P=1. 22 Yes. In fact, you don't even need the quadratic term: there are degree-1 polynomials giving arbitrarily long sequences of primes by the Green-Tao theorem. (Of course, you could also let P be a constant prime). 15 The Rational Root Test shows that the only possible rational solutions are \pm 1. Substituting gives that x = -1 is one (but x = 1 is not), so polynomial long division gives p(x) = -(x + 1) q(x) for some quintic q. Substituting x = -1 gives that -1 is not a root of q, so if q factors over \Bbb Q, it does so into an irreducible quadratic ... 3 The degree of q is also n (which is even), and the coefficient of the largest exponant is 1. Thus q(x) goes to +\infty when x approaches +\infty or -\infty. Therefore q has a lower bound. Let us denote m = \inf \limits_{x \in \mathbb{R}} q(x). You can easily prove that there exists a \in \mathbb{R} such that q(a) = m. We have q'(a) ... 1 The real problem here is to keep the computations at a minimum, in order to not get lost. So I suggest giving a symbolic name to the right-hand side:$$ w=-\frac{1}{2}+i\frac{\sqrt{3}}{2} $$So the equation becomes$$ \frac{z+i}{2z-i}=w $$Cross multiply and move terms around in the usual way$$ z-2wz=-wi-i $$that becomes$$ z=\frac{i(w+1)}{2w-1} $$Now ... 1 You didn't cross multiply the z correctly. Here is the right steps involved:$$ 2z + 2i = (-2z + i) + z * 2i\sqrt 3 + 1\sqrt 3 \\ 2z + 2z - z * 2i\sqrt 3 = i + \sqrt 3 - 2i \\ 4z - z * 2i\sqrt 3 = \sqrt 3 - i \\ z(4 - 2i\sqrt 3) = \sqrt 3 - i \\ z = \frac{\sqrt 3 - i}{4 - 2i\sqrt 3} \\ $$Now, to rationalize the denominator, we multiply and divide by the ... 0 a) 6 is a zero. (x-6) = (x+1). So \frac {x^3 +4x^2 +5x +2}{x+1} = (x+1)(x+2) b) x^2 + x +8 = x^2 -9x + 8 = (x-1)(x-8) = (x+9)(x+2). 5 We're in luck because this is a very small field. We can just test out all the values. Obviously 0 is not a root. f(1) = 1 + 3 + 5 = 9 = 2 \neq 0. f(2) = 8 + 6 + 5 = 19 = 5 \neq 0. f(3) = 27 + 9 + 5 = 41 = 6 \neq 0. f(4) = 64 + 12 + 5 = 81 = 4 \neq 0 f(5) = 125 + 15 + 5 = 145 = 5 \neq 0 f(6) = 216 + 18 + 5 = 239 = 1 \neq 0. So there are ... 2 No, x is not a unit; there is no polynomial p(x) such that xp(x)=1. This is because \deg xp(x)=1+\deg p(x)\ge 1. Actually, one shows that, for a polynomial of positive degree to be a units, it is necessary that its leading coefficient to be a zero-divisor, which implies n to be composite. A simple example: in \mathbf Z/4\mathbf Z[x], we have: ... 2 Much of this would be covered in "basic ring theory". Let's say that R is a commutative ring with unity (although more general settings can be accommodated, it seems like a good place to start). As you studied "polynomials" in calculus you were asked to consider them as functions of one or more variables. In abstract algebra there is more of an emphasis ... 1 Yes, I'm pretty sure that's definition is correct. Basically, this means that the ring has commutative addition, commutative multiplication, and there's no way you can multiply two non-zero polynomials and get 0. Any polynomial that has an x term in it can not be a unit. There is no way you can multiply something with an x in it and then get 1. It ... 2 Consider F=\Bbb Q[x]/(x^2-c). Then the image \overline{f}(x)\in F has as a factor (x-a-b\sqrt{c}). But then f(x) is divisible by the minimal polynomial for a+b\sqrt{c} over \Bbb Q which is (x-a-b\sqrt c)(x-a+b\sqrt c)= x^2 -2ax+(a^2-cb^2)\in\Bbb Q[x], which of course has a-b\sqrt c as a root. 0 Use that for the complex conjugate we have$$(1)\;\overline{z+w}=\overline z+\overline w\;,\;\;(2) ;\overline{wz}=\overline w\overline z\,,\;\;(3)\;\;\overline z=z\iff z\in\Bbb R$$So that$$p(\overline z)=\overline{p(z)}=\overline0=0$$1 Yes, x being a zero of f just means that f(x) = 0. As for what a polynomial ring R[x] is, it's simply creating a ring (in the usual way) by taking a base ring R, and then using it to create polynomial in x with coefficients from R. So the elements of R[x] are simply polynomials f(x) = \sum_{i=0}^{n} a_i .x^i, where n is some finite ... 2 \cos 3 y=4\cos^3 y-3\cos y. When \cos 3 y=-1/2 and x=\cos y, we have 8 x^3-6 x +1=0, with 3 solutions x=\cos [2\pi(1+3 n)/9] for n\in \{0,1,2\}. 2 Hint: for y=8x^3-6x+1 and y'=24x^2-6 . So ve have two stationary points for x=\pm1/2. Now you can see that there is a positive valued max at x=-1/2 and a negative valued min at x=1/2 and the function is negative for x=-1 and positive for x=1. Use continuity (intermediate value theorem) to show that there are three roots in [-1,1] 0 For f(x)=8x^3-6x+1 the gradient function is f'(x)=24x^2-6=6(4x^2-1). This leads to an observation about the ordinates of the stationary points. Then find f(-1) and f(1). The conclusion follows. 6 Say we consider Q(x)=(x+1)P(x)-x where P(x) is a polynomial of degree n. So we can say that Q(x)=0 for x=0,1,2,\ldots ,n, or in other words, we can say that Q(x) is a polynomial of degree n+1 with n+1 roots namely, 0,1,2,\ldots ,n. Hence we can write that$$Q(x)=c(x-0)(x-1)\ldots (x-n) where $c$ is the leading co-efficient. Therefore ...

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