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1

$$((9x)+(4iy^3))((9x)-(4iy^3))$$


1

You are using the law of large numbers, which says that if you pick the number $\frac kn$ with probability $\binom n k p^k (1-p)^{n-k}$, then with high probability, $\frac kn$ is close to $p$. Now you want to argue that with high probability, $f(\frac kn)$ is close to $f(p)$. And if $f$ isn't continuous, you can't make that last step. For example, if ...


0

HINT: Solve by substituting $x$ with $y+\frac{a}{y}$ and you will solve for $a$ later on.


0

Quick answer: Thanks to egreg, the elements in $\mathcal{R}$ commutes with $x$.


0

A try for a solution. Put $q=1-p$. You want to show $$\frac{(1-q^n)^2}{1-q} \geq 1+(n-1)q^{n-1}-nq^n$$ We have: $$(n-1)q^{n-1}\leq q+\cdots+q^{n-1}$$ and $$ q^n+\cdots+q^{2n-1}\leq nq^n$$ Hence: $$1+(n-1)q^{n-1}-nq^n\leq 1+(q+\cdots+q^{n-1})-(q^n+\cdots+q^{2n-1})=(1-q^n)(1+\cdots+q^{n-1})$$ and we are done.


5

Your equation can be rewritten $$\left(\frac x{x_A}\right)^5+\left(\frac x{x_B}\right)^3-1=0.$$ With $y=\dfrac x{x_A}$ and $r=\dfrac{x_A}{x_B}$, a single parameter remains: $$y^5+r^3y^3-1=0.$$ Then $$r=\sqrt[3]{\frac{1-y^5}{y^3}}=\frac1y\sqrt[3]{1-y^5},$$ can be compared to your approximation $$y=\frac1{\sqrt[4]{1+r^4}},$$i.e. ...


3

The answer given is wrong: $9vw^3$ and $-vw^2$ are not like terms, so they cannot be added to make a single term. Maybe the problem was supposed to be $9vw^2 - 7v^3w - vw^2$?


0

Assume $al_1+bl_2+cl_3=0$, where $0$ is the zero function. Then it is true that $$al_1(\lambda)+bl_2(\lambda)+cl_3(\lambda)=0$$ for every $\lambda$. Conclude by evaluating in $\lambda_i$ for $i=1,2,3$.


1

$$ P(p) = \sum_{N_1 \geq N_2+2} \binom{4}{N_1}\binom{0}{N_2} p^{N_1+N_2}q^{4 - N_1 -N_2 }$$ $\binom{0}{x}$ is 1 if $x=0$ and 0 otherwise so the only $N_2$ That can affect the sum is $N_2 = 0$. similarly the other binomial coefficient is zero if $4=n_1<N_1$ so we don't have an infinite sum, $$P(p) = \sum_{ k= 2}^{4} \binom{4}{k} p^k q^{4-k} $$ Typing on ...


5

This is the generating function for the number of partitions into distinct parts: http://mathworld.wolfram.com/PartitionFunctionQ.html


2

You can use the quo-function, like this: Q<x0,y0,z0> := quo<P | x^2+y, y*z+1>; This gives you a new ring $Q$, which is isomorphic to $P$ modulo the ideal generated by $x^2+y$ and $yz+1$. This is documented in the subsection "Affine Algebras" of the section "Commutative Algebra" in the MAGMA handbook. Changing the normal form: I think (but ...


1

When $a=1$, you can factor $x^2+bx+c=(x+\alpha)(x+\beta)=x^2+(\alpha+\beta)x+\alpha\beta$, and the rule is obvious: factor the independent coefficient $c=\alpha\beta$ and sum to get the linear coefficient $b=\alpha+\beta$. When $a\ne1$, we can modify the procedure by multiplying by $a$ to yield $$a^2x^2+abx+ac,$$ and solve for $z=ax$ instead of $x$: ...


0

$f(x)=ax^2 + bx + c = a(x^2 + \frac ba x + \frac ca)$ assuming $a\neq0$ Continuing.. $f(x)=a(x^2+\frac ba x+ \frac{b^2}{4a^2} -\frac{b^2}{4a^2}+\frac ca)$ $f(x)=a[(x+\frac{b}{2a})^2-(\frac{\sqrt(b^2-4ac)}{2a})^2]$ ...... (Milestone 1) Now any polynomial of the form $g(x,y)=x^2-y^2$ can be simplified as below - ...


0

Your statement, that l has to be between 1/6 and 1/4 for a real solution with t between 0 and 1, does not seem to be correct. As you wrote, let t = 1/2, a = -96*l + 24, b = 8 - 24*l, c = 96*l - 24, d = 72*l - 16 Then this is a solution no matter what l is. Furthermore, if t is arbitrary then this is a solution: l = (t - t^2)/3, a = (4*t + 2) / t^2 , ...


3

This method is just a clever way to set things up so that we can factor by grouping. If $a\ne0$ and we can write $ac=\alpha\beta$ such that $\alpha+\beta=b,$ then we have: $$\begin{align}ax^2+bx+c &= ax^2+(\alpha+\beta)x+\frac{\alpha\beta}a\\ &= ax^2+\alpha x+\beta x+\frac{\alpha\beta}a\\ &= ax\left(x+\frac\alpha a\right)+\beta\left(x+\frac\alpha ...


0

So the remainder when $ax^2+bx+c$ is divided by $(x-1)^2=x^2-2x+1$ is $2x+1$ and $$ax^2+bx+c-a(x^2-2x+1)=(b+2a)x+(c-a)=2x+1$$ So that $2a+b=2$ and $c-a=1$ Then set $x=3$ to obtain $9a+3b+c=15$. Now solve for $a,b,c$


1

Let the polynomial be $f(x)$ We have $$f(x)=(x-1)^2Q_1(x)+2x+1$$ and $$f(x)=(x-3)Q_2(x)+15$$ We also have $$f(x)=(x-1)^2(x-3)Q_3(x)+ax^2+bx+c$$ putting $x=1,$ $$f(1)=3=a+b+c$$ putting $x=3,$ $$f(3)=15=9a+3b+c$$ Therefore eliminating $c$, we have $4a+b=6$ Furthermore, $$f'(x)=2(x-1)Q_1+(x-1)^2Q'_1+2\Rightarrow f'(1)=2$$ and also, ...


0

HINT: Let $$f(x)=A(x-1)^2(x-3)+B(x-1)^2+C(x-1)(x-3)+D(x-3)$$ $$15=f(3)=B(3-1)^2\iff B=?$$ $$2x+1\equiv C(x-1)(x-1-2)+D(x-3)\pmod{(x-1)^2}$$ $$\equiv D(x-3)-2C=Dx-(2C+3D)$$ which needs to be $2x+1$


0

You multiply both sides by $x^{4-n}$ to get $$ x^{n+4-n} = 3 \cdot x^{n-4+4-n} $$ which we can simplify to obtain $$ x^4 = 3 \cdot x^0 = 3 $$


10

It makes a difference if you are considering irreducibility in $\mathbb{Z}[x]$ or in $\mathbb{Q}[x]$. A counterexample for the first case: $f(x) = x$ is irreducible in $\mathbb{Z}[x]$, but $f(2x) = 2\cdot x$ is not. In the second case your statement is true for all $n\neq 0$, since then $n$ is a unit in $\mathbb{Q}[X]$, so $f(x)\mapsto f(nx)$ is an ...


3

Let $n$ be the degree of the polynomial $p(x) = a_n x^n + \dots +a_1x+a_0$. Since $F$ is infinite, there exist $n+1$ distinct elements $b_0, b_1, \dots, b_n \in F$. Call $y_k=p(b_k) \in F$ for $k=0,\dots, n$. Now, you have $$\left( \begin{matrix} 1 & b_0 & b_0^2 & \dots & b_0^n \\ 1 & b_1 & b_1^2 & \dots & b_1^n \\ \vdots ...


0

A proof by contradiction: If $R$ is an integral domain, but $R[x]$ is not, there must exist some minimal counter-example, that is some $f$ of minimal $\text{deg}\ n \geq 0$, for which there exists $g \neq 0$ such that $fg = 0$. If $f(x) = a_0 + a_1x +\cdots + a_nx^n$ and: $g(x) = b_0 + b_1x +\cdots + b_mx^m$ it follows from $fg = 0$ that $a_nb_m = 0$. ...


0

A simple observation: assume $f,g \in R[x] \setminus \{0\}$ then $$f(x) = \sum_{n=0}^N a_n x^n \ g(x) = \sum_{n=0}^M b_n x^n$$ where $a_N,b_M \in R \setminus \{0\}$. The leading coefficient of $fg$, i.e. the coefficient of degree $N+M$ is $a_Nb_M$ which is not null since $a_N,b_M \ne 0$ and $R$ is a domain. As a consequence $fg$ as degree at least $N+M$ ...


0

This is a bit extreme, but this is also a consequence of the famed McCoy's Theorem. If $f(x)\in R[x]$ is a zero divisor, then there is a non-zero $r\in R$ such that $r\cdot f(x)=0$. Thus if $R[x]$ were not a domain, there would be an $f(x)$ which is non-zero and a zero-divisor. But that would give you an $r\in R$ that would kill all of the coefficients of ...


4

Suppose that neither $f$ nor $g$ is the zero polynomial. Then there exist non-negative integers $k$ and $l$ and ring elements $a_0,a_1,\dots, a_k$, with $a_k\ne 0$, and $b_0,b_1,\dots,b_l$, with $b_l\ne 0$, such that $$f=a_0+a_1x+\cdots+a_kx^k \quad\text{and}\quad g=b_0+b_1x+\cdots+b_lx^l.$$ The coefficient of $x^{k+l}$ in the product $fg$ is $a_kb_l$. ...


3

$\bf{Alternatively::}$ If $x=\alpha\;,x=\beta\;,x=\gamma$ are the roots of the equation $x^3+5x^2+7x+11=0.\;,$ Then $\alpha+\beta+\gamma = -5$ Now have to find an equation whose roots are $\displaystyle \frac{\alpha+\beta}{2}$ and $\displaystyle \frac{\beta+\gamma}{2}$ and $\displaystyle \frac{\gamma+\alpha}{2}$ Now Using above relation ...


3

Let $\alpha,\beta,\gamma$ are the roots of the equation $x^3+5x^2+7x+11=0\;,$ Then $\alpha+\beta +\gamma = -5$ and $\alpha\cdot \beta+\beta\cdot \gamma+\gamma\cdot \alpha = 7$ and $\alpha\cdot \beta \cdot \gamma = -11$ Now we have to calculate a polynomial equation whose roots are $\displaystyle ...


6

It is the product of positive and strictly increasing functions. Alternatively: The derivative of $p$ has degree $n-1$ and at least one root in each interval $(a_i,a_{i+1})$ (Rolle). Hence it has no root besides these, especially it does not change signs beyond $x=a_n$.


0

$$\left(x^m-\dfrac1{x^m}\right)\left(x^2+\dfrac1{x^2}\right)=x^{m+2}-\dfrac1{x^{m+2}}+x^{m-2}-\dfrac1{x^{m-2}}$$ $$\implies F_{m+2}=F_m\left(x^2+\dfrac1{x^2}\right)+F_{m-2}$$ where $F_r=x^r-\dfrac1{x^r}$ Now $x^2+\dfrac1{x^2}=\left(x-\dfrac1x\right)^2+2=F_1^2+2$ $$F_3=\left(x-\dfrac1x\right)^3+3\left(x-\dfrac1x\right)$$ ...


3

One has if we denote $Y_n=x^n-1/x^n$ and $Y_1=Y$ $$\begin{align} \left(x-{1\over x}\right)^5 &=x^5-5x^3+10x-{10\over x}+{5\over x^3}-{1\over x^5}\\ &=Y_5-5Y_3+10Y \end{align}$$ And $$\begin{align} \left(x-{1\over x}\right)^3 &=x^3-3x+{3\over x}-{1\over x^3}\\ &=Y_3-3Y \end{align}$$ We then derive from the second identity that ...


1

Try $P(x)=x^5+5x^3+5x $. The idea is to eliminate the highest order at each step.


2

Hint...write out the binomial expansions of $(x-\frac 1x)^5$ and $(x-\frac 1x)^3$ and rearrange terms


1

It's essentially due to Runge's Phenomenon for interpolating polynomials. Even without uncertainty, the behavior of higher order polynomials at their endpoints is very sensitive to the parameter values. Here is a link to a technical explanation that I will regurgitate here. It relates Runge's Phenomenon to regression. Below is a intuitive, but admittedly ...


1

Let $f(x)$ be this polynomial. Then $f'(x) = 5ax^4 + 3bx^2+c>0$ for any $x$ because all of the coefficients are $>0$ and $x^2,x^4 \ge 0$. So, $f$ is strictly increasing everywhere. With $\lim_{x \to -\infty}f(x) = -\infty$ and $\lim_{x \to +\infty} f(x) = +\infty$, we conclude that $f(x) = 0$ has a unique real solution. This of course doesn't mean ...


1

It is a polynomial in $y=x^{\frac 1{12}}$ namely $$p(y)=a_4y^3+a_3y^4+a_2y^6+a_1y^{12}+a_0$$and is thus an element of the polynomial ring obtained by adjoining a $12^{th}$ root of $x$ to the original polynomial ring.


4

There's no name in common use for the set of expressions in exactly the form you quote. One problem with the concept is that they are not closed under multiplication, like polynomials are. For example we have $$ (a_2\sqrt x + a_1x)(b_2\sqrt x + b_1 x) = a_2b_2 x + (a_1b_2+a_2b_1) x^{3/2} + a_1b_1 x^2 $$ where the middle term looks neither like $ax^n$ nor ...


2

You could call it a Puiseux polynomial.


1

To prove the indicated statement from scratch, we first sketch a proof of another result that's called Dickson's lemma: For any infinite sequence of distinct $\alpha_0, \alpha_1\in \mathbb{N}^n$, there exist $i < j$ such that $\alpha_i \leq \alpha_j$ (that is, each component $\alpha_i^k \leq \alpha_j^k$.) Assume the result holds for $(n-1)$-tuples. ...


1

As an addendum to Antonio Vargas's answer, let's prove that the roots of $S_p$ are indeed symmetrically distributed around $-1/2$, or in other words that if $r$ is a root, then so is $-1-r$. A somewhat more precise result is that $S_p(-1-x) = S_p(x)$ when $p$ is odd, and $S_p(-1-x) = -S_p(x)$ when $p$ is even (thus confirming that the polynomial $T(x) := ...


1

The ideals $J_0 \subset J_1 \subset J_2 \subset \cdots$, and $k[x_1, \dots, x_n]$ is Noetherian.


3

Yes. Assume that $P_1$ and $P_2$ are equal on some open interval and let the degrees of $P_1$ and $P_2$ be at most $n$. Then $P_1 - P_2$ is a polynomial of degree at most $n$ that has at least $n + 1$ distinct roots [since every number in the interval is a root]. But a nonzero polynomial of degree at most $n$ has at most $n$ roots.


1

Only if $P_1=P_2$ and that holds for any degrees


0

If your random permutation polynomials are of degree one, they have the form $T(x) = ax+b$ for some $a, b \in \mathbb{Z}_{p}$, $a \neq 0$. I'm not sure how many fixed inputs there are. For a single input $x_{1}$, $T(x_{1}) = ax_{1}+b = y_{1}$ whenever $b = y_{1}-ax_{1}$; there is precisely one $b$ for any choice of $a$, and $p$ choices for $b$, so the ...


5

If two polynomials agree on any open interval (which seems to be what you mean by "lay on each other over more than one point"), then they are identical polynomials. So the answer to your question is "no", because $deg(P_3) \ge 1 + deg(P_2)$, hence $P_3$ cannot be a constant multiple of $P_1$. Why must the two polynomials be equal? Consider points $x = ...


3

Suppose that $$a_n + a_{n-1} x + \ldots + a_0 x^n = \left( b_i + b_{i-1} x + \ldots + b_0 x^i \right) \left( c_j + c_{j-1} x + \ldots + c_0 x^j \right).$$ Then what is $\left( b_0 + b_1 x + \ldots + b_i x^i \right) \left( c_0 + c_1 x + \ldots + c_j x^j \right)$ ?


1

For small scale problems, simply using a global solver appears to work very well, at least for the data I tried. Here is some YALMIP code (MATLAB Toolbox, developed by me) to solve a small instance using YALMIPs global solver bmibnb. It is solved in a second or so if you have a good MILP solver installed. Similiar with scips global solver N = 10; M = 20; ...


1

You shouldn't need to worry about PIDs or factorization or anything like that. The proof that $I\subset A$ is maximal if and only if $A/I$ is a field is a short one, and therefore it is easy to convert your proof that $\mathbb Z[x]/(x,2)$ is a field into a direct proof that $(x,2)$ is a maximal ideal in $\mathbb Z[x]$. Let me show you how that's done. ...


1

We prove that if $I\subsetneq J$ is an ideal, then $J= \mathbb Z[X]$. Let $P(x) \in J \backslash I$ be any polynomial. Then you can do long division by $X$ and get $$P(x)=XQ(X)+m$$ Now, $XQ(X) \in I \subset J$ which implies that $m \in J$. If $m$ is even, then $XQ(X) \in I, m \in I$ would imply that $P(X) \in I$. Therefore $m=2k+1$ for some $k$. This ...


2

One might argue in this fashion, instead: Suppose that $J$ is an ideal such that $I\subsetneq J\subseteq\Bbb Z[x],$ and take any $y\in J\setminus I.$ By definition of $I,$ we know that $x$ doesn't divide $y,$ and so if $y=px+n$ for some $p\in\Bbb Z[x]$ and some $n\in\Bbb Z,$ then $n\ne 0.$ In fact, $n$ cannot be any even integer. (Why?) Thus, if $y=px+n$ for ...


4

Try it like this. Let $p(x)= a_0 + a_1x + a_2x^2 + \dots + a_nx^n$ (with each $a_k \in \mathbb{Z}$) be any element that is in $J$ but not in $I$. Now note that $a_1x + a_2x^2 + \dots + a_nx^n$ is already in $I$ (because it is a multiple of $x$), hence it must be true that $a_0 \in J$ but $a_0 \notin I$. Now $a_0$ can't be an even integer -- if it were, it ...



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