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0

I think the best way is just to make $$ f(x) = \epsilon - c$$ $\forall \; 0<c\leq\epsilon \;$ this shows why the lower bound is degree zero.


1

Yes. The lower bound is the degree of the zero polynomial.


1

Not only the product is an integer, but all the coefficients of your polynomial (that is, all the symmetric polynomials in the given $2^n$ numbers) are. The given numbers are in $\mathbb Q(\sqrt{p_1},\dots,\sqrt{p_r})$ where $p_1,\dots,p_r$ are the primes appearing in the decomposition of $2,\dots,n$. The $\mathbb Q$-automorphisms of $\mathbb ...


5

There is no upper bound, $f(x) = \epsilon x^n$ works for every $n$.


1

"Polynomial" is a precisely defined term. A polynomial is constructed from constants and variables by adding and multiplying. One could add "subtracting", but $x-y$ is $x+(-1)y$, so adding and multiplying are enough. "Alebraic expression" is not a precisely defined term. Algebraic expressions include many things that are not polynomials, including ...


0

Some difference is: Algebraic Expression: may not be a continuous function on $R = (-\infty,\infty)$, but Polynomial is. Example: $\dfrac{x}{x+1}$, and $x^2-1$. The former is not defined at $x=-1$, while the latter is continous throughout $R$.


1

Hint $ $ gcds in a PID D such as $\,\Bbb Q[x]\,$ persist in extension rings because the gcd may be specified by the solvability of (linear) equations over D, and such solutions always persist in extension rings, i.e. roots in D remain roots in rings $\rm\,R \supset D.\:$ More precisely, the Bezout identity for the gcd yields the following ring-theoretic ...


0

let $ f $ be a irreducible polynomial over finite field $\Bbb{F}_q$ and $ \alpha$ is a zero of $f$. let $ d=\mathrm{deg}(f)$. then degree of $\Bbb{F}_q(\alpha)$ is $d$ and the zero is also zero of $ x^{q^d}-x$. therefore all irreducible polynomial with degree $d$ is factor of $x^{q^d}-x$. If $f$ is not a factor of $x^{q^d}-x$, then $f$ is reducible.


0

The first one is easily proven to be irreducible since a polynomial of degree $\;\le 3\;$ is reducible over some field iff it has a root in that field. For the second one observe that $$x^4+x^2+1=(x^2+x+1)^2\in\Bbb F_2[x]$$


3

Hint: If $p(a)=0$, then you can write $p(x)=(x-a)q(x)$. Consider how many times you can do that. Solution:


0

Let $A = \begin{bmatrix}500^3&500^2&500&1\\1100^3&1100^2&1100&1\\2100^3&2100^2&2100&1\\2700^3&2700^2&2700&1\\3400^3&3400^2&3400&1\end{bmatrix}$, $x = \begin{bmatrix}a\\b\\c\\d\end{bmatrix}$, and $b = \begin{bmatrix}0.476\\1.038\\1.982\\2.557\\3.240\end{bmatrix}$. Then, we are trying to find a ...


2

We can rewrite this system as $$ \begin{bmatrix} 500^3 & 500^2 & 500 & 1 \\ 1100^3 & 1100^2 & 1100 & 1 \\ 2100^3 & 2100^2 & 2100 & 1 \\ 2700^3 & 2700^2 & 2700 & 1 \\ 3400^3 & 3400^2 & 3400 & 1 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \\ d \\ \end{bmatrix} = \begin{bmatrix} 0.476 \\ 1.038 \\ ...


1

(The OP suggested a connection to this post.) Because this question is too broad, it spreads thin and may be vague. I suggest it be limited so coefficients $a_i$ are rational, and $x,y$ are also rational. Having said that, two nice results are discussed in Kevin Brown's website. I. Deg 2: The sum of $24$ consecutive squares. $$F(x) = ...


1

As Derek Holt hinted, you should consider the polynomial $h=f-g$. By hypothesis $h\ne 0$. Then consider the associated polynomial function of $h$, denoted here by $h_K$. Suppose the contrary that $h_K=0$, that is, $h_K(a)=0$ for all $a\in K$. This means that all elements of $K$ are roots of $h$. But over a field a polynomial can't have more (distinct) roots ...


1

Since $p$ has even degree and positive leading coefficient we have $\lim_{|x| \to \infty} p(x) = \infty$. Hence $p$ has a global minimum at some point $x^*$. Then you must have $p''(x^*) \ge 0$ (if not, you can combine this with $p'(x^*)=0$ to contradict the fact that $x^*$ is a minimizer). Hence $p(x^*) \ge 0$ and the result follows.


6

Hint: Let $P_n(x)$ be your polynomial. Then show $P_{n+1}(x)=P_n(x-\sqrt{n+1})P_n(x+\sqrt{n+1})$, and show inductively that $P_n(x)$ always has only integer coefficients.


6

Consider $x = \frac{10}{12}$: $$\begin{align*} f\left(\frac{10}{12}\right) &= a\left(\frac{10}{12}\right)^2+b\left(\frac{10}{12}\right)+c\\ &= \frac{10}{144}\left(10a+12b+\frac{144}{10}c\right)\\ &= \frac{10}{144}\left(10a+12b+15c-\frac{6}{10}c\right)\\ &= -\frac{6}{144}c \end{align*}$$ And consider $x=0$: $$f(0) = 0^2a + 0b + c = c$$ If ...


2

From the definition of $P_k(n)$ we get $$ P_k(n) = \sum_{i=1}^n i^k \Rightarrow P_0(n) = \sum_{i=1}^n 1 = n. $$ We now use complete induction over $k$ to proof the statement $S(k)$ $$ P_k(x) = k \int\limits_0^x P_{k-1}(t) \, dt + C_k \, x \quad (*) $$ Base case For $k=1$ we have $S(1)$: $$ 1 \int\limits_0^x P_0(t) \, dt + C_1 \, x = \int\limits_0^x t \, ...


5

Since $P_k(x)-P_k(x-1)=x^k$, $$P_k'(x) - P_k'(x-1) = kx^{k-1} = k(P_{k-1}(x)-P_{k-1}(x-1))$$ Hence, integrating from $0$ to $x$, we find $$P_k(x)-P_k(x-1) - P_k(-1) = I_k(x) - I_k(x-1)$$ where $I_k(x) = k\int_0^x P_{k-1}(t) dt$. Both $P_k(x)$ and $I_k(x)$ are polynomials. Let $c_k = P_k(-1)$. Then we can rewrite the above as $$(P_k(x) - c_k x) - ...


0

Hm. This problem is weird, in that as defined $P_k(n)$ is only defined on the naturals. Although as you noted, you can find a closed form and evaluate it at an arbitrary point. Have you tried taking the derivative of both sides and using Fundamental Thm of Calculus? That would get rid of the integral, and turns the $C_{k+1}\cdot x$ term to just a constant, ...


3

Hint: Start by assuming $$a_1p_1 + a_2p_2 + a_3p_3 = 0$$ for some scalars $a_1, a_2, a_3$. You have to show that in fact $a, b, c$ are all zero. To do this remember that the equation above is an equality of polynomials. So the coefficients of $x^2$ on the left should equal the coefficient of $x^2$ on the right (which is zero!). Same for the coefficient ...


5

We can in fact give a stronger result: Let $P_N$ denote the set of monic polynomials of degree $n > 0$ in $\mathbb{Z}[x]$ whose coefficients all have absolute value $< N$. S. D. Cohen gave in The distribution of Galois groups of integral polynomials (Illinois J. of Math., 23 (1979), pp. 135-152) asymptotic bounds for the ratio in the above limit, and ...


3

Setting $(a,b,c,d) = (0,0,0,0)$ yields $4P(0)^2 = 2P(0)$, i.e. $P(0) = 0$ or $P(0) = \frac{1}{2}$. However, since all the coefficients of $P(x)$ are integers, $P(0)$ must be an integer. Hence, $P(0) = 0$. Then, setting $(a,b,c,d) = (x,0,y,0)$ yields $P(x)P(y) = P(xy)$ for all reals $x,y$. Also, setting $(a,b,c,d) = (0,x,0,y)$ yields $P(x)P(y) = P(-xy)$ ...


1

Starting from $P(0)=0$, now set $b=d=0$ to get $P(a)P(c)=P(ac)$ and $a=c=0$ to get $P(b)P(d)=P(-bd)$ so $P$ must be even. $a=d=1, b=c=0$ gives $P(1)^2=P(1)$, so $P(1)=0,1$ $a=b=c=d=1$ gives $4P(1)^2=P(2)$ and generally $a=b=c=d$ gives $4P(a)^2=P(2a^2)$ It sure looks like the only choices are $P(x)=0$ or $P(x)=x^2$ though I haven't proven that. If we plug ...


1

If this cubic factors over the rationals, by Gauss's lemma it factors over the integers, and it must factor as (linear)(quadratic), i.e. as $(x - r)(x^2 + a x + b)$ where $r,a,b$ are integers, and $r$ is a root of the polynomial. Since $rb = 4$, there aren't too many possibilities to try for $r$ ...


32

Here is a proof which does not require Galois theory. Write $$f(z)=z^n-x^n\quad\hbox{and}\quad g(z)=(z+1)^n-(x+1)^n\ .$$ It is clear that these polynomials have rational coefficients and that $x$ is a root of each; therefore each is a multiple of the minimal polynomial of $x$, and every algebraic conjugate of $x$ is a root of both $f$ and $g$. However, if ...


0

Yes, every polynomial is infinitely differentiable, since every derivative is again a polynomial. A potential source of confusion: a polynomial of degree $n-1$ has $(n-1)$ derivatives that are not identically zero, and the derivatives of higher orders are identically zero. (They are still perfectly valid derivatives.)


1

I claim that the set of automorphisms of $\mathbb{Z}[x]$ are the ring homomorphisms $\phi$ that satisfy $\phi(x) = \phi_0+\phi_1 x$ where $\phi_0 \in \mathbb{Z}$ is arbitrary and $\phi_1 \in \{\pm 1\}$. Suppose $\phi$ is an automorphism. As above, we have $\phi(n) = n$ for $n \in \mathbb{Z}$. Since $\phi$ is an automorphism, we can find an inverse for $x$. ...


1

If $f(x)=a_0+a_1x+\cdots+a_nx^n$, $a_n\ne0$, $n\ge1$, then $\phi(f(x))=a_0+a_1\phi(x)+\cdots+a_n\phi(x)^n$ (note that $a_i\in\mathbb Z$ and then $\phi(a_i)=a_i$) which has the degree equal to $\deg\phi(x)^n=dn\ge d$. Moreover, if $d\ge2$ then $\deg\phi(f(x))=dn\ge2n\ge 2$, so the image of $\phi$ consists of constants and of polynomials of degree $\ge 2$. ...


1

Hint From the vectors $(1,x)$ and using the Gram-Schmidt process construct an orthonormal basis $(e_1,e_2)$ The closest polynomial $p(x)$ to $f$ is it's orthogonal projection onto $\Bbb P^1$: $$p(x)=\langle f,e_1\rangle e_1+\langle f,e_2\rangle e_2$$


3

Call $ax+b$ the inverse. Force: $$(2x+1)(ax+b) = 1 \implies 2ax^2+(a+2b)x+b = 1.$$ This suggests $b = 1$. And $a$ must satisfy: $$a +2\equiv0 \pmod 4 \quad \mbox{and} \quad 2a \equiv 0 \pmod 4.$$ Notice that $a = 2$ does the job. I leave you to check that $(2x+1)(2x+1) = 1$...


0

In this case $L^2=\{f : \int_{0}^{1}|f(x)|^2dx<1\}$. By definition $\langle f,g\rangle=\int_{o}^{1}f(x)g(x)dx$. No, they're asking for such a polynomial $p(x)=a+bx^2$ that $\|f-p\|^2=\langle f-p,f-p\rangle$ is smallest possible. Find projection $f$ onto subspace $\text{span}\{1,t^2\}$, for example here you can find an algorithm.


1

There are no mistakes in Wolfram Alpha's calculation or your input. Excel is acting weirdly though. It seems it somehow doesn't recognize the x-values. This can already be seen by the fact that the x-axis on the graph is not to scale. I think it calculates the formula based on the points $x = 0,1,2,\ldots, 10$. Look at ...


1

are you sure you are not solving $f(x-x_s)$ rather than $f(x)$ i.e. the polynomial for the first one is $$-0.0102(x-x_s)^2 + 0.0584(x-x_s) + 1.9332$$, where $x_s$ is some characteristic shift..i.e if we use $x_s = 114.0$ then we yield 1.9332 which corresponds more closely to the solution. or better still $\bar{x} = \frac{x}{x_s}$ which leads to $$ ...


0

Let $Z=7$ and consider $y=W(2x+3x^2+e^{5x+x^2})$ where $W$ is a Lambert Omega function. In general you have $y=W(2x+3x^2+e^{5x+x^2}-7+Z)$ where $z$ is just a chosen number. I don't see any simplification after this point.


27

I will prove the following fact: Let $x$ be a real number such that $x^n$ and $(x+1)^n$ are rational. Then $x$ is rational. Without loss of generality, I assume that $x \neq 0$ and $n>1$. Let $F = \mathbb Q(x, \zeta)$ be the subfield of $\mathbb C$ generated by $x$ and a primitive $n$-th root of unity $\zeta$. It is not difficult to see that ...


0

It seems the following. We can prove that $x$ is rational provided $n$ is a power of an odd prime $p$. Denote $x^n=r$ and $(x+1)^n=s$ with $r$, $s$ rationals. If both $r$ and $s$ are $p$-th powers of rational numbers, then we descent from $n$ to its $p$-th root. So without loss of generality we can assume that one of the numbers $r$ and $s$ (for instance, ...


1

No particular rule for polynomials: they are elements of a vector space of dimension $3$. Since $\{1;x;x^2\}$ is obviously a basis for your space, you can simply show that the matrix $$ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ -1 & -1 & 1 \end{bmatrix} $$ has rank $3$, which is done by a simple elimination. Why is this true? ...


0

This does not immediately answer your question but here is how I would approach it. Let $X = 4 y$ and $Y = 2 x^2 + 1$. Then there exists integers $x, y$ satisfying $y^2 = 3 x^4 + 3 x^2 + 1$ if there exists an integer solution $(X, Y)$ to $$X^2 - 12 Y^2 = 4$$ satisfying $4 \mid X$ and $Y$ is odd of the form $2 x^2 +1$. We know the fundamental solution is $(X, ...


0

You need to show two things: (i) $S = \{p_1, p_2, p_3\}$ form a linearly independent set, and (ii) $\operatorname{span} S = \mathbb{P}^2$ The first part is relatively easy, let $\vec p_1, \vec p_2, \vec p_3$ be the coefficient vectors of $p_1, p_2$, and $p_3$ respectively. Then, $S$ is linearly independent if the system $$\begin{bmatrix} \vec p_1 & ...


1

Doing almost the same as mixedmath, I have very slightly different results since $$e^{x^2} = 1+x^2+\frac{x^4}{2}+\frac{x^6}{6}+\frac{x^8}{24}+O\left(x^9\right) $$ $$\ln(1 + x^2) = x^2-\frac{x^4}{2}+\frac{x^6}{3}-\frac{x^8}{4}+O\left(x^9\right)$$ $$\cos(2x)=1-2 x^2+\frac{2 x^4}{3}-\frac{4 x^6}{45}+\frac{2 x^8}{315}+O\left(x^9\right)$$ $$2x\sin x =2 ...


0

Working in $(\mathbb{Z}/3\mathbb{Z})[[x]]/\langle x^2 + x + 1 \rangle$, one (awful) way to do it is to notice that $2x + 1 = 1 - x$, so $$(2x+1)^{-1} = \frac{1}{1-x} = 1 + x + x^2 + x^3 + \dotsb. $$ But $1 + x + x^2 = -1 = 2$ and, since $x^2 = -x - 2 = 1-x$, $$x^3 + \dotsb = \frac{x^3}{1-x} = \frac{x^3}{x^2} = x,$$ giving $(2x+1)^{-1} = 2 + x$.


1

Assume $a(1-x^2) + b(x-x^2) + c(x+x^2) = 0, \forall x \in \mathbb{R}$, you need to establish that $a = b = c = 0$. Put $x = 1$, $2c = 0 \rightarrow c = 0$, and let $x = 0 \rightarrow a = 0$, finally put $x = 2 \rightarrow -2b = 0 \rightarrow b = 0$.


0

You can just let the unknown inverse be $ax+b$ and then multiply $(ax+b)(2x+1)=2ax^2+(a+2b)x+b.$ Equate this to $1$ in $Z_3[x]/(m)$ which means using $x^2=-x-2$ that the term $2ax^2$ becomes $2a(-x-2)=-2ax-4,$ which is then added to $(a+2b)x+b.$ Try from there. I'm pretty sure this method should work, though in a comment, the user Unit already gave the ...


1

As you suggested, let $m$ be the least common multiple of the denominators of the coefficients of $f$, and $f^* = mf$, so that $f^*$ has integer coefficients. Now let $d$ be the greatest common divisor of the coefficients of $f^*$, and set $df'=f^*$ where $f'$ is primitive, so you have $df' = mf$, and $f = \frac{d}{m} f'$. This shows $(2)$. Now suppose ...


2

$$=e^{\sin(x)}=1+\left(x-\dfrac{x^3}{6}\right)+\dfrac{\left(x-\frac{x^3}{6}\right)^2}{2}+\dfrac{(x-x^3/6)^3}{6}+\dfrac{(x-x^3/6)^4}{24}+O(x^5)$$ and this term $$\dfrac{(x-x^3/6)^4}{24}=\dfrac{x^4}{24}+o(x^5)$$ so $$e^{\sin{x}}=1+x+\dfrac{x^2}{2}+\left(-\dfrac{x^4}{6}+\dfrac{x^4}{24}\right)+o(x^5)$$


0

HINT: You have to consider all summands up to $O(x^5)$.


4

Then $f(x)=g(x)h(x)$ where $\deg g,\deg h< \deg f$. If $f(x)$ is a prime for infinitely many integers $x$, then either $g(x)$ or $h(x)$ is $1$ for infinitely many $x$, so...


4

Well, you can use $f(x-1)=f(x)+(2x)^2$ to compute values for $f(-1), f(-2)$, and $f(-3)$ starting with $f(0)$. Or you could use finite difference to start with $f(x)-f(x-1)=-(2x)^2$.



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