User i707107 - MathOverflow

Answer by i707107 for Any closed form for series like $F(x)=\Sigma_{i=p}^{\infty}x^p$,p is prime$?

2013-05-24T01:44:57Z

See this blog post: http://uniformlyatrandom.wordpress.com/tag/power-series/

contains a proof of the result by Fatou:

A function whose power series expansion has integer coefficients and radius of convergence 1 is either rational(in $\mathbb{Q}(x)$) or transcendental(over $\mathbb{Q}(x)$).

If $r$ is rational, then the decimal expansion will be eventually periodic. So we have rational function. (Indeed this can be done explicitly)

Otherwise, when $r$ is irrational, then the resulting function cannot be rational(plug in $1/10$, then you get irrational number). Thus, we have transcendence of $f$.

In particular, your functions $F$ in the beginning are transcendental. However, getting closed form will be extremely hard for those examples.

Ramification in Division field of Abelian Varieties

2013-05-16T20:13:34Z

This might be a very simple question, and that might be the reason that I could not find any reference on this.

My question is

Let $A$ be an abelian variety defined over a number field $k$, and $N$ the conductor. Let $m\geq 2$. Consider the division field $k(A[m])$. Let $\mathfrak{p}$ be a prime ideal in $k$ that divides $m$. Also suppose that $k(A[m])\neq k$. Then is it true that $$\mathfrak{p} \textrm{ is ramified in } k(A[m]) ? $$

If this is not true, then can anyone provide a counterexample?

I know that if $\mathfrak{p}$ does not divide $mN$, then it should be unramified.

Answer by i707107 for Efficient (divergent) summation for sum of zetas at negative arguments?

2013-04-19T22:51:07Z

This is a partial answer for $m=-1$, proving that Gottfried Helms' guess is correct.

What we need is the following:

When $$B_1(x)= \left(\frac 12- \frac 1x -{1 \over 1-\exp(x)}\right) \cdot \frac 1x, $$

$$\int_0^\infty \exp(-t) B_1(t) dt {=}1- \log( \sqrt{2 \pi}) $$

I have used the same idea for this question in MSE: http://math.stackexchange.com/questions/340718/references-to-integrals-of-the-form-int-01-left-frac1-log-x-frac/342072#342072

The idea is considering the following integral:

$$F(s)=\int_0^\infty \left(\frac 12- \frac 1t -{1 \over 1-\exp(t)}\right) \cdot t^{s-1} e^{-t} dt$$

This is well-defined if $\textrm{Re}(s)>-1$. In particular for $s=0$.

This integral can be treated term by term if we have $s$ with sufficiently large real part (absolute convergence).

The result is $$\frac 12 \Gamma(s) - \Gamma(s-1) + \Gamma(s)(\zeta(s)-1)$$

Since the integral defines analytic function on $\textrm{Re}(s)>-1$, the result of integral should be analytic continuation of the function $\frac 12 \Gamma(s) - \Gamma(s-1) + \Gamma(s)(\zeta(s)-1)$.

Now the result for $s=0$ follows if we take limit $s\rightarrow 0$ in that expression.

Ordinary Generating Function for Mobius

2013-04-03T19:55:27Z

Is there any information known for the Ordinary Generating Function for Mobius? $$ \sum_{n=1}^{\infty} {\mu(n)}x^n $$ I know that

It has radius of convergence 1.
Does not have limit as $x\rightarrow 1$.

My question is

Does it have limit as $x\rightarrow \textrm{exp}(i\theta)\neq 1$?
Is there a similar result for $\textrm{exp}(i\theta)\neq 1$ with $$ \sum_{n\leq x}\mu(n)=o(x)$$ i.e. is there a result of this type? $$ \sum_{n\leq x}\mu(n)\textrm{exp}(in\theta) = o(x)$$

EDIT1: Aside from my questions, there are some known results for the OGF for Mobius.

EDIT2: Found a reference for (III)

Reference: JASON P. BELL, NILS BRUIN, AND MICHAEL COONS "TRANSCENDENCE OF GENERATING FUNCTIONS WHOSE COEFFICIENTS ARE MULTIPLICATIVE" available at arXiv:1003.2221v2

P. Borwein, T. Erd´elyi, and F. Littman, Polynomials with coefficients from a finite set available at http://www.cecm.sfu.ca/personal/pborwein/PAPERS/P195.pdf

(I) This function has the unit circle as natural boundary.

(II) This function is a transcendental function.

(III) It is not bounded on any open sector($\{z:|z|<1, z=re^{i\theta}, \alpha<\theta<\beta\}$)of unit disc.

This result (III) together with Baire Category answers my question 1 with a dense set of $\theta$'s, but as Prof Tao mentioned, there can be a possibility that it is bounded for some values of $\theta$.

Answer by i707107 for $\prod_{n=1}^{\infty} n{}^{\mu(n)}=\frac{1}{4 \pi ^2}$

2013-03-22T22:37:15Z

This is a discussion following from David Speyer about Cesaro summability of the series. This series is not Cesaro summable in any order.

In "The General Theory of Dirichlet Series" by Hardy and Riesz. Chapter 4.

Let $\kappa\geq 0$. $0\leq \lambda _0<\lambda_1<\cdots<\lambda_n\rightarrow\infty$, and write $$ A_{\lambda}^{\kappa}(w) = \sum_{\lambda_n \textrm{<}w}(w-\lambda_n)^{\kappa}a_n. $$

\bf Summability $(\lambda,\kappa)$. \rm If $A_{\lambda}^{\kappa}(w)\sim Aw^{\kappa}$ as $w\rightarrow\infty$, then we say that $\sum a_n$ is summable $(\lambda,\kappa)$ to sum $A$.

The second consistency theorem(we do not need first consistency theorem here) states that If $\sum a_n$ is summable $(l,\kappa)$ where $l_n=e^{\lambda_n}$, then it is summable $(\lambda,\kappa)$ to the same sum. In particular, Cesaro summability of order $\kappa$ implies $(\log n, \kappa)$ summability to the same sum.

We consider the formula $$ \sum_{n=1}^{\infty}a_ne^{-\lambda_ns}=\frac{1}{\Gamma(\kappa+1)}\int_0^{\infty}s^{\kappa+1}e^{-s\tau}A_{\lambda}^{\kappa}(\tau)d\tau.$$

This is originally valid if $\sigma> max(0,\sigma_c)$ where $\sigma_c$ is the abscissa of convergence of the left series. On the other hand, the right allows an analytic continuation of the function represented by the series on the left up to $\sigma>0$.

What we use here is $a_n = \mu (n)\log n$ and $\lambda_n = \log n$ .

Hence, assuming Cesaro summability of any order $\kappa$, will allow $$-\frac{\zeta'(s)}{\zeta^2(s)}$$ to have an analytic continuation to $\sigma>0$ which contradicts the fact that $\zeta(s)$ has zeros on the critical line.

Answer by i707107 for Expression for the sum of square roots of zeros of a polynomial

2013-03-11T22:58:19Z

Let us consider this in the rational function field of $n$ variables $x_1,\cdots, x_n$, Let $f(x)=(x-x_1)\cdots (x-x_n)=x^n+c_1x^{n-1}+\cdots+c_{n-1}x+c_n$ Then the coefficients are elementary symmetric functions in $x_1,\cdots, x_n$. We see that $K=\mathbb{Q}(x_1,\cdots,x_n)$ is a Galois extension of $k=\mathbb{Q}(c_1,\cdots,c_n)$ with Galois group $S_n$.

Now we consider $r=\sqrt{x_1}+\cdots +\sqrt{x_n}$. Then $r$ has degree $2^n$ over $K$, so there is a polynomial $p$ of degree $2^n$ which is satisfied by $r$. Let $p(x)=x^{2^n}+f_1 x^{2^n-1} +\cdots + f_{2^n}$, where $f_i\in K$. Further, we have $f_i\in k$, indeed each $f_i\in\mathbb{Z}[c_1,\cdots,c_n]$.

Thus, it follows that $F=k(r)$ also has degree $2^n$ over $k$. Note that $F$ is not a normal extension of $k$. However, $KF=\mathbb{Q}(x_1,\cdots,x_n,r)$ is a normal extension of $k$. Indeed, it is the normal closure of $k(r)$ over $k$. So, the Galois group $Gal(KF/k)$ has $S_n$ as a quotient. $S_n$ is not solvable if $n\geq 5$. Therefore, if $n\geq 5$, then we cannot have "a root formula" for $r$ in terms of $c_1,\cdots c_n$.

Answer by i707107 for Invertibility of a certain matrix indexed by the Hamming cube

2013-03-07T09:53:03Z

After seeing very good proofs of this, I could not think other ways to prove than using induction. I read O. Selim's proof, and I think it is possible to simplify their induction argument.

We can associate each subset of $S$ to a binary expansion so that natural numbers from $0$ to $2^n-1$ will represent all subsets of $S$. The components of $2^n \times 2^n$ matrix $A_n$ where $|S|=n$ is then $$ A_{ij}= 1 \textrm{ if the binary expansions of $i$ and $j$ has 1 in common at some digit} $$ $$A_{ij}=0 \textrm{ otherwise} $$ where $i,j = 0, 1, \cdots , 2^n-1$. So, this matrix is basically one column and one row of zeros added to your original matrix, this does not change the rank.

Let $E_n$ be the $2^n\times 2^n$ matrix with all 1's. Then we have the following block matrix form $$ A_{n+1}=\begin{pmatrix}{A_n}&{A_n}\\ {A_n}&{E_n} \end{pmatrix}, \ E_{n+1}-A_{n+1}=\begin{pmatrix}{E_n-A_n}&{E_n-A_n}\\ {E_n-A_n}&{0}\end{pmatrix} $$ We assume our induction hypothesis $$ \textrm{rank}A_n=2^n-1, \ \textrm{rank}(E_n-A_n)=2^n $$ After elementary row and column operations, we have $$ \textrm{rank}A_{n+1}=\textrm{rank}\begin{pmatrix}{A_n}&{0}\\ {0}&{E_n-A_n} \end{pmatrix}, \\ \textrm{rank}(E_{n+1}-A_{n+1})=\textrm{rank}\begin{pmatrix}{0}&{E_n-A_n}\\ {E_n-A_n}&{0} \end{pmatrix} $$ Then we have $$ \textrm{rank}A_{n+1}=2^{n+1}-1, \ \textrm{rank}(E_{n+1}-A_{n+1})=2^{n+1} $$ Added) This method might also work for finding inverse matrix. Then we have to consider $n-1\times n-1$ minor of $A_n$ with row and column of all zeros deleted.

Answer by i707107 for Series defined by a fixed-point functional equation

2013-03-07T04:10:11Z

This maybe off-topic answer, but since OP asked

(1) Have you encountered other fixed-point functional equations of the described form?

I will put an example that I have seen.

$$ S(x)=x-S(x^2) $$ This gives the power series $$ S(x)=\sum_{k=0}^{\infty} (-1)^k x^{2^k}. $$ This is the famous example given by Hardy. Using Tauberian Theory, one can prove that this is not Abel summable. i.e. $$ \lim_{x\rightarrow 1-} S(x) \textrm{ does not exist.} $$

cf) G. H. Hardy, “On certain oscillating series”, Quart. J. Math. 38 (1907), 269–288.

Answer by i707107 for Smith normalform of a Matrix with -1 outside the diagonal

2013-03-04T01:39:50Z

This is a partial answer for the special case when all $a_i$ are distinct. I will work on complex(or algebraic closure of $\mathbb{Q}$) for convenience. Writing $M-\lambda I = D_{\lambda}+E$ where $D_{\lambda}$ is the diagonal matrix with diagonal entries $a_0-\lambda+1, \cdots , a_n-\lambda+1$. With this expression, it is easy to calculate the determinant, which will give the characteristic polynomial of $M$.

If $f(\lambda)=(a_0-\lambda+1)\cdots (a_n-\lambda+1)$, then we have $$ \textrm{det}(M-\lambda I) = f(\lambda)+f'(\lambda).$$ Considering the identity $(f(t)e^t)'=(f(t)+f'(t))e^t$, we can find the roots of characteristic polynomial by looking at the critical points of $f(t)e^t$. Also, by Mean Value theorem, we know that the critical points of $f(t)e^t$ are all distinct.

Therefore, if we let $\lambda_0, \cdots, \lambda_n$ be the critical points of $f(t)e^t$, then we have the following Smith Normal form of the matrix $M-\lambda I$ over the polynomial ring $\mathbb{C}[\lambda]$, (or $\overline{\mathbb{Q}}[\lambda]$): $$ \textrm{Diag}(1,\cdots, 1, (\lambda-\lambda_0) \cdots (\lambda-\lambda_n)).$$ Hence, we obtain the Smith Normal form of $M$ in this case: $$\textrm{Diag}(1,\cdots, 1,\lambda_0 \cdots \lambda_n).$$ There is a natural way of bringing this down to $\mathbb{Z}$, then we have to deal with the irreducible factors of $f(t)+f'(t)$. This is indeed $$\textrm{Diag}(1,\cdots, 1, f(0)+f'(0)).$$ Added) This method also works for the case below:

The cardinality of $\{ j: a_i = a_j \}$ is at most $2$, for any $i=0,\cdots, n$.

Added2) This gives the SNF of $M$ over $\mathbb{Q}$. Over $\mathbb{Z}$ will be certainly more difficult.

Answer by i707107 for questions on central simple algebras

2013-02-27T00:52:19Z

The double centralizer theorem (cf. Knapp, Anthony W. (2007), Advanced algebra, Cornerstones p 115 Theorem 2.43) implies that $$\textrm{dim}_F B \cdot \textrm{dim}_F C = \textrm{dim}_F A.$$ Thus we should have $$m^2\textrm{dim}_F \Delta_B \cdot k^2\textrm{dim}_F \Delta_C = n^2\textrm{dim}_F \Delta_A.$$ Furthermore, $C$ is simple, and $B$ is the centralizer of $C$ in $A$.

As a corollary of the double centralizer theorem, If $\Delta$ is a central finite dimensional division algebra over a field $F$, and if $K$ is the maximal subfield of $\Delta$, then $$\textrm{dim}_F\Delta=(\textrm{dim}_F K)^2.$$ So, for $\Delta_A$, we have $\textrm{dim}_F \Delta_A = (Ind_A)^2$. For $\Delta_B$ and $\Delta_C$, note that $Z(B)=B\cap C = Z(C)$. Thus, $$\textrm{dim}_F\Delta_B=\textrm{dim}_{B\cap C}\Delta_B\cdot\textrm{dim}_F(B\cap C)=(Ind_B)^2\cdot\textrm{dim}_F(B\cap C)$$ $$\textrm{dim}_F\Delta_C=\textrm{dim}_{B\cap C}\Delta_C\cdot\textrm{dim}_F(B\cap C)=(Ind_C)^2\cdot\textrm{dim}_F(B\cap C).$$

Matrix Transpose Similarility

2013-02-19T19:40:49Z

A famous problem in linear algebra is that "A $n\times n$ matrix $A$ over a field $\mathbb{F}$ is similar to its transpose $A^T$."

I know one proof using Smith Normal form. However, I want to know an elementary proof avoiding any concepts related to SNF.

My question is: "Is there an elementary way to prove this?"

Requirements are:

Do NOT use the structure theorem over PID.

Do NOT use the Smith Normal form (nor Jordan canonical form).

Do NOT use the concept of invariant factors.

Provide an explicit invertible matrix P such that $$A=PA^T P^{-1}.$$

Answer by i707107 for Riemann zeta at even integers

2012-04-12T17:37:12Z

These are the proofs that I have seen:

The proof using Fourier series: Reference: Stein Shakarchi "Fourier Analysis, the introduction" p 97 Exercise 4 The key ingredient of this proof is the following identity $$\sum_{n=1}^{\infty} \frac{1}{n^2-\alpha^2}=\frac{1}{2\alpha^2}-\frac{\pi}{2\alpha\tan{\alpha\pi}}$$ which can be proved by using Fourier series of $\cos(\alpha x)$. This allows an expression of $\zeta(2n)$ by Bernoulli numbers.

The proof using the functional equation of $\zeta(s)$: Reference: E.Titchmarsh "The Theory of the Riemann zeta function" p18 (2.4 Second method)

Chapter 2 of this book is entirely devoted to the proofs of the functional equation $$\pi^{-s/2}\Gamma(s/2)\zeta(s)=\pi^{-(1-s)/2}\Gamma((1-s)/2)\zeta(1-s).$$ Section 2.4 is one of the proofs, it uses te residue theorem of complex analysis, and derives the formula $$\zeta(1-2m)=\frac{(-1)^mB_{2m}}{2m}$$ for $m=1,2,\cdots$ The formula for $\zeta(2n)$ is now followd by the functional equation.

The result is: $$2\zeta(2n)=(-1)^{n+1}\frac{(2\pi)^{2n}}{(2n)!}B_{2n}$$ where $$\frac{z}{e^z-1}=\sum_{n=0}^{\infty} \frac{B_n}{n!}z^n.$$

Answer by i707107 for Linear algebra and Cayley Hamilton

2013-02-06T02:16:59Z

Let $T^{(i)}=A^{i-1}b$ for $i=1,\cdots n$. You can solve $\widehat{b}$ from the equation $T\widehat{b}=b$. By Cramer's rule,

$$ \widehat{b_1}=\frac{det[b, T^{(2)}, \cdots, T^{(n)}]} {det T} $$ $$ \widehat{b_2}=\frac{det[b, b, T^{(3)}, \cdots, T^{(n)}]} {det T} $$ $\cdots$ $$ \widehat{b_n}=\frac{det[b, T^{(2)}, \cdots, T^{(n-1)},b]} {det T} $$ Then you can see that $\widehat{b_1}=1, \widehat{b_i}=0$ for all $i=2,\cdots, n$.

Further for $\widehat{A}$, use Cramer's rule to each column vector of $\widehat{A}$, in the equation $T\widehat{A} = AT$.

Answer by i707107 for If $f$ is infinitely differentiable then $f$ coincides with a polynomial

2013-02-02T10:46:36Z

I remember solving this in one whole week, but after a while, forgot how I did. I actually tried to remember but couldn't, so I tried this again. Spending five days, I got the solution. Compared to other solutions posted here, mine is more brute force approach. Mine has the same line of argument with the proof of Baire Category Theorem.

Problem: $f\in C^{\infty}(\mathbb{R})$, for all $x\in \mathbb{R}$, there exists $n_x\in \mathbb{N}$ such that $f^{(n_x)}(x)=0$. Show that $f$ is a polynomial.

My solution: Suppose $f$ is not a polynomial.

Let $A_n = \{x\in\mathbb{R}|f^{(n)}(x) = 0\}$. Each $A_n$ is a closed set, so it can be decomposed as $A_n=P_n\cup C_n$, where $P_n$ is perfect set, $C_n$ is at most countable. Note that $\cup_n A_n = \mathbb{R}$, and $P_n\subset P_{n+1}$ for all $n$. We derive a contradiction by showing that $\cap_n P_n^c$ is uncountable. (This is a contradiction since $\cap_n P_n^c\subset \cup_n C_n$).

Let $(a,b)$ be any maximal interval of a $P_n^c$(which exists since we assumed $f$ is not a polynomial). Then $P_{n+1}$ cannot contain intervals $(a,s)$ or $(t,b)$, otherwise, $f^{(n)}$ be constant on those intervals, and the constant should be zero, which contradicts maximality of $(a,b)$.

Thus, we have either one of two cases:

$P_{n+1}^c$ has at least two maximal intervals inside $(a,b)$. Call one of them by $L$, and one of the others by $R$. (let all members of $L$ be less than any members of $R$)
$(a,b)$ remains a maximal interval of $P_{n+1}^c$.

Let $I_{n+1}$ be 'either $L$ or $R$' in Case 1, '$(a,b)$' in Case 2.

We continue finding maximal interval $I_{m+1}$ of $P_{m+1}^c$ inside $I_m$ where $m\geq n$.

Considering choices of $I_m$ for $m\geq n$, and taking intersections $\cap_{m\geq n} I_m$, we can generate uncountably many members of $\cap_n P_n^c$.

Remark:: If Case 1 occurs infinitely many times, consider $LR$ sequences that both have $L$ and $R$ infinitely many times.

If Case 1 only occurs finitely many, then the interval sequence $I_m$ is stationary.

Answer by i707107 for sum of $1/ \phi(n)^2$

2012-06-05T17:41:18Z

Theorem 2.14 of "Multiplicative Number Theory" I. Classical Theory, by Montgomery & Vaughan implies that $$\sum_{n\leq x} \left(\frac{n}{\phi(n)}\right)^2 = O(x)$$

Use this and partial summation method with $$\sum_{n\leq x} \left(\frac{n}{\phi(n)}\right)^2 \frac{1}{n^2}$$

Answer by i707107 for What is the probability that two numbers are relatively prime?

2012-05-15T20:19:47Z

This is a very standard counting problem in analytic number theory. Here's a rigorous proof: It is enough to derive an asymptotic formula for $$\sum_{a,b\leq n, (a,b)=1} 1 $$ This is $$\sum_{a,b\leq n, d|a, d|b} \mu(d) $$ $$=\sum_{d\leq n} \mu(d)\sum_{k\leq n/d , l\leq n/d} 1$$ $$=\sum_{d\leq n} \mu(d) ((n/d)^2 + O(n/d) ) $$ $$=n^2\sum_{d\leq n} \mu(d)/d^2 + O(n\log n)$$ $$=n^2\sum_{d=1}^{\infty} \mu(d)/d^2 + O(n) + O(n\log n)$$. $$=n^2 6/\pi^2 + O(n\log n).$$

Answer by i707107 for Some Questions about zero-dimensional subsets of the unit interval related to cantor set

2012-05-09T18:09:00Z

For Q2, consider a mapping $\phi:K\rightarrow \mathbb{P}$ defined by $$\phi( \sum_{n=0}^{\infty} \frac{2a_n}{3^n})=\sum_{n=0}^{\infty} \frac{a_n}{2^n}$$ where $a_n$ is a sequence entirely consisted of 0 and 1, and not periodic.

For Q3, the answer is yes. H is just intersection of $\mathbb{P}$ and the complement of cantor set. The complement of cantor set is union of countably many disjoint open intervals.

Answer by i707107 for Trig functions based on convex curves

2012-05-09T15:36:59Z

The sine function for the first one is $$f(\theta)=\frac{\sin \theta}{\sqrt{2} \sin( \frac{\pi}{4}+\theta)}$$ for $\theta\in [0,\frac{\pi}{2}]$.

The second one is $$g(\theta)=\frac{-\tan \theta+\sqrt{\tan^2\theta+4}}{2}\tan\theta$$ for $\theta\in [0,\frac{\pi}{2}]$.

Then extend this symmetrically for other values of $\theta$. Perhaps, you might be able to obtain desired identities using usual $\cos$ and $\sin$ identities, but judging from above formulas, I don't see the reason that there should be simple identities of this sort.

Answer by i707107 for Generalized multinomial coefficient

2012-05-04T03:54:00Z

We have $$(1+y+z)^{\alpha} = \sum_{t=0}^{\infty} \binom{\alpha}{t}(y+z)^t$$ The trinomial identity is just a rearrangement of the above by $$\binom{\alpha}{r,s}=\binom{\alpha}{r+s}\binom{r+s}{s}$$

To have convergence, we assume $|y+z|<1$, $|y|<1$, and $|z|<1$. The generalization follows similarly.

Answer by i707107 for Find generating function

2012-04-20T20:43:59Z

This can be done step by step. First note that $\binom{n}{s}/\binom{i}{s}$ can be written as $n(n-1)\cdots(n-s+1)/i(i-1)\cdots(i-s+1)$ Since we have the generating function (with assuming $a_i=0$ for $i< s$) $$a(x)x^{-s}=\sum_{i=s}^{\infty} a_ix^{i-s}$$ We obtain the following by integrating $s$ times. Let $A_0(x)=a(x)x^{-s}$, and $A_{k+1}(x)= \int_0^x A_k(t)dt$. Then $$A_s(x)=\sum_{i=s}^{\infty} a_i\frac{x^i}{i(i-1)\cdots (i-s+1)}$$ The generating function for $b_n/ n(n-1)\cdots (n-s+1)$ can be obtained from the product $$\left(\sum_{i=s}^{\infty} a_i\frac{x^i}{i(i-1)\cdots (i-s+1)}\right)\left(\sum_{j=1}^{\infty} x^j\right)$$

Now the generating function for $b_n$ follows from differentiating s times Again assuming $b_i =0$ for $i< s$, we have $$B_0(x)= A_s(x) \frac{x}{1-x}$$ $$B_{k+1}(x)=\frac{d}{dx} B_k(x)$$ $$\sum_{n=s}^{\infty} b_n x^n = B_s(x) x^s$$

Answer by i707107 for Integer polynomials mapping the unit disk into itself

2012-04-05T07:41:19Z

Since $P\in \mathbb{Z}[X]$, We have $P(0)\in \mathbb{Z}$. Suppose that $P(0)\neq 0$, then $|P(0)|\geq 1$. In that case $P(D)\subset D$ is not satisfied unless $P$ is constant by open mapping theorem. So, if $P$ is non-constant, then we must have $P(0)=0$.

Write $P(X)=XQ(X)$. Then $Q(X)=P(X)/X$. On a disk $D_r$ of radius $0< r<1$ centered at $0$, We have by Maximum modulus theorem that
$$|Q(X)|\leq 1/r$$

since $|P(X)|\leq 1$. Letting $r\rightarrow 1$, we have also that $|Q(X)|\leq 1$ whenever $|X|\leq 1$. By repeating the same argument for $Q(X)$, we obtain that $P(X)=\pm X^k$, when $k\in \mathbb{N}$ are the only polynomials in $\mathbb{Z}[X]$ satisfying the property.

Jacobson's theorem and further

2012-04-05T01:11:59Z

Jacobson's theorem states that If $R$ is a ring, and for every $x\in R$, there exists $n(x)\geq 2$ such that $x^{n(x)}=x$. Then $R$ is commutative.

I wonder if the following stronger assertion(in case $R$ has unity) is true.

Let $R$ be a ring with unity. For every $x$ in $R$, there exists $n(x)\geq 2$ such that $x^{n(x)} = x$. Then, $R$ is embedded in a product(possibly infinite) of fields $F_i$, where each $F_i$ is an algebraic extension of $F_{p_i}$ (prime field of $p_i$ elements).

If this is not true, then I am also interested in counterexample.

Thanks in advance.

Answer by i707107 for Sums of powers mod p

2012-04-03T03:36:02Z

For fixed $r>1$, use of Weil's bound will give the result. We refer to Theorem 5.1 from chapter 2 of W.M.Schmidt, Equations over Finite Fields: An Elementary Approach.

Theorem Let $f_1,\cdots, f_n$ be polynomials with coefficients in $\mathbb{F}_q$ and of degree $\leq m$. Put $\delta=\textrm{lcm}(d_1,\cdots,d_n)$, and $d=d_1d_2\cdots d_n$. Let X be a variable and let $\eta_1,\cdots,\eta_n$ be algebraic quantities with \begin{equation} \eta_1^{d_1}=f_1(X), \cdots, \eta_n^{d_n}=f_n(X).\ \end{equation} Suppose \begin{equation} [\overline{\mathbb{F}}_q(X, \eta_1,\cdots,\eta_n):\overline{\mathbb{F}}_q(X)]=d. \end{equation} Then if $q>100\delta^3m^2n^2$, the number $N$ of solutions $(x,y_1,\cdots,y_n)\in\mathbb{F}_q^{n+1}$ of the equations $y_1^{d_1}=f_1(X),\cdots,y_n^{d_n}=f_n(X)$ satisfies \begin{equation} |N-q|<5mnd\delta^{5/2}q^{1/2}. \end{equation} We will use this theorem to solve the problem for fixed $r>1$. We have $q=p$ prime number. Let $g$ be a primitive root modulo $p$. We put $n=r+1$, $m=1$, $d_1=\cdots=d_n=r$, $\delta=r$, $d=r^{r+1}$, $f_i(X)=g(X-g^{is})$ for $0\leq i \leq r-1$, and $f_r(X)=gX$. Then the condition is satisfied, and the number $N$ of solutions to the system $Y_i^r=f_i(X)$ ($0\leq i \leq r$) is \begin{equation} |N-p|<5(r+1) r^{r+1+5/2}p^{1/2}. \end{equation}

We look for the number $N^{*}$ of solutions with no $Y_i$ being zero. Then we have $$N^{*}\geq p-5(r+1) r^{r+1+5/2}p^{1/2}-(r+1)r^{r+1}.$$ This is in fact greater than zero for sufficiently large $p$. For any such solution $(X,Y_0,\cdots, Y_r)\in \mathbb{F}_p^{r+2}$, we obtain that $$X\in \mathbb{Z}_p-(A+B).$$ This proves the result.

Answer by i707107 for Extension of pointwise convergence of a sequence of uniformly continuous functions that converges on a dense set

2012-03-07T20:01:44Z

I don't know what example that you have in mind in the first statement, but you can find such example in your question on $X=[0,1]$ as pointed out by Peter, any continuous function on $X$ is uniformly continuous. Consider $f_n(x)=0$ on $x\in [\frac{1}{n},1]$, and $(-1)^nn(x-\frac{1}{n})$ on $x\in [0,\frac{1}{n}]$. This sequence is pointwise convergent to $0$ on $(0,1]$ which is a dense set in $[0,1]$.

Answer by i707107 for How to calculate Euler Totient Function up to a limit?

2012-02-17T05:37:19Z

Just a simple counting argument. It is $x\phi (n)/n +O(1)$. There is an error term $O(1)$ coming from the case when $x$ is not a multiple of $n$.

A raceway problem

2012-02-15T03:48:12Z

Let $f(x)=\sin x$, and $g(x)=\sin x + 1$. Consider a set $S={(x,y)| f(x)\leq y \leq g(x), x\in [0,2\pi]}$. This set $S$ can be considered as "Raceway" My question is finding the shortest path in $S$ such that initial point lies in ${0}\times [0,1]$, and the terminal point lies in ${2\pi}\times [0,1]$.

Question on division field of abelian variety

2012-02-02T21:54:54Z

I am wondering if the following holds or not.

Let A be an abelian variety of dimension $d\geq 1$ over $\mathbb{Q}$. Then there is a positive number c depending on d and A such that $[\mathbb{Q}(A[n]):\mathbb{Q}]\geq c^{w(n)} n^2$ where $w(n)$ is the number of distinct prime factors of n.

I know that it must hold for d=1(elliptic curve case). I guess that we can even have stronger inequality like $[\mathbb{Q}(A[n]):\mathbb{Q}]\geq c^{w(n)} |GSp_{2d}(Z/nZ)|$ ,but don't know how to prove it, and couldn't find any reference other than d=1 case.

Thanks, Sungjin Kim.

Comment by i707107

2013-05-24T15:40:54Z

Hold for series convergent with radius 1. Any series obtained from your setting, will have radius of convergence 1, or infinity. The case when radius of convergence infinity, is only when $r$ has finite decimal expansion.

Comment by i707107

2013-05-24T08:35:54Z

Yes. Weil pairing applies to $n$-torsion too.

Comment by i707107

2013-05-24T06:43:03Z

As I mentioned in my answer, $r$ can be either rational or irrational. When $r$ is rational, the function $f$ is rational, and when $r$ is irrational, the function $f$ is transcendental.

Comment by i707107

2013-05-24T02:02:07Z

This might be better expression: $$\sum_{p\textrm{ prime}} x^p$$

Comment by i707107

2013-05-16T21:09:34Z

Maybe I should have put, $m$ avoids all primes of bad reduction.

Comment by i707107

2013-05-16T20:50:11Z

Thank you. This is also very clear. So $\mathbb{Q}(E[2])=\mathbb{Q}(\sqrt{d})$, and primes ramify in $\mathbb{Q}$ are precisely the prime divisors of $d$.

Comment by i707107

2013-05-16T20:43:32Z

Thank you for clear counterexample. I'd like to know what happens when $m$ is a power of $p$.

Comment by i707107

2013-05-15T22:40:11Z

To clarify: $||v_i||\leq 1$ right? not $||v_i||=1$?

Comment by i707107

2013-05-09T20:06:28Z

What is $\tau$ in the last formula? Is it the "number of divisors" function?

Comment by i707107

2013-05-08T22:59:12Z

It is also strange that LaTeX is completely working on my iPhone, but not on PC.

Comment by i707107

2013-05-08T14:13:59Z

Or, are those separate problems?

Comment by i707107

2013-05-08T14:12:49Z

Putting $x=y=z$, you get $2X^2=1$, and $X^3=1$ where $X=f(x,x)$, which is nonsense.

Comment by i707107

2013-05-07T19:52:38Z

I think so too.

Comment by i707107

2013-05-07T02:53:11Z

It is also possible to obtain an asymptotic formula with main term $C\frac{1}{\log N}$ for some positive constant $C$.

Comment by i707107

2013-04-20T16:33:55Z

See this: wolframalpha.com/input/…