I have the following function, and I want to prove/disprove that it is $\Omega$(n$^2$): \begin{cases}     4[sum(n/2,n)],&amp; \text{if n is even } \\     2n-1+sum(n-3,n),              &amp; \text{otherwise} \end{cases} (it should be 2 n-1, so a space between 2 and n) where sum( j,k ) is a ‘partial arithmetic sum’ of the integers from j up to k, that is \begin{cases}     0&amp; \text{if j&gt;k } \\     j + (j+1) + (j+2) + ... + k,              &amp; \text{otherwise} \end{cases} e.g. sum(3,4) = 3 + 4 = 7, etc. Note that sum(j,k) = sum(1,k) – sum(1,j-1) I tried to solve it, however have not come across one before where it has conditions...
120555
I have the following function f(n), and I want to prove/disprove that it is O(n$^2$): \begin{cases}     4[sum(n/2,n)],&amp; \text{if n is even } \\     2n-1+sum(n-3,n),              &amp; \text{otherwise} \end{cases} (it should be 2 n-1, so a space between 2 and n) where sum( j,k ) is a ‘partial arithmetic sum’ of the integers from j up to k, that is \begin{cases}     0&amp; \text{if j&gt;k } \\     j + (j+1) + (j+2) + ... + k,              &amp; \text{otherwise} \end{cases} e.g. sum(3,4) = 3 + 4 = 7, etc. Note that sum(j,k) = sum(1,k) – sum(1,j-1) How do I prove or disprove that this is O(n$^2$)?
57506
I've seen the identity $\displaystyle \sum_{i \mathop = 0}^n \binom n i^2 = \binom {2 n} n$ used here recently.  I checked for proofs here   I couldn't have figured out the combinatorial proof by myself, and the inductive proof assumes you already know the answer...  So my question is : do you know how to prove directly through computation that $\displaystyle \sum_{i \mathop = 0}^n \binom n i^2 = \binom {2 n} n$ ?
74516
This is from Shreve's Stochastic Calculus for Finance II, Appendix A.3. Consider infinitely many independent coin tosses with outcomes $\omega_n$, $n = 1, 2, \dots$ such that $\omega_n$ can be either $H$ = heads or $T$ = tails. Define \begin{equation} Y_n\left(\omega\right) = \begin{cases}1 &amp; \text{ if }\omega_n = H \\ 0&amp; \text{ if } \omega_n = T\text{.} \end{cases} \end{equation}  Define $\displaystyle Y = \sum\limits_{n=1}^{\infty}\dfrac{2Y_n}{3^{n}}$. Shreve shows that $Y$ can only take values in the Cantor set, and since that particular set has measure zero, its Lebesgue measure is zero as well, thus it does not have a probability density function (since integrating such a density function would give a value of $0$). This part I understand, as well as my professor.  Now Shreve claims that $Y$ does not have a probability mass function.      If it did, then for some number $x \in C$ [the Cantor set] we would have $\mathbb{P}\left(Y=x\right) &gt; 0$. But $x$ has a unique base-three expansion    \begin{equation} x = \sum\limits_{n=1}^{\infty}\dfrac{x_n}{3^{n}}\text{,} \end{equation}   where each $x_n$ is either 0, 1, or 2 unless $x$ is of the form $\dfrac{k}{3^{n}}$ for some positive integers $k$ and $n$. In the latter case, $x$ has two base-three expansions... In either case, there at most two choices of $w \in \Omega_{\infty}$ [that is, the infinite coin-toss space] for which $Y(\omega) = x$. In other words, the set $\{\omega \in \Omega; Y(\omega) = x\}$ has either one or two elements. The probability of a set with one element is zero, and the probability of a set with two elements is $0 + 0 = 0$. Hence $\mathbb{P}\left\{Y = x\right\} = 0$.   My professor and I were able to follow this until the statement "The probability of a set with one element is zero." Is this a true statement? How so?   I thought it would be perhaps the fact that the Lebesgue measure of $[a, a] = \{a\}$ is zero, but I don't see any reason why Shreve would only use the Lebesgue measure, since it is not a general probability measure.
118851
Prove that $\mathbb{Z}[\omega]=\{ a+b \omega | a,b \in \mathbb{Z} \}$ is an integral domain, where $\omega^3=1$.   My solution:  Let $(a+b\omega)(c+d\omega)= (ac-bd)+(ad+bc-bd)\omega=0$  $\Rightarrow ac=bd \; \&amp; \; ad+bc=bd. $  If $a\neq  0$, then put $c= bd/a$ to solve the equation and get  \begin{equation} \begin{split} &amp;\quad d(a^2+b^2-ab)=0 \\  &amp;\quad \Rightarrow d= 0 \; \text{or} \; a^2+b^2=ab. \end{split} \end{equation}  The first case $d=0$ solves the problem, but I am not able to figure out, how to reject case 2, when $a^2+b^2=ab$.
30493
I'm trying to determine if 1317 and 116 can be written as the sum of three squares?  I have the condition that if it is not of the form $4^{\alpha}(8k+7)$ then it can be written as a sum of three squares, but how do I use this condition to work it out?  Thank you
826
Let $f: \Bbb C \to \Bbb C$ be an entire (analytic on the whole plane) function such that exists $\omega_1,\omega_2 \in \mathbb{S}^1$, linearly independent over $\Bbb R$ such that: $$f(z+\omega_1)=f(z)=f(z+\omega_2), \quad \forall\,z\in \Bbb C.$$Prove that $f$ is constant.   The intuition seems clear to me, we have the three vertices of a triangle given by $0$, $\omega_1$ and $\omega_2$. All points in the plane are one of the vertices of that triangle under a suitable parallel translation. The constant value will be $f(0)$, fine. Throwing values for $z$ there, I have found that $$f(n\omega_1) = f(\omega_1) = f(0)=f(\omega_2) = f(n\omega_2), \quad \forall\, n \in \Bbb Z.$$  I don't know how to improve the above for, say, rationals (at least). Some another ideas would be:   Checking that $f' \equiv 0$. I don't have a clue of how to do that. Write $w = a\omega_1+b\omega_2$, with $a,b \in \Bbb R$, do stuff and conclude that $f(w) = f(0)$. This approach doesn't seem good, because I only have a weak result with integers above. Finding that $f$ coincides with $f(0)$ on a set with an accumulation point. This seems also bad: the set on with $f$ coincides with $f(0)$ by which I found above is discrete.    Nothing works and this is getting annoying... And I don't see how analyticity comes in there. I'll be very thankful if someone can give me an idea.  (On a side note.. I know that this title is not informative at all. Feel free to edit if you come up with something better.)
70463
Assume that $k,n\in\mathbb{Z}^+$. Prove that the sum \begin{equation*} \dfrac{1}{k+1}+\dfrac{1}{k+2}+\dfrac{1}{k+3}+\ldots +\dfrac{1}{k+n-1}+\dfrac{1}{k+n} \end{equation*} is not an integer.   The case $k=0$ is proved in this question: ""
101702
Let $p$ be an odd prime and $\omega$ be a primitive $p$th root of unity. The question is to prove that:  $$(1-\omega)(1-\omega^2) \cdots (1-\omega^{p-1})=p$$    What I have done so far is:  I can see that this is true for $p=3$  $$(1-\omega)(1-\omega^2)=1-(\omega+\omega^2)+w^3=1-(-1)+1=3=p$$  I am not able to prove this in general....  $$(1-\omega)(1-\omega^2) \cdots (1-\omega^{p-1})=1-(\omega+\omega^2+\cdots+\omega^{p-1})+????+\omega^{\frac{p(p-1)}{2}}$$  I do not have any idea what that $????$ could be but all I can say is:   $(\omega+\omega^2+\cdots +\omega^{p-1})=-1$ $\omega^{\frac{p(p-1)}{2}}=1$   So, $(1-\omega)(1-\omega^2)\cdots (1-\omega^{p-1})=3+????$  I am not able to do more than this.  There could be some (hard) way of doing it by hand multiplying all those things but I am looking for a more theoretical idea.  Please help me to clear this up.
105571
I guess it has something to do with Riemann sums but this is new for me.  $\displaystyle\lim \limits_{n \to \infty}\sum \limits_{k=n}^{2n}\frac{k}{k^2+n^2}$  How do I start?
114161
How to prove that $$ \int_{t_0}^{t_0+T} \sin(m\omega t)\sin(n\omega t)\,\mathrm{d}t$$  will equal to $0$ when $m\ne n$ and $\frac{T}{2}$ when $m=n\ne 0$? Besides $$ \int_{t_0}^{t_0+T} \cos(m\omega t)\cos(n\omega t)\,\mathrm{d}t$$ will equal to $0$ when $m\ne n$ and $\frac{T}{2}$ when $m=n\ne 0$ and $T$ when $m=n=0$?
99990
Suppose $\Omega$ is a region,$f\in H(\Omega)$, and $f\not\equiv 0$.If for every positive integer n,there is $f_n \in H(\Omega)$,$(f_n)^n=f$,prove that there is $g\in H(\Omega),f=e^g$.  My conjecture :  $$g=lim  (nf_n - n)$$  But I don't know how to deal with it.  I will appreciate your help.
93100
I am having trouble with some simple ordinal arithmetic.  I am trying to figure out what $(\omega+2)\cdot \omega$ is.  If you represent $\omega + 2$ as: $\omega+2 = 0_0 &lt; 1_0 &lt; 2_0 &lt; ... &lt; \omega + 1 &lt; \omega + 2$  Then when I try to solve it out by writing out the representations, I get:  $ \begin{align}   &amp;0_0 &lt; 1_0 &lt; 2_0 &lt; ... &lt; (\omega+1)_0 &lt; (\omega+2)_0 \\ &lt; &amp;0_1 &lt; 1_1 &lt; 2_1 &lt; ... &lt; (\omega+1)_1 &lt; (\omega+2)_1 \\  &amp;... \end{align} $  Which if I just rename the terms becomes:  $ \begin{align}   &amp;0_0 &lt; 1_0 &lt; 2_0 &lt; ...  \\ &lt; &amp;0_1 &lt; 1_1 &lt; 2_1 &lt; ...  \\   &amp;... \end{align} $  Or simply: $\omega + \omega + \omega + ... = \omega^2$  Is this right? or am I thinking about this the wrong way
101737
To be specific, why does the following equality hold? $$ \prod_{0\lt n\lt\omega}n=2^{\aleph_0} $$
24782
I just wanted to evaluate  $$ \sum_{k=0}^n \cos k\theta $$  and I know that it should give  $$  \cos\left(\frac{n\theta}{2}\right)\frac{\sin\left(\frac{(n+1)\theta}{2}\right)}{\sin(\theta / 2)}   $$  I tried to start by writing the sum as  $$ 1 + \cos\theta + \cos 2\theta + \cdots  + \cos n\theta $$  and expand each cosine by its series representation. But this soon looked not very helpful so I need some clue about how this partial sum is calculated more efficiently ...
50713
Show that $\displaystyle\sum_{i=0}^l \dfrac{{l\choose i}}{{n\choose k + i}} = \dfrac{n+1}{(n-l+1) {n-l\choose k}}$ for all $k,l, n\geq 0$ with $k+l \leq n.$  The $k+l\leq n$ part doesn't seem that important. The claim clearly holds for $n=0,$ and in that case $k$ and $l$ must both be zero. I think the claim can be shown by induction, but I'm not sure how to do so; I'm not sure how to use the inductive hypothesis to prove the inductive step. The sum on the LHS is not always an integer, so finding a combinatorial argument doesn't seem very feasible. For $n = 1,$ we can have $(k,l) = (0,0), (1,0), (0,1),$ and in each case the sum holds. Perhaps multiplying both sides by a number (idk ${n-l\choose k}$) might be useful?
105419
How do you derive the formula for the sum of the series when $S_n = \sum_{j=1}^n j^2$?  The relationship $(n, S_n)$ can be determined by a polynomial in $n$. You are supposed to use finite differences to determine the degree and then use a calculator to derive the function. But how?
16510
Let $X_{1},\ldots,X_{n}\stackrel{\text{ i.i.d }}{\sim}N(\mu,\mu^{2})$.  $T=\left(\sum_{i=1}^{n}X_{i},\sum_{i=1}^{n}X_{i}^{2}\right)$ is a sufficient statistic for $\mu$. Also $T$ is minimal sufficient.  However, the question is: Prove that $T$ is not complete, hint: use $f(x,y)=x-2y^{2}/(n+1)$.  To prove that $T$ is not complete, I must find an example of a Borel function, $f$, for which $E_{\theta}(f(T))=0$ does not imply $f(T)=0$. However, I do not know how to interpret the 'hint', since I only can conclude that $f(T)=0$, whenever using $x=\sum_{i=1}^{n}X_{i}$ and $y=\sum_{i=1}^{n}X_{i}^{2}$ and calculating $E_{\theta}(f(T))=0$.