text stringlengths 83 79.5k |
|---|
H: How to solve this trigonometric equalities?
We have to solve $ \tan(x) = 2 \sin(x)$ on $0 \leq x \leq 2 \pi$.
I rewrote it to $\sin(x) = \sin(2x)$ using $2\sin(x)\cos(x) = \sin(2x)$.
But how can I now find the set of solutions by hand?
AI: You have already reached at $\sin2x=\sin x$
$\displaystyle\implies 2x=n\pi+(-1)^nx$ where $n$ is any integer
Deal the even and odd cases of $n$ separately |
H: What's wrong with my conjecture?
I was doing math homework, and I formulated the following conjecture from one of the questions:
If $f(x)$, $g(x)$ and $h(x)$ are continuous functions and the equations $f(x) = h(x)$ and $g(x) = h(x)$ both have only one root, then the equation $f(x) = g(x)$ has only one root. So can anyone find a counterexample? I can't think of one, but maybe that's because I haven't read Counterexamples in Analysis yet.
AI: You can just take $f(x) = g(x) = x$ and $h(x) = 0$.
If you want a less trivial example, take $f(x) = x + \sin x$, $g(x) = x + \cos x$, and $h(x) = 0$. See it here.
For a less trivial $h$, you can take $f(x) = x \sin x$, $g(x) = x \cos x$, and $h(x) = ax$ for any $a$ such that $|a| > 1$). See it here for $a=2$.
I would like to make another example, fundamentally different from the previous ones: $f(x) = -(x+1)^2$, $g(x) = (x-1)^2$, $h(x) = -4x$. Here, your conditions are also met, but $f(x) = g(x)$ has no real solutions. See it here. |
H: Piecewise continuous function
I'm trying to prove this, and here what I got so far
Let f be defined on A, $a=kπ$ is cluster point of A
Part 1: if $a=kπ$ for some k∈Z then $f$ is continuous at $a$
Take ε>0, let $δ=sin^{-1}(ε)$ st $0<|x-kπ|<δ$. When $x$ is a rational
$|f(x)-f(a)|=|sin|x|-sin|kπ||=|sin〖|x||〗<ε$
Observe that
|x-kπ|<δ
Now I'm stuck. I don't know hot link the last line to the inequality above it
AI: Consider any sequence of rational numbers converging to $k\pi$, $k \in \mathbb{Z}$, say $\{x_n\}$. So $\{f(x_n)\}$ converges to $0$. Take any sequence of irrational numbers say $\{y_n\}$ converges to $k\pi$. See $\{f(y_n)\}$ converging to $0$. So $f$ is continuous at $k\pi$.
For any other points $x$ construct any two sequences of rational and irrational numbers, converging to $x$ and see if the sequential criterion holds for the function $f$.
So you may conclude that $f$ is continuous only at $k\pi$. |
H: Finding numbers when their sum and LCM are given
The sum of two numbers is $2000$ and thier LCM is $21879.$ Find the numbers.
My attempt: Let two numbers be $x, 2000-x.$ Product of two numbers is equal to the product of their lcm and hcf. So, $x(2000-x)=21879*hcf.$
Now we have two variables and one equation. So I am stuck.
But the book simply considers $x(2000-x)=21879,$ thereby $x=1989,11.$
My question is - Is it implied anywhere in the question that hcf is $1$ or is the question incomplete?
More impotantly, can we have any other pair of numbers whose sum is $2000$ and lcm is $21879 (hcf <> 1)?$
AI: Note that $21879$ and $2000$ are coprime. So this means that $x$ and $2000-x$ are coprime (simply because their gcd must divide $x+(2000-x)=2000$). |
H: Conditional probability problem in Elementary Probability Theory with Stochastic Processes
There are two kinds of tubes in an electronic gadget. It will cease to
function if and only if one of each kind is defective. The probability
that there is a defective tube of the first kind is .1; the probability
that there is a defective tube of the second kind is .2. It is known
that two tubes are defective, what is the probability that the gadget
still works?
This is from Elementary Probability Theory with Stochastic Processes, K.L.Chung.
My solution: the probablity that the gadget won't work is $\frac{0.1\times 0.2}{0.1^2+0.2^2+0.1\times 0.2} = \frac{2}{7}$, assuming that the defections are independent in this gadget. So the answer shall be $\frac{5}{7}$.
But the correct answer is $\frac{5}{9}$. So I assume that the $0.1\times 0.2$ term above should actually be doubled. But I can't account for the factor of 2. Could someone explain this to me? Thanks!
AI: Let $X_{1}$be the first tube that breaks down and $X_{2}$ the second.
Then there are four possibilities:
(1) $X_{1}$ a tube of the first kind and $X_{2}$ is a tube of the
first kind. Unconditional probability: $0.01$
(2) $X_{1}$ a tube of the first kind and $X_{2}$ is a tube of the
second kind. Unconditional probability: $0.02$
(3) $X_{1}$ a tube of the second kind and $X_{2}$ is a tube of the
first kind. Unconditional probability: $0.02$
(4) $X_{1}$ a tube of the second kind and $X_{2}$ is a tube of the
second kind. Unconditional probability: $0.04$
Adding these probabilities gives $0.09$ and in the cases (1) and
(4) the gadget still works. This leads to conditional probability
$\frac{5}{9}$ |
H: "Preimage" of a binary relation
Consider the binary relation $R \subseteq X \times Y$. Is there a standard name and notation for the set $X' = \{x\ |\ (x, y) \in R\}$?
ProofWiki calls $X'$ the preimage of $R$, denoted as
$\operatorname{Im}^{-1}(R)$.
This site calls $R$ a correspondence, calls $X$ the predomain of $R$ and calls $X'$ the domain of $R$.
Both conventions seem to make sense, although $X$ is generally called the domain of $R$, which I prefer as well. But the notation $\operatorname{Im}^{-1}(R)$ is rather clunky and unwieldy - not to mention that I hardly see it elsewhere - and I would prefer to use something simpler like $\operatorname{dom} R$ or $\operatorname{pre} R$, if these are accepted.
Of course, I should follow the standard notation, but I can't seem to find a consensus online!
AI: I claimed in a comment that $\operatorname{dom}(R)$ is standard notation. I was asked by the OP to give out some references to support this claim. I present them below.
From D.J. Velleman's How to Prove It: A Structured Approach:
From Elements of Set Theory by H. B. Enderton:
And finally, from Thomas Jech's classic Set Theory: |
H: Find the expressions for the common difference and common ratio
The first 2 terms of a geometric progression (first term $a$ and common ratio $r$) are the same as the first 2 terms of an arithmetic progression (first term $a$ and common difference $d$). The third term of the geometric progression is twice as big as the third term of the arithmetic progression. Find 2 different expressions for $d$ and hence or otherwise find the 2 possible values of $r$.
So I've tried to tackle this problem, to no avail:
$$ ar = a + d$$
$$ar^2 = 2(a + 2d)$$
So using this I found $d = ar -a $ and $d = \dfrac{1}{4}ar^2 - \dfrac{1}{2}a$, so $$ ar -a =\dfrac{1}{4}ar^2 - \dfrac{1}{2}a$$
$$a(r^2-4r+2)=0$$
Here I knew I was wrong, because I'm sure that you're not supposed to be using the quadratic for this exercise (judging from how much time you have).
So where did I go wrong?
AI: I'm afraid that you've done nothing wrong. (There's an odd thing to say.)
It's worth noting that if $a=0,$ then we get $d=0$ immediately, but $r$ can take literally any value. Since the problem expects two possible values for $r,$ then we clearly must assume $a\neq0.$
Since $a\ne 0,$ then $r^2-4r+2=0,$ and from there, the quadratic formula will give you the two possible values for $r$. Alternately, you could note the following: $$r^2-4r+2=0\\r(r-4)+2=0\\(r-2+2)(r-2-2)+2=0\\(r-2)^2-2^2+2=0\\(r-2)^2-2=0$$
This is an end-around method of completing the square by using the difference of squares formula, from which you can readily conclude that $r=2\pm\sqrt2,$ if you'd rather avoid the quadratic formula. |
H: sequence of functions and its limit
Let $f_n (x) = 1 $ if $x \in [n, n+1] $. And $f_n(x) = 0$ otherwise. I want to find the
$$ \lim_{ n \to \infty } f_n $$
MY attempt: we know $ 0 \leq f_n \leq 1 $ for all $n$ So $f_n$ is bounded. IT is obviously monotone since $f_n \leq f_{n+1} $ So limit exists. But I have trouble trying to see whether the limit is $0$ or $1$. Can someone help me see this last part?
AI: The sequence is not monotone. For example $f_{2}(2)>f_{4}(2)$ the way the sequence is defined. This sequence converges point-wise to 0 (Hint: If you fix an $x$ then you can pick $n$ large enough so that $f_{m}(x)=0$ for all $m>n$. To see this fix an $x$. Then by the Archemedian property there is a natural number $n>x$. Now because of the way the sequence is defined $f_{m}(x)=0$ for all $m>n$. For example $f_{n}(2)=0$ for $n>2$, $f_{n}(3.5)=0$ for $n\geq{4}$). It does not converge uniformly to $0$ and hence doesn't converge uniformly ($\|{f_{n}-0}\|_{\infty}=1$ for any $n$ and the uniform limit and the point-wise limit has to agree (see the proof of the theorem uniform convergence implies limitwise convergence)). |
H: building Matrix "echelon form"
given is
A = $ \begin{pmatrix} a & b & 0 & b \\ b & a & b & 0 \\ 0 & b & a & b \\ b & 0 & b & a \end{pmatrix} $
I need the echelon form for this matrix:
Whenever I begin calculating, I end up with an entry like "a/b", which isn't defined for b = 0.
However, I am supposed to name the echelon form for real numbers a, b.
What am I doing wrong?
AI: First switch the last row with 3rd row
Then divide your first Row by a to introduce a leading 1 then all you need to do is perform operations to make all elements below the leading 1 ZERO .
e.g after doing 1st step perform $$ R_2 - bR_1 $$ then perform $$ R_3 - bR_1 $$
Repeat this with remaining rows |
H: Implication of $L^p$ convergence
Take $U$ as open subset of $\mathbb{R}^{n}$. If $u_{m} \rightarrow u$ in $L^{p}(U)$ then does it follow that $||u_{m}||_{L^{p}(U)} \rightarrow ||u||_{L^{p}(U)}$?
AI: Yes, by the triangle inequality (dropping the subscript on the norms):
$\|u\|=\|u-u_m+u_m\|\le\|u-u_m\|+\|u_m\|$, so $\|u\|-\|u_m\|\le\|u-u_m\|$.
Interchanging $u$ and $u_m$ in this and combining the two inequalities leads to $\bigl|\|u\|-\|u_m\|\big|\le\|u-u_m\|$, and the result follows. |
H: Why is $t=\frac 3 2$ in $1+t=\sqrt{4+t^2}$?
I am confused about solving $1+t=\sqrt{4+t^2}$.
When I solve it per hand I come to the conclusion that $t$ can be everything.
$$\begin{align*}
1+t =\sqrt{4+t^2}& \qquad | \cdot^2 \tag{1} \\
1+t^2 = 4+t^2 & \qquad | -t^2 \tag{2} \\
1 = 4 \tag{3}
\end{align*}$$
However wolfram alpha tells me the result is $\frac 3 2$.
What am I doing wrong?
AI: $1+t=\sqrt{4+t^2}
\implies (1+t)^2=1+2t+t^2=4+t^2
\implies 2t=3
\implies t=3/2$ |
H: Removable singularity when multiplied by linear factor
I was thinking about this situation: If $f(z)$ is holomorphic everywhere except at $a$ (where it has a pole), and has a removable singularity or a pole at $\infty$, does $(z-a)f(z)$ also have a removable singularity or a pole at $\infty$?
In the pole case, $f(z)/z^k$ is holomorphic in some neighborhood of $\infty$. Since $(z-a)$ is holomorphic, $(z-a)f(z)/z^k$ should also be holomorphic in that neighborhood of $\infty$. But can we conclude that? I'm not sure because $z-a$ is not defined at $\infty$.
In the removable singularity case, we can define $f(\infty)$ so that $f(z)$ is holomorphic in a neighborhood of $\infty$. Again, can we conclude anything about $(z-a)f(z)$?
AI: Yes, in that situation, $(z-a)f(z)$ has a pole at infinity, or, if $f$ had a zero at infinity, a removable singularity.
When $k$ is such that $f(z)/z^k$ is holomorphic in a neighbourhood of infinity, then
$$\frac{(z-a)f(z)}{z^{k+1}} = \frac{f(z)}{z^k} - a\frac{f(z)}{z^{k+1}}$$
also is holomorphic in a neighbourhood of $\infty$. |
H: Understanding of Analytic Branch
What does it mean for a function ''to have a branch that is analytic''? Can someone give me an example of it as well please?
thanks
AI: Some functions have multiple values, for example $\log z = \log |z| + i(\text{arg } z + 2\pi k)$, $k \in \mathbb{Z}$. Such a function "has an analytic branch" when we can define it to be single valued and analytic. We can do this for the logarithm by restricting the argument to be $-\pi < \theta < \pi$, which is called the principal branch. |
H: Proof that the Riemann-Integral satisfies $\int_A \lambda f = \lambda \int_A f$
Suppose $A\subset\mathbb{R}^n$ is a closed rectangle and $f:A\to \mathbb{R}$ is Riemann-Integrable on $A$. I want to show that $\lambda f$ is integrable and that
$$\int_A \lambda f =\lambda\int_Af $$
My approach was the following: suppose $\lambda \geq 0$, by integrability of $f$ we know that there is a partition $P$ of $A$ such that $U(f,P)-L(f,P)< \epsilon/\lambda$. Then, let $S$ be a subrectangle of this partition, we know that $m_S(f)\leq f(x)\leq M_S(f)$ for all $x\in S$. Since $\lambda \geq 0$ we have $\lambda m_S(f) \leq \lambda f(x)\leq M_S(f)$ for all $x \in S$, and hence, since $m_S(\lambda f)$ is the greatest lower bound and $M_S(\lambda f)$ is the least upper bound we have $m_S(\lambda f) \geq \lambda m_S(f)$ and $M_S(\lambda f)\leq \lambda M_S(f)$.
In this case, we have the relation for the upper and lower sums:
$$U(\lambda f,P) =\sum_{S\in P}M_S(\lambda f,P)v(S) \leq \lambda \sum_{S\in P}M_S(f)v(S) = \lambda U(f,P)$$
$$L(\lambda f,P)= \sum_{S\in P}m_S(\lambda f,P)v(S) \geq \lambda \sum{S\in P}m_S(f)v(S) = \lambda L(f,P)$$
and this gives us $U(\lambda f,P)-L(\lambda f,P) \leq \lambda(U(f,P)-L(f,P)) < \lambda \epsilon/\lambda = \epsilon$ which proves integrability.
In the case $\lambda < 0$, there is a partition $P$ of $A$ such that $U(f,P)-L(f,P)< -\epsilon/\lambda$. Now, if $S$ is a subrectangle of $P$, from $m_S(f)\leq f(x)\leq M_S(f)$ for all $x\in S$ we have $\lambda M_S(f) \leq \lambda f(x) \leq \lambda m_S(f)$ for all $x\in S$. Hence, by similar arguments as above, $m_S(\lambda f)\geq \lambda M_S(f)$ and also $M_S(\lambda f)\leq \lambda m_S(f)$. This gives us
$$U(\lambda f,P)=\sum_{S\in P}M_S(\lambda f,P)v(S) \leq \lambda \sum_{S\in P}m_S(f)v(S) = \lambda L(f,P)$$
$$L(\lambda f,P)= \sum_{S\in P}m_S(\lambda f,P)v(S) \geq \lambda \sum_{S\in P}M_S(f)v(S) = \lambda U(f,P)$$
and this gives $U(\lambda f,P)-L(\lambda f,P) \leq\lambda(L(f,P)-U(f,P)) < \lambda \epsilon/\lambda = \epsilon$ and therefore $\lambda f$ is integrable again.
Now, first of all: is this proof correct? I think it is correct, but I'm a little unsure on the case $\lambda < 0$. Also, I'm stuck now, I need a hint on how to show that
$$\int_A \lambda f = \lambda \int_A f.$$
Thanks very much in advance!
AI: Cleaned-up Version:
Let $\lambda \ne 0, \epsilon > 0$ be given and chose a partition $P$ of $A$ s.t.
$$U(f;P) - L(f;P) < \frac \epsilon {|\lambda|}$$
WLOG assume $\lambda > 0$ since $f$ is Riemann integrable on $A \Leftrightarrow -f$ is Riemann integrable on $A$ (shown?)
Then see that $\def\vol{\ {\rm vol}}$
$$U(\lambda f;P) = \sum_{P_i \in P} \max_{x\in P_i} \lambda f(x) \vol(P_i) = \sum_{P_i \in P} \lambda \max_{x\in P_i} f(x) \vol(P_i) = \lambda \sum_{P_i\in P} \max_{x\in P_i} f(x) \vol(P_i) = \lambda U(f; P)$$
$$L(\lambda f;P) = \sum_{P_i \in P} \min_{x\in P_i} \lambda f(x) \vol(P_i) = \sum_{P_i \in P} \lambda \min_{x\in P_i} f(x) \vol(P_i) = \lambda \sum_{P_i\in P} \min_{x\in P_i} f(x) \vol(P_i) = \lambda L(f; P)$$
So now
$$U(\lambda f; P) - L(\lambda f; P) = \lambda U(f;P) - \lambda L(f;P) = \lambda (U(f;P) - L(f;P)) < \lambda \frac\epsilon\lambda = \epsilon$$
There for $\lambda f$ is Riemann integrable and
$$\int_A \lambda f = \lim_{|P| \to\infty} U(\lambda f;P) = \lim_{|P|\to\infty} \lambda U(f;P) = \lambda \lim_{|P|\to\infty} U(f;P) = \lambda \int_A f$$
(the last part can also be formulated with $L(f;P)$ etc.) |
H: Determine $a<0$ such that $\int_a^0 f(x) dx = f(a)$
The function $f$ is given by
$$f(x)=\frac{e^{\frac 1 x}}{x^2}$$
where $x\ne 0$. Determine a number $a<0$ such that
$$\int_a^0 f(x) dx = f(a)$$
AI: Just go ahead and compute the integral. You have
$$
\int_a^0 f(x) dx=\lim_{\epsilon \uparrow 0} \int_a^\epsilon \frac{e^{1/x}}{x^2}dx=\lim_{\epsilon\uparrow 0}(-e^{1/\epsilon}+e^{1/a})=e^{1/a}.
$$
Here, $\epsilon<0$ and I used $u=1/x$ substitution to integrate. You should be able to finish from here. |
H: Directional Derivative (3 variables)
Find the derivative of the function $ u = xy + yz + zx$ at the point $M(2,1,3) $ in the direction from this point to the point $N(5,5,15)$
When it is 2 variables, I can easly find a vector that pass through the given points and is a unit vector. But how should I procedure in $\mathbb{R}^3$?
P.S: i've added the tag linear algebra since I think my problem is not, directly, the calculus.
Thanks!
AI: To find the relative vector between the points, you merely subtract the coordinates of $M$ from the coordinates of $N$.
$$N-M = (5,5,15) - (2,1,3) = (3, 4, 12)$$
This vector is not unit, but if you divide this vector by its magnitude, you will have a unit vector. Do you know how to find the magnitude of a vector? |
H: Example of a Subgroup That Is Not Normal
Can you kindly provide an example of a subgroup that is not normal? I have been told many times that, for coset multiplication to be defined, the subgroup must be normal. I have seen the proof and examples of quotient group multiplications. Now, I am trying to find out where the process will break down if the subgroup is not normal.
Suppose that $x \in a_1H = a_2H, y \in b_1H = b_2H$, where $H$ is not normal.
So $x = a_1h_1 = a_2h_2$ for some $h_1, h_2 \in H$, and $y = b_1h_1^* = b_2h_2^*$ for some $h_1, h_2 \in H$.
$xy = (a_1b_1)(h_1h_1^*) = (a_2b_2)(h_2h_2^*)$. However, both $h_1h_1^*, h_2h_2^*$ are still elements of $H$.
So I don't see how $a_1b_1H \neq a_2b_2H$.
AI: Note that in your proof, $xy = (a_1h_1)(b_1h_1^*)$. When you write that $xy = (a_1b_1)(h_1h_1^*)$ is assuming the group in question is abelian. But this assumption is unwarranted, because when commutativity does not hold, we cannot assume $xy = (a_1h_1)(b_1h_1^*)=(a_1b_1)(h_1h_1^*)$.
Indeed, if a group is abelian, then every one of its subgroups are normal, as you've shown to be the case, but this doesn't hold in nonabelian groups.
Consider, for example, the group $G = S_3$ and the subgroup $H\lt G$, $H = \{id, (12)\}$. This group is not normal in $S_3$. Consider other non-abelian groups, as well, to find many other counterexamples. |
H: Extend a linear bounded functional defined on dense subset of $C([0,1])$
Let $D \subseteq C([0,1])$ be a dense countable subset of the continuous functions from $[0,1]$ to $\mathbb{R}$. Let $L\colon D \rightarrow \mathbb R$ be a linear and bounded functional with norm $\leq 1$. How can I extend it to a continuous functional over all $C([0,1])$ with norm ${}\leq 1$?
For every point $x \not \in D$ I can define $L(x)$ as the limit of a sequence in $D$ converging to $x$. But is this limit well defined?
AI: Yes: we can check that this definition does not depend on the choice of the approximating sequence (if $x_n\to x$ and $y_n\to x$, consider $(L(x_1),L(y_1),L(x_2),L(y_2),\dots)$: this is a Cauchy sequence hence the subsequences $(L(x_{k}))$ and $(L(y_k))$ have the same limit. |
H: Limit of $\sum\limits_{i=1}^n\sin\left(\frac{i}{n^2}\right)$
Compute $$\lim_{n\rightarrow\infty}\sum_{i=1}^n\sin\left(\frac{i}{n^2}\right)$$
Using Taylor expansion for $\sin x$, I know that this is $$\lim_{n\rightarrow\infty}\sum_{i=1}^n\sum_{k=1}^{\infty}\frac{(-1)^k}{(2k+1)!}\left(\frac{i}{n^2}\right)^{2k+1}$$
How do I transform this into a Riemann integral?
AI: Let $S_n=\sum\limits_{i=1}^n\sin(i/n^2)$. For every $x\geqslant0$, $x-x^3\leqslant\sin x\leqslant x$ hence
$$
\frac1{n^2}T_n-\frac1{n^6}R_n\leqslant S_n\leqslant \frac1{n^2}T_n,
\qquad T_n=\sum\limits_{i=1}^ni,\quad R_n=\sum\limits_{i=1}^ni^3.
$$
Note that $T_n=\frac12n(n+1)$ and $R_n\leqslant\sum\limits_{i=1}^nn^3=n^4$, hence
$$
\frac12\frac{n+1}n-\frac1{n^2}\leqslant S_n\leqslant\frac12\frac{n+1}n,
$$
which implies that
$$
\lim\limits_{n\to\infty}S_n=\frac12.
$$ |
H: Find horisontal, vertical and slant asymptotes of this function...
$$y=\sqrt{x^2-1}$$
I found that this function has no asymptotes, but I have no idea if it's true.
AI: Hint: Presumably you're only defining the function $f(x) = \sqrt{x^2-1}$ for $|x|\geq 1$. There are no horizontal or vertical asymptotes for this function, but there is (at least one) slant asymptote. You should show that
$$
\lim_{x\to\infty}\frac{\sqrt{x^2-1}}{x}
$$
converges to a real number. (Which one?) You can use this to determine the slant asymptote of $y=f(x)$ as $x\to\infty$. Of course you should also determine if
$$
\lim_{x\to-\infty}\frac{\sqrt{x^2-1}}{x}
$$
exists, as well, and if so determine the slant asymptote of $y=f(x)$ as $x\to-\infty$. |
H: How to evaluate the series: $ \frac {20} {(5-4)(5^2-4^2)} + \frac {20^2} {(5^2-4^2)(5^3-4^3)} + \frac {20^3} {(5^3-4^3)(5^4-4^4)} + \ldots $
How to evaluate this infinte summation?
$$ \dfrac {20} {(5-4)(5^2-4^2)} + \dfrac {20^2} {(5^2-4^2)(5^3-4^3)} + \dfrac {20^3} {(5^3-4^3)(5^4-4^4)} + \ldots $$
Telescopic cancellation is the way to go, I feel, but I don't know how to get this into telescopic form.
Hints?
AI: Notice that:
$$\frac{20^k}{(5^k-4^k)(5^{k+1}-4^{k+1})}=\frac{4^k}{5^k-4^k}-\frac{4^{k+1}}{5^{k+1}-4^{k+1}},$$
in order to prove that:
$$\sum_{k=1}^{+\infty}\frac{20^k}{(5^k-4^k)(5^{k+1}-4^{k+1})} = 4.$$ |
H: Dirichlet's Approximation Theorem For Rationals
Dirichlet's Approximation Theorem implies that for all $x \in \mathbb{R} \setminus \mathbb{Q}$ we have
$$
\left|x-\frac{p}{q} \right| \leq \frac{1}{|q|^2} \quad \text{for infinitely many $(p,q) \in \mathbb{Z}^2$.}
$$
This fails for $x \in \mathbb{Q}$:
$$
\left|\frac{a}{b}-\frac{p}{q} \right| = \left|\frac{aq - bp}{bq}\right| \geq \frac{1}{|b||q|} \quad \text{if } \frac{a}{b} \neq \frac{p}{q}.
$$
Is there an $\epsilon > 0$ such that for all $x \in \mathbb{Q}$ we have
$$
\left|x-\frac{p}{q} \right| \leq \frac{1}{|q|^{1+\epsilon}} \quad \text{for infinitely many $(p,q) \in \mathbb{Z}^2$?}
$$
AI: For $x = \dfrac{a}{b}\in\mathbb{Q}$ and $x \neq \dfrac{p}{q}$, we have, as you wrote yourself in the question,
$$\frac{1}{\lvert b\rvert\cdot \lvert q\rvert} \leqslant \left\lvert x - \frac{p}{q}\right\rvert,$$
so an inequality
$$\left\lvert x - \frac{p}{q}\right\rvert \leqslant \frac{1}{\lvert q\rvert^{1+\varepsilon}}\tag{1}$$
can only hold if $$\lvert q\rvert^\varepsilon \leqslant \lvert b\rvert \iff \lvert q\rvert \leqslant \lvert b\rvert^{1/\varepsilon}.$$
So whatever positive $\varepsilon$ one chooses, $(1)$ can only hold for finitely many $q$. And for each of these $q$, only for finitely many $p$. |
H: Cover information theory 7.21 tall, fat people
I am stuck on Thomas Cover information theory 2nd edition, problem 7.21 Fat, tall people. The problem is like following:
7.21 Tall, fat people. Suppose that the average height of people in a room is 5 feet. Suppose that the average weight is 100 lb.
(a) Argue that no more than one-third of the population is 15 feet tall.
(b) Find an upper bound on the fraction of 300-lb 10-footers in the room.e
For (a), I think height should follow a Gaussian, so if the std is known, then it should not be hard to know 15 feet falls out of how many std away, thus its probability to occur. However, the std is not know?
For (b), I think it is a joint probability of height and weight (iid), so it should be a 2D Gaussian. So asking for upper bound sounds like estimating the std for both?
And this chapter (chapter 7 in "Elements of Information Theory") is particularly about channel capacity and joint typical sequence, so should I use joint typical sequence to solve this problem? But I don't see how is it related. I am really getting stuck, anyone can please give any pointers? Or anyone has a solution I can look at it? Thanks!
(b)
P(w≥300, h≥10)=P(w≥300)P(h≥10)≤(5/10)(100/300)=(1/2)*(1/3)=1/6 ?
AI: Don't think about a Gaussian, but of Markov's inequality: $P(X\geq c)\leq\mu_X/c$, so for (a) the answer is $P(X\geq15)\leq5/15$. Can you go from here to (b)? |
H: Prove if prime can be written as $2^n+1$, $n = 2^k$
Prove that if prime can be written as $\ 2^n + 1$ then $n = 2^k$, $\;\;n, k \in \mathbb N$.
I am pretty new in this part of math.
AI: If n is odd, then $2^n+1 = (2+1)\cdot {2^n+1 \over 2+1} $ where the fraction evaluates to an integer, because this is true even for general bases $b$ : we'll have $b^n+1 = (b+1)\cdot {b^n+1 \over b+1} $.
So for odd $n$, except trivially $n=1$, the expression cannot be prime.
Thus $n$ must be even, say $n=2m$ , assume with an odd $m$. Then our initial expression changes to $4^m+1 = (4+1)\cdot {4^m+1 \over 4+1} $.
We see, that we have the same problem as before: only if $m=1$ an odd $m$ allows the expression to be prime, so also $m$ must be even, say $m=2l$.
Repeat this to see that only numbers $n=2^k$ allow a prime solution of your exponential expression. |
H: Why is the empty set bounded?
Why is the empty set bounded below and bounded above? If it has no elements, how can you say that an upper or lower bound exists?
AI: Recall that implication has the property that when the assumption is false, the implication is true. In other words, if $P$ is false, then $P\implies Q$ is true.
Let $S$ be a set of real numbers. Then $M$ is an upper bound for $S$ if the following implication holds, $$s\in S\implies s \leq M.$$ Now let us examine the case for the empty set, $\emptyset$.
Proposition Let $M$ be any real number. Then $M$ is an upper bound for the empty set (of real numbers).
Proof:
Since the statement, $s\in\emptyset$ is false, the implication, $$s\in\emptyset\implies s \leq M$$ is true.
Note: An almost identical proof works for the lower bound case. A nice little slogan to remember here is
All things are true about the MEMBERS of the empty set. |
H: Evaluation of a limit
Here is a question on limits. I would like to ask help. Here it goes: $$\lim_{N\to\infty}\left(\frac{\sum_{j=0}^{N}\left(\frac{j}{N}\right)^{n+1}}{\sum_{j=0}^{N}\left(\frac{j}{N}\right)^{n}}\right)$$ I do not know where to start but so far, I think I have to use L'Hopital's Rule, maybe because it becomes $\frac{0}{0}$, which is indeterminate? I hope somebody can explain. I really can't figure out. Thanks.
P.S. Some texts simplified this as $$\frac{n+1}{n+2}$$Is this correct? Why?
AI: Each of the sums in your limit is a Riemann sum. Multiply up and down in the fraction by $1/N$, and use, e.g.,
$$\lim_{N \to \infty} \frac{1}{N}\sum_{j=0}^N \left (\frac{j}{N}\right )^n = \int_0^1 dx \, x^n = \frac{1}{n+1}$$ |
H: Recurrence relation $g(n) = g( \lfloor {n/2}\rfloor) + \lfloor{log_2{n}}\rfloor $
$g(n) = g( \lfloor {n/2}\rfloor) + \lfloor{log_2{n}}\rfloor \\ g(0) = 0$
Series is like this:
$0,0,1,1,3,3,3,3,6,6,6,6,6,6,6,6,10,...$ and it's changes similar as $\lfloor{log_2{n}}\rfloor $
$0,0,1,1,2,2,2,2,3,3,3,3,3,3,3,3,4,....$
But I can't find smart closed-formula, that's easy to proof.
$g(n) = g(\frac{n}{2}) + \lfloor{log_2{n}}\rfloor for \ even$ and
$g(n) = g(\frac{n-1}{2}) + \lfloor{log_2{n}}\rfloor for \ odd$
My guess was $g(n) = \lfloor log_2{\lfloor\frac{n}{2}\rfloor} \rfloor +\lfloor log_2n\rfloor $ but I am having problem proving this one, even though it's seems to be true :S Could you tell me if im on the right track?
Any help very appreciated
AI: Hint: Start by considering $g(N)$ for $N = 2^n$. Notice in this case that
$$
g(2^n) = g(2^{n-1}) + n = g(2^{n-2}) + (n-1) + n
= \ldots = g(0) + 1 + 2 + \ldots + n = \frac{n(n+1)}{2} = \frac{(\log_2 N)^2 + \log_2N}{2}.
$$
Now look at the general case. |
H: What is the difference between Taylor series and Laurent series?
Can someone intuitively describe what is the difference between Taylor series and Laurent series? Also, what is the most general formula for both?
AI: Laurent series allows for terms with negative power. Intuitively, this allows for singularities to occur. |
H: If A, B, C, D are non-invertible $n \times n$ matrices, is it true that their $2n \times 2n$ block matrix is non-invertible?
Is it true that
$ \left( \begin{array}{ccc}
A & B \\
C & D \\\end{array} \right)$
is non-invertible?
Assume that the matrix is over a field.
AI: Try the $2\times 2$ matrices that make up $$\begin{pmatrix}1&0&0&0\\
1&0&1&0\\
0&1&0&0\\
0&1&0&1\end{pmatrix}.$$
(This example works over all fields, no matter what characteristic). |
H: Tangents from a certain point, to a circle?
We have a circle with radius 2, centred on the origin. Find the equation of the lines passing through the point $(0,4)$ which are tangent to the circle.
So we have the circle $$x^2 + y^2 = 4$$
We need 2 lines $$y=ax+4$$
So if you fill that in the equation of the circle, we end up with $$a^2x^2 + 8ax + x^2 + 12 = 0 $$.
So we'd have to solve $$ a = \dfrac{-8x \pm \sqrt{64x^2-4x^2(x^2+12)}}{2x^2}$$
I must have done something wrong, since this causes the discriminant to be negative, and thus gives no answer. Is there a smarter way I overlooked?
AI: Suppose $A(x_1,y_1)$ is the point of the circle where the tangent is drawn.
Then the equation of the tangent is $xx_1+yy_1=4$. Since $(0,4)$ belongs to the tangent it satisfies her equation so $0x_1+4y_1=4\rightarrow y_1=1$.
Then subtitute this value to the circle equation
we get $x_1^2+y_1^2=4\rightarrow x_1^2+1^2=4\rightarrow x_1=\sqrt3$ or $x_1=-\sqrt3$.
Thus we have two tangents with equations:
$\sqrt3x+y=4$ and $-\sqrt3x+y=4$. |
H: How to solve equations of this form: $x^x = n$?
How would I go about solving equations of this form:
$$
x^x = n
$$
for values of n that do not have obvious solutions through factoring, such as $27$ ($3^3$) or $256$ ($4^4$).
For instance, how would I solve for x in this equation:
$$x^x = 7$$
I am a high school student, and I haven't exactly ventured into "higher mathematics." My first thought to approaching this equation was to convert it into a logarithmic form and go from there, but this didn't yield anything useful in the end.
My apologies if this question has been asked and answered already; I haven't been able to find a concrete answer on the matter.
AI: Equations like $x^x=7$ often don't have ''nice'' solutions. Whenever you see something like $x^x$, something that you should think about is the Lambert W function. This is the function $W(x)$, implicitly defined by $z=W(z)e^{W(z)}$.
In your case, taking natural logs gives $x \log x=\log 7$. So $e^{\log x} \log x=\log 7$, so $\log x=W(\log 7)$, and finally $x=e^{W(\log 7)}$. The Lambert W function has many different branches, which is something sort of akin to the fact that when you take a square root you can choose either the positive or negative square root. This means that there isn't a unique solution to your equation. You can get some families of solutions using Wolfram Alpha: http://www.wolframalpha.com/input/?i=x%5Ex%3D7. |
H: Using table of integrals to solve $\int y \sqrt{6+12y-36y^2}dy$
I'm supposed to use a table of integrals to solve the below equation:
$$\int y \sqrt{6+12y-36y^2}dy$$
I'm having trouble identifying the form to use because I guess my weakness in algebra shows in my inability to compete the square of the square root function.
Here is how I tried to complete the square (first rewriting the equation and factoring out a -12):
$$6-12(-y+3y^2)$$
And I assume the next step is dividing my $b$ by $2$ and then squaring the number which would give:
$$6-12(\frac{1}{4}-\frac{1}{2}y+3y^2)$$
But I don't think this is right and I don't know how to get it in the form of $\sqrt{a^2-u^2}$
AI: As $6+12y-36y^2=7-(6y-1)^2,$
Put $6y-1=\sqrt7\sin\theta$
$$\int y\sqrt{6+12y-36y^2}dy=\frac{\sqrt7\sin\theta+1}6\sqrt7\cos\theta\frac{\sqrt7\cos\theta}6 d\theta$$
$$=\frac7{36}\int(\sqrt7\sin\theta+1)\cos^2\theta d\theta$$
$$=\frac7{36}\sqrt7\int \sin\theta \cos^2\theta d\theta+\frac7{36}\int \cos^2\theta d\theta$$
For the first part, put $\cos\theta=u$
For the second, $$\int \cos^2\theta d\theta=\frac{(1+\cos2\theta)}2 d\theta=\frac{\theta}2+\frac{\sin2\theta}4$$ |
H: How many numbers from 1 through 60100 are divisible by none of the numbers from 2 through 6?
My thoughts on doing this problem:
total numbers is 60100
so from the total I subtract the numbers divisible by 2, 3, 4, 5, and 6.
Yet my answer
60100-30050-20033-15025-12020-10016
is a negative number. How to solve this problem?
AI: We need to consider the primes only.
The number of Numbers divisible by at least one of $2,3,5$ will be $$\left\lfloor\frac{60100}2\right\rfloor+\left\lfloor\frac{60100}3\right\rfloor+\left\lfloor\frac{60100}5\right\rfloor-\left\lfloor\frac{60100}{2\cdot3}\right\rfloor-\left\lfloor\frac{60100}{2\cdot5}\right\rfloor-\left\lfloor\frac{60100}{3\cdot5}\right\rfloor+\left\lfloor\frac{60100}{2\cdot3\cdot5}\right\rfloor$$ |
H: Dicrete Math Interesting question about Tromino
Prove that for a m$\times$n rectangle, if this rectangle can be covered completely by trominoes of the shape indicated in the picture, then mn is divisible by 3.
I came up with a tentative way to prove the statement above. There are two problems that are bothering me, first, I am not sure if I've covered all the cases in my proof; second, I am guessing there will be a more elegant way to prove this. I'd appreciate if you can help!
Assume that 3 divides m. Suppose that n is even, we can partition the rectangle into 3$\times$2 blocks(the following graph indicates how this would be done.) Suppose n is odd, clearly we can't have n=1. If m is even, the first three columns of the rectangle may be partitioned into 2$\times$3 blocks while the remaining can be partitioned into 3$\times$2 blocks. Supppose m is odd, obviously, we can't have m=3, or n =3, since we can only pack a 3$\times$n rectangle by partitioning it into 3$\times$2 blocks. But that requires n to be even. If m$\geq$9 and n$\geq$5, the first five columns of the first nine rows consist of a 9$\times$5 block(indicated below).
The remaining columns of these rows are partitioned into 3$\times$2 blocks. The remaining rows can be covered if and only if 3 divides mn. Provided that both m and b are greater than 1, and neither is equal to 3 when the other is odd.
AI: The tromino's area is a multiple of 3, so if there's a covering by them then the total area is a multiple of the tromino's area and hence a multiple of 3. Read the problem carefully.
This is an example of what's called a necessary but not sufficient condition. If the rectangle is to be tilable then we have to have 3 divides $mn$, but that fact alone won't guarantee the rectangle is tilable. |
H: The existence of "arbitrary large" connected compact sets in the plane
Studying some complex analysis I came up with the following hypothesis:
Let $\Omega \subseteq \mathbb C$ be a region (an open and connected set), and let $E \subset \Omega$ be a compact subset. There exists a connected compact set $F$ satisfying $E \subset F \subset \Omega$.
I believe that this is true and I want to prove it. My idea was considering $$F_n=\left\{z \in \Omega:|z| \leq n,\text{dist}(z, \partial\Omega) \geq \frac{1}{n} \right\} $$ for large integers $n$, since $\Omega$ itself is connected and the $F_n$ "tend" to $\Omega$.
Unfortunately, I couldn't prove $F_n$ works.
Is my statement correct? Could you please provide me with a proof?
Thanks!
AI: The sets $F_n$ are compact, but they need not be connected. Consider a region $\Omega$ consisting of the lower half plane, a disk with radius $\frac12$ and centre $k+i$ for all $k \in \mathbb{Z}$, and for each $k\in\mathbb{Z}$ a corridor with width $2^{-(4+\lvert k\rvert)}$ connecting the half plane with the disk. Then the corridors are too narrow, so you have islands of $F_n$ in the disks for large enough $\lvert k\rvert < n$ that aren't connected to the main body in the lower half plane.
The statement is however true (and from that it follows that $E$ is contained in one connected component of $F_n$ for all large enough $n$).
First, we have $\delta := \operatorname{dist}(E,\,\partial \Omega) > 0$ since $E$ is compact. Thus $E_1 := \{ z : \operatorname{dist}(z,\,E) \leqslant \delta/2\}$ is a compact subset of $\Omega$. Now, $E_1$ is a union of (closed) disks of radius $\delta/2$, hence it can have only finitely many connected components. $\Omega$ is open and connected, hence path-connected. Now you can connect all of the finitely many connected components of $E_1$ with paths in $\Omega$. |
H: Complex formula that is equivalent to $f(x) = x$
In a computer you can't store any real number that you want to because in the $[0.0;1.0]$ interval there are infinite numbers and computer memory is finite. I want to show it in examples, which is why I need some formulas that do some complex calculations and return the initial number. Logically it should be the same but computer will calculate and return a different number. I want to get maximal difference.
Sample expressions can be like $f(x) = x^2/x$ or $(x + x) / 2$
Of course these are too simple ones.
AI: In languages that use IEEE floating point, you can get counterintuitive results with even something as simple as
$$1.0-0.3-0.3-0.3-0.1 = -2.77556\times 10^{-17} \neq 0$$
and so for $f(x) = x + 1.0-0.3-0.3-0.3-0.1$, you will have $f(0)\neq 0$. |
H: Induction proof of $a^r \ge 1$
I understand induction with one variable well, however I am not sure what to do when there are 2 or more variables.
The problem I came across is following:
Prove that $a^r \ge 1$, where $r \in \mathbb{N}$ and $a \in \mathbb{R} \wedge a \ge 1$
My solution which I am not sure whether is right:
1) For $r=0$:
$a^0 = 1$, $1\ge1$
2) Now assuming $a^r \ge 1$, I try to prove that $a^{r+1} \ge 1$.
$a^{r+1} = a^r * a$
As assumed $a^r \ge 1$ and also $a \ge 1$, therefore $a^r * a \ge 1$.
Is this proof correct or do I have to do it for $a$ somehow as well?
AI: Your proof looks fine to me. For this proof, there's no need to induct on $a$.
The induction proof as is assumes that $a$ is just some arbitrary number for which $a \in R \wedge a \geq 1$ is true. This is enough to show that any $a$ that fulfills $a \in R \wedge a \geq 1$ will work (since you can freely substitute any $x$ such that $x \in R \wedge x \geq 1$ and the proof doesn't change). |
H: Double factorial identity
Does anyone know a strategy for proving
$$
2\cdot(2k-3)!!=\sum_{i=1}^{k-1}(2i-3)!!(2(k-i)-3)!!\binom{k}{i}
$$
for $k\geq 2$? Note that $(-1)!!=1$. Hints would be most appreciated. Full solutions not so much.
I have considered induction but whereas the left hand side is multiplied by the next odd number in the induction step the right hand side becomes one term longer and each term is multiplied by a different factor.
AI: First note that $$(2n-1)!!=\frac{(2n)!}{2^nn!}\;,$$ so that the desired identity can be written
$$\begin{align*}
\frac{2(2k-2)!}{2^{k-1}(k-1)!}&=\sum_{i=1}^{k-1}\left(\frac{(2i-2)!}{2^{i-1}(i-1)!}\cdot\frac{\big(2(k-i)-2\big)!}{2^{k-i-1}(k-i-1)!}\binom{k}i\right)\\\\
&=\frac1{2^{k-2}}\sum_{i=1}^{k-1}\frac{(2i-2)!(2k-2i-2)!k!}{i!(i-1)!(k-i)!(k-i-1)!}\\\\
&=\frac{k!}{2^{k-2}}\sum_{i=1}^{k-1}\left(\frac1i\binom{2(i-1)}{i-1}\cdot\frac1{k-i}\binom{2(k-i-1)}{k-i-1}\right)\\\\
&=\frac{k!}{2^{k-2}}\sum_{i=1}^{k-1}C_{i-1}C_{(k-2)-(i-1)}\\\\
&=\frac{k!}{2^{k-2}}\sum_{i=0}^{k-2}C_iC_{(k-2)-i}
\end{align*}$$
where $C_n$ is the $n$-th Catalan number. Now do a little simplification and apply a basic Catalan identity, and you’ll have it. |
H: Prove convergence using $\varepsilon$-$N$ definition
This is an example presented by professor in class. I understand the idea behind this kind of definition, but I'm having trouble following my professor's thought process.
We must prove that the $\displaystyle \lim_{n\to\infty} \frac{2n-1}{n^2} = 0$ using the $\varepsilon$-$N$ definition of a limit of a sequence.
So, the foregoing limit is true iff $\forall \varepsilon>0,\ \exists N(\varepsilon)>0 : \forall n \ ( n \geq N \Rightarrow |f(n) - 0| < \varepsilon)$.
Now, the way I understand it is that if we prove the conditional statement by choosing an appropriate $N$, we've proven the limit.
So first we fix $\varepsilon>0$ since it's given, and play around with $$\displaystyle \left| \frac{2n-1}{n^2} \right| < \varepsilon $$
$$|2n-1| < \varepsilon \ n^2 .$$
We note that $|2n-1|<|2n|$ which sits right with me.
However, my professor then reasons that $|2n-1|<|2n|<\varepsilon \ n^2$ and so $2n < \varepsilon \ n^2$.
How do we know that $2n < \varepsilon \ n^2$? If epsilon is really small, wouldn't there be a point when this inequality is no longer true? This is what is bothering me.
From that we conclude that $\displaystyle n>\frac{\varepsilon}{2}$ and choose $\displaystyle N=\frac{\varepsilon}{2}$ so that
$$n \geq \frac{\varepsilon}{2} \Rightarrow \left| \frac{2n-1}{n^2} \right| < \varepsilon$$
is always true. And that is true iff the limit is true, so the limit is true.
I'd like to know if my understanding of this definition is correct, and if anyone can explain to me the reasoning behind the inequalities above.
Thank you very much!
AI: I'll give you another way to think about it. Instead of putting $\epsilon$ there from beginning, my approach is always to simply find an upper bound for $|a_n - L|$, and then look at it to figure it out which conditions put over $n$ assure that this bound is less than $\epsilon$.
Your sequence is $(a_n)$ where $a_n = (2n-1)/n^2$. Now, we want to bound $|a_n - 0|$ and look at that bound to see what $n$ must be bigger than in order to make that less than $\epsilon$.
Since $n \geq 1$ we have that $2n - 1 > 0$ and so $|a_n| = a_n$, in other words, we can look just to $(2n-1)/n^2$. Now, this is the same as
$$\dfrac{2n-1}{n^2}=2\dfrac{1}{n}-\dfrac{1}{n^2},$$
Now, look that $-1/n^2 < 0$ always, so that
$$\dfrac{2n-1}{n^2} <\dfrac{2}{n}$$
Now, what must $n$ be greather than in order to make that less than $\epsilon$. Of course, if $n > 1/\epsilon$, then $1/n < \epsilon$, however there's a $2$ there, so that to take care of it we make $n > 2/\epsilon$, now when we divide by $n$ we get $1/n < \epsilon/2$, and hence the condition follows. So, given $\epsilon > 0$, if $n > 2/\epsilon$ we have that
$$\left|\dfrac{2n-1}{n^2}-0\right|=\dfrac{2n-1}{n^2}=\dfrac{2}{n}-\dfrac{1}{n^2}<\dfrac{2}{n} <\epsilon$$
and hence $a_n \to 0$ when $n\to \infty$. |
H: Heat Equation: Initial value boundary value problem
Given the initial-boundary value problem
$$ u_t −u_{xx} = 2, \ \ \ \ \ \ x ∈ [−1,1], t ≥ 0,$$
With initial and boundary conditions
$$u(x,0) = 0$$
$$ u(−1, t) = u(1, t)=0$$
Claim: the solution, $u(x,t)$, is such that
$$ u(x, t) ≤ −x^2 + 1$$
for all $x ∈ [−1,1], \ \ t ≥ 0$.
My idea: this is of course the heat equation and I am asked to prove that the solution cannot exceed a certain function, namely $-x^2+1$(which solves the heat equation), since the solution must "diffuse" in time. I was thinking that minimum principle should be involved here but I am not quite sure how to show the claim.
Thanks for the help.
AI: The solution consists of two parts: the stationary solution $-x^2+1$ plus transient part, which is simply the solution to the homogeneous heat equation with the same boundary conditions. This solution can be easily found in explicit form (note that the initial condition will change). Do this and show that the transient solution is always less than 0. |
H: How to find laurent series of $\exp(1/z)$
I want to find the laurent series of $f(z)=\exp(1/z)$.
I started with the formula for laurent series: $f(z)=\sum_{0}^{\infty} a_n (z-z_0)^n +\sum_{1}^{\infty} b_n (z-z_0)^{-n}$, but I don't know how to apply it.
Can someone help me find the series?
AI: (Posting a comment as an answer)
Do you know the series expansion for $f(w)=e^w$? If you do, write that out and then see what happens when $w=1/z$. |
H: Find the limit of $\sum\limits_{k=1}^n\left(\sqrt{1+\frac{k}{n^2}}-1\right)$
$$\lim_{n\rightarrow\infty}\sum_{k=1}^n\left(\sqrt{1+\frac{k}{n^2}}-1\right)$$
Note that $\forall x\ge 0, \sqrt{x}-1\le\sqrt{1+x}-1\le x$
Then
$$\sum_{k=1}^n\left(\sqrt{\frac{k}{n^2}}-1\right)\le S_n\le \sum_{k=1}^n\frac{k}{n^2}=\frac{1}{2}-\frac{1}{2n}$$
How do I evaluate the sum on the left?
AI: We have $$S_n=\sum_{1,n} \sqrt{1+k/n^2}-1=\sum_{1,n}\frac{k/n^2}{\sqrt{1+k/n^2}+1}$$
hence
$$\frac{n(n+1)}{2n^2 (\sqrt{1+1/n}+1)}=\sum_{1,n}\frac{k/n^2}{\sqrt{1+n/n^2}+1}\leq S_n\leq \sum_{1,n}\frac{k/n^2}{\sqrt{1}+1}=\frac{n(n+1)}{4n^2}$$
and so $\lim S_n =\frac{1}{4}$ by squeezing. |
H: Are strictly upper triangular matrices nilpotent?
An $n\times n$ matrix $A$ is called nilpotent if $A^m = 0$ for some $m\ge1$.
Show that every triangular matrix with zeros on the main diagonal is nilpotent.
AI: Its characteristic polynomial is $T^n$, so by Cayley-Hamilton, $A^n=0$. |
H: Star-shaped set
Definition.- A set $S\subset \mathbb{R}^n$ is called star-shaped if there exists a point $z_0$in $S$ such that the line segment between $z_0$ and any point $z$ in $S$ is contained in $S$, $z_0$ is called a center of $S$.
Let $S\subset \mathbb{R}^n$ be a star-shaped open set , I have a problem is the set $$ \color{blue}Z=\{z\in S:z \text{ is a center of S}\}$$ an open set ?
and if $S$ is bounded ?
Any hints would be appreciated.
AI: Take a right-angled triangle that is not isosceles. Fit four of them around a single point, like a pinwheel.
EDIT: Take the interior of the octagon with vertices (in order) $(0,2),(0,1),(2,0),(1,0),(0,-2),(0,-1),(-2,0),(-1,0)$ |
H: A question of H.G. Wells' mathematics
H.G Wells' short story The Plattner Story is about a man who somehow ends up being "inverted" from left to right. So his heart has moved from left to right, his brain, and any other asymmetries belonging to him. Then H.G Wells' goes on a slight metaphysical exposition:
There is no way of taking a man and moving him about in space, as ordinary people understand space, that will result in our changing his sides. Whatever you do, his right is still his right, his left his left. You can do that with a perfectly thin and flat thing, of course. If you were to cut a figure out of paper, any figure with a right and left side, you could change its sides by simply lifting and turning it over. But with a solid it is different. Mathematical theorists tell us that the only way in which the right and left sides of a solid body can be changed is by taking that body clean out of space as we know it, - taking it out of ordinary existence, that is, and turning it somewhere outside space. This is a little abstruse, no doubt, but anyone with a slight knowledge of mathematical theory will assure the reader of its truth. To put the thing in technical language, the curious inversion of Plattner's right and left sides is proof that he has moved out of our space into what is called the Fourth Dimension, and that he has returned again to our world.
This is leaving me a bit perplexed. A reflection is an odd isometry after all, and rotations/translations are direct. So it shouldn't be possible for the man to be reflected using continuous,direct transformations, regardless of how many dimensions he is using. But then again, his argument with the 2/3D analogue is convincing... Thoughts and explanations?
AI: I think the key here is that a flat object living purely in the $xy$-plane is not affected by reflection in $z$. Thus, an apparent reflection across the $x$-axis can be achieved by reflecting across both the $x$ and $z$ axes — this is a net rotation and hence can be achieved by continuous transformation without breaking orientability.
The same principle would apply to a 3D object embedded in $\mathbb R^4$. Having no width along the fourth dimension makes one invariant under reflections across that dimension. |
H: Is $\mathbb{Q}(5^{1/3})$ a Galois extension over $\mathbb{Q}$?
I am trying to prove or disprove that the simple extension $\mathbb{Q}(5^{1/3})$ is Galois over $\mathbb{Q}$.
I suspect that this extension is not Galois, because an extension if Galois over $\mathbb{Q}$ if and only if it is the splitting field of an irreducible polynomial over $\mathbb{Q}$. We know that $5^{1/3}$ is a root of $x^3-5$ over $\mathbb{Q}$, but $\mathbb{Q}(5^{1/3})$ is not the full splitting field of $x^3-5$. However, I am not sure how to prove that this extension is in fact not Galois.
AI: If an extension is Galois it must be normal and $K/E$ being normal is equivalent to saying that if an irreducible polynomial in $E[x]$ has one root in $K$, then all its roots must lie in $K$.
So, to show that $\mathbb{Q}(5^{1/3})/\mathbb{Q}$ isn't Galois, all you need to do is find an irreducible polynomial with one root in your extension field that doesn't split, which it sounds like you have done. |
H: We have $T(n) \leq T(\lceil \frac{n}{5} \rceil) + T(\lceil \frac{7n}{10} \rceil)$. Show that $T(n) < c'n$.
We have $T(n) \leq T(\lceil \frac{n}{5} \rceil) + T(\lceil \frac{7n}{10} \rceil)$. Show that $T(n) < c'n$ for all $n$ and for some constant $c'$.
Looks straightforward enough, but suprisingly I get stuck at the base case:
For $n=1$ we have:
$T(1)\leq 2T(1)+cn \Leftrightarrow -T(1) \leq cn $
Which doesn't help me much.
AI: Is $T(n)$ defined for $n\in\mathbb{N}$?
Show that $\lceil\frac{n}{5}\rceil\leq\frac{n+4}{5}$ and similar for the other term.
Find an $N$ for which,
$\,\,$ If $n>N$, and $T(k)\leq c\,'k$ for all $k\leq n$, then $T(n+1)\leq c\,'(n+1)$ as well.
Lastly, find a $c\,'$ that works for all $k=1..N$ I think it will depend on the values of $T(1), T(2)$ and $T(3)$ but on only a finite number of $T(n)$ |
H: How do you factor this using complete the square? $6+12y-36y^2$
I'm so embarrassed that I'm stuck on this simple algebra problem that is embedded in an integral, but I honestly don't understand how this is factored into $a^2-u^2$
Here are my exact steps:
$6+12y-36y^2$ can be rearranged this way: $6+(12y-36y^2)$ and I know I can factor out a -1 and have it in this form: $6-(-12y+36y^2)$
This is the part where I get really lost. According to everything I read, I take the $b$ term, which is $-12y$ and divide it by $2$ and then square that term. I get: $6-(36-6y+36y^2)$
The form it should look like, however, is $7-(6y-1)^2$
Can you please help me to understand what I'm doing wrong?
AI: we know that $$ (ay - b)^2 = a^2y^2 - 2ayb + b^2 \space \space \space \space (1) $$
we have $$ 6 + 12y - 36y^2 = -(36y^2 - 12y - 6) $$
we need to factor $$ 36y^2 - 12y - 6 $$
from (1) let $ a^2 = 36 \Rightarrow a = \pm 6, 2ab = 12 \Rightarrow b = \pm 1 $
thus $$ 36y^2 - 12y + 1 - 7 = (6y - 1)^2 \ or \ (-6y +1)^2 - 7 \Rightarrow 6 + 12y - 36y^2 = 7 - (6y-1)^2 $$ |
H: Write $y=-2\sin (x- \frac{\pi}{3})$ in the form $y= a\sin k(x-\beta)$, where a and k are nonnegative
I would like some help with this question please:
Write the following in the form $y= a\sin k(x-\beta)$, where a and k are nonnegative
$$y=-2\sin (x- \frac{\pi}{3})$$
I am confused over the effect the negative sign in front of the 2 will have on the expression in brackets. I know that $-\sin x = \sin (-x)$, so does$-2\sin x = 2 \sin (-x)$?
Following on from this, does:
$$y=-2\sin (x- \frac{\pi}{3}) = 2\sin (-x - \frac{\pi}{3})?$$
AI: $$y=-2\sin\left(x-\frac\pi3\right)=2\sin\left(\pi+x-\frac\pi3\right)=\cdots$$ as $\sin(\pi+x)=-\sin x $ |
H: Logically speaking, why can variables be substituted?
Suppose that
$$a^2+a+1=b$$
Suppose also that $a=5/4$. What makes it valid to substitute $5/4$ into the first equation? Is it because equality is transitive?
AI: It's just a basic principle of first-order logic with equality that if $a = b$ and $P(b)$ for a formula $P$, then $P(a)$. It's probably the most important aspect of equality that usually "goes without saying" in mathematics, except in logic courses. |
H: Give a push down Automata for this language: the length of is odd and it's middle symbol is 0
Give a push down automaton for this language:
{w| the length of w is odd and it's middle symbol is 0}
Here is the CFG I wrote for this language:
S --> 0|0S0|0s1|1s0|1s1
This what I have done for odd length part (I'm not using a stack for first part but I'm sure I need to use stack for second part):
I've spent days on this question but no luck,
AI: Broad Hint: You shouldn't have to check that the string has odd length. Note that a string $w$ is in the language $L$ described above iff there is an occurrence of the symbol $\mathtt{0}$ in $w$ such that there are exactly as many symbols in $w$ appearing before that occurrence of $\mathtt{0}$ as there are after. (For a hint that is a bit more direct, whenever you read a $\mathtt{0}$ from the string non-deterministically guess that this is the correct one, using the stack to check whether the above property holds.) |
H: Show that $\dot{n_s}=-\kappa_s t$
I found the question in a differential geometry textbook while studying. This question seems so intesting to me. So please help me solving it.
Show that, if $\gamma$ is a unit-speed plane curve, $$\dot{\mathbf{n}}_s=-\kappa_s\mathbf t.$$
I know that
$$\dot t =\kappa_s n_s$$ and $$\kappa =|\kappa_s|$$
AI: Check this out:
$\langle \mathbf n_s, \mathbf t_s \rangle = 0,\tag {1}$
whence
$\frac{d}{ds}\langle \mathbf n_s, \mathbf t_s \rangle = 0, \tag{2}$
so that
$\langle \dot{\mathbf n}_s, \mathbf t_s \rangle + \langle \mathbf n_s, \dot{\mathbf t}_s \rangle = 0, \tag{3}$
and using
$\dot {\mathbf t}_s =\kappa_s \mathbf n_s \tag{4}$
we get from (3)
$\langle \dot{\mathbf n}_s, \mathbf t_s \rangle = -\kappa_s \tag{5}$
since $\langle \mathbf n_s, \mathbf n_s \rangle = 1$. And since $\langle \mathbf n_s, \dot{\mathbf n}_s \rangle = 0$ and we are operating in $\Bbb R^2$, we can conclude that
$\dot {\mathbf n}_s = -\kappa \mathbf t_s. \tag{6}$
QED!
Hope this helps! Cheerio,
and as always,
Fiat Lux!!! |
H: Prove why this algorithm to compute all list permutations works
Note: this is not homework or for a class, as I'm no longer in school.
Let's say I have a list of characters {1,2,...,N} and I want to generate all permutations.
For example, if I had {1,2,3} the permutations would be
{1,2,3}
{1,3,2}
{2,1,3}
{2,3,1}
{3,1,2}
{3,2,1}
Of course we could come up with a number of algorithms do solve this that work for various reasons, but I'm interested in the combinatorial theory behind the following algorithm.
The Algorithm
In brief: I'll generate a round of subsets by swapping the ith value with the Nth for all existing subsets generated in previous rounds.
In the same round, I'll generate more subsets by swapping the ith value with the (N-1)th, and then with the (N-2)th etc. until (N-j) = i
Some definitions:
Let List be a container of characters.
List(List) is a container of containers of characters
Containers are 1 indexed,
i.e., List[1] gives the first element, and List[N] gives the last one
In our example, the original list {1,2,...,N} is a List, and the set of permutations is a List(List)
In pseudocode:
L = {1,2,...,N}
Permutations = {}
For(i=1 to N){
List(List) thisRound
For each permutation p in Permutations{
For(j=N; j > i; --j)
{
List t = p
Swap(t[i],t[j])
thisRound.Insert(t)
}
}
Permutations.AddAll(thisRound) //add all the permutations in thisRound
}
Full C++ Implementation
Assumes that we're generating permutations of strings
Simple I/O: Asks for a string of characters
Outputs each permutation as well as the total count
#include <iostream>
#include <string>
#include <vector>
using namespace std;
void GeneratePermutations(vector<string>& _permutations);
int main(){
string input;
cout << "Input a string: " << endl;
cin >> input;
vector<string> permutations;
permutations.push_back(input);
cout << "Generating all permutations..." << endl;
GeneratePermutations(permutations);
cout << "Permutations: " << endl;
size_t count = 0;
for (vector<string>::iterator vecIter = permutations.begin(); vecIter != permutations.end(); ++vecIter){
cout << *vecIter << endl;
++count;
}
cout << "Total: " << count << endl;
}
void GeneratePermutations(vector<string>& permutations)
{
size_t i = 0;
size_t length = permutations.front().size();
for(size_t i=0;i<length;++i){
vector<string> tmp;
for(vector<string>::iterator strIter = permutations.begin(); strIter != permutations.end(); ++strIter){
for(size_t j=(*strIter).size()-1; j > i; --j){
string nextStr = *strIter;
std::swap(nextStr[i],nextStr[j]);
tmp.push_back(nextStr);
}
}
permutations.insert(permutations.end(), tmp.begin(), tmp.end());
}
}
Example run
Input a string:
1234
Generating all permutations...
Permutations:
1234
4231
3214
2134
1432
1324
4132
4321
3412
3124
2431
2314
1243
4213
3241
2143
1423
1342
4123
4312
3421
3142
2413
2341
Total: 24
My Question
I have only a moderate intuition about why this method works.
First we generate all permutations of the first column, which gets reused over and over (4321,4321), and then we do something similar for the second column, but it excludes the values we generated the first time around.
I just can't come up with a real pattern/explanation for WHY it works. I've looked (briefly) online for similar implementations to generate string permutations, and I haven't found anything like this.
Can you determine why it works?
AI: Basically, the Swap command multiplies by the transposition $(i\ j)$ (I'm used to think of it as left-multiplication, but every other person would think of it as right-multiplication, so just pick what's comfortable to you and stick with it). So what we're doing is first including all $(1\ j)$'s, then adding all $(1\ j)(2\ k)$, and so on. In particular, pick a permutation and represent it as the product of disjoint cycles, and it becomes clear how it is generated. For example, $(1\ 3\ 4)(2\ 5)\in S_5$ is generated as follows:
$$(1\ 3)\to(1\ 3)(2\ 5)\to(1\ 3)(2\ 5)(3\ 4) = (1\ 3)(3\ 4)(2\ 5) = (1\ 3\ 4)(2\ 5).$$ |
H: Why does "imread" generate 3 images?
The MATLAB help for imread says:
A = imread(filename, fmt)
The return value A is an array containing the image data. If the file
contains a grayscale image, A is an M-by-N array. If the file contains
a truecolor image, A is an M-by-N-by-3 array.
I thought it would generate 3 matrices for the RGB (one with the Red, one with the Green and one with the Blue), so I did:
img = imread('myimg')
size(img) % ans = 720 960 3
imshow(img) % the original image
imshow(img(:,:,1)) % print grayscaled image
imshow(img(:,:,2)) % print grayscaled image
imshow(img(:,:,3)) % print grayscaled image
And, as commented, the 3 images are just the original but grayscaled. So, why are 3 images generated?
(using Matlab R2011b)
AI: the 3 images are just the original but grayscaled"
Uh, no. There are not 3 images, there is a single image represented as a 3D matrix, or, a stack of 3 matrices, one for each channel (R,G,B). When you do img(:,:,1) you are extracting the matrix corresponding to the red channel, and when you call imshow(img(:,:,1)) you tell Matlab to interpret that matrix as an image (and because that matrix has only one channel, it interprets it as a gray scale image). The three calls to imgshow should then show each channel as a grayscale image. They should not be equal, though; to be sure of that, use some image that has eg a pure red corner: that would display as white in the first channel, black in the others.
This question is more apt for stackoverflow, though. |
H: Is $P( A \cup B\, |\, C) $ the same as $P(A | C) + P(B | C) $ ( $A$ and $B$ are mutually exclusive)
Is $P( A \cup B \,|\, C)$ the same as $P(A | C) + P(B | C)$ ?
Here $A$ and $B$ are mutually exclusive.
AI: Yes.
Remember:
$$
\begin{align*}
P(A\cup B\mid C)&=\frac{P((A\cup B)\cap C)}{P(C)}\\
&=\frac{P((A\cap C)\cup (B\cap C))}{P(C)}\\
&=\frac{P(A\cap C)+P(B\cap C)}{P(C)}\\
&=\frac{P(A\cap C)}{P(C)}+\frac{P(B\cap C)}{P(C)}\\
&=P(A\mid C)+P(B\mid C).
\end{align*}
$$
(In the third line, we used the fact that if $A\cap B=\varnothing$, then $(A\cap C)\cap (B\cap C)=\varnothing$.) |
H: A student must answer five out of $10$ questions on a test, including at least two of the first five questions.
A student must answer five out of $10$ questions on a test, including at least two
of the first five questions. How many subsets of five questions can be answered?
$$\binom52\binom53+\binom53\binom52+\binom54\binom51+\binom55$$
if im correct , is there any other way to think about this problem ?
AI: What you’ve done is correct. You could also have calculated
$$\binom{10}5-\binom51\binom54+\binom55\;,$$
i.e., the total number of ways to choose $5$ questions minus the number of ways that include $1$ or $0$ questions from the first $5$. |
H: Is this derivative wrong?
Given the following in WebAssign: $\ y = 8x + {{7} \over {x }} $
Why is it that the derivative is shown to be: $\ {dy \over dx}= {8x^2 - 7 \over { x^2}} $
Yet, when I find the derivative the quotient rule being used on the $\ 7 \over x $ makes my derivative come out to: $\ {dy \over dx}= {8x^2 +x - 7 \over { x^2}} $
Am I doing something wrong?
AI: As Amzoti said the derivative of $y = 8x + \frac{7}{x}$ is $\frac{dy}{dx} = 8 - \frac{7}{x^2}$ and you just have to add up the values but if you insist on doing some application of the quotient rule you can add up the terms at first and then derivative.
\begin{eqnarray}
y & = & 8x + \frac{7}{x} \\
y & = & \frac{8x^2 + 7}{x}
\end{eqnarray}
Now apply the quotient rule and you should get the solution shown in your textbook or wherever you found that solution. |
H: Some explinations on the proof of a lemma from (Gamelin-Topology) book
In the following lemma, I need to understand why $g$ is continuous and where he used the property $|h(t)|\to \infty$ as $|t|\to \infty$.
My attempt: Let $h(t_n)\to h(t)$, then $g(h(t_n))=t_n-^?\to t=g(h(t))$. I'm not sure in which cases we can remove $(?)$.
AI: Okay, the $|h(t)|\to\infty $ as $ |t|→∞$ information is essential. It tells you that for each $M>0$ there is a $N>0$ such that whenever $|t|>N$ then $|h_i(t)|>N$ for $i=1,2$. This means that the preimage of a bounded set $B\subset\Bbb R^2$ is bounded. Since the preimage of a closed set is closed due to continuity, you can follow that preimages of compact sets are compact, i.e. the map $h$ is a so-called proper map.
Now, a proper map $f:X\to Y$ into a locally compact Hausdorff space $Y$ is closed (Try to prove it yourself or ask me if you'd like to see a proof; it is only required that each point in $Y$ has a compact Hausdorff neighborhood). Since this applies to $h:\Bbb R\to\Bbb R^2$, we conclude that $h$ is a closed map. Therefore it is a closed embedding. |
H: Finding the smallest positive integer $ n $ satisfying a modular identity.
Is there any good way of finding the smallest positive integer $ n $ such that
$$
3^{n} \equiv 1 \pmod{1000000007}?
$$
AI: First to make things clear, we are trying to find the multiplicative order of $3$ modulo $1000000007$
Now use the fact that $1000000007$ is a prime number and $(3,1000000007) = 1$. Then by Euler's Theorem and Fermat's Little Theorem we have:
$$3^{1000000007-1} \equiv 3^{1000000006}\equiv 1 \pmod {1000000007}$$
From Lagrange Theorem for group order we know that the order of one element in the group divides the order of the group. So let $G_{1000000007}$ denote the group of non-zero elements of the integers (mod ${1000000007}$) under multiplication, which has order $1000000007-1=1000000006$.
So we have: $ord_{1000000007}3 \mid 1000000006$
Because both numbers are natural numbers and $1000000006$, has $4$ divisors and those are:
$$ 1, \quad \quad 2, \quad \quad 500 000 003, \quad \quad 1000000006$$
That means that $ord_{1000000007}3$ must be one of them. It's easy to check that $ord_{1000000007}3=1$ and $ord_{1000000007}3=2$ isn't true. We now need to check whether $ord_{1000000007}3=500 000 003$.
From Euler's criterion we have:
$$\left(\frac{3}{1 000 000 007}\right) \equiv 3^{500 000 003} \pmod {1 000 000 007}$$
Where$\left(\frac{3}{1 000 000 007}\right)$ is Legendre symbol. From property of Legendre symbol we have:
$$\left(\frac{3}{1 000 000 007}\right) = (-1)^{\lfloor\frac{1 000 000 007+1}{6}\rfloor} = (-1)^{166666668} = 1$$
So this means that:
$$3^{500 000 003} \equiv 1 \pmod {1 000 000 007}$$
And $n=500 000 003$ is the smallest integer satisfying that congruence relation. |
H: Generating Functions and Catalan Numbers
The task is to evaluate
$$\sum_{k=0}^{n}\dfrac{1}{k+1}{2k\choose k}{2(n-k)\choose n-k} $$
This is what I've gotten so far.
$$\begin{align*}
A(x)&=\sum_{n\geq 0}\left(\sum_{k\geq 0}^{n}\dfrac{1}{k+1}{2k\choose k}{2(n-k)\choose n-k}\right)x^{n}\\
&=\sum_{n\geq 0}C_{n}x^{n}\sum_{n\geq 0}\binom{2n}{n}x^{n}\\
&=\dfrac{1-\sqrt{1-4x}}{2x}\cdot\dfrac{1}{\sqrt{1-4x}}\\
&=\dfrac{1}{2x\sqrt{1-4x}}-\dfrac{1}{2x}
\end{align*}$$
So we have
$$ A(x)=\dfrac{1}{2}\sum_{n\geq 1}\binom{2n}{n}x^{n-1}$$
So the answer would be $\dfrac{1}{2}\binom{2(n+1)}{n+1}$ but I know that is wrong because I got $\binom{2n-1}{n}$ when using a combinatorial arguement, which was verified.
AI: $\dfrac{1}{2}\binom{2n+2}{n+1}=\binom{2n+1}{n}$ I just counted the lattice paths to $(n+1,n+1)$. Don't know why I didn't think of that earlier. |
H: Simple Polynomial Interpolation Problem
Simple polynomial interpolation in two dimensions is not always possible. For example, suppose that the following data are to be represented by a polynomial of first degree in $x$ and $y$, $p(t)=a+bx+cy$, where $t=(x,y):$
Data: $f(1,1) = 3, f(3,2)=2, f(5,3)=6$
Show that it is not possible.
I've been trying to think of ways to use either Newton's Form or Divided Differences to prove this is not possible, but can't come up with how to work it out. Looking for any help :)
AI: Since you want a polynomial of degree $\leqslant 1$, you have the three equations
$$\begin{align}
a + b + c &= 3\tag{1}\\
a + 3b + 2c &= 2\tag{2}\\
a + 5b + 3c &= 6\tag{3}
\end{align}$$
Subtracting $(1)$ from $(2)$ yields $2b + c = -1$, and subtracting $(2)$ from $(3)$ yields $2b + c = 4$. These are incompatible. |
H: Why not keep $\epsilon$ in Proof : Any Convergent Sequence is Bounded
I am trying to understand the proof of the proposition:
Any convergent sequence is bounded.
In my textbook, the author uses the definition of convergence for a sequence $\{a_n\}\to l$ and fixes $\epsilon=1$ so that there is a natural number $N$ such that \begin{align*}n>N&\implies|a_n-l|<1\\&\implies |a_n|<1+|l|.\end{align*} Then $\{a_n\}$ is bounded by $\pm U$ where $U=\max\{|a_1|,|a_2|,|a_3|,\dots,|a_{N-1}|,|a_N|,1+|l|\}$.
What I don't understand about this proof is why do we have to fix $\epsilon=1$? Wouldn't it be enough if we had simply fixed some $\epsilon>0$ and then claimed that \begin{align*}n>N &\implies |a_n-l|<\epsilon\\&\implies |a_n|<\epsilon+|l|.\end{align*}In this case $\{a_n\}$ would be bounded by $\pm M$ where $M=\max\{|a_1|,|a_2|,|a_3|,\dots,|a_{N-1}|,|a_N|,\epsilon+|l|\}$. So, why does the author decide to fix $\epsilon$ to be 1 when it would be enough fix some $\epsilon$. For example, what if I had fixed $\epsilon=\pi$, would the proof be incorrect?
AI: Yes, you could choose an other positive value of $\epsilon$ and the proof is identical. The only reason for choosing $\epsilon = 1$ is that it's a slightly more convenient choice. |
H: expected value of $f(U)$ where $U$ has a uniform distribution on $[0,1]$ and $f$ is a measurable function on $L^1[0,1]$
Let $ (\Omega, F , P)$ a probability space. And let $(\mathbb R,B(\mathbb R))$ the real numbers and the borel sets. Let's consider $ U: (\Omega, F , P)\to (\mathbb R,B(\mathbb R))$ be a random variable that has uniformly distribution on $[0,1]$, let's consider a mesurable function $f:[0,1] \to \mathbb R$, such that $\int_0^1 |f(x)|dx <\infty$. I want to compute the expected value of the new random variable $f(U): (\Omega, F , P)\to (\mathbb R,B(\mathbb R))$ using the change of variable formula
$Ef(U)=\int_{\Omega}f(U(\omega))dP=\int_{\mathbb R}f(x)d\mu$
Where $\mu$ the the measure defined by $\mu(A)=P(U\in A)$
I want to compute that expected value but I don't know how to proceed. I think that the first step is to separate $\mathbb R = (-\infty,0)\cup[0,\infty)$ and since $\mu ((-\infty,0))=0$ then
$\int_{\mathbb R}f(x)d\mu=\int_{[0,\infty)}f(x)d\mu$
The second possible step it's to separate $[0,\infty)=[0,1]\cup (1,\infty)$ and use the fact that on $[0,1]$ $\mu$ is the lebesgue measure. But I'm very confused to work on $(1,\infty)$ Please help me!
AI: I think you are working a little too hard on this. The statement $U$ is uniformly distributed on $[0,1]$ means $\mu = 1_{[0,1]}d\lambda$ where $\lambda$ is the Lebesgue measure. Then we have $$Ef(U) = \int_\Omega f(U)dP = \int_{\mathbb{R}} f(u) d\mu(u) = \int_\mathbb{R} f(u) 1_{[0,1]}(u)du = \int_0^1 f(u)du.$$
The key point here is justifying the change of measure
$$
\int_\Omega f(X(\omega))dP(\omega) = \int_{X(\Omega)}f(x)dPX^{-1}(x)
$$
which is proved through simple functions and holds for $f\geq0$ measurable or $f\in L^1$.
Also I should point out that it is important for the $L^1$ case, $f \in L^1(PX^{-1})$, in general $\int_0^1|f(x)|dx < \infty$ is not enough, it is in this case since $\mu = 1_{[0,1]}d\lambda$ so the previous statement exactly says $f \in L^1(\mu)$. |
H: 4 Element abelian subgroup of S5.
I have a homework question from my intro to group theory class.
Question:
Find a 4 element abelian subgroup of $S_5$. Write it's table.
This is where I've gotten so far, but I don't even know if I'm on the right mental track.
We know $S_5 = \{1,2,3,4,5\}$.
Abelian means the subgroup of $S_5$ is commutative.
I am having trouble figuring out how to start this. Do I pick any 4 elements of $S_5$? Or are there particular elements that must be chosen?
Do I perform permutations on the group first?
Or do I chose my 4 elements, and then perform the permutations?
Thank you!
AI: $S_5 \neq \{1, \,2,\, 3,\, 4, \,5\}$, but rather, $S_5$ is the group of permutations of the 5 elements in $\{1,\, 2,\, 3,\, 4,\, 5\}.$
Hint: Pick an abelian subgroup of order $4$ (hence not just any four elements, but four elements in $S_4$ that together meet the criteria of a commutative subgroup of $S_5$.)
Such a group can easily be found by choosing a permutation, call it $\sigma$, of order $4$, and then determining the cyclic subgroup $H$ of $S_5$ that is generated by $\sigma$: $\;\; H =\{id,\, \sigma,\, \sigma^2, \,\sigma 3\}$. And so we have a cyclic, and therefore abelian, subgroup $H$ of $S_5$.
Additional hint: Let $\sigma$ be a $4$-cycle in $S_5$ (A $4$-cycle is cycle of length $4)$.
Take, for example, the $4$-cycle $\sigma = (1\,2\,3\,4)(5) = (1\,2\,3\,4)\in S_5.\,$ Then $\,\sigma^2 = (1\,3)(2\,4) \in S_5,\,$ and $\,\sigma^3 = (1\,4\,3\,2)\in S_5.\,$ Finally, $\sigma^4 = id$. Hence we have that $\langle \sigma\rangle = \{\sigma, \sigma^2\, \sigma^3, id\}$ is a cyclic (hence commutative) subgroup of order $4$ in $S_5$. |
H: An open set that contains $\mathbb{R}\setminus\mathbb{Q}$
If $A\subseteq\mathbb{R}$ is an open set such that $\mathbb{R}\setminus\mathbb{Q}\subseteq A$, then $A=\mathbb{R}$?
AI: Let $X\subseteq\Bbb Q$ a closed subset of $\Bbb R$, for example a singleton or $\Bbb Z$ or even far more complicated sets.
Then $A=\Bbb R\setminus X$ is an open set, and it contains $\Bbb{R\setminus Q}$ as wanted.
For example one can pick for every $k\in\Bbb Z$ a rational sequence $k_n$ such that $k_n\to k$ and $k_n\in(k-1,k)$. Then the set $X=\{k_n\mid k\in\Bbb Z,n\in\Bbb N\}\cup\Bbb Z$ is closed, but very very complicated. And one can do even more. Much more. |
H: Convergence of improper integrals in $\mathbb{R}^n$
This is probably an elementary result, but every time I need it I'm always confused.
I also looked for the solution but I could not find it, so I think this will serve well as a reference for the future!
Here are the problems: consider $A_n = B(0,1) \subset \mathbb{R}^n$ (the unit ball centered at the origin) and $B^n = \mathbb{R}^n \setminus B(0,1)$.
$1)$ find all values of $p$ for which
$$\int_{A_n}|x|^p\ dx < \infty;$$
$2)$ find all values of $q$ for which
$$\int_{B_n}|x|^q\ dx < \infty.$$
I think (hope) that looking at how you characterize such values of $p$ and $q$ will help me to remember them!
Thank you!
AI: Try the $n = 1$ case first. That is very well-known (it is done wherever improper integrals are treated). Then you want to reduce the higher-dimensional case to the $n = 1$ case. There are several ways to do this -- a slick way would be to do a generalized version of polar coordinates where you integrate a constant function over all the spheres of radius $r$. But if you just set up the integral and write it as an iterated integral (by Fubini's Theorem), then it seems to me that nothing particularly clever is required to evaluate it: you can just use the one-variable Fundamental Theorem of Calculus $n$ times.
Have you tried to evaluate these integrals? How far did you get? |
H: Prove a formula in terms of n:
$1+5+9+...+(4n+1)$
I HAVE to use induction, but I am new to induction, so I am a bit confused...
I believe I have to use the base case first: so $n=1$ is $4(1)+1=5$, but i get the second term in the sequence instead of the first term? is this ok? Next I tried to substitute k+1 in the formula and got: $4(k+1)+1=4k+5$. Is my approach feasible or what do I do next?
Thanks!
Edit: I realise the formula would be $n+4$... but how would I go about proving it and explaining it using induction?
AI: You may know the formula $1+2+\cdots+n=\frac{n(n+1)}{2}$. By this formula we get
$$
\begin{aligned}
1+(4+1)+(4\cdot 2+1)+\cdots + (4n+1)
&=4(1+2+\cdots +n )+(n+1)
\\&=2n(n+1)+(n+1)
\\&=(2n+1)(n+1).
\end{aligned}
$$
So desired formula is $(2n+1)(n+1)$. If you want to prove by induction, First you check $1+5=(2+1)(1+1)$ and you check $1+5+9+\cdots+(4n+1)=(n+1)(2n+1)$ implies $1+5+9+\cdots+(4n+1)+(4n+5)=(n+2)(2n+3)$. |
H: limsups and liminfs
Suppose we have a sequence $(x_n)$ in $\mathbb{R}$ such that $x_n \leq X $ for all $n$, then we have $$ \limsup x_n \leq X $$
and if $X \leq x_n$ for all $n \implies X \leq \liminf x_n $.
My try. Since $x_n \leq X$, then $X$ is an upper bound for the set $\{ x_n \}$. In particular $ \sup \{ x_n \} \leq X $ if we then pass to the limits we have $ \limsup x_n \leq X $ as required. Similarly for liminf .
Is this correct? thanks for your comments.
AI: Your try looks correct to me.
I apologize for using the "t" word, but this seems like a trivial consequence of the definitions of $\lim\sup$ and $\lim\inf$. Under your assumptions,
$$\lim\sup_{n\to\infty} x_n \equiv \lim_{n\to\infty}\left(\sup\{x_m \mid m \geq n\}\right) \leq X $$
because $x_m \leq X$ for all $m$. Maybe I'm missing something. The only thing that might not be obvious is the existence of $\lim\sup_{n\to\infty} x_n$. Clearly, $\lim\sup_{n\to\infty} x_n$ is the limit of a nonincreasing sequence so it exists (it is a real number or $-\infty$).
You use a similar argument for the $\lim\inf$. |
H: Fermat's Little Theorem and congruences.
I am taking number theory and have hit a roadblock taking the next logical step in one of my proofs. I am told that $n=195=3 \cdot 5 \cdot 13$. I am asked to show that $a^{n-2} \equiv_n a, \; \forall a \in \mathbb{Z}.$
The previous part of the problem was relatively straight-forward, but this one has me stuck. So far, I have established the trivial result by considering
$$
n | a \Rightarrow n|a^{n-2}
$$
thus,
$$
a^{n-2} \equiv_n 0 \equiv_n a.
$$
At which point, I'd be finished. To show that this holds for all results, it seems I need to consider the congruences mod $3, 5, \text{and } 13$ and show that $a^{n-3}=a^{192}$ is congruent to $1$ for each prime factor. The problem is that I can't assume that $p \nmid a$ for $p \in \{3,5,13\}$ to be able to invoke Fermat's without loss of generality.
Could someone please point me in the right direction? I feel like I'm missing an easy (albeit fundamental) step here. Thanks!
AI: To prove $$a^{193}-a\equiv 0\pmod{195}$$ it suffices to prove it mod $3,5,13$ then use the CRT. Prove this modulo each prime. Either $3|a$ or $3\nmid a$; in either case you should be able to prove $a^{193}-a\equiv 0\pmod{3}$. Etc. |
H: Normal distribution
I have this question:
A normal distribution is such that 16% of it is smaller than 13, and 2.5% of it is larger than 22. What's the mean of this normal distribution?
I know I should be using the 68-95-99.7 rule, but I have no clue how to start without the standard deviation being given. Any hints are appreciated, thank you.
AI: Hint:
You are supposed to come up with two equations, relating the unknown mean and unknown standard deviation to $13$ in the first case and $22$ in the second
Then solve these as simultaneous equations in two unknowns to find the mean; you could also find the standard deviation. |
H: Let $a,b$ nonnegative, $a\ge cb$ for every $c$ in $(0,1)$, Should $a \ge b$ be true?
In the proof of monotone convergent theorem, The above technique is used. I really appreciate if anybody can explain it to me.
AI: Suppose that $a<b$. Then choose $c=\frac{b+a}{2b}$ and clearly we have $c<1$. Moreover:
$$
cb=\frac{b+a}2> \frac{a+a}2=a.
$$
which is not possible because for all $c\in(0,1)$ we have $cb\leq a$. Therefore $a\geq b$. |
H: Use the relation of Laplace Transform and its derivative to figure out $L\left\{t\right\}$，$L\left\{t^2\right\}$，$L\left\{t^n\right\}$
If $F(s) = L\left\{f(t)\right\}$, then $F'(s) = -L\left\{tf(t)\right\}$
Use this relation to determine
$(a)$ $L\left\{t\right\}$
$(b)$ $L\left\{t^2\right\}$
$(c)$ $L\left\{t^n\right\}$ for any positive integer $n$.
I was able use the definition of Laplace Transform and integration to figure out $(a)$,$(b)$ and $(c)$.
Namely
$L\left\{t\right\} = \dfrac{1}{s^2}$
$L\left\{t^2\right\}=\dfrac{2}{s^3}$
$L\left\{t^n\right\} = \dfrac{n!}{s^{n+1}}$
But how I use the relation between Laplace Transform and its derivative to figure out $(a)$,$(b)$ and $(c)$?
AI: Induction. You know how to write down $F'(s)$. Now, $\mathscr L\{t\cdot t^n\}=\cdots$?
For example, $\mathscr L\{t^3\}=\mathscr L\{t\cdot t^2\}=\mathscr L\{tf(t)\} $ with $f(t)=t^2$. If you knew that $\mathscr L\{t^2\}=\dfrac{2!}{s^3}$ then $\mathscr L\{t^3\}=-F'(s)=-\dfrac{d}{ds}\left(\dfrac{2!}{s^3}\right)=\dfrac{3!}{s^4}$. |
H: easy calculus thing that i forgot
Suppose $(s_n)$ is a non- negative sequence in $R$. Suppose $m > n $, then is that true that we must have $ (s_n) \subseteq (s_m ) $ and hence we can conclude $ \inf s_n \geq \inf s_m $. but we know that $h_n = \inf_{ k \geq n } s_k $ is a non-decreasing sequence. What am i doing wrong?
AI: If $m>n$ then the other inclusion holds. That is, $$\{s_n,s_{n+1},\ldots,s_m,s_{m+1},\ldots\}\supset \{s_m,s_{m+1},\ldots\}$$ |
H: How many 9-digit numbers are there with twice as many different odd digits involved as different even digits.
How many 9-digit numbers are there with twice as many different odd digits
involved as different even digits (e.g., 945222123 with 9, 3, 5, 1 odd and 2, 4
even).
(5c1)(5c2)3^9 +(5c2)(5c4)6^9+(5c3)(5c6)9!
even:0 2 4 6 8
odd:1 3 5 7 9
pick out the even one first
then pick out odd , then order them
is that correct?
i think im missing the situation that the number cannot start with 0.
AI: You’re right in thinking that the number of distinct odd digits cannot be more than $3$. You’re also right in thinking that there are $\binom51\binom52$ ways to choose one odd digit and two even digits, $\binom52\binom54$ ways to choose two odd and four even digits, and $\binom53\binom56$ ways to choose three odd and six even digits, but you should notice that $\binom56=0$: you can’t pick $6$ different even digits when only $5$ are available. Thus, there are really only two cases to consider: one odd digit and two even digits, and two odd and four even digits. The real work, though comes in counting the ways to arrange the digits once you’ve chosen them.
Suppose that you’ve chosen an odd digit, $d$, and two even digits, $e_0$ and $e_1$. If you use $d$ only once, you can put it in any of $9$ positions in the number. Then you can distribute $e_0$ and $e_1$ in the other $8$ positions however you please, except that you cannot use just $e_0$ or just $e_1$. There are $2^8$ ways to distribute $e_0$ and $e_1$ among the $8$ open positions, but two of those use only one of the digits, so there are $2^8-2$ ways that use both digits. Thus, we get a total of $9(2^8-2)$ nine-digit numbers in this subcase. If you use $d$ twice, there are $\binom92$ pairs of positions in which you can place the two $d$’s, and by the same reasoning as before there are $2^7-2$ ways to distribute $e_0$’s and $e_1$’s amongst the remaining $7$ places, so there are $\binom92(2^7-2)$ nine-digit numbers in this subcase. Continuing in this fashion, it’s a little tedious but not hard to get the number of nine-digit numbers in this case.
If instead you’ve chosen odd digits $d_0$ and $d_1$ and even digits $e_0,e_1,e_2$, and $e_3$, you can reason similarly, though there’s a little more work involved. Suppose, for instance, that you’ve decided to put the odd digits into $4$ spots, leaving the other $5$ for the even digits. There are $\binom94$ ways to choose the $4$ positions for the odd digits. There are $2^4-2$ ways to distribute $d_0$ and $d_1$ amongst these $4$ positions, excluding the two ways that use only one of the two digits. It remains to count the ways to distribute the even digits amongst the other $5$ positions. Clearly one of them must appear twice and the other three once each. There are $4$ ways to pick the one that appears twice, and there are $\binom52$ ways to pick the slots for it. There are then $3!$ ways to permute the remaining $3$ even digits amongst the remaining three slots, for a grand total of $\binom94(2^4-2)\cdot4\binom52\cdot3!$ numbers. There are a few more subcases, depending on how many places you fill with odd digits, but you can use the same basic ideas for all of them. |
H: Does "Big Data" Have a Ramsey Theory Problem?
I'm erring on the side of conservatism asking here rather than MO, as it is possible this is a complex question.
"Big Data" is the Silicon Valley term for the issues surrounding the huge amounts of data being produced by the global IT structure. Advanced mathematics is starting to pay attention to this, with very early thoughts on topological approaches. For example, see the Wiki here.
But one obvious way to think about patterns in Big Data is as polychromatic colored complete graphs: let the vertices be your data, let the edges represent relations between data, and let the colors be specific relations (which are the objective of Big Data visualization), with some neutral color representing no relation.
By the very definition of Big Data and Ramsey Theory, this virtually guarantees the existence of monochromatic complete subgraphs which may be nothing but spurious relations that must exist because of Ramsey Theory.
I am NOT a graph theorist in any way. So what I am asking specifically is this:
Are there other techniques that can be overlayed on to a graph theoretical approach that add information that the monochromatic subgraphs are real "signal" and not Ramsey noise? Or, am I misunderstanding, and the Ramsey structures are not actually noise?
AI: As requested, I'll post the comment above as an answer:
The OP is right that there are inevitably patterns in large data sets, and in fact often ones of the sort that we want to find. Here are a couple of very common examples.
In statistics, traditionally you do "hypothesis testing" where you try to find evidence for or against the hypothesis that a parameter has a particular value, for example that the mean height of American males is 5'11''. You do this by measuring the mean height of a sample and then seeing if it is "significantly" different from 5'11''. The problem is, if your sample is big enough, it is always "significantly different", because significance, whatever that is, increases with the size of the sample.
Another example is finance, where people called technical analysts look for support and resistance patterns, which is where a price keeps declining after reaching a certain value (say \$20), and then bouncing back up again. This is evidence that people are selling when the price reaches \$20 and taking their profits. However, such patterns also very commonly appear in random walks, so it is often not clear whether they represnt anything real about the financial situation.
It is said that "If you torture the data for long enough, it will confess." Some people use the term Data Mining in a perjorative sense to refer to finding patterns which aren't really there, in a sense that all sufficiently large data sets should contain such patterns just by chance.
One of our main defences against this problem is to split the data randomly into subsets and look for patterns in one subset, then see whether they generalize to the other subsets. In machine learning, people say "training" instead of "fitting a statistical model" or "looking for patterns" and then talk about "validation". The idea is that the validation should show you how your model is likely to perform on unseen data. If your model learns a spurious pattern, then you hope that this will show up as a poor performance on the validation data. Such a model is said to be over-fitted.
Splitting up the whole data set into $k$ subsets and validating a model fitted to one of the subsets on the rest of the data, and doing this $k$ times, once for each subset, is called $k$--fold cross-validation and is a common way of estimating how your model will perform on new data. This is why the Stackexchange statistics site is called Cross Validated.
The motivation for this procedure is that we really want to see how our patterns generlize to unseen data. If they are real patterns, they should show up in unseen data as well. But if we don't have any unseen data yet, we just pretend that some of the data we have got is unseen, and use it to validate the model. And it makes perfect sense, because usually our reason for wanting to fit models/find patterns/learn is to make predictions about new data that we haven't seen yet. |
H: A statistical approach to the prisoners problem
Two days ago, I found this problem on reddit (I didn't have access to reddit when I did the math, so I did it with 24 instead of 23, and I decided the warden picked someone every day, not "whenever he feels like it"):
A prison warden tells 24 prisoners he has a "game" for them. Once per day, the warden chooses one prisoner at random and leads them to the Switch Room. The aptly named Switch Room has two switches, both at the Off position at first, that are connected to nothing. The called prisoner has to toggle exactly one switch.
At any time, any prisoner can go to the warden and tell him that the 24 prisoners all went to the Switch Room at least once. If that's true, they're all freed. If not, they're going to be made into sausages for the other inmates or something.
The prisoners can come up with a plan now but won't ever be able to communicate again until someone tells the warden.
EDIT After some debating with @alex.jordan, I wanted to clarify my intent. My perspective is that this is a metaphor for "how would you tell that N distinct events, uniformly and randomly happening, all happened at least once if you only had 2 bits to spare", and has nothing to do with an actual warden (who could be biased in his random choices or use a different distribution model to screw up the prisoners), or actual prisoners.
I am solely interested in an answer that assumes a prisoner is selected randomly and uniformly once every day (not "from time to time" as stated in the reddit riddle), so you can safely ignore any assumption that the warden is out there to grind the prisoners to sausages or selects people with a bias or just won't ever select anyone.
The "classical" solution is that the prisoners designate one leader. When a prisoner enters the Switch Room (except the leader), if they've never been there and find the first switch in the Off position, they turn it on. Otherwise, they toggle the other switch. When the leader enters the Switch Room, if he sees the first switch in the On position, he knows that someone who's never been there before has been, so he counts one and turns off the switch. When he has counted to 23, he knows that everyone has been there at least one.
The thing is, this solution sucks. Assuming 24 prisoners again, and knowing they're picked at random, we can represent the whole thing as a series of geometric distributions (this also assumes I'm doing it right, which I may not be):
$$
X: \text{number of days before the Leader goes to the Switch Room}\\
X \sim Geom\left(\frac{1}{24}\right)\\
E(X) = 24, Var(X) = 552\\
$$
$$
Y_n: \text{number of days before one of the n remaining prisoner}\\
\text{goes to the Switch Room for the first time (n from 1 through 23)}\\
Y_n \sim Geom\left(\frac{n}{24}\right)\\
E(Y_n) = \frac{24}{n}, Var(Y_n) = \frac{24 (24-n)}{n^2}
$$
Since expected values and variance can be linearly added, we can expect that the Leader will tell the warden after on average 642 days, with a standard deviation of 116, by assuming the leader will go $E(X)$ days after any given prisoner went for the first time as modeled by $E(Y_n)$:
$$
Z: \text{Number of days before the Leader announces everyone's been to the Switch Room}\\
E(Z) = \sum_{n=1}^{23}E(X) + E(Y_n) = 552 + \sum_{n=1}^{23}E(Y_n) \approx 552 + 89.6229 \approx 641.6229\\
Var(Z) = \sum_{n=1}^{23}Var(X) + Var(Y_n) = 12696 + \sum_{n=1}^{23}Var(Y_n) \approx 12696 + 833.3521 \approx 13529.3521\\
\sigma = \sqrt{13529.3521} = 116.3157
$$
Very simple maths tell us that after 642 days, every prisoner's been on average 26 times to the Switch Room. This looks like a horrible waste of time.
I'm pretty sure it's possible to calculate how many days you would need to wait before you have 99% chances (or higher) that each prisoner has been there. Problem is, I'm only halfway through my college stats class, and we've only seen easy distributions where successes are independent, so I'm not too sure how to tackle that.
How would you calculate the chances that each prisoner has been to the Switch Room after $Z$ days?
EDIT I just made a quick and dirty program to run "simulations", and it takes on average 90.6 days until every prisoner has visited the Switch Room, with a standard deviation of 28.5. Making that into a normal distribution, it should be around 157 days before we can say with 99% certainty that each prisoner has visited the Switch Room at least once, and 179 days for 99.9% certainty. Needless to say, you're pretty safe after 641 days...
It's an empirical technique and it doesn't really feel mathematically satisfying, so the question is still open for better answers.
AI: It is even worse than that. On day $1$ some prisoner turns the switch on (unless the leader goes first). Then all the prisoners that visit next can't turn it on, so those visits are wasted.
The page here has a good writeup. Your solution is Solution 6: Simple count and he claims with 100 prisoners you need 28 years. There are some strategies, and I found a very nice paper that I can't find again.
For your question, we can pretend that each prisoner is independent of each other (which is a much better approximation than it sounds). After $d$ days the prisoner has $(\frac {23}{24})^d$ chance of not having visited the room. To have a $99\%$ chance that they all have visited, we need $$(1-(\frac {23}{24})^d)^{24}=0.99
\\1-(\frac {23}{24})^d=0.99^{\frac 1{24}}\\(\frac {23}{24})^d=1-0.99^{\frac 1{24}}\\d=\frac {\log (1-0.99^{\frac 1{24}})}{\log \frac {23}{24}}\approx 182.76$$ You are right that this is much faster |
H: Proving any product of four consecutive integers is one less than a perfect square
Prove or disprove that : Any product of four consecutive integers is one less than a
perfect square.
OK so I start with $n(n+1)(n+2)(n+3)$ which can be rewritten $n(n+3)(n+1)(n+2)$
After multiplying we get $(n^2 + 3n)(n^2 + 3n + 2)$
How do I proceed from here to end up with something squared $- 1$?
AI: You might note that for any $N$ at all, $$(N-1)(N+1) = N^2-1$$ and so is one less than a perfect square. I presume you have seen this before.
Then you could take $N = M+1$ in the formula above, and get $$M\cdot(M+2) = (M+1)^2 - 1$$ is one less than a perfect square.
And here you have $M = n^2+3n$, don't you? |
H: Generating Function for the adjusted Fibonacci numbers
The task is to find another relation for the adjusted Fibonacci numbers. I've found there genertaing function
$$A(x)=\dfrac{1}{1-x-x^{2}}$$
Furthermore I've created the generating function in a different way and now want to grab the correct coefficient.
Since $\hat{F}_{n}$ is the number of ways to add one and two to get $n$,
$$A(x)=\sum_{n\geq 0}(x+x^{2})^{n}$$
Now I'm having trouble grabbing the $n^{th}$ coefficient.
It should be $\sum_{k=0}^{n}\binom{k}{n-k}$.
All I have so far is
$$A(x)=\sum_{n\geq 0}\sum_{k\geq 0}^{n} \binom{n}{k}x^{n-k}x^{2k}$$
Then I arrive at
$$A(x)=\sum_{n\geq 0}(1+x)x^{n}$$
Which does me no good. A hint would be very nice. Thank you.
AI: Hint: We factor $1-x-x^2=(1-\phi x)(1-\psi x)$ for appropriate real numbers $\phi,\psi$. Then solve the partial fractions problem $$\frac{1}{1-x-x^2}=\frac{A}{1-\phi x} + \frac{B}{1-\psi x}$$
Then you have the sum of two geometric series. |
H: Find the value of $\lim_{n\rightarrow\infty}\frac{a_n}{1\cdot 2}+\frac{a_{n-1}}{2\cdot 3}+\ldots +\frac{a_1}{n(n+1)}$
$\displaystyle \lim_{n\rightarrow\infty}\frac{a_n}{1\cdot 2}+\frac{a_{n-1}}{2\cdot 3}+\ldots +\frac{a_1}{n(n+1)}$ if $\lim_{n\rightarrow\infty}a_n=a$
Note that $\displaystyle\sum_{k=1}^n\frac{a_{n-k+1}}{n^2}\le S_n\le \sum_{k=1}^n\frac{a_{n-k+1}}{k^2}$
Can I bound $S_n$ by $a$ instead of $a_{n-k+1}$ in the numerator instead?
AI: We will show that the limit is $a$.
Let $\varepsilon>0$. First, $(a_n)$ is bounded, so $|a_n|<M$ for some $M>0$, for all $n\in\mathbb N$. There exists $n_0\in\mathbb N$ such that, for all $n\geq n_0$, $|a_n-a|<\frac{\varepsilon}{6}$. Also, there exists $n_1\in\mathbb N$ with $n_1>n_0$ such that, if $n\geq n_1$, then $\sum_{k=n_1}^{\infty}\frac{1}{k(k+1)}<\frac{\varepsilon}{6M}$. Finally, there exists $n_2\in\mathbb N$ such that, if $n\geq n_2$, we have that $\frac{1}{n+1}<\frac{\varepsilon}{3M}$. Then, if $n\geq n_0,2n_1,n_2$,$$\left|\frac{a_n}{1\cdot2}+\frac{a_{n-1}}{2\cdot3}+\dots+\frac{a_1}{n(n+1)}-a\right|=$$$$\left|\frac{a_n-a}{1\cdot2}+\frac{a_{n-1}-a}{2\cdot3}+\dots+\frac{a_1-a}{n(n+1)}-a+\frac{a}{1\cdot2}+\frac{a}{2\cdot3}+\dots+\frac{a}{n(n+1)}\right|\leq$$$$\left|\frac{a_n-a}{1\cdot2}+\frac{a_{n-1}-a}{2\cdot3}+\dots+\frac{a_1-a}{n(n+1)}\right|+\left|\frac{a}{1\cdot2}+\frac{a}{2\cdot3}+\dots+\frac{a}{n(n+1)}-a\right|.$$ The term in the second absolute value is equal to $$\left|a-\frac{a}{2}+\frac{a}{2}-\frac{a}{3}+\dots+\frac{a}{n}-\frac{a}{n+1}-a\right|=\left|-\frac{a}{n+1}\right|\leq\frac{M}{n+1}<\frac{\varepsilon}{3},$$ since $n\geq n_2$. For the first term, you have that $$\left|\frac{a_n-a}{1\cdot2}+\frac{a_{n-1}-a}{2\cdot3}+\dots+\frac{a_1-a}{n(n+1)}\right|\leq$$$$\left|\frac{a_n-a}{1\cdot2}+\frac{a_{n-1}-a}{2\cdot3}+\dots+\frac{a_{n_1}-a}{(n-n_1+1)(n-n_1+2)}\right|+$$$$\left|\frac{a_{n_1-1}-a}{(n-n_1+2)(n-n_1+3)}+\dots+\frac{a_1-a}{n(n+1)}\right|.$$ Since $n_1>n_0$, the first term here is bounded by $$\left|\frac{\varepsilon/6}{1\cdot2}+\dots+\frac{\varepsilon/6}{(n-n_1+1)(n-n_1+2)}\right|<\frac{2\varepsilon}{6}=\frac{\varepsilon}{3},$$ and the second term is bounded by $$2M\sum_{k=n-n_1+2}^\infty\frac{1}{k(k+1)}<\frac{2M\varepsilon}{6M}=\frac{\varepsilon}{3},$$ since $n-n_1+2\geq n_1$. After adding those inequalities, we get that this limit is equal to $a$. |
H: Integration of $(5^{3/x}-\sqrt[3]{x^8}+9)/(3x^2)$
Problem: evaluate $$\int\frac{5^{3/x}-\sqrt[3]{x^8}+9}{3x^2}\mathrm dx$$
Please provide me a hint. I tried to separate to 3 parts but don't know how to integrate $\frac{5^{3/x}}{3x^2}$.
AI: Hint: Let $u = \frac{3}{x}$, and then $du = \frac{-3}{x^2} dx$. Then
$$\int \frac{5^{3/x}}{x^2} dx = \int 5^u \frac{du}{-3}$$ |
H: Show that $\lim_{n \to \infty} \frac{k^n}{n!} =0$ for all $k$ in $\mathbb{R}$
$$\lim_{n\to\infty} \frac{k^n}{n!}=0 \, \forall\:k\in \mathbb{R}$$
I have tried to find an $N$ in term of epsilon in the definition of limit, but to no avail. I've tried log, but $\log (n!)=\log(1)+\log(2)+...+\log(n)$ does not seem to help here.
Could you help me with this problem?
Thanks.
AI: Consider the ratio
$$\frac{\frac{k^{n + 1}}{(n + 1)!}}{\frac{k^n}{n!}} = \frac{n!}{(n + 1)!} \frac{k^{n + 1}}{k^n} = \frac{k}{n + 1} \to 0$$
as $n \to \infty$. Hence by the ratio test, the sum
$$\sum_{n = 0}^{\infty} \frac{k^n}{n!}$$ converges for each $k \in \mathbb{R}$. In particular, the terms of the sequence being summed must tend to $0$. |
H: Discrete Math Combinatorics Homework Help
Find the value of
${n\choose0} + 3{n\choose1} + 9{n\choose2} + {27}{n\choose3} + \dots + 3^n{n\choose n}$
I know that ${n\choose0} = 1$, ${n\choose1} = n$ so $3{n\choose1} = 3n$, and ${n\choose n}=1$ so $3^n{n\choose n}=3^n$. But I don't realize the pattern going on here.
2.Color all the odd numbers in Pascals triangle red and all the even numbers blue. What pattern do you get? Describe it as precisely as you can.
For this one I realize you get this picture going on:
So you get like upside down decreasing in size triangles for the even numbers inside of the bigger triangles marked by the odd numbers. But I'm not sure exactly what I should be saying about the picture. Am I missing a crucial discovery here?
3.Let $k,l,m,n \in Z \geq 0$ be such that $n=k+l+m$. The trinomial coefficient ${n \choose k,l,m}$ is given by the rules:
for $k+l=n$, ${n\choose k,l,0} = {n \choose k,0,l} = {n \choose 0,k,l} = {n\choose k}$
${n\choose k,l,m} = {n-1 \choose k-1,l,m} + {n-1 \choose k,l-1,m} + {n\choose k,l,m-1}$
The following questions use this definition.
(a) What are all the trinomial coefficients for $n=1,2,3$?
(b) Describe the "triangle" of trinomial coefficients (Hint: Think three dimensional Pascal's triangle).
This one I just don't really understand because of "trinomial" talk. I can visualize a Pascal's "pyramid." One where the number below is a sum of the 3 above it. Something like that. But other than that, I'm not really sure what's going on.
AI: For the first, recognize the binomial theorem:
\begin{align*}
&{n \choose 0} + 3 {n \choose 1} + 9 {n \choose 2} + ... + 3^n {n \choose n} \\&= {n \choose 0} 1^n 3^0 + {n \choose 1} 1^{n - 1} 3^1 + {n \choose 2} 1^{n - 2} 3^2 + ... + {n \choose n} 1^0 3^n \\
&= \sum_{k = 0}^n {n \choose k} 1^{n - k} 3^k \\
&= (3 + 1)^n
\end{align*} |
H: Differential Equation $y(x)'=(y(x)+x)/(y(x)-x)$
can someone give me some tips on how to solve this differential equation.
I looked at the Wolfram solution which substituted $y(x)=xv(x)$. I'd know how to solve from there, but I have know idea why they did it in the first place, well why the algorithm did it in the first place. When do you substitue $y(x)=xv(x)$?
What are other ways on solving it?
AI: Approach 1
$$y = v x \rightarrow y' = v + x v'$$
Substitute into original ODE. Looks like you know how to do this.
Why does this work?
To solve the equation $\tag 1 \dfrac{dy}{dx} = \dfrac{f_1(x, y)}{f_2(x,y)},$
where $f_1(x,y)$ and $f_2(x,y)$ are homogeneous functions of the same degree in $x$ and $y$, we use the following approach. Let
$$f_1(x,y) = x^n \phi_1\left(\dfrac{y}{x}\right), ~f_2(x,y) = x^n \phi_2\left(\dfrac{y}{x}\right).$$
From $(1)$, we have
$$\dfrac{dy}{dx} = \dfrac{f_1(x, y)}{f_2(x,y)} = \dfrac{x^n \phi_1\left(\dfrac{y}{x}\right)}{x^n \phi_2\left(\dfrac{y}{x}\right)} = \dfrac{\phi_1\left(\dfrac{y}{x}\right)}{\phi_2\left(\dfrac{y}{x}\right)}$$
We can now write this as
$$\dfrac{dy}{dx} = f\left(\dfrac{y}{x}\right)$$
Now, if we say $v = \dfrac{y}{x} \rightarrow y = vx$ and we substitute, we get a separable equation of the form:
$$\dfrac{dv}{f(v) - v} = \dfrac{dx}{x}.$$
Approach 2
Let $x + y = z$.
We have $x + y = z \rightarrow 1 + \dfrac{dy}{dx} = \dfrac{dz}{dx} \rightarrow \dfrac{dy}{dx} = \dfrac{dz}{dx} - 1$.
For the numerator, we substitute $z$ and for the denominator, we substitute $y-x = z-2x$. Do you see how to get that last one?
Now, the ODE is separable.
Approach 3
Let $y-x = z$.
Follow what was done in approach 2. |
H: How can one show that $ f(0)\ln(\frac{b}{a})=\lim_{\epsilon\rightarrow 0}\int_{\epsilon a}^{\epsilon b} \frac{f(x)}{x}dx$?
Let $f:[0, 1] \rightarrow \mathbb{R}$ a continuous function. If $a>0$, show that:
$$ f(0)\ln(\frac{b}{a})=\lim_{\epsilon\rightarrow 0}\int_{\epsilon a}^{\epsilon b} \frac{f(x)}{x}dx$$
Tried using Riemann sum, but did not succeed.
AI: $$\int_{\epsilon a}^{ \epsilon b} \frac {f(x)}{x} dx = \int_{\epsilon a}^ {\epsilon b}\left[\frac {f(0)}{x} +\frac {f(x) - f(0)}{x}\right] dx = f(0)[\ln (\epsilon b) - \ln (\epsilon a)] + \int_{\epsilon a}^ {\epsilon b} \frac {f(x) - f(0)}{x}dx$$
The last integral goes to 0 as $\epsilon \to 0$ because $a > 0$ (and I am also assuming $b > 0$), so $x$ is not near $0$.
So the final answer is $f(0)[\ln (\epsilon) +\ln (b) -(\ln (\epsilon) + \ln(a))]$
and this goes to $f(0)\ln \dfrac{b}{a}$ as $\epsilon \to 0$. |
H: Compute $\int_0^1\int_0^1...\int_0^1\lfloor{x_1+x_2+...+x_n}\rfloor dx_1dx_2...dx_n$
Compute $\int_0^1\int_0^1...\int_0^1\lfloor{x_1+x_2+...+x_n}\rfloor dx_1dx_2...dx_n$ where the integrand consists of the floor (or greatest integer less than or equal) function.
The case $n=1,2,3$ all can be solved geometrically. Actually the case $n=3$ is pretty fun where you get to dissect the unit cube. I guess the cases $n\ge4$ don't have useful geometric interpretations... so how to approach them (well, you still get to divide the sum into different cases: sum between 0 and 1, between 1 and 2, ..., between $n-1$ and $n$, and multiply the floor of the sum with the measure of the corresponding region) ?
AI: Let $f$ be the integrand; let $\Omega_k$ be the region of the unit hypercube over which the integrand has value $k$: that is, the region over which the sum of the coordinates lies between $k$ and $k+1$.
Note that the unit hypercube is invariant under the transformation which sends each coordinate $x_i$ to $1-x_i$. It follows that $\Omega_k$ (the region over which $\sum x_i$ lies between $k$ and $k+1$) and $\Omega_{n-1-k}$ (the region over which it lies between $n-k-1$ and $n-k$) have equal measure.
Thus $$\int_{\Omega_k+\Omega_{n-1-k}}f = k \mu(\Omega_k)+(n-1-k)\mu(\Omega_{n-k-1})=(n-1)\mu(\Omega_k) \, ,$$
and so
$$\int_{[0,1]^n} f = \sum_{k=0}^{n-1} \int_{\Omega_k} f = \frac{1}{2}\sum_{k=0}^{n-1} \int_{\Omega_k+\Omega_{n-k-1}} f = \frac{n-1}{2} \sum_k \mu(\Omega_k) \, .$$
As the unit cube has measure $1$ and the $\Omega_k$ form a decomposition of it, it follows that the integral has value $\dfrac{n-1}{2}$.
Note: Let $\left<a\right>$ denote the fractional part of $a$. Then, by using the identity $\left<a\right>+\lfloor a \rfloor=a$, you can see that the above computation is equivalent to the following fact:
If $X_1,\dots, X_n$ are independent uniform random variables on $[0,1]$, the expected fractional part of their sum is $\frac{1}{2}$.
You could prove this directly by a similar symmetry argument (passing from the random variables $X_1,\dots,X_n$ to the random variables $1-X_1,\dots,1-X_n$), but I also think it's a more intuitively clear statement than the original one even before proof... |
H: How can I evaluating limits? $\lim_{x\rightarrow0^+}(xe^{2x}+1)^{5/x}$ and $\lim_{x\rightarrow{\pi / 2}}(1+\sec(3x))^{\cot(x)}$
Can someone please give me an idea how to do these two limits. I guess it is the L'Hopital's rule, not sure.
$$\lim_{x\rightarrow0^+}(xe^{2x}+1)^{5/x}$$
$$\lim_{x\rightarrow{\pi / 2}}(1+\sec(3x))^{\cot(x)}$$
Thank you.
AI: I'll suggest one way to do the first, and let you see about the second: Let $y = \left(xe^{2x} + 1\right)^{5/x}$. Then
$$\ln{y} = \frac{5}{x} \ln\left(xe^{2x} + 1\right)$$
By L'Hospital's rule,
$$\lim_{x \to 0^+} \frac{5 \ln\left(xe^{2x} + 1\right)}{x} = \lim_{x \to 0^+}\frac{5}{xe^{2x} + 1} \left(2xe^{x} + e^{2x}\right) = \frac{5(0 + 1)}{1} = 5$$
So $\ln{y} \to 5$, and $y \to e^{5}$. |
H: How can I get more intuition about CX?
Let $X$ be a topological space and define $CX$ as the quotient space $X \times I / X \times \{0\}$. From my understanding $X \times \{0\}$ defines some sort of equivalence relation, but I am unsure how this can be described. Is this assumption correct? If so, what is a better description of the relation?
AI: The name of this operation is the biggest source of intuition about it: it is called the "cone" over $X$.
Taking the quotient by $X \times \{0\}$ does in fact define an equivalence relation: it identifies all points in $x\times I$ with the second component (the one from $I$) equal to $0$.
Think about what happens when you apply $C$ to the circle: first you create $S^1 \times I$, a cylinder, and then you collapse one end of the cylinder (the one corresponding to $S^1\times \{0\}$) to a single point, forming a cone (topologically, a disk). In fact, this generalises to higher dimensions: $CS^n = D^n$, a fact whose proof could greatly help your intuition.
In general the operation $C$ is carried out (informally) by adjoining a single extra point to you space, and then forming lines between this point and every other point in the space. |
H: Evaluate $\oint_C\frac{dz}{z-2}$ around the circle $|z-2| = 4$
I don't completely understand how to approach these questions. I suppose the notation $\oint_C$ is something I'm not used to.
So far, I have $\oint_C\frac{dz}{z-2} = \log(z-2)$. From here, I suppose I could utilize $|z-2| = 4$ and $z-2 = re^{i\theta}$ somehow, but I'm not completely sure.
Is it the matter of that that $r = 4$, and $\theta = 2\pi$ and $\theta = 0$ (considering we're going over the entire region? Then, we have $\log(4e^{i2\pi}) - \log(4e^{i0}) = 2\pi i+\log(4) - (i0 + \log(4)) = 2\pi i $
I think I may be right. I know the answer is correct, but I don't know if my approach was necessarily correct.
AI: By definition,
$$\oint_C f(z) dz = \int_a^b f(\gamma(t)) \gamma'(t) dt$$
where $\gamma : [a, b] \to \mathbb{C}$ is a parameterization of $C$ (provided $C$ is nice, as occurs in this case). This is simply a matter of definition.
Now in order to compute the integral, finding an antiderivative won't really work as you've tried. The problem is that $\log{(z - 2)}$ cannot be made holomorphic (differentiable) away from $2$ - you'll still have to have a branch cut, so it's not even an antiderivative anymore. Note that $e^{2\pi i} = e^0$ in the complex plane, so if we have
$$\ln{e^{2\pi i}} = 2\pi i \text{ while } \ln{1} = 0$$
then $\ln$ isn't even a single-valued function anymore. This is why we'd have to take a branch cut or view the logarithm as multi-valued.
The general technique for solving these problems is to write down a parameterization and make it an integral over some real interval. A circle can be parameterized very nicely; let's take
$$\gamma : [0, 2\pi] \to\mathbb{C}$$
defined by $$\gamma(t) = 2 + 4e^{it}$$
Then
\begin{align*}
\oint_C f(z) dz &= \int_0^{2\pi} \frac{1}{(2 + 4e^{it}) - 2} 4ie^{it} dt \\
&= \int_0^{2\pi} \frac{4ie^{it}}{4e^{it}} dt \\
&= i \int_0^{2\pi} dt
= 2\pi i
\end{align*} |
H: Help with simple algebra
Can someone please explain what to do with x in the following equation, I have not done algebra for a long time so a bit lost:
$$f(x)= -5 x^5 + 69 x^2 - 47$$
AI: Given:
$$f(x)= -5 x^5 + 69 x^2 - 47$$
We have:
$f(0) = -5(0)^5 + 69 (0)^2 - 47 = -47$
$f(1) = -5(1)^5 + 69 (1)^2 - 47 = 17$
$f(2) = -5(2)^5 + 69 (2)^2 - 47 = 69$
$f(3) = -5(3)^5 + 69 (3)^2 - 47 = -641$ |
H: Partial derivatives of second order
Find all functions $f:\mathbb{R}^2\rightarrow \mathbb{R}$ of class ${\cal C}^2$, such that:
$\frac{\partial^2f}{\partial x\partial y} = 0$
$\frac{\partial^2f}{\partial x^2} = \frac{\partial^2f}{\partial y^2}$
(Separate questions)
For the first one I prove that $f(x,y) = h(x)+g(y)$ for some $h,g\in{\cal C}^2$, but I can't determine a condition for $f$ in the second part.
AI: The second equation is the one-dimensional wave equation which has solutions $f(x, y) = g(x+y) + h(x-y)$; this form is usually obtained via d'Alembert's solution which is explained in the link. |
H: Giving presents to 8 employees out of 10
Each of 10 employees brings one (distinct) present to an office part. Each present is given to a randomly selected employee by Santa (an employee can get more than one present). What is the probability that at least two employees receive no presents? How would you approach this?
AI: It is easier to find the probability that all get a present, or all but one get a present.
There are $10^{10}$ equally likely ways for the presents to be distributed. For we can line up the presents, and for each of the $10$, there are $10$ people it can go to.
There are $10!$ ways to distribute the presents so that everybody gets one.
We now count the ways to distribute so that exactly one person gets left out. The sad person can be chosen in $\binom{10}{1}$ ways. For each such choice, the lucky person who will get $2$ can be chosen in $\binom{9}{1}$ ways. The presents she gets can be chosen in $\binom{10}{2}$ ways. And the remaining $8$ presents can be distributed to the remaining $8$ people, one to each, in $8!$ ways. |
H: Determine the maximal ideals of $\mathbb{R}^2$.
Determine the maximal ideals of $\mathbb{R}^2$.
Well for any real number that is not divisible by another number other than 1 and itself generates a maximal ideal for $\mathbb{R}$. Is that right? Would it be the same as for $\mathbb{R}^2$?
AI: Not quite, since there aren't any primes in $\mathbb{R}$ - every non-zero element has a multiplicative inverse. Any non-trivial ideal in $\mathbb{R}$ is, in fact, $\mathbb{R}$. For if $r \in I$ and $r \ne 0$, then
$$\frac{1}{r} I \subseteq I \implies \frac{1}{r} r \in I \implies 1 \in I$$
And any ideal containing $1$ is the entire ring. So $0$ is actually the unique maximal ideal.
On the other hand, suppose that $(a, b) \in I$ for some ideal $I \subseteq \mathbb{R} \times \mathbb{R}$. Consider some cases:
If $a$ and $b$ are both non-zero, show that $0 \ne I = \mathbb{R} \times \mathbb{R}$ by constructing something analogous to $1/r$.
If every element of $I$ looks like $(0, b)$, conclude that $I = 0 \times \mathbb{R}$ by thinking about the element $(0, \frac{1}{b})$.
Likewise for $(a, 0)$. |
H: Conditional Joint Density Function $(Y | X) \sim \mathcal{N}(x, 1)$ and $X \sim \mathcal{N}(0,1)$ whats $f(y | x)$
Conditional Joint Density Function: $(Y | X) \sim \mathcal{N}(x, 1)$ and $X \sim \mathcal{N}(0,1)$ whats $f(y | x)$
Since $X \sim \mathcal{N(\mu, \sigma^2)}$ then $f(x) = \frac{1}{\sigma \sqrt{2 \pi}} \exp\{ -\frac{1}{2\sigma^2}(x - \mu)^2 \}$.
... I would think $(Y | X) \sim \mathcal{N}(x, 1)$ so $f(y|x) = \frac{1}{\sqrt{2 \pi}} \exp\{ -\frac{1}{2}(x - x)^2 \} = \frac{1}{\sqrt{2 \pi}} \exp\{ 0 \}$ but I believe this is not the case? In my lecture notes
$$f(x, y) = f(y|x)f(x) = \frac{1}{2\pi} \exp\{ -\frac{1}{2} ((y-x)^2 + x^2) \}$$ It seems like they did $f(y|x) = \frac{1}{\sqrt{2 \pi}} \exp\{ -\frac{1}{2}(\color{red}{y} - x)^2 \}$. So somehow $x$ became $y$ (colored red)?
AI: Notice that $f(y\,|\,x)$ is a function of $y$, so in the p.d.f. substitute the "dummy" $x$ with $y$ and $\mu$ with $x$, we get: $$f(y\,|\,x)=\frac{1}{\sqrt{2 \pi}} \exp\left(-\frac{(y - x)^2}{2}\right)$$ |
H: Hint is required for an integral
$$\int\frac{\mathrm dx}{\sqrt x \cos(1-\sqrt x)}$$
please provide a hint about the substitution. The website gives a long answer See here.
Wondering if there is a simplification.
AI: Hint:
$$u = 1 - \sqrt{x} \implies du = -\frac{1}{2 \sqrt{x}} dx$$
so the integral can be written as
$$\int \frac{1}{\cos(1 - \sqrt{x})} \frac{dx}{\sqrt{x}} = \int \frac{1}{\cos{u}} \frac{du}{-2} = -\frac{1}{2} \int \sec{u} du$$ |
H: What does $\text{poly}$ stand for in $O(\log^{10.5}n \cdot \text{poly}(\log \log n))$?
I posted this question on cstheory and found that "poly(f(n))" is shorthand for "polynomial in f(n)" or $f(n)^{O(1)}$, hence poly(log log n) is shorthand for $(log log n)^{O(1)}$. However, I don't understand the notation where the big oh is an exponent. I understand big oh notation but not when big oh itself is an exponent. Why have the word poly at all in the expression? Can anybody explain how this works or refer me to some articles?
AI: Putting the $O(1)$ in the exponent doesn't involve any special notation or hidden meaning that you haven't already seen.
As you probably already know, if something is in $O(1)$, it's bounded above by some constant $c$. By saying $f(n)^{O(1)}$, all they are saying now is that the exponent is bounded above by some constant $c$.
$poly(f(n))$ means that the complexity of whatever you are analyzing is upper bounded by some polynomial in $f(n)$. With regards to the $f(n)^{O(1)}$ bit, the constant would coincide with the degree of that polynomial. |
H: Trigonometry - 'Find all solutions to the equation'
I have 3 questions to solve for trigonometry, and I'm not sure of the process to solve them. They are:
Find all solutions to the equation:
$4 \sin 2x = 2.4$ $\forall$ $0 ≤ x ≤ 2\pi$
$2 \cos 2x = 1.3$ $\forall$ $0 ≤ x ≤ 2π$
$3 \tan 2(x + \frac{\pi}{3}) = 6$ $\forall$ $0 ≤ x ≤ 2\pi$
AI: HINT: The sine/cosine/tangent functions are periodic. This means that over a sufficiently large interval, you will have repeating y-values for different x-values. To get the intuition, you should plot/graph your functions, like $4\sin 2x$:
$2\pi$ is roughly equal to $6.3$. Look at the interval from 0 to $6.3$. Do you see how there are multiple x-values for the same y-values? For example, for $y=1$, there are three points that look like they'll satisfy the equation.
For actually obtaining the computational answers themselves, it's just two or three steps of rearranging and simplifying to find the first answer. Then use the period of the function to find out when it'll hit that point again. |
H: Probability on drawing colored balls
Here is another question from the book of V. Rohatgi and A. Saleh. I would like to ask help again. Here it goes:
An urn contains $r$ red and $g$ green marbles. A marble is drawn at random and its color noted. Then the marble drawn, together with $c > 0$ marbles of the same color, are returned to the urn. Suppose that $n$ such draws are made from the urn. Find the probability of selecting a red marble at any draw.
So here is what I have so far:
$$\text{P{a red ball in at least one of the $n$ draws}}=1-\text{P{green balls in all $n$ draws}}$$ I noted that $$\text{P{all green balls}}=\text{P{X$_1$$\;$=$\;$green}}\times{...}\times\text{P{X$_n$$\;$=$\;$green$\;$|$\;$X$_1$$\;$=$\;$${...}$$\;$=$\;$X$_{n-1}$=$\;$green}}$$ Thus, I calculated $\text{P{a red ball in at least one of the $n$ draws}}$ as $$1-\left(\frac{g}{r+g}\right)\left(\frac{g+c}{r+g+c}\right){...}\left(\frac{g+(n-1)c}{r+g+(n-1)c}\right)$$
I do not know if my way is correct or not. However, I feel like this is wrong since the text gave an answer of $$\frac{r}{r+g}$$ If that is the actual answer, can anyone help explain why? Thanks.
AI: The question is to get $$\text {The probability of selecting a red marble at any draw}$$
Not $$\text {The probability of selecting a a red marble in at least one of the $n$ draws}$$
The First time you draw a marble the Probability is:
$${\text{Red Marbles}\over\text{Total marbles}}={\frac{r}{r+g}}$$
The Second time you draw a marble the Probability is:
$${\frac{r}{r+g}}*{\frac{r+c}{r+g+c}}+{\frac{g}{r+g}}*{\frac{r}{r+g+c}}={\frac{r}{r+g}}$$
And so on...
As you can see the Probability does not depend on $\text{ n or c}$. |
H: Tallest bubble tower induction proof
A hemispherical bubble is placed on a spherical bubble of radius $1$. A smaller hemispherical bubble is then placed on the first one. This process is continued until $n$ chambers, including the sphere, are formed. (The figure shows the case $n = 4$.) Use mathematical induction to prove that the maximum height of any bubble tower with $n$ chambers is $1 + \sqrt{n}$.
I can find the relationship of the radii :$$\frac{\sqrt{2}}{2}(R_{n}) = R_{n+1}$$
But, I still do not see how to use the inductive hypothesis to make a proof.
Suppose $R$ is the radius of the $n$'th bubble and $x$ is the radius of the $(n+1)$'th bubble. And suppose $d$ is the height added when the $(n+1)$'th bubble is added to the top of the tower.
$$\sqrt{{R_n}^2 - {R_{n+1}}^2} + R_{n+1} - d = R_n$$
Let $d(x)$ take $R_{n+1}$ as the input, and outputs the added height to the tower, $d$.
$$d(x) = \sqrt{R^2 - x^2} + x - R$$
$$d'(x) = -\frac{x}{\sqrt{R^2 - x^2}} + 1 = 0$$
$$\sqrt{R^2 - x^2} = x$$
$$R^2 - x^2 = x^2$$
$$R^2 = 2x^2$$
$$\frac{\sqrt{2}}{2}R = x$$
$$\frac{\sqrt{2}}{2}R_n = R_{n+1}$$
AI: Suppose the hypothesis is true for towers with upto $n$ bubbles. Now consider a tower with $n+1$ bubbles. The bottom bubble has a sphere of radius 1. Let $r$ be the radius of 2nd bubble. Then, from hypothesis, bubbles $2$ to $n+1$ can have a maximum height of $r\sqrt{n}$ starting from the center of bubble number $2$. So maximum height of all $n+1$ bubbles is $1+\sqrt{1-r^2}+r\sqrt{n}$. First order condition for maxima is $r/\sqrt{1-r^2}=\sqrt{n}$ which yields $r^2=n/(n+1)$. Substituting this value of $r$, the maximum height of $n+1$ bubbles is $1+\sqrt{n+1}$. |
H: Proving that there is no values of z such that cosz = isinz
Here is my attempt:
Assume there are values of $z$ such that $\cos z = i \sin(z)$
Then
$\cos(z) = \dfrac{e^{iz} + e^{-iz}}{2}$
and
$i\sin(z) = \dfrac{i(e^{iz} - e^{-iz})}{2i}$
and so after doing some algebra we get
$4e^{-iz} = 0$
and this will never happen, since e^whatever will never be zero , that's a contradiction and hence there are no values of $z$ such that $\cos(z) = i\sin(z)$
AI: That's one way to do it, and is correct.
Alternative approach: If $\cos{z} = i \sin{z}$, then
$$\cos^2{z} + \sin^2{z} = -\sin^2{z} + \sin^2{z} = 0$$
contradicting that $\cos^2{z} + \sin^2{z} = 1$. |
H: algebraically determining if a number is irrational or not
Is it possible to use an algebraic formula, equation, concept, or principle to determine with perfect accuracy (or high precision, if not perfect) whether or not a number is rational?
An example number I have in mind is $\sqrt{937}$.
AI: I think you need to consult an expert on this, but I would boldly suggest the answer is unknown. The current method is to assume the number coming from a certain algebraic equation of degree $n$, then check the decimal part of the number and see if it become linearly dependent after taking the power a few times, etc. I think for a lot of constants like $\zeta(2n+1)$ or Euler's constant, we do not actually know whether they are rational or irrational, largely because the above method fails, which in fact only tells us the weaker result that the number should be algebraic by bounding its "height". For quadratics it is easy, but if the number is not a period or coming from a non-obvious limiting process, then the answer can be tricky and difficult.
Again, excuse me if I made any mistakes. A real expert in algebraic number theory would be better to answer your question. |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.