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H: convergence of test functions
Let $\phi\in C_0^\infty(\mathbf R^n)$ such that $0\leq \phi(x)\leq 1$ and $\phi(x)=1$ for $|x|\leq 1$ and $\phi(x)=0$ for $|x|\geq 2$. Defining $\phi_n(x)=\phi(x/n)$. How do we show that $\phi_n\rightarrow \phi$ pointwise as $n\rightarrow\infty$ to conclude that $\phi_nf\rightarrow\phi f$ pointwise for any integrable function $f$?
AI: With $\phi(x) = 1$ for $\lvert x\rvert \leqslant 1$, we have $\phi_n(x) = 1$ for all $x$ with $\lvert x/n\rvert \leqslant 1$, or, written in a different way, $\lvert x\rvert \leqslant n$. So, for every $x$, almost all $\phi_n$ have $\phi_n(x) = 1$, thus $\phi_n \to 1$ pointwise.
And therefore we also have $\phi_n\cdot f \to f$ pointwise for every function $f$. When $f \in L^p$, for $p <\infty$, then the convergence is not only pointwise, but also $L^p$-convergence, thus such a sequence of cut-off functions can serve to approximate any integrable ($L^p,\,p<\infty$) function by integrable ($L^p$) functions with compact support (and smooth, if $f$ is smooth). Such approximations are often useful, for example because many theorems can be proved more easily for functions with compact support, and then extended to the general case by continuity. |
H: Find an algorithm with O(n) for making Chairlifts on N mountains
Today, I encountered an interesting problem in my textbook. The problem is:
Utopia city has N mountains with height of $$h_{1}, h_{2}, h_{3}, ..., h_{n}$$. We want to make a chairlifts which pass from the top of k consecutive mountain. To do this, we have to make the height of all these mountains the same. If the secondary height of mountains are $$d_{1}, d_{2}, d_{3}, ..., d_{n}$$, the cost of building chairlifts from mountain #t to
the mountain #t+k-1 is:
$$\sum_{i=t}^{t + k -1} (h_{i} - d_{i})$$
We are looking for an algorithm to find the minimum cost of building the chairlifts in $$O(n).$$
in the book it is mentioned that it is easy to find the algorithm. But it doesn't mention the algorithm. I can't think of an O(n) algorithm as a solution.
Can anyone help please?
AI: Let $PS[i]$ be the prefix sum of $h$ (sum of $h_1,\ldots,h_i$). So we can calculate the sum of any interval in $O(1)$.
Do some preprocessing on $h$ in $O(n)$ time, so that we can answer Range Minimum Queries (RMQs) in $O(1)$. See Wiki page for RMQ.
Run through all possible starting positions, calculate the sum of the next $k$ mountains, and get the minimum height of those mountains, and calculate the cost of building the chairlift there (that is, $sum - minimum \times k$). Record the minimum cost.
Update:
It is important to note all ranges we run RMQ on are of length $k$. It is therefore possible to exploit this fact and find simpler algorithm to answer those RMQs. I'll try to explain such an algorithm in the following.
Consider the equivalent problem of finding the minimum of all length $k$ sub-arrays for every ending position $i$.
We maintain a monotone queue (actually deque) $Q$ of our active set, that is, only those elements that could still be the minimum of the current interval are stored in $Q$.
Consider e.g. $Q=[h_1=1,h_3=7,h_7=9,h_9=13]$, $i=10$ and $h_i=8$.
From this point on, we are sure that $h_9$ and $h_7$ can never be the minimum for any ending positions after $i$ (inclusively) because $h_i=8$ comes after them, they are in a sense "dominated" by $h_i$. Therefore we drop those elements at the end of the queue which are larger than or equal to $h_i$, and then append $h_i$ to the queue, resulting in a new monotone queue $Q'=[h_1=1,h_3=7,h_{10}=8]$.
Then, we remove all elements that are too far away from $i$, from the beginning of $Q'$. E.g. if $k=8$, then $h_1$ is too far away from $h_{10}$, resulting in $Q''=[h_3=7,h_{10}=8]$.
From this queue, we choose the first element. That will be the minimum of the length-$k$ sub-array ending at $i$.
The algorithm is $O(n)$, because every element can only be in the queue at most once, and be dropped at most once. The total cost is therefore $O(n)$, although it could also take up to $O(n)$ in a single position $i$. |
H: Name of shape with constant distance to a line segment
For a computer graphics problem I have a shape that is defined by a constant distance to a line segment:
I tried to find a name for this shape, but my Google skills have failed me. Does it have a dedicated name?
AI: Maybe it is a ciiiiiiiiircle?
Kidding aside, I've never seen a dedicated name for it, but in metric topology one can always define the set of points at distance $R$ from a given set.
I also ran across this question in another forum which appears to be asking the same thing. It is again a lot of "I don't know a specific word for it" but the words that came up were things like "race track", "truncated circle" and "stadium." (Note while some of the posters mention elliptical shapes, the OP there clearly stated he's asking about a circle split in half reconnected with straight sides to make the shape the OP is asking about here.)
Wikipedia also suggests that in common English, the "split circle connected by lines" is often called an oval, and that this name is applied to oval tracks.
Edit: Arthur noted in the comments that if you happen to follow the link to stadium, it suggests that stadium is exactly the word you're looking for. I guess if the line gets longer, you just call it a long stadium.
In my searches for terms, I came across this article on topological skeletons which seems to be a sort of "backwards" version of this problem where you start with a shape and then find points inside that are as "central" as possible.
I got the idea then to try to find answers under the heading of shape analysis, but I also ran out of luck with search terms. It seems like a promising clue, though. |
H: How find this value of $\prod_{1\le i
give the positive integer number $n$, and $w=\cos{\dfrac{2\pi}{n}}+i\sin{\dfrac{2\pi}{n}}$
where $i^2=-1$
find the vaule
$$\prod_{1\le i<j\le n}(w^i-w^j)^2$$
My try:note
$$w^n=1$$
$$\prod_{1\le i<j\le n}(w^i-w^j)^2=\prod_{1\le i<j\le n}(w^i-w^j)(w^i-w^j)$$
and I know this
$$\prod_{i=1}^{n-1}(1-w^i)=n$$
becasuse
$$1+x+x^2+\cdots+x^{n-1}=(x-w)(x-w^2)\cdots (x-w^{n-1})$$
$$\prod_{1\le i<j\le n}(w^i-w^j)^2=\prod_{1\le i<j\le n}w^{2i}(1-w^{j-i})^{2}$$
Then I can't works,Thank you
AI: Hint: First, find out the value of $$\prod_{i=1}^{n-1} (1-w^i)$$ |
H: Quadratic twist of an elliptic curve
I found this page:
http://en.wikipedia.org/wiki/Twists_of_curves#Quadratic_twist
which tells me $dy^2=x^3+a_2x^2+a_4x+a_6$ is equivalent to
$y^2=x^3+da_2x^2+d^2a_4x+d^3a_6$. Why is this equivalent (for $d$ as given on that page)?
AI: Put $y = y'/d^2$ and $x=x'/d$. Then
$$d^{-3}y'^2 = d^{-3}x'^3 + a_2d^{-2}x'^2 + a_4d^{-1}x' + a_6.$$
Multiply both sides by $d^3$. |
H: Prove $\sqrt{2} + \sqrt{5}$ is irrational
How do you prove that $\sqrt{2} + \sqrt{5}$ is irrational?
I tried to prove it by contradiction and got this equation: $a^2/b^2 = \sqrt{40}$.
AI: Use proof by contradiction. Assume that the sum is rationial, that is $$\sqrt2 +\sqrt5 = {a\over b}$$ where $a$ and $b$ are integers with $b\neq0$. Now rewrite this as $$\sqrt5={a\over b}-\sqrt2.$$ Squaring both sides of this equation we obtain $$5={a^2\over b^2}-2\sqrt2{a\over b}+2.$$ Now, carefully solve for $\sqrt2$ and obtain $$\sqrt2={-3b\over 2a}+{a\over 2b.}$$ This implies that $\sqrt2$ is a rational number which is a contradiction. Thus $$\sqrt2+\sqrt5$$ is an irrational number. |
H: Hypergeometric die toss
A fair die is tossed until a $2$ is obtained. If $X$ is the number of trials required to obtain the first $2$, what is the smallest value of $x$ for which $P(X \leq x) \geq \frac{1}{2}$?
My thoughts: The general formula for hypergeometric random variables is given by $p(k) = P(X=k) = \frac{C(n,k)C(N-n, r-k)}{C(n,r)}$. I need to find $x$ such that $P(x) + P(x-1) + \cdots + P(0) \geq \frac{1}{2}$. So, $k=x$. $N, n,$ and $r$ have been hard for me to find. Maybe $N=6^{X}$ (since there are $6^{X}$ possible outcomes given $X$ trials), and $n=6$, but I'd like some feedback because I'm not sure. Anyway, thanks in advance for your help.
Source: Finan's A Probability Course for the Actuaries
AI: Are you certain that it is the hypergeometric distribution that you have to use? This problem looks like it can be dealt with using the negative binomial distribution.
EDIT: Okay here is how I would do it using the negative binomial distribution. Since we are interested in 1 sucess, i.e. only the first two, the negative binomial reduces to the geometric distribution that has PMF: $$ p_X(x) = p \left( 1-p \right)^{x-1} $$ X is the number of trials needed to get the first success so $x-1$ is the number of failures prior to success. The Cumulative distribution function is given by $$ CDF(x)= 1-\left( 1-p \right)^x $$
Since you are looking for the median you need to find $ x $ such that the cumulative distribution above equals $ 0.5\ \text{and}\ 1-p=5/6 $ Solve using logarithms. |
H: Finding inradius given the heights
I'm given the heights of a triangle. Find the inradius. I know that inradius is area/semiperimeter. But then?
AI: Let the area of triangle be $T$, the heights $h_a, h_b, h_c$, and $s$ the semi-perimeter. Then:
$$2T = ah_a = bh_b=ch_c$$
$$a = \frac{2T}{h_a}, b = \frac{2T}{h_b}, c = \frac{2T}{h_c} $$
$$a+b+c= 2T(\frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c})$$
$$\frac{s}{T}=\frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c}$$
I hope you get it from here, the sum of the reciprocals of the heights is the reciprocal of the inradius itself. Nice little theorem. |
H: What is $\int \frac1{1+(a\tan x)^2}dx$?
What is $\int \frac1{1+(a\tan x)^2} \mathrm dx$?
This is a difficult integral. If you can, please give a step-by-step solution - I would be delighted.
AI: It's not that hard if you take the suggestion from Clayton:
$u=a \tan{x}$; $x = \arctan{(u/a)}$; $dx=a\, du/(u^2+a^2)$. Then the integral is
$$a \int \frac{du}{(u^2+a^2)(u^2+1)} $$
Assume $a \ne 1$. Use partial fractions:
$$\frac{1}{u^2+a^2}-\frac{1}{u^2+1} = \frac{1-a^2}{(u^2+a^2)(u^2+1)}$$
so the integral becomes
$$\frac{a}{1-a^2} \int du \left (\frac{1}{u^2+a^2}-\frac{1}{u^2+1} \right ) = \frac{a}{1-a^2} \left (\frac{1}{a} \arctan{\frac{u}{a}} - \arctan{u} \right )+C$$
Therefore
$$\int \frac{dx}{1+a^2 \tan^2{x}} = \frac{1}{1-a^2} \left (x - a \arctan{(a \tan{x})}\right )+C $$
When $a=1$, however, we get that the integral is
$$\int \frac{dx}{1+\tan^2{x}} = \int dx \cos^2{x} = \frac{x}{2} + \frac12 \sin{x} \cos{x} + C$$ |
H: Double embedding or double restriction
The following generalizes both embedding and restriction for sets $A$ and $B$:
$A \rightleftarrows B = ( A ; B ; \operatorname{id}_{A \cap B})$.
$A \rightleftarrows B$ is considered as a morphism of the category $\mathbf{Rel}$.
It is an embedding $A \hookrightarrow B = ( A ; B ; \operatorname{id}_A)$ if $A\subseteq B$.
It can be called a restrition if $B\subseteq A$ (in this case
$A \rightleftarrows B = ( A ; B ; \operatorname{id}_B)$.).
It is well known that embedding into a set $A$ and then embedding into a set $A'$ is the same as embedding into $A'$ (if $A\subseteq A'$). In simple words: Double embedding is embedding.
It is well known that restricting to a set $A$ and then restricting to a set $A'$ is the same as restricting to $A'$ (if $A'\subseteq A$). In simple words: Double restriction is restriction.
The following obvious theorem (enough conditions for $( B \rightleftarrows C) \circ ( A \rightleftarrows B) = ( A \rightleftarrows
C)$) generalizes it:
$( B \rightleftarrows C) \circ ( A \rightleftarrows B) = A \rightleftarrows
C$ if $C \subseteq B$.
$( B \rightleftarrows C) \circ ( A \rightleftarrows B) = A \rightleftarrows
C$ if $A \subseteq B$.
(Here composition of $\mathbf{Rel}$-morphisms is induced by the composition of binary relations: $g\circ f=\{(x;z) \,|\, \exists y:(xfy\wedge ygz) \}$.)
My question: What are necessary conditions for $( B \rightleftarrows C) \circ ( A \rightleftarrows B) = ( A \rightleftarrows
C)$?
My guess the condition is $A\subseteq B\vee C\subseteq B$.
Note: The above is for the category $\mathbf{Rel}$. I use it in my research of categories $C$ equipped with a functor $\mathbf{Rel}\rightarrow C$. Thus it is important.
AI: $( B \rightleftarrows C) \circ ( A \rightleftarrows B) = A \rightleftarrows C$ is equivalent to:
$(B;C;\operatorname{id}_{B\cap C})\circ(A;B;\operatorname{id}_{A\cap B}) = (A;C;\operatorname{id}_{A\cap C})$;
$(A;C;\operatorname{id}_{A\cap B\cap C}) = (A;C;\operatorname{id}_{A\cap C})$;
$A\cap B\cap C = A\cap C$;
$B\supseteq A\cap C$.
So the answer is $B\supseteq A\cap C$. |
H: $A \subset C$ implies $A \cap C \subset B \cap C$
I need help with proving this set of expressions in boolean algebra:
$$
A \subset B \implies A \cap C \subset B \cap C
$$
I tried to solve it like this:
$$
A \subset B \implies A \cap C \subset B \cap C\\
\lnot(A\lor B) \lor (\lnot(A\land B))\lor (B\land C)\\
(\lnot A\land\lnot B) \lor (\lnot A\lor \lnot B) \lor (B\land C)\\
...
$$
but it gets too complicated, and I am not sure I get the correct result.
I'd like to have some clue on how to solve it.
AI: Try element chasing:
$A \subset B$ means, by definition, $$\begin{align}(x \in A \rightarrow x\in B) & \implies [(x \in A \land x \in C) \rightarrow (x \in B \land x \in C)] \\ \\ & \iff (x \in A\cap C \rightarrow x \in B\cap C)\end{align}$$
And this means $A\cap C \subset B\cap C$. |
H: What is the moment generating function from a density of a continuous random variable?
Let X be a random variable with probability density function $$f(x)=\begin{cases}xe^{-x} \quad \text{if } x>0\\0 \quad \text{ } Otherwise.\end{cases} $$
Determine the mgf of X whenever it exists.
I know that $M(t) = E(e^{tx}) =\int e^{tx}f(x)~dx$ but not sure what to do from there.
Thanks for the help.
AI: We begin, as you indicated, with the integral
$\int_{-\infty}^\infty e^{tx}f(x)dx$ In order to find this integral, we may proceed as follows:
$$
\int_{-\infty}^\infty e^{tx}f(x)dx
= \int_0^\infty x e^{tx} e^{-x}dx
=\int_0^\infty x e^{(t-1)x}dx
$$
From there, use integration by parts. That is, we have $\int u\,dv = uv - \int v\, du$. For this problem, we choose $u = x$ and $dv = e^{(t-1)x}dx$. Applying the rule gives us
$$
\begin{align}
\int_0^\infty xe^{(t-1)x}dx &=
\left[\frac{1}{t-1}x e^{(t-1)x}\right]_0^\infty
- \int_0^\infty \frac{1}{t-1} e^{(t-1)x}dx\\
&= \frac{1}{t-1} \left[x e^{(t-1)x}\right]_0^\infty -
\frac{1}{(t-1)^2}\left[e^{(t-1)x}\right]_0^\infty\\
&= \frac{1}{t-1} \cdot 0 - \frac{1}{(t-1)^2}\cdot (-1)\\
&= \frac{1}{(t-1)^2}
\end{align}
$$ |
H: Find all $x,y,z$ satisfying $xy=z-x-y$ and cyclic permutations
Find all ordered pairs $(x,y,z)$ real numbers, which satisfy the following system of equations:
$$xy=z-x-y\\xz=y-x-z\\yz=x-y-z$$
AI: Hint:
$$xy=z-x-y \quad \iff \quad (x+1)(y+1)=z+1.$$
Hence $(x+1)(y+1)=z+1$, $(x+1)(z+1)=y+1$, $(z+1)(y+1)=x+1$. |
H: Show that $\{f_n^2\}\rightarrow f^2$ in measure.
From Royden's Analysis, 4th Edition, Chapter 5, section 2 Problem 7:
Let $E$ have finite measure, $\{f_{n}\}\rightarrow f$ in measure on $E$ and $g$
be a measurable function on $E$ that is finite a.e. on $E$. Prove that $\{f_{n}\cdot g\}\rightarrow f\cdot g$ in measure, and use this to show that $\{f_{n}^{2}\}\rightarrow f^{2}$ in measure.
The first part is complete, but I am having issues with the squared sequence. I understand that I can get a subsequence, $\{f_{n_k}\}\rightarrow f$ pointwise almost everywhere on $E$. Is this helpful for this problem? So, even with this subsequence, I don't have that $f$ is finite almost everywhere, so I can't really use the previous portion of the problem.
Any insight at all would be great! Thanks.
AI: A sequence converges in measure iff every subsequence has a further subsequence converging a.e. to the target function. Apply this principle and you will get it right away; in fact you can generalize the result. |
H: Using Compact Support to write function as integral
If we have $u \in C_{c}^{1}(\mathbb{R}^{n})$ then why does we require the compact support of $u$ in order to write:
$u(x) = \int_{-\infty}^{x_{i}}u_{x_{i}}(x_{1},…,x_{i-1},y_{i},x_{i+1},..,x_{n})dy_{i}$ ?
Thanks
AI: I think it might be required in order to apply the Fundamental Theorem of Calculus (FT1).
The Fundamental Theorem of Calculus requires that $f$ be continuous on a closed compact set $[a,b]$, since $u$ is continuous and has a compact support we can evaluate the integral over the support which is closed and compact using the Fundamental Theorem of Calculus. Just an idea. In fact since the $\text{supp}u$ is closed and bounded we can find such an $a$ and $b$ such that $\text{supp} u \subset [a,b]$ and then apply the first theorem of the fundamental theorem of calculus. |
H: Proof of $(\forall x)(x^2+4x+5 \geqslant 0)$
$(\forall x)(x^2+4x+5\geqslant 0)$ universe is $\Re$
I went about it this way
$x^2+4x \geqslant -5$
$x(x+4) \geqslant -5$
And then I deduce that if $x$ is positive, then $x(x+4)$ is positive, so it's $\geqslant 5$
If $ 0 \geqslant x \geqslant -4$, then $x(x+4)$ is also $\geqslant -5$.
If $ x < -4$, then $x(x+4)$ will be negative times negative = positive, so obviously $\geqslant -5$
I'm wondering if there's an easier to solve this problem? My way seems a little clunky.
AI: Assuming $x$ is real, $(x+2)^2\ge0\implies x^2+4x+4\ge0$ |
H: time speed distance
Two horses start simultaneously towards each other and meet after $3h 20 min$. How much time will it take the slower horse to cover the whole distance if the first arrived at the place of departure of the second $5 hours$ later than the second arrived at the departure of the first.
MY TRY::
Let speed of 1st be a kmph and 2nd be b kmph
Let the distance between A and B be d km
d = 10a/3 + 10b/3
and
d/a - d/b = 5
now i cant solve it. :(
Spoiler: The answer is $10$ hours.
AI: Let the distance from the meeting place of the departure of the first Horse $A$ is $a $ meter
and the distance from the meeting place of the departure of the first Horse $B$ is $b$ meter
So, the total distance is $a+b$ meter
So, the speed of the first horse is $\displaystyle\frac a{200}$ meter/minute and that of the second is $\displaystyle\frac b{200}$ meter/minute
So, the first horse $A$ will need to cover $b$ meter more which it will take $\displaystyle\frac b{\frac a{200}}=\frac{200b}a$ minute
So, the total time taken by $A$ will be $\displaystyle200+\frac{200b}a$ minute
Similarly, the total time taken by $B$ will be $\displaystyle200+\frac{200a}b$ minute
If $A$ is slower than $B,$ $\displaystyle200+\frac{200b}a-\left(200+\frac{200a}b\right)=300\implies 2b^2-3ab-2a^2=0\implies b=2a$ (why?)
The total time taken by $A$ will be $\displaystyle\frac{a+b}{\frac a{200}}$ minute |
H: About definition of "ordered semi-ring"
I need the definition of "ordered semi-ring". Can I use these properties:
$a \preceq b \to a + c \preceq b + c$
$0 \preceq a \wedge 0 ≤ b \to 0 \preceq a \cdot b$ (or: $a \preceq b \wedge 0 \preceq c \to a \cdot c \preceq b \cdot c $)
???
Thanks in advance!
AI: I found three slightly different definitions of ordered semiring. This one may be the oldest:
Definition: A semiring $R$ is ordered if there is a partial order $\preceq$ on $R$ such that
$a+c\preceq b+c$ whenever $a,b,c\in R$ and $a\preceq b$, and
$ac\preceq bc$ and $ca\preceq cb$ whenever $a,b,c\in R$, $0\preceq c$, and $a\preceq b$.
$R$ is positive if $0\preceq a$ for all $a\in R$.
Another definition removes the restriction that $0\preceq c$ from (2); the two are clearly equivalent for positive ordered semirings. The third adds to the second the requirement that $0\preceq a$ for all $a\in R$, so it’s equivalent to the notion of positive ordered semiring in either of the first two senses of ordered semiring.
All of these definitions explicitly include additive monotonicity, your first property. Your second property follows from (2) above, so it holds no matter which of these definitions is in use: if $0\preceq a$ and $0\preceq b$, then $0=0\cdot b\preceq a\cdot b$. Thus, under any of these definitions of ordered ring you can use both of the properties that you listed. |
H: Where's the mistake? Integral of a bounded non-analytic function over a closed curve.
Let $f(z)$ be a bounded non-analytic function defined on a closed curve $\gamma$. What is wrong with the following:
$$\left| \oint_\gamma f(z) dz\right| \le \oint_\gamma \left|f(z) \right| dz \le M \oint_\gamma 1 \cdot dz =0$$
because $1$ is analytic on $\gamma$.
Of course when one writes
$$\left| \oint_\gamma f(z) dz \right| = \left| \int_a^b f(z(t)) z'(t)dt \right| \le \int_a^b \left| f(z(t)) z'(t) \right| dt \le M \int_a^b|z'(t)|dt=M|\gamma|.$$
it works out as it should. What's wrong with the first one then? Which step is illegal?
AI: $dz$ is a complex measure/differential form (whichever interpretation you prefer), so $$\oint_\gamma \lvert f(z)\rvert\,dz$$ is in general not real, hence the inequality doesn't make sense.
You get $$\left\lvert\oint_\gamma f(z)\,dz\right\rvert \leqslant \oint_\gamma \lvert f(z)\rvert\, \lvert dz\rvert,$$
where $\lvert dz\rvert$ is the total variation of $dz$ in the measure interpretation, which is realised by $\lvert z'(t)\rvert\,dt$. |
H: Integral of Infinite Division
In my Calculus II class this question was posed. It is a definite integral, but the function integrated is infinitely dividing. I tried $u$ substitution, but cannot find the correct term to use as my $u$.
$$
\int\limits_0^6 \frac{xdx}{1-\frac{x}{1-\frac{x}{1-\frac{x}{\quad\ddots}}}}
$$
Any help? Thanks so much!
AI: Let $y$ be the continued fraction. Then
$$y = \frac{x}{1-y}$$
or
$$ y^2-y + x=0$$
which means that
$$y = \frac12 \pm \frac12 \sqrt{1-x}$$ |
H: Generalizations of pairing function
A pairing function is a one-to-one mapping from $N^2$ into $N$; for example the Cantor pairing function is given by:
$$J(x,y) = \frac{(x+y)(x+y+1)}{2} + 1$$
Another one is: $f(x,y)=2^x(2y+1)$.
They can be easily generalized to encode $n$-tuples (points in a $n$-dimensional space) ($J(x_1,x_2,...,x_n) = J(x_1,J(x_2,...,J(x_{n-1},x_n)))$).
Are there generalizations for an infinite dimension space? I.e. functions that are bijections between (integer) points in an infinite dimension space and $N$ ?
AI: We take $\mathbb{N}$ to include $0$. Minor modification will take care of things if we define $\mathbb{N}$ to start at $1$.
Consider the set $S$ of all sequences $(a_0,a_1,a_2,\dots)$ of natural numbers that are $0$ from some point on. Let $\psi$ be the mapping that takes the sequence $(a_0,a_1,\dots)$ to $\left(\prod_0^\infty p_i^{a_i}\right)-1$, where $p_i$ is the $i$-th prime. Then $\psi$ is a bijection from $S$ to $\mathbb{N}$. |
H: How far does she travel in her Journey?
Ok, I'm doing a Mock End of Unit Test Revision paper and I'm stuck on a few questions.
The first question is:
Susan completes the journey in $2$ stages of her journey. In stage 1 of her journey, she drives at an average speed of $80$km/h and takes $1$ hour and $45$ minutes.
(a) How far does she travel?
I know that answer is $140$ because:
$1$ hour $=$ $80$km/h
A quarter of $80$ ($15$ minutes) is $20$ and I need three-quarters as there's $45$ minutes which is $60$.
$80$ $+$ $60$ $=$ $140$. I know it's the right answer however how come on the Marking Paper the method of working it out is doing $80$ x $1.75$. Where do you get $1.75$ from?
Second question:
Altogether Susan travels $190$km/h and takes the total time of $2$ hours and $15$ minutes.
(b) What is her average speed, in km/h, in stage $2$ of her journey?
I know that you would start of by doing $190$ $-$ $140$ = $50$. I'm not sure what to do after that though. Thanks everyone, I know it seems quite easy.
AI: For question $a$, $1.75$ is how many hours it takes for the journey (note that $45$ minutes = $0.75$ hours). For question $b$, the average speed is $\frac{distance}{time} = \frac{190-140}{2.25-1.75} = \frac{50}{0.5} = 100$ km/h. In words, in the second part a distance of $50$ km is travelled in $30$ minutes, so the average speed is $100$ km/h. |
H: Induction with sets: $\forall n \ge 1: \overline{\bigcap_{i=1}^nA_i}=\bigcup_{i=1}^n \overline{A_i}$
I know how to do induction with equations, but for this thing with sets:
$$\forall n \ge 1: \overline{\bigcap\nolimits_{i=1}^nA_i}=\bigcup\nolimits_{i=1}^n \overline{A_i}$$
exactly I don't have an idea. If someone can I would appreciate it very much.
AI: The same way. First, look at the base case, which is $n=1$. It is easy. Now do $n=2$-can you verify that? Finally, assume it is true for values up to $n$. When you write the equation for $n+1$, note that associativity lets you group the first $n$ terms on each side into one set of parentheses, and you have the two element case. |
H: How do I prove inequalities and one-to-one function?
Can anyone please help me with these questions?
1.Given
x + 1 < 0
Prove that:
i) $2x - 1 < 0 $
ii) ${2x-1\over x+1} > 2$
2.For $g(x) = {kx + 8\over 4x - 5}$
i) Find k if gg(x) = x
Is it fine if I just let any value of x for this question?
ii) Find the value of k so that g(x) is not a one-to-one function.
Thank you very much.
AI: (1 i) $x+1<0\Rightarrow 2\left( x+1 \right)<2.0\Rightarrow 2x+2<0\Rightarrow 2x+2-3<-3\Rightarrow 2x-1<-3<0$
so $2x-1<0.$
(1 ii)$$\frac{2x-1}{x+1}=\frac{2\left( x+1 \right)-3}{x+1}=2-\frac{3}{x+1}$$
because $x+1<0$ given we can write $$-\frac{3}{x+1}>0\Rightarrow 2-\frac{3}{x+1}>2 \Rightarrow\frac{2x-1}{x+1}>2$$
(2 i) we know if $g\left( x \right)=\frac{ax+b}{cx+d}$ then ${{g}^{-1}}\left( x \right)=\frac{-dx+b}{cx-a}$.
$g\left( g\left( x \right) \right)=x\Rightarrow g\left( x \right)={{g}^{-1}}\left( x \right)$ this means $g(x)=\frac{kx+8}{4x-5}=\frac{5x+8}{4x-k}={{g}^{-1}}\left( x \right)$ so $k=5.$
(2 ii) similar way |
H: Real root of a complex equation.
I was working on a problem from Gamelin; where I was required to find out zeros of $2z^5+6z^1-1$ , in the unit disk (in $\mathbb C$). I applied Rouché's theorem and find out zeros in the unit disk and I got to know that there is only one zero inside it.
Further, I have to show that it has one zero inside $(0,1)$ : Following is my answer and I am not sure about it. please correct me if i am wrong.
This polynomial have only one zero inside the open unit disk. Therefore the only root that exist in unit disk must be a real one. Because complex roots exist only in pairs.
Now we have to show that this zero is positive. Is it right if I say that, there is 1 changes in sign of the function's coefficients, so there will be at most 1 positive roots (maybe less). And now put -z on the place of z then all coefficients of the polynomial will turn negative. Therefore there is no negative root. So the only root that we have in the unit disk must be real; so it will lie in $(0,1)$.
AI: Hint: $2\cdot 0^5 + 6\cdot 0-1 < 0 < 2\cdot 1^5+6\cdot 1-1$. |
H: Family with three children. Chances of at least one boy and girl?
Here's my exercise:
A family has three children. What's the probability of event $A \cup C$ where:
$A$- Family has children of both sexes.
$C$- Family has at most one girl.
Well I see two ways to look at this conundrum:
The first one would be to differentiate possibilities by triples, $B$ meaning boy, $D$ meaning girl. So we have eight possibilities: $BBB,BBD,BDB,DBB,BDD,DDB,DBD,DDD$. So the answer would be $P(A \cup C)=\frac{7}{8}$.
And this is correct according to solutions in my textbook. Yet my first solution was this, since they ask about if the family has two boys, and not about whether the first and third child are boys, then $BDB$ and $BBD$ are indistinguishable, then we have only four possibilities: one girl and two boys, three boys, three girls, one boy and two girls. Then $P(A \cup C)=\frac{3}{4}$.
Why was my first intuition wrong?
AI: The only order that fails $A \cup C$ is $DDD$, one out of eight, so $\frac 78$ is correct. Your other approach misses the fact that (unordered) $BDD$ and $BBD$ are three time more likely than $BBB$ and $DDD$ |
H: Divisibility by $9$
Suppose we have a natural number $N$ with decimal representation $A_kA_{k-1}\ldots A_0$. How do I prove that if the $\sum\limits_{i=0}^kA_i$ is divisible by $9$ then $N$ is divisible by $9$ too?
AI: Hint: $\overline{A_kA_{k-1}\ldots A_1A_0} = 10^kA_k + 10^{k-1}A_{k-1} + \ldots + 10A_1 + A_0$.
$(10^kA_k + 10^{k-1}A_{k-1} + \ldots + 10A_1 + A_0) - (A_k + A_{k-1} + \ldots + A_1 + A_0) = (10^k-1)A_k + (10^{k-1}-1)A_{k-1} + \ldots + (10-1)A_1$ is divisible by $9$. |
H: What is the answer to this syllogism? Why is option D incorrect?
Q. a. Some books are not reference books.
b. All books are encyclopedias.
A Some reference books are no encyclopedias
B No reference books are encyclopedias
C All reference books are encyclopedias
D None of the above
AI: All reference books are books, and hence encyclopedias. So $C$ is the correct answer.
Even without knowing the meaning of 'reference book' and 'encyclopedia', either $A$ or $C$ must be true. If $C$ is false, i.e. not all reference books are encyclopedias, then there must be a reference book that is not an encyclopedia, and $A$ is true. |
H: Probability of getting the product divisible by $3$.
What is the probability that the product of two randomly chosen (distinct) numbers between $1$ and $100$ (inclusive) is divisible by $3$?
Now, what I am doing is to find out the number of ways of selecting a multiple of $3$ out of $100$ and then selecting any of the other remaining numbers. This can be done in $^{33}C_1 \times ^{99}C_1$ ways. The whole process of selecting two numbers can be carried out in $^{100}C_2$ ways. That gives the probability as: $ \dfrac { 33 } { 100 } $. However this answer is wrong. What's wrong in what I am doing?
The given solution considers two separate cases. First case, you select both the numbers as multiples of $3$ and in the other case, you select one multiple and one non-multiple of $3$. Why do that?
AI: Can you tell how many times the pair (3,6) was considered in your solution?
Once, choosing a multiple of 3 like 3 and then one of the remaining numbers, 6.
Once again, choosing a multiple of 3 like 6 and then one of the remaining numbers, 3.
In you solution, the pair (3,6), just like all pairs with both numbers divisible by 3, was considered twice. That's why the solution split the problem in these cases. |
H: How to solve this class of problems?
I was presented with the following problem:
Ricardo, Rogério and Renato are brothers. One of the is a medic, the other one is a teacher and the other one is a musician. It is known that:
Ricardo is a medic or Renato is a medic.
Ricardo is a teacher or Rogério is a musician.
Renato is a musician or Rogério is a musician.
Rogério is a teacher or Renato is a teacher.
Then, their professions are respectivelly:
Teacher, Medic, Musician
Medic, Teacher, Musician
Teacher, Musician, Medic
Musician, Medic, Teacher
Medic, Musician, Teacher
And I know there may be some way to interpret the propositions and solve the problem in a fashion similar to the way we multiply numbers. But it's not very clear to me how to use this way, I've read about Propositional Calculus and the Method of Analytic Tableux but I'm still lost. I've also tried to think in a way to use Venn diagrams but I still don't have much clue on what to do. I'm looking for a general way of solving this class of problems. I've also tried some random explorations on Mathematica but couldn't achieve anything. Can you guide me?
I guess I have managed to make a solution for this (following the answers given by Vadim123 and Peter Smith, I've made a table with the propositions:
$$\begin{vmatrix}
{Ricardo Medic}&{Renato Medic}\\
{Ricardo Teacher}&{Rogério Musician}\\
{Renato Musician}&{Rogério Musician}\\
{Rogério Teacher}&{Renato Teacher}
\end{vmatrix}$$
And then another one to follow the consequences, then If Ricardo is a medic, I checkmark the corresponding option and mark the other professions that he could have with an X mark. If Ricardo is a medic, then it's impossible for him to be a teacher nor Renato be a musician due to the nature of the XOR operation.
$$\begin{vmatrix}
{\checkmark }&{X}\\
{X }&{ - }\\
{ -}&{ -}\\
{- }&{- }
\end{vmatrix}$$
Now I make a bet in the profession of the other brothers, I'll make a bet that Renato is a musician, which implies that he can't be a teacher and also that Rogerio can't be a musician. The only job left for Rogerio is the job of musician.
$$\begin{vmatrix}
{\checkmark }&{X}\\
{X }&{ X }\\
{ \checkmark }&{X}\\
{X }&{\checkmark }
\end{vmatrix}$$
Now I have two problems:
How to determine the number of solutions of the given exercise? How to determine that there is only one option that answers it? I know that I gave a list of answers, but ignore that it exists, how to know that Ricardo is a medic, Renato is a musician and Rogerio is a teacher is the only answer? I can only think of brute force and perhaps the truth table, but for the latter, I'm still not sure it works. I'm worried because using brute-force, I guess I've been able to find 3 solutions until now:
$$\begin{matrix}
{S_1}&{RicardoTeacher}&{RenatoMedic}&{RogerioMusician}\\
{S_2}&{RicardoMedic}&{RenatoMusician}&{RogerioTeacher}\\
{S_3}&{RicardoMedic}&{RenatoTeacher}&{RogerioMusician}\\
\end{matrix}$$
Are there ways to algebrize this problem? I've tried to make a truth table for it but I'm not sure it works. I'm interested in having a general approach for this kind of problem, in this case, there are $4$ propositions with the $XOR$ conector, what if the connector could be any conector? What if one of the propositions has a $XOR$, the other one has a $NOR$ and the next one has an $AND$?
AI: It is evident from 4, since only one of the brothers is a teacher, that Ricardo isn't a teacher, and 2 then tells you
Rogerió is musician.
So he isn't a teacher so 4 tells you
Renato is a teacher.
So he isn't a medic, so 1 tells you
Ricardo is a medic.
Job done!
OK: but this is an ad hoc solution to the particular problem. Can we mechanise the reasoning? In particular, the OP is interested whether we can use the propositional calculus. Well, yes ....
Use e, i, o for Renator, Ricardo, and Rogerió.
Use D, T, M for Doctor, Teacher, Musician.
So we have nine atoms we can write $De, Di, Do; Te, Ti, To; Me, Mi, Mo$.
The basic structure of the situation is that we have there is exactly one doctor
$(De \land \neg Di \land \neg Do) \lor (\neg De \land Di \land \neg Do) \lor (\neg De \land \neg Di \land Do)$
similarly for teacher and musicians,
and there there is exactly one profession for Renato
$(De \land \neg Te \land \neg Me) \lor (\neg De \land Te \land \neg Me) \lor (\neg De \land \neg Te \land Me)$
and similarly for the other two guys.
So that's six general "axioms": and then there are four specific axioms
$Di \lor Do$
and so on.
We've got nine atoms (so it's a 512 line truth-table to do!) and ten axioms. And we can then do a truth-table to test any axiom to see if it follows. Be my guest .... The principle of the thing is mildly interesting, but of course implementing it in practice would be stunningly boring! |
H: Rewriting $x^3-3xy^2+2xy+i(-y^3+3x^2y-x^2+y^2 )$ in terms of $z$, with $z=x+yi$
How do I write $f=u+iv$
with:
$u=x^3-3xy^2+2xy$ and
$v=-y^3+3x^2y-x^2+y^2 $
in terms of $z$ with $z=x+yi$?
AI: Have you heard of the binomial theorem? It tells us that (for example) $$(a+b)^2=a^2+2ab+b^2$$ and $$(a+b)^3=a^3+3a^2b+3ab^3+b^3.$$ See if you can use those to do the trick, bearing in mind that $i^2=-1$. |
H: About $S_n=\{(x,y)\mid \lfloor kx\rfloor=\lfloor ky\rfloor,k=1,2,\cdots n;~x,y\in [0,1]\}$
Let $$S_n=\{(x,y)\mid \lfloor kx\rfloor=\lfloor ky\rfloor,k=1,2,\cdots n;~x,y\in [0,1]\}$$
Here are the pictures of $S_1,S_2,\cdots S_{10}$:
We can see that $S_1$ has only one blue region, $S_2$ has $2$ different blue regions. Denote $R_n$ be the number of different blue regions of $S_n,$ then $R_n=1, 2, 4, 6, 10, 12, 18, 22, 28, 32, \cdots$
I find that $R_n$ is A002088: $$R_n=\sum_{d=1}^n \phi(d)\tag 1$$
Can you prove $(1)$ ? What's the total area of the blue regions of $S_n$?
AI: Proof of (1)
Let $S = \left\{ \dfrac{p}{q} \,|\; 1 \le p \le q \le n \right\}$. Then all the elements of $S$ would lie between $0$ and $1$. Arrange the elements into increasing order and let it be $a_1 < a_2 < \cdots < a_N$ = 1. Also, let $a_0=0$.
First note that $$N = \sum_{d=1}^{n} \phi(d)$$ since there are exactly $\phi(d)$ number of fraction which the denominator is $d$ in its reduced form.
If some $a_i = \frac{p}{q}$ lies between $x$ and $y$, then one of $\lfloor qx \rfloor$ and $\lfloor qy \rfloor$ should be smaller than $p$ and one greater or equal to $p$. Thus, for $(x,y)$ to be a element of $S_n$, they must both be in some interval $[a_i, a_{i+1})$.
On the other hand, if $x, y \in [a_i, a_{i+1})$, there cannot be any $q \le n$ such that there exists a integer $p$ between $qx$ and $qy$. Hence, there are no $q$ which $\lfloor qx \rfloor \neq \lfloor qy \rfloor$ and $(x,y) \in S_n$.
Now we know that $S_n$ only consists of squares $[a_i,a_{i+1}) \times [a_i,a_{i+1})$ for $i = 0, 1, \cdots, N-1$ and $(1,1)$. Therefore, the number of squares is $R_n = N$. |
H: "Moving" a filter from one set to an other set, two equivalent formulations
By filters I mean filters on sets (not on one fixed set but on arbitrary sets).
Informally saying, this question concerns "moving" a filter from one set to an other set.
I define rebase a filter $\mathcal{A}$ (on a set $\mathfrak{A}$) to a set $\mathfrak{B}$ by the formula
$$\mathcal{A} \div \mathfrak{B} = \left\{ X \in \mathscr{P} \mathfrak{B} \,|\,
\exists Y \in \mathcal{A} : Y \subseteq X \right\} .$$
I want to prove that
$$\mathcal{A} \div \mathfrak{B} = [\{ X\cap \mathfrak{B} \,|\, X\in\mathcal{A} \}]_{\mathfrak{B}}$$
where $[F]_{\mathfrak{B}}$ means the filter on $\mathfrak{B}$ induced by the set $F$.
I'd prefer not any arbitrary proof, but a clean "structured" proof, maybe as a series of lemmas.
AI: Without conditions on $\mathfrak{A} , \mathfrak{B}$ these families may not be equal.
Suppose that $\mathcal{A}$ is the principal (ultra)filter on $A$ generated by $a \in \mathfrak{A}$, and $\mathfrak{B}$ does not contain $a$. It it easy to show that $\mathcal{A} \div \mathfrak{B} = \varnothing$ (since no set in $\mathcal{A}$ can be a subset of any subset of $\mathfrak{B}$). However $\{ X \cap \mathfrak{B} : X \in \mathcal{A} \} = \mathscr{P} ( \mathfrak{A} \cap \mathfrak{B} )$. |
H: Problem applying Cauchy's estimate in proof
Let $f$ is an analytic map from the unit disk to itself.
Taylor series of $f$ centered at 0: $f=\sum_n a_n z^n$ ).
Prove that: $|a_n|\leq 1 \; \forall n$
I've started using Cauchy's estimate:
$|a_n|=\frac{f^{(n)}(0)}{n!}\leq \frac{M_r}{r^n}$ with $M_r=\max\{|f(z)|: |z|=r\}$
We see that: $M_r\leq 1 $ and $r=1$. But why is this ($M_r\leq 1 $ and $r=1$) the case ?
AI: You have $M_r \leqslant 1$ because by assumption $f$ maps the unit disk to itself, so $\lvert f(z)\rvert < 1$ for all $z$ with $\lvert z\rvert < 1$.
The assumptions don't allow you to set $r = 1$, since $f$ need not be defined, let alone continuous, on the boundary of the disk, the circle $\lvert z\rvert = 1$. But you have
$$\lvert a_n \rvert \leqslant \frac{M_r}{r^n} \leqslant \frac{1}{r^n}$$
for all $r < 1$, so you can take the limit $r \to 1$ to conclude $\lvert a_n\rvert \leqslant 1$. |
H: Prove $X - \bigcup_{C \in \scr{C}}C = \bigcap_{C \in \scr{C}}(X - C).$
Prove $X - \bigcup_{C \in \scr{C}}C = \bigcap_{C \in \scr{C}}(X - C).$
Proof. Assume $a \in X - \bigcup_{C \in \scr{C}}C$.
Then $a \in X$ and $a \not\in \bigcup_{C \in \scr{C}}C$.
Then $a \in X$ and $a \not\in \bigcap_{C \in \scr{C}}C$, since for all $C \in \scr{C}$ we have $a \not\in C$.
I don't think I am going the right path here. How do I continue?
AI: Some hints to get you going again.
To complete this direction, we have to prove that $a \in \bigcap\limits_{C \in\mathscr C} (X-C)$. By definition of intersection, this amounts to proving $a \in X -C$ for every $C \in \mathscr C$. It may be helpful to look at the definition of union, and think about what it means not to be an element of it.
For the converse direction, if $a \in X-C$ for every $C \in \mathscr C$, we need to prove $a \in X$ and $a \notin \bigcup\limits_{C \in\mathscr C} C$.
I'm confident you can handle it from here. |
H: Faithfulness of adjoint representation of Lie algberas
Are there any simple or useful conditions (necessary & sufficient) under which the adjoint representation lie algebra is faithful ? One sufficient condition is semisimplicity, but perhaps this is not necessary.
AI: The adjoint representation of a Lie algebra has a kernel which is very easy to describe: It is the center of the Lie algebra (this is easy to check directly from the definition).
So the adjoint representation is faithful iff the Lie algebra has a trivial center. |
H: Does a discrete set of points in $\mathbb{R}^{n}$ define a locally finite collection of hyperplanes?
Let $v_{1},v_{2},...$ be a discrete set of non-zero vectors in $\mathbb{R}^{n}$. By discrete, I mean that any $v_{i}$ is surrounded by an $\epsilon$-ball not containing any other point $v_{j}$. Equivalently, any compact subset $K$ of $\mathbb{R}^{n}$ contains only finitely many of the $v_{i}$. (EDIT: These two conditions are not equivalent, and I mean the latter condition. I don't want accumulation points.)
Define affine hyperplanes $H_{i}=\{x \mid \langle v_{i},x \rangle=1\}$. Here $\langle,\rangle$ is the standard inner product.
Is the set of hyperplanes $H_{i}$ locally finite?
Here, locally finite means that given a point $x$ in the union of the hyperplanes, we can find an $\epsilon$-ball around $p$ that intersects only finitely many of the hyperplanes. Equivalently, any compact subset $K$ intersects finitely many of the $H_{i}$. (EDIT: Two conditions needn't be equivalent. I want the latter condition.)
AI: In $\Bbb R^2$ let $v_k=\langle k,k\rangle$ for each $k\in\Bbb Z^+$; the set $V=\{v_k:k\in\Bbb Z^+\}$ is certainly discrete. However, $H_k$ is the line whose equation is $kx+ky=1$, the line of slope $-1$ with $x$- and $y$-intercepts at $\frac1k$. $\{H_k:k\in\Bbb Z^+\}$ clusters at every point on the line $y=-x$, so it is not locally finite in $\Bbb R^2$. In particular, the compact set
$$\{\langle 0,0\rangle\}\cup\left\{\left\langle\frac1k,\frac1k\right\rangle:k\in\Bbb Z^+\right\}$$
meets every $H_k$. This set of lines is locally finite in itself, however.
Added: Better yet, let $v_k=\langle 2^k,1\rangle$ for $k\in\Bbb N$; $\{v_k:k\in\Bbb N\}$ is a discrete subset of $\Bbb R^2$, but $H_k$ is the line $x+2^ky=1$ whose $x$-intercept is $1$ and whose $y$-intercept is $2^{-k}$, so $$\langle 1,0\rangle\in\bigcap_{k\in\Bbb N}H_k\;,$$ and $\{H_k:k\in\Bbb N\}$ isn’t even discrete in itself. |
H: If $A$ is a countable subset of $\omega_1$, then there is an $ \alpha < \omega_1$ such that $A \subseteq \alpha$
Prove : if $A$ is a countable subset of $\omega_1$. Then there exists $ \alpha < \omega_1$ with $ A \subseteq \alpha$.
I don't really know where to start, can anyone give a tip first?
AI: HINT: Recall that $\alpha=\bigcup A=\sup A$. Then $\alpha$ is the countable union of countable ordinals. |
H: Determination of a uniformly continuous, not locally Hölder-continuous function over an open set
Given an open interval $I=(a,b)$, I would like to exhibit a uniformly continuous function over $I$ that is not locally Hölder-continuous with exponent $\alpha$ for any $\alpha\in(0,1)$.
It is quite easy to exhibit an UC-function that is not globally Hölder-continuous for a given exponent, or an UC-function that, in a neighbourhood of a single point, is not Holder-continuous for any exponent, but how can we bring this type of "singularity" in any open neighbourhood belonging to $I$?
My first try was to consider UC-functions with "very fast local variation", i.e. nowhere or almost-nowhere differentiable functions, like the Weierstrass function. Unfortunately, the Weierstrass function, despite its weird behaviour, is Hölder-continuous.
AI: By considering that (thanks to Marco D'Addezio)
$$ f(x)=-\frac{1}{\log x} $$
is not $\alpha$-Holder continuous in a right neighbourhood of zero for any $\alpha\in(0,1)$ (as the inverse of the "really flat" function $\exp(-1/x)$), we can take:
$$ g(x)=\frac{1}{\log 4-\log \sin^2(\pi x)}$$
(defined as $0$ if $x\in\mathbb{Z}$) to have a 1-periodic uniformly continuous function over $\mathbb{R}$, that is not locally Holder-continuous in any neighbourhood of an integer, and is locally Lipschitz-continuous over $\mathbb{R}\setminus\mathbb{Z}$. By condensation of singularities, i.e. by considering the following sum of translates of $g(x)$ over the dyadic rationals in $(0,1)$ (any countable dense subset can play the same role):
$$h(x)=g(x)+\sum_{j=1}^{+\infty}\frac{1}{3^j}\sum_{k=1}^{2^{j-1}}g\left(x-\frac{2k-1}{2^j}\right)$$
we still have a periodic uniformly continuous function over $\mathbb{R}$, that is nowhere locally $\alpha$-Holder continuous for any $\alpha\in(0,1)$. |
H: proof by contradiction example
Any ideas on how I can use proof by contradiction to show that at least 3 of any 25 days chosen must fall in the same month of the year?
I don't even understand the question.
AI: Pick any $25$ different dates in $2013$ (or any other year), like $13$ May, $27$ June, etc. The claim is that no matter which $25$ dates you pick, at least three of them will be in the same month.
HINT: Suppose that you picked at most two dates from each month of the year; what’s the largest number of dates that you could possibly pick? |
H: Counterexample to in $\mathcal{Mod}_A$ colimits of filtered index categories are exact
This is not a true general fact for any abelian category, as Vakil points out in 1.6.12. He gives the following counterexample, which puzzled for it is in the category of abelian groups, and every abelian group is a module over $\mathbb Z$. So my question is: why is his counterexample not also a counterexample in the category $\mathcal{Mod}_\mathbb Z$?
⋆⋆
Here is a counterexample. Because the axioms of abelian categories are self-dual, it suffices to give an example in which a filtered limit fails to be exact, and we do this. Fix a prime $p$. In the category $\mathcal{Ab}$ of abelian groups, for each positive integer $n$, we have an exact sequence $\mathbb Z \rightarrow \mathbb Z/(p^n)\rightarrow 0$. Taking the limit over all n in the obvious way, we obtain $\mathbb Z\rightarrow \mathbb Z_p\rightarrow 0$, which is certainly not exact.)
**
AI: The category $\text{Mod-}A$ of right modules over a ring is known to be a Grothendieck category; in particular, filtered colimits are exact in this category.
This category is not “coGrothendieck”, though. Indeed it's not hard to show that if an abelian category is both Grothendieck and coGrothendieck, then it is the one object and one morphism category.
The self-dual nature of the axioms for an abelian category doesn't allow to conclude that for any property of an abelian category $\mathcal{A}$ then the dual property holds in $\mathcal{A}$. Exactness of filtered colimits in $\text{Mod-}A$ is an example.
What can be derived from the self-duality of the axioms is that if a property holds in all abelian categories, then also its dual property holds in all abelian categories. So the “counterexample” you present only shows that in some abelian categories filtered colimits are not exact (for instance in the opposite category of $\text{Mod-}\mathbb{Z}$). |
H: Proof of $|x^{\alpha} - y^{\alpha}| \le \alpha^{\alpha} |x-y|$ for $\alpha \ge 1, x,y\in [0,1]$
I want to prove
$$
|x^{\alpha} - y^{\alpha}| \le \alpha^{\alpha} |x-y|
$$
for $\alpha \ge 1$ and $x,y \in [0,1]$. For $\alpha \in \mathbb N$ I already got the proof by using the formulae
$$
(x^n - y^n) = (x-y)(x^{n-1} + x^{n-2}y + \ldots + xy^{n-2} + y^{n-1})
$$
but I have no idea how to generalize to arbitrary $\alpha \ge 0$, any hints?
AI: Hint:
$$(x^{\alpha} - \alpha^{\alpha} x)' = \alpha(x^{\alpha-1} - \alpha^{\alpha-1}) < 0.$$ |
H: $f: \omega_1 \to \omega_1$. $\forall$ $\alpha$ $\exists$ $\beta > \alpha $ with $f(\beta)=\beta $.
Prove: if $f: \omega_1 \to \omega_1$ is an increasing, continuous, unbounded, function, then $\forall$ $\alpha$ $\exists$ $\beta > \alpha $ with $f(\beta)=\beta $.
Can anyone give me a tip?
AI: The use of quantifiers in the expression ‘$\exists\,\forall\alpha$ a $\beta>\alpha$ with $f(\beta)=\beta$’ really isn’t correct: you mean ‘$\forall\alpha\,\exists\beta>\alpha$ such that $f(\beta)=\beta$’.
HINT: First prove that $f(\alpha)\ge\alpha$ for all $\alpha\in\omega_1$. Now let $\alpha\in\omega_1$ be arbitrary. Let $\alpha_0=\alpha+1$. Given $\alpha_n$ for some $n\in\omega$, let $\alpha_{n+1}=f(\alpha_n)$. Can you see how to use $\langle\alpha_n:n\in\omega\rangle$ to find a $\beta$ with the desired properties? |
H: Inductive definition with choice for sequence
In topology there is a very common way to define a sequence. This usually go something like:
"Define $\{z_{n}\}$ to be a sequence such that $z_{0}$ is <blah blah blah>, and $z_{n}$ is such that $R(z_{0},z_{1},\ldots,z_{n})$ is true. The sequence is well-defined since there always exist $z_{n}$ that satisfy that relationship because <blah blah blah>."
(here $R$ is some sort of relationship, such as for example $z_{n}\subset\bigcap\limits_{i=1}^{n-1}B_{\delta}(z_{i})$, though the particular is irrelevant for this question)
Sure it seems intuitively clear, and I use it all the time. My professor, despite being a constructivist, accept this reasoning. But I have been wondering out of curiousity: is there anyway to prove that such a sequence exist? Hope someone can shed a light on this. I have only seen a book (about Complex Analysis) that mention that and call this principle of inductive definition though fortunately for it, it never need to use choice anywhere. The name appeared to be made up for the book, since I cannot find anything with that exact name on the Internet.
Note that the issue mainly at play here are:
-We can prove any finite subsequence exist by induction. The question is about the existence of the infinite sequence.
-There is always choice needed to be made at each step ie. there are many $z_{n}$ that fit the bill. Since this is usually about general topological space, there is no good way to actually pick out an element specifically. Axiom of Countable Choice is allowed of course, but even given that, how do you prove that an infinite sequence exist. If Axiom of Countable Choice is not enough, then why, and how do you prove it using the full Axiom of Choice?
I was able to prove that a sequence exist when no choice is needed. I can also prove it assuming that the space is countable (basically reduce to non-choice version through well-ordering). However most topological space is not countable and choice are needed, so I'm stuck.
Thank you for your help.
AI: The principle used to define infinite sequences by induction is stronger than the axiom of countable choice and it is known as Dependent Choice, or $\sf DC$. The principle has many equivalent formulations, one of them is as follows:
Suppose that $S$ is a non-empty set and $R$ is a binary relation on $S$, whose domain is all $S$. Then there exists function $f\colon\Bbb N\to S$ such that $f(n)\mathrel{R}f(n+1)$ for every $n\in\Bbb N$.
The axiom of countable choice is sufficient when we can uniformly define the families from which we are choosing. If we want an arbitrary point whose distance from $x$ is at most $\frac1n$ then this is doable with the axiom of countable choice as we can choose $x_n$ from $B_{\frac1n}(x)$. But often we define $x_n$ to be an element definable from the previously chosen elements, in which case we are in fact using $\sf DC$.
Finally, to use the full axiom of choice one can usually do one of the two following things:
Well-order the space, then proceed to pick the least $x$ in the well-ordering which satisfies the needed requirements. Or, more directly,
Fix a choice function on all the non-empty subsets of the space (or just the relevant subsets, if we care about open sets or whatever), then define use induction to choose using our choice function. Using the choice function as a parameter the choice of element is no longer arbitrary and we can define the infinite sequence without difficulty. |
H: Inserting values left to right in a binary search tree
What does it mean to build a binary search tree by inserting values from left to right starting from an empty tree? The "left to right" part confuses me..I know how to build one by normally inserting values from the top..but what does this mean?
Could anyone please explain how to insert values from left to right with an example? An example would be really helpful..please and thank you.
AI: I would assume "left to right" refers to the order of processing a given key sequence.
E.g. given the key sequence $[1, 5, 2, 3]$, You just start with an empty tree and insert $1$, then $5$, then $2$ and at last $3$.
Other orderings include "in random order" and "in reversed order (or right to left)". |
H: Rooted Trees & Induction
So I am a little stumbled upon this question:
A full binary tree is a rooted tree where each leaf is at the same distance from the root and each internal node has exactly two children. Inductively, a full binary tree of depth 0 is the one node tree N, and a full binary tree of depth d+1 is a rooted tree whose two subtrees are each full binary trees of depth d. How many nodes and how many edges are in a full binary tree of depth d? Prove this answer by using the inductive defininition given in this problem.
Anything that could help me understand and get me started proving this would be awesome!
Thank you in advance!
AI: The full binary trees of depths $0,1$, and $2$ look like this:
0 1 2
* + o
/ \ / \
* * / \
+ +
/ \ / \
* * * *
I’ve used different symbols for the nodes to make it easier to see how they are built up inductively: the tree of depth $1$ has a root + from which descend two copies of the tree of depth $0$, and the tree of depth $2$ has a root o from which descend two copies of the tree of depth $1$. Similarly, the full binary tree of depth $3$ has a root from which descend two copies of the full binary tree of depth $2$. If $f(n)$ is the number of nodes in the full binary tree of depth $n$, you can see that $f(0)=1$, $f(1)=3$, and $f(2)=7$, and from the description above it’s clear that $f(3)=2f(2)+1=2\cdot7+1=15$. At this point we have the following data:
$$\begin{array}{rcc}
n:&0&1&2&3\\
f(n):&1&3&7&15
\end{array}$$
It’s not hard to extend this table further: the inductive construction implies that $f(n+1)=2f(n)+1$ for each $n$, since we get the full binary tree of depth $n+1$ by putting two copies of the full binary tree of depth $n$ side by side and adding a root at the top.
You’re supposed to conjecture a closed formula for $f(n)$ and then prove it by induction on $n$. If you don’t immediately see a plausible formula, try extending the table a little further; the sequence of numbers $f(n)$ is very close to a sequence that ought to be pretty familiar. |
H: I don't understand how sets can be closed, yet disjoint?
What are some closed, disjoint subsets $A, B$ in $R^2$ where $inf\{d(A, B) = 0 \forall a \in A \forall b \in B\}$?
AI: For example, $A=\{(x,y)\mid xy=1,x>0\}$ and $B=\{(x,y)\mid xy=-1,x<0\}$ |
H: Finding if $\sum\frac{1}{2+3^{-k}}$ divergent or convergent
How would one find if the following series is divergent or convergent.
$$\sum\frac{1}{2+3^{-k}}$$
I did the following
$$\sum\frac{1}{2+3^{-k}}<\sum\frac{1}{3^{-k}}$$
But I am not sure what test I should use the only ones I know are limit comparison and basic comparison test.
But what should I do.
AI: Hint: $$\lim_{k \to +\infty} \frac{1}{2+3^{-k}} = \frac12.$$ |
H: Finding if $\sum_{}^{}\frac{\ln k}{k}$ converges or diverge
How can I find whether the serie
$$\sum_{}^{}\frac{\ln k}{k}$$
converge or diverge using the basic comparison test or limit test.
The part that is confusing me is how to deal with $\ln$.
AI: You know that the harmonic series $\sum_{k = 1}^\infty \dfrac 1k$ diverges.
$\dfrac{\ln k}{k} > \dfrac 1k$ as $k$ grows, indeed, for all $k \gt 3$. You don't even need to know that when exactly, $\dfrac {\ln k}{k}$ exceeds $\dfrac 1k$, only that as $k$ grows large (indeed, very early on), $\dfrac{\ln k}{k} \gt \dfrac 1k.$
So you can use the comparison test for convergence/divergence.
Added for clarification about the harmonic series and its general term $a_n = \dfrac 1n$:
In a follow-up question below, it was asked why the harmonic series diverges even though the limit as $n \to \infty$ of the general term is zero?
Recall that with respect to the convergence behavior of a given series $\sum_{n = 1}^\infty a_n$,
"IF $\;\lim_{n\to \infty} |a_n| \neq 0,\;$ then it diverges."
So this theorem tells us definitely that any such series diverges. What it doesn't tell us is what happens when that same limit does approach $0.\;$ In those cases where $\lim_{n\to \infty} |a_n| = 0$, as in the case of the general term of the harmonic series, the associated series may converge or it may diverge. So that particular theorem is inconclusive in this case.
For a nice discussion about the divergence of the harmonic series, with proofs of its divergence (using the comparison test and one using the integral test), see the Wikipedia entry on the divergence of the harmonic series. |
H: How to construct co-equalizers in $\mathbf{Top}$?
How to construct co-equalizers in the category $\mathbf{Top}$?
Well, do co-equalizers in $\mathbf{Top}$ exist at all?
AI: If $f,g:X\to Y$ are two parallel morphisms, then the coequalizer $c:Y\to Z$ of $f,g$ will be the quotient map to $Z=Y/\sim$ where "$\sim$" is the smallest equivalence relation such that $f(x)\sim g(x)$. So it is the coequalizer in $\mathbf{Set}$ equipped with the quotient topology.
This is no coincidence. All colimits in $\mathbf{Top}$ have the same underlying set as the colimit of the underlying sets in $\mathbf{Set}$. In other words, the forgetful functor $U:\mathbf{Top}\to\mathbf{Set}$ preserves colimits. This is due to the fact that it has a right adjoint, namely the functor $\mathbf{Set}\to\mathbf{Top}$ which equips a set with the indiscrete topology.
Edit: Regarding the OP's comments, the adjointness only guarantees that IF the coequalizer exists, then its underlying set (and set map) is the coequalizer (diagram) in $\mathbf{Set}$.
To show that $Y/∼$ is the coequalizer in $\mathbf{Set}$, let $h:Y\to W$ be a function with $hf=hg$. Then the relation $h(y)=h(y')$ is an equivalence relation on $Y$ which contains the relation generated by $f(x)∼g(x)$ since $h(f(x))=h(g(x))$. Therefore the map $k:Y/∼\to W,\ k(c(y))=h(y)$ is well-defined and is obviously the only function $k:Y/∼\to W$ such that $kc=h$.
To show that $c$ is also the coequalizer in $\mathbf{Top}$, we just have to make sure that $c$ is continuous, and that the only possible function $k:Z\to W$ satisfying $kc=h$ is continuous. But $c$ is continuous because we equip $Z$ with the final topology, and this is also the reason why $k$ is continuous if $kc=h$ is. So $c:Y\to Z$ is the coequalizer in $\mathbf{Top}$.
Proving that $G:\mathbf{Set}\to\mathbf{Top},\ X\mapsto(X,\tau_{id})$ is right adjoint to $U$ is easy. Just note that $((X,\tau_{id}),\ 1:UGX\to X)$ is universal from $U$ to $X$: For each set map $f:Y\to X$ we have a continuous map $(Y,\tau_Y)\to(X,\tau_{id})$, namely $f$ itself. |
H: Give the transformation matrix that send $\vec i$ on $ [1,1/2,0]$
I'm asked to give the linear transformation that sends $\vec i$ on $[1,1/2,0]$ , $\vec j$ on $[-1/3,1,0]$ , $\vec k$ on $[1/4,1/4,1]$.
Is it a translation where I need to use the actual $i,j,k$ vectors or is it just changing the $i,j,k$ vectors to the new ones? It's in french and it says 'qui envoie $\vec i $ sur $[1,1/2,0]$'
What would be transformation matrix ?
AI: Each of the image vectors is one column of the corresponding matrix representation of the linear map.
To see this, consider the following, for coefficients $\alpha, \beta, \ldots$:
$$A(\vec i) = \begin{bmatrix} \alpha & \delta & \eta \\ \beta & \epsilon & \theta \\ \gamma & \zeta & \iota \end{bmatrix} \begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix} = \begin{bmatrix} \alpha \\ \beta \\ \gamma \end{bmatrix}$$
But you know the result should be
$$A(\vec i) = \begin{bmatrix} 1 \\ 1/2 \\ 0\end{bmatrix}$$ |
H: sum over primes less than 'x
is there a function $ f(x) $ so
$$ \sum_{p\le x}f(p)=S(x)$$
where $ S(x)=g(f(x), \pi(x) $
this means that the sum S(x) depends on the function $ f(x)$ but also on the prime number counting function
the only case is $ f(x)=0$ but can be another alternatives ?
for example, the integral of a function $ \int f(x)dx =f(x) $ means that $ f(x)=e^{x} $
but how about a sum over primes which is equal to itself for a non zero function ??
AI: Henry has a good example in the comments:
$$\sum_{p<x|\text{p is prime}}k = k\pi(x)$$
(where $\pi(x)$ is the prime counting function) |
H: For sets $A,B$, $A \cap (B \setminus A) \subseteq \varnothing$
I need help with this proof:
Let $A$ and $B$ be sets. Prove that $A\cap (B\setminus A)\subseteq \varnothing$.
My problem: Since $B\setminus A$ = $B\cap A^{c}$, we can say that if $x \in B$ then $x\in B\cap A^{c}$and $x \in A^{c}$ as well. So $x\notin A$.
Where do I go from here?
AI: Hint:
$$x\in A\cap (B\setminus A)\implies x\in A\;\text{ and also }\;x\in B\setminus A\implies$$
$$x\in A\;\text{and also}\;\left(x\in B\;\text {but}\;x\notin A\right)\implies\ldots$$ |
H: Using Gauss's lemma to show if p(a) = p(b) = p(c) = p(d) = 5, there is no integer k with p(k) = 8
I'm attempting to prove this:
Let p(x) be a monic polynomial with integer coefficients. Suppose that there are distinct integers a, b, c, d with p(a) = p(b) = p(c) = p(d) = 5. There is no integer k with p(k) = 8.
I have a hint saying to use Gauss's lemma, but I don't know how. Just plain following my nose gave me this:
Consider the polynomial $p'(x) = p(x) - 5$. $p'(a) = p'(b) = p'(c) = p'(d) = 0$. Therefore, $p'(x) = (x-a)(x-b)(x-c)(x-d)q(x)$, where $q(x)$ is another polynomial. Suppose to the contrary that there exists an integer $k$ so that $p(k) = 8$. Then $p'(k) - 3 = (k-a)(k-b)(k-c)(k-d)q(k) - 3 = 0$. So if $p''(x) = p'(x) - 3,$ then $p''(x) = (x-3)r(x)$ for some polynomial $r(x)$.
I'm looking for a contradiction, but I don't know where it is. Am I even on the right track with this?
AI: Hint: You have already got $(k-a)(k-b)(k-c)(k-d)q(k)=3$. Now $k-a, k-b, k-c, k-d$ are four distinct integers whose product is a factor of $3$. |
H: Simple tax puzzle
I recently saw some post on facebook whining about taxes. Simplifying it (and changing numbers, facts, etc.), this was saying:
For each dollar an employer wants to pay you:
20% go in taxes that your employer pays
out of the remaining 80%, you pay 40% of various taxes (from income taxes, to sales taxes etc.)
48% of the initial amount remains for you and 52% has gone in taxes.
Now, assume that you spend all of your money. This amounts to spending everything in paying people's salary : whether it is directly a taxi driver, or the set of people that are required to bring your tomato to the supermarket: cashier, CEO, farmer etc. (the tomato itself does not cost money as the earth is not asking us to bury dollar bills each time we harvest a tomato).
So, out of these salaries, as an (indirect) "employer", the 48% you spent will partially go in employer taxes (20% in our scenario). The facebook post concludes that in fact, out of the initial money that was offered to you, a total of 61.6% have gone into taxes.
I want to go further and say that, actually, the people that I indirectly employed also paid their 40% share of "various taxes", and with their remaining 48% they also used their money to buy things (hence employ people, that will buy things...), etc.
You recognize here a converging series... that converges to 100%.
Solutions to the paradox:
there is no paradox: if the government just absorbs taxes and does not employ itself people, there is indeed 100% of an initial amount of money that end up in taxes
the problem is that we should stop without entering the recurrence : you have 48% of the initial amount, you spent it, and bought stuffs with that. That's the value you got, that got transformed into goods (or services).
when I said a tomato costs nothing per se, and only work has a value in terms of salary, I was wrong. (why?).
I made a mistake and the series does not converge to 100% (why??).
I actually can't find the actual solution to this paradox I imposed myself. No politics please ;)
AI: I'm not sure if this is on topic for this site, but no one has voted the question down yet so I'll give a shot at an answer.
Assume that the money supply is constant and assume that every monetary transaction is taxed at a constant rate (whenever an amount of money changes hands then the government takes a fixed proportion of that amount.)
If we assume that money held by the government does not circulate, and that money held by everyone else circulates at a constant rate (called the velocity of money) then your argument shows that the proportion of the money supply that ends up in the hands of the government approaches 100% as time goes on.
To be precise, the amount of money not held by the government at time $t$ forms a geometric series that approaches zero as $t \to \infty$.
However, this situation does not reflect reality. The government spends money too, rather than holding on to it forever. For example, the government spends money on payroll, entitlement programs, and buying goods from the private sector.
Now if we assume instead that the government spends money like everyone else, a somewhat different thing happens. Every dollar keeps circulating forever, and is taxed an unlimited number of times, so eventually the total amount the government has collected will exceed the money supply, and indeed will eventually exceed any fixed amount. (But of course it will not have all this money at that time, because it will have already spent much of it.)
Eventually every dollar will have been involved in many transactions: sometimes being part of taxable transactions within the private sector, sometimes being paid to the government in taxes, and sometimes being paid back by the government to the private sector. |
H: Volume of $n$-ball in terms of $n-2$ ball
Let $V_n(R)$ be the volume of the ball with radius $R$ in $\mathbb{R}^n$. This page says
$$V_n(R)=\int_0^{2\pi}\int_0^RV_{n-2}(\sqrt{R^2-r^2})r\,dr\,d\theta$$
I don't really understand the explanation given in there. Could someone explain it in the case $n=3$ (so $n-2=1$) how this integral is derived?
AI: Let's do the $n=2$ case first. What is a zero-dimensional sphere, you ask? Well, it's nothing but a point, and $0$-dimensional volume is, I claim, just cardinality. So $V_0 (R) = \lvert\{\ast \}\rvert= 1$.
Now, how do we calculate $V_2 (R)$, the $2$-volume (i.e. area) of a $2$-ball (i.e. disc)? Well, we want to add up the volume of all those teensy $0$-dimensional balls which make up the disc, parametrized by the radius and the angle. First, imagine cutting the disc into very skinny rings. We will look at one ring with inner radius $r$ and outer radius $r+dr$. Next, we cut this ring into extremely thin sections. We will focus on one such section, that sweeps out a very small angle, $d\theta$.
Call the area of this little piece $dA$. Then the volume we want is exactly the sum of all those little pieces, which is exactly $\int dA$. There are very general principles that tell us how to compute $dA$, but here is how to work it out from scratch.
Our little chunk is very very close to being a rectangle when $d\theta$ is very small. It's so close, that we might as well pretend it is a rectangle (if this makes you uncomfortable, then try working out the error term and seeing how quickly it goes to zero, or else look at the rigorous proof of the multivariate change-of-variables formula).
And what are the side lengths of this rectangle? The radial one is obvious: it's just $dr$. The other one is a little more tricky. Whatever it is, if we add it up all around the circle, we should get the circumference. Since $\int_0^{2\pi} r d\theta = 2\pi r$, the length we want is $r d\theta$.
So we conclude that $dA = r \,dr\, d\theta$. Since we're interested in adding up these areas over all $r$ and $theta$ with $0\leq r \leq R$ and $0\leq \theta \leq 2\pi$, we have $V_2 (R) = \int dA = \int r\,dr\,d\theta = \int_0^{2\pi} \int_0^R r\,dr\,d\theta$.
If we want, we can actually compute this out as a sanity check, and get $2\pi \left.\frac{1}{2} r^2 \right|_0^R = \pi R^2$.
Now that we've done $n=2$, we're more than ready for $n=3$. Imagine that you're looking down on a sphere of radius $R$, so that it looks like a disc. Now we want to chop up the disc exactly as before, except now, instead of tiny near-rectangular bits, these are very thin bars that go all the way through the sphere (picture an apple slicer). Their volume is just the area $dA$ that we had before, times... hmm...
We need to break out the Pythagorean theorem to find the length of our rectangular bar. Say that the length is $2s$. Then, if you think for a minute, you should find that $s^2 + r^2 = R^2$, because the top of our bar lies on the sphere, which has radius $R$, and the line from the origin is the hypoteneuse of a right triangle that goes radially outwards (from our top-down perspective) a distance $r$, then up (towards us) a distance that we defined to be $s$. So the length we want is $2s$—which just so happens to be $V_1 (s)$, the "volume" of the $1$-sphere of radius $s$.
So we have $V_3 (R) = \int 2s \cdot r\,dr\,d\theta = \int V_1 (s) \cdot r\,dr\,d\theta$ $= \int V_1 (\sqrt{R^2 - r^2})r\,dr\,d\theta$ $= \int_0^{2\pi} \int_0^R V_1 (\sqrt{R^2 - r^2}) r\,dr\,d\theta$.
One again, we can do a sanity check, and actually compute the thing to get $\frac{4}{3} \pi R^3$.
Now the general case should be clear. We will dividing up the $n$-ball into pieces, making rectangular cuts. The rectangles will have side lengths $dr$ and $rd\theta$, and they will cut out a ball of dimension $n-2$ and radius $s$, with $s^2 + r^2 = R^2$. |
H: Strong induction definition clarification: is the hypothesis that it is true for all $k$ with $0 \le k < n$ or for all $k$ with $0 \le k \le n$?
I have a general question about strong induction:
Assuming that the base case is 0, if I let my inductive hypothesis be that for all 0 <= k < n some statement is true, and if I prove that that statement is true for n, does this mean that I've proved it for all natural numbers?
Some examples do something a little differently: Assuming that for all 0 <= k <= n is true which includes n, they prove it's true for n + 1..are these two equivalent? Or is only the second method valid? Thanks.
AI: They are equivalent. Note that both methods show that assuming the statement is true for $0\le k\le 16$, it is also true for $17$. The first method does this when $n=17$, the other when $n=16$. Since with both methods one shows the induction step for all $n$, the difeerence in notation does not matter.
Note however that the first variant (correctly) allows one to drop the base case:
The induction step for $n=0$ shows that, assuming the statement is true for all $k$ with $0\le k<0$ it also holds for $0$; but as $0\le k<0$ is impossible, the assumption is vacuously true, hence the case $n=0$ follows without special considereation.
This does not work with the second formulation because to do the same trick, we'd have to use $n=-1$, which we don't want to. |
H: Why is the gradient of F a constant multiple of a parallel vector?
Problem 14:
Solution:
I don't understand why the gradient of F is a constant multiple of a parallel vector. Why is the equation in the black box true? I understand why it would be true if the gradient of F is a stretched version of (27, 8, 1) but it is only parallel to (27, 8, 1). Doesn't that mean that there is a possibility that there is some sort of translation?
AI: The question is explicitly asking you to find where the gradient is parallel to the given vector. Two vectors are parallel if and only if they are multiples of each other. These are not "free vectors" but rather vectors considered in $\mathbb{R^3}$, so they all begin at the origin. |
H: $\limsup_{n\rightarrow\infty} a_n^{1/\log n}<1/e$ and $a_n>0$ then $\sum a_n$ converges
If $\limsup_{n\rightarrow\infty} a_n^{1/\log n}<1/e$ and $a_n>0$ then $\sum_{n=1}^{\infty} a_n$ converges.
$$0<a_n < e^{-\log n}=\frac{1}{n}$$
Also, $\exp\{\frac{1}{\log n}\log a_n \} \le \exp(-1)$ so $\displaystyle\frac{\log a_n}{\log n}\le -1$.
How does the limsup help here?
AI: You are close. Let the $\limsup$ be $\frac{1}{e^{1+\delta}}$. Then for large enough $n$ we have
$$\log a_n\lt -(1+\delta/2)\log n.$$
That means that $a_n \lt \frac{1}{n^{1+\delta/2}}$. Now Comparison does it. |
H: solutions of a linear equation system
The following matrix is given over $\mathbb R$
A=$\begin{pmatrix} 1 & -1 & -1 \\ 1 & 0 & a \\ 1 & a & 0 \end{pmatrix}$
The linear equation System $Ax=\begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}$ has exactly one solution for all $a$ except $a=-2$ which has no solution. Is it right?
AI: Almost correct.
At $a = -2$, we have no solution, so you are correct on that.
At $a = 0$, our augmented RREF matrix gives:
$$\begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & -1 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$
This means we have infinite solutions as:
$x = 1, y = -1 + z$, where $z$ is a free variable.
At all other $a$, we have a single unique solution. |
H: Find a L-sentence which is true in a structure $M$ iff the universe $A$ of $M$ consists of exactly two elements
Find a L-sentence which is true in a structure $M$ iff the universe $A$ of $M$ consists of exactly two elements, where the language L consist a unary function $S$ and $2$-ary predicate $<$.
This is an exercise $2.d$ in my text: A Friendly introduction to mathematical Logic by Leary.
My attempt to solve this is this formula:
$\phi = \mathop{\sim}[\ \forall a\ \exists b\ \exists c\ ([a\ne b] \wedge [ a\ne c] \wedge [b\ne c] \ )\ ] \wedge [\ \forall x \exists y (x\ne y)\ ] $
I chose this formula as it's of course a sentence (no free variables) and it says that the structure doesn't have more than two element and that it doesn't have one element so the only universe satisfies this property must have exactly two elements!
I wonder if this is a right choice and if yes, is there any better sentence we can provide?
AI: Yes, that works, although I found it a bit confusing. You could also say
$$ \exists x\,\exists y\,(x \ne y) \wedge \neg \exists x\,\exists y\,\exists z\,(x \ne y \wedge x \ne z \wedge y \ne z),$$
which says "there are two distinct elements and it is not the case that there are three distinct elements," or
$$ \exists x\,\exists y\,(x \ne y \wedge \forall z\,(z = x \vee z = y)),$$
which says "there are two distinct elements such that every element is equal to one of the two." |
H: which one? permutation or combination?
Let say we have a bookshelf that can fit 6 books, we want 4 computer science books and 2 physics books but computer books should be together and physics books also should be together, we have 8 computer and 6 physics book in total in how many ways we can do that?
I believe that the answer is like this:
c(8,4)*c(6,2) + c(6,2)*c(8,4)
but my instructor sloved it in this way(by p I mean permutation):
p(8,4)*p(6,2) + p(6,2)*p(8,4)
could u please tell me which one is right?
AI: It depends.
If the order of the books matter (as in, computer science book A needs to come before computer science book B, for example), then your instructor would be right.
If they don't, then you would be right. |
H: Elementary question regarding binomial coefficient
I would like to show that $ {n \choose j}$ is a multiple of $n$ if and only if $n$ and $j$ are coprime. Thanks for the help!
AI: You do get one implication: suppose $j$ and $n$ are coprime. Then: $$\binom{n}{j} = \frac{n!}{j!(n-j)!} = \frac{n (n-1) \ldots (n-j+1)}{j!}.$$ This is an integer (either combinatorially or because $j!$ divides the product of $j$ consecutive integers). So $j!$ divides $n(n-1)\ldots(n-j+1)$. But you can split this as: $$\frac{n}{j} \cdot \frac{(n-1)\ldots(n-j+1)}{(j-1)!}.$$ Again, the second part is an integer (because $(j-1)!$ divides the product of $j-1$ consecutive integers). So this reduces to: $$\frac{n}{j} \cdot k$$ for some $k$. Of course, $j$ is coprime to $n$, so for this product to be an integer, $j$ must divide $k$. Hence, the product is a multiple of $n$. |
H: Prove a function is continuous
Discuss the continuity of the function $f:\mathbb{R} \to \mathbb{R}$
if
$f(x) = 0$ if $x$ is not rational
$f(x) = \frac1q$ if $x = \frac{p}{q}$ where $p,q$ are integers with no
common divisors other than $\pm 1$.
I know that the function is not continuous so I must show that for every $\epsilon>0$ there exists $\delta >0$ such that if $x\in \mathbb{R}$ and $|x-x_0|<\delta$ then $|f(x) -f(x_0)|>\epsilon$, for some $x_0 \in \mathbb{R}$ but I don't know how to show this.
AI: Hint 1:
If $x \in \mathbb Q$, you can find some sequence $x_n \in \mathbb R \setminus \mathbb Q$ such that $x_n \to x$.
Then $x_n \to x$ but $f(x_n)=0$ and $f(x) \neq 0$. What does this tell you?
Hint 2: If $x \notin \mathbb Q$.
Pick an $\epsilon >0$. Let $n$ be so that $\frac{1}{n+1} < \epsilon \leq \frac{1}{n}$.
Prove that there are only finitely many rational numbers $\frac{p}{q} \in (x-1,x+1)$ so that $q \leq n$.
Pick a such number $r$ which is closest to $x$ and let $\delta= \frac{|x-r|}{2}$.
Now, if $y \in (x- \delta,x+ \delta)$ what is $f(y)$? Note that you have two cases to cover: $y$ rational and $y$ irrational. |
H: Inequality with condition similar to Schwarz lemma
Suppose $f(z)$ is holomorphic and $|f(z)|\leq 1$ for $|z|\leq 1$. Show that $$\frac{|f'(z)|}{1-|f(z)|^2}\leq \frac{1}{1-|z|^2}.$$
If I also have the condition $f(0)=0$, I would be able to use the Schwarz lemma to conclude that $|f(z)|\leq|z|$ and $|f'(0)|\leq 1$. But I don't know how I can imply the inequality above.
If $f(0)=0$, I want to define $g(z)=f(z)-f(0)$, so that $g(0)=0$, but then the condition $|g(z)|\leq 1$ for $|z|\leq 1$ is not true.
How can I get around those issues?
AI: I want to define $g(z)=f(z)−f(0)$
Put another way, you wanted to compose $f$ with the automorphism of $\mathbb C$ that sends $f(z)$ to $0$. This idea needs just one tweak to work: compose $f$ with the automorphism of the unit disk that sends $f(z)$ to $0$. Namely, the Möbius map
$$\phi(w) = \frac{w-f(0)}{1-w\overline{f(0)}}$$
The Schwarz lemma applies to $\phi\circ f$ and yields $|f'(z)|\le |\phi'(f(z))|^{-1}$, which is the desired inequality in disguise. |
H: Tiny question about the proof of $L^p \cap L^r \subset L^q$
I'm reading about $L^p$ spaces as a sort of self-studying thing. And usually when I read proofs, I'll try to fill in the steps that the author skips myself, but I'm having trouble with this one:
The proof is for: if $1 \leq p < q < r \leq \infty$, then $L^p \cap L^r \subset L^q$.
The author writes:
Let $\lambda \in [0,1]$, and for $f \in L^q$,
\begin{align}
\Vert f \Vert_{L^q} &= \left( \int_X \vert f \vert^q \right)^{\frac{1}{q}} \\
&= \left( \int_X\vert f \vert^{\lambda q} \vert f \vert^{(1-\lambda)q} \right)^{\frac{1}{q}} \\
&= \left( \Vert f \Vert_{L^p}^{\lambda q}\Vert f \Vert_{L^r}^{(1-\lambda)q} \right)^{\frac{1}{q}} \\
&= \Vert f \Vert_{L^p}^{\lambda }\Vert f \Vert_{L^r}^{(1-\lambda)}
\end{align}
I understand how this proof shows that the $L^q$ norm of the function is controlled by the $L^p$ and $L^r$ norm of the function, so that if it's in $L^q \cap L^r$ then it's in $L^q$.
What I'm not getting is a minor detail of the proof: I'm not getting how I should get from the second line to the third line. At first I thought it's an application of Holder's inequality, but if I apply that to the expression in the second line, I get
\begin{align*} \left(\int_X \vert f^{\lambda q} \vert^p\right)^\frac{1}{p}\left(\int_X \vert f^{(1-\lambda) q} \vert^r\right)^\frac{1}{r} \end{align*}
but the expression on the third line is equal to
\begin{align*}
\left( \int_X \vert f \vert^p \right)^\frac{\lambda q}{p}\left( \int_X \vert f \vert^r \right)^\frac{(1-\lambda q)}{r}
\end{align*}
and I'm not sure how I can pull the $\lambda q$, $(1-\lambda)q$ out of the integral. I thought about it, but I don't think Jensen's inequality applies here.
Am I missing something obvious? Or do I need more sophisticated analysis machinery?
Thanks in advance!
AI: Let $\lambda$ be such that $\frac{\lambda}{p} + \frac{1 - \lambda}{r} = \frac{1}{q}$ and apply Holder with exponents $\frac{p}{\lambda q}$ and $\frac{r}{(1 - \lambda)q}$.
:) |
H: Defining matrices in MATLAB
I'm completely new to MATLAB, and I can't figure out how to do the following: I have
A=[-1,0.4,0.8;1,0,0;0,1,0];
b=[0;0.3;6];
What I want to do is define, for $N$ fixed but arbitrary, a $3\times N$ matrix C by having the $i$th column of C be $A^{(N-i)}b$. Without doing the calculations separately beforehand, how can I have MATLAB compute and set C as this matrix?
AI: A=[-1,0.4,0.8;1,0,0;0,1,0];
b=[0;0.3;6];
N=10;
Ai=eye(3); % Identity 3 x 3 matrix
C=zeros(3,N);
for i=1:N
C(:,N-i+1)=Ai*b;
Ai=Ai*A;
end
Note that the first column of C will be equal to $A^{N-1}b$, while the last column will be equal to $b$.
If you wanted them to be $A^Nb$ and $Ab$ respectively then replace the fourth line of the code with
Ai=A;
Edit: The code in littleO's answer is much faster. Mine is an example of what you shouldn't do (at least when you care about efficiency)! |
H: Contingency table on Bayers Rule
The question is this:
A medical test for the disease StatsLove will correctly state that a diseased student suffers from StatsLove 95% of the time. The medical test correctly states that a non-diseased student does not suffer from StatsLove 99% of the time.
From prior terms it is known that approximately 80% of students suffer from StatsLove.
I didn't understand at all. Please help me to build a contingency table.
My understanding is:
STATSLOVE HEALTHY
Infected *95% 80%
Not Infected *99% 20%
The *95% is $80\%\times95\%$, means the positive percentage; similarily, *99 is $99\%\times20\%$.
Any other thoughs? Thanks.
AI: It looks as if you essentially know what to do. If you prefer to use percentages, you may do so. But where you have written $\ast95\%$, and indicate that you know it is $95\%\times 80\%$, I would write $(0.95)(0.8)$, and perhaps even compute the product as $0.76$.
The entry next to the right is for the probability that the person is diseased but is diagnosed as healthy. Given that you are diseased, the probability that you are falsely called healthy is $1-0.95$, that is, $0.05$. So the probability that you are diseased but declared healthy is $(0.8)(0.05)=0.04$. That is the required entey. Note that the two entries $0.76$ and $0.04$ add up to $0.80$, so once you knew the $0.76$, you could have computed the $0.04$ as $0.8-0.76$.
For the bottom row, note that the probability that someone who is not diseased is falsely called diseased is $1-0.99=0.01$. The entry you have labelled $\ast 99\%$ is indeed $(0.2)(0.99)=0.198$.
The entry at the bottom right should therefore be $0.2-0.198=0.002$, or equivalently can be computed as $(0.2)(0.01)$.
Remark: The "heart" of the table is:
$$\begin{matrix} 0.76 & 0.04 \\0.002& 0.198
\end{matrix}
$$
However, you may want also to have the marginal sums, both for the rows and the columns, like this:
$$\begin{matrix} 0.76 & 0.04&0.8 \\0.002& 0.198&0.2\\ 0.762 &0.238 &
\end{matrix}
$$
If you have your accounting hat on, you may want to insert a $1$ in the bottom right corner. |
H: Number of roots of $x^n+ax+b$
Let $P(x) = x^n + ax+b$ with $n\geq 2$ and $a,b\in\mathbb{R}$. Then one of these is true:
the number of distinct real roots of $P$ can be any number between $0$ and $n$.
the number of distinct real roots of $P$ is less than or equal to $3$.
the number of distinct real roots of $P$ is at least $1$.
$P$ cannot have multiple roots.
$P$ has always at least one multiple root if $n\geq 3$.
Is easy to notice that 3. and 4. are false, but I cannot decide between 1.,2. and 5.
Any hint?
AI: I think 5 is false. For example, $x^{3}+x=x(x^2+1)$ has roots 0,i, -i and no multiple root.
May be statement 2 is correct.
If a is not equal $0$
$x^n + ax + b = 0$
$x^n = -ax -b$
Graph functions $y = x^n$ and line $y= -ax - b$.
X-coordinates of their intersections would be real roots of P.
Maximum number of intersections is 3 if n is odd.
If n is even maximum number of intersections is 2.
Statement 2 is true.
If $a=0$
$x^n= - b$
There are $0$ or $1$ or $2$ distinct real roots.
So statement 2 is true and statement 5 is false. |
H: Question about a proof that The Cantor set is uncountable.
I am reading a proof from a paper I found online, and it goes like this: We want to show that there exists a surjection $f$ from the cantor set $\mathfrak{C}$ to the interval $[0,1]$ then we can show that $|\mathfrak{C}|\geqslant [0,1]$, but since $\mathfrak{C}\subseteq [0,1]$ it follows that $|[0,1]|=|\mathfrak{C}|$. Since the unit interval is uncountable, the Cantor set must be uncountable as well.
I am stuck on one word here:
After showing that for each $p\in \mathfrak{C}$ none of its digits can equal 1, we then define a function $f : C \to [0, 1]$ by taking the number not
consisting entirely of the digits $\{0, 2\}$ (why not entirely?) and replacing each occurrence of $\{2\}$ by $\{1\}$ in its representation. To show that $f$ is surjective, consider any element in $a\in[0, 1]$.
Represent $a$ in its binary form, and then replace each occurrence of $\{1\}$ digit by
a $\{2\}$. This new number, which we label $b$, satisies $f(b) = a$. As a result, $f$ is
surjective. And then it follows that follows that $|[0,1]|=|\mathfrak{C}|$.
I can't seem to see the necessity of the word entirely I can't see why the numbers $1=0.2222...$ and $0=0.000...$ would hurt anything, and $0,1\in\mathfrak{C}$, so if $f$ doesn't consider those numbers then it is an inconsistent mapping from $\mathfrak{C}$ to $[0,1]$. (by inconsistent I mean that $\text{Dom}f\neq \mathfrak{C}$).
AI: I think what “taking the number not consisting entirely of the digits $\{0,2\}$ and replacing each occurrence of $\{2\}$ by $\{1\}$ in its representation” means it the following. Take any number $b\in\mathfrak C$. Then, the ternary representation of $b$ consists entirely of the digits $\{0,2\}$ (i.e., it does not contain any digit $1$). Let $a$ be the number constructed from $b$ in such a way that each digit $2$ in $b$ is replaced by $1$ in $a$ (and you leave the zeros alone). Define $f(b)=a$.
Then, take any $a\in[0,1]$. The aim to show that $a=f(b)$ for some $b\in\mathfrak C$. Consider the binary representation of $a$. Replace all digits $1$ in the binary representation in $a$ by $2$ and view this new number as a ternary representation. Call this new number $b$. Since $b$ consists entirely of the digits $\{0,2\}$, $b\in\mathfrak C$, and $f(b)=a$ by the construction of $f$. Hence, $f$ is surjective.
Upon a closer look, I might have misunderstood the source of the confusion (metaconfusion :-)). You may just need to replace “not consisting” by “consisting.” That is, you just consider those numbers whose ternary representations consist entirely of $\{0,2\}$ The set of such numbers is $\mathfrak C$, issues with uniqueness of representations aside. |
H: Meaning and Intuition behind product $\sigma$ algebras , Folland Definition
I am currently studying Real Analysis using Folland's book by the same name, but I have troubles in understanding the product $\sigma$-algebras definition which is as follows:
Let $\{ X_{\alpha} \}_{\alpha \in A}$ be an indexed collection of nonempty sets, $X = \prod_{\alpha \in A}X_{\alpha}$, and $\pi_{\alpha}: X \rightarrow X_{\alpha}$ the coordinate maps. If $\mathcal{M}_{\alpha}$ is a $\sigma$-algebra on $X_{\alpha}$ for each $\alpha$, the $\textbf{product $\sigma$-algebra}$ on $X$ is the $\sigma$-algebra generated by :
$$\{ \pi^{-1}_{\alpha}(E_{\alpha}):E_{\alpha} \in \mathcal{M}_{\alpha}, \alpha \in A \}.$$
First I don't know what is the definition for the coordinate maps and then what is the intuition or how can I understand more practically product $\sigma$-algebras ?
Thank you!
AI: The coordinate maps are also called projection maps. They send an element of a product set into the $\alpha$-th "coordinate" space. For instance, if $X = A\times B$, then $\pi_A:X\to A$ would be defined by $\pi(a,b) = a$ for all $(a,b)\in A\times B$. Countable products can be thought of similarly, since you can write elements as infinite sequences $(a_1, a_2, \ldots)$. Uncountable products are a little bit more unwieldy.
Let's also use this example to get some feel for the product $\sigma$-algebra. For a finite product $X = A\times B$, where now $A, B = \mathbb{R}$ and each is equipped with the Borel $\sigma$-algebra. Then the product $\sigma$-algebra on $X$ is generated by
$$\{\pi_A^{-1}(E): E\in \mathscr{B}_A\}\cup\{\pi_B^{-1}(F):F\in\mathscr{B}_B\}.$$
For any $E\in\mathscr{B}_A$, we have $$\pi_A^{-1}(E) = E\times\mathbb{R}.$$ Similarly $$\pi_B^{-1}(F) = \mathbb{R}\times F.$$
Since the product $\sigma$-algebra is closed under finite intersections, this means $E\times F \in\mathcal{M}_{A\times B}$. So the product $\sigma$-algebra contains all sets of the form $E\times F$, where $E\in\mathcal{M}_A$ and $F\in\mathcal{M}_B$. (These $\sigma$-algebras are the Borel $\sigma$-algebras in this example.) In particular, taking $A$ and $B$ to be intervals (open, half-open, closed, whatever) we see that $\mathcal{M}_{A\times B}$ contains all rectangles. Since the Borel $\sigma$-algebra on $\mathbb{R}^2$ can be generated by half-open rectangles, this shows you that the product $\sigma$-algebra contains the Borel $\sigma$-algebra. This is one of the things that we expect out of the Lebesgue $\sigma$-algebra on $\mathbb{R}^2$, once we get around to defining it.
This isn't a complete description of the product $\sigma$-algebra. In general, there are many more sets in the product $\sigma$-algebra than just the Cartesian products of measurable sets. But I hope this example gives you some insight into how the product $\sigma$-algebra works - it is not as intimidating as Folland makes (arguably everything) seem. |
H: $f^{\prime}(x)\rightarrow 0$ as $x\rightarrow\infty$. $g(x)=f(x+1)-f(x)$, then $g(x)\rightarrow 0$ as $x\rightarrow\infty$
If $f$ is a defined and $\forall x>0, f^{\prime}(x)\rightarrow 0$ as $x\rightarrow\infty$. $g(x)=f(x+1)-f(x)$, then $g(x)\rightarrow 0$ as $x\rightarrow\infty$.
Note that
$$g(x)=\frac{f(x+1)-f(x)}{(x+1)-x}=\frac{f(x+1)-f(x)}{1}$$
$$\lim_{n\rightarrow\infty} f'(x)=0\implies \forall\varepsilon >0, \exists N>0 \space\space\text{s.t.}\space\space\space |f'(x)|<\varepsilon \space\space\forall x>N$$
$$\left| \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}\right| < \varepsilon \space\space\forall x>N$$
How can I somehow link the $h$ and $1$ together so that $g(x)$ is bounded by $\varepsilon$?
AI: Hint: Mean Value Theorem.${}{}{}{}{}$ |
H: If $P$ is a prime ideal in a commutative ring $R$, is the ideal $P\times P$ a prime ideal in $R \times R$
If $P$ is a prime ideal in a commutative ring $R$, is the ideal $P\times P$ a prime ideal in $R \times R$
AI: As Matemáticos Chibchas points out, it will not be a prime ideal (assuming $R$ has a multiplicative identity), since $$(1,0)\cdot(0,1)=(0,0)\in P\times P,$$ but $1\notin P,$ so $(1,0),(0,1)\notin P\times P$.
However, $P\times R$ and $R\times P$ will be prime ideals of $R\times R$ whenever $P$ is a prime ideal of $R.$ |
H: How to prove $\cos ^6x+3\cos ^2x\space \sin ^2x+\sin ^6x=1$
Prove the following equation.
\begin{eqnarray}
\\\cos ^6x+3\cos ^2x\space \sin ^2x+\sin ^6x=1\\
\end{eqnarray}
I can't prove it by many methods I use.
Please give me some hints.
Thank you for your attention
AI: It seems one sign in your equality is wrong. One actually has
\begin{align}
1=\left(\cos^2x+\sin^2x\right)^3&=\cos^6x+3\cos^4x\sin^2x+3\cos^2x\sin^4x+\sin^6x=\\
&=\cos^6x+3\sin^2x\cos^2x\underbrace{\left(\cos^2x+\sin^2x\right)}_{=1}+\sin^6x=\\
&=\cos^6x+3\sin^2x\cos^2x+\sin^6x.
\end{align} |
H: Idea of how to approach Exponential Functions Questions Please
Can someone help me on how to approach this question?
AI: The model is, to put it nicely, implausible. It assumes that the seeds have been hidden in sets of $100$, perhaps by a mathematically minded squirrel. It also assumes that searching for Type 1 and for Type 2 cannot be done simultaneously. And the answer is obvious, since the good stuff is easiest to find. But let us continue, but not without lodging a complaint at this parody of applied mathematics.
The expected caloric find if we spend the proportion $x_1$ searching for Type 1, and $x_2$ for Type 2, is
$$500x_1^3+300x_2^5.$$
Substituting $1-x_1$ for $x_2$, we arrive at the function
$$V(x_1)=500x_1^3+300(1-x_1)^5.$$
Differentiation yields
$$1500(x_1^2-(1-x_1)^4)=0.$$
So we want to solve $x_1^2=(1-x_1)^4$. Since everything is non-negative, we get equivalently $x_1=(1-x_1)^2$, or equivalently $x_1^2-3x_1+1=0$. This has the solution $x_1=\frac{3-\sqrt{5}}{2}$, and therefore $x_2=\frac{-1+\sqrt{5}}{2}$. Calculation shows that this is a horrible strategy. By using the second derivative, or by examining the first derivative, we see that in fact this gives an absolute minimum caloric intake.
So we need to look at the endpoints. It is clear that best strategy is to look for Type 1 only. |
H: Convergence of a sequence whose even and odd subsequences converge
Suppose $\{a_n\}$ is a sequence such that the subsequences $\{a_{2n−1}\}$ and $\{a_{2n}\}$ converge to the same limit, say $a$. Show that $\{a_n\}$ also converges to $a$.
AI: Hint: Take any $\epsilon>0.$ Since $a_{2n-1}\to a,$ then there is an $N_1$ such that $|a_{2n-1}-a|<\epsilon$ whenever $n>N_1.$ Likewise, there is an $N_2$ such that $|a_{2n}-a|<\epsilon$ whenever $n>N_2.$ How can we use $N_1$ and $N_2$ to come up with an $N$ such that $|a_n-a|<\epsilon$ whenever $n>N$? |
H: Solution of $ 1+ 5 * 2^m =n^2$ Equation and some other question from this equation
I want to find all integer Solutions of $ 1+ 5 * 2^m =n^2$ Equation . From this eqn I have to answer the following answer .
I have to find an expression for $ n^2-1 $
Are $ (n+1) $ and $ (n-1) $ both even or both odd or is one even and the other odd ?
Let $ a = \frac{n-1}{2} $ . I have to find an expression for $ a(a+1) $ .
If a is odd , is $ (a+1) $ even or odd ?
From parts 3 and 4 , is it possible for $ a(a+1) =1 $ or $ a=1 $ ?
I have to find the pnly possible values a can take and then have to find what m and n should be ?
I want to have the solution of the above equation . Then I will answer the above question .
AI: I am considering the case $m\ge 2$
Observe that $5\cdot2^m=n^2-1$ is even and so are $n\pm1$
and $(n-1,n+1)=(n-1,n+1-(n-1))=(n+1,2)=2$
alternatively, if integer $d>0$ divides both $n+1,n-1, d$ will divide $n+1-(n-1)=2$
$\implies (n-1,n+1)|2\implies (n-1,n+1)=2$ as $n\pm1$ are even
So, we have
$$5\cdot2^{m-2}=\frac{n+1}2\cdot \frac{n-1}2$$ with $\left(\frac{n+1}2,\frac{n-1}2\right)=1$
So, either $\displaystyle\frac{n+1}2=5; \frac{n-1}2=2^{m-2}$
or $\displaystyle\frac{n-1}2=5; \frac{n+1}2=2^{m-2}$
or $\displaystyle\frac{n-1}2=5\cdot2^{m-2} ;\frac{n+1}2=1$
or $\displaystyle\frac{n+1}2=5\cdot2^{m-2} ;\frac{n-1}2=1$ |
H: Show that if x divides a power of 2, then x is a power of 2
I'm trying to prove that if $x$ divides $2^a$ for some integer $a \geq 0$, then $x = 2^b$, where $a \geq b$. In other words, if $x$ divides a power of 2, then $x$ is a power of 2. This makes sense, since the all the factors of a power of 2 are also powers of 2, e.g. the factors of 256 are 1, 2, 4, 8, 16, 32, 64, 128, 256. But I'm not so sure how to word this in a rigorous fashion.
A related question: Does this extend to powers of any integer, instead of just 2? If $x$ divides $k^a$, does $x = k^b$?
AI: In answer to the related question, no. Let $k=4$ and $x=8$. $8$ divides $4^2$, but $8$ is no integer power of 4.
This does extend to any prime $k$, however. |
H: How to solve $e^{ax}+e^{bx}+e^{cx}+d=0$
How to solve an equation like $e^{ax}+e^{bx}+e^{cx}+d=0$ (i.e. to write $x=...$) where $a,b,c,d$ are fixed non-zero real numbers.
I have tried assuming that $x=ln(y)$ for $y>0$ but it goes nowhere.
$$e^{ln(y)a}+e^{ln(y)b}+e^{ln(y)c}+d=y^a+y^b+y^c+d=0$$
AI: $\newcommand{\+}{^{\dagger}}%
\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
\newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
\newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
\newcommand{\dd}{{\rm d}}%
\newcommand{\isdiv}{\,\left.\right\vert\,}%
\newcommand{\ds}[1]{\displaystyle{#1}}%
\newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
\newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
\newcommand{\ic}{{\rm i}}%
\newcommand{\imp}{\Longrightarrow}%
\newcommand{\ket}[1]{\left\vert #1\right\rangle}%
\newcommand{\pars}[1]{\left( #1 \right)}%
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\pp}{{\cal P}}%
\newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
\newcommand{\sech}{\,{\rm sech}}%
\newcommand{\sgn}{\,{\rm sgn}}%
\newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
\newcommand{\ul}[1]{\underline{#1}}%
\newcommand{\verts}[1]{\left\vert #1 \right\vert}%
\newcommand{\yy}{\Longleftrightarrow}$
$\large\tt Hint:$
We can set $a \leq b \leq c.\quad$ $d$ must be negative
($d \equiv -\expo{\mu}$). The equation becomes
$$
{\expo{ax} \over \verts{d}}
+
\verts{d}^{b/a - 1}\pars{{\expo{ax} \over \verts{d}}}^{b/a}
+
\verts{d}^{c/a - 1}\pars{{\expo{ax} \over \verts{d}}}^{c/a}
=
1\,,\qquad a \not= 0\,,\quad d < 0
$$
$$
z + z_{b}\,z^{\beta} + z_{c}\,z^{\gamma} = 1
\quad\mbox{where}\quad
\left\vert%
\begin{array}{rcl}
z & \equiv & {\expo{ax} \over \verts{d}}
\quad\yy\quad x = {\ln\pars{\verts{d}z} \over a}
\\[1mm]
z_{b} & \equiv &\verts{d}^{b/a - 1}\,,\quad\beta\equiv {b \over a}
\\[1mm]
z_{c} & \equiv& \verts{d}^{c/a - 1}\,,\quad\gamma\equiv {c \over a}
\\[2mm]
\quad a \not= 0\,,&& d < 0
\end{array}\right.
$$
Since the equation solution requires $\quad 0 < z < 1,\quad$
$0 < z_{b}z^{\beta} < 1\quad$ and $\quad 0 < z_{c}z^{\gamma} < 1,\quad$ it's obvious that a numerical candidate is the
bisection method.
When $a = 0$ (with $b \leq c$), we can repeat the above procedure. Whenever $a = b = 0$ and $c \geq 0$ the analysis is trivial. |
H: Possibility of making diagonal elements of a square matrix 1,if matrix has only 0 or 1
Let $M$ be an $n \times n$ matrix with each entry equal to either $0$ or $1$. Let $m_{i,j}$ denote the entry in row $i$ and column $j$. A diagonal entry is one of the form $m_{i,i}$ for some $i$. Swapping rows $i$ and $j$ of the matrix $M$ denotes the following action:we swap the values $m_{i,k}$ and $m_{j,k}$ for $k = 1, 2 ..... n$. Swapping two columns is defined analogously We say that $M$ is re-arrangeable if it is possible to swap some of the pairs of rows and some of the pairs of columns (in any sequence) so that, after all the swapping, all the diagonal entries of $M$ are equal to $1$.
(a) Give an example of a matrix $M$ that is not re-arrangeable, but for which at least one entry in each row and each column is equal to $1$.
(b) Give a polynomial-time algorithm that determines whether a matrix $M$ with $0-1$ entries is re-arrangeable.
I tried a lot but could not reach to any conclusion please suggest me algorithm for that.
AI: Do a search for Hall's Marriage Theorem. Imagine each row is a man, each column is a woman, each $1$ represents a compatible couple, and the question is whether you can pair off the men and women into compatible couples. Hall's Marriage Theorem gives a simple necessary and sufficient condition.
Then think about $$\pmatrix{1&1&1\cr1&0&0\cr1&0&0\cr}$$ |
H: $S$ is a closed subset of $E'$, then $f^{-1}(S)$ is a closed subset of $E$
Let $E,E'$ be metric spaces, $f: E\to E'$ a continuous function. Show that if $S$ is a closed subset of $E'$, then $f^{-1}(S)$ is a closed subset of $E$.
I need to show that $C(f^{-1}(S))\subset E$ is open. Since $S$ is a closed subset of $E'$ then $C(S)$ is open in $E'$. Since $f$ is continuous then we know that $f^{-1}(C(S))\subset E$ is open. But we need to show that $C(f^{-1}(S))\subset E$ is open, so does $C(f^{-1}(S))= f^{-1}(C(S))$? Are they both equal?
AI: Let $S^c$ be the complement of $S$. Then $S^c$ is open, and therefore $f^{-1}(S^c)$ is open. We will be finished once we show that
$$f^{-1}(S)=(f^{-1}(S^c))^c.\tag{1}$$
The relation (1) is indeed true for any function.
For let $x\in f^{-1}(S)$. Then $f(x)=z$ for some $z\in S$. Thus $x\not\in f^{-1}(S^c)$, since $f$ cannot simultaneously map $x$ to an element in $S$ and to an element in $S^c$. It follows that $x\in (f^{-1}(S^c))^c$.
A similar argument shows that if $x\in (f^{-1}(S^c))^c$, then $x\in f^{-1}(S)$. |
H: $f$ defined on $[1,\infty )$ is uniformly continuous. Then $\exists M>0$ s.t. $\frac{|f(x)|}{x}\le M$ for $x\ge 1$.
$f$ defined on $[1,\infty )$ is uniformly continuous. Then $\exists M>0$ s.t. $\frac{|f(x)|}{x}\le M$ for $x\ge 1$.
I know f uniformly continuous $\implies \forall\varepsilon > 0s.t.\forall x,y\in [1,\infty)$,
$$|f(x)-f(y)|<\varepsilon \space\space\forall |x-y|<\delta$$
By Mean value theorem, $\exists c\in (x,y)\forall x,y\in [1,\infty)$ such that
$$\frac{f(x)-f(y)}{x-y}=f'(c)$$
$$\frac{|f(x)-f(y)|}{|x-y|}=|f'(c)|<M$$ for some $M>0$
$$|f(x)-f(y)|<M|x-y|<M\delta $$
How can I get an expression for $|f(x)|/x$?
AI: Assume that $f(1)=0$ for the moment. Let $\epsilon=1$. Then there is $\delta $ such that
$$|f(x) - f(y)| < 1$$
whenever $|x-y|\leq\delta$. Let $x\in [1, \infty)$. The idea is that you chop the interval $[1, x]$ into small intervals so that you can apply the above inequality. Now there is $n\in \mathbb N$ such that
$$1 + n\delta \leq x < 1 + (n+1)\delta,$$
(Just chopping the intervals) Then write $x_m = 1+ m\delta$,
$$\begin{split} |f(x)| &= |f(x)-f(1)| \\
&\leq |f(x) - f(x_{n})| + |f(x_{n}) - f(x_{n-1})| + \cdots \\
&\ \ \ +|f(x_2) - f(x_1)|+ |f(x_1) - f(1)| \\
&\leq 1 + 1 + \cdots + 1 = n+1 <2n\\
&< \frac{2}{\delta}(1+ \delta n)\\
&\leq Mx\ .
\end{split}$$
By letting $M = \frac{2}{\delta}$ (note that $\delta $ is fixed). Thus
$$\frac{|f(x)|}{x} \leq M\ .$$
In general, if $f(x) \neq 0$, then letting $g(x) = f(x) - f(1)$, then $g$ is also uniformly continuous. Thus there is $M$ such that
$$\frac{|f(x) - f(1)|}{x} = \frac{|g(x)|}{x} \leq M$$
which implies
$$\frac{|f(x)|}{x} \leq M+ \frac{|f(1)|}{x} \leq M + |f(1)|\ .$$ |
H: Calculating sample size
A city Humane Society wishes to determine the life expectancy of adopted dogs from their shelter.
How many of the dog owners do they need to contact in order to be 90% confident of being within 0.29 of the true mean life expectancy? A preliminary survey indicates that the ages are approximately normally distributed with a standard deviation of 1.9.
I've interpreted this question for the following variables:
$$\sigma = 1.9\\
z_c = 1.645\\
E = 0.29$$
And using the below formula:
$$\begin{align}
n & = (\frac{z_c\sigma}{E})^2 \\
& = (\frac{(1.645)(1.9)}{0.29})^2 \\
& \approx (10.78)^2 \\
n & \approx 116.2
\end{align}$$
This is incorrect, however. Why? Is this the incorrect formula to use?
AI: The way I see it, you have a confidence interval:
$(\mu-(1.645)(1.9)/(n)^{1/2}, \mu+(1.645)(1.9)/(n)^{1/2} $, and you want
the radius of the interval to be 0.29 |
H: Solve $\lfloor{x}\rfloor+\lfloor2x\rfloor+\lfloor4x\rfloor+\lfloor16x\rfloor+\lfloor32x\rfloor=12345$
Solve for $x$
$$\lfloor{x}\rfloor+\lfloor2x\rfloor+\lfloor4x\rfloor+\lfloor16x\rfloor+\lfloor32x\rfloor=12345$$
I tried to put $x$=$I$+$f$ where $I$ is integer part and $f$ is fractional part but that didn't work.
AI: As $y-1 < \lfloor y \rfloor \leq y$ you get
$$x+2x+4x+16x+32x -5< \lfloor{x}\rfloor+\lfloor2x\rfloor+\lfloor4x\rfloor+\lfloor16x\rfloor+\lfloor32x\rfloor \\ \leq x+2x+4x+16x +32 x$$
Thus
$$55x-5 < 12345 \leq 55x$$
or
$$11x-1 < 2469 \leq 11x$$
This implies
$$ 224.\bar{45} \leq x \leq 224.\bar{54}$$
Now all you need is to see the intervals for $2x, 4x, 16x, 32x$ and see how many choices you have for each. |
H: Convergence of exponential matrix sum
Let $A$ be an $n\times n$ matrix. Consider the infinite sum $$B=\sum_{k=1}^\infty\frac{A^kt^k}{k!}$$ Each term $\dfrac{A^kt^k}{k!}$ is an $n\times n$ matrix. Does the sum $B$ always converge? (i.e. does the sum for each of the $n^2$ entries always converge?)
AI: This is just $\exp(tA)-I$, the matrix exponential, except that you forgot to start summation at$~0$. The sum always converges for the same reason the ordinary exponential does, namely the $k!$ in the denominator beats the at most exponential growth of each entry of $(tA)^k$. To be precise, each entry of $(tA)^k$ is bounded in absolute value by $(nm)^k$ where $m$ is the maximal absolute value of all entries of $tA$; this can be shown by an easy induction on $k$. |
H: Polynomial convergence to zero
Let $f_k $ be a series of n degree polynomials, that converges to $0$ uniformly in $[-M, M] $ for every $M$.
Say $f_k = a_{(0,k)} + a_{(1,k)}x +... + a_{(n, k)}x^n$
Prove that for every i, $a_{(i, k)} $ converges to 0.
It seems very very obvious to me, but I can't prove it formally.. . Any help will be appreciated!
AI: Let $V$ be the space of polynomials of degree $n$ on $[-M,M]$ with the norm
$$
\|f\|_{\infty} := \sup_{x\in [-M,M]}|f(x)|
$$
Now consider a different norm
$$
\|\sum_{i=0}^n a_ix^i \|_1 := \sum_{i=1}^{n} |a_i|
$$
Since $V$ is finite dimensional, any two norms are equivalent, and so $\exists c >0$ such that
$$
\|f\|_1 \leq c\|f\|_{\infty}
$$
Thus, if $f_n \to 0$ uniformly, then $\|f_n\|_1 \to 0$, which implies that
$$
|a_{i,n}| \leq \|f_n\|_1 \to 0 \quad\forall i
$$ |
H: Conditional Probability - Two methods
There are two methods, A and B, to finish a work.
Method A succeeds with probability $1/3$, but if it fails one tries method B with probability $3/4$ or method A again with probability $1/4$.
Method B succeeds with probability $1/4$, but if it fails one tries method A with probability $2/3$ or method B again with probability $1/3$.
Successive choices are made according to the scheme above until success.
What is the probability that success will be made by method A if
a) one starts with method A ?.
b) one starts with method B ?.
I thought the problem was dealing with a markov chain, but i think it's more related to conditional probability now. Still stuck!
Any help will be greatly appreciated. Thank you very much!
AI: Let $a$ be the probability one will (ultimately) succeed with Method $A$, given that one starts with Method $A$, and let $b$ be the probability one will succeed with Method $B$, given that one starts with Method $B$.
We assume without proof that if we start with $A$, then with probability $1$ we ultimately succeed, and the same holds if we starts with $B$. We have then
$$a=\frac{1}{3}+ \frac{2}{3}\left(\frac{1}{4}a +\frac{3}{4}(1-b)\right),$$
and
$$b=\frac{1}{4} +\frac{3}{4}\left(\frac{2}{3}(1-a) +\frac{1}{3}b\right).$$
Solve for $a$ and $b$. We get $a=\frac{2}{3}$ and $b=\frac{5}{9}$.
Remark: It is a Markov chain problem, but does not require us to deploy the machinery. The two key equations come from conditional probability considerations. For the first equation, suppose we start with Method $A$. There is immediate success with probability $\frac{1}{3}$. With probability $\frac{2}{3}$ we continue. If (probability $\frac{1}{4}$) we use Method $A$, then the probability that (ultimate) success comes with Method $A$ is $a$. If (probability $\frac{3}{4}$) we switch to Method $B$, then the probability success still comes from Method $A$ is $1-b$. |
H: Prove that $(A,*)$ is a monoid
$F$ is a field.
$A$ is defined as the group of all the functions from $\mathbb{N}_0=\{0,1,2,3,...\}$ to $F$.
Let's define a binary operation $*$ on $A$ as follows:
For all $f,g\in A$ the function $f*g\in A$ is defined by: $(n)f*g=\sum_{k=0}^n (k)f(n-k)g$ for every $n\in \mathbb{N}_0$.
Please help me prove that $(A,*)$ is a monoid.
Second part (added lately):
Let us say that a function $f\in A$ has a finite support if the set $\{n \mid (n)f\ne 0\}$ is finite. Let $B$ be the set of all the functions in $A$ with a finite support.
Prove that $B$ is a subsemigroup of $A$.
AI: HINT: We can think of elements of $A$ as infinite sequence of elements of $\mathbf{F}$. If $x=\langle x_n:n\in\Bbb N\rangle$ and $y=\langle y_n:n\in\Bbb N\rangle$, then $x*y=\langle z_n:n\in\Bbb N\rangle$, where
$$z_n=\sum_{k=0}^nx_ky_{n-k}\;.$$
Clearly $A$ is closed under $*$, and it’s not hard to see what the identity element is. HINT: Its first term is $1_{\mathbf{F}}$.
The hardest work is proving that $*$ is associative.
Let $x=\langle x_n:n\in\Bbb N\rangle$, $y=\langle y_n:n\in\Bbb N\rangle$, and $z=\langle z_n:n\in\Bbb N\rangle$ be arbitrary elements of $A$, and let $x*y=\langle u_n:n\in\Bbb N\rangle$ and $y*z=v=\langle v_n:n\in\Bbb N\rangle$. Then the $n$-th term of $(x*y)*z=u*z$ is
$$\sum_{k=0}^nu_kz_{n-k}=\sum_{k=0}^n\left(\sum_{i=0}^kx_iy_{k-i}\right)z_{n-k}\;,$$
the $n$-th term of $x*(y*z)=x*v$ is
$$\sum_{k=0}^nx_kv_{n-k}=\sum_{k=0}^nx_k\left(\sum_{i=0}^{n-k}y_iz_{k-i}\right)\;,$$
and the problem is to show that they are equal. To do this, show that both double summations are equal to
$$\sum\{x_iy_jz_k:0\le i,j,k\le n\text{ and }i+j+k=n\}\;.$$
Added: For each sequence $x=\langle x_n:n\in\Bbb N\rangle\in B$ there is a minimal $m_x\in\Bbb N$ such that $x_n=0$ for all $n\ge m_x$. To show that $B$ is closed under $*$, show that $m_{x*y}\le m_x+m_y$. |
H: Proof of A is orthogonal $\Leftrightarrow \|Ax\|=\|x\|$
I want to proof this property.
A is orthogonal $\Leftrightarrow \|Ax\| = \|x\|$
I tried to elaborate from this, but cannot see how to get any further:
$\|A\| \times \sqrt{<x,x>}$
I really appreciate your answer!!!
AI: You should add to your equivalence
$$
A \quad \text{orthogonal} \qquad \Longleftrightarrow \qquad || Ax|| = ||x|| \quad \text{for all} \quad x \ .
$$
Said this, one implication is clear: if $A$ is orthogonal, then
$$
||Ax||^2 = (Ax)^t(Ax) = x^t(A^tA)x = x^tx = ||x||^2 \ , \quad \text{for all} \quad x \ .
$$
And, since on both sides you have positive numbers, you can delete both squares.
For the other implication, you can take advantage of the computations already done: $||Ax|| = ||x||$ is the same as $x^t(A^tA)x = x^tIx$. Of course, this wouldn't mean that, necessarily $A^tA = I$, if the equality was true just for one or some $x$: you can't just cancel $x$'s on both sides. But, if it's true for all $x$, then you can say more:
First, for any matrix $B$, what's $e_i^tBe_j$? (Here $e_i^t = (0, \dots, 1, \dots, 0)$: just one $1$ in the $i$th coordinate, and the rest are zeros.)
Then you should try to prove this equality (probably you already know it):
$$
x^ty =\frac{1}{2} \left( ||x+y||^2 - ||x||^2 - ||y||^2 \right) \ .
$$ |
H: To derive an inequality of the form $f(r) \leq cr^\alpha$
Let $f$ be a continuous positive function on $[0,1]$ which satisfies
$$ f(s/2) \leq \theta f(s)$$
for all $s\in [0,1]$, where $\frac{1}{2} < \theta < 1$. Can we show that
$$f(s) \leq Cs^\alpha \ \ \forall s\in [0.1]$$
for some positive $C$ and $\alpha\in (0,1)$?
AI: Then $f(1/2^n)\leqslant\theta^nf(1)$ for every $n\geqslant0$.
$(\ast)$ Assume that $f$ is nondecreasing and that $\theta\lt1$.
Then, for every $s$ in $(0,1)$ there exists some $n\geqslant0$ such that $s\leqslant1/2^n\lt2s$, hence $f(s)\leqslant f(1/2^n)\leqslant\theta^nf(1)$. And $2^{n+1}\gt1/s$ hence $\theta^{n+1}\lt\theta^{-\log_2(s)}$, that is, $\theta^{n+1}\lt s^{-\log_2\theta}$.
One gets $f(s)\leqslant Cs^\alpha$ with $C=\theta^{-1}f(1)$ and $\alpha=-\log_2\theta$. Finally, if $\theta\gt1/2$, then $\alpha\lt1$.
Edit: The hypotheses put on $f$ in the question seem to vary a lot (let us hope they finally stabilize), hence one could mention that the same approach works for every function, nondecreasing or not, provided:
$(\ast\ast)$ Assume that $f$ is bounded on $[1/2,1]$ and that $\theta\lt1$.
To see why, replace the step $f(s)\leqslant f(1/2^n)\leqslant\theta^nf(1)$ of the proof above by $f(s)\leqslant\theta^{n}f(2^{n}s)\leqslant\theta^{n}M$ where $M=\sup\limits_{[1/2,1]}f$, valid since $1/2\lt2^{n}s\leqslant1$.
Then $\alpha$ stays the same and $C$ becomes $C=\theta^{-1}M$.
Anyway, the continuity of $f$ is not needed, except to guarantee $(\ast\ast)$. |
H: Find the coefficient of a power of x in the product of polynomials - Link with Combinations?
I came across a new set of problems while studying combinatorics which involves restrictions to several variables and use of multinomial theoram to evaluate the number of possible combinations of the variables subjected to those restrictions.For example
If we were to pick 6 balls out of a bag which contained 3 red , 4 green and 5 white identical balls then the method i've been taught is as follows :
Suppose
$a$ = red balls , $b$= green balls and $c$=white balls .
Now,
$a +b+c =6$ where $0<=a<=3,0<=b<=4,0<=c<=5.$
Now i am supposed to find the coefficient of $x^6$ in the following product :
$(1+x^1+x^2+x^3)(1+x^1+x^2+x^3+x^4)(1+x^1+x^2+x^3+x^4+x^5)$ and that is the solution.
I want to understand what's actually happening here since finding a coefficient seems to have no link with the possible combinations. THANKS.
AI: Consider for a moment the product
$$\left(x^0+x^1+x^2+x^3\right)\left(y^0+y^1+y^2+y^3+y^4\right)\left(z^0+z^1+z^2+z^3+z^4+z^5\right)\;.$$
When you multiply it out, you get a total of $4\cdot5\cdot6=120$ terms, each of the form $x^ay^bz^c$, where $0\le a\le 3$, $0\le b\le 4$, and $0\le c\le 5$.
Now change the $y$’s and $z$’s back to $x$’s to get the original product; when multiplied out, it’s still a sum of $120$ terms, which are now of the form $x^ax^bx^c=x^{a+b+c}$, where $0\le a\le 3$, $0\le b\le 4$, and $0\le c\le 5$. The lowest power of $x$ in the product is clearly $x^0$, and the highest is $x^{3+4+5}=x^{12}$.
Suppose that $0\le n\le 12$; what’s the coefficient of $x^n$ in this product when we collect like terms? It’s simply the number of products $x^ax^bx^c$ for which $a+b+c=n$. In other words, the coefficient of $x^n$ in this product is the number of solutions to the equation $a+b+c=n$ such that $a,b$, and $c$ are integers satisfying the inequalities $0\le a\le 3$, $0\le b\le 4$, and $0\le c\le 5$. In your ball problem that’s the number of distinguishable combinations of $n$ red, green, and white balls that can be drawn from that particular bag. |
H: How find this limit $I=\lim_{n\to\infty}(1+\sin{(\sqrt{4n^2+1}\cdot\pi)})^n$
find this limit
$$I=\lim_{n\to\infty}(1+\sin{(\sqrt{4n^2+1}\cdot\pi)})^n$$
My try: let
$$I=e^{\lim_{n\to\infty}n\sin{(\sqrt{4n^2+1}\pi})}$$
AI: $$\sin((\sqrt{4n^2+1})\pi)=\sin(2\pi n(\sqrt{1+\frac{1}{4n^2}}))=\sin(2\pi n+2 \pi n\frac{1}{8n^2}+O(1/n^3))$$
$$=\sin(\frac{\pi}{4n}+O(1/n^3))=\frac{\pi}{4n}+O(\frac{1}{n^3})$$ |
H: If $a^3 + b^3 +3ab = 1$, find $a+b$
Given that the real numbers $a,b$ satisfy $a^3 + b^3 +3ab = 1$, find $a+b$.
I tried to factorize it but unable to do it.
AI: There are a continuum of solutions to
$$
a^3+b^3+3ab=1
$$
Suppose that
$$
x=a+b
$$
then
$$
\begin{align}
1
&=a^3+(x-a)^3+3a(x-a)\\
&=x^3-3ax^2+3a^2x+3ax-3a^2
\end{align}
$$
which means
$$
\begin{align}
0
&=(x-1)\left(x^2+(1-3a)x+3a^2+1\right)
\end{align}
$$
So either $x=1$ irregardless of $a$, or
$$
x=\frac{3a-1\pm(a+1)\sqrt{-3}}{2}
$$
Thus, other than $x=1$, the only real $x$ is $-2$, which comes from $a=-1$.
That is, the only two real values of $a+b$ are $1$ and $-2$. |
H: Number of models for some theory
Let $\mathcal L = \{ E(\_,\_) \}$ and $T$ be the $\mathcal L$-theory that says that $E$ is an equivalence relation with an infinite number of infinite classes. (I find this statement not clear, because it does not seem to specify the number of finite classes. In the following I assume that all classes are infinite)
I'd like to know how many models with given cardinality there are. I think there is only one model of $T$ of cardinal $\aleph_0$, because there is $\aleph_0$ classes of cardinal $\aleph_0$, hence any two models would be isomorphic by simply identifying the classes. This shows that $T$ is $\aleph_0$-categorical and hence complete.
Now if I look at models $\mathcal M$ of cardinal $\aleph_1$, then the "isomorphism type" of $\mathcal M$ (I don't know if this is a valid terminology) is given by the number $\kappa$ of equivalence classes and for all $i<\kappa$ by the cardinal $\lambda_i$ of the $i$th class. Since $Card(\mathcal M) = \aleph_1$, we have that $\sum_{i<\kappa} \lambda_i = \aleph_1$. If all $\lambda$'s are equal, I know that this sum is $\kappa\times\lambda = \max\{\kappa,\lambda\}$ and there are 3 non-isomorphic models of $T$. But if the $\lambda$'s are different, I don't know how to evaluate the infinite sum of cardinals.
Could you help me finish this computation? Any reference for cardinal arithmetic would be appreciated too!
AI: Let $M$ be the underlying set of $\mathscr{M}$, and assume that $|M|=\omega_1$. For a partition $\mathscr{P}$ of $M$ into infinitely many infinite blocks let $\kappa(\mathscr{P})$ be the number of blocks of cardinality $\omega_1$, and let $\lambda(\mathscr{P})$ be the number of blocks of cardinality $\omega$.
Clearly $\kappa(\mathscr{P})$ can assume any value in $K=\{0,1,2,\ldots,\omega,\omega_1\}$.
If $\kappa(\mathscr{P})=0$, then clearly each block of $\mathscr{P}$ has cardinality $\omega$, so $\lambda(\mathscr{P})=\omega_1$.
If $\kappa(\mathscr{P})>0$, $\lambda(\mathscr{P})$ can assume any value in $K$.
The isomorphism type of $\mathscr{P}$ is completely determined by $\left\langle\kappa(\mathscr{P}),\lambda(\mathscr{P})\right\rangle$. |
H: meaning of math symbol in probability expression?
anyone knows the meaning of the marked symbol in the pic and its Latex syntax?
AI: It is an indicator function that is 1 if its argument is true and 0 if it is false. The latex syntax is $\mathbb{I}$ (\mathbb{I}). |
H: Proving that roots of a quadratic lie between two values
To prove that one of the roots of a quadratic $f(x) = ax^2 + bx + c$ with real coefficients lies between two values $x_1, x_2$ is it enough to prove that:
$$f(x_1) < 0 < f(x_2) $$
Can this be generalized to a polynomial with an arbitrary degree?
AI: Any polynomial, $f$, is a continuous function on $\mathbb{R}$. If there exists an $x_1$ such that $f(x_1)<0$ and an $x_2$ such that $f(x_2)>0$, (and wlog assume that $x_1<x_2$) then it follows by the intermediate value theorem that there exists an $x\in (x_1, x_2)$ such that
\begin{equation*}
f(x)=0.
\end{equation*}
Thus, not only is your statement true for any polynomial, it is true for any continuous function on $\mathbb{R}$. |
H: How to find this integral $I=\int_{-\pi}^{+\pi}\frac{x\sin{x}\arctan{e^x}}{1+\cos^2{x}}dx$?
Find the integral
$$I=\int_{-\pi}^{+\pi}\dfrac{x\sin{x}\arctan{e^x}}{1+\cos^2{x}}dx$$
My try:
let
$$I=\int_{-\pi}^{0}\dfrac{x\sin{x}\arctan{e^x}}{1+\cos^2{x}}dx+\int_{0}^{+\pi}\dfrac{x\sin{x}\arctan{e^x}}{1+\cos^2{x}}dx=I_{1}+I_{2}$$
for $I_{2}$,we let $x=-t$
AI: Putting $x=-t,$
$$I_1=\int_{-\pi}^0\frac{x\sin x\arctan(e^x)}{1+\cos^2x}dx$$
$$=\int_{\pi}^0\frac{(-t)\sin(-t)\arctan(e^{-t})}{1+\cos^2(-t)}(-dt)$$
$$=-\int_{\pi}^0\frac{t\sin t\arctan(e^{-t})}{1+\cos^2t}dt$$
$$=\int_0^{\pi}\frac{t\sin t\arctan(e^{-t})}{1+\cos^2t}dt$$
Now, $\arctan \frac1x=\text{arccot} x=\frac\pi2-\arctan x$
$$\implies I_1=\int_0^{\pi}\frac{t\sin t\left(\frac\pi2-\arctan(e^t)\right)}{1+\cos^2t}dt $$
$$\implies I=I_1+I_2=\frac\pi2 \int_0^{\pi}\frac{t\sin t}{1+\cos^2t}dt $$
$$\text{Applying }\int_a^bf(x)dx=\int_a^bf(a+b-x)dx\text{ on }I \text{ with }a=0,b=\pi$$
$$ I=\frac\pi2\int_0^{\pi}\frac{t\sin t}{1+\cos^2t}dt=\frac\pi2\int_0^{\pi}\frac{(\pi-t)\sin(\pi- t)}{1+\cos^2(\pi-t)}dt=\frac\pi2\int_0^{\pi}\frac{(\pi-t)\sin t}{1+\cos^2t}dt$$
$$\implies I=\frac{\pi}2\left(\int_0^{\pi}\frac{\pi\sin t}{1+\cos^2t}dt-I\right)$$
Can you take it from here? |
H: Is $\mbox{lcm}(a,b,c)=\mbox{lcm}(\mbox{lcm}(a,b),c)$?
$\newcommand{\lcm}{\operatorname{lcm}}$Is $\lcm(a,b,c)=\lcm(\lcm(a,b),c)$?
I managed to show thus far, that $a,b,c\mid\lcm(\lcm(a,b),c)$, yet I'm unable to prove, that $\lcm(\lcm(a,b),c)$ is the lowest such number...
AI: Let the highest power of prime $p$ in $a,b,c$ be $A,B,C$ respectively.
The highest power of prime $p$ in lcm$(a,b,c)=$max$(A,B,C)$
The highest power of prime $p$ in lcm$(a,b)=$max$(A,B)$
The highest power of prime $p$ in lcm$($lcm$(a,b),c)=$max$($max$(A,B),C)$
Can you see max$($max$(A,B),C)=$max$(A,B,C)$ ?
This holds true for any prime that divides at least one of $a,b,c$ |
H: Local sections of Hopf fibration $(S^3,\pi,S^2)$
In the lecture we showed the local triviality of the Hopf fibration $(S^3,\pi,S^2)$ as a principal-$S^1$-bundle by constructing local sections
$$s_1:S^2\setminus\{\infty\}\cong\mathbb{C}\to S^3,\qquad s_1(z):=\frac{1}{\sqrt{1+\lvert z\rvert^2}}\left(z,1\right)$$ and $$s_2:S^2\setminus\{0\}\to S^3,\qquad s_2(z):=\frac{1}{\sqrt{1+\lvert z\rvert^2}}(\lvert z\rvert,\lvert z\rvert).$$
At least that is what I have written in my notes, but in general $s_2(z)\notin S^3$. So, what is a correct section $s_2$?
AI: We want $s_{1}$ and $s_{2}$ to be compatible with each other under "inversion," i.e. $z\mapsto 1/z$, so the right formula for $s_{2}(z)$ would be just $s_{1}(1/z)=\frac{1}{\sqrt{1+|z|^{2}}}(\frac{|z|}{z},|z|)$. |
H: a basic question on continuity
Suppose $f:\Bbb R \to \Bbb R$ which is not injective and continuous. I need to find an open set in the domain which does not get mapped to an open set in the range. how ?
AI: Consider a value $y_0 \in \mathbb{R}$ that gets mapped to by several $x\in \mathbb{R}$. Now one of the following is certainly true: Either there is an interval $[a,b]$ in the domain where all the $x \in [a , b ]$ get mapped to $y_0$. Then we can use $(a,b)$ as the open set we're looking for.
Or there are $x_1 < x_2 \in \mathbb{R}$ such that $f(x_1) = f(x_2) = y_0$ and $f(x) \neq y_0$ for all $x_1 < x < x_2$. Then $f$ either takes a minimum or a maximum on $[x_1,x_2]$ which is not $y_0$ and therefore, $f( (x_1,x_2))$ will also have a maximum or minimum, so it is not open. |
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