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H: Pascal's Triangle Problem?
Guys I have been trying to solve this problem for a long time but cant seem to come up with anything. Problem is essentially a Pascal triangle but I cant figure out how to sum up a column for a specific time, t. Please help. Following is the problem:
Consider a bacterial cell constrained to a one-dimensional environment (some sort of tube, for example). This bacteria reproduce by binary fission: Every hour, each cell divides into two equal daughter cells. Immediately after the division, two daughter cells move away from each other and stop at points x + 1 and x − 1, where x is the original position of their parent cell before the fission.
The process starts with a single cell located at x = 0. We set the clock to t = 0 in the beginning of the experiment. After one hour, at time t = 1, there are two cells: at x = −1 and at x = 1, respectively. The process continues indefinitely.
Let N(t,x) be the number of cells at the position x at the time t. Derive a formula for N(t,x).
Thanks for any help !
AI: It’s clear that $N(t,x)=N(t,-x)$ for all $t\ge 0$ and all $x$. The recurrence is straightforward:
$$N(t+1,x)=N(t,x-1)+N(t,x+1)\;,$$
with initial value $N(0,0)=1$ and $N(0,x)=0$ for $x\ne 0$. The first few values:
$$\begin{array}{c|cccc}
t\backslash x&-4&-3&-2&-1&0&+1&+2&+3&+4\\ \hline
0&&&&&1\\
1&&&&1&0&1\\
2&&&1&0&2&0&1\\
3&&1&0&3&0&3&0&1\\
4&1&0&4&0&6&0&4&0&1
\end{array}$$
Empirically it’s clear that this is just Pascal’s triangle with zeroes inserted between the entries and the horizontal indexing shifted. Fix $t\ge 0$: as $x$ runs from $-t$ to $t$ in steps of $2$, we get the binomial coefficients $\binom{t}k$ for $k=0,\ldots,t$. In other words, we can see that
$$N(t,-t+2k)=\binom{t}k$$
for $k=0,\ldots,t$ and that $N(t,x)=0$ for all other values of $x$. This can be rephrased a bit more nicely: since you get $k$ from $-t+2k$ by adding $t$ and dividing by $2$, we can see that
$$N(t,x)=\begin{cases}
\binom{t}{\frac{x+t}2},&\text{if }x+t\text{ is even}\\\\
0,&\text{otherwise}\;.
\end{cases}$$
To finish the problem, you need only prove by induction on $t$ that this is actually correct. |
H: In graph theory, what is the difference between a "trail" and a "path"?
I'm reading Combinatorics and Graph Theory, 2nd Ed., and am beginning to think the terms used in the book might be outdated. Check out the following passage:
If the vertices in a walk are distinct, then the walk is called a path. If the edges in a walk are distinct, then the walk is called a trail. In this way, every path is a trail, but not every trail is a path. Got it?
On the other hand, Wikipedia's glossary of graph theory terms defines trails and paths in the following manner:
A trail is a walk in which all the edges are distinct. A closed trail has been called a tour or circuit, but these are not universal, and the latter is often reserved for a regular subgraph of degree two.
Traditionally, a path referred to what is now usually known as an open walk. Nowadays, when stated without any qualification, a path is usually understood to be simple, meaning that no vertices (and thus no edges) are repeated.
Am I to understand that Combinatorics and Graph Theory, 2nd Ed. is using a now outdated definition of path, referring to what is now referred to as an open walk? What are the canonical definitions for the terms "walk", "path", and "trail"?
AI: You seem to have misunderstood something, probably the definitions in the book: they’re actually the same as the definitions that Wikipedia describes as the current ones. |
H: Probability of getting 499000–501000 heads if a fair coin is flipped $10^6$ times
A fair coin is flipped $10^6$ times.
What's the probability that the number of heads is at least 499000 and at most 501000?
I'm not sure how to even go about starting this. Does it involve the $Q$ function?
AI: By the central limit theorem, the distribution may be well approximated by a normal distribution. The mean $\mu = 10^6 (1/2) = 500000$, and the variance is $\sigma^2 = 10^6 (1/2) (1/2)$ so that the standard deviation is $\sigma = 500$. You are then asked the probability of being within $\pm 2$ standard deviations of the mean. |
H: $\frac{\partial f(x,y)}{\partial x}=\frac{\partial f(x,y)}{\partial y}$ implies $f(x,y)=g(x+y)$
Assume $f(x,y) \in C^{(1)}(\Bbb{R}^2)$,If $$\frac{\partial f(x,y)}{\partial x}=\frac{\partial f(x,y)}{\partial y}$$ for all $(x,y) \in \Bbb{R}^2$. Show that there exists a function $g(t)$,such that $f(x,y)=g(x+y)$.
AI: Your given condition implies that the directional derivative in the $(1,-1)$ direction is always zero. So $f(x,y)$ is constant on each line $x + y = c$... |
H: Variaton of parameters
Find a particular solution to the differential equation
$-6 y'' + y' + y = t^2 - t - 5 e^{-4 t} $.
This is my work so far.
I know I should be setting things to match up with the right hand side of the differential but I am stuck. I feel that my guess of at^2 + bt + c (e^(-4t)) is incorrect.
Any help would be appreciated. Thanks.
AI: The form of the solution should be $a t^2 + b t + c + d e^{-4t}$. The point being that ($0$ not being a root of the characteristic polynomial) to handle a polynomial of degree $2$ on the right side you need a polynomial of degree $2$ in the solution, potentially with all possible terms. Since $-4$ is also not a root, to handle the the $e^{-4t}$ term you need a constant times $e^{-4t}$. Note that those terms are added, not multiplied.
EDIT: By the way, this is "undetermined coefficients", not "variation of parameters". Variation of parameters would start with finding the basic solutions $e^{t/2}$ and $e^{-t/3}$ of the homogeneous system, and write the solution as
$u_1(t) e^{t/2} + u_2(t) e^{-t/3}$. |
H: How to show that this process is a Markov chain?
This question is from DEGROOT's "Probability and Statistics".
Question:
Suppose that a coin is tossed repeatedly in such a way
that heads and tails are equally likely to appear on any
given toss and that all tosses are independent, with the
following exception: Whenever either three heads or three
tails have been obtained on three successive tosses, then
the outcome of the next toss is always of the opposite type.
At time $n (n ≥ 3)$, let the state of this process be specified
by the outcomes on tosses $n − 2$, $n − 1$, and $n$. Show that
this process is a Markov chain with stationary transition
probabilities and construct the transition matrix.
The answer at the back of the book :
$$\begin{array}{c|lcr}
& \text{HHH} & \text{HHT} & \text{HTH} & \text{THH} & \text{TTH} & \text{THT} & \text{HTT} & \text{TTT} \\
\hline
HHH & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
HHT & 0 & 0 & \frac{1}{2} & 0 & 0 & 0 & \frac{1}{2} & 0 \\
HHH & 0 & 0 & 0 & \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\
THH & \frac{1}{2} & \frac{1}{2} & 0 & 0 & 0 & 0 & 0 & 0 \\
TTH & 0 & 0 & 0 & \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\
THT & 0 & 0 & \frac{1}{2} & 0 & 0 & 0 & \frac{1}{2} & 0 \\
HTT & 0 & 0 & 0 & 0 & \frac{1}{2} & 0 & 0 & \frac{1}{2} \\
TTT & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
\end{array}$$
But I don't know how to show that this process is a Markov chain.
Actually, I can't understand the meaning of the sentence: At time $n (n ≥ 3)$, let the state of this process be specified
by the outcomes on tosses $n − 2$, $n − 1$, and $n$.
Please help me showing this a Markov chain. Thank you.
AI: Say you've tossed the coin ten times and the result is
$$
HHTHTTTHTH
$$
Then view it like this:
$$
HHTHTTT\,\overbrace{HTH}
$$
At that point the state of the Markov chain is $HTH$, i.e. the state is given by the last three tosses. If you then get $T$ on the next toss, then the whole history is
$$
HHTHTTT\,\overbrace{HTH}\,T
$$
and you look at the last three tosses:
$$
HHTHTTTH\,\overbrace{THT}
$$
and the state becomes $THT$. If you get $H$ instead, then the state becomes $THH$. You have probability $1/2$ of either of those outcomes.
In other words, if you've toss the coin $11$ times, so that $n=11$, then $n-1=10$ and $n-2=9$, and the state of the system is determined by the $9$th, $10$th, and $11$th tosses.
Thus from the state $HTH$ one can transition into either of the two states $THH$ or $THT$ with equal probabilities. But if the state is $TTT$ then one necessarily (thus with probability $1$) goes next to the state $TTH$, and if the state is $HHH$ then the next state is necessarily $HHT$. |
H: Convert Parametric form of plane convert to Cartesian, to determine if it is a subspace
I found a linear algebra problem asking to determine if different structures form subspaces.
In one of the examples, it gives a parametric equation of a plane.
It has one constant vector followed by the two parameters, something like this:
Plane: $(x,y,z)= (2,3,4) + t_1(1,1,1) + t_2(1,2,3)$
The answer at the back says that this is a subspace.
This equation does satisfy the condition with $t_1=-1$ and $t_2=-1$, to contain the zero vector, which is the first condition of a subspace.
But I know a theorem or corollary that says that planes of the Cartesian form:
$Ax+By+Cz=0$, go through the origin and are subspaces.
So if I convert my above parametric equation to Cartesian form, I have doubt that
it will be this form, due to the constant.
(NOTE: I put one constraint on this problem, you can't use Cross Product to find the
Normal vector.)
So I guess my question is, if a parametric equation of a plane has the constant vector
$(2,3,4)$ does that imply that it has a constant $D$ in the Cartesian form:
$Ax+By+Cz + D =0$. (which means it does not pass through the origin)
But I can find parameters to make the zero vector.
So what is the relationship between $D$ and the constant vector, if there is any.
AI: But the plane does go through the origin! Take $t_1=t_2=-1$. Yes, $D=0$. |
H: Proving Multiplication by Positive Element
I need help proving this simple fact.
If $a,b,c \in \mathbb{Z}$, all $a,b,c>0$, and $a>b$, then $ca>b$.
How do I prove this?
AI: Observe $cb \geq b$ iff $cb - b \geq 0$ iff $b(c-1) \geq 0$.
This last inequality holds, since $b$ and $c-1$ are both nonnegative.
Similarly, we have $ca > cb$ iff $ca - cb > 0$ iff $c(a - b) > 0$.
This last inequality holds, since $c$ and $a-b$ are both positive (the latter from $a>b$).
Putting it all together: $ca > cb \geq b$, hence $ca > b$. QED |
H: Expected value calculation.
Let $E(h,k,i,j)$ , $i \leq h$,$j\leq k$ be the expected number of the common numbers of two independently chosen subsets of a set of $h$ elements and $k$ elements respectively, where the first and second subsets respectively contain $i$ and $j$ elements. How to calculate $E(h,k,i,j)$ for two possible case here $h\leq k$ and $k\leq h$? Here is one example shown in figure .
In this scenario how can I calculate expected number of overlap or common numbers which are indicated by T,T' in the figure? Here T is chosen from (5*9 = 45) grid and T' is chosen (7*13=91) grid.
AI: If we assume that $H=\{1,2,\dots,h\}$ and $K=\{1,2,\dots,k\}$, and if $i$ elements are chosen (uniformly and without replacement) from $H$ and $j$ elements are chosen (uniformly and without replacement) from $K$, and if $E(h,k,i,j)$ is the expected number of elements common to both samples, then $E(h,k,i,j)=\dfrac{ij\min(h,k)}{hk}=\dfrac{ij}{\max(h,k)}$; this is obtained by adding, for each $n\in\{1,2,\dots,\min(h,k)\}$, the probability that $n$ is chosen in both samples.
Is that what you meant? |
H: Find two vector b such that Ab = b
I have a question which I am unable to solve.
I have this matrix:
$
A = \begin{bmatrix}
1 & -1 & 1 & -1\\
1 & 1 & 1 & 1\\
1 & 2 & 4 & 8\\
1 & 3 & 9 & 27
\end{bmatrix}
$
I have to find two vector $\bf{b}$ such that $A\bf{b}=\bf{b}$
I know that the vector $0$ is one of the answer but I don't know how to find a non zero vector.
Is it possible for you to lead me in the right direction and help me out?
AI: $
A = \begin{bmatrix}
1 & -1 & 1 & -1\\
1 & 1 & 1 & 1\\
1 & 2 & 4 & 8\\
1 & 3 & 9 & 27
\end{bmatrix}
$
Let $
b = \begin{bmatrix}
p \\
q \\
r \\
s
\end{bmatrix}
$
Then,we get
$
\begin{bmatrix}
1 & -1 & 1 & -1\\
1 & 1 & 1 & 1\\
1 & 2 & 4 & 8\\
1 & 3 & 9 & 27
\end{bmatrix}
$
$
\begin{bmatrix}
p \\
q \\
r \\
s
\end{bmatrix}
$
=$
\begin{bmatrix}
p \\
q \\
r \\
s
\end{bmatrix}
$
You will get equations as
$-q+r-s=0$
$p+r+s=0$
$p+2q+3r+8s=0$
$p+3q+9r+26s=0$
Solve the equations and find unknowns
It has trivial solution i.e $p=q=r=s=0$
Hence $
b = \begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}
$ |
H: Question about the essential supremum
Say $f: E \to \bar{\mathbb{R}} $ is measurable function. Define the Essential Supremun $esssup f$ as $ \inf \{z : f \leq z \; \text{a.e.} \} $
a.e. means almost everywhere which means everywhere except on sets of measure zero.
MY question is: While studying the concept I have made the following remarks based of the definition. Are they correct? thanks in advanced for any generous response.
THe $esssup f $ can be $\infty$ and this happens if $\{z : f \leq z \; \text{a.e.} \} = \varnothing $, which occurs when $ f > z$ a.e. Also, $esssup f$ is $- \infty$ if $\{z : f \leq z \; \text{a.e.} \} = \bar{\mathbb{R}} $ which occurs when $f \leq z$ everywhere.
AI: Your remarks are ill-founded: In "$\{z:f\leq z \mbox{ a.e.}\}=\emptyset$, which occurs when $f>z$ a.e.", you need to quantify the second $z$. Also, the negation of $f\leq z$ a.e. is not $f>z$ a.e., it's $f > z$ on a set of measure greater than zero.
You should instead say something like,
The essential supremum of $f$ can be $\infty$, and this happens if $\{z:f\leq z\mbox{ a.e.}\} = \emptyset$, which occurs if for every $z$, there exists a set $E$ with $\mu(E)>0$ and $f|_E > z$.
Similar remarks hold for your second point. |
H: Determining whether a subspace of the metric space of real sequences is separable
Let $$X=\left\{(a_n)_{n \in \mathbb N} \in \mathbb R^N : \exists n_0 \in \mathbb N\, \forall n\ge n_0 \big(a_n\le \sqrt{n}\big)\right\}$$ with the metric $$d\big((a_n)_{n \in \mathbb N},(b_n)_{n \in \mathbb N}\big)=\sup_{n \in \mathbb N}\frac{|a_n-b_n|}{n}$$
Determine if $(X,d)$ is separable
Prove that for every Cauchy sequence in $X$ there exists a sequence of real terms such that the Cauchy sequence converges to that one but that $X$ is not complete.
Can anyone give me a hint?
AI: With this definition of $X$, the function $d$ isn’t defined on all of $X\times X$: for example, if $z$ is the zero sequence and $x=\langle -n^2:n\in\Bbb N\rangle$, then $z,x\in X$, but $$\sup_{n\in\Bbb N}\frac{|0-(-n^2)|}n=\sup_{n\in\Bbb N}n$$ doesn’t exist. I’m going to assume that the last condition in the definition of $X$ was supposed to be that $|a_n|\le\sqrt{n}$. Then if $x=\langle x_n:n\in\Bbb N\rangle,y=\langle y_n:n\in\Bbb N\rangle\in X$, then there is an $m\in\Bbb N$ such that $$\frac{|x_n-y_n|}n\le\frac{2\sqrt{n}}n=\frac2{\sqrt{n}}$$ for all $n\ge m$, and $d(x,y)$ is defined.
Consider rational sequences that are eventually $0$.
For $n,k\in\Bbb N$ let $$x_k^{(n)}=\begin{cases}2\sqrt{k},&\text{if }k\le n\\0,&\text{if }k>n\;,\end{cases}$$ and let $x^{(n)}=\left\langle x_k^{(n)}:k\in\Bbb N\right\rangle$. Show that $\left\langle x^{(n)}:n\in\Bbb N\right\rangle$ is a Cauchy sequence in $X$ that has no limit in $X$. There is a sequence $x\in\Bbb R^{\Bbb N}\setminus X$ such that $d\big(x^{(n)},x\big)$ is defined for each $n\in\Bbb N$ and converges to $0$ as $n\to\infty$; what is it?
More generally, show that if $\left\langle x^{(n)}:n\in\Bbb N\right\rangle$ is any Cauchy sequence in $X$, where $x^{(n)}=\left\langle x_k^{(n)}:k\in\Bbb N\right\rangle$ for each $n\in\Bbb N$, then $\left\langle x_k^{(n)}:n\in\Bbb N\right\rangle$ is a Cauchy sequence in $\Bbb R$ for each $k\in\Bbb N$. Use this observation to show that there is an $x\in\Bbb R^{\Bbb N}$ such that $d\big(x^{(n)},x\big)$ is defined for each $n\in\Bbb N$ and converges to $0$ as $n\to\infty$. |
H: Calculus remark I forgot.
Suppose we have a set $X$. Let $a = \inf X$. then for all $n$, $a + \frac{1}{n} \in X$.
maybe this is wrong anyway I am trying to show it. So, we know $a \leq x $ for all $x \in X$. By archimidean, can find $n$ such that $a + \frac{1}{n} \leq x$. Im stuck here. Any help would be greatly appreciated. thanks in advance
Suppose $X \neq \varnothing$ and $X \subseteq \mathbb{R}$
AI: If $X$ is a nonempty set of real numbers with a lower bound, then $a=\inf X$ satisfies:
For all $x\in X$, $a\leq x$.
For each positive integer $n$, there exists $x\in X$ such that $x<a+\frac{1}{n}$.
However, you cannot conclude that $a+\frac{1}{n}$ is in $X$, only that it is not a lower bound for $X$, which follows from the definition of the infimum as the greatest lower bound of $X$. For example, $X$ might only contain one element. |
H: Big O estimate of simple while loop
Give a big-O estimate for the number of operations, where an operation is an addition or a multiplication, used in this segment of an algorithm (ignoring comparisons used to test the conditions in the while loop).
i := 1
t := 0
while i ≤ n
t := t + i
i := 2i
My attempt:
n = 1 i=2
n = 2 i=4
n = 3 i=8
n = 4 i=16
relationship of i to iteration is i = 2^n
How many iterations(n’) until 2^(n’) > n (basically solving for n')
n’ > log2(n) thus the big O estimate is:
O(log_2(n)) (read as log base 2 of n)
However, the book says it's O(log(n)) - why isn't it base 2?
AI: $O(\log_2(n))$ and $O(\ln{n})$ are the same thing, since $\log_2$ and $\ln$ are related by the formula
$$\log_2{n} = \frac{\ln{n}}{\ln{2}} \approx 1.44 \ln{n}$$
The multiplicative constant is irrelevant for the Big O notation.
More precisely, we have the relations
$$1.44 \ln{n} \le \log_2{n} \le 1.45 \ln{n}$$ |
H: how to construct a homeomorphism between $T$ and $S$?
$S=\{x^2+y^2\le 1\}$, $T=\{|x|+|y|\le 1\}$, could anyone tell me how to construct a homeomorphism between $T$ and $S$?
AI: Hint: We can consider both $S$ and $T$ in $\mathbb C$ as $S=\{z:|z|\leq 1\}$ and $T=\{z:z=a+ib,|a|+|b|\leq 1\}$, then define $f:T\rightarrow S$ by $f(x)=re^{i\theta},$ where $r$ is the unique $r\in [0,1]$ such that $x\in \{z:z=a+ib,|a|+|b|=r\}$, and $\theta$ is the argument of $x$. |
H: Is a space where only finite subsets are compact sets always discrete?
If in a topological space only finite subsets are compact sets, is it then the discrete topological space?
Thank you.
AI: A topological space is anticompact if all compact sets are finite. In this answer Stefan H pointed out that an uncountable set with the cocountable topology is an example of an anticompact $T_1$-space which is not discrete. The real line, with the topology generated by the usual open sets and the cocountable sets, is an example of an anticompact Hausdorff space which is not discrete.
The set $\mathbb N$ of natural numbers, with the topology consisting of the sets not containing $1$ and the sets of asymptotic density $1$, is an example of a countable zero-dimensional Hausdorff space which is anticompact but not discrete. This is the Appert space.
P.S. In the last example, instead of taking the sets of asymptotic density $1$ as neighborhoods of the point $1$, we could take the sets $U$ such that $\sum_{n\notin U}\frac1n\lt\infty$; maybe this example is slightly easier or more familiar. |
H: upper bound for a function f
Let $f(z)=z^3e^{z^2}-e^{3z^2}$.
For $R>0$, I need an upper bound for
$$\max_{z\in\bar{B(0,R)}}|f(z)|,$$
which will look nice if I take its natural logarithm, that is
$$\ln|\{\max_{z\in\bar{B(0,R)}}|f(z)|\}|.$$
Hope you can help. Thanks!
AI: Note that if $a$ and $b$ are positive numbers, then $e^a+e^b<e^{a+b}+1$, and $a<e^{a}-1$. Thus
$$R^3e^{R^2}+e^{3R^2}=e^{R^2}(R^3+e^{2R^2})<e^{R^2}(e^{R^3}+e^{2R^2}-1)<e^{R^2}e^{R^3+2R^2}=e^{R^3+3R^2}.$$
This can be improved using $1+\dfrac{x^3}{3!}<e^x$ for $x>0$, so that $\left(\dfrac{x}{\sqrt[3]{6}}\right)^3<e^x-1$, which implies that $x^3<e^{\sqrt[3]{6}x}-1$. Thus another bound that works is $e^{3R^2+\sqrt[3]{6}R}$. |
H: When is $\sum_{k=1}^{\infty}a(k) \sum_{k=1}^{\infty}b(k)\ge \sum _{k=1}^{\infty}a(k)b(k)?$
Given two convergent series $\sum_{k=1}^{\infty}a(k)$ and $\sum_{k=1}^{\infty}b(k)$ under what conditions is valid the inequality:
$$\sum_{k=1}^{\infty}a(k)\sum_{k=1}^{\infty}b(k)\ge\sum_{k=1}^{\infty}a(k)b(k)$$
Thanks.
AI: A sufficient condition is for the "cross-terms" $a(i)b(j)$, $i\ne j$ to be positive, which means that $\text{sgn}(a(i))=\text{sgn}(b(j))$, for all $i,j$. |
H: Integral over X as supremum of integrals over finite subsets of X.
I am trying to prove that
$$\int f d\mu = \sup \left\lbrace \int_E f d \mu, E \in S, \mu(E)< \infty \right\rbrace,$$
given that $\int f d\mu < \infty$. $f$ is a positive measurable function $f:X\rightarrow \mathbb{R}$. $S$ is $X$'s $\sigma$-algebra.
Attempt:
Since for all $E \in S$, it is true that $E \subset X$, then we get $f\chi_E\le f$, from which $$\int_E f \le \int f .$$
It follows that: $$\sup \left\lbrace \int_E f d \mu, E \in S, \mu(E)< \infty \right\rbrace \le \int f d\mu.$$
I want to show that:
$$\int f d\mu \le \sup \left\lbrace \int_E f d \mu, E \in S, \mu(E)< \infty \right\rbrace.$$
By exhibiting $E\in S$ satisfying: $\mu(E)<\infty,$ such that $\int f \le \int_E f$, as I would then have both inequatities, proving the equality.
I am having trouble trying to use the hypothesis: $\int f d\mu < \infty$. I tried constructing a sequence of functions $f_n$ whose limit-inferior was $f$ in order to apply Fatou's lemma, but I don't know if $\mu(X)<\infty,$ so this approach lead me nowhere.
Anyone got any ideas on how to proceed?
Thanks in advance.
AI: You can work directly from the definition.
Since $f$ is integrable, we have a sequence $s_n$ of simple measurable functions such that $0 \le s_n \le f$ and $\int s_n \to \int f$.
Each $s_n$ has the form $s_n = \sum \alpha_k 1_{A_k}$, where $\alpha_k > 0$ and the$A_k$ are measurable, $\mu A_k < \infty$ and, without loss of generality, disjoint. Let $E_n = \cup A_k$. Then $\mu E_n < \infty$. Then we have $s_n \le f \cdot1_{E_n}$, and hence $\int s_n \le \int f \cdot1_{E_n} = \int_{E_n} f \le \int f$. The result follows, since $\int s_n \to \int f$.
(Note: The fact that $\mu A_k < \infty$ follows from $\int s_n = \sum_k \alpha_k \mu A_k \le \int f < \infty$, and $\alpha_k >0$.)
Addendum: We have $\sup \{ \int_E f | \mu E < \infty \} \ge \int_{E_n} f$. Since $\int_{E_n} f \to \int f$, we have $\sup \{ \int_E f | \mu E < \infty \} \ge \int f$, as required. |
H: Trouble finding the Laurent Expansion .
I'm having trouble progressing through (in my experience) the tedious calculations required to obtain a Laurent Expansion of a complex function.
The problem arises in finding the series within the annulus. Thank you in advanced for your help.
AI: Hint for (a): where are the singularities of $f(z)$?
Hint for (b): partial fractions. |
H: How to calculate total profit in a set amount of days with a 2% daily gain?
I want to know if there is a formula that would give me the total amount at the end of a 365 day period with a 2% daily gain. Adding 365 days up would take for ever.
AI: Yeah, it's called compound interest. Note that if you have some amount of money $x$, and it grows by 2%, that's the same as multiplying: $x \cdot 1.02$. If it does that for 365 days in a row, then you multiply $1.02$ by itself 365 times = $1.02^{365}$ and multiply the end result by $x$.
In short: if you start with $x$ amount of money and your pile of money grows by $2\%$ a day, then at the end of the year, you'll have $x \cdot 1.02^{365} = x \cdot 1377.$
EDIT: If you want to know the pure profit, you will have to subtract from that result the amount of money you started with: so you'll have made a profit of $(x \cdot 1377) -x$. |
H: How is $\frac{\big(\frac{3}{2}\big)^{99}-1}{\big(\frac{3}{2}\big)^{100}-1}\approx\frac{1}{\big(\frac{3}{2}\big)}$
I read somewhere that $$\frac{\big(\frac{3}{2}\big)^{99}-1}{\big(\frac{3}{2}\big)^{100}-1}\approx\frac{1}{\big(\frac{3}{2}\big)}$$I don't know how to have it. Please let me know how this is approximated.
AI: Another point of view:
$$(3/2)\left((3/2)^{99}-1\right)=(3/2)^{100}-(3/2)=(3/2)^{100}-1-0.5$$ |
H: Help finding the interior of a set
Let $$S=\{(x,y,z) \in \mathbb{R}^3|\,\,\, 0<x<1,\,\, y^2+z^2 \le1 \}$$
I know this might seem very simple but I don't understand the geometric representation of it my 3D graphing is really rusty right now. My guess is that the interior would be $$int(S)=\{(x,y,z) \in \mathbb{R}^3|\,\,\, 0<x<1,\,\, y^2+z^2 <1 \}$$
Is this the right way to approach it? Thanks in advance.
AI: I’m not going to prove anything, but I’ll try to describe the region in such a way that you can sort it out.
Ignore $x$ for a moment: the equation $y^2+z^2=1$ gives you the unit circle in the $yz$-plane, so the inequality $y^2+z^2\le 1$ gives you the closed unit disk in the $yz$-plane. If you now allow $x$ to be arbitrary, you get the solid closed infinite cylinder of radius $1$ whose axis is the $x$-axis in $\Bbb R^3$. You’re chopping that off at the planes $x=0$ and $x=1$, which are not included, so you’re getting the part of the cylinder lying strictly between those planes. The cylinder’s ends are not included, but the lateral (cylindrical) side is; it’s the only part of the solid that isn’t in the interior. |
H: Convergence to a delta distribution
Is it Okay to say that when I have a probability density $P(x;\mu,\sigma)$, with $\mu$ the first moment of the probability density and $\sigma$ the square root of the second central moment of the probability density that the following relation holds:
$\lim\limits_{\sigma\rightarrow0}P(x;\mu,\sigma) = \delta(x-\mu)$, where $\delta(x-\mu)$ is my delta-distribution centered around $\mu$?
AI: Indeed, $P[A;\mu,\sigma]\to1$ if $\mu$ is in $A$ and $P[A;\mu,\sigma]\to0$ if $\mu$ is not in $A$. If this is how you define $\delta$, the limit holds. |
H: Is this an accurate way to represent n! using Π?
I recently learned of the $\Pi$ symbol, and was wondering if the following is an accurate way to represent $n!$:
$\Pi_{i=0}^{n-1} n - i$
AI: Yes, apart from the missing parentheses: make it $$\prod_{i=0}^{n-1}(n-i)\;,$$ and you’ll be fine. As $i$ runs from $0$ up through $n-1$, $n-i$ runs from $n$ down through $n-(n-1)=1$, which is exactly what you want. |
H: How to get the transformation?
Let T be the transformation of the xy-plane that reflects each vector through the x-axis and then doubles the vector's length. If A is the 2*2 matrix such that
then A=. I don't know how to get this answer.Please help me, thank you.
AI: You remember the first week of linear algebra when they told you that linear transforms are the type where $f(ax + by)$ = $a\cdot f(x) + b \cdot f(y)$ ? Well that's how you find the matrix that represents a linear transform.
Suppose $e_1=(1, 0)$ is your unit vector on the x-axis and $e_2=(0, 1)$ is your unit vector on the y-axis. Then any vector $(x, y)$ is the same as $xe_1 + ye_2$.
So your transform is $T(\begin{bmatrix} x \\ y \end{bmatrix}) = \text{flipped then scaled}$. Assume for now that it's a linear transform, so that means it has a matrix.
$T(\begin{bmatrix} x \\ y \end{bmatrix}) = T(xe_1 + ye_2) = xT(e_1) + yT(e_2) = \begin{bmatrix} T(e_1) & T(e_2) \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}$
So what happens to $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ when you flip it and double it? You get $\begin{bmatrix} 2 \\ 0 \end{bmatrix}$.
What happens to $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ when you flip and double it? You get $\begin{bmatrix} 0 \\ -2 \end{bmatrix}$.
So $T(\begin{bmatrix} x \\ y \end{bmatrix}) = \begin{bmatrix} 2 & 0 \\ 0 & -2 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}$. |
H: 3 Intersecting lines point, triangulation.
I currently have 3 circles that intersect each other. At these points I create a line, although I am stumped at how I can find the point of which all 3 lines intersect?
I know the coordinate center of each circle as well as its radius and each of the lines start and end coordinate points.
I have tried following some other intersecting line posts and maths but haven't managed to fully understand the principles behind them.
Hope someone could find a solution and explain the process please?
Thanks,
Marc.
AI: Subtracting the equations of two circles gives a linear equation. |
H: Are these two surfaces topological equivalents?
I believe they are not the same since after deforming both, we get a surface with 5 holes in in the first picture, and a picture with 4 holes in the second picture. What do you think?
AI: The first seems to be the connected sum of five tori ($S^1\times S^1$), while the second of four. |
H: Can an infinite permutatation be decomposed into finite number of infinite cycles?
Let $\sigma \in Perm(\mathbb{N})$ the set of permutations on the naturals. Then can $\sigma$ be written as a finite composition of possibly infinite disjoint cycles?
AI: Not necessarily. Consider the permutation given by
$$\sigma(n)=\begin{cases}
n+1,&\text{if }n\text{ is even}\\
n-1,&\text{if }n\text{ is odd}\;;
\end{cases}$$
it’s $(01)(23)(45)\ldots\;$. |
H: For each of the following functions, what do the first- and the second-order optimality conditions say about wether 0 is a minimum on $\mathbb{R}$.
For each of the following functions, what do the first- and the second-order optimality conditions say about wether 0 is a minimum on $\mathbb{R}$.
$f_1(x)=x^2$
$f_2(x)=x^3$
$f_3(x)=x^4$
$f_4(x)=-x^4$
So the dazzling thing is that the second derivative for all these functions gives $0$ when $0$ is input, i.e. $f_1''(0)=f_2''(0)=f_3''(0)=f_4''(0)=0$. Since the Hessian is given by the second order derivatives, and since these derivatives determine wether it is positive definite, negative definite, indefinite or singular. And since these properties determine wether a critical point is a minimum, maximum, saddle point or some other pathological situation. I get confused.
So all functions are neither a maximum, minumum or saddle point at $0$? Why would the book ask a silly question? (Im have a strong guess that I am doing things wrong here).
AI: You'll need do consider higher-order derivatives, if the Hessian is indefinite. In this case
but $$f_1'' (0) = 2 \neq 0$$
So you made a mistake there.
For $f_3, f_4$ you have
$$f_3''(0) = f_3'''(0) = f_4''(0) = f_4'''(0) = 0$$
And
$$f_3^{(4)}(0) = 24 = -f_4^{(4)}(0)$$
So you get local (global) minima / maxima there but nothing can be derived from $f_3''(0)$ and $f_4''(0)$ |
H: Negate the following sentence: $\forall x, |x−a| < δ \Rightarrow |f(x)−L| < \varepsilon$
Negate the sentence: $\forall x, |x−a| < δ \Rightarrow |f(x)−L| < \varepsilon$
For my negation I got:
$\exists x, |x-a| < δ \Rightarrow |f(x)-L| \geq \varepsilon$
Would that be correct?
AI: You are correct that $\lnot(\forall x P(x))$ is $\exists x \lnot P(x)$, but your negation of the implication is incorrect. The negation of $P \implies Q$ is $P \land \lnot Q$. See for example this truth table:
P Q P ⇒ ¬Q P ∧ ¬Q
============================
T T F F
T F T T
F T T F
F F T F
In this case, the negation should read in plain English as something like "there is an $x$ such that $x$ is $\delta$-close to $a$ and $f(x)$ is at least $\varepsilon$ away from $L$." |
H: Filling 4l, 5l bottles from two 10l bottles
There are two bottles of 10litre each filled with water.
Now two persons having empty bottles of 4litre and 5litre want to take 2litres of water each from the previous 10litre bottles..
Now you have to pour the water without wasting or throwing it.
This is an question in a previous placement paper. I was not getting how to proceed.
Can anyone give me some hint....
AI: Call containers by $A=10,B=10,C=5,D=4$.
Start by pouring $A$ into $C$, then $B$ into $D$. Pour $D$ into $A$, then $C$ into $B$, then $C$ into $D$. Now you have $1$ in $D$. Pour from $A$ into $C$, then from $C$ into $D$, leaving $2$ in $C$. Now pour from $D$ into $A$, then $B$ into $D$, then $D$ into $A$ again, leaving $2$ in $D$.
Alternate recipe (among many), uses "front loading":
Pour $A$ into $D$, $D$ into $C$, $A$ into $D$, $D$ into $C$, $C$ into $A$, $D$ into $C$, $B$ into $C$, $C$ into $A$ (note $C$ contains $2$), $A$ into $D$, and $D$ into $B$ and done.
The strategy in the first example was a sort of guess and check. In the second example, I focused on building appropriately-sized "holes" in the larger containers, which made it very easy. |
H: $K$ is a three digit number such that the ratio of the number to the sum of its digits is least.Find the difference between the hundreds and the tens…
$K$ is a three digit number such that the ratio of the number to the sum of its digits is least. What is the difference between the hundreds and the tens digits of $K$
a) 9
b) 8
c) 7
d) None of these
I was able to solve this question by using hit and trial but could not think of a proper way to solve it.
AI: We want to minimise $\cfrac {100a+10b+c}{a+b+c}$
This is equal to $$1+\frac{99a+9b}{a+b+c} $$ and this is clearly least when c is greatest. So we have $c=9$.
We then rewrite the fraction as $$100-\frac{90b+99c}{a+b+c}$$ which is least when $a$ is least. So we have $a=1$.
Isolating $b$ as above gives $$10+\frac {90a-9c}{a+b+c}$$ Since $a$ is non-zero the numerator is positive and the minimum is obtained when $b=9$. |
H: A function satisfy $x\frac{\partial f(x,y)}{\partial x}+y\frac{\partial f(x,y)}{\partial y}=0$ in a convex domain implies it is a constant
Assume $D \subset \mathbf{R}^2$ is a convex domain which contain original point. $f \in C^{(1)}(D)$,if $$x\frac{\partial f(x,y)}{\partial x}+y\frac{\partial f(x,y)}{\partial y}=0, \left((x,y)\in D\right)$$
show that $f(x,y)\equiv \text{constant}$
AI: Given $(x,y)\in D$, it suffice to show that $f(x,y)=f(0,0)$. Define
$$g:[0,1]\to \Bbb R,\quad g(t)=f(tx,ty).$$
Then for every $t\in (0,1)$,
$$t\cdot g'(t)=tx\cdot\frac{\partial f}{\partial x}(tx,ty)+ty\cdot\frac{\partial f}{\partial y}(tx,ty)=0\Longrightarrow g'(t)=0,$$
so $g$ is constant on $[0,1]$. The conclusion follows. |
H: What figure does one obtain from a Möbius band if one shrinks the boundary circle to a point?
'Im trying to solve the following problem:
What figure does one obtain from a Möbius band if one shrinks the boundary circle to a point?
I don't really quite understand the problem. What does it mean the boundary circle of Möbius band? Can someone explain me what is this question asking for? thanks in advance.
AI: If you make a cylinder with a strip of paper, there will be two edges, a top and bottom. If you make a Möbius band, these two edges will just be a single contiguous edge, which is what is meant by "boundary circle".
The easiest way to see what happens when you contract this circle is to draw a rectangle with arrows to indicate how two opposite edges will be glued together. Now squeeze the two sides that aren't being identified, until you have a circle with two marked points (though they are identified in the gluing), and two arrows on either side going in opposite directions.
So now it's clear: you get a disc, with its boundary glued along the antipodal map. In other words, you get the real projective plane. |
H: Sinusoid sum of cosine and sine
I am studying Fourier series right now. I asked a question before of math.statckexchange regarding Fourier series. This question is related and hopefully quite simple:
Generally Fourier series works because a sinusoid can be recomposed from a linear combination of a cosine and sine. In a book (Fourier Analysis Stein & Shakarchi) that was kindly pointed to me by someone here on Stack there is this definition:
If you can't access the image, I will reproduce the formula:
$a \cos(ct) + b \sin(ct) = A\cos(ct - \phi)$
What I like is that it says this can be easily verified. $\phi$ is the phase, A the amplitude. However considering I am trying to learn maths myself, I was wondering if someone could confirm my findings (at least I tried to "verify" it myself).
I used trigonometry identity:
$A\cos(ct - \phi) = A\cos(ct) * \cos(\phi) + A\sin(ct) * \sin(\phi)$
Because $\phi$ is a constant then we can write $a = A * cos(\phi)$ and $b = B * sin(\phi)$.
I would like to know if this is correct? Also can this be used as a proof (?) that any sinusoid can be recomposed from a combination of cosine and sine functions?
Thanks a lot.
AI: What you do is correct so far as it goes -- if you know $A$ and $\varphi$, you can find corresponding $a$ and $b$ in the way you describe.
Note, however, that this is not what the part you quote from the book does. (And what you quote is not a definition, by the way). The book says you can go in the other direction: If you already know $a$ and $b$, it is always possible to find matching $A$ and $\varphi$. That is not much more difficult to prove, though it is slightly trickier because $\varphi$ will not be uniquely determined. |
H: General solution to a recursive equation
What is the general solution of the following recursive equation?
$$N(t)=(1+f)\cdot\left(N(t-1)-N(t-T)+N(t-T-1)\right)$$
By "general solution" I mean an equation where $N(t)$ stands alone on the left hand-side and is expressed only in terms of f and T only.
This recursive equation has been written for a case-study in biology where $N(0)=1$. Also, I guess that we should consider that $N(any.negative.number)=1$ (although it makes no sense for my study-case to have a negative value in the N(..) function) so that when t is lower than T, the equation simplifies into $N(t)=(1+f)\cdot N(t-1)$
Let me know if my question makes sense and if all informations to solve this equation are given.
AI: Here is a "constructive" attempt. Sorry if it is long.
First of all, let
$$X(k)=\left[\begin{array}{c}N(t)\\N(t+1)\\\vdots\\N(t+T)\end{array}\right], X(k+1)=\left[\begin{array}{c}N(t+T+1)\\N(t+T+2)\\\vdots\\N(t+2T + 1)\end{array}\right]$$
The idea is that we can write your equation in a matricial form
$$X(k+1) = MX(k)$$, where M is a $(T+1) \times (T+1)$ matrix. To do this, let's pose $g = 1 + f$. Then:
$$ N(t+T+1) = g(N(t) - N(t+1) + N(t+T)) \Rightarrow $$
$$X_1(k+1) = g(X_1(k) - X_2(k) + X_{T+1}(k)) $$
$$ N(t+T+2) = g(N(t+1) - N(t+2) + N(t+T+1)) \Rightarrow $$
$$X_2(k+1) = g(X_2(k) - X_3(k)) + g X_{1}(k+1) $$
$$\cdots$$
$$ N(t+2T) = g(N(t+T-1) - N(t+T) + N(t+2T-1)) \Rightarrow $$
$$X_{T}(k+1) = g(X_{T}(k) - X_{T+1}(k)) + g X_{T-2}(k+1) $$
$$ N(t+2T+1) = g(N(t+T) - N(t+T+1) + N(t+2T)) \Rightarrow $$
$$X_{T+1}(k+1) = gX_{T+1}(k) + g(-X_{1}(k+1)+X_{T}(k+1)) $$
This means that we have the following:
$$X(k+1) = AX(k) + BX(k+1)$$
where $A$ and $B$ are $(T+1)\times(T+1)$ matrices, and then
$$X(k+1) = (I-B)^{-1}AX(k)$$
with $M = (I-B)^{-1}A$
$$A=\left[\begin{array}{ccccccc}g & -g & 0 & 0 & \cdots & 0 & g\\0 & g & -g & 0 & \cdots & 0 & 0\\\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots\\0 & 0 & 0 & 0 & \cdots & g & -g\\0 & 0 & 0 & 0 & \cdots & 0 & g\end{array}\right]$$
$$B=\left[\begin{array}{ccccccc}0 & 0 & 0 & \cdots & 0 & 0 & 0\\g & 0 & 0 & \cdots & 0 & 0 & 0\\0 & g & 0 & \cdots & 0 & 0 & 0\\\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots\\0 & 0 & 0 & \cdots & g & 0 & 0\\-g & 0 & 0 & \cdots & 0 & g & 0\end{array}\right]$$
Given $f$ and $T$, you can build $A$, $B$ and then $M$. At this point, you can evaluate the solution at any time. That is
$$X(k+h) = M^h X(k)$$
where
$$X(k+h)=\left[\begin{array}{c}N(t+h(T+1))\\N(t+h(T+1)+1)\\\vdots\\N(t+h(T+1)+ T)\end{array}\right]$$ |
H: Subgroups of $U(n)$ isomorphic to a direct sum of cyclic groups
Let $U(n)$ to be the set of all positive integers less then $n$ and relatively prime to $n$. Then $U(n)$ is a group under multiplication modulo $n$.
A. Find integer $n$ such that $U(n)$ contains subgroup isomorphic to $Z_5\oplus Z_5$.
B. Show that there is an $U(n)$ containing a subgroup isomorphic to $Z_3 \oplus Z_3$, but not isomorphic to $Z_4 \oplus Z_4$.
AI: (A) If $n$ is prime, then you know that $U(n)$ is cyclic, and so it cannot have a non-cyclic subgroup. So your goal is to find a composite $n$.
The simplest such would be $n = pq$, where $p, q$ are prime and $p\neq q$. Now
$$
U(n) \cong U(p) \times U(q) \cong \mathbb{Z}_{p-1} \times \mathbb{Z}_{q-1}
$$
To find a subgroup of the form $\mathbb{Z}_5\times \mathbb{Z}_5$, one can look for two primes $p$ and $q$ such that
$$
5\mid (p-1), \text{ and } 5\mid (q-1)
$$
So you list down numbers of the form $(5k+1)$ until you hit two distinct primes : 11, and 31.
Hence, $U(341) \cong \mathbb{Z}_{10}\times \mathbb{Z}_{30}$ has a subgroup isomorphic to $\mathbb{Z}_5\times \mathbb{Z}_5$.
Now can you try the same thing for (B)? |
H: Calculating the instantaneous position of an object constant speed and varying direction
Given an initial position for an object that's moving at constant speed but with a direction that varies in a linear fashion over time, how can I calculate the instantaneous position of the object?
For example, if an object is moving a 2 km/h, with an initial position of (0,0), an initial direction of 0° and a final direction of 90° over a period of 1 hour, how can I get it's position at any point during that hour?
Please bear in mind that I'm a programmer, not a mathematician! With that in mind, I have no idea what to tag this question with....
AI: Assuming you mean that the angle goes from start angle to final angle at a constant rate, the object traces out a circular arc.
You can prove this by writing out the differential equations and solving them. I won't give a proof, however, because it doesn't seem like you are necessarily looking for one. Instead, I'll just give some general details about the arc traced by the object.
Suppose $\theta_0$ is the initial angle, $\theta_1$ is the final angle, $T$ is the total time between the two angles, and $V$ is the constant velocity. Then:
The angle at any given time $t$ is $\theta(t) = \theta_0 + \frac{t}{T} (\theta_1 - \theta_0)$.
Assuming the angles are in radians, the radius of the circle traced out is $R = \frac{VT}{\theta_1 - \theta_0}$.
This follows from that the distance traveled is $R(\theta_1 - \theta_0)$, so the velocity must be $V = \frac{R\theta_1 - \theta_0}{T}$; then just solve for $R$.
If the object starts at $(0,0)$, the center of the circle is at $(R\cos(\theta_0 + 90^\circ), R\sin(\theta_0 + 90^\circ))$. (Adding $90^\circ$ to the angle $\theta_0$ gives us the direction from the object to the circle's center.)
This simplifies to $(-R\sin \theta_0, R \cos \theta_0)$.
The displacement from the center of the circle to the object at any time is $(-R \cos(\theta + 90^\circ), - R \sin(\theta +90^\circ))$, where as above $\theta$ depends on $t$. This simplifies to $(R \sin \theta, - R \cos \theta)$.
To get the position of the object (in $(x,y)$ coordinates) at a general time $t$ just add the center of the circle and the displacement together:
$$
\Big(x(t), y(t)\Big) = ( R\sin \theta - R\sin \theta_0, -R\cos \theta + R\cos \theta_0 )
$$
You can of course factor out the $R$ if you want.
Clarification
Regarding the limit as $\theta_1 - \theta_0 \to 0$, consider the $x$-coordinate:
$$
R (\sin \theta - \sin \theta_0) = VT \frac{\sin \theta - \sin \theta_0}{\theta_1 - \theta_0}
= VT \frac{\sin \left((\theta_0 + \frac{t}{T} (\theta_1 - \theta_0)\right) - \sin \theta_0}{\theta_1 - \theta_0}
$$
Write $\theta_1 - \theta_0 = \Delta$.
The limit as $\Delta$ goes to zero may be found by recognizing the definition of the derivative:
$$
x(t) = \lim_{\Delta \to 0} VT \frac{\sin \left((\theta_0 + \frac{t}{T} \Delta\right) - \sin \theta_0}{\Delta} = VT \left( \frac{t}{T} \cos(\theta_0) \right) = Vt\cos\theta_0
$$
Similarly one may compute $y(t) = Vt\sin\theta_0$. Hence the object is just traveling in a straight line in direction $\theta_0$ with velocity $V$ (as expected). |
H: Taking a power of a polynomial to make it symmetric
Suppose I have a non-symmetric multi-variable polynomial in $n$ variables $P(x_1,x_2,...,x_n)$. For example $P$ might be $x_1^2+x_2$ or $x_1-x_2$
Under what conditions will some power $m$ of $P$ (that is $P(x_1,x_2,...,x_n)^m$) be a symmetric polynomial in $x_1,x_2,...,x_n$?
For example it seems like there is no such $m$ in the case of $x_1^2+x_2$.
But there is such an $m$ in the case of $x_1-x_2$ since $(x_1-x_2)^2=x_1^2-2x_1x_2+x_2^2$.
AI: First of all: this will happen if and only if every permutation of the variables multiplies $P$ by some root of unity.
If $P\circ\sigma = \zeta_\sigma P$ for every permutation $\sigma$, then $P^n$ will be symmetric, where $n$ is the least common multiple of the orders of the $\zeta_\sigma$, because $P^n \circ \sigma = (P\circ \sigma)^n = (\zeta_\sigma P)^n = \zeta_\sigma^n P^n = P^n$.
On the other hand, if $P^n$ is symmetric, then, for any permutation $\sigma$, we have $(P\circ \sigma)^n = P^n$, so $P\circ \sigma$ and $P$ have the same irreducible factors with the same multiplicity, hence $P\circ \sigma = aP$ for some constant $a$ with $a^n = 1$.
But this cannot really happen if the full symmetry group is acting!—since a symmetry group is generated by transpositions, we can have at most order $2$ behavior. And since two transpositions generate a symmetry of order $3$, there are only two possibilities: either $P$ is symmetric already, or it is antisymmetric in every pair of variables (e.g. $(a-b)(a-c)(a-d)(b-c)(b-d)(c-d)$).
These examples are fairly interesting, actually—they are Vandermonde determinants, which pick out the sign of the acting permutation. Specifically, the antisymmetric polynomials are those which are the product $\prod_{i<j} (x_i -x_j)$ times any symmetric polynomial.
If you require only cyclic symmetry, then you will get interesting examples of every degree, however, e.g. $(a+ib-c-id)^4$ is invariant under the transformation $a\mapsto b\mapsto c\mapsto d\mapsto a$. |
H: Fastest way to check if $x^y > y^x$?
What is the fastest way to check if $x^y > y^x$ if I were writing a computer program to do that?
The issue is that $x$ and $y$ can be very large.
AI: If both $x$ and $y$ are positive then you can just check: $$ \frac{\log(x)}{x} \gt \frac{\log(y)}{y}$$
so if both $x$ and $y$ are greater than $e \approx 2.7183$ then you can just check: $$x \lt y$$ |
H: What am I doing incorrectly; logarithms?
We have an increasing number of books on a bookshelf. Every year, 2 books are added and each book is twice as long as the previous book. At the beginning of 1935 the volume was 1 cm thick. We define the 'velocity of the front cover' as the thickness of that volume divided by the number of seconds in six months.Determine the year when the front cover of the volume stacked will exceed the velocity of light. I tried to do it like this:
$$ v = \dfrac{ \text {thickness volume}}{ \text {number of seconds in 6 months}}$$
$$ \text{thickness volume} = 3 \times 10^8 \cdot 15552000 = 4.6656 \times 10^{15}$$
$$ \log_{10}(ans) \approx 15.7$$
So that's the thickness of the book which exceeds the velocity of light. Since the volumes become four times as big every year, if we calculate in logs, every year the books become $\log_{10}(4)$ bigger. So I get $$\dfrac{15.66..}{\log_{10}{4}} \approx 26.02 $$
So my answer would be that in the year $26+1935 = 1961$ the velocity would exceed that of light. Which is a wrong answer, but why?
AI: Your measurement of the speed of light is in meters per second, but your books are moving in speeds of centimeters per second. |
H: How to prove that $f$ is injective if $\exists h \in H: |f^{-1}(\{h\})| = 1$?
Let $G,H$ be groups and $f: G \to H$ a group homomorphism. I want to show that the following statements are equivalent:
$f$ is injective.
$\ker f = \{e_G\}$.
$\exists h \in H: |f^{-1}(\{h\})|=1$.
The standard solution suggests to show the equivalence of 1 and 2, and then the equivalence of 2 and 3 by proving both directions in each case. Therefore, it proves four implications. When I started to prove this, I tried to show only three implications, namely $1 \implies 2 \implies 3 \implies 1$. The first two implications were pretty easy, however I am stuck with the last implication.
Can anyone give me a hint?
AI: Hint: Let $g_1, g_2 \in G$ with $f(g_1) = f(g_2)$. Let $h$ as in 3 and $f^{-1}(\{h\}) = \{g\}$. Then $f(gg_1g_2^{-1}) = h$, so $gg_1g_2^{-1} = g$. |
H: Find an efficient algorithm to calculate $\sin(x) $
Suggest an efficient algorithm to determine the value of the
function $ \sin(x) $ for $ x \in [-4\pi, 4\pi] $.
You can use only Taylor series and $ +, -, *, /$.
I know, that $$\sin(x)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n + 1)!}$$
but I can't find an efficient algorithm.
Thank for your help.
AI: Hint: The coefficients of the series are related to one another by the simple relation:
$$
t_{n+1}=\frac{x^{2(n+1)+1}}{(2(n+1)+1)!}=\frac{x^2}{(2n+2)(2n+3)}\frac{x^{2n+1}}{(2n+1)!}=
\frac{x^2}{(2n+2)(2n+3)}t_n
$$
(Here $t_n$ represents the coefficient of $x^{2n+1}$, multiplied by $x^{2n+1}$. So $\sin x=\sum_{n=1}^\infty t_n$.)
This means that you don't need to work out each coefficient separately: once you've worked out $t_n$, you've done most of the work you need to do to work out $t_{n+1}$.
The other thing you'll need to do is to work out how many terms of the series you'll need to ge a good enough approximation on $[-4\pi,4\pi]$. |
H: Is $x^8+1$ irreducible in $\mathbb{R}[x]$
Question is to check if :
$x^8+1$ is irreducible over $\mathbb{R}[x]$.
even before this I tried to see $x^4+1$ and $x^2+1$.
for $x^2+1$, it does not have a root in $\mathbb{R}$ So, it is irreducible.
for $x^4+1$, checking for roots does not imply anything.
So I tried to solve for $a,b,c,d \in \mathbb{R}$ in $x^4+1=(x^2+ax+b)(x^2+cx+d)$ and concluded that $x^4+1$ is reducible.
It is becoming more difficult when the power of $x$ is getting bigger.
The only thing I can say about $x^8+1$ is it does not have a factor of degree $1,3,5,7$ (if not, it would have a real number as a root which is not possible)
I am not able to proceed further.. please help me
AI: Do you know that $\mathbb C$ is the algebraic closure of $\mathbb R$ and it has dimension $2$ as $\mathbb R$-vector space?
This implies that every irreducible polynomial over $\mathbb R$ has degree less or equal to $2$.
More details: every polynomial $p \in \mathbb R[x]$ has root in $\mathbb C$ and the $\mathbb R[\alpha]$, the subring of $\mathbb C$ generated by $\alpha$ a root of $p$, is field and so a subfield of $\mathbb C$. Its dimension as $\mathbb R$-vector space is equal to the degree of the irriducible polynomial having root $\alpha$, but such dimension must also be less or equal then the dimension of $\mathbb C$, which is $2$. So the irriducibile polynomial of $\alpha$ can either have degree $1$ (the element belong to $\mathbb R$) or $2$. |
H: Arithmetic Progressions containing integers close to a power of 2
Consider an arithmetic progression of the form $\{kq: k \in \mathbb{Z}\}$, where $q$ is an odd integer. Do such APs always contain a number of the form $2^n \pm 1$?
I was initially interested in the largest $K(q)$ for which $\{kq: 0 \leq k < K(q)\}$ didn't contain a number of the form $2^n \pm 1$. Then I realized that I couldn't even prove $K(q)$ is always finite.
I wrote a small python code to check the $K(q)$ numbers for the first 30 odd numbers. The distribution is interesting.
1 1
3 1
5 1
7 1
9 1
11 3
13 5
15 1
17 1
19 27
21 3
23 89
25 41
27 19
29 565
31 1
33 1
35 117
37 7085
39 105
41 25
43 3
45 91
47 178481
49 42799
51 5
53 1266205
55 19065
57 9
59 9099507
61 17602325
63 1
65 1
K(67) however is at least 100 million.
AI: The question can be rephrased as follows:
Given $q\in\mathbb{N}$, $q$ odd, does there exist $n\in\mathbb{N}$ such that
$q$ divides $2^n+1$ or $2^n-1$?
The answer is yes. Let $q\in\mathbb{N}$ be odd. Consider the set
$$
P=\{2+1,2^2+1,\dots,2^q+1\}.
$$
If $q$ divides some element in $P$ there is nothing to prove. If not, by the Pigeon Hole Principle, there exist $1\le m<n\le q$ such that $2^m+1$ and $2^n+1$ have the same remainder when divided by $q$, that is,
$$
2^m+1\equiv 2^n+1\pmod q.
$$
Then $q$ divides $2^n-2^m=2^m(2^{n-m}-1)$. Since $q$ is odd, $q$ must divide $2^{n-m}-1$. |
H: If $U$ is open and $B\subset U$ is a closed ball, is there an open set $V\subset U$ such that $V\supset B$ and $V,\partial V$ are Jordan Measurable?
I'm studing Evans PDE book, especifically, the proof of theorem 6 here.
Let $U\subset\mathbb{R}^n$ be an open set and $f:U\to\mathbb{R}$ a locally integrable function. In the proof is taken an open set $V\subset\mathbb{R}^n$ such that $B\subset V\subset\overline{V}\subset U$ (where $B$ is a closed ball that contains the support of $\eta_\varepsilon$).
As $f$ is locally integrable, we can conclude that $f$ is integrable over $\overline{V}$. However I need $f$ integrable over $V$. My analysis book says that if $X,Y\subset\mathbb{R}^n$ are both Jordan measurable and $f:X\cup Y\to\mathbb{R}$ is integrable then $f$ is integrable over $X$ and over $Y$.
So, my question is: can we ensure that $V$ and $\partial V$ are both Jordan measurable? Or, to be more precise, given an open set $U\subset\mathbb{R}^n$ and a closd ball $B\subset U$, is there an open set $V\subset U$ such that (i) $V\supset B$ and (ii) $V$ and $\partial V$ are both Jordan Measurable? (if so, I think we can conclude that $f$ is integrable over $V$ because it's integrable over $\overline{V}=V\cup\partial V$).
Thanks.
AI: Evans is not concerned with Jordan measurability.
It seems that your analysis book does not provide adequate background for reading Evans. You should find a book that does Lebesgue integration. (Real Analysis by Folland is one of such.) Every Borel set is Lebesgue measurable. |
H: Prove the straight line as a tangent to a parabola.
I was going through some past exam papers and I came across this problem and I'm bit puzzled on how to approach this, could someone please help me out?
Equation of the parabola $y^2-7x-8y+14=0$, prove that the straight line given by $7x+6y=13$ is a tangent and find the point of contact.
AI: given parabola : $y^2-7x-8y+14=0$,
given line : $7x+6y=13$
point of contact : $y^2-7x-8y+14=y^2+(6y-13)-8y+14=0$ i.e., $y^2-2y+1=0$
i.e., $y=1$ i.e., $(1,1)$ is "the" point of contact
when "a" point of contact is "the" point of contact.. It must be tangent... |
H: Limit of sequences ($\lim x_n = a > 0 \Rightarrow \lim x_n ^{1/k} = a^{1/k}$)
I need to show that if $(x_n)_{n \in \mathbb{N}}$ is a sequence such that $\lim x_n = a>0$, then $\lim \sqrt[k]{x_n} = \sqrt[k]{a}$.
It was suggest to use the equality $(x-a) = (x^{1/k} - a^{1/k})\left(\sum_{i=0}^{k-1} x^{i/k} a^{-i + 1/k} \right)$ and I was thinking to use the definition to this proof (since I don't know yet what is $ \lim x_n^{p/q} $). But, if a get a particular $x_j$ such that $x_j <0$, if k is pair $x_j^{1/k}$ is not well defined in $\mathbb{R}$ and I don't know, in this case, if I can say that
$n \geq n_{\epsilon} \Rightarrow |x_n - a| < \epsilon \leq \epsilon.|\sum_{i=0}^{k-1}(...)| $
So, what should I do, in this case? Change the proof to "eliminate" the problem or we must not consider this condition? I know that the sequence must have at least one positive number because it's limit is positive.
thanks for the help!
AI: Since you only want the limit for large values of $n$, you can just start after all of the negative values of $x_n$. Say for $n>N$. But how can you be sure that the $x_n$ are (after a while) always positive? Check the definition of limit. |
H: Writing a permutation as products of transpositions
If a can write a permutation $\sigma$ as a product like $\Delta \alpha \beta$, where $\Delta$ is a product of transpositions (in fact, anything) and $\alpha$ and $\beta$ are two disjoint transpositions, so the symbols moved by $\alpha$ and $\beta$ belong to the support of different cycles in disjoint cycle decomposition of $\sigma$?
Is this true? If so, does somebody have some clue of where to find a proof for that?
Many thanks for any help...
Luiz
AI: No they don't: $(12)(13)(24) = (2413)$. |
H: Where to take Real Analysis and Linear Algebra?
I am undergraduate in economics. As you may know, most prestigious departments in economics now require their aspirants to have taken Real Analysis (and Linear Algebra, too) before entering their programs.
I have time to take these courses but don't know whether any college will give admission in their math program to an econ undergrad. I don't know whether distance learning courses will help me.
Can anyone suggest me how to take math courses?
Your replies will oblige me.
AI: I am a little confused by the question. When you ask whether college mathematics departments will accept an undergraduate economics major, what do you mean? The two options I am imagining are (a) you want to be a mathematics major later on and you're concerned the econ background will not help you, or (b) you feel you need to be admitted to graduate work in mathematics, complete that work, and then apply to economics graduate departments.
If it is option (a) then to some extent you are right. Many of the most prestigious mathematics programs will not look at the mathematical preparation of an economics major as sufficient for graduate study. It varies by school certainly, but more often than not you will need many rigorous advanced math classes and lots of extra-curriculars (e.g. doing well on the Putnam exam) to demonstrate that even though your coursework focused on economics, you have gathered the needed mathematical maturity along the way.
Some exceptions would be applied mathematics graduate programs. In that case, if you have learned programming skill or applied modeling, these can be viewed as assets that differentiate you from applicants who have done only theoretical mathematical work. It often depends on the professors who have availability to accept students and whether they see your background as a good fit for their lab's work or their on-going projects.
If it is option (b) then I see no need to formally enroll in graduate study in mathematics. Most modern universities offer linear algebra and real analysis as a basic part of the undergraduate curriculum. If your university does not, you may look into what options you have with Coursera or commuting as a special student to a nearby university (possibly as an online student) to take the classes elsewhere.
Rigorous undergraduate courses in linear algebra and real analysis will often be sufficient for entry into an economics program. The textbooks you become familiar with might also help you. This is by no means a comprehensive list, but some of the books that are often used in graduate real analysis include: Folland, Stein and Shakarchi, and Rudin.
For two very readable and gentle introductions to the subject at an undergraduate level, check out Abbott and Saxe.
If you take a course in analysis and do well -- and separately make a study of some of these books -- then you may list these textbooks when you apply to graduate school and it should often be a sufficient signal to the faculty that your mathematical maturity is high enough to do rigorous economics work.
If you are interested in a graduate-level economics book which covers the standard subjects from economics that require a lot of real analysis, then check out the first five or so chapters of Mas-Colell, Whinston, and Green. |
H: Formal proof $\binom{n}{k}$ is an integer
In mathematics one defines:
$\left(\begin{array}{c}n\\k\end{array}\right)=\displaystyle\frac{n!}{k!\cdot (n-k)!}$
This is the number of combinations of $k$ elements from a collection of $n$ elements.
I was wondering if it is possible to prove that the result is an integer in a formal way. It's easy to see that $k!$ in the denominator divides the numerator, but how do you prove that the denominator $k!$ also divides the product of all elements between $n-k+1$ and $n$?
AI: $$
\frac{n!}{k!\cdot (n-k)!}
=\frac{n(n-1)\cdots(n-k+1)}{k!}
=\frac{(n-k+k)(n-1)\cdots(n-k+1)}{k!}\\
=\frac{(n-1)\cdots(n-k)}{k!}+
\frac{(n-1)\cdots(n-k+1)}{(k-1)!}
$$
Now use (double) induction.
This is the relation that defines Pascal's triangle:
$${n \choose k} = {n-1 \choose k} + {n-1 \choose k-1}.$$ |
H: Is any differentiable function $f : (0,1)\rightarrow [0,1]$ is uniformly continuous
Question is to check if :
any differentiable function $f : (0,1)\rightarrow [0,1]$ is uniformly continuous.
I know that any continuous function on compact subset of $\mathbb{R}$ is uniformly continuous.
As $(0,1)$ is not compact, we can not say anything at this time.
Now, as it is given that $f$ is differentiable, if its derivative $f'$is bounded then $f$ is uniformly continuous.
So, I am trying to look for differentiable functions $f$ on $(0,1)$ such that $f'$ is unbounded.
i am not very familiar with large number of differentiable functions with unbounded derivatives.
I know $f(x)=\sqrt{x}$ has unbounded derivative, but $\sqrt{x}$ is uniformly continuous....
So, I would like someone to help me out with some hint.
P.S : I have just now saw one example
$$f(x)=x^2\sin{\frac{1}{x^2}}$$
which is differentiable but is not bounded.
I see that $\sin(\frac{1}{x^2})$ is bounded by $1$ and if $x\in (0,1)$ then so is $x^2$ and so is $x^2\sin{\frac{1}{x^2}}$
So, $f(x)=x^2\sin{\frac{1}{x^2}}$ is from $(0,1)$ to $[0,1]$ whose derivative is unbounded.
Now, the problem reduces to show that $f(x)$ is not uniformly continuous... :(
AI: Consider the function $f(x)=\frac12(1+\sin\frac1x)$ on $(0,1)$. It's obviously differentiable. However, $f(\frac{1}{(2n-\frac12)\pi})=0$ and $f(\frac{1}{(2n+\frac12)\pi})=1$, but $\frac{1}{(2n-\frac12)\pi}$ and $\frac{1}{(2n+\frac12)\pi}$ can be arbitrary close (if you take large $n$). |
H: Are complete intersection prime ideals of regular rings regular ideals?
Let $(R, \mathfrak{m})$ be a regular local ring and let $\mathfrak{p}$ be a prime ideal of $R$ which is a complete intersection, i.e. the minimal number of generators of $\mathfrak{p}$ equals its height $h$. Then by Macaulays theorem there is a system of parameters (or equivalently - a regular sequence) $\{a_{1},\dots, a_{h}\}$ which generates $\mathfrak{p}$.
Is it then also true that $\mathfrak{p}$ can be generated by elements $\{b_{1}, \dots, b_{h}\}$ which can be extended to a regular system of parameters for $R$?
Phrased differently, I am asking whether every complete intersection prime ideal in $R$ is regular (in the sense that $R/ \mathfrak{p}$ is regular).
I am asking this question being interested in the situation where $R = \mathbb{C}\{x_{1},\dots, x_{n}\}$ is the ring of convergent power series.
AI: Take $R=\mathbb{C}[x,y]_{(x,y)}$, and take $\mathfrak{p}=(x^2-y^3)$. $R/\mathfrak{p}$ is not a regular local ring, since it isn't integrally closed in its field of fractions. |
H: When f is absolutely integrable and contiunous prove that $\sqrt{f}$ is absolutely integrable.
If $f: (a ,b) \rightarrow [0,\infty)$ is continuous and absolutely integrable on (a,b), then prove that $\sqrt{f}$ is absolutely integrable on (a,b).
I have that $\sqrt{f}$ is locally integrable. I am working on $|\sqrt{f}|$ is improperly integrable. I think I should do something by comparison, but I am having trouble coming up with what to compare it to.
AI: Hint: $\sqrt f \le \sup\{f, 1\}$. |
H: Showing the limit of a flow must be an equilibrium point under certain restrictions.
I'm stumped on how to approach this one:
Consider the autonomous ODE $\dot{x} = f(x)$, $x \in \mathbb{R}^n$ with initial condition $x(0) = x_0$ and $f : \mathbb{R}^n \rightarrow \mathbb{R}^n$
(at least) continuously differentiable. Let $\phi(t; x_0)$ be a solution such that
$$\lim_{t \rightarrow\infty}\phi(t; x_0) = a$$
for a certain $a \in \mathbb{R}^n$. Prove that $a$ must be a critical point (equilibrium) of the system.
Now intuitively I get that since $\phi(t,x_0) \rightarrow a$ we in a sense have that $\dot{x} \rightarrow 0$ as $x \rightarrow a$ since the speed at which our solution approaches $a$ becomes slower and slower. But how to rigorously put this, and how $f$ being continuously differentiable has got me stuck.
Can anyone help me along the right path?
AI: We have
$$
f(x)=f(a)+O(|x-a|)
$$
Given $\varepsilon_0>0$, we have
$$
|\phi(t,x_0)-a|<\varepsilon_0 \;\;\mbox{ if } t>t_0
$$
and
$$
|\phi'(t)-f(a)|=|f(\phi(t,x_0))-f(a)|\le M|\phi(t,x_0)-a|<M\varepsilon_0\; \mbox{ if } t>t_0
$$
Now if we take $t_1$ and $t_2>t_0$, using the previous inequalities, we obtain
$$
2\varepsilon_0>|\phi(t_2,x_0)-\phi(t_1,x_0)|=\left|\int_{t_1}^{t_2}\phi'(t)\,dt\right|\ge\left|\int_{t_1}^{t_2}f(a)\,dt\right|-\left|\int_{t_1}^{t_2}\phi'(t)-f(a)\,dt\right|\\ >|f(a)(t_2-t_1)|-M\varepsilon_0|t_2-t_1|
$$
Then
$$
|f(a)|<\frac{2\varepsilon_0+M\varepsilon_0|t_2-t_1|}{|t_2-t_1|}
$$
Since we can take $\varepsilon_0$ so small as we want, we conclude that $f(a)=0$. |
H: marching band conductor
Let $f(x)$ be the unique polynomial that satisfies:
$f(n)=\sum_{i=1}^{n} i^{101}$, for all positive integers $n$.
The leading coefficient of $f(n)$ can be expressed as $\frac {a}{b}$, where $a$ and $b$ are positive co prime integers. What is the value of $a+b$?
AI: If $f_k(n)$ is the polynomial defined by $f_k(n)=\sum_{i=1}^n i^k$, then the leading coefficient is $\frac{1}{k+1}$ (easy to show by induction).
So we have $a=1$ and $b=102$, and hence $a+b=103$. |
H: Prove the following statement...
So, how can I prove this limit: $$\lim_{x \to x_0} \frac{1}{(x-x_0)^2}=+\infty$$
What I've tried is to multiply both numerator and denominator with (x+x0), but I guess that's wrong, what can I do to solve this? Please help...
AI: What happens to the denominator as $x \to x_0$?
Since $(x - x_0)$ is squared, it is positive regardless of whether we approach $x_0$ from the left or right. Hence, the denominator shrinks towards zero (becoming incredibly small, yet remaining positive) as $x\to x_0,\;$ and thus the resultant function approaches $+\infty$. |
H: How can these be the weights of the adjoint representation?
This is perhaps a stupid question. We consider $G =\text{SU}(3)$ and $\pi : G \to \textrm{GL}(\mathfrak{g})$ the adjoint representation that sends $g \in G$ to $Ad_g$ that acts on the Lie algebra $\mathfrak{g}$ be the following formula. For $X \in \mathfrak{g}$,
$$Ad_g(X) = gXg^{-1}.$$
Now I am trying to find the weight vectors for this representation. I get that they are the elementary matrices $E_{ij}$ but there is one problem: These $E_{ij}$ technically don't live in $\mathfrak{g}$! This is because an elementary matrix $E_{ij}$ does not satisfy the relation $X + X^\ast = 0$. What am I misunderstanding here?
AI: Yes, absolutely, the issue is that eigenvectors (weight vectors) often lie only in the complexified Lie algebra. Here, a useful, compatible model of the complexification of $\frak su$$(n)$ is $\frak sl$$(n,\mathbb C)$, making the relevance of @TobiasKildetoft's comments perhaps clearer. Indeed, the question does turn into asking about the action of the complexified algebra on itself. |
H: Matlab - finding gradient
Given the function:
$$v(x,y) = x + e^{-((x-1)^2 + (y-1)^2)}$$
I am supposed to calculate the gradient of this expression in Matlab for x defined in the interval -1:0.1:0.9 and y defined in the interval -2:0.1:1.9. My task is to find the absolute value of the gradient of this function, and I'm supposed to do this two ways - first by calculating the gradient analytically by myself, and then by using Matlab's built'in gradient-function. However, I get quite different results when I do this. For the analytical part I did as follows:
[x,y] = meshgrid(-1:0.1:0.9, -2:0.1:1.9);
analytic_partialx = 1 - 2*(x-1).*exp(-((x-1).^2 + (y-1).^2));
analytic_partialy = -2*(y-1).*exp(-((x-1).^2 + (y-1).^2));
abs_gradient = sqrt(analytic_partialx.^2 + analytic_partialy.^2);
When using Matlab's built-in function I did as follows:
[x,y] = meshgrid(-1:0.1:0.9,-2:0.1:1.9);
v = x + exp(-((x-1).^2 + (y-1).^2));
[partialx,partialy] = gradient(v);
abs_gradient = sqrt(partialx.^2 + partialy.^2);
However, in the first case I get a matrix primarily containing values in the range of 1-1.5, whereas in the secand case I get a matrix primarily containing values in the range of 0-0.2. So one of my approaches is wrong here since they should have been almost completely similar. If anyone can help me by seeing what I do wrong here, then I would be extremely grateful!
AI: I am not near Matlab at the moment. Does the gradient command have an option where you input all three of x,y,v?
Otherwise, you should divide your Matlab answer by 0.1, which is the value of dx and dy in the grid. |
H: Question about the complement
If $E=\{(x,y):x\in\mathbb{R},y\in(-\infty,0)\}$, what is the complement of $E$ in $\mathbb{R}^2$?
Is $E^c=\phi\times[0,+\infty)$?Is this an empty set?
AI: I will try to give a bit more detail than DonAntonio's answer contains.
Recall the definition of complement in a given universe of discourse $U$:
$$E^c \stackrel{\rm def}= \{u \in U: u \notin E\}$$
In our case, $U = \Bbb R^2$, so we're looking to describe:
$$E^c = \{(x,y) \in \Bbb R^2: (x,y)\notin E\}$$
Now, the definition of $E$ stipulates that $(x,y) \in E$ precisely when $x \in \Bbb R$ and $y \in (-\infty,0)$.
So $(x,y) \in E^c$ when not both $x \in \Bbb R$ and $y \in (-\infty,0)$. That is, when $x \notin \Bbb R$ or $y \notin (-\infty,0)$. (Note how and became or by negating the statement.)
Since $x \in \Bbb R$ by definition, it remains to conclude that:
$$(x,y) \in E^c \text{ precisely when } y \notin (-\infty,0)$$
That is to say:
$$E^c = \{(x,y)\in\Bbb R^2: y \ge 0\} = \{(x,y)\in\Bbb R^2: y \in [0,\infty)\}$$ |
H: comparing bit lengths of binary numbers
Suppose I have two binary numbers x and y that have bit lengths of nx and ny which are unknown. I'm looking for a fast method of comparing their bit lengths without computing their bit lengths; I think I found one using bitwise OR, and a pair of left shifts, but I'm not sure:
def comparelen(x, y):
'''return -1 if bitlength(x)<bitlength(y),
return 1 if bitlength(x)>bitlength(y),
return 0 if bitlength(x)==bitlength(y)
'''
xy = x | y # bitwise OR
if (x << 1) <= xy:
return -1
if (y << 1) <= xy:
return 1
return 0
If x and y are the same length, it's easy to see that this will return the correct result.
If nx < ny - 1, or ny < nx - 1, this will return the correct result.
What I can't seem to figure out is if my function is correct for bit lengths that differ by one. Can anyone help me?
AI: It does work. Suppose $x$ has length $n$ and $y$ has length $n+1$. Then $x|y$ begins with two leading ones, so if $2x \geq x|y$, then $x$ must begin with two ones. But then $x|y$ begins with three ones so $x$ must begin with three ones, and so forth. You finally conclude that $x$ is all ones, but in that case you still have $2x < x|y$ because $x|y$ is a string of $n+1$ ones whereas $2x$ is of length $n+1$ and ends in a zero. Thus $2x < x|y$ is guaranteed if the length of $x$ is less than the length of $y$. Nice algorithm you found :) |
H: What is the anti derivative of $ f(x) = \int\frac{1}{1+e^{-x}}dx $?
I know u-substitution and integration by parts, but I don't know how to solve this one.
AI: Use
$$\int\frac{f'(x)}{f(x)}dx=\log f(x)+C$$
and
$$\frac1{1+e^{-x}}=\frac{e^x}{e^x+1}$$ |
H: Are all prime numbers up to Mersenne Prime 48 known?
In January this year the biggest prime number so far has been found and it is a Mersenne Prime. $$2^{57885161}-1$$.
My question:
Are all prime numbers from 0 up to $2^{57885161}-1$ found, or not?
If you only look for prime numbers which are Mersenne Primes, you skip quite a few.
AI: As you may know, GIMPS is conducting a search. They give each user a different prime number to test. Then they give that number to a different user to check.
They only test numbers of the form $2^p-1$, where $p$ is also prime. So they have only checked a few million possible values of $p$, that is a few million possible values of $2^p-1$.
The link www.mersenne.org/report_milestones/ shows that they have checked all primes
$2^p-1$ if $p$ is below 44,576,437, and have double-checked primes $2^p-1$ for values of $p$ up to $p=26,187,517$.
There are around $(2^{57885161})/57885161$, or around $2^{57885135}$ unknown primes below the current record. |
H: Equivalence of Definitions for $T_1$
A topological space $(X,T)$ is $T_1$ (or Fréchet) if and only if for every $x\in X$, $\{x\}$ is the intersection of all neighborhoods $N\in N_x$
I have to use the following definition:
$(X,T)$ is $T_1$ if for $x,y\in X$ exists a neighborhood $ U_x$ of $x$ not containing $y$ and a neighborhood $ U_y$ of $y$ not containing $x$.
Thank you.
AI: If $\langle X,\mathcal{T}\rangle$ is a T$_1$ space, then for all $x\in X$ and all $y\in X\setminus\{x\},$ there is some $U\in\mathcal T$ such that $x\in U$ and $y\notin U$. That is, there is some $U\in\mathcal N_x$ such that $y\notin U$, so that $y\notin\bigcap\mathcal N_x.$ This holds for all $y\in X\setminus\{x\},$ so since $x\in\bigcap\mathcal N_x,$ what can you conclude?
Suppose that $\langle X,\mathcal T\rangle$ is not a T$_1$ space. Then there is some $x\in X$ and some $y\in X\setminus\{x\}$ such that for all $U\in\mathcal T$ with $x\in U$ we have $y\in U$. That is, $y\in U$ for all $U\in\mathcal N_x,$ so $y\in\bigcap\mathcal N_x,$ and so $\{x\}\neq\bigcap\mathcal N_x.$ |
H: How does $7\log(8x) = 7\ln8x$?
I was working on some math homework with a program called scientific notebook. I was check that I was writing something correctly.
The original equation is $(\log(x^4)+\log(x^5))/\log(8x)=7$
I then converted it to $\log(x^{(4+5)})=7\log(8x)$
I was expecting to get $\log((8x)^7)$ when I entered the $7\log(8x)$ into the program to evaluate. Can someone please explain this to me?
AI: In higher mathematics it is customary to take $\;\log=\ln=\log_e\;$ , and that's what most mathematicians mean when writing that.
Added: You had
$$\frac{\log x^4+\log x^5}{\log 8x}=7\iff \log x^9=7\log 8x=\log(8x)^7\implies$$
$$x^9=8^7x^7\implies x^2=8^7\ldots$$ |
H: Recurrence $T(n) = T({2n\over5}) +n$ using Master Theorem
Solve the recurrence
$$T(n) = T\left({2n\over5}\right) +n$$
My attempt:
$a=1$,$\ b=\frac 52$, $f(n)=n$
For the most part I believe that is correct. Now I was wondering if my math is correct in this next step. $n^{\log_b a}$ if $a=1$ and $b=\frac 52$ then:
$n^{\log_b a} = n^{\log_{5/2} 1} = n^0 = 1$ (Let me know if this is incorrect)
$f(n) \ \ \ \ \text{vs.} \ \ \ \ n^{\log_b a} \\
\ \ \ n \ \ \ \ \ \ \text{vs.} \ \ \ \ \ \ \ 1$
Assuming $n\ge 1$ this is case 3.
$n = \mathcal O(n^{1+\epsilon}) \quad \text{ for }\epsilon = ? $
Does/can epsilon equal $0$?
I can figure out the regularity condition, I just want to make sure these steps are correct, before I move on.
AI: You want to know if $T(n)$ is comparable to $n$. Just try the obvious, $T(n) \sim Cn$ and find $C$:
$Cn = \frac{2}{5}Cn + n$, so that $C = \frac{5}{3}$ and in fact this solves the recurrence.
$ T(n)= 5n/3$ is a particular solution (for the smoothed recurrence that allows rational $n$ as arguments instead of rounding $2n/5$)
and $E(n) = T(n) - 5n/3$ satisfies
$E(n) = E(2n/5)$ which for rational-$n$ solutions bounded near $0$ would mean $T(n) = 5n/3 + O(1)$.
If you really meant the unsmoothed, integer recurrence, the recurrences for $T$ and $E$ are solved up to bounded error $O(1)$ and compute the value at $n$ in $O(\log n)$ steps that add up those errors, so the result will be
$T(n)=\frac{5n}{3} + O(\log n)$. |
H: Superposition of Sine and Cosine functions
I was wondering this:
Let a and b be real numbers. Is it always possible to find real numbers c, d, and e such that $$a\sin(x)+b\cos(x)=c\cos\left(\frac{x+d}e\right)$$
Why is this the case?
Thanks.
AI: $$a\sin(x)+b\cos(x)=\sqrt{a^2+b^2}\left(\dfrac{a}{\sqrt{a^2+b^2}} \sin{x} + \dfrac{b}{\sqrt{a^2+b^2}} \cos{x} \right)\\
=\sqrt{a^2+b^2}\left(\sin{\varphi} \sin{x} + \cos{\varphi} \cos{x} \right)=\sqrt{a^2+b^2}\cos{(x-\varphi)},$$ where
$$\sin{\varphi} =\dfrac{a}{\sqrt{a^2+b^2}}, \\
\cos{\varphi} =\dfrac{b}{\sqrt{a^2+b^2}}.
$$ |
H: Integrate the following.
Given
$$\dfrac{dN}{N}=4\pi\left(\dfrac{m}{2\pi K_b T}\right)^{3/2}v^2e^{mv^2/2K_bT}dv $$
How to integrate this from $v_0$ to infinity.?
AI: You want
$$\int_{v_0}^{\infty} dv \, v^2 \, e^{- a v^2} $$
where $a = m/(2 K_b T)$. (I know this because I know this is the Maxwell-Boltzmann distribution.) Integrate by parts to get
$$-\frac{1}{2 a} \left [v \, e^{- a v^2} \right]_{v_0}^{\infty} + \frac{1}{2 a} \int_{v_0}^{\infty} dv \, e^{- a v^2} = \frac{v_0}{2 a} e^{-a v_0^2} + \frac14 \sqrt{\frac{\pi}{a^3}} \text{erfc}\left (\sqrt{a} v_0 \right )$$ |
H: proof that convergence in mean implies convergence in probability
I'm attempting to understand a proof, but I am failing to see how a step is pulled off.
Claim: $\text{if } f_n \longrightarrow_{L_p} f$ then $f_n \longrightarrow_{P} f$
Proof: Let $\epsilon > 0$. Then
$P\left( \left\lbrace \omega: |f_n(\omega) - f(\omega)| > \epsilon \right\rbrace \right) = P \left( \left\lbrace \omega: |f_n(\omega) - f(\omega)|^p > \epsilon^p \right\rbrace \right) \\$
$ \leq \frac{1}{\epsilon^p} \int_\Omega |f_n(\omega) - f(\omega)|^pdP(\omega) \longrightarrow_n 0$
I understand that this would show convergence in probability since by assumption the integral converges to zero. Its the step before that (moving from equality to inequality) that I am miffed by. This is my attempt:
$P \left( \left\lbrace \omega: |f_n(\omega) - f(\omega)|^p > \epsilon^p \right\rbrace \right) = \int_\Omega I[|f_n(\omega) - f(\omega)|^p > \epsilon^p]dP(\omega) \\ = \frac{\epsilon^p}{\epsilon^p} \int_\Omega I[|f_n(\omega) - f(\omega)|^p > \epsilon^p]dP(\omega) \\ = \frac{1}{\epsilon^p} \int_\Omega \epsilon^p I[|f_n(\omega) - f(\omega)|^p > \epsilon^p]dP(\omega) \\ \leq \frac{1}{\epsilon^p} \int_\Omega |f_n(\omega) - f(\omega)|^pdP(\omega) $
The last being from the fact that $\epsilon^p$ times the indicator will be $\epsilon^p$ or 0 unless the indicator is satisfied, which is less than $|f_n(\omega) - f(\omega)|$ for those $\omega$ since if the indicator is 1, $|f_n(\omega) - f(\omega)| > \epsilon^p$. But this feels way to loose (especially since we haven't changed the fact that were integrating over $\Omega$). Any help making this more clear/rigorous would be great.
Thanks!
AI: Since $\varepsilon>0$ we have
$$
|f_n-f|^p>\varepsilon^p\;\;\iff\;\;g_n:=\frac{|f_n-f|^p}{\varepsilon^p}>1,
$$
and hence
$$
\int_\Omega \mathbf{1}_{\{|f_n-f|^p>\varepsilon^p\}}\,\mathrm dP=\int_\Omega \mathbf{1}_{\{g_n>1\}}\,\mathrm dP.
$$
But $0\leq \mathbf{1}_{\{g_n>1\}}(\omega)\leq g_n(\omega)$ for all $\omega$ and hence
$$
\int_\Omega \mathbf{1}_{\{g_n>1\}}\,\mathrm dP\leq\int_\Omega g_n\,\mathrm dP=\frac{1}{\varepsilon^p}\int_{\Omega}|f_n-f|^p\,\mathrm dP.
$$ |
H: Solving $\lim_{n \to \infty}{\frac{n^a}{\log\left(\left| \log(n^a)\right|\right)}}$
I haven't practiced limits for years, now I need them to solve an exercise and I don't know whether I have come up with the right solution.
$$\lim_{n \to \infty}{\frac{n^a}{\log\left(\left| \log(n^a)\right|\right)}}$$
where $a$ is a fixed constant.
Since I have the form $\frac{\infty}{\infty}$, I apply the De L'Hopital theorem, so I derive both numerator and denominator, so:
$$\lim_{n \to \infty}{\frac{an^{a-1}}{\frac{a}{n\log(n^a)}}} = \lim_{n \to \infty}{n\log(n^a)} = \lim_{n \to \infty}{a n\log(n)} = \infty$$
Can you please give me any feedback?
AI: The expression is defined only for $a\ne0$, because for $a=0$ the denominator would have $\log 0$.
So, assume $a>0$; then you can rewrite the denominator as
$$
\log(|\log n^a|)=\log(|a\log n|)=\log(|a|\log n)=\log|a|+\log\log n
$$
(at least for $n>1$, which is implied).
Since both numerator and denominator go to infinity, you can apply l'Hôpital's theorem:
$$
\lim_{n\to\infty}\frac{n^a}{\log|a|+\log\log n}=
\lim_{n\to\infty}\frac{an^{a-1}}{\frac{1}{\log n}\frac{1}{n}}=
\lim_{n\to\infty}an^{a-1}\cdot n\log n=
\lim_{n\to\infty}an^a\log n
$$
provided this last limit exists. Does it?
Don't forget to apply the chain rule when differentiating $\log\log n$.
For $a<0$ you have a slightly different, but easier, situation. |
H: Selecting 1 ball of each type from the box
There are n White balls and n Black balls in a box. You repeatedly withdraw 2 balls simultaneously till the box is empty. Find the probability such that every withdrawal consists of 1 White ball and 1 Black ball?
EDIT:-let's just find the probability for the first simultaneous withdrawal of 2 balls such that 1 is W and 1 is B.
AI: Let's try a long winded way of doing this through classical probability. I think the below is correct.
Let number of remaining white balls be $w_i$ and similarly for black balls $b_i$ where $i$ is the ordinal number denoting the draw, e.g. $i=1$ for the first draw. Initially, there are a total of $2n$ balls in the box and that $i=0$, the number of white balls is $w_0=n$ and number of black balls is $b_0=n$.
At the first draw, i.e. $i=1$, the probability for drawing one white and one black ball is given by the probability of drawing a white first then black, denoted $P(b_1|w_1)$ plus that of drawing a black first then black, denoted $P(w_1|b_1)$. They are given by:
$P(b_1|w_1)=\frac{b_1}{(b_1+w_1)}\times\frac{w_1}{(b_1+w_1-1)}$
$P(w_1|b_1)=\frac{w_1}{(b_1+w_1)}\times\frac{b_1}{(b_1+w_1-1)}$
Notice that they are equal to each other. Hence,
$P(b_1,w_1)=P(b_1|w_1)+P(w_1|b_1)=\frac{2w_1b_1}{(b_1+w_1)(b_1+w_1-1)}=\frac{2n^2}{2n(2n-1)}=\frac{n}{(2n-1)}$
Now we have to generalize for second draw, third draw and so on. To do so, note that since there were $n$ white balls initially, the number of white balls $w_i$ after the $i^\text{th}$ draw is given by $w_i=n-i+1$. For example, if we had 10 white balls, at the second draw we would have $10-2+1=9$ white balls left. We can apply the same logic for the black balls and obtain the identical formula $b_i = n-i+1$. One then obtains the probability of drawing 1 white and 1 black ball at the $i^\text{th}$ draw (provided one have done so similarly up to this point in previous draws) as:
$P(b_i,w_i)=\frac{2w_ib_i}{(b_i+w_i)(b_i+w_i-1)}=\frac{2(n-i+1)^2}{(2(n-i+1))(2(n-i+1)-1)}=\frac{(n-i+1)}{2(n-i+1)-1}$
The probability $P(d)$ for a chain of consecutive draws of one black and one white happening, until box is empty (this is when $n$ draws have happened s.t. $i=n$) is simply the cumulative product of $P(b_i,w_i)$. We obtain:
$\prod_{i=1}^{i=n}{\frac{(n-i+1)}{2(n-i+1)-1}}=\frac{(2^{-n}(-n)_n)}{(0.5-n)_n}$
where $(x)_n$ is the falling factorial such that $(x)_n=x(x-1)(x-2)...(x-n+1)$. A table of values for a box with $2n$ balls, i.e. $n$ white and $n$ black balls and the probability of drawing one of each consecutively 2 at a time, until the box is empty, is given by:
$n$$P(d)\!\:$
1$\;\;\; 1\;\;\;$
2$\;\:\!\;\frac{2}{3}\;\:\;$
3$\;\:\!\;\frac{2}{5}\;\:\;$
4$\;\!\;\frac{8}{35}\;\:\;$
5$\;\!\;\frac{8}{63}\;\:\;$
6$\;\frac{16}{231}\;\!\;$ |
H: How to determine whether critical points (of the lagrangian function) are minima or maxima?
$f(x,y) = 2x+y$ subject to $g(x,y)=x^2+y^2-1=0$. The Lagrangian function is given by
$$
\mathcal{L}(x,y,\lambda)=2x+y+\lambda(x^2+y^2-1),
$$
with corresponding
$$
\nabla \mathcal{L}(x,y,\lambda)= \begin{bmatrix} 2 + 2\lambda x \\ 1+2\lambda y \\ x^2+y^2-1 \end{bmatrix}.
$$
From the latter we can see that $x=\frac{-1}{\lambda}$ and $y=\frac{-1}{2\lambda}$, which we can substitute into $x^2+y^2=1$ to obtain $\lambda = \pm\sqrt{\frac{5}{4}}$. Meaning that $x = \pm \dfrac{2}{\sqrt{5}}$ and $y= \pm \dfrac{1}{\sqrt{5}}$. We find the critical points $(\frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}})$ and $(\frac{-2}{\sqrt{5}}, \frac{-1}{\sqrt{5}})$.
I am confused on how I should proceed to check wether these points or minima or maxima? I know the Hessian is involved, but which one?
AI: In general when optimizing $f(x)$ subject to $g(x)=0$, you solve the problem $\nabla f(x)=\lambda \nabla g(x)$ and the critical points can be checked by the bordered Hessian matrix:
$$H=\begin{pmatrix}
0 & g_x & g_y\\
g_x & f_{xx}+\lambda g_{xx} & f_{xy}+\lambda g_{xy}\\
g_y & f_{yx}+\lambda g_{yx} & f_{yy}+\lambda g_{yy}
\end{pmatrix}.$$
Define $h:=\det(H)$. If $h>0$ at the critical point, you're at a maximum and if $h<0$ you're at a minimum. Equivalently, you can replace all instances of $g$ in the above Hessian with $-g$ so that the derivatives in the bottom right square become those of $f-\lambda g$. Indeed, consider expanding $f-\lambda g$ around each critical point taking into account that you are moving along $g$. |
H: Prove that $u \circ f $ is plurisubharmonic on $\Omega_1$
I'm trying to show that the theorem in my book:
Let $f: \Omega_1 \to \Omega_2$ be a holomorphic map between open sub - sets $\Omega_1, \Omega_2$ of $\Bbb C^n$. If $u$ is plurisubharmonic on $\Omega_2$ ($u \in PSH(\Omega_2)$) then $u \circ f $ is plurisubharmonic on $\Omega_1$ ($u \in PSH(\Omega_1)$).
======================================================
Assume that $u \in C^2(\Omega_2)$. I tried to show that $\forall z \in \Omega_1, \omega \in \Bbb C^n$ then $$\sum_{j,k=1}^n \dfrac{\partial^2 (u\circ f)}{\partial z_j \partial \overline{z_k}}(z)\cdot \omega_j \overline{\omega_k} \ge 0 \tag I$$
Since $u \in PSH(\Omega_2)$ we have $$\sum_{j,k=1}^n \dfrac{\partial^2 u}{\partial z_j \partial \overline{z_k}}(z)\cdot \omega_j \overline{\omega_k} \ge 0 \tag 1$$
======================================================
How can we show that $(I)$?
Any help is always appreciated! Thanks!
AI: For $u$ twice differentiable, it's a simple consequence of the chain rule. Since the $f_m$ are holomorphic, the general
$$\frac{\partial (u\circ f)}{\partial z_j} = \sum_{m}\left(\frac{\partial u}{\partial w_m}\circ f\right)\cdot \frac{\partial f_m}{\partial z_j} + \sum_m \left(\frac{\partial u}{\partial \overline{w_m}}\circ f\right)\cdot \frac{\partial \overline{f_m}}{\partial z_j}$$
simplifies to
$$\frac{\partial (u\circ f)}{\partial z_j} = \sum_{m} \left(\frac{\partial u}{\partial w_m}\circ f\right)\cdot \frac{\partial f_m}{\partial z_j}.$$
No $\frac{\partial u}{\partial \overline{w}_m}$ occur since the components $f_m$ are holomorphic, so $\frac{\partial \overline{f_m}}{\partial z_j} = 0$. Then differentiating with respect to $\overline{z_k}$ yields
$$\frac{\partial^2(u\circ f)}{\partial z_j\partial \overline{z_k}} = \sum_{m,n} \left(\frac{\partial^2 u}{\partial w_m \partial \overline{w_n}}\circ f\right)\cdot \frac{\partial f_m}{\partial z_j}\cdot \overline{\frac{\partial f_n}{\partial z_k}},$$
where the holomorphy of $f$ causes all terms $\frac{\partial^2 u}{\partial z_j\partial z_k}$ to be annihilated, and for the matrix of the Levi form, you get
$$L_{u\circ f} = J_f^\ast\cdot L_u\cdot J_f$$
(or with $\overline{J_f}$ instead of $J_f$, depends on which is the row and which the column index). If $L_u$ is positive semidefinite, any $A^\ast\cdot L_u\cdot A$ is also positive semidefinite. |
H: Inductive step in the induction: $\sum^{n}_{i=0} q^i = \frac {1-q^{n+1}}{1-q}\times2$
I am trying induction for the following formula:
$$\sum^{n}_{i=0} q^i = \frac {1-q^{n+1}}{1-q}\times2$$
I have done the initial step which gives me for $n=1$ for both sites $1+q$
In the inductive step I wrote:
$$(\sum^{n}_{i=0} q^i ) \times \frac {1-q^{n+1}}{1-q}\times2 =\frac {1-q^{n+1}}{1-q}\times2 \times \frac {1-q^{n+2}}{1-q}\times2 = 4 \times \frac {-q*q^n - q^2 q^n + 1 + q^3 q^{2n}}{1-q}$$
However, here I am stuck...
I really would appreciate your answer!!!
AI: Beginning note: all operations below assume that $q\ne 1$. To avoid other pitfalls, also assume that $q\ne 0$.
For induction on a sum, you would use the following process:
$$\sum^{n}_{i=0} q^i = \frac {1-q^{n+1}}{1-q}\times2$$
Show for $n=1$ (or $n=0$, if desired)
Assume statement is true for an arbitrary $n=k\gt 1$
Show statement holds for $n=k+1$
For the base case, use $n=1$ as the point where you started. Then we have:
$$\sum^{1}_{i=0} q^i = \frac {1-q^{1+1}}{1-q}\times2 \implies 1+q=\frac {(1-q)(1+q)}{1-q}$$
$$=\frac {1-q^2}{1-q}\ne \frac {1-q^2}{1-q}\times 2$$
Already, we have failed to show the required induction. At this point I would start again, ensuring that your formula is correct.
Continuing forward as if we had succeeded at the base step, we would have the following:
Show that:
$$\sum^{k+1}_{i=0} q^i = \frac {1-q^{k+2}}{1-q}\times2$$
To do this, we would use
$$\sum^{k}_{i=0} q^i = \frac {1-q^{k+1}}{1-q}\times2$$
(our assumption step) and add the $k+1$ term to both sides, like so:
$$\sum^{k}_{i=0} q^i + q^{k+1} = \frac {1-q^{k+1}}{1-q}\times2 + q^{k+1}$$
$$\implies \sum^{k+1}_{i=0} q^i = \frac {1-q^{k+1}}{1-q}\times2+\frac{(1-q)q^{k+1}}{1-q}$$
$$=\frac {2-q^{k+1}-q^{k+2}}{1-q}\ne \frac {1-q^{k+2}}{1-q}\times2$$
The problems we found at the base case continue through the entire induction process and show again that the original formula as stated cannot be correct.
A related form of your formula looks like this in the final steps:
$$\sum^{k+1}_{i=0} q^i = \frac {1-q^{k+1}}{1-q}+\frac{(1-q)q^{k+1}}{1-q}$$
$$=\frac {1-q^{k+1}+(1-q)q^{k+1}}{1-q}=\frac {1-q^{k+2}}{1-q}$$ |
H: Are all values of $\sin(x)$ algebraic.
Can we prove that for all $x$ in $(0,2\pi)$ $\sin(x)$ is an algebraic number?
I have seen people express various values of $\sin(x)$ like $\sin(3)$ and $\sin(30)$ using radicals so I suspect that all values of $\sin(x)$ must be algebraic. Is that correct? Can we prove it?
AI: Quick answer: no. Consider that $\sin$ is continuous over the range $[-1,1]$, and thus can take on values such as $e^{-1}, \frac 2\pi$, etc. |
H: Prove that $\lim_{n\to\infty} a_k$ is nonnegative for a convergent sequence of nonnegative terms $a_k.$
Suppose we have a convergent sequence $(a_k)$ such that $a_k\ge 0 $ for all $k\ge 1.$
Show that $\lim_{n\to\infty} a_k\ge 0$.
I have to prove by contradiction. This is the first time I've dealt with a question like this. Can someone give me help?
AI: Hint: Take $x<0$ and put $\epsilon=-x=|x|>0.$ What can you say about $|a_k-x|$ for all $k$, given your hypotheses? |
H: Evaluate the limit of $(\sqrt{5-x}-2)/(\sqrt{2-x}-1)$ as $x\to 1$
Can you help me with it and explain the steps
$$\lim_{x \to 1} \frac{\sqrt{5-x}-2}{\sqrt{2-x}-1}$$
I tried to multiply at conjugate expression but I failed.
AI: Hint: $$\frac{\sqrt{5-x}-2}{\sqrt{2-x}-1}=\frac{(\sqrt{5-x}-2)(\sqrt{5-x}+2)(\sqrt{2-x}+1)}{(\sqrt{2-x}-1)(\sqrt{2-x}+1)(\sqrt{5-x}+2)}.$$ |
H: Where does log(x) / x take maximum value?
If the base of the logarithm is e, one can say log(x)/x takes maximum at e.
If the base of the logarithm is 10, one can say log(x)/x takes maximum at 10.
But log10(x)/x is nothing but (loge(e)/loge(10))/x.
The two functions are just a constant multiple (1/loge(10)) of each other. Shouldn't they have the same maxima?
AI: Your last sentence is correct, whatever base $b$ we take (as long as $b\gt 1)$, the function $\frac{\log_b x}{x}$ reaches a maximum at $x=e$. The reasoning that led to that conclusion is clearly stated.
The assertion in the second sentence that $\frac{\log_{10} x}{x}$ reaches a maximum at $x=10$ is not true. |
H: Check whether W is a T-invariant subspace of V
$V$ = $C([0,1])$
$T(f(t))$ =[ $\int\limits_0^1f(x)dx$]$t$
$W$ = {$f \in V $ : $f(t)$ = $at+b$ $for$ $some$ $a$ $and$ $b$}
I got $T(f(t))$ for $f(t)$ = $at+b$ is,
[ $\int\limits_0^1(ax+b)dx$]$t$ = $(a/2 + b)$$t$ ,which does not belong to $W$.
Hence $W$ is not a $T$-invariant subspace of $V$.
But my book says otherwise i.e. $W$ is a $T$-invariant subspace of $V$.
I can't figure out where i went wrong ?
AI: Why doesn't $\left(\frac a2+b\right)t$ belong to $W$? It is of the form $ct+d,$ where $c,d\in\Bbb R$, yes? |
H: $\sigma$-algebra generated by open sets coincides with $\sigma$-ring generated by open sets.
Under the topic of Metric spaces in my measure theory book I came across this definition that says:
"Denote by $B$ the $\sigma$-ring generated by the class of all the open sets of X. The sets of $B$ are called Borel-sets."
"Note that $B$ coincides with the $\sigma$-algebra generated by the open sets of X."
I saw a nice example about a sigma-algebra being generated, but this note I don't really get, also maybe because I'm not sure how to generate a sigma-ring.
Thanks for any hint or tip!
AI: Let $B$ denote the $\sigma$-ring generated by the open sets and $C$ the $\sigma$-algebra generated by them. Obviously $B \subseteq C$, it suffices to show that $X \in B$, as a $\sigma$-ring containing the base set is a $\sigma$-algebra. But $X$ is open. |
H: Find $\lim_{x\to 2} \frac{x - \sqrt{3 x - 2}}{(x^2 - 4)}$
Can you help me with this limit?
What do I have to do ?
I tried to multiply by the conjugate expression but it didn't work
$$
\lim_{x\to 2} \frac{x - \sqrt{3 x - 2}}{(x^2 - 4)}.
$$
Thanks
AI: Hint: After multiplying by the conjugate note that $x^2-(3x-2)=x^2-3x+2=(x-1)(x-2)$. |
H: Column Space of AA' is equal to column of A.
This is my question. How to show that the column space of matrix A
is just equal to the column space of AA'?.. A' represents the transpose of A.
I know that the column space of AA' is a subset of the column space of A
which is just trivial. But the other way around I still used inclusion
but it seemed that it is going nowhere. Maybe it can be done by some manipulation
or the barbaric way of doing this. I just want to obtain a simple proof. Anyone?
AI: Hint. If you can show that $A'x=0$ whenever $AA'x=0$, then $\operatorname{nullity}(AA')\le\operatorname{nullity}(A')$ and hence $\operatorname{rank}(AA')\ge\operatorname{rank}(A')=\operatorname{rank}(A)$, i.e. the dimension of the column space of $A$ does not exceed the dimension of the column space of $AA'$.
If you really want a super-duper simple proof, use singular value decomposition. |
H: relation between Jordan form and Jordan decomposition
How to get the Additive Jordan Decomposition of a matrix from its Jordan Canonical Form?
I tried with few matrices and its Jordan Canonical Forms, but I could not understand. So please explain the relation.
Thanks in advance.
AI: Let $A = S J S^{-1}$ be a Jordan decomposition. Note that $J$ has a form $J = D + N$, where $D$ is diagonal, and $N$ is strictly upper triangular with zeroes everywhere except some positions of the main superdiagonal, where it has ones. In other word,
\begin{align*}
J &= \begin{bmatrix} J_{ij} \end{bmatrix}, \quad J_{ij} = \begin{cases} \text{$0$ or $1$}, & i = j-1, \\ \lambda_i, & i = j, \\ 0, & \text{otherwise}. \end{cases}, \\
D &= \operatorname{diag}(\lambda_1,\dots,\lambda_n), \\
N &= \begin{bmatrix} N_{ij} \end{bmatrix}, \quad N_{ij} = \begin{cases} \text{$0$ or $1$}, & i = j-1, \\ 0, & \text{otherwise}. \end{cases}.
\end{align*}
So,
$$A = S J S^{-1} = S (D + N) S^{-1} = \underbrace{S D S^{-1}}_{A_s :=} + \underbrace{S N S^{-1}}_{A_n :=}.$$
Is this what you had in mind? |
H: show that a measure is complete
If $(X,\mathcal{M},\mu)$ is a measure space and $\mathcal{\overline{M}}:=\{E\cup F:E\in\mathcal{M}\text{ and }F\subset N\text{ for some }N\in \mathcal{N}\}$ is a completion of $\mathcal{M}$ with respect to $\mu$ where $\mathcal{N}:=\{N\in\mathcal{M}:\mu(N)=0\},$ then $\mathcal{\overline{M}}$ is a $\sigma$-algebra.
If now $\overline{\mu}(E\cup F):=\mu(E)$ how to show that $\overline{\mu}$ is a complete measure?
I suspect that we need to show that any subset of a null-set is measurable.
AI: Complete just means that if $A$ is a measurable set with $\overline{\mu} A = 0$, and $B \subset A$, then $\overline{\mu} B = 0$ also.
Suppose $A \in \overline{M}$ such that $\overline{\mu} A = 0$. Then $A = E \cup F$,
where $E \in M$ and $F \subset N$, with $N \in {\cal N}$. Since $\overline{\mu} A = 0$, we have $\mu E = 0$ also. In particular, $A \subset E \cup N$, and $\mu(E \cup N) = 0$.
Now suppose $B \subset A$. The we can write $B = \emptyset \cup B$, and $B \subset E \cup N$, where $E \cup N \in {\cal N}$. Hence $B \in \overline{M}$, and $\overline{\mu} B = \mu \emptyset = 0$. |
H: derivative of an integral which has a bound with multiple variables?
I want to find the derivative with respect to x of:
$$\int_0^{{\frac{x}{\sqrt4t}}} {e}^{-s^2}\,\mathrm{d}s$$
where t and x are both independent variables.
I thought you should use the fundamental theorem of calculus. However, since the upper bound of the integral is in terms of t and x, does this complicate the question?
Do I need to somehow use the chain rule to get it in terms of just x?
Any help would be greatly appreciated.
AI: You can use the Leibiz integral rule to calculate the answer. http://en.wikipedia.org/wiki/Leibniz_integral_rule |
H: Ways to prove the Zsigmondy's theorem
How to prove the Zsigmondy's Theorem with cyclotomic polynomials?
Is there a proof which does not rely on cyclotomic polynomials? How many ways are there?
AI: There are several proofs available for Zsigmondy's theorem: Zsigmondy (1892), Birkhoff and Vandiver (1904), Dickson (1905), Artin (1955), Hering (1974) and Lüneburg (1981).
All of them use cyclotomic polynomials (which is natural).
If there is some proof without it, it might be just artificial.
A reference for a proof is (among many other sources) the article "On Zsigmondy primes" by
M. Roitman (1997). |
H: How to determine whether a line is at the clockwise side of another line?
This is my first post in math-overflow , I am trying to implement an algorithm where , i am given 2 lines , two lines have one point in common , i need to determine
if one line is at the clockwise side of another line ? My points are in 2D.
Let , p is a point not actually in any of the x-axis or y-axis .
There is ray(Line) from pw3
Now I need to determine The edges incident to w3 => w3w1 and w2w3 , are they in the clockwise or anticlockwise side of the Ray ?
Like in this scenario both the edges are at anti clockwise direction to pw3
But in scenario 2, when the Ray is passing through pw2, edge w2w3 is at clockwise direction whereas w1w2 is at anti-clockwise direction
How to possibly determine this ?
AI: For points $P$, $A$ and $B$ in the $xy$ plane, the sign of the $z$ component of the 3-dimensional cross product of $\vec{PA} \times \vec{PB}$ is positive (negative) if $B$ is on the left-hand side (right-hand side) of the ray from $P$ towards $A$, and zero if $B$ is exactly on the line $PA$. See e.g. the illustrations in the Wikipedia page Cross product. You can use this to determine on which side the endpoints of the line segments, and thus the line segment, are. |
H: Can someone help me solve this limits question?
$$\begin{align}\lim x → ∞\end{align}$$
$$\begin{align}
f(x) = {\frac{2^{x+1}+{3^{x+1}}}{2^x + 3^x}} \\
\end{align}$$
I tried using L Hopitable but that gives the same expression. Also tried using substitution but I didn't get anywhere. Help would be appreciated.
AI: To begin with
$$f(x)=\frac{2\cdot(\frac23)^x+3}{(\frac23)^x+1}$$
it follows that
$$\lim_{x\to+\infty}f(x)=3.$$
However
$$f(x)=\frac{2+3\cdot(\frac32)^x}{1+(\frac32)^x}$$
in such a case, we get that
$$\lim_{x\to-\infty}f(x)=2.$$ |
H: Best way to learn from textbooks that have the Theorem-Proof format?
I've always loved the theorem-proof format in textbooks, e.g. Hardy and Wright's Introduction to the theory of numbers. However, the problem is that I can't remember anything I read in this format because usually books of this kind don't have exercises at the end (not all, but usually). What do you think is the best way to actually learn from theorem-proof type books? Should I read a section, write down all the theorems and then close the book and try proving them myself a week later or so? Should I just take notes as I read and construct my own problems to practise with as I go along?
Thanks! :)
AI: I think no textbook is really only Theorem Proof, next Theorem, next Proof. Watch out for the motivation for theorems, look at the examples, try to consider the history. My suggestion is basically to not only look at one book. It won't help you in mathematical research to know a list of theorems. The theorems are just the cornerstones in a theory and what you really want is to understand the entire theory. This understanding comes from working within and with the theory, for instance by trying to prove for yourself new theorems in this area.
One concrete hint: You can always make your own exercises by checking why all the requirements of any given theorem are necessary. |
H: Uniqueness of map by dot product
I know that for a map on a complex vector space we have that if $\langle Ax,x \rangle = 0$
then $A = 0$ via the standard polarization trick. But what is the case if we are talking about real vector spaces? Is this then also true?
AI: No, consider the rotation by $90^\circ$ in $\mathbb{R}^2$. |
H: Laplace Equation -> Boundary Conditions
Given $\nabla^2 \phi=0$,
With B.C.
$\Gamma_1 =-V$ (Left side of rectangle)
$\Gamma_2 = V$ (right side of rectangle)
$\Gamma_0 = 0$ (top and bottom sides of rectangle)
Can Separation of Variables be applied to this problem? Will the superposition principle have to be applied to the second ODE because there are two non-zero BC?
AI: Absolutely. First write $\phi(x,y) = \phi_1(x,y)+\phi_2(x,y)$, where:
$$\nabla^2 \phi_1 = 0$$
$$\phi_1(x,0) = 0$$
$$\phi_1(x,d)=0$$
$$\phi_1(0,y) = -V$$
$$\phi_1(w,y) = 0$$
and
$$\nabla^2 \phi_2 = 0$$
$$\phi_2(x,0) = 0$$
$$\phi_2(x,d)=0$$
$$\phi_2(0,y) = 0$$
$$\phi_2(w,y) = V$$
We may write $\phi_1(x,y) = X(x) Y(y)$ to find that
$$\frac{X''}{X} = -\frac{Y''}{Y} = k^2$$
Then
$$Y(y) = A \cos{k y} + B \sin{k y}$$
The boundary conditions in $y$ imply that $A=0$ and $k = n \pi/d$. For the $x$ equation, we may write
$$X(x) = C_n \cosh{\frac{ n \pi (w-x)}{d}} + D_n \sinh{\frac{ n \pi (w-x)}{d}}$$
I wrote the solution this way so that we can have $C=0$ from the boundary condition at $x=w$. The general solution is then
$$\phi_1(x,y) = \sum_{n=1}^{\infty} D_n \sinh{\frac{ n \pi (w-x)}{d}} \sin{\frac{ n \pi y}{d}}$$
We find the $D_n$ from the B.C. at $x=0$. By orthogonality, we have
$$D_n = \frac{2}{d \sinh{\frac{ n \pi w}{d}}} \int_0^d dy (-V) \sin{\frac{ n \pi y}{d}} = -\frac{2 V}{n \pi \sinh{\frac{ n \pi w}{d}}} \left ( 1-\cos{n \pi}\right )$$
We find $\phi_2$ similarly, except the expansion is now given by
$$\phi_2(x,y) = \sum_{n=1}^{\infty} C_n \sinh{\frac{ n \pi x}{d}} \sin{\frac{ n \pi y}{d}}$$
We find the $C_n$ from the B.C. at $x=w$:
$$D_n = \frac{2}{d \sinh{\frac{ n \pi w}{d}}} \int_0^d dy (V) \sin{\frac{ n \pi y}{d}} = \frac{2 V}{n \pi \sinh{\frac{ n \pi w}{d}}} \left ( 1-\cos{n \pi}\right )$$
Putting this altogether, I get as the sought-after solution:
$$\phi(x,y) = \frac{4 V}{\pi} \sum_{k=0}^{\infty} \frac{\sinh{\frac{(2 k+1) \pi x}{d}}-\sinh{\frac{(2 k+1) \pi (w-x)}{d}}}{(2 k+1) \sinh{\frac{(2 k+1) \pi w}{d}}} \sin{\frac{(2 k+1) \pi y}{d}} $$ |
H: "$x$ times as many as" versus "$x$ times more than"?
I keep coming across such questions during my GRE preparation:
A is how many times of B versus A is how many times greater than B
A is what percentage of B versus A is what percentage greater than B
Some websites treat both phrasings in each pair as meaning the same, however I believe that "$A$ is __times greater than $B$" is asking for $$\frac AB-1.$$ As such, I think that these four options are all correct:
Let A=9 and B=3.
1: A is 3 times of B.
2: A is 2 times greater than B.
3: A is 300% times of B.
4: A is 200% times greater than B.
Yet the author of the following question would pick only options 1 and 3 above:
What is the right convention to interpret these phrasings?
AI: The phrase "A is what % of B" should be written as $A=x\cdot B$. And now solve for x, and then multiply by 100.
Example 1a: If A is 100, and B is 50, then $100=x\cdot 50$, means that $x = 2$, and A is 200% of B.
The phrase "A is what % greater than B," should be written as $A=x\cdot B$, just as before. But now, when you solve for x, and multiply by 100, you want to take the additional step of subtracting 100. Notice that this will only work if A is actually greater than B.
Example 1b: In the above example, A would be 100% greater than B.
Example 2: if A is 150, and B is 100, then solving for x in $A=x\cdot B$, would give us $x = 1.5$, and so A is 150% of B. But A is 50% greater than B. |
H: Cauchy-Schwarz inequality in $\mathbb{R}^3$
Use the Cauchy-Schwarz inequality on the euclidean space $ \mathbb{R}^3 $ (usual inner product) to show that, given 3 strictly positive numbers $a_1, a_2, a_3$ we have
$$ (a_1 + a_2 + a_3) \left(\frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3}\right) \geq 9 $$
My attempt was to expand this inequality and check if I could find 2 vectors of $\mathbb{R}^3 $ to prove the inequality. We have:
$ 1 + 1 + 1 + \frac{a_1}{a_2} + \frac{a_1}{a_3} + \frac{a_2}{a_1} + \frac{a_2}{a_3} + \frac{a_3}{a_1} + \frac{a_3}{a_2} \geq 9 \Leftrightarrow$
$a_1^2a_2 + a_1^2 a_3 + a_2^2 a_1 + a_2^2 a_3 + a_3^2 a_1 + a_3^2 a_2 \geq 6 a_1 a_2 a_3 \Leftrightarrow$
$ (a_1 + a_2)^3 + (a_1+a_3)^3 + (a_2+a_3)^3 \geq 2a_1(a_1^2 + a_2a_3) + 2a_2(a_2^2 + a_1a_3) + 2a_3(a_3^2 + a_1a_2)$
so I tried $v=(2a_1, 2a_2, 2a_3) $ and $u=(a_1^2 + a_2a_3, a_2^2 + a_1a_3, a_3^2 + a_1a_2)$ but I had no sucess (well, I couldn't verify the left side of the inequality). Maybe a should put the coefficients not like this
Is there a better way to solve it? Maybe using a consequence of the cauchy-schwarz inequality
Thanks for the help!
AI: If you want it using Cauchy-Schwarz, set $b_i=\sqrt{a_i}$. Then CS tells you that $$\left(\sum (\frac{1}{b_i})^2\right)\left(\sum b_i^2\right)\ge \left(\sum b_i\frac{1}{b_i}\right)^2=3^2=9$$ |
H: domain of composition of continuous functions
Let $f$ and $g$ be continuous functions from $R^2$ to $R^2$.
$f$ and $g$ both have open domains. That is, $f$ and $g$ are both defined on open sets of $\mathbf{R}^2$.
Is it true that the domain of $f(g)$ is also open?
By domain of $f(g)$, I mean the set of all $x \in \mathbf{R}^2$ such that $f(g(x))$ is well defined.
How to prove this?
AI: Edit: In fact, we have more conditions than we need. It suffices that $g$ is continuous and that $\operatorname{dom}(f),\operatorname{dom}(g)$ are open, so that $$\begin{align}\operatorname{dom}(f\circ g) &= \{x\in\Bbb R:x\in\operatorname{dom}(g)\text{ and }g(x)\in\operatorname{dom}(f)\\ &= \{x\in\operatorname{dom}(g):g(x)\in\operatorname{dom}(f)\}\\ &= g^{-1}[\operatorname{dom}(f)]\end{align}$$ is relatively open in $\operatorname{dom}(g)$ as the preimage of an open set under a continuous function, and so open in $\Bbb R^2$ since $\operatorname{dom}(g)$ is open.
Now, if we want $f\circ g$ to be continuous, then we will in general need $f$ to be continuous, as well. |
H: Common direction between two vectors
I have two vectors with the same origin and I need to find the common direction between them, that is the vector perpendicular to the line that join them. For instance, referring to this image I need the vector that points perpendicularly from $A$ to the line joining $B$ and $D$.
I am very rusty at algebra, can someone help me?
AI: From the picture that you link to the line that you want is the orthogonal projection of the vector $v = \overrightarrow{AB}$ onto the vector $u = \overrightarrow{BD}$. What this means is you want the vector which is $v$ minus the component of $v$ that points in the direction of $u$. The vector that you want is given by
$\displaystyle v - \frac{\langle v, u\rangle}{\langle u, u\rangle}u$
where $\langle\ \cdot\ ,\ \cdot\ \rangle$ is the inner product.
To see where this comes from consider that we have two arbitrary vectors $u,\ v \in \mathbb{R}^{2}$. Consider that $v$ can be represented as
$v = \alpha u + u^{\perp}$
for $u^{\perp} \perp u$ (i.e. $u^{\perp}$ is perpendicular to $u$). Now consider that perpendicular vectors have zero inner product, such that
$\langle u, u^{\perp}\rangle = 0$
and so
$\langle v, u \rangle = \langle \alpha u + u^{\perp}, u \rangle \\
\quad \quad\ = \alpha\langle u , u \rangle$
from which
$\displaystyle \alpha = \frac{\langle v, u\rangle}{\langle u, u \rangle}$
implies that
$\displaystyle u^{\perp} = v - \alpha u = v - \frac{\langle v, u \rangle}{\langle u, u \rangle}u.$
Since $u^{\perp}$ was defined to be perpendicular to $u$, we have the orthogonal vector to $u$.
If you want to brush up on your linear algebra then 'Linear Algebra and its Applications' by Strang is a very good book |
H: Solving system of equations by using simple iteration method
I have a problem:
$$\begin{cases}
\sin(x) + 2y = 2 \\
\cos(y - 1) + x = 0.7
\end{cases}
$$
with margin of error 0.00001
And I need to solve this by using Fixed-point iteration method.
Can someone help me with that? Algorithm for full solution might be the best thing ever, since I gave to solve this in Maple later..
Thanks is advance.
P.S. one more question. What is q in this method and how to find it?
AI: Try this iteration:
$
\qquad x_{n+1}=0.7-\cos(y_n-1)
$
$
\qquad y_{n+1}=(2-\sin(x_n))/2
$
Taking $x_0=y_0=0$ works fine for me. I get that
$\qquad x^*=-0.28980932884903, \quad y^*= 1.1428847552227$
is a fixed point of $F(x,y)=(0.7-\cos(y-1),2-\sin(x))/2)$
after 27 steps,
but it only takes 11 steps to get an error below $10^{-6}$. The error is measured by $\| F(x,y)-(x,y) \|_1$, where $\|(x,y)\|_1 = \max(|x|,|y|)$. |
H: common between family of lines
suppose that there is a question like this :
i just want to make sure that i have understand everything,maybe it is meant by this question that graph structure is common,i mean their graph is straight line right?because let us take several value of $m$
$1.m=1$
$f(x)=1+(x+3)=x+4$
$2.m=2$
$f(x)=1+2*(x+3)=2*x+7$
$3. m=0$
$f(x)=1$
we see that for different $m$,we have different set of linear forms,so common is their graph right?others like $x$ intersection or $y$ intersection clearly is not common,as well slopes,am i wrong?
AI: Observe that the point $\;(-3,1)\;$ is common
to them all:
$$1=1+m(-3+3)\ldots$$ |
H: Showing a set is open in $\mathbb{R}^2$
Show that $\mathbb{R}^2 \setminus \{(0,0)\}$ is open in $\mathbb{R}^2$.
I'm not sure if this is obvious I can't give enough details. Every point in $U=D(v,||v||)$ has a $\epsilon$-neighbourhood contained in U such that $v=(a,b)\ne (0,0)$.
Can someone please help me on describing the details? Than you.
AI: Have you proved that open disks are topologically open? If you have (or if you can), then it follows that $D(v,\lVert v\rVert)$ will be an open neighborhood of $v$ contained in $\Bbb R^2\setminus\{(0,0)\}$ for any $v\in\Bbb R^2\setminus\{(0,0)\}.$ |
H: Complex integral of exponential and power
Let $n$ be an integer. Compute $$\int_{|z|=1}e^zz^{-n}dz.$$
If I parametrize $z(t)=e^{it}$ for $t\in[0,2\pi]$, this becomes $$\int_0^{2\pi}e^{e^{it}}e^{-nit}ie^{it}dt = i\int e^{e^{it}-(n-1)it} dt$$ and this looks too complicated.
Also I thought about using Cauchy's formula, but the function $e^zz^{-(n-1)}$ might not even be defined at $z=0$.
How can I find this integral?
AI: The Cauchy Integral formula tells you that the integral is
$$ f^{n-1}(0) = \frac{(n-1)!}{2 \pi i} \int_{|z|=1} \frac{f(z)}{z^n}dz$$
from which you get $\displaystyle \frac{2 \pi i}{(n-1)!}$
The other way as AlexR mentioned in comment, we look for the residue of the integrand in the domain $|z| \le 1$.
$$z^{-n} e^z = z^{-n} \sum_{k=0}^\infty \frac{z^{k}}{k!}$$
The residue is $\displaystyle \frac{1}{(n-1)!}$ which is coefficient of $z^{-1}$. |
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