text stringlengths 83 79.5k |
|---|
H: Prove that if a function $f: X\to Y$ continuous then its graph is closed
The graph of $f$ is $G(f) = \{(x,f(x)) : x\in X\} \subseteq X\times Y$
$X$ and $Y$ are metric spaces.
a) Suppose $f$ is continuous and prove that $G(f)$ is a closed set.
b) Suppose that $G(f)$ is compact and prove that $f$ is continous
For a), the definition of a closed set that comes to my mind is a set that contains all its limit points (or was it accumulation points?), is there another equivalent definition that may b more helpful to prove a)? Is it possible to prove this directly? Because at first glance the only way I could imagine to prove this is by contradiction or contrapositive.
I imagine that the proof of b) will be immediately derived from a).
AI: a) Let $(z_n)=(x_n,f(x_n))$ be a convergent sequence of $G(f)$. If $(x,y)$ is its limit, show that $y=f(x)$.
b) Let $x\in X$ and $(x_n)$ a convergent sequence with limit $x$. You have to prove that $(f(x_n))$ is convergent in $Y$ with limit $f(x)$. Use the sequence $z_n=(x_n,f(x_n))$ and use the fact that $G(f)$ is compact to prove that $(f(x_n))$ has $f(x)$ as an accumulation point.
Then prove that any subsequence of $(f(x_n))$ has $f(x)$ as an accumulation point. |
H: Structure constants for $\mathbb{F}_q$ as an $\mathbb{F}_p$-algebra
Let $k'=\mathbb{F}_q$ and $k=\mathbb{F}_p$, where $q=p^d$. We may regard $k'$ as a $d$-dimensional $k$-algebra via $k\hookrightarrow k'$. For any choice of $k$-basis $\{e_1,\ldots,e_d\}$ we obtain structure constants $a_{ij}^l$ giving the multiplication of basis elements:
$$e_i\cdot e_j=\sum_{l=1}^da_{ij}^le_l$$
I'm trying to define this algebra structure on $\mathbb{F}_q$ in Magma, so I need an explicit description of the $a_{ij}^l$. Is there a way to pick a particularly nice basis so that we can arrive at a formula for the $a_{ij}^l$?
I tried starting with $e_1$, a generator of the multiplicative units of $k'$, and considering the basis of the images of $e_1$ under the Galois group, that is, $e_i=e_1^{p^i}$. But it is not clear to me how to express $e_1^{p^i+p^j}$ as a linear combination of the $e_l$.
AI: Eltseq, when applied to, e.g., an element of GF(q) will give you the coefficients of an element of $\mathbb{F}_q$ relative to the power basis. I'm not sure if you have to construct the field in a special way to get this to work.
Are you sure whatever you're trying to do can't be done with magma's finite field functionality instead? |
H: Is it possible to remove a variable from an expression?
Given the Expression:
$$\dfrac{N(N+1)(2N+1)}{6} + \dfrac{6N}{6}− \dfrac{3N(N+1)}{6}$$
Is it possible to remove N from the third expression to give -3 / 6, giving a final expression of:
$$\dfrac{N(N+1)(2N+1-3)+6N}{6}$$
If so, what is this called. And how can I learn more about when to do it.
Many thanks
AI: You're thinking of the distributive property. And yes you can do this with it. One way to think of it is to say $\frac{N(N+1)}{6}$ is a thing. I have 2N+1 or them over here, and -3 of them over here. They are being added(or subtracted) together, so I can put them together. |
H: Solving a PDE - do you have an idea?
Do you have an idea how to solve
$$
v_{\xi\eta}=\frac{1}{2} v_{\xi}\cdot\xi?
$$
First I thought of using
$$
v_{\xi\eta}=v_{\eta\xi},
$$
substituting $z:=v_{\xi}$ and then getting
$$
\frac{dz}{d\eta}=\frac{1}{2}z\cdot\xi.
$$
First I thought that I can solve this by separation of variables, but it is not the needed form to do so...
AI: \begin{align}
v_{\xi \eta} &= \frac 12 v_\xi \xi \\
\frac {v_{\xi \eta}}{v_\xi} &= \frac 12 \xi \\
\left( \ln v_\xi \right )_\eta &= \frac 12 \xi \\
\ln v_\xi &= \frac 12 \xi \eta + C_1(\xi) \\
v_\xi &= e^{C_1(\xi) + \frac 12 \xi \eta} \\
v &= \int e^{C_1(\xi) + \frac 12 \xi \eta} d\xi + C_2 (\eta)
\end{align} |
H: Showing that $A(p)\circ A(q)=p \circ q$ for every $p,q\in \mathbb{R}^3$
Let $\{x_A,y_A,z_A\}$ be an orthonormal set of vectors in $\mathbb{R}^3$. We call the coordinate system by these vectors $A$. Let $v\in \mathbb{R}^3$. We define $ A(v)$ to be the unique vector with components $a,b,c$ such that $v=ax_A+by_A+cz_A$ (uniqueness follows because $\{x_A,y_A,z_A\}$ is a base) . It follows easily from orthogonality relations that $a=x_A\circ v,b=y_A\circ v,c=z_A \circ v$ .
Now I believe that the following is true:
For any two vectors $p,q\in \mathbb{R}^3$, $A(p)\circ A(q)=p \circ q$.
Things get messy when I try to prove it. Is there a smart way to see this ?
Thank you
Note: I am currently taking a robotics course, to which knowledge of rotation matrices is useful. I never learnt anything about rotation matrices before. I only took a weak non-rigorous engineering linear algebra course. Thus, I would like to see a proof without assuming knowledge about rotation matrices. My robotics book introduces rotation matrices in a non-rigorous, and I'd like to make things very rigorous. Thanks for your help.
AI: By definition, we have
$$p = (x_A \circ p)\cdot x_A + (y_A\circ p)\cdot y_A + (z_A \circ p)\cdot z_A,$$
and $$A(p) = \begin{pmatrix}x_A\circ p\\y_A\circ p\\z_A\circ p \end{pmatrix},$$
and similar for $q$.
So we have
$$\begin{align}
A(p)\circ A(q) &= (x_A\circ p)\cdot (x_A\circ q) + (y_A\circ p)\cdot(y_A\circ q) + (z_A\circ p)\cdot(z_A\circ q)\\
&= ((x_A\circ p)\cdot x_A)\circ q + ((y_A\circ p)\cdot y_A)\circ q + ((z_A\circ p)\cdot z_A)\circ q\\
&= ((x_A \circ p)\cdot x_A + (y_A\circ p)\cdot y_A + (z_A \circ p)\cdot z_A)\circ q\\
&= p\circ q
\end{align}$$
by the bilinearity of the inner product. |
H: Proving Wilson's theorem for $n=11$
Wilson's theorem establishes that if a $n$ number is prime then:
\begin{align}
(n-1)! &\equiv -1\ \textrm{mod}\ (n)
\end{align}
I have probed the theorem for the particular case where $n = 7$ like this:
I first consider the set $\{2,..,n-2\}$, in my case, $\{2,3,4,5\}$ and then I take the pair of numbers $a$ and $a¹$ where:
\begin{align}
2.4 &\equiv 1\ \textrm{mod}\ (7)
\end{align}
\begin{align}
3.5 &\equiv 1\ \textrm{mod}\ (7)
\end{align}
Then,
\begin{align}
2.3.4.5 &\equiv 1\ \textrm{mod}\ (7)
\end{align}
So,
\begin{align}
6 &\equiv -1\ \textrm{mod}\ (7)
\end{align}
And finally,
\begin{align}
6! &\equiv -1\ \textrm{mod}\ (7)
\end{align}
How can I prove the theorem where $n=11$? How can I retrieve $a$ and $a¹$?
AI: Answer to Old Question
You can't prove the statement when $n=8$ because it's false. Note $7!$ is $0$ mod $8$.
Wilson's theorem is in fact an if and only if statement. The proposition $(n-1)! \equiv -1 \pmod n$ is true if and only if $n$ is prime.
Answer to Edit
To answer your new question, use the pairs $(1,10)$, $(2,6)$, $(3,4)$, $(5,9)$, and $(7,8)$. |
H: "Nested" recursion preserves primitive recursive functions
Problem: Assume the functions $f$, $\pi$, and $g$ are given. They take one, two, and three arguments respectively. Prove a unique function $h$ exists such that:
$$h(0,y)\cong f(y)$$
$$h(x+1,y)\cong g(x,y,h(x,\pi(x,y)))$$
Furthermore, prove that if the functions $f$, $\pi$, and $g$ are primitive recursive, then so is $h$.
Attempted solution:
We have
$h(0,y)=f(y)$;
$h(1,y)=g(0,y,h(0,\pi(0,y))=g(0,y,f(\pi(0,y))$;
$h(2,y)=g(1,y,h(1,\pi(1,y))=g(1,y,g(0,\pi(1,y),f(\pi(0,\pi(1,y))))$;
$h(3,y)=g(2,y(h(2,\pi(2,y)))=g(2,y,g(1,\pi(2,y),g(0,\pi(1,\pi(2,y)),f(\pi(0,\pi(1,\pi(2,y))))))$
Let us define $P$ by:
$P(0,x,y)=y$
$P(n+1,x,y)=\pi(x \dot - (n+1),P(n,x,y))$
Now I need a new function $Q$ such that
$Q(0,x,y)=P(x,x,y)$
How should I continue? How should I go about the uniqueness of $h$? Any help would be appreciated.
AI: For uniqueness, suppose that there are $h_1$ and $h_2$ that both satisfy your conditions. If they are not equal, then there must be a minimal $x$ such that $h_1(x,y)\ne h_2(x,y)$ for some $y$. Argue that because $x$ is minimal $h_1(x,y)$ must equal $h_2(x,y)$, which is a contradiction.
For "primitive recursive", it looks like you're going in the right direction with your $P$, but if would be easier to rearrange things such that $n$ (which you sometimes call $s$?) is the first argument. Then $P(n,x,y)$ is easily seen to be primitive recursive.
I don't think you need the $Q$. Instead you need to rewrite the equations for $h$ such that the varying $y$ argument is computed anew from the original $y$ using $P$ for each iteration:
$$ H(0,x_0,y_0) = f(P(x_0,x_0,y_0)) $$
$$ H(x+1,x_0,y_0) = g(x,P(x_0\dot-(x+1), x_0, y_0), H(x,x_0,y_0))$$
$H$ is then clearly primitive recursive, and you should be able to prove by appropriate induction arguments and under appropriate conditions on the variables (unless I've made a fencepost error somewhere, which I probably have, in which case find and fix it!) that
$$ P(n+1,x_0+1,y) = P(n,x_0,\pi(x_0,y)) $$
and then
$$ H(x,x_0+1,y) = H(x,x_0,\pi(x_0,y)) $$
and then
$$ h(x,y) = H(x,x,y) $$
which gives an alternative definition for $h$ that shows it to be primitive recursive. |
H: Integral of quartic function in denominator
I'm sorry, I've really tried to use MathJaX but I can't get integrals to work properly.
indefinite integral
$$\int {x\over x^4 +x^2 +1}$$
I set it up to equal
$$x\int {x\over x^4 +x^2 +1} - \int {x\over x^4 +x^2 +1}$$
$$\text{so } (x-1)\int {1\over x^4 +x^2 +1}$$
OKAY, now I set the denominator to $(X^2 +.5)^2 + \frac 34$
So I multiplied the top and bottom by $\frac 43$
then I absorbed it into the squared quantity by dividing it (within the parenthesis) by $\sqrt 3\over 2$
so
$${1\over \left({X^2 +.5\over {\sqrt 3\over 2}}\right)^2 + 1}$$
so $\arctan\left({X^2 +.5\over{\sqrt 3\over 2}}\right)$
FINAL ANSWER: $(x-1)\arctan\left({X^2 +.5\over{\sqrt 3\over 2}}\right) + C$
Thanks for reading, I have no way of checking this work... tutors are always have too many people wanting help.
AI: HINT:
As $\displaystyle x^4+x^2+1=(x^2+1)^2-x^2=(x^2+x+1)(x^2-x+1)$ and $\displaystyle (x^2+x+1)-(x^2-x+1)=2x$
$$\frac x{x^4+x^2+1}=\frac12\frac{(x^2+x+1)-(x^2-x+1)}{x^4+x+1}=\frac12\left(\frac1{x^2-x+1}-\frac1{x^2+x+1}\right)$$
Now as, $\displaystyle x^2+x+1=\frac{(2x+1)^2+(\sqrt3)^2}4$ put $2x+1=\sqrt3\tan\theta$
and similarly for $\displaystyle\int\frac{dx}{x^2-x+1}$ |
H: Trigonometric problem in triangles.
I need your help. I'm studying physics, but I have a trigonometric problem. I attached a figure where depicts the angles and the unknown $x$. The idea that I want to understand is how to express $x$ in terms of $m$, $n$, $a$ and $b$. Because the solution is $x=m\cdot\sin a + n\cdot\cos b$.
Please, show me the clues to get the solution.
AI: It seems that the radius of the circle is $1$. Assuming that, realize that $\sin b=n$ and $\cos b = m$. We have that
$$x=\sin (a+b)=\sin a \cos b+\sin b \cos a=m\sin a+n\cos a$$ |
H: From $ \sum^\infty_{\lfloor \log n \rfloor + 1}n/{2^r} $ to $ \sum^\infty_{r=0}1/2^r $?
$$
E[h] = E[\sum^\infty_{r=1}I_r]
= \sum^\infty_{r=1}E[I_r]
$$
$$
= \sum^{ \lfloor \log n \rfloor}_{r=1}E[I_r] + \sum^\infty_{\lfloor \log n \rfloor + 1}E[I_r]
$$
$$
\leq
\sum^{ \lfloor \log n \rfloor}_{r=1}1 + \sum^\infty_{\lfloor \log n \rfloor + 1}n/{2^r}
$$
$$
\leq
\log n + \sum^\infty_{r=0}1/2^r
$$
$$
= \log n + 2
$$
I'm trying to understand how you go from $ \sum^\infty_{\lfloor \log n \rfloor + 1}n/{2^r} $ to $ \sum^\infty_{r=0}1/2^r $
AI: Assuming $\log$ here means the base-$2$ logarithm (since the inequality doesn't generally hold if the base is larger), we have
$$n < 2^{\lfloor \log n\rfloor+1},$$
and therefore
$$\sum_{r=\lfloor \log n\rfloor+1}^\infty \frac{n}{2^r} < \sum_{r=\lfloor \log n\rfloor+1}^\infty \frac{2^{\lfloor \log n\rfloor+1}}{2^r} = \sum_{r=\lfloor \log n\rfloor+1}^\infty \frac{1}{2^{r-(\lfloor\log n\rfloor+1)}} = \sum_{k=0}^\infty \frac{1}{2^k}.$$ |
H: Prove that if $A$ is compact and $B$ is closed and $A\cap B = \emptyset$ then $\text{dist}(A,B) > 0$
Let $X$ be a metric space. For nonempty subsets $A,B\subseteq X$. Define $\text{dist}(A,B) := \inf\{d(x,y) : x\in A, y\in B\}$
a) Prove that if $A$ is compact and $B$ is closed and $A\cap B = \emptyset$ then $\text{dist}(A,B) > 0$
b) Suppose that $X = \mathbb{R}^n$, with the standard metric. Under the same assumptions of (a), prove that $\exists x_0\in A$ and $\exists y_0\in B$ such that dist$(A,B) = ||x_0 - y_0||$.
For a), it seems to me that the crucial point is that $A\cap B = \emptyset$. I can immediately see how this would imply that $\text{dist}(A,B) > 0$. However, I do not understand why it's necessary for $A$ to be compact, and $B$ to be closed. Could you explain why those conditions are necessary?
AI: a) You need $B$ to be closed so that for any $x\notin B, d(x,B) > 0$. You need $A$ to be compact, so that you can restrict your attention to only finitely many $x$'s as follows :
For every $x \in A, \epsilon_x := d(x,B) > 0$, so there is a $\delta_x > 0$ such that
$$
d(x,y) < \delta_x \Rightarrow d(y,B) > \epsilon_x/2
$$
Now the balls $B(x,\delta_x)$ cover $A$. Take a finite subcover $\{B(x_i,\delta_{x_i})\}$, and choose $\epsilon$ to be the minimum of the corresponding $\epsilon_{x_i}$'s. Now check that
$$
d(A,B) > \epsilon/2 > 0
$$
b) Note that
$$
x \mapsto d(x,B)
$$
is a continuous function on $A$, and so there is $x_0 \in A$, at which the infimum is attained. ie.
$$
d(x_0, B) = \inf\{\|x-y\| : y \in B\} = \inf\{\|x_0 - y\| : y \in B\}
$$
Now there is a sequence $(y_n) \in B$ such that
$$
d(x_0,B) = \lim \|x_0 - y_n\|
$$
Check that $(y_n)$ must be a bouned sequence, and so it must have a convergent subsequence (since you are in $\mathbb{R}^n$). This subsequence must converge to a point $y_0 \in B$. Then
$$
d(A,B) = d(x_0,B) = \|x_0 - y_0\|
$$ |
H: nitpicking the definition of a polynomial function
A textbook I'm reading says that $f(x)=0$ is NOT a polynomial function, yet $g(x)=8$ IS a polynomial function since $g(x)=8=8x^0$ which satisfies the non-negative integer degree requirement. Yet, it's still a monomial! Or, can it be considered $g(x)=8=8x^0=8x^0+0$ which is now technically a POLYnomial ?
Also, $f(x)=0$ is not a poly by the same logic since it CAN be written as $f(x)=0=0x^{-3}$ which doesn't fit the poly definition above. (Counterexample)
Edit: The book says the zero function is not assigned a degree, while the nonzero constant function has a degree of 0.
AI: The problem with your book's choice of definition is that, by that definition, the sum of two polynomial functions may not be a polynomial function. This is, generally speaking, a property one would really like to hold. I'm not going to go so far as to say the book is wrong (any author can define things however they like), but it seems a very unfortunate choice.
Edit: Most agree that the zero function is a polynomial function. The matter of degree is another story. It may be considered to have no degree or to have a degree of $-\infty$. The latter choice makes some theorems easier to state. |
H: Uniform Convergence of Sequence of Functions Doubt
For the sequence of functions $$f_n={nx\over 1+n^3x^3},x\in[0,1]$$ If we wish to check the Uniform Convergence. I tried doing it in the following two ways
Method 1 :
$\lim f_n=0=f(x)$ , for a given $\epsilon>0$
$|f_n(x)-f(x)|=|{nx\over 1+n^3x^3}-0|={nx\over 1+n^3x^3}<{nx\over n^3x^3}={1\over n^2x^2}<\epsilon$
so we have , the result holds good for $n>{1\over x \sqrt{\epsilon}}$but, as $x\rightarrow 0, m\rightarrow\infty $. Tis means that $f_n$ is not uniformly convergent
Method 2: Using $M_n-test$
let $y={nx\over 1+n^3x^3},y'={n(1-n^3x^2)\over (1+n^3x^2)^4}$. Setting $y'=0, y={1\over n^{3/2}}$ This is the maximum value as confirmed by $y''$. As $M_n\rightarrow 0 \ as \ n\rightarrow\infty $, The function must be Uniformly convergent.
So where am i doing/understanding it wrong ?
AI: HINT: The functions converge pointwise to $0$, and $$ f_n\left( \frac{1}{n} \right) = \frac{1}{2}.$$ |
H: Nonzero subspace that is invariant under any operator cannot be proper?
This is not for homework, and I would just a like a hint please. The question asks
Prove or give a counterexample: If $U$ is a subspace of $V$ that is invariant under every operator on $V$, then $U = \{ 0 \}$ or $U = V$.
In the question, $V$ is finite dimensional. My gut feeling is that this claim is true, so I tried to begin a proof. I began by assuming $U \neq \{ 0 \}$. I want to try to exploit the fact that $U$ is invariant under every operator to try and force every basis vector of $V$ into $U$. I would like just a hint please!
AI: Hint: assume $U\ne\{0\}$ and $U\ne V$; then if $\{u_1,\dots,u_k\}$ is a basis for $U$, you can extend it to a basis $\{u_1,\dots,u_k,u_{k+1},\dots,u_n\}$ of $V$. You can define a linear map $V\to V$ by telling the action on a basis. Can you make $f$ that shows $U$ is not $f$-invariant?
Note: you don't need finite dimensionality, but of course the proof in the non finite dimensional case requires the axiom of choice.
Details (added after accept)
We know that $0<\dim U=k<\dim V=n$, so in the argument we do have $u_1$ and $u_{k+1}$. Thus the linear map $f\colon V\to V$ defined by
$$
f(u_i)=\begin{cases}
u_{k+1} & \text{if $i=1$}\\
0 & \text{if $1<i<n$}
\end{cases}
$$
shows $U$ is not $f$-invariant.
The same idea can be used in the general case: take a non-zero vector $u\in U$ and extend it to a basis $B$ of $U$; now take $v\notin U$ and extend $B\cup\{v\}$ to a basis $C$ of $V$ (this is possible using Zorn's lemma). Then define the linear map $f\colon V\to V$ by $f(u)=v$ and $f(w)=0$ for $w\in C$, $w\ne u$.
Another possibility is to define $f(u_i)=u_{i+1}$ for $1\le i<n$ and $f(u_n)=u_1$. Since $f(u_k)=u_{k+1}\notin U$, $U$ is not $f$-invariant. |
H: Is it a Transitive Set?
Is $\{\{0\},\{\{0\}\}\}$ a transitive set? Or only $\{0,\{0\},\{\{0\}\}\}$ is transitive?
if the first isnt a transitive set, can someone give me an example of a transitive set which does not contain urelements?
Thanks!
AI: We have $0\in\{0\}$ and $\{0\}\in\{\{0\},\{\{0\}\}\}$, but $0\notin\{\{0\},\{\{0\}\}\}$, so no, it is not transitive.
$\{\}$ is transitive and has no urelements. |
H: Verify that an ellipse has four vertices.
Verify that an ellipse has four vertices.
The ellipse is given by $$ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
And I took $$x=a\cos t$$ and $$y=b \sin t$$ for $t\in [0,2\pi]$
Please can someone help me how to verify this?
AI: Hint:
Consider the values $t=\frac {k\pi}2$ for $k\in \mathbb Z$ and the functions $\cos t,\sin t$. Are there any values of $k$ such that $\cos \frac {k\pi}2=0$? What about $\sin \frac {k\pi}2=0$?
Further hint:
We can separate the two partial derivatives $\partial x\over \partial t$ and $\partial y\over \partial t$ and get values satisfying either one of them according to the definition you are claiming, so consider the following:
$$\cos \frac \pi 2=\cos \frac {(2k+1)\pi}2=0$$
This covers two of your four "ellipse vertices". Where are the other two? |
H: Deciding whether a subset of a regular language is regular
So let's have $2$ regular languages $R_1$ and $R_2$. Then we have language $L$ where $R_1 \subseteq L \subseteq R_2$. Decide whether $L$ is always regular language or not.
So my approach was this. Let's have
$$R_1=\{a^mb^l \;|\;m,l\ge0\}$$
$$L=\{a^nb^n\;|\; n \ge0\}$$
$$R_2=\{a^mb^l \;|\;m,l\ge0\}$$
So we can easily prove that $R_1,R_2$ are regular by an automaton, and we can prove that $L$ is not regular by Pumping lemma. So we have found languages for which that does not hold $\implies$ it is not always regular.
Is this correct? Or have I made mistake somewhere?
AI: A quick hacky answer is that regular languages don't have to be infinite, so we just take $R_1 = \{ab, aabb\}$ and show it's a subset of both $R_2$ and $L$.
If you want a more damning answer, let $R_1$ be the set of all possible strings you can form from the characters $($ and $)$. Let $L$ be the set of all possible strings of balanced parenthesis, ie $(()(())) \in L$, $((($ and $())($ are $\notin L$. Finally, let $R_2 = \{(), ()(), ()()(), ...\}$. Can you show that $R_2$ is regular and $L$ is not? |
H: Prove that $\mathbb{F}_5[X]/(X^2+3)$ is a field with 25 elements
Prove that the ring shown above is a field containing 25 elements.
Research effort:
$$\mathbb{F}_5[X]/(X^2+3)\cong (\mathbb{Z}[X]/5\mathbb{Z}[X])/ \overline{(X^2+3)} \cong \mathbb{Z}[X]/(5,X^2+3)$$
Modifying the term $3$, I tried to find $k$ such that $5k -3 = n^2$ for some $n \in \mathbb{N}$, in order to find a zero for the polynomial $X^2+3$. But there is no $k \in \{1, 2 , \cdots, 9 \}$, so that $k^2$ ends with a two or a seven,so we won't find a $k\geq 10$ that ends on $7$ or $2$ either, since $(10k +m)^2 = 10(10k^2+2km) +m^2$.
Maybe it's possible to modidy the other terms of the polynomial to find a root. In that case we could use a substitution homomorphism to prove that it's isomorphic to $\mathbb{F}_5$.
May intuition says that this is not possible however. Can you provide me a hint for this?
This means that we can not modify the polynomial to show that it has a root.
AI: You probably saw a theorem that states that for any field $F$ and irreducible polynomial $P(X)\in F[X]$, the quotient ring $F[X]/(P(X))$ is a field. So, you need to establish that your polynomial $X^2+3$ is irreducible over the field $F=\mathbb F_5$. Since the polynomial has degree $2$, all you need to do is verify that the polynomial has not roots in the field. Since there are only $5$ elements in the field, it's quite easy to check them one by one. |
H: Solving a recurrence relation using repeated substitution
So, basically I am having a big issue with this recurrence relationship:
$$T(n) = T(n-1)+n, T(1) = 0$$
using repeated substitution I get down to:
$$i=1, T(n-1) + n$$
$$i=2, T(n-2) + 2n - 2$$
$$i=3, T(n-3) + 3n - 3$$
$$T(n) = T(n-i) + i*n - i$$
Base case: $n-i = 1, i.e. i = n - 1$
$$T(n) = T(1) + n(n-1) - (n-1)$$
$$T(n) = n^2 - 2n + 1 $$
But, wolframalpha gives me:
$$T(n) = 0.5*n^2 + 0.5*n + c$$
And with $T(1) = 0$:
$$T(n) = 0.5*n^2 + 0.5*n - 1$$
Am I doing something wrong??
Please help!!
AI: You seem to be off by 1
$$T(n) = T(n-1) + n = T(n-2) + n -1+n = T(n-2) + 2n-1$$
And you have $2n-2$. The next step gives
$$T(n) = T(n-3) + n-2 + 2n-1 = T(n-3) + 3n - 3$$
And one more gives
$$T(n) = T(n-4) + n-3 +3n-3 = T(n-4) + 4n -6$$
In geeneal the constant term will be the sum from $1$ to $k-1$ of $k$. In this case $6 = 1+2+3$. I'm sure you can find the right formula. |
H: Problem of color painting on a circle.
Is it possible to color each point on a circle either red, yellow, or blue in such a way that no three points of the same color or totally different colors from the vertices of an isosceles triangle?
I think we must be able to find at least one of such an isosceles triangle. My guess is to use pigeonhole principle. What should I do?
AI: Hint: Consider any regular pentagon on the circle. Then any triple of vertices forms an isosceles triangle. |
H: Find the next four largest 4-combinations
Find the next four largest 4-combinations of the set 1,2,3,4,5,6,7,8 after 1,2,3,5.
Not sure how to do this, need some help~!
AI: If you want the next ones, increment the last digit as far as you can go, then go back and increment the next to last, bringing the last up as close as you can. |
H: Let $I$ be a prime ideal. Show that $f^{-1}(I)$ is a prime ideal of $R$. Is this also true for maximal ideals?
I'm solving some exercises to prepare for my ring theory exam.
Let $f:R→R'$ be a ring homomorphism, with $f(1)=1$, and $R,R'$
commutative rings with $1$. Let $I$ be a prime ideal. Show that
$f^{-1}(I)$ is a prime ideal of $R$. Is this also true for maximal
ideals? Prove or disprove.
I was able to prove this for prime ideals. But I don't know how to prove anything for maximal ideals. I think it is not true. But I find it hard to think of a counterexample.
AI: $I$ a prime ideal $⇒$ $f^{-1}(I)$ a prime ideal
Suppose $I$ is a prime ideal. Then we know that $I≠R'$. We need to prove that $f^{-1}(I)≠R$. Suppose $f^{-1}(I)=R$. Then $f(1)=1 \in f^{-1}(I)$. So then $1 \in I$. And then $I=R'$.
Let $x,y \in R$ with $xy \in f^{-1}(I)$. Then $f(xy) \in I$. Then $f(x),f(y) \in I$. Then $x,y\in f^{-1}(I)$.
$I$ a maximal ideal $\not ⇒$ $f^{-1}(I)$ a maximal ideaal
Take the ring homomorphism $f : ℤ →\mathbb{Q} : x ↦ x$. As $\mathbb{Q}$ is a field, therefore $(0)$ is a maximal ideal. But $f^{-1}((0))=(0)$ is not a maximal ideal because $ℤ$ is not a field. |
H: For any given set of 13 distinct real numbers, prove we can always find two numbers $x$ and $y$ that $0<\frac{x-y}{1+xy}\leq 2-\sqrt{3}$.
For any given set of 13 distinct real numbers, prove we can always find two numbers x and y that $0<\dfrac{x-y}{1+xy}\leq 2-\sqrt{3}$.
I knew we can always make $0<\dfrac{x-y}{1+xy}$ happen. Since x and y are distinct, we can just switch the order of x and y to change the sign of $\dfrac{x-y}{1+xy}$, but what should I do for the right hand side?
AI: Put $x = \tan(u)$ and $y = \tan(v)$ for some $u, v$ considered modulo $\pi$ (since $\tan$ has period $\pi$). Then $\frac{x-y}{1+xy}$ is $\tan(u-v)$.
So the key is to recognize that two of the angles must be within $\pi/12$ of each other. This is by a pigeonhole principle where we are trying to slot 13 numbers into 12 sectors given by $[0, \pi/12]$, $[\pi/12, 2\pi/12]$, etc. |
H: Finite field extension
Let $K\supset F$ be a finite field extension of degree 2 and the characteristic of $F$ is not 2. Show that there exists an isomorphism of rings $\sigma:K\to K$ such that
$$F=\{x\in K:\sigma(x)=x\}.$$
In this case, I didn't see why we need the condition char $F\neq 2$. The extension has degree 2 this means this extension is algebraic. Then I am thinking of using the polynomial ring over $F$. Then I get stuck. Any hints?
AI: Hint: Is there a $w\in K\setminus F$ such that $w^2 \in F$? (here you will need the hypothesis ${\text {char}}(F)\neq 2$ and $[K:F]=2$). Now, the last hint is: think about the automorphisms of ${\mathbb Q}(\sqrt{2})$, relate it to your problem. |
H: How does this factor to 6(1-ln6)
-6(ln6)+6
I got....
6(-1(ln6)
shouldn't the 1 be negative not the ln6
AI: You forgot the part of the expression when we add $6$; when you factor out $6$ from each term, you're left with:
$$-6\cdot (\ln (6)) + 6 = 6\cdot\Big ((-1)\ln(6) + 1\Big) = 6\cdot \Big(-\ln(6) + 1\Big) = 6\cdot\Big(1 - \ln(6)\Big)$$ |
H: Poisson Process: probability
Find the chance that the fifth waiting time of the length at least T for an arrival, occurs at the time of tenth arrival.
{N(t), t>0} with rate lampda
AI: The probability that the waiting time between two arrivals is $\ge T$ is $e^{-\lambda T}$. Call this number $p$.
We want the probability that there is a wait of $\ge T$ between the $9$-th and $10$-th arrival, and that this kind of wait has occurred exactly $4$ times in the $9$ arrivals before that.
Call a wait of $\ge T$ a success (it gives workers a much needed break). We want the probability of exactly $4$ successes in $9$ trials, followed by success.
The required probability is therefore
$$\binom{9}{4}p^4(1-p)^5p.$$ |
H: Matrix representation of a co-domain restriction of a linear operator
Consider the finite-dimensional linear operator:
$\mathcal{A}:\mathbb{R}^{3}\rightarrow\mathbb{R}^{3},$ with $Ax=y,$
$A=\left[\begin{array}{ccc}
1 & 0 & 1\\
1 & -2 & -1\\
0 & 1 & 1
\end{array}\right].$
Let $\mathcal{A}_{1}:\mathbb{R}^{3}\rightarrow\mathcal{R}\{A\}$ be
the co-domain restriction of $\mathcal{A}$ to $\mathcal{R}\{A\}$.
Give the matrix representation $A_{1}$ of $\mathcal{A}_{1}.$
I have gotten $\mathcal{R}\{A\}$, which is 2-dimensional. So I know the matrix representation of the restricted operator will be a 2x3 matrix, which maps $\mathbb{R}^{3}$ to the plane parametrized by the 2 vectors which span the range. I am unsure of how to obtain the restricted operator. Any advice would be greatly appreciated. Thank you!
AI: The matrix representation you are asking is not unique, it will depend on the basis of $\mathcal{R}\{A\}$ you are considering. To find a basis for $\mathcal{R}\{A\}$, find $Ae_1$ and $Ae_2$: you get that $$Ae_1=\left[\begin{array}{ccc}
1 & 0 & 1\\
1 & -2 & -1\\
0 & 1 & 1
\end{array}\right]\left[\begin{array}{c}
1 \\0\\0
\end{array}\right]=\left[\begin{array}{c}
1 \\1\\0
\end{array}\right],\,\,\,\,\,Ae_2=\left[\begin{array}{ccc}
1 & 0 & 1\\
1 & -2 & -1\\
0 & 1 & 1
\end{array}\right]\left[\begin{array}{c}
0 \\1\\0
\end{array}\right]=\left[\begin{array}{c}
0 \\-2\\1
\end{array}\right].$$ Those are independent and $\mathcal{R}\{A\}$ is two dimensional, so they span it. Now, for the general $z=(a,b,c)$: $$Az=\left[\begin{array}{ccc}
1 & 0 & 1\\
1 & -2 & -1\\
0 & 1 & 1
\end{array}\right]\left[\begin{array}{c}
a \\b\\c
\end{array}\right]=\left[\begin{array}{c}
a+c \\a-2b-c\\b+c
\end{array}\right]=\left[\begin{array}{c}
a+c \\a+c\\0
\end{array}\right]+\left[\begin{array}{c}
0 \\-2b-2c\\b+c
\end{array}\right]=$$$$(a+c)Ae_1+(b+c)Ae_2.$$ So, a matrix representation (with respect to the usual basis of $\mathbb R^3$ and the basis $Ae_1, Ae_2$) is $$\left[\begin{array}{ccc}
1 & 0 & 1\\
0 & 1 & 1
\end{array}\right].$$ |
H: Calculus position/velocity question.
Disclaimer: I am not a student trying to get free internet homework help. I am an adult who is learning Calculus. I am deeply grateful to the members of this community for their time.
Here is the question (Not sure of the best way to format this)
$$s(t)=t^3-9t^2+24t-6$$
Q: The speed of the particle is decreasing for:
1) t<1
2) t>2
3) t<3
4) t<1 and t>2
5) all t
So, I took the derivative of position to get velocity, and set it to zero.
$$v(t)=3t^2-18t+24=0$$
Now, I have a parabola with roots at (2,0) and (4,0) and vertex (3,-3)
I understand that Speed is DECREASING when either:
a) velocity is (+) and slope is (-)
(I'm moving forward, but velocity is becoming less positive)
b) velocity is (-) and slope is (+)
(I'm moving backwards, but velocity is becoming less negative)
So, MY answer is not in the list $t<2$ and $3<t<4$
$t<2$ is where the parabola meets criterion (a)
$3<t<4$ is where the parabola meets criterion (b)
Yet, the answer key says choice (3) ??
What am I doing wrong?
=====================
Update:
There was another question in the same set that is equally perplexing!
Q: The minimum value of the speed is:
1) -3
2) -2
3) -1
4) 0
5) 1
MY answer would be 0. (choice 4), since speed = |v|
Yet, the answer key says choice 3
How could speed OR velocity be equal to -1 ?
Minimum velocity is -3 (choice 1)
AI: It appears the answer key is using velocity instead of speed. Following your definition of speed as absolute value of velocity, you are correct that it is decreasing on $t \lt 2$, but it is also decreasing on $3 \lt t \lt 4$ (your case b). That is still not one of the choices. It looks like you understand the issue, though. |
H: Limit as $x$ approaches $0$ with constant $a$
Find the limit where $a$ is a constant
$$ \lim_{x \to 0}\frac{\left [ \cos(a+x)-\cos(a-x) \right ]^2}{\tan^2(3x)} $$
I don't know what to do. At first I thought I could replace $a$ with an arbitrary number and then solve the limit but then I got stuck as to how to use the squeeze theorem on this question. Any help?
AI: Hint: $$ \frac{\left[ \cos(a+x)-\cos(a-x) \right]^2}{\tan^2(3x)} = \frac{4\sin^2 a \sin^2 x}{\tan^2(3x)} = \frac{4\sin^2 a }{9}\frac{(3x)^2}{\tan^2(3x)} \frac{\sin^2 x}{x^2}.$$ |
H: Partial derivative with respect to a function?
I have $f(x, y)$ is a class $C^2$ function where $x = u + v$, and $y = u - v$
How do I get $$\frac{\partial^2 f}{\partial x^2}$$
Well before we even talk about that one, I don't even know how to get the first partial. I'm confused about how to go about solving it when its with respect to x. Any hints?
Edit: I don't have a specific equation for $f(x, y)$. The full question was to show
$$\frac{\partial^2f}{\partial{u}\ \partial{v}} = \frac{\partial^2f}{\partial{x^2}} - \frac{\partial^2f}{\partial{y^2}}$$
AI: You can show this by using the chain rule twice. First,
$$\frac{\partial{f}}{\partial{v}} = \frac{\partial{f}}{\partial{x}}\frac{\partial{x}}{\partial{v}} + \frac{\partial{f}}{\partial{y}}\frac{\partial{y}}{\partial{v}} \\
= \frac{\partial{f}}{\partial{x}} - \frac{\partial{f}}{\partial{y}}$$
and then
$$\frac{\partial^2 f}{\partial{u}\ \partial{v}} = \frac{\partial}{\partial{u}}\frac{\partial{f}}{\partial{x}} - \frac{\partial}{\partial{u}}\frac{\partial{f}}{\partial{y}} \\
= \left(\frac{\partial^2 f}{\partial{x}^2}\frac{\partial{x}}{\partial{u}} + \frac{\partial^2 f}{\partial{y}\ \partial{x}}\frac{\partial{y}}{\partial{u}}\right) - \left(\frac{\partial^2 f}{\partial{x}\ \partial{y}}\frac{\partial{x}}{\partial{u}} + \frac{\partial^2 f}{\partial{y}^2}\frac{\partial{y}}{\partial{u}}\right) \\
= \left(\frac{\partial^2 f}{\partial{x}^2} + \frac{\partial^2 f}{\partial{y}\ \partial{x}}\right) - \left(\frac{\partial^2 f}{\partial{x}\ \partial{y}} + \frac{\partial^2 f}{\partial{y}^2}\right) \\
= \frac{\partial^2 f}{\partial{x}^2} - \frac{\partial^2 f}{\partial{y}^2}$$
The last equality is valid because $$\frac{\partial^2 f}{\partial{y}\ \partial{x}} = \frac{\partial^2 f}{\partial{x}\ \partial{y}}$$ when $f$ is of class $C^2$. |
H: How to solve for $n$?
e.g.
$$\int_1^n \frac{1}{x} dx = \pi$$
I want to know the method used to solve it for any constant $c$ of any integral.
AI: We have $\int_1^n \frac{1}{x} dx = \ln (n) = \pi$ therefore $e^{\ln n} = e^{\pi} = n.$ |
H: Meta Theory when studying Set Theory
What exactly serves as the meta theory in the study of set theory (for example in Kunen's text). It seems to involve a certain amount of number theory and a certain concept of what it means to be finite.
Also: We usually use $\bigcup_{\alpha\in{ON}}R_{\alpha}$ as our set theortic universe $V$ (in the sense we believe that all sets we are interested in live there) where $R_{0}=\emptyset$, $R_{\alpha+1}= \mathcal{P}(R_{\alpha})$ and $R_{\alpha}=\bigcup_{\beta<\alpha}R_{\beta}$ when $\alpha$ is a limit ordinal. The study of independence results then seem to be based on relatavizing notions with this $V$ as our set theoretic universe. It seems that it is implicitly assumed that the $\omega$ in $V$ has no non-standard elemements (in the sense that there is no $n>a$ in $\omega$ where $a$ is obtained be repeatedly applying the successor operation to $0$.) Is this true and can it be formalized or is it something that is taken for granted?
e.g. If the above were true then $ZFC\vdash{\neg{Con(\ulcorner{ZFC}\urcorner)}}$ would imply that $ZFC$ is inconsistent since we may look at the $ZFC$ axioms $\phi_{1},...,\phi_{n}\vdash\neg{Con\ulcorner{ZFC}\urcorner}$, then look at a ctm these axioms (which exists by reflection and downward Lowenheim Skolem) and recover the proof of the contradiction.
AI: There are generally two accepted approaches:
You can use some arithmetical theory, e.g. $\sf PA$, or even a fragment which is sufficient to develop first-order logic and syntactic manipulation of proofs. Then one can define the language of set theory, write the axioms and proofs and so on. In fact $\rm Con(\sf ZFC)$ is in fact a statement about natural numbers rather than a statement about sets and models.
This is even true if one wants to introduce forcing. And I ran into a recent masters thesis in which this (usually folklore, I believe?) result is given in details.
You can use $\sf ZFC$ itself. Then you have some universe of set theory (usually $\sf ZFC+\rm Con(\sf ZFC)$ and even more), but you are in fact working inside a set model of set theory within that universe. In that case you are free to use all sort of fun model theoretical tools, and forcing is done directly and so on.
However in many many instances we in fact omit the meta theory, and we just care about it sufficient to develop first-order logic. We often work within the universe. So there is no real model of set theory, there is a universe and we work with that. We can do forcing using the universe because we can define Boolean-valued models and prove independence results using that, and so on.
Let me quote from a Ph.D. thesis written by VanLiere,
Since these questions all have to do with first-order provability, we could take as our metatheory some very weak theory (such as Peano arithmetic) which is sufficient for formalizing first-order logic. However, as is customary in treatises about set theory, we take as our metatheory $\sf ZF$ plus the Axiom of choice in order to have at our disposal the infinitary tools of model theory. We will also use locutions such as ... which are only really justifiable in some even stronger metatheory with the understanding that they could be eliminated through the use of Boolean-valued models or some other device.
This sums up quite nicely, I think, the very utilitarian approach to set theory. We really just need a weak theory sufficient for first-order logic, but it's easier and nicer to work with much stronger metatheories because they make our life so much easier.
About the second part, we often take the universe to have the same integers as the metatheory. This has some serious implications, e.g. the validity of $\rm Con(\sf ZFC)$ is inherited from the meta-theory. This is not always the case, and sometimes it is worth looking at non-standard models, or even models which do not agree on the integers with the universe or with the metatheory.
However whenever you see the term "transitive model" then you immediately know that the model and the universe agree on the integers, which has some strong implications on the truth values of some statements in the model. This is why arguing about transitive model is not enough to invoke the completeness theorem and have provability; but it is enough to establish independence.
This is again a utilitarian point. Transitive models are nice, and they have very nice properties that they inherit from the universe. When we are interested in independence, then working with transitive models is generally enough. So we do that, because we are utilitarians.
I should probably add another thing about utilitarianism here. Of course one can make all sort of crazy assumptions "because they would make life easier", but philosophically one is likely to agree that the consistency of $\sf ZFC$ is true if one is interested in investigating $\sf ZFC$ and its consequences, and so it's reasonable to add this assumptions to your list of axioms in the metatheory, and even more. Even if you end up working just with $\sf ZFC$ itself.
As the quote points out, some assumptions that we often use can only be justified by much stronger theories, but they are "philosophically sound" (i.e. working with $\sf ZFC+\lnot\rm Con(\sf ZFC)$ as a metatheory for investigating $\sf ZFC$ is really weird), and we can remove these extraneous assumptions by using slightly more complicated devices anyway. |
H: How to obtain $y$
The question was written with dark-blue pen. And I tried to solve this question. I obtained $x$ as it is below. But I cannot obtain $y$ Please show me how to do this.
By the way, $\gamma (t)$ may not be clearly readable. So, I wrote again.
$$\gamma (t)=( \cos ^2 (t)-1/2, \sin(t)\cos (t), \sin (t))$$
Thanks for helping.
-sorry for not writing with MathJax. -
AI: From your work,
$$x^2+y^2=\frac{1}{4} \Rightarrow y^2=\frac{1}{4}-x^2,$$
and
$$x=\cos^2 t -\frac{1}{2}.$$
Substituting the latter into the former produces
\begin{align*}
y^2 &=\frac{1}{4}-x^2 \\
&=\frac{1}{4}-\left( \cos^2 t - \frac{1}{2} \right)^2 \\
&=\frac{1}{4}-\left( \cos^4 t - \cos^2 t + \frac{1}{4} \right) \\
&=\cos^2 t - \cos^4 t \\
&=\cos^2 t\left( 1-\cos^2 t \right) \\
&=\cos^2 t \sin^2 t \\
\Rightarrow y &= \sin t \cos t,
\end{align*}
as was your intention. |
H: Stuck because of possible error in exercise of "How to Prove It" by Velleman
I am self-studying "How to Prove It" by Velleman, and I believe there must be an error on exercise 3.3 #14.
I'll show the question here, and where I think the error is, and then I'd love to find out if you believe I'm correct that there is an error, or if not, where am I missing the boat?
The exercise:
"Suppose $ \{A_i \mid i \in I \} $ is an indexed family of sets. Prove that $ \bigcup_{i \in I} \mathbb{P}(A_i) \subseteq \mathbb{P}(\bigcup_{i \in I} A_i) $."
Where I believe the problem to be:
It seems to me that the left-hand side of the subset symbol would be a "set" (in other words, "flat" if you will), while the right-hand side would be a "set of sets." To me, it appears that the definition of subset makes it impossible for a "set" to be a subset of a set-of-sets.
Is the book wrong, or am I? (I suspect I am wrong, but after approaching it from several different angles, I simply DO NOT SEE WHERE).
Thanks.
AI: Suppose $A_1=\{1,2\}$ and $A_2=\{2,3\}$. Then $\mathbb P(A_1) = \{\varnothing,\{1\},\{2\},\{1,2\}\}$ and $\mathbb P(A_2) = \{\varnothing,\{2\},\{3\},\{2,3\}\}$, so
$$
\mathbb P(A_1)\cup\mathbb P(A_2) = \{\varnothing,\{1\},\{2\},\{3\},\{1,2\},\{2,3\}\}
$$
and that is a subset of
$$
\mathbb P(A_1\cup A_2) = \{ \varnothing,\{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\},\{1,2,3\} \}.
$$
Of course this doesn't prove the result, but it shows how your objection to the conclusion misses something. |
H: How to calculate the minimal and maximal distance between these two objects?
Suppose that an object $\mathcal{O}$ travels on the $xy$ plane following a path with respect to the time $t$ of the form $\mathcal{O}(t)=(2\cos(t), 2\sin (t))$ and another object $\tilde{\mathcal{O}}$ at the same time follows the path $\tilde{\mathcal{O}}(t)=(2\cos (t+\delta)+1, \sin(t+\delta))$, $t \in \Bbb R$, where $\delta$ is a real number. I need to know how to calculate the minimal and maximal distance between these two objects as a function of $\delta$. I aready made a graph of the two paths, but I need to calculate this distance analytically. I calculated
$$||\mathcal{O}(t)-\tilde{\mathcal{O}}(t)||=\sqrt{9+8[\cos(t)\cos(t+\delta)-\sin(t)\sin(t+\delta)]+4[\cos(t)-\cos(t+\delta)]},$$
but I don't what else to do. Actually, I don't know if this is the correct and/or easiest way to solve this problem. I'd appreciate any ideas to solve this. Thanks in andvance. I leave you the graph of the paths:
AI: First of all notice that if $f(t)$ is always positive, then finding the maximum and minimum of $\sqrt{f(t)}$ is the same as finding them for $f(x)$. This is obviously your case, where
$$f(t) = \big(1+2\cos(t+\delta)-2\cos(t)\big)^2+\big(\sin(t+\delta)-\sin(t)\big)^2$$
Now find the extrema as usual, by solving
$$\frac{\partial}{\partial t}f(t) = 0$$
and then checking the second derivative to determine which are maxima and which are minima. |
H: Prove that $I= \{a+bi \in ℤ[i] : a≡b \pmod{2}\}$ is an maximal ideal of $ℤ[i]$.
I'm making some exercises to prepare for my ring theory exam:
Prove that $I= \{a+bi \in ℤ[i] : a≡b \pmod{2}\}$ is an maximal ideal
of $ℤ[i]$.
I know that $a+2l=b$ with $l\in ℤ$ (or should I say $l \in ℤ[i]$)? Therefore I can write $a+(a+2l)i=a+ai+2li$. So I don't have $0+i$ in my ideal. For even real parts I have even complex parts, and for odd real parts I have odd complex parts. But I don't see what I can conclude now.
AI: Hint: (1) Check that $I$ is an ideal of $\mathbb Z[i]$, that is an additive subgroup with $\mathbb Z[i]\cdot I \subseteq I$.
(2) To show it is maximal, let $a+bi \in \mathbb Z[i] \setminus I$. Then $a-1 + bi \in I$ (why?). What can you conclude about the ideal $J = (I, a+bi)$, noting $a+bi - (a-1+bi) = 1$? |
H: How can I optimize this? Finding someone using several factors
I have 100 students, and they all need colored pencils.
Each of them needs the same colors of pencils, however, they can have different shades of the color.
What's the least amount of color and shade combinations I need to purchase so that each student has a unique combination of shades?
So for example, if I decide 10 colors and 10 shades, Student 42 will have (I assigned the shades randomly)
Color Green Shade A
Color Red Shade J
Color Blue Shade C
Color Orange Shade A
Color Yellow Shade E
Color Indigo Shade B
Color Purple Shade D
Color Violet Shade A
Color Light Blue Shade F
Color Magenta Shade G
And I can be 100 percent sure that if given those color / shade combos, I will be able to tell you it's student 42.
But how can I optimize this so I need less color and shade combos (what's the least I need)? And once I have that smallest amount, how do I go about distributing it so it still has a 100% chance of finding the student?
AI: Find some numbers $s_1,s_2,\dots,s_r$ that multiply to at least 100; then you can get by with $r$ colors, $s-1$ shades of the first color, $s_2$ shades of the second color, and so on, to $s_r$ shades of the last color.
For example, $5\times5\times4=100$, so it works to have 5 shades of red, 5 shades of blue, and 4 shades of green.
I suppose you could use 2 shades of orange, and 50 shades of gray (if you count gray as a color). |
H: Prove the limit related to a recurrence
For the following sequence $\{a_n\}_{n=1}^{\infty}$, we define $a_1=\alpha\in(0,1)$, and for any $n\geq 2$, $a_{n+1}=a_n(1-a_n)$.
Prove: $\lim_{n\rightarrow\infty}{na_n} = 1$.
AI: Hint
Prove that $(a_n)$ is decreasing sequence bounded below by $0$ so it's convergent to $0$
Prove that
$$\frac{1}{a_{n+1}}-\frac{1}{a_{n}}\sim 1$$
Use the Cesaro theorem to find the desired result. |
H: Largest box fitting inside an ellipsoid
Find the volume of the largest box with sides parallel to the $xy$, $xz$, and $yz$ planes that can fit inside the ellipsoid $(x/a)^2 + (y/b)^2 + (z/c)^2 = 1$.
My answer: We want to maximize $f(x,y,z) = xyz$ subject to $(x/a)^2 + (y/b)^2 + (z/c)^2 \leq 1$. So $\nabla f $ parallel to $\nabla g$, i.e.
$$(yz, xz, xy) = \lambda \left( \frac{x}{a}, \frac{y}{b}, \frac{z}{c} \right)\\ \Rightarrow \frac{ayz}{x}=\frac{bxz}{y}= \frac{cxy}{z}\\ \Rightarrow ay^2 = bx^2 ; bz^2 = cy^2 ; az^2 = cx^2 \\ \Rightarrow (x,y,z) \text{ parallel to } \left( 1, \sqrt{\frac{a}{b}}, \sqrt{\frac{a}{c}}\right)$$
so we need to find $t$ such that $\left( t, t\sqrt{\frac{a}{b}}, t\sqrt{\frac{a}{c}}\right)$ is on the ellipsoid, i.e.
$$\left( \frac{t}{a\sqrt{a}}\right)^2 + \left( \frac{t}{b\sqrt{b}}\right)^2 + \left( \frac{t}{c\sqrt{c}}\right)^2 = 1.$$
So the volume is $$8t^3 \frac{1}{\sqrt{abc}}\\ = 8\frac{1}{\left(\sqrt{ \frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{c^3} }\right)^3}\cdot \frac{1}{\sqrt{abc}}$$
But the answer which the T.A. has indicated is $\frac{8abc}{3\sqrt{3}}$ (of course there could be a mistake). Where is my error?
AI: It is not difficult to verify that the largest box in the unit sphere is the cube with sides $2/\sqrt3$, whence for such a sphere you get the volume $8/3\sqrt3$. Now notice that $f:\mathbb R^3\to\mathbb R^3$, $(x,y,z)\mapsto(ax,by,cz)$ maps unit sphere to your ellipsoid, and $V(f(X))=abc V(X)$ where $V$ is the volume and $X$ is any set. Moreover, $f$ brings your box to a box again. Therefore the maximum is necessarily attained for $f(X)$ where $X$ is the cube. And you get $V(f(X))=8abc/3\sqrt3$.
Your mistake is that $\partial_x (x/a)^2 = 2x/a^2$ and not $2x/a$. |
H: Binary representation of 2-adic integers
I would like some examples of the binary representation of 2-adic integers that are not standard integers. What is the 2-adic expansion of $1/3$? Of $-1/3$? What number does $...010101$ represent?
AI: To answer your last question first, $$...010101 = 1 + 4 + 16 + \cdots + 4^n + \cdots = 1/(1-4) = -1/3.$$ This also answers your second question. As for your first question, $$1/3 = \dfrac{1}{2}( 1 - 1/3) = ...0101011.$$ |
H: Compact Subsets of $C[a,b]$
Consider the set $G = \lbrace f \in C\left[a,b\right] : |f(x)| \le |g(x)|,\ \forall x \in [a,b] \rbrace$
Find all values of $g$'s for which $G$ is a compact subset of $C[a,b]$ with the max norm.
Attempts:
I understand the fact that it suffices to just show that the set $G$ is closed because we know that $C[a,b]$ with the infinity norm is in fact a compact subspace. However, for which value of $g$ will these sets be dense? My first intuition on the argument would be that the only necessary conditions on $g$ would be that it ins in $C[a,b]$ and that any function would do.
However, my only real justification is that given any function $f \in G$ we know that if $O$ is any neighborhood of $f$ $\exists p(x) \in G$ such that $p(x) \in O$ by the density of polynomials in $C[a,b]$. However, first hand I don't know whether or not this is necessarily true, I also have no intuition on showing such a function exists.
Any ideas? Or is showing that $G$ is closed not the right choice for the proof.
Edit: Problem in inequality
Edit2: Now knowing that $C[a,b]$ is not compact.
AI: Edit: this is the answer after the question has been edited.
We claim that $g\equiv 0$. If not, there exists $x_0$ such that $g(x_0)\neq 0$. $g$ is continuous, so there exists $\varepsilon>0$ so that $|g(x)|>\varepsilon$ close to $x_0$. So, $|g(x)|>\varepsilon$ in an interval of the form $(c,d)$. Let $k$ be the middle point of $(c,d)$, and consider the sequence of functions $$f_n(x)=\left\{\begin{array}{cl}0,& x\leq k-\frac{1}{n}\,\,\text{or}\,\,x\geq k+\frac{1}{n}\\ \frac{n\varepsilon}{2}(x-k)+\frac{\varepsilon}{2},&k-\frac{1}{n}\leq x\leq k \\ -\frac{n\varepsilon}{2}(x-k)+\frac{\varepsilon}{2},&k\leq x\leq k+\frac{1}{n}\end{array}\right.$$ Those functions belong to $G$, and $G$ is compact, so $(f_n)$ should have a convergent subsequence. But $f_n$ converges pointwise to a discontinuous function, which gives a contradiction.
So, $g\equiv 0$. |
H: Reference for Deligne-Mumford
What is a good reference for someone new to the theory of Deligne-Mumford stacks, other than the original Deligne-Mumford paper itself? The paper itself seems readable with some effort; but the fear is that the reader will miss out on whatever happened in the later years. Is there a good introduction at the graduate level?
AI: There is the book of Laumon and Moret-Bailly.
There is of course the stacks project.
There are also Artin's paper on algebraic spaces (his survey on the implicit function theorem in algebraic geometry from the Bombay conference, and also his two papers on algebraization of formal moduli). When I was recently trying to learn this material, I found Artin's papers very helpful, even though they are about algebraic spaces rather than stacks proper. |
H: Kind of a silly question regarding compact operators
If $T \in \mathcal{B}(X)$ is a compact operator, how about $-T$, i.e. it's additive inverse? Does this notation even make any sense with regard to the minus sign? Is it trivial?
Further, suppose we have another compact operator $S \in \mathcal{B}(X)$. I know that $S+T$ is compact, but how about $S-T$ or $T-S$?
AI: The set of compact operators on a Banach space is itself a Banach space, with the supremum norm. So linear combinations and limits of compact operators are again compact. |
H: Prove that a function is irreducible
Let $F$ be a field. Let $\varphi : F[x] \rightarrow F[x]$ be an isomorphism such that $\varphi(a)=a$ for every $a \in F$.
Prove that $f \in F[x]$ is irreducible if and only if $\varphi(f)$ is.
How will I be able to start this proof? Any help will be greatly appreciated. Thanks in advance!
AI: If $R \to S$ is any ring isomorphism, then it maps irreducible elements of $R$ to irreducible elements of $S$. This can be derived directly from the definitions. Note that this is - in some sense - trivial a priori since isomorphisms are required to preserve the whole ring structure and irreducibility only refers to this structure.
It doesn't make any sense to restrict to polynomial rings here. The proof doesn't simplify in this special case, and this restriction obscures what is really going on. Also the assumption that the isomorphism is the identity on constants is not necessary. Of course this special case is useful etc., but if you have this extra assumption you are somehow invited to use it in your proof, but then your proof will be either a) wrong, or b) too complicated. |
H: Improper integral divergence
Use the graph of 1/x and the sum of areas of rectangles to show that $\int _{ 1 }^{ \infty }{ \frac { 1 }{ x } dx }$ = +$\infty$.
Would the sum of rectangles just be:
1 + 1/2 + 1/3 + 1/4 +....+1/n + = +$\infty$.
AI: You will need to draw a picture. Then maybe use:
First rectangle: base $[1,2]$, height $1/2$;
Second rectangle: base $[2,4]$, height $1/4$;
Third rectangle: base $[4,8}$, height $1/8$;
Fourth rectangle: base $[8,16]$, height $1/16$;
And so on.
Note that each rectangle has area $1/2$, and the union of the rctangles lies in the region "below" $y=1/x$ and "above" the $x$-axis. |
H: Let $V$ be $n$ dimensional real vector space. Show that of $T[V]=\ker(T)$, then $n$ is even
Let $V$ be an $n$-dimensional real vector space and $T:V\rightarrow V$ a linear transformation from $V$ to itself. Suppose that $T[V]=\ker(T)$. Show that $n$ is even.
I'm really lost. Since $\operatorname{Im}[T]=\ker(T)$, does it mean that there has to be an even number of $T(\vec v)$ such that half of them are additive inverse of the other half, and $\operatorname{Im}[T]=\ker(T)=0$?
AI: Hint: By the rank-nullity theorem,
$$n = \dim V = \dim T[V] + \dim \ker(T)$$ |
H: Find the volume of the solid obtained by rotating the region bounded by the curves $y=x^2$, $x=4$ and $y=0$ about the $x$-axis
I do not understand how they want the volume between $y=x^2$ and $y=0$? I don't understand how to do the problem. Please help. I have other similar homework problems and would like to LEARN how to do them properly.
AI: The first step is to find the area between the given curves. I like to draw a picture. Here we have a parabola $y=x^2$ and two lines $x=4$, which is a vertical line, and $y=0$, which actually is just the x axis. Think now of rotating this area around the x axis. You get a sort of bell or horn shaped thing. What if we cut this horn into really thin disks that are circles with radius $x^2$ going perpendicular to the x axis? We could find their volume, add it up, and get the total volume.
This should make you think of an integral.
The volume of each thin disk I mentioned is $\pi r^2$ times the disk thickness "dx", which we could also write as $\Delta x$. The radius r of each disk is $x^2$. So, en total we can form an integral $\int_0^4 \pi (x^2)^2 dx=\int_0^4 \pi (x^4)dx$.
Maybe consult a good calculus text to see more methods. |
H: Differentiation/Integration continuous function (metric spaces)
I have the following two questions:
Is differentiation, $f(x) \mapsto f'(x)$ a continuous function from
$C^1[a,b] \longrightarrow C[a,b]$
and
Is integration, $f(x) \mapsto \int_a^x f(t) \ dt$, a continuous
function from $C[a,b] \longrightarrow C[a,b]$
$C[a,b]$ denotes the space of continuous real-valued functions on $[a,b]$ with the sup metric and $C^1[a,b]$ denotes the space of continuously differentiable functions on $[a,b]$
So far, I've attempted the first one and I'm trying to use the theorem that a map is continuous if and only if the inverse map applied to some set is closed in $f(x)$ whenever the set is closed in $f'(x)$ but haven't been able to get anywhere (or think of a counterexample)
AI: You need to specifiy a norm on these spaces to discuss continuity. Usually, one gives $\mathcal{C}[a,b]$ the sup-norm. In this case, integration is continuous, since if we define $$T(f)(x) = \int_0^x f(t)\,dt, \qquad t\in[a,b]$$
we have
$$|T(f)(x)| \le \left| \int_0^x f(t)\, dt\right| \le \int_0^x |f(t)| dt
\le \|f\|(b - a),$$
where the norm denotes the uniform norm. This map $T: \mathcal{C}[a,b]\to
\mathbb{R}$ is continuous.
If you make the $\mathcal{C}^1$ norm $f\mapsto \|f\| + \|f'\|$, then differentiation is continuous in that norm. |
H: Decomposition of $S^2(\wedge^2 E)$
Consider bianchi map $$ b(T)(x,y,z,t) = \frac{1}{3}(T(x,y,z,t)+T(y,z,x,t) + T(z,x,y,t))$$ where $T\in S^2(\wedge^2 E)$
I already checked that $b(b(T))=b(T)\in S^2(\wedge^2 E)$
But how can we derive the following ?
$$ S^2(\wedge^2 E) = {\rm Ker}\ b \oplus {\rm Im}\ b$$ and
$${\rm Im}\ b = \wedge^4 E$$
AI: The splitting $V = \mathrm{ker\ } T \oplus \mathrm{im\ } T$ occurs for any projection operator (i.e. idempotent) $b$ on a finite-dimensional vector space $V$. To see this note that any $v\in V$ can be written $v = (v-b v) + b v$ and apply the rank-nullity theorem.
To show that $\mathrm{im\ }b = \bigwedge^4 E$, you just need to show that $3b(T)_{ijkl} = T_{ijkl} + T_{jkil} + T_{kijl}$ is completely antisymmetric when $T \in S^2 \bigwedge^2 E$. Since you already know $b(T)$ has some antisymmetries (it is still in $S^2 \bigwedge^2 E$), you really just need to check that you pick up a sign change when you swap one of the first two indices with one of the last two. (This shows $\mathrm{im\ }b \subset \bigwedge^4 E$; for the other direction just show that $b$ acts as the identity on $\bigwedge^4 E$.) |
H: Is sum of tail probability always less than integral of tail probability?
I'm working through some Probability and Measure Theory, and frequently we have been using the fact that for $X_i$ iid
$\sum\limits_{k = 1}^{\infty} P(|X_1| > k) \leq \int\limits_0^{\infty}P(|X_1| >t)dt$
Intuitively this makes sense but other than seeing a graph, I haven't been able to convince myself why this is always true, nor have I been able to come up with a counter example. Any insights on how to prove/disprove for a generic iid r.v?
Thanks!
AI: You have $\int_0^1 P(X_1\gt t) dt \ge \int_0^1 P(X_1\gt 1)dt=P(X_1 \gt 1)$ and similarly $\int_n^{n+1}1 P(X_1\gt t) dt \ge \int_0^1 P(X_1\gt n)dt=P(X_1 \gt n)$, so each interval in the integral dominates the element in the sum at the end. |
H: What's wrong with my answer? $\int \frac{3x^2-2}{x^2-4x-12} \mathrm dx$
Here is the original problem: $\int \frac{3x^2-2}{x^2-4x-12}\ \mathrm dx$
After doing polynomial division and factoring the denominator I got this:
$$\int 3 + \frac{12x+36}{(x-6)(x+2)}\ \mathrm dx$$
Then using partial fraction decomposition I got the following:
$$\int 3\ \mathrm dx+ \frac{27}{2}\int \frac{\ \mathrm dx}{x-6} -\frac{3}{2}\int \frac{\ \mathrm dx}{x+2}$$
For the final answer I got this:
$$3x+\frac{27}{2}ln|x-6|-\frac{3}{2}ln|x+2|+C$$
But it says my answer is incorrect. Can you all spot my error or should I provide more details?
AI: You did the calculation wrong.
$$\frac{3x^2-2}{x^2-4x-12} = \frac{3(x^2-4x-12)-2+12x+36}{x^2-4x-12} = 3+\frac{12x+34}{x^2-4x-12}$$
Now, try to evaluate your integral |
H: Inner Product Space on linear transformation on itself
So $V$ is an inner product space and $T : V \to V$ is a linear map such that $$||T(v)|| = ||v||$$ for all $v \in V$. Prove that $$\langle T(v), T(w)\rangle = \langle v, w\rangle$$ for all $v,w \in V$.
However, I missed the class where they talked about this, and reading up in my book hasn't been completely helpful. I don't know what my first step should be in trying to prove this.
AI: $$\langle v,v\rangle +\langle w,w \rangle +2\langle v,w \rangle=
\langle (v+w),(v+w) \rangle $$
$$=\langle T(v+w),T(v+w) \rangle =\langle T(v),T(v) \rangle+2\langle T(v),T(w) \rangle +\langle T(w),T(w) \rangle $$
So $$2\langle v,w \rangle=
2\langle T(v),T(w) \rangle $$ |
H: $S_n = \sum_{k=1}^n \frac{(-1)^{k+1}}{k !}$ for each $S_n$ in terms of $n$?
How can we compute
$S_n = \sum_{k=1}^n \frac{(-1)^{k+1}}{k !}$ for each $S_n$ in terms of $n$?
I tried to compute 2 items together but that didn't work for me. I also tried to find a recurrence relation between $n$ and $n-1$ but that also didn't work. Does anyone know how to solve this?
AI: The limit as $n \to \infty$ is $1-\frac 1e$, which can be derived by looking at the series for $e^{-1}$. I suspect you won't do better for finite sums than just adding them up. |
H: question on Brownian Motion stopping time and end state
I came across this equation in my lecture notes, which states:
$P(T_a < t , W_t \ge a) = P(W_t \ge a)$ where $T_a = \min\{t \ge 0, W_t \ge a\}$.
I'm really confused by this equation: as far as I understand it, the R.H.S $P(W_t \ge a)$ is the probability that at the end of duration $t$, the Brownian Motion hits level $a$ and it doesn't really care about what happened within the $t$ time frame (meaning the Brownian Motion may or may not hit level a before the end of $t$). Whereas the L.H.S $P(T_a < t , W_t \ge a)$ clearly requires the the Brownian Motion hits a before the end of t. So shouldn't $P(T_a < t , W_t \ge a)$ be strictly less than $P(W_t \ge a)$? since R.H.S also includes such processes where $a$ is not hit a single time before the end of $t$.
Could someone please enlighten me on this? Thanks a lot in advance.
AI: Remember that Brownian motion is (almost surely) continuous. So if it is at or above level $a$ at time $t$, then by the intermediate value theorem, it (almost surely) must have hit level $a$ at some time at or prior to $t$. (This assumes $W_0 = 0$ and $a > 0$.) That gives you $P(T_a \le t, W_t \ge a) = P(W_t \ge a)$. To replace $T_a \le t$ with $T_a < t$, note that if $T_a = t$ then $W_t = a$ which happens with probability 0. |
H: How do I prove that the limit as one function goes to infinity is equal to another function?
I was playing around with the integral $\int x^ndx$ and noticed that it is always $\frac{x^{n-1}}{n-1}+c$ except for the singular case $n=-1$. So I could pick $n$ arbitrarily close to $-1$, and the formula works, but as soon as I hit $-1$, it breaks (so to speak)!
Here's my work thus far:
$$\int x^{-(n+1)/n}dx=-nx^{-1/n}+c$$
Working with $n=100$, with the initial condition $y(1)=0$, I got:
$$-100x^{-1/100}+c = 0$$
$$-100+c=0$$
$$c=100$$
So I figure the following could be true:
$$-nx^{-1/n}+n \approx \ln x$$
How can I prove that as I take $n\rightarrow \infty$, the two functions are equivalent?
$$\lim_{n\rightarrow \infty}-nx^{-1/n}+n=\ln x$$
Or equivalently,
$$e^{\lim_{n\rightarrow \infty}-nx^{-1/n}+n}=x$$
AI: The key to this is to be careful with the integral, and in particular, the limits of integration. Note that $$\int^t_1 x^n=\frac{t^{n+1}-1}{n+1}, n\neq 1$$ So now we want to compute $$\lim_{n\to -1}\frac{t^{n+1}-1}{n+1}.$$ By making a change of variables, $n\mapsto n-1$, we get $$\lim_{n\to 0}\frac{t^{n}-1}{n}$$ which is known to be the natural logarithm.
A similar type of thing happens with the integral $$\lim_{n\to 1}\int\sin(nx)\sin(x)dx.$$ |
H: Boundary of a set
Can boundary of any subset of $\mathbb R$ under the usual topology be $[0,1]$? I think it is not possible. If $\partial A=[0,1]$, then $[0,1]\subset \overline{A}$ and $[0,1]\cap A^{0}=\emptyset$. Then what to do? Please give some hint.
AI: Well, the boundary of a dense set with empty interior is its closure (Can you prove this?).
So, in this case, you can take $A = \mathbb{Q}\cap [0,1]$. |
H: Factoring with 5 terms
I'm doing some factoring and have arrived at the point where the book says that:
$$(k+1)/30(6k^4 + 39k^3 + 91k^2 + 89k + 30)$$
factors to:
$$(k+1)/30(k+2)(2k+3)(3(k+1)^2+3(k+1)-1)$$
I cannot derive the latter using the former. Can someone help? This must be a tough cookie to crack because even my solutions manual doesn't actually do the factoring, it just assumes it by 'knowing what we expected the simplified expression to be'. I'd like to know how to do it on my own, seeing how a lot of the advanced courses involve heavy-duty factoring. Thanks.
AI: It looks like a good application of the rational roots theorem. $-2$ and $-\frac 23$ would be in the list of possibilities. What is left is $3k^2+9k+5$. Finding the expression with $(k+1)$ would be a bit of inspection or inspiration |
H: Cylindrical Shell Volumes Problem
Use the method of cylindrical shells to find the volume generated by rotating the region bounded by $y=3+2x−x^2$ and $x+y=3$ about the y-axis.
I have already turned $x+y=3$ into $y=3-x$. However I don't know what to do with the polynomial to continue into graphing them and using the cylindrical shell method $dV=2pirht$. I don't understand how to continue this problem. I'd appreciate any help.
AI: $3+2x-x^2=(3-x)(1+x)$, so $y=3+2x-x^2$ is a parabola opening down with $x$-intercepts at $-1$ and $3$; meets the line $x+y=3$ at the points $\langle 0,3\rangle$ and $\langle 3,0\rangle$. The region between the two curves therefore lies in the first quadrant, above the straight line and below the parabola. When you slice it into vertical strips, you get a strip at each $x$ in the interval $[0,3]$.
If $0\le x\le 3$, the top end of the strip at $x$ lies on the parabola, so it has $y$-coordinate $3+2x-x^2$. The bottom end of the strip lies on the straight line, so it has $y$-coordinate $3-x$. The length of the strip, and therefore the height of the cylindrical shell that it generates, is
$$\text{top}-\text{bottom}=(3+2x-x^2)-(3-x)=3x-x^2\;.$$
The radius of the shell is the distance from $x$ to the axis of revolution, which in this problem is the $y$-axis; that distance is $|x-0|=x$, since $x\ge 0$. The thickness of the shell is $dx$, so its volume is
$$dV=2\pi x(3x-x^2)\,dx\;.$$
Can you finish it from there? |
H: Limit of a multivariable function approaching to $(0,0)$
So I have
$$\lim_{(x,y) \to (0,0)}\frac{x\sin(xy)}{y}$$
And I was wondering if I'm allowed to do this:
$$= \lim_{x \to 0} x * \lim_{(x,y)\to(0,0)} \frac{\sin(xy)}{y} \\=0 * \lim_{(x,y)\to(0,0)} \frac{\sin(xy)}{y} \\ = 0$$
AI: \begin{align*}\lim_{(x,y) \to (0,0)}\frac{x\sin\ (xy)}{y}&= \lim_{(x,y) \to (0,0)}\ x^2\bigg(\frac{\sin\ (xy)}{xy}\bigg)\\&=\lim_{x \to 0}\ x^2 \cdot \lim_{(x,y)\to(0,0)} \frac{\sin(xy)}{xy} \\&=0 \cdot 1=0 \end{align*} |
H: Showing a set of functions $F$ is bounded
I have a set of functions given by;
$$F = \{f:[0,1]\rightarrow\mathbb{R}|\int_0^1 f(x)dx = 0, |f(x)-f(y)|\leq|x-y|, x,y\in[0,1]\}.$$
I have a solution for the question so my questions are about the solution.
Let $f\in F \implies \int_0^1 f(x)dx = 0$.
$\implies$ there exists $x_0\in [0,1]$ such that $f(x_0) = 0$.
$\implies |f(x)| = |f(x)-f(x_0)|\leq |x-x_0|\leq 1$.
Therefore $F$ is bounded.
My questions are:
How can we deduce that there exists an $x_0\in [0,1]$ such that $f(x_0)=0$?
How do we see that $|x-x_0|\leq 1$? I know the Lipschitz constant is less than or equal to one but how does that imply $|x-x_0|\leq 1$?
Thank you for any help and comments.
AI: Since $|f(x) - f(y)| \le |x - y|$, $f$ is continuous (and Lipschitz continuous with constant at most $1$) as you've noted. By the mean value theorem, such an $x_0$ must exist: The MVT for integrals states that for some $x_0$,
$$f(x_0) = \frac{1}{1 - 0} \int_0^1 f(t) dt = \frac{0}{1}$$
Then since $x_0 \in [0, 1]$ and $x \in [0, 1]$, it is immediate that $|x_0 - x| \le 1$; they are elements of the same interval of length $1$. |
H: Integrate $\int \cos x\cos^5(\sin x)dx$
I set $u$ to $\sin x, du$ to $\cos x$, then had $\cos^5udu$, then replaced $(\cos^2u)^2$ with $(1-\sin^2u)^2$ to get $\cos u(1-\sin^2u)^2$, then set $w$ to $\sin u$ and $dw$ to $\cos u$ and got $(1-w^2)^2dw$ then integrated to get $w- 2/3w + w^5/5$ then plugged back in $\sin u$ into the $w$'s and $\sin x$ into the $u$'s to get: $\sin(\sin(x)) - 2/3\sin^3(\sin(x)) + \sin^5(\sin(x))/5$... but wolfram alpha says $5/8\sin(\sin(x))+5/48\sin(3 \sin(x))+1/80\sin(5\sin(x))$ ... I think I'm leaving out a step.
AI: You left out the $+C$. The rest is fine. On to the next question!
Wolfram Alpha seems to have been programmed to prefer linear combinations of $\sin kx$ and $\cos kx$. Because of the endless trigonometic identities, integrals of trigonometric functions can be given in various ways, sometimes not obviously equivalent. |
H: Detailed proof that no essential singularity at infinity implies polynomial
Suppose $f(z)$ is holomorphic in the whole plane, and that $f(z)$ does not have an essential singularity at $\infty$. Prove that $f(z)$ is a polynomial.
I've tried following the hint given in this question. Since $f(z)$ has a nonessential singularity at $\infty$, so $g(z)=f(1/z)$ has a nonessential singularity at $0$. There are two cases:
1) $g(z)$ has a removable singularity at $0$. This means $\lim_{z\rightarrow 0}zg(z)=0$.
2) $g(z)$ has a pole at $0$. This means $g(z)=h(z)/z^k$ for some analytic function $h(z)$ such that $h(0)\neq 0$.
How can I finish each of these cases?
AI: 1) Since $g$ is continuous, we can bound $g(z)$ inside some interval of $z$. Since $g(z)$ is defined as $f(1/z)$, it turns out that from the bound on $z$ we deduce that $f$ is a constant function:
There exists $M, \varepsilon >0$ such that $|g(z)|\leq M$ for all $|z| \in (0,\varepsilon)$. Hence $|f(z)|\leq M$ for all $|z| > 1/\varepsilon$. Letting $\varepsilon \to \infty$ $f$ is bounded and entire. It follows by Liouville's Theorem that $f$ is a constant function.
2) We use that $g$ has a Laurent expansion at $0$ and $f$ has a Taylor expansion since $f$ is entire. We can 'invert' the Laurent expansion so to say and by uniqueness of the Laurent expansion see that $f$ must be a polynomial.
Since the pole is of order $m$, the Laurent expansion of $g$ at $0$ is $$g(z) = \sum_{k=-m}^{\infty} a_k z^k$$ for $|z|\in (0,\varepsilon)$. We can invert this to get $f$: $$f(z) = \sum_{k = -\infty}^{m} a_{-k}z^k$$ for $|z|>1/\varepsilon$. The Taylor expansion of $f$ around $0$ given by $$f(z)=\sum_{k=0}^{\infty}b_kz^k$$ and the Laurent expansion must be equal by uniqueness, so $$f(z)=\sum_{k=0}^{m} b_k z^k$$ where $a_{-k}=b_k$ for all $k$. So $f$ is a polynomial.
See the following pdf: math.berkeley.edu/~mjv/Math185hw8.pdf |
H: sketching the graph of functions
Good morning!
How to draw the graph of the following function and discuss the increasing and decreasing intervals of y.
$y = $x$ + \sqrt[]{x^2}$
Also, I would like to know free online graphing calculator
Thanks in advance.
AI: Hint: Recall that $\sqrt{x^2} = |x|$, so
$$f(x) = x + |x| = \left\{\begin{array}{lr} 2x &: \text{ if } x \ge 0 \\ 0 &: \text{ if } x < 0 \end{array}\right.$$
For an online graphic calculator, I would suggest Wolfram Alpha. |
H: Is $\max(f)$ well defined if $f$ is finite?
Say I have a function that's finite almost everywhere within an interval $[a,b]$. Does that mean that it has an upper bound if I ignore those points on which it is infinite? i.e. is:
$$\text{max}_{[a,b]}(f)$$
well defined?
AI: No, for one of two possible reasons:
1) The function may still be unbounded in $[a,b]$. Bounded and finite are not the same concept. Bounded means there is a number $M<\infty$ such that $|f(x)|<M$ for $x\in[a,b]$, whereas finite means that $|f(x)|<\infty$ for $x\in[a,b]$. This doesn't change if you say these properties hold almost everywhere.
2) Even if the function is finite a.e. in $[a,b]$, it may not achieve its supremum, in which case it has no maximum. There are conditions on $f$ that will guarantee it achieves its supremum (e.g. defined everywhere on a compact set and continuous), but under your minimal assumptions this cannot be guaranteed.
However, in the case that $f$ is unbounded on $[a,b]$ minus a null set and achieves $\infty$ on this null set, then if you allow yourself the extended real numbers you can take $\max_{[a,b]} f = \infty$. |
H: How should I grade my students?
Anyone who is interested in my experience as a grader in the past can check this thread.
This semester I was assigned a grader for a certain class, which runs two sections. There are two professors, Professor $A$ and Professor $B$. Both of them using Professor $A$'s personal written textbook and taught on the same schedule using the same set of problems to evaluate the students. As I am the grader, they hand in their midterms to me on Wednesday. They told me because it is not worth upsetting the students with a low grade, I should grade fairly but cut off no more than five points per problem. This gave an effective lower bound of 75 out of 100 for the students. Obviously I did not have a choice but to comply, because they often modify my grading in the past and complain I graded too harshly. I thought at 25 point scale the grade can still reflect the student's performance.
However, I was shocked when I hand in the graded midterms Professor B told me I still graded too harshly. For example, at a certain step in proving $\frac{2n-1}{5n+1}\rightarrow \frac{2}{5}$ the student computed for $n$ large $|f(n)-L|$ equal$$\frac{1}{4n-2}<\frac{1}{4n},\forall n\ge 2$$etc. I cut off all the five points available and gave a mark $\frac{15}{20}$ for this problem. But Professor $B$ claim that this is a inconsequential arithemetic mistake, 2-3 points should be the maximum. In this way he systematically raised up the students' grades to over $90$ unless the students did extremely badly. He told me that if the students receive a mark over 90, then they would be happy. Moreover since no one will complain if he taught the students well, he has no moral hazard by giving free $A$s to most of the students. When I questioned him if the students would learn anything after such a grading process, he admitted that is questionable. But "who cares?". He told me in order to get tenure one has to care about teaching references.
Since there is nothing I can do to Professor $B$'s grade inflation, I graded Professor $A$'s section of students as I did before, only with $15/20$ lower bound on each problem. Professor $A$ asked me how is Professor $B$'s class. After hearing my story, Professor $A$ exhibited deep sympathy with the situation. But to my surprise he proposed to let me inflate his section's grades as well so the students would not consider themselves to be treated unfairly. To be more specific, now about 2/3 of the students can get over $90/100$. However, there is an exception to this rule. There is a student who honestly written he/she cannot finish the test in time, and has to leave the last problem in blank. Since I cannot give a blank problem $15/20$, her final grade around $70$ was the lowest in the class even though her performance is not the lowest judged by the test paper. There are plenty of student written nonsense like the above equation and I have to grade them $15/20$ at first, and now maybe $18/20$ with Professor $A,B$'s personal policy.
I want to know (without disclosing more detailed personal information), if there is anything I can do to avoid behaving dishonestly like this in future. I know I have to compromise to live in the academia, but I found the whole inflation scenario to be totally absurd. When I think about it, the obvious answer is I should refuse to grade with $75/100$ at first, then maybe the next string of things would not happen. However doing that would not change the grade inflation at all since the Professors have the final say on the grades. I proposed to be the TA for their sections for free as I felt morally awkward at this moment. I want to ask for others with similar experience if this is a normality to be expected in any ranking $\approx 100$ university or a sign of academic corruption singularly at here.
Also, why the honest person should suffer while others being inflated? To me it is not someone's fault that he or she tried hard but still learned the subject very slowly. I know my department's funding policy depends on the students' performance. My guess is with a new $70$ on her midterm, her academic position is endangered significantly. But there is nothing I can do to change it. I felt somehow guilty even though I am not responsible for her performance as I am not the teaching for the class.
update:
My proposal of setting up personal office hour was accepted. However it seems I have to inflate my grades in future as well for Professor $B$. I felt very complicated.
update:
After taking a look at my grading results, Professor $A$ considered the inflation I had done is not enough to bring it to the level of Professor $B$'s inflations. As a result he asked me to cancel my office hours and spend more time on grading (he suppose if I spend more time on the grade inflation thing, I would do it better). Since I have to "match" Professor $B$'s inflation scale, I asked him personally and he said he is effectively grading on 5 point scale. So as long as student write something, no one will be cut off more than 3 points.
This is one of the most disappointing events in my short budding career.
AI: "I want to know (without disclosing more detailed personal information), if there is anything I can do to avoid behaving dishonestly like this in future."
Two things. You can tell the chair and the graduate coordinator that you do not want to be assigned to either of those two professors any more and explain why (I hope that it is not the universal attitude towards grading at your college; if it is, you are just in a wrong place at a wrong time to save your moral integrity). When you become a professor yourself, grade in the way consistent with your own moral principles and make sure that you are a good enough researcher to make the department worry about losing you, not you worry about losing job at any particular place. Moral integrity of a slave is an oxymoron, and the choice between living on knees and dying upright can be postponed for a worthier crusade. |
H: combinations possible of distributing 8 apples to 4 people
I saw a problem where you need to distribute 8 apples to 4 different people where there is also a possibility of one or more of them getting 0 apples too. How do you compute in combinatorics the number of total combinations. The answer is given as 165. I am unable to deduce a combinatoric term for the answer.
AI: First, you need to state the problem carefully. It sounds like the people are labeled-then this becomes a stars and barsproblem. Since you seem to allow that some people don't get any apples, distribute $12$ apples to $4$ people, requiring that each gets at least one. Now you have to choose four places of eleven to place dividers, so the answer is ${11 \choose 3}=165$ but I am not confident I am answering the correct problem. |
H: Find a closed expression for the sum of the entries of the Pascal triangle inside the upper n x n rhombus.
Find a closed expression for the sum of the entries of
the Pascal triangle inside the upper $n \times n$ rhombus.
For example, for $n = 3$, you need to sum the entries:
\begin{array}{cccccccc}
&&&&1\\
&&&1&&1\\
&&1&&2&&1\\
&\cdot&&3&&3&&\cdot\\
\cdot&&\cdot&&6&&\cdot&&\cdot
\end{array}
I found this problem was really interesting but I was struggling to solve this problem.
I found that the next number below $6$ is $2(1+3+6)$, double the sum of the numbers of the edge of the rhombus. Also, $6=2(1+2)$. I also found that the $m$th number on the edge of the $n \times n$ rhombus is the sum of the first $m$ numbers on the edge of the $(n-1) \times (n-1)$ rhombus. However, I couldn't figure out how to sum them up.
AI: $\newcommand{\cb}{\color{brown}}$You can get it from the hockey stick (or Christmas stocking) identity, which is identity $(8)$ here when $j=k$ in that identity. You have
$$\begin{align*}
\sum_{k=0}^{n-1}\sum_{m=0}^{k+n-1}\binom{m}k&=\sum_{k=0}^{n-1}\binom{k+n}{k+1}\\\\
&=\sum_{k=0}^{n-1}\binom{k+n}{n-1}\\\\
&=\sum_{k=n}^{2n-1}\binom{k}{n-1}\\\\
&=\binom{2n}n-\binom{n-1}{n-1}\\\\
&=\binom{2n}n-1\;.
\end{align*}$$
For example, when $n=3$ it’s $\binom63-1=19$, as can be seen from the figure in the question.
Geometrically this is quite elegant: just subtract $1$ from the binomial coefficient directly below the bottom point of the rhombus.
Added: It may be easier to see why the sum of the entries in the rhombus is the double summation above if you square up Pascal’s triangle:
$$\begin{array}{c|cc}
m\backslash k&0&1&2&3&4&5&6\\ \hline
0&\cb1&0&0&0&0&0&0\\
1&\cb1&\cb1&0&0&0&0&0\\
2&\cb1&\cb2&\cb1&0&0&0&0\\
3&1&\cb3&\cb3&1&0&0&0\\
4&1&4&\cb6&4&1&0&0\\
5&1&5&10&10&5&1&0\\
6&1&6&15&20&15&6&1
\end{array}$$
I’ve colored the $3\times 3$ rhombus brown. |
H: Can a complex function be complex-differentiable at a point and not in a neighborhood?
Is it possible for a function $f:\mathbb{C} \to \mathbb{C}$ to be complex-differentiable at a point $z_0\in \mathbb{C}$ without being analytic in a neighborhood of $z_0$? How can we prove this?
AI: Yes; try $f(z)=|z|^2$ ; then Cauchy-Riemann:
$u_x=2x; u_y=2y; v_x=v_y=0$, is satisfied only at $(x,y)=(0,0)$ , tho maybe that is not
exactly what you wanted? |
H: The Hahn-Banach theorem for Hilbert spaces follows from Riesz's theorem
How does the Hahn-Banach theorem for Hilbert spaces follow from Riesz's representation theorem?
AI: Yes. Given a subspace $M$ of a Hilbert space $H$, and a continuous linear functional $f : M \to \mathbb{C}$, you can extend $f$ to a unique continuous linear functional $g : \overline{M} \to \mathbb{C}$.
Now, $\overline{M}$ is a Hilbert space, to there is $y \in \overline{M}$ such that
$$
g(x) = \langle x, y\rangle
$$
for all $x\in \overline{M}$.
Now write $h : H \to \mathbb{C}$ to be $x \mapsto \langle x, y\rangle$ and this extends $f$. |
H: A particular version of Gronwall's inequality
We have this theorem (Gronwall's inequality):
Let $f$, $g$ and $h$ be continuous nonnegative functions defined for $t\ge t_0$. If$$f(t)\le h(t)+\int_{t_0}^{t}g(s)f(s)\,ds\>,$$then$$f(t)\le h(t)+\int_{t_0}^{t}g(s)h(s)e^{\int_s^tg(u)\,du}\,ds$$
How do I prove another version of the theorem, letting $h(t)=k$ , where $k$ is any nonnegative constant, i.e If$$f(t)\le k+\int_{t_0}^{t}g(s)f(s)\,ds\>,$$then$$f(t)\le ke^{\int_{t_0}^tg(s)\,ds}\,ds$$
I worked out the following
\begin{align}
f(t)&\le k+\int_{t_0}^{t}g(s)ke^{\int_s^tg(u)\,du}\,ds\\\\&=k\left(1+\int_{t_0}^{t}g(s)e^{\int_s^tg(u)\,du}\,ds\right)
\end{align}
How should I proceed?
AI: HINT
Let $w(t)=v(t)\cdot\exp \left(- \int_{t_0}^{t}g(s)\mathrm{d}s\right)$
We have: $w'(t)<0, \forall t \ge t_0 \ge 0$, where $v(t)=k+\int_{t_0}^{t}g(s)f(s)\,\mathrm{d}s$.
Whence $w(t) \le w(t_0), \forall t \ge t_0 \ge 0$ |
H: Generating all solutions for a negative Pell equation
How to get all solutions for a negative Pell equation? For example, the equation $x^2 - 2 y^2 = -1$ has two solutions - $(7, 5)$ and $(41, 29)$, and the $(7, 5)$ is the fundamental one, right? How to get the $(41, 29)$ solution from the fundamental one?
AI: The fundamental solution of the equation $x^2-2y^2=-1$ is $(1,1)$. We get all positive solutions by taking odd powers of $1+\sqrt{2}$. The positive solutions are $(x_n,y_n)$, where $x_n+y_n\sqrt{2}=(1+\sqrt{2})^{2n-1}$.
One can alternately obtain a recurrence for the solutions. If $(x_n,y_n)$ is a positive solution, then the "next" solution $(x_{n+1},y_{n+1})$ is given by
$$x_{n+1}=3x_n+4y_n,\qquad y_{n+1}=2x_n+3y_n.$$
Note that your solution $(7,5)$ is the case $n=2$, and $(41,29)$ is the case $n=3$.
Similarly, the positive solutions of the equation $x^2-dy^2=-1$ (if they exist) are obtained by taking the odd powers of the fundamental solution. |
H: integrate $(t\sin^2(t))dt$... why is my answer coming out wrong?
I tried setting $u = t, dv = \sin^2t\mathrm{d}t$, then converting $dv$ to $1/2(1-\cos 2t)\mathrm{d}t$
with half angle identity,
Then $du = 1dt$ and $v = 1/2 - 1/4(\sin2t)$, then $$t\sin^2tdt = t(1/2 - 1/4 \sin2t) - (1/2 - 1/4 \sin2t)dt$$ then after evaluating I got $-(t\sin2t)/4 - (\cos2t)/8$
...
But the solutions manual says: $(1/4)t^2 -(t\sin2t)/4 - (\cos2t)/8$...
Did I miss a step?
AI: Hint: We have:
$$\int t \sin^2 t ~ dt = \int \dfrac{1}{2} t(1 - \cos 2t)~dt = \dfrac{1}{2}\left(\int t~dt - \int t ~ \cos 2t ~ dt\right)$$
For the second integration, use integration by parts.
Spoiler
$\dfrac{1}{8} \left( 2 t (t - \sin 2t) - \cos 2t \right) + c$ |
H: Why is the $0$th power mean defined to be the geometric mean?
Mentioned in the wikipedia article, the $0$th power mean is defined to be the geometric mean. Why is this? I understand that a convenient consequence is that the means are ordered by their exponent. But is there an intuitive reason why the $0$th mean should be the geometric mean? Is it true that
$$\lim_{r\downarrow 0} \left(\frac{a_1^r + \dotsb + a_n^r}{n}\right)^{1/r}= \left( a_1 \dotsb a_n\right)^{1/n}?$$
If I take the logarithm and use L'Hospital's rule I get
$$\ln L = \lim_{r\downarrow 0} \frac{a_1^r + \dotsb + a_n^r}{n\left( a_1^r \ln{a_1} + \dotsb + a_n^r \ln{a_n} \right)}.$$
How might we evaluate this?
AI: Use the fact that as $r \rightarrow 0$, for all $a>0$:
$$a^r = 1 + r \log a + o(r)$$
So the sum becomes:
$$\frac{1}{n}\sum_{i=1}^n a_i^r=\frac{1}{n}\sum_{i=1}^n(1 + r \log a_i + o(r))=1+r\frac{1}{n}\sum_{i=1}^n\log a_i + o(r)$$
Furthermore $\lim_{r \rightarrow 0} (1+rx + o(r))^{1/r}=e^x$.
$$\lim_{r \rightarrow 0} \left(\frac{1}{n}\sum_{i=1}^n a_i^r\right)^{1/r}=\lim_{r \rightarrow 0} \left(1+r\frac{1}{n}\sum_{i=1}^n\log a_i + o(r)\right)^{1/r}=e^{\left(\frac{1}{n}\sum_{i=1}^n\log a_i\right)}= \left( \prod_{i=1}^n a_i\right)^{1/n}.$$ |
H: Vector space and vector subspace
We know that a vector space V is a set of vectors on which two operations are defined, sum, which satisfies the following properties:
\begin{equation}
u+v=v+u
\end{equation}
\begin{equation}
u+(v+w)=(v+u)+w
\end{equation}
\begin{equation}
u+0=0+u=0
\end{equation}
\begin{equation}
u+(-u)=(-u)+u=0
\end{equation}
(for each element u,v,w in V, -u opposite element, 0 neutral element) and product by scalar, which satisfies the following properties:
\begin{equation}
a(u+v)=av+au
\end{equation}
\begin{equation}
(a+b)u=au+bu
\end{equation}
\begin{equation}
(ab)u=a(bu)
\end{equation}
\begin{equation}
1 u=u
\end{equation}
(for each element u,v in V; a,b real scalar; 1 neutral element of product). Not all sets of vectors are vector spaces. For example, space $R^2$ with the following operations:
\begin{equation}
(x,y)+(x',y')=(x+x',y+y')
\end{equation}
\begin{equation}
a(x,y)=(ax,0)
\end{equation}
is not a vector space because $1(x,y)=(x,0)\neq (x,y)$.
A vector subspace, instead, is a set S of vectors included in the vector space V, which in turn is a vector space. In other words, S is a vector subspace if
\begin{equation}
0\in S
\end{equation}
\begin{equation}
u+v\in S, \ \ \ u,v\in S
\end{equation}
\begin{equation}
av\in S, \ \ \ \ v\in S, a \in R
\end{equation}
For example:
\begin{equation}
S=\left\{(x,y)\in R^2 | y=2x \right\}
\end{equation}
is a vector subspace. In fact, the generic element of S is thus formed: $(x,2x)$. If $x=0$, we have the neutral element $(0,0)\in S$. My doubt is involved in product by scalar, because: $av=a(x,2x)$ and the next step should be, of course, $(ax,2ax)$. Namely: $a(x,2x)=(ax,2ax)$. But this step is legitimate? We are trying to prove that S is a subspace, then we do not know if $a(x,2x)=(ax,2ax)$. For example, it may happen that $a(x,2x)=(ax,0)$ (see example above on vector space). Perhaps, after considering: 1) S included in V, 2) V vector space; we are assuming that: V vector space with the following operations:
\begin{equation}
(x,y)+(x',y')=(x+x',y+y')
\end{equation}
\begin{equation}
a(x,y)=(ax,ay)
\end{equation}?
I hope to be able to explain. Thank you very much.
AI: You are correct; $S\subset\mathbb{R}^2$, as given, may not be a subspace of $\mathbb{R}^2$ if the action of $\mathbb{R}$ is not compatible. And similarly with addition—how do we know that $(x,2x) + (x',2x') = (x+x',2x+2x')$, unless we have specified the addition law on $S$?
The answer is that these kinds of questions and statements usually take for granted that $S$ inherits certain properties from its parent, $\mathbb{R}^2$. So if $a\in\mathbb{R}$, $v\in S$, we calculate $av$ according to the multiplication rule on $\mathbb{R}^2$, namely $av=a(v_0,v_1)=(av_0,av_1)$. And when we add two vectors, we add them as elements on $\mathbb{R}^2$.
So yes, $S$ is only given as a set, and a set is not a vector space without some additional structure, and it is only a subspace when that structure is compatible with the structure on $\mathbb{R}^2$. It is by convention, however, that we always assume that $S$ is granted exactly the structure that makes it compatible. Your conclusion at the very end is exactly the solution; we do assume these things, and this is a good point to recognize. |
H: A problem of minimizing distance.
A power house, P, is on one bank of a straight river $200$ m wide,
and a factory, F, is on the opposite bank $400m$ downstream from P.
The cable has to be taken across the river,
under water at a cost of Rs $6/m$.
On land the cost is Rs $ 3/m$.
What path should be chosen so that the cost is minimized?
I first looked at the extreme cases and then tried to find the solution. But then too there was a problem
AI: Draw a picture. Let $A$ be the point directly across the river from $P$.
Suppose we go underwater directly to a point $x$ metres downstream from $P$. Then by the Pythagorean Theorem, we will have a $\sqrt{x^2+200^2}$ underwater stretch. Then we need to travel $400-x$ overland to $F$. The cost $C(x)$ is given by
$$C(x)=6\sqrt{x^2+200^2}+3(400-x).$$
To minimize, find the derivative $C'(x)$ and set it equal to $0$. We have
$$C'(x)=\frac{6x}{\sqrt{x^2+200^2}}-3.$$
Solving $C'(x)=0$ is easy. We get $2x=\sqrt{x^2+200^2}$. Square both sides.
When you have done this, and found a candidate $x$, you need to check whether this really gives the minimum cost.
Remark: There are various other ways to solve the problem. We chose the most basic beginning calculus approach. |
H: distribution of chocolate problem?
I have 6 packet where , 10 20 30 35 50 60 are chocolate in that 6 packet.I have to distribute one packet to k children. where k=3 .Then unfair is calculated by
find=min(|sum(Xi-Xj)|) where Xi and Xj are set of number chocolate .ex:-
10 20 30 we get unfair distribution as
|30-20|+|30-10|+|30-20|=40
10 30 35 we get unfair distribution as
|35-30|+|35-10|+|35-30|=30
...similar for 30 35 50 as I get 40 and 35 50 60 as I get 50
So minimum of them is 30 (10 30 35)
My question is there any way to determine the minimum faster rather than performing all the calculation.
AI: For three packets, the unfairness is max minus min, times two: If $a\le b\le c$, then $$|a-b|+|a-c|+|b-c|=(b-a)+(c-a)+(c-b)=2c-2a.$$ |
H: Is there any weighted graph which smallest eigenvalue of its adjacency matrix is greater than 1?
Is there any weighted or unweighted graph which smallest eigenvalue of its adjacency matrix is greater than 1?
AI: If the graph has no loops, then the answer is "no" since the sum of the eigenvalues is equal to the trace of the adjacency matrix which is zero. Hence there must be at least one eigenvalues that is $\leq 0$.
If the graph has loops, then yes: just take a single vertex with two loops. |
H: Using counting formula to get |G| = |kernel φ||image φ|
The counting formula I am saying :
Let S be a finite set on which a group G operates, and let Gs and Os be the stabilizer and orbit of an element s of S. Then
|G|=|Gs||Os|
or
(order of G)=(order of stabilizer)(order of orbit)
then I have the following question:
Suppose that φ : G → G′ is a homomorphism between finite groups. Explain how the familiar formula |G| = |kernel φ||image φ| follows from the counting formula.
I was told that if G acts on the set of left coset of Kernel φ, i.e. G/K,
then by counting formula, use |kernel φ|=|stabilizer| and |image φ|=|orbit|
we can get |G| = |kernel φ||image φ|
but I can not rationalize it, I think it is due to my poor understanding of coset. Can someone explain it a little? Thanks in advance.
AI: (Left) cosets are subsets of $G$ of the form $gK$ (here with $K=\ker \phi$).
For $g,h\in G$ we have either $gK=hK$ (even if possibly $g\ne h$ of $gK\cap hK=\emptyset$, and the union of all cosets is of course all of $G$ (as $g\in gK$).
How does $G$ act on $G/K$? For $g\in G and $A\in G/K$ (say, $A=hK$ with $h\in G$) the set $gA=ghK$ is also a left coset - and that's it.
Let us pick a specific coset in $G/K$ and determine ists stabilizer. For simplicity, pick $K$ itself.
For which $g\in K$ is $gK=K$?
Since $e\in K$, this specifically implies $g\in K$, i.e. $g\in\ker \phi$ by our choice of $K$. On the other hand $g\in K$ also gives us $gK=K$. So the stabilizer of $K$ is just $\ker \phi$ itself.
How can we relate $\operatorname{im}\phi$ with $G/K$?
If $g'\in G'$ is in the image of $\phi$, then $\phi^{-1}(\{g\})$ is a left coset (if $\phi(g)=g'$ and $k\in G$ then $\phi(gk)=g'$ iff $k\in\ker\phi$) and of course for different $g'$ we obtain different cosets. (On the other hand if $g'$ is not in the image, then $\phi^{-1}(\{g\})=\emptyset$)
Thus $\phi$ induces a bijective map between the sets $G/K$ and $\operatorname{im}\phi$ (in fact this is a group isomorphism, but we don't need that here; one could do this part with an isomorphism theorem anyway) |
H: Finding the characteristic polynomial of a linear transformation
Let $T: M_n(\Bbb{R}) \to M_n(\Bbb{R})$ be a linear transformation defined by $T(A)=A^t+A$. What is the characteristic polynomial of $T$?
If I use the basis $E_{ij}$, I get $T(E_{ij})=E_{ij}+E_{ji}$, but don't know how to compute the characteristic polynomial.
AI: Let $E_{ij}$ be the matrix with a $1$ at the $(i,j)$-th entry and zero elsewhere. Then all the $E_{ij}$s form a basis of $M_n$. Now, order this basis as follows: $E_{11},E_{22},\ldots,E_{nn}$ go first, followed by every pair of $\{E_{ij},E_{ji}\}$ with $i\ne j$. The order of the pairs are unimportant. What is crucial is that each $E_{ij}$ is adjacent to $E_{ji}$ in the ordered basis.
The matrix of $T$ with respect to this basis is then $\operatorname{diag}(2I_n, J, J, \ldots, J)$, where there are $\frac{n(n-1)}{2}$ copies of $J=\pmatrix{1&1\\ 1&1}$. You can calculate the characteristic polynomial of this block-diagonal matrix easily. |
H: Calculate unique Integer representing a pair of integers
I have a pair of positive integers $(x, y)$, with $(x, y)$ being different from $(y, x)$, and I'd like to calculate an integer "key" representing them in order that for a unique $(x, y)$ there is an unique key and no other pair $(w, z)$ could generate the same key.
I don't mind if we can/can't guess the pair values based on the key.
The key must be < 2,147,483,647 and the value of $x$ and $y$ as high as possible.
Any solution ?
AI: You could use something like $\text{key}(x,y) = 2^{15}x + y$, where $x$ and $y$ are both restricted to the range $[0, 2^{15}]$. In programming terms, this just means stuffing $x$ into (roughly) half of your 31-bit space and $y$ into the other half.
It sounds like you are using a 32-bit integer variable in some programming language to represent the key. This is the reason for the limit of 2,147,483,647, I suppose. This means you really only have 31 bits available, because one bit is reserved for the sign of the integer. But this is slightly wasteful. If you use an unsigned integer variable (which is available in most programming languages), then you'll have 32 bits available. You can then use 16 bits for $x$, and 16 bits for $y$.
Using this sort of approach, recovering $x$ and $y$ from a given key is easy to code and extremely fast -- you just grab the relevant bits from the key variable. I can write the code, if you don't know how.
If you still want to use an int variable for the key (k), then the code (in C#) is:
int k = 32768*x + y;
With this approach, we require $x \le 65535$ (16 bits) and $y \le 32767$ (15 bits), and k will be at most 2,147,483,647, so it can be held in a 32-bit int variable.
If you are willing to use an unsigned int variable for the key (k), instead, as I suggested, then the code (again in C#) is:
uint k = 65536*x + y;
With this approach, we require $x \le 65535$ (16 bits) and $y \le 65535$ (16 bits), and k will be at most 4,294,967,295, which will fit in a 32-bit unsigned int variable. |
H: How to evaluate this using algebra?
We have $$1.001^6 - 1.001^5$$
How do I evaluate this? Normally I would use algebra to rewrite it, but I don't know how to cleverly rewrite $a^6 - a^5$ to a simpler form.
AI: How about (let $x = 0.001$ for simpler writing)
\begin{align*}
(1+x)^6 - (1+x)^5 &= (1+x)^5 \cdot (1+x - 1)\\
&= x \cdot (1+x)^5\\
&= x^6 + 5x^5 + 10x^4 + 10x^3 + 5x^2 + x\\
&= 0.001\,005\,010\,010\,005\,001
\end{align*} |
H: Nonnegative measurable function on measure zero set
Let $f$ be a nonnegative measurable function from $X$ to $R$.
Then, for $\mu$$(A)=0$, $\int_A f d\mu$=0?
I think by using increasing simple function to $f$ and Monotone convergence theorem,
it's true but i'm not sure...
AI: Let $s=\sum_{i=1}^na_i\chi_{A_i}$ be a simple function with $0\leq s\leq f$. Then, $$\int_As\,d\mu=\sum_{i=1}^na_i\mu(A\cap A_i)=0.$$ Since this holds for all those $s$, you have that $\displaystyle\int_Af\,d\mu=0$. |
H: Fact about PSD matrices
I'm reading something where $C$ is a positive semi-definite matrix. Also, the centring matrix is $H = (I - (1/n)11^{T})$.
My notes say that $C$ is p.s.d. $\implies HCH$ is p.s.d. I'm just not seeing seeing why.
AI: A matrix $C$ is called positive semidefinite if $y^\ast Cy\ge0$ for all vector $y$. When $C$ is positive semidefinite, $H^\ast CH$ is positive semidefinite for every matrix $H$, because for any vector $x$, if we put $y=Hx$, we get $x^\ast(H^\ast CH)x=(Hx)^\ast C(Hx)=y^\ast Cy\ge0$.
Now your $H$ is real symmetric. Hence $H^\ast=H$ and $HCH=H^\ast CH$. So, the statement in the previous paragraph is applicable. |
H: Unbounded function with uniform bounds on Integrals
I'm looking for examples of unbounded functions $f\colon\Bbb R\rightarrow\Bbb R$ with the property that for all $L$ exists some $C(L)$ with
$$\int_x^{x+L} f(t)dt \le C$$
for all $x$.
AI: We can choose $f$ non-negative, continuous, integrable and unbounded, for example
a piecewise linear function for which $f(k)=k$, $f(k-2^{-k})=f(k+2^{-k})=0$.
Instead of a continuous function, one can choose a smooth one, scaling adequately a bump function in order to get a smooth non-negative integrable bounded function.
In these cases $C$ will not depend on $L$. |
H: Area between curve $x^n$ and other lines
Find the area between the curve $y = |x^n|$, where $n$ is a positive constant, the line defined by $y=-2$, and the lines defined by $|x| = 2$.
I have answered this question, but I don't know if my answer is correct, so that's why I'm posting it here.
We want $\displaystyle \int_{-2}^{2} (y = |x^n| -2)$
$$ [\space |\dfrac{x^{n+1}}{n+1}| -2x ]_{-2}^{2}$$
Evaluate it and get $$ | \dfrac{2^{n+1}}{n+1} | - | \dfrac{-2^{n+1}}{n+1} | + 8$$
I believe the first 2 cancel out because of the absolutes, so $8$ is the answer.
Is this correct?
AI: The final answer cannot be independent of $n$ - it is clearly different when $n=1$ and $n=2$. You are making a mistake integrating $|x|$ :
Since the function is even, you get
$$
A = 2\int_0^2 (x^n - 2)dx = 2\left [ \frac{2^{n+1}}{n+1} -4 \right ]
$$ |
H: A notation problem of partial derivatives
I have a notational problem of partial derivatives:
Let $z=f(x,\phi(x))$,and let $\phi(x),f$ be a differentiable functions.What is the difference between $\frac{\partial z}{\partial x}$ and $\frac{\partial f}{\partial x}$?In my eye,they ought to have no difference,but in my text book,the author writes:
$$\frac{\partial z}{\partial x}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial \phi(x)}\frac{\partial \phi(x)}{\partial x}.$$
Thanks for any clarifications.
AI: Abuse of notation. You have a function $f(u,v)$ of two real (for the sake of argument) variables. You also have a function $\Phi$ of one real variable with values in $\mathbb{R}^2$, namely $\Phi(x) = (x,\phi(x))$.
Then you define $z = f \circ \Phi$. And the chain rule yields
$$\frac{\partial z}{\partial x}(x) = \frac{\partial f}{\partial u}(\Phi(x))\cdot \frac{\partial \Phi_1}{\partial x}(x) + \frac{\partial f}{\partial v}(\Phi(x))\cdot \frac{\partial \Phi_2}{\partial x}(x).$$
Now note that $\Phi_1(x) = x$, hence $\frac{\partial \Phi_1}{\partial x} \equiv 1$, and $\Phi_2(x) = \phi(x)$, so what remains is
$$\frac{\partial z}{\partial x}(x) = \frac{\partial f}{\partial u}(x,\phi(x)) + \frac{\partial f}{\partial v}(x,\phi(x))\cdot \frac{\partial \phi}{\partial x}(x).$$
Then commit the abomination of writing $\frac{\partial f}{\partial \phi(x)}$ for $\frac{\partial f}{\partial v}$. |
H: Probability with average of measures
I am trying to solve Problem 2 from this problem set.
Let $\mathbf{Y}$ be the avreage of $5$ independent measurements.
For a single measurement one have $\sigma^2 = 0.060^2$ and $\mu = 6.8$.
$\textbf{b})$ What is the probability that $\mathbf{Y}$ deviates more than $0.06$ from $\mu$ ?
I tried to use the probability that one measurement deviated more than $0.06$ from $\mu$. And then the total probability should be $1 - P(\text{all are below } 0.06)$.
But I see that this is wrong.
I also tried calculating
$$
1 - P\left( Z < \frac{5\cdot 6.8- (5\cdot 6.8 + 0.06)}{\sigma/\sqrt{5}} \right)= 1 - P(Z<\sqrt{5\ \!})
$$
Which also turns out to be false. The correct answer should be
P = $0.026$, but I can not quite get there. I am not experienced with dealing with
the average of independent measurements, so any literature, hints or tips is very welcome.
AI: The questions says "results which are assumed to be independent and normally distributed", which is important to note.
In effect you are being asked to find the distribution of $Y$ (which is a mean, not a sum as in your expression).
What is the expected value of $Y$? (It is not $5\cdot 6.8$.)
What is the standard deviation of $Y$? (As you seem to say, it is $\sigma/\sqrt5$.)
Now, what is the probability of $Y$ deviates from $\mu$ by more than $0.06$? Remember that $Y$ might be greater than or less than $\mu$, and the deviation here probably means the absolute value of the difference. |
H: Proving sequential compactness from open cover compactness.
Let $(\mathcal M,d)$ be a metric space and $A\subset\mathcal M$. The following types of compactness are equivalent:
(i) Each open cover of $A$ contains a finite subcover.
(ii) $A$ is sequentially compact, i.e. each sequence in $A$ contains a subsequence which converges in $A$.
In the lecture, I've seen proofs for (i) $\Rightarrow$ (ii) and (ii) $\Rightarrow$ (i). I have a problem with the first one, (i) $\Rightarrow$ (ii), which reads as follows:
If an (infinite) sequence $(x_k)_{k\in\mathbb N}\subset A$ does not have any cluster points, then $\forall\ y\in A$ we can find a ball of radius $r_y>0$ around $y$ which contains only finitely many elements of $(x_k)_k$.
We can use these balls to construct an (infinite) open cover of $A$, $A\subset\bigcup_{y\in A}B_{r_y}(y)$. By (i), there exists a finite subcover, i.e.
$\exists\ y_1,\dots,y_n\in A$, such that $A\subset \bigcup_{i=1}^nB_{r_{y_i}}(y_i)$.
Since $(x_k)_k\subset A$ has infinitely many elements but each of the finitely many balls contains only finitely many elements of $(x_k)_k$, we have a contradiction and, thus, $(x_k)_k$ has to have at least one cluster point. So far, I agree.
Now the argument is that, $(x_k)_k$ has to contain a subsequence which converges and, thus, we've shown (ii). My point is: we are required to show that the subsequence should converge in $A$ but I don't seen why the cluster point of the original sequence should be in $A$. For this to hold, $A$ would have to be closed. In the lecture, we had a lemma stating that $A$ is closed if it is compact, but we used the equivalence of (i) and (ii) in the process.
AI: The contradiction shows that there is at least one $y \in A$ such that every ball with centre $y$ contains infinitely many $x_k$. Then there is a subsequence converging to $y$.
Alternatively, we can show that a compact set is closed without the equivalence:
Let $A\neq \overline{A}$. Let $x \in \overline{A}\setminus A$. Then the open cover
$$\{ \mathcal{M}\setminus \{ x : d(x,y) \leqslant 2^{-n}\} : n \in \mathbb{N}\}$$
of $A$ has no finite subcover. |
H: Sum of sines $\sum_{k=0}^{n} \sin(\phi +k\alpha)$
I've got the following problem. I'd like to prove that
$$\sum_{k=0}^{n} \sin(\phi +k\alpha) = \frac{\sin\left(\frac{n+1}{2}\right)\alpha + \sin\left(\phi + \frac{n\alpha}{2}\right)}{\sin\frac{\alpha}{2}}$$ Can you help me? I tried writing sines using complex numbers (that $\sin x = \frac{e^{ix}-e^{-ix}}{2i}$), but it only made the calcuations more difficult... May somebody show me how to solve this problem? I'd be grateful.
AI: I might be wrong but I think there might be a mistake in your first post : see my answer below. The main idea here is $\Im(e^{i\theta})=\sin(\theta)$ ($\Im(z)$ denotes the imaginary part of $z$). Using this idea, you get :
$$ \sum_{k=0}^{n} \sin(\phi+k\alpha) = \Im \Big( \sum_{k=0}^{n} \exp\big( i\phi + ik \alpha \big) \Big) \; . $$
Let's work on $\displaystyle \sum_{k=0}^{n} \exp(i\phi + ik\alpha)$. Let's assume that $\alpha \notin 2\pi \mathbb{Z}$. You have :
$$
\begin{eqnarray*}
\sum_{k=0}^{n} \exp \big( i\phi + ik\alpha \big) & = & \exp(i\phi) \sum_{k=0}^{n} \exp(ik\alpha) \\[2mm]
& = & \exp(i\phi) \frac{1 - \exp \big( i(n+1)\alpha \big)}{1 - \exp(i\alpha)} \\[2mm]
& = & \exp(i\phi) \frac{\exp \big( i\frac{n+1}{2}\alpha \big) \left[ \exp \big( -i \frac{n+1}{2} \alpha \big) - \exp \big( i\frac{n+1}{2} \alpha \big) \right] }{\exp \big( i\frac{\alpha}{2} \big) \left[ \exp \big( -i\frac{\alpha}{2} \big) - \exp \big( i\frac{\alpha}{2} \big) \right] } \\[2mm]
& = & \exp(i\phi) \exp \big( i\frac{n\alpha}{2} \big) \frac{\sin \big( \frac{n+1}{2}\alpha \big)}{\sin \big( \frac{\alpha}{2} \big)} \\[2mm]
& = & \exp \big( i\phi + i\frac{n\alpha}{2} \big) \frac{\sin \big( \frac{n+1}{2} \alpha \big)}{\sin \big( \frac{\alpha}{2} \big)} \; . \\
\end{eqnarray*}
$$
So,
$$
\begin{eqnarray*}
\Im \Big( \sum_{k=0}^{n} \exp \big( i\phi + ik \alpha \big) \Big) & = & \Im \left[ \exp \big(i\phi + i\frac{n\alpha}{2} \big) \frac{\sin \big( \frac{n+1}{2} \alpha \big)}{\sin \big( \frac{\alpha}{2} \big)} \right] \\
& = & \sin \big( \phi + \frac{n\alpha}{2} \big) \frac{\sin \big( \frac{n+1}{2} \alpha \big)}{\sin \big( \frac{\alpha}{2} \big)} \; .\\
\end{eqnarray*}
$$
Therefore, if $\alpha \notin 2\pi \mathbb{Z}$,
$$\sum_{k=0}^{n} \sin(\phi + k\alpha) = \sin \big( \phi + \frac{n\alpha}{2} \big) \frac{\sin \big( \frac{n+1}{2} \alpha \big)}{\sin \big( \frac{\alpha}{2} \big)} \; .$$
And if $\alpha \in 2\pi\mathbb{Z}$,
$$ \sum_{k=0}^{n} \sin(\phi+k\alpha) = \sum_{k=0}^{n} \sin(\phi) = (n+1)\sin(\phi) \; . $$ |
H: Isomorphism between partially ordered sets - What is wrong with my argument?
I'm trying to show
Isomorphism between two partially ordered sets is an equivalence relation.
Suppose $M$ and $M^{\prime}$ are two partially ordered set and $f:M\to M^{\prime}$ is isomorphism between them. To show reflexivity, let $a\in M$ then since $M$ is partially ordered $a\leq a$, so $f(a)\leq f(a)$.
If $f(a)\leq f(b)$ and $f(b) \leq f(c)$ imply $a \leq b$ and $b \leq c$. $M$ is partially ordered set so, $a\leq c$, hence, $f(a)\leq f(c)$.
I have problems to show symmetry. If $f(a)\leq f(b)$ I can't show $f(b)\leq f(a)$. Either my argument is completely wrong or I am missing something important. Thanks for any help!
AI: I assume what you mean by isomorphism is that $f$ preserves the structure of $M$
when mapping to $M^{\prime}$ in the following sense:
$f$ is a bijection
$x\leq_{M}y$ iff $f\left(x\right)\leq_{M^{\prime}}f\left(y\right)$
I also assume you define equivalence as follows: $M$ and $M^{\prime}$
are equivalent (written $M\sim M^{\prime}$) if there exists an isomorphism
(as defined above) between them. Then
Taking $f$ to be the identity map, $M\sim M$
Suppose $M\sim M^{\prime}$. Then there exists an isomorphism $f_{1}\colon M\rightarrow M^{\prime}$.
Taking $f\equiv f_{1}^{-1}$ ($f_{1}$ is a bijection), we get $M^{\prime}\sim M$.
Suppose $M\sim M^{\prime}$ and $M^{\prime}\sim M^{\prime\prime}$.
Then there exist isomorphisms $f_{1}\colon M\rightarrow M^{\prime}$
and $f_{2}\colon M^{\prime}\rightarrow M^{\prime\prime}$. Taking
$f\equiv f_{2}\circ f_{1}$, we can show that $f$ is an isomorphism.
Suppose $a,b\in M$ with $a\leq_{M}b$. Then $f_{1}\left(a\right)\leq_{M^{\prime}}f_{1}\left(b\right)$,
and as such $f_{2}\left(f_{1}\left(a\right)\right)\leq_{M^{\prime\prime}}f_{2}\left(f_{1}\left(b\right)\right)$.
Edit: The notation $a \leq_M b$ refers to the partial order on $M$ explicitly. The subscript is often omitted. |
H: Which is bigger $x^{log(x)}$ OR $(log(x))^x$
I'm trying to find out which is bigger $x^{log(x)}$ OR $(log(x))^x$
As $x \to \infty$
I tried to take the log of both but I didnt't reach any where.
AI: You can try to take $\lim\limits_{x\rightarrow +\infty}x^{\log(x)}/\log(x)^x$. To do that, see that:
$$\begin{aligned}
\lim\limits_{x\rightarrow +\infty} \frac{x^{\log(x)}}{\log(x)^x} &=\lim\limits_{x\rightarrow +\infty}
\frac{\exp\left(\log(x)\log(x)\right)}{\exp\left(x\log(\log(x))\right)} \\
&= \lim\limits_{x\rightarrow +\infty} \exp\left(\log(x)\log(x)-x\log(\log(x))\right) \\
&= \exp\left(\lim\limits_{x\rightarrow +\infty}(\log(x))^2-x\log(\log(x))\right)
\end{aligned}$$
and so you have to compare $(\log(x))^2$ and $x\log(\log(x))$. So take
$$\lim\limits_{x\rightarrow +\infty}(\log(x))^2-x\log(\log(x))$$
This equals
$$\lim\limits_{x\rightarrow +\infty}x\log(\log(x))\left(\frac{(\log(x))^2}{x\log(\log(x))}-1\right)$$
which is a much easier limit to analyse. Applying l'Hôpital's rule to the fraction inside the parentheses we have
$$\lim\limits_{x\rightarrow +\infty}\frac{2\log(x)/x}{\log(\log(x))+1/\log(x)}$$
As x approaches infinity, the numerator approaches 0 and the denominator approaches infinity, so the whole limit approaches 0. Therefore our original limit can be found because
$$\begin{aligned}
\lim\limits_{x\rightarrow +\infty} (\log(x))^2-x\log(\log(x)) &= \lim\limits_{x\rightarrow +\infty}x\log(\log(x))\left(\frac{(\log(x))^2}{x\log(\log(x))}-1\right) \\
&= +\infty(-1) \\
&= -\infty
\end{aligned}$$
Now, this was just the exponent of $e$ in the definition of our original limit. Thus:
$$\begin{aligned}
\lim\limits_{x\rightarrow +\infty}\frac{x^{\log(x)}}{\log(x)^x} &= \exp\left(\lim\limits_{x\rightarrow +\infty}(\log(x))^2-x\log(\log(x))\right)\\
&= \exp(-\infty)\\
&= 0
\end{aligned}$$
So we reach the conclusion that, as $x \rightarrow +\infty$, $\log(x)^x$ grows faster than $x^{\log(x)}$. |
H: $2$-dimensional Hyperbolic space with fundamental group ${\bf Z}$ and constant curvature $-1$
$$ d\rho^2 + \cosh^2\rho\ d\theta^2$$
Only one ? Is there any other example ?
AI: You have essentially two choices: either the quotient of the hyperbolic plane by (the cyclic group generated by) a hyperbolic isometry or the quotient of the hyperbolic plane by (the cyclic group generated by) a parabolic isometry. |
H: Solve recurrence relation $ a_{n+1} = (n+1)a_n + 1 $
$a_0 = 1 \\ a_{n+1} = (n+1)a_n + 1 $
Could you help me solve this?
And maybe someone know good source explaining how to solve recurrence relations?
AI: as @Did said in the comments we can put
$$ n!b_n = a_n \Rightarrow (n+1)!b_{n+1} = (n+1)!b_n + 1 $$
$$ \Rightarrow (b_{n+1} - b_n)(n+1)! = 1 \Rightarrow b_{n+1} - b_n = \frac{1}{(n+1)!} $$
you can do the following because you'll get a telescoping sum $$ \sum_{n=0}^k (b_{n+1} - b_n) = \sum_{n=1}^{k+1} \frac{1}{n!} \Rightarrow b_{k+1} - b_0 = \sum_{n=1}^{k+1} \frac{1}{n!} $$
$$ a_0 = 1 = 0!b_0 = b_0 \Rightarrow b_{k+1} = 1 + \sum_{n=1}^{k+1} \frac{1}{n!} \Rightarrow b_n = 1 + \sum_{k=1}^{n} \frac{1}{k!} $$
$$ \Rightarrow a_n= n! \left(1 + \sum_{k=1}^{n} \frac{1}{k!} \right) $$ |
H: How to guess that $ \sum_{i=1}^{n}3^i = \frac{3}{2}(3^n - 1)$
As in title how do you guess that $ \sum_{i=1}^{n}3^i = \frac{3}{2}(3^n - 1)$?
I have homework about solving recurrence relations and using iterate method I can find that http://www.wolframalpha.com/input/?i=RSolve%5B%7Ba%5Bn%5D+%3D+a%5Bn-1%5D+%2B+3%5En%2C+a%5B1%5D+%3D+3%7D%2Ca%5Bn%5D%2Cn%5D series shown in this example is equal to $ \sum_{i=1}^{n}3^i$, but I am curious how you change it into $ \frac{3}{2}(3^n - 1)$. I know how to prove it's true, but I don't know how to guess it on my own. Anyone could explain?
AI: Observe that this is a Geometric Series of $n$ terms, the first Term being $=3$ and common ratio $=3$
we can safely use the formula of summation of $n$ terms |
H: Continuous irrational function proof
Let $f,g$ be two defined and continuous function on $[a,b]$ suppose that $f(x)=g(x)$ for all rational $x∈[a,b]$. Prove that that $f(x)=g(x)$ for all real $x∈[a,b]$
Since this is true for rational I know that I need to prove it also true for irrational, but I don't know how to do it.
AI: Define $h(x)=f(x)-g(x)$ for all $x\in [a,b]$. The function $h$ is continuous on $[a,b]$ and $h(x)=0$ for all rational $x\in [a,b]$. Let $y\in [a,b]$ be an arbitrary irrational, then there exists sequence $\{x_{n}\}$ of rational elements of $[a,b]$ such that $x_{n}\rightarrow y$, now since $h$ is continuous on $[a,b]$, so $h(x_{n})\rightarrow h(y)$ but $h(x_{n})=0$ for every natural $n$, therefore $h(y)=0$ and the proof is done. |
H: Flat space Minkowski metric
I am having some problem understanding the why in Minkowski spacetime, the continuity equation is written as $$\partial_\mu J^\mu=0.....................(*)$$
Physically, I know that $$\partial_t \rho=-\nabla\cdot\vec j$$
but isn't the Minkowski metric diag$\{1,-1,-1,-1\}$, such that $(*)$ reads
$$\partial_t \rho=\nabla\cdot\vec j$$ ? What have I done wrong?
Thank you.
Ref: http://en.wikipedia.org/wiki/Continuity_equation
AI: In (pseudo)Riemannian manifold, divergence, in most cases, is no long the usual form $\text{div}X=\partial_iX^i$. The divergence is defined as
$$\text{div}X=\frac1{\sqrt g}\partial_i(\sqrt gX^i)$$
where $g$ is the determinant of metric matrix. So for Minkowski metric, $g=-1$ and thus
$$\text{div}J=\frac1i\partial_\mu(iJ^\mu)=\partial_\mu J^\mu=\frac{\partial \rho}{\partial t}+\nabla\cdot j=0$$ |
H: $\prod_{k=1}^n (1+ \frac{z}{k})$ converges to $0$ when $\Re (z)<0$
We have a complex number $z$ such that $\Re (z)<0$ and the sequence $ z_n = \prod_{k=1}^n (1+ \frac{z}{k})$. Prove that $\lim_{n \rightarrow \infty} z_n = 0$.
How to do it? I guess it will be easier to prove that $\lim_{n \rightarrow \infty} |z_n|^2 = 0$, it's a trick that usually helps, but even though I thought about if for a longer while, I don't know how to prove it. May somebody help?
AI: Hint: For every $w$ in $\mathbb C$, $\left|1+w\right|^2=1+2\Re(w)+|w|^2\leqslant\mathrm e^{2\Re(w)+|w|^2}$, hence, for every $n$,
$$
|z_n|^2\leqslant\exp\left(2\Re(z)\sum_{k=1}^n\frac1k+|z|^2\sum_{k=1}^n\frac1{k^2}\right)\leqslant C(z)\cdot n^{2\Re(z)}.$$ |
H: Working out modulo without a calculator
How does one determine modulo without a calculator in cases like this:
$$15^7 - 13^5(\mod14)$$
Normally I would simply divide what is given by the modulo number and take the decimal output and times it by the modulo number. How can I work out $15^7 - 13^5(mod14)$ without the use of a calculator?
Now what I am thinking is:
$$15 \cong 1 \mod 14 $$
$$15^7 \cong 1 \mod 14$$
$$13 \cong -1 \mod 14$$
$$13^2 \cong 1 \mod 14$$
$$13^5 \cong -1 \mod 14$$
$$[15^7 - 13^5(\mod 14)] = 1 (\mod 14) + 1 (\mod 14) = 2\mod 14$$
Is that right?
AI: Yes, that's correct! $$[15^7 - 13^5]\pmod{14} \;\;= \;\;2\pmod {14}$$ |
H: Integrating $\int_\gamma \frac{\cosh z}{2 \ln 2-z} dz$
I need to solve the following integral using Cauchy Integral Formula:
$$\int_\gamma \frac{\cosh z}{2 \ln 2-z} dz$$
with $\gamma$ defined as:
$|z|=1$
$|z|=2$
With $|z|=2$ I've solved already, as it is quite easy. All one needs to do is re-write the denominator as $z-2 \ln 2$ and since $2 \ln 2$ is inside $\gamma$, then one can apply easily the formula.
With 1 I'm not sure how to re-write the integral so I can use the Cauchy Integral formula (Can I even do it?).
AI: For part 1, we can use the fact that the integrand
$$\frac{\cosh z}{2\ln 2 - z}$$
is analytic in $\mathbb{C}\setminus \{2\ln 2\}$, in particular, in a neighbourhood of the closed unit disk, since $\cosh z$ is an entire function. Thus by the Cauchy integral theorem, the integral is $0$.
An alternative roundabout way is to consider the function
$$f(z) = \frac{z\cosh z}{2\ln 2 - z},$$
and note that the integral is
$$\int_{\lvert z\rvert = 1} \frac{f(z)}{z}\,dz = 2\pi i f(0).$$ |
H: If $k^2-1$ is divisible by $8$, how can we show that $k^4-1$ is divisible by $16$?
All is in the title:
If $k^2-1$ is divisible by $8$, how can we show that $k^4-1$ is divisible by $16$?
I can't conclude from the fact that $k^2 - 1$ is divisible by $8$, that then $k^4-1$ is divisible by $16$.
AI: Hint: $$k^4 - 1 = (k^2 - 1)(k^2 + 1) = (k^2 - 1)\Big((k^2 - 1) + 2\Big)$$
ADDED per comment: So yes, we have that if $(k^2 - 1)$ is divisible by $8$, then $$k^4 - 1 = (k^2 - 1)(k^2 + 1) = 8b(k^2 + 1)$$ for some integer $b$.
And now, if $k^2 - 1$ is divisible by 8, it is even, then so is $k^2 + 1$.
That is, $k^2 + 1 = (k^2 - 1) + 2 = 8b + 2$. So $$(k^2 - 1)(k^2 + 1) = 8b(8b + 2) = 16b (4b + 1)$$ |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.