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H: Finding number of roots of the polynomial $\,\,x^3-x\,$ in $\,\,\Bbb Z /6 \Bbb Z$
The number of roots of the polynomial $x^3-x$ in $\Bbb Z /6 \Bbb Z$ is : $1,2,3,6.$
I have to pick the correct choice.
Since $\,\,\Bbb Z /6 \Bbb Z=\{\overline 0,\overline 1,\overline 2,.....,\overline 5\}$ and we see $\overline 0,\overline 1$ satisfies $x^3-x=0$ and so the number of roots the polynomial $x^3-x$ in $\Bbb Z /6 \Bbb Z$ is $2$. Am I right?
Thanks in advance for your time.
AI: $$\overline{2}^3 = \overline{8} = \overline{2}$$$$\overline{3}^3 = \overline{27} = \overline{3}$$$$\overline{4}^3 = \overline{64} = \overline{4}$$$$\overline{5}^3 = \overline{125} = \overline{5}$$
So the correct answer is 6. |
H: What if $(a) \land (b) \iff (c)$ and $(c)$ is false?
My question is about the logical inverse of this Lemma:
Lemma: Let $g$ be a real analytic function. Then we have the equivalence $((a)∧(b))⇔(c)$, where the statements $(a),(b)$ and $(c)$ are given by:
(a) $g$ has infinitely many real zeros and the set of those zeros is unbounded in both directions.
(b) $g$ assumes arbitrarily large and arbitrarily small values, i.e., for all $K>0$, there are $s_1,s_2$ with $g(s_1)<-K$ and $g(s_2)>K$,
(c) The fiber $g^{-1}(w)$ is infinite for all $w\in \Bbb R$.
In particular, what if for some $w \in \Bbb R$, the fiber $g^{-1}(w)$ is finite?
AI: The lemma establishes that $(c)$ and the conjunction of $(a)$ and $(b)$ are logically equivalent.
In particular, if $(c)$ is false, so must $(a) \land (b)$ be. Concretely, this means $(a)$ or $(b)$ must be false.
That is, we infer that at least one of the following must be the case:
(The negation of $(a)$, that is:) The set of zeroes of $g$ is bounded in at least one direction;
(The negation of $(b)$, that is:) $g$ is either bounded above or bounded below.
I suggest you think for a moment on why these are indeed the negations of $(a)$ and $(b)$, respectively. |
H: Calculating a Fraction's Reciprocal
Is there any way or equation that allows me to calculate the reciprocal of any fraction?
I mean if i have 5/6 and i need it's reciprocal by using a formula or an equation to calculate it.
Is there or not?
thanks all
AI: Flip it, ${}{}{}{}{}{}{}$ (provided the numerator is not $0$.).
The reciprocal of $\;\dfrac xy,\;\; x, y \neq 0\;\;$ is given by $\;\dfrac 1{\left(\frac xy\right)} = \dfrac yx$.
So for your fraction, the reciprocal of $\dfrac 56$ is given by $ \dfrac 65$. |
H: Taylor's theorem for vector valued functions
I'm reading about linear and nonlinear programming and on one page I have the following statment (I have highlighted the areas where I have problems and drawn questions for them in the bottom of it):
Proposition $5$. Let $f\in C^2$. Then $f$ is convex over a convex set $\Omega$ containing an interior point if and only if the Hessian matrix $\textbf{F}$ of $f$ is positive semidefinite throughout $\Omega$.
Proof. By Taylor's theorem we have
$$\color{#48BE6B}{\boxed{\,\,\,\displaystyle\color{black}{f(\mathbf{y})=f(\mathbf{x})\color{#ED1C24}{\boxed{\color{black}{=}}}\boldsymbol\nabla f(\mathbf{x})(\mathbf{y}-\mathbf{x})+\frac12(\mathbf{y}-\mathbf{x})^T\mathbf{F}\color{#FF7F27}{\boxed{\color{black}{(\mathbf{x}+\alpha(\mathbf{y}-\mathbf{x}))}}}(\mathbf{y}-\mathbf{x})}\,\,\,}}\tag{12}$$
for some $\alpha, 0\leqslant\alpha\leqslant1$. Clearly, if the Hessian is everywhere positive semidefinite, we have
$$f(\mathbf{y})\geqslant f(\mathbf{x})+\boldsymbol{\nabla}f(\mathbf{x})(\mathbf{y}-\mathbf{x}).\tag{13}$$
which in view of Proposition $4$ implies $f$ is convex.
Now suppose the Hessian is not positive semidefinite at some point $\mathbf{x}\in\Omega$. By continuity of the Hessian it can be assumed, without loss of generality, that $\mathbf{x}$ is an interior point of $\Omega$. There is a $\mathbf{y}\in\Omega$ such that $(\mathbf{y}-\mathbf{x})^T\mathbf{F}(\mathbf{x})(\mathbf{y}-\mathbf{x})<0$. Again by the continuity of the Hessian, $\mathbf{y}$ may be selected so that for all $\alpha, \,0\leqslant\alpha\leqslant1$.
$$(\mathbf{y}-\mathbf{x})^T\mathbf{F}(\mathbf{x}+\alpha(\mathbf{y}-\mathbf{x}))(\mathbf{y}-\mathbf{x})<0.$$
This in view of $\text{(12)}$ implies that $\text{(13)}$ does not hold; which in view of Proposition $4$ implies that $f$ is not convex. $\blacksquare$
$\begin{align}\color{#ED1C24}{\blacksquare}\,\,\,& \text{Is this a mistake? Should there be a + instead of =?}\\
\color{#FF7F27}{\blacksquare}\,\,\,& \text{I just don't understand this... Why is the alpha there?}\\
\color{#48BE6B}{\blacksquare}\,\,\,& \text{In general what is the Taylor theorem for vector valued}\\
& \text{functions? The formula in Wikipedia looks different than}\\ &\text{the formula here.}\\
\end{align}$
reference on wikipedia:
http://en.wikipedia.org/wiki/Taylor%27s_theorem#Taylor.27s_theorem_for_multivariate_functions
Any help much appreciated `=)
AI: Yes there should be a plus.
You see this Taylor expansion uses "=" which means exactly the same. We call the orange part Lagrangian form of the remainder, which also can be found on the same page of wikipedia.
The general Taylor expansion is exactly what wiki writes. And the theorem in this book, the author takes the first order approximation, which is the simplest case of Taylor expansion. You can also expand the function to higher order according to the extend how precise is the approximation. |
H: Matrix powers and recurrence relations
The nth Fibonacci number can be found by raising the matrix $\begin{pmatrix}1 & 1 \\ 1 & 0 \end{pmatrix}$ to the nth power. Are there other recurrence formulas that can be solved like this? This yields faster algorithms for computing them.
AI: Yes - all constant coefficient homogeneous linear recurrence relations can be solved this way:
Let $a_n=\alpha_{1}a_{n-1}+...+\alpha_{r}a_{n-r}$ be a recurrence relation with initial conditions - given $a_0,...,a_{r-1}$. Then, observing that
$$\begin{pmatrix}a_{n}\\a_{n-1}\\a_{n-2}\\\vdots\\a_{n-r+1}\end{pmatrix}=\begin{pmatrix}
\alpha_1&\alpha_2&\alpha_3&... &\alpha_{r-1}&\alpha_r\\
1&0&0&...&0&0\\
0&1&0&...&0&0\\
\vdots& &\ddots& &&\vdots\\
0&0&0&...&1&0
\end{pmatrix}
\begin{pmatrix}a_{n-1}\\a_{n-2}\\a_{n-3}\\\vdots\\a_{n-r}\end{pmatrix}$$
Denote that matrix by $A$. Then we see that
$$\begin{pmatrix}a_{n}\\a_{n-1}\\a_{n-2}\\\vdots\\a_{n-r+1}\end{pmatrix}=A^{n-r}
\begin{pmatrix}a_{r-1}\\a_{r-2}\\a_{r-3}\\\vdots\\a_{0}\end{pmatrix}$$
So you need to raise $A$ to the $n$th power to find $a_n$. |
H: Solution verification: $\lim\limits_{x\rightarrow \infty} \frac{\sqrt{x}+x^2}{2x-x^2} = -1$
I am trying to find the following limit
$$\lim_{x\rightarrow \infty} \frac{\sqrt{x}+x^2}{2x-x^2}$$
and I did the following steps:
\begin{align}
\require{cancel}
&\lim_{x\rightarrow \infty} \frac{\sqrt{x}+x^2}{2x-x^2} \\
&\lim_{x\rightarrow \infty} \frac{x^2\left(\frac{\sqrt{x}}{x^2}+1\right)}{x^2\left(\frac{2x}{x^2}-1\right)} \\
& \lim_{x\rightarrow \infty} \frac{\cancel{x^2}\left(\frac{\sqrt{x}}{x^2}+1\right)}{\cancel{x^2}\left(\frac{2\cancel{x}}{\cancel{x^2}}-1\right)}\\
& \lim_{x\rightarrow \infty} \frac{\left(\frac{\sqrt{x}}{x^2}+1\right)}{\left(\frac{2}{x}-1\right)} \\
\end{align}
Now here, the top portion goes to $0$ because the there is a larger power of $x$ in the denominator leaving only a $+1$ on top. On the bottom, the same thing happens, $\frac{2}{x}$ goes to $0$ and we left with $-1$ in the denominator. Therefore
$$\lim_{x\rightarrow \infty} \frac{\sqrt{x}+x^2}{2x-x^2} = -1$$
Is my solution correct and did I take the right steps with the correct logic?
Thanks!
AI: Excellent work. You've made a correct and tight case for the limit being $-1$.
Just replace
Now here, the top portion goes to $0$ because the there is a larger power of $x$ in the denominator leaving only $(+1)$ on top.
With
Now here, the top portion goes to $1$ because in the first term of the numerator there is a larger power of $x$ in the denominator leaving only the term of $(+)1$ on top. On the bottom...
You could also argue that in the numerator, $\dfrac {\sqrt x}{x^2} = \dfrac 1{x^{3/2}} \to 0$ as $x \to \infty$. |
H: Solve the following Differential Equation $x \ln x\ \mathrm{d}x+(y-\ln x\ \mathrm{d}y)=0$
I want to solve the following equation:
$$x \ln x\ \mathrm{d}x+(y-\ln x\ \mathrm{d}y)=0$$
How can i solve it? Thanks.
AI: It's strange that there is no $dx$, I am assuming $dy$ just means $y' = dy/dx$ really.
So we must solve
$$
\begin{split}
0 &= y' x \ln x + y - y' \ln x = y + y' (x-1) \ln x \\
y &= y'(1-x) \ln x \\
\frac{y'}{y} &= \frac{1}{(1-x) \ln x} \\
\int \frac{dy}{y} &= \int \frac{dx}{(1-x) \ln x}
\end{split}
$$
and you can finish the arithmetic. |
H: Proper method for solving quadratic equations with exponents
$(\sqrt {x^2-5x+6}+\sqrt{x^2-5x+4})^{x/2}$ + $(\sqrt {x^2-5x+6}-\sqrt{x^2-5x+4})^{x/2}$ = $2^{(x+4)/4}$
I have found out, by trial and error method, that $x=0$ and $x=4$ satisfy this equation. But is there a proper way to solve this equation and get the solutions?
AI: Let's start by putting $$y_1=\left(\sqrt {x^2-5x+6}+\sqrt{x^2-5x+4}\right)^{x/4}$$ and $$y_2=\left(\sqrt {x^2-5x+6}-\sqrt{x^2-5x+4}\right)^{x/4}.$$ Now, observe that $$\begin{align}2 &= (x^2-5x+6)-(x^2-5x+4)\\ &= \left(\sqrt {x^2-5x+6}+\sqrt{x^2-5x+4}\right)\left(\sqrt {x^2-5x+6}-\sqrt{x^2-5x+4}\right),\end{align}$$ and so $$2^{(x+4)/4}=2\cdot 2^{x/4}=2y_1y_2.$$ Thus, the given equation becomes equivalent to solving $$y_1^2+y_2^2=2y_1y_2,$$ or $$(y_1-y_2)^2=0.$$ Hence, we need only solve the equation $$y_1=y_2,$$ so since both $y_1,y_2>0$ (why?), then we face the equivalent task of solving $$y_1^2=y_2^2.$$ As you've seen, $x=0$ readily does the trick, but if $x\ne 0,$ then the only way that we can have $a^x=b^x$ for $a,b>0$ is if $a=b.$ Hence, it remains only to solve $$\sqrt {x^2-5x+6}+\sqrt{x^2-5x+4}=\sqrt {x^2-5x+6}-\sqrt{x^2-5x+4},$$ which I leave to you. |
H: Prove that the centralizer subgroup is normal in the normalizer subgroup
To my dear friends with gratitude.
I want to get help proving centralizer of a nonempty subset of a group is a normal subgroup in the normalizer of that set in the mentioned group.symbolically:
$C_G (S)\trianglelefteq N_G (S)$
AI: Let $z \in C_G(S)$ and let $n \in N_G(S)$. We want to show that $nzn^{-1} \in C_G(S)$.
If $s \in S$, then $(nzn^{-1})s(nzn^{-1})^{-1}=nz(n^{-1}sn)z^{-1}n^{-1}$.
Now $n^{-1}sn \in S$ and so $nz(n^{-1}sn)z^{-1}n^{-1}=n(n^{-1}sn)n^{-1}=s$. |
H: Sequence of the form $pn^2+qn+r$
A sequence 192, 360, 576 is formed by multiplying the corresponding two different arithmetic progression. What is the eighth term of the sequence?
Solution says that answer will be of the form $pn^2+qn+r$. Why is this, and how can I find it?
AI: Let's let the first arithmetic progression be given by $x_n=a+nc$ and the second by $y_n=b+nd,$ so that your sequence is given by $$z_n=x_ny_n=bdn^2+(ad+bc)n+ab.$$ Put $r=ab,$ $q=ad+bc,$ and $p=bd$ for simplicity (since we aren't actually interested in the original sequences, anyway). In particular, we are given the following system: $$\begin{cases}192=r & \\360=p+q+r & \\576=4p+2q+r\end{cases}$$
Solving the above system for $p,q,$ and $r,$ what can we then do? |
H: Limit of functions
I need to give a counterexample to the following statement:
If $ \lim_{x \to 0} \left( \frac {f(x)}{g(x)} \right) = 1 $, then $ \lim_{x \to 0} \left( f(x) - g(x) \right) = 0 $.
The problem is I think this statement is correct for any functions because:
$$ \lim \left( \frac {f(x)}{g(x)} \right) = \dfrac {\lim \left( f(x) \right)}{\lim \left( g(x) \right)}, $$ so if it equals $1$< then both functions have the same limit so, if we subtract them, the limit is $0$.
I tried to find a counterexample for hours! And found nothing. Thanks for the help.
AI: Converting njguliyev's comment to an answer, consider: $$ \begin {eqnarray*} f(x) &=& \dfrac {x+1}{x^2}, \\ g(x) &=& \dfrac {1}{x^2}. \end {eqnarray*} $$ |
H: Why does this function not have any extrema?
Why does the following function not have any extrema: $z=(xy-1)^2 +x^2$? I calculated that $\dfrac{dz}{dx}=2((xy-1)y+x)$ and $\dfrac{dz}{dy}=2(xy-1)x$ which are both zero when $x$ and $y$ are zero. What's my mistake?
AI: It is possible that the function has a saddle point there. Check by using the second partial derivative test. |
H: Haar measure of $SO(3)$ obtained from $SU(2)$
I am reading 'Analysis on Lie groups, an introduction' by Faraud and don't understand the following statement
… the image by the map Ad of the Haar measure $\mu$ of $SU(2)$ is equal to the Haar measure $\nu$ of $SO(3)$.
We know that $Ad:$ $SU(2)\rightarrow SO(3)$ is a covering of order 2. I also know the Haar measure on $SU(2)$, but cannot see why the above statement holds.
I appreciate any hint, comment or answer.
AI: Trying to clarify my brief comment: on a compact (Hausdorff) topological group, up to scalar multiples there is a unique left-and-right translation-invariant (regular, Borel) measure. Thus, any translation-invariant measure on $SO(3)$ is necessarily a multiple of the Haar measure on $SO(3)$. Ok, given a set $E\subset SO(3)$, let $E'$ be its inverse image in $SU(2)$, and say that the measure of $E$ in $SO(3)$ is the measure of $E'$ in $SU(2)$. This measure is invariant: given $k\in SO(3)$, choose any $k'$ in $SU(2)$ mapping down to $k$. The inverse image of $k\cdot E$ is $k'\cdot E'$, which has the same measure as $E'$. And the total mass is $1$ if the measure on $SU(2)$ was normalized in this fashion. |
H: Find probability that only one event will occur
I have a problem with such simple task:
Probabilities of two independent events $A_1$ and $A_2$ are equal repectively $p_1$ and $p_2$. Find the probability that only one of the events will occur.
The answer given in the book is $P=p_1+p_2-2 \cdot p_1 \cdot p_2$
However I get the different answer. The result from above is got when such way of solving is applied:
$$P((A_1 \cap A_2') \cup (A_1' \cap A_2))=P(A_1 \cap A_2') + P(A_1' \cap A_2)$$
and so on. I don't udnerstand why there is no substraction of common part of events, as there is not mentioned that those events are mutually exclusive. Shouldn't the first step be:
$$P((A_1 \cap A_2') \cup (A_1' \cap A_2))=P(A_1 \cap A_2') + P(A_1' \cap A_2)- P((A_1 \cap A_2') \cap (A_1' \cap A_2))$$
?
AI: The events $A_1\cap A_2'$ and $A_1'\cap A_2$ are mutually exclusive:
$$(A_1\cap A_2')\cap(A_1'\cap A_2)=(A_1\cap A_1')\cap(A_2\cap A_2')=\varnothing\cap\varnothing=\varnothing\;$$
You can certainly subtract $\Bbb P\big((A_1\cap A_2')\cap(A_1'\cap A_2)\big)$, but that’s $\Bbb P(\varnothing)=0$, so there’s no need to do so.
Since the events are independent, you then have $$\Bbb P(A_1\cap A_2')=p_1(1-p_2)$$ and $$\Bbb P(A_1'\cap A_2)=(1-p_1)p_2\;,$$ from which the result is a matter of a little algebra. |
H: What exactly is a linear space?
I often stumble upon books using the term "linear space" (outside of Linear Algebra) and I have never been totally comfortable with this. Perhaps I am over complicating this, but my intuition says that a "linear space" is one which has some sort of basis, say $\{a, b\}$, and all the other elements can be thought of as linear combinations of the elements in the basis like $j=\alpha a + \beta b$ (for some $\alpha, \beta$). This would further exclude the possibility of some sort of non-linear combinations like $j=\alpha a + \gamma a^2 + \beta b$ from our linear space.
Is my intuition on the right track? Should I even bother trying to define "linear space" to myself? Is "linear space" synonymous with "vector space"?
AI: Formally, you need an underlying field $F$ under your linear space. The space (let's say $L$) is then whatever set of whatever object, such that you have two maps: Sum as a map from $L\times L\mapsto L$ and multiplication as a mup from $L\times F\mapsto L$. They have to satisfy some axioms of course (sum is commutative, both are associative, the distibutive law is valid).
So for example take the set of real functions $\mathbb{R}\to\{0,\dots,6\}$, and define $(a+b)(x)=a(x)+b(x)\mod 7$ and $(ta)(x)=ta(x) \mod 7$. Since $\mathbb{Z}_7$ is a field, this is a linear space, by definition. No intuition about lines and bases works here.
The elements of your $L$ can be potatoes, as long as you're able to find a field and two operations that satisfy the definition.
However, in general if the literature says that something "is a linear space", they in general mean that it's a set closed under addition and multiplication by numbers from some field, where the defition of the operations as well as the underlying field are usually clear from the context. |
H: Weierstrass... thing
There is in my maths text-book this property/theorem given under the name of Weierstrass property/theorem:
Let $ (a_n) $ be a sequence of real numbers.
a)If $ (a_n) $ is monotonically increasing and has an upper bound, then $ (a_n) $ is convergent.
b)If $ (a_n) $ is monotonically decreasing and has a lower bound, then $ (a_n) $ is convergent.
When searching on Google, I can't find anything related to this, so my first question is:
What is the name of this thing?
What I want to do is to attempt to prove this property/theorem, I would like some hints on what to start with and what should I end with.
Thanks!
AI: HINT: Use the fact that $\Bbb R$ has the least upper bound property: if $\varnothing\ne A\subseteq\Bbb R$, and $A$ is bounded above, then $A$ has a least upper bound. If your increasing sequence is $\langle x_n:n\in\Bbb N\rangle$, let $x$ be the least upper bound of $\{x_n:n\in\Bbb N\}$, and show using the definitions of least upper bound and limit of a sequence that $x=\lim\limits_{n\to\infty}x_n$.
The proof for the other result is similar. Alternatively, if $\langle x_n:n\in\Bbb N\rangle$ is decreasing and bounded below, then $\langle-x_n:n\in\Bbb N\rangle$ is increasing and bounded above, so by the first result it converges to some $x$. Now show that $\langle x_n:n\in\Bbb N\rangle$ converges to $-x$.
Forgot to say: in my experience it’s usually called the monotone convergence theorem. |
H: Orthogonal Matrix statements/proofs
I am currently learning about Orthogonal matrices and have three statements that I can not figure out.
Here are three Proofs that I have written down but can not figure how to prove them:
If $Q$ is an orthogonal matrix, then $Q^{-1}$ is orthogonal.
If $Q$ is an orthogonal matrix, then $Q^T$ is orthogonal.
If $Q_1$ and $Q_2$ are Orthogonal matrices, then $Q_1Q_2$ is orthogonal.
I am having trouble understanding these. Can anyone clear these up?
Thanks!
For the first statement, I have:
$Q$ = $Q^T$
($Q$ - 1) $Q$ = $Q^T$ ($Q$ - 1)
($Q$ - 1)$Q$ = I
AI: Suppose that $Q$ is orthogonal so that the rows and columns of the matrix are orthonormal. Then the $i^{th}$ row of $Q$ is the $i^{th}$ column of $Q^T$ so that $Q^T$ has orthonormal rows and columns as well. Hence it's orthogonal. The $ij^{th}$ entry of $QQ^T$ is the dot product of the $i^{th}$row of $Q$ with the $j^{th}$ column of $Q^T$. If $i=j$ this dot product is 1 by orthonormal if $i \neq j$ this dot product is 0 by orthonormal. Thus $QQ^T=I$. Since inverses are unique $Q^T=Q^{-1}$. Last $(Q_1Q_2)(Q_1Q_2)^T = Q_1Q_2Q_2^TQ_1^T =Q_1IQ_1^T=I$ showing that the product is and orthogonal matrix. |
H: Proving that $\Bbb{R}_\ell$ is finer than $\Bbb{R}$.
Let us take the two topologies $\Bbb{R}_\ell$ and $\Bbb{R}$. The book "General Topology" by Munkres says that $\Bbb{R}_\ell$ is finer than $\Bbb{R}$. This article says that every open set of $\Bbb{R}$ has to be an open set in $\Bbb{R}_\ell$ for the latter to be finer than the former.
The basis of $\Bbb{R}_\ell$ consists of sets of the form $[a,b)$, where $a,b\in \Bbb{R}$. Similarly, the basis of $\Bbb{R}$ consists of sets of the form $(a,b)$.
How can any open set in $\Bbb{R}_\ell$ ever be of the form $(a,b)$? I have tried both all possible unions of the basis sets, and they all seem to be of the form $[a,b)$.
Thanks in advance!
AI: The basic open sets aren’t the only open sets: every union of basic open sets is open as well, and
$$(a,b)=\bigcup_{a<x<b}[x,b)$$
is a union of basic open sets in $\Bbb R_\ell$. |
H: If $f^n$ is mixing then $f$ is mixing?
Let $(X,\mathcal{A},\mu)$ be a probability space and $f:X\to X$ be a measurable map that preserves $\mu$. Fix $n\in \mathbb{Z}^+$.
It's not hard to see that $f$ ergodic does not necessarily imply $f^n$ ergodic. For example, take $X=\mathbb{Z}/4\mathbb{Z}$ with the uniform probability measure, $f:x\mapsto x+1$ and $n=2$.
Is it true that if $f$ is (strong) mixing then $f^n$ is mixing?
EDIT: Yes, it is, since a subsequence of a real convergent sequence is convergent and has the same limit. What I meant to ask is:
Is it true that if $f^n$ is (strong) mixing then $f$ is mixing?
The result is easily seen to be true if we replace "mixing" by "ergodic".
AI: "The map $f$ is strong mixing" means that for each $A,B\in\mathcal A$, we have
$$\lim_{N\to\infty}\mu(A\cap T^{-N}(B))=\mu(A)\mu(B).$$
Since the sequence $(\mu(A\cap (T^{n})^{-k}(B))-\mu(A)\mu(B),k\geqslant 1)$ is a subsequence of a sequence which converges to $0$, it converges itself to $0$.
For the converse, fix $A,B\in\mathcal A$ and define
$$c_k:=\mu(A\cap T^{-k}(B))-\mu(A)\mu(B).$$
We have that $c_{nk}\to 0$ by assumption. Using $T^{-i}B$ instead of $B$ in the , we get
$c_{nk+i}\to 0$ as $k\to\infty$, for each $0\leqslant i\leqslant n-1$. |
H: Does this follow from the definition of the LambertW function?
The LambertW function $W(s)$ also called ProductLog seems to satisfy this relation:
$$-W(s) = \underbrace{-s e^{-s e^{\cdot^{\cdot^{-s e^{-W(s)}}}}}}_n$$
Or truncated:
$$-W(s) = -se^{-s e^{-s e^{-s e^{-s e^{-s e^{-s e^{-s e^{-s e^{-s e^{-s e^{-s e^{-s e^{-W(s)}}}}}}}}}}}}}$$
for some complex number $s$.
Does this follow from the definition of the LambertW function?
Mathematica:
s = ZetaZero[1]
N[-s*Exp[-s*
Exp[-s*Exp[-s*
Exp[-s*Exp[-s*
Exp[-s*Exp[-s*
Exp[-s*Exp[-s*
Exp[-s*
Exp[-s*Exp[-s*Exp[-ProductLog[s]]]]]]]]]]]]]], 30]
N[-ProductLog[s], 30]
which outputs:
-1.88420341024267232833723733662 - 1.03367078404796759147577622553 I
and
-1.88420341024267232833723733662 - 1.03367078404796759147577622553 I
Excel spreadsheet formula (European dot-comma) apparently giving $\frac{W(n)}{n}$:
=IF(OR(ROW()=1; COLUMN()=1);0; IF(ROW()>=COLUMN();EXP(-SUM(INDIRECT(ADDRESS(ROW()-COLUMN()+1; COLUMN(); 4)&":"&ADDRESS(ROW()-1; COLUMN(); 4); 4)));0))
by using the relation above.
Excel output:
0 | 0,56714329 | 0,426302751 | 0,349969501 | 0,300547296 | 0,26530177
which is the same as the Mathematica command:
N[Table[LambertW[n]/n, {n, 1, 12}]]
AI: Yup, it follows from $W(s)e^{W(s)} = s$. Rearranging yields
$$
W(s) = se^{-W(s)} = se^{-se^{-W(s)}} = \cdots
$$ |
H: Proof about boundedness of $\rm Si$
$\def\Si{{\rm Si}}$
I want to prove the boundedness of
$$\Si(x) := \int_0^x \frac {\sin \xi} \xi d\xi$$
as part of a homework (about the non-surjectivity of $\mathcal F : L^1(\mathbb R) \to C_0^0(\mathbb R)$). As a first step, I found (by means of plotting)
$$\frac{\sin x}{x} \leq \cos \frac x 2 \qquad \forall\ x \in (0, \pi)$$
but couldn't easily prove that. If that is proved, we had, including some other arguments
$$\Si(x) \leq \int_0^\pi \frac {\sin \xi} \xi d\xi \leq \int_0^\pi \cos\frac\xi 2 d\xi = 2$$
And I'd be done.
How can I prove
$$\frac{\sin x}{x} \leq \cos \frac x 2 \qquad \forall\ x \in (0, \pi)$$
in a simple way, or do you have any alternative proof for my objective?
Thanks in advance
Okay with the hints provided by @HaraldHanche-Olsen in his answer, it became very clear:
$$\frac{\sin x} x \stackrel{\text{addition thm.}}{=} \frac{2 \sin\frac x 2 \cos\frac x 2}{x} = \underbrace{\frac{\sin\frac x 2}{\frac x 2}}_{\leq 1} \cos \frac x 2 \leq \cos\frac x 2$$
AI: Here are two hints for the price of one: $$\frac{\sin x}{x}\le 1$$ (I am sure you know that one) and $$\sin x=2\sin\frac x2\cdot\cos\frac x2.$$ |
H: How find the limit $I=\lim_{m\to 0,n\to 0}(m^2-2n)^n$
find the limit
$$I=\lim_{n\to 0}\lim_{m\to 0}(m^2-2n)^n$$
my try:
$$I=\lim_{n\to 0}(0^2-2n)^n=\lim_{n\to 0}(-2n)^n$$ is not exsit,
my try is true? Thank you
AI: The limit is not well-defined, i.e. the value depends on the choice of a sequence $a_j = (m_j, n_j)$ with $\lim_{j\to\infty} a_j = (0,0)$.
For example, $a_j = (\frac 1 j, 0)$ yields $\lim_{j\to\infty} F(a_j) = 1$ whereas $a_j = (0, \frac 1 j)$ produces undefined values and thus an undefined limit. |
H: Suppose that $0 \le f(n) \le 1$, why $\lim_{n \to \infty} (1 - f(n))^n = 0 \iff \lim_{n \to \infty} f(n)n = \infty$?
Suppose that $0 \le f(n) \le 1$ and consider two eqauation:
$$ \lim_{n \to \infty} (1 - f(n))^n = 0 \tag{A} $$
$$ \lim_{n \to \infty} f(n)n = \infty \tag{B} $$
It seems that A and B are equivalent. But how we can show it formally?
Maybe you know a reference for this result?
Thanks in advance
AI: Note that $1-x\leqslant\mathrm e^{-x}$ for every $x$, hence $(1-x)^n\leqslant\mathrm e^{-nx}$ for every $x\leqslant1$, in particular $(1-f(n))^n\leqslant\mathrm e^{-nf(n)}$. Thus, $nf(n)\to\infty$ implies $(1-f(n))^n\to0$.
In the other direction, if $nf(n)$ does not converge to infinity, there exists infinitely many $n_k$ and some finite $C$ such that $n_kf(n_k)\leqslant C$, hence $(1-f(n_k))^{n_k}\geqslant(1-C/n_k)^{n_k}$ for every $k$ large enough. Since $\lim\limits_{x\to+\infty}(1-C/x)^x=\mathrm e^{-C}$ and $n_k\to\infty$, this yields $(1-f(n_k))^{n_k}\geqslant\mathrm e^{-C}/2\gt0$ for every $k$ large enough, in particular $(1-f(n))^n$ does not converge to $0$. |
H: Proof that $\bigcap_{n=1}^\infty J^n=0$ in commutative noetherian ring
If we let $R$ be a commutative noetherian ring. Then $\bigcap_{n=1}^\infty J^n=0$ where $J$ is the jacobson radical of $R$
Proof.
Denote $X=\bigcap_{n=1}^\infty J^n=0$. Then let $XJ=Q_1\cap \ldots \cap Q_n$ be a primary decomposition of $X$. We then fix $i$.
Then if $ X \not\subset \sqrt{Q_i}$ then $J\subset \sqrt{Q_i}$ and so $J^k\subset Q_i$ for some $k$.
This then gives that either $X\subset Q_i$ or $J^k\subset Q_i$, this gives $X\subset XJ$ and applying Nakayamas Lemma gives $X=0$
My question is why is this part true? :
Then if $X\not\subset \sqrt{Q_i}$ then $J\subset \sqrt{Q_i}$
I'm sure that I am missing something simple.
Thanks for any help.
AI: I think it's supposed to say "If $X\not\subseteq\sqrt{Q_i}$."
Since $Q_i$ is primary, $\sqrt{Q_i}$ is prime. By definition of a prime ideal, $XJ\subseteq P$ and $X\nsubseteq P$ implies $J\subseteq P$. |
H: Compute homotopy classes of maps $[T^{2},T^{2}]$
How to compute $[T^{2},T^{2}]$ the set of homotopy class of continuous maps $f:T^{2}\longrightarrow T^{2}$?
Thanks.
AI: Since $T^2=S^1\times S^1$, you get
$$
[T^2,T^2]=[T^2,S^1]\times [T^2,S^1].
$$
So, the problem reduces to finding $[T^2,S^1]$. Since $S^1$ is the first Eilenberg-Maclane space, we have that
$$
[T^2,S^1]\cong H^1(T^2;\mathbb{Z})\cong \mathbb{Z}\oplus\mathbb{Z}.
$$ |
H: number of automorphisms for group in order 169
Let $G$ be a group with order 169. Prove number of automorphisms is at least 143.
I thought that 169 is 13 squared so maybe G isomorphic to $ Z_{169} $ but I dont have any idea. How can I solve it?
AI: Let $p$ be prime and $G$ a group of order $p^2$.
If $G$ is cyclic, there are $\phi(p^2)=p(p-1)$ automorphisms.
So assume $G$ is not cyclic, hence each element $\ne 1$ has order $p$.
$G$ acts on itself by conjugation. The length of the orbit of an element $g\in G$ is either $1$ or $p$ or $p^2$. In fact $p^2$ is excluded beacuse $1$ is not conjugate to any other element.
As the $p^2-1$ nontrivial alements of$G$ cannot be partitioned into orbits of size $p$, there must exist some $g\in G\setminus\{1\}$ with orbit length $1$, i.e. $N:= \langle g\rangle $ is a nontrivial normal subgroup of $G$.
Then $N\cong G/N\cong Z_{p}$ and $$1\to N\to G\to G/N\to 1$$
splits (just pick a preimage of a generator of $G/N$) thus making $G$ a semidirect product of $Z_{p}$ with itself. As $Z_{p}$ has only $p-1$ automorphisms, it has no automorphism of order $p$ and we conclude that the product is in fact direct. But $G\cong Z_{p}\times Z_{p}$ has at least $(p-1)^2$ automorphisms (from picking one per factor; we could easily find more by switching the factors). |
H: Fields of positive characteristic
Let $\mathbb F$ be an infinite field of characteristic $p>0$. It is true that every element of $\mathbb F$ is algebaric over the prime subfield $\mathbb F_p$ of $p$ elements?
AI: Consider ${\mathbb F}_p (x)$, the field of rational functions over ${\mathbb F}_p$. Is $x$ algebraic over ${\mathbb F}_p$ ? |
H: $\epsilon$-$\delta$ Verification of a Lipschitz Function
A function $f:D\rightarrow \mathbb{R}$ is said to be a Lipschitz function provided that there is a nonnegative number $C$ such that
$|f(u)-f(v)|\le C|u-v|$ for all $u,v\in D$.
Show that a Lipschitz function satisfies the $\epsilon$-$\delta$ criterion on $D$.
In order for this to be true, for $\epsilon>0$, we must find a $\delta>0$ such that
$|f(x)-f(x_0)|<\epsilon$ $\hspace{10pt}$ if $\hspace{10pt}$ $|x-x_0|<\delta$.
If we let $u=x$ and $v=x_0$, then we have
$|f(x)-f(x_0)|\le C|x-x_0|$ for all $x\in D$, but I'm not sure where to go after this. Any suggestions?
AI: Let $\delta={\varepsilon \over C}$.
Then $|x - x_0| < {\varepsilon \over C}$, which means $C|x - x_0| < {\varepsilon}$.
And since $|f(x)-f(x_0)|\le C|x-x_0|$, $|f(x)-f(x_0)|< {\varepsilon}$. |
H: Consider the equation $\,\,x^{2007}-1+x^{-2007}=0.\,$
I am stuck with the following problem:
Consider the equation $\,\,x^{2007}-1+x^{-2007}=0.\,$Let $\,m$ be the number of distinct complex non-real roots and $\,n$ be the number of distinct real roots of the above equation. Then $\,m-n\,$ is
1.$\,0$
2.$\,2006$
3.$\,2007$
4.$\,4014$
Can someone explain? Thanks in advance for your time.
My Attempt: $y=x^{2007}$ gives $y^2-y+1=0 \implies y=\frac12 \pm \frac{\sqrt 3i}{2}.\,\,$Now, I am not sure how to progress.
AI: HINT:
Putting $x^{2007}=a$ we get $a-1+\frac1a=0\iff a^2-a+1=0$
Observe that the discriminant of the last quadratic equation $<0$
Hence the values of $a$ are complex
As $x^{2007}(=a)$ is complex, so will be all the values of $x$
for if $x$ is real, so will be $x^n$ for any integer $n$
Do you know the number of roots of $$x^{2007}-1+x^{-2007}=0$$ |
H: Last removed number in Sieve of Eratosthenes
I want to find the last deleted number in Sieve of Eratosthenes when applied on numbers below 1000. How can I find it?
AI: HINT: If $n<1000$ is composite, it will be removed when its smallest prime factor is processed. $31<\sqrt{1000}<32$, so the smallest prime factor or $n$ can be at most $31$. |
H: What's wrong with that proof?
What wrong with this proof?
$(-1)=(-1)^{\frac{2}{2}}=(-1)^{2\times \frac{1}{2}}=\sqrt{1}=1$ then $1=-1$
AI: $x^{\frac{1}{2}}$ is a multiple-valued "function", since in general $x$ has two square roots. One could also write:
$$\sqrt1=-1$$ |
H: Proving something with Wilson theorem
I need to prove that $x^2\equiv -1\pmod p$ if $p=4n+1$.
($p$ is prime of course...)
I need to use Wilson theorem.
AI: We have, $p-r\equiv-r\pmod p$
Putting $r=1,2,\cdots,\frac{p-1}2$ and multiplying we get $$\prod_{1\le r\le \frac{p-1}2}(p-r)\equiv(-1)^{\frac{p-1}2}\prod_{1\le r\le \frac{p-1}2} r$$
Multiplying either sides by $\prod_{1\le r\le \frac{p-1}2} r$
$$(p-1)!\equiv(-1)^{\frac{p-1}2}\left(\prod_{1\le r\le \frac{p-1}2} r\right)^2\pmod p$$
Wilson theorem says: $(p-1)!\equiv-1\pmod p$
If $p=1+4n, \frac{p-1}2=2n$ which is even $\implies (-1)^{\frac{p-1}2}=\cdots$ |
H: What is the probability that the number $5$ comes up on exactly two of three loaded dice?
I roll three different loaded dice. For the first die, the probability of getting a $5$ is $0.7$, for the second die the probability of getting a $5$ is $0.48$, and for the third die the probability of getting a $5$ is $0.38$.
What is the probability that the number $5$ comes up on exactly two of the three dice?
AI: If $E_i$ is the event that dice $i$ ($i=1,2,3$) comes up a $5$, then the event of getting exactly two 5s is
$$
\big(E_1 \cap E_2 \cap E_3^c\big)\cup\big(E_1 \cap E_2^c \cap E_3\big)\cup(E_1^c \cap E_2 \cap E_3\big)
$$
Each of the pieces in parentheses is disjoint from the others (why!?) so using the properties of the probability $P$,
$$
P(\text{exactly two } 5s) = P\big\{\big(E_1 \cap E_2 \cap E_3^c\big)\cup\big(E_1 \cap E_2^c \cap E_3\big)\cup(E_1^c \cap E_2 \cap E_3\big)\big\} =
P\big(E_1 \cap E_2 \cap E_3^c\big) + P\big(E_1 \cap E_2^c \cap E_3\big)+P\big(E_1^c \cap E_2 \cap E_3\big).
$$
Now, assuming that the $E_i$ are independent,
$$
P\big(E_1 \cap E_2 \cap E_3^c\big) + P\big(E_1 \cap E_2^c \cap E_3\big)+P\big(E_1^c \cap E_2 \cap E_3\big) = P(E_1)P(E_2)P(E_3^c) + P(E_1)P(E_2^c)P(E_3) +P(E_1^c)P(E_2)P(E_3)
$$
and now you can figure out the right side of the last equality with the information you're given. |
H: How to construct a contractible space but not locally path connected?
I am looking for a space which is contractible and not locally path connected.
I know the cone $CX$ of every space $X$ is contractible. Besides, it seems that if $X$ is locally path connected, so is $CX$. Therefore, I need to find a space which is not locally path connected.
Therefore, I tried $\sin(1/x)$-space and considered its cone.
Since $X$ can be topologically embedded in $CX$, $\sin(1/x)$-space is a subspace of its cone. It seems that open subsets of a locally path connected space are again locally path connected. However, $X$ is not open in $CX$.
Therefore, I am not sure whether it is locally path connected or not.
Any help will be appreciated. (I prefer some simple examples of such spaces, since I am a beginner in algebraic topology.)
AI: If $X$ is not locally path-connected, then there is a point $x$ and a neighborhood $U$ of $x$, which does not contain a path-connected neighborhood. Now choose $0<\epsilon<1$. The subspace $U\times[0,ϵ)$ of $X\times I$ is also a subspace of the cone $C(X)$. If it had a path-connected neighborhood $V$ of $(x,0)$, then the projection of $V$ onto $X\times\{0\}$ would be a path-connected neighborhood of $x$ within $U$ (since the projection is continuous and open). That's why the cone isn't locally path-connected either.
On the other hand, if $X$ is locally (path-)connected, then so is $X\times I$ as $I$ is locally path-connected. This property then carries over to the cone $C(X)$ since quotient maps preserve local (path-)connectedness. |
H: conditional probability that randomly chosen
In a certain village sports club, 46 % of members play football, 36 % of members play cricket, and 17 % of members play both games. What is the probability (between 0 and 1) that a randomly chosen member does not play football given that he/she plays cricket?
Give your solution accurate to 4 decimal places.
AI: Let $F$ be the event that the person plays football and $C$ be the event they play cricket. Then, what you want is
$$
P(F^c | C) = \frac{P(F^c \cap C)}{P(C)}.
$$
Now, $P(F^c \cap C) = P(\text{the person plays cricket, but not football})$, which you can calculate, since you know that 17% play both cricket and football. |
H: Find the probability of two random real numbers $x$ and $y$ between $0$ and $2$, where $\min(x,y) < 2/3$
Here is a picture of what I did so far.
http://sdrv.ms/HhxIvu
I got a result of $\frac59$, because the total area is $4$, and I'm subtracting the square with side of $\frac43$.
Can anyone confirm that this is the correct answer?
Thank you.
AI: Yes, you are right.
$\left(\frac{4}{3}\right)^2 = \frac{16}{9} $ is the area of empty subsquare.
Probability of being in empty subsquare is $\frac{16/9}{4} = \frac{4}{9}$.
The probabiliy of being in filled area is equal to $1-\frac{4}{9}$.
So, you are right. |
H: Why are singleton sets connected?
Why are singleton sets connected? I know that a subset $S$ of a metric space $X$ is connected if and only if given subsets $U$ and $V$ of $X$:
i. $U$ and $V$ are open
ii. The intersection of $U$, $V$, and $S$ equals the empty set
iii. $S$ is a subset of the union of $U$ and $V$.
AI: Because there exist no such sets U and V such that both $U\cap S\neq\emptyset$ and $V\cap S\neq\emptyset$.
A set $S$ is connected if there exist no open sets $U$ and $V$ such that $U\cap V\cap S=\emptyset$, $U\cap S\neq\emptyset$, $V\cap S\neq\emptyset$, and $S\subseteq U\cup V$. |
H: if neither f nor g is differentiable at x=a. is $f+g$ differentiable at $x=a$?
Suppose that $f$ and $g$ are defined on R and that neither f nor g is differentiable at x=a. prove or disprove: f+g is not differentiable at x=a.
I know how to show if f and g are differentiable at x=a.then f+g is differentiable at x=a. But in this problem I don't know how to use definition of differentiable function to prove it. I can't find counterexample to disprove it either.
AI: Hint: If $f$ is not differentiable at $a$, then $-f$ is also not differentiable at $a$. |
H: Prove that f is a constant function
Let $f$ be a function defined on R and suppose that there exists $M>0$ such that for any $x,y∈R$, $|f(x)-f(y)|≤M|x-y|^2$. Prove that $f$ is a constant function.
I don't even know how to start, I know that I need to show that $f(x)$ equal to some number , zero for instance. I think that $M$ is the upper bound of $f$ but I don't think it help.
AI: For any $x,y$, the difference quotient of $f$ obeys
$$\bigg| \frac{f(x) - f(y)}{x-y} \bigg| \leq M|x-y|.$$
In particular $f$ is differentiable and its derivative is zero everywhere (both by the squeeze theorem).
Since the derivative is everywhere zero, what can you say about $f$? |
H: Simplification of a trilogarithm of a complex argument
Is it possible to simplify the following expression?
$$\large\Im\,\operatorname{Li}_3\left(-e^{\xi\,\left(\sqrt3-\sqrt{-1}\right)-\frac{\pi^2}{12\,\xi}\left(\sqrt3+\sqrt{-1}\right)}\right)$$
where
$$\large\xi=\frac{\sqrt[3]3}6\sqrt[3]{27+\sqrt3\,\sqrt{243-\pi^6}}$$
and $\Im\,\operatorname{Li}_3(z)$ denotes the imaginary part of the trilogarithm.
AI: Step 1: Note that the trilogarithm argument
$$z=e^{2\pi i x}=-e^{\xi\,\left(\sqrt3-\sqrt{-1}\right)-\frac{\pi^2}{12\,\xi}\left(\sqrt3+\sqrt{-1}\right)}$$
lies on the unit circle.
Step 2: Use that
$$\operatorname{Li}_n(e^{2\pi i x})+(-1)^n\operatorname{Li}_n(e^{-2\pi i x})=-\frac{(2\pi i)^n}{n!}B_n(x),$$
where $B_n(x)$ is the $n$th Bernoulli polynomial. Of course, here $n=3$.
Step 3: Simplify.
I am pretty sure this will give you the answer $\frac12$ of @VladimirReshetnikov . |
H: Application of Main Homomorphism Theorem
This is related to this question. I just didn't want a prolonged discussion in the comments.
Let $\phi: G \to G'$ be a homomorphism. Let $G$ be a finite group. Let $K \leq G$ be the kernel of $\phi$. Let $I \leq G'$ be the image of $\phi$.
Let $H' \leq G'$. Find a formula relating the order of $\phi^{-1}(H')$ in terms of $H', I, K$.
Attempt at a solution:
$|\phi^{-1}(H')|=|H'\cap I|\cdot |K \cap \phi^{-1}(H')|$.
My justification for this is as follows. $\phi^{-1}(H')/(K \cap \phi^{-1}(H'))\cong H' \cap I$
AI: As long as you can justify the isomorphism in question, you should be fine (and as Daniel said, you can simplify by noting that $K\subseteq\phi^{-1}(H')$). Also, you want to make sure you're assuming that $G$ is a finite group, otherwise the formula might not make sense (if $G$ is infinite, $\left|G/H\right|\neq\left|G\right|/\left|H\right|$, because the right side is not well defined). |
H: What can we conclude from "f is not little-o of g"?
Given two functions $f$ and $g$, what does "$f$ is not $o(g)$" mean ? What can we conclude from this statement ?
I know "$f$ is $o(g)$" means the limit at infinity of $\frac fg$ is zero.
So does "$f$ is not $o(g)$" mean the limit at infinity of $\frac fg$ is different from zero or does it mean that this limit doesn't exist ?
AI: $f = o(g)$ does mean that the $\lim_{x \to \infty} \frac{f(x)}{g(x)} = 0$. Notice that saying $\lim_{x \to \infty} \frac{f(x)}{g(x)} = 0$ is actually the statement "the limit of $\frac{f(x)}{g(x)}$ exists as $x \to \infty$ AND the limit is $0$". The negation of this, "$f$ is not $o(g)$", means that we negate the previous statement resulting in "the limit does not exist OR if it does exist, it is not $0$." |
H: Can an uncountable group be generated from a single element?
First question : can an uncountable group be cyclic?
Ok so my though is if $G$ is generated by i then for $x\in G$ we have $x=i^n$ for integer n, so then it must be countable. Is there a way to generate an uncountable group in some way from a single element?
Second question: can an uncountable group be generated by a countable set?
Something related to my question is that for any positive $k: \mathbb Q^k$ is smaller than $\mathbb R$, however since we can approach any number infinitely using a binary system it makes sense to me that if we build $(N,b_1,b_2,b_3....)$ where $ N$ is the whole part of the number and b is 0 or 1 then this should have the same cardinality as $\mathbb R$.
My question is the following: can an uncountable group be generated from a countable group?
I believe that the set $2^{-n}$ for n a non negative integer does the trick in the group $\mathbb R $under addition. Am I right?
Regards
AI: If a group is generated by one element, then it is cyclic. However, all cyclic groups are isomorphic to either $\Bbb Z/n\Bbb Z$ for some $n$ or $\Bbb Z$. $\Bbb Z$ is countable, so any (infinite) group generated by one element must also be countable. We can also see this directly by simply noting that by definition, $G = \langle r\rangle = \{r^n\mid n\in\Bbb Z\}$, so there is an immediate bijection between $G$ and $\Bbb Z$.
As for the second paragraph, I'm not entirely sure what you mean. If you mean for each $b$ to be the same, then there's a bijection between $\{(N,b,b,b,\ldots)\mid N\in\Bbb Z, b\in\{0,1\}\}$ and $\Bbb Z\times\{0,1\}$ (which is countable). If you mean for each $b$ to be possibly different, just notice that if you disregard the $N$, you can put $\{(b_1,b_2,\ldots)\mid b_i\in\{0,1\}\}$ in bijection with $[0,1]$ (which is uncountable) by sending $(b_1,b_2,\ldots)\mapsto .a_1a_2\ldots$, where $.b_1b_2\ldots$ is the base $2$ decimal expansion of $.a_1a_2\ldots$, so you obtain a bijection between $\{(N,b_1,b_2,\ldots)\mid b_i\in\{0,1\}, N\in\Bbb Z\}$ and $\Bbb Z\times [0,1]$ (still uncountable).
The set $\{2^{-n}\mid n\in\Bbb Z\}$ will not generate all of $\Bbb R$ as a normal group - it won't even generate all of $\Bbb Q$ (you'll always have powers of $2$ in the denominator)! In fact, it is known in general that if a group is countably generated, it is countable. However, if you want to talk about topological groups, then you can get uncountable groups from countable generating sets, because you take the topological closure in the end. $\Bbb R$ does have a topology, so if you consider the generating set $\{10^n\mid n\in\Bbb Z\}$ as generating a topological group with the usual topology (induced by the Euclidean metric), you can indeed generate all of $\Bbb R$ (think about limits and infinite decimal expansions). However, the key here is that in $\Bbb R$ you can make sense of things like infinite sums and limits. See here for some more commentary. |
H: Graph Theory Complements
Let G be a simple graph with n vertices. What is the relation between the number of edges of G and the number of edges of the complement G'?
In the example below, I noticed that by adding the vertices and edges of G and the edges of G' you get a total of 10. Then the number of edges in G' would be 3. It is the same for every graph with four vertices. Is there a formula for calculating the number of edges of G' based on the number of vertices and edges in G?
AI: Assuming that you’re talking about simple graphs (i.e., graphs without loops or multiple edges), there is a nice, straightforward relationship. If there are $n$ vertices, there are $\binom{n}2=\frac12n(n-1)$ pairs of distinct vertices, so there are $\binom{n}2$ possible edges. If $G$ has $m$ edges, then $G'$ must have every one of the possible edges that $G$ doesn’t have, so $G'$ has $\binom{n}2-m$ edges.
If there are $4$ vertices, for example, there are $\binom42=6$ possible edges, so if $G$ has $m$ edges, then $G'$ has $6-m$ edges. |
H: are any two vector spaces with the same (infinite) dimension isomorphic?
Is it true that any 2 vector spaces with the same (infinite) dimension are isomorphic? I think that it is true, since we can build a mapping from $V$ to $\mathbb{F}^{N}$ where the cardinality of $N$ is the dimension of the vector space - where by $\mathbb{F}^{N}$ I mean the subset of the full cartesian product - where each element contains only finite non zero coordinates?
AI: Your argument is quite correct. |
H: Trying to prove that the sequence: 3, 3, sin(1), 3, 3, sin(2), 3, 3, sin(3), 3, 3, sin(4), $\ldots$ does not converge to 3
$\textbf{Proof}$:
I need to show that $\exists \epsilon > 0$, such that, $\forall N \in \mathbb{N},\ \exists n \ge N$, such that, $\ \left | a_n - 3 \right | \ge \epsilon$
Let $\epsilon = 2$, then if you take $n = 30$, we have that $a_{30} = sin(10) \approx -0.544 \ldots$
So we have that $ \left | sin(10) - 3 \right | \ge 2$
Therefore I have showed that, $\exists \epsilon > 0$, such that, $\forall N \in \mathbb{N},\ \exists n \ge N$, such that, $\ \left | a_n - 3 \right | \ge \epsilon$, where $\epsilon = 2$ and $n = 30$.
I am wondering if this proof is right.
AI: You have to find $\varepsilon>0$ such that, for all $N\in\mathbb N$, there exists $n\geq N$ such that $|a_n-3|\geq \varepsilon$. What you did will only work for $N\leq 30$, but, if for example $N=31$, you have to find an $n\geq 31$ such that $|a_n-3|\geq\varepsilon$.
Instead, you could do the following: given $N\in\mathbb N$, there exists $n\geq N$ such that $a_n=\sin k$ for some $k\in\mathbb N$. So, for that $n$, $|a_n-3|=|\sin k-3|\geq\left||3|-|\sin k|\right|\geq 3-|\sin k|\geq 2$. |
H: Continuous real valued functions and inner product space?
Let $V$ be the space of all continuous real valued functions on the interval $[1,4]$ with the inner product defined by:
$$\langle f,g\rangle = \int_1^{4} f(t)g(t)\,dt.$$
(i) Find an orthonormal basis of the space $W$ of polynomials of degree less than or equal to $4$.
Here is my attempt:
You should use the [Gram–Schmidt process][1]: Take any three linearly independent polynomials, for example $1,x,x^2$. Now apply the process to this set.
$$\|1\|^2=\langle1,4\rangle=\int_1^4 dt=4-1\quad\Rightarrow\quad p_0(x)=\frac{1}{\|4\|}$$
$$proj_{p_0}(v_1)=\langle v_1,p_0 \rangle p_0=p_0\cdot\langle x,p_0 \rangle=\frac{1}{\|4\|}\int_1^4dt$$
and then $u_1=v_1-proj_{p_0}(v_1)$ and $p_1=\frac{u_1}{\|u_1\|}$ and so on.
(ii) Find the polynomial of degree less than or equal to 4 that is closest to $\log t$.
Can you please help me with this? I saw this in a numerical analysis book and was trying to solve this problem the other day. This is what I tried to do but I know this is probably wrong. I have come to the point of exhaustion trying to teach myself this. I tried following other examples but they are not the same. Sorry for not being able to do much.
AI: First of all, you should know how the Gram-Schmidt proccess works (see Wikipedia, for example).
Let $\left\{p_1,\ldots,p_5\right\}$ be a basis for the set $W$ of polynomials (in $[1,4]$) of degree $\leq 4$. For example, you can take $p_i(x)=x^{i-1}$. Apply the Gram-Schmidt process as explained in Wikipedia. You'll obtain 5 polynomials, say $q_1,\ldots,q_5$ s.t. $<q_i,q_j>=\delta_{ij}=\begin{cases}1&\text{, if }i=j\\ 0&\text{ otherwise}\end{cases}$. Remember that, when applying the Gram-Schmidt process, you'll have to calculate some polynomial integrals, which can take quite some time...
The set $\left\{q_1,\ldots,q_5\right\}$ is an orthonormal basis for $W$.
Now, to find polynomial in $W$ that is closest to $\log$, you should already know that this is given by
$$p=\sum_{i=1}^5<q_i,\log>q_i$$
that is, the polynomial $p$ is given by
$$p(t)=\sum_{i=1}^5\left(\int_1^4q_i(s)\log(s)ds\right)q_i(t)$$
for every $x\in[1,4]$. |
H: lagrange multiplier slope
I was reading the link given in the thread's last comment. I understood initial part. I understand that in case of the hill, if we take any point on that hill, the gradient of the original function will always point towards the peak of the mountain. But I am a bit confused about the gradient of the constraint. Why would gradient of the constraint at constraint's topmost point will point in the direction of hill's peak? I thought that it will be 0 as we have reached the topmost point of the constraint. any explanation?
AI: You are moving on a level set of the constraint -- $g(x) = 0$ -- and since the gradient of $g$ is always perpendicular to its level sets, $\nabla g$ is always perpendicular to $g(x)=0$.
When you are at the constrained maximum of $f$, $\nabla f$ is also perpendicular to $g(x)$. Therefore $\nabla f$ and $\nabla g$ are parallel, i.e. $\nabla f = \lambda \nabla g$ and since $\nabla f$ points uphill, so does $\nabla g$. But that is a consequence of the fact that the two gradients are parallel, and only true at the top of the constrained hill.
EDIT: No, the gradient is never zero. (In fact, the method of Lagrange multipliers requires zero to be a regular value of $g$, i.e., $\nabla g \neq 0$ whenever $g=0$.)
Here's a concrete example:
$$\max\, x \quad \textrm{s.t.}\quad x^2+y^2=1$$
In other words, you are trying to maximize $x$ over the unit circle. This has obvious solution $x=1, y=0$.
Now the constraint function is $g(x,y) = x^2+y^2-1.$ The gradient is
$$\nabla g = (2x, 2y).$$
At the top of the constrained hill, $(1,0)$, this gradient points further uphill -- in the direction $(2,0)$ -- but is nowhere near zero. In fact it is easy to check that when $g=0$, the gradient of $g$ is never zero:
$$\|\nabla g\| = \sqrt{4x^2+4y^2} = 2\sqrt{g+1} = 2.$$ |
H: I'm a beginner in Abstract algebra and am trying to show the homomorphism
You know that map where G goes to H and then we also know a surjective group homomorphism G to G/N exists. How do I connect the G/N to H? As in, how do I show that there is a surjective map between them? Also, how come showing that a map is well-defined does not imply injective?
AI: The first isomorphism theorem for groups says that if $\phi:G\to H$ is a homomorphism of groups, then $G/\ker\phi\cong \phi(G)\subseteq H$. I assume that in your question, $N$ is meant to be the kernel of your homomorphism. If the original map is not surjective, there is not necessarily a surjective map from $G/\ker\phi\to H$. However, if the original $\phi$ is surjective, then the first isomorphism theorem tells you there is a surjection (as there is an isomorphism).
As for the second part of your question, a function $f:X\to Y$ is well-defined if you obtain the same element of the codomain no matter how you represent an element in the domain (different representations of the same element give the same result under the mapping). This has nothing to do with injectivity. For example, the function $f : \Bbb Q\to\Bbb Z$ given by $f(q) = n - m$ (with $q = n/m$) is not well-defined, because $2 = 2/1 = 4/2$, but $2 - 1 = 1\neq 2 = 4 - 2$, while the function $f : \Bbb Q\to \Bbb Z$ given by $f(q) = 0$ is well-defined (writing $q = n/m = n'/m'$ in two different ways will not change $f(q)$), but very far from being injective. |
H: Given identity map $U:\ell^n_a\rightarrow \ell^n_b$ for $\mathbb{R}^n$, how to computer operator norm $\forall a,b$?
If $U:\ell^n_a\rightarrow \ell^n_b$ is the identity map of the underlying vector space $\mathbb{R}^n$, then how do you compute the operator norm $U$ for all possible values of $a$ and $b$?
AI: Supposing I guessed right:
For $a \leqslant b$, we have $\lVert U\rVert = 1$. The standard basis vectors have $\lVert e_k\rVert_a = \lVert e_k\rVert_b = 1$, so $\lVert U\rVert \geqslant 1$. But if $\lVert x\rVert_a = 1$, then $\lvert x_k\rvert \leqslant 1$ for all $k$, so
$$\sum_{k=1}^n \lvert x_k\rvert^b \leqslant \sum_{k=1}^n \lvert x_k\rvert^a = 1,$$
and $\lVert x\rVert_b \leqslant 1$, so $\lVert U\rVert \leqslant 1$.
For $b < a$, we have $\lVert U\rVert = n^{1/b-1/a}$. For the vector with all components $1$, we have $\lVert \mathbb{1}\rVert_a = n^{1/a}$, so that shows $\lVert U\rVert \geqslant n^{1/b-1/a}$. On the other hand, by Hölder's inequality,
$$\sum_{k=1}^n \lvert x_k\rvert^b \leqslant \left(\sum_{k=1}^n 1^{a/(a-b)}\right)^{(a-b)/a}\cdot \left(\sum_{k=1}^n \lvert x_k\rvert^a\right)^{b/a},$$
so $\lVert x\rVert_b \leqslant n^{1/b-1/a}\lVert x\rVert_a$. |
H: What is a polynomial and how is it different from a function?
I have a problem that asks me to find a polynomial $P(x)$ so that $P(3)$ is 9.
Now I can say with certainty that $P(x)$ can be $x^2$. This is a second degree polynomial.
But what about functions such as $\frac{1}{x^2+3}$, are these not polynomials? If not, why?
P.S. I seem to be having troubles with the math tags. Some help on that would be appreciated as well.
AI: A polynomial (in one variable, over the real numbers) is by definition an expression of the form
$f(x) = a_0 + a_1x+a_2x^2 + \cdots + a_nx^n$
for some non-negative integer $n$ and real numbers $a_i,\ i=0,1,\dotsc, n$.
Note that the variable $x$ must have non-negative powers; things like $r(x) = \frac{1}{x^2+3}$ are, by definition, not polynomials in $x$ since they include negative powers of $x$. (Properly speaking, it contains a negative power of the polynomial $x^2-3$, and hence cannot be written in the same form as $f(x)$ above.)
Of course, the numerator and denominator of $r$ are both polynomials; expressions of this form $\frac{p}{q}$ where $p,q$ are polynomials are called rational functions. |
H: Hamming Code Error Detection
I am learning few things about hamming code and error detection so my question may sound stupid. So i know that lets i ahve (7,4) hamming code and i made transpose of parity check matrix H(t). Now say my code word was Y="1001000" now i need to find the error i know the procedure that you compute syndrome by Y.H(t) and i get 110 now i know from the table my error is "e=001000" and i can go on to correct the third bit my question is how do you compute "e" if you cannot have a look at the syndrome table in other words what is the formula to compute "e" when you only have your code word and your H(t) matrix and you cannot correspond to the syndrome table? If it is stupid pardon me.
Thanks anyways.
AI: There is no such thing as a stupid question, and this question is actually an important question in coding theory. Your problem is in fact equivalent to the "decoding problem," i.e. you want to find the transmitter codeword $x$ given that you have received $y$ (As once you know $e$, you can get $x = y+e$ and vice versa). The optimal way of doing this (to minimize the block error probability) is to choose the most likely transmitted vector $x = Gs$ given that what you have received is $y$. Due to its various symmetries, this becomes a rather straightforward task in the case of a Hamming code. In a general code with a general parity check matrix, there is unfortunately no easy way of doing it: You will have to calculate the probabilities $p(s_i|y)$, where $s_i$ are your possible data-words, and choose the $s_j$ that maximizes this probability. In fact, the only reason as to why there are different (suboptimal) decoding algorithms (such as the message passing algorithms for LDPC codes) is that the optimal decoding procedure described above is a very difficult (computationally expensive) task. |
H: Why it does not produce a Klein bottle?
I cannot understand why the action $\mu : (\mathbb{Z}\oplus \mathbb{Z})\times \mathbb{R}^2 \longrightarrow \mathbb{R}^2 $ given by $\mu((m,n), (x, y)) = (x+ m, (-1)^m(y + n))$ does not produce the Klein bottle $K$. If $X = \mathbb{R}^2 / (\mathbb{Z}\oplus \mathbb{Z})$, then $X \cong K$ would imply that $\pi_1(K,p) \cong \mathbb{Z}\oplus \mathbb{Z}$, however $\pi_1(K,p) \cong \mathbb{Z} \rtimes \mathbb{Z}$. What's is wrong here?
Thanks in advance.
AI: The problem is that this action is not an action of $\mathbb Z\oplus \mathbb Z$. Note that it doesn't behave well with respect to the group law on $\mathbb Z\oplus\mathbb Z$:
\begin{align*}
(m_1+m_2,n_1+n_2)\cdot (x,y) &=(x+m_1+m_2, (-1)^{m_1+m_2}(y+n_1+n_2)), \text{ but}\\
(m_2,n_2)\cdot[(m_1,n_1)\cdot (x,y)]&=(m_2,n_2)\cdot(x+m_1, (-1)^{m_1}(y+n_1))\\
&=(x+m_1+m_2, (-1)^{m_2}((-1)^{m_1}(y+n_1)+n_2)\\
&= (x+m_1+m_2, (-1)^{m_1+m_2}(y+n_1)+(-1)^{m_2}n_2)
\end{align*}
So the two sides do not agree and this is not an action.
Another problem is that $\pi_1(K)\neq \mathbb Z\oplus\mathbb Z_2$! Indeed the fundamental group of the Klein bottle is not Abelian:
$$\pi_1(K)=\langle a,b\,|\,abab^{-1}=1\rangle.$$
You can make your action well-defined by using this group instead, and everything works out the way it should. |
H: Non-elementwise Matrix Derivatives
Let A,B,C,D,X be matrices.
I'd like to perform a Gradient Descent minimization to the loss functin
$$ tr[(AXBX^TC-D)^T(AXBX^TC-D)] $$
My question is, how to take the gradient efficiently w.r.t. $B$?
I have came up with the following:
$$ \frac {\partial} {B_{ij}} tr[(AXBX^TC-D)^T(AXBX^TC-D)] =$$
$$ tr[(\frac {\partial} {B_{ij}}(AXBX^TC-D)^T)(AXBX^TC-D)]+$$
$$tr[(AXBX^TC-D)^T\frac {\partial} {B_{ij}}(AXBX^TC-D)] =$$
recalling that $trX^T=trX$:
$$ 2tr[(\frac {\partial} {B_{ij}}(AXBX^TC-D)^T)(AXBX^TC-D)]=$$
$$ 2tr[((AX(\frac {\partial} {B_{ij}}B)X^TC)^T)(AXBX^TC-D)]=$$
$$ 2tr[((AXJ^{ij}X^TC)^T)(AXBX^TC-D)]$$
where $J^{ij}$ is all zero except that the $i,j$ element is 1.
The problem is that I need to have an expression for whole $B$ rather than only $B_{ij}$ so I won't need to do all this multiplication as the amount of $B$'s elements when I update $B$ by Gradient Descent.
I have read many material (including the matrix cookbook) and still very confused.
Thanks for your help!!!
AI: You can always absorb $X$ into $A$ and $X^\top$ into $C$, so that $X$ is eliminated in your objective function. Now, using the cyclic property of the trace of matrix product, you can rewrite the objective function as
\begin{align*}
f(B)&=\operatorname{tr}\left((ABC-D)^\top(ABC-D)\right)\\
&=\operatorname{tr}\left(B^\top A^\top ABCC^\top\right)
- 2\operatorname{tr}\left(B^\top A^\top DC^\top\right)
+\operatorname{tr}\left(D^\top D\right)\tag{1}
\end{align*}
In general, if $M$ is some constant matrix, the derivative of $\operatorname{tr}(B^\top M)$ with respect to $B$ is $M$. Using the product rule to handle the first summand in $(1)$, we see that the derivative of $f(B)$ with respect to $B$ is $2(A^\top ABCC^\top - A^\top DC^\top)$ for the new $A$ and $C$. So, with the original $A$ and $C$, the derivative should be
$$2(X^\top A^\top AXBX^\top CC^\top X - X^\top A^\top DC^\top X).$$ |
H: What function can achieve the following?
So I have some values which are computed linearly. But I want to stress the middle range more so I want the values to be "transformed" into something like this:
So basically, say for $x$ between $0$ and $100$, it should start steeper, $f(x) - x$ should be biggest in the middle at $50$ and cross $f(x) = x$ at $100$. After $100$ it should have a rather slow slope.
Note that the plot above is just a rough sketch, the real function should of course be smooth and have no saltus.
Can you show me a form of a function which has these attributes?
AI: There are many ways to meet the conditions you sketch, which can have quite different behavior for $x\gg 100$.
One natural approach, since you're concerned about the difference between $x$ and $f(x)$ reaching a maximum for a particular $x$, is to design this difference instead of $f$ itself. In that case we're looking for a $g$ such that
$g(0)=g(100)=0$,
$g(x)$ has a maximum at $x=50$, and
$g(x)$ never slopes downward so much that $f(x)=x+g(x)$ would have a downward slope (you don't say this explicitly, but it's implicit in your sketch).
The third condition means that we cannot use a simple parabola as $g$. But we can use a hyperbola instead: $g(x)=b-\sqrt{(x-50)^2+a}$ for appropriate constants $a$ and $b$. This works out to
$$ f(x) = x + \sqrt{2500+a} - \sqrt{(x-50)^2+a} $$
where the positive constant $a$ determines (indirectly) how much larger than $x$ $f(x)$ gets at $x=50$; you should select an $a$ that suits your application by trial and error.
This solution has the property that $f(x)$ will approach a finite limit for large $x$. If you want $f(x)$ to continue increasing with a nonzero slope (but smaller than 1), multiply $g$ by $(1-c)$ where $c$ is your desired limiting slope:
$$ f(x) = x + (1-c)\sqrt{2500+a} - (1-c)\sqrt{(x-50)^2+a} $$ |
H: Largest Circle in a Polygon
My polygon is given by $P=$$\left\{x\geq 0, y\geq 0, 3x-4y\leq 2, 4x+3y\leq 12\right\}$
Now trying to find the largest circle inscribed inside these half-planes. But whenever I formulate it as an LP problem, the answers don't make sense. I'm using the method of Chebyshev Center and these notes. So then I get:
Maximize r
Subject to the constraints:
$3x_c + 5r \leq 2$
$-4x_c + 5r \leq 2$
$4x_c + 5r \leq 12$
$3x_c + 5r \leq 12$
Plugging into maple I get $(x_c, r) = (1/4, 1/4)$ but that doesn't really make sense. Is there something wrong with my formulation?
AI: In your notes, $x_c$ is a 2-d vector. If you write it out, a circle is represented by the center $(x,y)$ and radius $r$, so you need to maximize $r$ subject to
$4x+3y+5r \leq 12$,
$3x-4y+5r \leq 2$,
and $x \geq r$ and $y \geq r$ with $r \geq 0$,
which is indeed a LP. Solve it using Wolfram Alpha to get $(x,y,r) = (0.9, 1.3, 0.9)$ |
H: Markov-chain properties
I have some questions about a Markov-chain $(X_n)$ on a finite state-space $S$ with transition matrix $P$. A function $f:S\rightarrow\mathbb R$ is a columns vector and $Pf$ therefore a matrix multiplication. Now there are three points I am not aware of:
$\mathbb E^x$ means that $X_0=x\in S$. Then it holds that $[Pf](x)=\mathbb E^xf(X_1)$
I honestly do not see where this comes from.
Consider an arbitrary function $g:S\rightarrow\mathbb R$. Then $g(X_n)$ is $\sigma(X_n)$-measurable and $g(X_n)$ is $\mathcal F_n$-measurable where $\mathcal F_n=\sigma(X_k:0\le k\le n)$
For fixed $f:S\rightarrow \mathbb R$ the process $M_n=f(X_n)-f(X_0)-\sum_{k=0}^{n-1}[(P-I)f](X_k)$ is a martingale with respect to the filtration $\mathcal F_n$
I do not see why this should be true because if $Pf\le f$ then $\mathbb E[f(X_{n+1})|\mathcal F_n]$ is a supermartingale
AI: (1) is obvious. Think about it. write out the matrix $P$ and $f$ and multiply them. what is the expression for each term of the result?
I think for the row corresponding to $x$, it is $\sum_y P_{xy}f(y)$
Convince yourself, this is that the same as $E^xf(X_1)$.
(2) why dont you just check the martingale property the same way you check a simple symmetric random walk on Z is a martingale? Can you not see how part 1 is connected to this? |
H: differentiability with complex numbers
Let $f:\mathbb{C}\rightarrow \mathbb{C}$ defined by $f(z)=z^{3}$. Prove that does not exist a point $z_0$ for the line segment that joins $z_1=1$ and $z_2=i$ such that
$f(z_2)-f(z_1)=f'(z_0)(z_2-z_1)$
...any idea how to start, please?
AI: Hint: This is really equivalent to asking whether there exists a $z_0$ on this line segment for which
$$-i - 1 = f'(z_0) (i - 1)$$
or even better,
$$f'(z_0) = i$$
Now try showing that for any $z_0$ on this line segment, $3z_0^2 \ne i$. Try considering absolute values. |
H: Can I use this trick for my proof?
I am given the following problem, with a hint that states that I should use the General Lebesgue Dominated Convergence Theorem.
Let $\{f_n\}$ be a sequence of integrable functions on $E$ for which $f_n \rightarrow f$ a.e. on $E$ and $f$ is integrable over $E$. Show that $\int_E \lvert f - f_n \rvert \rightarrow 0$ if and only if $\lim_{n \rightarrow \infty} \int_E \lvert f_n \rvert = \int_E \lvert f \rvert$.
($\Rightarrow$) Let $g_n = k_n \lvert f - f_n \rvert$ where $k_n \in \mathbb{R}$ and $\lvert f_n \rvert \leq g_n$ for all $n$. If we let $k = \text{sup}\{ k_n \}$, then $\lvert f_n \rvert \leq \frac{k}{k_n}g_n$
It seems like I should be able to do something like this using the Archimedean property and the fact that it does not change the value of the limit since $k \cdot lim_{n \rightarrow \infty} \int_E \lvert f - f_n \rvert = k \cdot 0 = 0$.
I finish the forward direction of the proof by stating: since $g_n$ converges, and $f_n \rightarrow f$, it follows by the General Lebesgue Dominated Convergence Theorem that
$$
\lim_{n \rightarrow \infty} \int_E f_n = \int_E f
$$
as desired.
AI: Yes, but, how would you use the dominated convergence theorem there? You have to talk about the limit of the integrals of the absolute values of $f_n$, not $f_n$ themselves.
This implication is a consequence of the inequality $$\left|\int_E|f_n|-\int_E|f|\right|=\left|\int_E(|f_n|-|f|)\right|\leq\int_E||f_n|-|f||\leq\int_E|f_n-f|,$$ and the last term goes to $0$. This is also one way of getting this at the proof of dominated convergence theorem.
Edit: For the other direction, you can do the following: first, you have that $\left||f_n-f|-|f_n|\right|\leq|f_n-f-f_n|=|f|$ and $f$ is integrable. Since $f_n\to f$ almost everywhere, you have that $|f_n-f|-|f_n|\to -|f|$ almost everywhere. Therefore, from the dominated convergence theorem, $$\int_E(|f_n-f|-|f_n|)\to-\int_E|f|,$$ hence $$\int_E|f_n-f|=\int_E(|f_n-f|-|f_n|)+\int_E|f_n|\to-\int_E|f|+\int_E|f|=0.$$ |
H: Confused by a proof in Strichartz' book on Fourier Transforms
Hi I'm confused by a proof on page 53 in Strichartz book on Fourier Transforms. Specifically, in the first equation on page 53, why is it valid to interchange the action of the distribution with the integral? I know that distributions are linear, but integrals are in general infinite sums. I have tried looking earlier in the book to find a justification for this, but I could not find anything. I would really appreciate some help with understanding this.
AI: welcome to SE.
In the above, the author is not justifying his calculation. In fact, he says `formally.' Here is one sketch of proof:
Let $O_\phi = \{y : supp(\phi_y) \subset \Omega\}$. We can take $\psi^\delta = \eta_\delta * \psi$, where $\eta_\delta$ is a mollifier.
In particular, we have that $\psi^\delta * \phi \rightarrow \psi * \phi$ uniformly as $\delta \rightarrow 0$, and $\int_{O_\phi} |\psi^\delta - \psi| < \delta$. By continuity of $T$, the former bound gives $T(\psi^{\delta} * \phi) \rightarrow T(\psi * \phi)$. These two facts yield
\begin{align*}
\left|T(\psi*\phi) - \int_{O_\phi} \psi(y) T(\phi_y) dy\right| &\leq |T(\psi*\phi)-T(\psi^\delta*\phi)| + \left|\int_{O_\phi}(\psi(y)-\psi^{\delta}(y))T(\phi_y) dy\right| \\
&\leq \epsilon + \sup|T(\phi_y)|\int_{O_\phi} |\psi(y)-\psi^{\delta}(y)|dy
\end{align*}
Hence, we prove the claim for $\psi \in C^\infty$. Since $\phi \in C^\infty$ (and so $\phi_y \in C^\infty$), we can approximate the integral by Riemann sums. It follows that
\begin{align*}
\int_{O_\phi} \psi(x)T(\phi_y) dx &\leq \sum_{i=1}^m \Delta_i \psi(x_i)T(\phi_y) + \epsilon \\
&= T\left(\sum_{i=1}^n \Delta_i \psi(x_i)\phi(x_i-y)\right) + \epsilon
\end{align*}
where we approximated with lower Riemann sums. We can also approximate with upper Riemann sums to get the reverse inequality and find that in the limit the two expressions must be equal. |
H: Using an Interpretation, Prove Two Equations Are Not Equivalent
I have been working on this problem for hours and cant seem to understand how to go about doing it. The question is to prove that
Is not equivalent to
by giving an interpretation which is a model for one, but not for the other.
I don't see how simply switching the position of the existential quantifier makes a difference. Can someone please give me some pointers? Thank you.
AI: Hint: Let $g(w,x,y,z)$ hold if $x\lt y$, domain the reals. |
H: What is a 0-ball?
I'm reading a paper that says $\bigcap V_{T,X}$ is either empty or a closed $l$-ball where $T \subset S$ is a subset of points $S$ and $\operatorname{card}{T} = m + 1 - l$ where $m$ is the dimension of the smooth manifold $\Sigma$ that the points $S$ are sampled from. My question is what happens when $l=0$, meaning that $\operatorname{card}{T} = m+1$. In this case the intersection must be a closed $0$-ball. How is a $0$-ball defined and what does it mean for an intersection of two topological spaces to be a $0$-ball?
AI: The closed $0$-ball $\bar B(x,0)$ is the singleton $\{x\}$; the open $0$-ball $B(x,0)$ is empty. (Note that in this case $\bar B$ is just notation, since it is not the closure of $B(x,0)$.)
Edit: If the designation $0$ is for the dimension and not the radius, we generalize the definition of the $n$-ball: $B^n=\{x\in\Bbb R^n:|x|=\sqrt{\sum_i x_i^2}<1\}$ (where the closed ball has weak inequality) to dimension $0$. $\Bbb R^0$ is the set of all $0$-tuples, of which there is only one, $\langle\rangle=\emptyset$ (the empty sequence). we have $\sum_i x_i^2=0$ (the empty sum), so both $\bar B^0$ and $B^0$ contain this point (since $0\le1$ and $0<1$). Thus $\bar B^0=B^0=\{\emptyset\}=\Bbb R^0$. |
H: Rotation of 2D polar graph in a 3D space along some fixed axis?
Does there exist some systematic way of rotating a 2-D polar graph $r=f(\theta)$ around some axis in a 3D space?
For example: $f(\theta)=cos(\theta)$ in 2-D looks like:
If we want to rotate the above plot along the y-axis (in 3D of-course) the plot should look like donut, as shown below:
The Question is how to get the mathematical equation of the above "donut", either in rectangular, spherical coordinate system, or cylindrical system?
Thanks !
AI: In spherical coordinates, your $\theta_{2D}$ is given by:
$$\theta_{2D} = \pi/2 - \theta$$
And you have $r = f(\theta)$.
So for your graph you'ld have:
$$r = \cos(\pi/2 - \theta) = \sin(\theta)$$
$\quad\quad\quad\quad\quad\quad\quad$ |
H: Is This Interpretation A Model Of This Formula
I have a formula
and an interpretation I:
Where D is the domain, and R1 is the set of relations.
I am trying to prove or disprove that I is a model for A. I believe that it is. However, I am unsure how I would go about proving it. Couldn't any random assortment of two-tuples from the domain be in R1 and this interpretation would still be a model?
AI: Hint: rewrite the part $\lnot A\lor\lnot B\lor C$ of the formula as $(A\land B)\to C$, and draw the given relation $R_1$ (observe that only the numbers $1,2,3,5,8$ are used). |
H: Proof of $gcd(f_{n},f_{n+2})=1$ for natural numbers
I'm going to use the Principle of Mathematical Induction to prove the above statement.
Base cases:
$(n=1)$ $f_{1}=1, f_{3}=2$ so $gcd(1,2)=1$
$(n=2)$ $f_{2}=1, f_{4}=3$ so $gcd(1,3)=1$
Assume that $gcd(f_{n}, f_{n+2})=1$ holds for some natural numbers. I want to show that $gcd(f_{n+1}，f_{n+3}=1)$
Let $d=gcd(f_{n+1}, f_{n+3})$, then $d|f_{n+1}$ and $d|f_{n+3}$
Therefore, $d|f_{n+3}-f_{n+1}$, so $d|f_{n+2}$, and by the inductive hypothesis...
I don't know how to proceed from then on. Can anyone shed some light on continuing the proof from here?
AI: Do it like this. Suppose $d\gt 1$ divides $f_{n+1}$ and $f_{n+3}$. Argue exactly as you did that $d$ divides $f_{n+2}$. Then argue in basically the same way that $d$ divides $f_n$. But by the induction hypothesis this is false, for $\gcd(f_n,f_{n+2})=1$. |
H: Proof of if $A \times B = A \times C$, and $A \neq \varnothing$, then $B=C$
Proof: suppose $A \times B = A \times C$
Then $\frac{A \times B}{A} = \frac{A \times C}{A}$
Therefore $B=C$
Is this proof valid?
AI: Your argument, while suggestive, does not mean anything because there is no division operation defined for sets. If in addition to assuming that $A$ is nonempty, we assume that $A$, $B$, and $C$ are all finite, we do have something like what you wrote for the cardinalities of the sets:
$$ \frac{\left|A \times B\right|}{\left|A\right|} = \left|B\right|.$$
Even so, you are asked to prove that $B$ and $C$ are equal, not just that they have the same cardinality. So you will need a different approach. |
H: How to approach factoring problems?
Generally speaking, how should I approach a problem involving factoring? I usually don't have a problem with the more typical forms, but sometimes I just don't know what to do.
My calc2 question is this:
The given curve is rotated about the y-axis. Find the area of the resulting surface.
$$y = \frac{1}{4}x^2- \frac{1}{2}\ln x,\quad 1 \le x \le 3$$
So the first thing I did was this:
$$\frac{dy}{dx}= \frac{1}{2}x-\frac{1}{2x}$$
And then this:
$$1+\left(\frac{dy}{dx}\right)^{\!2}= 1 + \frac{1}{4}x^2-\frac{1}{2x}+\frac{1}{4x}$$
I was then able to simplify it down to this:
$$\frac{x^2}{4}+\frac{1}{2}+\frac{1}{4x^2}$$
But leaving it in this form doesn't make it an easy integral to deal with. My solution manual shows them simplifying it to this:
$$\left(\frac{x}{2}+\frac{1}{2x}\right)^{\!2}$$
And I have no idea how to get from the form I was in to the form above. Is there any strategy, technique, books or videos I can read to strengthen my factoring ability?
Lastly, how did they get to that form above?
AI: You have
$$\frac{x^2}{4}+\frac{1}{2}+\frac{1}{4x^2} = {x^4 + 2x^2 + 1\over 4x^2}.$$
The denominator is a perfect square; now factor
$${x^4 + 2x^2 + 1\over 4x^2} = {(x^2 + 1)^2\over 4x^2} = \left(x^2 + 1\over 2x\right)^2. $$
Does this make it less opaque? |
H: $\sum_{n=0}^{14}\tan(12n+1^\circ)$
I often fail to find trigonometric sums such as the one in the question shown in the following.
When I tried the question, I first led $z=e^{i\pi/180}$.
After simple calculations, I obtained
$\sum_{n=0}^{14}\tan(12n+1)=\sum_{n=0}^{14}\dfrac{z^{24n+2}-1}{z^{24n}+z^2}$
How can I proceed calculation further?
AI: This is a special case of the more general identity:
$$\sum_{k=0}^{n-1} \tan\left(\theta + \frac{k\pi}{n}\right) = -n\cot \left(\frac{n\pi}{2} + n\theta \right)$$
In your case of course $\theta = 1^\circ$ and $n=15$. A proof of this identity may be found as an answer to a previous question on this site, here. |
H: Surfaces of genus g
The problem: give maps $f:\Sigma_{g}\longrightarrow\Sigma_{h}$ not homotopic to a constant map with $0<g<h$.
Any idea would be helpful.
AI: The existence of such a map is guaranteed by $K(G,1)$ theory because $\Sigma_g$ is aspherical for $g\geq 1$. To construct a map explicitly, try to come up with a map from the torus to $\Sigma_2$, which is not homotopic to the constant map. For a hint, try flattening the torus to an annulus first, then map the annulus into $\Sigma_2$ in a homotopically non-trivial way. |
H: What is the minimum value of $(1 + a_1)(1 + a_2). . .(1 + a_n)$?
Suppose $a_1, a_2,\dots , a_n$ are $n$ positive real numbers with $a_1a_2 \dots a_n = 1$.
Then what is the minimum value of $(1 + a_1)(1 + a_2). . .(1 + a_n)$ ?
I think $(1 + a_1)(1 + a_2). . .(1 + a_n)$ takes its minimum value when $a_1=a_2=\dots=a_n=1$ and thus the minimum value is $2^n$.
I don't know how to prove it. Please help.
AI: No need for calculus here; by AM-GM we have $1+a_1\ge2\sqrt{a_1},\ldots, 1+a_{n}\ge 2\sqrt{a_n}$ and multiplying them yields $$\prod_{1\le i\le n} (1+a_i)\ge 2^{n}\sqrt{a_1a_2\cdots a_n}=2^n,$$ with equality when $a_1=\cdots = a_n = 1$. In general if $a_1a_2\cdots a_k = M$ for some positive real $M$, then we can give a lower bound on the minimum value, which is similarly computed to be $2^n\cdot \sqrt{M}$, with equality at $a_1=\cdots = a_n = \sqrt[n]{M}$. The absolute minimum, as discussed below, is $(1+\sqrt[n]{M})^n$. |
H: Clear explanation about uniform continuity.
Can anyone explain the uniform continuity clearly with picture if possible?? I have read the section on this topic in my text book but I am still not clear on this. Thanks.
--edit
In my text book, it gives two definitions of a continuous function. One with sequence in the domain of a function and $\varepsilon$-$\delta$ definition. I think I am good with a continuous function at a point. The definition of the uniform continuity in my text book is this.
Let $f$ be a real-valued function defined on a set $S \subseteq R.$ Then $f$ is uniformly continuous on $S$ if $\forall \varepsilon >0,\exists \delta > 0\text{ s.t } x,y \in S \text{ and } |x-y| < \delta \Rightarrow \left|f(x)-f(y)\right|<\varepsilon.$
From this definiton the points $x,y$ are not fixed points in the domain. So for given $\varepsilon >0$ we find $\delta >0$ such that the distance between $x$ and $y$ in the domain is less than $\delta$ implies the distance between $f(x)$ and $f(y)$ less than given $\varepsilon.$ Then the function is uniformly continuous. This is as far as I know about the uniform continuity.
AI: First, uniform continuity is about an interval, not just a point. You can say f(x) is continuous on the interval $(a,b)$ means: $\lim_{x \rightarrow c} f(x) = f(c)$ for all $c$ in $(a,b)$. Then you have to say what you mean by a limit, and you get the old $\varepsilon$-$\delta$ statement: for every $\varepsilon$ there exists a $\delta$ such that $|x-c| < \delta \Rightarrow |f(x) - f(c)| < \varepsilon$.
Now what they never tell you is that $\delta$ depends on $x$. For some functions $\delta$ has to get smaller and smaller to get $|f(x) -f(c)| < \varepsilon$. An example of this is the function $f(x) = 1/x$ on $(0,1)$. It's continuous there everywhere, but as $x$ approaches $0$ the function gets steeper and steeper. That means you have to take your $x$ and $c$ closer and closer together -- i.e $\delta$ smaller and smaller, to get $|f(x) -f(c)| < \varepsilon$.
What uniform continuity means is that the $\delta$ does NOT depend on x. You can find a single $\delta$ that depends only on $\epsilon$ for the entire interval.
Intuitively uniform continuity means your function can't get infinitely steeper on $(a,b)$ the way $1/x$ does on $(0,1)$. Steepness doesn't only mean that $f$ may go off to infinity in your interval. It could also oscillate around in some unpleasant way. Look for example at $f(x) = \sin(1/x)$ on $(0,1)$. |
H: Short Prove or Disprove: $R_1 \cap R_2$ is an equivalence relation
Suppose $R_1, R_2$ are both equivalence relations defined on nonempty set $A$. Prove or disprove: $R_1 \cap R_2$ is an equivalence relation.
What method (if any) would you take to prove this in as few sentences as possible?
AI: Let $R=R_1\cap R_2.$ You must show the following:
For any $a\in A,$ $\langle a,a\rangle\in R.$ ($R$ is reflexive on $A$.)
For any $a,b\in A$, if $\langle a,b\rangle \in R,$ then $\langle b,a\rangle\in R.$ ($R$ is symmetric on $A$.)
For any $a,b,c\in A$, if $\langle a,b\rangle,\langle b,c\rangle\in R,$ then $\langle a,c\rangle\in R$. ($R$ is transitive on $A$.)
Each step can be proved very directly and straightforwardly, since $R_1,R_2$ are equivalence relations on $A$. Let me prove symmetry to give you a taste of it.
Suppose that $a,b\in A$ such that $\langle a,b\rangle\in R.$ Then $\langle a,b\rangle\in R_1,$ so since $R_1$ is symmetric on $A$, then $\langle b,a\rangle\in R_1.$ Likewise, $\langle b,a\rangle\in R_2,$ so $\langle b,a\rangle\in R_1\cap R_2=R.$
Alternately, if you're not used to thinking of binary relations as sets of ordered pairs, we can proceed as follows.
We still let $R=R_1\cap R_2.$ By this, we mean that $a\:R\:b$ if and only if $a\:R_1\:b$ and $a\:R_2\:b.$ Then we need to show the following:
For any $a\in A,$ $a\:R\:a.$
For any $a,b\in A$, if $a\:R\:b,$ then $b\: R\: a.$
For any $a,b,c\in A$, if $a\: R\: b$ and $b\:R\:c,$ then $a\:R\:c$.
Each step is again straightforward, and very similar to the approach that would be taken above. Once again, I'll prove symmetry.
Suppose that $a,b\in A$ such that $a\:R\:b.$ Then $a\:R_1\:b,$ so since $R_1$ is symmetric, then $b\:R_1\:a.$ Likewise, $b\:R_2\:a,$ so $b\:R\:a$ by definition of $R=R_1\cap R_2.$ |
H: Is my proof correct? (Conformal equivalence of two circular annuli)
I want to show that the two annuli $$A=\{r<|z-z_0|<R\} $$
$$A'=\{r'<|z-z_0'|<R'\} $$
are conformally equivalent (i.e. there exists a biholomorphic map between the two) iff $$\frac{R}{r}=\frac{R'}{r'}. $$
Sufficiency:
Suppose the ratios of the radii are the same. A suitable scaling (by $\frac{r'}{r}$) will make $A$ congruent to $A'$, and a translation will map the scaled $A$ onto $A'$. This is clearly a conformal homeomorphism.
Necessity:
The annulus $A$ has the following set of periods
$$\left\{ \pm \frac{2 \pi}{\log (R/r)} \right\} $$
(this step required some computations which I will omit)
Similarly $A'$ has
$$\left\{ \pm \frac{2 \pi}{ \log(R'/r')} \right\}$$
as its set of periods. Since the set of periods is a conformal invariant. The ratios of the radii must indeed by the same.
Is this proof correct?
Thanks!
AI: You did not present much of actual proof here, but yes, the logic is correct. The periods of a domain are determined by the set of harmonic functions it supports, and conformal maps preserve harmonicity. Hence, the periods are conformal invariants of a domain.
For a completely different proof, see When can we find holomorphic bijections between annuli?
And for further results, Conformal maps of doubly connected regions to annuli. |
H: A question on the isomorphism induced by a homotopy equivalence.
Now, I am learning a proof that a homotopy equivalence induces an isomorphism. However, since I am a beginner in algebraic topology, I cannot fully understand the proof.
Suppose $\varphi:X\to Y$ is a homotopy equivalence. To prove $\varphi_\ast:\pi_1(X,x_0)\to\pi_1(Y,\varphi(x_0))$ is an isomorphism for every $x_0\in X$, consider the following diagram:
$$\pi_1 (X, x_0) \xrightarrow{\varphi_\ast} \pi_1 (Y, \varphi(x_0)) \xrightarrow{\psi_\ast} \pi_1 (X, \psi(\varphi(x_0))) \xrightarrow{\varphi_\ast} \pi_1 (X, \varphi(\psi(\varphi(x_0))))$$
where $\varphi\circ\psi\sim 1_Y$ and $\psi\circ\varphi\sim 1_X$.
Then there are three steps:
1) It follows that $(\varphi\circ\psi)_\ast$ is an isomorphism. Hence, $\varphi_\ast$ is surjective.
2) It follows that $(\psi\circ\varphi)_\ast$ is an isomorphism. Hence, $\varphi_\ast$ is injective.
3) Conclude that $\varphi_\ast$ is an isomorphism.
My Question is:
In 1), we actually prove that $\varphi_\ast:\pi_1 (X, x_0)\to\pi_1(Y,\varphi(x_0))$ is surjective.
In 2), we actually prove that $\varphi_\ast:\pi_1(X,\psi(\varphi(x_0)))\to \pi_1 (X, \varphi(\psi(\varphi(x_0))))$ is injective.
They are not the same function, although both are denoted by $\varphi_\ast$. So how to prove $\varphi_\ast:\pi_1(X,x_0)\to\pi_1(Y,\varphi(x_0))$ is injective?
Perhaps this question is really silly. I must miss something. Could anyone point it out? Thanks!
AI: Since $(\psi\circ\varphi)_*=ψ_*\circφ_*$ is an isomorphism, $ψ_*$ is surjective. And since $(φ∘ψ)_*$ is an isomorphism, $ψ_*$ must be injective. So $ψ_*$ is an isomorphism. It follows that $φ_*=ψ_*^{-1}∘(ψ∘φ)_*$ is an isomorphism, as well. |
H: Limit evaluation: very tough question, cannot use L'hopitals rule
I found a very tough limits question online. The question asks you to evaluate the limit $$\lim_{x \to 0}\frac{(x+4)^\frac{3}{2}+e^{x}-9}{x}$$ without using L'Hôpitals rule.
I tried to treat the top as a radical expression with the $e^x-9$ grouped and the other in root form to try to attempt rationalization. It did not work because you still get $\frac 0 0$.
I tried a trick of double rationalization but that did not work, got back to the starting.
Second attempt I tried to let $x=z-4$, a substitution, but it still did not lead to something that could remove a zero from the numerator.
Then I tried to break this up into three fractions, by dividing $x$ into each term in the numerator, and I basically got $+\infty$, then can't do $e^x/x$ and then $-\infty$.
So I have exhausted all the algebraic tricks I can think of.
Anybody out there think they they can crack this one?
Hope someone can.
Sincerely,
Palu
AI: Let $f(x) = (x+4)^\frac{3}{2}+e^{_{x}}-9$. Your limit can be written as
$$\lim_{x \to 0}\frac{f(x)- f(0)}{x - 0}$$
Which is the definition of $f'(0)$. Thus the answer is ${3 \over 2}(0 + 4)^{1 \over 2} + e^0 = 4$. |
H: Question on Proof of the Contraction Mapping Theorem
Contraction Mapping Theorem
If $T\colon X\to X$ is a contraction mapping on a complete metric space $(X,d)$ then there is exactly one solution $x\in X$.
Proof:
Let $x_0$ be any point in $X$. We define a sequence by
$$x_{n+1}=Tx_n, \qquad \text{for } n\geq 0.$$
Denote the $n$th iterate of $T$ by $T^n$, so that $x_n = T^n x_0$. First, we show that $(x_n)$ is a Cauchy Sequence. If $n\geq m\geq 1$ then
$$d(x_n,x_m) = d(T^n x_0, T^m x_0) \leq c^md(T^{n-m} x_0,x_0).$$
This is only some of the proof but my question is about the last inequality in particular the $T^{n-m}$. How is this obtained?
Thank you for any help and comments.
AI: I assume from context that $c$ is defined to be a constant for which
$$d(Tx, Ty) \le c d(x, y)$$
for all $x$ and $y$ (and in particular, $c < 1$ since $T$ is a contraction). Then
$$d(T^n x_0, T^m x_0) \le c d(T^{n - 1} x_0, T^{m - 1} x_0) \le c^2 d(T^{n - 2} x_0, T^{m - 2} x_0)$$
and so on. Apply the condition a total of $m$ times. |
H: Representing a double for-loop as a series.
What would be the best way to represent the sum formula (series) of this for-loop:
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j+=i)
count++
What process is best to determine this?
EDIT:
Sum in this sense: $\sum \limits_{k=1}^N k^2$
AI: Notice that the inner loop increments count when $j=1, j=i+1, j=2i+1, \ldots$
If $k$ is the largest integer with $ki+1 \leq n$, then count will be incremented $k+1$ times. Therefore, the inner loop increments count by $\lfloor \frac{n-1}{i}\rfloor+1.$
Therefore the full formula for the loop is
$$\sum_{i=1}^n \left\lfloor \frac{n-1}{i}\right\rfloor+1$$
or
$$n + \sum_{i=1}^{n-1} \left\lfloor \frac{n-1}{i}\right\rfloor.$$
I wouldn't be surprised if there is a nice closed form for that sum, although I couldn't find one with a quick search through tables of floor function identities. Maybe another answer will continue the calculation. |
H: Solving trigonometry equation
Please help me understand how to solve this for $0\leq x\leq360 $
I seem to have a problem with equations with powers.
$$3\sin^2 x-3\cos^2x+\cos x-1=0 $$
thinking that I would start by simplifying:
$$3 (\sin^2 x- \cos^2x)+\cos x - 1=0 $$
How I wish the equation in the bracket was in form $\sin^2 x+ \cos^2x$ which is equal to 1.
I also tried to substitute $\sin^2 x=1- \cos^2x$
AI: Well since you know $sin^2{x} = 1 - cos^2{x}$ then let's try putting it into the equation:
\begin{eqnarray} 3(1 - 2cos^2{x}) + cosx - 1& & \end{eqnarray}
Now try to substitute $u = cos(x)$, we get then $3(1 - 2u^2) + u - 1 = -6u^2 + u + 2$ now solve for $u$. Don't forget that $u = cos(x)$, so $x = arccos(u)$ . |
H: Calculate the complex integral $\oint_{|z|=1}\sin{\frac{1}{z}} dz$
How do I calculate this complex integral?
$$\displaystyle\oint_{|z|=1}\sin\left ({\displaystyle\frac{1}{z}}\right ) dz$$
I made the Taylor series for this:
$$\displaystyle\sum_{n=0}^\infty \displaystyle\frac{(-1)^n*(1/z)^{2n+1}}{(2n+1)!}$$
But now I don't know what to do.
AI: You can either apply the Residue Theorem or use the power series of $\sin$. Using the former we have
\begin{eqnarray}
\int_{|z|=1}\sin\left(\frac{1}{z}\right)\,dz&=&\int_{|z|=1}\frac{1}{z}\,dz+\sum_{n=1}^\infty\frac{(-1)^n}{(2n+1)!}\int_{|z|=1}\frac{1}{z^{2n+1}}\,dz\\
&=&\int_{|z|=1}\frac{1}{z}\,dz=\int_0^{2\pi}\frac{ie^{it}}{e^{it}}\,dt=2i\pi.
\end{eqnarray} |
H: Define $f$ on $\Bbb R$ by $f(x)=x^3$ for $x\ge0$ and $f(x)=0$ for $x\lt 0$. Find all $n\in\mathbb N$ such that $f^{(n)}$ exists on all of $\Bbb R$.
Define $f$ on $\Bbb R$ by $f(x)=x^3$ for $x\ge0$ and $f(x)=0$ for $x\lt 0$. Find all $n\in\mathbb N$ such that $f^{(n)}$ exists on all of $\Bbb R$.
I am studying for Analysis midterm.
I saw this question and I am not sure what exactly f^(n) means.
Is it talking about a derivative? or something else?
I am not 100% sure. It looks pretty simple.
any tips or suggestions would be great!
AI: $f^{(n)}(x)$
means the $n$-th derivative of $f$
at $x$.
For your problem, $f(x)$
is defined on two disjoint domains
as a different function on each domain.
These separate functions
are "nice" in their own domains,
so the only problem can be
where the domains meet,
which is at zero.
So what you need to do
is to see what the derivatives
of each function are at zero
and check if they agree or disagree.
The rest is up to you.
I may have said too much. |
H: The choice of the eigenfunction of Laplacian
$M$ is a closed Riemannian manifold and $\lambda_1>0$ is the first nontrivial eigenvalue of $\Delta$. Can we find a eigenfunction $f$ of $\lambda_1$ such that $\mathop {\sup }\limits_M f - \mathop {\inf }\limits_M f = 2$ and $\mathop {\inf }\limits_M f \geq-1$?
AI: Yes. Let $f$ be a eigenfunction of $\Delta$ with respect to $\lambda_1$. By multiplying a constant, we can assume that $\sup f - \inf f = 2$ (It can be found as $f$ is nonconstant). If $\inf f \geq -1$, we are done. If $\inf f < -1$, then $\sup f < 1$, and $h=-f$ satisfies $\inf h \geq -1$. |
H: If $f(x)=\int_0^x x^2 \sin {t^2}~dt $, find $f'(x)$.
Stumbled with this problem
If $f(x)=\int_0^x x^2 \sin {t^2}~dt $, find $f'(x)$.
How do you solve problems like this?
AI: HINT: $f(x) = x^2 \cdot g(x)$, where $g$ is the integral after the $x^2$ has been taken out.
So $f'(x) = (x^2)'\cdot g(x) + g'(x) \cdot(x^2)$ by the product rule.
The derivative of $g(x)$ is exactly the Fundamental Theorem of Calculus.
BONUS: What if the upper limit of the integral is $3x$? |
H: Trig equations solution
Solving the following for x:
$$
\frac{3\cos(2x)+5\cos(x)-1}{\sqrt{-\cot(x)}}=0
$$
The solution says that the answer is $x=-\frac{\pi}{3}+2\pi k$ where $k$ is an integer.
I am not sure why there is the minus sign in front. Can someone please help me out?
AI: First off, there is a domain issue with the function on the left-hand side of the equation, since the denominator is undefined wherever $ \ \cot x \ \ge \ 0 \ . $ Since the tangent function is positive in the first and third quadrants, so is the cotangent function; also, $ \ \cot x \ = \ 0 \ $ wherever $ \ \cos x \ = \ 0 $ and undefined where $ \ \sin x \ = \ 0 $ . Putting this all together tells us that there are only solutions to the equation for $ \ (\frac{2n-1}{2}) \cdot \pi \ < \ x \ < \ n \cdot \pi \ , $ the second and fourth quadrants without their boundaries.
The numerator is zero for
$$ \ 3 \ \cos(2x) \ + \ 5 \cos x \ - \ 1 \ = \ 3 \ (2 \cos^2 x \ - \ 1 ) + \ 5 \cos x \ - \ 1 $$
$$ = \ 6 \cos^2 x \ + \ 5 \cos x \ - \ 4 \ = \ (3 \ \cos x \ + \ 4 ) \ \cdot \ ( 2 \ \cos x \ - \ 1) \ = \ 0 . $$
The first factor is never zero, but the second one is for $ \ \cos x \ = \ \frac{1}{2} \ \Rightarrow \ x \ = \ \pm \frac{\pi}{3} + 2k\pi \ , $ as Oliver has said. However, the $ + \frac{\pi}{3} + 2k\pi \ $ "family" of solutions is barred since those are in the first quadrant. Thus, the permissible family of solutions is $ - \frac{\pi}{3} + 2k\pi \ . $ |
H: Determinant from matrix entirely composed of variables
I don't want the answer, but I'd love to kick in the right direction. I'm really not sure how to approach this question.
$$\begin{align}
& -6 = det\begin{bmatrix}
a & b & c \\
d & e & f \\
g & h & i \\
\end{bmatrix} \\
& x = det\begin{bmatrix}
a & b & c \\
2d & 2e & 2f \\
g+3a & h+3b & i+3c \\
\end{bmatrix}
& \text{Solve for }x
\end{align}$$
I believe if I set $a=1$, $e=2$, and $i=3$ (all other variables $0$), the determinant of the first matrix is $6$, and then for the second matrix is $12$. These were arbitrary variable initializations and can be any number. The relationship between the two (a scalar multiple of 2) will be the same irrespective of what I set the variables to.
I can then infer that the determinant of the second matrix is $2*det[A]$ or $-12$ because I'm investigating the relationship between the two matrices rather than actually calculating anything. I imagine this is a cheap way out, though. But, from this method I do get $x = -12$, which I believe is the correct answer.
What is the proper way of solving this? I don't want the answer, but I'd like to know the process.
AI: Your answer will be closely related to the $-6$ that was given. You are NOT supposed to find values of $a,b,c,d,...$ that "work". You are supposed to use pure theory.
A) when you add a multiple of one row to another row, the determinant stays the same.
B) multiplying a row by a constant multiplies the determinant by __. |
H: Cyclotomic Polynomial Evaluated at 1
I noticed that when trying to evaluate the $n$th cyclotomic polynomial at $1$, $\Phi_n(1)$, we run into an issue while using the explicit formula $\Phi_n(x)=\prod_{d|n}(x^d-1)^{\mu(n/d)}$. In particular, $$\Phi_n(1)=\prod_{d|n}(1^d-1)^{\mu(n/d)}=\prod_{d|n}0=0?$$ This is certainly not true, since $\Phi_n(1)$ is $1$ for all integers $n>1$ such that $n$ is not a power of a prime and $\Phi_n(1)=p$ if $n=p^k$ (Proved as a lemma here). Can someone explain what's going on here? Any help is appreciated.
AI: The answer lies in $\mu(n/d)$. In particular $\mu(p) = -1$ for any prime p. Since n/d is certainly a prime number for some choice of d, we have that $\mu(n/d) = -1$ for some choice of d|n. Then we have a rational function whose denominator evaluates to 0 at 1, so $x = 1$ isn't in the domain of this function.
Hope that helps. |
H: Does the principle of complete induction imply the well-ordering principle
Proof: Assume the PCI. Let $T$ be a nonempty subset of $\mathbb{N}$. Then $T$ has some element $x$. Then $\{1,2,...,x-1\}$ is a subset of $\mathbb{N} - T$. By the PCI, $x$ is an element of $\mathbb{N}-T$. This is a contradiction, because $x$ is an element of $T$ and $x$ is an element of $\mathbb{N}-T$. Therefore $T$ has a smallest element.
Is this proof valid?
AI: The proof is not valid. You use the Well-Ordering principle to prove the Well-Ordering Principle. In particular you choose an x, and then you assume that $\mathbb{N} - T$ contains all elements less than x, which can only happen if x was the smallest element to begin with. |
H: Proof that changing a finite number of terms in a series does not change where or not it converges
I want to prove the following theorem:
Changing a finite number of terms in a series does not change whether or not it converges, although it may change the value of its sum if it does converge
I don't know how to begin as I am not sure how to translate that statement into symbols. Furthermore, I am not sure what "Changing a finite number of terms" means. Does it mean changing the value of a particular term or omitting it entirely or something else?
Please advise.
AI: Suppose $x_n \to x$. This means for all $\epsilon>0$ there exists some $N$ such that if $n \ge N$, then $|x_n-x| < \epsilon$.
Now suppose $x'_n$ is a sequence such that for $n \ge M$, then $x'_n = x_n$.
Let $\epsilon>0$, there exists some $N$ that 'works' for the original sequence $x_n$. Now take $N'= \max(N,M)$. Then if $n \ge N'$, we have $|x'_n -x| < \epsilon$. Hence $x'_n \to x$.
Now consider a convergent series $\sum_n x_n$. If we let $s_n = x_1+...+x_n$, then we have $s_n \to s$.
Now consider the series $\sum_n x'_n$, where for $n \ge M$, then $x'_n = x_n$.
Let $s'_n = x'_1+...+x'_n$. Note that for $n \ge M$, we have $s'_n -s'_{M-1} = x'_M+...x'_n = x_M+...x_n$, and so $s'_n -s'_{M-1} = s_n -s_{M-1} \to s -s_{M-1}$.
Hence $s'_n \to (s -s_{M-1}+s'_{M-1})$. |
H: Drawing a lattice for a set partially ordered by divisibility
I am a little bit confused as to how I would draw a lattice for the set $S= \{1,2,3,4,6,9,12,18\} $when it is partially ordered by divisibility.
I was able to prove that it has a greatest lower bound, and a least upper bound for all elements $x,y \in S$. I believe my proof is correct, and by having these two bounds, the set is therefore a lattice.
I am just a bit confused as to how I would draw it.
AI: Just how are you confused about drawing the lattice? Here's how I did it. I started by putting 1 at the bottom, since it's the least element of the lattice, it divides everything.
Since 2 and 3 are divisible by 1, but not by each other, I put 2 one cm northwest of 1, I put 3 one cm northeast of 1, and draw lines from 2 to 1 and from 3 to 1.
Since 6 is divisible by everything so far, I put 6 one cm NE of 2 (and one cm NW of 3) and draw lines from 6 to 2 and from 6 to 3. Of course I don't draw a line from 6 to 1.
I now have a diamond, with 1 at the bottom, 2 and 3 on the sides, 6 at the top.
Next I look at 4: divisible by 1 and 2, not divisible by 3 or 6. So I put 4 one cm NW of 2, and draw a line from 4 to 2.
12 is divisible by everything so far. I put 12 one cm NE of 4 (and NW of 6), and draw lines from 12 to 4 and from 12 to 6.
9 is divisible by 1 and 3, but not divisible by 2, 4, 6, or 12. I put 9 one cm NE of 3, and draw a line from 9 to 3.
18 is divisible by 6 and 9, and therefore also by 1, 2, and 3; but 18 is not divisible by 4 or 12. I put 18 one cm NE of 6 (NW of 9) and draw lines from 18 to 6 and 9.
That's the diagram. I'd draw it for you if I knew how.
Looking at the diagram, it's easy to see that it's a lower semilattice, meaning that any two elements have a greatest lower bound.
It's also easy to see that it's not an upper semilattice (and therefore not a lattice); the pairs {4,9}, {4,18}, {9,12}, and {12,18} do not have least upper bounds.
By the way, what did you think was the least upper bound of 12 and 18? Maybe there was a typo in your question, and 36 was supposed to be an element of S? |
H: The limit of a sequence when at $n-1$
Suppose $\sum\limits_{n=1}^{\infty} a_n$ is a series that converges.
Therefore, $\lim\limits_{n \to \infty} S_n$ exists, where $S_n$ is the sum of the first $n$ terms of the series.
So, let $\lim\limits_{n \to \infty} S_n = L$.
How do I formally justify that $\lim\limits_{n \to \infty} S_{n-1} = L$ too?
I think this is true because if we draw a number line and mark $L$ on it and $x_n$ as the sum of the first $n$ terms in the series, then as $n$ gets large, $x_n$ will get closer and closer to $L$. So, for really large values of $n$, $x_n$ will approach $L$ and so does $x_{n-1}$ too.
But I don't know how to show this mathematically.
AI: $S_n$ converges to $L$ if and only if for every $\epsilon > 0$ there exists $N$ such that if $n > N$ then we have $|S_n - L| < \epsilon$.
The previous statement is also true for $N + 1$, hence for $n - 1 > N$.
EDIT: If a property holds for all $n > N$ then it holds for all $n > N+1$ necessarily, since if there exists $n_0 > N + 1$ such that the property (i.e. that $|S_{n_0} - L| < \epsilon$) does not hold, then (being $n_0 > N + 1$ and hence bigger than $N$) this would imply that the property was not true for all $n >N$. Is it clearer?
Also note that this is just a way to rewrite your intuition using the definition of limits |
H: Does this hold: $(1+\sqrt[n]{M})^n=2^n\cdot\sqrt M$
$(1+\sqrt[n]{M})^n=2^n\cdot\sqrt M $ (TRUE/FALSE)
My try: Using binomial theorem, I got $$(1+\sqrt[n]{M})^n=\sum_{k=0}^n\binom{n}{k}\big(\sqrt[n]{M}\big)^{n-k}=\sum_{k=0}^n\frac{n!}{(n-k)!\cdot k!}\big(M^{\frac{1}{n}}\big)^{n-k}.$$
I don't know what to do next. Please help.
AI: Assuming $M \ge 0$ (so there are no ambiguities about branches of complex numbers), if $t = M^{1/(2n)}$ your equation says
$$ (1 + t^2)^{n} = 2^n t^n $$
and thus
$$ 1 + t^2 = 2 t$$
which is only true for $t = 1$ (and thus $M = 1$). |
H: All Partial sums of two given sequences are bounded by a positive constant
Let $\theta \in \mathbb R$ be a non-integer multiple of $2\pi$. Prove that the sequences $(\sin(n\theta))_{n \in \mathbb N}$ and $(\cos(n\theta))_{n \in \mathbb N}$ verify $|S_N|\leq K$ where $K>0$ and $S_N=a_1+...+a_N$ for a given sequence $(a_n)_{n \in \mathbb N}$.
I began with the sequence involving the cosine, I suppose that the other case is analogue. I tried to express $\cos(n\theta)$ in its exponential form. Then $\sum_{n=1}^N \cos(n\theta)= \sum_{n=1}^N \frac {1} {2}(e^{in\theta}+e^{-in\theta})$. The second member of the equation can be separated into $\frac {1} {2} (\sum_{n=1}^N e^{in\theta}+ \sum_{n=1}^N e^{-in\theta})$. Both of these are the first $N$ terms of two geometric series. So $\frac {1} {2} (\sum_{n=1}^N e^{in\theta}+ \sum_{n=1}^N e^{-in\theta})=\frac {1} {2} (\frac {e^{i(N+1)\theta}-e^{i\theta}} {e^{i\theta}-1} + \frac {e^{-i(N+1)\theta}-e^{-i\theta}} {e^{-i\theta}-1})$. Well, I know that the denominator is never $0$ by the hypothesis we have on $\theta$. I've been fighting with this last term but I don't get to something nice. Am I doing something wrong?
AI: Note that
$$S_N=\Re\left(\sum_{n=1}^N\mathrm e^{\mathrm in\theta}\right)\quad\text{or}\quad S_N=\Im\left(\sum_{n=1}^N\mathrm e^{\mathrm in\theta}\right),
$$
and that
$$
\sum_{n=1}^N\mathrm e^{\mathrm in\theta}=\frac{\mathrm e^{\mathrm i\theta}-\mathrm e^{\mathrm i(N+1)\theta}}{1-\mathrm e^{\mathrm i\theta}}.
$$
Furthermore, for every complex number $z$, $|\Re(z)|\leqslant|z|$ and $|\Im(z)|\leqslant|z|$, hence
$$|S_N|\leqslant\left|\frac{\mathrm e^{\mathrm i\theta}-\mathrm e^{\mathrm i(N+1)\theta}}{1-\mathrm e^{\mathrm i\theta}}\right|\leqslant K_\theta,
$$
with
$$
K_\theta=\frac2{|1-\mathrm e^{\mathrm i\theta}|}.
$$ |
H: Maclaurin series of: $ f(x) = {x + 5\over1-x^2}$.
I'm trying to get the Maclauren series of:
$ f(x) = {x + 5\over1-x^2}$.
I am sure there is some trick here, the result according to Mathematica is:
$5 + x + 5x^2 + x^3 + 5x^4 + x^5 + 5x^6 + \ ...$
I defined:
$f^n_m = {x^n\over(1-x^2)^m}$
$f^n_m = 0$ for $n < 0 $ (just a definition necessary to make the next formula work for $n = 0$)
I've computed:
$(f^n_m)' = nf^{n-1}_m + 2mf^{n+1}_{m+1}$.
We have $f(x) = f^1_1(x) + 5f^0_1(x)$
I tried going from here, I didn't try induction as I want to see how I could get this result without assuming it first.
If I keep taking derivatives I get more and more terms of the form $f^s_r$.
But didn't find the pattern, I can see in the Mclauren series we evaluate the derivatives at 0 and thus only terms of the form $f^0_r$ contribute.
I tried also to just take some derivatives brute force and looking at the result, but they just keep getting more and more terms and I cannot ignore any of them because eventually as I differentiate they will all contribute at some point to the evaluation at 0.
Any advice will be helpful but I would like to see how I could have arrived at the result, so proving it by induction if it assumes the result won't be of much help.
Thank you very much!
AI: Notice we all know the following
$$ \sum^{\infty} x^n = \frac{1}{1-x} $$
for $|x| < 1 $. Now, applying this
$$ \frac{1}{1 - x^2} = \sum x^{2n} \implies\frac{x+5}{1-x^2} = (x+5) \sum x^{2n} = \sum x^{2n + 1} + \sum 5x^{2n} =$$
$$ = 5 + x + 5x^2 + x^3 + 5x^4 + x^5 + 5x^6 + \ ... $$ |
H: Complex Numbers and Transformations
If a transformation t acts by rotating every point of the plane around the origin by $\pi/5$ clockwise and then proceeds to translate it by vector $v$ = $(1,2)$.
How do I describe this transformation by complex numbers? (Define the function t(z) so that the image of any point z is t(z) after this sequence of transformations.)
So do I go with... $t(z) = z + (1+2i)$?
AI: A rotation is multiplying by a root of unity and in this case, it is $e^{-i\pi/5}$. A translation is addition, in this case, it is $1+2i$.
So putting everything together, we have $T(z)=z\cdot e^{-i\pi/5} +(1+2i)$. |
H: Logial Entailment vs. Material Conditional: binding free variables?
I think that now I DO understand basic logic manipulations.
I DO understand why (camels have feathers) -> (Michigan has a lot of great lakes) is true.
Nevertheless, we can still write $\mathscr{A}$ for (camels have feathers) and $\mathscr{B}$ for (Michigan has a lot of great lakes), then why is it NOT true that
$$ \mathscr{A} \implies \mathscr{B} $$
Or maybe it is true. That's why I'm confused.
QUESTION: What does this have to do with "binding free variables" and "substitutions"? And how do we know when a logical sentence needs "bindings" or "substitutions"?
AI: Before turning to the symbols, think first about the difference between something being true, plain and simple, and something being logically true (true as a matter of logic).
It is plain true that I'm less than six foot tall. But that's not a matter of logic: it's just a contingent fact about how things have turned out. (A few growth supplements when I was small and things might have turned out differently!) Compare the proposition that I'm not both less than six foot tall and taller than six foot. That uninformative tautology is true as a matter of logic. However things go with the contingent facts it will remain the case that I can't be both less than and taller than six foot tall.
Similarly, it is true that if I press this switch the light will go on. But that's contingent, it just happens that the wiring goes like that, there isn't a power cut and so forth. It's not a matter of logic that light-switches work. Compare: it is logically true that if I press the light switch and turn on the coffee machine then I press the light switch. There is no logically possible way that the antecedent of that conditional can be true and the consequent false.
Similarly again here. Define the so-called material conditional $A \to C$ to be equivalent to $\neg A \lor C$. [And let's not tangle now with the question of quite how the material conditional relates to the "if ... then ..." of ordinary language, as this isn't the key thing that is being asked for here.] Then with $A$ for camels have feathers and $C$ for Michigan has a lot of great lakes, it is plain true that $\neg A \lor C$ since both conjuncts are true and $\lor$ is inclusive, and hence -- trivially, by definition -- $A \to C$ is true too. But $A \to C$ is not logically true. We could imagine a world I guess where camels evolve to have a feathery coat, and the topography of Michigan is different so it has a lot of little lakes instead.
In logic we are concerned with what follows from what; so we are going to be interested in cases when, if a given $A$ is true, then $C$ has to be true as well as a matter of logic. Let's use $A \Rightarrow C$ to express this strong relation between $A$ and $C$. Then, at least as a first approximation, we have $A \Rightarrow C$ just when it is logically true that either we don't have $A$ or we $C$ together with $A$, which is equivalent to
$A \Rightarrow C$ just in case it is logically true that $A \to C$.
Now, with the camels/lakes readings for $A$ and $C$, as we saw, although it is plain true $A \to C$ is not logically true. So $A \Rightarrow C$ comes out false.
Of course, we'll want to fancy up our definition of $A \Rightarrow C$ so as not to rely on the intuitive notion of logical truth, so we'll perhaps want to define it in terms of their being no interpretation/model on which $A$ comes out true and $C$ false, where we give a fancy account of interpretations. Or we might give a definition along the lines of, for every substitution for the non-logical vocabulary in $A$ and $C$, $A \to C$ remains true. But that's fine tuning. We've already said enough to see how $A \Rightarrow C$ and $A \to C$ [in one use of the notations] can peel apart. |
H: In a cylic group of order $12$, we can find an element $g \in G$ such that $x^2 = g $ has no solutions.
$$ \textbf{PROBLEM} $$
If $G = \{ g^n : 0 \leq n \leq 11 \} $. Then we can find an element $a
\in G$ such that the equation $x^2 = a$ has no solutions.
$$ \textbf{ATTEMPT} $$
My claim is that the multiples of $3$ would do. So, take $a = g^3$. We claim that $x^2 = g^3$ has no solutions. Suppose $b \in G$ is a solution. $b$ must be of the form $g^k$ for some $0 \leq k \leq 11 $. Therefore, we have
$$ (g^k)^2 = g^3 \implies g^{2k} = g^3 \implies g^{2k - 3} = e \implies |g| \; \; \text{divides} \; \; 2k - 3$$
In other words, $|g|n_0 = 2k - 3 $ for some $n_0 \in \mathbb{Z}$. But I dont see how I can get a contradiction from here. Any suggestion will greatly be appreaciated. thanks
AI: Your group is isomorphic to the additive group $\mathbb{Z}/12\mathbb{Z}$. Thus, transposing to an additive notation, your problem is to solve the equation $x + x = a$ in this group. It should be clear now that your equation has no solution for $a = 1$. |
H: Extended Pigeonhole Principle: How to prove it?
A version of the pigeonhole principle is:
(1) If m objects are put in n boxes and n < m, then at least one box
contains at least ceil(m/n) objects
An alternate (more generalized) version is:
(2) For a nonempty finite collection of integers (not necessarily
distinct), the maximum value is at least the average value.
How do I prove (2) starting from (1)? Every description I have read leaves this as obvious but I can't seem to make the link so easily.
AI: Assume you have $n$ integers $a_1,\dots,a_n$. Assign them $n$ "boxes" $A_1,\dots,A_n$, such that for all $i$, $A_i$ is filled with $a_i$ "objects". The total number of objects will be $m:=a_1+\dots+a_n$. (1) applies and tells, that there is a box $A_i$ which contains at least $ceil(m/n)$ objects. Now $a_i$ is the number of objects of $A_i$, so:
$$a_i\geq ceil(m/n)\geq m/n = \frac{a_1+\dots+a_n}{n}$$ |
H: Proving a Sequence Does Not Converge
I have a sequence as such:
$$\left( \frac{1+(-1)^k}{2}\right)_{k \in \mathbb{N}}$$
Obviously it doesn't converge, because it alternates between $0,1$ for all $k$. But how do I prove this fact?
More generally, how do I prove that a sequence does not converge? Are there any neat ways other than "suppose, for contradiction, that the sequence converges. Then..."?
AI: A sequence of real numbers converges if and only if it's a Cauchy sequence; that is, if for all $\epsilon > 0$, there exists an $N$ such that
$$n, m \ge N \implies |a_n - a_m| < \epsilon$$
If you haven't shown or seen this, I'd strongly suggest trying to prove it (or look it up in pretty much any basic analysis book).
No matter what we choose for $N$ here, however, just choose $n = N + 1$ and $m = N$, with $\epsilon = \frac{1}{2}$. Then $|a_n - a_m| = 1 > \epsilon$, a contradiction.
Alternatively, show that if a sequence is convergent, then every subsequence is convergent to the same limit. Then choose appropriate subsequences. |
H: Help with solving $\int_0^n \exp(-rt)\exp\left(\frac sre^{-rt}\right).dt$
Given
$$\int_0^n \exp(-rt)\exp\left(\frac sre^{-rt}\right).dt$$
Can you please show the step(s) involved to reach this next line in the textbook:
$$\left[-\frac 1s\exp\left(\frac sr e^{-rt} \right) \right]_{t=0}^{t=n}$$
Is it done (or can it be done) with integration by parts? It's the third $e$ within the second $e$ that completely throws me on how to approach it, any tips on how to deal with that would be really useful, thanks.
AI: Notice $$ \frac{d( e^{-rt} )}{-r} = e^{-rt} dt \implies \frac{s}{r} \frac{d( \frac{r}{s}e^{-rt} )}{-r} = e^{-rt} dt$$
$$ \therefore \int e^{-rt}e^{\frac{s}{r}e^{-rt}}dt = \frac{-s}{r^2}\int e^{\frac{s}{r}e^{-rt}} d( \frac{r}{s}e^{-rt} ) = \frac{-s}{r^2} e^{\frac{s}{r}e^{-rt}} + C = F(t)$$
$$ \therefore F(n) - F(0) = \frac{-s}{r^2} e^{\frac{s}{r}e^{-rn}} -
\frac{-s}{r^2} e^{\frac{s}{r}} $$ |
H: Induction: $2^n = \sum_{v=0}^{n} \binom{n}{v}$
I have to prove the following identity for $n \in \mathbb{N}$:
$\displaystyle 2^n = \sum_{v=0}^{n} \binom{n}{v}$
Is there a way to show it through induction? Or is there a easier way? My steps so far:
$\displaystyle n=2: 2^2=\sum_{v=0}^{2} \binom{n}{v} \Rightarrow 2 \cdot 2 = \binom{0}{2} + \binom{1}{2} + \binom{2}{2} \Rightarrow 4 = 1 + 2 +1 \checkmark$
Now we suppose that the idenity is true for $n \in \mathbb{N}$, then it has to be true for $n+1 \in \mathbb{N}$, too:
$\displaystyle n \to n+1: 2^{n+1} = \sum_{v=0}^{n+1} \binom{n}{v} \Rightarrow 2^n \cdot 2 = \sum_{v=0}^{n} \binom{n}{v}+\binom{n+1}{v} \Rightarrow \ldots ?$
AI: You can use the Newton's binomial that states that $(a+b)^n=\sum_{k=0}^n \binom{n}{k}a^{k}b^{n-k}$. Thus you have that $2^n=(1+1)^n=\sum_{k=0}^n \binom{n}{k}1^{k}1^{n-k}=\sum_{k=0}^n \binom{n}{k}$ |
H: Prove: $\lim\limits_{n\to \infty} a_n ⋅ b_n = \infty$
How do I prove:
Let
$\lim\limits_{n\to \infty} a_n = \infty$
and
$\lim\limits_{n\to \infty} b_n = \infty$
Prove: $\lim\limits_{n\to \infty} a_n ⋅ b_n = \infty$
Thank you
AI: Notice can find and $\alpha $ such that $0 < \alpha < \lim b_n$. For large $n$, in particular, $b_n > \alpha. $(proof?)
Let $M > 0$ . Therefore, by hypothesis we can find $N$ such that $a_n > \frac{M}{\alpha} $ for all $n > N $
Therefore,
$$ a_nb_n > \alpha \frac{M}{ \alpha} > M $$
$$ \therefore (a_nb_n) \to \infty $$ |
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