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H: Proving that the Fourier coefficients of a functional determine it
Proving that the Fourier coefficients of a functional determine it
I have the following exercise, taken from old homework of a functional
analysis course:
Let $\mu\in C(\mathbb{T})^{*}$. Define the Fourier coefficients of
$\mu$ by $$ \hat{\mu_{n}}=\mu(e^{2\pi int}) $$
Prove that if $\mu_{1},\mu_{2}$ have the same Fourier coefficients
then $\mu_{1}=\mu_{2}$
I have tried to prove this by calculating $\mu_{1}(f)$ for $f\in C(\mathbb{T})$
and tried to prove that it is the same as $\mu_{2}(f)$.
My problem is that I have used
$$
f=\sum_{-\infty}^{\infty}\hat{f}(n)e^{2\pi int}
$$
which I was told that is not necessarily true in $C(\mathbb{T})$,
but rather in $L^{2}$.
I was also told that I should of used the Stone-Weierstrass theorem,
but I didn't understand how.
Can someone please explain how can I use S-W in order to solve the
question ?
AI: Letting $\mu = \mu_1-\mu_2$, the question boils down to showing that if $\hat{\mu}_n = 0$ for all $n$, then $\mu = 0$.
Let me use the notation $e_n(t) = e^{i 2 \pi n t}$.
Suppose $\mu(f) > 0 $ for some $f \in C(\mathbb{T})$. Let $p_n = \sum_i \alpha_n^i e_i$ be a sequence of trigonometric polynomials converging uniformly to $f$. Since $\mu(p_n) = \sum_i \alpha_n^i \mu(e_i)$, we see that for sufficiently large $n$, we have $\mu(p_n) > 0$. In particular, we must have $\hat{\mu}_k=\mu(e_k) \neq 0$ for some $k$.
Hence if $\mu$ is non zero, the $\hat{\mu}_k$ cannot all be zero. |
H: If the transitive closures of two sets are equal can they be disjoint?
I came across a problem today that I can't get out of my head: if $\mathcal{TC}(A)=\mathcal{TC}(B)$, can $A$ and $B$ be disjoint? My guess is no, but I can't quite show it. A direct proof seems unsuccessful, while a proof by contradiction going such as suppose $A$ and $B$ disjoint and $\mathcal{TC}(A)=\mathcal{TC}(B)$ gives a hypothesis that is difficult to work with. Any input is appreciated, thanks in advance!
AI: Consider $A=\{0,2,4,6,8,\cdots\}$ and $B=\{1,3,5,7,9,\cdots\}$. It is obvious that $A\cap B$ is empty. However the transitive closure of $A$ and $B$ are $\omega$. |
H: How can I find the domain and the range of a function?
I'm working on a task which has the following question:
What is the domain and the range to the function g?
Here is the function g:
Let A - $\{$1, 2, 3, 4$\}$, B - $\{$a, b, c, d$\}$
Let g : B $\rightarrow$ A be defined so that g - $\{$$\lt$a,
1$\gt$,$\lt$b, 2$\gt$, $\lt$c, 4$\gt$, $\lt$d, 4$\gt$$\}$
Since the question is asking for the domain to the function g and since B is the domain to the function g and A is the codomain to the function g, would the answer; B - {a, b, c, d} is the domain, hold as a correct answer? Or would I have to add something more to it?
Also how can I find the range of the function g?
Would appreciate some help, thanks alot!
AI: The domain here is $B=\{a,b,c,d\}$ as it's the set of elements which you feed in to the function. The range of $g$ is the set of elements hit by $g$ which is a subset of the codomain (which in this example is $A$), and so the range of $g$ is $\{1,2,4\}$ because there does not exist any element $x$ in $B$ such that $g(x)=3$.
In your notation, the last sentence would read "because there does not exist any element $x$ in $B$ such that $\langle x,3\rangle \in g$". |
H: Independence of two (discrete) Random Variables
We know that for two discrete random variables $X$ and $Y$ to be independent, the joint density of random vector $Z=(X,Y)$ has to be equal to product of marginal densities of $X$ and $Y$.
Now, this means:
a) To establish independence of $X$ and $Y$, we gotta test the above equality for every vector of euclidean plane.
b) To establish independence of $X$ and $Y$, we can test the above equality for only one vector of euclidean plane, and this suffices.
Thanks.
AI: For discrete random variables taking on finite sets of values, the joint probability mass function can be viewed as a matrix $P_{X,Y}$ with, say $m$ rows and $n$ columns, while the marginal mass functions can be viewed as $1\times m$ and $1\times n$ matrices (row vectors) $P_X$ and $P_Y$. Independence requires you to check whether $P_{X,Y} = P_X^TP_Y$, and this requires you to verify this matrix equality (that is, check whether the entries are all the same on both sides). So, $m\times n$ tests. |
H: Convergence of series implies convergence of Cesaro Mean.
Proof. Let $\sum_{k = 0}^N c_k \rightarrow s$, let $\sigma_N = (S_0 + \dots + S_{N-1})/N$ be the $Nth$ Cesaro sum where $S_K$ is the $Kth$ partial sum of the series. Then $s - \sigma_N \\= s - c_0 - c_1(N-1)/N + c_2(N-2)N +\dots+c_{N-1}/N \\ =c_1/N + c_2 2/N + \dots + c_{N-1}(N-1)/N + c_N + \dots$
Where do I go from here?
AI: You don't need the fact that it's a series, which is maybe why this is confusing for you.
Suppose $S_n \to S$ is a converging sequence. Then $\frac 1n \sum_{k=1}^n S_k \to S$ also. Roughly speaking, if you take $n$ large enough, then all the big terms (big index, not big in value) are close to $S$ ; all the small terms (small index) will get killed when $n$ goes to infinity.
Non-roughly speaking,
$$
\left| \left( \frac 1n \sum_{k=1}^n S_k \right) - S \right| = \frac 1n \left| \sum_{k=1}^n (S_k - S) \right| \le \frac 1n \sum_{k=1}^n |S_k - S| = \frac {\sum_{k=1}^{\ell} |S_k - S|}{n} + \frac {\sum_{k=\ell+1}^n |S_k - S|}{n}
$$
Let $\varepsilon > 0$, and choose $\ell$ such that for all $k > \ell$, $|S_k - S| < \varepsilon/2$ by convergence of $S_k$ to $S$. Now that $\ell$ is fixed, choose $N$ large enough so that for all $n > N$,
$$
\frac{\sum_{k=1}^{\ell} |S_k - S|}{n} < \varepsilon / 2.
$$
(Note that the numerator does not depend on $n$ so we still have freedom.) It follows that for all $n > N$,
$$
\frac {\sum_{k=1}^{\ell} |S_k - S|}{n} + \frac {\sum_{k=\ell+1}^n |S_k - S|}{n} \le \frac {\sum_{k=1}^{\ell} |S_k - S|}{n} + \frac{(n-\ell) (\varepsilon/2)}n \le \varepsilon.
$$
For your particular problem, put $S_n = \sum_{k=0}^n c_k$.
Hope that helps, |
H: Probability distribution of $X_N$, $N = min\{n \geq 2: X_n = $ second largest of $ X_1, \ldots, X_n\}$
Given a sequence $X_1, X_2, \ldots$ of independent, continuous random variables with the same distribution function $F$ and density function $f$, let $N = min\{n \geq 2: X_n = $ second largest of $ X_1, \ldots, X_n\}$.
Denote $X_N$ as the first random variable which, at the time of observation, is the second largest of those observed so far. Find the probability density function $f_{X_N}(x)$.
I'm extremely lost on this problem; all I've come up with thus far is to look at random variables $X_M$, the maximum of all observed so far, and $X_{M-1}$, the 'second-to-maximum' but I have been unable to find their respective distribution/density functions.
Thanks in advance for any and all help.
AI: Let $f(x)$ be the density function of the $X_i$, and $F(x)$ their cdf.
Let $Z$ be the maximum of the $X_i$. For completeness we find the distribution of $Z$, though that is likely familiar to you.
The event $Z\le z$ happens precisely if all the $X_i$ are $\le z$. This has probability $(F(z))^n$. Thus
$$F_Z(z)=(F(z))^n.$$
For the density function of $Z$, differentiate. We get $f_Z(z)=nf(z)(F(z))^{n-1}$.
Let $Y$ be the second largest of the $X_i$. We give a highly informal derivation of the density function $f_Y(y)$ of $Y$.
Let $dy$ be "small." We find the probability that $Y$ lies between $y$ and $y+dy$.
This will be approximately $f_Y(y)\,dy$.
Neglecting terms in higher powers of $dy$, the probability that the second largest lies between $y$ and $y+dy$ is the probability that some $X_i$ lies in this interval times the probability that $n-2$ of the $X_i$ lie below $y$ and $1$ lies above $y+dy$.
The $X_i$ that lies between $y$ and $y+dy$ can be chosen in $n$ ways. The probability it lies in the interval is approximately $F(y)\,dy$.
The probability that $n-2$ of the remaining $X_i$ lie below $y$, and $1$ lies above, is $\binom{n-1}{1}(1-F(y))^1 (F(y))^{n-2}$. "Thus,"
$$f_Y(y)=n\binom{n-1}{1}f(y)(1-F(y))(F(y))^{n-1}.$$
For the cdf $F_Y(y)$, integrate from $-\infty$ to $y$. The integral is in principle easy, make the substitution $u=F(y)$. |
H: How to integrate this function?
Evaluate
$$\int\sqrt{x^3+x^2}\;dx$$
What I have tried
Using substitution (which I believe was applied incorrectly) I get:
$$\frac{(x+2)\sqrt{4x+4}}{4x+4}$$
How can this integral be evaluated?
AI: For the first integral, suppose that $x\gt 0$. Let $u^2=1+x$. Then
$$\sqrt{x^3+x^2}=x\sqrt{1+x}=(u^2-1)(u).$$
Since $2u\,du=dx$, we end up with the integral
$$\int (2u)(u^2-1)(u) \,du,$$
which is easy. |
H: Heat Equation - Similarity Solution
Let $ s = xt^{-1/2} $ and look for the solution to the heat equation $ u_{t} = u_{xx} $ which is of the form: $ u(x,t) = t^{-\frac{1}{2}}f(s) $, which satisfies the condition $ \int_{-\infty}^{\infty} u(x,t) \; \mathrm{d}x = 1.$
This is what I've tried so far:
$ u = t^{-\frac{1}{2}}f(xt^{-\frac{1}{2}})$
$ u_{t} = -\frac{1}{2}t^{-\frac{3}{2}}f(xt^{-\frac{1}{2}}) + t^{-\frac{1}{2}}f'(xt^{-\frac{1}{2}})(-\frac{1}{2}xt^{-\frac{3}{2}})$
$ u_{t} = -\frac{1}{2}t^{-\frac{3}{2}} \left[ (f(s) + sf'(s) \right]$
$ u_{x} = t^{-\frac{1}{2}}f'(xt^{-\frac{1}{2}}) \cdot t^{-\frac{1}{2}} $
$ u_{xx} = t^{-\frac{1}{2}}f''(xt^{-\frac{1}{2}}) \cdot t^{-\frac{1}{2}} \cdot t^{-\frac{1}{2}} = t^{-\frac{3}{2}}f''(s)$
$ u_{t} = u_{xx}$
$ -\frac{1}{2}t^{-\frac{3}{2}} \left[ (f(s) + sf'(s) \right] = t^{-\frac{3}{2}}f''(s) $
$ f''(s) + \frac{s}{2}f'(s) + \frac{1}{2}f'(s)=0$
Basically, this is where I get stuck.
AI: $$f^{′′}(s)+\frac{s}{2}f^′(s)+\frac{1}{2}f(s)=0$$
$$f^{''}(s)+{(\frac{s}{2} f(s))}^{'} = 0 $$
$$f^{'}(s)+\frac{s}{2}f(s) = A $$
$$ (f(s).\exp(\frac{s^2}{4}))^{'} = A\exp(\frac{s^2}{4}) $$
$$f(s).\exp(\frac{s^2}{4}) = f(0)+\int_{u=0}^s A\exp(\frac{u^2}{4}) \:du$$
you can integrate at any point you want , i chose the 0 but you can choose another one |
H: continuity of a map on $M(\mathbb{R}^n)$
Let $M:=M(\mathbb{R}^n)$ be the space of probability measures on $\mathbb{R}^n$ with respect to the Borel $\sigma$-algebra. Let $K\subset M$ be a compact convex subset. $K$ carries a natural topological structure, i.e. the weak topology induced by the bounded continuous functions. I have a continuous function $f:\mathbb{R}^n\to\mathbb{R}$ given such that
$$|f(x_1,\dots,x_n)|\le K(1+\sum_{i=1}^n|x_i|)$$
for some constant $K$. I want to verify the continuity of $F:K\to\mathbb{R}$ on $K$, where
$$F(\mu):=\int_{\mathbb{R}^n}f d\mu$$
We know that every $\mu\in K$ has marginals $\rho_1,\dots,\rho_n$ with finite first moments. I have two questions:
From the finite first moment it should follow: $\int_{\mathbb{R}^n\backslash[-a,a]^n}fd\mu\to 0$ uniformly in $\mu\in K$. Why is this the case? I've never heard the term finite first moment for a measure (just for r.v.)
This should prove the continuity of $F$. How exactly?
AI: The measure $\nu$ on the Borel subsets of the real line has a finite first moment if $$\int_{\mathbb R}|x|\mathrm d\nu<\infty.$$
It's equivalent to say that the real valued random variable $X$ has a finite first moment and $\mu_X$, the measure associated with $X$, has a finite moment.
Using the inequality about $f$, it's enough to show that
$$\lim_{a\to \infty}\sup_{\mu\in K}\int_{\mathbb R^n\setminus[-a,a]^n}|x_j|\mathrm d\mu=0, \quad j\in\{1,\dots,n\}.$$
Using the fact that each element of $K$ has marginals $\rho_j$, we are reduced to prove that
$$\lim_{a\to \infty}\int_{\mathbb R\setminus [-a,a]}|x|\mathrm d\rho_j=0.$$
It's the case by monotone convergence.
In this context, weak convergence is metrizable, so we only have to check sequential continuity. Take $\{\mu_k\}\subset K$ which converges in distribution to $\mu$ (an element of $K$). Fix $\varepsilon>0$ and $a$ such that $\sup_k\left|\int_{\mathbb R^n\setminus[-a,a]^n}f\mathrm d\mu_k\right|<\varepsilon$. Consider $\eta\colon\mathbb R^n\to \mathbb R$ a continuous function with compact support such that $\eta(x)=1$ if $x_j\in [-a,a]$. Then
$$\left|\int_{\mathbb R^n}f(x)\mathrm d\nu_k-\int_{\mathbb R^n}f(x)\mathrm d\nu\right|\leqslant 2\varepsilon+\left|\int_{\mathbb R^n}f(x)\eta(x)\mathrm d\nu_k-\int_{\mathbb R^n}f(x)\eta(x)\mathrm d\nu\right|,$$
and we conclude taking $\limsup_{k\to \infty}$ (the last term vanishes because we now have a continuous bounded function). |
H: Solutions of triangles, homework
I have 2 questions in which I have doubt :-
Q1. prove that:- a cosBcosC + b cosAcosC + c cosBcosA = ar(ABC)/R
A1. I have used cosine rule and have put the values of all cosines here and after adding them, this is what I get:-
LHS= [1/abc]* $[2b^2c^2+2a^2c^2+2a^2b^2-($a^4+b^4 +c^4$)]$
I can't go ahead:(.
Q2. If D is the midpoint of CA in triangle, then show that:-
tan(angleADB)=4*ar(ABC)/$(a^2-c^2)$
A2. AD=b/2
BD=1/2*square root of$(2a^2+2b^2-c^2)$
BE=2*ar(ABC)/b
ED=some complex term which I am unable to solve.
diagram:- http://prntscr.com/1v8psh
dia http://prntscr.com/1v8psh
Thanks in advance.
AI: For the first use $a=b\cos C+c\cos B$ to get
$$a\cos B\cos C+b\cos C\cos A+c\cos A\cos B=a\cos B\cos C+\cos A(b\cos C+c\cos B)$$
$$=a\cos B\cos C+a\cos A=a\{\cos B\cos C+\cos(\pi-\overline{B+C})\}$$
$$=a\{\cos B\cos C-\cos(B+C)\}\text{ as }\cos(\pi-x)=-\cos x$$
$$=a\{\cos B\cos C-\cos B\cos C+\sin B\sin C)\}$$
$$=a\sin B\sin C$$
But from the sine law $a=2R\sin A$ etc. and we know $\displaystyle\triangle =\frac{abc}{4R}$
For the second, using Law of cosines,
in
$\displaystyle \triangle CDB, a^2=BD^2+\left(\frac b2\right)^2-2BD\cdot \frac b2\cos\angle CDB$
$\displaystyle a^2=BD^2+\left(\frac b2\right)^2-2BD\cdot \frac b2\cos(\pi-\angle ADB)$
$\displaystyle\implies a^2=BD^2+\left(\frac b2\right)^2+2BD\cdot \frac b2\cos\angle ADB \ \ \ \ \ (1)$ as $\cos(\pi-x)=-\cos x$
in
$\displaystyle \triangle ADB, c^2=BD^2+\left(\frac b2\right)^2-2BD\cdot \frac b2\cos\angle ADB\ \ \ \ (2)$
$\displaystyle(1)-(2)\implies a^2-c^2=2BD\cdot b\cos\angle ADB\ \ \ \ (3)$
Now we need to eliminate $BD$ from $(3)$
So, applying sine law in $\displaystyle \triangle ADB,$ we get $\displaystyle\frac{BD}{\sin A}=\frac c{\sin\angle ADB}$
Putting the value of $BD$ is $\displaystyle(3), a^2-c^2=2\frac{c\sin A}{\sin\angle ADB}\cdot b\cos\angle ADB$
$\displaystyle\implies \frac{\sin\angle ADB}{\cos\angle ADB}=\frac{2bc\sin A}{a^2-c^2}=\frac\triangle{a^2-c^2}$ |
H: I can't solve this limit...
I tried to solve it as difference of two squares. But I guess I can't move any longer from that. Please help...
AI: Rationalize the numerator to get $$\lim_{x \to \infty} \frac{ \sqrt{x + \sqrt{x + \sqrt{x}}}}{\sqrt{ x + \sqrt{x + \sqrt{x + \sqrt{x}}}} + \sqrt{x}}$$ You can eliminate the indeterminacy by multiplying top and bottom by $\frac{1}{\sqrt{x}}$. |
H: How do you find out what the function $g(f(2))$ and $f(g(2))$ is?
I'm trying to find what the *g*$(f(2))$ and the f $(g(2))$ is.
Here are the functions for f and g:
Let A - $\{$1, 2, 3, 4$\}$ and B - $\{$a, b, c, d$\}$
Let f : A $\rightarrow$ B be defined so that f - $\{$$\lt$1, b$\gt$,$\lt$2, c$\gt$, $\lt$3, d$\gt$, $\lt$4, a$\gt$$\}$
Let g : B $\rightarrow$ A be defined so that g - $\{$$\lt$a, 1$\gt$,$\lt$b, 2$\gt$, $\lt$c, 4$\gt$, $\lt$d, 4$\gt$$\}$
This is what I've done so far, but i'm not sure if it's done correctly, please feel free to correct me.
$g(f(2))$ - We have $f(2)$ in g which is $f(2) = c$ and $g(c)$ which is $g(c) = 4$.
Is this done correctly or I'm I missing something out on this?
I'm not sure how i can solve the f $(g(2))$ would appreciate some help.
Thanks a lot
AI: You did $g(f(2))$ correctly. You can't do $f(g(2))$ because $g(2)$ isn't defined since $2$ isn't in the domain of $g$. |
H: $S\subset \mathbb{R}^2$ with one and only one limit point in $\mathbb{R}^2$ such that no three points in $S$ are collinear
Question as in the title, but here it is re-typed just in case not all of the title is visible on your screen (you're welcome):
I am interested if there is a set $S\subset \mathbb{R}^2$ with one and only one limit point in $\mathbb{R}^2$ such that no three points in $S$ are collinear. I also require the limit point of $S$ to not be a member of the set $S$.
Either proof by construction or a less direct proof are both adequate.
Thanks.
AI: How about $S = \{(\frac{1}{n}, \frac{1}{n^2}) \mid n \in \mathbb{N}\}$? This set has only one limit point $(0, 0)$ and no three points in $S$ are collinear. |
H: Convergence of sequence (write a proof)
I need to prove the following affirmation: If $ \lim x_{2n} = a $ and $ \lim x_{2n-1} = a $, prove that $\lim x_n = a $ (in $ \mathbb{R} $ )
It is a simple proof but I am having problems how to write it. I'm not sure it is the right way to write, for example, that the limit of $(x_{2n})$ converges to a:
$ \forall \epsilon > 0 \; \exists n_1 \in \mathbb{N} $ such that if $n \in \mathbb{N}$ and $n \geq n_1$, then (*) $ | x_{2n} - a| < \epsilon $
About the (*) step, is that correct to write $ x_{2n} $? Or should I use another notation?
Thanks for the help!
AI: You can do so. As a matter of fact,you consider the sequence $(y_n)$ given by $y_n=x_{2n}$ and could write - as you may be more accustomed to - that $|y_n-a|<\epsilon$ for all $n>n_1$. But as $y_n=x_{2n}$ you really get back what you (correctly) wrote: $|x_{2n}-a|<\epsilon$. |
H: Sum-to-one constraint
This is a general question, but I am asking it since I am not able to find any good material online. Can someone please explain what's meant by a "sum-to-one constraint"?
Thanks.
AI: There should be some context to suggest the meaning. Evidently, some things are constrained (perhaps in an optimization problem) to have a sum of $1$. |
H: Series Convergence
What does this series converge to?
$$ \sqrt{3\sqrt {5\sqrt {3\sqrt {5\sqrt \cdots}}}} $$
and also this?
$$ \sqrt{6+\sqrt {6+\sqrt {6+\sqrt {6+\sqrt \cdots}}}} $$
And, generally speaking, how should one approach these kind of questions?
AI: Here's for
$$
\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{\cdots}}}}.
$$
First define a sequence recursively by $a_1 := \sqrt{6}$ and $a_n := \sqrt{6+a_{n-1}}$ ($n \geq 2)$. Then we wish to compute $\lim_{n \to \infty} a_n$. In order to find this limit, we need, of course, to know that it exists. But instead of trying to show right away that it does, we will assume that it does and try to obtain candidates for its value. This will help us to show convergence (as it is often the case... in fact, most of the time, when we want to show that a sequence converges to some limit $L$, we need to first guess the value of $L$, which is not always simple... Cauchy sequences are useful justly because we don't have to guess a limit in order to prove convergence).
So, assuming the limit exists, put $\lim_{n \to \infty} a_n := L$. Then by definition of $a_n$ we have
$$
\lim_{n \to \infty} a_{n+1} = \lim_{n \to \infty} \sqrt{6+a_n},
$$
which gives
$$
L = \sqrt{6+L}.
$$
In this case W.W. showed that $L \in \{-2, 3\}$ and correctly observed that $L = -2$ is impossible, since $a_n \geq 0$ $\forall n$. Thus, the only candidate for $L$ is $3$.
Now it is easily seen that $a_{n+1} \geq a_n$, so the sequence is increasing. In order to show convergence we know it suffices to exhibit an upper bound for $a_n$. But we might as well try to show that $a_n \leq 3$, in view of our candidate for $L$. This we can do by induction :
Base step : Clearly $a_1 = \sqrt{6} \leq \sqrt{9} = 3$ is true.
Induction step : Assume $a_{n-1} \leq 3$ for some $n \geq 2$. Then $a_n = \sqrt{6+a_{n-1}} \leq \sqrt{6+3} = 3$.
By the principle of mathematical induction, we have shown that $a_n \leq 3$ $\forall n$, which is what we were seeking.
We may conclude that $a_n$ converges and that its limit is 3. |
H: Genus of Edwards curve
Let us work over a field $\Bbbk$ of characteristic not equal to two. Let $d\in\Bbbk\setminus\{0,1\}$. It is said in the wikipedia article about Edwards curves that the plane quartic defined by the equation
$$x^2+y^2 = 1 + d\cdot x^2y^2$$
is birationally equivalent to a curve in Weierstraß form, i.e. a plane cubic. However, the genus of a plane curve of degree $d$ is equal to
$$g(d)=\frac{(d-1)(d-2)}{2}.$$
Furthermore, the genus is a birational invariant and $g(4)=3\ne 1=g(3)$. How can it be that the Edwards curve is birationally equivalent to a cubic?
AI: The expression you have written down is the genus of a nonsingular plane projective curve. The projective closure of the Edwards curve in $\mathbf P^2$ is not smooth! |
H: Best applications-oriented introductory calculus textbooks?
Note: I've edited this question on October 9th, after establishing a bounty on it.
What are the best introductory calculus textbooks that
explain why calculus is important in a broad intellectual and scientific context, justifying its inclusion in a liberal education. Necessarily, this means it puts great emphasis on applications to science and engineering, with specifics;
are non-dogmatic, i.e. they justify what they say, in most cases in a serious way, i.e. heuristic rather than rigorous (logical rigor in mathematics, a beautiful thing in some contexts, can approach sarcasm in other contexts);
Are at a level that can be understood by students adequately prepared in prerequisites but primarily interested in other things than mathematics and not inclined to develop their mathematical ability beyond what is needed to become liberally educated and to know that such a field as mathematics exists (in a way in which most educated people currently do not know there is such a field).
Such a book would not say "This differential equation arises in the study of fluid flow", but rather "This section will explain how to derive this differential equation from these physical principles, and some of the exercises are on understanding steps in this and related derivations, and some of them are on applying the techniques." That's a part of the "non-dogmatic" nature of the book. It would also explain why, or at least that, a particular equation is consequential beyond mathematics.
Such a book would necessarily be ruthless about refusing to include some topics that are part of the (far too) ritualistic standard course.
PS: Could answers describe the books' contents and say why they are judged to answer the points above?
AI: Try looking at Agnew's book, which now seems to be freely available on the internet:
Ralph Palmer Agnew, Calculus. Analytic Geometry and Calculus, with Vectors, McGraw-Hill Book Company, 1962, xiv + 738 pages.
Agnew's book is probably not as "strongly oriented towards applications in science and engineering" as some books (still, it is probably above average in this regard, being written back during a time when such applications were more thorough than is the case today), but the writing is remarkably fresh and the exercises are among the best you can find in an elementary calculus text.
Another book worth looking at is below. This one is a bit more strongly oriented towards applications and has some novel approaches to certain topics.
James Callahan, et al, Calculus in Context. The Five College Calculus Project, W. H. Freeman, 1995/2008, xxi + 845 pages. [This also is online. See .pdf file for Chapters 1 through 6 and .pdf file for Chapter 7 through 12.] |
H: $f\in C^1$ and $K$ compact, prove that $f:K \to ℝ^p$ is Lipschitz continuous
I try to prove Corollary 2.5.5:
Corollary 2.5.5. Let $f: {\bf R}^n \to {\bf R}^n$ be a $C^1$ mapping and let $K \subset {\bf R}^n$ be compact. Then the restriction $f|_K$ of $f$ to $K$ is Lipschitz continuous.
Obviously, I somehow need to use this theorem:
Theorem 2.5.3 (Mean Value Theorem). Let $U$ be a convex open subset of ${\bf R}^n$ and let $f: U \to {\bf R}^p$ be a differentiable mapping. Suppose that the derivative $Df: U \to \operatorname{Lin}({\bf R}^n,{\bf R}^p)$ is bounded on $U$, that is, there exists $k>0$ with $\lVert Df(\xi)h\rVert \leq k\lVert h\rVert$, for all $\xi \in U$ and $h \in {\bf R}^n$, which is the case if $\lVert Df(\xi)\rVert_{\text{Eucl}} \leq k$. Then $f$ is Lipschitz continuous on $U$ with Lipschitz constant $k$, in other words
$$ \lVert f(x) - f(x')\rVert \leq k\lVert x-x'\rVert \qquad (x,x' \in U). \tag{2.17} $$
Here are my thoughts about this problem:
If $x,x'\in U(x,x')$ where $U(x,x')$ is a convex open subset of $ℝ^n$ and the derivative $Df$ is bounded on $U(x,x')$, then $\|f(x)-f(x') \| ≤ k(x,x') \|x-x'\|$.
Surely, I can find for every $x,x'\in K$ such an $U(x,x')$ and because $K$ is compact I can even find finite many of them, so that I can take $k$ as the maximum of $k(x,x')$.
The problem in my reasoning is that if make the open cover $∪U(x,x')$ finite, then I have for single $x\in K$ a $U(x,x')$, but I don't think I can find for every pair $(x,x')$ where $x,x'\in K$ a set $U(x,x')$ in my finite cover that contain both.
Edit: The answer of @Umberto P. makes sense to me, but when I look at the official solution, I don't see why they would make it so difficult then:
http://mathematics.discoursehosting.net/uploads/db1409/484/12492d22695e2e3a.png
AI: If $K$ is compact it is in particular bounded. You can cover $K$ with a single convex open set $U$ (for instance a ball). |
H: equivalence of matrices and the image
I was working on a problem and I would like to have someone give me an advice. The question was,
Let $A$ and $B$ be two $n \times m$ matrices. Show that the $img(A)$ contains the $img(B)$ if and only if there is an $m \times m$ matrix $X$ so that $AX = B$.
My answer so far is if $A$ and $B$ are two matrices and if $X$ is a matrix so that $AX =B$, then $img(B)$ is contained in the $img(A)$ because $b \in img(B)$ if and only if there is a vector so that $By = b$ since $AX = B$, $b = By = (AX)y$
Does this make sense? Is there anyway to phrase better?
AI: Suppose ${\cal R} B \subset {\cal R} A$. Then for all $e_k$ (vector of zeroes with one in the $k$th position), there is some $x_k$ such that $A x_k = B e_k$, for $k=1,...,m$. Let $X = \begin{bmatrix} x_1 & \cdots & x_m\end{bmatrix}$. Then $AX = \begin{bmatrix} B e_1 & \cdots & B e_m\end{bmatrix} = B$, as required.
Now suppose there is some $X$ such that $AX=B$. Clearly ${\cal R} B = {\cal R} (AX) \subset {\cal R} A$. |
H: travelling salesman understanding constraints
I am trying to program TSP problem in R. From wikipedia page section "Integer linear programming formulation", I was able to understand all the constraints except the last one.
Need help to understand the last constraint...what are the variable U/artificial varaibles? are they slack variables for earlier constraints?
AI: This constraint is introduced to avoid disconnected routes of the salesman. Indeed, if a route consists of more than one cycle, then summing the last constraints along the cycle with the length $k<n$, we obtain $nk\le n(k-1),$ which is impossible.
Addendum: Let we begin to solve a TSP by a method of the integer programming. On the first step we choose anywise the unknowns $x_{ij}=0,1$ and $u_i$. Thanking to the last constraints we necessarily get a connected route. The same thing happens on the second step etc. So we will only deal with connected routes. |
H: Finding a dual basis
This is one of my homework questions - I'm pretty sure I understand part of it.
Let $V=\Bbb R^3$, and define $f_1, f_2, f_3 \in V^*$ as follows:
$$f_1(x,y,z) = x - 2y;\quad f_2(x,y,z) = x + y + z; \quad f_3(x,y,z) = y - 3z.$$
Prove that $\{f_1, f_2, f_3\}$ is a basis for $V^*$ (they are linearly independent, so this part is true), and then find a basis for $V$ for which it is the dual basis. The textbook does a horrible job as explaining dual bases in general. Can someone explain me the methods behind formulating the dual basis here?
Thanks.
AI: You need to find vectors
$$
e_1 = (x_1,y_1,z_1), e_2 = (x_2,y_2,z_2), e_3 = (x_3,y_3,z_3),
$$
so that
$$
f_i(e_j) = \begin{cases}
1 & i = j \\
0 & i \neq j
\end{cases}
$$
Write down what this means for $e_1$:
$$
x_1 - 2y_1 = 1 \\
x_1 + y_1 + z_1 = 0 \\
y_1 - 3z_1 = 0
$$
and solve for $x_1,y_1,z_1$. Then do the same for $e_2$ and $e_3$. |
H: Probability of secretary making 4 or more errors on a page
I have this problem, and I want to figure out how to do it, or at least figure out the subject that it deals with.
A secretary who only does word processing makes $2$ errors per page when typing. What is the probability that in the next page she makes $4$ or more errors?
Thank you!
AI: You mean that the secretary makes an average of two errors per page.
Usually, errors of this type are modelled using the Poisson distribution with parameter $\lambda$ equal to the mean number of errors per "unit," in this case page.
So if $X$ is the number of errors in a given page, our Poisson model gives
$$\Pr(X=k)=e^{-2}\frac{2^k}{k!}.$$
The probability of $4$ or more errors is $1$ minus the probability of $3$ or fewer errors. And
$$\Pr(X\le 3)=\sum_{k=0}^3 e^{-2}\frac{2^k}{k!}.$$
Remark: At best, the Poisson model will fit reality only modestly well. |
H: Intuition for dense sets. (Real analysis)
I have been having problems with dense sets as my lecturer didn't really develop an intuition for dense sets in my class. So can any of you please help me with that? And can you please tell me (the general case) how I should go about proving that a set is dense in R.
AI: A set $D$ is dense in $\Bbb R$ if every non-empty open interval of $\Bbb R$ contains a point of $D$; in symbols, $D$ is dense in $\Bbb R$ if and only if $D\cap(a,b)\ne\varnothing$ whenever $a,b\in\Bbb R$ and $a<b$. The set $\Bbb Q$ of rationals is dense in $\Bbb R$: if $a$ and $b$ are real numbers, and $a<b$, there is always a rational number between $a$ and $b$, so $\Bbb Q\cap(a,b)\ne\varnothing$. There is also always an irrational number between $a$ and $b$, so $\Bbb R\setminus\Bbb Q$, the set of irrationals, is also dense in $\Bbb R$. And of course $\Bbb R$ itself is dense in $\Bbb R$.
Another example of a dense subset of $\Bbb R$ is $\Bbb R\setminus\Bbb Z$, the set of real numbers that are not integers: you can easily prove that if $a<b$, the interval $(a,b)$ contains a non-integer. Similarly, $\Bbb R\setminus F$ is dense for any finite $F\subseteq\Bbb R$.
Here are a couple of less obvious examples. Let $$D=\left\{\frac{2m+1}{2^n}:n\in\Bbb N\text{ and }m\in\Bbb Z\right\}\;;$$ the elements of $D$ are the dyadic rationals, the rational numbers whose denominators in lowest terms are powers of $2$. This set $D$ is dense in $\Bbb R$; you might try to prove that $D\cap(a,b)\ne\varnothing$ whenever $a<b$. HINT: For a fixed $n$, the dyadic rationals with denominator $2^n$ are $\frac1{2^n}$ apart. Finally, let $C$ be the middle-thirds Cantor set; then $C$ does not contain any non-empty open interval, so $\Bbb R\setminus C$ intersects every non-empty open interval and is therefore dense in $\Bbb R$. |
H: Galois group of $\mathbb{Q}[\sqrt{3},\sqrt{2}]$
I am trying to compute the Galois group of $\mathbb{Q}[\sqrt{3},\sqrt{2}]/\mathbb{Q}$ in the following way:
First, $\mathbb{Q}[\sqrt{3},\sqrt{2}]/\mathbb{Q}$ is a Galois extension of the separable polynomial $(x^2-2)(x^2-3)$ (separable because $\text{char}(\mathbb{Q})=0$).
Write $G=\text{Gal}(\mathbb{Q}[\sqrt{3},\sqrt{2}]/\mathbb{Q})$. Every element of $G$ sends roots $x^2-2$ to roots of $x^2-2$ and roots of $x^2-3$ to roots of $x^2-3$ (because both polynomials are irreducible). I would like to show that all combinations are possible.
Take the identity automorphism $\mathbb{Q}\rightarrow\mathbb{Q}$. There are two ways to extend it to an automorphism $\mathbb{Q}[\sqrt{2}]\rightarrow\mathbb{Q}[\sqrt{2}]$. I would now like to extend it to an automorphism of $\mathbb{Q}[\sqrt{3},\sqrt{2}]$. The minimum polynomial of $\sqrt{3}$ over $\mathbb{Q}[\sqrt{2}]$ divides $x^2-3$. If it equals $x^2-3$ then we are done.
How do I show that the minimum polynomial of $x^2-3$ over $\mathbb{Q}[\sqrt{2}]$ is $x^2-3$.
One way to go is to show that $\mathbb{Q}[\sqrt{2}]$ does not contain a square root of $3$ by writing $(a+b\sqrt{2})^2=3$ and deriving a contradiction. I did not manage to do that, and anyway, I am hoping for a cleaner way.
(another approach: if we assume that the minimal polynomial of $\sqrt{3}$ over $\mathbb{Q}[\sqrt{2}]$ is quadratic, then $[\mathbb{Q}[\sqrt{2},\sqrt{3}:\mathbb{Q}]=4$, and then by Galois theory we have $|\text{Aut}(\mathbb{Q}[\sqrt{2},\sqrt{3}/\mathbb{Q})|=4$, meaning that all 4 possibilies must define automorphisms. This is the approach suggested in the answers to this related question, but they don't explain the part I have problems with).
Note: If you think that an entirely different approach to compute this Galois group is cleaner, please do show it!
AI: Your approach looks good. And you only have to prove that $\sqrt{3} \notin \mathbb{Q}[\sqrt{2}]$. Well, otherwise $(a+b \sqrt{2})^2=3$ for some rationals $a,b$. Expanding and using that $1,\sqrt{2}$ are linearly independent, it follows that $a^2+2b^2=3$ and $2ab=0$. Hence $a=0$ and $2b^2=3$, or $b=0$ and $a^2=3$. But you probably already know that $\sqrt{3}$ and $\sqrt{3/2}$ are irrational.
More generally, if $m_1,\dotsc,m_r$ are coprime square-free integers, then $\mathbb{Q}[\sqrt{m_1}\,,\,\dotsc,\,\sqrt{m_r}\,]$ has degree $2^r$ over $\mathbb{Q}$, and the Galois group is isomorphic to $(C_2)^r$. See here. |
H: How to find a perpendicular line from a point to a line.
I have something i am trying to take a consistent measurement of, but the issue is that the sample is not always consistent with rotation.
However from the sample I have a point, and 2 parallel lines.
Given A as the reference point, | | are lines B & C respectively, and D is the measurement point, it would look something like this.
B C
| |
A-------| D |
| |
I need to find a way to make sure that the point D in-between lines B & C is always consistent, regardless of rotation, using the reference point A.
So my thought is to make a line that intersects Line B from Point A.
This Line must be made to always be perpendicular.
I then take the middle point of this new line that falls in-between lines B & C which will always give me a consistent location for D.
Only how do I figure out the slope to give the line from point a to Line B so that the line is perpendicular to Line B and C?
AI: The slope of a perpendicular is the negative reciprocal of the slope of the line. So if the slope of line B is $20$, the slope of $AB$ is $\frac {-1}{20}$. If $B$ and $C$ are parallel, they have the same slope and this will be perpendicular to $C$ as well. |
H: Why does the central limit theorem imply that the standard deviation approaches $\frac{\sigma}{\sqrt{n}}$?
According to the central limit theorem, if one takes random samples of size $n$ from a population of mean $\mu$ and standard deviation $\sigma$, then as $X$ gets large, $X$ approaches the normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$.
$\frac{\sigma}{\sqrt{n}}$ doesn't make sense to me. Lets look at the extreme case. Say my sample consists of the entire population. Then, shouldn't my standard deviation be just $
\sigma$ instead of $\sigma/\text{(population size)}$?
AI: If $X$ represents here the sample mean $\bar X_n$, then the Central Limit Theorem says that the quantity
$$Z = \sqrt n(\bar X_n-\mu)$$ tends in distribution to $N(0,\sigma^2)$ as $n$ tends to infinity, and then by abusing notation and asymptotics, we write
$$ \bar X_n = \frac{1}{\sqrt n}Z + \mu$$ which gives us that $\bar X_n \approx N(\mu,( \frac{\sigma}{\sqrt n})^2) $.
...which in a sense holds for some "intermediate range" of $n$ - because if $n$ truly passes over to infinity, then the distribution collapses to a single point, since the variance goes to zero (which is as it should). |
H: An extension of a question about no 3 points in $\mathbb{R}^2$ being collinear
I am interested to see if we can use any of the answers here - or other methods - to answer the following question(s):
Does there exist a set $S\subset \mathbb{R}^2$ whose set of limit points in $\mathbb{R}^2$ contains the line $[0,1] =$ {$(x,0): 0 \le x \le 1$} such that no 3 points in S are collinear?
Update: I've deleted a silly follow-on question from the OP to simplify things for readers.
AI: To produce our set $S$, we "fly over" the interval $[0,1]$ again and again from left to right. At the $n$-th stage, we add points with $x$-coordinates $\frac{k}{2^n}$, where $k$ ranges over the odd numbers from $0$ to $2^n$. The $y$-coordinates of the points added during the $n$-th stage are positive and have absolute value $\lt \frac{1}{2^n}$.
At any time, we have only chosen a finite number of points, which determine a finite number of lines. So when we are choosing the $y$-coordinate of the point in $S$ corresponding to $x$-coordinate $\frac{k}{2^n}$, only finitely many points need to be avoided.
Take any real $r$ in the interval $[0,1]$. For any $n$, there is a point of our set whose $x$-coordinate differs from $r$ by at most $\frac{1}{2^n}$, and therefore whose distance from the point $(r,0)$ is positive and $\lt \frac{\sqrt{2}}{2^n}$. So there is a sequence of points in $S$ with limit $(r,0)$. |
H: Prove that $x_{n}$ is convergent if $|x_{n+1}-x_n|\leq(\frac45)^n$
The question says; prove the sequence $x_{n}$ given by:
$\left | x_{n+1} - x_{n}\right| \leq (\frac{4}{5})^n \ \forall \ n \in \mathbb{N} $
is convergent.
Here is how I approached the problem:
Since the difference between the terms applies to all $n \in \mathbb{N}$, then we know that it is a Cauchy sequence (specifically, a contractile sequence) where we can assume that $\left| x_2 - x_1 \right|=1 $ then $K = \frac{4}{5} <1$
Also since for large values of $n$, $\lim_{n\rightarrow \infty} (\frac{4}{5})^n = 0$ so the constant $(\frac{4}{5})^n$ can replace $\varepsilon$ in the criteria for contractive sequence:
$\left | x_{n+1} - x_{n}\right| \leq \varepsilon $
Then we can have, $\lim_{n\rightarrow \infty} \left|x_{n+1} - x_{n} \right| = 0$
which is a convergent sequence using the contraction principle.
However, it seems like I should have used another method that uses similar idea but explicitly shows that the difference between $\left|x_{n+1}-x_{n} \right| \leq \frac{4}{5}\left|x_{n}-x_{n-1} \right|$ and so on until I get to $\left | x_{n+1} - x_{n}\right| \leq (\frac{4}{5})^n \left| x_2 - x_1 \right|$.
I would like to get some feedback and the what are the major flaws in my solution. Suggestions are very welcome
AI: No, what you are saying is not enough. For example, the sequence $$ x_n=1+\frac12+\dots+\frac1n $$ satisfies that $ |x_n-x_{n+1}|=\frac1{n+1}$, which approaches $0$, and yet the sequence diverges.
(There is also the obvious typo that $\bigl(\frac45\bigr)^n$ is not a constant.)
You are not given nearly enough information to find the limit of the sequence, so really the most direct way (the only way?) of verifying convergence is to argue that the sequence is Cauchy. (I assume you are working on the reals, or at least in some complete metric (normed?) space, so the Cauchy criterion ensures convergence. If this is not the case, one can produce easy counterexamples.)
OK, to check that the sequence is Cauchy, we need to bound $|x_n-x_m|$ (not just $|x_n-x_{n+1}|$), in a way that ensures that $|x_n-x_m|$ can be made arbitrarily small, as long as we restrict $n,m$ to sufficiently large values.
The obvious attempt (really, again, the only thing we can do) is to use the triangle inequality. For concreteness, suppose $n<m$. We then have that
$$ |x_n-x_m|\le |x_n-x_{n+1}|+|x_{n+1}-x_{n+2}|+\dots+|x_{m-1}-x_m|. $$
Using the bounds we are given, this gives us that
$$ |x_n-x_m|\le\left(\frac45\right)^n+\left(\frac45\right)^{n+1}+\dots+\left(\frac45\right)^{m-1}. $$
It is here that the specific numbers $\bigl(\frac45\bigr)^n$ become useful. If instead all we had is that $|x_n-x_{n+1}|\to0$, we would not be able to control $|x_n-x_m|$. However, in this case, we have that
$$ |x_n-x_m|\le \left(\frac45\right)^n\frac{1-\left(\frac45\right)^{m-n}}{1-\frac45}\le\left(\frac45\right)^n\frac1{1-\frac45}=5\left(\frac45\right)^n. $$
The latter expression can be made as small as desired as long as $n$ is large enough. That is: For any $\epsilon>0$ there is an $N$ such that, provided that $N<n\le m$, we have that $|x_n-x_m|<\epsilon$. This is precisely what it means for the sequence to be Cauchy.
Of course, if you already have established a result ensuring the convergence of a sequence satisfying $|x_{n+1}-x_{n+2}|\le K|x_n-x_{n+1}|$ for all $n$ and some constant $K<1$, then you can readily conclude that the sequence converges. Anyway, if you have such a lemma, most likely its proof was along the lines of the argument I gave. |
H: Linear Combination of Vectors simple problem
So I've embarked on teaching myself Linear Algebra with G. Hadley's book which I found in my mom's library. The first chapter teacher vectors, linear dependence, and define subspaces and bases but I can't solve this simple problem in the problems section:
Express x=[4,5] as a linear combination of a=[1,3], b=[2,2].
So by my understanding this is asking what scalars can you multiply these vectors by and then add them so you get [4,5] as a solution. But it seems to me logically that no matter how you manipulate these values you can't get the difference between them to be only 1, since [1,3] has a difference of 2 and [2,2] just scales both values up equally. I'm sure I'm missing something simple and fundamental. Thanks for any help!
AI: Solve the linear system
$$\begin{cases}x\;\,+2y=4\\3x+2y=5\end{cases}$$
and you'll find the coefficients $\;x,y\;$ that give you the linear combination. As observed in the comments, these coefficients do not have to be integers, and also check the above system has a unique solution (why?) |
H: Integral $\int_{-1}^{0}\dfrac{1}{x(x^2+1)}$
Suppose I have to compute $\int_{-1}^{0}\dfrac{1}{x(x^2+1)}$. I use partial fractions to get $\int_{-1}^{0}\left(\dfrac1x-\dfrac{x}{x^2+1}\right)$, which integrates to $\log(x)-\log(x^2+1)$. Now, the $\log$ is not defined for $x\in[-1,0)$. What do I do in this case?
AI: First: I would point out that your antiderivative should be
$$
\int\left(\frac{1}{x}-\frac{x}{x^2+1}\right)\,dx=\ln\lvert x\rvert-\frac{1}{2}\ln\lvert x^2+1\rvert+C,
$$
which takes care of the problems on $(-1,0)$. However, you still have a problem at $0$ itself.
Remember, this is an improper integral, and so you actually have to compute it as
$$
\lim_{A\rightarrow0^{-}}\int_{-1}^A\left(\frac{1}{x}-\frac{x}{x^2+1}\right)\,dx.
$$
Try doing it this way, and check to see whether or not the limit converges. If it doesn't, then this integral is divergent, and you can't put a value to it. |
H: spherical geometry
A mobile, on the surface of the earth, is at a point A. Travels 200 km south arriving at a point B. Later moves 200 km west arriving at a point C. Finally moves over 200 kilometers to the north, back to point A. Assuming that the surface of the earth is a perfect sphere, which the geometric place of the points where the mobile could be?
AI: Just so we have an answer, one possibility is the North pole. There are more near the South pole, where the Westward trip circles the pole any whole number of times. |
H: T/F: $v_1,\dots,v_k\in\mathbb{R}^n$ linearly independent, $u\in\mathbb{R}^n,u\neq0$, then the matrices $uv_1^T,\dots,uv_k^T$ are linearly independent.
If $v_{1},v_{2},\dots,v_{k}$ are linearly independent vectors in $\mathbb{R}^{n}$ and if $u$ is any nonzero vector in $\mathbb{R}^{n}$, then the matrices $uv_{1}^{T}, uv_{2}^{T}, \dots, uv_{k}^{T}$ are linearly independent.
Is this true or false? Please help me.
AI: Hint: Note that $a_1uv_1^T + \dots a_kuv_k^T = u(a_1v_1^T+\dots a_kv_k^T)$. Also, if $u, v \in \mathbb{R}^n\setminus\{0\}$, $uv^T \neq 0$. |
H: Solving a definite integral using Integration by parts [simple]
The problem is to evaluate this integral:
$\displaystyle\int_{0}^1 ye^{-3y} dy$
I have uploaded an image of my work as well as my attempt to make my writing clearer in Paint.
!
Thanks so much for any help, not sure what I've done wrong.
AI: You did the "hard part" correctly. You erred at the very end, after evaluating the integral at the bounds of integration. The following is correct:
$$-\frac{1}{3e^3} - \left(\dfrac 1{9e^3} - \dfrac 19\right)$$
But, you need to distribute the negative over the parenthetical expression, to obtain $$-\dfrac{4}{9e^3} + \dfrac 19$$ |
H: rank and the span of column matrix
Can someone look over my answer for this question?
If $A$ and $B$ are conformable matrices for the product
$AB$ and if the rank of $AB$ equals the rank of $A$, then the span
of the columns of $AB$ equals the span of the columns of $A$.
Here's my answer:
Span of the columns of $AB$ equals the span of the columns of $A$ is equivalent as $Img(AB)=Img(A)$. We also know that $rank(A)=dim(Img(A))$. If $rank(A)=rank(AB)$, then $rank(A)=rank(AB)=dim(Img(AB)=dim(Img(A))$ Therefore we can conclude that if the rank of $AB$ equals the rank of $A$, then the span of the columns of $AB$ equals the span of the columns of $A$.
Does my answer makes sense?
AI: You're missing the main point. Indeed two subspaces can have the same dimension without being equal: think to the spans of $[0\;1]^T$ and of $[1\;0]^T$.
Hint: the image of $AB$ consists of the vectors of the form $ABv$; the image of $A$ consists of the vectors of the form $Aw$. Therefore …
Because of this, since the two subspaces have the same dimension, they are equal. |
H: Tangential Acceleration Proof
Given $$a = a_TT + a_NN = (\frac{d^2s}{dt^2})T + \kappa(\frac{ds}{dt})^2 N$$ prove that $$a_T = a \cdot T = \frac{a \cdot v}{||v||}$$
Where $a_T$ is tangential velocity, $a$ is an acceleration function, and $s$ is an arc length parametrization.
Where do I begin?
AI: Hint: Try computing $\renewcommand{\vec}[1]{\mathbf{#1}}\vec{a}\cdot\vec{T}$, using your assumption that $\vec{a}=a_{\vec{T}}\vec{T}+a_{\vec{N}}\vec{N}$. You may find it helpful to recall that $\vec{v}\cdot\vec{v}=\lvert\vec{v}\rvert^2$ for any vector $\vec{v}$, and that the unit tangent vector $\vec{T}$ and the principal unit normal vector $\vec{N}$ are orthogonal. |
H: proof of a limit by using epsilon delta definition
I need to prove that $$\lim_{x\to \infty}\frac{x^2}{2^x}=0$$ using only the formal definition of limit. Can anyone help?
AI: First prove that $2^x>x^3$ for all $x$ in some interval $[M,\infty)$, $M$ possibly large. Then use that $$\lim_{x\to\infty}x^{-1}=0$$
A possible way would be to look at $x\log 2>3\log x$, and use that $$\frac{\log x}x\to 0$$ |
H: Lebesgue integral in two dimensions over fraction
Let $X=[-1,1],Y=[0,1]$, and $$f(x,y)=\dfrac{xy}{(x^2+y^2)^2}$$ for $x\in X,y\in Y$. Let $\mu$ be the Lebesgue measure. Does the following integral exist:
$$\int_{X\times Y} f(x,y)d(\mu\times\mu)$$
It seems very hard to follow the definition of Lebesgue integrals using simple functions here. To do that we must find simple functions that are lower bounds for the function $f$. But the form of $f$ is quite complicated. What would be the way to compute this integral?
AI: Hint: A Lebesgue integral exists if and only if its positive part and negative part both exist and are finite.
So, why not start by considering these separately: the positive part is precisely the part over $[0,1]\times[0,1]$, and the negative part is the part over $[-1,0]\times[0,1]$. If you can show that either does not exist, then you are done; otherwise, you will have shown that both exist, and are also done!
Think about
$$
\int\limits_{[0,1]\times[0,1]}\frac{xy}{(x^2+y^2)^2}\,d(\mu\times\mu).
$$
In Riemann integration theory, we would consider this as an improper integral because of problems as we approach $(0,0)$; we would further find that the integral does not exist in that context, because it shoots of to $+\infty$ to quickly as we approach $(0,0)$. Can you use this sort of inuition to find some step functions that demonstrate that this integral will be $+\infty$? |
H: Working out the difference in earnings
I'm mathematically impaired/ignorant and trying to figure out the difference in earnings between my partner and I to work out a fair split of the bills.
So; I earn £2060 per month and partner earns £1650. As a percentage, how much more than her do I earn?
Therefore; If we had a mortgage payment of £850, by what percentage should we split the figure by so it’s proportionate to the difference in our earnings.
Any help sincerely appreciated,
M.
AI: Your combined earnings are $2060+1650$. So if you will split the rent in proportion to your earnings, you should pay
$$850\times \frac{2060}{2060+1650}$$
and your partner should pay
$$850\times \frac{1650}{2060+1650}.$$
The amounts are (to $2$ decimal places) $471.97$ and $378.03$ respectively.
As to your question about percentage, you earn
$$\frac{2060-1650}{1650}\times 100\%,$$ that is, about $24.85\%$ more than your partner. I believe that this gives a less clear path to seeing what your rent contributions should be.
Remark: There is a good argument that the contributions should be proportional to after tax income rather than income. That argument may not please your partner. |
H: Calculating the limit $\lim_{x\to\infty}\frac{\sin^2\left( \sqrt{x+1}-\sqrt{x}\right)}{1-\cos^2\frac{1}{x}}$
I'm having a really tough time trying to evaluate the limit for this expression
$\lim_{x\to\infty}\frac{\sin^2\left( \sqrt{x+1}-\sqrt{x}\right)}{1-\cos^2\frac{1}{x}}$
The only hint I was given is that $\lim_{x\to\infty}\frac{\sin(x)}{x} = 0$.
By using trigonometric identities I can see that this leads to $\lim_{x\to\infty}\frac{\sin^2\left( \sqrt{x+1}-\sqrt{x}\right)}{\sin^2(\frac{1}{x})}$. I proceeded then to do multiply everything by $\frac{1}{x}$ which leaves me with $\lim_{x\to\infty}\frac{x^{-1}\sin^2\left( \sqrt{x+1}-\sqrt{x}\right)}{1}$ (according to the hint I was given).
However, now I don't know how to simplify the expression in the numerator. Any tips would be really appreciated.
Thanks
AI: Hint: In the argument of sine, multiply by $$\frac{\sqrt{x+1}+\sqrt{x}}{\sqrt{x+1}+\sqrt{x}}.$$ |
H: Prove that $\lim_{n\to\infty}a_n^2=a^2$ when ${a_n}$ is a sequence in $\mathbb{R}$ with $\lim_{n\to\infty}a_n=a$
Question:
Let ${a_n}$ be a sequence in $\mathbb{R}$ with $\lim_{n\to\infty}a_n=a$
Prove that
$\lim_{n\to\infty}a_n^2=a^2$
Attempt:
Without using any properties of limits (this is a question in a section of the book that came before any list of any limit theorems)
By definition:
$|a_n^2-a^2|<\epsilon$
$|a_n^2-a^2|$ $\leq$ $|(a_n+a)(a_n-a)|$ $\leq$ $|(a_n+a)| |(a_n-a)|$ $\leq$
$(|a_n | + |a|)(|a_n -a|)$ $<$ $\epsilon$
And since I know that $\lim_{n\to\infty}a_n=a$, I'll subsitute that in for $|a_n|^2$ as $a^2$
$\therefore$ $a^2-|a|^2 < \epsilon$ which is true since $a^2-|a|^2$ in my case.
I wonder if this is a possible solution?
AI: You have written down the intuition towards a formal proof.
You are given an $\epsilon \gt 0$, and want to show there is an $n$ such that if $n\gt N$ then $|a_n^2-a^2|\lt \epsilon$.
By your calculation,
$$|a_n^2-a^2|\le (|a_n|+|a|)(|a_n-a|).$$
We want to make the left side "small." That will be the case if $|a_n-a|$ is small, unless $|a_n|+|a|$ is big. So we first want to show that after a while, $|a_n|+|a|$ cannot be big.
There are two cases, $a\ne 0$ and $a=0$.
If $a\ne 0$, there is an $N_1$ such that if $n\gt N_1$ than $|a_n-a|\lt \frac{|a|}{2}$. Thus if $n\gt N_1$, then $|a_n|\lt \frac{3|a|}{2}$, and therefore
$$|a_n|+|a|\lt \frac{5|a|}{2}.$$
Since the sequence $(a_n)$ has limit $a$, there is an $N_2$ such that if $n\gt N_2$, then
$$|a_n-a|\lt \frac{2\epsilon}{5|a|}.$$
Let $N=\max(N_1,N_2)$. If $n\gt N$ then $|a_n^2-a^2|\lt \epsilon$.
The argument for the case $a=0$ goes along similar lines, but is technically simpler. We leave it to you.
Remark: We gave the full "$\epsilon$-$N$" argument. More informally, the sequence $(|a_n|+|a|)$ is bounded, and $|a_n-a|\to 0$, so $|a_n^2-a^2|\to 0$. |
H: Clarification about negation in propositional logic
I am a little stumped on the concept of resolution, and want to clarify that something is correct, primarily negation.
if an expression in CNF is ${x = (a \lor b) \land (\lnot a \lor \lnot b)}$
It's negation should be: ${\lnot x = (\lnot a \land \lnot b \land a \land b)}$
I got that as follows:
${\lnot x = \lnot(a \lor b) \land \lnot(\lnot a \lor \lnot b)}$
${\lnot x = (\lnot a \land \lnot b) \land (a \land b)}$
${\lnot x = (\lnot a \land \lnot b \land a \land b)}$
Is this correct, because I am using this as the basis for something in a larger assignment. Which involves something I read here: http://logic.stanford.edu/classes/cs157/2009/notes/chap05.html about halfway down the page is this example:
${\frac{(p, q)\\(\lnot p, \lnot q)}{(p, \lnot p)\\(q, \lnot q)}}$
where () denotes a cluase and the --- denotes the resolution from above to below. From my understanding, they have taken a pair of clauses above that are satisfiable, and made a pair of clauses that are valid (satisfiable under any assignment) below, have I misunderstood this?
Thanks.
AI: Your initial negation is incorrect: the negation of
$${x = (a \lor b) \land (\lnot a \lor \lnot b)}$$
is
$$\begin{align*}
\neg\Big((a\lor b)\land(\neg a\lor\neg b)\Big)&\equiv\neg(a\lor b)\lor\neg(\neg a\lor\neg b)\\
&\equiv(\neg a\land\neg b)\lor(a\land b)\;.
\end{align*}$$
You failed to apply De Morgan’s law $\neg(p\land q)\equiv\neg p\lor\neg q$ correctly in your first step.
You can check this by informal reasoning. The original sentence $x$ says that at least one of $a$ and $b$ is true and at least one is false, so it says that exactly one of $a$ and $b$ is true. The final sentence in my derivation of the negation says that $a$ and $b$ are both false or both true, which clearly is the negation of the assertion that exactly one of them is true.
Added: The resolution
$$\begin{array}{}
\{p,q\}\\
\{\neg p,\neg q\}\\ \hline
\{p,\neg p\}\\
\{q,\neg q\}
\end{array}$$
is just an abbreviation for a pair of resolutions,
$$\begin{array}{}
\{p,q\}\\
\{\neg p,\neg q\}\\ \hline
\{p,\neg p\}
\end{array}$$
via elimination of $q$ and $\neg q$, and
$$\begin{array}{}
\{p,q\}\\
\{\neg p,\neg q\}\\ \hline
\{q,\neg q\}
\end{array}$$
via elimination of $p$ and $\neg p$. It amounts to saying that from $p\lor q$ and $\neg p\lor\neg q$ you may derive either $p\lor\neg p$ or $q\lor\neg q$, which is surely true, if not very interesting! |
H: Question about the Geometric Sequence Theorem
I am reading my professor's notes, and there is something that I'm not understanding about this. The Geometric Sequence Theorem states that if $r$ is a real number such that $|r| < 1$, then $\lim_{n\to\infty} r^n = 0$. I understand that well enough, but then she gives us an example that asks whether $(2^n/3^{n+1})$ is convergent.
She ends up writing:
\begin{align*}
\lim_{n \to infty} (a_n)
&= \lim_{n\to \infty}\frac{2^n}{3^{n+1}} \\
&= \frac 1 3 \lim_{n\to \infty}\frac{2^n}{3^n}
\end{align*}
and I'm wondering how in the WORLD she took a $1/3$ out of there.
Also, is the geometry sequence theorem basically stating that if $r$ is a fraction, then $r^n = 0$?
AI: Hint --I don't have enough points for a comment:
$3^{n+1}=3(3^n)$ , so that $\frac {2^n}{3^{n+1}}$ =$\frac {2^n}{3.3^n}=....$
And, be careful; if r is a fraction with ratio between -1 and 1 (non-inclusive), then
$r^n\rightarrow 0$ as $n \rightarrow \infty$ |
H: For any positive integer $n$, what is the value of $t^*$ that maximises the following expression?
For any positive integer n, what is the value of t* that maximises the following expression?
$$\displaystyle \sum_{j=1}^{n-t^*}\left(\frac{t^*-j+2}{t^*+j}\right)$$
where $t^*$ is some integer in the set $\{0,1,2...,n-1\}$.
Clearly $t^*$ = $f(n)$ but I am unable to find what this function is? E.g. $t^*=\frac{3}{5}(n-1)$ rounded to the nearest whole number is a decent estimate but not exact. Perhaps there is no closed-form solution. Thanks for your help!
AI: Reindex with $j\mapsto j-t^*$ and you have $$\sum_{j=1+t^*}^n\frac{2t^*-j+2}{j}$$ or just $$\sum_{j=1+t^*}^n\left(\frac{2t^*+2}{j}-1\right)$$ which is $$(2t^*+2)\sum_{j=1+t^*}^n\frac{1}{j}-(n-t^*-1)$$
The difference between this expression and the same expression with $t^*$ replaced by $t^*-1$ is $$(2t^*+2)\sum_{j=1+t^*}^n\frac{1}{j}-(n-t^*-1)-2t^*\sum_{j=t^*}^n\frac{1}{j}+(n-t^*)$$ which simplifies to $$2\sum_{j=1+t^*}^n\frac{1}{j}-1$$ The critical value for $t^*$ will happen when this difference is $0$. Either the floor or the ceiling of that number is the answer you are seeking. Representing $H(m)$ as the $m$th partial Harmonic summ, we are trying to solve $$2H(n)-2H(t^*)=1$$ or rather $$H(t^*)=H(n)-\frac{1}{2}$$ If you can find an inverse function for $H$, you have your answer. For large $n$, $H(n)\approx \log(n)+\gamma$, where $\gamma$ is the Euler-Mascheroni constant, so if we take that and run with it, $$t^*\approx\exp\left(\log(n)-\frac{1}{2}\right)=\frac{1}{\exp(1/2)}n\approx0.6065n$$ agreeing with your empirical estimate of $3/5$. This is more accurate the larger the value of $n$. |
H: Measure $\mu_f$ for function $f(x)=x^2$
Let $X=[0,1]$ and $\mu$ the Lebesgue measure. Describe the measure $\mu_f$ for the function $f(x)=x^2$.
By definition of $\mu_f$: For any Borel subset $A\subseteq R$, we have $$\mu_f(A)=\mu(f^{-1}(A)) = \mu(\{x\in[0,1]\mid x^2\in A\})$$
Is there anything that can/should be said about $\mu_f$? I'm not sure because of the vague problem statement.
AI: Perhaps you could say that
$$
\mu_f((a,b]) = \sqrt{b} - \sqrt{a}
$$
for any interval $(a,b] \subset [0,1]$; or that
$$
\mu_f(A) = \int_A \frac{1}{2\sqrt{x}}dx
$$
for any Borel set $A \subset [0,1]$? |
H: schwarz class and $L^2(\mathbb{R})$
Schwarz and $L^2$ both have the property that the Fourier transform is defined and bijective as a self-map of these spaces. Are they related in anyway or is this coincidence? (i.e. dual in some sense, or perhaps both arising from some more general construction.)
AI: $L^2$ is where it all starts; from it new such spaces (including $\mathcal S$, for Schwartz with the t) can be constructed in a routine way. Let $w$ be a locally integrable function on $\mathbb R$ such that $\inf w>0$. The space
$$S_w=\{f: wf\in L^2 \ \text{ and } \ w\hat f\in L^2\}$$
is a subset of $L^2$ on which the Fourier transform $\mathcal F$ is a bijection.
If $w\equiv 1$, we have $L^2$ itself.
Another example is given by $w(x)=(1+x^2)^{p/2}$, for any $p>0$.
The property of being mapped onto itself by $\mathcal F$ is preserved under intersection of function spaces. Taking the intersection of $S_w$ with $w(x)=(1+x^2)^{p/2}$ over all $p>0$, we get $\mathcal S$. (To see that this agrees with the usual definition, observe that derivatives remain in this intersection, and that the integral norm of a sufficiently high derivative controls the supremum of a function.)
We get something smaller than $\mathcal S$ with $w=\exp{|x|}$; this space is still nontrivial (contains the Gaussian function). I suspect that beyond $w=\exp(x^2)$ the space $S_w$ becomes trivial, but don't have a proof. |
H: Goldilocks Packing type problem
This is a resource allocation problem I am attempting to formulate myself, so bear with me this isn't from the 12th edition of some math book.
A miner is selecting 'rocks' from amongst his mine to haul back to the top. There are three components, or attributes, that define a rock; mineral_1, mineral_2, and mineral_3, location.
Example Rock A) 50 units mineral_1, 20 units mineral_2, 0 units mineral_3
Example Rock B) 10 units mineral_1,10 units mineral_2,10 units mineral_3
Example Rock C) 0 units mineral_1,0 units mineral_2, 95 units mineral_3
His cart can hold about** 300 units of rock, and at the top he wants to have about** 100 units of each rock component.
Example Cart 1)
Contains Rock A and C.
Holding 165 units total.
50 units mineral_1,
20 units mineral_2,
95 units mineral_3
His rocks are generally scattered about the caverns of the mine. Their removal comes at a cost of manpower. The manpower required to move a particular rock can be easily calculated as a (,say, linear) function of distance from the location of the cart.
One rock 100 yards away requires 100 rupees of man power, 99 yards, 99 rupees ... 98, 98...[nevermind other things like how much the rock weighs for example.]
**Now about the "about." Think of the cart like a grocery cart. There really is no hard limit on how much it can be stacked. For each case, choose an arbitrary limit "L" and an arbitrary minimum "M" > 0.
*For the individual components, there is a limit "l" for too many of a type of mineral, and a minimum "m" < 0 for too few of a type of mineral.
* The goal being always to get as close to the ideal 100,100,100 composition.
I want to know, given full knowledge of the composition of every rock in the cave, and its location, what is the best choice of rocks to haul back to the surface on this trip which minimizes cost.
Help translate to math? I want to avoid a subjective element, meaning I don't want to have to consider any more variables or elements, particularly as it pertains to a propensity to choose 'closeness' over cost. If this is unavoidable, I need help capturing its essence concisely.
The algorithm I've roughed out it Like this:
Divide each element of each rock by the cost associated with obtaining the whole rock. Now I have the rocks in terms of their elements per unit cost, calling them "Rc"
Assume no one rock alone can overflow any element. Meaning, no rock contains more than 100 of any mineral.
Cart = [mineral_1,mineral_2,mineral_3] = [0,0,0]
Choose the Rc with the greatest sum, and add it to the cart.
Cart = [0+Rc1,0+Rc2,0+Rc3]
Here is a difficulty in choosing. Simply adding the previous rock again will most likely (though not certainly) lead to an unbalance.
...
So I decided to create a scaler.
scaler = [(100-Cart1)/100,(100-Cart2)/100,(100-Cart3)/100]
To decide on the next rock to choose, I examine each Rc and do the dot product with scaler. I choose the Rc/rock that has the greatest dot product in this case.
And so on until I get reasonably close to the goal.
One thing I am issues with in this case is respecting both the cart boundary and the element boundaries.
AI: What you need to define is a utility function that incorporates the returned quantities and the cost to obtain them. How many rupees better is $100,100,100$ than $120,80,60$? Once you have that function and your constraints, this is a knapsack style problem-find the set of rocks that maximize utility. More commonly, there would be no maximum on any mineral, there would just be a price for each one, but that is a constraint like putting a maximum weight on the cart. The utility function sounds like it will be harder and more important than the optimization problem. |
H: Having trouble understanding this proposition from my textbook.
I'm seeing this perplexing proposition in my optimization textbook:
Suppose an LP
$$\max\{z(x)=\vec{c}^{T}x+\bar{z}:A\vec{x}=\vec{b},\vec{x}\geq\vec{0}\}$$
and a basis $B$ of $A$ are given. Then, the following linear program is an equivalent linear program in canonical form for the basis $B$,
$$\max\hspace{1em}z(x)=\vec{y}^{T}\vec{b}+\bar{z}+(\vec{c}^{T}-\vec{y}^{T}A)\vec{x}
$$subject to $$A_{B}^{-1}A\vec{x}=A_{B}^{-1}\vec{b}$$$$\vec{x}\geq\vec{0}$$where $\vec{y}=A_B^{-T}\vec{c}_B$.
Apparently, in this case "basis" means a set of columns in a vector which are linearly independent (rather than anything doing with spanning a vectorspace). $M_B$ denotes an ordered matrix of the columns of $M$ which are members of $B$.
My main trouble is two points:
What on earth is $A_B^{-T}$?
Why is $\vec{c}^{T}-\vec{y}^{T}A$ well defined? It seems very possible for $y$ and $c$ to have different lengths.
The book did not give a good proof of this, just a worked example with real numbers that happens to fit this pattern. Is there an approachable resource I can use to understand optimization of linear models better? Googling only gives resources way above my head, and the textbook for our course is badly written.
AI: $A_B^{-T}$ is a shorthand for $(A_B^T)^{-1}$ or equivalently, $(A_B^{-1})^T$.
Yes, $\vec{y}$ and $\vec{c}$ may have different lengths, but $\vec{c}^{T}$ and $\vec{y}^{T}A$ have identical lengths and so one can calculate their difference. For instance, you can compute $(1,2)-(3,4,5)\pmatrix{1&2\\ 3&4\\ 5&6}$ although $(1,2)$ is a 2-vector and $(3,4,5)$ is a 3-vector.
Multiplying $A_{B}^{-1}A\vec{x}=A_{B}^{-1}\vec{b}$ by $A_B$ on both sides, the constraint is equivalent to $A\vec{x}=\vec{b}$. Hence the new linear program has the same constraints as the old one. In addition,
\begin{align*}
\vec{y}^{T}\vec{b}+\bar{z}+(\vec{c}^{T}-\vec{y}^{T}A)\vec{x}
&=\vec{y}^{T}(\vec{b}-A\vec{x})+\bar{z}+\vec{c}^{T}\vec{x}\\
\end{align*}
Under the constraint $A\vec{x}=\vec{b}$, we have $\vec{y}^{T}(\vec{b}-A\vec{x})=0$ and the objective function in the new linear program is equal to the old one. Hence the two linear programs are equivalent. |
H: Finite cover with zero symmetric difference
The original problem is following: Let J be a finite subinterval in $\mathbb{R}$ and A be a measurable subset of J. Then for any $\epsilon > 0$, there exists a finite union of intervals B such that $d(A,B) = \mu^*((A - B) \cup (B - A)) < \epsilon$
I have solved this problem (posted below if you need hints). I try to put more restriction on B, that is, $B \supset A$. I have tried this problem for a while but without success. Could anyone help me with that problem? I feel I come really close to the solution, but somehow I still miss some details.
AI: It doesn't work with the restriction $B\supset A$. For example, if $A=\mathbb Q\cap[0,1]$, then for every finite union of intervals $B$ such that $B\supset A$, you can show that $d(A,B)=\mu(B)\geq 1$. |
H: Distance from the origin to a plane... where is my logic wrong here?
I'm going through MIT's multivariable calc online course, and came to the following recitation question:
Compute the distance from P = (0,0,0) to the plane with equation $2x+y-2z=4$
The correct answer is $\frac{4}{3}$, and the TA solves this by picking a random point on the plane, drawing a vector from the origin to that point, then computing the dot product between that vector and the normal vector $<2, 1, -2>$ to find the projection in the direction of the normal vector. Makes sense, okay.
But the way I did it originally still seems right to me, even though I'm getting the wrong answer. The shortest path to the plane is a perpendicular vector that starts at $(0,0,0)$ and ends on the plane. So I solved the simple equation below:
$$2\alpha + \alpha - 2\alpha = 4$$
No matter what $\alpha$ is, the left side of the equation will be parallel to the normal vector. In this case, $\alpha = 4$, and so I get the vector $<8, 4, -8>$. If you start it at the origin, it does indeed land on the plane with the correct normal vector direction. Then, distance = $\sqrt{8^2 + 4^2 + (-8)^2} = 12$. I'm clearly thinking about this incorrectly. But how?
AI: Taking a vector in the normal direction from the origin gives $\langle 2\alpha,\alpha,-2\alpha\rangle$. The condition that this vector ends on the plane is $2(2\alpha) + \alpha -2(-2\alpha) = 4$, i.e., $9\alpha = 4$, or $\alpha = \frac{4}{9}$. Thus the vector is $\langle \frac89, \frac49, \frac{-8}{9}\rangle$, with length $\frac{4}{3}$. |
H: Matrix math syntax in wolfram alpha
I'm having trouble getting Wolfram Alpha to do my bidding with regard to matrix manipulations. I am trying to take the derivative of the following matrix expression with respect to m, and was hoping WolframAlpha could be used:
-(x-m)^T * E^-1 * (x-m)
...However, I cannot discover how to input matrices as variables, and do not know what to use for the transpose operator. (^T was used in this case. x and m are vectors, E is a symmetric matrix.)
Give a man a fish: How can this derivative be solved?
Teach a man to fish: How can I input such expressions into Wolfram Alpha?
If it helps, I know the solution should look something like (but may be off by some scalar constant):
E^-1(x-m)
AI: You probably want:
$-$Transpose[x-m].Inverse[E].(x-m)
Although WolframAlpha might automatically think E is the number 2.71828.... |
H: How do I prove $2^{n+1} + 2n + 1 = 2^{n+2} - 1$
I am attempting to prove using induction:
$\sum_0^n 2i = 2^{n + 1} - 1$
I have gotten to the point where I need to show:
$2^{n+1} + 2n + 1 = 2^{n+2} - 1$
How do I prove this? Or should I be proving the initial question a different way
AI: You missed the power of $2$ in the Left hand side
and the range of $i$ should start from $0$ instead of $1$
We find $\displaystyle\sum_0^1 2^i =1+2=3 $ and $ 2^{1 + 1} - 1=4-1=3$
So, $\displaystyle\sum_0^n 2^i = 2^{n + 1} - 1$ is true for $n=1$
Let $\displaystyle\sum_0^n 2^i = 2^{n + 1} - 1$ is true for $n=m$
$$\implies \sum_0^m 2^i = 2^{m + 1} - 1$$
$$\implies \sum_0^{m+1}2^i =\sum_0^m 2^i +2^{m+1}= 2^{m + 1} - 1 +2^{m+1}=2^{m+1}(1+1)-1=2^{m+2}-1$$ |
H: Find the derivative of $f(x)=\frac{1}{\ln x}$ and approximate $f'(3.00)$ to 4 decimals.
Find the derivative of $f(x)=\frac{1}{\ln x}$ and approximate $f'(3.00)$ to 4 decimals.
I've been having trouble with this one for a while, I've been using quotient rule but I keep getting 3 when the answer should be a decimal answer
How do I solve it?
Please Help!!
AI: Noting that $f(x) = \left(\ln{x}\right)^{-1}$, we see that
$$f'(x) = -\left(\ln{x}\right)^{-2} \cdot \frac{1}{x} = -\frac{1}{x(\ln{x})^2}$$
Hence $$f'(3) = \frac{-1}{3(\ln{3})^2}$$ |
H: Isometries on the Banch Space M([0,1]) of regular Borel Measures
I'm trying to define an isometric isomorphism $T:M([0,1])\to M([0,1])$ that is not weak-star continuous (by $M([0,1])$ I mean the Banach space of regular Borel measures). How I can build one? One observation is that $T$ is not weak-star continuous if the adjoint mapping $T^*$ doesn't map $C([0,1])$ onto itself.
AI: Perhaps one could try the following: Take $\varphi:[0,1]\to [0,1]$ which permutes two points and leaves the others fixed. Given $\mu\in \mathcal{M}[0,1]$ and $A$ Borel, define $T\mu(A)=\mu(\varphi(A))$. One can check $w^*$ discontinuity by recalling the strong relationship between the topology of $[0,1]$ and the $w^*$ topology on the Dirac point evaluation functionals. |
H: If $\aleph_\lambda>2^{\aleph_0}$ is a limit cardinal, then $\aleph_\lambda^{\aleph_0}=\aleph_\lambda$?
I know that for successor cardinals, the result holds; i.e. if $\aleph_\alpha$ is a successor cardinal, then because this implies that it is a regular cardinal, then we can use this to show that if $\aleph_\alpha\geq 2^{\aleph_0}$, then $\aleph_\alpha^{\aleph_0}=\aleph_\alpha$. The proof of this can be found in For every $n < \omega$, $\aleph_n^{\aleph_0} = \max(\aleph_n,\aleph_0^{\aleph_0})$, and depends entirely on the fact that $\aleph_\alpha$ is regular.
Clearly, if $\aleph_\lambda$ is a limit cardinal, then the proof in the above cannot be used. It doesn't seem like assuming the the set of all limit ordinals for which the result is false is non-empty and then picking the least one is fruitful in finding a contradiction, since the fact that it is a limit ordinal stops us from easily looking elements smaller than it, though looking at ordinals smaller than the cofinality seems to be similar in concept to that of the successor cardinal case. However, to me it seems that in order to use this in much the same manner as in the above proof, the cofinality would need to be strictly larger than $\aleph_0$, which seems like a strong assumption to make.
Is this identity true, and if so, what approach should/could be used to prove it?
AI: This is not true. König's theorem tells us that for all cardinals $\kappa$, $\kappa < \kappa^{\text{cf}(\kappa)}$. So for any $\kappa$ with $\text{cf}(\kappa) = \aleph_0$, we have $\kappa^{\aleph_0} > \kappa$. And there are plenty of cardinals greater than $2^{\aleph_0}$ with cofinality $\aleph_0$. For an explicit example, if CH holds, take $\aleph_{\omega}$. |
H: Can someone help me please?
$A^+A=I$ where $A^{+}=(A^TA)^{-1}A^T$, $A_{m \times n}$
I have tried with $(A^TA)^{-1}=A^{-1}{A^T}^{-1}$ but the matrix is not squared
AI: Notice
$A^{+}A=((A^{T}A)^{-1}A^{T})A=(A^{T}A)^{-1}(A^{T}A)$ |
H: Prove $\sum\limits_{i=1}^{n}i\binom{n}{i}=n2^{n-1}$ using induction.
I have already derived the formula $\sum_{i=1}^{n}i\left(\begin{array}{c}n\\i
\end{array}\right)=n2^{n-1}$ directly just by doing some algebraic manipulations to the summand, which is indeed proves the validity of the formula. However, for the sake of a challenge I thought that I might try to prove this using induction, and it has turned into a nightmare! After checking the base cases and establishing the hypothesis $\sum_{i=1}^{k}i\left(\begin{array}{c}
k\\
i
\end{array}\right)=k2^{k-1}$ for all $k>1$ I get the following:
Step 1:
$$\sum_{i=1}^{k+1}i\left(\begin{array}{c}
k+1\\
i
\end{array}\right)=\sum_{i=1}^{k+1}i\left[\left(\begin{array}{c}
k\\
i-1
\end{array}\right)+\left(\begin{array}{c}
k\\
i
\end{array}\right)\right]$$
Step 2:
$$\sum_{i=1}^{k+1}i\left(\begin{array}{c}
k\\
i-1
\end{array}\right)+\sum_{i=1}^{k+1}i\left(\begin{array}{c}
k\\
i
\end{array}\right)=\sum_{i=1}^{k+1}i\left(\begin{array}{c}
k\\
i-1
\end{array}\right)+\sum_{i=1}^{k}i\left(\begin{array}{c}
k\\
i
\end{array}\right)+(k+1)\left(\begin{array}{c}
k\\
k+1
\end{array}\right)$$
Step 3:
$$\sum_{i=1}^{k+1}i\left(\begin{array}{c}
k\\
i-1
\end{array}\right)+\sum_{i=1}^{k}i\left(\begin{array}{c}
k\\
i
\end{array}\right)+(k+1)\left(\begin{array}{c}
k\\
k+1
\end{array}\right)=\sum_{i=1}^{k+1}i\left(\begin{array}{c}
k\\
i-1
\end{array}\right)+k2^{k-1}$$
Step 4: let $i=j+1$
$$\sum_{i=1}^{k+1}i\left(\begin{array}{c}
k\\
i-1
\end{array}\right)+k2^{k-1}=\sum_{j=0}^{k-1}(j+1)\left(\begin{array}{c}
k\\
j
\end{array}\right)+k2^{k-1}$$
Step 5:
$$\sum_{j=0}^{k-1}j\left(\begin{array}{c}
k\\
j
\end{array}\right)+\sum_{j=0}^{k-1}\left(\begin{array}{c}
k\\
j
\end{array}\right)+k2^{k-1}=\sum_{j=1}^{k-1}j\left(\begin{array}{c}
k\\
j
\end{array}\right)+k\sum_{j=0}^{k-1}\left(\begin{array}{c}
k-1\\
j
\end{array}\right)+k2^{k-1}$$
Step 6:
$$\left(\sum_{j=1}^{k}j\left(\begin{array}{c}
k\\
j
\end{array}\right)-k\left(\begin{array}{c}
k\\
k
\end{array}\right)\right)+k\sum_{j=0}^{k-1}\left(\begin{array}{c}
k-1\\
j
\end{array}\right)+k2^{k-1}=2^{k-1}-k+k2^{k-1}+k^{k-1}$$
I can't get what in need $(k+1)2^{k}$ from this, where did I go wrong?
AI: At step $4$ you got the wrong upper limit on the summation when you shifted the index. I make it:
$$\begin{align*}
\sum_{i=1}^{k+1}i\binom{k+1}i&=\sum_{i=1}^{k+1}i\left(\binom{k}{i-1}+\binom{k}i\right)\\
&=\sum_{i=1}^{k+1}i\binom{k}{i-1}+\sum_{i=1}^{k+1}i\binom{k}i\\
&=\sum_{i=1}^{k+1}i\binom{k}{i-1}+\sum_{i=1}^ki\binom{k}i\\
&=\sum_{i=0}^k(i+1)\binom{k}i+k2^{k-1}\\
&=\sum_{i=0}^ki\binom{k}i+\sum_{i=0}^k\binom{k}i+k2^{k-1}\\
&=\sum_{i=1}^ki\binom{k}i+2^k+k2^{k-1}\\\\
&=k2^{k-1}+2^k+k2^{k-1}\\\\
&=2k2^{k-1}+2^k\\\\
&=k2^k+2^k\\\\
&=(k+1)2^k
\end{align*}$$ |
H: Help with a step in Diestel's proof of Tutte's theorem in Graph Theory
The proof is given on page 8 of the pdf here which has page number 42.
We let $G=(V,E)$ be an edge-maximal graph without a 1-factor—that is, if we add any edge to $G$, the resulting graph has a 1-factor. We need to show that a set $S$ exists such that all components of $G-S$ are complete and every vertex $s\in S$ is adjacent to all the vertices of $G-S$.
Let $S$ be the set of all vertices of $G$ which are adjacent to every other vertex in $G$. Suppose that $S$ does not satisfy the above condition. Then there are two nonadjacent vertices $a,a'$ in some component of $G-S$. This implies there are vertices $a,b,c,d$ with $ab\in E$, $bc\in E$, but $ac\notin E$ and $bd\notin E$. Then there is a 1-factor $M_1$ of $G+ac$ and a 1-factor $M_2$ of $G+bd$. Now let $P=d\dots v$ be a maximal path in $G$ with its first edge in $M_1$ and all subsequent edges alternately in $M_2$ and $M_1$.
So far so good, but now we get to, "if the last edge of $P$ is in $M_1$ then $v=b$, since otherwise we could continue $P$". I just can't for the life of me see how this is true. How must it be true that $b$ is the only vertex incident with a matched edge in $M_1$ and no matched edge in $M_2$?
AI: $M_2$ is a 1-factor of $G+bd$, and $G$ does not have a 1-factor, so $bd$ must be an edge of $M_2$.
This means that the edges of $M_2$ that are in $G$ hit all vertices except $b$ and $d$.
So if we cannot continue $P$, it must end with either $b$ or $d$.
We also know that $P$ begins and ends with an edge of $M_1$.
Since $M_1$ only has independent edges, the last one cannot be incident with the first one,
in other words, the last one cannot hit $d$, since $P$ begins with $d$.
So $P$ must end with $b$. |
H: Find all primes of the form $n^n + 1$ less than $10^{19}$
Find all primes of the form $n^n + 1$ less than $10^{19}$
The first two primes are obvious: $n = 1, 2$ yields the primes $2, 5$. After that, it is clear that $n$ has to be even to yield an odd number.
So, $n = 2k \implies p = (2k)^{2k} + 1 \implies p-1 = (2k)^{k^2} = 2^{k^2}k^{k^2}$. All of these transformations don't seem to help. Is there any theorem I can use? Or is there something I'm missing?
AI: As you noted, we only have to check even $n\gt 2$. Note that $16^{16}$ is (barely) too big to be considered.
So we are only worried about $4, 6,\dots, 14$.
Let $n=km$ where $k$ is odd and $\gt 1$. Then $n^n+1=(n^m)^k+1$. This is divisible by $n^m+1$.
We have therefore eliminated all the even numbers in our range except $4$ and $8$. Your turn! |
H: Please explain Monte Carlo method
Generally I understand the idea of the Monte Carlo method.
However, when I read articles about it, there is always shown an example of calculating pi using a square, into which we insert 1/4th of a circle. Then we start putting randomly points into the square, and because the area of a 1/4 of the circle is pi/4, and the area of the square is exactly 1, we get that pi/4 = number of points in the circle / number of points in the square.
This simple example had been obvious for me, I'd liked it much, but one day I realized, that this gives correct (with some error of course) results because we know exactly what we want to get. So in this example we know that this square should be 1-sized.
Let us consider we know nothing about the function (in this case the circle) so we don't know which borders to choose. If we mistakenly take a 2x2 square, we won't get good result.
Ok, so we should first find the square. To find a square we need the function values, at least max and min. So I need to search them. But searching increases complexity. If I check each value for each x why just don't perform numerical integration in the same loop?
I understand the example of searching pi is trivial, however for me this does not make sense as we know how to fit data to get desired result.
(Please note I'm not a mathematician, but an engineer so I have some basis in maths, but I'm not good in advanced problems)
AI: This yields a binomial distribution for the points inside the circle. The probability that a point is inside the circle is
$p = \left(\pi/4\right)/\left(1\times 1\right) = \pi/4$. Then, the average number of points inside the circle is $\overline{n} = Np = N\pi/4$ where $N$ is the total number of points. That means
$$
\pi = 4\,{\overline{n} \over N}
$$
In a computer, you generate $N$ points inside the square $\left(~N \gg 1~\right)$
and counts how many are inside de circle ( that is $\overline{n}$ ). This little script ( in $\tt C++$ ) generates a Montecarlo $\pi$:
// piForSE0.cc
#include <cstdlib>
#include <iostream>
using namespace std;
typedef unsigned long long ullong;
const ullong N=1000000ULL; // One million
const double RANDMAX1=double(RAND_MAX) + 1.0;
inline double rand01()
{
return rand()/RANDMAX1;
}
int main()
{
double x,y;
ullong pointsInCircle=0;
for ( ullong n=0 ; n<N ; ++n ) {
x=rand01();
y=rand01();
if ( (x*x + y*y)<1.0 ) ++pointsInCircle;
}
cout<<"PI = "<<(4.0*(double(pointsInCircle)/N))<<endl;
return 0;
} |
H: Showing $[a, \infty) = \bigcap (a - \frac{1}{n}, \infty) $
I am trying to show: $[a, \infty) = \bigcap_{n=1}^{\infty} (a - \frac{1}{n}, \infty) $
If we take $x \in [a, \infty)$, then $x > a > a - \frac{1}{n} $ for all $n$. Hence, $x$ is in the intersection.
Similarly, if $x$ is in the intersection, then $x > a - \frac{1}{n} \implies x = a \implies x \geq a \implies x \in [a , \infty )$
Is this a valid solution? Can someone help me? Thanks
AI: The first half of your argument, showing that $[a,\infty)\subseteq\bigcap_{n\ge 1}\left(a-\frac1n,\infty\right)$, is correct, but the second half is not. It is certainly not true that $x>a-\frac1n$ implies that $x=a$, which is what you’ve written, though probably not what you meant. What is true is that if $x>a-\frac1n$ for all $n\in\Bbb Z^+$, then $x\ge a$; this is what you need, but it’s easier to prove it indirectly.
It’s probably easiest to show that $\bigcap_{n\ge 1}\left(a-\frac1n,\infty\right)\subseteq[a,\infty)$ by showing that if $x\notin[a,\infty),$, then $x\notin\bigcap_{n\ge 1}\left(a-\frac1n,\infty\right)$, so suppose that $x\notin[a,\infty)$; then $x<a$, so $a-x>0$. By the Archimedean property there in an $n\in\Bbb Z^+$ such that $\frac1n<a-x$, and hence $x<a-\frac1n$, and I expect that you can easily finish it from here. |
H: Prove the indication in Integration
Any body can help me to prove this problem in simple function?
Let if $g_n \geq 0$, $g_n \rightarrow g$ and $\int g_n d\mu \leq E \leq \infty$ then $\int g_n d\mu \leq E$
with $(S, \Sigma, \mu)$ as measurable function and $E\in \Sigma$.
Hope that this is will be clear, thanks..
AI: By Fatou's Lemma we have that since $g_{n}\ge0$ and $g_{n}\to g$ then
$\int gd\mu\le\lim\inf_{n\to\infty}\int g_{n}d\mu\le E$.
(I have assumed that you wanted to prove $\int gd\mu\le E$ since otherwise it follows directly from the statement of the question. Also I think you meant $E\in[0,\infty)$) |
H: Does every non-trivial subgroup of $S_9$ containing an odd permutation necessarily contain a transposition?
Does every non-trivial subgroup of $S_9$ containing an odd permutation necessarily contain a transposition?
Here $S_9$ denotes the group of all permutations (i.e. bijections with itself) of the set $\{1,2, \ldots, 9 \}$ under the binary operation of composition of functions.
By a transposition is meant a permutation that leaves all elements except two fixed, mapping either of these two onto the other.
A permutation can only be expressed either as a product of an even number of transpositions or as a product of an odd number of transpositions (but not both) and as such is defined to be even or odd, respectively.
Reference: Fraleigh p. 95 Question 23d. in A First Course in Abstract Algebra
AI: No. Take $\sigma=(1\ 2)(3\ 4)(5\ 6)$ and $H=\langle\sigma\rangle$. |
H: Power of a number in the difference of two factorials
What is the highest power of 3 available in $58! - 38!$ ( ! stands for factorial)
I can take $38!$ out as common to get $38! ( \frac{58!}{38!} - 1).$
I know how to find out the power of 3 in $38!$ But it is the difference term inside the brackets which I am not able to handle. What power of 3 will be contained in that term?
Is my approach correct in the first place? If yes then how to proceed further and if not then what approach should I take.
AI: Hint: $$\frac{58!}{38!} = 3\cdot 13\cdot\frac{58!}{39!}.$$ |
H: Example of an infinite complete lattice which is distributive but not complemented
Which is an example of an infinite complete lattice which is distributive but not complemented? Is the set of natural numbers under the relation divides an example? Also is the set of natural numbers under the usual $<=$ (less than or equal to) an example?
AI: Every complete lattice is necessarily bounded, since the set of all elements must have a join, and the empty set must have a meet. Your second example has no maximum element, so it’s not complete. If your $\Bbb N$ includes $0$, your first example is a bounded lattice, with $0$ as its maximum element; otherwise it’s not. (Note also that quite apart from the question of completeness, if you’re going to talk about lattices being complemented or not, you really ought to restrict yourself to bounded lattices anyway, so that complements are at least possible.)
You can turn $\Bbb N$ under the usual ordering into a bounded lattice by adding a top element $\top$: let $L=\Bbb N\cup\{\top\}$, where $n<\top$ for each $n\in\Bbb N$. Every linearly ordered lattice is distributive, so $L$ is distributive, and clearly no element of $L$ besides $0$ and $\top$ has a complement: if $n\in\Bbb Z^+$, then $n\land x=0$ if and only if $x=0$, while $n\lor x=\top$ if and only if $x=\top$. $L$ is also complete; the possible difficulty is with the join of an infinite set, and that’s always $\top$.
If your $\Bbb N$ does not $0$, you can simply add a top element, as I did above. The resulting lattice is complete, but it’s not complemented; e.g., there is no $n$ such that $2\land n=1$ and $2\lor n=0$. It is distributive, however, so it’s also an example. |
H: Arrangements of 1, 2, 2, 4, 6, 6, 6 greater than 3,000,000
How many numbers greater than 3,000,000 can be formed by arrangements of 1, 2, 2, 4, 6, 6, 6?
Any approach on how to solve this?
AI: HINT: You have only seven digits available, and $3,000,000$ is already a seven-digit number, so you’ll have to use all seven digits.
There are only two possibilities for the first digit; what are they?
Note that every seven-digit number that you can form with one of those first digits will be large enough. For each of the two possible first digits, calculate the number of distinct ways of arranging the other $6$ digits. |
H: Let $S$ be a set and $p$ a prime number and $m\in N$ such that $p$ does not divide $m$. Show $p$ does not divide $|S|$.
I'm currently doing an exercise where this should be shown:
Let $S$ be a set and $p$ a prime number and $m\in N$ such that $p$ does not divide $m$.
Suppose $|S| = p^rm(p^rm-1)...(p^rm-p^r+1)/{p^r(p^r-1)...1}$ is the size of $S$.
Show $p$ does not divide $|S|$.
Could someone tell me why this is true ?
Thanks in advance,
AI: Well, in fact
$$|S|=\frac{m(p^rm-1)(p^rm-2)\cdot\ldots\cdot(p^rm-p^r+1)}{(p^r-1)(p^r-2)\cdot\ldots\cdot 1}$$
Now just prove the easy
$$\forall\;k\in\Bbb N\,,\,1\le k\le p^r-1\;;\;\;p\mid(p^rm-k)\iff p\mid (p^r-k)$$
and deduce that $\;|S|\;$ has no factor equal to any non-zero positive power of $\;p\;$ . |
H: Solving system of linear eqaution in special cases
I have to solve for $Ax=B$. Here the diagonal elements of $A$ are $-1$ and all other elements are $1$. $A$ is $n \times n$ matrix . In this special case can we solve for $x$ quickly?
EDIT: quick is in terms of asymptotic complexity.
AI: Note that $A= e e^T -2I$, where $e=(1,1,...,1)^T$.
It is easy to check that $\sigma(A) = \{n-2,-2\}$, hence for $n \neq 2$, $A$ is invertible.
I will assume $n \neq 2$ subsequently.
Suppose $b = \alpha e + v$, where $v \bot e$. Then if we let $x = \frac{\alpha}{n-2} e - \frac{1}{2} v$, it is easy to check that $Ax = b$.
It is trivial to compute $\alpha = \frac{1}{n} \langle b , e \rangle$, and $v = b-\alpha e$ to get
$$ x = \frac{\langle b , e \rangle}{n(n-2)} e - \frac{1}{2} (b - \frac{1}{n} \langle b , e \rangle e) = \frac{\langle b , e \rangle}{2(n-2)} e - \frac{1}{2} b$$
I count $3n+2$ flops. |
H: Why does my calculator show $2^{-329} = 0?$
On my calculator, I usually get a $0$ when I divide something by $2$, a lot of times if that makes sense, but I was just wondering why does $2^{-329} = 0?$
AI: The precise name for the feature is underflow, which means it's less than the smallest number the registers can hold. It's kind of like overflow. |
H: Is $\text{fix}_\Omega(G_\alpha)$ a block of imprimitivity when $G$ is infinite?
Let $G$ be an infinite transitive permutation group acting on a set $\Omega$. Is $\text{fix}_\Omega(G_\alpha)$ a block of imprimitivity for $G$ in $\Omega$?
$G_\alpha$ is the set of elements of $G$ that fix $\alpha \in \Omega$
$\text{fix}_\Omega(G_\alpha)$ is the set of points in $\Omega$ fixed by $G_\alpha$.
A block of imprimitivity for $G$ in $\Omega$ is a set $B\subseteq \Omega$ such that for any $g \in G$, $B^g=B$ or $B^g \cap B = \emptyset$
I can show this in the finite case where $\text{fix}_\Omega(G_\alpha) = \alpha^{N_G(G_\alpha)}$, the orbit of $\alpha$ under the normalizer of $G_\alpha$, but in the infinite case I'm not sure if this is still true.
Thank you for your help.
AI: This is false in general, because there are infinite groups $G$ containing a subgroup $H$ and an element $y$ such that $K := y^{-1}Hy$ is a proper subgroup of $H$. An example is the Baumslag-Solitar group $\langle x,y \mid y^{-1}xy = x^2 \rangle$ with $H = \langle x \rangle$. (This can also be defined as a $2 \times 2$ group of matrices over ${\mathbb Q}$.)
Consider the transitive permutation action by right multiplication on the right cosets of $H$ in $G$. Let $\alpha=H$ and $\beta=Hy$. Then $G_\alpha=H$ and $G_\beta = y^{-1}Hy=K$, so $G_\beta$ is a proper subgroup of $G_\alpha$. Hence ${\rm Fix}(G_\alpha)$ is a proper subset of ${\rm Fix}(G_\beta)$ (it contains $\alpha$ but not $\beta$). But $y^{-1}$ maps ${\rm Fix}(G_\beta)$ to ${\rm Fix}(G_\alpha)$, so ${\rm Fix}(G_\beta)$ cannot be a block. |
H: What are the possible values of $a$ such that $f(x) = (x + a)(x + 1991) + 1$ has two integer roots?
What are the possible values of $a$ such that $f(x) = (x + a)(x + 1991) + 1$ has two integer roots?
$(x + a)(x + 1991) + 1 = x^2 + (1991 + a)x + (1991a + 1)$
This is of the form $ax^2 + bx + c$. Applying the quadratic formula $\left(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\right)$, we get, that the rooots must equal:
$\frac{-(1991 + a) \pm \sqrt{(1991 + a)^2 - 4(1991a + 1)}}{2} = \frac{-(1991 + a) \pm \sqrt{1991^2 + a^2 + 2\times1991a - 4\times1991a -4}}{2} = \frac{-(1991 + a) \pm \sqrt{(1991 - a)^2 - 2^2}}{2} = \frac{-(1991 + a) \pm \sqrt{(1991 - a - 2)(1991-a+2)}}{2}$
For the formula to yield an integer, the discriminant must be a perfect square or $0$. Clearly, the discriminant will be zero if $a = 1993$ or $a = 1989$. The only thing that remains to be shown is that these are indeed the only possible solutions.
I couldn't think of any way to do so. Are there any alternate solutions, perhaps some that involve more number theory and less algebra?
AI: If $(1991-a)^2-2^2=b^2$, say, then $(1991-a)^2-b^2=4$, and you have two squares that differ by $4$. Of course $2^2$ and $0^2$ work, as do $(-2)^2$ and $0^2$; these are the solutions that you already have. And they’re the only ones: if $x$ and $y$ are integers, and $x^2-y^2=4$, then $(x-y)(x+y)=4$. The factors $x-y$ and $x+y$ have the same parity, so they must both be even, and therefore either both are $2$, or both are $-2$. In either case one of $x$ and $y$ is $0$ and the other is $2$, giving you the solutions that you already have. |
H: Why is the restriction map $\mathcal{O}_X(U) \to \mathcal{O}_X(V)$ a flat morphism?
I am reading page 255 of Qing Liu and he claims that if $U,V$ are affine open subsets of a scheme $X$, then $\mathcal{O}_X(U) \to \mathcal{O}_X(V)$ is a flat morphism. Why is this necessarily the case? There are no finiteness assumptions or anything right now on $X$.
If $V = \operatorname{Spec} A_f$ and $U = \operatorname{Spec} A$ then the restriction map is just the canonical map $A \to A_f$ which is flat by exactness of localization. What about the general case of $V$ not necessarily being a basic open set?
AI: While reading page 255, your question is answered on page 136. Namely, we have the usual definition 3.1 of flatness. Then Proposition 3.3 tells us that open immersions are flat (which is trivial) and that a morphism of affine schemes $X \to Y$ is flat iff $\mathcal{O}(Y) \to \mathcal{O}(X)$ is a flat homomorphism (which is already proven on page 11 in Corollary 2.15). Hence, if $X \to Y$ is an open immersion of affine schemes, then $\mathcal{O}(Y) \to \mathcal{O}(X)$ is flat.
Remark: If $X \to Y$ is an open immersion between arbitrary schemes, then $\mathcal{O}(Y) \to \mathcal{O}(X)$ doesn't have to be flat. |
H: Whats the name of this function?
I read this function in an exercise. It looks quit familiar to me, however I do not know its name.
Whats the name of the $\rho_n$ function and who brought it up first?
AI: Here $\rho_{n}=B\left(\frac{1}{2},\frac{1}{2}n-1\right)^{-1}$. I
do not know a special name for it. |
H: How can a unit step function be differentiable??
Recently, I am taking a Signal & System course at my college. In all of the signal & system textbooks I have read, we see that it is written " When we differentiate a Unit Step Function, we get an Impulse function. " But as far as I have read, a unit step function is a piece-wise linear function as well as it is a continuous function but it is non differentiable. My question is that how can a non differentiable function be differentiated to obtain the impulse function?? Where is the contradiction.....
I am confused...would be happy if anybody helps...
For more info,you can see this link
AI: The derivative of the unit step function (or Heaviside function) is the Dirac delta, which is a generalized function (or a distribution). This wikipedia page on the Dirac delta function is quite informative on the matter.
One way to define the Dirac delta function is as a measure $\delta$ on $\mathbb{R}$ defined by
$$
\delta(A) = \begin{cases}
0 &: \text{ if } 0 \notin A \\
1 &: \text{ if } 0 \in A
\end{cases}
$$
Then one can write down precisely what is meant by the expression
$$
\int fd \delta = f(0)
$$ |
H: $f: [0,1]\rightarrow \mathbb{R}$ be an injective function, then :
Question is :
$f: [0,1]\rightarrow \mathbb{R}$ be a one one function, then which of the following statements are true?
$(a)$ $f$ must be onto
$(b)$ range of $f$ contains a rational number
$(c)$ range of $f$ contains an irrational number
$(d)$ range of $f$ contains both rational numbers and irrational numbers
I can see that $f$ need not be onto function by considering :
$f : [0,1]\rightarrow \mathbb{R}$ with $f(x)=x$ is one-one , but not onto.
Now, I should not take a continuous function to show a contradiction for $(b)/(c)/(d)$ because, any continuous function takes intervals to intervals and thus, Range of $f$ will have both rational and irrational numbers.
I recalled all functions i have thought could be discontinuous but those are not helping me at all.
I would be thankful if some one can help me to crack this problem and similar kind of problems.
THank You.
AI: Consider $f : [0, 1] \to \mathbb R$ defined as:
$$
f(x) = \begin{cases}
x &: x \text{ irrational} \\
x + \sqrt{2} &: x \text{ rational}
\end{cases}
$$
It is easy to show that this is an injective function from $[0, 1]$ into $\mathbb R - \mathbb Q$. |
H: Limits of integral: $\iiint_{D} \frac {\mathrm{d}x\mathrm{d}y\mathrm{d}z}{(x + y + z + 1)^3}$ , where $D =\{ x > 0 , y > 0 , z > 0 , x + y + z < 2\}$
What are the limits of the integral:
$$\iiint_{D} \frac {\mathrm{d}x\mathrm{d}y\mathrm{d}z}{(x + y + z + 1)^3}$$ where
$$ D =\{ x > 0 , y > 0 , z > 0 , x + y + z < 2\}$$
I have previously done double integrals and was wondering if someone could help me derive the limits of this triple one.
Help is appreciated!
AI: Hint: $(x,y,z)\in D$ if and only if $0<x<2$, $0<y<2-x$, and $0<z<2-x-y$. Then, $$\underset{D}{\iiint}\frac{\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z}{(x+y+z+1)^3}=\int_{x=0}^{x=2}\left[\int_{y=0}^{y=2-x}\left(\int_{z=0}^{z=2-x-y}\frac{1}{(x+y+z+1)^3}\mathrm{d}z\right)\mathrm{d}y\right]\mathrm{d}x.$$ Be careful: since the limits depend on other variables in some of the integrals on the right-hand side, the order of integration must not be changed! That this trick works is a consequence of Fubini's famous theorem.
Detailed calculations: $$\int_{z=0}^{z=2-x-y}\frac{1}{(x+y+z+1)^3}\mathrm{d}z=\left[-\frac{1}{2(x+y+z+1)^2}\right]_{z=0}^{z=2-x-y}=-\frac{1}{18}+\frac{1}{2(x+y+1)^2}.$$
\begin{align*}\int_{y=0}^{y=2-x}\left(-\frac{1}{18}+\frac{1}{2(x+y+1)^2}\right)\mathrm{d}y=&\left[-\frac{y}{18}-\frac{1}{2(1+x+y)}\right]_{y=0}^{y=2-x}\\=&\,\frac{x-2}{18}-\frac{1}{6}+\frac{1}{2(1+x)}\end{align*}
\begin{align*}\int_{x=0}^{x=2}\left(\frac{x-2}{18}-\frac{1}{6}+\frac{1}{2(1+x)}\right)\mathrm{d}x=&\left[\frac{\dfrac{x^2}{2}-2x}{18}-\frac{x}{6}+\frac{\log (1+x)}{2}\right]_{x=0}^{x=2}\\=&\,-\frac{2}{18}-\frac{2}{6}+\frac{\log 3}{2}-0=-\frac{4}{9}+\frac{\log3}{2}.\end{align*} |
H: Is this graph edge-connected?
I would like to know the following: is the following graph still connected even if every edges of two specific colors are eliminated, and is this true no matter how these two colors are chosen? If this graph doesn't satisfy the above condition, would you have some advice to make this graph to satisfy it?
The above graph has seven colors. If this graph doesn't satisfy the above condition, I think the graph with eight colors which my friend constructed will be the best graph which we can have. Yet I wanted to try a graph with seven colors.
Do you know any way to check this sort of edge-connectivity fast?
AI: Look at the long dark blue edge on the left.
Let's check if its end points are always connected: To avoid connection one of the forbidden colours must be dark blue and the other must be light blue or light green (path throuigh the bottom left). For both cases you easily find an explicit path.
Now check the lower left vertex. It suffices to show that it is connected to either of the two vertices we just considered (as they are always connected). Therefore we need only consider the case that light blue and light green are forbidden. You easily find a path for this case.
You can continue along this line of thought: To disconnect the lower right vertex, you need to forbid red and either dark blue or black, but then it is connected to the top vertex.
To disconnect the other vertex at the bottom, you'd now need to forbid black and dark blue and then can easily find a path to the left, for example. Similarly, for the remaining vertex on the perimeter, you'd need to forbid yellow and light blue and again find a path to "known territory".
Once you have such a firm base, it becomes more and more easy to extend vertex by vertex until you have checked the condition for the complete graph. |
H: union of countable many positive sets has a positive signed measure
Let $P_{1},P_{2},....$ be positive sets for $\nu$, then $P=\displaystyle\bigcup_{n=1}
^{\infty}P_{n}$ is a postive set for $\nu$
Proof:
Let $Q_{1}=P_{1},Q_{2}=P_{2}\setminus P_{1}$....$Q_{n}=P_{n}\setminus\displaystyle\bigcup_{j=1}^{n-1}P_{j}$ So all $Q_{n}'s$ are disjoint and $P=\bigcup_{j=1}^{\infty}Q_{j}$ since $Q_{j}\subset P_{j}$ for each $j$ then each $Q_{j}$ is postivie set for $\nu$.
So now we want to show that any subset of $P$ is positive for $\nu$.
let$ E\subset P$ then $\nu(E\cap Q_{n})>0$ and $\nu(E)=\sum_{j=1}^{\infty}\nu(E\cap Q_{j})\ge 0$
how do we get the last line? why is $\nu(E\cap Q_{n})>0$ is it because $Q_{n}$ is positivie for arbitrary $n$ and $E$ is a measurable set and a measure of any set is non-negative?
AI: $\nu(E\cap Q_n)>0$ in the last line clearly cannot be guaranteed without further information on $E$. For example, if $E$ is empty, then $\nu(E\cap Q_n)=0$. I'm pretty sure it's a typo.
The proof can be expanded on as follows. Let $E\subseteq P$ be a measurable set. Then, $$E=E\cap P=E\cap\left(\bigcup_{j=1}^{\infty} Q_j\right)=\bigcup_{j=1}^{\infty} (E\cap Q_j).$$ Since the $Q_j$'s are disjoint, so are the $E\cap Q_j$'s, so that $$\nu(E)=\nu\left(\bigcup_{j=1}^{\infty} (E\cap Q_j)\right)=\sum_{j=1}^{\infty}\nu(E\cap Q_j)\geq0,$$ where the inequality follows from the fact that $Q_j$ is a positive set for $\nu$ for all $j$ and $E\cap Q_j\subseteq Q_j$, so that $\nu(E\cap Q_j)\geq0$ for all $j$. |
H: how to find the elements of additive Group - $\mathbb{Z_7^+}$
I am given this additive Group G=$\mathbb{Z_7^+}$
I tried to find all its elements and I did:
$$gcd(1,7) = 1 \\ gcd(2,7) = 1 \\ gcd(3,7) = 1 \\ gcd(4,7) = 1 \\
gcd(5,7) = 1 \\ gcd(6,7) = 1 \\ gcd(7,7) = 7 \\ $$
Then I took all which gives $1$. so $1,2,3,4,5,6$ are all elements of this Group $G$. Am I right?
is it the only and good way of finding elements of given Group? regardless whether multiplicative or additive?
thanks a lot
AI: The gcd is irrelevant when you are looking for elements of $\mathbb{Z}_7$. The elements of this group are in reality subsets of $\mathbb{Z}$ : You say that two numbers $n$ and $m$ in $\mathbb{Z}$ are equivalent iff $7\mid (n-m)$. Then equivalence class of $n$ is
$$
[n] = \{m \in \mathbb{Z} : 7\mid (n-m)\}
$$
This is a subset of $\mathbb{Z}$. There are exactly 7 such subsets, because for any $n \in \mathbb{Z}$, $n$ is equivalent to its remainder upon division by 7. (For instance, $10$ is related to $3$, and so
$$
[10] = [3]
$$
The elements of $\mathbb{Z}_7$ are these equivalence classes
$$
\{[0], [1], \ldots, [6]\}
$$ |
H: Is $\Pr(X
Given that $X$ and $Y$ are non-negative random variables, and $a$ is a non-negative constant, is $\Pr(X<Y|X<a)\geq \Pr(X<Y)$? I mean, is that $X<a$ gives useful informaiton on the guess of $X<Y$?
Thanks a lot in advance.
AI: Not necessarily. It holds of course if $X$ and $Y$ are independent.
But if they are not independent, anything can happen. Suppose for example that $Y=X^2$ holds and $a=1$ ...
In the independent case, note that
$$ \mathrm P(X<Y \mid X<a) ~\ge~ \mathrm P(Y\ge a\mid X<a) ~=~ \mathrm P(Y\ge a\mid X\ge a) ~\ge~ \mathrm P(X<Y \mid X\ge a) $$
where the middle equals sign is because $X$ and $Y$ are independent.
Since $\mathrm P(X<Y)$ is a weighted average of the two extremes of this inequality, it cannot be larger than the left-hand side. |
H: Quadratic Manipulation
Is there some easy way to transform a quadratic equation like $ax^2+bx+c$ into a quadratic equation of the form $d(x+s)^2+e(x+s)$ where $a,b,c,d,e$ are constants and $a,d,e>0$
$-s$ is a root of the function
Thanks in advance
AI: NOTE: This is an answer to the problem OP asked before editing.
let $ax^2+bx+c=d(x+s)^2+e(x+s)$, compare coefficients and get the desired form
$ax^2+bx+c=d(x+s)^2+e(x+s)=d(x^2+s^2+2xs)+ex+es$ which gives the equations
$a=d$
$b=2sd+e$
$c=ds^2+es$ |
H: Constant functions are measurable
Let $f = C$ a constant, and I want to show $f$ is measurable.
In other words, if we take $(a, \infty)$, we show $f^{-1} (( a, \infty) )$ is measurable.
But $ f^{-1}((a, \infty))$ is just the real line $\mathbb{R}$, and therefore measurable? Am I right?
AI: That's one of the two parts.
But bear in mind, if $a \ge c$, then $f^{-1}[(a,\infty)] = \varnothing$. Of course, this doesn't alter the conclusion, since $\varnothing$ is also measurable. |
H: How to solve the limit $\lim_{n \to \infty} \sqrt[8]{n^2+1} - \sqrt[4]{n+1}$?
How do I show that this limit is zero?
$$\lim_{n \to \infty} \sqrt[8]{n^2+1} - \sqrt[4]{n+1} = 0$$
I've done the multiply by conjugates thing, which seems to lead nowhere:
$$\lim_{n \to \infty} (\sqrt[8]{n^2+1} - \sqrt[4]{n+1}) (\frac{\sqrt[8]{n^2+1} + \sqrt[4]{n+1}}{\sqrt[8]{n^2+1} + \sqrt[4]{n+1}})$$
$$=$$
$$\lim_{n \to \infty} \frac{\sqrt[4]{n^2+1} + \sqrt[2]{n+1}}{\sqrt[8]{n^2+1} + \sqrt[4]{n+1}}$$
I also considered using the squeeze theorem, as $\sqrt[8]{n^2+1} - \sqrt[4]{n+1} \leq \sqrt[8]{n^2+1} - \sqrt[4]{n}$, but I'm not sure how to bound it from below.
What's the correct approach to solving this limit?
AI: From identity
$$\sqrt[8]a-\sqrt[4]b=\frac{a-b^2}{(\sqrt[8]{a}+\sqrt[4]{b})(\sqrt[4]{a}+\sqrt{b})(\sqrt{a}+b)}$$
for $a=n^2+1$ and $b=n+1$ we have that
$$\lim_{n\to\infty}\sqrt[8]{n^2+1}-\sqrt[4]{n+1}=$$
$$=\lim_{n\to\infty}\frac{1}{\sqrt[8]{n^2+1}+\sqrt[4]{n+1}}\frac{1}{\sqrt[4]{n^2+1}+\sqrt{n+1}}\frac{n^2+1-{(n+1)^2}}{\sqrt{n^2+1}+n+1}=$$
$$=\lim_{n\to\infty}\frac{1}{\sqrt[8]{n^2+1}+\sqrt[4]{n+1}}\frac{1}{\sqrt[4]{n^2+1}+\sqrt{n+1}}\frac{-2}{\sqrt{1+1/n^2}+1+1/n}=0\cdot0\cdot(-2)=0$$ |
H: Sum of closed and compact set in a TVS
I am trying to prove: $A$ compact, $B$ closed $\Rightarrow A+B = \{a+b | a\in A, b\in B\}$ closed (exercise in Rudin's Functional Analysis), where $A$ and $B$ are subsets of a topological vector space $X$. In case $X=\mathbb{R}$ this is easy, using sequences. However, since I was told that using sequences in topology is "dangerous" (don't know why though), I am trying to prove this without using sequences (or nets, which I am not familiar with). Is this possible?
My attempt was to show that if $x\notin A+B$, then $x \notin \overline{A+B}$. In some way, assuming $x\in\overline{A+B}$ should then contradict $A$ being compact. I'm not sure how to fill in the details here though. Any suggestions on this, or am I thinking in the wrong direction here?
AI: Suppose $x\notin A+B$, then for each $a\in A$, $x \notin a+B$, which is a closed set (since $v \mapsto a+v$ is a homeomorphism). Since every TVS is regular, there are open sets $U_a$ and $V_a$ such that
$$
x\in U_a, \quad a+B \subset V_a, \quad \text{ and } U_a\cap V_a = \emptyset
$$
Now,
$$
V_a - B = \cup_{b\in B} (V_a - b)
$$
is open and contains $a$. Hence, $A\subset \cup_{a\in A}(V_a - B)$. Since $A$ is compact, there is a finite set $\{a_1, a_2, \ldots a_n\}$ such that
$$
A \subset \cup_{i=1}^n (V_{a_i} - B)
$$
Let $U = \cap_{i=1}^n U_{a_i}$, then $U$ is a neighbourhood of $x$.
We claim that $U\cap (A+B) = \emptyset$. If not, then $y = a+b \in U\cap(A+B)$, then
$$
y \in V_{a_i} \quad\text{ for some } i
$$
and $y \in U_{a_i}$, which is a contradiction. |
H: Variance and covariance function of stochastic process
Let $X_t = Z_t + \theta Z_{t-1}$ where $ \left\{ Z_t \right\} \approx WN(0, \sigma^2)$. Find variance $ VarX_t$ and covariance function.
Of course we have $EX_t = 0$. Then $VarX_t = EX_t^2 = E(Z_t^2 + 2 \theta Z_t Z_{t-1} + \theta^2Z_{t-1}^2) = \sigma^2 + \theta^2 \sigma^2$
I can't understand the last equal.
I suppose that $E(Z_t^2 + 2 \theta Z_t Z_{t-1} + \theta^2Z_{t-1}^2) = EZ_t^2 + 2 \theta EZ_tZ_{t-1} + \theta^2 EZ_{t-1}^2 = E(Z_t - EZ_t)^2 + 2 \theta EZ_tZ_{t-1} + \theta^2 E(Z_{t-1} - EZ_{t-1})^2 = \sigma^2 + 2 \theta EZ_tZ_{t-1} + \theta^2 \sigma^2 $.
Hence I suppose that $EZ_tZ_{t-1} = 0$ but why? $Z_t$ and $Z_{t-1}$ are independent?
If you explain me that, maybe i will know how find covariance function.
Thanks in advance!
AI: The W in WN$(0,\sigma^2)$ stands for "white" as in "white noise" and means that the process $(Z_t)$ is independent. Since $(Z_t)$ is also centered, yes, $E[Z_tZ_{t-1}]=0$ and yes, $E[Z_t^2]=\sigma^2$. |
H: Evaluating $\int{\frac{1}{\sqrt{x^2+y^2}}\mathrm dx}$
Attempting to calculate $\displaystyle \int{\dfrac{1}{\sqrt{x^2+y^2}}\mathrm dx}$,
$$\int{\dfrac{1}{\sqrt{x^2+y^2}}\mathrm dx}=\int{\frac{1}{\sqrt{(y\tan\theta)^2+y^2}}y\sec^2\theta \mathrm d\theta}=\int{\sec\theta d\theta}=\ln(\sec\theta +\tan\theta)=\ln\left(\sqrt{\left(\frac{x}{y}\right)^2+1}+\frac{x}{y}\right)=\ln\left(\frac{1}{x}\left(\sqrt{x^2+y^2}+x\right)\right),$$
where $x=y\tan\theta$
However Wolfram Integrator somehow returns
$$\ln\left(2\left(\sqrt{x^2+y^2}+x\right)\right)$$
as the answer. Where did I go wrong? Many thanks.
AI: Hint: after you fix the mistake in the last step, differentiate the function $$x\mapsto \ln\left(\frac{1}{y}\left(\sqrt{x^2+y^2}+x\right)\right)-\ln\left(2\left(\sqrt{x^2+y^2}+x\right)\right)$$ with respect to $x$. It's easy to do so in your head. Also do not forget that you should consider the absolute value appropriately ($\int \frac 1x\mathrm dx)=\ln (|x|)$) and to add an arbitrary constant when you integrate. |
H: Infinite Dihedral Group.
Let $D_{\infty}= \langle x,y \mid x^2=y^2=1\rangle$ be the infinite dihedral group. Are the following statements true?
Since $G$ is not torsionfree, $\mathbb{Q}[G]$ is not a domain.
$D_{\infty}$ is an infinite subgroup of $G= \langle x,y \mid x^2=y^2\rangle$.
The infinite dihedral group has a free abelian subgroup $F$ generated by $\langle x y \rangle$ of rank one and index $2$, and thus normal. $F$ is also subgroup of $G$.
AI: Yes, it is not a domain.
No, it is a factor group.
Yes. |
H: Are coefficients and values for x in F[x] in the same set?
I'm trying to understand the construction of $F[x]$ where $F$ is a field. As far as I understand it now, all coefficients and roots for all polynomials $f(x) \in F[x]$ lies in $F$. But what about the domain for the polynomials? The set $F[x]$ contains polynomials, but as functions when evaluating any polynomial, do we care how $f$ is defined or must it be that $f:F \to F$ for the field F ?
For example, say $F=Q$. A polynomial $f(x) = x^2-2$ that has a root $\sqrt{2}$, is that polynomial excluded from $F[x]$ or is it just that we really don't care about what $f(x)$ evaluates to?
AI: No. You don't exclude $x^2-2$ from $\mathbb{Q}[x]$. You only need the coefficients to be in $\mathbb{Q}$, not the roots.
The polynomials in $F[x]$ are merely symbols. When you start thinking of them as functions, then the question of domain pops up - this is where field extensions play a role (you can enlarge the domain of a polynomial so that it may now have a root in that new domain).
However, until then, treat them as symbols ($F[x]$ is a vector space over $F$, if you will, which basis elements that can be multiplied together) |
H: Proving a Logic Equation
I have two information.
$x+y = 1$ and $xy = 0$.
Now,I need to prove this equation : $xz + x'y + yz = y + z$
What I tried:
$z(x+y) + x'y = z + x'y$
Thats all
What do you think?
AI: xz+x′y+yz = xz+ xy + x′y+yz .... (adding xy=0)
= xz + y(x + x') +yz
= xz + y + yz .....( x+x' always = 1)
= z(x+y) + y ... rearranging terms
= z + y ........... as x + y = 1 is given. |
H: Fourier Series: going from $a_n$ and $b_n$ to $c_n$
I sort of understand the principle of the Fourier series, but when I watch the wiki page I don't understand how to get from:
${a_0 \over 2} + \sum_{n=1}^N[a_n cos({2\pi n x \over P}) + b_n sin({2\pi n x \over P})]$
to
$\sum_{n=-N}^N c_n e^{i{2\pi n x \over P}}$
To be clear what I don't understand is the transition from using the coefficients $a_n$ and $b_n$ to $c_n$. I am familiar with the Euler's formula. I noticed that the sum goes from -N to N in the second equation. In the article they write:
$\begin{align}
a_n & = c_n + c_{-n} \\[12pt]
b_n & = i(c_n - c_{-n}) \\[12pt]
c_n & = \begin{cases}
\frac{1}{2}(a_n - i b_n) & \text{for } n \ne 0, \\[12pt]
\frac{1}{2}a_0 & \text{for }n = 0.
\end{cases}
\end{align}$
So to get an intuition I tried to take a simple example in which N = 4 for example. I assume if I was taking n=-1 and n=1 for example, summing the terms using the above relationships, etc. I would get back to $a_1 cos(...) + b_1 sin(...)$. But I didn't succeed. I realise it probably takes quite some time to show how to get from one to another, but I would be grateful if someone could show me or at least put me on the right track.
Thank you.
AI: We need consider only one $n > 0$ (the case $n = 0$ is easily verified). In the exponential form, we consider the two terms with indices $n$ and $-n$,
$$\begin{align}
c_n e^{2\pi inx/P} + c_{-n} e^{-2\pi inx/P} &= c_n (\cos (2\pi nx/P) + i\sin (2\pi nx/P)) + c_{-n}(\cos (2\pi nx/P) - i \sin (2\pi nx/P))\\
&= (c_n + c_{-n})\cos (2\pi nx/P) + i(c_n - c_{-n})\sin (2\pi nx/P),
\end{align}$$
so $a_n = c_n + c_{-n}$, and $b_n = i(c_n - c_{-n})$. In the trigonometric form, we consider
$$\begin{align}
a_n \cos (2\pi nx/P) + b_n\sin (2\pi nx/P) &= \frac{a_n}{2}\left(e^{2\pi inx/P} + e^{-2\pi inx/P}\right) + \frac{b_n}{2i}\left(e^{2\pi inx/P} - e^{-2\pi inx/P}\right)\\
&= \frac{a_n - ib_n}{2}e^{2\pi inx/P} + \frac{a_n + ib_n}{2}e^{-2\pi inx/P}
\end{align},$$
so we get $c_n = \frac{a_n-ib_n}{2}$ and $c_{-n} = \frac{a_n + ib_n}{2}$. |
H: Prove that there exist no positive integers $m$ and $n$ for which $m^2+m+1=n^2$
The problem: Prove that there exist no positive integers $m$ and $n$ for which $m^2+m+1=n^2$.
This is part of an introductory course to proofs, so at this point, the mathematical machinery should not be too involved. This is supposed to be proven by contradiction. I've been messing around with this for a bit and can't help but feel that I'm missing something completely obvious. My first instinct was to evaluate this case by case based off of combinations of even and odd for m and n, which led to contradictions in the case that both m and n are even or that m is odd and n is even (the contradiction being that zero equates to an odd number). The problem comes when trying to find a contradiction where n is a positive odd number and m is either even or odd.
I then tried to approach it by showing that if $n^2=m^2+m+1$ and n is a positive integer, that $(m^2+m+1)^{\frac{1}{2}}$ must be a positive integer, and that this leads to a contradiction. It certainly looks like this will be the case, since, for the first few positive integer values of m, we get $3^{\frac{1}{2}}$,$7^{\frac{1}{2}}$,$13^{\frac{1}{2}}$,$21^{\frac{1}{2}}$, none of which are positive integers, but at this point at least, I don't know how to demonstrate this with a proof.
A gentle nod in the correct direction would be greatly appreciated. Just starting off with this stuff, so any help/insight would be great.
AI: Hint: If such $n$ existed, then it would have to be between $m$ and $m+1$ since $$m^2<m^2+m+1<m^2+2m+1=(m+1)^2.$$ |
H: Difference between parallel and orthogonal projections
i would like to understand what is a difference between parallel and orthogonal projection?let us consider following picture
We have two non othogonal basis and vector A with coordinates($7$,$2$),i would like to find parallel projection of this vector to these basis,i am studying Covariant and Contrivant components,so i would like to understand how to find parallel projection and also orthogonal projection?according to wikipedia, orthogonal projection is defined by
http://en.wikipedia.org/wiki/Vector_projection
what about parallel projection?please help me
AI: In geometric terms ...
In a parallel projection, points are projected (onto some plane) in a direction that is parallel to some fixed given vector.
In an orthogonal projection, points are projected (onto some plane) in a direction that is normal to the plane.
So, all orthogonal projections are parallel projections, but not vice versa. A parallel projection that is not an orthogonal projection is called an "oblique" projection.
This could all be translated into the language of linear algebra, I suppose, but I don't think that would make it any clearer. |
H: show that function is convex
Let $f:\mathbb{R}\to\overline{\mathbb{R}}$. Show that
$$f\left(x\right)=\begin{cases}
+\infty & \mbox{ if }x\in\left(0,\infty\right)\\
0 & \mbox{ if }x=0\\
-\infty & \mbox{ if }x\in\left(-\infty,0\right)
\end{cases}$$
is convex.
AI: define $+\infty−\infty=0$
In that case the function is not convex. We have
$$\infty = f\left(\frac13\right) = f \left(\frac23\cdot 1 + \frac13\cdot (-1)\right),$$
but
$$\frac23f(1) + \frac13 f(-1) = \frac23\infty + \frac13(-\infty) = \infty - \infty = 0 < \infty$$
with that definition. For a convex function, we'd have $f(1/3) \leqslant \frac23 f(1) + \frac13 f(-1)$. |
H: Finding normal vector of point $(x,y)$ given $f(x, y)$
Ignoring the codes, this is more or less a calculus question. Given an equation of a surface $z = x^2 + y^2$, I need to find normal vector of a point. I am not good at maths ... so ...
For (a), I believe the $\frac{\partial z}{\partial x}$ is equation of tangent of curve created by plane $y=c$ (probably wrong strictly speaking but you get the idea) with surface $f(x, y)$. Similarly for $\frac{\partial z}{\partial y}$ is equation of tangent of curve created by plane $x=c$ with surface $f(x, y)$
For (b) is tangent vector the same as the equations found in (a)?
For (c), do I do a cross product of the partial derivatives? If so, what I don't understand is how do I map the equations to a "matrix notation form" that I am familiar to do cross product?
AI: For (a), your thinking is correct.
For part (b), the tangent vector corresponding to $\frac{\partial z}{\partial x}$ is
$$
z_x = \left(1, 0, \frac{\partial z}{\partial x} \right)
$$
Similarly, the tangent vector corresponding to $\frac{\partial z}{\partial y}$ is
$$
z_y = \left(0, 1, \frac{\partial z}{\partial y} \right)
$$
For part (c), your idea of taking a cross product is correct. The surface normal is $z_x \times z_y$. You may have to unitize this vector before passing it to OpenGL.
In computer graphics, it's more common for a surface to be given by parametric equations $\mathbf{S} = \mathbf{S}(u,v)$, rather than in the form $z=f(x,y)$. Then the surface normal is simply the cross product of the partial derivative vectors:
$$
\mathbf{N} = \frac{\partial \mathbf{S}}{\partial u} \times
\frac{\partial \mathbf{S}}{\partial v}
$$
What we have here can be considered a special case where $u=x$, $v=y$, and $\mathbf{S}(x,y) = (x, y, z(x,y))$. |
H: category of linear maps
Let $V,W$ be vector spaces. Let's define a category whose objects are linear map $f:V\to W$ and morphisms from $f$ to $g$ are pair of linear maps $(\alpha,\beta)$ where $\alpha:V\to V,\beta :W\to W$ such that $g \circ\alpha = \beta \circ f$.
Does this category have some special name?
If $f_i$ are indexed family of objects, is there product of them in this category?
AI: As for this category having cartesian products: it's a very strange question, but the answer is 'no', unless both $V$ and $W$ are trivial ($0$-dimensional).
The basic problem can be illustrated by asking: does this category have a terminal object? Suppose $f: V \to W$ is terminal. This means that for any $h: V \to W$, there is exactly one pair of morphisms $g_1: V \to V$, $g_2: W \to W$ such that $g_2 \circ h = f \circ g_1$. Of course, by taking $g_1$ and $g_2$ both to be zero maps, there is always at least one such pair. But suppose we consider when $h: V \to W$ is the zero map. In that case every pair $(0, g_2)$ works. Thus we get uniqueness of the pair only if $W$ is trivial. But then $f = 0$, and now every pair $(g_1, 0)$ works, and we get uniqueness only if also $V$ is trivial.
The basic principle is that this category of linear maps admits products insofar as the full subcategory consisting of the single object $V$ (and similarly, the one for $W$) has products. For example, the category of linear maps has binary products if and only if we have isomorphisms $\phi: V \to V \times V$ and $\psi: W \to W \times W$, which is the case iff $V$ and $W$ are infinite-dimensional (let's agree to ignore cases where $V$ or $W$ is trivial). I.e., the composites
$$p_i = (V \stackrel{\phi}{\to} V \times V \stackrel{\pi_i}{\to} V), i = 1, 2$$
give product projections $p_1, p_2: V \to V$ in the full subcategory determined by $V$. If $q_1, q_2: W \to W$ are similar product projections for $W$, then the pairs $(p_1, q_1)$, $(p_2, q_2)$ furnish product projections for the category of linear maps. This is not hard to check. Of course, by cardinality considerations, we cannot have arbitrary products for any full subcategory determined by $V$ (again, except for the trivial case), since there will be no isomorphism $V \to V^S$ for a sufficiently high power $S$, and this carries over to the category of linear maps as well. |
H: Is this a justified expression for $\int \lfloor x^2 \rfloor \, \mathrm{d}x$?
By observing patterns in Riemann sums for the following integral, I'm convinced that $$\int_0^\sqrt{n} \lfloor x^2 \rfloor \, \mathrm{d}x = (n-1)\sqrt{n}-\sqrt{n-1}-\sqrt{n-2}-\cdots-\sqrt{1},$$ with $n$ a positive integer. (We choose $\sqrt{n}$ as the upper limit because $\lfloor x^2 \rfloor$ changes its $y$ at $x=\sqrt{n}$.)
Is this a correct assertion? Can this be used to find $\int_0^n \lfloor x^2 \rfloor \, \mathrm{d}x$?
AI: I think it's easier to just evaluate the integral directly, noting that there is a step at each square root of an integer:
$$\int_0^{\sqrt{n}} dx \, \lfloor x^2 \rfloor = \sum_{k=0}^{n-1} k \left ( \sqrt{k+1}-\sqrt{k}\right)$$
The stated result follows from summation by parts:
$$\sum_{k=0}^{n-1} k \left ( \sqrt{k+1}-\sqrt{k}\right) = n \sqrt{n} - \sum_{k=0}^{n-1} \sqrt{k+1}$$ |
H: Related Rates question from Pure Mathematics 1 by Hugh Neil
There is a question in my Pure Maths book which seems to confuse me. Ive done the rest but somehow this question seems to confuse me I think enough info is not given.
A water tank has a rectangular base 1.5m by 1.2m. The sides are vertical and water is being added to the tank at a constant rate of 0.45m^3 per minute. At what rate is the depth of water in the tank increasing?
AI: You don't need calculus for this.
Every minute $0.45 \text{ m}^3$ of water is added. Since $1.5\times1.2 \text{ m}^2$ is the area of the base, $0.45 \text{ m}^3$ of water will have height $\dfrac{0.45 \text{ m}^3}{1.5\times1.2 \text{ m}^2}=0.25 \text{ m}$. Thus the required rate is $0.25 \text{ m min}^{-1}$. |
H: Divisors of zero in $ \mathbb Z_{p^k}$
Let $p$ be a prime numer and let $k$ be a natural numer such that $k\geq 2$. I wish to descripe all zero's divisors in $\mathbb Z_{p^k}$. Obviously elements of the form
$np$, where $n=0,...,p^{k-1}-1$, are zero divisors, because $p^k|np^k$.
Are there others?
AI: Suppose $ab=0$, and $p\nmid a$. Well $p^k|ab$, so $p^k|b$, so $b=0$. Hence you found them all. |
H: Let $f,g$ be two distinct functions from $[0,1]$ to $(0, +\infty)$ such that $\int_{0}^{1} g = \int_{0}^{1} f $.
Let $f,g$ be two continuous, distinct functions from $[0,1]$ to $(0, +\infty)$ such that $\int_{0}^{1} g = \int_{0}^{1} f $.
Given $n\in \mathbb{N},$ let $y_n = \int_{0}^{1} \frac{f^{(n+1)}}{g^{(n)}} $ How do I show that $(y_n)$ is increasing and divergent?
Appreciate all advice. Thank you.
AI: Let $A = \{ x \in [0,1] : f(x) > g(x)\}$, $B = \{ x \in [0,1] : f(x) < g(x)\}$ and $C = [0,1] \setminus (A\cup B)$. Then since $f \neq g$ and $\int_0^1 f(x)\,dx = \int_0^1 g(x)\,dx$, we have
$$\int_A f\,dx + \int_B f\,dx = \int_A g\,dx + \int_B g\,dx \iff \int_A (f-g)\,dx = \int_B (g-f)\,dx,$$
and $\int_A (f-g)\,dx > 0$. Now,
$$\begin{align}
y_{n} - y_{n-1} &= \int_0^1 \frac{f^n(f-g)}{g^n}\,dx\\
&= \underbrace{\int_A \left(\frac{f}{g}\right)^n(f-g)\,dx}_{a_n} - \underbrace{\int_B \left(\frac{f}{g}\right)^n(g-f)\, dx}_{b_n}.
\end{align}$$
On $A$, we have $\frac{f}{g} > 1$, and on $B$, we have $\frac{f}{g} < 1$, so $a_n > a_0 = b_0 > b_n$ for $n > 0$. Hence $y_n > y_{n-1}$. By the monotone convergence theorem. $\lim\limits_{n\to \infty} b_n = 0$ and $\lim\limits_{n\to\infty} a_n = \infty$, hence $\lim\limits_{n\to\infty} (y_n-y_{n-1}) = \infty$. |
H: What's the fastest way to solve this inequality?
$|x^2-3x+2|>|x|+1$
Thanks in advance.
AI: $x^2-3x+2=(x-1)(x-2)$.
Thus
If $x \in (-\infty, 0]$ the inequality becomes....
If $x \in (0, 1]$ the inequality becomes....
If $x \in (1, 2]$ the inequality becomes....
If $x \in (2, \infty]$ the inequality becomes....
A faster way would be to draw the two graphs, since they are easy to draw. |
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