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H: Volume and surface area of a sphere
The volume of a spherical balloon increases by $1cm^3$ every second. What is the rate of growth of the radius when the surface area of the balloon is $100cm^2$
The surface area of a sphere is $4\pi r^2$, and its volume is $\dfrac{4}{3}\pi r^3$.
The answer sheet states that $\dfrac{dV}{dt} =1$, and we need to find $\dfrac{dr}{dt}$, but I don't understand this, can anyone explain?
AI: $\dfrac{dV}{dt}$ is the rate of change of the volume. In particular, $\dfrac{dV}{dt}=1$ cm per second. $\dfrac{dr}{dt}$ is the rate of change of the radius, which is what we want to know.
Now, the surface area of the sphere is actually $4\pi r^2.$ Observe, then, that $$\frac{dV}{dt}=\frac{d}{dt}\left[\frac43\pi r^3\right]=\frac43\pi\frac{d}{dt}\left[r^3\right]=\frac43\pi\cdot 3r^2\frac{dr}{dt}=4\pi r^2\frac{dr}{dt}.$$ Can you take it from there? |
H: Confustion with Fourier Transform of harmonic functions
Note: this is homework.
Say we assume a time-varying momentum is of the form
$p(t) = \Re \{p(w) e^{-iwt} \}$
Now we would like to know the solution of the following equation, in frequency space:
$ \tfrac{d}{dt} p(t) = -\tfrac{1}{\tau}p(t) - eE(t) $
The part where I get confused is the fourier transform of $ \tfrac{d}{dt} p(t) $. I thought the Fourier transform of the derivative, ie $\tfrac{d}{dt} p(t)$, is $ (iw) p(w) $. However, in the solutions to the problem, it is $(-iw)p(w)$. Further more, Wolfram Alpha outputs the same.
Is this because of how we assume $p(t)$ to look like? Or does my class and Wolfram Alpha define the Fourier Transformation differently than what I'm used to?
AI: This is most likely due to a sign convention mismatch in the definition of your Fourier transform and your reference's definition. Some like to define the Fourier transform as
$$
\mathcal{F}[f](w)=\frac{1}{\sqrt{2\pi}}\int_\mathbb{R}e^{-i\omega t} f(t)dt
$$
and others as
$$
\mathcal{F}[f](w)=\frac{1}{\sqrt{2\pi}}\int_\mathbb{R}e^{i\omega t} f(t)dt,
$$
which will change the sign of the formula for the transform of the derivative of a function. |
H: Sum the following series, where $e^y < 1$
Sum the following series: $$ 1 + e^y + e^{2y} + e^{3y} + .. $$ where $e^y < 1$
So $e^y$ is less than one, that means $y$ is a negative number, which means that each term will be contributing less. But I find this to be a very vague question. Can you actually say that the sum is just $1$?
AI: This is a geometric series:
$$
\frac{1}{1-e^y} = 1+e^y+e^{2y}+\cdots
$$
which converges uniformly if $e^y<1$. |
H: Solving nested summation $\sum_{i=0}^{n-1} \sum_{j=i}^{n-1}p(A_i)p(B_j) $
I am having trouble solving the following nested summation:
$$\sum_{i=0}^{n-1} \sum_{j=i}^{n-1}p(A_i)p(B_j) $$
where $p(A_i) = \frac{1}{n}$, and $n$ is a constant (length of an array). Same goes for $p(B_i) = \frac{1}{n}$.
I tried rewriting it into this form:
$$\sum_{i=0}^{n-1} \left(p(A_i)\sum_{j=i}^{n-1}p(B_j)\right) $$ as $p(A_i)$ doesn't depend on the index j, however I still have no idea what to do with the inner sum.
AI: First, $\sum_{j=i}^{n-1}p(B_j)$ means you're adding up $n-i$ $\frac{1}{n}$'s, so this sum becomes just $\frac{n-i}{n}$. So your desired sum is \begin{eqnarray*} \sum_{i=0}^{n-1}\frac{n-i}{n^2} &=& \frac{1}{n^2}\sum_{i=0}^{n-1}(n-i)\\ &=& \frac{1}{n^2}\sum_{i=0}^{n-1}n - \frac{1}{n^2}\sum_{i=0}^{n-1}i\\ &=&\frac{1}{n^2}n(n)-\frac{1}{n^2}\frac{(n-1)n}{2} \\ &=& 1-\frac{n-1}{2n} \\ &=&\frac{1}{2}+\frac{1}{2n}.\end{eqnarray*}
PS, A slightly quicker way: In the first step, notice $\sum_{i=0}^{n-1}(n-i)=\sum_{i=1}^{n}i = \frac{n(n+1)}{2}$... |
H: Find $2$ unit vectors orthogonal to $(2,1)$
I understand that in order to find an orthogonal vector to $(2,1)$ I will solve this :
$\langle(x,y),(2,1)\rangle = 0 $
but I don't understand how this is related to $2$ unit vectors.
AI: As you did:
$$2x+y=1\iff y=1-2x\,,\,\,\text{so for example}\;\;u:=\binom{\;\;1}{-1}
$$
is one possibility and, in fact, all the solutions are given by (assuming we're in $\;\Bbb R^2\;$)
$$S:=\left\{\;\binom {\;\;x}{\!\!-2x}\;;\;\;x\in\Bbb R\right\}+u=\text{Span}\,\left\{\;\binom{\;\;1}{-2}\;\right\}+\binom{\;\;1}{-1}$$
Note that
$$||u||=\sqrt{1+1}=\sqrt2\implies \frac u{\sqrt2}\;\;\text{is a unit vector, and thus}$$
$$\frac {ku}{\sqrt2}=\binom{\;\frac k{\sqrt2}}{\!\!-\frac k{\sqrt2}}$$
is a vector of length $\;k\;$ and orthogonal to what you want |
H: Odd and even function properties...
Does it mean when the function is even it's in 100% cases y-axis symmetric, and when it's odd it's in 100% cases origin symmetric?
AI: Yes, that is the geometric interpretation of even or oddness. Also, if a function has no lines of symmetry in the plane, then it cannot be even or odd. |
H: Show: $C^1(\Omega)\subset C^{0,1}(\Omega)\subset C^{0,\lambda}(\Omega)\subset C^0(\Omega)$.
Show that
$$
C^1(\Omega)\subset C^{0,1}(\Omega)\subset C^{0,\lambda}(\Omega)\subset C^0(\Omega)~~~~~~~\forall0<\lambda\leq 1.
$$
Hello, I have some problems to show these inclusions! In order to get some help, I wrote down all the definitions that we had (please see below).
So consider $u\in C^1(\Omega)$. Concerning to the second definition of 1.), then $u$ is continious on the domain $\Omega$ and the first partial derivatives are continious on $\Omega$.
Now I have to show, that $u\in C^{0,1}(\Omega)$.
Which definition do I have to use resp. what do I have to show?
This are the definitions that we wrote down: (unfortunately rather many)
1.) $\Omega\subset\mathbb{R}^n$ a domain and $k\in\mathbb{N}_0$
$$
C(\Omega):=C^0(\Omega):=\left\{u\colon\Omega\to\mathbb{R}|u\mbox{ continious on }\Omega\right\}
$$
$$
C^k(\Omega):=\left\{u\in C(\Omega)|\forall\alpha\in\mathbb{N}_0^n\mbox{ with }0\leq\lvert\alpha\rvert\leq k: D^{\alpha} u(x)\in C(\Omega)\right\}
$$
$$
C^{\infty}(\Omega):=\bigcap\limits_{k=0}^{\infty}C^k(\Omega)
$$
2.) $\Omega\subset\mathbb{R}^n$ limited domain and $k\in\mathbb{N}_0$
$$
C^k(\overline{\Omega}):=\left\{u\in C^k(\Omega)|\forall\alpha\in\mathbb{N}_0^{n}\mbox{ with }\lvert\alpha\rvert\leq k\mbox{ is }D^{\alpha}u\mbox{ uniformly continious on }\Omega\right\}
$$
3.) $\Omega\subset\mathbb{R}^n$ a (not necessarily limited)) domain and $k\in\mathbb{N}_0$
$$
C^k(\overline{\Omega}):=\left\{u\colon\overline{\Omega}\to\mathbb{R}|\forall~R>0: u\in C^k(\overline{B_R(0)\cap\Omega})\right\}
$$
4.) $\Omega\subset\mathbb{R}^n$ limited domain, $\lambda\in ]0,1]$ and $k\in\mathbb{N}_0$
$$
C^{k,\lambda}(\overline{\Omega}):=\left\{u\in C^k(\overline{\Omega})|\forall\alpha\in\mathbb{N}_0^n\mbox{ with }\lvert\alpha\rvert\leq k: H_{\alpha,\lambda}(u)<\infty\right\},
$$
whereat
$$
H_{\alpha,\lambda}(u):=\sup\limits_{x,y\in\overline{\Omega}, x\neq y}\left\{\frac{\lvert D^{\alpha} u(x)-D^{\alpha} u(y)\rvert}{\lVert x-y\rVert^{\lambda}}\right\}
$$
5.) $\Omega\subset\mathbb{R}^n$ a (not necessarily limited) domain, $\lambda\in ]0,1]$ and $k\in\mathbb{N}_0$
$$
C^{k,\lambda}(\Omega):=\left\{u\in C^k(\Omega)|\forall\overline{B}_R(x)\subset\Omega: u\in C^{k,\lambda}(\overline{B}_R(x))\right\}
$$
$$
C^{k,\lambda}(\overline{\Omega}):=\left\{u\colon\overline{\Omega}\to\mathbb{R}|\forall R>0: u\in C^{k,\lambda}(\overline{B_R(0)\cap\Omega})\right\}
$$
AI: I - $C^1(\Omega)\subset C^{0,1}(\Omega)$
Take any $B_R(x)$ such that $\overline{B_R(x)}\subset \Omega$. If $u\in C^1(\Omega)$ then $u\in C^1(\overline{B_R(x)})$, which implies that $\|u'(x)\|\leq C$, where $C>0$ is a constant.
Can you concude from here?
Hint: Apply Mean Value Theorem.
II - $C^{0,1}(\Omega)\subset C^{0,\lambda}(\Omega)$
Take any $B_R(x)$ such that $\overline{B_R(x)}\subset \Omega$ and $z,y\in \overline{B_R(x)}$. Note
\begin{eqnarray}
\frac{\|u(z)-u(y)\|}{\|z-y\|^\lambda} &=& \frac{\|u(z)-u(y)\|}{\|z-y\|}\frac{\|z-y\|}{\|z-y\|^\lambda} \nonumber \\
&=& \frac{\|u(z)-u(y)\|}{\|z-y\|}\|z-y\|^{1-\lambda}\nonumber
\end{eqnarray}
Can you conclude from here?
III - $C^{0,\lambda}(\Omega)\subset C^0(\Omega)$
This is immediate from definition. |
H: See if "7<4 implies 7 is ..." Is the following conclusion valid?
For my homework I need to see if the following conclusion is correct.
$$
7<4 \implies 7\ \text{is not a prime number}\\
\lnot(7<4)\\
-----------------\\
\text{7 is prime number}\\
$$
To tell you the truth, I have no idea how to start this, letalone how to finish it, so any help is welcome.
Thanks!!!!
AI: Let , P be " $ 7<4$ " and Q be " $ 7\ \text{is not a prime number} $ "
So actually you want to know whether the following identity holds
$ P \implies Q \\
\lnot P \\
-----------------\\
\lnot Q \\$
Well . Actually $ P \implies Q $ can be written as $ (\lnot P \lor Q) \\ $ . If this is true and $ \lnot P $ is true you can not certainly tell that $ \lnot Q $ is true . Here a fault remains .
But according to Modus Tollens the following identity is correct .
$\lnot Q \\
P \implies Q \\
-----------------\\
\lnot P \\$
Hope this helps . |
H: Automorphisms of order $2$ of the multiplicative group of a field
Let $k$ be any field, finite or infinite, (even though I'm more interested in the infinite case) of $\operatorname{Char}(k)\neq2$. Let $\varphi$ be an automorphism of order $2$ of the multiplicative group $k^\times$ of $k$ ($\varphi$ is not assumed to be a field automorphism). Assume further that $\varphi$ fixes precisely $2$ points: $\pm 1$.
The only example of such $\varphi$ that I have is $\varphi(x)=x^{-1}$.
Can anyone show another example of such $\varphi$?
Alternatively, can one show that those conditions force $\varphi$ to be the inverse?
NB: I came across this in my research (which is not exactly in the theory of fields), and I'm looking for conditions to force $\varphi$ to be the inverse.
AI: Let $x \in k^\times$. Then $\phi(x\phi(x)) = \phi(x)\phi^2(x) = \phi(x)x = x\phi(x)$, so $x\phi(x)$ is a fixpoint of $\phi$. Under your assumption, you get $x\phi(x) = 1$ or $x\phi(x) = -1$.
So you get a group morphism $k^\times \to \{-1;+1\}$ given by $x \mapsto x\phi(x)$.
Its kernel is either $k^\times$ or a subgroup $H$ of index $2$ containing $-1$.
Conversely, for every subgroup $H$ of index $2$ of $k^\times/\{\pm 1\}$, there is a corresponding involution $\phi$ defined by $\phi(x) = 1/x$ for $x\in H$ and $\phi(x) = -1/x$ for $x\notin H$
For example, let $k = \Bbb Q$. $\Bbb Q^\times /\{\pm 1\} \cong \bigoplus_{p} \Bbb Z$ (where for prime $p$, $\Bbb Z \to \Bbb Q$ is given by $n \mapsto p^n$).
For example, for each prime $p$ there is the subgroup $H_p$ consisting of rationals whose $p$-valuation is even. It has index $2$ and contains $-1$, and they correspond to a non-trivial involution $\phi_p(x) = (-1)^{v_p(x)}/x$ |
H: Help with an implication of a really basic question in algebraic geometry
I'm starting to study algebraic geometry and I'm trying to prove this implication:
For a homogeneous ideal $\mathfrak a\subset S$, show that $Z(\mathfrak
a)=\emptyset \implies \sqrt {\mathfrak a}=\text{either}\ S \ \text{or
the ideal}\ S_+=\bigoplus_{d\gt0}S_d$.
I have seeing somewhere that this is true because of these implications:
$Z(\mathfrak a)$ is empty $\implies$ in $\mathbb A^{n+1}$, $Z(\mathfrak a)$ must be empty or $(0,\ldots 0) \implies \sqrt{\mathfrak a}=S$ or $\sqrt{\mathfrak a}=\bigoplus_{d\gt 0} S_d$.
I don't understand both implications, I would appreciate if someone could help me, I really need help.
Thanks.
AI: I assume that $S=k[x_0,...,x_n]$, where $k$ is an algebraically closed field. This is exactly your setup, no?
Anyways, the homogeneous ideal $\mathfrak{a} \subset S$ defines a projective algebraic set $Z^{proj}(\mathfrak{a}) \subset \Bbb P^n$. But $\mathfrak{a}$ is also an honest ideal of the polynomial ring $k[x_0,...,x_n]$, hence it also defines an affine algebraic set
$$
Z^{aff}(\mathfrak{a}) \subset \Bbb A^{n+1}
$$
Note that by construction $\Bbb P^n$ is a quotient of $\Bbb A^{n+1} \setminus \{0\}$. Denote by $[P] \in \Bbb P^n$ the residue class of $P \in \Bbb A^{n+1} \setminus \{0\}$.
Now if $P \in Z^{aff}(\mathfrak{a})$ and $P \neq 0$, then $[P] \in Z^{proj}(\mathfrak{a})$, so if $Z^{proj}(\mathfrak{a})$ is empty then either $Z^{aff}(\mathfrak{a})$ is empty as well, or $Z^{aff}(\mathfrak{a})=\{0\}$.
The second implication follows from the well-known order preserving bijection between radical ideals and affine algebraic sets over an algebraically closed field, which in turn follows from Hilbert's Nullstellensatz. In particular, if $Z^{aff}(\mathfrak{a})=\emptyset$ then
$$
\sqrt{\mathfrak{a}}=\mathfrak{a}=S
$$
and if $Z^{aff}(\mathfrak{a})=\{0\}$ then
$$
\sqrt{\mathfrak{a}}=(x_0,...,x_n)=\bigoplus_{d>0}S_d
$$ |
H: How is $\sqrt{\frac{\sqrt{3}+2}{4}} = \frac{\sqrt{2}(\sqrt{3}+1)}{4}$
How is $\sqrt{\frac{\sqrt{3}+2}{4}} = \frac{\sqrt{2}(\sqrt{3}+1)}{4}$?
(Prove by using algebraic manipulation not by calculation)
I've tried to come up with something myself but I can't find a solution, I must be missing something.
AI: Well, note that $$\frac{\sqrt{2}(\sqrt{3}+1)}4$$ is positive, and that $$\left(\frac{\sqrt{2}(\sqrt{3}+1)}4\right)^2=\frac{2(\sqrt3+1)^2}{16}=\frac{2\sqrt3+4}8=\frac{\sqrt3+2}4,$$ so we're done. |
H: Converting Vector Valued Function
I am having trouble turning this vector function into something like a $y=mx +b$ equation.
$$r(t) = 2cos(t)^3 \hat{i} + 2sin(t)^3 \hat{j}$$
Normally I would say $x = 2cos(t)^3$ and $y=2sin(t)^3$ and either rearrange and solve x for t or use trigonometric identities to simplify things. But I don't see what can be done here.
AI: Clues:
$\sin(t)^2 + \cos(t)^2 = 1$ |
H: Right adjoint of forgetful functor from Top
How to prove this?
The forgetful functor $U:\mathbf{Top}\to\mathbf{Set}$ has a right adjoint, namely the functor $\mathbf{Set}\to\mathbf{Top}$ which equips a set with the indiscrete topology and left adjoint which equips a set with the discrete topology.
AI: If $X$ is a topological space and $S$ a set, it should be quite clear that a map from $U(X)$ to $S$ is "the same" as a continuous map from $X$ to $S$ equipped with indiscrete topology, and that a map from $S$ to $U(X)$ is "the same" as a continuous map from $S$ with discrete topology to $X$.
Note that this uses essentially that all maps from a discrete or to an indiscrete space are continuous. |
H: does $\int ^1_0 \frac {\ln x}{1-x^2}\ \mathrm dx$ - converge?
Question:
$\int ^1_0 \frac {\ln x}{1-x^2}dx$ - converges or diverges?
What we did:
We tried to compare with $-\frac 1x$ and $-\frac 1{x-1}$ but ended up finding that these convergence tests fail. Our book says this integral diverges, but Wolfram on the other hand says it converges. How come?
AI: There are two potential sources of divergence if the integral were to diverge (it doesn't): at $x=0$ and $x=1$. At $x=0$, the integral behaves as $\ln{x}$, which has antiderivative $x \ln{x}-x$. You may show that the limit of this expression as $x \to 0$ is $0$ using e.g.,L'Hopital. So the integrand is integrable at $x=0$.
At $x=1$, you may show that
$$\lim_{x\to 1} \frac{\ln{x}}{1-x^2} = -\frac12$$
so that the integrand is integrable here as well. Thus, the integral converges. |
H: Probability of getting Required Sum
What is the probability of when four dice rolled together once,and getting a sum of Thirteen
If we do by just calculating all possible values of sum, then it will take more time; so we can solve the above problem as Multinomial Coefficents of sum, i.e.:
$$
x_1+x_2+x_3+x_4 = 13\;\text{ where },1\leq x_i \leq 6\;\;\forall\;1\leq i \leq 4.
$$
I got stuck that form the above equation how can we obtain $N_{\text{ways to get $\sum=13$}}$ or is there any easy method to find this kind of problems?
AI: Consider the polynomial $u(x)=(x+x^2+\cdots+x^6)^4$. The number of ways to get $13$ is $N=[x^{13}]u(x)$, that is, the coefficient of $x^{13}$ in $u(x)$.
To compute $N$, note that $u(x)=x^4(1-x^6)^4v(x)$ where $v(x)=(1-x)^{-4}$ and that $x^4(1-x^6)^4$ equals the sum of $x^4-4x^{10}$ and of some terms of degree at least $16$.
Hence $N=[x^9]v(x)-4[x^3]v(x)$. Now, $6v(x)$ is the third derivative of the series $(1-x)^{-1}=\sum\limits_{n}x^n$ hence, for every $n$, $[x^n]v(x)={n+3\choose 3}$.
Finally, the probability to get $13$ is
$$
\frac1{6^4}N=\frac1{6^4}\left({12\choose 3}-4{6\choose 3}\right)=\frac{35}{324}.
$$ |
H: Obvious inequality
Suppose $f$ and $g$ are Lebesgue integrable functions on $E$. Then:
$$ \int\limits_E | f + g | \leq \int\limits_E ( |f| + |g| ) $$
Does this follow easily? I can see we have to use the triangle inequality, but does it work even with integrals? I'm not seeing this inequality.
AI: Of course $|f+g| \leq |f|+|g|$ and integral is monotone, so yeah
$$\int\limits_E | f + g | \leq \int\limits_E ( |f| + |g| )$$ |
H: Need help with Sigma-algebra
I am confused on how to determine a Sigma-algebra.
The following partitions of a set are given:
$$
A1 = \{1,3\} $$ $$
A2 = \{2,4,6,8\} $$ $$
A3 = \{5,7,9\}$$
And Omega is $$
\Omega = \{1,2,3,4,5,6,7,8,9\}
$$
I know that at least Omega and the empty set has to be part of the Sigma-algebra as well as the single partitions and their complements:
$$
\sigma(A1,A2,A3) = \{\emptyset,\Omega,\{1,3\},\{2,4,5,6,7,8,9\},\{2,4,6,8\},\{1,3,5,7,9\},\{5,7,9\},\{1,2,3,4,6,8\}\}
$$
Additionally, there has to be the union of the elements inside and this is the point I am not sure about. Are the following elements part of the Sigma-algebra as well?
$$ A1 \cup A2 = \{1,2,3,4,6,8\}$$
$$ A2 \cup A3 = \{2,4,5,6,7,8,9\}$$
$$ A1 \cup A3 = \{1,3,5,7,9\} $$
$$ A1 \cup A2 \cup A3 = \Omega $$
I think I can ignore all of them since they are already in my Sigma-algebra.
Is my Sigma-algebra fullfinished and where my thoughts about creating the algebra correct? I know that there have to be 2^n elements inside the algebra and due to the fact of there are 8 elements this might be a correct solution.
Another question:
In case a union (e.g. A1 U A2) would not have been in the algebra. Do I just have to add A1 U A2 to the algebra or also its complement?
AI: You're right on target. The ($\sigma$-)algebra on $\Omega$ generated by the partition $\{A_1,A_2,A_3\}$ is precisely $$\{\emptyset,A_1,A_2,A_3,A_1\cup A_2,A_1\cup A_3,A_2\cup A_3,\Omega\}.$$ More generally, if you're given a partition of any set $\Omega$ into $n$ subsets, the ($\sigma$-)algebra on $\Omega$ generated by that partition will consist of $2^n$ subsets, consisting of all finite unions of the partition elements (including the empty union).
When you aren't dealing with partitions, it is less simple. At that point, you'll need to worry about complements. Roughly speaking, take all the countable unions, take all the complements, and repeat until you're done. |
H: Recurrence Relation.
I was searching the internet when I came a across a question, and just couldn't solve it. I kept rearranging and substituting but kept going around in loops.
"For $n:= 1,2,3,.....,$ Let $$ I_n = \int_0^1 \frac{x^{n-1}}{2-x}.dx $$
Writing $x^n =x^{n-1}(2-(2-x))$, show that this sequence of numbers satisfies the recurrence relation $$I_{n+1} = 2I_n - \frac{1}{n} $$
Extension
The value taken for $I_1 = \ln{2}$ is $0.6931 $ if the recurrence relation is now used to calculate successive values for $I_n$ we find $I_{12}=-0.0189$ (you are not required to confirm the calculation). Explain carefully both why this cannot be correct and the source of the error, given that all the intermediate arithmetical operations have been performed exactly.
AI: Well, $$\begin{align}I_{n+1} &= \int_0^1\frac{x^n}{2-x}\,dx\\ &= \int_0^1\frac{x^{n-1}\bigl(2-(2-x)\bigr)}{2-x}\,dx\\ &= \int_0^1\frac{2x^{n-1}-(2-x)x^{n-1}}{2-x}\,dx\\ &= \int_0^1\left(\frac{2x^{n-1}}{2-x}-\frac{(2-x)x^{n-1}}{2-x}\right)\,dx.\end{align}$$ Can you take it from there? |
H: Algebra rules when multiplying matrices with unknown variables
I'm about to compute the determinant of a given matrix, this matrix however contains variables instead of actual values, so I'm a bit uncertain what to do here.
I have the 3x3 matrix A:
$$A = \left(\begin{matrix} a-b-c & 2a & 2a\\ 2b & b-c-a & 2b\\ 2c & 2c &c-a-b\end{matrix}\right).$$
When computing the determinant i start off by taking the first element and multiply this with the determinant of the $2\times 2$ matrix and here I encounter the problem as I'm not sure how multiplying work out in this scenario:
$$\det(A) = (a-b-c)\left((b-c-a)(c-a-b) - (2b)(2c)\right). $$
Do I start off by multiplying $b$ with $c$, then $-a$ and then $-b$ or do I try to simply the expressions, like $b$ cancels out $-b$ etc.?
This is my attempt for the determinant of the $2\times 2$ matrix for the first part:
\begin{eqnarray*} (b-c-a)(c-a-b) &= &bc - ab -b^2 - c^2 + ac + bc - ac + a^2 + ab \\&=& a^2 - b^2 + 2ab - c^2.\end{eqnarray*}
AI: Since the determinant of a matrix is defined in terms of sums of products, which entails the use of the standard operations of addition and multiplication, the "algebra" needed for computing the determinant is simply the standard arithmetic manipulation on variables and scalars.
First, note that you have not obtained the full expression for $\det (A)$ when expanding along the first column.
$$\begin{align} \det(A) & = (a-b-c)\Big((b-c-a)(c-a-b) - (2b)(2c)\Big) \\
& - \;\;2b\Big(2a(c-a-b) - (2a)(2c)\Big) \\
& + \;\;2c\Big((2a)(2b) - 2a(b-c-a)\Big)\end{align}$$
Also, for the product within the first term of the sum, we have (note the bold-face term, which is in contrast to your computation):
\begin{eqnarray*} (b-c-a)(c-a-b) &= &bc - ab -b^2 - c^2 + ac + bc - ac + a^2 + ab \\&=& a^2 - b^2 + {\bf 2bc} - c^2.\end{eqnarray*}
Then for the complete first term of the sum, we need to compute:
$$(a - b - c)\Big((a^2 - b^2 + 2bc - c^2)- 4bc\Big) = (a - b - c)(a^2 - b^2 -c^2 - 2bc)$$
The second and third term in $\det(A)$ should be a bit less tedious to compute. |
H: last two digits of $9^{1500}$ (Dummit Foote -Abstract Algebra preliminaries $0.3.5$)
Question is to find last two digits of $9^{1500}$ (No Euler totient theorem please)
What i have done so far is :
$9^2\equiv 81\pmod{100}$
$9^4 \equiv 61\pmod{100}$
$9^8\equiv 21\pmod{100}$
$9^{16} \equiv 41\pmod{100}$
$9^{32} \equiv 81\pmod{100}$
higher powers of $9$ namely $64,128,256,512,1024$ will be in repeated pattern as above.
$9^{64} \equiv 61\pmod{100}$
$9^{128} \equiv 21\pmod{100}$
$9^{256} \equiv 41\pmod{100}$
$9^{512} \equiv 81\pmod{100}$
$9^{1024} \equiv 61\pmod{100}$
Now,I want to split power of $9$ i.e., $1500$ to powers which i have noted down above. i.e,
$9^{1500}=9^{1024}.9^{476}$
$9^{1500}=9^{1024}.9^{256}.9^{220}$
$9^{1500}=9^{1024}.9^{256}.9^{128}.9^{92}$
$9^{1500}=9^{1024}.9^{256}.9^{128}.9^{64}.9^{28}$
$9^{1500}=9^{1024}.9^{256}.9^{128}.9^{64}.9^{16}.9^{12}$
$9^{1500}=9^{1024}.9^{256}.9^{128}.9^{64}.9^{16}.9^{8}.9^4$
When you multiply two positive integers, the last digit in the product depends on those two integers only through their last digits.
So, I will look only for last two digits of $9$ in above powers.
$9^{1500}=9^{1024}.9^{256}.9^{128}.9^{64}.9^{16}.9^{8}.9^4$
$\equiv 61.41.21.61.41.21.61 \pmod{100} $
$\equiv (61.61).(21.21).(41.41).61\pmod{100}$
(we have already seen above $61.61\equiv 21 \text{mod}100$ and similarly for other cases). So, we would be left with :
$\equiv (21).(41).(81).(61)\pmod{100}$
$\equiv (61)(41)\pmod{100}$
$\equiv (01) \pmod{100}$
I would be happy if someone can verify the procedure I have done and I would be thankful if some one can help me to make this less laborious and more efficient.
AI: You have seen that $9^2 \equiv 9^{32} \equiv 81$ mod $100$. As gcd$(81,100) = 1$, this implies $9^{30} \equiv \frac{9^{32}}{9^2} \equiv \frac{81}{81} \equiv 1$, and thus $9^{1500} = (9^{30})^{50} \equiv 1$.
Your method also works, but it is longer. |
H: Convergence of the series $ \sum_{n=1}^\infty \frac 1{n!} $ using the Cauchy convergence criterion
Study the convergence of the following series, using the Cauchy Convergence criterion:
$$ \sum_{k=1}^n \frac 1{k!} $$
Following the $ \sum_{k=n}^{n+p} \frac 1{k!} <\epsilon $, I must show that $ \frac 1{(n+1)!}+\frac 1{(n+2)!}+...+\frac 1{(n+p)!}<\epsilon $.
Because $ \frac 1{(n+1)!}< \frac 1{n!} $ and $ \frac 1{(n+p)!}< \frac 1{n!} $ and so on, I get that $ \sum_{k=n}^{n+p} \frac 1{k!}<\frac p{n!} $.
Now I must show that there exista a $ N(\epsilon) $ such that $ \frac p{n!}<\epsilon \ \forall \ n > N(\epsilon) $.
I get two cases:
$ \epsilon > \frac p{n!} $ so the inequality holds (this being the simple case).
$ \epsilon \leq \frac p{n!} $ in this case I should show that there exists a $ N(\epsilon) $ such that the inequality $ \epsilon \leq \frac p{n!} $ does not hold.
I am stuck on finding that $ N(\epsilon) $.
AI: According to the Cauchy criterion you need to show that $ \sum_{k = n}^{m} \frac{1}{k!} $ is less that any positive given number if n is n is big enough. So for given $\epsilon$ $$ \sum^{m}_{k = n} \frac{1}{k!} \leq \sum^{m}_{k = n} \frac{1}{n! n^{k-n}} \leq \sum^{m}_{k = n} \frac{1}{n! 2^{k-n}} \leq \frac{2}{n!}. $$
It approaches zero and so for some big n it will be less than this $\epsilon$. |
H: How to write this conversion algorithm
I have to calculate Grade Average from the scale of 1 to 5 such that 1 being the best (100%) and 5 being the worst (0%). So we have,
1 = 100% (Best)
2 = 75%
3 = 50%
4 = 25%
5 = 0% (Worst)
For a given subject, the student scored 40 out of 100. So we have 40%.
Now how do I convert the 40% into the scale of 1 to 5. I know it must be between 4 (25%) and 3 (50%).
Would it be 3.some decimal (3.3) or 4.some decimal (4.3)? I mean, should I travel with the decimals upwards from the 3 or 4?
I am making a formula in Excel.
AI: A score of $40\%$ would give a grade of $3.4$. In general, a score of $x\%$ corresponds to the grade $5 - \frac{x}{25}$. |
H: identity for squared binomial coefficient
I was wondering if there is an identity for squaring a binomial coefficient.
I know there is one with converting it to a linear equation, but I am looking to stay at a "coefficient" level. something like:
$${n \choose k}^2={n \choose K}$$
where K is a function of k. Did not find anything like it, sorry if this is a re-post.
Thanks
AI: If you take $n=2k>0$ there is no $K$ satisfying the equation:
$${n \choose k}^2={n \choose K}$$
because ${n \choose k}$ is maximum when $k=n/2$.
So, in general, your equation cannot be satisfied. |
H: Mapping cone not homotopy equivalent to quotient space
My question is about a "non-example" to theorem 1.6 in chapter VII in Bredon.
We have an inclusion $i: A \to X$, with $A = \{0\} \cup \{1/n | n = 1,2,...\}$, and $X = [0,1]$. Then $X/A$ is a one-point union of an infinite sequence of circles with radii going to zero. $C_i$, the mapping cone, contains homeomorphs of circles joined along edges.
Bredon claims that because the circles in $C_i$ do not tend to a point, any prospective homotopy equivalence $X/A \to C_i$ would be discontinuous at the image of $\{0\}$ in $X/A$.
I don't find this last argument very compelling as it stands (although I "intuitively" see it's correct). Could anyone expand on it? Or perhaps supply a more algebraic proof showing that homology or fundamental groups are different?
AI: Let's consider the fundamental group $\pi_1(C_i)$. Let $a_0$ be the result of collapsing $A$ in the mapping cone $C_i$. Let $f : I \to C_i$ be a loop at $a_0$. Since $I$ is compact, $f$ is uniformly continuous. Put $\epsilon < 1/2$ and choose the corresponding $\delta > 0$. Cover $I$ with a finite collection of intervals $\{I_j\}$ of length less than $\delta$. The image of each such interval $f(I_j)$ is confined to an open ball of radius $\epsilon < 1/2$. Thus, it cannot be a nontrivial loop (any nontrivial loop has to travel through $a_0$ and $X$, which involves a straight distance larger than $\epsilon$). Since $\{I_j\}$ is finite, it follows that $f$ must consist of finitely many loops. Hence $\pi_1(C_i)$ is countable.
$X/A$, on the other hand, is the Hawaiian earring. Its fundamental group is uncountable. To see why, let $X/A = \bigcup_{n=1}^\infty B_n$, a union of circles of decreasing radius. We have retractions $r_n : X/A \to B_n$ resulting from collapsing all but one circle to the origin. Each $r_n$ induces a surjection $\rho_n : \pi_1(X/A) \to \mathbb Z$. The product $\rho$ of all $\rho_n$ gives a homomorphism from $\pi_1(X/A)$ to $\prod_{n=1}^\infty \mathbb{Z}$. This homomorphism is surjective since for every sequence $\{k_n\} \in \prod_{n=1}^\infty \mathbb{Z}$ there is a loop $g$ in $X/A$ that goes around the circle $B_n$ for $k_n$ times during the interval $[1 - 1/n, 1 - 1/(n+1)]$. $g$ is clearly continuous at points other than the origin. The fact the circles $B_n$ become closer and closer to the origin is crucial for showing that $g$ is continuous at the origin. Every neighborhood of the origin contains all but finitely many circles. This makes it possible to show that $g$ is continuous at the origin, and hence everywhere. |
H: Derivative of an integral? $f(y) = \frac{d}{dy} F(y) = \color{red}{\frac{1}{\sqrt{y}}}\Phi'(\sqrt{y})$
Am I right to say if I differenciate an integral, I get back the thing inside the integral?
$$\frac{d}{dx} \int f(x) \, dx = f(x)$$
Then why is it in the below question,
The last line marked by the arrow ...
$$\frac{d}{dy} F(y) = \frac{d}{dy} (2 \Phi(\sqrt{y})-1) = \frac{d}{dy} (2(\int^{\sqrt{y}}_{-\infty} \frac{1}{\sqrt{2\pi}} e^{-u^2/2}
\; du)) = 2 \frac{1}{\sqrt{2\pi}} e^{-y/2}$$
How do I get the extra $\frac{1}{\sqrt{y}}$ part?
AI: $F(y)=2\Phi (\sqrt y)$. By the chain rule, you get:$$f(y)=F'(y)=2\cdot \Phi '(\sqrt y) \cdot \frac{d}{dy}[\sqrt y]=\frac{\Phi'(\sqrt y)}{\sqrt y}.$$
The issue with your proof is that, when you take the derivative of a function of the form $$H(x)=\int _a ^{f(x)} g,$$
you are considering a composition: $$h(t)=\int _a ^t g , \quad H(x)=(h\circ f)(x),$$
so the derivative of $H$ is:
$$H'(x)=(h'\circ f) (x) \cdot f'(x)=g(f(x))f'(x).$$ |
H: What is the use of Euler Totient or Phi Function?
What is most motivating way of introducing this function? Does it in itself have any real life applications that have an impact. I can only think of a^phi(n)=1 (mod n) which is powerful result but is this function used elsewhere.
AI: RSA, or public-key cryptography is one of them. |
H: Why $x^{p-1}+x^{p-2}+\cdots+1$ is irreducible over $\mathbb{Q}$?
I'm tring to know why $x^{p-1}+x^{p-2}+\cdots+1$ is irreducible over $\mathbb{Q}$.
Can you help me with a proof or showing me some references, please.
AI: The usual trick is to look instead at $f(x+1) = \frac{(x+1)^p-1}{x}$. Since the binomial coefficients are divisible by $p$, this polynomial is Eisenstein at $p$, and therefore irreducible. Thus $f$ is also irreducible. |
H: If a function $f$ is continuous in $[a,∞)$ and finite $\lim_{x→+∞}f(x)$ exists, then it's uniformly continuous in $[a,+∞)$.
Prove that if $f$ is defined and continuous in $[a,+∞)$ and if there exists a finite limit $\lim_{x→+∞}f(x)$, then $f$ is uniformly continuous in $[a,+∞)$
I know that since there exists a finite limit $\lim_{x→+∞}f(x)$, so $f(x)$ is convergent. How can I use this to prove it's uniformly continuous?
AI: The idea is to reduce the proof to the compact case.
Let $L= \lim_{x\to\infty} f(x)$. Choose $\epsilon>0$, and $M>0$ such that if $x>M$, then $|f(x)-L| < \frac{1}{2}\epsilon$.
$f$ is continuous, hence uniformly continuous on $[a,M+1]$. Choose $0<\delta<1$ so that if $x,y \in [a,M+1]$ and $|x-y| < \delta$, then $|f(x)-f(y)| < \epsilon$.
Now choose $x,y \in [a,\infty)$ such that $|x-y| < \delta$. Then either $x,y \in [a,M+1]$ or $x,y \in (M, \infty)$. In the first case we have $|f(x)-f(y)| < \epsilon$, in the second case, we have $|f(x)-f(y)| \le |f(x)-L|+|f(y)-L|< \frac{1}{2}\epsilon + \frac{1}{2}\epsilon = \epsilon$. |
H: $T_2$ spaces and isolated points
Is there a topological Hausdorff space with an infinite number of isolated points such that any infinite set of isolated points have an infinite number of limit points !? (Of course it would be impossible if a limit point is a limit of a sequence.)
AI: Yes: $\beta\Bbb N$, the Čech-Stone compactification of $\Bbb N$, which is a compact Hausdorff space, has that property. In fact, every infinite subset of $\Bbb N$ has $2^{\mathfrak{c}}$ limit points. (Every point of $\Bbb N$ is isolated.) |
H: Calculate $10,000e^{-\int_2^{10}\left(0.05+0.01/(t+1)\right)\,dt}$
This equation is used as an example in a text book with a given answer of $\approx$ 6,617
I cannot get to this solution as somewhere along the way I must be making an error. If it is a problem with the integration, please can you point out any fundamental things I'm doing wrong?
Sorry for all the fractions in the exponent, I tried to write it as clear as I could.
$$\begin{align}
& 10,000e^{-\int_2^{10}\left(0.05+\frac{0.01}{t+1}\right)\,dt} \\ \\
& 10,000e^{-\left[0.05t+0.01\frac{1}{0.5t^2+t}\ln|t+1|\right]_2^{10}} \\ \\
& 10,000e^{-\left[\left(0.05(10)+\frac{0.01}{0.5(10)^2+10}\ln11\right)-\left(0.05(2)(+\frac{0.01}{0.5(2)^2+2}\ln3\right)\right]} \\ \\ \\
& 10,000e^{-\left[\left(0.5+\frac{0.01}{60}\ln11\right)-\left(0.1+\frac{0.01}{4}\ln3\right)\right]}\\ \\
\approx& 6,719
\end{align}$$
AI: To integrate $\frac{1}{t+1}dt$, use the substitution $u=t+1, du=dt$. The integral becomes $\int\frac{1}{u}du=\ln|u|+C=\ln|t+1|+C$. In context, $\int 0.05+\frac{0.01}{t+1}dt = \int 0.05dt + 0.01\int\frac{1}{t+1}dt=0.05t+0.01\ln |t+1|+C$. |
H: Simplification of boolean algebra from "not s and p" to "not s"
I am trying to learn more about "Rules of Inference" and their application, but one thing always confuses me, and that is simplification "not s and p" to "not s".
I have looked at some examples:
http://www.site.uottawa.ca/~lucia/courses/2101-10/lecturenotes/04InferenceRulesProofMethods.pdf page 18
http://www2.cs.siu.edu/~nojoumian/CS215/Files/Lec06_CS215.pdf page 16
And I simply dont understand how is it possible to reduce expression.
Any help is welcome.
Thanks.
AI: $\lnot s \not\equiv (\lnot s \land p),\;$ but it is the case that $\;\lnot s\;$ follows from $\;\lnot s \land p$.
"$\lnot s \land p$ is true" means
$\lnot s$ is true, and $ p $ is true.
So it certainly follows that
"$\lnot s$ is true,"
just as it follows that
$p$ is true.
More simply put, we have
$\lnot s$ AND $p$.
Therefore $\lnot s$.
Therefore $p$. |
H: Does the sequence $1,-1,1,1,-1,1,1,1,-1,1,1,1,1,-1,1,1,1,1,1,-1,\ldots$ have a closed form?
Question : Can we represent the following sequence $\{a_n\}\ (n\ge 0)$ as a closed form?$$a_n : 1,-1,1,1,-1,1,1,1,-1,1,1,1,1,-1,1,1,1,1,1,-1,\ldots$$ Suppose that there exist ${(i+1)}$ $1_s$ between the $i_{th}$ $(-1)$ and ${(i+1)}_{th}$ $(-1)$ for $i=1,2,\cdots$.
Though I've thought about this, I'm facing difficulty. Can anyone help? A simpler form is better if closed forms exist.
Motivation : I've known that a periodic sequence can be represented as a closed form. For example, if a sequence $\{b_n\}\ (n\ge0)$ is defined as
$$b_n : 1,1,1,-1,1,1,1,1,1,-1,1,1,1,1,1,-1,\ldots,$$
(Suppose that $b_n=-1\ (n\equiv 4),b_n=1\ (n\not\equiv 4)\ ($mod$ 6$))
then we can represent $\{b_n\}$ as
$$b_n=\frac 13\left(2+(-1)^n+2\cos\frac{n\pi}{3}-2\cos\frac{2n\pi}{3}\right).$$
Then, I got interested in the above sequence $\{a_n\}$. I'm facing difficulty because $\{a_n\}$ is not periodic.
AI: $$a_n=-(-1)^{\left\lceil\frac{-3+\sqrt{8n+9}}{2}\right\rceil-\left\lfloor\frac{-3+\sqrt{8n+9}}{2}\right\rfloor}.$$
or
$$a_n=-\cos\pi\left( {\left\lceil\frac{-3+\sqrt{8n+9}}{2}\right\rceil-\left\lfloor\frac{-3+\sqrt{8n+9}}{2}\right\rfloor} \right).$$
The basic idea of find this form is find a pattern of place of $-1$ occurs. By simple calculation, we can find that $-1$ occurs when $n=\frac{1}{2}k(k+1)+k$ for some $k$.
Let we find a function $f$ satisfy that $f(\frac{1}{2}(k^2+3k))$ is odd and otherwise $f(n)$ is even. If we can find this function we can define a sequence $a_n=(-1)^{f(n)}$. But we know that
$$
\lceil x\rceil -\lfloor x\rfloor =\begin{cases} 0 & \text{if $x$ is integer }\\ 1 &\text{otherwise}\end{cases}
$$
and $\frac{1}{2}(n^2+3n)=x\Longleftarrow x= \frac{-3+\sqrt{8x+9}}{2}$. In that case, $x$ is integer iff $x=\frac{1}{2}(k^2+3k)$ for some $k$. So if we define $f(n)=\lceil x\rceil -\lfloor x\rfloor+1$ then we get desired function. |
H: Giving an explicit example of a vector that is perpendicular to $v$
Let $v\in\mathbb{R}^3$ be a unit vector. It is possible to show that there exists vectors $\{w_1,w_2\}$ such that $\{v,w_1,w_2\}$ is orthonormal by applying the Gram-Schmidt process, but can we do so continuously?
Question: Is it possbile to give these vectors explicitly
? More precisely: Can someone give examples of continuous functions $f,g:\mathbb{S}^2\rightarrow \mathbb{S}^2$ such that for every unit vector $v$ in $\mathbb{R}^3$ the set $\{v,f(v),g(v)\}$ is orthonormal ?
To solve the above question, it suffices to find a continuous function $h:\mathbb{R}^3-\{0\} \rightarrow \mathbb{R}^3-\{0\}$ such that for every nonzero vector $v\in \mathbb{R}^3$, we have $h(v)\cdot v=0$. Once one finds this function we can set $f(v)=\frac{h(v)}{||h(v)||},g(v)=\frac{v\times h(v)}{||v\times h(v)||}$. I spent some time to find such a function, but I didn't find any. In case this impossible, I am guessing it would be due to something about the topology of $\mathbb{R}^3$. Thus, I am tagging this question as general topology as well.
Thank you
AI: Do you know the 'hairy ball theorem', that you can't have a continuous non-zero vector field on the surface of the sphere? The vector pointing from $v$ to $f(v)$ would be that field. |
H: What is the standard notation for $\arcsin$
I found a lecture notes that claims the following. Is this standard?
The notation $\overline{\text{arc}}\text{ sin }x$ is the inverse function of $\sin x$ restricted to $\left [ -\frac{\pi}{2},\frac{\pi}{2}\right ]$ and $\text{arc sin }x $ mean all those $y$ satisfying $\sin x=y.$
AI: The most common notation used is either $\,\arcsin x\,$ or $\,\sin^{-1}x$.
When the desired value of $\,f(x) = \arcsin x\,$ is restricted to those values lying in $[-\pi/2, \pi/2]$, this is usually stated explicitly. I presume the lecturer introduced $\overline{\text{arc}}\sin x$ to spare the need from restricting the range of solutions repeatedly. |
H: Mayer-Vietoris for a cover without triple intersections
Let $M = \bigcup_i U_i$ be a cover with open sets $U_i$ such that for
for distinct $i,j,k$ we always have $U_i \cap U_j \cap U_k = \emptyset$.
I would like to show the existence of the following exact sequence
$$\rightarrow \bigoplus_{i < j} H_q(U_i \cap U_j) \rightarrow \bigoplus_{i} H_q(U_i) \rightarrow H_q (M) \rightarrow \bigoplus_{i < j} H_{q-1}(U_i \cap U_j) \rightarrow$$
using only the Eilenberg–Steenrod axioms and without using CW-approximation.
For a finite cover I thought one could somehow induce on the Mayer-Vietoris
but I failed.
I heard that for generic covers (without the tripel intersection property) there is a spectral sequence allowing to compute the homology of $M$ with the homology of the $U_i$. Sadly I don't know
spectral sequences and so I am hoping for a proof without them.
AI: You can find this in the book Bott, Tu "Differential forms in algebraic topology". Section 8 "Generalized Mayer-Vietoris principle" is exactly what you are asking about.
They actually implicitely do use spectral sequences, but they don't call it this way, and the explanation is very clear and elementary. |
H: Modulo operation, the remainder of division of one number by another
The equation is:
$241 \equiv_{N} 35$
I have no clue how to get value of $N$, any ideas?
AI: Hint: In order to find $N$ you need to remember that if $a = b \pmod N$ then $N \mid a-b$, in your case $N \mid 241-35$, can you take over from here?
In your case $241-35 = 206$, so $N=206$. But as I told you before you need to find the $N$ values that divide $241-35$ and therefore $206$ and those are $103$ and $2$.
In conclusion $241 \equiv_{N} 35$ if $N=1, 2, 103, 206$. |
H: In which the real number system that sum of geometric progression involve?
I want to know about sum of geometric progression
a and r
Are they real number it integer .. Etc ?
AI: The formula
$$(1-r)(a+ar+ar^2+....+ar^n)=a(1-r^{n+1})$$
hold in any ring. Thus it is true in integers, rationals, reals and complex numbers.
Now, if you want to divide by $1-r$, you need "$\frac{1}{1-r}$" to make sense in the numbers you have. In particular, that formula holds in any field, as long as $r\neq 1$, and it also holds in Integral domains, as long as $1-r$ divides the RHS.
Hence.
$$a+ar+ar^2+....+ar^n=a \frac{1-r^{n+1}}{1-r}$$
is true in rationals, real and complex numbers. Moreover, is $a,r$ are integers, since the formula holds in rationals, and the LHS is integer, it follows that both sides are integers and equal. Thus formula also holds in integers. |
H: Are the numbers 1, 0, -1 necessarily in every field?
Do the numbers 1, 0, and -1 belong to every field? To me, this seems a fairly obvious conclusion of the field axioms, though I haven't seen it stated like so in any textbooks.
AI: $0$ is in $F$, because $F$ is a group w.r.t. addition, it must have an identity element, we denote it as $0$.
$1$ is in $F$, because $F\backslash\{0\}$ is a group w.r.t. multiplication, it must have an identity element, we denote it as $1$.
$-1$ is in $F$, since $-1$ is the additive inverse of $1$ in the group $F$.
However, those elements can be the same, e.g. $-1=1$ in $\mathbb{Z}_2$. |
H: How to calculate the integer part of the value of the following equation?
How to calculate the integer part of the value of the following equation?
$$y=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+\ldots+\frac{1}{\sqrt{1000000}}$$
It should be calculated in a special way, after all the equation is so long.
AI: We have from the decreasing of the function $x\mapsto\frac{1}{\sqrt x}$
$$\sum_{k=1}^n\int_{k}^{k+1}\frac{dx}{\sqrt x}\le\sum_{k=1}^n\frac{1}{\sqrt k}\leq\sum_{k=2}^n\int_{k-1}^k\frac{dx}{\sqrt x}+1$$
hence
$$ 2(\sqrt{n+1}-1)=\int_1^{n+1}\frac{dx}{\sqrt x}\le\sum_{k=1}^n\frac{1}{\sqrt k}\leq\int_{1}^n\frac{dx}{\sqrt x}+1= 2(\sqrt n-1)+1$$
Numerically for $n=1000000$ the integer part is $1998$ |
H: Prove that $1/f$ is uniformly continuous on ...
I need hints for this particular question:
Prove that if a function $f$ is uniformly continuous on $A\subseteq \mathbb{R}$ and $|f(x)|\geq k>0$ for all $x\in A$, then the function $\frac{1}{f(x)}$ is also uniformly continuous on $A$.
My attempt: From the rough work given that $f(x)$ is uniformly continuous on $A$ for all $\epsilon> 0$, there exists a $\delta$ such that $|f(x)-f(y)|<\epsilon k^2$ when $x\in A$ and $|x-y|< \delta$, which impiles $\frac{1}{k^2} |f(x)-f(y)|< \epsilon$ for all $\epsilon >0$ and $|x-y|<\delta$.
Taking the hint from Ayman Hourieh into consideration we have using the same $\delta$ from rough work,
$\left |\frac{1}{f(x)} - \frac{1}{f(y)} \right | $ = $\left |\frac{1}{f(x)f(y)} \right | \left | f(x) - f(y) \right |\leq \frac{1}{k^2} \left | f(x)-f(y) \right |< \epsilon$ when $\left | x-y \right |<\delta$.
Is this proof okay?
AI: Let $x_1, x_2 \in A$. We have:
$$
\left|\frac{1}{f(x_1)} - \frac{1}{f(x_2)}\right| = \frac{\left|f(x_2) - f(x_1)\right|}{\left|f(x_1)f(x_2)\right|}
$$
Now use the inequality you have to make an estimate... |
H: $\int_{-\infty}^{0}xe^{x}$ diverge or converge
How would one find whether the following improper integral converge or diverge.
$\int_{-\infty}^{0}xe^{x}$
I did the following.
$t\rightarrow\infty$
$\int_{t}^{0}xe^x$
I did the integration by parts.
$u=x$
$dv=e^x$
$xe^x-\int 1e^x$
$xe^x-1xe^x$
$(0)(e^0)-e^0(0)-te^t-te^t$
$0-\infty-\infty$
would this mean there is a divergence.
AI: You've made an error in your integration by parts. $$xe^x-\int 1e^x=xe^x-e^x=(x-1)e^x$$
It can be shown that $$\lim_{x\to-\infty}(x-1)e^x=0,$$ so the improper integral converges.
As a side note, we also have that $$\lim_{x\to-\infty}xe^x=0,$$ so even if your integration by parts hadn't been erroneous, your evaluation of the limit would have been. |
H: tangent planes and linear approximations and partial derivatives
I have to study tangent planes and linear approximations, there is this theorem :
THEOREM: if the partial derivatives $f_x$ and $f_y$ exist near $(a,b)$ and are continuous at $(a,b)$, then $f$ is differentiable at $(a,b)$
Actually, it's foggy in my head about what I have to do to show their existences, because everytime I just calculating the partial derivatives straightforward.
Could you, please, give to me a short example where one of the partial derivative doesn't exist
AI: How about $f(x,y) = |x|y$? Or, for an example where the partial derivatives exist everywhere except at a single point, the cone $f(x,y) = \sqrt{x^2+y^2}$. |
H: Some questions on convex sets.
Are all bounded closed convex sets in a metric space $(M,d)$ compact? or if not are they complete?
The positive definite matrices form a convex set (Why does a positive definite matrix defines a convex cone?), do they also form a metric space? If so what is the metric? Are they complete or compact?
P.S.
I am trying to understand if interior point methods can be applied to non Euclidean convex sets.
Thanks a lot.
AI: The set of positive definite matrices of a given size is convex. To see this, let $\mathbf A$ and $\mathbf B$ be two positive definite matrices in $\mathbb R^{n\times n}$, where $n$ is a positive integer. Let $\mathbf x\in\mathbb R^n$ be any nonzero vector and $\lambda\in[0,1]$. Then
\begin{align*}
\mathbf x^T\left[\lambda\mathbf A+(1-\lambda)\mathbf B\right]\mathbf x=\lambda \mathbf x^T\mathbf A\mathbf x+(1-\lambda)\mathbf x^T\mathbf B\mathbf x>0,
\end{align*}
since both quadratic forms are strictly positive (because $\mathbf A$ and $\mathbf B$ are positive definite and $\mathbf x$ is not zero) and at least one of $\lambda$ and $1-\lambda$ is positive.
For a given $n$, you can conceive the set of positive definite matrices of size $n$ as a subspace of $\mathbb R^{n^2}$ after rearranging the elements of the matrices into vectors. You can use the Euclidean metric on $\mathbb R^{n^2}$ on the subspace of vectors thus extracted from positive definite matrices. This set is not compact, because there exist positive definite matrices with arbitrarily large elements (say, $m\mathbf I$, where $m>0$ and $I$ is the identity matrix), nor is it complete, as $(1/m)\mathbf I$ is positive definite for all $m\in\mathbb Z_+$, but the limit of such a sequence is the zero matrix. (However, the set of positive semi-definite matrices, view as a subspace, is complete, because it is closed.)
To see a counterexample for a bounded, closed, convex set that is not compact, consider any infinite set $M$ that is also a vector space, endowed with the discrete metric $d$. That is, for any $x,y\in M$, let $$d(x,y)=\begin{cases} 0&\text{if $x=y$,}\\1&\text{otherwise.}\end{cases}$$ Then, $M$ is bounded (because the distance between any two points is at most $1$), convex (since $M$ is also a vector space), and closed (the whole space is always closed in any metric space), but $\{m\}_{m\in M}$ is an infinite collection of open sets (hint: in a discrete metric space, every set is open) whose union is $M$, but there is no finite subcollection whose union is still $M$ (because $M$ is infinite by assumption). Hence, $M$ is not compact. |
H: History of complex numbers
I'm interested in the history of complex numbers - their origin and their subsequent development.
I'd be very interested if anyone can provide references for finding out more about this topic.
AI: The book An Imaginary Tale: The Story of $i$ by Paul Nahin is very nice, and has a fair amount of historical material. |
H: Smallest such $n \in \mathbb{N}$ that $2^{n} \equiv 1 \pmod{5\cdot 7\cdot 9\cdot 11\cdot 13}$
Can anybody give me a hint about how to find smallest such $n \in \mathbb{N}$ that $2^{n} \equiv 1 \pmod{5\cdot 7\cdot 9\cdot 11\cdot 13}$?
I thought that I will find it piece by piece with help from my friend Fermat's Little Theorem, so :
$2^{n_{1}} \equiv 1 \mod 5$, so $n_{1}=4$.
$2^{n_{2}} \equiv 1 \mod 7$, so $n_{2}=6$.
$2^{n_{3}} \equiv 1 \mod 3$, so $n_{3}=2$.
$2^{n_{4}} \equiv 1 \mod 11$, so $n_{4}=10$.
$2^{n_{5}} \equiv 1 \mod 13$, so $n_{5}=12$.
So, since I know that $x \equiv y \mod mz \Leftrightarrow x \equiv y \mod m$ when $z \neq 0$, so my lucky was to take the lowest common multiple of $4,6,2,10,12$, which is $60$ and lo and behold it fits the bill. But is it the smallest such $n$? If so, how to explain it?
AI: You are looking for the smallest $n_1,..,n_5$, FLT only gives you some $n$. But you should know that the order (the smallest such $n_i$) divides any other $n$.
So in each line, the smallest one must be a divisor of the one given by FLT.
You know $2^{n_{1}} \equiv 1 \mod 5$ and $n_1$ is a divisor of $4$. Since $1,2$ don't work $n_1=4$.
You know $2^{n_{2}} \equiv 1 \mod 7$ and $n_1$ is a divisor of $6$. We test $1,2,3,6$ and the smallest one which works is....
Continue.
At the end, the smallest number which works must be divisible by all of them, thus the lcm is the right choice. |
H: Counter examples to inscribed squares conjecture
Can the counter examples found by me qualify as a counter proof for the "inscribed squares problem" (the Toeplitz' conjecture) ?
I ask this here because the problem stands as unsolved for 100 years, and the counterexamples below appear too simple. It is hard to believe that I am the first to find them. So, is there something I missed out?
Here are the curves:
Solution http://automaton.opa.ro/wp-content/uploads/2013/09/solution_4_s30.png
Solution http://automaton.opa.ro/wp-content/uploads/2013/09/solution_1_s30.png
Solution http://automaton.opa.ro/wp-content/uploads/2013/09/solution_3_s30.png
AI: So no, it's not a counterexample |
H: help find my error in a statistics transformation computation
This is a low-priority question, but it has been bugging me so I thought I'd ask. In my stats homework I have the following exercise:
Exercise. Suppose $X_1$ and $X_2$ are iid observations from the pdf $f(x\mid\alpha)=\alpha x^{\alpha-1}e^{-x^\alpha}$, $x>0$, $\alpha>0$. Show that $(\log X_1)/(\log X_2)$ is an ancillary statistic.
This is easy to do by showing that if $X\sim X_i$ then $(\log X)$ is a scale family and then applying the Theorem which states that if $U$ is a scale family and $U_1,\cdots,U_n$ are iid observations from $U$ then each $U_i/U_j$, $i\neq j$, is ancillary. However I attempted a different method which led to a curious problem. Observe.
Let $U=\frac{\log X_1}{\log X_2}$ and $V=\log X_2$, then consider the transformation $(X_1,X_2)\mapsto(U,V)$. Then $x_1=e^{uv}$ and $x_2=e^v$, giving us the Jacobian
\begin{equation*}J=\left|\begin{array}{cc}\frac{\partial x_1}{\partial u}&\frac{\partial x_1}{\partial v}\\\\\frac{\partial x_2}{\partial u}&\frac{\partial x_2}{\partial v}\end{array}\right|=e^{(u+1)v}.\end{equation*}
Thus by independence of $X_1$ and $X_2$ we have
\begin{multline*}f_{U,V}(u,v)=f_{X_1,X_2}(e^{uv},e^v)|J|\\\\=[\alpha (e^{uv})^{\alpha-1}e^{-(e^{uv})^\alpha}][\alpha (e^v)^{\alpha-1}e^{-(e^v)^\alpha}]e^{(u+1)v}\\\\=\alpha^2\exp[{(u+1)v\alpha}-e^{uv\alpha}-e^{v\alpha}].\end{multline*}
Hence by changing variables $z=v\alpha$ we get $dv=dz/\alpha$, and then
\begin{multline*}f_U(u)=\int_{-\infty}^\infty\alpha^2\exp[{(u+1)v\alpha}-e^{uv\alpha}-e^{v\alpha}]\;dv\\\\=\alpha\int_{-\infty}^\infty\exp[{(u+1)z}-e^{uz}-e^z]\;dz=:\alpha g(u),\end{multline*}
where $g(u)$ is independent of $\alpha$. However this is impossible!
This leads to my question: Where is my error?
AI: The error is in the computation of the Jacobian which equals $J=v \exp\left((u+1) v\right)$
$$
\begin{equation*}J=\left|\begin{array}{cc}\frac{\partial x_1}{\partial u}&\frac{\partial x_1}{\partial v}\\\\\frac{\partial x_2}{\partial u}&\frac{\partial x_2}{\partial v}\end{array}\right| = \left|\begin{array}{cc} v \exp(u v) & u \exp(u v)\\\\0&\exp(v)\end{array}\right| =v e^{(u+1)v}.\end{equation*}
$$ |
H: How to read a contour plot?
I am taking machine learning, and I have seen a few contour plots in the course.
It seems that I can't understand how to read this plot, I have tried looking it up in Wikipedia, but I don't even understand the first example (the figure on the right).
Another example is from the course:
The remark in the brackets say that it is easier to see that the function plotted is convex from the counter plot. but I don't see it at all.
What are those circles ? what is the meaning that the red x is in the inner circle (or even near its middle) ?
I would appreciate it if someone could please explain what information I should be able to read from the right drawing (the contour plot) and how ?
AI: The circles you see are level sets: the function $J$ is equal where the colors are the same, i.e. around each circle. The red X is, presumably, the point where $J$ attains its minimum.
If you've ever seen a topographic map, it is also a form of contour plot, where the function being plotted is the height of the ground.
From the plot you should be able to see that the function $J$ is "bowl-shaped" and therefore convex. Moreover, from the spacing you can see that the function increases quickly starting from the X and moving up-right or down-left, and slowly moving up-left or down-right. |
H: Solve $\left(x^{2010}+1\right)\left(1+x^2+x^4+x^6+.......+x^{2008}\right)=2010x^{2009}$
Solve for $x$
$\left(x^{2010}+1\right)\left(1+x^2+x^4+x^6+.......+x^{2008}\right)=2010x^{2009}$
solution should be by hand
AI: First it is clear that $x \geq 0$.
By AM-GM inequality we have:
$$x^{2010}+1 \geq 2 x^{1005}$$
$$1+x^2+x^4+x^6+.......+x^{2008} \geq 1005 \sqrt[1005]{1x^2...x^{2008}}=1005 x^{\frac{2+4+6+...+2008}{1005}}=1005x^{1004}$$
Multiplying you get
$$\left(x^{2010}+1\right)\left(1+x^2+x^4+x^6+.......+x^{2008}\right) \geq2010x^{2009}$$
You get equality if and only if you have equality in the above. Thus
$$x^{2010} =1$$
and
$$1=x^2=x^4=...=x^{2008}$$ |
H: Describing the ideal in polynomial ring with $n$ indeterminates
Let $R$ be a commutative ring and let $m$ be a natural number. Describe the ideal $(X_1,X_2,...,X_n)^m$ of the ring $R[X_1,X_2,...,X_n]$ of polynomials over $R$ in indeterminates $X_1,...,X_n$.
I know that, if $K[X,Y]$ is the ring the generators of ideal $(X,Y)^2$ are $\{X^2,XY,Y^2\}$.
Generators of ideal $(X_1,X_2,...,X_n)^m$ are like $\{X_1^m,X_1^{m-1}X_2,...\}$. I know but it looks like very complex. I can not arrange the generators in a good way. Maybe I am totally wrong.
Thanks for help.
AI: The set of generators is the set of homogenous polynomials of degree $m$. It can be described as
$$\{X_1^{t_1} \cdots X_n^{t_n} \ | \ t_1 + \cdots + t_n = m, t_i \in \mathbb{N} \}$$. |
H: 4D to 3D projection
Im trying to calculate the position of 4D point in 3D world. I started with 2D and tried to extend it to the 3D and then to 4D. Firstly, I found out that its easy to calculate the projected position of 2D point on the line.
Whoops, there should be () in the first equation: x/(a+y)
Now I figured out that the same will apply in the 3D world if I split the P(X,Y,Z) to the P1(X,Z) and P2(Y,Z), calcualte their Q and then build a point of P'(Q1,Q2) (Assuming Im looking Z axis positive infinity from C(0,-a) point and rendering to the XY plane).
nx = (a*x)/(a+z);
ny = (a*y)/(a+z);
Then I thought its just as simple as adding next point P3, and came up with
nx = (a*x)/(a+z);
ny = (a*y)/(a+z);
nw = (a*w)/(a+z);
I felt it was weird, becouse W (new axis) actually affects only Z of the last point, and referring to the tesseract it should affect all dimensions...
This isn't working, so I'd like to ask if you can possibly provide some details of what Im doing wrong. Im pretty sure that its the "point splitting" problem, and the equation should be more complex. Please, don't attack me with matrixes and quaternions. I just want to have a simple static camera at (0,-1) looking at (0,0)...
Thanks for any help!
AI: Let's say you have a point ${\bf p} \in \mathbb{R}^4$ that you want to project on to a three-dimensional hyperplane that is sat inside $\mathbb{R}^4$. Let ${\bf q} \in \mathbb{R}^4$ be the point from which you project and assume that the plane has the equation $a_1x_1+a_2x_2+a_3x_3+a_4x_4=c$, where the $a_i$ and $c$ are fixed numbers.
You can parametrise the ray from ${\bf q}$ to ${\bf p}$ and beyond by ${\bf x} = (1-t){\bf q}+t{\bf p}$, where $t \ge 0$. When $t=0$ we're at the point of projection ${\bf q}$. When $t=1$, we're at the point to be projected ${\bf p}$. For all other $t \ge 0$ we're on the straight line starting at ${\bf q}$, passing through ${\bf p}$ and carrying on forever.
This line hits the plane when $x_i= (1-t)q_i+tp_i$ solve the equation $a_1x_1+a_2x_2+a_3x_3+a_4x_4=c$. All of the $p_i$ and $q_j$ will be number that you know, as will all of the $a_k$ and $c$. You will get a linear equation in $t$. Solve that equation to give, say $t=\tau$ then plug that back into the vector equation ${\bf x} = (1-t){\bf q}+t{\bf p}$. |
H: Simplifying this Further
$$2x(x^2-3)^{10} + 20x(x^2+3)(x^2-3)^9$$
I would like to double check my answer (if anyone can double check this please)
Please simplify the top
AI: $$2x(x^2-3)^{10} + 20x(x^2+3)(x^2-3)^9=2x(x^2-3)^9(x^2-3+10(x^2+3))=2x(x^2-3)^9(11x^2-27)$$ |
H: Are the solutions to the equation $f(n \cdot x)=x$ always expressible in closed form?
Are the solutions to the equation:
$$f(n \cdot x)=x$$
always expressible in closed form?
$$n=1,2,3,4,5,...$$
AI: Depends what you mean by closed form, but if you mean in terms of elementary functions, then no. Consider, for instance, $f(x) = e^x-2$. There is no closed form for $n=1$ much less the other values of $n$. |
H: How do I find the supremum and infimum of this set
$$A=\left\{\dfrac{mn}{1+m+n} \mid m,n \in \mathbb N \right\}$$
I'm relatively new to this whole infimum and supremum proving. I've tried for a long time to prove the infimum of this set (which I believe to be 1/3). Can someone please help me and provide a proof for that?
AI: Hint: $$0 < \dfrac1{mn} + \dfrac1{m} + \dfrac1{n} \le 3, \qquad m,n \in \mathbb{N}.$$ |
H: Prove that $\sqrt{2n^2+2n+3} $ is irrational
I have proven this by cases on $n$.
I would like to see a neater proof. One similar to the proof of the fact that $\sqrt{2}$ is irrational.
AI: Hint: $2n^2 + 2n + 3 = 2n(n+1) + 3 \equiv 3 \pmod 4$. |
H: Definition of Vector Space
What is the meaning of objects in a vector space?
Definition: A vector space is a nonempty set V of objects, called vectors, on
which are defined two operations, called addition and multiplication
by scalars, subject to the ten axioms...
Can entries of a vector be anything other than real or complex numbers?
AI: A vector is an object. It does not need to have "entries"; this is a property that $\mathbb{R}^n$ has, but not most other vector spaces.
For example, the set of all real polynomials in the variable $t$, denoted $P(t)$, is a vector space. There are no entries; a vector looks like $1+2t-3t^5$. |
H: free groups , a question of listing elements and drawing multiplication table
I'm requested to list the elements and draw the multiplication table for the group $\langle a, b : |a| = 2 = |b|\rangle$ without any more details. But hence this group is infinite isn't it ?
while listing the elements i found $ = \{ a , b , ab , ba , aba , bab , abab , baba, \dots \}$
and so on ! I see that we must have an information about the order of $ab$ or $ba$?
Am I wrong ? Can anybody help me?
AI: Why the tag "free-groups"? Yours is not a free group but a free product, namely $\;C_2*C_2\;$ = the infinite dihedral group.
By the general theory, $\;C_2*C_2\;$ is an infinite group and the only elements with finite order are those who are conjugates to one of the elements in either factor.
Thus, if we put for the first factor $\;\langle a\rangle=\{1,a\}\;$ , and for the second one $\;\langle b\rangle=\{1,b\}\;$ , the normal form of an element in this group is of the form
$$abababa\ldots\;,\;\;\text{or}\;\;bababa\ldots$$
each of the two forms above being a finite word in those two letters, and the finite order elements are those of the form $\;g^{-1}ag\;,\;\;g^{-1}bg\;,\;\;g\in C_2*C_2\;$ , for example
$$ababa=(ab)a(ba)=(ab)a(ab)^{-1}\;,\;\;abababa=(aba)b(aba)=(aba)a(aba)^{-1}\;\ldots etc.$$ |
H: Integral convergent calculating
I have to calculate if this integral is convergent for any "$a$":
$$ \int_2^\infty \frac{1}{x\log^a(x)} dx $$
(I made sure to not make any mistake when writing the integral from paper into this forums!)
I really don't what is meant by this..
AI: Changing variable to $t=\log x$, one has
$$
\int_2^{+\infty}\frac{1}{x\log^{\alpha} x}\textrm{d}x=\int_{\log 2}^{+\infty}\frac{1}{t^{\alpha} }\textrm{d}t,
$$
so the integral converges if $\alpha>1$, diverges if $\alpha\le 1$. |
H: How do I find the area in this question?
Find the area bounded by
$$ f(x) = x + 6 $$
$$ g(x) = x^3 $$
$$ h(x) = -\frac{x}{2} $$
Edit Fixed simple error on h(x)
I already drew the grew, although it is very hard to really tell where they seem to intersect.
It looks like on the left side of the graph f(x) intersects with h(x) at -12. However going toward the right side of the graph it looks like I'll have to integrate base on f(x) and g(x).
I'm thinking this is a two part integration problem, but I'm unsure of what my bounds should be.
AI: Here is the graphs to help you. Decompose your region in two parts: one for $-4\leq x\leq 0$ and the other for $0\leq x\leq 2$. Then you just use integrals to compute the area.
First interval (left): below red minus below green
Second interval (right): below red minus below blue |
H: Convergence of average of sums
Let $a_1,a_2,\ldots$ be a sequence of real numbers, and define $s_n=a_1+\ldots+a_n$ for all $n$. Define $t_n=\dfrac{s_1+s_2+\ldots+s_n}{n}$. There is a theorem that if $\{s_n\}$ converges, then $\{t_n\}$ converges (to the same limit, I believe).
Does the converse hold? That is, if $\{t_n\}$ converges, then does $\{s_n\}$ converge?
AI: No. Choose $s_n = (-1)^n$; then
$$t_k = \left\{\begin{array}{lr} -\frac{1}{n} &: n \text{ odd} \\ 0 &: n \text{ even} \end{array}\right.$$ |
H: Evenness of Fourier coefficients
Let $f\in L^1(\mathbb{R}/2\pi\mathbb{Z})$ and let $F(n)$ denote its Fourier coefficients $$F(n)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{-inx}dx$$ I want to prove that $f$ is even if and only if $F(n)=F(-n)$ for all $n$.
Suppose $f$ is even. Then I have to prove $\int_{-\pi}^\pi f(x)e^{-inx}dx=\int_{-\pi}^\pi f(x)e^{inx}dx$. This is true because the value on the left-hand side at $a$ is $f(a)e^{-ina}$, while the value of the right-hand side at $-a$ is $f(-a)e^{-ina}=f(a)e^{-ina}$.
What about the converse? Suppose $F(n)=F(-n)$ for all $n$. How can I show that $f(x)=f(-x)$ for all $x$?
AI: Changing variables, we have
$$
F(n)=F(-n)\rightarrow \int_{-\pi}^\pi f(x) e^{inx}dx=\int_{-\pi}^\pi f(x) e^{-inx}dx
=\int_{-\pi}^\pi f(-x) e^{inx}dx,
$$
so
$$
\int_{-\pi}^\pi[f(x)-f(-x)]e^{inx}dx = 0.
$$
If $\mathcal{F}$ denotes the Fourier transform, then we have that
$$
0=\mathcal{F}[f(x)-f(-x)]\rightarrow\mathcal{F}[f(x)] = \mathcal{F}[f(-x)].
$$
Applying the inverse Fourier transform should give the result. |
H: Find $\vec u$ such that $D_\vec uf(x^2+y^2)$ is maximum, minimum, and zero at $(1,2)$
I understand how to find the maximum and minimum, but I'm having trouble understanding how to find it when it's zero. My professor said that I need to set $\phi=\pi/2$ in $D_uf(x,y)=\|\nabla f\|\cos{\phi}$. But $\cos{\pi/2}=0$, so wouldn't it make the entire expression equal to $0$?
My calculations for finding maximum and minimum:
$\nabla f(x,y)=\langle 2x,2y\rangle$
$\nabla f(1,2)=\langle 2,4\rangle$
$\|\nabla f(1,2)\|=2\sqrt{5}$
To get maximum, let $\vec u=\frac{\nabla f(1,2)}{\|\nabla f(1,2)\|}$, which yields $\bigl\langle\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}}\bigr\rangle$. To get minimum, I just negate the vector, which yields $\bigl\langle-\frac{1}{\sqrt{5}},-\frac{2}{\sqrt{5}}\bigr\rangle$.
How do I find $\vec u$ when it's equal to $0$?
AI: Since your function is differentiable, I'd rather use the formula
$$D_uf(x,y)=\nabla f(x,y)\cdot u\;,\;\;u=\text{a unit vector in the wanted direction}\implies$$
$$\nabla f(1,2)\cdot (u_1,u_2)=(2,4)\cdot (u_1,u_2)=2u_1+4u_2=0\iff u_1=-2u_2$$
and we're done...! |
H: showing existence and uniqueness of solution of $y'(t)=\frac1{1+|y(t)|}$
Given \begin{align*}
y'(t)&=\frac1{1+|y(t)|},&y(0)=y_0&&\textrm{for }t\in[a,b]
\end{align*}
I want to show that this IVP has a unique solution
My attempt: We get $f(t,x)=\frac1{1+|x|}$. If $f$ is continuous there exists a solution on $[a,b]$ by Peano. If $f$ is Lipschitz-continuous, the solution is unique.
Since fractions and $|\cdot|$ are continous and $|\cdot|\geq0$, $f$ is continuous and so there exists a solution by Peano, right?
Now showing Lipschitz-continuity: $$|f(t,x)-f(t,y)|=\left|\frac1{1+|x|}+\frac1{1+|y|}\right|=\left|\frac{1+|y|+1+|x|}{(1+|x|)(1+|y|)}\right|$$
but now I am stuck. How do you get $\leq L|x-y|$ ?
AI: You wrote $+$ instead of $-$ in the expression for $f(t,x)-f(t,y)$. It should be
$$|f(t,x)-f(t,y)|=\left|\frac1{1+|x|}-\frac1{1+|y|}\right|=\frac{\left||x|-|y|\right|}{(1+|x|)(1+|y|)} \le \left||x|-|y|\right| \le |x-y|.$$ |
H: A question on limit of functions
Let $\{f_n\}$ be a sequence of continuous functions such that$ f_n \to f$ uniformly on $\mathbb R$. Suppose that $x_n\to x_0$. Prove that $\lim_{n\to\infty} f_n(x_n)=f(x_0)$.
Let me know if I should format it better! This question seems so straightforward that I don't know what definition I'm supposed to employ...
AI: Let $\;\epsilon >0\;$:
$$|f_n(x_n)-f(x_0)|=|f_n(x_n)-f(x_n)+f(x_n)-f(x_0)|\le$$
$$|f_n(x_n)-f(x_n)|+|f(x_n)-f(x_0)|$$
For the first summand in the second line above there exists $\;N_1\in\Bbb N\;$ s.t. it is less than $\;\frac\epsilon2\;$ for any $\;x_n\;$ as long as $\;n>N_1\;$, and for the second summand there exists $\;N_2\in\Bbb N\;$ s.t. it is less than $\;\frac\epsilon2\;$ as long as $\;n>N_2\;$ by continuity (why?) ...and we're done. |
H: Which volume formula do I use for this problem?
Find the volume generated by revolving about the $x$-axis the region bounded by the curves:
$$f(x)= x^2 ,\ g(x) = 2-x^2.$$
I drew the graph and if we are revolving around the $x$-axis it looks like we are going to have some sort of gap inside part of the shape (washer)?
So I went ahead and got the bounds: $-1,\ +1$.
I'm thinking I should be integrating using this formula:
$$ V = \int_{-1}^{1} \pi [ (g(x)^2 - (f(x))^2 ]\, dx $$
Does that seem correct?
AI: Yes, you're "spot on." Your assessment of the problem is correct, as is your integral. |
H: A book/text in Stochastic Differential Equations
Somebody know a book/text about Stochastic Differential Equations? I'm in the last period of the undergraduate course and I have interest in this field, but my university don't have a specialist in this area. So, I want a book that can introduce me in this field without many difficulty and that permite me study still others more advanced texts. Thanks!
AI: Here is a free book by Evans:
https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=0CC4QFjAA&url=http%3A%2F%2Fmath.berkeley.edu%2F~evans%2FSDE.course.pdf&ei=AhtsUraRJ87qkAfSp4Bo&usg=AFQjCNE53z56RnKcJrcmImNp4BT4egSFag&sig2=xo7gynJelChMeL0-An8adg
This is also a good text:
http://www.amazon.com/Stochastic-Differential-Equations-Introduction-Applications/dp/3540047581/ref=sr_1_1?ie=UTF8&qid=1382816460&sr=8-1&keywords=stochastic+differential+equations |
H: If a function is integrable, then it is bounded
Probably a simple question, but I wonder about the following.
I know that if a function $f : \mathbb{R} \rightarrow \mathbb{R} $ is (Riemann)integrable, then it is bounded. I wonder if I can generalize this to functions on $\mathbb{R}^3$ (now for an ingral over a volume).
It seemed logical to me that, because this theorem is true on $\mathbb{R}$, that it should be true on $\mathbb{R}^n$. But apparently it isn't, because the integral in theorem 10.1 of the document http://people.maths.ox.ac.uk/kirchhei/section_1008.pdf converges according to the theorem. The function under the integral is not bounded however. So can't I generalize the theorem that a integrable function should be bounded?
AI: It is not a theorem, it is part of the definition. The Riemann integral is defined for bounded functions on a bounded domain. If the function, the domain or both are unbounded, then the integral may exist as an improper integral. |
H: Evenness of Fourier coefficients implies even function
Let $f\in L^1(\mathbb{R}/2\pi\mathbb{Z})$ and let $F(n)$ denote its Fourier coefficients $$F(n)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{-inx}dx$$Assume that the Fourier coefficients determine an $L^1$ function.
How can we prove, without using Fourier transform, that if $F(n)=F(-n)$ for all $n$, then $f(x)=f(-x)$ for all $x$?
AI: Let $g(x) = f(x) - f(-x)$. Then $\hat{g}(n) = \hat{f}(n) - \hat{f}(-n) = 0$ for all $n$. So by the uniqueness theorem $g(x) = 0$ a.e.
I'm not sure what you mean about not using the Fourier transform since the statement itself involves Fourier coefficients, but the usual proof of this uses convolutions and taking limits of them, with not much emphasis on the Fourier transform side. |
H: the table at the end of Theoretical Computer Science Cheat Sheet
Theoretical Computer Science Cheat Sheet, created by Steve Seiden, is a hodgepodge of well-known mathematical theorems and notions.
I can understand (or guess at least) many of them, but I'm not sure about this 10-by-10 table at the end of the document. What is this matrix? The document has no explanation at all, and I'm wondering why the author put in a cheat sheet.
Is there any special meaning in computer science or math that this matrix stands for?
AI: Arranging the integers from $0$ to $99$ in a $10\times10$ square in such a way that no two in the same row or column have the same first digit or the same second digit.
It can be seen as a pair of orthogonal latin squares of order $10$: the first
digits form one square, the second digits the other.
Source: quick google search for the first row. See this PDF, 23.5.1. |
H: Linear Differential Equation $y'''−3y′+2y=\cos t+e^t$
I'm trying to find the solution to this non-homogenous third-order linear differential equation.
I know the solution is supposed to be:
$$c_1e^t+c_2te^t+c_3^{-2t}+\frac{e^tt^2}{6}-\frac{\sin(t)}{5}+\frac{\cos(t)}{10}$$
So far I've solved the left side of the equation to get the first half of the answer:
$$c_1e^t+c_2te^t+c_3^{-2t}$$
I don't know how to get the solutions from the right-hand side though. Thanks.
AI: You can try the method of undetermined coefficients by setting $y_p= Ae^t+Bte^t+Ct^2e^t+D\sin t+E\cos t$. You take the derivatives of $y_p$ and substitute into the equation. By equating the coefficients of same terms you can determine constants $A, B, C, D$ and $E$. |
H: Did I correctly set up this volume problem?
The question:
Find the volume of the solid generated by revolving about the y-axis the region bounded by:
$$y = x^2$$
$$x-axis$$
$$x = 2$$
First, I believe we want to do this problem in terms of y. If this is correct, we would then set it up like the following:
$$ V = \int_0^4 \pi [(2)^2 - (y^\frac{1}{2})^2] dy $$
$$ V = \int_0^4 \pi [4 - y] dy $$
$$ V = \pi [4y - \frac{1}{2} y^2 ] $$ from 0 to 4
$$ V = \pi [16 - 8] $$
I feel like I messed up with the problem, the $2^2$ doesn't seem right.
AI: Here's one way of solving this problem:
So you notice that the volume you get by rotating the region around the y-axis sort of resembles a shot glass, or maybe a cylinder with a bullet-shaped object carved out. Either way, it might not be the ideal shape to work with. But you notice that you can "fill the shot glass" or "fit the bullet inside" (whichever analogy you go with) to make a simple cylinder of radius $2$ and height $4$. So now it comes to finding the volume of the cylinder and subtracting the volume of the "drink" or "bullet".
The volume of the cylinder is easy: $\pi r^2 h = \pi \cdot 2^2 \cdot 4 = 16 \pi$.
Now, for the "bullet". You note that translating or rotating an object does not change its volume, so now let's instead consider the object that you get by rotating $y = 4 - x^2$ around the y-axis. By cutting this object into cylindrical "shells" of width dx, we get that each "shell" has the volume of $2\pi y dx = 2\pi (4-x^2) dx$. Integrating this from $0$ to $2$, we get
$\displaystyle \int_0^2 2\pi (4-x^2)dx = 2\pi \left(4x - \frac{x^3}{3}\right) = 2\pi \left(4\cdot2-\frac{2^3}{3}\right) = 2\pi \left(8-\frac{8}{3}\right) = \frac{32\pi}{3}$.
Finally, going back to the original question, we subtract the "bullet" from the cylinder and get $\displaystyle 16\pi - \frac{32\pi}{3} = \frac{16\pi}{3}$.
Of course, it is very possible to do the cylindrical shell method on the original object as well. Each shell has a volume of $2\pi y dx = 2\pi x^2 dx$, and integrating that from $x = 0$ to $x = 2$ gives us $\displaystyle \int_0^2 2\pi x^2 dx = 2\pi \int_0^2 x^2 dx = 2\pi \left(\frac{x^3}{3}\right) = \frac{16\pi}{3}$. And in fact, in this case it was actually easier to calculate it this way.
There's more than one way to go around solving problems like these. But either way, dealing with square roots tends to get a little ugly, so I avoided it by solving it in terms of $x$ instead of $y$. It is, of course, very possible to do it the way you were thinking of it, by solving the integral in terms of $y$. That's the beauty of math - there are multiple ways of solving a problem, but there's only one right answer - but the process can determine which method is the preferred one. |
H: Finding a marginal PDF of a joint probability distribution
I understand the idea of how to do it, but I'm currently getting a constant as my marginal PDF, which doesn't make sense.
The overall distribution is as follows:
$f(x,y) = 5ye^{-xy}$ for $0 < x, 0.2 < y < 0.4$
I'm trying to find the probability that $0 < x < 2$ given that $y = 0.25$, which should be $\frac{f(0<x<2,y=0.25)}{f_y(0.25)}$. As a minor double-check, should the top be a single integral evaluated with $y=0.25$?
The bulk of my question is that I'm finding $f_y$ to be a constant, which doesn't make sense. What am I doing wrong?
$f_y = \int_{0}^{\infty} 5ye^{-xy} dx = 5[-e^{-xy}]_{0}^{\infty} = 5[0-(-1)]=5$
AI: What am I doing wrong?
Answer: you are forgetting the support of the density.
A sure way to stop making this mistake is to write the densities rigorously, for example using indicator functions. In your case the density $f$ is the function defined on the whole set $\mathbb R^2$, by
$$
f(x,y)=5y\mathrm e^{-yx}\mathbf 1_{x\gt0}\mathbf 1_{0.2\lt y\lt0.4}.
$$
Hence, the second marginal of $f$ is the function $f_Y$ defined on the whole set $\mathbb R$, by
$$
f_Y(y)=\int_\mathbb R f(x,y)\mathrm dx=5\,\mathbf 1_{0.2\lt y\lt0.4}\int_0^\infty y\mathrm e^{-yx}\mathrm dx=5\,\mathbf 1_{0.2\lt y\lt0.4}.
$$
Thus, $f_Y$ is a constant on the interval $(0.2,0.4)$, which makes perfect sense. Once again, note that $f_Y(y)$ is well defined, for every $y$ in $\mathbb R$, for example $f_Y(3)=0$.
Edit: By definition, the conditional density of $X$ conditionally on $Y$ is defined for every $x$ and every $y$ such that $f_Y(y)\ne0$ by
$$f_{X\mid Y}(x\mid y)=f(x,y)/f_Y(y).
$$
(If $f_Y(y)=0$, one can set $f_{X\mid Y}(x\mid y)=0$.) Then, for every $y$ such that $f_Y(y)\ne0$,
$$
P[0\lt X\lt 2\mid Y=y]=\int_0^2f_{X\mid Y}(x\mid y)\mathrm dx.
$$
Try this computation and you will get a result for $P[0\lt X\lt 2\mid Y=y]$ between $0$ and $1$, as it should. |
H: How to get the equation of tangent plane when the point is an unknown?
Problem:
Solution:
I don't understand how they got the equation in the black box.
Here is my attempt:
$$\overrightarrow { \nabla } F({ x }_{ 0 },{ y }_{ 0 },{ z }_{ 0 })=(2x-2y)\widehat { i } +(6y-2x)\widehat { j } +8z\widehat { k } \\ \overrightarrow { \nabla } F({ x }_{ 0 },{ y }_{ 0 },{ z }_{ 0 })\cdot (x-{ x }_{ 0 },y-y_{ 0 },z-z_{ 0 })=0\\ (2x_{ 0 }-2y_{ 0 },6y_{ 0 }-2x_{ 0 },8z_{ 0 })\cdot (x-{ x }_{ 0 },y-y_{ 0 },z-z_{ 0 })=0\\ (2x_{ 0 }-2y_{ 0 })\cdot (x-x_{ 0 })+(6y_{ 0 }-2x_{ 0 })\cdot (y-y_{ 0 })+8z_{ 0 }(z-z_{ 0 })=0$$
AI: The gradient at $(x_0,y_0,z_0)$ is a normal to the surface. The equation that you've boxed is simply the gradient at $(x,y,z)$ from the previous line but with $(x,y,z) = (x_0,y_0,z_0)$ (and simplified slightly).
To complete the exercise, you can set up equations to solve for $P = (x_0,y_0,z_0)$ based on the known form of normal to planes parallel to the coordinate planes. For example, in part (a), note that the tangent plane at $P$ is parallel to the $xz$-plane if and only if it has the same normal vectors as the $xz$-plane. So its normal vectors are all of the form $c \widehat j$ for $c \neq 0$. So $\overrightarrow{\nabla}F(x_0,y_0,z_0) = 2(x_0-y_0)\widehat i + 2(3y_0-x_0)\widehat j + 8z_0 = c\widehat j$ for some $c\neq0$, from which we get (equating components)
\begin{align*}
2(x_0-y_0) &= 0\\
2(3y_0 - x_0) &= c\\
8z_0 &= 0.
\end{align*}
The rest of the solution is algebra (solving for $P$ in terms of $c$ and then resubstituting into the equation of the surface to solve for $c$). |
H: Evaluating $ \lim_{x\rightarrow \infty}e^{-x } + 2\cos(3x)$
Find the limit or prove that it does not exist by $\varepsilon-\delta$ approach:
$$ \lim_{x\rightarrow \infty}e^{-x } + 2\cos(3x)$$
Note:I found this question when I was doing exercise from the book Calculus:Early Transcendentals. The book just need me to show it does not exist, but I think it would be interesting to strictly prove it by $\varepsilon-\delta$ language.
Update: This question is not duplicate to the other question at all, since I am asking a strictly proof by $\varepsilon-\delta$ approach here rather than just show it does not exist. I also emphasized the requirement in my Note when I post this question. Although those two questions hold the same functions, but they have different requirements.Hence, It actually totally different from another one.
AI: No limit, since the limit along the sequence of general term $2n\pi$ is $2$ and the limit along the sequence of general term $2n\pi+\pi$ is $-2$. The asymptotics is best described by saying that the limit set is $[-2,2]$ (this is the set of possible limits).
Edit: To prove that the function $f:x\mapsto\mathrm e^{-x}+2\cos(3x)$ has no limit when $x\to+\infty$ using what you call the epsilon-delta approach, one can show:
$$
\forall N\gt0,\exists x\gt N,\exists y\gt N,|f(x)-f(y)|\gt1.
$$
Now that we know that $x$ and $y$ are to be looked for near odd and even multiples of $\pi$, the rest should follow. |
H: Properties of Natural Logarithm I need help finding the Derivative
$y=\ln(x)^2$
I am not sure why the answer would be $\frac{2\ln(x)}{x}$
I used this property "power rule" "$\ln(x^n) = n\ln(x)$
So i got $2\ln(x) $
the derivative of that using the constant multiplier rule i got
$\frac{2}{x}$
can I use the other chain rule to $y=f(u)$ and $g=g(x)$
Am i not supposed to bring that 2 in front becuase the whole expression is getting raised not the $x$?
Any help would be great,
AI: Apply the chain rule with $\;f(x):=x^2\;,\;\;g(x):=\log x$ :
$$(f(g(x)))'=f'(g(x))\cdot g'(x)$$
So
$$(\log x)^2=2\log x\cdot \frac1x=\frac{2\log x}x$$ |
H: Does $m(E)>0$ imply that $E$ must contain a nondegenerate interval?
Does $m(E)>0$ imply that $E$ must contain a nondegenerate interval?
$E\subset\mathbb{R}.$ $m$ refers to Lebesgue measure. $I$ refers to a nondegenerate interval.
AI: No, try any fat Cantor set. $ $ |
H: Limit of e with imaginary number
Important part:
$\lim\limits_{x\to\infty} e^{-ix} - e^{-x} $
This is suppose to approximate to "$1$" but the way I see it we have $0 - 0$ ...
AI: Ok, so the limit of a complex sequence (function) exists iff the limit of its real and imaginary parts exists:
$$e^{-ix}-e^{-x}=\cos x-e^{-x}-i\sin x\;,\;\;\text{and since}\;\lim_{x\to\infty}\sin x$$
doesn't exist the original limit doesn't, either.
Added: If you wanted instead
$$\lim_{x\to\infty}|e^{-ix}-e^{-x}|=\lim_{x\to\infty}\sqrt{\cos^2x-2e^{-x}\cos x+e^{-2x}+\sin^2x}=$$
$$=\lim_{x\to\infty}\sqrt{1-2e^{-2x}\cos x+e^{-2x}}=\sqrt{1-0+0}=1$$ |
H: Wolfram-Alpha's choice of $k$ in a complex logarithm
I'm puzzling on this complex integral:
$$ \int \frac{2ie^{it}}{2e^{it} - 1}dt = \log(2e^{it} -1)$$
The numerator is the derivative of the divisor, so the primitive is the log of the divisor.
When you ask Wolfram-Alpha to compute the integral over the range $0..2\pi$:
$$\left. \log(2e^{it} -1) \right|_0^{2\pi}$$
The answer is $2i\pi$. And that's probably a valid answer:
$$\log(2e^{i2\pi}-1) - \log(2e^{i0}-1) = \ln(1) - \ln(1) + 2k\pi i = ..., -2\pi i, 0, 2\pi i, ... $$
Still in this case, the canonical answer should be the one where $k=0$. When you ask Wolfram-Alpha for $log(1)$, it returns $0$, and not $2\pi i$.
Why does Wolfram-Alpha return $2 \pi i$ and not $0$ for the integral?
AI: Wolfram's answer is correct, and the only correct answer here. If you consider the integral as a function of the upper bound,
$$I(x) = \int_0^x \frac{2ie^{it}}{2e^{it}-1}\,dt,$$
you have a continuous function of $x$, and indeed we have
$$I(x) = \log (2e^{ix} - 1)$$
for some value of the logarithm, but that value depends continuously on $x$. When $x$ increases from $0$ to $2\pi$, the curve described by $2e^{it}-1$ winds once around the origin, so the imaginary part of the logarithm increases from $0$ to $2\pi i$, and $I(2\pi) = 2\pi i$. |
H: Condition on Fourier coefficients for real-valued function
Let $f\in L^1(\mathbb{R}/2\pi\mathbb{Z})$ and let $F(n)$ denote its Fourier coefficients $$F(n)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{-inx}dx$$ I want to prove that $f$ is real-valued if and only if $F(n)=\overline{F(-n)}$ for all $n$.
We have $$\overline{F(-n)}=\frac{1}{2\pi}\int_{-\pi}^{\pi}\overline{f(x)}e^{-inx}dx$$
If $f(x)$ is real, then $f(x)=\overline{f(x)}$, and clearly $F(n)=\overline{F(-n)}$.
If $F(n)=\overline{F(-n)}$, then $$\int_{-\pi}^{\pi}f(x)e^{-inx}dx=\int_{-\pi}^{\pi}\overline{f(x)}e^{-inx}dx$$
How can I conclude that $f$ is real-valued?
AI: Define $g(x):=f(x)-\overline{f(x)}$: we have to show that $g(x)=0$ for all $x$. We have $\int_{-\pi}^\pi g(x)e^{-inx}\mathrm dx=0$ for all $n$. Function of the form $\sum_{j=-N}^Na_je^{ijx}$ are dense in $C[-\pi,\pi]$, hence
$$\int_{-\pi}^\pi g(x)h(x)\mathrm dx=0$$
for each $h\in C[-\pi,\pi]$. By an other approximation argument, we get that for each closed subset of $[-\pi,\pi]$, $\int_Fg(x)\mathrm dx=0$. |
H: How can the length of a normal vector matter?
Problem:
Solution:
Graph:
For part a, I understand mathematically why the value of c matters. What I don't understand is how it can possibly matter intuitively.
I get that $\overrightarrow { \nabla } F(x_{ 0 },y_{ 0 },z_{ 0 })=(0,c,0)$ and therefore, c can't be any value. My question is: how can a tangent plane possibly depend on the length of the normal vector? If the vector is normal, then the length shouldn't matter because the vector will always be normal for all values for c except 0.
AI: You want to find all points on the surface where the normal vector points in a specified direction (along the $y$-axis), but the easiest way to calculate the normal vector (using gradient) not only gives you a direction, but full-fledged vector. In order to prepare for this, you need to look for places where the gradient goes out in a specific direction, without specifying from the start what length of vector you're looking for in said gradient.
Solving the equation $\nabla F(x_0, y_0, z_0) = (0, c, 0)$ gives you a curve parametrized by $c$ (in this case $(c/4, c/4, 0)$) for which the gradient of $F$ always has direction along the $y$-axis. Then you solve $F(c/4, c/4, 0) = 16$ in order to figure out where that curve intersects with your surface. |
H: Integration for Fourier coefficients of $x$
To compute the Fourier coefficients of $x$, I was trying to integrate $$\int_{-\pi}^\pi xe^{-inx}dx$$ How to integrate this? (I already did it, and I'm posting my answer just to keep the records and possibly help others in the future.)
AI: So I used integration by parts. $$\int_{-\pi}^\pi xe^{-inx}dx=x\frac{e^{-inx}}{-in}-\int_{-\pi}^\pi\frac{e^{-inx}}{-in}dx$$
The first term is $\dfrac{i}{n}(\pi(\cos(-n\pi)+i\sin(-n\pi)+\pi(\cos(n\pi)+i\sin(n\pi))=\dfrac{2\pi i}{n}\cos(n\pi)$
The second term is $-\dfrac{1}{n^2}(\cos(-n\pi)+i\sin(-n\pi)-\cos(n\pi)-i\sin(-n\pi))=\dfrac{2i}{n^2}\sin(n\pi)$
EDIT: As N.S. helpfully pointed out, when $n=0$ this should be done separately. So it's $\int_{-\pi}^\pi xdx=0$. Also, for other $n$ the answer is $\dfrac{2\pi i}{n}\cos(n\pi)-\dfrac{2i}{n}\sin(n\pi)=\dfrac{2\pi i(-1)^n}{n}$ |
H: Matrix representation of Automata
Is anyone know if there is any tutorial for the matrix representation of automata?? I am taking a theoritical computer science in this semester and the professor uses the matrix in his lecture. I gonna have test next week so I have to study for it. I tried to find any tutorial on good but there was not useful tutorial.. doe anyone know any tutorial??
AI: Let $A = \begin{pmatrix} a & b \\ c & d\end{pmatrix} $ where the entries are alphabet letters and $A_{ij} = a \iff$ there's an $a$ transition from state $i$ to state $j$ labeled $a$. Then $A^2 = \begin{pmatrix} a^2 + bc & ab + bd \\ ca + dc & cb + d^2\end{pmatrix}$, where $+$ means language union. Each entry is the language of all $2$-length substrings of your automata from state $i$ to state $j$. A similar statement holds for $A^k$.
Take the language $b^* a$ which has automata $0 \to^a 1$, $0 \to^b 0$, where $0$ is the starting node and $1$ the final node. Its matrix looks like
$$
A =
\begin{pmatrix}
b & a \\
\varnothing & \varnothing
\end{pmatrix}
$$
$$
A^k = \begin{pmatrix}
b^k & b^{k-1} a \\
\varnothing & \varnothing
\end{pmatrix}
$$
Exercise: prove the above characterization.
Let $f(1), \dots f(k)$ be fhe final states of an automaton $A. \ $ Then the language accepted by $A$ can be calculated using its matrix, also called $A$:
Let $B = A^0 + A^1 + \dots \ $ Then $L(A) = B_{1f(1)} + B_{1f(2)} + \dots + B_{1f(k)}$, where $1$ is the starting state.
Let's consider the boolean automata
$1 \rightarrow^0 2 \circlearrowright^1\\
\downarrow^1 \nearrow_0\\
3$
Directly represent it as the matrix $A$ where $A_{ij}$ holds the strings of length $1$ from state $i$ to state $j$:
$$
A = \begin{pmatrix}
\varnothing & 0 & 1 \\
\varnothing & 1 & \varnothing \\
\varnothing & 0 & \varnothing
\end{pmatrix}
$$ |
H: Proof with mathematical induction that $ (\frac{n}{n+1})^2 + (\frac{n+1}{n+2})^2 + ... + (\frac{2n - 1}{2n})^2 \le n - 0.7 $
Proof with mathematical induction.
I have the following induction problem:
$ (\frac{n}{n+1})^2 + (\frac{n+1}{n+2})^2 + ... + (\frac{2n - 1}{2n})^2 \le n - 0.7 $
This property applies to all $n \ge 1$. I've shown that it is true for $n = 1$ and $n = k$, however I can not show for $n = k+1$.
This is how I start for $n = k+1$:
$ (\frac{k+1}{k+2})^2 + (\frac{k+2}{k+3})^2 + ... + (\frac{2k - 1}{2k})^2 + (\frac{2k}{2k + 1})^2 + (\frac{2k + 1}{2k + 2})^2 \le k + 1 - 0.7 $
Any ideas?
AI: You are almost there. At the end, you wrote the difference between the two sums as a single complicated fraction. Maybe you have miscalculated something there, maybe not, anyway, it is not easy to see through. It becomes much easier if we proceed in steps, pairing like things with like things. The difference is
$$\begin{align}
\left(\frac{2n}{2n+1}\right)^2 + \left( \frac{2n+1}{2n+2}\right)^2 - \left(\frac{n}{n+1}\right)^2
&= \left(\frac{2n}{2n+1}\right)^2 + \left( \frac{2n+1}{2n+2}\right)^2 - \left(\frac{2n}{2n+2}\right)^2\\
&= \left(\frac{2n}{2n+1}\right)^2 + \frac{(2n+1)^2-(2n)^2}{(2n+2)^2}\\
&= 1 - \frac{(2n+1)^2 - (2n)^2}{(2n+1)^2} + \frac{(2n+1)^2-(2n)^2}{(2n+2)^2}\\
&= 1 - (4n+1)\left(\frac{1}{(2n+1)^2}-\frac{1}{(2n+2)^2}\right)\\
&< 1.
\end{align}$$ |
H: Triangle bisection on non-right angle triangle with known angle and two sides
I'm creating 3D road intersections and to create the corner points I detect when the edge vectors of the road intersect. It's extremely accurate.
However, I would like to pre-calculate the corner positions instead. The interior angles of the intersection are all known and are never 90 degrees.
I've boiled it down to this 2D problem where side a and b are always known and the angle specified is random but always known. A and B will always equal each other. Everything else is unknown:
How do I get the length of line D (bisect)?
AI: If the triangle is isosceles, then the intersection of the lines D and the line C forms a right angle and so
$$ D = B \cos\left(\dfrac{\theta}{2}\right) $$ |
H: The derivative of $e^x$ using the definition of derivative as a limit and the definition $e^x = \lim_{n\to\infty}(1+x/n)^n$, without L'Hôpital's rule
Let's define
$$
e^x := \lim_{n\to\infty}\left(1+\frac{x} {n}\right)^n, \forall x\in\Bbb R
$$
and
$$
\frac{d} {dx} f(x) := \lim_{\Delta x\to0} \frac{f(x+\Delta x) - f(x)} {\Delta x}
$$
Prove that
$$
\frac{d} {dx} e^x = e^x
$$
using the definition of $e^x$ and derivation above, without using L'Hôpital's rule or the "logarithm trick" and/or the "inverse function derivative trick".
$$
\left( \frac{d} {dx} f^{-1}(x)= \frac{1} {\left(\frac{d}{d(f^{-1}(x))}f(x)\right)(f^{-1}(x))}\right)
$$
Or equivalently prove that the following two definiton of $e$ are identical
$$
1)\space\space\space\space e =\lim_{n\to\infty}(1+\frac{1} {n})^n
$$
$$
2) \space\space\space\space e\in\Bbb R,\space\space(\frac{d} {dx} e^x)(x=0) = 1
$$
What I've got is
$$
\frac{d}{dx}e^x=e^x \lim_{\Delta x\to0}\frac{e^{\Delta x} - 1} {\Delta x} =
e^x \lim_{\Delta x\to0}\lim_{n\to\infty}\frac{\left(1+\frac{\Delta x}{n}\right)^{n}-1}{\Delta x} =
e^x \lim_{\Delta x\to0}\frac{e^{0+\Delta x}-e^0}{\Delta x}
$$
If i assume that $n\in\Bbb N$ I could use binomial theorem but I didn't got much out of it. Wolframalpha just uses L'Hospital rule to solve it, but I am looking for an elementary solution. What I'm interested in is basically is the equivalence of the two definition of $e$ mentioned above. And I'd like to get a direct proof rather than an indirect(I mean which involves logarithms or the derivatives of invers functions).
I look forward getting your aswers.
AI: Define the function sequence $f_n(x):=\left(1+\frac{x}{n}\right)^n$. Then all $f_n$ are differentiable, and you can easily show by using the definition of the derivative that
$$
f'_n(x)=f_n(x)\left(1+\frac{x}{n}\right)^{-1},
$$
hence $f_n'(x)\to e^x$ as $n\to\infty$. Moreover, one can show that $f_n$ converges locally uniformly to $e^x$ and hence $f_n'$ converges locally uniformly to $e^x$ as $n\to\infty$ (show that $f_n\leq f_{n+1}$, i.e. $f_n$ is monotonically increasing, and use Dini's theorem). Finally we conclude, that $(e^x)'=\lim_{n\to\infty}f_n'(x)=\lim_{n\to\infty}f_n(x)=e^x$. |
H: the value of $\int_{0}^{2\pi}f(e^{it})\cos t dt$?
$f$ be analytic function then what is the value of $\int_{0}^{2\pi}f(e^{it})\cos t dt$?
AI: Let $z=e^{it}$ the the integral becomes $$\int_{|z|=1}f(z)({z+z^{-1}\over 2}){dz\over iz}=0={1\over 2i}\int_{|z|=1} f(z)dz+{1\over 2i}\int_{|z|=1}{f(z)\over z^2}dz=\pi f'(0)$$. |
H: Using DeMorgan's Laws to complement a function
Using DeMorgan's Law, write an expression for the complement of $F$ if:
$F(x,y,z) = x(y' + z)$.
$F=x'+(y'+x)'$
$F(x,y,z) = xy + x'z + yz'$
$F=(xy)'(x'z)'(yz')'$
$F(w,x,y,z) = xyz' (y'z + x)' + (w'yz + x' )$.
$F=[(xyz')'+(y'z+x)](w'yz+x')'$
My answers are underneath the numbered questions. Is everything correct? I'm not 100% sure as to what I'm exactly supposed to do. I just took all the ANDs, negated them and made them ORs and vice-versa.
AI: You don't have to stop there. Continue using De Morgan until you get to these simpler forms.
$F^c = x' + y z'$
$F^c = (x'+y')(x + z')(y'+z)$
$F^c = x(w+y' +z')$
Since you asked, let me briefly explain how I got (3). At some point during the reduction you will come across a long formula like this:
$((x' + y' + z) + (y'z + x))((w + y' + z')x)$
Now, noticing that disjunction (+) is associative we can collapse the parens on the left to get:
$(x' + y' + z + y'z + x)((w + y' + z')x)$
And here we immediately realize that we're dealing with the tautology $(x + x')$:
$(\color{blue}{x' +} y' + z + y'z + \color{blue}{x})((w + y' + z')x)$
This means that the entire formula on the left is materially equivalent ($\equiv$) to $\top$:
$\color{blue}{\top}((w + y' + z')x)$
Now since $(\top \land \phi) \equiv \phi$, for all $\phi$, we cancel it, getting:
$(w + y' + z')x$
Lastly, since conjunction is commutative, I just put the x in front to obtain:
$x(w + y' + z')$ |
H: Solution of $y''+4y=\cos^2t$
I'm trying to find the solution, which is supposed to be
$y(t)=c_2\sin(2t)+c_1\cos(2t)+\frac{t\sin(2t)}{8}+\frac{\cos^2(t)}{4}$,
but I'm doing something wrong along the way and I can't figure out what I did. I was hoping someone would be able to find my mistake for me. I'll show what I've gotten:
First I got the complementary solution $y_c(t)=c_1\cos(2x)+c_2\cos(2x)$
I then used the particular solution $y_p=A+B\cos(2t)+C\sin(2t)$ - I'm guessing this is where my error occurs, but I don't see what's wrong with it.
I derived it two times and got $y_p''=-4B\cos2t-4C\sin(2t)$
I plugged it into the original equation (assuming $\cos^2t$ is $\frac{1}{2}+\frac{\cos(2x)}{2}$) and then tried to solve the problem, at which point I found myself running in circles:
$\sin2t(-4C-4C)+\cos2t(-4B+4B)+4A=\frac{1}{2}+\frac{\cos(2x)}{2}$.
Can anyone help me find my mistake(s)? Thanks a lot.
AI: You figured out the issue with:
$$y_c(t) = c_1 \cos 2t + c_2 \sin 2t$$
For the particular solution, some care is needed given the homogeneous (complementary) solution. We expand the $\cos^2 2t$ term as:
$$\cos^2 2t = \dfrac{1}{2}\left(1 +\cos 2t\right)$$
Now, we notice that this has a common term with the homogeneous term, so we multiply the particular solution by $t$ to account for that. Thus, we choose a particular solution as:
$$y_p(t) = a + b~ t \cos 2t + c~ t \sin 2t$$
Now, substitute and solve for $a$, $b$ and $c$.
You should get:
$$a = \dfrac{1}{8}, b = 0, c = \dfrac{1}{8}$$
Your final solution will be:
$$y(t) = y_c(t) + y_p(t)$$ |
H: Complex integral of an exponent divided by a linear ($\int \frac{e^u}{u-1}$)
Here is the question I'm working on:
Evaluate the following integral:
$$ \oint_{|z+1|=1} \frac{\sin \frac{\pi z}{4}}{z^2-1}dz$$
I've tried along the following line. Substitute $sin(z) = \frac{e^z-e^{-z}}{2i}$:
$$ \frac{1}{2i} \int \frac{e^{\pi z/4}-e^{-\pi z/4}}{z^2-1}dz$$
The contour is a circle of radius 1 around $-1$, so substitute $z=-1+e^{it}$, with $dz = ie^{it}dt$:
$$ \frac{1}{2i} \int_0^{2\pi} \cfrac{e^{\pi/4(e^{it}-1)}-e^{-\pi/4(e^{it}-1)}}{(e^{it}-1)^2-1}ie^{it}dt = \frac{1}{2} \int \cfrac{e^{\pi/4(e^{it}-1)}-e^{-\pi/4(e^{it}-1)}}{e^{it}-2}dt$$
A repeat of the same substitution $u = e^{it} - 1$ with $du = ie^{it}dt$ gives:
$$\frac{1}{2} \int \cfrac{e^{\pi u/4}-e^{-\pi u/4}}{u-1}du$$
At this point I've run out of ideas. Any hint or pointer would be appreciated.
AI: Use the Residue Theorem:
$$\oint\limits_{|z+1|=1}\frac{\sin\frac{\pi z}4}{z^2-1}dz=\frac12\left(\overbrace{\oint\limits_{|z+1|=1}\frac{\sin\frac{\pi z}4}{z-1}dz}^{=0}-\oint\limits_{|z+1|=1}\frac{\sin\frac{\pi z}4}{z+1}dz\right)=$$
$$=\left.-\frac{2\pi i}2\sin\frac{\pi z}4\right|_{z=-1}=\frac{\pi i}{\sqrt2}$$ |
H: Problem with permutations
The problem says:
We have strings formed by two letters, followed by two digits and then followed by three letters. In each group repetitions are not allowed, but the last group of three letters can contain up to one of those used in the first group. If the number of letters available is 12 how many different strings can be formed?
A) 23522400
B) 980100
C) 7840
Well, my solution is this:
Considering that the first letter is repeated, the letters and digits in order:
(12*11)*(10*9)*(12*10*9)=12830400
Considering that the second letter is repeated, the letters and digits in order:
(12*11)*(10*9)*(11*10*9)=11761200
So, by the sum of the two cases: 12830400+11761200=24591600
The result of my solution is not in the list of choices, but is close to A.
I noticed that A can be obtained with: 11761200+11761200=23522400, wich i dont understand why that would be correct, this is like saying 12830400+12830400 is ok too...
I think A is wrong, but the others are wrong too, so maybe I am doing something wrong.
AI: There are two cases, depending on whether the last group repeats one letter of the first group or not.
If all five letters are different, there are $(12\cdot11)\cdot(10\cdot9)\cdot(10\cdot9\cdot8)=8,553,600$ possible strings: $12$ choices for the first letter, $11$ for the second letter, $10$ for the first digit, $9$ for the second digit, $10$ for the third letter (because it has to be different from the first two), $9$ for the fourth letter, and $8$ for the last letter.
If the last group of letters has one letter in common with the first group, there are $12$ ways to choose that letter. Then we have to decide whether it’s the first or the second letter in the first group; we can make that choice in $2$ ways. We also have to decide whether it’s the first, second, or third letter of the last group, and we can make that choice in $3$ ways. Then there are $11$ choices for the other letter of the first group, $10$ for the first digit, $9$ for the second digit, $10$ for the first of the two remaining letters in the last group, and $9$ for the second of the two remaining letters in the last group. Altogether, then, we have $$12\cdot2\cdot3\cdot11\cdot10\cdot9\cdot10\cdot9=6,415,200$$ possible strings.
The final total is $8,553,600+6,415,200=14,968,800$; none of the answers in your list is correct. |
H: Prove that $n^9 \equiv n \pmod{30}$ if $(30,n) > 1$
I'm looking at the following number theory problem:
Prove that $n^9 \equiv n \pmod{30}$ for all positive integers $n$ if $(30,n) > 1$.
It is easy to show that $n^9 \equiv n \pmod{30}$ if $(30,n)=1$ by using Euler's Theorem. Is there any way to prove the above easily when $(30,n)$ are not relatively prime? I'm looking for some easy realization, and not factoring $n^9-n=n(n^8-1)=n(n^4+1)(n^4-1)=n(n^4+1)(n^2+1)(n-1)(n+1)$ and then deducing divisibility. Is there a simple realization for this?
Thanks!
AI: If $p$ is one of $2,3,5$ then $n^9\equiv n\pmod{p}$. This is true by Fermat's Theorem if $n$ and $p$ are relatively prime, and trivially true if $p$ divides $n$. |
H: Examples of complex functions with infinitely many complex zeros
What are some examples of complex functions with infinitely many complex zeros? There are no particular restrictions on the functions I am just curious and having a hard time finding examples. Also what can be said about a complex function with infinitely many complex zeros, must they have any special properties?
AI: There is nothing special about functions with infinitely many zeros. In fact, it is the norm.
If $f(z)$ is any entire function which isn't a polynomial, then $\infty$
is an essential signularity of $f(z)$. By Big Picard's theorem, $f(z)$ takes on all possible values of $\mathbb{C}$, with at most a single exception, infinitely often.
This means if your $f(z)$ is entire, not a polynomial with finitely many zeroes, then $f(z) + \alpha$ for any constant $\alpha \ne 0$ has infinitely many zeros. |
H: Rotation Matrix inverse using Gauss-Jordan elimination
I'd like to calculate the inverse of a rotation matrix, let take the simplest case which is a $2$ x $2$ rotation matrix:
$R =\begin{bmatrix} \cos \theta & -\sin \theta \\[0.3em] \sin \theta & \cos \theta \end{bmatrix}$
I know that the inverse is the following
$R^{-1} =\begin{bmatrix} \cos \theta & \sin \theta \\[0.3em] -\sin \theta & \cos \theta \end{bmatrix}$
and I know that I can calculate it using the transpose method as such:
$R^{-1}=R^T$
but I fail to calculate the inverse using $Gauss-Jordan$ elimination, that is I don't know how to substract $\cos \theta$ from $\sin \theta$ in the second row. It all gets a bit complicated; I've looked around and nobody has a full step method using $G.-J.$ only the solution or the transpose method. Could someone provide me a full-step solution using $G.-J.$?
AI: $$
\begin{bmatrix}
\cos t&-\sin t&|&1&0\\ \sin t&\cos t&|&0&1
\end{bmatrix}
\xrightarrow{\frac1{\cos t}R1}
\begin{bmatrix}
1 &-\frac{\sin t}{\cos t}&|&\frac1{\cos t}&0\\ \sin t&\cos t&|&0&1
\end{bmatrix}
\xrightarrow{R2-\sin t\,R1}
\begin{bmatrix}
1 &-\frac{\sin t}{\cos t}&|&\frac1{\cos t}&0\\ 0&\cos t+\frac{\sin^2t}{\cos t}&|&-\frac{\sin t}{\cos t}&1
\end{bmatrix}=
\begin{bmatrix}
1 &-\frac{\sin t}{\cos t}&|&\frac1{\cos t}&0\\ 0&\frac1{\cos t}&|&-\frac{\sin t}{\cos t}&1
\end{bmatrix}
\xrightarrow{\cos t\,R2}
\begin{bmatrix}
1 &-\frac{\sin t}{\cos t}&|&\frac1{\cos t}&0\\ 0&1&|&-\sin t&\cos t
\end{bmatrix}
\xrightarrow{R1+\frac{\sin t}{\cos t}R2}
\begin{bmatrix}
1 &0&|&\frac1{\cos t}-\frac{\sin^2t}{\cos t}&\sin t\\ 0&1&|&-\sin t&\cos t
\end{bmatrix}
=\begin{bmatrix}
1 &0&|&\cos t&\sin t\\ 0&1&|&-\sin t&\cos t
\end{bmatrix}
$$
This is a terrible method to calculate the inverse of any $2\times 2$ matrix.
Edit: of course this does not work when $\cos t=0$; but this is a much easier case: you simply divide by $\sin t$ and permute the rows. |
H: An example where $E[X_1 X_2] - E[X_1]E[X_2] = 0$ for functions $X_1$ and $X_2$
I computed $E[X_1 X_2] - E[X_1]E[X_2]$ using $X_1 = 3x+1$ and $X_2 = 2x+5$ the following way:
$E[(3x+1)(2x+5)] - E[3x+1]E[2x+5] = E[6x^2 + 17x + 5] - (3E[x]+1)(2E[x]+5) = 6(E[x^2] - E^2[x]) = 6\sigma^2$
When I was done I asked myself when does $E[x^2] = E^2[x]$ for some functions $X_1$ and $X_2$? But I couldn't think of a good example. I also wondered why I didn't get zero since $X_1$ and $X_2$ appear to be independent.
AI: I am assuming you mean $X$ is a random variable (most people would use a capital letter here), and $X_1$ and $X_2$ are the random variables defined by $X_1 = 3X+1$ and $X_2=2X+5$. The answer to your question is that $E(X^2)-E(X)^2 =0$ if and only if $X$ is a constant. This is because $E(X^2)-E(X)^2$ is the variance of $X$. And if $X$ is a constant, $X_1$ and $X_2$ are constants, and if you look carefully at the definition of independence, you'll find this implies $X_1$ and $X_2$ are independent.
So, if $E(X^2)-E(X)^2 \neq 0$, then $X_1$ and $X_2$ are not independent. You wouldn't expect them to be, anyway (unless $X$ is constant), since one is an affine function of the other. |
H: Should I be able to prove Law of Cosines, Half Angle formula, etc?
This is more of a general question then the title suggests, but the laws in the title are what I'm currently studying. I can read the proofs of both and understand them after a while, but I could never produce such a proof unless I committed it to memory which is really my question: should I be concerned about needing to prove something like the Half Angle formula for the sine function on my own is or is it enough to be able to read the proof and understand it?
My long term goals are to read Apostol's calculus or Spivak's but right now I'm just studying Pre-Calc. In the case of Spivak's book I know he gives a proof of why $-a*b = -(a*b)$ for positive $a$ and $b$ and I can follow it but I could never have come up with it on my own. Should I be discouraged by that or does that just mean I need to study harder. Anyways, Cheers!
AI: All the trig identities become easy to remember and easy to prove once you learn the connection between complex exponentiation and trig functions.
For me personally, I did not learn about that connection until sometime in college, after or during a calculus course on sequences and series. Until then, I was aware that trig identities existed and was comfortable looking them up when I thought they might help. |
H: Real projective $n$ space
We define $\sim$ on $\mathbf{R}^n - \{0\}$ by $x \sim y$ if $x = \lambda y$ for some $\lambda \in \mathbf{R}$. We define projective $n$ space by $X = (\mathbf{R}^n - \{0\})/{\sim}$.
I am having trouble showing that $X$ is an $(n-1)$-dimensional topological manifold. The definition of a topological manifold that I'm working with is the following: $X$ is a topological $n$-manifold if, for all $x \in X$, there is an open nhood $U$ of $x$ such that $U$ is homeomorphic to an open subset of $\mathbf{R}^n$.
I understand that one can take $U$ to be homeomorphic to any open ball in $\mathbf{R}^n$, or any open nhood ball in $\mathbf{R}^n$.
I am also aware that the open nhoods of $\overline{x} \in X$ will look like a half-sphere (at least I think so; I think my work was OK here).
AI: Hint: If $[x_0:\cdots:x_n]\in X$ corresponds to the line $\lambda(x_0,\ldots,x_n)$ and $x_i\neq0$, then $[x_0:\cdots:x_n]=[x_0/x_i:\cdots:x_{i-1}/x_i:1:x_{i+1}/x_i:\cdots:x_n/x_i]$.
Now on the open set $U_i=\{[x_0:\cdots:x_n]\in X:x_i\neq0\}$, take the chart $U_i\to\mathbb{R}^{n}$ where $[x_0:\cdots:x_n]\mapsto(x_0/x_i,\ldots,x_n/x_i)$ (with the $i$-th coordinate removed).
EDIT I realize I used $\mathbb{R}^{n+1}$ mod the action of $\mathbb{R}$, not $\mathbb{R}^n$ as in the question above. Sorry! |
H: fundamental complex integral( in Conway's book)
I am reading Conway's Functions of One Complex Variable. I have some trouble in doing some exercise.
Find $\int_\gamma(z^2-1)^{-1}dz$, where $\gamma$ is a path. $\gamma(t)=1+e^{it}$ for $0\leq t\leq 2\pi$.
Find $\int_\gamma(z^2-1)^{-1}dz$, where $\gamma$ is a path. $\gamma(t)=2e^{it}$ for $-\pi\leq t\leq \pi$.
These two are similar. They are in Chapter 4 Section where Riemann-Stieltjes integral is just defined. Since Cauchy theorem is introduced in later chapter, I guess we can solve these without using Cauchy theorem or its corollaries.
I know
$$
\int_\gamma\frac 1{z^2-1}=\frac12\int_\gamma\frac 1{z-1}-\frac1{z+1}dz=\frac 12\int_\gamma\frac 1{z-1}dz-\frac 12\int_\gamma\frac1{z+1}dz
$$
What should I do next?
I know the question I ask is very fundamental but I am stuck. Thanks in advance.
AI: The definition of a complex integral like this is
$$\int_{\gamma} f(z) dz = \int_a^b f(\gamma(t)) \gamma'(t) dt$$
where $a$ and $b$ are the endpoints of the interval parameterizing $\gamma$. So here, $\gamma(t) = 1 + e^{it}$ and $a = 0$, $b = 2\pi$, so
\begin{align*}
\int_{\gamma} \frac{1}{z^2 - 1} dz &= \int_0^{2\pi} \frac{1}{(1 + e^{it})^2 - 1} ie^{it} dt \\
&= \int_0^{2\pi} \frac{ie^{it} dt}{2e^{it} + e^{2it}} \\
&= \int_0^{2\pi} \frac{i dt}{2 + e^{it}}
\end{align*}
I'll leave it to you to finish evaluating the integral. The other one is done similarly. |
H: Let V denote the Klein 4-group. Show that $\text{Aut} (V)$ is isomorphic to $S_3$
After a week in my Abstract Algebra class, the professor proposed this as a problem. I'm not entirely sure where to begin. $ V = \{ e, \tau, \tau_1, \tau_2 \}$, so I'm not sure exactly what is meant by $\text{Aut} (V)$. Is it simply saying that the only way to have an automorphism on $V$ is to rearrange the order of the elements of $V$ and therefore $\text{Aut} (V) \cong S_3$?
AI: Any automorphism $\phi$ of $V$ sends its identity element $e$ to itself. What is interesting is the way that $\phi$ rearranges the other three elements. The claim here is that any of the six possible bijections from $\tau, \tau_1, \tau_2$ to itself is in fact a group automorphism of $V$. This will give an isomorphism from $\operatorname{Aut}(V) \cong S_3$ that sends an automorphism $\phi \in \operatorname{Aut}(V)$ to its corresponding permutation of the three elements $\tau, \tau_1, \tau_2$. |
H: How do I get an answer of $14$ using simpsons rule for $\frac{152e}{180n^4}<.0001$
I must have the algebra wrong somewhere but here is the original equation:
$$\frac{152e}{180n^4}<.0001$$
If I then multiply like this:
$$152e<.0001(180)n^4$$
Which then gives:
$$152e < .0018n^4$$
And then dividing:
$$\frac{152e}{.0018} < n^4$$
And then taking the fourth root:
$$\sqrt[4]{\frac{152e}{.0018}}< \sqrt[4]{n^4}$$
And I get about $n = 46.338$
But the book gives a solution of $14$. Can someone explain where I went wrong?
AI: $\dfrac{152 \mathrm{e}}{180n^4}<0.0001$
$152 \mathrm{e} < 0.018n^4$
$\dfrac{152 \mathrm{e} }{0.018} < n^4$
$n > \sqrt[4]{22954.38}$
$n > 12.31$
This still doesn't explain why the solution is listed as $14$, though. |
H: What's the probability that nine people were born in the same two months (but not the same month)?
Find the probability that nine people were born in the same two months (but not
all in the same month).
No clue how to approach this. I was thinking well you have to choose 8 out of the 9 people and then match them up to one month. Then take the remaining person and match him up to another month. How do i do that?
AI: HINT: Imagine that you pick your $9$ people and arrange them in alphabetical order, say. Since there are $12$ months, there are $12^9$ possible sequences of birth months, and you’re intended to assume that they are equally likely. You need to find $N$, the number of sequences that meet the requirement that exactly two months appear in the sequence; the desired probability will then by $\frac{N}{12^9}$.
You cannot assume that $8$ people were born in one month and $1$ in another: it’s quite possible, for instance, that $3$ were born in one month and $6$ in another. I’ll get you started on one possible approach to counting these sequences.
Let $m_1$ be the birth month of the first person in line. There has to be exactly one other birth month represented in this group; Call it $m_2$. Let $S$ be the set of people among the remaining $8$ who were born in $m_1$; $S$ cannot be the entire set of $8$ people, since at least one of them has to have been born in $m_2$, but that’s the only restriction. Note that once we know $S$, $m_1$, and $m_2$, we know the entire sequence: the first person and everyone in $S$ were born in $m_1$, and every one of the $8$ who is not in $S$ was born in $m_2$.
How many ways are there to choose $m_1$?
How many ways are there to choose $m_2$?
How many ways are there to choose $S$?
Finally, how should you combine the answers to the first three questions in order to get $N$? |
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