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Real matrix with the property that every nonzero vector in $\mathbb{R}^n$ is an eigenvector of $A$. so I'm supposed to let $A$ be a square real matrix with the property that every nonzero vector in $\mathbb{R}^n$ is an eigenvector of $A$. And I'm supposed to show that $A=\lambda I$ for a constant $\lambda$ in $\mathbb{R}$.
But I am having a lot of trouble with how to start this proof... Does it have anything to do with $Av=\lambda v$? I'm really confused. Any help would be appreciated.
|
If $n=1$, there's nothing to prove, as all matrices are scalar multiples of the identity.
If $n\ge 2$, take two linearly independent vectors $v_1,v_2$ with corresponding eigenvalues $\lambda_1$, $\lambda_2$. Since every vector is an eigenvector, so is $v_1-v_2$ with corresponding eigenvalue $\lambda_3$. So then $A(v_1-v_2) = \lambda_3 (v_1 -v_2) = \lambda_3v_1 - \lambda_3v_2$.
Distributing $A$ however gives us $A(v_1-v_2) = Av_1-Av_2 = \lambda_1v_1 - \lambda_2v_2$, and so $\lambda_1v_1 - \lambda_2v_2 = \lambda_3v_1 - \lambda_3v_2$. Combining like terms on both sides, we see $(\lambda_1-\lambda_3)v_1 + (\lambda_3-\lambda_2) v_2 = 0$, so then as $v_1$ and $v_2$ are linearly independent, $\lambda_1-\lambda_3 = 0$ and $\lambda_3-\lambda_2 = 0$. Thus $\lambda_1=\lambda_2=\lambda_3$.
Since this works for any two vectors, namely it works for any pair of basis vectors. So letting $\{ v_1,\cdots, v_n\}$ be a basis for $\mathbb{R}^n$, we see that the above shows that every basis vector is sent to a scalar multiple of itself, and all those scalars are the same, say $\lambda$. Thus for any $v \in \mathbb{R}^n$, we can write $v = a_1v_1+\cdots +a_nv_n$ where each $a_i \in \mathbb{R}$, and: $$Av = A\displaystyle\sum_{i=1}^n a_iv_i = \sum_{i=1}^n a_iA(v_i) = \sum_{i=1}^n a_i\lambda v_i = \lambda\sum_{i=1}^n a_iv_i = \lambda v.$$ Since this holds for any $v \in V$, we have $A = \lambda I$.
Notably, we used nothing about the fact that we were using a matrix, real or not, and any property of $\mathbb{R}^n$. This fact actually holds for any linear operator on a finite dimensional vector space satisfying this property. It may work in infinite dimensions too, but I haven't done much infinite dimensional linear algebra.
|
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Limit to zero of the $p$-norm I have the $p$-norm defined as $$\|x\|_p=\left(\sum_{k=1}^n|x_k|^p\right)^\frac{1}{p}.$$ I am trying to find the limit as $p\to0^+$ of $\|x\|_p$. I've seen it defined as $\{x_k:x_k\neq0\}$. Why is this true? Is there a way to prove this?
|
Let $m=\min\{|x_i|:\ x_i\neq0\}$, and $k$ the number of non-zero $x_i$. Then
$$||x||_p=m\cdot\left(\sum_{i=1}^{n}\left|\frac{x_i}{m}\right|^{p}\right)^{1/p}\to\begin{cases}0&\text{ if }k=0\\m&\text{ if }k=1\\\infty&\text{ if }k>1\end{cases}$$
The reason for the value when $k>1$ is that the $\sum$ above has $k$ terms $\geq1$, and $1/p\to\infty$.
|
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multivariable limit problem I have a confusion regarding this problem.
Problem: $\displaystyle f(x,y)=\frac{\sin^2|x+2y|}{x^2+y^2}$ is continuous for all $(x,y)\neq (0,0)$. True or false?
I think that the limit does not exist so the function is not continuous.
How to prove that limit does not exist?
Any help will be appreciated. Thanks in advance
|
Try approaching the limit by the line $y=k\cdot x$.
Applying the standard limit $$\lim_{t\rightarrow 0} \frac{\sin(t)}{t}$$
You can prove that the limit depends on the constant K, hence does not exist
|
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A digraph is a graph where every edge is directed. How many digraphs on $n$ vertices are there? So far I have that between any two vertices (say $j$ and $k$) there are 3 options.
*
*there is no edge between $j$ and $k$
*there is an edge directed from $j$ to $k$
*there is an edge directed from $k$ to $j$
And so from this, we can see that there must be $3^{\binom{n}{2}}$ digraphs on $n$ vertices.
I'm not sure if it should be $\binom{n}{2}$ or $\binom{n}{3}$? Can anybody clarify this for me?
|
In a directed graph, there are $n(n-1)$ distinct edges, each of these can be present or not. Thus the number of digraphs with $n$ vertices should be
$$2^{n(n-1)}$$
Note that this allows two vertices to be connected by directed edges in both ways. This coincides with the definition on wikipedia (see the example image).
If instead you want to probhibit situations like $v\leftrightarrow v'$ (double-connectedness), you are correct in your enumeration, we pick two vertices in $\binom n2$ ways and assign one of three cases to it independently of the others:
$$3^{\binom n2} = 3^{\frac12 n(n-1)}$$
If the same method was used do allow the fourth case, we would get the original result:
$$4^{\binom n2} = (2^2)^{\frac12 n(n-1)} = 2^{n(n-1)}$$
|
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Chinese Remainder Theorem for non prime-numbers. Let's say I want to find x such that x leaves remainder 2 when divided by 3 and x leaves remainder 3 when divided by 5.
x % 3 = 2
x % 5 = 3
We break down the problem to:
x % 3 = 1
x % 5 = 0
Therefore,
5k % 3 = 1
2k % 3 = 1
k = 2
10, when remainder = 1
20, when remainder = 2
Now x % 3 = 0
and x % 5 = 1
3k % 5 = 1
k = 2
6, when remainder = 1
18, when remainder = 3
Therefore, final solution is 20 + 18 = 38.
38 is a solution
LCM of 3,5 = 15.
38 - 15 - 15 = 8.
8 is the least number, that is the solution.
But now if I have
x % 7 = 3
x % 4 = 2
How do I solve the question ?
|
Exactly the same way. The equation $x\equiv 3 \mod 7$ tells you that $x=3+7y$ .
Plugging this into the second equation gives you $3+7y\equiv 2 \mod 4$, that is $-y\equiv -1 \mod 4$, so $y=1+4z$, and $x=10+28z$, i.e. $x\equiv 10 \mod 28$.
|
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Pattern Recognition - How to solve this problem? A general term of the sequence
$$
-8,-7,-10,-1,-28,53,\ldots
$$
can be expressed as $\dfrac{a-b^{n-1}}{4}$, where $a$ and $b$ are integers.
What is the value of $ab$?
Teach me how to solve this problem.
|
If we call this sequence $x_n$, then $a-4x_n$ is the geometric sequence $b^{n-1}$. So the first element should be $1$, meaning $a-4(-8)=1$, and the second element should be $b$, so $b=a-4(-7)$.
|
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Show that $\mathrm{gcd}(x+4,x-4)$ divides $8$ for all integers $x$. I want to prove that $\mathrm{gcd}(x-4,x+4)$ divides $8$ for all $x\in \mathbb{Z}$
Since they are both polynomials of degree $1$, it suggests that the $\mathrm{gcd}$ is a constant.
Using Euclidean Algorithm, I get: $(x+4) = 1(x-4) + 8$, so $\mathrm{gcd}(x-4,x+4)=\mathrm{gcd}(x-4,8)$ thus it will always divide $8$.
Is this the correct approach / use of EA for polynomials?
Thanks.
|
Your approach is correct. Another way of proving it is the following:
$$
d\mid x-4\text{ and }d\mid x+4\implies d\mid(x+4)-(x-4)=8.
$$
|
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|
Adjoint Transformations and Self-Adjoint Operators I don't quite understand the whole adjoint and self adjoint thing. I know their definitions:
Given a linear transformation $A:\mathbb{R}^d \to \mathbb{R}^m$, its adjoint >transformation, $A^*:\mathbb{R}^m \to \mathbb{R}^d$ is defined by $$(Ax,y)=(x,A^*y).$$
A linear operator $B:\mathbb{R}^k \to \mathbb{R}^k$ is self-adjoint if $B=B^*$.
So, $B$ is defined by $(Bx,y)=(x,By)$, right? Also, could someone please give me some examples of both $A$ and $B$? Thanks in advance.
|
We can't say that any $B$ is defined by $(Bx,y) = (x,By)$. However, if $B$ is self-adjoint, then $B$ has the property that $(Bx,y) = (x,By)$ for every $x,y \in \Bbb R^k$.
Here are some examples with matrices: recall that the usual inner product for $x,y \in \Bbb R^n$ is given by
$$
(x,y) = x^Ty
$$
Define the transformation $T_A: \Bbb R^3 \to \Bbb R$ by $A = \pmatrix{1&2&3}$. That is,
$$
T_A(x) = Ax = x_1 + 2x_2 + 3x_3
$$
The adjoint $T_A^*$ (with matrix $A^*$) is defined by the property
$$
(Ax,y) = (x,A^*y)
$$
for all $x \in \Bbb R^3$ and $y \in \Bbb R$. Since $A^* \in \Bbb R^{3 \times 1}$, it suffices to compute $T_{A^*}(1) = A^*1 = A^*$. We note that for all $x$, we have
$$
x^TA^* = (x,A^*1) = (Ax,1) = x_1 + 2x_2 + 3x_3
$$
So, in particular, we must have
$$
A^* = \pmatrix{1\\2\\3}
$$
Define the transformation $T_B:\Bbb R^2 \to \Bbb R^2$ by the matrix
$$
B = \pmatrix{0&1\\1&2}
$$
For any $x,y \in \Bbb R^2$, we compute
$$
(Bx,y) = \left( \pmatrix{x_2\\x_1 + 2x_2}, \pmatrix{y_1\\y_2} \right)
= x_2 y_1 + (x_1 + 2x_2)y_2 = x_1 y_2 + x_2y_1 + 2x_2 y_2
$$
Similarly,
$$
(x,By) = \left( \pmatrix{x_1 \\ x_2}, \pmatrix{y_2\\y_1 + 2y_2} \right)
= x_1 y_2 + x_2(y_1 + 2y_2) = x_1 y_2 + x_2y_1 + 2x_2 y_2
$$
So, we say that $B$ is self-adjoint.
|
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Check whether or not $\sum_{n=1}^{\infty}{1\over n\sqrt[n]{n}}$ converges. Check whether or not $\sum_{n=1}^{\infty}{1\over n\sqrt[n]{n}}$ converges.
I tried few things but it wouldn't work out. I would appreciate your help.
|
$\require{cancel}$
Using the limit comparison test with the harmonic series we see
$$\lim_{n\to\infty} {\cancel{(1/n)}\cdot 1\over \cancel{(1/n)}\sqrt[n]{n}}=1$$
hence the series diverges since the harmonic series does.
|
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Is this theorem equivalent to "existential instantiation" rule? In Enderton's, There is a theorem called "existential instantiation", it says:
Assume that the constant symbol $c$ does not occur in $\alpha ,\beta , \Gamma$ and that:
$$ \Gamma\cup\{\alpha^x_c\}\vdash\beta$$
Then
$$\Gamma\cup\{\exists\alpha\}\vdash\beta$$
I've googled the name of the theorem but found that on Wikipedia here for example, It says something else namely, Given $\exists x\alpha$ we can infer $\alpha^x_c$ for some contant symbol $c$ that has not been used before in the deduction.
It seems that what wikipedia states is the converse of the EL theorems in Enderton's and I don't see how both forms coincede.
Could anyone clarify for me how they state the same fact?
|
If you can infer $A$ from $B$, and you have $\Gamma \cup \{A\} \vdash \beta$, then you can conclude that $\Gamma \cup \{B\} \vdash \beta$. (The left-side of a turnstile is like the left-hand side of an implication: strengthening an assumption preserves validity.) Hence there is no conflict between Enderton's rule that operates on an assumption and the rule you found in the Wikipedia that operates on a conclusion.
|
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Can you recommend a book with techniques for solving hard algebra/arithmetic problems? I'm a university student who never really studied maths in high school (I did the basic courses, but because I'm dyslexic I was to embarrassed to try the harder courses) now I'm getting back into it, and, while I'm perfectly fine with understanding the advanced concepts and grasping the theoretical side of things, I have a lot of trouble with the arithmetic in the harder questions. A lot of this is due to my not really knowing a lot of the techniques used in manipulating terms in complex fractions and just generally not having a good set of problem solving tools to fall back on.
Can you recommend any books that deal with arithmetic techniques for more advanced problems?
just to give you an idea, here are some examples of question formats that I would like to practice:
number 1
$$
\frac{x^{2}}{y}+\frac{y^{2}}{x}
$$
where $x = 2 + \sqrt{3}$ and $y = 2 - \sqrt{3}$
Number 2
factorise $54x^3 + 16$
number 3
given that $$a^{2}+b^{2} = 7ab$$
express $$\left(\frac{a+b}{3}\right)^{2}$$
in terms of a and b. Hence show that $$\log\left (\frac{a+b}{3}\right )=\frac{1}{2}\left(\log a+\log b \right)$$
if would be especially great if it had lots of practice questions.
Thanks.
|
Have a look at 'How to solve it - George Polya'. It's more general but quite helpful.
|
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Area enclosed by cardioid using Green's theorem Let $$\gamma(t) = \begin{pmatrix} (1+\cos t)\cos t \\ (1+ \cos t) \sin t \end{pmatrix}, \qquad t \in [0,2\pi].$$ Find the area enclosed by $\gamma$ using Green's theorem.
So the area enclosed by $\gamma$ is a cardioid, let's denote it as $B$. By Green's theorem we have for $f=(f_1, f_2) \in C^1(\mathbb{R}^2, \mathbb{R}^2):$
$$\int_B \text{div} \begin{pmatrix} f_2 \\ -f_1 \end{pmatrix} d(x,y) = \int_{\partial B} f \cdot ds$$
So if we choose $f(x,y) = \begin{pmatrix} -y \\ 0 \end{pmatrix}$ for example, we get
$$\begin{eqnarray}
\text{Area of $B$} &=& \int_{\partial B} f \cdot ds
\\&=& \int_{\gamma} f(\gamma(t)) \cdot \gamma'(t) dt
\\&=& \int_0^{2\pi} \begin{pmatrix} -(1+ \cos t) \sin t \\ 0 \end{pmatrix} \cdot\begin{pmatrix} \sin t ( 1 - 2 \cos t) \\ \cos^2t - \sin^2t + \cos t \end{pmatrix} dt ,\end{eqnarray}$$
which I guess we can evaluate but if I keep going, this will become very nasty and tedious.
There must be a nicer way to do this. Please help me see it.
|
As a rule of thumb, it's better to use $$\frac12\int_{\partial B}(-y\,dx+x\,dy)$$ whenever trig functions are involved. That will give a simpler expression once you use standard trig identities.
Comment: It's more straight forward in this particular case to do the area integral directly in polar coordinates: $$\int_0^{2\pi}\int_0^{1+\cos\theta} r\,drd\theta.$$
|
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Prove $f(x) = \sum_{n=1}^\infty \frac{\sin nx + \cos nx}{n^3}$ is well-defined and $C^1$.
Prove $f(x) = \sum_{n=1}^\infty \frac{\sin nx + \cos nx}{n^3}$ is well-defined and $C^1$.
First of all I need to prove that $f(x)$ is well-defined. I'm not so sure what does it mean.
Basically I can claim that:
$$ f(x) = \sum_{n=1}^\infty \frac{\sin nx + \cos nx}{n^3} \le \sum_{n=1}^\infty \frac{2}{n^3} < \infty$$
So, by Weierstrass M-test we get that $f(x)$ converges uniformly to some continuous $f(x)$.
Hence, we may take the derivative of the term inside the sum and get:
$$f'(x) = \sum_{n=0}^\infty \frac{n\cos nx -n\sin nx + 3n^2(\sin nx + \cos nx)}{n^6} \le \sum_{n=0}^\infty \frac{2n + 6n^2}{n^6} < \infty$$
So again $f_n'(x)$ continuous and converges uniformly to $f'(x)$ which is also continuous by Weierstrass M-test.
I'd be glad if you could examine my solution and tell me if it right/rigorous.
|
Here's some remarks:
*
*You said that "$f(x)$ converges uniformly to some continuous $f(x)$" which is misspoke and we should write "the series converges uniformly..."
*You said "... Hence, we may take the derivative of the term...": I don't see the implication: how if $f$ is continuous then you may take the derivative?
Finally, notice that to prove $\sum_n f_n(x)$ is well defined and $C^1$ you just need to show:
*
*the series $\sum_n f_n(x)$ is point-wise convergent on $\Bbb R$;
*the functions $f_n$ are $C^1$ on $\Bbb R$;
*the series $\sum_n f'_n(x)$ is uniformly convergent on $\Bbb R$.
|
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Proof that these two definitions are equivalent
Where can I find a proof of or how can I prove that these two definitions are equivalent?
Definition 1: The Lie algebra of a Lie group $G \subset GL_n$ is the tangent space at $I$.
Definition 2: The Lie algebra of a Lie group $G \subset GL_n$ is all elements $g\in G$ with the property that for all $t \in \mathbb R$ the element $e^{t g}$ is also in $G$.
|
Definition 2 is incorrect: In general, the Lie algebra of $G\subset GL_n$ is not a subset of $G$ (or even of $GL_n$). For example, the zero matrix is an element of the Lie algebra, but it's not an element of $GL_n$.
In place of Definition 2, you could write
Definition 2$'$: The Lie algebra of a Lie group $G \subset GL_n$ is the set of all $n\times n$ matrices $g$ with the property that for all $t \in \mathbb R$ the element $e^{t g}$ is also in $G$.
The equivalence of this and Definition 1 should be proved in most elementary books about Lie groups. You can also find it in my Introduction to Smooth Manifolds (chapters 7 and 20).
|
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How can I prove every Riemann sum of x^2 of [a,b] is the integral? I find that to prove $\int_a^b x\,dx = (b^2-a^2)/2$ (using the $\epsilon - \delta$ definition of Riemann integrable) is pretty straightforward, and after manipulating some sums, I end up with an expression like
$\left| \sigma - \frac{b^2 - a^2}{2} \right| < \epsilon$ if $||P||<\delta = \frac{2\epsilon}{b-a}$.
Now for $\int_a^b x^2\,dx$ I start: consider a Reimann sum over $P$ with $n$ subintervals
$\sigma = \sum_{i=1}^n c_i^2 (x_i - x_{i-1})$ where $x_{i-1} \leq c_i \leq x_i$. Note that we can write
$c_i = \frac{x_i+x_{i-1}}{2} + d_i$ where $|d_i| < \frac{x_i - x_{i-1}}{2}$ (first term is midpoint, second term is max difference)
Then I expand this out, and since (intuitively) $d_i$ can get arbitrarily small by making $||P|| < \delta$, from the following expression I can 'remove' the $d_i$ terms from this sum:
$\sum_{i=1}^n \left[ \frac{x_i^3 - x_{i-1}^3}{4} + \frac{x_i^2x_{i-1} - x_i x_{i-1}^2}{4} + d_i^2(x_i - x_{i-1}) + d_i(x_i^2 - x_{i-1}^2) \right]$
Which results in me 'proving' that the sum gets arbitrarily close to
$\frac{b^3-a^3}{4} + \frac{b^2a - a^2b}{4}$ which does not seem right at all :)
Where am I going wrong here?
|
Let $\{I_i\}_{i=1}^n$ be a partition of $[a,b]$ with $n$ sub intervals. Let us choose the tag of the interval $I_i=[x_{i-1},x_i]$ to be the mid point $q_i=\dfrac{1}{2}(x_{i-1}+x_i)$.
Then the contribution of this term to the Riemann sum corresponding to the tagged partition
$Q=\{(I_i,q_i)\}_{i=1}^n$ is
$h(q_i)(x_i-x_{i-1})=\dfrac{1}{2}(x_i^2-x_{i-1}^2)$
So, $S(h; Q)=\dfrac{1}{2}\sum_{i=1}^n(x_i^2-x_{i-1}^2)=\dfrac{b^2-a^2}{2}$
If $P=\{(I_i,y_i)\}_{i=1}^n$ is arbitrary tagged partition of $[a,b]$ with $\|P\|<\delta$ so that $x_i-x_{i-1}<\delta$ for $i=1,....,n$.
Also let $Q$ have the same partition points, but where we choose the tag $q_i$ to be the midpoint of the interval $I_i$. Since both $t_i$ and $q_i$ belong to
this interval, we have $|t_i - q_i|\lt \delta $. Using the Triangle Inequality, we deduce
$|S(h,P)-S(h,Q)|\lt \delta \implies |S(h,P)-\dfrac{1}{2}|\lt \delta$
So we led to take $\delta_\epsilon \le \epsilon$. If we take $\delta_\epsilon = \epsilon$ then we can retrace the arguement to conclude that h is Riemann Integrable in $[a,b]$ so that $\int_a^b x^2 dx=\dfrac{1}{2}(b^2-a^2)\quad \square$
|
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Euler's formula, is this true? *I've changed this question as below.
Let me have a function such as $ f(k) = \exp(j 2 \pi k ) $, where $k$ is real value.
Using Euler's formula, we can write $f(k)$ as below,
$$ f(k) = \exp(j 2 \pi k ) = \cos(2\pi k)+j \sin(2\pi k).$$
If $k$ is integer, this always goes to 1.
Up to here, nothing is weird and makes sense.
However, if the equation goes to
$$f(k) = \exp(j2\pi k)=e^{j2\pi k}=(e^{j2\pi})^k = (\cos(2\pi)+j\sin(2\pi))^k=1^k=1$$
What I want to know is this above equation makes sense or not.
Thank you in advance.
|
Following on from Yves Daoust:
$$\left(z^x\right)^y$$
$$=\left(e^{\log \left(z^x\right)}\right)^y=\left(e^{x\log \left(z\right)}\right)^y$$
$$=e^{\log \left(\left(e^{x\log \left(z\right)}\right)^y\right)}=e^{y\log \left(e^{x\log \left(z\right)}\right)}$$
$$= e^{y\left(x\log \left(z\right)+i\left(2nπ\right)\right)}$$
$$=z^{xy} ⋅ e^{i\left(2nyπ\right)}$$
|
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A property of the fourier series Show that any periodic function $f(x)$ with period $2\pi$ which is both odd and satisfies $f(\pi-x)=f(x)$ has $b_{n}=0$ for $n$ even and so has a fourier series of the form $$f(x) = \sum^{\infty}_{m=0} c_{m}\sin{(2m+1)x}.$$
So we know $f$ is odd and periodic with period $2\pi$ thus it has a fourier series of the form $$f(x) = \sum_{n=1}^{\infty}b_{n}\sin{(nx)}\quad \text{where}\quad b_{n}=\frac{2}{\pi}\int^{\pi}_{0}f(x)\sin{(nx)} \ {\rm d}x.$$
Integrating $b_{n}$ by parts yields
\begin{align}b_{n} &= \frac{2}{\pi}\left[-\frac{1}{n}\Big[f(x)\cos{(nx)}\Big]^{\pi}_{0}+\frac{1}{n}\int^{\pi}_{0}f'(x)\cos{(nx)} \ {\rm d}x\right]\\ &= -\frac{2}{n\pi}\Big[f(\pi)\cos{(n\pi)}-f(0)\Big]+\frac{2}{n\pi}\int^{\pi}_{0}f'(x)\cos{(nx)} \ {\rm d}x\end{align}
Notice however $f(x) = f(\pi-x)$ thus $f(\pi) = f(0)$ so $$b_{n} = -\frac{2f(0)}{n\pi}\Big[(-1)^{n}-1\Big] + \frac{2}{n\pi}\int^{\pi}_{0}f'(x)\cos{(nx)} \ {\rm d}x$$
Now from this point onwards I'm not sure what to do with the second integral since we don't know $f(x)$ it will never cancel. Is what I have done going on the right track and what would be the next steps?
|
$$f(\pi-x) = \sum_{n=1}^{\infty} b_n \sin{n(\pi-x)} = \sum_{n=1}^{\infty} (-1)^{n+1} b_n \sin{n x}$$
The only way $f(\pi-x)=f(x)$ is when $b_n=0$ when $n$ is even.
|
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|
Understanding graphically the convergence of alternating harmonic and divergence of harmonic I understand the rules of convergence of a series so that I know that $\sum \frac 1 n$ (the harmonic series) diverges and $\sum \frac 1 {n^2}$ squared converges. It doesn't make sense graphically to me. One just gets to $0$ a little faster. What graphically sets that borderline?
For that matter, what makes the harmonic series diverge by the integral test and the alternating harmonic series converge by the alternating series test. That also seems counterintuitive graphically.
|
The Cauchy condensation test may help here. For a non-increasing sequence $(f(n))_{n\in\mathbb{N}}$ of non-negative real numbers we have (from Oresme's proof of the divergence of the harmonic series) :
\begin{align}
\tag{1}\sum_{k=1}^\infty f(k)&\le \underbrace{f(1)}_{f(1)}+\underbrace{(f(2)+f(2))}_{2\times f(2)}+\underbrace{(f(4)+f(4)+f(4)+f(4))}_{4\times f(4)}+\cdots\\
&\le\sum_{n=0}^\infty\;2^n\,f(2^n)\\
\end{align}
The same way
\begin{align}
\sum_{n=0}^\infty\;2^n\,f(2^n)&= f(1)+(f(2)+f(2))+(f(4)+f(4)+f(4)+f(4))+(f(8)+\cdots\\
&\le(f(1)+f(1))+(f(2)+f(2)+f(3)+f(3))+(f(4)+f(4)+\cdots\\
&\le 2\sum_{k=1}^\infty\;f(k)\\
\end{align}
so that $\;\sum_{k=1}^\infty\;f(k)\;$ will converge iff $\;\sum_{n=0}^\infty\;2^n\,f(2^n)$ converges.
Since $\,f(k):=\dfrac 1{k^{\,p}}\,$ satisfies the required criteria for any fixed non-negative power $p$ we deduce that $\;\sum_{k=1}^\infty\;\dfrac 1{k^{\,p}}\;$ will converge iff $\;\sum_{n=0}^\infty\;2^n\,\dfrac 1{2^{n\,p}}=\sum_{n=0}^\infty\;2^{n\,(1-p)}\;$ that is if and only if $\,p>1$.
The following illustration may clarify this. The picture at the middle shows the inequality in $(1)$ : the total area can't exceed the sum of the areas of all the $2^n$-width grey rectangles. In the specific case $p=1\,$ each of these rectangle will have an area of $1$ (the actual area can't differ from that by a factor larger than $2$) while for larger values of $p$ the areas will decrease exponentially and thus be finite.
The same reasoning applies for $\int \frac 1{x^p}\,dx$ from the integral test.
For alternative and possibly more convincing proofs of the divergence of the harmonic series see Kifowit's answers in "The Harmonic Series Diverges Again and Again" and "More Proofs of Divergence of the Harmonic Series".
|
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|
Can the null space be empty? I was reading a proof of the theorem that the range of a linear map $T$ is always a subspace of the target space, and when the author was showing that the $0$ vector was included in the range, he made an appeal to a previous theorem which says that the null space of $T$ is always a subspace of $T$.
In other words, he says that because the nullspace is a subspace, $0$ is always in the nullspace, and therefore since $T(0) = 0$, then $0$ is in the range of $T$.
That makes sense, but is it possible that the nullspace is empty? My feeling is no. Because $T$ acts on a vector space $V$, then $V$ must include $0$, and since we showed that the nullspace is a subspace, then $0$ is always in the nullspace of a linear map, so therefore the nullspace of a linear map can never be empty as it must always include at least one element, namely $0$.
|
Let $T:V\to W$ be a linear map. Then
$$
T(\mathbf{0})=T(\mathbf{0}-\mathbf{0})=T(\mathbf{0})-T(\mathbf{0})=\mathbf{0}
$$
This proves that $\mathbf 0$ is always in the nullspace of $T$. Hence the nullspace of $T$ cannot be empty.
|
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|
Are $\emptyset$ and $X$ closed, open or clopen? It is indeed a very basic question but I am confused:
(1) In an 2013 MSE posting under general topology here, I was told that $\emptyset$ is an open set and therefore I assume $X$ must be open too.
(2) But in Wikipedia page on clopen set here, it says "In any topological space $X$, the $\emptyset$ and the whole space $X$ are both clopen."
(3) And yet in another Wikipedia page on closed set here, "The $\emptyset$ is closed, the whole set is closed."
I must have missed something. Can you help me with a supreme verdict, once and for all, as sure as the sun rises from the east each morning, if $X$ and $\emptyset$ are open, closed or clopen. Of course I am talking about topology, thanks for your time.
|
By the first axiom in the definition of a topology, $X$ and $\emptyset$ are open. However, closed sets are precisely those whose complements are open, by definition. Hence the empty set and $X$, being each others complements, are also closed. So, they are clopen.
|
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Integration inequality question help: Sketch the curve y=1/u for u > 0... Sketch the curve $y=\frac{1}{u}$ for $u > 0$. From the diagram, show that $\int\limits_1^{\sqrt{x}}\frac{du}{u}< \sqrt x-1$, for x > 1. Use this result to show that $0 < \ln(x) < 2(\sqrt{x}-1)$ and hence that $\frac{\ln x}{x}\to0$ as $x\to\infty$. Deduce that $x\ln(x)\to0$ as $x\to0$.
Basically, I figured out the first part in terms of it being from the diagram. From 'use this result' in the question, I don't know what to do next. Can anyone help me out?
Thank you!
|
Big Hint:
We know by the fundamental theorem of calculus that
$$\int_{1}^{\sqrt{x}} \frac{du}{u} = \frac{ln(x)}{2}.$$
However,
$$\int_{1}^{\sqrt{x}} \frac{du}{u} < \sqrt{x} -1.$$
Thus, we can say $$0<\frac{ln(x)}{2} < \sqrt{x} -1 \text{ or } 0<lnx<2(\sqrt{x} -1).$$
Now, if we divide by $x$ and apply Squeeze Theorem, one will quickly find what they asked for.
|
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|
Does $A\setminus B = A\setminus C$ imply $B=C$? Let $A, B, C$ be sets with $B \subset C$ and $C \subset A$. Does $A\setminus B = A\setminus C$ imply $B=C$?
I am not sure what the "\" means, so I don't know how to solve this.
|
Yes it does, since, if $B\subset A$, then $A\setminus (A\setminus B)=B$. So $A\setminus B = A\setminus C$ implies $A\setminus (A\setminus B) = A\setminus (A\setminus C)$, i.e. $\,\,B=C$.
|
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|
Proving that $\sin^7\theta + \cos^7\theta <1$ using basic trigonometry and identities How do I prove $\sin^7\theta + \cos^7\theta < 1$ for an angle between $(0,\pi/2)$?
|
Hint: If $\theta \in (0, \pi/2)$, we have $$0 < \sin \theta < 1 \text{ and }0 < \cos \theta < 1,$$ which gives us that $\sin^7 \theta < \sin^2\theta$ and $\cos^7\theta < \cos^2\theta$. So...
|
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|
How to connect a line between 4 randomly placed points on a plane such that the line does not cross itself You get 4 coordinates of points on a plain. You need to connect them all with a line. The line must not cross itself.
What's your strategy?
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Let $${\bf z}_k=(x_k,y_k)\qquad(1\leq k\leq 4)$$
be the four points, arranged such that
$$x_1\leq x_2\leq x_3\leq x_4\ .$$
This means that the leftmost point gets number $1$, the point with next-larger $x$-coordinate gets number $2$, and so on. The rightmost point gets number $4$.
It is assumed that no three of the four given points are collinear. In particular $x_1<x_4$, and the line $g$ connecting ${\bf z}_1$ with ${\bf z}_4$ is not vertical.
As an auxiliary tool we now introduce the affine linear function
$$\phi(x,y):=(x_4-x_1)(y-y_1)-(y_4-y_1)(x-x_1)\ .$$
It is designed such that $\phi(x,y)=0$ when the point $(x,y)$ is on the line $g$. When the point $(x,y)$ is lying above $g$ then $\phi(x,y)>0$, and when $(x,y)$ is lying below $g$ then $\phi(x,y)<0$. Using this $\phi$ compute the test quantities
$$t_2:=\phi(x_2,y_2), \qquad t_3:=\phi(x_3,y_3)\ .$$
The signs of these two quantities (they cannot be $=0$) decide how the subsequent connections should be drawn. Proceed as follows:
If $t_2\>t_3>0$ (meaning that ${\bf z_2}$ and ${\bf z}_3$ are on the same side of $g$) draw $\ {\bf z}_1\to{\bf z}_2\to{\bf z}_3\to{\bf z}_4\to{\bf z}_1$.
If $t_2\>t_3<0$ (meaning that ${\bf z_2}$ and ${\bf z}_3$ are on opposite sides of $g$) draw $\ {\bf z}_1\to{\bf z}_2\to{\bf z}_4\to{\bf z}_3\to{\bf z}_1$.
It helps to draw a figure in order to understand that this works in all nondegenerate cases. If three points on a line are allowed more work is necessary.
|
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|
How many 0's are in the end of this expansion? How many $0's$ are in the end of:
$$1^1 \cdot 2^2 \cdot 3^3 \cdot 4^4.... 99^{99}$$
The answer is supposed to be $1100$ but I have absolutely NO clue how to get there.
Any advice?
|
There as many zeros as the number of the factors of 5 in the product:
So we need to check the subproduct
$$
5^5\cdot 10^{10}\cdots 95^{95}
$$
We have $5+10+15+\cdot+95=5(1+2+\cdots+19)=950$. And we have missed the case where we have two factors: $25,50,75$. So we have another $25+50+75=150$. Altogether $950+150=1100$.
|
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How can I see that the space curve given by $\gamma(s) = (\frac 4 5 \cos s, 1 - \sin s, - \frac 3 5 \cos s)$ forms a circle in space ? Let $\gamma: \mathbb R \rightarrow \mathbb R^3$ be a space curve given by $\gamma(s) = (\frac 4 5 \cos s, 1 - \sin s, - \frac 3 5 \cos s)$.
How do I see that $\gamma$ has image in $\mathbb R^3$ that forms a circle ? Also how can I determine its radius and determine the plane it lies in ?
I've worked with circles in $\mathbb R^2$ given by equations of the form $(x-a)^2 + (y-b)^2 = r^2$. Can I utilize this knowledge here ?
I see the $x$-coordinate goes between $\pm \frac 4 5$, the $y$-coordinate between $0$ and $1$ and the $z$-coordinate between $\pm \frac 3 5$. Intuitively, this points towards a circle in space, but how can I determine this rigorously ?
|
Since you seem to see what values the coordinates "go between", then you can already make an educated guess: what about the circle with center $\;(0,1,0)\;$?
$$x^2+(y-1)^2+z^2=\frac{16}{25}\cos^2x+\sin^2s+\frac9{25}\cos^2s=1$$
and there you go.
|
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|
How to prove this: $a^4+b^4+2 \ge 4ab$? How to prove this: $a^4+b^4+2 \ge 4ab$?
$a$ and $b$ are reals.
|
Another possible method is generation of a function and calculating its minimum:
$$f(a,b)=a^4+b^4+2-4 a b$$
$$D_a f(a,b)=4 a^3-4 b = 0$$
$$ a^3=b$$
$$ f(a)=a^4+a^{12}+2-4a^4=a^{12}+2-3a^4$$
$$D_a f(a)=12a^{11}-12a^3=0$$
$$a^3 (a^8-1)=0$$
Minimum must be at one of:
$$f(0,0)=2$$
$$f(1,1)=0$$
$$f(-1,-1)=0$$
so
$$f(a,b)=a^4+b^4+2-4 a b\geq0$$
|
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Number theory: show that ${ 1^2, 2^2, 3^2,... , m^2}$ cannot be a complete residue system Is this an acceptable answer?
Question: show that ${ 1^2, 2^2, 3^2,... , m^2}$ cannot be a complete residue system.
Since the above has $m$ elements, one must show it cannot be a complete residue system
modulo $m$. Consider the complete residue system modulo m:
{1,2,3,... ,$m$}. Now between the two the first element $1^2$ and $1$ matches. But
what about the second last element $ (m-1)^2$. Here we have $(m-1)^2$ = $m^2 -2m +1$
which is clearly congruent to the first element, $1$, modulo m.
So there exist two elements congruent modulo $m$ to each other in ${ 1^2, 2^2, 3^2,... , m^2}$ so it cannot be a complete residue system.
|
Let's start with an example. Below is a congruence table modulo &11&.
\begin{array}{rrr}
k & k^2 & k^2 \pmod{11} \\
\hline
0 & 0 & 0\\
1 & 1 & 1\\
2 & 4 & 4\\
3 & 9 & 9\\
4 & 16 & 5\\
5 & 25 & 3\\
6 & 36 & 3\\
7 & 49 & 5\\
8 & 64 & 9\\
9 & 81 & 4\\
10 & 100 & 1\\
\hline
\end{array}
Notice that the last column does not contain all eleven of the integers from $0$ to $10$ because, of the eleven numbers, there are repeated numbers: $k^2 \equiv (11-k)^2 \pmod{11}$.
This is true for all moduli, N.
$$(N-k)^2 \equiv N^2 - 2kN + k^2 \equiv k^2 \pmod N$$
|
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|
How to solve this ordinary differential equation? I am just trying to find general solution
$$\frac{dy}{dx} = 1 + \sqrt{1 - xy}$$
|
Let $u=-\sqrt{1-xy}$ ,
Then $y=\dfrac{1-u^2}{x}$
$\dfrac{dy}{dx}=\dfrac{u^2-1}{x^2}-\dfrac{2u}{x}\dfrac{du}{dx}$
$\therefore\dfrac{u^2-1}{x^2}-\dfrac{2u}{x}\dfrac{du}{dx}=1-u$
$\dfrac{2u}{x}\dfrac{du}{dx}=u-1+\dfrac{u^2-1}{x^2}$
Approach $1$:
$\dfrac{2u}{x}\dfrac{du}{dx}=\dfrac{(u-1)x^2+(u+1)(u-1)}{x^2}$
$(x^2+u+1)\dfrac{dx}{du}=\dfrac{2ux}{u-1}$
Let $v=x^2$ ,
Then $\dfrac{dv}{du}=2x\dfrac{dx}{du}$
$\therefore\dfrac{x^2+u+1}{2x}\dfrac{dv}{du}=\dfrac{2ux}{u-1}$
$(x^2+u+1)\dfrac{dv}{du}=\dfrac{4ux^2}{u-1}$
$(v+u+1)\dfrac{dv}{du}=\dfrac{4uv}{u-1}$
This belongs to an Abel equation of the second kind.
Let $w=v+u+1$ ,
Then $v=w-u-1$
$\dfrac{dv}{du}=\dfrac{dw}{du}-1$
$\therefore w\left(\dfrac{dw}{du}-1\right)=\dfrac{4u(w-u-1)}{u-1}$
$w\dfrac{dw}{du}-w=\dfrac{4uw}{u-1}-\dfrac{4u(u+1)}{u-1}$
$w\dfrac{dw}{du}=\dfrac{(5u-1)w}{u-1}-\dfrac{4u(u+1)}{u-1}$
Approach $2$:
$\dfrac{2u}{x}\dfrac{du}{dx}=u-1+\dfrac{u^2-1}{x^2}$
$u\dfrac{du}{dx}=\dfrac{u^2}{2x}+\dfrac{xu}{2}-\dfrac{x}{2}-\dfrac{1}{2x}$
Let $u=\sqrt{x}v$ ,
Then $\dfrac{du}{dx}=\sqrt{x}\dfrac{dv}{dx}+\dfrac{v}{2\sqrt{x}}$
$\therefore\sqrt{x}v\left(\sqrt{x}\dfrac{dv}{dx}+\dfrac{v}{2\sqrt{x}}\right)=\dfrac{xv^2}{2x}+\dfrac{x\sqrt{x}v}{2}-\dfrac{x}{2}-\dfrac{1}{2x}$
$xv\dfrac{dv}{dx}+\dfrac{v^2}{2}=\dfrac{v^2}{2}+\dfrac{x\sqrt{x}v}{2}-\dfrac{x}{2}-\dfrac{1}{2x}$
$xv\dfrac{dv}{dx}=\dfrac{x\sqrt{x}v}{2}-\dfrac{x}{2}-\dfrac{1}{2x}$
$v\dfrac{dv}{dx}=\dfrac{\sqrt{x}v}{2}-\dfrac{1}{2}-\dfrac{1}{2x^2}$
Let $t=\dfrac{x^\frac{3}{2}}{3}$ ,
Then $\dfrac{dv}{dx}=\dfrac{dv}{dt}\dfrac{dt}{dx}=\dfrac{\sqrt{x}}{2}\dfrac{dv}{dt}$
$\therefore\dfrac{\sqrt{x}v}{2}\dfrac{dv}{dt}=\dfrac{\sqrt{x}v}{2}-\dfrac{1}{2}-\dfrac{1}{2x^2}$
$v\dfrac{dv}{dt}=v-\dfrac{1}{\sqrt{x}}-\dfrac{1}{x^\frac{5}{2}}$
$v\dfrac{dv}{dt}=v-\dfrac{1}{\sqrt[3]{3t}}-\dfrac{1}{(3t)^\frac{5}{3}}$
This belongs to an Abel equation of the second kind in the canonical form.
Please follow the method in https://arxiv.org/ftp/arxiv/papers/1503/1503.05929.pdf or in http://www.iaeng.org/IJAM/issues_v43/issue_3/IJAM_43_3_01.pdf
|
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|
Closure, interior and boundary of $(0, 1)$ with Zariski topology If we consider $\mathbb{R}$ together with the Zariski topology, what is the closure, interior and boundary of $(0,1)$?
A set is closed iff it is either finite or $\mathbb{R}$ under this topology, so since the closure of $(0,1)$ must be closed am I right in thinking that this must be either $\{0,1\}$, $\emptyset$ or $\mathbb{R}$? I think if I can correctly understand the closure of this set I would be able to get the latter parts of my question.
|
The closure is the smallest closed superset, so in the case is $\Bbb R$. The interior is the greatest oper subset, so in this case is... And the border is...
|
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|
$d(n)$ is odd if and only if $n = k^2$ Can someone help me prove that $d(n)$ is odd if and only if $n = k^2$ for some integer $k$?
For reference:
$d(n)$ gives the number of positive divisors of $n$, including $n$ itself.
|
Hint : A number with the prime factorization $p_1^{a_1}\times ...\times p_n^{a_n}$ has
$(a_1+1)\times ...\times (a_n+1)$ divisors. When is the latter product odd ?
|
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|
Show that the arithmetic mean is less or equal than the quadratic mean I tried to solve this for hours but no success.
Prove that the arithmetic mean is less or equal than the quadratic mean.
I am in front of this form:
$$
\left(\frac{a_1 + ... + a_n} { n}\right)^2 \le \frac{a_1^2 + ... + a_n^2}{n}
$$
With rewriting the inequality in other forms I had no luck.
I think maybe induction would be OK, but I have no idea, how to do it in this case. Do you know a good proof for this?
Thanks!
|
This is kind of a simplistic and brute-force approach:
$$\bigg(\frac{\sum_{i = 1}^n a_i}{n}\bigg)^2 \leq \frac{\sum_{i=1}^n a_i^2}{n} \iff %
\bigg(\sum_{i = 1}^n a_i\bigg)^2 \leq n \sum_{i=1}^n a_i^2$$
Now open the bracket on the left of the inequality and simplify and note that the result is equivalent to:
$$\sum_{1 \; \leq \; i \; < \; j \; \leq \; n} (a_i - a_j)^2 \geq 0$$
This is obvious.
|
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|
Find $\int_{-1}^3xf(x)\,dx$ where $f(x)=\min(1,x^2)$ Find:
$$\int_{-1}^3xf(x)\,dx,$$
where $f(x)=\min(1,x^2)$.
I thought about solving it like this:
$$\int_{-1}^1 x^3\,dx + \int_{1}^3x\,dx = \cdots = 4.$$
But the solution is $\frac{26}{3}$
and I don't understand how they got it.
|
My attempt::
Assuming $f:\Bbb R \to \Bbb R$,
$$f(x) = \min\Big(1\ ,\ x^2\Big) =
\begin{cases}
x^2\quad,\ x \in [-1,1] \\
1 \quad,\ x \in (-\infty,-1)\cup(1, \infty)
\end{cases}$$
The graphs of $ y = \min(1, x^2)$ and $y = x\min(1,x^2)$ are respectively as follows:
Now,
$$\int x\cdot f(x)\ \mathrm dx =
\begin{cases}
\displaystyle \int x^3 \mathrm dx = \frac{1}{4}x^4 \color{lightgray}{+ \mathcal C}\ , x \in [-1,1]\\
\displaystyle \int x \mathrm dx = \frac{1}{2}x^2 \color{lightgray}{+ \mathcal C} \ ,\ x \in \Bbb R \sim [-1,1]\end{cases}$$
Hence,
$$ \int_{-1}^{3}xf(x)\mathrm dx = \int_{-1}^{1}xf(x)\mathrm dx + \int_{1}^{3} xf(x)\mathrm dx\\
= \frac{1}{4}x^4 \Big]_{-1}^{+1} + \frac{1}{2}x^{2}\Big]_{1}^{3} \\
= \frac{1}{4}\left[\Big((+1)^4 - (-1)^4\Big) + 2\Big(3^2 - 1^2\Big)\right]\\
= \frac{0 + 16}{4} = 4$$
So, yes, you're absolutely right =) The textbook was wrong. There is no room for doubt.
$$\boxed{\displaystyle \int_{-1}^{3}xf(x)\mathrm dx = 4}$$
|
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|
How many $6$ digit numbers there are, where first and last digits and also consecutive digits are not equal? We have a 6 digit number
$a_1a_2a_3a_4a_5a_6$ and $a_1 \neq 0$
also $a_1 \neq a_2, a_2 \neq a_3, a_3 \neq a_4, a_4 \neq a_5, a_5 \neq a_6, a_6 \neq a_1$
All of numbers where $a_1 \neq a_2, a_2 \neq a_3, a_3 \neq a_4, a_4 \neq a_5, a_5 \neq a_6$ are $9^6$
I'm having troubles with the last case.I know that this is generalization of $a_1a_2a_3$ for which there are $8*9^2$ numbers but cannot figure out how they are related
and this is the source file which counts all those digits.
bool foo(int num)
{
char str[15] = {};
int i=0;
while(num)
{
str[i]=num%10;
num/=10;
i++;
}
str[i]=str[0];
for(int j=0;j<i;j++)
if(str[j]==str[j+1])
return false;
return true;
}
int main()
{
int sum=0;
for(int i=100000;i<=999999;i++)
if(foo(i))
sum++;
cout<<sum;
return 0;
}
Please note that this program produces answer: 478305, which is correct
Also note that the answer 8*9^5 misses some cases
|
Here's another approach. We will do cases based on the digits in positions $1$, $3$, and $5$.
Case 1:
$a$, ___ , $a$, ____, $a$, ____.
There are $9$ ways to select the digit $a$, and $9$ ways to fill each blank. So $9^4$ numbers of interest in this case.
Case 2.1, 2.2, 2.3:
$a$, ___ , $a$, ____, $b$, ____
and $a$, ___ , $b$, ____, $b$, ____
and $b$, ___ , $a$, ____, $a$, ____
For each of three subcases, there are $9\cdot 9$ ways to choose $a$ and $b$, and $9\cdot 8\cdot 8$ ways to fill the blanks. So there are a total of $3\cdot 9^3\cdot 8^2$ numbers of interest in these three subcases combined.
Case 3:
$a$, ___ , $b$, ____, $c$, ____.
There are $9\cdot 9\cdot 8$ ways to choose $a, b,c$, and $8$ ways to fill each blank. So there are a total of $9^2\cdot 8^4$ numbers of interest in this case.
Grand total: $9^4+3\cdot 9^3\cdot 8^2+9^2\cdot8^4=478305$ numbers of interest.
|
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|
Show that if the sum of an diverges, no discrete probability space can contain independent events Suppose that $0\leq p_n\leq 1$, and put $a_n= \min \{p_n, 1-p_n\}$. Show that if
$\sum a_n$ diverges, then no discrete probability space can contain independent events $A_1, A_2, \ldots$ such that $A_n$ has probability $p_n$.
|
Let's do not consider the $P_n=0$ cases, because if the empty set is independent with any sets.
So , consider the discrete probability space, contains at most countable many points with each point assigned with a positive probability.
All the probabilities sum up to 1.
Now $\sum a_n$ diverge,
suppose $A_n$ are the events,
they are independent, therefore $P(\bigcap_n A_n)=\prod_n a_n=0$, here because $p_n$ are at most $\frac{1}{2}$,
then $\bigcap_n A_n$ is simply empty in our discrete setting.
(*) indeed, if we choose any infinite $A_n$ their intersection will be empty.
Use Borel-Cantelli lemma, $\sum a_n$ diverge then $\limsup_{n\rightarrow\infty} A_n$ will have measure $1$, in our discrete setting this means $\limsup_{n\rightarrow\infty} A_n$ is the whole set.
This would imply that any point in our set will be contained in infinitely many $A_n's$
Contradict with (*)
The whole point is that in discrete probability space, measure 0 set is empty and measure 1 set is the whole set. (of course, the points with 0 probability we can simply delete them without changing anything).
Hope it will help. :D
|
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|
How to prove the cofactor of a matrix is related to it's minor? How do I prove $A_{ij}$, the cofactor of $a_{ij}$ of a matrix $D$, has the property $A_{ij}=(-1)^{i+j}M_{ij}$, where $M_{ij}$ is the minor obtained by deleting the $i$th row and $j$th column?
|
For the case of $i=j=1$, $A_11$ will be the sum of the terms $(1)^{N(\alpha_1=1,\alpha_2,\dots,\alpha_n)} a_{\alpha_2 2} \cdots a_{\alpha_n n}$. Observe that there is no transposition (or inversion) that contains $a_{11}$. So $N(\alpha_1=1,\alpha_2,\dots,\alpha_n)=N(\alpha_2,\dots,\alpha_n)$. Then we have the sum of terms like $(-1)^{N(\alpha_2,\dots,\alpha_n) } a_{\alpha_2 2} \cdots a_{\alpha_n n}$. The sum of these term, however, is the determinant of the following matrix
$$
\begin{pmatrix}
a_{22} & \cdots & a_{2 n} \\
\vdots & \vdots & \vdots \\
a_{n 2} & \cdots & a_{nn} \\
\end{pmatrix}
$$
and clearly this matrix is the minor $M_{11}$ of the original matrix. Thus $A_{11}=M_{11}$. This shows the claim $A_{ij}=(-1)^{i+j} M_{ij}$ when $i=j=1$. In the general case you just move the element from $i$th row $j$th column to the first row first column, and follow a similar argument.
|
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|
Can this be solved for $f(n)$? While working upon a partial sum formula for the harmonics, I came across a necessity for the function defined below
$f\left(\frac{n(n-1)}{2} +1\right) = n!$
Can it be solved for $f(n)$?
|
Let your formula define $f(t)$ for all integer $t$ on the form $t=\frac{n(n-1)}{2}+1$ with $n$ integer. For other $t$ define $f$ to be whatever you want. The resulting function will satisfy your equation. This means that there is no unique function satisfying your equation.
However, there is a 'natural' function. Solving $\frac{n(n-1)}{2}+1 = t$ for $t$ we get
$$n = \frac{1 \pm \sqrt{8t-7}}{2}$$
and by using $n! = \Gamma(n+1)$ for integer $n$ where $\Gamma$ is the gamma function we get (by taking the positive branch)
$$f(t) = \Gamma\left(\frac{1 + \sqrt{8t-7}}{2} + 1\right)$$
which is a real function for all $t\geq \frac{7}{8}$.
|
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|
A nonmetrizable image of a metrizable space It is well known that a hausdorff continuous image of a compact metric space is metrizable. What is a counterexample for noncompact case?
|
Let $X$ be an infinite set, let $\mathcal{T}_\text{d}$ be the discrete topology on $X$, and $\mathcal{T}$ any non-metrizable Hausdorff topology on $X$. Then the identity mapping $\operatorname{id}_X : X \to X$ is a continuous function from $( X , \mathcal{T}_{\text{d}} )$ onto $( X , \mathcal{T} )$, and the discrete topology is metrizable (by the discrete metric).
|
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|
$ A_n $ =$[\frac{n}{n+1},\frac{n+1}{n+2}] $ be closed subsets find $\bigcup_{n=1}^\infty A_n $ Let $ A_n =\frac{n}{n+1},\frac{n+1}{n+2}] $, $n=1,2,3...$ be closed subsets of real line R. Then $\bigcup_{n=1}^\infty A_n $ is
*
*(1/2,1)
*[1/2,1)
*(1/2,1]
*[1/2,1]
My attempt : think it could be [1/2,1) since $\lim \frac{n+1}{n+2}$=1 . I am not sureit is just a guess.any hints plz.
|
Hint: Since $f(x)=\frac{x}{x+1}$ is monotonically increasing, we have that for any $0\le x\lt1$,
$$
n=\left\lfloor\frac{x}{1-x}\right\rfloor\iff n\le\frac{x}{1-x}\lt n+1\iff\frac{n}{n+1}\le x\lt\frac{n+1}{n+2}
$$
Thus, for any $\frac12\le x\lt1$, if $n=\left\lfloor\frac{x}{1-x}\right\rfloor$, then $n\ge1$ and $x\in\left[\frac{n}{n+1},\frac{n+1}{n+2}\right]$.
Furthermore, note that for all $n\ge0$, $\frac{n+1}{n+2}\lt1$.
|
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|
Does $\sum_{n\ge0} \cos (\pi \sqrt{n^2+n+1}) $ converge/diverge? How would you prove convergence/divergence of the following series?
$$\sum_{n\ge0} \cos (\pi \sqrt{n^2+n+1}) $$
I'm interested in more ways of proving convergence/divergence for this series.
My thoughts
Let
$$u_{n}= \cos (\pi \sqrt{n^2+n+1})$$
Let's first check A necessary condition for the convergence of that seires is that the limit $$\lim\limits_{n \to \infty} u_n$$ should exist and be equal $0$
but i have answer in my old book but i can't remember how they got that
$$u_n=\cos \left(n\pi+\frac{\pi}{2}+\frac{3\pi}{8n}+o(\frac{1}{n^2})\right)=\frac{(-1)^{n+1}.3\pi}{8n}+o(\frac{1}{n^2})$$
since $|\frac{(-1)^{n+1}.3\pi}{8n}|$ decreasing. and converge $0$
then by compraison $u_n$ is convergent
any help would be appreciated
|
A hint:
$$\sqrt{n^2+n+1}-\left(n+{\textstyle{1\over2}}\right)={3/4 \over
\sqrt{n^2+n+1}+\left(n+{1\over2}\right)}=:\ \alpha_n\searrow\ 0\qquad(n\to\infty)\ .$$
|
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|
Why do we pretend + and x not to be addition and multiplication while eventually they are considered so while explaining groups, rings and fields? Aren't commutative, associative properties result of addition and multiplication? Then why most of the definitions of Groups and Fields contain + and x but don't accept that they are addition and multiplication?
|
As an exemple of how '' addition'' and ''multiplication'' can be different from the usual operations in $\mathbb{R}$, take the ring $(M_2(\mathbb{R},\oplus, \odot)$ of $2\times 2$ matrices with real entries.
Here the $\oplus$ is defined as
$$
A\oplus B=
\left(
\begin{array}{ccccc}
a_1&a_2 \\
a_3&a_4
\end{array}
\right) \oplus
\left(
\begin{array}{ccccc}
b_1&b_2 \\
b_3&b_4
\end{array}
\right) =
\left(
\begin{array}{ccccc}
a_1+b_1&a_2+b_2 \\
a_3+b_3&a_4+b_4
\end{array}
\right)
$$
and we can easly see that this operation is associative, commutative, has a neutral element (the matrix with all $0$ entries, that we call $0$) and that every matrix $A$ has an ''inverse $X$'' (that we call ''opposite'') such that $A \oplus X=0$ ( and we call this opposite $-A$), i.e. $(M_2(\mathbb{R}),\oplus)$ is a commutative group.
Is all that sufficient to call $\oplus$ a ''sum'' in analogy whit the sum in $\mathbb{R}$? NO, because also the product in $\mathbb{R}$ has the same formal properties.
But we also define the operation $\odot$ as:
$$
A\odot B=
\left(
\begin{array}{ccccc}
a_1&a_2 \\
a_3&a_4
\end{array}
\right) \odot
\left(
\begin{array}{ccccc}
b_1&b_2 \\
b_3&b_4
\end{array}
\right) =
\left(
\begin{array}{ccccc}
a_1b_1+a_2b_3&a_1b_2+a_2b_4 \\
a_3b_1+a_4b_3&a_3b_2+a_4b_4
\end{array}
\right)
$$
and we see that such opertion is associative, but not commutative, there is a neutral element, the matrix
$$
I=
\left(
\begin{array}{ccccc}
1&0 \\
0&1
\end{array}
\right)
$$
but there are matrices $M \ne 0$ that don't have an ''inverse'', as:
$$
M=
\left(
\begin{array}{ccccc}
0&m \\
0&0
\end{array}
\right)
$$
So $(M_2(\mathbb{R}),\odot)$ is not a group but only a monoid.
So the first operation $\oplus$ has the same formal properties of $+$ and $\cdot$ in $\mathbb{R}$, but the second operation $\odot$ has not these properties. Why ve call the first a ''sum'' and the second a ''product''? Because there is another important property: $\odot$ is distributive over $\oplus$, i.e.:
$$
A\odot(B\oplus C)=a\odot B \oplus A\odot C
$$
And here $\oplus$ and $\odot$ have the same behavior as $+$ and $\cdot$.
So $\oplus$ is very similar to the $+$ operation on $\mathbb{R}$ and we can call it a ''sum''. Finally we chose to call $\odot$ a ''product'' ( or ''multiplication'') only to save words, but we we must be careful that this ''product'' is not commutative and not invertible.
|
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|
How do i find P(2)? Let $P(x)$ be the polynomial of least degree with real coefficients such that $P(i)=P(1+i)=0$ and $P(1)=4$. Find $P(2)$.
Here i is the complex number. I have done such questions with real numbers but I don't know why I am not getting it.
|
Hint: $P$ has real coefficients. So, if $P(i) = 0$, then $P(-i) = 0$. If $P(1+i) = 0$, then $P(1 - i) = 0$.
Can you construct the smallest complex polynomial that has these zeros?
|
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|
Weird integration issue: $\ln(x+1)=\ln(2x+2)$ ?! Weird integration issue: Using $(\ln[f(x)])'=\frac {f'(x)}{f(x)}$ we get that $\int \frac{2\,dx}{2x+2}=\ln(2x+2)$. Yet, $\int \frac{2\,dx}{2x+2}= \int\frac{dx}{x+1}=\ln(x+1)$ using the same rule as earlier.
What is wrong here?
|
Just add a constant term and also notice that $\ln2$ is also a constant.
|
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|
dimension of inverse image Let $f : V \to W$ be a linear transformation between two vector spaces. Let $E$ be a subspace of $V$.
Assume $V$ to be finite dimensional. Is there a way to obtain the dimension of $f^{-1} (f(E))$ in terms of the dimension of $E$, the dimension of $ \mathrm{im}f$ and the dimension of $\mathrm{ker}f$ ?
|
One has $f^{-1}(f(E))=E+\ker(f)$, which sum need not be direct, so the general formula applies
$$
\dim(f^{-1}(f(E)))=\dim(E+\ker(f))=\dim(E)+\dim(\ker f)-\dim(E \cap\ker(f)).
$$
So apart from the dimension of $E$ and of the image of $f$ (from which you can deduce that $\ker f$), you need to know hoe $E$ intersects $\ker f$.
|
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Why is a circle 1-dimensional? In the textbook I am reading, it says a dimension is the number of independent parameters needed to specify a point. In order to make a circle, you need two points to specify the $x$ and $y$ position of a circle, but apparently a circle can be described with only the $x$-coordinate? How is this possible without the $y$-coordinate also?
|
That is a bad definition of dimension. By this definition, the (filled-in) unit square in the real coordinate plane is one-dimensional, since you only need one number to describe a point $(x,y)$. An easy way to do this is to write out the decimal expansions of $x$ and $y$:
$$x=0.a_1a_2a_3a_4a_5\dots$$
$$y=0.b_1b_2b_3b_4b_5\dots$$
Now the point $(x,y)$ can be described with just one number:
$$0.a_1b_1a_2b_2a_3b_3\dots$$
But suppose we overlook this shortcoming of your definition, and we apply it to a circle. Then we can use the polar coordinate map
$$(x,y) \rightarrow (r,\theta)$$
This lets us write every point $(x,y)$ in terms of just one variable: $\theta$ (since $r$ is constant).
|
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|
Can the cohomology ring of the two-fold torus be calculated abstractly? In our lectures, we are given an unusual definition of cohomology and cup products which makes explicit calculations a bit tedious (that is, even more tedious than usual).
For the $n$- and $m$-spheres $S^n$ and $S^m$, I know that there’s an isomorphism of rings
$$H^•(S^n×S^m) \overset \simeq \longrightarrow H^•(S^n) \otimes H^•(S^m).$$
Can I use merely this fact, and other basic facts about cohomology like the Mayer–Vietoris sequence, Poincaré duality and the functoriality of cohomology to actually calculate the cohomology of the two-fold torus? What other facts would I need?
|
I think Qiaochu Yuan's post (https://qchu.wordpress.com/2013/10/12/the-cohomology-of-the-n-torus/) is quite relevant. In particular the first part gives a derivation of the cohomology (group) of the $n$-torus using Mayer–Vietoris . Of course you need more tools to compute the ring structure, but I think Qiaochu did a quite good job at elaborating it as well. So I feel an answer by me might be simply duplicating some parts of what he did. Of course you can use Poincare duality, but the computation of the homology groups takes some effort if you just use Mayer–Vietoris or long exact sequence.
|
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|
Do you know any almost identities? Recently, I've read an article about almost identities and was fascinated. Especially astonishing to me were for example $\frac{5\varphi e}{7\pi}=1.0000097$ and $$\ln(2)\sum_{k=-\infty}^{\infty}\frac{1}{\left(\sqrt{2}+\frac{1}{\sqrt{2}}\right)^k}=\pi+5.3\cdot10^{-12}$$ So I thought it would be nice to see a few more. Therefore, my question is: Do you know a fascinating almost identity? Can you, in some sense, prove it?
|
There is a substantial list on Wikipedia:
http://en.wikipedia.org/wiki/Mathematical_coincidence
Some of the more interesting (imo) examples are:
\begin{equation}
\pi\approx\frac{4}{\sqrt{\varphi}}\\
\pi^4+\pi^5\approx e^6\\
\frac{\pi^{(3^2)}}{e^{(2^3)}}\approx10\\
e^{\pi}-\pi\approx20
\end{equation}
There's also the claim made in the April 1975 Scientific American (more specifically, the April Fool's claim) that Ramanujan had predicted that $e^{\pi\sqrt{163}}$ is an integer. (It isn't, but it is extremely close.)
Oh, and lest we forget, $\pi=3.2$.
|
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|
$X+Y$ is closed $\Leftrightarrow$ $\|x\|\leq c\|x+y\|$ for all $x\in X$ and all $y \in Y$. The problem says
Let $(Z,\|\cdot\|)$ be a Banach space. Let $X$ and $Y$ be two closed subspaces of $Z$ such that $X\cap Y=\{0\}$. Prove that $X+Y$ is closed if, and only if, there exists $c\geq 0$ such that
$$\|x\|\leq c\|x+y\|,\quad \forall\ x\in X,\quad\forall \ y \in Y.\tag{1}$$
My attempt: Suppose that condition $(1)$ holds. Let $(w_n)$ be a Cauchy sequence in $X+Y$. Then, for all $n$, we have $w_n=x_n+y_n$ with $x_n\in X$ and $y_n\in Y$. By $(1)$ we have
$$\|x_n-x_m\|\leq c\|w_n-w_m\|,\quad \forall \ n,m\in\mathbb{N}.$$
It follows that $(x_n)$ and $(y_n)$ are Cauchy sequences. Since $X$ and $Y$ are complete, $(x_n)$ converges to some $x\in X$ and $(y_n)$ converges to some $y\in Y$. Taking $z=x+y$ we see $(z_n)$ converges to $z\in X+Y$. So, $X+Y$ is complete and thus closed.
1) In my argumment, the hypothesis $X\cap Y=\{0\}$ wasn't used. It's correct? There are any alternative approach?
2) Please, hints for the converse.
Thanks.
|
For the inverse,
consider
$X+Y$ is closed, but (1) is not sure.
This means
$\forall c\geqslant 0$ we can find a $x\in X$ and $y\in Y$ such that,
$\Vert x\Vert>c\Vert x+y\Vert $
divided by $\Vert x\Vert$ we can assume that $x$ has norm 1.
then for any $\frac{1}{c}>0$ we can find
$\Vert x+y\Vert<\frac{1}{c}$,
now
$\vert\Vert x\Vert-\Vert y\Vert\vert\leqslant\Vert x+y\Vert$.
So,
$\Vert y\Vert$ can be also assumed to be bounded.
notice,
$\Vert x+y\Vert$+$\inf_{x'\in X}\Vert y-x'\Vert\geqslant\inf_{x'\in X}\Vert x+y+y-x'\Vert$=$\inf_{x'\in X}\Vert 2y-x'\Vert$=$d(2y,X)$.
also,
$\inf_{x'\in X}\Vert 2y-x'\Vert$=$2\inf_{x'\in X}\Vert y-x'\Vert$=$2d(y,X)$
which means,
$2d(y,X)\leqslant d(y,X)+\Vert x+y\Vert$
so,
$d(y,X)$ can be any small.
denote $Y_M^m$=$\{y\in Y\vert m\leqslant\Vert y\Vert\leqslant M\}$, since we assume $y$ to be bounded we can find such $M$ that fits our situation.
$d(Y_M^m,X)$ can not be $0$ by the closeness of those sets.
$d(y,X)$ can be any small means we can choose $y\notin Y_M^m$ but the norm is bounded by $M$, therefore by the arbitrary nature of smallness of $d(y,X)$ we have to conclude that $y=0$ but then $x$ will have to be $0$, not as we required to be norm $1$.
Ok, in the whole discuss we can choose $\frac{1}{c}$ to be $\frac{1}{2^n}$ and denote the sequence $x_n,y_n$ and do the above argument.
|
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|
Find a simple expression for $\sum_{k=0}^{n} k(k-1)\binom{n}{k}$ for $n\geq0$ Find a simple expression for $\sum_{k=0}^{n} k(k-1)\binom{n}{k}$ for $n\geq0$.
My intuition is to somehow make use of the fact that $\binom{n}{k} = \frac{n}{k}\binom{n - 1}{k - 1}$. Can I simply replace the $\binom{n}{k}$ with this identity and solve from there somehow?
|
Let $X \sim \operatorname{Bin}(n,\tfrac{1}{2})$ be a binomial random variable. You are after $$2^n\mathbb{E}[X(X-1)] = 2^n(\mathbb{E}[X^2]-\mathbb{E}[X]) = 2^n(\mathbb{V}[X] + \mathbb{E}[X]^2 - \mathbb{E}[X]).$$ Since expectation and variance are additive, it is easy to calculate $\mathbb{E}[X] = \tfrac{n}{2}$ and $\mathbb{V}[X] = \tfrac{n}{4}$. Therefore
$$2^n\mathbb{E}[X(X-1)] = 2^n\left(\frac{n}{4} + \frac{n^2}{4} - \frac{n}{2}\right) = 2^{n-2}n(n-1).$$
|
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|
Differential calculus: integrate $\frac{1}{x \log^3 (x)}$ I would like a step by step description of how to integrate $$\frac{1}{x \log^3 (x)}$$
*
*I know that the answer is - $\frac{1}{2\log^2(x)}$
*and that the integral of $\frac{1}{\log^2(x)}$ is $\frac{1}{\log(x)}$
*and that the integral of $\frac{1}{\log(x)}$ is $\log(\log(x))$
But I don't see how these are obtained.
|
For this problem:$$\int\frac{1}{x\log^3(x)}dx\tag{1}$$we can use the substitution:$$u=\log^3(x)\tag{2}$$$$\therefore \log(x)=u^{\frac{1}{3}}\tag{3}$$which leads to:$$du=3\log^2(x)\times\frac{1}{x}dx$$$$=3(u^{\frac{1}{3}})^2\times\frac{1}{x}dx=\frac{3u^{\frac{2}{3}}}{x}dx$$$$\therefore \frac{dx}{x}=\frac{du}{3u^{\frac{2}{3}}}\tag{4}$$Now we substitute this into (1) to get:$$\int\frac{1}{x\log^3(x)}dx=\int\frac{1}{3u^{\frac{2}{3}}\times u}du=\frac{1}{3}\int\frac{1}{u^{\frac{5}{3}}}du$$$$=\frac{1}{3}\int u^{-\frac{5}{3}}du=\frac{1}{3}\left(\frac{u^{-\frac{2}{3}}}{-\frac{2}{3}}\right)=-\frac{1}{2}\times\frac{1}{u^{\frac{2}{3}}}=-\frac{1}{2\log^2(x)}$$
|
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|
proof $z \mid b$ and $w\mid b$ Question I'm working on: Let $a,b$ be integers with $b$ not equal to $0$. suppose $x^2+ax+b=0$ and $x=z,w$. If $z,w$ are integers, show that $z\mid b$ and $w\mid b$.
Is it sufficient for me to show that $x^2+ax+b=(x-z)(x-w)$ so $b=zw$?
|
Here is a better approach:
We need to show $z|b$ and $w|b$. What does it mean for $z$ to "divide" $b$? It means there is some integer $k$ such that $z \cdot k = b$. Well, what are we given? We are given that if $x = z$ or $x = w$, then $x$ satisfies $x^{2} + ax + b = 0$.
But that means $z^{2} + az + b = 0$, or $z^{2} + az = -b$. But then that means $z(z + a) = - b$. Since $z + a$ is an integer (because $z$ and $a$ are integers), then we have found an integer $k$ such that $z \cdot k = -b$, so $z$ divides $-b$. Is it clear that if $z$ divides $-b$, then $z$ divides $b$?
You can use the same argument for $w$. Does this argument make sense to you? Do you understand every step?
|
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|
Partial Derivatives and the Fundamental Theorem of Calculus I am being asked to evaluate the 1st-order partial derivatives $-$ $f_{x}$(x,y) and $f_{y}(x,y)$ $-$ of the following multi-variable function: $f(x,y) = \int_{y}^{x} cos(-1t^2 + 3t -1) dt$. Any help is appreciated.
|
$$
f(x,y)=\int_y^x g(t) dt=G(t)|_y^x=G(x)-G(y) \text{ where } G'=g
$$
so differentiating w.r.t $x$ on both sides of $f(x,y)=G(x)-G(y)$ gives
$$
f_x(x,y)=(G(x)-G(y))_x$$
and by difference rule that is
$$
f_x=(G(x))_x = g(x).
$$
|
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|
Using two coins to select a person fairly. Good evening, I would like to know if the solution to this problem, I know it can be solved because it is from a Hungarian Olympiad. The problem is as follows: You need to fairly select a person between $n$ persons using two unfair coins for which you get to decide the odds.
This is simple if you label the people $1$ through $n$ in binary and then flip $\lceil \log_2 n \rceil$ coins to get a number in binary. If the number corresponds to a person that person gets selected, otherwise repeat the process again.
The problem here is that we need to give a bound for the number of flips before hand. I don't really know how to solve the problem now.
Thank you very much in advance.
Regards.
|
Here's a quite straightforward method with two coins, bounded number of throws, and both coins having rational values. First coin is unbiased; second coin has the following probability for a Head:
$$p = \frac{2^m}{n!}; m = \lfloor{\log_2(n!)}\rfloor$$
With the unbiased coin you can emulate a biased coin which has $\frac{k}{2^m}$ probability for a Head ($k$ is less than $2^m$). So, using the two coins we can emulate a coin which has probability $\frac{1}{r}$ for a Head with $1\leq r \leq n$ (just emulate with the unbiased coin a biased coin with probability $\frac{n!}{r\cdot 2^m}$ for a Head). That would be enough to select uniformly from the set of $n$ people.
Example (simplified): $n=7$. Second coin $p=\frac{8}{21}$. With $3$ throws of the unbiased coin emulate a coin of $\frac{3}{8}$; using that and one additional throw of the biased coin we emulate a coin of:
$$\frac{3}{8} \cdot \frac{8}{21} = \frac{1}{7}$$
Similarly, with 3 throws of the unbiased coin and a single throw of the biased coin we can emulate a coin of:
$$\frac{7}{8} \cdot \frac{8}{21} = \frac{1}{3}$$
So, we actually have the following 'coins' to use:
$$\frac{1}{2}, \frac{1}{3}, \frac{1}{7}$$
And these are enough to choose uniformly from a set of size $7$.
|
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|
Doubts on solving an initial value problem The problem is : $y(\ln y)'= \frac{2t}{y}$ with $y(0)=0$. How to obtain $y'$ ?
|
$y(\ln y)'=y(\dfrac{1}{y}\dfrac{dy}{dt})=\dfrac{dy}{dt}.$
So the problem is $\dfrac{dy}{dt}=\dfrac{2t}{y}\implies ydy=2t dt$
So, $\dfrac{y^2}{2}=t^2+C,$ thus with the initial condition $y(0)=0$, we get $C=0$
ie, $y^2=2t^2\implies y=\pm \sqrt2t \square$
|
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|
Binomial expansion (sort of ) rearrangement Let
$$
x_n=\sum_{i=1}^n\binom{n}{i}y_iz_{n-i} \qquad n=1,\ldots,k
$$
For general $k$, can you find an explicit expression for $y_k$ only in terms of $x_1,\ldots,x_k$ and $z_1,\ldots,z_k$?
For example, if $k=3$, then
\begin{align*}
x_3 &=y_3+3y_2z_1+3y_1z_2+z_3 \\
x_2 &=y_2+2y_1z_1+z_2 \\
x_1 &=y_1+z_1
\end{align*}
So
\begin{align*}
y_3&=x_3-3y_2z_1-3y_1z_2-z_3 =x_3-3(x_2-2y_1z_1-z_2)z_1-3(x_1-z_1)z_2-z_3 \\
&=x_3-3(x_2-2(x_1-z_1)z_1-z_2)z_1-3(x_1-z_1)z_2-z_3 \\
&=x_3-3x_2z_1+6x_1z_1^2-6z_1^3-3x_1z_2+6z_1z_2-z_3 \\
\end{align*}
For general $k$, I have tried to recursively substitute in and get a very complicated expression.
Can anyone help me find a general (preferably simple) formula for $y_k$? Thanks.
Edit: Here's my lengthy (I think correct) solution:
\begin{align*}
& y_k=x_{k}-\sum_{k_{1}=2}^{3}\binom{k}{k-k_{1}}z
_{k_{1}}x_{k-k_{1}} -\sum_{k_{1}=4}^{5}\binom{k}{k-k_{1}}\left( z
_{k_{1}}-\sum_{k_{2}=2}^{k_{1}-2}\binom{k_{1}}{k_{2}}z _{k_{1}-k_{2}}z
_{k_{2}}\right) x_{k-k_{1}} \\
& -\sum_{k_{1}=6}^{7}\binom{k}{k-k_{1}}\left( z
_{k_{1}}-\sum_{k_{2}=2}^{k_{1}-2}\binom{k_{1}}{k_{2}}z _{k_{1}-k_{2}}z
_{k_{2}}+\sum_{k_{2}=4}^{k_{1}-2}\binom{k_{1}}{k_{2}}z
_{k_{1}-k_{2}}\sum_{k_{3}=2}^{k_{2}-2}\binom{k_{2}}{k_{3}}z
_{k_{2}-k_{3}}z _{k_{3}}\right) x_{k-k_{1}}-\cdots \\
& -\sum_{k_{1}=2l}^{k}\binom{k}{k-k_{1}}\left( z
_{k_{1}}-\sum_{k_{2}=2}^{k_{1}-2}\binom{k_{1}}{k_{2}}z _{k_{1}-k_{2}}z
_{k_{2}}+\sum_{k_{2}=4}^{k_{1}-2}\binom{k_{1}}{k_{2}}z
_{k_{1}-k_{2}}\sum_{k_{3}=2}^{k_{2}-2}\binom{k_{2}}{k_{3}}z
_{k_{2}-k_{3}}z _{k_{3}}-\cdots \right. \\
& \left. +\left( -1\right) ^{l-1}\sum_{k_{2}=2l-2}^{k_{1}-2}\binom{k_{1}}{
k_{2}}z _{k_{1}-k_{2}}\cdots \sum_{k_{l-1}=4}^{k_{l-2}-2}\binom{k_{l-2}}{
k_{l-1}}z _{k_{l-2}-k_{l-1}}\sum_{k_{l}=2}^{k_{l-1}-2}\binom{k_{l-1}}{k_{l}
}z _{k_{l-1}-k_{l}}z _{k_{l}}\right) x_{k-k_{1}}
\end{align*}
I needed to program this in and it wasn't as bad as I first thought.
If someone can find a more elegant expression (or simplification), I would accept it as a solution.
|
Given a sequence $s=(s_1,s_2,\ldots)$, it is useful to consider the exponential generating function
$$
f_s(T):=1+\sum_{n\geq 1}s_n\frac{T^n}{n!}.
$$
The relationship between the $x=(x_n)$, $y=(y_n)$, and $z=(z_n)$ can be expressed as
$$
f_x(T) =f_y(T)\cdot f_z(T).
$$
We want to solve for $y$ in terms of $x$ and $z$, so we write
$$
f_y(T)=\frac{f_x(T)}{f_z(T)}.
$$
The constant term of $f_z(T)$ is $1$, so we can invert as follows:
\begin{align*}
\frac{1}{f_z(T)}&=\sum_{k\geq 0}(-1)^k\left(\sum_{n\geq 1}z_n\frac{T^n}{n!}\right)^k\\
&=\sum_{k\geq 0}(-1)^k\sum_{n_1,\ldots,n_k\geq 1}\,z_{n_1}\ldots z_{n_k}\frac{T^{n_1+\ldots+n_k}}{n_1!\ldots n_k!}\\
&=\sum_{m\geq 0}\sum_{k\geq 0}(-1)^k\sum_{\substack{n_1,\ldots,n_k\geq 1\\n_1+\ldots+n_k=m}}z_{n_1}\ldots z_{n_k}\frac{T^m}{n_1!\ldots n_k!}.
\end{align*}
Finally, we multiply this by $f_x(T)$ and extract the coefficient of $T^n$ to obtain
\begin{align*}
y_n&=n!\sum_{i=0}^n \frac{x_{n-i}}{(n-i)!}\sum_{\substack{n_1,\ldots,n_k\geq 1\\n_1+\ldots+n_k=i}}(-1)^kz_{n_1}\ldots z_{n_k}\frac{1}{n_1!\ldots n_k!}\\
&=\sum_{\substack{n_1,\ldots,n_k\geq 1\\b\geq 0\\n_1+\ldots+n_k+b=n}}\hspace{-5mm}(-1)^k \binom{n}{n_1,\ldots,n_k,b}x_b z_{n_1}\ldots z_{n_k}.
\end{align*}
Here $\binom{n}{n_1,\ldots,n_k,b}$ is a multinomial coefficient, and we adopt the convention (as it looks like you did) that $x_0=1$.
|
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|
Determining whether equation $\ln x=\frac{1}{3}x$ has at least 3 solutions I'm trying to say whether
$$\ln x=\frac{1}{3}x$$
Has at least 2 solutions, and after that the problem asks us to tell if the third solution exists...
So I managed to find 1 solution: $f(x)=\ln x - \frac{1}{3} x$, $f(e)>0, f(1)=-\frac{1}{3} < 0$, so by the mean value theorem we have $f(y)=0$ for some $y\in [1,e]$.
What to do about the other two?
UPDATE: I haven't covered derivatives yet in my course...
|
If you look at the graph of the given equation, the plot of ln(x) bends towards X-axis after the second point of intersection while the plot of $\frac13x$ continues straight above. So, there won't be a third solution of the equation.
|
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|
Bisector and symmetric point I can't solve the following problem and need help.
Given an acute-angled triangle ABC.CL is the bisector of $\angle C$ where $L \in AB$. CL instersects the circumcircle of the triangle at point D. O is the center of the circumcircle and I is the incenter of the triangle. If $P$ is the symmetric point of D with respect to $AB$, prove that $\angle LOI =\angle CPI$.
|
Since $\widehat{ACD}=\widehat{DCB}$, $D$ is the midpoint of the arc $AB$ in the circumcircle $\Gamma$ of $ABC$, hence $D,P,O$ are collinear and, obviously, $OC=OD$.
Moreover $\widehat{DOC}=\widehat{BOC}+\widehat{BOD}=2\widehat{A}+\widehat{C}$, so $CD = 2R\sin(\widehat{A}+\widehat{C}/2) $.
Now it is easy to check that:
$$ DP\cdot DO = DL\cdot DC, $$
hence $L,P,O,C$ lie on the same circle, so $\widehat{LPC}=\widehat{LOC}$ and we just need to prove that $\widehat{LPI}=\widehat{IOC}$. Since $DA=DB=DI$, we have that the circle $\Omega$ with center $D$ through $I$ is orthogonal to the circle through $L,P,O,C$, so:
$$ DP\cdot DO = DL\cdot DC = DI^2 $$
gives that the circular inversion with respect to $\Omega$ switches $P,O$ and $L,C$, giving $\widehat{LPI}=\widehat{IOC}$ as wanted.
|
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|
Number of digits $d$ in $d^k$ The title says it all really. For example, how many occurences of $6$ are there in $6^k$? It starts $6, 36, 216, \dots$ so $1, 1, 1,\dots$ The question can now be generalized into any digit or group of digits.
|
Here is a general formula for base $10$ and $d\in[1,10-1]$:
$$\sum\limits_{n=1}^{k}1-\left\lceil\left(\frac{\left\lfloor\frac{d^k}{10^{n-1}}\right\rfloor-10\left\lfloor\frac{d^k}{10^n}\right\rfloor-d}{\left\lfloor\frac{d^k}{10^{n-1}}\right\rfloor-10\left\lfloor\frac{d^k}{10^n}\right\rfloor+d}\right)^2\right\rceil$$
Here is a general formula for base $B$ and $d\in[1,B-1]$:
$$\sum\limits_{n=1}^{k}1-\left\lceil\left(\frac{\left\lfloor\frac{d^k}{B^{n-1}}\right\rfloor-B\left\lfloor\frac{d^k}{B^n}\right\rfloor-d}{\left\lfloor\frac{d^k}{B^{n-1}}\right\rfloor-B\left\lfloor\frac{d^k}{B^n}\right\rfloor+d}\right)^2\right\rceil$$
|
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|
Minimum value of trigonometry function How can I get the minimum value of function
$$f(x) = (2 + \sin x)(5 - \sin x)$$
I have used the differential ways but the answer was not match with the key answer.
By the way the key answer is $6$.
|
You can use the change of variable $t=\sin x$ and remember that $-1\le t\le 1$.
Then
$$(2+t)(5-t)=10+3t-t^2$$
has the derivative $3-2t$, which cancels at $t=\frac32$, out of the domain of $t$.
The minimum can only arise at one of the ends of that domain,
$$f(-1)=\color{green}6$$or $$f(1)=12.$$
No trigonometry !
|
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|
connected and compact subset of $\mathbb{C}$ is either a singleton or has uncountable boundary Let's say that $A\subseteq\mathbb{C}$ is connected and compact but not a singleton. I want to conclude that $\partial A$ is uncountable. Can anyone help me find a reference to cite?
I can see how to prove this in a routine way, by picking distinct elements $a$ and $b$ in $A$, which we may assume without loss of generality have distinct real parts, and then finding an element in $\partial A$ with any real part strictly between the real parts of $a$ and $b$. However, I would much rather just give a reference, since the argument is uninteresting.
|
The boundary of such a set is a perfect set (closed with no isolated points). Such set in $\Bbb R^k$ is uncountable. See Rudin's Principles of Mathematical Analysis Theorem 2.43.
|
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|
Modularity and prime number sequence I tried to solve this modular equation involving first $n$ prime numbers. And it is:
$$2^{3+5+7+11+13+.....+p_{n-3}+p_{n-2}}\equiv p_{n-1}\ \left(\text{mod }p_{n}\right),$$
where $p_{n}$ is the $n$-th prime number.
I couldn't find any solution for this equation until first $300$ primes. Is there any solutions for $n$?
|
For what is worth, next solutions $\{n,p_n\}$ are:
$$
\{306\,311,4\,353\,467\}\\
\{859\,825,13\,174\,621\}\\
\{1\,700\,098,27\,291\,793\}\\
$$
There are no more with $n\le10^7$.
|
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|
matrix multiplication by column vectors If $A$ is an $n \times n$ matrix, and $x$ and $y$ are column vectors in $R^n,$ show that $(A x)\cdot y=x \cdot (A^t y),$ and then that $(A x) \cdot (A y)= x(A^t A y).$ I tried to take $2 \times 2$ or $3 \times 3$ matrices and it worked but i did not know how to generalize it. thank you
|
hint. here are things you need to know:
(a) $x \cdot y = x^Ty = y^Tx,$
(b) $(Ax)^T = x^TA^T$
(c) $A(BC) = (AB)C$
i will do the first one. $Ax \cdot y = (Ax)^Ty = x^TA^Ty = x.A^Ty$
can you see how to do the second one?
|
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|
Show that the series $\sum_{k=0} ^\infty (-1)^k \frac{x^{2k+1}}{2k+1}$ converges for $|x|<1$ and that it converges to $\arctan x$ Show that the series $\sum_{k=0} ^\infty (-1)^k \dfrac{x^{2k+1}}{2k+1}$ converges for $|x|<1$ and that it converges to $\arctan x$
I tried using the ratio test but I got that it equals 1, so it is inconclusive. How can I show that it converges, and that it converges to arctan x? I think I'm trying to show that the Legrange error term is 0 but I'm not sure how to proceed with it - the nth derivatives of arctan x keep changing so...
|
Modifying Jack D'Aurizio's answer,
since
$\sum_{k=0}^n x^k = \dfrac{1-x^{n+1}}{1-x}$,
putting $-x^2$ for $x$ we get
$\sum_{k=0}^n (-1)^kx^{2k}
= \dfrac{1-(-1)^{n+1}x^{2n+2}}{1+x^2}
= \dfrac1{1+x^2}+\dfrac{(-1)^{n}x^{2n+2}}{1+x^2}
$
or
$\dfrac1{1+x^2}
=\sum_{k=0}^n (-1)^kx^{2k} -\dfrac{(-1)^{n}x^{2n+2}}{1+x^2}
$.
Integrating from $0$ to $y$,
$\begin{array}\\
\arctan(y)
&=\int_0^y \dfrac{dx}{1+x^2}\\
&=\int_0^y \left(\sum_{k=0}^n (-1)^{k-1}x^{2k}-(-1)^{n}\dfrac{x^{2n+2}}{1+x^2}\right)dx\\
&=\sum_{k=0}^n (-1)^{k-1}\int_0^y x^{2k}dx-(-1)^{n}\int_0^y \dfrac{x^{2n+2}dx}{1+x^2}\\
&=\sum_{k=0}^n \dfrac{(-1)^{k-1}x^{2k+1}}{2k+1}-(-1)^{n}\int_0^y \dfrac{x^{2n+2}dx}{1+x^2}\\
\end{array}
$
so
$\arctan(y)-\sum_{k=0}^n \dfrac{(-1)^{k-1}x^{2k+1}}{2k+1}
=(-1)^{n+1}\int_0^y \dfrac{x^{2n+2}dx}{1+x^2}
$
so that
$\begin{array}\\
|\arctan(y)-\sum_{k=0}^n \dfrac{(-1)^{k-1}x^{2k+1}}{2k+1}|
&\le\int_0^y \dfrac{x^{2n+2}dx}{1+x^2}\\
&\le\int_0^y x^{2n+2}dx\\
&=\dfrac{x^{2n+3}}{2n+3}\\
&\to 0
\qquad\text{as } n \to \infty\\
\end{array}
$
|
{
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|
if Y1 and Y2 are independent, does it follow that U and X are independent, where U = f1(Y1, Y2) and X= f2(Y1, Y2) If X, Y are iid rvs, and U and Z are r.v.s that can each be written in terms of X and Y, does that mean that U and Z are independent?
|
No. Consider i.i.d. standard normal random variables, $\,X,Y\sim\mathcal{N}\left(0,1\right)\,.$ Let $\,\,U{}={}a_1X+a_2Y\,\,$ and $\,\,Z{}={}b_1X+b_2Y$ for non-zero real numbers $a_1, a_2, b_1$ and $b_2$.
Then, see that
$$
\mathbb{C}ov\left(U,Z\right){}={}a_1b_1+a_2b_2\ne0\,,
$$
therefore $U$ and $Z$ can't be independent.
|
{
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|
Why must a locally compact second countable Hausdorff space be second countable to imply paracompactness? The textbook version of the result I've seen states: A locally compact second countable Hausdorff space is paracompact. Is the property of being second countable needed, or have I missed something?
My thinking:
*
*If the space is locally compact then each point has a compact neighborhood.
*For this compact neighborhood each covering has a finite sub-covering.
*The finite sub-covering is a locally finite refinement.
Thanks in advance.
|
Without the second countable condition the result is not true. The long line is locally compact Hausdorff and is not paracompact because it is locally metrizable but not metrizable.
|
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|
Intersection of context-free language and its reversal I know that intersection of two context-free languages is not always context-free and the following problem:
Given two context-free languages A and B, is $A \bigcap B \neq \emptyset$ ?
is undecidable. But is that true in particular case when we know that $B = \{ w^{R} | w \in A \}$?
|
If $A$ contains a palindromic string $w$ then $w^R=w$. So in this case, $A \cap B \neq \varnothing$. But if $A$ doesn't contain any palindromic string, we cannot find a common element of $A$ and $B$.
|
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|
Show that Möbius transformations that preserve the unit disk are of the matrix form $\tiny \begin{bmatrix}a & b \\ \bar{b} & \bar{a} \end{bmatrix}$ Show that Möbius transformations that preserve the unit disk are of the matrix form $$\begin{bmatrix}a & b \\ \bar{b} & \bar{a} \end{bmatrix},$$ where $|a|^2 - |b|^2 = 1$ and $a,b \in \mathbb{C}$.
I tried approaching this by noting that these transformations must first and foremost preserve the unit circle. So I looked at whether I'd get anything useful out of seeing what a random matrix $\begin{bmatrix}a & b \\ c & d \end{bmatrix}$ would need to satisfy in order for, say, $1, -1$ and $i$ to stay on the unit circle. However, I got nothing useful out of that.
I looked at some threads here already (for example, Möbius Transforms that preserve the unit disk), but none of them seem to help in getting me this form.
Help would be greatly appreciated.
|
First note that your statement of the result is not quite accurate. For example, the Mobius transformation
$$f(z)=\frac{az+b}{cz+d}=\frac{iz+0}{0z+i}=z$$
clearly preserves the unit disc even though the corresponding matrix
$$\pmatrix{i&0\cr0&i\cr}$$
does not have the form you state. However, multiplying $a,b,c,d$ by a common factor does not alter $f$, and so we may assume that $ad-bc=1$. If we suppose that this is the case, then the result is true.
Now note that if $\beta$ is real then the inequation
$$\def\o#1{\overline#1}
z\o z-\alpha z-\o\alpha\o z+\beta<0$$
can be written as
$$|z-\o\alpha|^2<|\alpha|^2-\beta\ ;$$
this is the open disc with centre $\o\alpha$ and radius $\sqrt{|\alpha|^2-\beta}$ if $\beta<|\alpha|^2$, the empty set otherwise.
A bit of algebra shows that the inverse image of $|f(z)|<1$ is
$$z\o z-\alpha z-\o\alpha\o z+\beta<0$$
with
$$\alpha=\frac{c\o d-a\o b}{a\o a-c\o c}\ ,\quad
\beta=\frac{b\o b-d\o d}{a\o a-c\o c}\ .$$
Setting $|\alpha|^2-\beta=1$ and simplifying shows that this is the unit disc if and only if
$$c\o d-a\o b=0\ ,\quad b\o b-d\o d=-1\ .$$
We can therefore write
$$a=a(d\o d-b\o b)=\o d(ad-bc)+b(c\o d-a\o b)=\o d\ ,$$
and similarly $b=\o c$.
|
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|
Interpreting a group homomorphism $f: \mathbb{Z}_{12} \to \mathbb{Z}_{3}$ visually I am having a hard time studying and I am a visual learner. How could I visually imagine a (group) homomorphism
$$\mathbb{Z}_{12} \to \mathbb{Z}_3?$$
Also, if the question states that the map $f$ is a group homomorphism such that $f(1)=2$, how could I find the kernel $K$ of $f$?
|
I find it useful to imagine the group $\mathbb{Z}_n$ a group of rotations (of, e.g., the plane) by multiples of $\frac{1}{n}$ revolution, that is, by multiples of $\frac{2\pi}{n}$ radians, so that $[k] \in \mathbb{Z}_n$ corresponds to a (say, anticlockwise) rotation by $\frac{2 \pi k}{n}$ radians.
Since $\mathbb{Z}_n$ is cyclic, any group homomorphism $f: \mathbb{Z}_n \to H$ is determined by $f(1)$. This leaves three candidate maps, namely
- the map defined by $f([1]) = [0]$, which is just the zero homomorphism $f([n]) := 0$,
- the map defined by $f([1]) = [1]$,
- the map defined by $f([1]) = [2]$.
Per the above mnemonic, we can regard $[1] \in \mathbb{Z}_{12}$ as an anticlockwise rotation by $\frac{1}{12}$ of a revolution, or $\frac{\pi}{6}$ radians, and, e.g., $[2] \in \mathbb{Z}_{3}$ as an anticlockwise rotation by $\frac{2}{3}$ of a revolution, or $\frac{4\pi}{3}$ radians. So, using that $f$ is a group homomorphism, so that $f([n]) = n \cdot f([1])$ we can think of $\phi$ as the map that takes a given rotation in $\mathbb{Z}_{12}$ and applies it $\frac{\frac{2}{3} \text{ rev}}{\frac{1}{12} \text{ rev}} = 8$ times, which by construction is always a rotation in $\mathbb{Z}_3$ so imagined.
(As a word of warning, this is a convenient way to think about finite cyclic groups, but if one wants to think about $\mathbb{Z}_3$ sitting inside $\mathbb{Z}_{12}$ the way we did here, one must specify that one is using this mnemonic; indeed, there is more than one way to put $\mathbb{Z}_3$ into $\mathbb{Z}_{12}$ in a way that respects group multiplication.)
Thinking of a group as we did here as a set of (linear) transformations of some vector space, by the way, is a (the) central idea of representation theory, which has proved to be an immensely powerful too for understanding groups.
|
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|
Equation of normal vector pointing away from ellipse Assuming that I have an ellipse, centered at $(h,k)$ of type:
$$\left(\frac {x-h}{a}\right)^2 + \left(\frac {y-k}{b}\right)^2 = 1$$
The gradient of the normal is:
$$\frac{a^2(y-k)}{b^2(x-h)}$$
From this I can get a vector representation of the normal, but my question is, is there a way to get the normal vector such that it is always pointing away from the center?
|
I think the simplest way is look at the coordinate $(x,y)$ on the ellipse.
For example, if $x-k > 0$, then choose your $x$-component to be positive. Same for $y$-component
explicitly, $(\frac{x-h}{a^2},{\frac{y-k}{b^2}})$
For gradient, do you mean
$$\nabla\left(\left(\frac{x-h}{a}\right)^2+\left(\frac{y-k}{b}\right)^2\right)=2(\frac{x-h}{a^2},{\frac{y-k}{b^2}}) ?$$
Usually for any function $f(x,y)$, the normal direction of the contour $f(x,y)=c$ is easily obtained by taking gradient of $f(x,y)$ as you did. If you want to obtain the normal pointing outward, just choose the right sign in front of $f(x,y)$ such that the value of the function increases as you go outward. Then take the gradient and the normal will be automatically pointing outward.
(I would like to make it a comment but I don't have enough reputation. XP )
|
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|
Prove or disprove absolute convergence Prove or Disprove:
if $\sum_{n=1}^{\infty} a_n$ is absolutely convergent, then $\sum_{n=1}^{\infty} max\{a_n,\frac{1}{n^2}\}$ is also absolutely convergent.
|
ok, I think I just missed something pretty simple:
The sum of two convergent series is also convergent.
$\sum_{n=1}^{\infty} max\{a_n,\frac{1}{n^2}\} < \sum_{n=1}^{\infty} |a_n| + \sum_{n=1}^{\infty} |\frac{1}{n^2}| $
therefore by the first comparison test, the original series is absolutely convergent.
|
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|
Prove $|f|$ is constant implies $f$ is constant Let $f$ be an entire function (differentiable everywhere over $\mathbb{C})$. Suppose that $|f|$ is constant. Prove that $f$ is constant.
Hint: $|f|\equiv c$ implies that $u^2+v^2\equiv c^2$. Take partial derivatives and apply the Cauchy-Riemann equations.
Proof: Suppose $|f|=c$ for some $c \in \mathbb{C}$. Writing $f(z)=f(x+iy)=u(x,y)+iv(x,y)$, this means
\begin{equation}
u(x,y)^2+v(x,y)^2=c^2
\end{equation}
Please correct me if I'm not differentiating correctly, but taking the partial with respect to $x$ I find that
\begin{equation}
2u(x,y) u_x(x,y)+2v(x,y)v_x(x,y)=0
\end{equation}
Simiarly, taking the partial with respect to $y$ shows that
\begin{equation}
2u(x,y) u_y(x,y)+2v(x,y)v_y(x,y)=0
\end{equation}
Is this correct so far?
Edit: Thanks to AlexR for his tip. I continue the proof below.
Substituting the C-R equations ($u_x=v_y$ and $v_x=-u_y$), the first equation gives two possible equations:
\begin{equation}
2\left(u(x,y) v_y(x,y)+v(x,y)v_x(x,y)\right)=0
\end{equation}
or
\begin{equation}
2\left(u(x,y) u_x(x,y)-v(x,y)u_y(x,y)\right)=0
\end{equation}
Similarly, depending on the choice of substitution, the second equation gives two possible equations:
\begin{equation}
2\left(v(x,y)v_y(x,y)-u(x,y)v_x(x,y)\right)=0
\end{equation}
or
\begin{equation}
2\left(u(x,y) u_y(x,y)+v(x,y)u_x(x,y)\right)=0
\end{equation}
(does it matter which equations we choose to use?)
|
A short, but different proof can be given by using the fact that a non-constant holomorphic function is an open map: By assumption $f(\mathbb C)$ is contained in a circle in $\mathbb C$ with radius $c$. Any non-empty subset of such circle is not open in $\mathbb C$, hence $f$ is constant, q.e.d.
|
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Is it possible for a relation to be transitive and symmetric but not reflexive with only one element? E.g. On the set $A = \{1,2,3,4,5,6\}$, is the relation set $R = \{(1,1)\}$ a transitive and symmetric relation but not reflexive?
|
Yes. The relation $R$ that you have given is transitive and symmetric, but not reflexive. To see this, note that $(2,2)\not\in R$, but by definition, $R$ is reflexive if $(a,a)\in R$ for all $a\in A$. It's easy to check that $R$ is symmetric and transitive. For example, note that for all $(a,b)\in R$, we also have $(b,a)\in R$, since $R=\{(1,1)\}$.
|
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Using first order sentences, axiomize the $\mathcal{L}$-theory of an equivalence relation with infinitely many infinite classes. Problem
Let $\mathcal{L} = \{E\}$ where $E$ is a binary relation symbol. Let $T$ be the $\mathcal{L}$-theory of an equivalence relation with infinitely many infinite classes.
Current Solution
First, the axioms for equivalence class:
\begin{alignat*}{3}
&\forall x &&E(x,x),\\
&\forall x \forall y &&(E(x,y) \rightarrow E(y,x)),\\
&\forall x \forall y \forall z &&(E(x,y) \wedge E(y,z) \rightarrow E(x,z))).\\
\end{alignat*}
Then to express infinitely many equivalence classes, we add infinitely many senteneces $\phi_n ~(n \ge 2$):
$$\phi_n = \exists x_1 \ldots \exists x_n ~ \bigwedge_{i < j \le n} \neg E(x_i,x_j).$$
That is we have sentences $\phi_2,\phi_3,\ldots$. Finally, we add infinitely many sentences $\psi_n ~(n \ge 1)$
$$\psi_n = \forall x \exists x_1 \ldots \exists x_n ~ \bigwedge_{i < j \le n} x_i \neq x_j \wedge \bigwedge_{i=1}^n E(x,x_j)$$
to axiomize each class is infinite.
Problem
I am not confident that $\{\phi_n\}$ and $\{\psi_n\}$ are not contradicting each other
|
You have axiomatized the theory asserting that $E$ is an equivalence relation with infinitely many classes, ALL of which are infinite. Your statement $\psi_n$ says that every class has at least $n$ elements.
But the theory mentioned in the title problem was just that $E$ should have infinitely many infinite classes (and perhaps also some finite classes). It turns out that this problem is impossible.
Theorem. The collection of equivalence relations with infinitely many infinite classes is not first-order axiomatizable in the language of one binary relation with equality.
Proof. Let $E_0$ be an equivalence relation on a set $X$ with infinitely many classes of arbitrarily large finite size, and no infinite classes at all. So $\langle X,E_0\rangle$ is not one of the desired models. Let $T$ be the elementary diagram of $\langle X,E_0\rangle$, plus the assertions with infinitely many new constants $c^n_i$, that $c^n_i\mathrel{E} c^n_j$ and $c^n_i\neq c^n_j$ and $\neg (c^n_i\mathrel{E} c^m_j)$, whenever $n\neq m$ and $i\neq j$. This theory is finitely consistent, and so it is consistent. Any model of $T$ will be an expansion of an elementary extension of the original model $\langle X,E_0\rangle$, but it will now have infinitely many infinite classes.
Thus, the property of having infinitely many infinite classes is not first-order expressible, since the original model did not have that property but the new model did, even though they were elementary equivalent. QED
|
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Moving from Finitely Additive to Countably Additive I have a question on basic measure theory here:
Let $(X, \mathscr A)$ be measurable space and $\mu$ is a non-negative set function that is finitely additive and such that $\mu (\emptyset) = 0$. Suppose that whenever $A_i$ is an increasing sequence of sets in $\mathscr A$, then $\mu(\cup_i A_i) = \lim_{i \to \infty} \mu(A_i)$. Show that $\mu$ is a measure.
Here are the questions I would like you to help:
(1) First, background question: In this problem, "finitely additive" means
$$\mu(\bigcup_{j=1}^n A_j) = \sum_{j=1}^n \mu(A_j), \quad \text{where }A_j \text{'s are pairwisely disjoint.}$$
Correct me if I am wrong here.
(2) Another background: The phrase "Suppose that whenever $A_i$ is an increasing sequence of sets..." means
$$\text{if } A_i \uparrow \text{ then } \mu(\cup_i A_i) = \lim_{i \to \infty} \mu(A_i), \quad \text{where } \ldots A_{i-1}\subset A_i \subset A_{i+1} \ldots$$
Again, correct me if I am wrong here.
(3) Done with the background. To solve this problem, my understanding is to "move" from finitely additive as given in (1) to countably additive as given in a measure's definition, but I do not how to proceed. Do please let me know the friendliest way of solving this problem.
Thank you for your time and effort.
|
The idea is to turn a countable disjoint union into an increasing union. To do that, you'll need to use different sets. So suppose $\{ A_n \}_{n=1}^\infty$ is a disjoint collection of measurable sets. Consider $\{ B_m \}_{m=1}^\infty$ defined by $B_m = \bigcup_{n=1}^m A_n$. This is an increasing union, so it has the limit property which was assumed. Use this to conclude the limit property you want for the measure of the union of the $A_n$.
|
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|
An example of an infinite open cover of the interval (0,1) that has no finite subcover I've been having a hard time solving this problem that I was given in class. The problem states
" Give an example of an infinite open cover of the interval (0,1) that has no finite subcover."
I know that the set has to be compact and that both 0 and 1 are limit points. Aside from those two known factors I'm at a complete loss as to how to go about solving this problem.
|
You can take the open cover $\{(0+\frac{1}{n}, 1-\frac{1}{n})\mid n>2\} $. It has no finite subcover because if it does simply take the largest $n$ and $\frac{1}{2n}$ is not included.
|
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|
Prove or disprove : if $x|y^2 $then $x|y$ How can I prove such statement?
I think that if $x|y^2$ then $x|(y*y)$ so $x|y$ or $x|y$ which means that in any case $x|y$. Am I correct?
I ask this question as such template because I think that it doesn't matter which $x$ or $y$ represents.
However, the original question is: if $6|n^2$ then $6|n$ and if $12|n^2$ then $12|n$.
|
The claim $x \mid y^2 \implies x \mid y$ is clearly false. Take $x = y^2 > 1$ a perfect square and see for yourself. What is true is: $$\begin{cases} x \mid yz \\ \gcd(x,y)=1\end{cases} \implies x \mid z.$$
Having this in mind, if $\gcd(x,y)=1$, and taking $y = z$...
|
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How to solve $y''+Ay'+By=a+bt+ct^{2}$? I know that the solution of this second-order non-homogeneous will be of the form $y = y_{p} + y_{c}$.
First I find the solution to the homogeneous DE: $y''+Ay'+By=0$, with the characteristic equation $r^{2}+Ar+B=0$. However, it's hard to actually get the roots since I don't know what $A$ and $B$ are.
Any ideas on how to overcome this problem?
EDIT: By using the quadratic formula: $r = \frac{-A\pm \sqrt{A^{2}-4B}}{2}$.
So we don't really know about the form of the complementary solution, $y_{c}$.
Now, we find the particular solution of $y''+Ay'+By=a+bt+ct^{2}$.
Let $y_{p} = A_{1}t^{2}+B_{1}t+C_{1}$.
So $y'_{p} = 2A_{1}t+B_{1}$ and $y''_{p} = 2A_{1}$.
We substitute $y_{p}$ and it's first and second derivatives back into the non-homogeneous DE to obtain:
$t^{2}(A_{1}B)+t(2AA_{1}+BB_{1})+(2A_{1}+AB_{1}+BC_{1})=ct^{2}+bt+a$
So $$A_{1}B=c;$$
$$2AA_{1}+BB_{1}=b;$$
$$2A_{1}+AB_{1}+BC_{1}=a.$$
From the above equations I obtain $A_{1} = \frac{B}{c}$, $B_{1} = \frac{cb}{B}-2A$, and $C_{1} = \frac{a - \frac{B}{c} - \frac{ACb}{B}+2A^{2}}{B}$.
So the particular solution, $$y_{p} = (\frac{B}{c}) t^{2} + (\frac{cb}{B}-2A)t + \frac{a - \frac{B}{c} - \frac{ACb}{B}+2A^{2}}{B}$$.
Then how do I write down expression for the general solution, $y$.
Is it just $$y = (\frac{B}{c}) t^{2} + (\frac{cb}{B}-2A)t + \frac{a - \frac{B}{c} - \frac{ACb}{B}+2A^{2}}{B} + y_{c}?$$
|
to find the particular soultion,we assume
$$y_p=A_1t^2+B_1t+C_1$$
$$y'_p=2A_1t+B_1$$
$$y''_p=2A_1$$
substitute in D.E to get
$$2A_1+A(2A_1t+B_1)+B(A_1t^2+B_1t+C_1)=a+bt+ct^2$$
$$BA_1=c$$
$$A_1=c/B$$
so the $c$,$B$ are known
$$2AA_1+BB_1=b$$
$$B_1=(b-2A(c/B))/B$$
$$2A_1+AB_1+BC_1=a$$
$$C_1=(a-2A_1-AB_1)/B$$
If you want to find the complementary solution
$$r^2+Ar+B=0$$
$$r=0.5A\mp \sqrt{0.25A^2-B}$$
$$r=\alpha \pm \beta $$
$$y_c=K_1e^{\alpha + \beta}+K_2e^{\alpha - \beta}$$
$$y_c=e^\alpha(K_1e^\beta+K_2e^{-\beta})$$
If the $\beta$ is complex part, you can change the exponential to $sin$ and $cos$
|
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|
How do I prove, using the definition, that the nth root is a continuous function? Stuff I've tried:
\begin{align}
|x^{\frac 1 n}-a^{\frac 1 n}| &< \epsilon
\\
|(x^{\frac 1 n}-a^{\frac 1 n})^n| &< \epsilon ^n
\\
\left|\sum_{i=0}^{n}C(n,i)x^{\frac in}(-a)^{1-\frac i n}\right| &< \epsilon ^n
\end{align}
(it looks like a one but the power on $x$ and $a$ it's actually $\frac i n$)
if we take $\delta \leq 1$
$$
|x| < |a| + 1
\\
\left|\sum_{i=0}^{n}C(n,i)x^{\frac in}(-a)^{1-\frac i n}\right| \leq 2^n(|a| + 1)
$$
I would love to be able to say that $\delta = \min \left\{1, \frac \epsilon{2^n(|a|+1)} \right\}$ but this doesn't seem to be the case.
I also tried using the $$x^n - y^n = (x-y)(x^{n-1} + ... + y^{n-1})$$ formula but it didn't help much either.
The way I used this formula for $n=3$ for example was:
$$|\sqrt[3]x - \sqrt[3]a| < \epsilon
\\
|x-a| < \epsilon ( \sqrt[3]{x^2} + \sqrt[3]xa + \sqrt[3]{a^2})
$$
if we use the same trick $\delta \leq 1$
$$( \sqrt[3]{x^2} + \sqrt[3]xa + \sqrt[3]{a^2}) < 3\sqrt[3]{(|a|+1)^2}$$
but even if we set $\delta = \min \left\{1, 3\epsilon \sqrt[3]{(|a|+1)^2}\right\}$
it can't be shown that $|x-a|< \delta$ implies $|\sqrt[3]x - \sqrt[3]a| < \epsilon$
Prove continuity for cubic root using epsilon-delta
|
With some accessible work, you can also show the inequality
$$
\left|x^{1/n}-y^{1/n}\right|\le|x-y|^{1/n}
$$
showing Hölder continuity and uniform continuity on $[0,\infty)$.
The proof starts with considering $0\le y\le x$, so that $x^{1/n}=y^{1/n}+h$, $h\ge0$ and uses only the binomial theorem,
$$
x-y=(y^{1/n}+h)^n-y=((y^{1/n})^n+...+h^n)-y\ge h^n
\\
\implies\sqrt[n]{x-y}\ge x^{1/n}-y^{1/n}
$$
|
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|
Solving natural logarithms with absolute value Question from my text: $e^{4x-2014} - 7 = |-3|$. I've never seen this before and my text is useless! Thank you!
|
Hint: First of all, what is $|-3|$? Second, sum $7$ on both sides of the equation. Third, apply $\ln$ in both sides of the equation. Fourth: sum $2014$ on both sides of the equation. And finally divide both sides of the equation by $4$. Carry the steps carefully and you'll see what's happening.
|
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|
Finding a paper/reference where a solution also exists I have already calculated the following sum:
$$\sum_{k=0}^\infty{B_2(k)}=\frac{\pi^2}{3}-2\approx1.2898681336964$$
where $B_2(k)$ is the 2nd hypergeometric Bernoulli number of order $k$. These numbers are defined with the following generating function:
$$\frac{x^N/N!}{e^x-1-x-...-(x^{N-1}/(N-1)!)}=\sum_{k=0}^\infty\frac{B_k(N)}{k!}x^k$$
My question is this: Is there a way to find a catalog of numbers such as this to see if this number is the result of other calculations, summations, products, integrals, etc? I checked OEIS for the decimal sequence but is was not there but didn't know if other such sites/references existed.
While the below answer is helpful for another way to get the closed form of the sum, I do not need that. Hopefully someone else has the answer I am looking for. Thanks.
|
Observe that we are trying to compute the sum
$$2\sum_{k=0}^\infty
[x^2] \frac{x^k/k!}{e^x-1-x-\cdots-x^{k-1}/(k-1)!}$$
which is
$$2\sum_{k=0}^\infty
[x^2] \frac{x^k/k!}{\sum_{q=k} \frac{x^q}{q!}}.$$
Use the Cauchy Product to extract the coefficient:
$$\frac{x^k}{k!}
= \left(\sum_{q\ge k} \frac{x^q}{q!}\right)
\times\left(a_0 + a_1 x + a_2 x^2 + \cdots\right).$$
This gives $a_0 =1$ and
$$a_0 \frac{1}{(k+1)!} + a_1 \frac{1}{k!} = 0 $$
so that $a_1 = - \frac{1}{k+1}.$
Furthermore
$$a_0 \frac{1}{(k+2)!} + a_1 \frac{1}{(k+1)!}
+ a_2 \frac{1}{k!} = 0$$
or
$$a_0 + a_1 (k+2)
+ a_2 (k+1)(k+2) = 0$$
so that $a_2 = \frac{1}{(k+1)^2(k+2)}.$
Hence we are computing the sum
$$\sum_{k\ge 0} \frac{1}{(k+1)^2(k+2)}$$
which is by partial fractions
$$\sum_{k\ge 0}
\left(\frac{1}{k+2}-\frac{1}{k+1} +\frac{1}{(k+1)^2}\right).$$
which evaluates to
$$-1 + \zeta(2).$$
Multiply by two to get
$$2\zeta(2) - 2 = \frac{\pi^2}{3} - 2.$$
|
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|
Find a solution: $3(x^2+y^2+z^2)=10(xy+yz+zx)$ I'm something like 90% sure that this diophantine equation has nontrivial solutions:
$3(x^2+y^2+z^2)=10(xy+yz+zx)$
However, I have not been able to find a solution using my calculator. I would greatly appreciate if someone could try to find one using a program. Or maybe you can just guess one that happens to work?
Thanks!
EDIT: By nontrivial I mean no $0$'s. (Credits to Slade for reminding me to define this)
EDIT2: In fact, you are free to find a nontrivial solution to $(3n-3)(x^2+y^2+z^2)=(9n+1)(xy+yz+zx)$ where $n\equiv 1\pmod 5$ is a positive integer. The one I posted above is the case $n=5(2)+1$, but you will make my day if you can find a nontrivial solution for any $n=5k+1$.
|
As far as I understand - this is the site for solving the problem. Programming and calculation using the computer is not mathematics. If you want to calculate - there is a special section. https://mathematica.stackexchange.com/questions
Here it is necessary to solve the equations.
For the equation:
$$3(x^2+y^2+z^2)=10(xy+xz+yz)$$
The solution is simple.
$$x=4ps$$
$$y=3p^2-10ps+7s^2$$
$$z=p^2-10ps+21s^2$$
$p,s - $ any integer which we ask.
Why make a program? What's the point? For what?
|
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|
First examples for topology of non-Hausdorff spaces I have absolutely no intuition about non-Hausdorff spaces. I would like to understand the topology of non-Hausdorff spaces (in particular spaces obtained by "bad" group actions).
As a first example, I would like to calculate the fundamental group of $\mathbb R^2 / \mathbb R \cong \mathbb C / \mathbb R$, where a real number $q$ acts on $r \operatorname e^{i \theta}$ by multiplication. This is like projective space "without taking out the zero".
Of course the origin is a deformation retract of the plane. We can choose a deformation retraction (scaling the radius), which is "equivariant" w.r.t. the scaling action of $\mathbb R$. Does this show that the fundamental group of $\mathbb R^2 / \mathbb R$ is trivial?
I would also welcome examples of other non-Hausdorff spaces and their fundamental groups and/or (co)homology groups.
|
Yes, it is trivial.
To get some intuition for non-Hausdorff spaces, consider this fact: if you take a simplicial complex and quotient every open simplex to a point, you get a space that has the same homotopy groups as the original space. For instance, you can take the boundary of a triangle (which is homeomorphic to a circle) and quotient out the interior of each edge. This gives you a non-Hausdorff space with 6 points, three of which are open (the three squished open intervals) and three of which are closed. This space has fundamental group equal to $\mathbb{Z}$ and its universal cover is the quotient of the real numbers obtained by squishing each open interval between integer points.
To get a 2-sphere (up to homotopy), take the boundary of a tetrahedron and send each open face to a point and each edge to a point. The resulting space is simply connected but with non-trivial $\pi_2$.
|
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|
Integral Power Rule Step-by-step Real quick, two things: I'm sorry if my notation or terminology is incorrect, and I know what I'm asking isn't strictly necessary for my studies, but writing something out step by step helps me to understand math intuitively.
I have spent a while thinking about this and cannot figure out where to go next.
I am taking Calc I and I understand differentiation intuitively and can derive functions using the various rules provided, while also understanding why they work and being able to do the steps manually.
So, for the power rule, I can show you the why of:
$$f(x) = x^n$$
$$f'(x) = nx^{n-1}$$
I also understand that to take the antiderivative, you reverse the process. I also know why the antiderivative provides a sum. Since I can't write it out step by step, though, I am not satisfied.
How can I show, algebraically, that
$$F(x^n) = \frac{x^{n+1}}{n+1}$$
So far (and I don't know how right I am, but I think I am close), I've gotten this far on my own:
$$f(x) = x^n$$
$$\lim_{n \rightarrow \infty}\sum_{i=0}^n \left[\frac{x}{n} \cdot f\left(i\cdot\frac{x}{n}\right)\right] = F(x)$$
How do I evaluate this limit? How do I cross this threshold to reach the final form of the antiderivative? I have looked through google and this site on my own for a solution, but I can't seem to come across how to proceed.
Thanks for your time.
|
Here is what the problem boils down to: You need to have a closed formula for summing $$\sum_{i=1}^ni^k$$ I know of such a formula for $k=1,2,3$ and a buddy of mind derived one for $k=4$, but beyond that I don't know that there is a general formula for any $k$. For $k=1$ we know $$\sum_{i=1}^ni^1 = \frac{n(n+1)}{2}$$ So if you had been interested in finding the antiderivative of $f(x)=x$, you evaluate the sum as you correctly outlined: $$\lim_{n \rightarrow \infty}\left[\sum_{i=0}^n \frac{x}{n} \cdot f\left(i\cdot\frac{x}{n}\right)\right] = \lim_{n \rightarrow \infty}\left[\sum_{i=0}^n \frac{x}{n} \cdot \frac{xi}{n}\right] \\ = \lim_{n \rightarrow \infty}\left[\frac{x^2}{n^2}\sum_{i=0}^n i\right]$$ It is valid to move the $\frac{x^2}{n^2}$ outside the sum since we are summing over $i$. Now use the closed formula above to get $$\lim_{n \rightarrow \infty}\left[\frac{x^2}{n^2}\sum_{i=0}^n i\right] = \lim_{n \rightarrow \infty}\left[\frac{x^2}{n^2}\frac{n(n+1)}{2}\right] \\ = \frac{x^2}{2}\lim_{n \rightarrow \infty}\left[\frac{n(n+1)}{n^2} \right] \\ = \frac{x^2}{2}$$ which is clearly the desired result. Again, you can find closed form sums for $k=2,3$ and maybe $4$ somewhere out there, but unless you know a general formula for an arbitrary positive integer $k$, your integral power rule options are limited.
|
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Using Euler's formula for cos/sin 12 degrees and cos/sin 48 degrees $(\cos12^\circ+i\sin12^\circ+\cos48^\circ+i\sin48^\circ)$. Using Euler's Formula, turn this into exponential form (i.e. something like $e^{i\frac{5\pi}{12}}$).
Would I need to use the $\cos$ and $\sin$ sum and difference formulas? I tried doing that and it became messy very quickly. Is there an alternative?
|
Observe
\begin{align*}
\cos 12^{\circ}=\cos(30^{\circ}-18^{\circ})&=\cos30^{\circ}\cos18^{\circ}+\sin30^{\circ}\sin18^{\circ} \\
\sin 12^{\circ}=\sin(30^{\circ}-18^{\circ})&=\sin30^{\circ}\cos18^{\circ}-\cos30^{\circ}\sin18^{\circ}\\
\cos 48^{\circ}=\cos(30^{\circ}+18^{\circ})&=\cos30^{\circ}\cos18^{\circ}-\sin30^{\circ}\sin18^{\circ} \\
\sin 48^{\circ}=\sin(30^{\circ}+18^{\circ})&=\sin30^{\circ}\cos18^{\circ}+\cos30^{\circ}\sin18^{\circ}
\end{align*}
Then, from trigonometric identities, we can reduce it as follows
\begin{align*}
\cos12^{\circ}+i\sin12^{\circ}+\cos48^{\circ}+i\sin48^{\circ}&=2\cos30^{\circ}\cos18^{\circ}+i2\sin30^{\circ}\cos18^{\circ}\\
&=2e^{i\frac{\pi}{6}}\cos\left(\frac{\pi}{10}\right)
\end{align*}
|
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|
Showing some Function $fg$ is Integrable when $g(x)\le e^{-\vert x\vert }$ and given some Conditions on $f$. Suppose $\int_a^b \vert f \vert <\infty$ for all real $a,b$ and that $$\int_{-r}^r \vert f \vert \le (r+1)^a$$ for all real $r$ some real $a$, and that $$g(x)\le e^{-\vert x \vert}$$ I want to show $$f(x)g(tx)\in L^1(\mathbb{R})$$ when $t\not =0$.
My biggest issue is finding a decent upper bound for $\vert f \vert$. My idea is to find an upper bound for $\vert f \vert$ on $[-r,r]$, use this upper bound to find an upper bound for $\vert f(x)g(tx)\vert$ such that when I integrate from $-r$ to $r$ and take a limit as $r\to \infty$ I have convergence. The problem is that although the integral of $f$ is bounded on $[-r,r]$, $f$ can assume arbitrarily large values over this interval. I blindly used $(r+1)^a$ as an upper bound for $\vert f\vert$ but when I do that I get $\int _{-r}^r \vert f(x)g(tx) \vert \le \int _{-r}^r (r+1)^a\cdot e^{-\vert xt\vert}\le (r+1)^a(\frac{e^{tr}-e^{-tr}}{t})$ if $t>0$, but this tends to $\infty$ as $r\to \infty.$
How can I find a sharper upper bound for $\vert f\vert$ or show that $fg\in L^1(\mathbb{R})$?
|
Motivation for this solution: we're given information about the integral of $f$, so we want to bring that function into the bounds somehow. That suggests integration by parts.
Let's just work on $[0,\infty)$ (the other half is similar) and assume $t>0$. Define $F(x) = \int_0^x |f(y)|\,dy$. Then
$$
\int_0^r |f(x)g(tx)|\,dx \le \int_0^r |f(x)| e^{-tx}\,dx = F(x)e^{-tx}\big]_0^r + t \int_0^r F(x) e^{-tx}\,dx.
$$
Since $F(0)=0$ and $F(x) \le (x+1)^a$, this gives
$$
\int_0^r |f(x)g(tx)|\,dx \le (r+1)^ae^{-tr} + t \int_0^r (x+1)^ae^{-tx}\,dx;
$$
taking $r\to\infty$ yields
$$
\int_0^\infty |f(x)g(tx)|\,dx \le t \int_0^\infty (x+1)^ae^{-tx}\,dx,
$$
which is a convergent integral.
|
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|
Let $x,y,z>0,xyz=1$. Prove that $\frac{x^3}{(1+y)(1+z)}+\frac{y^3}{(1+x)(1+z)}+\frac{z^3}{(1+x)(1+y)}\ge \frac34$ Let $x,y,z>0$ and $xyz=1$. Prove that $\dfrac{x^3}{(1+y)(1+z)}+\dfrac{y^3}{(1+x)(1+z)}+\dfrac{z^3}{(1+x)(1+y)}\ge \dfrac34$
My attempt:
Since it is given that $xyz=1$, I tried substituting $x=\dfrac{a}{b},y=\dfrac{b}{c},z=\dfrac{c}{a}$. But the expansion looked really ugly and I didn't think I could make out anything out of it.
So, I made another attempt, if each element was greater than $\dfrac{1}{4}$, we could have a solution, so, treating that way, I get $4x^3\ge 1+x+y+xy, 4y^3\ge 1+x+z+xz, 4z^3\ge 1+x+y+xy$. Using AM-GM I get an equality.
So, please help. Thank you.
|
As another approach, you could prove it as follows.
Due to Hölder's inequality, we have:
$$
\left(\sum_{cyc} \frac{x^3}{(1+y)(1+z)}\right)\cdot\left(\sum_{cyc} (1+y)\right)\cdot\left(\sum_{cyc} (1+z)\right)\ge(x+y+z)^3\iff \sum_{cyc} \frac{x^3}{(1+y)(1+z)}≥\frac{(x+y+z)^3}{\left(\sum_{cyc} (1+y)\right)\cdot\left(\sum_{cyc} (1+z)\right)}=\frac{(x+y+z)^3}{\left(3+x+y+z\right)^2}
$$
So it remains to prove that:
$$
\frac{(x+y+z)^3}{\left(3+x+y+z\right)^2}≥\frac34
$$
Setting $x+y+z=a$, this is equivalent to:
$$
\frac{a^3}{\left(3+a\right)^2}≥\frac34\iff 4a^3≥27+18a+3a^2\iff 4a^3-3a^2-18a-27≥0
$$
We have equality at $a=3$, so we can factor out $a-3$:
$$
(a-3)\left(4a^2+9a+9\right)≥0
$$
But since
$$
a=x+y+z≥3(xyz)^{\frac{1}{3}}=3\iff a-3≥0
$$
due to AM-GM, this is always true, so we're done.
|
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|
Requirement for Hall-$\pi$-subgroups to be conjugate (Wielandt) I am reading through Isaacs Theory of finite groups and towards the end of chapter 3 on "split extensions" there is mention of a theorem of Wielandt which states that if a finite group $G$ contains a nilpotent Hall-$\pi$-subgroup then all the Hall-$\pi$-subgroups of $G$ are conjugate. However I cannot find an exact statement and proof of this theorem in any text regarding the matter. Does anyone know of any such texts which could be of use?
|
You can look at the Theorem and proof in Derek J.S. Robinson's "A course in the Theory of Groups" Chapter 9 (p. 258-259). It says: Let the finite group $G$ possess a nilpotent Hall $\pi$-subgroup $H$. Then every $\pi$-subgroup of $G$ is contained in a conjugate of $H$. In particular all Hall $\pi$-subgroups of $G$ are conjugate. (The proof is easy using a lemma of O.J. Schmidt, and I do not know any other that does not uses it). I hope I have helped you.
|
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Algebraic characterization of commutative rings with Krull dimension=1,2, or 3 A commutative ring $R$ (with $1$) is $0$-dimensional if and only if $R/\sqrt 0$ is von Neumann regular. Besides this result, there is a wealth of information about zero-dimensional rings.
I could not find any information about rings of higher Krull dimension. If the problem for general commutative rings is too hard, how about the case when $R$ is Noetherian and/or an integral domain?
|
What about the following result due to Coquand and Lombardi
(A short proof for the Krull dimension of a polynomial ring.
Amer. Math. Monthly 112 (2005), no. 9, 826–829.)
Let $R$ be a commutative unital ring. For $N\in \mathbb N$,
the following assertions are equivalent:
(1)The Krull dimension of $R$ is at most $N$.
(2) For all $(a_0,\dots, a_{N})\in R^{N+1}$ there exists $(x_0,\dots, x_{N})\in R^{N+1}$ and $(n_0,\dots, n_{N})\in \mathbb N^{N+1}$ such that
$$
a_0^{n_0}\Biggl( a_1^{n_1} \Bigl(\cdots \bigl( a_{N}^{n_{N}}(1 +a_{N} x_{N})+\cdots\bigr)\Bigr) +a_0x_0\Biggr)=0.
$$
|
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Finding circle of a sphere through two points We have two points $P_1, P_2$ on a sphere $S$ of radius $R$. Suppose for $r \ll R$, the distance between $P_1$ and $P_2$ is less than $2r$.
Then, $P_1$ and $P_2$ both lie on exactly two radius-$r$ circles of the sphere. This is clear: of the set of planes that contain $P_1$ and $P_2$, there are two of those planes that intersect $S$ to form a circle with radius $r$.
Given $P_1$, $P_2$, $R$ and $r$, how can I calculate the centers of these two circles?
I would prefer to do as little messy spherical geometry as possible :).
|
Here is a partial answer, if you don't mind a bit of linear algebra.
Let $u_1$ and $u_2$ be the vectors from the center of the sphere to $P_1$ and $P_2$ respectively. These vectors have length $R$. The points you are looking for will each correspond to vectors of length $R$ as well; let us call either of them $x$.
So you want a vector $x$ of length $R$ so that $$u_i\cdot x=R^2\cos\theta$$ for $i=1,2$. Here $\theta$ is the angle of the cone defined by the center of the spehere and the circle of radius $r$. (The exact value depends on whether $r$ is measured along the surface of the sphere or in the plane spanned by the circle.)
Writing $x=x_1u_1+x_2u_2+x_3u_3$, where $u_3$ is a suitable vector orthogonal to $u_1$ and $u_2$ (for example their cross product), you get two linear equations from the above equation. These involve only $x_1$ and $x_2$, so you get a unique solution. Then $x_3$ can be found from the requirement that the length of $x$ be $1$. There will be two solutions to this, a simple quadratic equation.
While carrying out the detailed calculations, don't forget that $u_1$ and $u_2$ are not orthogonal.
|
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Non-linear ODE $1+y'^2=yy'^2$, $y(0)=b,y(a)=c$ How to solve
$$1+y'^2=yy'^2?$$
The initial condition is $y(0)=b,y(a)=c$.
It is a non-linear ODE.
Some hint is also helpful! Thanks for your help!
|
We have $$1+(y')^2=y(y')^2$$
Divide by $(y')^2$ to get $$\frac{1}{(y')^2}=y-1$$
Let us solve for $x=x(y)$ instead of for $y=y(x)$. Then we get the equation:
$$x'=\sqrt{y-1}\ \ \ \ \ \ \text{We need to do the other sign of the square root too.}$$
Separate variables now: $$\begin{align}dx&=\sqrt{y-1}dy\\x+C&=\frac{2}{3}(y-1)^{3/2}\\y&=\left[\frac{3}{2}(x+C)\right]^{2/3}+1\end{align}$$
And we need to check which of these solutions do satisfy the original equation.
|
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Prove that $4^n > n^4$by induction for $n\ge 5$ That is a simple question and I can't start a simple desenveloment.
Just $k=5$ we have $4^5 = 1024 > 5^4 = 625$
for $k+1$: $4^{k+1} > (k+1)^4\Rightarrow 4^k > (k+1)^4/4$
And How can i proceed after that?
Best,
|
Hint:
For $k \geq 5$ the following is true:
$$k\sqrt2 > k+1$$
As:
$$5(\sqrt2 -1)>5\times \frac25 =2> 1$$
We can show that $\sqrt2 -1 >\frac25$ as such:
$$\sqrt2>\frac75$$
Now as both $\sqrt2$ and $\frac75$ are bigger than zero, the inequality is persevered by squaring.
$$2=\frac{50}{25}>\frac{49}{25}$$
|
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How to prove there always exists a two variables or one variable equation corresponding to a curve drawn on a 2D plane? How to prove there always exists a two variables or one variable equation corresponding to a curve drawn on a 2D plane?
For example, a circle 's equation is $x^2+y^2=r^2$.
|
Since Victor didn't specify what he means by "a curve drawn ...", I'll assume that
this is a parametric curve $p = \gamma(t)$, $0 \le t \le 1$, with $\gamma: [0,1] \to \mathbb R^2$ continuous. The image $C = \gamma([0,1])$ is then a compact set in the plane, and $F(p) = \text{dist}(p, C)$ (i.e. the distance from $p$ to $C$) is a continuous function on $\mathbb R^2$ such that $C = \{p: F(p) = 0\}$, i.e.
$F(p)= 0$ can be taken to be the "equation" of the curve.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Closed and Bounded but not compact Let $(C[0,1];d_{\infty})$ the metric space with $C[0,1]$ the continuous functions on $[0,1]$ and $d_{\infty}(f,g)=\max_{x \in[0,1]}|f(x)-g(x)|$
Prove that $$S=\{f:|f(x)| \leq 1\}$$ is closed and bounded but no compact.
My attempt: I've already proved that is bounded using a ball with radius $2$. But I can't figure out how can I prove that is closed and is not compact, I tried with the limit points and the complement but I'm lost.
|
It is closed because the uniform limit of continuous functions is continuous, and the limit function will belong to S, because it is also the pointwise limit.
For non-compactness, since we are in a metric space, it is enough to find a sequence in S which (you can show) has no convergent subsequence. $\{x^n\}_{n\geq 1}$ will do the trick.
|
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|
Sum of entries made one Assume that I have performed some loop and summed values of $x_i$ and the sum is as follows $$S= \sum_{i=1}^n (x_i *0.0001) = 0.9999$$
However I would like to have the sum to be equal to 1 i.e,
$$\sum_{i} (x_i *0.0001)= 1$$
How much should I add to each of the entires $x_i$ so that I end with sum of values equal to one.
Is the solution the increment to each $x$
$$increment = \frac{\frac{0.9999}{0.0001} - \frac{1}{0.0001}}{n}$$
Thanks
|
$$I_1=\sum _{i=1}^n{x_i}*u=X_1$$
$$I_2=\sum _{i=1}^n{(x_i+ \delta)}*u=X_2$$
$$I_2-I_1=\sum _{i=1}^n{(\delta)}*u=X_2-X_1=\Delta$$
$$ n*\delta={\Delta \over u}$$
$$\delta = { \Delta \over (nu)}$$
For $\Delta=0.0001$ and $u=0.0001$
$$\delta = { 1 \over n}$$
|
{
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|
Does the convergence of $\sum_{n = 0}^{\infty} |a_{n}| \Rightarrow |a_{n}| \rightarrow 0$ Consider the power series $\sum_{n = 0}^{\infty} a_{n} x^{n}$.
If $\sum_{n = 0}^{\infty} |a_{n}|, n \in \mathbb{Z}$ is absolutely convergent, does that necessarily imply $|a_{n}| \rightarrow 0$?
When I think about it, it seems true. I'm not asking for a proof or anything I just want to get an intuition of why that is true (if it is to begin with).
|
Suppose $|a_n|$ converged to some nonzero number $L>0$. That implies that for any $\epsilon>0$, there is a natural number $N$ such that $L-\epsilon<|a_n|$ for all $n\ge N$. But then we would have
$$\sum_{n=0}^{\infty}|a_n|\ge \sum_{n=N}^{\infty}|a_n|>\sum_{n=N}^{\infty}(L-\epsilon)=\infty.$$
Contradiction.
|
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|
Solving for one of variables locally when the Implicit Function Theorem does not apply I'm having trouble deciding whether certain functions can be locally solved. I have some examples:
*
*Can $xye^{xz} - z\log y =0$ be locally solved in $(0,1,0)$ for x? y? z?
In this case, I used Implicit Function Theorem to answer affirmatively that $x$ can be solved locally in function of $y$ and $z$. As $d/dy$ and $d/dz$ are equal zero, I can't use the theorem in those cases. I tried proving that $x$ has a local maximum at $(0,1,0)$ and could prove the first derivatives of $x$ with respect to $y$ and $z$ are zero, but I couldn't go further.
*
*If $f(x,y) = \frac{\sin(xy^2)}{xy}$ if $x\neq y$ and $f(x,y)=0$ otherwise, can $f(x,y) = 0$ be locally solved for $y=y(x)$ around $(0,0)$?
I couldn't apply implicit function theorem again and I have a clue that we can solve this locally for $y$, using $sin$ power series.
Is there a general procedure for problems like that? Any help is appreciated.
|
When implicit function theorem does not apply, the approach is usually ad hoc, based on the details of the function. Two things to try:
*
*Change the variables. Some substitutions, like $u=x^3$, can reduce the function to a form in which the implicit function theorem applies.
*Restrict $f$ to some line parallel to the coordinate of interest. If the zero set has multiple points, one can't solve for that coordinate.
First example
Restriction to lines works here:
*
*The line $x=0=z$ is in the zero set, which shows you can't solve for $y$
*The line $x=0$, $y=1$ is in the zero set, which shows you can't solve for $z$
Second example
The above tricks are not needed: the structure of particular function helps. Indeed, $f(x,y)$ turns into zero precisely when $\sin(xy^2)=0$. And since we are in a neighborhood of $(0,0)$, that means $xy^2=0$, which is equivalent to $x=0$ or $y=0$. Now the local structure of the set $f=0$ is clear.
|
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|
Prove $\int_t^{\infty} e^{-x^2/2}\,dx > e^{-t^2/2}\left(\frac{1}{t} - \frac{1}{t^3}\right)$ How to formally prove the following inequality -
$$\int_t^{\infty} e^{-x^2/2}\,dx > e^{-t^2/2}\left(\frac{1}{t} - \frac{1}{t^3}\right)$$
|
For a better lower-bound you may use the following proof by @robjohn:
$$x\int_x^\infty e^{-t^2/2}\,\mathrm{d}t \le \int_x^\infty e^{-t^2/2}\,t\,\mathrm{d}t =e^{-x^2/2}$$
Integrate both sides of the preceding:
$$
\begin{align}
\int_s^\infty e^{-x^2/2}\,\mathrm{d}x
&\ge\int_s^\infty x\int_x^\infty e^{-t^2/2}\,\mathrm{d}t\,\mathrm{d}x\\
&=\int_s^\infty\int_s^txe^{-t^2/2}\,\mathrm{d}x\,\mathrm{d}t\\
&=\int_s^\infty\frac12(t^2-s^2)e^{-t^2/2}\,\mathrm{d}t\\
\left(1+\frac12s^2\right)\int_s^\infty e^{-x^2/2}\,\mathrm{d}x
&\ge\frac12\int_s^\infty t^2e^{-t^2/2}\,\mathrm{d}t\\
&=-\frac12\int_s^\infty t\,\mathrm{d}e^{-t^2/2}\\
&=\frac12se^{-s^2/2}+\frac12\int_s^\infty e^{-t^2/2}\,\mathrm{d}t\\
\left(s+\frac1s\right)\int_s^\infty e^{-x^2/2}\,\mathrm{d}x
&\ge e^{-s^2/2}
\end{align}
$$
and note that $\displaystyle \left(s+\frac{1}{s}\right)^{-1} = \frac{s}{1+s^2} > \frac{1}{s} - \frac{1}{s^3}$ for $s > 0$,
which is equivalent to $\displaystyle s^4 > s^4 - 1$.
Expalanation for integartion by parts:
\begin{align*}
&\int_x^{\infty} e^{-t^2/2} \mathrm dt\\
=& \int_x^{\infty} \color{red}{\frac{1}{t}} .\color{blue}{te^{-t^2/2}} \mathrm dt\\
=& \left[\color{red}{\frac{1}{t}} .\int\color{blue}{te^{-t^2/2}}\,dt\right]_x^{\infty}
- \int_x^{\infty} \left( \color{red}{\frac{1}{t}} \right )' \left (\int\color{blue}{te^{-t^2/2}}\,dt\right)\mathrm dt\\
=& \frac{e^{-x^2/2}}{x} - \int_x^{\infty} \frac{e^{-t^2/2}}{t^2} \mathrm dt.
\end{align*}
Now, for the second integral that we obtained in the previous line we employ similar idea as done above:
\begin{align*}
&\int_x^{\infty} \frac{e^{-t^2/2}}{t^2} \mathrm dt\\
=& \int_x^{\infty} \color{red}{\frac{1}{t^3}} .\color{blue}{te^{-t^2/2}} \mathrm dt\\
=& \left[\color{red}{\frac{1}{t^3}} .\int\color{blue}{te^{-t^2/2}}\,dt\right]_x^{\infty}
- \int_x^{\infty} \left( \color{red}{\frac{1}{t^3}} \right )' \left (\int\color{blue}{te^{-t^2/2}}\,dt\right)\mathrm dt\\
=& \frac{e^{-x^2/2}}{x^3} -3 \int_x^{\infty} \frac{e^{-t^2/2}}{t^4} \mathrm dt.
\end{align*}
Combining these together we have $$\int_x^{\infty} e^{-t^2/2} = e^{-x^2/2}\left(\frac{1}{x} - \frac{1}{x^3}\right)+3 \int_x^{\infty} \frac{e^{-t^2/2}}{t^4} \mathrm dt$$
Since, the integrand $\dfrac{e^{-t^2/2}}{t^4}$ is always positive, we get the desired inequality:
$$\int_x^{\infty} e^{-t^2/2} > e^{-x^2/2}\left(\frac{1}{x} - \frac{1}{x^3}\right)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Confusion on how to calculate mean value I've done this type of thing in multiple classes over multiple years since high school, and still, when it's presented to me, I fumble around like a dope.
Consider a gas of $N_0$ non-interacting molecules enclosed in
a container of volume $V_0$. Focus attention on any sub-volume, $V$, of this
container and denote by $N$ the number of molecules located within this
sub-volume. Each molecule is equally likely to be located anywhere within
the container; hence the probability that a given molecule is located within
the sub-volume $V$ is simply equal to $\frac{V}{V_0}$.
(a) What is the mean number, $\bar{N}$, of molecules located within $V$? Express
your answer in terms of $N_0$, $V_0$, and $V$.
So, the probability of a given molecule being in $V$ is just $\frac{V}{V_0}$. These molecules are independent of one another; thus, the probability of two given molecules being in $V$ should be $\left( \frac{V}{V_0} \right)^2$, the probability of three given molecules in $V$ should be $\left( \frac{V}{V_0} \right)^3$, etc. ... right?
The expected value of $N$ should be the sum of each outcome's probability multiplied by the outcome: $\bar{N} = \sum_{N = 1}^{N_0} \left( \frac{V}{V_0} \right)^NN$.
But this doesn't make sense; it should be the case that $\bar{N} \rightarrow N_0$ as $V \rightarrow V_0$. Clearly that does not happen in the above summation. I think, "Well, the probability is for a given particle. There are many ways that $V$ could have, say, 3 particles—there must be some issue with including duplicates."
So, I divide each term in the summation by $N!$, since there are $N!$ ways of creating a particular result. $\bar{N} = \sum_{N = 1}^{N_0} \left( \frac{V}{V_0} \right)^N \frac{1}{(N-1)!}$. This is definitely not right.
Aside from all the different little things I've tweaked, by far, the biggest problem is that all of these things make complete and total sense to me. There's a very fundamental intuition that I have that's both inconsistent with other "intuitions" on the same thing, and very wrong.
Could someone offer some help?
|
The probability of ONLY two molecules being in the volume is
$p^2(1-p)^{N-2}$ times the combinatorial (n, 2). Where $p$ is V/V0.
Etc Etc
|
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|
How can I solve a difference of inverse sines? I'm trying to solve the equation
$$\arcsin(\lambda_1 / d) - \arcsin(\lambda_2 / d) = \pi / 6$$
for $d$, where $\lambda_1$ and $\lambda_2$ are known.
I don't really know where to start here.
I don't know any identities for inverse trigonometric functions akin to the ones like $\sin(u + v) = \sin u \cos v + \cos u \sin v$.
I see via Wikipedia that these have complex logarithmic forms, but I don't see how that might help (and I don't know any complex analysis).
Mathematica and WolframAlpha each give me a closed-form expression, but refuse to show how:
What's the best/cleanest way to go about solving this?
|
Note that if $\sin x=y$, then since
$$\sin^2 x+\cos^2 x=1$$
we have that $\cos x=\pm \sqrt{1-y^2}$. Use this together with the sum formula.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Asymptotic Notation Analysis Problem I'm new here. I have some question on asymptotic analysis
I am trying to calculate the Big-O of these five functions and rank them up:
a: $$2^{\log(n)}$$
b: $$2^{2\log(n)}$$
c: $$n^{5\over2}$$
d: $$2^{n^2}$$
e: $$n^2\log(n)$$
These are my approaches:
a: $$O(n)$$ because I log the both sides and cancel out the log
b: $$O(n)$$ same as the approach of a
c: $$O(n^{5\over2})$$
d: $$O(2^{2n})$$
e: $$n^2\log n$$
So d is definitely the largest.
The second largest would be c because $$n^{5\over2}$$ is just $$n^2\times n^{1\over2}$$ and $$n^{1\over2}$$ is larger than $$\log n$$ so the third would be e.
Then a and b are the same.
So to rank from the lowest growing rate, I get "abecd". But my answers are wrong, can someone explain, please?
|
I don't see anything wrong if it is $\log_2$. In general $2^{\log_b(n)} = n^{\log_b(2)} = n^{\log_2(b)^{-1}}$ and so the asymptotic behaviour would depend on $b$.
Wait your reasoning for (d) is wrong. $2^{n^2} = 2^{(n^2)} \ne (2^n)^2$... Though it really is the biggest, much bigger than what you thought.
|
{
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"url": "https://math.stackexchange.com/questions/1136083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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