Find the power set $P(S)$ for $S=\{\emptyset, \{\emptyset\}, \{\emptyset \{\emptyset\}\}\}$ Find the power set $P(S)$ for $S=\{\emptyset, \{\emptyset\}, \{\emptyset \{\emptyset\}\}\}$

OK this problem confuses me for many reasons, but here is what I know.  The cardinality of a set is $2^n$ where $n$ is the number of elements in the set.  In this problem, however, how can the empty set be an element?

If I were just going to say this set has $2^n$ elements that would mean the set has $2^3$ or 8 elements, but I don't know what those elements would be other than empty sets.

Any help in understanding this problem is greatly appreciated!

Let's assume the actual question is to find the power set of $S=\{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\}$. Using Von Neumann's definition of the natural numbers, this is equivalent to finding the power set of $S=\{0,1,2\}$, where $0=\emptyset$, $1=\{0\}$, and $2=\{0,1\}$. The power set is then

$$\{\emptyset,\,\{0\},\,\{1\},\,\{2\},\,\{0,1\},\,\{0,2\},\,\{1,2\},\,\{0,1,2\}\}$$

Just change the $2$'s to $\{0,1\}$, then the $1$'s to $\{0\}$, then the $0$'s to $\emptyset$--and you are done!

You will end up with a royal mess, of course. I found it difficult enough to format the easier form of the power set above.

Every set has the empty set as an element. However #(S) should be zero where #(S) is the cardinality of set S.

The power set of the empty set or sets of empty sets should be simply the empty set.

Remember however that the power set of a set is not a cardinality. It is a set constructor that builds a set from the elements of set S. So technically your answer should be the set which is the factorial combination of each element ie if set S = { a, b, c} then P(S) = {a, b, c, {a,b}, {a,c}, {b, c}, {a, b, c} }

IMHO of course !


Limits without L'Hopitals Rule Evaluate the limit without using L'hopital's rule

a)$$\lim_{x \to 0} \frac {(1+2x)^{1/3}-1}{x} $$

I got the answer as $l=\frac 23$... but I used L'hopitals rule for that... How can I do it another way?

b)$$\lim_{x \to 5^-} \frac {e^x}{(x-5)^3}$$

$l=-\infty$

c)$$\lim_{x \to \frac {\pi} 2} \frac{\sin x}{\cos^2x} - \tan^2 x$$

I don't know how to work with this at all

So basically I was able to find most of the limits through L'Hopitals Rule... BUT how do I find the limits without using his rule?

$$L_1=\lim_{x \to 0} \frac {(1+2x)^{1/3}-1}{x}$$

Using binomail thorem: $$(1+2x)^{1/3}=1+\frac13.2x+\text{terms containing other higher powers of x}$$
Putting back in you'll get the limit as $2/3$

  $$L_2=\lim_{x \to 5^-} \frac {e^x}{(x-5)^3}=-\infty$$

Because denominator $\to0^-$ and the numerator is finite.

  $$L_3=\lim_{x \to \pi/2} \frac{\sin(x)}{\cos(x)^2} - \tan(x)^2$$

Since,
$$\frac{\sin(x)}{\cos(x)^2} - \tan(x)^2=\frac{\sin x-\sin^2 x}{\cos^2x}=\frac{\sin x(1-\sin x)}{(1-\sin x)(1+\sin x)}=\frac{\sin x}{1+\sin x}$$
So, obviously limit is $\frac 12$

EXPAND(1+2x)^3/2 and the limit will reduce to 2/3  HOWZ THAT


Prove $uvu=u$, $v$ unique implies $u$ invertible in a ring $R$ Suppose $R$ is a ring and $u,v \in R$ are so, such that $uvu=u$ and $v$ is unique. How do I show that $u$ is invertible? R is assumed to have multiplicative identity.
So far I have been able to prove that $u,v \neq 0$. I have also proven that $u$ cannot be a non-invertible element with a one-sided inverse.
Can I get a few more hints?

Given that $uvu=u$ for some unique $v$, note that $u^2vu=u^2$. Consider the quantity : $u(v+uv-1)u$. Note that $u(v+uv-1)u = uvu + u^2vu - u^2 = u + u^2 - u^2  = u$. But then, $v$ was the unique element with this property, hence $v=v+uv-1$ hence $uv=1$.

In fact, there is another generalization form of this lemma. For an element $u$ in the ring $R$, the sufficient and necessary condition for its invertibility is either (1) $uvu=u$, $vu^2v=1$ or (2) $uvu=u$ and such $v$ is unique.

The necessity is easy to verify and part 2 of the sufficiency has been given. Therefore, I would like to provide the proof of part 1. Since the condition (1) implies that the semigroup $(R,\cdot)$ contains an identity element and $vu^2v=(vu)(uv)=1$, we have $uv=(vu)^{-1}$. Note that $u = uvu = u(vu)^{-1} = u^2v$. Then we have $vu=vu^2v=1$, which implies that $u$ is invertible and furthermore, $v = u^{-1}$.


Why is $\ln N - \ln(N-1) = \frac1N$ for large $N$? May I ask why is $\ln N - \ln(N-1) = \frac1N$ for large $N$?

Thank you very much.

You can get quite far with just algebra:

$$\begin{align}
\ln N - \ln (N-1) & = \ln N - ( \ln N + \ln (1-1/N)) \\
& = -\ln (1-1/N)
\end{align}$$

using the laws for addition of logarithms. Now you can use the Taylor expansion of the natural logarithm:

$$-\ln(1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots$$

to get

$$-\ln(1-1/N) = \frac{1}{N} + \frac{1}{2N^2} + \cdots$$

so that $\ln N - \ln (N-1)$ is, for large $N$, equal to $1/N$ plus a correction term of order $O(1/N^2)$.

$\displaystyle \lim_{n \to \infty} (\ln n-\ln(n-1))=\displaystyle \lim_{n \to \infty}\left(\ln \frac{n}{n-1}\right)=\ln\left(\displaystyle \lim_{n \to \infty} \frac{n}{n-1}\right)=\ln 1=0 $

$\displaystyle \lim_{n \to \infty} \frac{1}{n}=0$

Hence , for large $n$ both $\ln n-\ln(n-1)$ and $\frac{1}{n}$ tends to zero .


Why is $\arctan(\tan(y))=y+2k \pi$? Why is $\arctan(\tan(y))=y+2k \pi$?

When the cycle of both $\arctan$ and $\tan$ are $\pi$?

It's actually wrong. The correct answer is actually
$$\arctan(\tan(y))=y+k\pi\quad k\in\mathbb{Z}\text{ s.t. } |y+k\pi|<\pi/2$$

if the standard $\arctan:\mathbb{R}\to(-\pi/2,\pi/2)$ function is used as an inverse.

because the function tan(x) have period $2\pi$.
I edited the answer.


Solving $\sinh z = 2i$ This is my attempt at the question (I stopped early because it starts to not work out...)
$$\sinh z = 2i \\ e^{iz} - e^{-iz} = 4i \\ = e^{2iz} - 4ie^{iz} - 1 = 0 $$
solving the quadratic gives
$$e^{iz} = i(2\pm \sqrt{3}).$$
I stop here to check:
$$\sinh z = \frac{e^{iz} - e^{-iz}}{2}= \frac{i(2\pm\sqrt{3}) + i(2\pm \sqrt{3})}{2}= i(2\pm \sqrt{3}) \neq 2i...$$

I'm not sure if I'm doing something really wrong... because I've redone this a few times and I can't see it.

Use $\sinh^{-1} w = \ln(w+\sqrt{1+w^2})$

$$z = \sinh^{-1} 2i=\ln(2i+\sqrt{1-4})=\ln(2i\pm i\sqrt3)=\ln i + \ln(2\pm\sqrt3)$$

Since $\ln i = \ln e^{i\frac\pi2 + i2n\pi} = i\frac\pi2 + i2n\pi$, the solutions are

$$z = i\frac\pi2 + i2n\pi + \ln(2\pm\sqrt3)$$

Actually, correct version of the last expression is

$$ \sinh(z)=\frac{e^{iz}-e^{-iz}}{2}=\frac{i(2\pm\sqrt{3})+i(2\mp\sqrt{3})}{2}=2i $$

Use $z=(2+\sqrt{3})i$ or $z=(2-\sqrt{3})i $ uniformly in the all part of expression


Does every invertible complex matrix have a non-zero eigenvalue? Let $M$ be an $n \times n$ matrix over the field of complex numbers. Additionally assume that $M$ is invertible. Now let $E$ be the set of eigenvalues that is

$$E = \{\lambda \in \mathbb{C}: \exists v \in \mathbb{C}^n\setminus\{0\}, Mv=\lambda v\}$$

Now is it true, that $E\cap\{0\}^c \neq  \emptyset$, i.e. does $M$ always have a non zero eigenvalue.

My thought are that by the fundamental theorem of algebra, I know that every complex polynomial, i.e. every polynomial with complex coefficients has at least one solution, and thus I can conclude that $E\neq \emptyset$. Now the general statement is clearly false if we don't assume that $M$ is invertible, since the zero matrix has only zero as an eigenvalue. So is the condition of invertibility sufficient?

Since every polynomial has a root over $\mathbb{C}$, the characteristic polynomial of any complex matrix must have a root, say $\lambda$. Then $\lambda$ is an eigenvalue of the matrix at hand. Since the matrix is assumed to be invertible, we have $\lambda \neq 0$.

Regarding the last statement, if $M$ has $0$ as eigenvalue, there is some non-zero eigenvector $x$: $Mx=0$, and $M$ is not invertible.

Let $A\in \Bbb C^{n\times n}$ be invertible $\iff |A|≠0$. Also $|A|=\lambda_1...\lambda_n$, therefore...


Is there is a limit to: $\lim\limits_{n \to \infty} \left ( 1 + \sin \frac1{n^2} \right )^{n^2} $ I need to calculate the limit or prove that the limit does not exists:

$$
\lim_{n \to \infty} \left ( 1 + \sin \frac{1}{n^2} \right )^{n^2}
$$

Can someone guide how to solve this?

And please, don't just write the answer, it's my homework and I need to learn, not to copy.

Yes this can be solved ! when n tends to infinity try inverting everything with using an h=1/x where x tends to 0 and you will be easily solving it for the power thing for such type of limits use logarithm .


If the real part of $f$ is bounded then $f$ is constant It isn't too hard to show that if $f:\mathbb{C}\to\mathbb{C}$ holomorphic everywhere (entire) and $\Re (f)$ is bounded, then $f$ is constant: it suffices to consider $\exp f$, which is entire, and by $|\exp f|=\exp\Re (f)$ bounded and therefore constant. That question has been asked on this site before.

I am now asked to show that this holds if $f$ is only defined on $U\subseteq\mathbb{C}$, $U$ path-connected and non empty, and the bound is attained in the interior of $U$. Because of the domain I can't use Liouville anymore, so I am not sure how to proceed.

Is it true that there exists a differentiable bijection $\phi: U\to\mathbb{C}$ for instance? That would resolve the issue... Otherwise some help would be welcome.

I think that the result is not true.

For example, if $U$ is the unit open disk centered at the origin, and $f(z)=\exp(z)$, then we have that $\exp(z)=\exp(x)\cos(y)+i\exp(x)\sin(y)$, for $z=x+iy$. We have that the absolute value of the real part, $|\exp(x)\cos(y)|$ is bounded by $\exp(1)$ for all $z\in U$, but we know that $\exp(z)$ is not constant in $U$.

If the bound is attained in the interior of $U$, the you can use the Maximum modulus principle as Sangchul Lee says.


How find this $\lim_{n\to\infty}a_{n}$ let sequence $\{a_{n}\}$,such  $a_{1}=2\pi-6$,
and
$$a_{n}=\left\lceil\dfrac{2\pi}{a_{n-1}}\right\rceil\cdot a_{n-1}-2\pi$$

Find the
$$\lim_{n\to\infty}a_{n}$$

where $\left\lceil\dfrac{2\pi}{a_{n-1}}\right\rceil$is the smallest integer not less than $\dfrac{2\pi}{a_{n-1}}$

My try: since
$$a_{n}>0,\dfrac{a_{n}}{a_{n-1}}=\lceil\dfrac{2\pi}{a_{n-1}}\rceil-\dfrac{2\pi}{a_{n-1}}\le 1$$
and then I can't Find this limit,and maybe other nice methods,

Thank you  very much

$a_1 > 0$ and $\bigg\lceil\frac{2\pi}{a_n}\bigg\rceil \ge \frac{2\pi}{a_n}$. By induction you can easily prove that $a_n >0 \, \forall \, n \in \mathbb N$.

As you proved, if $a_n >0$, then the sequence is strictly decreasing.

Remains to show that the actual limit is $0$, a reasonable guess.

Using the fact that the sequence is strictly decreasing and strictly positive, you know that the sequence has a unique limit.

The limit will solve the equation $a = \big\lceil\frac{2\pi}{a}\big\rceil a - 2\pi$ which can be rewritten as : $$\bigg\lceil\frac{2\pi}{a}\bigg\rceil - \frac{2\pi}{a} = 1$$

This has no solution for any $a_1 > a > 0$ therefore the limit is not strictly positive.

So the limit is $0$.

Perhaps easier (just thought about it), $$\lim_{a_n\to0} \bigg\lceil\frac{2\pi}{a_n}\bigg\rceil a_n  - 2\pi= 0$$
therefore $0$ is a limit. Since the limit is unique, it is $\underline{\text{the}}$ limit.

Another way I thought of doing this is to use the squeeze theorem. I'm sure there's a way to do it.

If there is a (nonzero) limit, it should satisfy
$$a = \lceil\frac{2\pi}{a}\rceil a - 2\pi,$$ which should have a solution (which I am having some trouble finding).


Question on Markov property for Itô diffusion : does it implies time homogeneity as well? Let $(X_t)$ a diffusion process. I denote $\mathbb P^x$ the measure : $$\mathbb P(C(t_1,...,t_n,A_1,...,A_n)\mid X_0=x),$$
for all cylinders $$C(t_1,...,t_n,A_1,...,A_n)=\{X_{t_1}\in A_1,...,X_{t_n}\in A_n\}.$$
So, I know that $(X_t)$ has Markov property if for all $h>0$, $$\mathbb P\{X_{s+h}\in A\mid \mathcal F_s\}=\mathbb P\{X_{s+h}\in A\mid X_s\},$$
for all Borel set $A$. In my lecture, it's denoted by $$\mathbb P^0(X_t\in A\mid \mathcal F_s)=\mathbb P^{X_s}(X_h\in A),\tag{1}$$
which I interpret as $$\mathbb P^0\{X_{s+h}\in A\mid \mathcal F_s\}(\omega )=\mathbb P\{X_h\in A\mid X_0=X_s(\omega )\}\tag{2}$$
The thing it's that $(1)$ and $(2)$ looks rather to be time homogeneity+Markov property than Markov process only. Indeed,
$$\mathbb P(X_{s+h}\in A\mid \mathcal F_s)(\omega )=\mathbb P(X_{s+h}\in A\mid X_s=X_s(\omega ))=\mathbb P(X_h\in A\mid X_0=X_s(\omega )),$$
where the first equality is Markov property and the second one is time homogeneity.
Q1) Am I right ?
Q2) For Itô diffusion, does Markov property means Markov property+ Time homogeneity ?

This looks like the strong Markov property. Under certain natural conditions, having the Markov property and time homogeneity implies the strong Markov property: see Exercise 13.1 of this; it seems you also gave a proof.
The answer to your question (2) is no if you define Markov property without the time homogeneity assumptions. There's just no reason that all Itô diffusion processes with the Markov property must be time homogeneous.


Eigenvalues of $AB$ from eigenvalues of $A$ and $B$ Is it possible to find the eigenvalues of $AB$ if we know the eigenvalues of $A$, say $\lambda_1, \lambda_2,...,\lambda_n$ and those of $B$ say $\lambda_1, \mu_2,...,\mu_n$ and given that $A$ and $B$ are positive semi/definite symmetric complex valued matrices. Even if not possible can we build a relation of magnitude of the eigenvalues?

Thank you.
Related to https://math.stackexchange.com/questions/492697/possible-determinant-inequality-det-leftiaaib-right-1-leq-det-l

$A\left\vert\alpha\right\rangle = \alpha\left\vert\alpha\right\rangle.\quad$
$B\left\vert n\right\rangle = n\left\vert n\right\rangle$

$$
\left\langle m\left\vert AB\right\vert n\right\rangle
=
n\left\langle m\left\vert \left(A%
\sum_{\alpha}\left\vert\alpha\right\rangle\left\langle\alpha\right\vert\right)
\right\vert n\right\rangle
=
n\sum_{\alpha}\alpha\,
\left\langle m\right\vert\left.\alpha\right\rangle
\left\langle\alpha\right\vert\left.n\right\rangle
\equiv
C_{mn}\,,
\qquad
C = AB
$$
$C_{mn}$ are matrix elements of $C = AB$ in the base of eigenvectors of $B$. Now,
you can expand a $C$ eigenvector $\left\vert\Psi\right\rangle$ in the same base:
$$
C\left\vert\Psi\right\rangle = \lambda_{\rm i}\left\vert\Psi\right\rangle
\qquad\Longrightarrow\qquad
\sum_{m}C_{mn}\Psi_{n} = \lambda_{\rm i}\Psi_{m}\,,
\quad
\Psi_{m}
\equiv
\left\langle m\right\vert\left.\Psi\right\rangle
$$


$|a-b| \geq | |a| - |b| | \Longleftrightarrow |a-b| \geq |a| - |b| $? Does the following always hold?

$|a-b| \geq | |a| - |b| | \Longleftrightarrow  |a-b| \geq  |a| - |b| $

Certainly, the second inequality implies the first, but the converse isn't so clear to me.

If it does hold, then why do we define the 'reverse triangle inequality' as the first inequality in the above?

The if and only if holds if you clarify "for all $a,b$".

The first inequality, for fixed $a,b$, implies the second inequality. The second inequality, on the other hand, does not imply the first inequality if you consider it  for fixed $a,b$; you need it to hold for $b,a$.

In summary, the first inequality contains more information.

I think the question that you're trying to ask is "how do we use the triangle inequality (which can be rearranged to $|a-b| \geq  |a| - |b|$) to prove the "reverse triangle inequality, namely $|a-b| \geq | |a| - |b| |$.

Here's one way to do it: note that if we exchange the roles of $a$ and $b$ in the first inequality, we find that
$$
|b-a| \geq  |b| - |a|
$$
However, we know that $|b - a| = |-(a - b)| = |a - b|$.  So, the above implies that $|a - b| \geq  |b| - |a|$.  That is, we know that
$$
|a - b| \geq |a| - |b| \quad \text{AND} \quad |a - b| \geq |b| - |a|
$$
To put it more succinctly, we have
$$
|a - b| \geq \max\{|a| - |b|, |b| - |a|\} = \max\{|a| - |b|, -(|a| - |b|)\} = ||a| - |b||
$$
which is the desired result.


How to solve the problem by using separation of variables? A tank in the form of a right circular cone of height $1.6$ meters and radius $1$ meter with its vertex below the base is filled with water. A $1$ centimeter hole at the vertex causes water to leak out. Determine how long it takes for the tank to be empty. Assume that the rate of the water flow out through the hole is given by $0.6 \, \sqrt{2gh}$. Where $h$ is the instantaneous height of the water.

Based on the problem I've came up of the differential equation:
$$\frac{π}{1.6^2}h^2\frac{dh}{dt}=-0.0001\pi[0.6(\sqrt{2gh})]$$
Which i think can be solved by seperation of variables, my current weakness in solving DE.
Thank you.

We have: $h^2\dfrac{dh}{dt} = -0.0001\cdot (1.6)^2\cdot 0.6\cdot \sqrt{2g}\cdot \sqrt{h}= C\sqrt{h}\to h^{1.5}dh = Cdt$. You can continue at this point...


Sequence of continuous functions whose pointwise limit is discontinuous Give an example of a sequence of continuous functions $f_n:\mathbb{R} \to \mathbb{R}$ such that the pointwise limit $f:\mathbb{R} \to \mathbb{R}$ exists but is discontinuous.
My idea is $f_n(x)=\displaystyle\frac{x}{n}.$ Then the pointwise limit is $0$ but $f$ is discontinuous on $\mathbb{R}.$ Is that correct?
I'm having a hard time understanding the definition of pointwise convergence. I have this definition:
$\displaystyle\lim_{n\to \infty} f_n=f$ pointwise iff $\displaystyle\lim_{n\to \infty} f_n(x)=f(x)$
But then what is $f(x)$? Where does the $n$ go? In the example I gave, for instance,
$\displaystyle\lim_{n\to \infty} f_n(x)=\displaystyle\lim_{n\to \infty} \frac{x}{n}=0.$ But does this equal $f(x)$? Is $f(x)$=0 a discontinuous limit?

The example you give has a point-wise limit of 0, which is not a discontinuous function.  Some standard examples of these kinds of functions are ones that kind of smoothly interpolate between two values, with the interpolating region getting smaller and smaller. For example, consider the sequence:

$$
f_n(x) = \tan^{-1}(nx)
$$

This is plotted for some various values of $n$ below, on the region $x\in [-1,1]$

Along with Suugaku's answer, this should give you the intuition you need about point-wise limits and continuity.

Yes, $f(x) = 0$ according to what you wrote. However, your answer is not correct because $0$ is a continuous function. An example of what you want are the functions:
$$
f_n(x) = \arctan(n x),
$$
which converges pointwise to
$$f(x) = \frac{\pi}{2}\mbox{sgn}(x) = \left\{\begin{array}{ccc}
-\pi/2 & \quad & x < 0\\
0 & \quad & x = 0\\
\pi/2 & \quad & x > 0
\end{array}\right.$$
a discontinuous function.


How can we say that the area of any circle or circular based shape is finite? I am saying this because the area of the circle is pi * radius * radius. We know that pi is a never ending value. So, if some one says I need a circle of 3 metre square area to make a rim of the wheel. How can I make so? The value of pi is non-terminating - not a practical value. Also, does it mean that area of a square is finite but area of circle is not finite. But, if area of circle is not finite, then how can I see/visualize it.

Non terminating doesn't mean infinite.  pi is a finite number of a value between 3 and 3 1/4.  The decimal expansion is infinite but the value is finite.  pi is between 3 and 4.  Go one more place value and we see pi is between 3.1 and 3.2.  Go another place value and we see pi is between 3.14 and 3.15, and so on forever.  The precision of pi is infinite but it's value is simply a finite number like any other.

To get an area of 3 square meters you do:

$ A = \pi*r^2 = 3 $

$r^2 = \frac{3}{\pi} $

$ r = \sqrt{\frac{3}{\pi}}$

That number is a finite number. According to my calculator it is approximately 0.97720502380583984317276924567669...  Like $\pi$, this number is non-terminating so I can't express it exactly.  So I'll say r = 0.977 and that is within a millimeter of what we need.  If I want to measure it within the diameter of a helium atom I round it off to 10 place values.  So r = 0.977205024 meters minus about 1/5 of a helium atom.

All space (area) is infinite.  Measurements express our understanding of such via finite rounding.  That is to say, in measuring we express a point as finite - when in fact a point has an area made up of an infinite number of "points."


Solve $f^2(x)=\frac{k}{f''(x)}$ While studying physics I have many time stumbled upon forces that are directly affected by the position of a particle. Moreover given $f(x)$ how could we approach solving the equation:
$$f^2(x)=\frac{k}{f''(x)}$$
I am new to differential equations and I would really appreciate if someone could explain the thinking process and the steps needed to reach a solution. Thanks in advance.

Well, we have that (multiply both sides by $\text{f}\space'\left(x\right)$):

$$\text{f}\left(x\right)^2=\frac{\text{k}}{\text{f}\space''\left(x\right)}\space\Longleftrightarrow\space\int\text{f}\space'\left(x\right)\text{f}\space''\left(x\right)\space\text{d}x=\int\text{f}\space'\left(x\right)\cdot\frac{\text{k}}{\text{f}\left(x\right)^2}\space\text{d}x\tag1$$

We get, when we use:

$$\int\text{f}\space'\left(x\right)\text{f}\space''\left(x\right)\space\text{d}x=\frac{\text{f}\space'\left(x\right)^2}{2}+\text{C}_1\tag2$$
$$\int\text{f}\space'\left(x\right)\cdot\frac{\text{k}}{\text{f}\left(x\right)^2}\space\text{d}x=\text{C}_2-\frac{\text{k}}{\text{f}\left(x\right)}\tag3$$

So:

$$\frac{\text{f}\space'\left(x\right)^2}{2}=\text{C}_3-\frac{\text{k}}{\text{f}\left(x\right)}\space\Longleftrightarrow\space\int\frac{\text{f}\space'\left(x\right)}{\sqrt{\text{C}_3-\frac{2\text{k}}{\text{f}\left(x\right)}}}\space\text{d}x=\int\pm1\space\text{d}x=\text{C}_4\pm x\tag4$$

For the intergal, substitute $\text{u}=\text{f}\left(x\right)$:

$$\int\frac{\text{f}\space'\left(x\right)}{\sqrt{\text{C}_3-\frac{2\text{k}}{\text{f}\left(x\right)}}}\space\text{d}x=\int\frac{1}{\sqrt{\text{C}_3-\frac{2\text{k}}{\text{u}}}}\space\text{d}\text{u}\tag5$$

Putting $y = f(x)$, we have
$$y^{\prime\prime} = \frac{k}{y^2}.$$
Now multiplying both sides by $2y^\prime$, we get
$$2y^\prime y^{\prime\prime} =  \frac{2ky^\prime}{y^2},$$
which can be rewritten as
$$\left( \left( y^\prime \right)^2 \right)^\prime  =  \frac{2ky^\prime}{y^2} = \frac{\mathrm{d}}{\mathrm{d} x} \left( - \frac{2k}{y} \right),$$
which upon integration of both sides gives
$$ \left( y^\prime \right)^2 = c -\frac{2k}{y},$$
where $c$ is an arbitrary constant of integration. So
$$ y^\prime = \left( c -\frac{2k}{y} \right)^{\frac{1}{2}},$$ and therefore
$$ \frac{y^\prime  }{ \left( c -\frac{2k}{y} \right)^{\frac{1}{2}} } = 1,$$
which we can rewrite as
$$ \frac{ y^\frac{1}{2} y^\prime }{\left( cy - 2k \right)^\frac{1}{2} } = 1, $$
or $$ \frac{1}{c^\frac{1}{2}} \left( \frac{y}{ y - \frac{2k}{c} } \right)^\frac{1}{2} y^\prime = 1.$$
Can you take it from here on?


Find $\lim_{x \to 0^{+}} \frac{\ln(x+1)}{x}$ without using L'Hopital's Rule? Without using L'Hopital's Rule, how does one find $\lim_{x \to 0^{+}} \frac{\ln(x+1)}{x}$?

I was hoping to find a way using basic calc I, pre-differentiation knowledge and not knowing the definition of $e$--much like you can prove $\lim_{x\to\infty} \frac{\sin x}{x} = 1$ using a geometric argument and the squeeze rule/theorem.

If you know that $\lim_{x\to\infty}\left(1+\frac1x\right)^x=e$, either because this is how $e$ was defined or because you’ve already proved it, you can begin by letting $L=\lim_{x\to 0^+}\frac{\ln(1+x)}x$. The exponential function is continuous, so

$$e^L=e^{\lim_{x\to 0^+}\frac{\ln(1+x)}x}=\lim_{x\to 0^+}e^{\frac{\ln(1+x)}x}=\lim_{x\to 0^+}(1+x)^{1/x}=\lim_{x\to\infty}\left(1+\frac1x\right)^x=e\;,$$

and hence $L=\ln e=1$.

You could use the series expansion: $$\ln(x+1)=x-x^2/2+x^3/3-x^4/4\ .\ .\ .$$


Consider Linear transformation Given the linear transformation $ f: \mathbb{R}^{4\times 4} \to  \mathbb{R}^{4\times 4}$ defined by $A \to f(A)= A + A^T$

How do you determine the dimensions of $\text{Ker}(f)$ and $\text{Img}(f)$?

First, you need to prove that the image of $f$ is precisely the space of symmetric matrices: the matrices $B$ for which $B=B^T$, or in other words those of the form
$$\begin{bmatrix}b_1&b_2&b_3&b_4\\b_2&b_5&b_6&b_7\\b_3&b_5&b_8&b_9\\b_4&b_6&b_9&b_{10}\end{bmatrix}$$

The dimension of the space of symmetric $4\times 4$ matrices is $10$: They are uniquely determined by the upper triangular part (including the diagonal), which contains 10 entries ($b_1,\ldots,b_{10}$ above). To find a basis of this space, for each $i=1,\ldots,10$, set $b_i=1$ and all $b_j=0$, for $j\neq i$ in the matrix above.

In other words, $\dim(im(f))=10$.

By the rank-nullity theorem $\dim(\ker(f))=\dim(\mathbb{R}^{4\times 4})-\dim(im(f))=16-10-6$.

Alternatively, you can notice that $A\in\ker(f)$ iff $A=-A^T$, so $A$ has the form
$$\begin{bmatrix}0&a_1&a_2&a_3\\-a_1&0&a_4&a_5\\-a_2&-a_4&0&a_6\\-a_3&-a_5&-a_6&0\end{bmatrix}.$$

So the dimension of $\ker(f)$ is $6$. Again by the rank-nullity theorem, $\dim(im(f))=10$.

Clearly $Ker(f)=\{A\in M(4,4,\mathbb{R})\,|\, A^T=-A\}$, the space of antisymmetric matrix that has dimension $6$(Skew-symmetric matrix). The image is clearly contained in the space of symmetric matrix, that has dimension $10$, and so by a famous theorem(Dimension Theorem) we can conclude by dimension argument


Simplify the inequality $2<\frac{10}{x}<3$ to an inequality for $x$. I'm not sure what to google search for this.  Let's say I have the expression:
$$ 2 < \frac{10}{x} < 3 $$
We need to see if x is positive or negative.  If positive, we have:
$$ 2x < 10 < 3x $$
If negative we have:
$$ 2x > 10 > 3x $$
It can't be negative since $2x$ can't be greater than $10$.  So $x$ must be positive.
Now, what do I do with $ 2x < 10 < 3x $ to get $x$ in the middle so that it is $x$ not $1/x$.  I am not sure what rule or term to google for.
Thanks!

Note that the inequality $2x<10<3x$ is actually two inequalities:
$$2x<10\qquad\text{ and }\qquad 3x>10.$$
Dividing them by $2$ and $3$, respectively, shows that
$$x<\frac{10}{2}\qquad\text{ and }\qquad x>\frac{10}{3},$$
which can be written more concisely as $\tfrac{10}{3}<x<\tfrac{10}{2}.$

More abstractly, for the function $f(x)=\tfrac1x$ we can write the inequalities as
$$f(\tfrac12)<f(\tfrac{x}{10})<f(\tfrac13).$$
Because $f(x)$ is positive if and only if $x$ is positive, we see that also $\tfrac{x}{10}$ is positive, so $x$ is positive. Moreover, because $f$ is strictly decreasing on the positive numbers, it follows that
$$\tfrac12>\tfrac{x}{10}>\tfrac13,$$
yielding the bounds $\tfrac{10}{3}<x<\tfrac{10}{2}$.

Solution without cases.
We need to solve $$\frac{10-3x}{x}<0$$ and $$\frac{2x-10}{x}<0,$$ which gives
$$x\in(-\infty,0)\cup\left(\frac{10}{3},+\infty\right)$$ and $$x\in(0,5),$$ which gives the answer:$$\left(\frac{10}{3},5\right).$$


$k$ with an even sum of digits for all multiples of $k$? Is there a number $k\in\mathbb{N}$ such that $k\cdot n$ has an even sum of digits for all $n\in\mathbb{N}$?

I would be grateful for any ideas of how to attack this problem...

There is no such $k \in \mathbb{N}$ : given any $k$, we can calculate a value $n$ such that $kn$ has an odd sum of digits.

Factorize $k$ as $k = 2^a 5^b m$. Then consider the following integer

$$q = 10^{\max(a, b)} \sum_{i=1}^{m} {\left(10^{\phi(m)}\right)}^{i}$$

where $\phi(m)$ is Euler's totient function. We claim that $q$ is divisible by $k$. Since $\gcd(2^a 5^b, m) = 1$, if we can show that $2^a 5^b \mid q$ and $m \mid q$ we will be done.

The left term in the product is divisible by $2^a 5^b$ by construction; now we consider the right term.

Since $\gcd(m, 10) = 1$, Euler's theorem tells us that $10^{\phi(m)} \equiv 1 \pmod m$. Additionally, for any power of $i$, $1^i \equiv 1 \pmod m$. So all $m$ summands have remainder one, meaning that the right term is equivalent to $1 \cdot m \equiv 0 \pmod m$, and $m$ divides $q$.

Consequently, $q$ is divisible by $k$, which we can write as $q = kn$ for some integer $n$.

What is the sum of the digits in $q$? The right term is a sum of $m$ distinct powers of ten, whose digits will sum to $m$. Multiplying by the left term, a power of ten, doesn't change the sum, and so the sum remains $m$ – which is odd by definition.

There is no such number. All the number crunching in the world is no proof, but when someone crunches up to a largish number, it looks very unlikely.

The way I would start to attack this problem is to show that for every positive $k$ there are infinitely many $kn$ with odd digital root. It's of course possible for $kn$ to have an odd digital root but an even digit sum (e.g., $99$).


Prove that sup $f(x)$ $\le$ inf $g(y)$. (proof verification) I saw this proof, but I don't understand one point.

$f(x) \le g(y)$ for $x,y \in D$

$\rightarrow$ sup$f(x)$ $\le$ $g(y)$

$\rightarrow$sup$f(x)$ $\le$ inf$g(y)$

I think from the second line to third line of proof, it assumes that inf$g(y)$ $\in$ $g(D)$.

But, how can we prove this?

Thanks in advance.

The property that they use is let $a \in \mathbb{R}$. Then $\forall y, a \le g(y) \implies  a \le \inf_y g(y)$.

$a$ is a lower bound for the set $\{ g(y): y \in D\}$, but remember that the infimum is the greatest lower bound, hence we have $a \le \inf_y g(y)$.

Here $a$ is $\sup f(x)$.

Since $f(x)\le g(y)$ for $x$,$y\in D$

$g(y)$ is an upper bound for $f(x)$, so the least upper bound is less than or equal $g(y)$

Thus  $\sup f(x) \le g(y)$.
That implies $\sup f(x)$ is a lower bound for $g(y)$ so the greatest lower bound  of $g(y)$ is greater than  or equal to sup $f(x)$.

That is:

$\sup f(x) \le \inf g(y)$


Why can't polynomials have negative exponents or division by a variable Why can't:

$$2x^{-3} - 3x$$

or

$$\frac{1}{2x}$$

be polynomials too?

Why have a definition that excludes these algebraic forms?

Polynomials are defined as they are for a few distinct reasons: (1) because polynomials as functions have certain properties that your 'polynomials with division' don't have, and (2) because there are other terms for more generalized algebraic forms.

First, the properties of polynomials: unlike e.g., $2x^{-3}+3x$, polynomials have no poles; there's no place where a polynomial 'blows up', where it goes to infinity.  In particular, this means that the domain of a polynomial as a function is all of $\mathbb{R}$ — or, alternately, all of $\mathbb{C}$).  In fact, polynomials in the complex plane have even nicer properties - they're analytic functions, which means that not only are they well-defined everywhere, but all of their derivatives are well-defined everywhere.  This means that polynomials have a lot of structural properties that make them 'nice' objects of study in ways that your expressions aren't.

Secondly, there's also a notion that (roughly) corresponds to your 'extended' version of polynomials, with its own set of nice properties: the ring of rational functions, which includes not just items like $x^2+\frac1x+5$ but also terms like $\dfrac{x^3+2}{x+5}$ that your formulation doesn't at least at first glance suggest.  These represent a useful extension of polynomials for study because they're closed under the operation of division; for any two rational functions $f(x)$ and $g(x)$, the function $h(x) = \frac{f(x)}{g(x)}$ is also a rational function.  This property doesn't hold for your 'Lambert polynomials', because there's no finite expression in positive and/or negative powers of $x$ that corresponds to the function $\frac1{x+1}$.

What's more, there's also an object occasionally studied that more directly corresponds to your notion: the notion of Laurent Polynomial.  As that article suggests, they're of particular importance and interest for their connections with the field of Hopf Algebras (and by extension, quantum groups).  I would opine that the reason that they're not the primary object of study is because they're not the 'simplest' structure of interest among any of their peers, and fundamentally the most important structures in mathematics tend to be the simplest structures exhibiting some given property.

I would have added this as a comment to Newb's answer but for some reason I can't add a comment today. But in answer to Quora Fea's question in the comments of that answer - is it possibly because a 'polynomial' with negative exponents could be converted into a valid one with positive exponents by multiplying by some x^n. The roots would be the same (with the exception of an extra root at x=0 in the converted polynomial).


Help with function question $f(2x + 1) = 2f(x) + 1$ I've been working on this question for a few hours but I can't figure out the answer.

$f(2x + 1) = 2f(x) + 1$ for all $x$ and $f(0) = 2$ what does $f(3)$ equal?

iteration speed of outside function is the same as inside function just they don't start with the same initial values F"Inside" starts with initial value of 0 and f"outside" starts with the initial value of 2 so you just solve for first order non homogenous recurrence relation of the form F(c+1)=2F(c)+1, with the seeds FI(0)=0, FO(0)=2. FI(t)=2^t-1, FO(t)=3*2^t-1, putting it in the form of parametric curve we get [FI(t),FO(t)], and solving for it as form of y(x). FI inverse of tI(x) =
log base 2(x+1). 2^log base 2(x+1)=x+1, 3(x+1)-1=3x+2.
so your final answer is 3x+2. checking 3(2x+1)+2=6x+3+2=6x+5.
2(3x+2)+1=6x+4+1=6x+5.


How can we calculate $\int \:\frac{1}{\sqrt{x^2-1}}dx$ Solve the following integral: $$\int \:\frac{1}{\sqrt{x^2-1}}dx$$

I attempted to solve it intergradation by parts by doing a

$$\int \:1\:\frac{1}{\sqrt{x^2-1}} \, dx$$

and set $u$ be $\frac{1}{\sqrt{x^2-1}}$ and $dv/dx$ be $1$:

but as I start doing, it gets more complicated. What is the right direction to solve this?

For $x^2-1$ in the radical use $x=\sec\theta$ as $\sec^2\theta-1=\tan^2\theta$

For $x^2+1$ in the radical use $x=\tan\theta$  for the same reason

For $1-x^2$ in the radical use $x=\sin\theta$ as $1-\sin^2\theta=\cos^2\theta$

the solution is $= \operatorname{Argch(x)}$

because $\operatorname{Argch}^{\prime}(x)= \frac{1}{\sqrt{x^2-1}}$.


Testing for the Convergence/Divergence of a series Given only that the series $\sum a_n$ converges, either prove that the series $\sum b_n$ converges or give a counterexample, when we define $b_n$ by,

i ) $\frac{a_n}{n}$

ii) $a_n \sin(n ) $

iii) $n^{\frac{1}{n}} a_n$

Is there any general approach to such questions?

Counterexample (ii):

The series

$$\sum_{n=1}^\infty a_n = \sum_{n=1}^\infty\frac{\sin n}{n},$$

converges by the Dirichlet test. Using $\cos 2n = 1 - 2 \sin^2 n$ we have

$$\sum_{n=1}^m a_n \sin n = \sum_{n=1}^m\frac{\sin^2 n}{n} = \sum_{n=1}^m\frac{1}{2n}- \sum_{n=1}^m\frac{\cos 2n}{2n}$$

The first sum on the RHS diverges (harmonic series) and the second sum converges by the Dirichlet test. Hence, the sum on the LHS diverges.

I'll leave it to you to attempt part (i) and (iii) where a standard series test applies.  Hint -- the sequence  $1/n$ is monotone decreasing and $n^{1/n}$ is   eventually monotone decreasing.  Case (ii) stands apart because $\sin n$ is not monotone.

Supplement:  Deriving $\sum_{n=1}^N \sin n = \frac{\sin (N/2) \sin( (N+1)/2)}{\sin (1/2)}$

We will use the identities $\sin a = - \cos (\pi/2 + a)$ and $\sin(a+b) - \sin(a-b) = 2 \sin b  \cos a. $

We have

$$\begin{align}2 \sin(1/2) \sum_{n=1}^N \sin n &= -\sum_{n=1}^N [2 \sin(1/2) \cos(\pi/2 +n)] \\ &=  -\sum_{n=1}^N [ \sin(\pi/2 + n + 1/2) - \sin(\pi/2 +n - 1/2)] \end{align}.$$

Since the series on the RHS is telescoping, we get

$$\begin{align}2 \sin(1/2) \sum_{n=1}^N \sin n &= - [\sin(\pi/2 + N + 1/2) - \sin(\pi/2 + 1/2)] \\ &=  -[ \sin( N/2 + 1/2 + \pi/2 + N/2 ) - \sin(N/2 + 1/2 + \pi/2 - N/2)] \end{align}$$

Using $\sin(a+b) - \sin(a-b) = 2 \sin b  \cos a $ with $a = N/2 + 1/2 + \pi/2$ and $b = N/2$,

$$\begin{align}2 \sin(1/2) \sum_{n=1}^N \sin n &= - 2\sin(N/2) \cos(\pi/2 + (N+1)/2) \\ &= 2\sin(N/2) \sin((N+1)/2) \end{align}$$

Hence,

$$\sum_{n=1}^N \sin n = \frac{2\sin(N/2) \sin((N+1)/2)}{\sin(1/2)}$$

For (iii) consider the sequence to be $n^{-n}$


Primes and 32 - where did this pattern come from? If we position all the natural numbers into a 'periodic table' with the period equal to $32$, we get the following pattern for primes.

The primes are colored according to their last digit. I did not color the primes past the first 'full diagonal', but the pattern continues indefinitely. The numbers on the 'prime diagonals' that are composite are in bold.

  How can this pattern be explained? I'm genuinely surprised by it.

Edit The so called 'pattern' was explained to me as trivial in the answers but what about the bold numbers $[49,77,91,119,121,133,143,...]$? Is there something special about this sequence? Some of them are squares of primes, some of them are the product of two primes, but maybe there is something else?

Actually, this is a little misleading.  If you ignore the primes and mark off every number by its last digit (but only consider those that end in 1, 3, 7, or 9) you will get the exact same pattern.  Except now the lines will be solid without breaks in them.

This isn't a mystery.  Each column we add 32 to each coresponding cell so by going over 1 column and up 2 rows you will always get a digit that has the same last digit and the lines with the same last digits will always form by going over 1 and up 2.  (If we colored all digits we'd get a confusing checkerboard like pattern but by restricting to odd values we can see lines.)

Filtering by primes (or any other condition) will put blank spots in the lines but will not change the basic existence of the lines.

Basically the lines exist because $32 \equiv 2 \mod 10$.  If we used 31 the lines would go over one up one.  If we used 30, the lines would be horizontal.  If we used 33 the lines would slant down.  Etc.

NOTHING to do with primes.

====

But what is surprising, is Ulam's spiral.  Google it.  (Which has NOTHING to do with what you have done--- which is utterly trivial.)

There's what you talking about https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes


Evaluate $h(t) = t + \frac{5}{t}$ Evaluate $$h(t) = t + \frac{5}{t}$$

$$h(-1)=$$

My answer was 7/-1 (Just saw my mistake I added the ones (1) instead of multiplying them)

However the answer is -6. Why is that?

Both denominators needed negative denominators and that caused the numerators for both terms to have positive values, thus I added them. What do I not understand?


Doubt in the definition of a compact set It's said a set $A$ is compact if for every finite cover $U$ of $A$ there exists a subset of $U$ which also covers $A$, let's say $U_1$.
Assuming $A$ is a compact set, we must be able to find a subcover of $U_1$, $U_2$ where $U\supset U_1\supset U_2$.
With this in mind, my questions are:

  Can this procedure can be done $n$-times?

If so, does some "infimum set" $U_f$ exists, such that from this set there cannot be obtained a subcover?
Sorry for the dull question, but I've been troubling with this for a long time.
Thanks, hope you all have a wonderful day

Once you fix your definition of compact, I think what you are asking is:

  Does every compact space satisfy, "Every cover of me has a minimal finite subcover?"

Here "minimal" means "containing no proper subcover", not "contained in every subcover."

The answer is yes: let $U$ be a cover of a compact space $X$. By compactness, $U$ has a finite subcover, $U_1=\{V_1, V_2, . . . , V_n\}$. Now let $U_i=\{V_i, V_{i+1}, V_{i+2}, ... , V_n\}$, and let $k$ be the largest number such that $U_k$ is an open cover of $X$ (there is such a largest number since $U_1$ is a finite cover); then $U_k$ is a minimal subcover of $U$.

Note that there is no reason for every open cover to have a unique minimal subcover.

This property can fail for noncompact spaces, even if we drop the word "finite." For example, take $\mathbb{R}$ with the usual topology and let $U=\{(-n, n): n\in\mathbb{N}\}$. Then $U$ has no minimal subcover.

Interestingly, it is not clear to me whether the property "Every open cover has a minimal subcover" is equivalent to compactness . . .

EDIT: OK, this is neat: it turns out the property "Every open cover has a minimal subcover" is indeed equivalent to compactness! The proof is cute, so I can't resist putting it here:

Suppose $X$ is a topological space such that every open cover of $X$ has a minimal subcover, and suppose $X$ is not compact. Let $\kappa$ be the least cardinality of some counterexample to compactness, and let $U=\{V_\eta: \eta<\kappa\}$ be a counterexample to compactness.
Let $W_\eta=\bigcup_{\beta<\eta}V_\beta$. Each $W_\eta$ is open, and clearly $U'=\{W_\eta: \eta<\kappa\}$ is an open cover of $X$. So $U'$ has a minimal subcover, $C$.
What could $C$ be? Well, it's easy to see that $C$ must have exactly one element: if $W_\eta, W_\theta\in C$ with $\eta<\theta$, then $C\setminus\{W_\eta\}$ would be a strictly smaller subcover.
OK, so $C=\{W_\eta\}$ for some $\eta$. Then - by definition of $W_\eta$ - we have that $U''=\{V_\beta: \beta<\eta\}$ is a subcover of $U$. But $\vert\eta\vert<\kappa$, so $U''$ has size $<\kappa$ - so by assumption on $\kappa$, $U''$ has a finite subcover.

"It's said a set A is compact if for every finite cover U of A there exists a subset of U which also covers A, let's say U1."

Well, as $U \subset U$ that is trivial.  This idea that a finite subcover must have strictly smaller subcover is absurd.

Consider that $[0,1]$ is compact.  Consider {$(-1,2)$} is a finite cover; it only has one set it in.  Can't get much more finite than that.  But it has no smaller subcover.

Your definition is wrong.  For every open cover $U = $ {$U_{\alpha}$} of A [A cover $U = ${$U_{\alpha}$} is a collection of sets so that $A \subset \cup_{\alpha} U_{\alpha}$] has a finite subcover.

Example: $[0,1]$ is compact. let $U = \{(a,b)| a \in \mathbb R; b \in \mathbb R; a < b\}$.  As $[0,1] \subset \cup_{a,b} (a,b)$, $U$ is a an open cover.  $U$ has uncountably many intervals $(a,b)$.  But as $[0,1]$ is compact there must be a finite subcollection that covers them.  And indeed $[0,1] \subset (-1,1) \cup (0,2)$ is a finite subcollection.  Indeed $[0,1] \subset (-1,2)$ is a subcollection with only one set.

"Assuming A is a compact set, we must be able to find a subcover of U1,U2. Where U>U1>U2"

Not at all!  For example, for $[0,1]$ the subcovers $\{(-1,1),(0,2)\}$ or $\{(-1,2)\}$ don't have any strictly sub covers.  But any infinite open cover must have a finite subcover.

"If so, does some "infimum set" Uf exists, such that from this set there cannot be obtained a subcover?"

Obviously.  But it's not accurate to call it "infimum".  Theres nothing unique or infinimum about it.  I've given several examples above.  For $[0,1]$.  {(-1,2)} is a minimal subcover.  As is {(-1,1),(0,2)}.  As is {(-.1,1/3),(1/4,2/3),(1/2,1.1)} etc.


Prove $x^6-6x^4+12x^2-11$ is irreducible over $\mathbb{Q}$ Extracted from Pinter's Abstract Algebra, Chapter 27, Exercise B1:

  Let $p(x) = x^6-6x^4+12x^2-11$, which we can transform into a polynomial in $\Bbb{Z}_3[x]$:
  \begin{align*}
x^6+1
\end{align*}
  Since none of the three elements $0,1,2$ in $\Bbb{Z}_3$ is a root of the polynomial, the polynomial has no factor of degree 1 in $\Bbb{Z}_3[x]$. So the only possible factorings into non constant polynomials are
  \begin{align*}
x^6+1 &= (x^3+ax^2+bx+c)(x^3+dx^2+ex+f)
\end{align*}
  or
  \begin{align*}
x^6+1 &= (x^4+ax^3+bx^2+cx+d)(x^2+ex+f)
\end{align*}
  From the first equation, since corresponding coefficients are equal, we have
  \begin{align}
x^0:\qquad & cf &= 1 \tag{1} \\
x^1:\qquad & bf + ce &= 0 \tag{2} \\
x^2:\qquad & af + be + cd &= 0 \tag{3} \\
x^3:\qquad & c + f + bd + ae &= 0 \tag{4} \\
x^5:\qquad & a + d &= 0 \tag{5} \\
\end{align}
  From (1), $c = f = \pm1$, and from (5), $a + d = 0$. Consequently, $af + cd = c(a + d) = 0$, and by (3), $eb = 0$. But from (2) (since $c = f$), $b + e = 0$, and therefore $b = e = 0$. It follows from (4) that $c + f = 0$, which is impossible since $c = f = \pm1$. We have just shown that $x^6 + 1$ cannot be factored into two polynomials each of degree 3.

For the second equation, however, $x^6+1=(x^2+1)^3$ in $\Bbb{Z}_3[x]$.  So we cannot say $p(x)$ is irreducible over $\Bbb{Q}$ because $x^6+1$ is irreducible over $\Bbb{Z}_3$.  What am I missing here?

For the second equation, however, $x^6+1=(x^2+1)^3$ in $\Bbb{Z}_3[x]$.  So we cannot say $p(x)$ is irreducible over $\Bbb{Q}$ because $x^6+1$ is irreducible over $\Bbb{Z}_3$.  What am I missing here?

You are entirely correct here. What you are missing is that the preceding argument shows that $p$ is not a product of two cubics in $\Bbb{Q}[x]$. After all, if it were, then in such a factorization
$$p=(x^3+ax^2+bx+c)(x^3+dx^2+ex+f),$$
all coefficients are integers by Gauss' lemma, and hence this reduces to a factorization into cubics in $\Bbb{F}_3[x]$. But you have just shown that no such factorization exists.

So it remains to show that $p$ is not the product of a quadratic and a quartic in $\Bbb{Q}[x]$, and there are infinitely many other primes to try. For example, in $\Bbb{F}_7[x]$ you have
$$x^6-6x^4+12x^2-11=x^6+x^4+5x^2+3.$$
A quick check shows that this polynomial has no roots in $\Bbb{F}_7$. Now proceed as before, expanding
$$x^6+x^4+5x^2+3=(x^4+ax^3+bx^2+cx+d)(x^2+ex+f),$$
to show that no such factorization exists in $\Bbb{F}_7[x]$.

If you are comfortable with a little more abstract algebra, here is an approach that does not require such ad hoc calculations. First it is easy to see that in $\Bbb{F}_3[x]$ the polynomial $p$  factors as
$$p=x^6+1=(x^2+1)^3,$$
where $x^2+1\in\Bbb{F}_3[x]$ is irreducible. It follows that every irreducible factor of $p$ in $\Bbb{Q}[x]$ has even degree. Now note that $p=h(x^2)$ where $h:=x^3-6x^2+12x-11\in\Bbb{Q}[x]$. A quick check shows that $h$ has no roots in $\Bbb{F}_7$, and therefore it is irreducible in $\Bbb{F}_7[x]$. This means the subring of the quotient ring $\Bbb{F}_7[x]/(p)$ generated by $x^2$ is a cubic field extension of $\Bbb{F}_7$, and therefore $p$ has an irreducible cubic or sextic factor. In the latter case $p$ is irreducible in $\Bbb{F}_7[x]$, and hence in $\Bbb{Q}[x]$ and we are done.

If $p$ has an irreducible cubic factor in $\Bbb{F}_7[x]$, then this is the reduction of an irreducible factor of $p$ in $\Bbb{Q}[x]$. As we saw before, the degree of this factor is even, so it is either quartic or sextic. Again, if it is sextic then $p$ is irreducible in $\Bbb{Q}[x]$ and we are done. If it is quartic then its reduction in $\Bbb{F}_7[x]$ is the product of a cubic and a linear factor. But $p$ has no roots in $\Bbb{F}_7$ because $p=h(x^2)$ and $h$ has no roots in $\Bbb{F}_7$, a contradiction.


Fourier Transform of a shifted & scaled rectangular pulse I'm trying to find the Fourier Transform of the following rectangular pulse:

$$ x(t) = rect(t - 1/2) $$

This is simply a rectangular pulse stretching from 0 to 1 with an amplitude of 1. It is 0 elsewhere. I tried using the definition of the Fourier Tranform:

$$ X(\omega) = \int_0^1 (1)*e^{-j\omega*t}dt $$

However carrying out the relatively simple integration and subbing in the bounds results for me in this:

$$ X(\omega) = \frac{1}{j\omega}[e^{-j\omega} - 1] $$

& unfortunately wolfram alpha has a different answer when I use it to compute this fourier transform. It's got the sinc function;
I'd appreciate any help on this, if I've got some giant conceptual error. I have an exam on this stuff in a bit less than a week :/

Edit: also realized I used j; it's the same with i (the imaginary #)

They are equal to each other. Rewrite it using the following:

$$
( e^{-j\omega} - 1 ) = e^{-j\omega/2}( e^{-j\omega/2} - e^{j\omega/2} ).
$$

You are doing a little mistake buddy, the Fourier transform of shifted rectangular function (given as ques) will be,
$X(w)=\frac1{-jw}\left(e^{-jw} - 1\right)$. Then solve.


Is Bayes' Theorem really that interesting? I have trouble understanding the massive importance that is afforded to Bayes' theorem in undergraduate courses in probability and popular science.
From the purely mathematical point of view, I think it would be uncontroversial to say that Bayes' theorem does not amount to a particularly sophisticated result. Indeed, the relation
$$P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{P(B\cap A)P(A)}{P(B)P(A)}=\frac{P(B|A)P(A)}{P(B)}$$
is a one line proof that follows from expanding both sides directly from the definition of conditional probability. Thus, I expect that what people find interesting about Bayes' theorem has to do with its practical applications or implications. However, even in those cases I find the typical examples being used as a justification of this to be a bit artificial.

To illustrate this, the classical application of Bayes' theorem usually goes something like this: Suppose that

1% of women have breast cancer;
80% of mammograms are positive when breast cancer is present; and
10% of mammograms are positive when breast cancer is not present.

If a woman has a positive mammogram, then what is the probability that she has breast cancer?
I understand that Bayes' theorem allows to compute the desired probability with the given information, and that this probability is counterintuitively low. However, I can't help but feel that the premise of this question is wholly artificial. The only reason why we need to use Bayes' theorem here is that the full information with which the other probabilities (i.e., 1% have cancer, 80% true positive, etc.) have been computed is not provided to us. If we have access to the sample data with which these probabilities were computed, then we can directly find
$$P(\text{cancer}|\text{positive test})=\frac{\text{number of women with cancer and positive test}}{\text{number of women with positive test}}.$$
In mathematical terms, if you know how to compute $P(B|A)$, $P(A)$, and $P(B)$, then this means that you know how to compute $P(A\cap B)$ and $P(B)$, in which case you already have your answer.

From the above arguments, it seems to me that Bayes' theorem is essentially only useful for the following reasons:

In an adversarial context, i.e., someone who has access to the data only tells you about $P(B|A)$ when $P(A|B)$ is actually the quantity that is relevant to your interests, hoping that you will get confused and will not notice.
An opportunity to dispel the confusion between $P(A|B)$ and $P(B|A)$ with concrete examples, and to explain that these are very different when the ratio between $P(A)$ and $P(B)$ deviates significantly from one.

Am I missing something big about the usefulness of Bayes' theorem? In light of point 2., especially, I don't understand why Bayes' theorem stands out so much compared to, say, the Borel-Kolmogorov paradox, or the "paradox" that $P[X=x]=0$ when $X$ is a continuous random variable, etc.

You are mistaken in thinking that what you perceive as "the massive importance that is afforded to Bayes' theorem in undergraduate courses in probability and popular science" is really "the massive importance that is afforded to Bayes' theorem in undergraduate courses in probability and popular science." But it's probably not your fault: This usually doesn't get explained very well.
What is the probability of a Caucasian American having brown eyes? What does that question mean? By one interpretation, commonly called the frequentist interpretation of probability, it asks merely for the proportion persons having brown eyes among Caucasian Americans.
What is the probability that there was life on Mars two billion years ago? What does that question mean? It has no answer according to the frequentist interpretation. "The probability of life on Mars two billion years ago is $0.54$" is taken to be meaningless because one cannot say it happened in $54\%$ of all instances. But the Bayesian, as opposed to frequentist, interpretation of probability works with this sort of thing.
The Bayesian interpretation applied to statistical inference is immune to various pathologies afflicting that field.
Possibly you have seen that some people attach massive importance to the Bayesian interpretation of probability and mistakenly thought it was merely massive importance attached to Bayes's theorem. People who do consider Bayesianism important seldom explain this very clearly, primarily because that sort of exposition is not what they care about.

First see the comments following this answer, especially the last few comments.  I was totally unaware that Bayes Theorem is simply a consequence of axioms around the definition of Conditional Probability.  Based on this assertion, I can't refute the idea that the following problem can be solved without Bayes Theorem.

Hard to imagine attacking a conditional probability problem without it.  Imagine traveling back in time 1000 years.  You are the captain of a ship.  You have two sailors, A and B that you independently use to predict rain.
A is right 90% of the time and B is right 80% of the time.
A says it will rain today, and B says it won't rain today.
Absent Bayes Theorem, and absent any info on how often (in general) it rains, how do you (intuitively) determine the chance that it will rain today?  Clearly, the problem is well defined, so the problem has a meaningful answer.  Absent Bayes Theorem, or anything like it, how do you compute the answer?


Technique for using direct and limit comparison tests? Suppose you have an infinite series $$\sum_{n=1}^{\infty} \frac{1}{n^2 + 9}$$

Using the direct comparison test, the sequence can obviously be compared to $\sum_{n=1}^\infty\frac{1}{n^2}$ since it is smaller, and then by the p-series test, since $\sum\frac{1}{n^2}$ converges $\sum\frac{1}{n^2+9}$ also converges. This makes sense since when $\lim{n\to \infty}$, the $9$ is negligible.

In other cases though, the series is not simple. Take for example: $$\sum_{n=1}^\infty\frac{\ln^4(x)}{n}$$ The end behavior at $\infty$ is not as simple as the first one. How do you know what sequence to compare $\frac{\ln^4(n)}{n}$ to? How do you know what term is dominant? Take the derivative and compare? Is there a technique?

Another example: $$\sum_{n=2}^\infty\frac{n}{n^3-1}$$
I know the sequence compares to $\frac{1}{n^2}$, but is there some logical technique that allows us to make that comparison?

Essentially, my question is how do you figure out the comparison sequence?  Before, I have been taught to take the most dominant term in the numerator and denominator and use those as a comparison, but I feel this technique is prone to error. I realize these questions are a bit elementary, but I think the basic logic would be useful.

Thank you.

Excuse me, but $log^4n \geq 1$ for $n \geq 3$, so $\sum \frac{log^4 n}{n}$ is bigger than $\sum \frac{1}{n}$, which diverges. Did you get it?
Basically, you learn about simple sequences, then you learn about simples series and after that, you deal with series more complicated. The same you do with integrals, you know what I mean?

Alternative method:
$$\sum_{n=1}^{\infty} \frac{n}{n^3 - 1}=\sum_{n=1}^{\infty} \frac{1/n^2}{1 - 1/n^3} \rightarrow 0$$


Does $\mathbb{N} \cap \mathbb{N}^2 = \varnothing$? Definitions. A cardinal $\mathfrak{a}$ is said to be finite if $\mathfrak{a} \ne \mathfrak{a} + 1$. A finite cardinal is also called a natural integer. The natural integers form a set, denoted by $\mathbb{N}$. The set $\mathbb{N}^2$ is the Cartesian product of $\mathbb{N}$ with itself. In other words, $\mathbb{N}^2$ is the set of ordered pairs
$$(m,n) = \bigl\{\{m\},\{m,n\}\bigr\}$$
where $m$ and $n$ are natural integers.

If you use this definition:

$$\begin{alignat}{2}
   0 & {} = \{\} && {} = \emptyset,\\
   1 & {} = \{0\} && {} = \{\emptyset\},\\
   2 & {} = \{0,1\} && {} = \{\emptyset,\{\emptyset\}\},\\
   3 & {} = \{0,1,2\} && {} = \{\emptyset,\{\emptyset\},\{\emptyset,\{\emptyset\}\}\}
 \end{alignat}$$

then every element of $\mathbb{N}$ is either $\emptyset$ or contains $\emptyset$. This is false for the elements of $\mathbb{N}^2$ using the standard definition of ordered pair:

$$(a, b) \triangleq \{\{a\}, \{a,b\}\}$$

since $\{a\}\neq\emptyset$ and $\{a, b\}\neq\emptyset$. So $\mathbb{N}\cap\mathbb{N}^2=\emptyset$.

If you use an alternative definition, like $(a, b) \triangleq \{a, \{a, b\}\}$, then we can have $a = b = \emptyset \Rightarrow (a, b) = \{\emptyset, \{\emptyset,\emptyset\}\} = \{\emptyset, \{\emptyset\}\} = 2 \neq \emptyset$. Under this definition of ordered pair, we have $\mathbb{N}\cap\mathbb{N}^2 \neq \emptyset$.

Note: No, you absolutely cannot take $$(\forall x)(\forall x^\prime)(\forall y)(\forall y^\prime)\big((x,y) = (x^\prime,y^\prime) \implies x = x^\prime \text{ and } y = y^\prime\big).$$ as the definition of ordered pairs! The above is a property of ordered  pairs, but it does not specify what $(x,y)$ is. In fact  there's a standard definition of $(x,y)$ in set theory.

Hint: If $x$ and $y$ are sets then $(x,y)\ne\emptyset$ and also every element of $(x,y)$ is nonempty.

(Of course the answer depends on the definition of $\Bbb N$.)


Calculus Proof Problem Let $f$ be a continuous function whose domain includes $[0,1]$, such that $0 \le f(x) \le 1$ for all $x \in [0,1]$, and such that $f(f(x)) = 1$ for all $x \in [0,1]$. Prove that $\int_0^1 f(x)\,dx > \frac34$.

I've been stuck on this question for a while now without any idea on how to get started. Is it possible for someone to help me?

Thanks!

While not a general proof of the proposition, it can be seen as true of a certain class of functions satisfying the conditions.

Let $f(x)$ be a continuous non-decreasing function defined on the interval $[0,1]$ with $a\in[0,1]$ as shown in the diagram and satisfying

$f(0)=a$
$f(a)=1$ for $x\ge a$

Then the integral $\int_0^1 f(x)\,dx$ must be greater than $a^2+(1-a)$, the sum of the two shaded rectangles.

But the minimum area of the two rectangles is $\frac{3}{4}$ and occurs when $a=\frac{1}{2}$.

It remains to be determined whether there is any other class of function satisfying the conditions.

Addendum: Actually we can drop the non-decreasing part and merely require that $f(x)\ge a$ on $[0,a)$.

Edit: This proof does not work if f (x) does not have range [0,1] for all x in [0,1]. The simple way to prove this:

We have function f(x) which is continuous along the interval [0,1].

f (x) is defined such that                                                              f(x) has domain [0,1] , and range [0,1].                  (a)

f(f (x)) has domain [0,1] and range [1].                 (b)

--Note: The domain of f(f(x)) comes from the range of f(x) along the interval [0,1].--

Now we want to prove that the integral of f(x) over [0,1] is > 3/4

In order to prove this we can break the problem into two parts. First we can show that f(f(x)) and f(x) have the same range for domain [0,1]. Second we can show that because the function f's range is [1] for all x in [0,1] that the integral over [0,1] must be greater than 1.

First Part:

Since f is a function it cannot have two different outputs for one input, so (a) and (b) cannot have contradictory ranges for the same domain. From (a) we know that the range of f over domain [0,1] is 0<=f(x)<=1, from (b) we also know the range over domain [0,1] is f(x)=1. So (a) and (b)'s range do not contradict each other  in so much as f(f(x))=1 is within the range of f(x). But since f is a function it must be the case that f(x)=1 for all x in the interval [0,1]. Otherwise the function f would have multiple outputs for the same inputs and would not be a function. Therefore it must be the case that for all input for f in the interval [0,1], f has an output of 1, and thus for the function f the domain [0,1] has range [1].

The Second Part:
Since the function f on domain [0,1] has range [1], then any integral within [0,1] of inputs must be greater than or equal to 1. Therefore since the integral is greater >=1 and 1 is > 3/4, the integral is > 3/4.

Proof complete.


bijective function from [a,b] to [c,d] Im trying to think about bijective function from the closed interval [a,b] to the closed interval [c,d]. When $a,b,c,d \in \mathbb{R}$ and $a < b,\;c < d$.
Is there such a function?

The idea is to construct a line whose domain is $[a, b]$ and whose range is $[c, d]$

Hence, two points on the line will be
$$
p_0 = (a, c) \\
p_1 = (b, d)
$$

Since we want $a \to c$, $b \to d$, and a straight line between them.

The slope of such a line will be
$$ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{d - c}{b - a}$$

Using point slope form of a line,

$$
y - y_1 = m \left ( x - x_1 \right) \\
y - c = \frac{d - c}{b - a} \left (x - a\right)
$$

Is the required equation.

$f(x)=(d-c)\frac{x-a}{b-a}+c$   is the (linear) function


invert multinomial logit link with three unknown I am attempting to invert the multinomial logit link with three variables.  I can do it with two variables, but I do not know how to do it with three.

A multinomial logit function for three states, i.e., three probabilities, $a, b, c$ is written as follows:

$a = \frac{e^{x}} {(1 + e^{x} + e^{y})}$

$b = \frac{e^{y}} {(1 + e^{x} + e^{y})}$

$c = 1 - a - b$

These three probabilities are defined by the parameters $x$ and $y$.  If we know $x$ and $y$ we can obtain $a$, $b$ and $c$.

However, given $a, b, c$, how do we obtain $x$ and $y$?  One way is to use multinomial logistic regression.  However, there should be a closed form solution in which $x$ and $y$ are obtained using basic algebra.  I can obtain the closed form solution for two parameters, $x$ and $y$:

$x = \log(\frac{a  (1 - b) + (a  b)}{ (1 - a)  (1 - b) - a  b})$

$y = \log(\frac{b  (1 - a) + (b  a) }{ (1 - b)  (1 - a) - b  a})$

Which simplifies to:

$x = \log(\frac{a}{1 - a - b})$

$y = \log(\frac{b}{1 - a - b})$

How can I obtain the closed form solution when there are three parameters $x, y, z\;$?

$a = \frac{e^{x}} {(1 + e^{x} + e^{y} + e^{z})}$

$b = \frac{e^{y}} {(1 + e^{x} + e^{y} + e^{z})}$

$c = \frac{e^{z}} {(1 + e^{x} + e^{y} + e^{z})}$

$d = 1 - a - b - c$

For every fixed number of variables, you are considering, for every $i$,

  $$a_i=\frac{e^{x_i}}{1+s}$$

where

  $$s=\sum_ie^{x_i}$$

and

  $$z=\frac1{1+s}$$

and you are asking how to invert this system, that is, how to deduce the collection $(x_i)$ from the collection $(a_i)$ and $z$, or even, from $(a_i)$ only.

To solve this, consider that $$z+\sum_ia_i=1$$ hence, for every $i$, $$e^{x_i}=a_i\cdot(1+s)=\frac{a_i}z$$ that is,

  $$x_i=\log a_i-\log z=\log\left(\frac{a_i}{1-\sum\limits_ka_k}\right)$$

Solve each variable $x$, $y$ and $z$ as a function of the other two.

$x = \log(\frac{(a + a  e^{y} + a  e^{z})}  {(1 - a)})$

$y = \log(\frac{(b + b  e^{x} + b  e^{z})}  {(1 - b)})$

$z = \log(\frac{(c + c  e^{x} + c  e^{y})}  {(1 - c)})$

Then:

Substitute $x$ into the equation for $y$.

Substitute $x$ into the equation for $z$.

Substitute $y$ into the equation for $x$.

Substitute $y$ into the equation for $z$.

Substitute $z$ into the equation for $x$.

And substitute $z$ into the equation for $y$.

Each parameter $x$, $y$ and $z$ is now expressed as a function of just one of the other two.

To express each parameter as a function of itself, substitute, for example, the equation expressing $x$ as a function of $y$ into the function expressing $y$ as a function of $x$.

Once these three equations are simplified we have:

$x = \log(-(a * c - a) / (c^2 + (b+a-2)*c + (a - 1)*b - a + 1 - a * b))$

$y = \log(-(b * c - b) / (c^2 + (b+a-2)*c + (a - 1)*b - a + 1 - a * b))$

$z = \log(((1-b)*c) / (((b+a-1)*c+b^2+(a-2)*b-a+1) - (a*c)))$


Can eight circles be constructed from three circles? Given three sufficiently spaced circles in a plane, is it possible, using a straight edge and compass, to construct the eight circles that are tangent to all three given circles?

yes, one outside all of them, one inside all of them, three inside one of them, and three inside two of them. As the question was edited, previously the three circles were of the same size and their centroids were concentric. the general case of three circles in space of arbitrary size and position is not always possible.

four of these circles are of significance, the rest can be obtained by rotations. One circle that is tangential to the outside of the circles, on that is tangential to the inside of all the circles. these two cannot be rotated to obtain more permutations. if you construct a circle that is tangential to the inside of ONE of the circles and the outside of the other two circles it can be rotated in space to obtain two more. if you construct a circle that is tangential to the inside of TWO of the circles and tangential to the outside of one, this can be rotated to obtain the remaining two.


Set-theoretic proof for $A\setminus(B\setminus C) = (A\setminus B) \cup C$, provided that $C\subseteq A$ I'm trying to prove $A\setminus(B\setminus C) = (A\setminus B)\cup C$ using the fact that $C\subseteq A$. Somehow, I can use the relation $(A\setminus B)^c = A^c \cup B$. However, I can't really see how to start the proof.

Any help on how to start/structure this proof would be appreciated

A funny way to prove set-theoretic equality is to use indicator functions. Given $A, B \subset X$, we have $\mathbb{1}_{A\setminus B} = \mathbb{1}_{A}(1 - \mathbb{1}_B)$. Then we have the following equalities:

$\mathbb{1}_{A\setminus (B\setminus C)} = \mathbb{1}_{A} - \mathbb{1}_{A}\mathbb{1}_{B} + \mathbb{1}_{A}\mathbb{1}_{B}\mathbb{1}_{C}$

$\mathbb{1}_{(A \setminus B) \cup C} = \mathbb{1}_{(A \setminus B)} + \mathbb{1}_{C} - \mathbb{1}_{(A \setminus B)\cap C} = \mathbb{1}_{A} - \mathbb{1}_{A}\mathbb{1}_{B} + \mathbb{1}_{C} - \mathbb{1}_{A}\mathbb{1}_{C} + \mathbb{1}_{A}\mathbb{1}_{B}\mathbb{1}_{C}$

Because $C \subset A$, $\mathbb{1}_{C} - \mathbb{1}_{A}\mathbb{1}_{C} = 0$, so that both indicator functions are equal.

Does $C\subseteq A$ is an assumption? The identity is false.
Take $A=B=\varnothing$ and $C=\{1,2,3\}$


Finding the basis of a null space I am trying to understand why the method used in my linear algebra textbook to find the basis of the null space works. The textbook is 'Elementary Linear Algebra' by Anton.

According to the textbook, the basis of the null space for the following matrix:

$A=\left(\begin{array}{rrrrrr} 1 & 3 & -2 & 0 & 2 & 0 \\ 2 & 6 & -5 & -2 & 4 & -3 \\ 0 & 0 & 5 & 10 & 0 & 15 \\ 2 & 6 & 0 & 8 & 4 & 18 \end{array}\right)
$

is found by first finding the reduced row echelon form, which leads to the following:

$(x_1,x_2,x_3,x_4,x_5,x_6)=(-3r-4s-2t,r,-2s,s,t,0)$

or, alternatively as

$(x_1,x_2,x_3,x_4,x_5,x_6)=r(-3,1,0,0,0,0)+s(-4,0,-2,1,0,0)+t(-2,0,0,0,1,0)$

This shows that the vectors

${\bf v_1}=(-3,1,0,0,0,0),\hspace{0.5in} {\bf v_2}=(-4,0,-2,1,0,0),\hspace{0.5in} {\bf v_3}=(-2,0,0,0,1,0)$

span the solution space.

It can be shown that for a homogenous linear system, this method always produces a basis for the solution space of the system.

Question

I don't understand why this method will always produce a basis for $Ax=0$. Could someone please explain to me why this method will always work? If it helps to explain, I already understand the process of finding the basis of a column space and row space. I also understand why elementary row operations do not alter the null space of a matrix.
What specific properties of matrices or vector space that I need to be aware of in order to understand why this method works?

The null space of $A$ is the set of solutions to $A{\bf x}={\bf 0}$.  To find this, you may take the augmented matrix $[A|0]$ and row reduce to an echelon form. Note that every entry in the rightmost column of this matrix will always be 0 in the row reduction steps. So, we may as well just row reduce $A$, and when finding solutions to $A{\bf x}={\bf 0}$, just keep in mind that the missing column is all 0's.

Suppose after doing this, you obtain
$$
\left[\matrix{1&0&0&0&-1 \cr 0&0&1&1&0 \cr 0&0&0&0&0 \cr  0&0&0&0&0 \cr }\right]
$$

Now, look at the columns that do not contain any of the leading row entries. These columns correspond to the free variables of the solution set to $A{\bf x}={\bf 0}$  Note that at this point, we know the dimension of the null space is 3, since there are three free variables. That the null space has dimension 3 (and thus the solution set to $A{\bf x}={\bf 0}$ has three free variables) could have also been obtained by knowing that the dimension of the column space is 2 from the rank-nullity theorem.

The "free columns" in question are 2,4, and 5. We may assign any value to their corresponding variable.

So, we set $x_2=a$, $x_4=b$, and $x_5=c$, where $a$, $b$, and $c$ are arbitrary.

Now solve for $x_1$ and $x_3$:

The second row tells us $x_3=-x_4=-b$ and the first row tells us $x_1=x_5=c$.

So, the general solution to $A{\bf x}={\bf 0}$ is
$$
{\bf x}=\left[\matrix{c\cr a\cr -b\cr b\cr c}\right]
$$

Let's pause for a second. We know:

1) The null space of $A$ consists of all vectors of the form $\bf x  $ above.

2) The dimension of the null space is 3.

3) We need three independent vectors for our basis for the null space.

So what we can do is take $\bf x$ and split it up as follows:

$$\eqalign{
{\bf x}=\left[\matrix{c\cr a\cr -b\cr b\cr c}\right]
&=\left[ \matrix{0\cr a\cr 0\cr 0\cr 0}\right]+
  \left[\matrix{c\cr 0\cr 0\cr 0\cr c}\right]+
  \left[\matrix{0\cr 0\cr -b\cr b\cr 0}\right]\cr
&=
a\left[ \matrix{0\cr1\cr0\cr 0\cr 0}\right]+
c\left[ \matrix{1\cr 0\cr 0\cr 0\cr 1}\right]+
b\left[ \matrix{0\cr 0\cr -1\cr 1\cr 0}\right]\cr
}
$$
Each of the column vectors above are in the null space of $A$. Moreover, they are independent. Thus, they form a basis.

I'm not sure that this answers your question. I did a bit of "hand waving" here.  What I glossed over were the facts:

1)The columns of the echelon form of $A$ that do not contain leading row entries correspond to the "free variables" to $A{\bf x}={\bf 0}$. If the number of these columns is $r$, then the dimension of the null space is $r$ (again, if you know the dimension of the column space, you can see that the dimension of the null space must be the number of these columns from the rank-nullity theorem).

2) If you split up the general solution to $A{\bf x}={\bf 0}$ as done above, then these vectors will be independent (and span of course since you'll have $r$ of them).

The software Mathematica can find a null-space spanning set for Matrices given with exact coefficients. It can find exactly the answer your book gives:

NullSpace[{{1, 3, -2, 0, 2, 0}, {2, 6, -5, -2, 4, -3}, {0, 0, 5, 10, 0, 15}, {2, 6, 0, 8, 4, 18}}]

gives

{{-2, 0, 0, 0, 1, 0}, {-4, 0, -2, 1, 0, 0}, {-3, 1, 0, 0, 0, 0}}

I suggest you start here if you are interested in how this is done in a computationally efficient and of course provably correct way. That Wikipedia section for instance points out, that in practice the so called Bareiss algorithm is more useful than Gaussian elimination.


What does it mean to take the exclusive OR of the three numbers $0, 2, 1$? This question asks about the exclusive-OR'ing of several numbers, each of which can assume more than two values (that is, more than $0$ and $1$).

What does this mean?  I did a Google search and found information on the ternary exclusive-OR, but that still involves $true$ and $false$ values, not $0, 1, 2, 3, ...$

Any ideas?

Taking them bitwise, which is probably what is intended, you have $00 \oplus 10 \oplus 01=11$

Exclusive-OR, or XOR in short, is an operation which will return true if an odd number of its inputs is true. It iwll return false if an even number of its inputs is false.

For example, an input of TTFTFFT will give false since there are 4 T's.

an input of TTTFFFFFFFF will be true since there are 3 T's.

Hope this helps. :)


Find whether : $\sum_{ I \subset \mathbb{N}} e^{-\sqrt{S(I)}}$ converges Does the following sum converges : $$\sum_{ I \subset \mathbb{N}} e^{-\sqrt{S(I)}}$$
where : $$ S(I) =\sum_{ i \in I} i$$

I don’t know how to approach this problem. Nevertheless, maybe it’s possible to have some intuition about the problem.
If $\mid I \mid = n$ then we know that the $S(I)$ which are going to give an important weight to the sum are in $O(n^2)$. Hence maybe luckily :
$$\sum_{I \subset \mathbb{N}, S(I) = O(\mid I \mid ^2)} e^{-\sqrt{S(I)}}$$
has the same nature of our sum.
In this case the square root makes it easy and the sum converges.
Yet it’s possible that this intuition is false since there are a lot of $S(I) \ne O(\mid I \mid ^2)$ (infinitely many actually) so it might give a lot of weight to the sum... I don’t really know.
Note that in order that $S(I)$ makes sens, we have : $I < \infty$.

To make things precise, when $|I| = \infty$, we redefine the value of $S(I)$ as $+\infty$ and $e^{-\sqrt{S(I)}} = 0$.

After this change, the summand inside the sum $\sum\limits_{I\subset \mathbb{N}} e^{-\sqrt{S(I)}}$ will be non-zero for countably many $I$. Since all summands are non-negative, the sum is well defined and takes value in $[0,\infty]$. Furthermore, we can compute it by enumerating those $I$ with $|I| < \infty$ in arbitrary order and get the same result. As a result,

$$\begin{align}\sum_{I\subset \mathbb{N}} e^{-\sqrt{S(I)}}
\stackrel{def}{=} \sum_{I\subset \mathbb{N}, |I| < \infty} e^{-\sqrt{S(I)}}
&= 2\sum_{I\subset \mathbb{Z}_{+}, |I| < \infty} e^{-\sqrt{S(I)}}
= 2\sum_{n=0}^\infty \sum_{I \subset \mathbb{Z}_{+}, S(I) = n} e^{-\sqrt{n}}\\
&= 2\sum_{n=0}^\infty q(n) e^{-\sqrt{n}}
\end{align}
$$

where $q(n) = | \{ I \subset \mathbb{Z}_{+} : S(I) = n \} |$ is the number of partitions
of integer $n$ into distinct parts.

The OGF of $q(n)$ equals to

$$\sum_{n=0}^\infty q(n) z^n = \prod_{k=1}^\infty ( 1 + z^k )$$

The closed form of $q(n)$ is not known. However, we do know for large $n$,${}^{\color{blue}{[1]}}$

$$q(n) \sim \frac{3^{3/4}}{12 n^{3/4}} \exp\left(\pi\sqrt{\frac{n}{3}}\right) $$

Since $\alpha \stackrel{def}{=} \frac{\pi}{\sqrt{3}} - 1 > 0$, the sub-sum over those $I$ with $S(I) = n$ blows up like $e^{\alpha\sqrt{n}}$. From this, we can deduce

$$\sum_{I\subset \mathbb{N}} e^{-\sqrt{S(I)}} = \infty$$

Refs

$\color{blue}{[1]}$ -
Philippe Flajolet, Robert Sedgewick
Analytic Combinatorics,
Cambridge University Press; (1st ed., 2009). Formula found at VIII.6 Saddle-point asymptotics / Integer partitions.

That sum, or just any sum with more than countably many positive summands cannot converge:

Let  $X$ be a set and $f\colon X\to\Bbb R$ a map such that $f(x)> 0$ fore all $x\in X$. Assume that  $\sum_{x\in X}f(x)$ converges, say $\sum_{x\in X}f(x)=S\in \Bbb R$. Then for $Y\subseteq X$, clearly $\sum_{x\in Y}f(x)\le \sum _{x\in X}f(x)$.
For $n\in \Bbb N$, let $X_n=\{\,x\in X\mid f(x)>\frac 1n\,\}$. Then $$M\ge \sum_{x\in X_n}f(x)\ge \sum_{x\in X_n}\frac 1n=\frac 1n|X_n|$$
and in particular, $X_n$ must be finite. As $X=\bigcup_{n\in\Bbb N}X_n$, we conclude that $X$ is countable.


Big O Notation / landau symbols I want to write these two in the big O notation:
(it's $h\rightarrow0$)

$f(h)=\sqrt{h^3}$
$f(h)=h\cdot \log h $

But I don't have any idea how to do this.

Thanks for helping!

What you ask makes no sense, really. You have two functions; they are fine as they stand, no big O is needed to understand them. Big O is a notation to related a given function to something else, presumably simpler, or just to write an estimate in a form where you don't want to bother with the implied constants. You could certainly write $\sqrt{h^3}=O(h)$, for example, but if it makes sense to do so depends on context. Ditto for the other one, which is (among many things) $h\cdot\log(h)=O(h^{1+\varepsilon})$ for any $\varepsilon>0$. But again, whether it makes sense to write it so depends on context.

I don't really why $h\rightarrow 0$, i use this notation when $h\rightarrow\infty$ for look the complexity of algorithm. So, may be you want to use a substitution with $u=1/h$ and see it like me.

$O(h^{1/3})$
$O(n\cdot \log n)$ look that is Big O, is like that Complexity for Quicksort Algorithm in the average case.

Big-O is for me like a function boundary for the function and take the most important for the result function. Then, we forget things like coefficient constants.
For Example:
$O(4x^4+ x^2)=O(4x^4)= O(x^4)$

Regards,


If in a normed space $a_n$ converges to $a$, prove that $||a_n||$ converges to $||a||$ If in a normed space an converges to $a$, prove that $||a_n||$ converges to  $||a||$

I have seen this proof: for all $\varepsilon>0$ there exists and $N$ s.t. $n>N$ implies
that the $||a_n-a||<\varepsilon$

by the reverse triangle inequality for the same $N \varepsilon > ||an-a|| > |\ ||an||-||a||\ |$

and therefore it converges.

But this is the absolute value of the difference of norms.  Don't you need the norm of the difference of norms to be less than $\varepsilon$?

As in

$||\ ||an|| - ||a||\ || < \varepsilon$

not

$|\ ||an|| - ||a||\ | < \varepsilon$

Or am I confused

Thanks!

The expressions $||a_n||$ and $||a||$ are real numbers, rather than elements of your normed vector space (unless the normed vector space is $\mathbb{R}$ with the usual absolute value, in which case $||\cdot||=|\cdot|$), so you need to look at the absolute value of the difference, rather than the norm of the difference.

Let your normed space be $X$. The map $||\cdot|| : X \to \mathbb R$ is continuous. Since $a_n \to a$, $||a_n|| \to ||a||$, as the norm is continuous, and continuous functions preserve sequential convergence, limits being sent to limits.


Number of solutions to $a_1 + a_2 + a_3 + \cdots + a_n = N;\; 1\leq a_i \leq K$ How can I find the number of solution to this -
$$a_1 + a_2 + a_3 + \cdots + a_n = N;\; 1\leq a_i \leq K$$
I know stars and bard theorem. I can solve if it was said that $a_i \geq K$. Then we can set $a_i' = a_i - K$ and convert that to $$a_1' + a_2' + a_3' + \cdots a_n' = N + Kn;\; a_i \geq 0$$
Then the number of solutions will be $$\binom{N+Kn + n - 1}{n-1}$$
But how can I handle this when $a_i \leq K$

It is the coefficient of $X^N$ in the expression $\prod_{i=1}^N(1+X+X^2+\dots +X^K)$

Consider the 3 problems below:

Problem A: $$x_1+\cdots+x_n=N\\s.t.~x_i\geq0~\forall i, ~\text{and}~x_1\geq k+1$$
Problem B: $$x_1+\cdots+x_n=N\\s.t.~x_i\geq0~\forall i, ~\text{and}~x_j\geq k+1~\text{for some}~j$$
Problem C: $$x_1+\cdots+x_n=N\\s.t.~x_i\geq0~\forall i$$
Problem D: $$x_1+\cdots+x_n=N\\s.t.~0\leq x_i\leq k~\forall i$$
Problem E: $$x_1+\cdots+x_n=N-k-1\\s.t.~x_i\geq0~\forall i$$
Let $N_a.N_b,N_c,N_d$ be the number of solutions to Problem A,B,C,D respectively. Then $$N_d=N_c-N_b$$$$N_b=n!\times N_a$$$$N_a=N_e$$


Average in a rating between 1 and 5 A user can vote on these numbers:

1, 2, 3, 4, 5

Then it's displayed as groups like this:

1 - 5 votes
2 - 3 votes
3 - 1 vote
4 - 17 votes
5 - 2 votes

Now I want the avarage rating score, like 3.9 or whatever it will be.

What I tried and failed:

Multiply all rows in the groups like:

1 - 1 * 5 votes
2 - 2 * 3 votes
3 - 3 * 1 vote
4 - 4 * 17 votes
5 - 5 * 2 votes

and then a sum of them where I get 52. Then I divide it with 15 which is the sum of all the valid scores. 52/15 = 3.46.

If I do that with the same votes on each, I can see how this is not correct:

1 - 1 * 1 votes
2 - 2 * 1 votes
3 - 3 * 1 vote
4 - 4 * 1 votes
5 - 5 * 1 votes

The result with my way would be 15/15 and that's 1 but it should be 3 in this case.

Any ideas?

Everybody is showing you how to compute the weighted average, but not why.

What you have is 28 votes. The average of 28 values is the sum of those values, divided by 28. In this case, that means:

$$\frac{1+1+1+1+1+2+2+2+3+4+\cdots+4+5+5}{28}$$

This, we see, is the same as:

$$\frac{1\cdot 5 + 2\cdot 3+ 3\cdot 1 + 4\cdot 17+5\cdot 2}{28}$$

Your scenario : -
1 - 1 * 5 votes
2 - 2 * 3 votes
3 - 3 * 1 vote
4 - 4 * 17 votes
5 - 5 * 2 votes

Your to total number of votes casted are :-
5+3+1+17+2 = 28

so my friend to calculate the correct average divisor has to be 28
Now just add all up - 1*5 + 2*3 + 3*1 + 4*17 + 5*2 = 92

So you average rating is 92/28


$A_4 \oplus Z_3$ has no subgroup of order 18 Here my solution: Suppose there exists and $H \leq A_4 \oplus Z_3$ such that order of H is 18. Now, notice index of H in $A_4 \oplus Z_3$ is 2. therefore, H is normal, and therefore, the $A_4 \oplus Z_3 / H$ exists. Now, for every $\alpha$ in $A_4 \oplus Z_3$, $\alpha^2H = (\alpha H)^2 = H$ since the quotient group has order 2. But notice, $A_4$ has only 3 elements of order 2, therefore, $A_4 \oplus Z_3$ have only 3 elements of order 2, which implies H has 3 elements of order 2 and the rest must have order 1, and this is an absurd.

Is this a correct solution? Do you guys have any other solution?

thanks,

The image of $H$ under the projection $A_4\oplus \mathbf Z_3\to A_4$ would be a subgroup of $A_4$ isomorphic to $H/(H\cap \mathbf Z_3)$, which is of order $18/\#( H\cap \mathbf Z_3)$. But $A_4$ neither has any subgroup of order $18$ (obviously) nor of order $6$ (requires only slightly more reflection).

$\operatorname{Order}(g1+g2)=\operatorname{lcm}(\operatorname{Order}(g1),\operatorname{Order}(g2))$.

But for any $g_1$ from $A_4$, $\operatorname{Order}(g_1)\leq4$ and for any $g_2$ from $Z_3$, $\operatorname{Order}(g_2)\leq3$.

So, $\operatorname{Order}(g_1+g_2)\leq12$.


If $f(x+1/n)$ converges to a continuous function $f(x)$ uniformly on $\mathbb{R}$, is $f(x)$ necessarily uniformly continuous? Suppose $f:\mathbb{R}\mapsto\mathbb{R}$ is a continuous function. Define $f_n(x)=f(x+1/n)$ for all $n\in\mathbb{N}^+$. If $f_n(x)\to f(x)$ uniformly on $\mathbb{R}$, can we conclude that $f$ is actually uniformly continuous? If not, can you give a counterexample of $f$ to be not uniformly continuous but satisfies all above conditions?
I know that if $f_n(x)$'s are all uniformly continuous and $f_n(x)\to f(x)$ uniformly, then $f(x)$ must be uniformly continuous. However, here we do not assume $f(x)$ to be uniformly continuous, and $f_n(x)$ has specific structure related to $f(x)$, so at least you cannot directly conclude they are uniformly continuous. I guess the statement above is wrong but it's hard for me to find a counter-example. Can anyone help me?

Updated answer:
No, $f$ need not be uniformly continuous, as shown by the example below. The example is tailored to the sequence $\{1/n\}_{n \in \mathbb{N}}$, and does not shed light on the general question of whether the condition that $f(x + \epsilon_n) \to f(x)$ uniformly implies that $f$ is uniformly continuous, for a given sequence $\{\varepsilon_n\}$ converging to zero.
First, define a sequence recursively by $a_1 = 3$ and $a_{n+1} = (a_n)!$ for $n \geq 1$. Let $h$ be the periodic function with period $1$ such that $h(x) = 2x$ for $x \in [0, 1/2]$ and $h(x) = 2-2x$ for $x \in [1/2, 1]$. In particular, $h(k) = 0$ for each integer $k$, and $h$ is continuous and Lipschitz with constant $2$. Now using the sequence, for each $n$ we define the function $g_n$ by
$$g_n(x) = \begin{cases} 0 & x < n \\ \frac{1}{n} h(a_n x) & x \geq n\end{cases}$$
so $g_n$ is continuous and Lipschitz with constant $2a_n/n$. Finally, define $f(x) = \sum_{n=1}^\infty g_n(x)$, where $f$ is continuous since each $g_n$ is continuous, and all but finitely many $g_n$ vanish on $(-\infty, b)$ for each $b$.
We will now show that $f_m \to f$ uniformly, where $f_m(x) = f(x + 1/m)$. Fix $m$, and let $k$ be the smallest index such that $a_k \geq m$. In particular, this means that $m$ divides $a_n$ for all $n \geq k+1$, and thus for each such $n$, $1/m$ is a multiple of $1/a_n$, so $g_n(x + 1/m) = g_n(x)$ for all $x$ not in $(n-1, n)$. Therefore for any $x$, there is at most one $n \geq k+1$ for which $g_n(x + 1/m) \neq g_n(x)$, and this $n$ necessarily satisfies $|g_n(x + 1/m) - g_n(x)| \leq \frac{1}{n} \leq \frac{1}{k}$. It follows that for sufficiently large $m$ we have
\begin{align*}
|f(x + 1/m) - f(x)|
&= \left|\sum_{n=1}^\infty g_n(x + 1/m) - g_n(x)\right| \\
&\leq \frac{1}{k} + \sum_{n=1}^k |g_n(x + 1/m) - g_n(x)| \\
&\leq \frac{1}{k} + \frac{1}{k} + \frac{1}{k-1} + \sum_{n=1}^{k-2} |g_n(x + 1/m) - g_n(x)| \\
&\leq \frac{4}{k} + \sum_{n=1}^{k-2} \frac{2a_n}{n} \cdot \frac{1}{m} \\
&\leq \frac{4}{k} + \frac{2ka_{k-2}}{m} \\
&\leq \frac{5}{k}
\end{align*}
where the last inequality follows from $m \geq a_{k-1} \geq 2k^2 a_{k-2}$, which clearly holds for sufficiently large $k$. Then if we define $k_m$ to be the smallest $k$ for which $a_k \geq m$, we see that $k_m \to \infty$ as $m \to \infty$, so since $|f(x + 1/m) - f(x)| \leq 5/k_m$ uniformly for sufficiently large $m$, it follows that $f(x + 1/m)$ converges to $f(x)$ uniformly.
However, $f$ is not uniformly continuous. Note that $\int_0^1 h(a_nx) \,dx = 1/2$, so for any positive integer $k$, $\int_0^1 f(k + x) \,dx = \frac{1}{2}\sum_{n=1}^k \frac{1}{n}$, and thus there is some $x_k \in (0, 1)$ with $f(k + x_k) \geq \frac{1}{2}\sum_{n=1}^k \frac{1}{n}$. On the other hand $f(k) = 0$ always. But if $f$ were uniformly continuous, there would be some $M$ for which $|f(x) - f(y)| \leq M$ whenever $|x - y| \leq 1$, which clearly does not hold, so $f$ cannot be uniformly continuous.
Original answer:
Too long for a comment:
Whatever the answer is, it might depend on the properties of the sequence $\{1/n\}_{n \in \mathbb{N}}$. Below is an example which shows that if we instead define $f_n(x) = f(x + 3^{-n})$, then the condition that $f_n \to f$ uniformly on $\mathbb{R}$ does not imply that $f$ is uniformly continuous. I was unable to construct a similar example for the sequence $\{1/n\}_{n \in \mathbb{N}}$.
Let $h(x)$ be the "triangle wave" function, such that $h(x)$ is periodic with period $1$ and has $h(x) = 2x$ for $x \in [0, 1/2]$ and $h(x) = 2 - 2x$ for $x \in [1/2, 1]$. In particular, $h$ is continuous and has $h(0) = h(1) = 0$, $h(1/2) = 1$, and $|h(x) - h(y)| \leq 2|x - y|$ for all $x, y$. Next, for each $n$, define $g_n(x)$ so that $g_n(x) = 0$ for $x < n$ and $g_n(x) = \frac{1}{n} h(3^n x)$ for $x \geq n$. With this definition, $g_n$ is continuous, and satisfies $|g_n(x) - g_n(y)| \leq \frac{2}{n} \cdot 3^n|x - y|$ for all $x, y$. Finally, define $f(x) = \sum_{n=1}^\infty g_n(x)$. Note $f$ is continuous since each $g_n$ is continuous, and all but finitely many $g_n$ vanish on $(-\infty, a)$ for each $a$.
Now we will show that $f_m \to f$ uniformly, where $f_m(x) = f(x + 3^{-m})$. For $n \geq m$, $g_n(x + 3^{-m}) = g_n(x)$ so long as $x \not \in (n - 3^{-m}, n)$, meaning $g_n(x + 3^{-m}) \neq g_n(x)$ only in $(n-1, n)$. Thus for any $x$, there is at most one $n$ with $n \geq m$ and $g_n(x + 3^{-m}) \neq g_n(x)$, and this $n$ necessarily satisfies $|g_n(x + 3^{-m}) - g_n(x)| \leq \frac{1}{n} \leq \frac{1}{m}$. Thus we have
\begin{align*}
|f(x + 3^{-m}) - f(x)|
&= \left| \sum_{n=1}^\infty (g_n(x + 3^{-m}) - g_n(x)) \right| \\
&\leq \frac{1}{m} + \sum_{n=1}^{m-1} |g_n(x + 3^{-m}) - g_n(x)| \\
&\leq \frac{1}{m} + \sum_{n=1}^{m-1} \frac{2}{n} \cdot 3^{-(m - n)} \\
&\leq \frac{C}{m}
\end{align*}
for some sufficiently large $C$ not depending on $m$. Thus $|f_m - f| \leq C/m$, so $f_m \to f$ uniformly.
On the other hand, $f$ is not uniformly continuous. Consider $x$ of the form $k + \frac{1}{2}$ for $k$ a positive integer. For $n \leq k$ we have $g_n(x) = \frac{1}{n}$ since $3^n x = 1/2$ modulo $1$, so $f(k + \frac{1}{2}) = \sum_{n=1}^k \frac{1}{n}$. In particular, letting $k \to \infty$ we see $f(k + \frac{1}{2}) \to \infty$, while $f(k) = 0$ always, so $f$ is not uniformly continuous.

By assumption, for every $\epsilon >0$, there is an $n_0 \in \Bbb N$ such that for every $x \in \Bbb R$
$$n \geq n_0 \implies \left| f(x) - f \left(x + \frac{1}{n} \right)  \right| < \epsilon$$Now if you repeatedly substitute $x = \pm \frac{1}{n}$ in the above inequality, we can conclude that for any two rationals $a,b$ of the form(say $*$-type) $\frac{m}{n}$, where $m,n \in \Bbb Z, n \geq n_0$, we have
$$|f(a)-f(b)| < \epsilon$$
Now observe that the rationals of the form $m,n \in \Bbb Z, 0<n < n_0$ are nowhere dense in $\Bbb R$. Therefore rationals of $*$-type are dense in $\Bbb R$(we are using the fact that the rationals are dense in $\Bbb R$). Since $f$ is continuous, we obtain that for any two real numbers $x,y$
$$|f(x)-f(y)| \leq \epsilon$$
But since $\epsilon$ was arbitrary, we can conclude that $f$ must be a constant function which is trivially uniformly continuous.


What is the probability that there is at least one partygoer who ends up selecting his or her own gift? Suppose that there are 10 people in a party. Each person brings along a gift
for exchange. The gifts are put in a pile and labeled with number 1 ~ 10.
Each person in the party will randomly select one gift by picking slips of
papers with numbers identifying the gifts.
What is the probability that there is at least one partygoer who ends up
selecting his or her own gift?
My solution

Consider the complement of P(At least one)
We want to find P(None)=$\frac{9}{10}*\frac{8}{9}*\frac{7}{8}*\frac{6}{7}*\frac{5}{6}*\frac{4}{5}*\frac{3}{4}*\frac{2}{3}*\frac{1}{2}$

I am stuck. Help please!!!

This is called sampling without replacement. You already know that

$
P($at least 1 selects his/her own gift$)=1-P($no-one selects his/her gift$)
$

Since the slips are not returned to pile, you need to account for this, hence

$
P($no-one selects his/her gift$) = \frac{9}{10} \cdot \frac{8}{9} \cdots \frac{1}{2}=\frac{1}{10}
$
Hence,

$
P($at least 1 selects his/her own gift$)=1-\frac{1}{10}=\frac{9}{10}
$

In case sampling was with replacement, i.e. tickets got returned to the pile, this probability converges to (for large number of guests)
$$
\big(1-\frac{1}{n}\big)^n \approx e^{-1}
$$


Intuitive Explanation of Bessel's Correction When calculating a sample variance a factor of $N-1$ appears instead of $N$ (see this link ).  Does anybody have an intuitive way of explaining this to students who need to use this fact but maybe haven't taken a statistics course?

http://en.wikipedia.org/wiki/Bessel%27s_correction

The Wikipedia article linked to above has a section (written by me) titled "The source of the bias".  It explains it via a concrete example.

But note also that correcting for bias, when it can be done, is not always a good idea.  I wrote this paper about that: http://arxiv.org/pdf/math/0206006.pdf

Here's a really simple example:

Say that you have a list of the heights of all the players in a sports league.  Call that your "population" and calculate the variance.  Now pick a sample of three players at random and calculate the variance of your sample.  Chances are that your sample won't pick up the most extreme values, thereby causing the variance of the sample to be lower than the variance of the overall population.

In order to correct for that, you need to adjust your sample variance upward.  That's why you use 1/(n-1) instead of 1/n.  At small values of n (such as 3) the difference between these two will be pretty big.  As your sample size increases, the difference gets smaller.  This is because your sample is more likely to capture the extreme values in the population and therefore your sample variance will approach the population variance.


Solving the logarithimic inequality $\log_2\frac{x}{2} + \frac{\log_2x^2}{\log_2\frac{2}{x} } \leq 1$ I tried solving the logarithmic inequality:
$$\log_2\frac{x}{2} + \frac{\log_2x^2}{\log_2\frac{2}{x} } \leq 1$$
several times but keeping getting wrong answers.

Let $\log_2 x=A$, then $\log_2 x^2=2\log_2 x=2A$ and $\log_2\frac{2}{x}=\log_22- \log_2x=1-A$. So the given inequality becomes:
$$(A-1)+\frac{2A}{1-A} \leq 1.$$
Consequently we get
$$\frac{4A-A^2-1}{1-A} \leq 1.$$
Furthermore you get
$$\frac{5A-A^2-2}{1-A} \leq 0.$$
Hopefully you can solve from here.


Find the directional derivative in the point $p$ in the direction $\vec{pp'}$ Find the directional derivative in the point $p$ in the direction $\vec{pp'}$. In other words, if $\vec{v_0}$ is the unitary vector in the direction of $\vec{pp}$. you need find $D_{\vec{v_0}}f(p)$.

$f(x,y,z)=x^2+3xy+y^2+z^2$ with $p=(1,0,2)$ and $p'(-1,3,4)$

I dont understand the exercise, is confused his statement. Can someone explain me how solve this type of exercise?

Hint

If you don’t know what the gradient is, you should use directly the limit of the directional derivative. Which means

Compute $v_0 = \frac{pp^\prime}{\Vert pp^\prime\Vert}= (a,b,c)$
Compute the limit
$$\lim\limits_{h \to 0} \frac{f(1+a.h,0+b.h,2+c.h)-f(1,0,2)}{h}$$

$$ D_{\vec{v_0}}f(\vec p)= \frac{\vec \nabla f(\vec p)\cdot (\vec {p}' - \vec p)}{||\vec {p}' - \vec p ||}   $$


How can I show $f*g = g*f$ for Lebesgue integral? I want to show that
$(f*g)(x) = (g*f)(x)$ for almost every $x\in\mathbb R^n$ in  Lebesgue integration case, where  * is convolution on $\mathbb R^n$, that is $$
(f*g) (x)=\int_{\mathbb R^n} f(x-y)g(y) dy
$$

Try to substitute $x-y=u$. This makes a trick.

If $G$ is locally compact group then $L^1(G)$ is commutative if and only if $G$ is abelian group.[ Hewitt-Ross, Abstract harmonic analysis, 20.24 theorem]. Now since $R$ is locally compact abelian group so $L^1(R)$ is cummotative.


Proof that if a set of vectors is linearly independent over R the it is linearly independent over C I need to prove this for an exercise but I don't seem to understand why the first implication is true.

$$
Let \hspace{2mm} v_1,...,v_n \in {\Bbb R}^{n}. Prove \ that \hspace{2mm} \{v_1,...,v_n \} \ is \ linearly \ independent \ over \ {\Bbb R} \iff \{v_1,...,v_n \} \ is \ linearly \ independent \ over \ {\Bbb C}.
$$

I was able to prove the second implication. What I did was :

Let's suppose that $\{v_1,...,v_n\}$ is linearly independent over ${\Bbb C}$ then $$0=a_1*v_1+...+a_n*v_n$$
with $a_k=0 \hspace{2mm} \forall \hspace{1mm} 1\leq k\leq n$ and $a_k \in {\Bbb C}$. This is the only possible solution for the $a_k$ over ${\Bbb C}$ and also over ${\Bbb R}$, because ${\Bbb R} \subset {\Bbb C}$.

I'd like you to help me prove the first implication. A hint or the complete proof are ok. Thank you :D

As the set of vectors $v_i, 1\le i\le n$ is LI over $\mathbb R$, so

$\sum_{i=1}^n c_iv_i=0$

$\implies c_i=0\in \mathbb R \forall i$

or  $ c_i=0+i.0\in \mathbb C \forall i$

hence $v_i, 1\le i\le n$ is LI over $\mathbb C$.

For converse part you can proceed backwards.


What is $48\div2(9+3)$? There is a huge debate on the internet on the value of $48\div2(9+3)$.
I believe the answer $2$ as I believe it is part of the bracket operation in BEDMAS. Mathway yields the same answer. I also believe that if $48\div2\times(9+3)$ was asked it would be $288$ which Mathway agrees with as well.
However, WolframAlpha says it is $288$ either way.
A friend of mine (who is better at math) told me that there is no such thing as 'implicit multiplication', only shorthand so that is in fact done after the division (going left to right, not necessarily because division occurs before multiplication. But he didn't explicitly give a reason)
What is the answer and why?

There is no Supreme Court for mathematical notation; there were no commandments handed down on Sinai concerning operational precedence; all there is, is convention, and different people are free to adhere to different conventions. Wise people will stick in enough parentheses to make it impossible for anyone to mistake the meaning. If they mean, $(48\div2)(9+3)$, they'll write it that way; if they mean $48\div\bigl(2(9+3)\bigr)$, they'll write it that way.

There is no order difference between implicit and explicit multiplications. Purplemath suggests that implied multiplication outside of parentheses also gets parenthetical order priority over all other multiplication(division). So they would interpret the implied multiplied parenthetical as $48\div(2\times(9+3)) = 2$.

Alternatively, all implicit-capable calculators I've tried give the same results as Wolfram Alpha which interprets the implication as $(48\div2)\cdot(9+3) = 288$

Some confusion also seems to be the $\div$ division symbol itself as $48/2(9+3)$ visually supports the mutual parenthetical implied multiplication of Wolfram Alpha: $\frac{48}{2}\cdot(9+3) = 288$

The short answer is that the formula as written is too easily misinterpreted and the author should clarify it to ensure its proper calculation.

PS: to further furrow your brow - employing the parenthetical as a variable yields different results. So, $48\div2c$ where $c=9+3$ yields $2$ - but this conflates the distinction from parenthetical to coefficient-variable syntax. $48\div2\cdot c$ where $c=9+3$ yields $288$.


Center of an irreducible faithful representation is cyclic. I read on many references that : given $G$ a finite group which has an irreducible faithful representation, then its center $Z(G)$ is cyclic. I want to prove it over $\mathbb{C}$. Any help ?

The $A$ group of automorphisms of the representation $f:G\rightarrow Gl(V)$ is contained in the group of homothetic map $(h(x)=cx, x\in\mathbb{C})$ since the representation is irreducible (Schur's lemma), you deduce that for every $g\in Z(G)$, $g(x)=c_gx$, $|c_g|=1$ since $G$ is finite. A finite subgroup of $S^1$ is cyclic.

The answer is here : Proof. I will let my post in case of other people ask the question.


Is the complex function $f(z)=e^{-z^{-4}}$ analytic at $z=0$? Is the complex function $$f(z)=\begin{cases}
e^{-z^{-4}} & z\ne0\\
0 & z=0
\end{cases}$$ analytic at $z=0$ ?

I am able to show that the Cauchy-Riemann equations are satisfied.

The other answers are using thermonuclear weapons to conclude...

The function is not even continuous at zero. For example, there exists in $\mathbb C$ a sequence $(w_n)_{n\geq0}$ such that $w_n\to\infty$ and $\exp w_n^4=1$ for all $n\geq0$. It follows that $1/w_n\to0$ as $n\to\infty$ and $f(1/w_n)=\exp(-(1/w_n)^{-4})=1$ for all $n$, and this together with $f(0)=0$ imply that $f$ is not continuous at $0$.

Since you already know the function is continuous, we can conclude directly that it is analytic from Riemann's theorem (about removable singularity).


Direct Proof Writing Technique I've been having this proof writing hangup, and the following example covers the essence of it.

In the beginning of my analysis book one of the exercises is:

" If $a,b \in \mathbb{R}$ and $a + b = 0$, then $b = -a$. "

Here's the method I used -- and frequently use -- to prove it. I take the given information and derive the desired result.

Proof
\begin{align}
a+b &= 0 &&\text{given}\\
(-a)+a + b &= (-a)+0  &&\text{definition of equality}\\
\left((-a)+a\right) + b &= (-a)  &&\text{associativity of addition/zero element}\\
0+b &=-a &&\text{additive inverse}\\
b &= -a &&\text{zero element}\\
\end{align}

But the author uses this other (and much more common) proof style where you start from the left side of the desired result, use some given information, then get the right side.

Proof

$$b = 0+b = (-a + a) + b = -a + (a+b) = -a + 0 = -a$$

While I get the aesthetic, I just don't understand why it wouldn't be done the my way, which seems a lot easier. Is one of these styles more correct than the other, or preferred? And if so, why?

P.S. I tagged this as a soft-question in case the answer is subjective.

The author's style has the benefit of explicitly exhibiting the axioms being used at each step. In fact that is probably the point of the exercise.
Your method leaves out many axioms, esp. in the last step which implicitly implies that $(a+b)-a = b  $ which needs something like commutativity, associativity, inverse, and identity axioms in that order, a sequence of four steps of the kind done by the author.
Your method is probably a good method of discovering the proof, but the author's method is better at showing the reasoning in this case. It is too bad that author's don't take the time to exhibit the difference.
If you claim that you don't need to go thru that axiomatic detail, then you really don't need to do this problem.