What does $e^{\mu}$ mean for a measure $\mu$? I have seen the notation $\int_M fe^{\mu}$ in some geometry books and I cannot even guess what $e^{\mu}$ might mean for a measure/form $\mu$ on the (symplectic) manifold $M$.

Any clarifications are appreciated.

Edit

I found this remark in a book by Guillemin, Ginzburg, and Karshon:

    It is convenient to work with the differential form (of mixed degree)
    $$\exp \omega=1+\omega+\frac{1}{2!}\omega\wedge\omega\dots .$$
    With the convention that $\int_M\beta=0$ if the degree of $\beta$ is different than the dimension of $M$, Liouville measure is given by integration of $\exp \omega$.

This shows that my guess in the above comments was correct! However, I would appreciate if an expert would clarify why it is "convenient" to work with this differential form of mixed degrees.

If $e^{\mu}=d\lambda$ for some measure $\lambda$ in $M$ then, it seems that  $\mu=\ln(d\lambda)$. But here is not clear whether $\ln(d\lambda)$ is a measure.


calculate the Probability density function of the absolute difference of two random variable If $X$ and $Y$ are two independent random variables with probability density functions $f$ and $g$, respectively, then the probability density of the difference $Y − X$ is given by the cross-correlation . In contrast, the convolution f * g gives the probability density function of the sum $X + Y$.

what i would like to have is the probability density functions of $|y-x|$ or $(y-x)^2$. Is that possible ?

is that the equivalent as doing the following with two histogram ?

for (bin b1 in Histogram1)
  for (bin b2: Histogram2)
        prob = b1.probability * b2.probability
        distance += prob * abs(b1.value - b2.value)

As you said, the density of $X-Y$ is the cross-correlation $h(t) = \int_{-\infty}^\infty f(t+y) g(y) \, dy$.  The density of $|X-Y|$ is $f_{|X-Y|}(t) = h(t) + h(-t)$ for $t > 0$, $0$ for $t < 0$.  The density of $(X-Y)^2$ is $f_{(X-Y)^2}(t) = (h(\sqrt{t}) + h(-\sqrt{t}))/(2 \sqrt{t})$ for $t > 0$, $0$ for $t < 0$.

It is a very big question.

Yes you can find the pdf of any combination of variables, there are many ways.

The one which works most of the time use the characteristic function:

http://en.wikipedia.org/wiki/Characteristic_function_(probability_theory)#Basic_manipulations_of_distributions

You have many complex and exotic tools to exploit the properties of CF.

Anyhow this process greatly depends from the type of variable. E.g.: It is much easier with exponential variables than with non exponential variables.


You cannot separate out the $\cos$ function as you have done in step two.

You can remember this identity.

$$\cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B)$$

Here $\arccos(\frac{3}{5})$ is an angle ( i.e., approximately $53$ degrees)

Using thus result you should get the desired answer.

Cos^-1(3/5) is 37°. So we get cos(37°+60°) which is equal to -0.39 same as the term you have mentioned as the answer.


Numerical Solution of $\frac{x}{1-e^{-x}} -5 = 0$ I am working on a problem at the moment which cuts down to the following question:

How do I get a numerical solution for:

$$\frac{x}{1-e^{-x}} -5 = 0?$$

I've been thinking about using Newton's method here but I am not quite sure how to get a reasonable starting point. Is there any other approach or maybe even just a method that provides me a suitable starting point?

Thank you very much!
FunkyPeanut

Without getting into the theory, you can often setup a fixed point iteration in lieu of the more powerful Newton's method. What I mean is consider the sequence
$$ x_{n+1} = 5(1-\mathrm{e}^{-x_n}). $$

If it converges, it will converge to your answer. And we know from another comment that $x\approx 5$, so that makes an excellent starting guess -- $x_0=5$. Here are the first few terms in the sequence:
$$ \{5, 4.96631, 4.96516, 4.96512, 4.96511\} $$

After only 4 iterations (easily done in a calculator, btw), we have to 5 decimal places the correct answer. Not bad.

It's easy to write it in another way

$$x-5=-5\cdot e^{-x}$$

and draw two graphs. Then you see, that you have two solutions, first near $x\in[-1,0]$, second near $5$.


Prove $ \{(p \lor q) \land (p \implies r) \land (q \implies r) \} \implies r$ is a tautology using logical properties I spent quite a bit of time on this and have little to no ideas on how to proceed.

Using the conditional laws and De Morgan's law, I got to
$$( \sim p \land \sim q) \lor (p \land  \sim r) \lor(q \land \sim r) \lor r$$.

Hint: the path is clearer if you leave the $\neg(p\vee q)$ as is when you use de Morgan's Law on everything else.

$\begin{array}{l|ll}
1: & \big((p\vee q)\wedge (p\to r)\wedge (q\to r) \big) \to r
\\ & \Updownarrow \text{Implication equivalence, thrice}
\\ 2: & \neg\big((p\vee q)\wedge (\neg p\vee r)\wedge (\neg q\vee r) \big) \vee r
\\ & \Updownarrow \text{de Morgan's Negations}
\\ 3: & \neg (p\vee q)\vee (p\wedge \neg r)\vee (q\wedge\neg r) \vee r
\\ & \Updownarrow \vee \text{ Idempotent Introduction, Commutivity}
\\ 4: & \neg (p\vee q)\vee (p\wedge \neg r)\vee r\vee (q\wedge\neg r) \vee r
\\  & \Updownarrow \text{Distribution, Complementation, }\wedge\text{ Identity: } (a\wedge\neg b)\vee b= a\vee b
\\ 5: & \neg (p\vee q)\vee (p\vee r)\vee (q\vee r)
\\  & \Updownarrow \text{Association, Commutation, }\vee\text{ Idempotence }
\\ 6: & \neg (p\vee q)\vee (p\vee q) \vee r
\\  & \Updownarrow \text{Complementation: } A\vee \neg A=\top
\\ 7: & \top \vee r
\\  & \Updownarrow \vee \text{ Annihlation: } \top\vee r=\top
\\ 8: & \top
\end{array}$

At this point in time you can't prove this as a tautology.  Tautologies get defined as well-formed formulas.  It is not clear if you mean the tautology (in Polish or Lukasiewicz notation)

C(K(K(A(p,q),C(p,r)),C(q,r)),r).

or if you mean the tautology

C(K(A(p,q),K(C(p,r),C(q,r))),r).

The proofs of those well-formed formulas are distinct.


If the derivative is 0, prove $f$ is constant without connectedness Given a continuous function $f$: U $\to\mathbb{R}$ where U is an open subset of $\mathbb{R}^n $and suppose that $\nabla $$f$ =0, it is well known that if U is connected then $f$ is constant, and connectedness of U is necessary, for example, if U=(0,1)$\cup $(1,2), define $f$ to be 1 on (0,1) and 0 on (1,2), then$ \nabla $$f$=0 on U but $f$ is not constant. My question is that without the connectedness of U and instead if we add another condition that $f$ is 0 on$ \partial$U and that $f$ is continuous on $\bar{U}$, does $f$ constant on U now? I think this is correct and managed to give a proof of this and I want to know if my result and the proof is right. Any comment is appreciated and thank you all in advance!
Here is my proof:
we first suppose U is connected then $f$ is constant on U,say, $f$$\equiv a$. Since $f$ is continuous near $\partial$U, $a$ must be 0. Next observing that U is locally connected, the boundary of a connected component is in fact contained in the boundary of U.Thus $f$ is 0 on the boundary of a connected component of U, so by the first step, $f$ is 0 on a connected component. Finally, if U is not connected, we can write U as a union of its connected components, and since $f$ is 0 on every connected component of U,  $f$ is 0 on U.

One objection could be that if $U$ does not have a boundary. Think of $(0,1) \cup (2,3)$ as a subset of the topological space $(0,1) \cup (2,3)$. Then you counterexample works, but your proof does not. Although
I admit it is a bit pathological.


Complete the square in the form $(px+q)^2+r, p > 0$ I'm going over some completing the square questions and I need to express, in the form:
$(px+q)^2+r, p > 0$

the quadratic equation is $16x^2-8x+11$

I know how to get it in the form $p(x+q)^2+r$

So can someone show me how to get from $ax^2+bx+c$ to $(px+q)^2+r, p > 0$


$x^{x^{x^{\dots}}}=2$, what is the value of $x$? I came up with this little simple exercise, stating:

$x^{x^{x^{\dots}}}$ infinite times is equal to $2$, find $x$.

As we're dealing with infinity, we can just separate the first $x$ and get $x^2=2 \implies x=\sqrt2$

But then I thought what if we change $2$ to $4$? Now we get $x^4=4 \implies x=\sqrt[4]4=\sqrt2$

Something went wrong, can someone explain?

One way to define the expression is $$y = x^{x^{x^{\cdots}}} $$

$$\Rightarrow y^x = x$$

A slightly different definition is to define the sequence $a_0 = x$ and $a_n = x^{a_{n-1}}$. The problem is then to find $$\lim_{n\rightarrow \infty} a_n$$

It can be shown that the interval of convergence for the power tower is $$[e^{-e}, e^{1/e}]$$


Combinatorics problem involving series summation Can you give a combinatorial proof of $1\cdot1! + 2\cdot2! ....+ (n-1)(n-1)! = n!-1$

All permutations of $\{1,2,\ldots,n\}$ objects (except the identity) can be obtained first swapping $k$ with its desired value and then permuting the lower $k-1$ objects, where $k$ (with $2\le k\le n$) is uniquely determined as the highest non-fixed position.


A cat chases a rat. For every $5$ leaps of the rat , the cat takes $3$ leaps problem A cat chases a rat. For every  $5$ leaps of the rat , the cat takes $3$ leaps, but the  $2$ leaps of the cat are the same as $3$ leaps of the rat. Compare the speeds of the cat and the rat

a)$4:5$

b)$12:13$

c)$16:15$

d)none of these

solution

clearly the  $2$ leaps of the cat are the same as $3$ leaps of the rat

so let distance of rat leap be $x$

then distance of cat's leap =$ \frac {3x}{2} $

then speed ratio = $ \frac {4.5x}{5x} $

$ \frac {9}{10} $

Am I right $?$

Quicker an simple
Assume 2 leaps of cat = 3 leaps of rat=6meter(lcm of 2;3)
NowNow 3 leap of cat is 9 m
Rat 5 leap is 10 meter
Ratio = distance /time
Time is same
So ratio 9/10


If $\left( {\begin{array}{*{20}{c}} A & B \\ B & A \\ \end{array}} \right) \ge 0 \Rightarrow A \ge B$ Let $A,B\in M_n$ and $\left( {\begin{array}{*{20}{c}}
   A & B  \\
   B & A  \\
\end{array}} \right) \ge 0$

Why does $A \ge B$ ?

Let $x$ be given. Then
$$
0\le \pmatrix{x\\-x}^T\pmatrix{A&B\\B&A}\pmatrix{x\\-x}
 = 2x^TAx - 2x^TBx,
$$
This implies
$$
x^T(A-B)x\ge0
$$
for all $x$, hence $A\ge B$.

If you take Determinant of the above matrix, it becomes,

$A^2-B^2 \ge 0$

Square of any number is positive and to yield the number greater than zero, A should be always greater than B.

Hence $A\ge B$


Entire function assume real values for $z=x^2+ix$ Let $f$ be an entire function such that $f(z)$ is real for $z=x^2+ix$.
Is there exists such a function which is not constant?

Previously I thought that  $\displaystyle\int_{0}^{z}\left(\sqrt{t-1/4}-i\right) dt \ $ would work but I realized that this function isn't continous so I cannot guarantee that this integral is entire.

So, to reach at this answer I saw some facts:

1)$\cos(\sqrt{z})$ is an entire function - we can see this intuitively by the fact that cosine is even and with this we avoid the problem to define square root in $(-\infty,0]$ - which is a question here on Stack Exchange

2) $\cos2 \pi ix=\cosh2 \pi x$

So, to conclude I used what I think that was expected to be used, $x^2 +ix-1/4=(x+i/2)^2$

$f(z)=\cos2 \pi \sqrt{1/4-z}$ is a entire function and for $w=x^2 +ix$ we have
$f(w)=\cos2 \pi \sqrt{-(x+i/2)^2}=\cos2 \pi i(x+i/2)=\cos(2 \pi ix -\pi)=-\cos2 \pi ix=-\cosh2\pi x$ which is a real number for every $x\in \mathbb{R}$

First we notice that the area above the given parabola is mapped eithe to the lower half plane or the upper half plane. Using conformal mapping we see that f(g) takes the upper half plane to itself with boundary to boundary. This needs not be constant for example z^3.


Geometry - length of triangle question [![enter image description here][1]][1]
Could anyone please help me with this question
I can't seem to solve for the ?.
The answer provided for ? is 1.
Thanks!

When you see concurrent lines, Ceva's Theorem and Menelaus's Theorem are often the intended way.

Consider the grey triangle, applying Ceva's Theorem yields $$\frac{3}{x}\cdot\frac{a}{b}\cdot\frac{d}{c}=1$$
Again, consider the grey triangle with the black transversal line, applying Menelaus's Theorem yields $$\frac{5+x}{2}\cdot\frac{a}{b}\cdot\frac{d}{c}=1$$
Equating both equations gives $x=1$.

PS: I'm not sure why using other lines and triangles (for instance, the line between the line segments with lengths $3$ and $x$) doesn't work. If anyone would like to expand on this answer, please do. Thanks in advance!

Let us guess $\Delta ABC$  is an equilateral triangle with side length of 6.  F and D are the mid points of AC and BC respectively, CD=DB=AF=FC=3. Find a point E between D and B with  DE=1.
If we can prove  EF,AD and CH  intercept at the same point G, then   DE=1 is the solution.
Set some coordinates:
A(0,0),B(6,0),C(3,3$\sqrt3$), D(4.5,1.5$\sqrt3$), E(5,$\sqrt3$),F(1.5,1.5$\sqrt3$)
AE: y=$\frac{\sqrt3}{5}x$ (I)
AD:$\frac{\sqrt3}{3}x$  (II)
FB:  $-\sqrt3/3x$+2$\sqrt3$ (III)
EF:    $-\sqrt3/7x$+12$\sqrt3/7$ (IV)
Solve (I) and (III) to get the intercept point of AE and FB at H($\frac{15}{4},\frac{3\sqrt3}{4}$), then
HC:   $-3\sqrt3x+12\sqrt3$  (V)
Solve (II) and (V) to get the intercept point of AD and HC  at G$_1$(3.6,$\frac{6\sqrt3}{5}$)
Solve (II) and (IV) to get the intercept point of  AD  and EF intercept at G$_2$(3.6,$\frac{6\sqrt3}{5}$)
We  proved AD,EF and HC intercept at the same point  $G_1=G_2=G$.
Now  one can conclude the answer is indeed  DE=1.


Prove that if $n$ is a positive integer then $2^{3n}-1$ is divisible by $7$. I encountered a problem in a book that was designed for IMO trainees. The problem had something to do with divisibility.

  Prove that if $n$ is a positive integer then $2^{3n}-1$ is divisible by $7$.

Can somebody give me a hint on this problem. I know that it can be done via the principle of mathematical induction, but I am looking for some other way (that is if there is some other way)

Hint: Note that $8 \equiv 1~~~(\text{mod } 7)$.
So,
$$2^{3n}=(2^3)^n=8^n\equiv \ldots~~~(\text{mod } 7)=\ldots~~~(\text{mod } 7)$$
Try to fill in the gaps!

Solution:
Note that $8 \equiv 1~~(\text{mod } 7)$. This means that $8$ leaves a remainder of $1$ when divided by $7$. Now assuming that you are aware of some basic modular arithmetic,
$$2^{3n}=(2^3)^n=8^n\equiv 1^n ~~(\text{mod } 7)=1~~(\text{mod } 7)$$
Now if $2^{3n}\equiv 1~~(\text{mod } 7)$ then it follows that,

$$2^{3n}-1=8^n-1\equiv (1-1)~~(\text{mod } 7)~\equiv 0~~(\text{mod } 7)\\ \implies 2^{3n}-1\equiv 0~~(\text{mod } 7)$$
Or in other words, $2^{3n}-1$ leaves no remainder when divided by $7$ (i.e. $2^{3n}-1$ is divisible by $7$). As desired

Use Proof by induction.It's easy to show that if this holds for "n" ,then it holds for "n+1" and of course it's true for n=1


Difference between $\lim_{n \to + \infty} f(n)$ and $\lim_{x \to + \infty} f(x)$ Let $n \in \mathbb{N}$ and $x \in \mathbb{R}$. Let $f \in \textbf{C}^{0}(\mathbb{R},\mathbb{R})$ and suppose its limit exists in $+ \infty$.
I know that if $\lim_{x \to +\infty} f(x) = l$, therefore $\lim_{n \to +\infty} f(n) =l$ (it follows from the the sequential characterization of limits). But is it true that if $\lim_{n \to +\infty} f(n) =l$ therefore $\lim_{x \to +\infty} f(x) = l$? What if $f$ isn't supposed continuous anymore?

The reverse implication is not true even if $f$ is continuous. Consider
$$f(x) =\cos 2\pi x$$
Then the integer limit is $1$ but there's no limit when you quantify over all reals.

The statement $\lim_{n \to +\infty} f(n) =l\implies\lim_{x \to +\infty} f(x) = l$ is not true in general, when it holds we need some bounding trick to prove it.

As a classical example from

$$a_n=\left(1+\frac1n\right)^n \to e$$

since for any real $x\to \infty, \quad \exists n \quad x\in[n,n+1)$, then we conclude that also  $f(x)=\left(1+\frac1x\right)^x\to e$ indeed

$$\left(1+\frac1{n+1}\right)^n \le \left(1+\frac1x\right)^x\le \left(1+\frac1n\right)^{n+1}$$

and both RHS and LHS tends to $e$.


What are subspaces of $A$? Can anyone please tell me what are the subspaces of $A$ ?

My Thoughts:  $\{\alpha I + \beta J : \alpha , \beta \in \mathbb R , \beta > 0 \}$ is not a vector space and it does not contain any  non-zero vector spcae as well.
Can anyone help me ?

$A$ is not a set, it's a matrix. Asking for the "subspaces of $A$" probably means the eigenspaces associated to each eigenvalue of $A$. (The question should be clearer about that)

The eigen values of A are $\alpha\pm i\beta$. Since beta>0, the eigen values are distinct and complex conjugates and hence A is invertible. So kernel is trivial and range is whole codomain.


Does the sequence $\{\sin(en)\}$ converge or diverge? Is it known if $\{\sin(en)\}$ converges or diverges?

Also, I have a more general question.  For almost every rational $r$, I think we can say that $\{\sin(rn)\}$ diverges.  Does that statement hold if we only assume $r$ is a real number?

The sequence $(\sin rn)_{n\in\mathbb N}$ converges if and only if $r$ is a multiple of $\pi$ (in which case it is the constant sequence $0,0,0,\ldots$).

If $r$ is any rational-but-not-integral multiple of $\pi$, then the sequence is periodic but not constant, and therefore doesn't converge.

If $r/\pi$ is irrational, then values in the sequence are dense in $(-1,1)$ -- in particular there are infinitely many elements in $(-1,-\frac12)$ and also infinitely many elements in $(\frac12,1)$, so it cannot converge in this case either.

It does not convergence for all $e$ and therfore diverges. We can easily see this if we pick $e = \frac{\pi}{2}$. For $n \in \mathbb{N}$ we now see that $\sin(\frac{\pi}{2} n) =  0$  if $n$ is even. While if $n$ is odd, this is either $-1$ or $1$. This is not convergent and thus divergent.


Calculate the largest possible floating-point value: formula? If I have a 32-bit representation of a floating-point where 1 bit is the sign, 8 bits are the exponent, and 23 bits make up the mantissa. The exponent notation is Excess-127.

32=1+8+23

And, where the "1." of the mantissa is implied and not stored in the binary sequence.

It makes sense to me to calculate the largest, possible, positive floating-point using this formula:

(implicit "1." value) + (largest possible bit sequence) × (largest possible exponent)

... which in this case is:

(1) + ( { 2^(23) - 1 } x {2^128} )

The textbook I'm studying from uses this formula:

+ { 1 - 2^(-24) } x 2^(128)

The only part; in this formula; that makes sense to me, is 2^(128). I don't understand why they are using 2^(-24). I don't understand why they are subtracting 2^(-24) from 1 either.

Is the formula correct? And if so, how does it work?
If anyone could point me to resources that explain this I would really be grateful.

Your formula is slightly off.  Your expression (2^23)-1 represents the 'all 1s' of the maximum mantissa - but you have forgotten that this is a fraction.  To make it a fraction you must scale it by a factor of 2^-23 before adding 1.  If you work out the math, you will see that then it will be equivalent to the textbook formula.

I also believe you and the textbook are both wrong about the exponent - it should be at most 2^127 (assuming the 8-bit value is signed twos-complement, which has the value range [-128..127]).


How to express the meaning I mention in one formula? There are two points: $(x_1,y_1),(x_2,y_2)$, if $|y_2-y_1|>|x_2-x_1|$ then  $\tan(A)=\frac{|y_2-y_1|}{|x_2-x_1|}$  else if $|y_2-y_1|<|x_2-x_1|$ then  $\tan(A)=\frac{|x_2-x_1|}{|y_2-y_1|}$

My question is, how to express the above meaning in one formula? what kind of Mathematical symbol should I use? thanks.


Could someone remind me why is incorrect to switch an infinite sum and an integral? Could someone jog my memory on this?

The order of operation between an $\int$ and $\sum_{n\in \mathbb{N}}$ is not always interchangable? Note that the sum is an INFINITE sum

Why is it that $\int \sum_{n \in \mathbb{N}} \neq \sum_{n \in \mathbb{N}} \int$

Is the reason because the integral itself is a sum and the order of "summing" actually matters? (I think it's Multivariable calculus related stuff now)

It is easier and more instructive to give a counterexample using sequences: For example $$\lim_{n\to\infty}\int_0^1 n^2x^n(1-x)\,dx=1$$ even though the integrand goes to zero everywhere in [0,1].
To understand what is going on here, note that the function in the integrand has a graph which is a tall, thin peak getting taller and taller and thinner and thinner as $n\to\infty$, while approaching $x=1$ from the left.

Taking differences, you can easily realize the sequence as partial sums of a series, thus providing the counterexample you seek. To be precise, consider $$\lim_{n\to\infty}\int_0^1\sum_{k=0}^{n-1} \bigl((k+1)^2x^{k+1}(1-x)-k^2x^k(1-x)\bigr)\,dx,$$ which is just a difficult way to write the limit above.

Switching an limit and an infinite sum constitutes the interchange of limit processes.  Said interchanges often yield unexpected results.


Does $f:\mathbb{R}\to\mathbb{R}$ mean that $f$ maps to all reals? This is a quick question. When people write $f:I\to J$ for instance, does $J$ need to be the range of $f$ or can it be any set containing the range of $f?$ For example, is $g(x)=\pi$ an $\mathbb{R}\to\mathbb{R}$ function?

Usually, $f:A\to B$ means the domain of $f$ is $A$ and $\forall a\in A$, $f(a)\in B$. The map does not have to be injective or surjective. The range of $f$ is usually denoted as img$(f)$ or the image of $f$, which in this case is a subset of $B$.

condensed from comments received :

$$f:X \to Y \equiv \forall x \in X \quad \exists y \in Y \text{ such that } f(x)=y$$
$$f:X \to Y \land f_\text{ onto/surjective}   \equiv  \forall y \in Y, \, \exists x \in X  \text{ such that } f(x)=y$$
$$f:X \to Y \land f_\text{ 1-1/bijective}   \equiv \forall x \in X \quad !\exists y \in Y \text{ such that } f(x)=y$$


Finding remainders using modulo Determine the remainder of $2014^{2015} \cdot 2016^{2017} + 2018^{2019}$ divided by 13.

I can't figure out how to manipulate the 2018 part to get it to some form of 13. Any suggestions?

2014 is $(-1)$ modulo 13. 2016 is 1 modulo 13. 2018 is 3 modulo 13, but then $2018^3$ is $3^3$ which is 27 which is 1 modulo 13. After this, you can work with the powers very easily, just remember to take the third power of 2018 into account.


Does a Galois group being $S_3$ correspond to the extension being the splitting field of a cubic? If $f(x)$ is an irreducible cubic, then $\operatorname{Gal}(f(x))\cong S_3$ or $A_3$. But what about the converse? That is, if $\operatorname{Gal}(K/F)\cong S_3$, is it necessarily true that $K$ is the splitting field of some irreducible cubic in $F[x]$?

Yes. To see this, note that if $K/F$ is a Galois extension, then by the Fundamental Theorem of Galois Theory there is an intermediate subfield $L$ with $[L : F] = 3$, corresponding to the fixed field of $\langle
(1,2) \rangle$. By the Primitive element theorem, which in particular holds for any Galois extension, $L = F[\theta]$ for some $\theta \in L$. Let $p(x)$ denote the minimal polynomial of $\theta$. Then $degree(p(x)) = 3$.

Now let $r \in K$ be another root of $p(x)$. If $p \in L$, then $L/F$ would also be a Galois extension--which contradicts the fact that the group $\langle
(1,2) \rangle $ is not a normal subgroup of $S_3$. Thus claim $r \notin L$, which gives that $K$ is the splitting field of $p(x)$.

A quick addition to the post that I thought will be useful for someone else who need a little more clarification:

$K$ being a Galois extension implies that it is Separable.
Since $p(x)$ has a root, $\theta$ in $L\subset K$, it says that $p(x)$ should linearly split in $K$ (By definition of  separability ).
Notice that for a splitting field, having all the roots is not enough. It should also be generated by those roots

Now as it is proved that other remaining two roots cannot be in the intermediate field, and from Galois correspondence as we know that there are no intermediate fields between $L$ and $K$, it must have that $F[\theta,r,s]=K$


Resolution on 3-SAT instance yields in polynomial many resolvents A SAT instance in CNF with $n$ variables has at most $2^n$ resolvents, therefore the resolution method is not in polynomial time.

Considering a 3-SAT instance, we have at most $n^3 + n^2 + n$ many resolvents. Therefore, we can decide 3-SAT in polynomial time by resolution.

I am sure, my second statement is wrong, because otherwise we would have $P = NP$. But I can not see my mistake. Can someone bring clarification here?

There are only that many clauses but that only gives an upper bound for the time to test a single candidate solution.  There are still $2^n$ candidates to test.

Using some algorithms like DPLL or CDCL SAT solvers leads to much less than $2^n$ candidate testing but still an exponential function.

Also, in local search algorithms best bounds are about $1.333^n$ by Schöning's 1999 paper.


Why is tangent infinite sometimes? I am confused by the definition of tangent. The math textbook says it is because when cosine is 0 tangent becomes infinite. The intuition I have is that when one of adjacent or opposite is 0, then tangent is infinite. But isn't cosine only adjacent over hypotenuse? So is tangent theta only infinite if adjacent side is 0? What if adjacent side is theoretically infinite?

You can imagine the tangent in the unit circle as a line that tangency it in the start: $n\cdot 2\pi : n \in R$

That animation will give you the intuition for the asymptotes (where it goes to negative and positive infinity):

You can see that when you go the the angle $\frac{\pi}{2}$ for example, you are in a point in the circle, that creating a line from the origin to it, you have the same inclination as the tangent line, and that's why your graph is tending to $\infty$ and you get that asymptote in the graph.

Actucally the tangent to a curve means the value  of dy/dx i.e. tan() at some pt (x,y) on the curve so when the denominator in the equation of dy/dx becomes zero the tangent is infinite of tan() is infinite , if signifies that curve is straight line at that point.


Can this gravitational field differential equation be solved, or does it not show what I intended? This is the equation I'm having trouble with:

$$G \frac{M m}{r^2} = m \frac{d^2 r}{dt^2}$$

That's the non-vector form of the universal law of gravitation on the left and Newton's second law of motion on the right. I assume that upon correctly modeling and solving this, I will have a function of time that gives the distance from a spherical mass in space (e.g. distance from the Earth from an initial condition of $r(0) = 10,000 \mathrm{km}$).

However, WolframAlpha gives a hell of an answer, which leads me to believe that I'm modeling this equation completely wrong. Can anyone shed some light on this problem?

Yes you're modeling it wrong. Gravity is an attractive force, so if the mass is at r = 0, and r is nonnegative, then the gravity force should point towards zero, i.e. negative.

$$ - \frac{GMm}{r^2} = m\ddot r$$

(This assumes $m\ll M$ so we don't need to consider reduced mass. See the other answers for the more general case.)

But now Alpha even refuses to solve it. That's fine, because Mathematica can't solve many things. The usual approach is to notice that

$$ \frac{d^2r}{dt^2} = \frac{dv}{dt} = \frac{dv}{dr}\frac{dr}{dt} = v \frac{dv}{dr} $$

to convert that 2nd-order ODE to a 1st-order ODE:

$$ -GM \frac{dr}{r^2} = v \,dv $$

This gives the solution v(r).

$$ \frac{GM}r + \text{constant} = \frac{v^2}2 \qquad (*) $$

But recall $v = dr/dt$. So we have another 1st-order ODE to solve, which gives r(t). The solution should look like that “hell of answer” because of the square root.

Note: If you rearrange the terms you should see that Equation (*) is just conservation of energy.

If you model gravity correctly, that is with negative acceleration, the answer to this is very simple:
is just the "CUBIC ROOT OF t SQUARED", up to suitable constants.

This solution describes Motion with Parabolic Velocity ( ie. zero speed at infinity ), the other two speeds are given by the general solution.


Checking if an element of tensor product is zero I am trying to understand tensor product.

Let $R$ be a ring and $R_0$ its subring. Then $R$ is a right $R_0$-module. Let $M$ be a left $R_0$-module.

  Is the element $1\otimes m$ of $R\otimes_{R_0} M$ nonzero for a nonzero $m\in M$?

I tried to express $(1,m)$ of $R\times M$ as a linear combination of elements of the form
$(r_1+r_2,m)-(r_1,m)-(r_2,m)$,
$(r,m_1+m_2)-(r,m_1)-(r,m_2)$,
$(rr_0,m)-(r,r_0m)$
or to see that such an expression is not possible, but failed at both.

I know how to prove this in the affirmative for vector spaces (if $R_0$ is a field) by choosing a basis, and I also know that in some tensor products $1\otimes m$ can be zero, e.g. in  $\mathbb Z_m\otimes \mathbb Z_n$ for $(m,n)=1$, but it does not help me to solve this problem.

Not necessarily! The tensor product $\mathbb Q\otimes_{\mathbb Z}\mathbb Z/n\mathbb Z$ is zero.

The element $1\otimes m$ is $0$ precisely if $m$ is annihilated by an element of $R_0$. Thus the answer to your question is yes if and only if $\operatorname{Ann}_{R_0}(M)\not=0$.


Why does Euler's formula have to be $e^{ix} = \cos(x) + i\sin(x)$ In part one of this youtube video the uploader goes on to explain the calculus proof for Euler's Formula.

The Formula
$$e^{ix} = \cos(x) + i\sin(x)$$
Differentiate
$$ie^{ix} = f'(x) + i g'(x)$$
Multiply original formula by $i$
$$ie^{ix} = if(x) - g(x)$$
Equate the differentiation and the multiplied version
$$f'(x) + ig'(x) = if(x) - g(x)$$
Equate real and imaginary (and cancel the i)
$$f'(x) = -g(x) \qquad g'(x) = f(x)$$

Then he goes on to explain $f(x) = \cos(x)$ and $g(x) = \sin(x)$. My question is why can't $f(x) = \sin(x)$ and $g(x) = -\cos(x)$? Can further proof be added to this proof to eliminate $f(x) = \sin(x)$ and $g(x) = -\cos(x)$?

Because we know the initial condition $e^{i0}=1$ holds. As with most differential equations, there's an family of answers that you need to use the initial condition to find the correct one for.

Let z=cos(x)+isin(x)
When you 8derive it (in terms of x),
 z'=-sin(x)+icos(x)
And since we are in the complex world, -1=i×i
=>z'=i×isin(x)+icos(x)=i(cos(x)+isin(x))=iz
This pretty much makes it a first order differential equation.
After dividing by z and multiplying by dx (z'=dz/dx) both on both sides  and integrating both sides, you get:
ln(z)=ix+c
ln(cos(x)+isin(x))=ix+c
If we set x=0, cos(0)=1,sin(0)=0 and ln(1)=0
=> ln(1)=c=>c=0
=> ln(cos(x)+isin(x))=ix
e^ix=cos(x)+isin(x).
Hope this helps.


Ranking athletes by nationality In an international track competition, there are five United States athletes, four Russian athletes, three French athletes, and one German athlete. How many rankings of the 13 athletes are there when

b) Only nationality is counted and all the U.S athletes finish ahead of all the Russian athletes?

Would the answer by C(13, 9) * P(8; 4, 3, 1)?

HINT: Rankings for this competition are simply arrangements of the 13 athletes. Since only nationality is counted, there are $\frac{13!}{5!4!3!}$ possible rankings (and 13 possible ranks). There are $5!$ possible rankings among the Americans, and $4!$ among the Russians. How can you find the number of possible rankings if all the U.S. athletes beat all the Russian athletes?       My advice is to simplify the problem and then generalize the mechanism (this helps with a lot of combinatorics problems). In other words, change the problem so that you have, say, 7 possible ranks, and fewer athletes. And think of the ranks as boxes into which you can put the athletes, subject to the constraint that some of them (the American ones) are always ahead of others (the Russian ones). After you do this, you should discover how it works, and you can then apply it to your original problem.

Think of it as the number of ways of arranging 13 balls, 3 of which are green, 5 are red, 4 are blue and 1 is white, such that none of the blue balls come before any red ball.

this is 4[C(8,4) + 5.C(7,4) + C(6,4).C(6,4) + C(7,3).C(5,4) + C(8,4)] .


Proof that $x_1=x_2$ maximizes $f(x_1)+f(x_2)$, when its derivatives are unequal until $x_1=x_2$. Given some function $f: \mathbb{R} \rightarrow \mathbb{R}$. define $g: \mathbb{R}^2 \rightarrow \mathbb{R}$ as

$$g(x_1,x_2)=f(x_1)+f(x_2)$$

Assume that $f$ is decreasing in argument. Define $X$ as the sum of $x_1+x_2$. $X$ is assumed to be some finite constant. Hence, whenever $x_1$ increases, the value of $x_2$ decreases by the same amount. Claim: $g(x_1,x_2)$ is maximized when $x_1=x_2$. I argue this using the following fact (that I know independently to be true):

$$\frac{\partial f(x_1)}{\partial x_1}>\frac{\partial f(x_2)}{\partial x_2} \iff f(x_1)>f(x_2).$$

Which (I believe) states that increasing $x_1$ and decreasing $x_2$ increases  $g(x_1,x_2)$, whenever $f(x_1)>f(x_2)$. Note that $f(x_1)>f(x_2)$ $\iff$ $x_1< x_2$ since $f$ is decreasing in its argument. Hence, $g(x_1,x_2)$ can be increased by transferring value from $x_2$ to $x_1$ until $f(x_1)=f(x_2)$. Conversely, I also know that

$$\frac{\partial f(x_1)}{\partial x_1}<\frac{\partial f(x_2)}{\partial x_2} \iff f(x_1)<f(x_2).$$

And we can increase $g$ by transferring value from $x_1$ to $x_2$ whenever $x_1>x_2$. Since this holds everywhere for $g(x_1,x_2)$ whenever $x_1 \neq x_2$ we can increase $g$ by transferring value between $x_1$ and $x_2$. Hence $g$ is maximized when $x_1=x_2$. Does this argument work?

A thought rather than an answer:

This looks as if it's closely related to the standard ideas of an optimum on the production-possibility curve in Economics. The proofs usually given in textbooks implicitly assume that the PP curve is convex, so that there's a unique maximum. You may be making a similar assumption here. As written, I don't actually believe the claim you're trying to prove, although I confess I haven't read it carefully -- it's too muddled for my taste.


Rewrite formula in terms of This is a homework question but I don't want the answer. I just want to know if I'm right or wrong to see if I understand what I'm doing.

Need to rearrange this in terms of d

$a = b \times c + b \times d$

My answer is:
$\frac{(a - bc)}{b} = d$

Is that correct?

If not any hints or resources would help.  I've used KhanAcademy which seems good.

Thanks!

To close this question off with an answer
$$
a = bc +bd = bd + bc
$$
so lets try to make $d$ the subject by bringing everything to one side with the exception of the terms with $d$ in it.
$$
a - bc = bd + bc - bc = bd + 0 = bd
$$
now we have a term on the right hand side $b\times d$, and what is the opposite of multiplication? it is division so we can divide both sides by $b$
$$
(a-bc)\div b = (b\times d)\div b = d
$$
and then you are done
$$
\frac{a-bc}{b} = d
$$
now this is the result you require (and you correctly obtained so well done!!)
Now from my experience some students have an issue of seeing that
$$
\frac{a-bc}{b} = d
$$
is the same or equivalent to
$$
d = \frac{a-bc}{b}
$$
(just in case)


Long Inequality problem for $a, b, c $ positive real numbers $$ \left( a+\frac{1}{b} -1\right) \left( b+\frac{1}{c} - 1\right) +\left( b+\frac{1}{c} -1\right) \left( c+\frac{1}{a} -1\right) +\left( c+\frac{1}{a} -1\right) \left( a+\frac{1}{b} -1\right) \geq 3$$
How we can prove the inequality   above.  Actually it take long time to prove it but I couldn't complete.  How we prove it? . Thanks for help

Maybe this could help:

If you denote $x=a+\frac{1}{b}>0, \, y=b+\frac{1}{c}>0, \, z=c+\frac{1}{a}>0$
you will have $x+y+z=a+1/a +b+1/b+c+1/c\ge 2+2+2=6$ and the equality is achieved iff $a=b=c=1$

You get $(x-1)(y-1)+(y-1)(z-1)+(z-1)(x-1)\ge 3$ which is equivalent to
$$xy+yz+zx-2(x+y+z)\ge 0$$
So you have to prove the last one, having in mind that $x,y,z>0$ and $x+y+z\ge 6$

Hint:

$a,b,c$ are positive real numbers.

So,$a+\frac 1b>0$

So,$a+\frac 1b-1>-1$.

Similarly all the values are  greater than $-1$.


What is the shortest way to find inverse of a matrix? I know two methods to find the inverse of a matrix already:-

Row and Column transformations
$A^{-1}= \frac{Adj(A)}{Det(A)}$

I want to know if there's any shorter method to do so because these two methods feel very lengthy.

I always use the method where you use Gauss row operations against the identity matrix to find the inverse.
Basically you put the matrix you want to know the inverse of, called A, on "the left side" and the identity matrix, called I, "on the right side". Then you use regular Gauss row operations to transfer the identity matrix to the left side and the one you end up with on the right side is your inverse.
Here is an example:
Where A is the matrix and A^-1 is it's inverse.


Does $\zeta(3)$ have a connection with $\pi$? The problem

Can be $\zeta(3)$ written as $\alpha\pi^\beta$, where ($\alpha,\beta \in \mathbb{C}$), $\beta \ne 0$ and $\alpha$ doesn't depend of $\pi$ (like $\sqrt2$, for example)?

Details

Several $\zeta$ values are connected with $\pi$, like:

$\zeta$(2)=$\pi^2/6$

$\zeta$(4)=$\pi^4/90$

$\zeta$(6)=$\pi^6/945$

...

and so on for all even numbers.

See this mathworld link to more details: Riemann Zeta Function

So the question is, could $\zeta(3)$ be written as:

  $$\zeta(3)=\alpha\pi^\beta$$
  $$\alpha,\beta \in \mathbb{C}$$
  $$\beta \ne 0$$
  $$\alpha \text{ not dependent of } \pi$$

See $\alpha$ not essencially belongs $\mathbb{Q}$ and $\alpha,\beta$ could be real numbers too.

When I wrote $\alpha$ is not dependent of $\pi$ it's a strange and a hard thing to be defined, but maybe $\alpha$ can be written using $e$ or $\gamma$ or $\sqrt2$ or some other constant.

Edit:

Maybe this still a open question. If

$ \sum_{k = 0}^{2} (-1)^{k} \frac{B_{2k} \ B_{2- 2k + 2}}{(2k)! \ (2  - 2k + 2)!}$

in $-4 \sum_{k = 0}^{2} (-1)^{k} \frac{B_{2k} \ B_{2- 2k + 2}}{(2k)! \ (2  - 2k + 2)!}\pi^3$ be of the form $\frac{\delta}{\pi^3}$ with $\delta$ not dependent of $\pi$

and $- 2 \sum_{k \geq 1} \frac{k^{-3}}{e^{2 \pi k} - 1}$ not dependent of $\pi$ too, this question still hard and open.

Edit 2:

I discovered a result, but later I've seen that this is something already known, either way, it is an interesting one to have it here.

  $$\zeta(3)=-4\pi^2\zeta'(-2)$$

but, if $\zeta'(-2)$ is of the form $\frac{\alpha}{\pi^2}$, with $\alpha$ not dependent of $\pi$, then this still remains as a hard and an open question.

I have a conjecture that $\zeta'(-2)$ will not cancel the $\pi^2$ term, but since I wasn't able to prove it and I can't use it here.

We can express $\zeta$ of odd numbers with $\zeta'$ in a easy way, with a "closed" form like this one.

The question is whether or not $\zeta(3)$ is connected with $\pi$. The answer is yes. Moreover, $\zeta(3) = \beta \pi^{\alpha}$ for some complex $\alpha, \beta$. Take $\alpha = 3$ and $\beta = 0.0387682...$. It is not known whether $\beta = 0.0387682...$ is algebraic or transcendental. It is known, however, that $\zeta(3)$ is irrational as shown by Apery.

Ramanujan conjectured and Grosswald proved that the following holds. If $\alpha, \beta > 0$ such that $\alpha \beta = \pi^{2}$, then for each non-negative integer $n$,
\begin{align}
 \alpha^{-n} \left( \frac{\zeta(2n+1)}{2} + \sum_{k \geq 1} \frac{k^{-2n-1}}{e^{2 k \alpha} - 1} \right) & = (- \beta)^{-n} \left( \frac{\zeta(2n+1)}{2} + \sum_{k \geq 1} \frac{k^{-2n-1}}{e^{2 k \beta} - 1} \right) -
\end{align}
\begin{align}
\qquad 2^{2n} \sum_{k = 0}^{n+1} (-1)^{k} \frac{B_{2k} \ B_{2n- 2k + 2}}{(2k)! \ (2n  - 2k + 2)!} \alpha^{n - k + 1} \beta^{k}.
\end{align}
where $B_n$ is the $n^{\text{th}}$-Bernoulli number.

For odd positive integer $n$, we take $\alpha = \beta = \pi$,
\begin{align}
\zeta(2n+1) = -2^{2n} \left( \sum_{k = 0}^{n+1} (-1)^{k} \frac{B_{2k} \ B_{2n- 2k + 2}}{(2k)! \ (2n  - 2k + 2)!} \right) \pi^{2n+1} - 2 \sum_{k \geq 1} \frac{k^{-2n-1}}{e^{2 \pi k} - 1}.
\end{align}

In particular, for $n = 1$,
\begin{align}
\zeta(3) = -4 \left( \sum_{k = 0}^{2} (-1)^{k} \frac{B_{2k} \ B_{2- 2k + 2}}{(2k)! \ (2  - 2k + 2)!} \right) \pi^{3} - 2 \sum_{k \geq 1} \frac{k^{-3}}{e^{2 \pi k} - 1}.
\end{align}

Observe that the coefficient of $\pi^{3}$ is rational, however, nothing is known about the algebraic nature of the infinite sum. This is a current topic of research. Indeed, it is conjectured that $\frac{\zeta(3)}{\pi^{3}}$ is transcendental.

Update: Recently, Takaaki Musha claims to have proved that $\frac{\zeta(2n+1)}{(2 \pi)^{2n+1}}$ is irrational for positive $n \geq 1$. However, some objection has since been raised (read comments below).

I'm hoping it'll turn out to be related  somehow to OEIS A104007

$\zeta(3)=c \cdot \pi^3/60$

$\zeta(5)=c \cdot \pi^5/378$

$\zeta(7)=c \cdot \pi^7/2700$

where $c$ is either a constant or a function of $n$. The sequence 6,60,90,378,945,2700 are the denominators of coefficients in the expansion of $\frac{x^2}{(1-e^{-2x})^{2}}$.


Is there a statement whose undecidability is undecidable? We know there are statements that are undecidable/independent of ZFC. Can there be a statement S, such that (ZFC $\not\vdash$ S and ZFC $\not\vdash$ ~S) is undecidable?

At least one version of this question has a nearly-trivial answer; if you want to know if there's some statement G such that ZFC (or your PA-compatible logic system of choice) doesn't prove G, ZFC doesn't prove ~G, and we can't prove that ZFC doesn't prove G or not G, then any undecideable statement works!

Specifically, it's impossible for ZFC to prove that it can't prove something, because such a statement is tantamount to the consistency of ZFC; if ZFC were inconsistent then it could prove everything, so proving that there's something that can't be proved is equivalent to proving the consistency of ZFC, and of course this is forbidden by the second incompleteness theorem.

Yes. This is true for any first order statement by Church's theorem, not just statements with respect to ZFC. First order logic is semidecidable, in that every first order statement that can be proven has a sound, complete proof calculus; Valid statements will always be decidable, and can always be proven, though this may take quite a bit of computation.

Satisfiability is undecidable in general however. If a statement is satisfiable, sometimes it can be shown to be satisfiable with finite models, or forcing. If the statement itself and it's negation is demonstrably satisfiable, then you've just proved the statement isn't a universally valid statement. However, one of the result's of Church's theorem proves that there's no algorithm that you can rely on to terminate trying to check for satisfiability of an arbitrary first order statement, because otherwise it would solve the Entscheidungsproblem.


Infinite dimensional representation such that every subrepresentation is reducible Let $V$ be a nonzero finite dimensional representation of an algebra $A$.

  a) Show that
  it has an irreducible subrepresentation.

  b) Show by example that this does not always hold for infinite dimensional representations.

I did not have any problems with part a), but I'm struggling to find an example for part b). Any help?

Let $A=\mathbb{C}[x]$ and consider the regular representation of $A$ acting on itself. Every nonzero element generates a subrepresentation isomorphic to $A$, so this definitely does not have any irreducible subrepresentations.

I'm not quite sure if this contraption is correct, but I'll give it a go, because I'm also interested in this question.

b) Consider an algebra $A = \mathbb{Z}$ and its infinite-dimensional representation $V\subset\mathbb{Z^\infty}$ defined by recurrence relation $c_{i+1} = c_{i} + 1$ together with a homomorphism of algebras $\rho: A \rightarrow EndV$ defined by an operator $T$ such that:

$$\rho(a) = T_a: (c_i,c_{i+1},c_{i+2},\dots) \rightarrow (c_{i+a},c_{i+a+1},c_{i+a+2},\dots)$$

Irreducibility of this representation follows immediately from the observation that every $\rho(a)$, $a \in A$ has an infinite image.


Finitely generated R-module Let $R$ be an integral domain and also Noetherian ring. Let $f$ be an element in Quot$R$ (field of fractions of $R$). Let $R[f]$ be the subring of Quot$R$ generated by $R$ and $ \{ f \}$. Let suppose that $R[f] \subseteq M$ where $M$ is an $R$-submodule of Quot$R$ and $M$ is finitely generated as $R$-module. Is it true that $R[f]$ is also finitely generated as $R$-module?

If is it true, can you give me a rigorous and possibly elementary proof?

The following conditions are equivalent for the commutative ring $R$:

$R$ satisfies the ascending condition on ideals
every ideal of $R$ is finitely generated
every submodule of a finitely generated submodule is finitely generated

If $R$ satisfies one (hence all) of these properties, it is said to be Noetherian.

Here is a quick outline at a pretty basic level. First, a finitely generated module over $R$ is also Noetherian (acc on modules). You can do that by induction on the number of generators and the fact (pretty easy to verify) that if $M$ is an $R$-module, $M'$ a submodule of $M$, then $M$ is noetherian if and only if $M'$ and $M''$ are and take $M'$ to be the submodule generated by one element of your generating set for $M$). Now use this same fact applied to $R[f]\subseteq M$.


Probability of Broken Computers A school orders  7 new computers for a classroom but they are told 3 will not work properly when they recieve them.  The school begins to turn each computer on, one by one, to figure out which computers do not work.
There are quite a few questions but most of them I am just having trouble setting up and I feel mostly are similar in set up.

What is the probability that no more than five computers need to be turned on to find the three computers that don't work?

For this question N is the event the computer does not work and W means it works:
\begin{align}&\color{white}=P(NNN)+P(NWNN)+P(NWWNN)\\&=\left(\frac{3}{7}\cdot\frac{2}{6}\cdot\frac{1}{5}\right)\left(3\cdot\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{4}{5}\cdot\frac{1}{4}\right)\left(4\cdot\frac{3}{7}\cdot\frac{2}{6}\cdot\frac{4}{5}\cdot\frac{3}{4}\cdot\frac{1}{3}\right)\\
&=\frac{12}{42875}\end{align}
The next couple questions are very similar I just don't know how to set up.  These all don't have specific order of which computers are picked but are told the computers are turned on in a couple possible spots.  These are the remaining questions if someondy can explain how these can be set up that be appreciated.

Given that exactly one of the computers not working was found within the first three computers, what is the probability that the other two computers that aren't wokring are found within the next three computers turned on?
Given that exactly two of the computers that don't work were found within the first three computers, what is the probability the last computer that doesn't work is found within the next two computers turning on?
Given that exactly two of the computers that don't work were found within the computers 1, 3, 5, what is the probability the other computer not working was found on tests 6 or 7?
Given the last computer doesn't work was found within the last two tests, what is the probability that the first two computers that don't work were found within the first three computers?

For the first question, it helps to consider the opposite event. What is the probability that more than five computers need to be turned on to find the three that do not work? In other words, what is the probability that after knowing the state of five computers, you still don't know the state of the two remaining computers?

Note that it really doesn't matter in which order the first five computers work or do not work.

For the first, we could use the fact that if the three computer that doesn't have been found it the give first attempts, the the last two are working. Your problem is equivalent to "find the probability that the last two computers are working." The order in which the non-working computers are found is irrelevent.  This probability is
$$\frac47\times \frac36=\frac27$$

EDIT

Has pointed out in the comments, There is also the possibility that the last two computer are non-working. Since it is known there is three non-working computers, if we found only one in the first five attempts, the last two are non-working. The probability that the last two are non-working is
$$\frac37\times\frac26=\frac17$$
Then the probability that we need only five attempts to find the non-working computers is
$$\frac27+\frac17=\frac37$$

For the other problems, we could do it with conditiinal probability or consider what is left and do it like a new problem.

  Given that exactly one of the computers not working was found within the first three computers, what is the probability that the other two computers that aren't wokring are found within the next three computers turned on?

After three attempts, we found two working and one non-working. We are left with four computers where two are not working. To found the two non-working within the next three attempts, the last remaining computer must be working. There are two working computers in the four we have left. The probability is
$$\frac24=\frac12$$

  Given that exactly two of the computers that don't work were found within the first three computers, what is the probability the last computer that doesn't work is found within the next two computers turning on?

This case is similar to the previous one. Consider that the out of the last four, the last two need to be working.
$$\frac34\times\frac23=\frac12$$

  Given that exactly two of the computers that don't work were found within the computers 1, 3, 5, what is the probability the other computer not working was found on tests 6 or 7?

It left computers tested on 2, 4, 6 and 7. Three of them work, and one doesn't. The one that doesn't work need to be tested on 6 or 7. It is the same as saying 2 and 4 need to work.
$$\frac34\times\frac23=\frac12$$

  Given the last computer doesn't work was found within the last two tests, what is the probability that the first two computers that don't work were found within the first three computers?

This one is a bit trickier since we don't know if there is exactly one non-working computer in the last two tests. We will have to do it with conditional probability. Let
$$A=\text{two non-working on first three attempts}$$
$$B=\text{at least one non-working in last two attempts}$$
$$P(A|B)=\frac{P(A\cap B)}{P(B)}$$
Event $B$ will be true unless the last two are working so
$$P(B)=1-\frac47\times\frac36=\frac57$$
It left to evaluate the events $A$ and $B$ will be true together. We have to do it the hard way.
$$P(A\cap B)=3\times\frac37\times\frac26\times\frac45\times2\times\frac14\times\frac33=\frac6{35}$$
Finally, the answer is
$$P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{\tfrac6{35}}{\tfrac57}=\frac6{25}$$


Why hasn't GCH become a standard axiom of ZFC? I've never seen a text that includes GCH in the ZFC axioms. I presume this means that GCH has not achieved widespread acceptance. This seems surprising to me, given that:

The cardinal numbers encountered in "ordinary" mathematics are the beth numbers $\beth_\alpha$, and
ZFC can't prove much about the beth numbers without at least GCH or something stronger, and
ZFC+GCH has been shown equiconsistent with ZFC.

So my question is, why hasn't GCH achieved more widespread acceptance (or has it)?

Unfortunately, I do not feel the arguments you mention are strong enough to make a convincing case. Before addressing $\mathsf{GCH}$ proper, note that in fact:

$\omega_1$ and other cardinals do appear in mathematics. For example, transfinite iterations of length $\omega_1$ are not uncommon in analysis (a simple instance of this is that the Borel sets can be presented as the union of an increasing chain of length $\omega_1$, and this representation is useful beyond simply saying that they are the $\sigma$-algebra generated by the open sets). The cardinal $\omega_1$ appears naturally in descriptive set theory (tied up to the boldface $\Pi^1_1$ sets) and other cardinals (not necessarily beth numbers) appear also in this context.
$\mathsf{ZFC}$ can prove a great deal about beth numbers without any restrictions in the cardinal arithmetic. For example, as pointed out in this answer, the beth numbers appear naturally when studying the spectrum problem in model theory. That answer also mentions one of Shelah's best known results in cardinal arithmetic, that $\aleph_\omega^{\aleph_0}$ is either the continuum, or less than $\aleph_{\omega_4}$. These results are invisible under strong cardinal arithmetic restrictions. Cardinal arithmetic in fact has a lot to say when we do not "trivialize" it by assuming $\mathsf{GCH}$ or the like. For example, there are results that are known to hold from $\beth_\omega$ on but may fail for an initial segment of the cardinals. The fact that we do not have as extensive a list of results without strong cardinal arithmetic restrictions is simply because such restrictions were typically adopted in classical results, and only relatively recently we have been exploring the landscape without them. There is also the obvious fact that the study of cardinal invariants of the continuum disappears if we adopt $\mathsf{CH}$; there is significant structure here, provably in $\mathsf{ZFC}$, but $\mathsf{CH}$ hides it.
Equiconsistency does not seem a good argument for the adoption of an axiom. Replacement makes $\mathsf{ZFC}$ strictly stronger than $\mathsf{ZC}$, and it is this strength that makes it a desirable axiom to adopt. Large cardinal axioms go beyond $\mathsf{ZFC}$, and are routinely taken for granted when studying the projective hierarchy or determinacy. In fact, the ladder of consistency strength hierarchy is one of the most remarkable mathematical objects we have uncovered. One could argue instead that extensions of $\mathsf{ZFC}$ should be as strong as possible to accommodate all mathematical discourse, that $\mathsf{ZFC}$ is in strength too weak to do this properly.

With that out the way, we need to consider what it is that we expect to accomplish with extensions of $\mathsf{ZFC}$. For example, we may want a rich mathematical universe, with as many examples as possible, even at the level of the reals. $\mathsf{CH}$ certainly gives us that: Under this assumption, there are many non-isomorphic linear orders of size $\omega_1$, there are discontinuous homomorphisms between Banach algebras, etc. Analysts tend to adopt $\mathsf{CH}$ for this reason. When we want instead a very orderly universe, with strong classification results, alternative axioms are adopted, usually in the form of forcing axioms. I elaborate this to some extent, and provide some references, here.

It is true that quite a few set theorists see untamed cardinal arithmetic as a pathology rather than a virtue. (Shelah is a notorious exception.) Even so, $\mathsf{GCH}$ seems unforgivably restrictive, typically because it is fairly easy to violate it with mild forcing. For this reason, $\mathsf{SCH}$, the singular cardinal hypothesis, is considered a more reasonable statement to desire. (I am not doing this here, but) one could even make the case that a version of this statement should be part of whatever extension of $\mathsf{ZFC}$ we end up agreeing upon in the future. Unlike $\mathsf{GCH}$, we have that $\mathsf{SCH}$ is typically preserved by forcing. It is in fact seen as "cofinally true" in the universe, since it holds beyond strongly compact cardinals. It is implied by a variety of reflection principles and forcing axioms, so it is somewhat expected to hold under the heuristic that the universe of sets should be both as tall and as wide as possible.

On the other hand, one might expect that the universe of sets should be close to a fine-structural model. Models possessing a fine structure satisfy strong versions of $\mathsf{GCH}$, so this is not by any means an unreasonable axiom to aspire to. In fact, Woodin's recent "ultimate $L$" program aims at developing a theory of the universe that would be compatible with all large cardinals and would possess fine-structural features. If this program succeeds, we would have a template from which we could prove the consistency of any reasonable statement by means of forcing, so the universe could be seen as a forcing extension of a very well-behaved fine structural core. Any other competing "completion" is expected to be mutually interpretable with this alternative.

Now, the mention of mutual interpretability brings us to an interesting point. The argument can be made that there may not be a "distinguished" completion of $\mathsf{ZFC}$, that any two theories are equally desirable if they are mutually interpretable. (A multiverse view, if you wish.) This recent paper by John Steel elaborated on this view. Under this approach, the question of whether $\mathsf{GCH}$ should be adopted is moot.

Even if we do not follow that route, I do not think we have strong evidence to advocate fully for $\mathsf{GCH}$. However, one can make the case that $\mathsf{CH}$ is a desirable axiom to have. Woodin has proved a maximality result, showing that $\mathsf{CH}$ "decides" all $\Sigma^2_1$ statements. (Somewhat more formally, under appropriate large cardinals, any such statement that can be forced actually holds in the presence of $\mathsf{CH}$. Note that $\mathsf{CH}$ itself is a $\Sigma^2_1$ statement.)

On the other hand, a few years ago, Woodin presented a program that seemed to conclude on fairly general grounds that $\lnot\mathsf{CH}$ was the reasonable option. Oversimplifying, the argument was: From large cardinals, we have a "complete" theory of $H(\omega_1)$ and, in particular, its theory cannot be modified by set forcing under large cardinal assumptions. Similarly, we can have a "complete" theory of $H(\omega_2)$; for example, Woodin introduced a principle $(*)$ that implies this, and implies the negation of $\mathsf{CH}$ (again, under large cardinals). In contrast, it was claimed that no such "complete" theory exists compatible with $\mathsf{CH}$. However, a flaw was found in the proof of this claim, so the argument for $\lnot\mathsf{CH}$ on these grounds has lost strength.

Nowadays, advocates of the negation of $\lnot\mathsf{CH}$ are willing to give up the "conditional maximality" that $\mathsf{CH}$ provides. But typically one does not argue for $\lnot\mathsf{CH}$ directly, but rather for principles that (usually for technical reasons) end up implying $\lnot\mathsf{CH}$.

My take on $\mathsf{GCH}$, such as it is, is that first of all, we are far from having enough mathematical evidence to make a decision. And second, whatever we end up adopting should not be directly this axiom (or its negation), but whether it holds or not should be a consequence of deeper structural principles, and the precise nature of those principles is not yet developed to the point where we can venture a guess. I confess I am partial to large cardinals, and to strong reflection principles, so I am not opposed to $\mathsf{SCH}$. (But note that I am not advocating for large cardinals or forcing axioms or the like here, I am just pointing out that arguments in their favor have been advanced, and I find them more compelling and encompassing than the arguments I am aware of for their negation.) I find the structure of the continuum provided by forcing axioms somewhat more appealing than the wild untamed riches brought about by $\mathsf{CH}$.

But, again, as I said, whether strong cardinal arithmetic restrictions end up being part of the accepted axiomatization of set theory, I expect that this will be a consequence of other principles rather than just adopted as an axiom on its own.

And here is a more light-weight broad discussion from Scientific American.Dispute over Infinity Divides Mathematicians

And this one is a little bit of circumstantial evidence for CH in that a statement that was expected to witness ~CH ended up disproved Mathematicians Measure Infinities and Find They’re Equal


In how many ways can we choose 6 candies from 8 brands that are available? (It is assumed that repetition of a brand is allowed) I have the solution and I know it is $\ \binom{13}{7} $ but I think it should be $\ 8^6 $ because:
We need to choose 6 candies and there are 8 brands to choose from for each candy. So there are $\ 8×8×8×...×8=8^6 $ ways to choose them.
Please help me understand why my logic is wrong.

The correct answer as you have mentioned is $\binom{13}{7}$ and this is a stars and bars problem.

Now lets see why $8^6$ is wrong.
Lets say the brands are $A,B,C,D,E,F,G,H$

Then a selection of $6$ candies can be represented as a string.

$ABCAAF$ where $3$ candies from $A$ and $1$ candy from $B,C,F$

Consider another set of candies $AAABCF$ this has a similar configuration as the first one $3$ from $A$ and $1$ from $B,C,F$. This string was already counted and thus there is some overcounting involved.

Similarly there are many examples of overcounting in $8^6$

Can't we do it by taking the  combination of 48 taken 6 at a time? Here I am imagining that there are 8 empty buckets and each has a capacity to hold six candies.So , it would be 48!/{(42!)6!}. What is the mistake in this technique?Please correct me.


What is the exact definition of a reflexive relation? Well... The three definitions I got for reflexive, symmetric, and transitive relations are:

R is said to be reflexive on A if $\forall x\in A : xRx$.

R is symmetric if $\forall x,y\in A : xRy \rightarrow yRx$.

R is transitive if $\forall x,y,z\in A : (xRy \wedge yRz) \rightarrow xRz$

So, if I understand it correctly, the empty relation {} is symmetric and transitive because the xRy will always be false, and so the implication is true. It's also not reflexive over A unless A is the null set.

What I'm wondering is: if I had a relation, say $R=\{(x,y)\in \mathbb{R}\times\mathbb{R} | x>0 \wedge x<10 \wedge x=y \}$, it's obviously not reflexive over $\mathbb{R}$, but would it be called "reflexive" in general?

I'm asking this because I have a homework problem that says R is a relationship over A, and R is reflexive. I'm not sure if that means R is reflexive over all of A, or in other words, if A is necessarily a subset (or equal to) the domain and range of R.

I would say that a relation should not be called "reflexive", without further qualification, unless the underlying set which the relation is on is specified (explicitly or implicitly).

For example your set of ordered pairs $R$ is a subset of $\mathbb{R} \times \mathbb{R}$ and hence is a relation on $\mathbb{R}$.  It is not reflexive on $\mathbb{R}$ because there are real numbers $x$ with the property that $(x,x)$ is not in $\mathbb{R}$.  But your set $R$ is also a subset of $(0, 10) \times (0,10)$ and hence is a relation on the open interval $I = (0,10)$, and it is is a reflexive relation on this set.  To abstract this somewhat, the lesson is that whether a set of ordered pairs is reflexive or not, depends on more than just the set of ordered pairs.  It depends on the set $A$ that you think of the set of ordered pairs as being a subset of, and this is not encoded in the set $R$ at all.

(This is a little bit like how the codomain of a function is not determined by the set of ordered pairs that represents the graph of a function.)

As you have defined them the notions of "symmetric" and "transitive" also depend on the underlying set, but this dependence is less sensitive, in the following sense: if $S$ is a set and $R$ is a subset of $S \times S$, and $R$ is symmetric (or transitive) as a relation on $S$, then for any set $T$ satisfying $T \supseteq S$ (so you have that $R$ is also a subset of $T \times T$, and hence a relation on $T$), $R$ will also be symmetric (or transitive) as a relation on $T$.  As the above example shows, reflexivity does not behave like this.  So symmetry and transitivity are far less sensitive to what set you consider a set of ordered pairs to be relation on, than reflexivity is.  So you should not be casual about saying that a relation is "reflexive" (without further qualification) unless a reader knows from context what set you have in mind.  (In actual mathematical practice this is never an issue because relations are not constructed arbitrarily for the purpose of testing knowledge of concepts like "reflexivity".  But in classroom exercises it is a good practice to follow.)

If I may veer into metamathematics for a moment.  Speaking very informally, reflexivity is fundamentally different in nature from the symmetry and transitivity because it is the only property that actually requires specific ordered pairs to be in a relation.  (The other two only require that if some ordered pairs are in there, then other ordered pairs must be in there also.)  In textbook exercises often people think of reflexivity as a silly and trivial property because it is easy to verify in many examples whether or not a relation is reflexive (on a given set).  This is more an artifact of what happens in textbooks than it is an artifact of actual mathematical practice.  In actual mathematical practice, it is often the case that reflexivity (when it occurs) is the most difficult property of these three to verify--- precisely because it is the only one that requires specific ordered pairs to be in a relation, it is the only one where in proving it you might be required to actually do something with the definition of the relation and not just use its formal properties.

As an illustration of this idea, suppose I let $A$ denote the set of simple closed curves in $\mathbb{R}^3$ and say that $C \sim D$ for $C, D \in A$ iff $C$ there is a surface $S$ in $\mathbb{R}^3$ satisfying [a long list of geometrical constraints] with the property that the boundary of $S$ is the disjoint union of a curve homeomorphic to $C$ and another curve homeomorphic to $D$.  In checking that $\sim$ is symmetric and transitive, I only need to think about what happens to my long list of geometrical constraints if I permute the order of the data I feed into it, or if I piece two surfaces satisfying these constraints together.  To prove that $\sim$ is reflexive I actually have to construct, for every curve $C$, a surface satisfying [long list of constraints] whose boundary is a disjoint union of two copies of $C$.  A totally different kind of problem.

Firstly we know about identity relation . identity relation is a relation if in all ordered pairs first entry and second entry is same.
Symbolically identity relation is defined as
R={ [X,X]:for all X belongs to A }

Reflexive Relation : it is said to be reflexive when it contains whole identity part.
Identity relation is always reflexive converse may not be true.


What's the connection between the Laplace transform and the Fourier transform? Both the Laplace transform and the Fourier transform in some sense decode the "spectrum" of a function. The Laplace transform gives a power-series decomposition whereas the Fourier transform gives a harmonic (or loop-based) decomposition.

Are there deep connections between these two transforms? The formulaic connection is clear, but is there something deeper?

(Maybe the answer will involve spectral theory?)

I don't know what answer you are looking for but for example both Laplace and Fourier transform are a so called Gelfand Transform.

You can find good introduction to Gelfand Transform in nice book Functional analysis for probability and stochastic processes: an introduction, A. Bobrowski. Look into Chapter 6.

Fourier transform does not exist for every signal application.So by introducing the region of convergence in Fourier transform which is known as Laplace Transform one may have indirectly the Fourier transform of signal.


How to calculate scratch card probability A scratch ticket has 15 cells.

How do you calculate the probability of scratching 6 winning images before scratching 9 losing images?

I'm trying to build a simple online game and struggling with the probabilities :(

I assume that we know beforehand that the card will have exactly 6 winning images amongst the 15 on the card.

There are 6 winning images in the 15 images. When choosing the first image to reveal, there is a $\frac{6}{15}$ probability of it being a winning one.

There are now 5 winning images left in the remaining 14 images. When choosing the second image to reveal, there is a $\frac{5}{14}$ probability of it being a winning one.

There are now 4 winning images left in the remaining 13 images. etc.

The probability of going through all 6 reveals successfully is therefore $\frac{6}{15}\cdot \frac{5}{14}\cdot \frac{4}{13}\cdot \frac{3}{12}\cdot \frac{2}{11}\cdot \frac{1}{10}$.

Direct enumeration is another possibility. All possible drawings are obtained by the following Mathematica codes:

ticket = Join[Table[w, 6], Table[l, 9]];
ords = Permutations[ticket];
Length[ords] (* output 5005 *)

The count is $5005$.

Drawing 6 w's before the 9 l's requires that the first 14 scratches contain 6 w's. The number of drawings which satisfy this is

Length[ Select[ #[[1 ;; 14]] & /@ ords, Count[#, w]==6 &]]

and this gives $3003$. The required probability is $3003/5005$.


Double orthogonal complement of any closed subspace is it self Let $H$ be a pre-Hilbert space such that any closed sub space $M \subset H$ has the property $M^{\bot \bot}=M$. Prove that $H$ is a Hilbert space (ie, prove that $H$ is complete)

My attempt:

As $H^*$ ($L(H,\mathbb{K})$) is complete then I want to prove that $H \cong H^*$. But I don't know how to construct an isometry between them.

I will appreciate highly who can give me some ideas

Thank in advance!

That space seems to be the same space here:
A counterexample to theorem about orthogonal projection
and it doens´t seems to have the property $M = M^{\perp \perp}$ for all closed subspaces $M$.

I found the same exercise in the book "Functional Analysis" of George Bachman and Lawrence Narici

It says:
If X is an inner product space and if $M=M^{\perp \perp}$ for every closed subspace M of X, show that X is a Hilbert space. [Hint Use the mapping T mentioned in Eq. (12.23)]

That mapping is:

$T:X \rightarrow \overset{\sim}{X}$ such that

$y \rightarrow Ty = f $

where the bounded functional $f$ is, for any given $x \in X$, given by:

$(Ty)(x)=f(x)=(x,y)$

Edit:
I think I was able to prove it:

Let's be $T:X \to \overset{\sim}{X}$ such that
$T(y)=f_{y}$ where $f_{y}(x)=(x,y) \forall x \in X$

And $\overline{T}$ to the same function but with domain $\overline{X}$  (the completation of X).

It's clear that the conjugate space of $X$ and the conjugate space of $\overline{X}$ are the same, in fact the function $\overline{T}$ is an isometry, is aditive and homogeneuos conjugate (it's almost lineal and it's almost a congruency).

Let's suppose, to arrive a contradiction, that there exists $x \in \overline{X} - X$

It's clear that $x_{0} \neq 0$ given that $0 \in X$ because $X$ it's a vector space.

Let $f=\overline{T} (x_{0})$

We have that $N_{f}= \{ z \in X / f(z)=0 \}$ is a closed subspace of $X$, because it's the inverse image of $0$ under the bounded functional (so, a continous functional) $f$

As $x \neq 0$ we have that $f \neq 0$ so $N_{f} \neq X$

Let's see that $N_{f}^{\perp} = \{ 0 \}$ so we can arrive to a contradiction because $N_{f}=N_{f}^{\perp \perp} = \{ 0 \}^{\perp} = X$ (where the first equality is given by hyphotesis)

We have $ N_{f}^{\perp} = \{ x \in X / (y,x) = 0 \forall y \in X \}$

Then we have $x \in N_{f}^{\perp}$ which is the same than saying $(y,x) = 0 \forall y \in X$ then $T(x)(y)=0 \forall y \in X$ so $T(x)=\lambda f=\lambda T(x_{0})$

As $\left [ x_{0} \right ] \cap X = \{ 0 \}$ given that $x_{0} \notin X$, we arrive the desired contradiction.

Then $X=\overline{X}$, in other words $X$ is a Hilbert Space.

Sorry for the bad redaction, English isn't my first language.

This statement appears to be false:

Let $H=\ell^2(\mathbb R)$ be the Hilbert space with the usual orthonormal basis sequence $(e_n)_{n \in \mathbb N}$. Let $L$ denote the linear span of $e_n$.

Then $L$ is an (incomplete) inner product space: $$ s_n =\sum_{k=0}^n {1\over 2^k}e_k$$

converges to the sequence $s = (1,{1\over 2}, {1\over 4},\dots)$ which is not in $L$ but $s_n \in L$.

Since $e_n$ are orthonormal it is easy to see that if $M$ is any subspace of $L$ then $M^{\bot \bot}=M$.


Convergence of Power Series with $z^{2^k}$ So i got this  power series $$\sum_{k=0}^\infty 3^k \cdot z^{2^k}$$

Can i do the convergence test like this:

$$\text{ratio test} \Longrightarrow \frac{3^{k+1}\cdot z^{2^{k+1}}}{3^k \cdot z^{2^k}} = \frac{3 \cdot z^{2^{k+1}}}{z^{2^k}} = 3 \cdot z^2 \lt 1 \Longleftrightarrow |z| < \frac{1}{\sqrt3}$$

If you want to see your series as a power series, observe that

$$\sum_{k=0}^\infty 3^k z^{2^k}=1\cdot z+3z^2+9z^4+\cdots+3^kz^{2^k}+\cdots$$

observe then that the sequence of coefficients is

$$a_n:=\begin{cases}0,&n\neq 2^k\\{}\\3^k,&n=2^k\end{cases}$$

Thus, the convergence radius is given by (the sequence above is non-negative):

$$\frac1R=\lim\sup_{n\to\infty}\sqrt[n]{a_n}=\lim_{n=2^k\to\infty\iff k\to\infty}3^{k/2^k}=1\implies R=1$$

The Ratio test gives the limit $ L=\lim_{k\to +\infty} 3|z|^{2^k}$.

$|z|<1\implies L=0\implies R\geq 1$
$|z|=1 $ and $ \sum 3^k $ diverges $\implies R\leq 1$

thus, the radius of convergence is $R=1$.


Finding the nth non square free number We all know what squarefree numbers are. I was more interested in checking an efficient way to find nth non square free number.

Let $S(n)$ be the count of squarefree numbers less than or equal to $n$. Note that each integer $n$ has a unique "squarefree decomposition" $n = a^2b$ where $b$ is squarefree.

One approach is to start with $n$ and then remove the counts we don't want, i.e. anything with $a>1$. Using inclusion-exclusion, this suggests a recursion:

$$S(n) = n - \sum_{a=2}^{\lfloor\sqrt{n}\rfloor} S\left(\left\lfloor\frac{n}{a^2}\right\rfloor\right)$$

This isn't the most efficient approach as-stated, but it's a good starting point for additional insights and optimizations. Can you take it from here?

The $n$-th non square integer is given by the formula $n + [ \sqrt{n} + 0.5],$ where $[•]$ denotes floor function. E.g. $n=1$ yields $2$, $n=2$ yields $3$, $n=3$ yields $5$,...and so on. Is this what you are looking for?


Why is this element in the center? (Part two) Suppose that $A$ is an abelian group, and that $G$ is a group. Suppose we have a group homomorphism $$f: A \times G \to G,$$ and suppose further that $f(a,f(b,g)) = f(ab,g)$ for all $a,b \in A, g \in G.$  I have seem the claim that this implies that $f(a,1)$ is in the center of $G.$ Why is this true?

Related Why is this element in the center?

This is a left $A$-act on $G$ with the action given by $f$. Denoting $G= (G,\cdot)$ and $A = (A,+)$.

Assuming $G$ is unitary i.e $f$ surjective we would have $g = f(s,h)$
\begin{align*}
f(a,1)\cdot g = f(a,1)\cdot f(s,h) &= f(a+s, 1\cdot h) = f(s+a,h\cdot 1)\\ &= f(s,h)\cdot f(a,1) = g\cdot f(a,1)
\end{align*}

which does not require the 'associativity' of the action only that $f$ is a homomorphism.

Not sure atm what do if $f$ is not surjective.


MIT PRIMES Question, polynomial satisfying conditions There was an MIT PRIMES application problem that goes like this: (don't worry, the admission ended on Dec 1, so I'm not cheating or anything)

For all $d\geq 0$, determine if there is a polynomial $p(x)$ with degree $d$ such that for all $n$ in $1\ldots 99$ inclusive satisfies $$p(n)=n+\frac1n$$ and if so, what is $p(x)$ and find the value of $p(100)$.

In English terms: find a polynomial with degree $d$ such that $$p(1)=1+\frac11=2$$ $$p(2)=2+\frac12=2.5$$ and so on. What values of $d$ work? What are the polynomials for each of those $d$? What is $p(100)$ for each of those polynomials?

My solution was to brute force it by solving a matrix (see Finding polynomials from huge sets of points), but Java wasn't good enough to do all that math (couldn't contain huge numbers with good enough precision).

  Is there a better, more elegant way to do the problem? If not, how would I brute force it?

If $p(x)$ satisfies the given conditions, then
$$ f(x) = xp(x)-x^2-1 $$
is a polynomial of degree $d+1$ (assuming $d\geq 1$) with zeros at $x = 1,2,\ldots,99$.

Note that $f$ satisfies $f(0) = -1$, so it cannot be the polynomial that is identically zero-valued. Thus $f$ can have at most $d+1$ distinct zeros,
so we must have $d\geq 98$.

If $d = 98$, then the location of zeros of $f(x)$ implies
$$f(x) = k(x-1)(x-2)\cdots(x-99)$$
for some constant $k$.
But since
$$ -1 = f(0) = k\cdot(-1)^{99}(99!),$$
we may solve for $k = \frac1{99!}$. In this case $f(100) = k\cdot(99!) = 1$, so then
$$p(100) = \frac1{100} (f(100) + 100^2 + 1) = 100 + \frac2{100}.$$

If $d\geq 99$, then we have
$$f(x) = \frac1{99!}(x-1)(x-2)\cdots(x-99)g(x)$$
for some (arbitrary) polynomial $g(x)$ of degree $d - 98$, normalized to $g(0) = 1$.
In this case $f(100) = g(100)$, and
$$ p(100) = \frac1{100} (g(100) + 100^2 + 1 ) = 100 + \frac{1 + g(100)}{100}.$$

From the given condition $nP(n)=n^2+1$ for $n=1,2,\cdots,99$. So the polynomial $P^{\star}(x)=xP(x)-x^2-1$ of degree $d+1$, has 99 zeros. If $d<98$ then $P^{\star}(x)$, despite of being of degree $d+1$, would have $>d+1$ roots, and hence should be identically equal to zero for every $x$. Then $P(100)$ would be simply $\frac{P^{\star}(100)+(100^2+1)}{100}=100+\frac1{100}$.

Now if $d=98$ then we can represent
$$P^{\star}(x)=(x-1)(x-2)\cdots(x-99)$$
assuming $P$ to be a monic polynomial. Then $P(100)=\frac{P^{\star}(100)+(100^2+1)}{100}=\frac{99!+10001}{100}$.

If $d>98$ then we can represent
$$P^{\star}(x)=(x-1)(x-2)\cdots(x-99)g(x)$$
With $g(x)$ being a polynomial of degree $\ge1$. But then due to indeterminacy of $g(x)$ we can't calculate the exact value of $P(100)$.


Proof of a lower bound $\lambda(n)$ of the smallest number of multiplications $\ell(n)$ needed to compute $a^n$ for an integer $a$ Let $\ell(n)$ be the smallest number of multiplications needed to compute $a^n$ for any integer $a$. Here, a multiplication is $a_i := a_j \cdot a_k$ for $j, k < i$ and $a_0 := a$, e.g. $\ell(8) = 3$. I want to show the following inequality:

$$\lambda(n) \leq \ell(n) \leq \lambda(n) + v(n) + 1,$$

where $\lambda(n)$ is one less than the number of significant bits of $n$ in the binary representation and $v(n)$ is the number of $1$s in the binary representation of $n$.

I tried to prove this inequality using induction and the upper bound for $\ell(n)$ is quite easy to show, however I don't see how I can prove the lower bound. The only thing about $\ell(n+1)$ I know is that it is less or equal to $\ell(n) + 1$. But if we for example look at $n=7$, we see that $\ell(n+1) < \ell(n)$, so $\ell(n)$ is not monotonically increasing... Can anyone give me a hint how to show the lower bound or how $\ell(n)$ is bounded?

Thanks for any help in advance.

For the lower bound, we argue by induction that $a_i \leq a^{2^i}$ for all $i$. In order to get $a_t = n$, we must have $a^{2^t} \geq a_t = a^n$ or $2^t \geq n$. It follows that
$$t \geq \log n \geq \lfloor \log n \rfloor = \lambda(n).$$

Look at it the other way and try to put an upper bound on the exponents of the computable powers knowing the number of multiplications : for any given integer $k$, what is the greatest power of $a$ you can compute with only $k$ multiplications ?


If $Y$ is a nonnegative absolutely continuous random variable and $E[X|Y]=Y/2$, is $E[X|Y=-1]=-1/2$? Is $E[X|Y=2]=1$? One of the definitions I learned for $E[X|Y=y]$ is the following:
$$ E[X|Y=y]=\int_{\mathbb{R}} x\,P_{X|Y=y}(dx), $$
where $P_{X|Y=y}$ a probability verifying
$$ P(X\in A, Y\in B)=\int_B P_{X|Y=y}(A)\,P_Y(dy), \;\;(*)$$
for all $A,B\in\mathcal{B}(\mathbb{R})$.

This probability $P_{X|Y=y}$ is unique in the following sense: if $Q_{X|Y=y}$ is another probability satisfying $(*)$, then $P_{X|Y=y}(C)=Q_{X|Y=y}(C)$ for all $C\in\mathcal{B}(\mathbb{R})$ and $y\notin N$, with $P_Y(N)=0$.

Let $Y$ be a nonnegative absolutely continuous random variable with $E[X|Y]=Y/2$. Let $N=(-\infty,0)\cup \{2\}$. We have $P_Y(N)=0$, so in principle $P_{X|Y=y}$ may be any probability on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$ for $y\in N$, right? Then $E[X|Y=y]$ could be any value for $y\in N$, right?

I am totally confused. Intuitively, $E[X|Y=2]$ should be $1$ and $E[X|Y=-1]$ should not be defined, since $Y$ is nonnegative.

As you can see in this question, in Lemma 5.22 of the book An Introduction to Computational Stochastic PDEs, it is stated that $E[W(t)|W(1)]=t\,W(1)$ implies that $E[W(t)|W(1)=0]=0$, where $W$ is a Brownian motion and $0\leq t\leq 1$. But $P(W(1)=0)$, so $E[W(t)|W(1)=0]$ is not uniqueley defined, it may be any value. I mean, $y\mapsto E[W(t)|W(1)=y]$ is defined $P_{W(1)}$-a.s., so at $y=0$ there is not a unique definition. It is like stating something about $f(0)$ when $f$ is a real function defined a.e.

I will state and prove a general theorem which formally justifies why a Brownian bridge can be thought of as the same object as a Brownian motion conditioned on its terminal value (maybe this answer will make a bit more sense if one is familiar with regular conditional probabilities).

So let $B=(B_t)_{t \in [0,1]}$ be a Brownian motion on $[0,1]$. For $x \in \Bbb R$ I will define the Brownian bridge ending at $x$ to be the process $\tilde B_t^x := B_t-t(B_1-x)$, and I will show that this naive definition coincides with your conditional probability definition.

Let $\mu$ denote the law of $B$, and let $\mu^x$ denote the law of $\tilde B^x$, considered as probability measures on $C[0,1]$ equipped with its Borel $\sigma$-algebra. In other words, $\mu(A):=P(B \in A)$ and $\mu^x(A) := P(\tilde B^x \in A)$, for Borel sets $A \subset C[0,1]$.

  Theorem: All of the following are true:

  The Law of $\tilde B^x$ is the law of $B$ conditional on $B_1=x$. Formally, this means that for any bounded measurable function $f:C[0,1] \to \Bbb R$, we have that $$\Bbb E[f(B)  |B_1] = \int_{C[0,1]} f(\varphi) \; \mu^{B_1}(d\varphi), \;\;\;\;\; a.s.$$ where the RHS is meant to be interpreted as a random variable $\omega \mapsto \int f(\varphi) \;\mu^{B_1(\omega)}(d\varphi) $. Another way of putting this is that the map $(x,A) \mapsto \mu^x(A)$ is a regular conditional probability for $B$ with respect to $B_1$.
  $B$ can be decomposed as the independent sum $B = \tilde B^0 + tB_1$. Equivalently, for any measurable subset $A \subset C[0,1]$, $$\mu(A) = \frac{1}{\sqrt{2\pi}} \int_{\Bbb R} \mu^x(A) e^{-x^2/2} dx$$
  For any continuous function $f:C[0,1] \to \Bbb R$ and any $x \in \Bbb R$, $$\lim_{\epsilon \to 0} \Bbb E\big[f(B) \;\big| \; |B_1-x| < \epsilon \big] = \Bbb E[f( \tilde B^x) ] =\int_{C[0,1]} f(\varphi) \; \mu^x(d\varphi)$$ If $f$ is merely assumed to be bounded and measurable, the same conclusion still holds for almost every $x \in \Bbb R$.
  In terms of density functions, we have that $$f_{\tilde B^x_{t_1},...,\tilde B_{t_n}^x}(y_1,...,y_n) = \frac{f_{B_1,B_{t_1},..., B_{t_n}}(x,y_1,...,y_n)}{f_{B_1}(x)}$$

Proof: We will first prove 2, then 1, then 3, then 4.

Proof of 2: We will use the following fact: If $(X_t,Y_t)_{t \in [0,1]}$ is a jointly Gaussian process such that $cov(X_t,Y_s)=0$, for all $s,t$, then $X$ and $Y$ are independent. Indeed, this would imply independence of the finite-dimensional distributions, which (via the $\pi$-$\lambda$ theorem) implies independence with respect to the product $\sigma$-algebra on $\Bbb R^{[0,1]}$, which coincides with the Borel $\sigma$-algebra when restricted to $C[0,1]$. Using this fact together with the fact that $cov(B_t-tB_1,B_1)=0$ for all $t\in [0,1]$, we see that $(B_t-tB_1)_{t\in[0,1]}= \tilde B^0$ is independent of $B_1$. This proves the first part of the claim 2, and for the second part, we use independence of $\tilde B^0$ and $B_1$ to say $$\mu(A) = P( B \in A) = P\big((\tilde B^0_t+tB_1)_{t\in[0,1]} \in A\big) = \int_{\Bbb R} P\big((\tilde B^0_t+tx)_{t\in[0,1]} \in A\big) \; \Bbb P^{B_1}(dx) $$$$=\int_{\Bbb R} P(\tilde B^x_t \in A) \; \Bbb P^{B_1}(dx)=\frac{1}{\sqrt{2\pi}}\int_{\Bbb R} \mu^x(A)e^{-x^2/2}dx$$which proves the formula in 2.

Proof of 1: As a corollary of the formula in 2, if
$f: C[0,1] \to \Bbb R$ is any bounded measurable function, then $$\int_{C[0,1]} f(\phi) \mu(d\phi) = \frac{1}{\sqrt{2\pi}}\int_{\Bbb R} \bigg[\int_{C[0,1]} f(\phi) \mu^x(d\phi)\bigg] e^{-x^2/2}dx$$ Cosequently, if $A \subset \Bbb R$ is any Borel measurable set, then $$\Bbb E[f(B)1_{\{B_1 \in A\}}] = \int_{C[0,1]} f(\phi) 1_A(\phi(1)) \;\mu(d\phi)$$ $$=\frac{1}{\sqrt{2\pi}}\int_{\Bbb R} \bigg[\int_{C[0,1]} f(\phi)1_A(x) \mu^x(d\phi)\bigg] e^{-x^2/2}dx $$$$= \frac{1}{\sqrt{2\pi}}\int_{A} \bigg[\int_{C[0,1]} f(\phi) \mu^x(d\phi) \bigg]e^{-x^2/2}dx$$ $$=\Bbb E \bigg[ \bigg(\int_{C[0,1]} f(x) \; \mu^{B_1}(dx) \bigg) \cdot 1_{\{B_1 \in A\}} \bigg]$$ which proves the formula in 1.

Proof of 3: Let $f:C[0,1] \to \Bbb R$ be continuous, and fix some $x \in \Bbb R$. Mimicking the proof of 1, we find that $$ \Bbb E[f(B) \cdot 1_{\{|B_1-x|< \epsilon\}}] = \frac{1}{\sqrt{2\pi}} \int_{x-\epsilon}^{x+\epsilon} \bigg[ \int_{C[0,1]} f(\phi) \mu^u(d\phi)\bigg] e^{-u^2/2}du$$ Therefore, $$\Bbb E\big[f(B) \;\big|\;|B_1-x|<\epsilon\big] := \frac{\Bbb E[f(B) \cdot 1_{\{|B_1 -x|<\epsilon\}}]}{P(|B_1-x|<\epsilon)}$$$$= \frac{\int_{x-\epsilon}^{x+\epsilon} \bigg[ \int_{C[0,1]} f(\phi) \mu^u(d\phi)\bigg] e^{-u^2/2}du}{\int_{x-\epsilon}^{x+\epsilon} e^{-u^2/2}du}$$Now letting $\epsilon \to 0$ and using continuity of $f$ (together with weak continuity of $u \mapsto \mu^u$) we get that $$\lim_{\epsilon \to 0}\Bbb E\big[f(B) \;\big|\;|B_1-x|<\epsilon\big]=\int_{C[0,1]}f(\phi)\mu^x(d\phi) = \Bbb E[f(\tilde B^x)]$$ This proves the claim for continuous $f$. If $f$ is merely bounded and measurable, the claim follows similarly but we apply Lebesgue's differentiation theorem at the last step.

Proof of 4: We wish to show that $f_{B_1}(x)f_{\tilde B^x_{t_1},...,\tilde B_{t_n}^x}(y_1,...,y_n)=f_{B_1,B_{t_1},..., B_{t_n}}(x,y_1,...,y_n)$, i.e., that for bounded measurable $g:\Bbb R^n \to \Bbb R$ and Borel sets $A \subset \Bbb R$, $$\int_A\bigg[\int_{\Bbb R^{n}} f_{\tilde B^x_{t_1},...,\tilde B_{t_n}^x}(y_1,...,y_n)g(y_1,...,y_n)dy_1 \cdots dy_n\bigg]f_{B_1}(x)dx = \Bbb E[g(B_{t_1},...,B_{t_n})\cdot 1_{\{B_1 \in A\}}]$$ But the LHS may be rewritten as $$\frac{1}{\sqrt{2\pi}} \int_{A} \Bbb E \big[ g(\tilde B^x_{t_1},...,\tilde B^x_{t_n})\big] e^{-x^2/2}dx$$ $$=\frac{1}{\sqrt{2\pi}} \int_{A}\bigg[ \int_{C[0,1]} g(\phi(t_1), \dots, \phi(t_n)) \mu^x(d\phi) \bigg] e^{-x^2/2}dx$$ $$=\int_{C[0,1]} g(\phi(t_1),...,\phi(t_n))\cdot 1_A(\phi(1)) \mu(d\phi)$$ $$=\Bbb E[g(B_{t_1},...,B_{t_n})\cdot 1_{\{B_1 \in A\}}]$$ where we used the formula from 2 in the third equality. This proves the claim.


Equation for a smooth staircase function I am looking for a smooth staircase equation $f(h,w,x)$ that is a function of the step height $h$, step width $w$ in the range $x$.

I cannot use the unit step or other similar functions since they are just one step. I have been experimenting with various sigmoid curves and while I can get a single smooth step I cannot get to realize the staircase shape. The closest staircase function I have found is given in this paper in equation (18) and depicted in Fig. 4 and it is a close example of what I want (i.e generate a staircase in the range $x$ for arbitrary step heights and widths) but it is not smooth at all.

Regarding smooth steps, a likely starting point I found is here but it gives a smooth function of just a single step. I have been unable to modify the equation to make it into a staircase. I would like to specify arbitrary step heights and widths and generate a smooth staircase in the range $x$ specified.

Edit (Extra info):
The smooth function I mention above has the problem that the upper, horizontal line is not equal in length to the lower, horizontal line which is why I have been unable to adapt it into a staircase function

Edit 2
Including some pictures

Edit 2
Plot of $s$ with a steep slope showing a different width on the first horizontal line

Here is an example based on Math536's answer: Wolfram link

$$f(h,w,a,x) = h \left[\frac{\tanh \left( \frac{ax}{w}-a\left\lfloor \frac{x}{w} \right\rfloor-\frac{a}{2}\right)}{2\tanh\left(\frac{a}{2}\right) } + \frac{1}{2} + \left\lfloor \frac{x}{w} \right\rfloor\right]$$

Where h is the step height, w is the period, and a is the smoothness

Stair cases of Height=H and Width =W may be represented by:

$$S\left(x\right)=H\left(\frac{x}{W}+\frac{1}{2}-\frac{\arctan\left(\tan\left(\pi\left(\frac{x}{W}+\frac{1}{2}\right)\right)\right)}{\pi}\right)$$

Here you can observe https://www.desmos.com/calculator/nxmj1yqrph


Graded rings: what does $\mathbb{Z}[y]/(y, y^{2n + 1})$ with $y$ of degree $2$ mean? I do not understand what $\mathbb{Z}[y]/(2y, y^{2n + 1})$ with $y$ of degree $2$ means.

If I read the Wikipedia page right, the graded ring $\mathbb{Z}[y]$ is the set of all polynomials in $y$ with coefficients in $\mathbb{Z}$ with the natural multiplication, $(2y, y^{2n + 1})$ is the ideal generated by $(2y, y^{2n + 1})$.

Then $\mathbb{Z}[y]/(y, y^{2n + 1})$ is $\bigoplus(R_i + I)/I$ with the canonical multiplication, where $R$ is all polynomials of the form $z y^i$, $I$ is the ideal $(y, y^{2n + 1})$ and $R_i + I = \{a + b \ | \ a \in R_i, b \in I\}$.

Question 2: Wouldn't that mean that $(R_i + I)/I \cong R$ because $[a + b] = [a]$ in $(R_i + I)/I$?

Question 3: What does $y$ of degree 2 mean?

Edit: A bit more context may help:

I need to find an isomorphism of graded rings

$$H^*(\mathbb{R}P^{2n}; \mathbb{Z}) \cong \mathbb{Z}[y]/(2y, y^{2n + 1})$$

Of course, $y^{2n+1}$ lies in the principal ideal generated by $y$, and hence is a redundant (or superfluous) generator. Also, the corresponding quotient ring is clearly isomorphic to $\mathbb{Z}$.


Prove that the set of continuous real-valued functions on $[0,1]$ is a subspace of $\mathbb{R}^{[0,1]}$ I know this question has been asked here and here.

I see the reason why there are three parts because we need to prove additive identity, closed under addition, and closed under scalar multiplication. That is:

Take $f_0 : [0,1] \to 0$, show that $f_0 + g = g$
$f, g\in V$, show that $f+g\in V$
$f\in V$, and $a\in F$, show that $af\in V$

I saw some answer used $\epsilon-\delta$ Definition of a Limit to prove this problem. I don't understand why the proof of continuous function is a subspace of $\mathbb{R}^{[0,1]}$ need to prove the functions are continuous. It seems to me that the focus is not to prove subspace but continuity.

If possible, can you also work out the proof here?

After reading this and this as well as Epsilon-Delta Definition of a Limit, I understand now.

In the question "Prove the set of continuous real-valued functions on the interval [0,1] is a subspace of $\mathbb{R}^{[0,1]}$", if I write $U$ to be the set of continuous real-valued functions, and $u,w \in U$,

Then the question (only close under addition) can be translated to prove that if you add two continuous functions, are you guaranteed to get a continuous function. i.e.$u + w \in U$ where $U$ is a set of continuous functions.

That's why we need to use Epsilon-Delta Definition of a Limit to prove. Similarly in closed under scalar multiplication.

This had me for a loop as well, but since the approach here is correct I had to rethink the strategy it implies. Doing so helped me understand, and so might help others.

The approach presented here does two things:

You need to test for a subspace using the definition: additive identity, closed under addition, closed under multiplication.
You need to enforce that the thing you are testing is in fact a continuous real valued function.

So in my simple view of things, the solution is like a filter that tests these two points. I think this is what John meant when he said the question could be "translated" to mean that the Epsilon-Delta Limit would satisfy both (subspace and continuous tests).

Feedback welcome, I am new to this and sticking my neck out, but its the fastest way to learn.


Are there other accumulation functions that holds $a(n-t)={a(n) \over a(t)}$? This might be a beginner's question regarding accumulation methods and their functions, but so far I have learned that compound interest satisfy

$$a(n-t)={a(n) \over a(t)}$$

Which allows nice results such as

$$s_{\bar{n}\rceil} = (1+i)^n a_{\bar{n}\rceil}$$

I also understand that the first property does not hold for simple interest.  But out of curiosity, are there accumulation functions that are not compound interests that holds the same property?  If so, does it show up in real life actuarial problems?

You can rearrange your equation by multiplying through to get $a(n-t) a(t) = a(n)$.  Let $n = s+t$ and you get $a(s) a(t) = a(s+t)$.   So you want to find a function satisfying $a(s) a(t) = a(s+t)$ for all choices of $s$ and $t$.

Now, you can rewrite this to get $a(s) a(s) = a(2s)$, and then $a(s) a(2s) = a(3s)$ so $a(3s) = a(s)^3$, and in general $a(ms) = a(s)^m$ for any real number $s$ and integer $m$.  If you let $s = 1/m$, then you have $a(1) = a(1/m)^m$ and so $a(1/m) = a(1)^{1/m}$.  Therefore, for any integers $p$ and $q$, you have $a(p/q) = a(1/q)^p = (a(1)^{1/q})^p = a(1)^{p/q}$.  Thus if we know $a(1)$, we know $a(p/q)$ for any rational number $p/q$.  If you assume that the function $a$ is continuous then you can prove that any function satisfying your initial rule satisifes $a(x) = a(1)^x$ -- which, to go back to your initial question, is just the rule for compound interest.

The only real rule of accumulation functions is that a(0)=1.  So any function that you can think of for which f(n-t) = f(n)/f(t) qualifies provided you define a(0)=1.  An easy one would be a(t)=e^k(t)^2.  I'm just a student so I don't know what comes up in practice though.


Finding distribution of $X^2+Y^2$ where $X,Y\sim N(0,1)$ Assume I have two random independent standard normal variables $X,Y\sim N(0,1)$, How can I find the distribution of $Z=X^2+Y^2$?

I thought integrating the convolution, i.e $F_Z(z)=\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f_{X^2}(t)f_{Y^2}(x-t)dt$. If X is a normal variable, does also $X^2$ is a normal variable (then I could find the density by simply substituting)?

Your distribution is the $\chi^2$ distribution with $2$ degrees of freedom, aka the exponential distribution with mean $2$.  You should be able to calculate the cdf of $X^2+Y^2$ directly by switching to polar coordinates.

Or else you can use your convolution idea. The random variable $X^2$ is not normally distributed, but you can find its distribution in various ways, such as the method of transformations. Or else you can express the probability that $X^2\le w$ as an integral of the normal density function, and then differentiate under the integral sign (fundamental theorem of calculus).

First, both $X^2$ and $Y^2$ are known distributions, namely a chi-squared distribution.  See https://en.wikipedia.org/wiki/Chi-squared_distribution.

If they are independent, to add them just convolve the pdf's.  See: https://en.wikipedia.org/wiki/Convolution_of_probability_distributions


need orthogonal basis for R3. i'm given one of them. how do i find the rest? I need orthogonal basis for R3. I am given v1 = (1,1,1), so I need so I need other two vectors in this basis but how do i find the other two?

at first i thought i would use gram schmidt but that doesn't seem plausible with just one vector.

Pick any vector $\mathbf u$ that is not a multiple of $\mathbf v_1$.
Set $\mathbf v_2 = \mathbf v_1 \times \mathbf u$.
Set $\mathbf v_3 = \mathbf v_1 \times \mathbf v_2$.

Just pick two more vectors arbitrarily and apply Gram-Schmidt. Gram-Schmidt preserves the direction of the first input vector, so you'll get a normalized copy of $(1, 1, 1)$.


Prove that if n is a perfect square, then n+2 is not a perfect square I'm working on proving the following statement:

If n is a perfect square, then n+2 is not a perfect square.

I also need to state this in first order logic with arithmetic, but have no idea what that looks like.

The only start I have so far in terms of the proof is:

$n$ = $a^2$

$n+2$ = $b^2$

But I don't know how to proceed from here? I've seen solutions to this already but do not understand how they actually prove anything.

Let's assume w.o.l.g that $a, b >= 0$

$b^2 = n+2$

$a^2 = n$

So $b^2 - a^2 = 2$, which means $(b-a)(b+a) = 2$. So $(b+a)$ is a divisor of 2 (either 1 or 2, since $b+a>0$).

Now only a few cases remain

$a = 0, b=1$
$a = 1, b=0$
$a = 0, b=2$
$a = 1, b=1$
$a = 2, b=0$

No case has $b^2 - a^2 = 2$

Let us assume $n+2$ is a perfect square:
then, since $n$ is a perfect square, $n=a^2$ (where $a$ is a positive integer).
Since according to our assumption, $n+2$ is a perfect square, $n+2=b^2$ (where $b$ is a positive integer)
Then, we have ,
$$
a^2 +2=b^2
$$
Since the sum of a perfect square (here, $a^2$) and an imperfect square (here, $2$) is always an imperfect square, we know that the left hand side of the equation is an imperfect square and the right hand side of the equation is a perfect square (according to our assumption). So, a clear contradiction arises. This contradiction is due to our false assumption. Thus, $a^2+2$ i.e. $n+2$ is an imperfect square. Hence proved


$f$ twice continuously differentiable on $[a, b], f^\prime(c) = 0$ for some $c \in[a, b], f''(c) > 0$ implies various statements Assume $f$ is twice continuously differentiable on $[a, b]$ and there's some $c \in [a, b]$ such that $f'(c) = 0$. Suppose $f''(c) > 0$. Prove that:

there exists $\delta > 0$ such that $f''(x) > 0$ for all $x \in (c - \delta, c + \delta)$.

(I was able to prove this already, but I'm leaving it here in case it's helpful for 2. and 3., which I haven't had any luck with.)

$f'(x) < 0$ for $x \in (c - \delta, c)$ and $f'(x) > 0$ for $x \in (c, c + \delta)$.

What I'm confused about here is finding a way to apply $\epsilon-\delta$ continuity (if that is even the right way to approach this part).

$f(x) > f(c)$ for $x \in (c - \delta, c + \delta)$ such that $x \neq c$.

Thanks in advance.

For part c, I first used the Mean Value Theorem that: for every interval $(c-\delta, c)$, we can find an $x_1$ and (likewise and $x_2$ for the interval $(c, c+\delta)$), such that:
$$f'(x_1) = (f(c) - f(c-\delta))/\delta$$ and $$f'(x_2) = (f(c+\delta) - f(c))/\delta$$

But we know from part b that $f'(x_1) <0 \implies f(c-\delta) > f(c)$ and $f'(x_2) >0 \implies f(c+\delta) > f(c)$ .

(*) Now we can choose smaller intervals within given interval, e.g. ($c-\delta_i, c+\delta_i$), where $\delta_i < \delta $, ($\delta_i>0$), and attain all possible values of f(x) = $f(c-\delta_i)>f(c)$ or $ f(c+\delta_i) >f(c)$ $ \implies f(x) >$ f(c).

I'm a little unsure whether my argument (*) works. I'm saying that for all x in the given interval, $f(x)> f(c)$ because any x in that interval can be found to equal a $c+\delta_i$ or $c-\delta_i$, which we found from part b is greater than f(c).

There is a counter example to 1)
Take $f(x)= x^2sin(1/x)+ \frac{x}{2} \forall x\in(-\infty,0)\cup(0,+\infty)$
and $f(x)=0$ for $x=0$.
Take a look