How to prove that the function $f$ is continuous on $\mathbb{R}$ Let
$$f:\mathbb{R}\to \mathbb{R} : f(x)=\begin{cases} a^x & \text{if $x \in \mathbb{Q}$} \\ \sup\{ f(y):y<x \text{ and }y \in \mathbb{Q} \} & \text{if $x \in \mathbb{R} \setminus \mathbb{Q}$}\end{cases}$$ where $a>0$.

Prove that $f$ is continuous on $\mathbb{R}$.

Hint: Use $f(x+y)=f(x)f(y)$ for $x,y\in\Bbb Q$ and $f(x)\approx 1$ for $0\approx x\in\Bbb Q$.

If $a \geq 1$ there is nothing to prove as $f(x)=a^x$, $\forall x$ by conitnuity of $a^x$ and the fact that $\mathbb{Q}$ is dense in $\mathbb{R}$.
So let $a <1$. If $a < 1$ then $f(x) = \infty$ for $x > 0$ and $x \in \mathbb{R} \setminus \mathbb{Q}$ and $f$ is not continuous as let $a=1/2$. $\sup\{(1/2)^y : y < x,y \in \mathbb{Q}\} \geq \lim_{n \rightarrow \infty} (1/2)^{(-n)} = \infty$.


Analyticity of roots of a polynomial in terms of coefficients Suppose that $f(z,w)$ is a non-constant polynomial in $z,w$ with coefficients in $\mathbb{C}$.
Fix $z$, we define $p(w)=f(z,w)$. From Liouville's theorem, we know that
$p(w)=0$ is solvable for $w$, and we choose a root $w=G(z)$.

It is known that $w=G(z)$ is a continuous function of $z$ because
any root of a polynomial over  $\mathbb{C}$ is continuous in terms of the coefficients.

I want to know: is it true that $w=G(z)$ is analytic in simply connected open regions excluding possible singularities and branch points ?

Yes, it is. Our numerical analysis teacher used to allude to it again and again. I think its name is after one the giant guy's name: Euler's theorem, Gauss theorem, ...
Google it!


Compute the limit of $n\cdot\left(\frac{\arccos\left(3/n^3\right)}{\arccos\left(3/(n+1)^3\right)}-1\right)$ when $n\to\infty$ How to solve limit like this?

$$\lim_{n\to\infty}{n\left(\frac{\arccos\frac3{n^3}}{\arccos\frac3{(n+1)^3}}-1\right)}$$

$$=\lim_{n\to\infty}{n\left(\frac{\frac\pi2}{\frac\pi2}-1\right)}$$

$$=\lim_{n\to\infty}{n(1-1)} = 0$$
but $\infty*0$ is not define, how to solve it?

This is just a naive thought, but have you tried the first order approximations $\arccos\left(\frac{3}{n^3}\right)\approx\frac{\pi}{2}-\frac{3}{n^3}$ and $\arccos\left(\frac{3}{(n+1)^3}\right)\approx\frac{\pi}{2}-\frac{3}{(n+1)^3}$ for $n\rightarrow\infty$? In this way the whole expression becomes
$$-\frac{18n^2+18n+6}{\pi n^5+3\pi n^4+3\pi n^3+(\pi+3)n^2},$$
whose limit is $0$ (as it should be).

Suppose $A\to \infty$ and $B\to 0$ as $n\to \infty$. If you want to solve $\lim_{n\to\infty} AB$, you can rewrite it as $\lim_{n\to\infty} \frac{A}{1/B}$ to get the indeterminate form $\frac{\infty}{\infty}$ in the limit, or as $\lim_{n\to\infty}\frac{B}{1/A}$ to get the indeterminate form $\frac{0}{0}$ in the limit. Either of these can then be solved using L'hopital's rule.


Uniform convergence of ${(1-\frac{x^2}{n})^n}$ Uniform convergence of: $$f_n(x):={(1-\frac{x^2}{n})^n}$$
(a) In a closed interval $[-L, L]$ as $L>0$.
(b) In $\mathbb{R}$.

My try:
There is a pointwise convergence to $e^{-x^2}$.
Obviously, there is no uniform convergence in $\mathbb{R}$ as $\lim\limits _{x\to \infty }|f_n(x)|=\infty$.$\;$ Thus, $\sup \limits _{x\in\mathbb{R}}|f_n(x)-e^{-x^2}|= \infty$
Can anyone give a direction for a closed interval?

Using the basic inequalities $e^y \geqslant 1 +y$ and $e^{-y} \geqslant 1 - y$ for $0 \leqslant y \leqslant 1$ we can show that for $x^2 < n$,

$$\left(1- \frac{x^2}{n}\right)^n \leqslant e^{-x^2}\leqslant \left(1+ \frac{x^2}{n}\right)^{-n}.$$

Applying the Bernoulli inequality $(1 - x^4/n^2)^n \geqslant 1 - x^4/n,$ we get

$$0 \leqslant e^{-x^2} - \left(1- \frac{x^2}{n}\right)^n = e^{-x^2}\left[1 - e^{x^2}\left(1- \frac{x^2}{n}\right)^{n}\right]\\ \leqslant e^{-x^2}\left[1 - \left(1+ \frac{x^2}{n}\right)^{n}\left(1- \frac{x^2}{n}\right)^{n}\right]\\= e^{-x^2}\left[1 - \left(1- \frac{x^4}{n^2}\right)^{n}\right]\leqslant e^{-x^2}\frac{x^4}{n}.$$

For $x \in [-L,L]$, the RHS is bounded by $L^4/n$, which proves uniform convergence.

To find $\sup \limits _{x\in\mathbb{[-L,L]}}|f_n(x)-e^{-x^2}|$; supremum will be exist for some $ {x\in\mathbb{[-L,L]}}$.Suppose $x=a$ then $\lim\limits_{n\rightarrow\infty} \sup |f_n(x)-e^{-x^2}|=0$.Hence uniformly convergent


Find PDF of $Y=-2\ln X$ given that $X$ is uniform on $(0,1)$ Let $X$ have a uniform distribution with p.d.f. $f(x) = 1$, $x$ is in $(0, 1)$, zero elsewhere.
Find the p.d.f. of $Y = -2 \ln X$.

I don't think this is a very difficult question, I just don't really understand what it is asking or where to start. Any help would be very much appreciated. Thank you!

Update: I did
$$F(y)= P(Y \le y) = P(-2\ln x \le y) = P(\ln x \ge -y/2) = P(x \ge e^{-y/2}).$$
Then $ x=e^{-y/2}$ and $dx/dy =-1/2e^{-y/2}$
Is this all I need to do? Also, I'm not 100% sure why I am using inequalities here- can someone give me a quick explanation?

you are asked to find the probability distribution of the random variable $Y$ that is related to the random variable $X$ by the relation $Y=-2\ln X$, being $X$ uniform in the interval $[0,1]$. You can solve it considering a change of variable applied to the cumulative function $F(x)$:
$$F(x)=\int_0^x f(x)dx = \int_\infty^y f(y)\Big|\frac{dx}{dy}\Big|^{-1}dy$$
so that
$$f(y) = f(x)\Big|\frac{dx}{dy}\Big|_y$$
and in your situation $x=\exp(-y/2)$, so $\Big|\frac{dx}{dy}\Big|_y = \frac{1}{2}\exp(-y/2)$, so $$f(y)= \frac{1}{2}\exp(-y/2)$$

If you try to integrate $f(y)$ between $\infty$ and $0$ you can verify that it gives you $1$.

It was such a long time ago I did this, but this seems to work too. At least in my computer simulations. Maybe it is just a coincidence it works for this particular function, if that is the case, please show me why.

$Y = -2 \ln(X)$ means that if a sample of X is x, then the corresponding sample of Y is $y = -2\ln(x)$

$y = \ln(x^{-2})$ and then $e^y = x^{-2}$ and $e^{-y/2} = x$. We see that since x is between 0 and 1, y will be between 0 and infinity. Now we just need to normalize, i.e. find $k$ so that $\int_0^\infty ke^{-y/2} dy = 1$ and that is an easy exercise in calculus.


what does it mean if $\text{Cov}(X,Y)$ = $\text{Cov}(Y,Y)$ = $\text{Var}(Y)$ If $\text{Cov}(X,Y)$ = $\text{Cov}(Y,Y)$ = $\text{Var}(Y)$

then what can be said about X and Y?

My rusty math skills took me this far:

$$\text{Cov}(X,Y) = E[(X-\mu_X)(Y-\mu_Y)]$$

so we get

$$E[(X-\mu_X)(Y-\mu_Y)] = E[(Y-\mu_Y)(Y-\mu_Y)]$$

Does this mean that $(X-\mu_X) = (Y-\mu_Y)$ ?

How should I interpret that?

Intuitive answers are also more than welcome.

We know that covariance is a bilinear function. So
 $$Cov(X,Y)=Cov(Y,Y)$$
$$\Longleftrightarrow Cov(X,Y)-Cov(Y,Y)=0$$
$$\Longleftrightarrow Cov(X-Y,Y)=0$$

Therefore, $X-Y$ and $Y$ are uncorrelated.

EDIT
Dividing both sides by the product of standard deviations of $X$ and $Y$, we get
$$Cor(X,Y)=\frac{SD(Y)}{SD(X)}$$

Therefore, we observe two more things:

1) X and Y are positively correlated.

2)$SD(Y)<SD(X)$

Hope it helps:)

Obviously, X and Y are not independent. So $(X-\mu_X) = (Y-\mu_Y)$ is uncorrect.


Modular aritmetic and fields I'm studying the concept of field applied to modular aritmetic.
Is it correct to say that, if the dimension is a prime number $p$ then field properties are satisfied for the integers $\bmod p$ ?
And that, if the dimension is a power of a prime number $p$ (that is $p^{n}$), not all integers $\bmod p^{n}$ form a field, but only a set of $p^{n}$ integers ?

The integers modulo $p$ for a prime $p$ does form a field.

The integers modulo $p^n$ for $n>1$ doesn't form a field. And there isn't a subset of that which forms a field either.

There are fields with $p^n$ elements for any prime $p$ and natural number $n$. However, for $n>1$ this field does not coincide with the ring of integers modulo $p^n$. For instance, the additive group of such a field is $(\Bbb Z_p)^n$.


Prove that $I=\{x \in R : ax=0\}$ is a subring of $R$ To do so we just need to prove that I is a ring itself.  By requirements needed to be a ring,

$(I,+)$ is an abelian group: Let $x,y \in R$ then we have that $ax=0=ay$.  So $ax+ay=0=ay+ax$.
$\ast$ is associative: Let $x,y, z \in R$, which means $ax=ay=az=0$.  So $(ax*ay)az=ax(ay*az) =0$
The left and right distributive laws hold:  Let $a,b,c \in R$.  So we see that $a(b+c)=(ab)+(ac)=0$ and $(b+c)a=(ba)+(ca)=0$

Is there anything wrong with this or can anyone suggest a better way to write this?  My instructor is ridiculously picky.

To show that the structure is a sub-ring you need to show that it is closed under the relevant operations (the other restriction are inherited from the ambient ring and do not need to be checked, as mentioned in a comment).

For addition you need to show: $0 \in I$, for all $x, y \in I$ you have $x+y \in I $ and $-x \in I$ (or combied $x-y \in I$).

For multiplication you need to show: for all $x, y \in R$ you have $xy \in I$ (as mentioned in a comment). [It appears your rings do not need to have 1, otherwise you would need to assert it's in I; but this would not be possible in this case.]

So how to do this for addition. Well $a0= 0$ so $0 \in I$ that is fine. And, for $x,y \in I$ you have $ax= ay = 0$ as you said, so $0 = ax + ay = a(x+y)$ and $x + y \in I$.
I leave the argument for $-x$ to you.


Show that this integral of a hyperbolic equals a certain equation Show that $$ \int_0^\infty e^{-ax}\sinh bx \,\mathrm dx = \frac{b}{a^2 - b^2}\,,\quad|b| < a $$ by expanding $\sinh bx$ in a Taylor series at $x = 0$ and integrating term by term.

The instructions confuse me somewhat. Expanding $\sinh bx$ in a Taylor series at $x = 0$ is the McLaurin series right? If this is so, I get this relationship:
$$\sinh bx =  \sum_{n = 0}^\infty \frac{{(bx)}^{2n+1}}{{(2n+1)}!}$$
Integrating this series term by term (skipping some steps) I find a new series:
$$\sum_{n=0}^\infty b^{n+1}\frac{x^n}{n!} = b^{n+1}e^x$$

Notice

Using Laplace transform of $\sinh(bx)$, we get  $$\int_{0}^{\infty}e^{-ax}\sin h(bx)dx$$$$=L[\sin h(bx)]_{s=a}$$
$$=\left[\frac{b}{s^2-b^2}\right]_{s=a}$$
$$=\left[\frac{b}{a^2-b^2}\right]$$$$=\color {red}{\frac{b}{a^2-b^2}}$$
$\forall \ \ |b|<a$


Laplace transform of product of $\sinh(t)$ and $\cos(t)$ If I have a function $f(t)=\sinh(t)\cos(t)$ how would I go about finding the Laplace transform? I tried putting it into the integral defining Laplace transformation:
$$
F(s)= \int_0^\infty \mathrm{e}^{-st}\sinh(t)\cos(t)\,\mathrm{d}t
$$
But this integral looks very hairy to me. Can $\sinh(t)\cos(t)$ be rewritten as something more manageable perhaps?

= L [ { (( e^(-t) - e^(t) ))/2 }.{ cost } ]

Taking Constant outside and separating Laplace to different terms,

= (1/2).{ L[ e^(-t).cost ] - L[ e^(t).cost ] }

Separating solving,

L[ e^(-t).cost ],

first solve ---->  L[ cost ] = s/ (s^(2) + 1)

Therefore for L[ e^(-t).cost ],

Using shifting property,

L[ e^(-t).cost ] = (s+1)/( (s+1)^2 + 1 )  ...........(1)

similarly,

L[ e^(t).cost ] = (s-1)/( (s-1)^2 + 1 )   ............(2)

Subtracting (1) by (2),
will be Answer.


Probability of going into a room randomly There are 6 different rooms and 3 persons. Every person chooses a random room blindfolded. a) What is the probability that all of them go to different rooms? b) what is the probability that they go to the same room?

a) Is it wrong to say that there is a total number of 6*5*4 = 120 ways that this can happen? person 1 has 6 choices, person 2 has 5 choices, person 3 4 choices. Or rather 6^3 = 216?

b) Would it be right to say since there are 6 rooms, so there is only 6 different ways to pick. so P = 6/216?

I am a bit confused here. Can somebody clear this up please and the above question?

Thanks

Probability here can be found by finding the number of different ways your event can occur and dividing by total number of possible outcomes, which is $6^3=216$.

You are right, for (a), the number of ways this can happen is $6 \times 5 \times 4$ -- now compute the relevant probability.

For (b), there is obviously exactly 6 ways this can happen. Can you compute the probability here as well?


Prove divergence of sequence $a_n=1+n\sin\left(\frac{n\pi}{2}\right)$

So the above sequence is obviously not convergent because of the sine, but how could one prove this? Thanks for any pointers.

Tom

Sequence $a_n=1+n\sin\left(\frac{n\pi}{2}\right)$ has a subsequence
$$b_0=a_1,\ b_1=a_5,\ b_2=a_9,\ \ldots,\ b_k=a_{4k+1},\ \ldots$$
of a form $b_k = 4k+2$ which is divergent. Hence sequence $\{a_n\}$ is divergent.

UPD:
$$b_k = a_{4k+1} = 1 + (4k+1)\sin\left(\frac{(4k+1)\pi}{2}\right)=
1 + (4k+1)\sin\left(2k\pi + \frac{\pi}{2}\right) =
$$
$$
= 1 + (4k+1)\sin\left(\frac{\pi}{2}\right) = 1 + 4k+1 = 4k+2
$$

We know that a sequence if diverging to $\infty$ iff $\forall G \in \mathbb{R}$, no matter how large it is, $\exists n \in \mathbb{N}$ such that $a_k>G \; \forall k \geq n$. Well, we can easily find any odd number $>m$ and then the property will hold for some of the terms (all the terms $a_k, k>G, k \equiv 1 \; (\mathrm{mod}\;2))$. So, we have a diverging subsequence. But, we have another subsequence $\{1,1,1,...\}$ which converges to $1$. So, the sequence is neither converging nor diverging to $\infty$, it's infinitely oscillating.

BUT, if we consider divergence as the negation of convergence, it's definitely diverging as it has a diverging subsequence.


Equivalent definitions of $\beta$-smoothness Suppose that the gradient of $f:\mathbb{R}^{n}\rightarrow\mathbb{R}$ is $\beta$-Lipschitz, for some $\beta \geq 0$, i.e.
\begin{equation}
      \|\nabla f(x)-\nabla f(y)\|_2\leq \beta\|x-y\|_2,
\end{equation}
for all $x,y\in\mathbb{R}^n$.
Show that this is equivalent to
\begin{align*}
        f(\theta x+(1-\theta)y)&\geq \theta f(x)+(1-\theta)f(y)-\tfrac{\beta}{2}\theta(1-\theta)\|x-y\|^2,\\
        f(\theta x+(1-\theta)y)&\leq \theta f(x)+(1-\theta)f(y)+\tfrac{\beta}{2}\theta(1-\theta)\|x-y\|^2.
\end{align*}
for all $x,y\in\mathbb{R}^n$ and all $\theta \in [0,1]$,
Moreover, show that it is equivalent to
\begin{align*}
        f(y)&\geq f(x)+ \nabla f(x)^{T}(y-x) - \tfrac{\beta}{2}\|x-y\|^2,\\
        f(y)&\leq f(x) + \nabla f(x)^{T}(y-x) + \tfrac{\beta}{2}\|x-y\|^2,
\end{align*}
for all $x,y\in\mathbb{R}^n$.

Since the gradient of $f$ is $\beta$-smooth over $\mathbb{R}^n$, we have that by the Descent Lemma $$f(y)\leq f(x)+\nabla f(x)^T(y-x)+\frac{L}{2}\|x-y\|^2$$ for all $x,y\in\mathbb{R}^n$. Then can further write
$$f(x)\leq f(\theta x+(1-\theta) y)+\nabla f(\theta x+(1-\theta) y)^T(x-(\theta x+(1-\theta) y))+\frac{L}{2}\|x-(\theta x+(1-\theta) y)\|^2 \\
f(y)\leq f(\theta x+(1-\theta) y)+\nabla f(\theta x+(1-\theta) y)^T(y-(\theta x+(1-\theta) y))+\frac{L}{2}\|y-(\theta x+(1-\theta) y)\|^2$$
which is the same as
$$f(x)\leq f(\theta x+(1-\theta) y)+(1-\theta)\nabla f(\theta x+(1-\theta) y))^T(x-y))+\frac{L(1-\theta)^2}{2}\|x-y\|^2 \\
f(y)\leq f(\theta x+(1-\theta) y)+\theta\nabla f(\theta x+(1-\theta) y))^T(y-x)+\frac{L\theta^2}{2}\|x-y\|^2$$
Now multiply the first inequality by $\theta$ and the second by $(1-\theta)$ and add them together to get the desired result for the first inequality you're trying to prove. The second inequality proof is similar.


Finding $f(x)$ when given a composite function? CONTEXT: Uni homework question made up by lecturer

If, for $n=0$, $1$, $2$, $\ldots$ you're given $f_0(x)=\frac{1}{2-x}$ and $f_{n+1}=f_0(f_n(x))$, how do you prove that your formula for $f_n(x)$ is correct by mathematical induction?

UPDATE

I have computed the first few terms:

$f_1(x)=\frac{2-x}{3-2x}$

$f_2(x)=\frac{3-2x}{4-3x}$

$f_3(x)=\frac{4-3x}{5-4x}$

$f_4(x)=\frac{5-4x}{6-5x}$

Then, from observation of these I developed:

$f_n(x)=\frac{n+1-nx}{n+2-(n+1)x}$

But I don't see how you're supposed to prove this with mathematical induction as it has variable $x$ in it as well as the $n$.

Said in very broad terms, induction is the process of solving some problem for all values of $n$ when you can solve it for some $n_0$ and can derive the solution of the problem $n+1$ knowing the solution for $n$.

In your case, the problem is to "find the symbolic expression of $f_n(x)$", which you managed perfectly.

$$f_0(x)=\frac{n+1-nx}{(n+2)-(n+1)x}=\frac{1}{2-x}$$ is true.

And

$$f_n(x)=\frac{n+1-nx}{(n+2)-(n+1)x}\implies f_{n+1}(x)=f_0(f_n(x))=\frac{n+2-(n+1)x}{(n+3)-(n+2)x}$$
is true as well, as is shown by reworking the expressions.

Just calculate $f_{n+1}$ from $f_{n}$, with both depending on x and check if the pattern holds. The variable x is irrelevant in this case, as you perform your induction on n.

Or, if you prefer it in these words for better understanding, the "pattern" has to still be true for every value of x


Distance between two skew lines. I am to show that the lines

$$x = y = z$$ and $$x + 1 = \frac{y}{2} = \frac{z}{3}$$

are skew and to find the distance between them.

My attempt: Let $$L_1: x=t \:\:\:\:\:\: y=t \:\:\:\:\:\: z=t$$
and

$$L_2: x=s-1 \:\:\:\:\:\: y = 2s \:\:\:\:\:\: z = 3s.$$

$t=s-1 \Rightarrow s = t+1 \Rightarrow t = 2t+2 \:\: \mbox{and} \:\: t = 3t+3$ which is impossible. So the lines do not have an intersection point and are not parallel because $\langle1,1,1\rangle$ and $\langle1,2,3\rangle$ are not parallel. Therefore the lines are skew.

Then, I have to find the distance between the lines. I sketched the following to do so:

$$\vec{m} = \vec{v_1} \times \vec{v_2} = \langle -1,2,-1 \rangle$$
$$\vec{n} = \vec{m} \times \vec{v_2} = \langle8,2-4\rangle$$

$$\Downarrow$$

$$P: 4x+y-2z+4 = 0$$

$$\Downarrow$$

$$P \cap L_1 : (-\frac{4}{3},-\frac{4}{3},-\frac{4}{3})$$

$$\Downarrow$$

$$L: x = -\frac{4}{3} -t \:\:\:\:\:\: y = -\frac{4}{3} +2t \:\:\:\:\:\: z =-\frac{4}{3} -t$$

$$\Downarrow$$

$L \cap L_2 $ does not exist! But it must be.

Where did I go wrong?

You have correctly identified the vector perpendicular to both lines, i.e. $$\left(\begin{matrix}1\\-2\\1\end{matrix}\right)$$

Now find the vector joining any two point on the respective lines, for example, that joining $(0,0,0)$ and $(-1,0,0)$, therefore $$\left(\begin{matrix}1\\0\\0\end{matrix}\right)$$

and calculate the projection of this vector onto the normal.

Therefore the required distance is $$\frac{\left(\begin{matrix}1\\0\\0\end{matrix}\right)\cdot\left(\begin{matrix}1\\-2\\1\end{matrix}\right)}{\sqrt{1+4+1}}=\frac{1}{\sqrt{6}}$$

From what I have found, using normal geometry extended to three different planes, as per your drawing, was:
L1:
y=x
y=z
z=x
L2:
y=2x+2
y=2/3*z
x=1/3*z-1
Then, I solved for each equation to find the three closest points:
{y=2x+2,y=x} » (-2,-2)
{y=2/3*z,y=z} » (0,0)
{x=1/3*z-1,x=z} » (-3/2,-3/2)
Then, I used the distance formula between each point: in (x,y,z) form
(-2,-2,any),(any,0,0)» distance 2
(-3/2,any,-3/2),(any,0,0)» distance 3/2
(-2,-2,any),(-3/2,any,-3/2)» distance 1/2
The distance is √(2^2+3^2/2^2+1/2^2)≈2.5495097568


Is there a sequence producing almost all primes? Is there a sequence which contains only primes and $$\lim_{n \to \infty} \frac{a_n}{p_n} = 1$$ (where $p_n$ is the $n$th prime)?

Edit: sigh. I meant a sequence produced using 0, 1, +, -, *, / used a constant amount of times, powers to a constant and any constant number of the previous elements.

Apart from the trivial solution mentioned in the comments, if you exclude a finite number $k$ of primes, then $a_n=p_{n+k}$ for $n\geq N$ for some $N$ and even in this case you have
$$ \lim_{n\to \infty} p_{n+k}/p_n=1$$
(See Cipolla's asymptotic expansion).

If you exclude primes in arithmetic progression, i.e. $a_n=p_{rn}$ for some $n$, then the limit is $r$.

On the other hand I would be surprised if the sequence $a_n$ of all primes but skipping Mersenne's primes doesn't verify what you want. Or all primes skipping twin primes. Or all primes skipping any set of rare primes.

Or if $a_n$ allow repetitions, you can find a lot more examples. For instance any set of primes in where you replace a subset of zero density of primes by the prime "2".

With the additional constraints on $a_n$ you gave, in any case solution $a_n$ to your problem has to match Cipolla's asymptotics, which are precise. I don't see how to do that with the given operations (they generate rational and exponential functions).

Primes become less common as integers become larger. This includes primes of certain forms, polynomial values and sequences. The Mersenne Numbers $2^n-1$ with $n$ prime have the condition that each divisor of $2^n-1$ must be of the form $8kn+1$ or $8kn-1$, raising the probability of them being prime compared to a random integer $N$ of similar size, however, primes of the form $2^n-1$ are still extremely rare.


Do two right triangles with the same length hypotenuse have the same area? I watched computer monitors and I asked myself, do two monitors with the same display diagonal have the same display area?

I managed to find out that the answer is yes, if two right triangles with the same length hypotenuse have the same area. The answer is trivial if the two triangles are identical, and according to Thales' theorem I know that there are cases when two right triangles have the same length hypotenuse, but they are not identical (the legs length and the angles are different).

So my final question:

  Do two, not identical right triangles with the same length hypotenuse have the same area?

I guess that the answer is yes, but I can't give a proof that I'm 100% sure.

No. Consider the following picture:

$\hspace{90pt}$

All the triangles have the same hypothenuse, but one of them has area of $r^2$ while others might have it arbitrarily small.

I hope this helps $\ddot\smile$

A basic algegra approach lets us not only demonstrate that there can be multiple different values of area for a given hypotenuse, it lets us work out the range of possible values for that area and calculate the corresponding lengths of the other two sides.

Define $A$ as area of the triangle, $c$ as the length of the hypotonuse and $a$ and $b$ as the length of the other two sides. $A > 0$, $a > 0$, $b > 0$, $c > 0$

$$a^2 + b^2 = c^2$$
$$A = \frac{ab}{2}$$
$$2A = ab$$
$$b = \frac{2A}{a}$$

Substituting

$$a^2 + 4\frac{A^2}{a^2} = c^2$$

$$a^4 + 4A^2 = a^2c^2$$

$$a^4 - c^2a^2 + 4A^2 = 0$$

Solving the quadratic in $a^2$.

$$a^2 = \frac{c^2\pm\sqrt{c^4-4*4A^2} }{2}$$

$$a = \sqrt{\frac{c^2\pm\sqrt{c^4-16A^2} }{2}}$$

If $c^4 < 16A^2$ (i.e. if $c^2 < 4A$) there are no soloutions.
If $c^4 = 16A^2$ (i.e. if $c^2 = 4A$) there is one soloution.
If $c^4 > 16A^2$ (i.e. if $c^2 > 4A$) there are two soloutions.


I am looking to show that $n$, $n + 1$, or $n + 2$ are divisible by 3 I am seeking to prove the following:

If $n \in \mathbb{Z}$, then exactly one of the following is true: $\frac{n}{3} \in \mathbb{Z}, \frac{n + 1}{3} \in \mathbb{Z}, \frac{n + 2}{3} \in \mathbb{Z}$.

I believe this is pretty self evident, and hence have no idea where to start. Any tips would be greatly appreciated!

By the Division Algorithm $\, n+2 = 3q+r,\,$ for some $\, r\in\{0,1,2\}$

Therefore we conclude that  $\ n+\! \underbrace{r'}_{\large 2\,-\,r}\!\! = 3q,\, $ for some $\,r'\in \{0,1,2\}$

If $n \in \mathbb{Z} \to$ $n = 3k$ or $3k+1$ or $3k+2$, hence one of the statements must be true.


Formula for the floor function I found the following formula for the floor function:

$$\lfloor x \rfloor = -\frac12+x+\frac{\arctan(\cot\pi x)}{\pi}$$

for all $x$ not an integer.

My question is where I can find the proof of this formula.

It is well-known that the cotangent function has period $\pi$, so that the cotangent of $\pi x$ has period $1$.

With the usual definition of the arc tangent, you get a number between $-\frac\pi2$ and $\frac\pi2$, or, after division by $\pi$, a number between $-\frac12$ and $\frac12$.

Hence,

$$\frac{\arctan\cot\pi x}\pi=\frac{\arctan\tan\left(\dfrac\pi2-\pi x\right)}\pi=\frac12-x\bmod1$$

Now,

$$-\frac12+\frac12+x-x\bmod1=x-\{x\}=\lfloor x\rfloor.$$

Anyway, the formula is virtually useless, as it is undefined for integer $x$.

The arctangent function is multivalued - when thought of as a function of a complex variable, its domain is a Riemann surface with multiple sheets joined at branch cuts in the complex plane. So your formula is not well-defined unless you specify which branch you are considering to be the principal branch. The most natural way to specify the usual principal branch of the arctangent function basically uses the idea of the floor function anyway, so your formula "for" the floor function is correct but somewhat circular.


ODE with singular coefficients I started with an ODE (first ODE) :

$-(1-x^2)y''(x) +x y'(x) -  q(x) y(x) = \lambda y(x).$

Then I got a more sophisticated differential equation ( second one)
and is given by
$$-(1-x^2)y''(x) +x y'(x) -  q(x)y(x) + \frac{1}{(1-x^2)}y(x) = \lambda y(x).$$
What I want is to understand the following:

Given a solution of equation (1), how can I generalize this one to a solution of equation (2).

With convenient change of variable, the ODE can be transformed to another ODE on a simplified form :  $\frac{y’’(t)}{y(t)}=P_2(\cosh(t))$ where $P_2$ is a polynomial of the second degree.

It is known that some particular ODE’s of this form can be solved in terms of MathieuS and MathieuC functions: http://mathworld.wolfram.com/MathieuDifferentialEquation.html
For example, this is the case if one of the coefficient is null and the two others not null : in the case $\gamma=0$ or in the case $\alpha=0$. But not in the case $\lambda=0$. Other cases are solvable in case of particular relationship between the coefficients, in terms of Mathieu funcions or in terms of other special functions of lower level for particular values of the coefficients of the polynomial.

In the general case, i.e. $\alpha, \gamma, \lambda$ not null and not related one to another, a closed form for the solutions would certainly require special functions of higher level, which are not referenced as standard functions. So, I think that the ODE on its general form cannot be analytically solved in that sense.

The second EDO with one more term becomes on the reduced form :

A fortiori, in the general case, the solutions cannot be expressed with a finite number of standard functions. But closed forms of solutions could exist in some cases of particular values of the coefficients $\alpha, \gamma, \lambda, \beta$

Adding a term of different nature to an ODE generally changes a lot the form of the solution. A simplified example is shown below : The solution of the ODE which was expressed in terms of elementary functions becomes much more complicated because requiring special functions (associated Legendre) :

Another example which involves even more complicated functions :

A solution on the exponential form exists in case of particular relationships between the parameters $\gamma, \alpha, \lambda$ :

With the additional term, an approximative solution on exponential form can also be found  in case of particular relationships between the parameters $\gamma, \alpha, \lambda, \beta$ :

But this is of low interest for practical application because it cannot be generalized to ODEs which parameters are not related on the same way.

All this draw to think that the analytic method is not convenient to tackel this problem in case of a so complicated ODE. Other approaches might be more convenient, especially systematic studies with numerical methods.

This isn't an answer but it's too long for a comment. Let $t=x^2$ and note that
$$ x \frac{d}{dx}=x \frac{dt}{dx}\frac{d}{dt}=x\cdot 2x\frac{d}{dx}=2t\frac{d}{dt},\\
   x^2\frac{d^2}{dx^2} = x \frac{d}{dx}\left(x \frac{d}{dx}-1\right)=2 t \frac{d}{dt}\left(2t\frac{d}{dt}-1\right)=4t^2 \frac{d^2}{dt^2}+2t \frac{d}{dt}.$$

If we then multiply your differential equation by $x^2$, we can thus use these identities to rewrite as $$-4t^2(1-t)y''(t)+2t(2t-1)y'(t) \\- 4 (2t-1) \left( \alpha  + \gamma (2t-1) \right) y(t)+\frac{4u^2-1}{4t(1-t)}y(t) = \lambda \, t \, y(t)$$

This may be easier to approach, though it's still quite busy.


When solving for eigenvector, when do you have to check every equation? For example, suppose we want to solve for the eigenvectors of:

$$A = \begin{bmatrix} 0 & 1 \\ 2 & -1 \end{bmatrix} $$

We quickly find the eigenvalues are $1, -2$ i.e. $\sigma(A) = \{1, -2\}$

Then $Av_1 = \lambda_1 v_1$

$$ \begin{bmatrix} 0 & 1 \\ 2 & -1 \end{bmatrix} \begin{bmatrix}v_{11} \\ v_{12} \end{bmatrix} =  \begin{bmatrix} -2v_{11} \\ -2v_{12} \end{bmatrix}$$

We get two equations:

$v_{12} = -2 v_{11}$

$2v_{11} - v_{12} = -2 v_{12}$

Solving either equation will yield equivalent eigenvectors i.e. $v_1 = [1 -2]^T$.

What account for this property? Why is it that we only need to check for one equation in this case?

Are there conditions or criteria for when we should check all equations instead just a single one?

Your question traces back into how one finds an eigenvector of a square matrix ${\bf{A}}$. Suppose that we are looking for nonzero vectors ${\bf{x}}$ called eigenvectors of ${\bf{A}}$ such that

$${\bf{Ax}} = \lambda {\bf{x}}\tag{1}$$

rewrite the equation in the form

$${\bf{Ax}} = \lambda {\bf{Ix}}\tag{2}$$

where ${\bf{I}}$ is the identity matrix. Now, collecting terms to the left side we can get

$$\left( {{\bf{A}} - \lambda {\bf{I}}} \right){\bf{x}} = {\bf{0}}\tag{3}$$

As you see, this is a system of linear algebraic equations. What is the requirement for this system to have nonzero solutions? Yes! the determinant of coefficient matrix should vanish which means

$$\det ({\bf{A}} - \lambda {\bf{I}}) = 0\tag{4}$$

this is the main equation that you find the eigenvalues from it. So, eigenvalues set the determinant of  ${{\bf{A}} - \lambda {\bf{I}}}$ to zero. When the determinant of this matrix is zero, it implies that the equations in (3) are linearly dependent. In your example, your two equations are linearly dependent as you can easily verify that they are really the same

$$\left\{ \begin{array}{l}
{v_{12}} =  - 2{v_{11}}\\
2{v_{11}} - {v_{12}} =  - 2{v_{12}}
\end{array} \right.\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\left\{ \begin{array}{l}
2{v_{11}} + {v_{12}} = 0\\
2{v_{11}} - {v_{12}} + 2{v_{12}} = 0
\end{array} \right.\,\,\,\, \to \,\,\,\,\,\left\{ \begin{array}{l}
2{v_{11}} + {v_{12}} = 0\\
2{v_{11}} + {v_{12}} = 0
\end{array} \right.\tag{5}$$

In conclusion, I might say that you will never use all equations in $(2)$ or $(3)$. If ${\bf{A}}$ is a $n \times n$ matrix, then you have to use $n-1$ or less equations.

Your answer is simple. If $v$ is eigenvector of an operator $A$ correspond to an eigenvalue, $cv$ is also eigenvector of operator $A$ correspond to same eigenvalue for all nonzero $c$, (actually each eigenvector introduce a subspace of domain) so for your question in 2 dimension it's not possible that your final two equations be independent and find only one vector.


Simplify the cube root How would one simplify the cube root of $x^{64}$? Please show the solution step by step if possible. Been a while since I took math so please forgive the simplicity of this question... Thank you in advance.

You need to be careful.  Assuming that $x$ is real, one has $x^{64}$ is always a non-negative real number, and the cube root of a non-negative real number will again be a non-negative real.

$\sqrt[3]{x^{64}}=|x|^{\frac{64}{3}}=|x^{21}\cdot \sqrt[3]{x}|$

Without such care, one can arrive at incorrect results such as $\sqrt{x^2}=x$, which would imply things such as $1=-1$ (by plugging in $-1$ in for $x$)


Showing that the loopspace $\Omega S^{\infty}$ is homotopic to $S^{\infty}$. Showing that the infinite dimensional sphere $S^{\infty}$ is contractible is rather easy by constructing an explicit contraction (Hatcher gives a nice one). I thought it might be a nice exercise to try and show this using loop spaces and the fact that $\pi_1(S^{\infty})$ is trivial.

Let the loopspace $\Omega X$ be the space of base pointed loops in $X$. For now, assume that the loopspace $ΩS^{\infty}$ is homotopy equivalent to $S^{\infty}$. Then, $\pi_2(S^{\infty})=\pi_1(\Omega S^{\infty})=\pi_1(S^{\infty})=0$. Continuing by induction, we get that every homotopy group of $S^{\infty}$ is trivial and so $S^{\infty}$ is contractible because $S^{\infty}$ is a CW-complex (by Whitehead). As far as I can tell, this argument has no holes in it.

My problem now is in proving the assumption that $\Omega S$ is homotopy equivlanet to $S$. I haven't had too much experience with loopspaces or function spaces in general so I'm struggling somewhat.

One approach might be to find a fibration $F\rightarrow\Omega S^{\infty}\rightarrow S^{\infty}$ and then playing about with long exact sequences in homotopy, but I'm not sure if such a fibration exists. Maybe something like 'If $f\colon [0,1]\rightarrow S^{\infty}$ is a loop, let $p\colon \Omega S^{\infty}\rightarrow S^{\infty}$ be given by $p(f)=f(1/2)$' would work? Is p a fibration? And if so, what do the fibers look like? The best case scenario would be for some fibration to exist with contractible fiber. However, I'm worried that showing such a fiber is contractible would amount to showing that $S^{\infty}$ is contractible, which defeats the object of the exercise.

You might be able to get something using the Freudenthal theorem (though this is a bit overkill admittedly), since it studies the connectedness of the natural map to $\Omega \circ S$. Supposing the assertion that the suspension of $S^\infty$ is again $S^\infty$ (we have to use the limiting behavior of this space somewhere), then if $S^\infty$ is $n$-connected, the map

$\pi_k(S^\infty) \to \pi_k(\Omega S^\infty) = \pi_{k+1}(S^\infty)$

is an isomorphism for $k \le 2n$. Then this is certainly true for $k = n$, which gives that $\pi_{n+1}(S^\infty)$ is trivial as well, and so $S^\infty$ is at least $(n+1)$-connected -- we can continue on to hit all the homotopy groups.

Since $\pi_1(S^\infty)=0$, it is rather easy to prove that $S^\infty$ and $\Omega S^\infty$ are homotopy equivalent : set $\rho : S^\infty \to \Omega S^\infty$, $\rho(p)(t)=p$ for every $t\in[0,1]$ (since $\rho(p)$ is a path, it makes sense to write $\rho(p)(t)=p$) and $\tau : \Omega S^\infty \to S^\infty$, $\tau(\gamma)=\gamma(0)$. Now $\tau\circ\rho$ is the identity, and $\rho\circ\tau$ is isotopic to the identity, since $\pi_1(S^\infty)=0$, and so every loop is homotopy equivalent to the constant loop on its base point : choose an isotopy $F_\gamma$ from each loop $\gamma$ to its base point, then set $F:\Omega S^\infty \times [0,1]\to\Omega S^\infty$ by $F(\gamma,t)=F_\gamma(t)$, and this is a homotopy between the identity and $\rho\circ\tau$.


Limit of infinitely small oscillating functions I have a question about limit of "small" oscillating functions.
Let's say I have the two functions of the image (really don't know what they could be). Is the limit for x that goes to infinity of the one in "black" 1 as the one in "red"?

Because if we have a function like f(x)=sin(x)*(1/x) it is clear that the limit for x that goes to infinity is 0, as it goes up and down of always smaller amounts around 0.
But of the function i drawn things are a bit different, as it goes only above y=1 to then "touch" y=1 back again. So I'm not sure if the limit is definite.

Yes it’s possible.... let assume a function f(x) which is the upper bound of our oscillating function h(x).we see that 1<=h(x) <=f(x)
So here if f(x) convergence to 1 then h(x) must have to converge at 1. But if the oscillating function h(x) convergence to 1 we can’t say anything about f(x).
Thank you


Status of The Triangle Book I am interested in finding out about the current status of the planned book: The Triangle Book by John H. Conway and Steve Sigur. I understand that Steve Sigur died some time back. I got no reply from Prof. Conway; is there someone who knows the fate of this project. Is it abandoned or is it still under preparation? Here is some information about the planned book.

The Triangle Book is on hold (Conway died this year) but there are people trying to revive it trust me.
Colm Mulcahy

On a little googling , i find that the book is for sale

http://www.amazon.com/Triangle-Book-John-H-Conway/dp/1568811659


prove $f\in L^2([0,1])$ Assume $f:[0,1]\to R$ is a measurable function such that $fg\in L^1([0,1])$ for all $g\in L^2([0,1])$.prove that $f\in L^2([0,1])$.

My opinion: if I can find a function g such that f+g is in $L^2([0,1])$, then use $$\int(f+g)^2-\int2fg+g^2$$to get $\int f^2$, then we can prove f is in $L^2([0,1])$. Is it right?

We may assume $f\geq 0$. We need to show that
$$ \Lambda_f : g\in L^2 \mapsto \int f g \; dx $$
is a bounded linear functional on $L^2$. If this is the case then
by Riesz there is $h\in L^2$ so that $\Lambda_f(g)=\int hg \; dx$ and one sees that $f=h$ a.s.

So assume that $\Lambda_f$ is not bounded and let $g_n\geq 0$ be a sequence with $\|g_n\|_{L^2}=1$ so that $\Lambda_f(g_n) \rightarrow \infty$. Extracting a subsequence we may assume that $\Lambda_f(g_n) \geq 4^n$. Then
$$ g= \sum_{n\geq 0} \frac{1}{2^n} g_n$$ has $L^2$ norm not greater than $2$ but by monotone convergence $\Lambda_f(g)=+\infty$.


A lighthouse problem (Calculus). Is the problem as easy as it seems? The problem is supposed to be a modified version of the lighthouse problem.

Description:

A lighthouse (in point N) is located 4 km away from the shore at sea. A bar is located on the shoreline, 4 km away from point W. Point W is located on the shore such that it is the shortest and the most direct path from the lighthouse to the shore. The shoreline is perpendicular to WN.
We know that a man who is now in the lighthouse wants to go the bar. Also, we know that he rows at the rate of 4 km/h and walks at the rate of 5 km/h. What path should the man choose in order to get to the bar in shortest period of time?

My question is: is the problem as easily solvable as I understand? It seems that no knowledge of calculus is needed to solve it. Or I am missing something?

My solution: we get a triangle which has a perpendicular. We calculate NK, where K is the bar and N is the lighthouse. So now we have 2 paths: NK and NWK. We use the rowing and walking rates to calculate how much time each path will take. Then we compare the results.

As a result, the shortest path is rowing straight from the lighthouse to the bar.

You are correct. The straight line solution is the correct solution. Here is why.

He will row along some straight line path to the shore and land at a point $x$ which is between $0$ and $4$ kilometers from the bar. We want to minimize the total time it takes. The total time will consist of his time on the water and his time on the shore. Making use of the distance formula $d=rt$, where $d$ is distance, $r$ is rate and $t$ is time, and the fact that the total is the sum of its parts, we get

$$t=\frac{\sqrt{16+x^2}}{4}+\frac{4-x}{5}$$ as the total time.

The first term in the sum is the time on the water and the second term represents the time on land.

Taking the derivative with respect to $x$ and setting it to $0$  gives $$\frac{dt}{dx}=\frac{x}{4\sqrt{16+x^2}}-\frac{1}{5}=0\implies x=\frac{4}{5}\sqrt{16+x^2}\implies x^2=\frac{16}{25}(16+x^2)\implies x^2=\frac{256}{9}$$ which gives $x=\frac{16}{3}$

But $\frac{16}{3}\gt 5$ and this expression only works for $x$ between $0$ and $4$.

Moreover, for $x$ between $0$ and $4$, $\frac{dt}{dx}\lt 0$ which means more $x$, less time. Therefore, the minimum time occurs when $x=4$ and the minimum time is $t=\sqrt{2}$ which is what you originally had.

You are right.Walking 8 km at 5 kph takes 1.6 hrs.Rowing $4\sqrt 2$ km at $4$ kph takes $\sqrt 2$ hrs. And $\sqrt 2<1.6 \iff 2<1.6^2=2.56$. And no calculus.Which is good because he should not drink and calculus.


Formal negation of $((p\rightarrow q) \vee (q \leftrightarrow r)) \rightarrow q$ Can someone give me an outline for how I can negate the following?

$((p\rightarrow q) \vee (q \leftrightarrow r)) \rightarrow q$

You want to work from the inside out. Call the inside of the outer parentheses $s$, so $s=(p\to q)\vee (q\leftrightarrow r)$.

Then, the desired negation is $\lnot (s\to q)$. How do we evaluate that? Well, what actually is $\to$? $s \to q$ means that if $s$ is true, then $q$ is true. If $s$ is false, it doesn't say anything. Thus, $(s\to q)$ is only false if $s$ is true and $q$ is false, or in other words, $\lnot (s\to q)=s \wedge (\lnot q)$

Another way to see this is with truth tables:

$$\begin{array}{c|c|c|c}s& q & s\to q&\lnot(s\to q)\\
T&T&T&F\\
T&F&F&T\\
F&T&T&F\\
F&F&T&F\end{array}$$

Expanding $s$ back out again gives $$((p\to q)\vee (q\leftrightarrow r))\wedge(\lnot q)$$

¬((p → q) ∨ (q ↔ r)) → ¬ q
¬p → ¬q ∧ ¬q ↔ ¬r → ¬ q


An integral of a rational function with high degrees: Evaluate $\int_{-\infty}^{\infty}\frac{(x^4 + 1)^2}{x^{12} + 1}dx$. Calculate: $$\int_{-\infty}^{\infty}\frac{(x^4 + 1)^2}{x^{12} + 1}dx.$$

What I have tried is to divide both numerator and denominator with $x^4 + 1$ and then get two following integrals, because of parity of the integrand (nothing else worked for me): $$2\int_0^\infty \frac{x^4 + 1}{(x^4 - \sqrt3x^2 + 1)(x^4 + \sqrt3x^2 + 1)}dx = $$ $$ = \int_0^\infty \frac1{x^4 - \sqrt3x^2 + 1}dx + \int_0^\infty\frac1{x^4 + \sqrt3x^2 + 1}dx = $$ $$ = \int_0^\infty \frac1{(x^2 - \frac{\sqrt3}2)^2 + \frac14}dx + \int_0^\infty \frac1{(x^2 + \frac{\sqrt3}2)^2 + \frac14}dx.$$
I don't see what would be continuation of this. Any help is appreciated.
Thank you for any help. Appreciate it.

Hint: For integrals of rational functions of even polynomials over the real line, the following substitution can come in handy:

$$\begin{align}
\mathcal{I}
&=\int_{-\infty}^{\infty}\mathrm{d}x\,\frac{\left(x^{4}+1\right)^{2}}{x^{12}+1}\\
&=\int_{-\infty}^{\infty}\mathrm{d}x\,\frac{\left(x^{4}+1\right)^{2}}{\left(x^{4}+1\right)\left(x^{8}-x^{4}+1\right)}\\
&=\int_{-\infty}^{\infty}\mathrm{d}x\,\frac{x^{4}+1}{x^{8}-x^{4}+1}\\
&=2\int_{0}^{\infty}\mathrm{d}x\,\frac{x^{4}+1}{x^{8}-x^{4}+1}\\
&=2\int_{0}^{\infty}\mathrm{d}x\,\frac{\left(x^{2}+x^{-2}\right)}{x^{2}\left(x^{4}+x^{-4}-1\right)}\\
&=2\int_{0}^{\infty}\mathrm{d}x\,\frac{\left(x^{2}+x^{-2}\right)}{\left(x^{4}+x^{-4}-1\right)};~~~\small{\left[x\mapsto\frac{1}{x}\right]}\\
&=\int_{0}^{\infty}\mathrm{d}x\,\frac{\left(x^{2}+x^{-2}\right)}{x^{2}\left(x^{4}+x^{-4}-1\right)}+\int_{0}^{\infty}\mathrm{d}x\,\frac{\left(x^{2}+x^{-2}\right)}{\left(x^{4}+x^{-4}-1\right)}\\
&=\int_{0}^{\infty}\mathrm{d}x\,\frac{x^{2}+1}{x^{2}}\cdot\frac{\left(x^{2}+x^{-2}\right)}{\left(x^{4}+x^{-4}-1\right)}\\
&=\int_{-\infty}^{\infty}\mathrm{d}y\,\frac{2\left(4y^{2}+2\right)}{\left(16y^{4}+16y^{2}+1\right)};~~~\small{\left[\frac{x^{-1}-x}{2}=y\iff x=\sqrt{y^{2}+1}-y\right]}.\\
\end{align}$$

Note that we have cut the degrees of the polynomials in half and still have all integer coefficients. This could be considered an advantage over the brute-force partial fraction method if you're looking to avoid messy factorizations.

More a comment than an answer:

According to Wolfy,
the indefinite integrals
involve arctan and
complex values.
But the definite integrals
come out like this:

$\int_0^∞ \dfrac{dx}{x^4 - \sqrt{3} x^2 + 1}
= \frac12 \sqrt{2 + \sqrt{3}} π≈3.0345
$
and
$\int_0^∞ \dfrac{dx}{x^4 + \sqrt{3} x^2 + 1} dx = \dfrac{π}{2 \sqrt{2 + \sqrt{3}}}
≈0.8131
$


Use a convergence theorem to evaluate the integral I am working with the Monotone, Dominant, and Bounded Convergence Theorems. I am attempting to evaluate the integral

lim$_{n\rightarrow\infty} \int_{0}^{1} \big(1+\displaystyle\frac{x}{n}\big)^{-n} \text{cos}\big(\frac{x}{n}\big) dx$

I have that $f_n(x) = \big(1+\displaystyle\frac{x}{n}\big)^{-n} \text{cos}\big(\frac{x}{n}\big)$ which is a continuous function on $[0,1]$ therefore it is Riemann Integrable.

$|f_n| = |\big(1+\displaystyle\frac{x}{n}\big)^{-n} \text{cos}\big(\frac{x}{n}\big)| \leq (1+0)^{-n} \cdot \text{cos}(0) = 0$

I am stuck on the rest. Now I have to show that $f_n\rightarrow$ to something almost everywhere.

It is a standard fact that $(1+\frac x n)^{n} \to e^{x}$ as $n \to \infty$. Hence $(1+\frac x n)^{-n} \to e^{-x}$ as $n \to \infty$. Since $\cos (\frac x n ) \to 1$ we get $f_n (x) \to e^{-x}$. To apply Dominated Convergence Theorem just note that $|f_n(x)| \leq 1$ (since $1+\frac x n \geq 1$ and $|\cos \frac x n| \leq 1$) and the constant function 1 is integrable. Hence the required limit is $\int_0^{1} e^{-x}dx=1-\frac 1 e$.

Hint: Observe
\begin{align}
\left(1+\frac{x}{m} \right)^m  \leq \left(1+\frac{x}{n} \right)^n
\end{align}
whenever $m\leq n$.

   Additional Hint: Take $m=2$.

Here's the spoiler:
Observe we have

   \begin{align}\left( 1+\frac{x}{n}\right)^{-n} \leq \left(1+\frac{x}{2}\right)^{-2}\end{align}

for all $n\geq 2$. Hence we have that

   \begin{align} \left|\left(1+\frac{x}{n} \right)^{-n}\cos\frac{x}{n} \right| \leq \left(1+\frac{x}{2}\right)^{-2}. \end{align}

Since $(1+x/2)^{-2}$ is integrable then by Dominated Convergence Theorem, we have that

   \begin{align} \lim_{n\rightarrow \infty}\int^1_0\left(1+\frac{x}{n}\right)^{-n}\cos\frac{x}{n}\ dx = \int^1_0 e^{-x}\ dx = 1-e^{-1}.\end{align}


Formal symbol for the integer division operation The integer division is a common and useful operation in Computer Science. It comes up in many domains, as in the manipulation of matrices and grids.

Is there any formal symbol for this operation? Or at least a widely recognisable symbol that can be easily differentiated from the standard division (i.e. inverse of multiplication)?

$\lfloor a/b \rfloor$ (@barakmanos comment suggestion's) is probably the only formal way to express this, without creating new definitions.

Alternatively, you could follow @Henry 's suggestion to explicitly define a symbol to do the integer division operation. The common ones are \ (backslash, as opposed to / forward slash used for regular division), or div.

The advantages of using $\lfloor a/b \rfloor$ are that:

you don't need a new definition
you often need to combine integer division with +1 for various indexing problems. This can be easily achieved with $\lceil a/b \rceil = \lfloor a/b \rfloor + 1$ (edit:) for $a/b \notin \mathbb{Z}$

The edit suggested in the comment (@Tonyk) arguably invalidates the second advantage, since $a/b$ will likely also be an integer in indexing problems. I would personally still prefer the $\lfloor a/b \rfloor$ notation, simply for the first point.

I often use $\text{div}$, which is complementary with $\text{mod}$.

There are many alternatives (some used in programming languages) such as $/,//,\backslash,\div,\%$ but these don't go without saying.

Also $\left\lfloor\dfrac\cdot\cdot\right\rfloor$.


Can you solve for the identity matrix where A is non-invertible? If you have an $n\times m$ matrix $A$, and we assume that those dimensions are something like $3\times6$, is there a matrix $D$ such that $AD = I$?

You can prove the following result:
Claim Let $A$ be a $2 \times n$ matrix. Then, there exists a $n \times 2$ matrix $B$ so that $AD=I_2$ if and only if $\mbox{Rank}(A)=2$.
More generally, for an $m \times n$ matrix $A$, there exists a $n \times m$ matrix $B$ so that $AD=I_m$ if and only if $\mbox{Rank}(A)=m$.
This can be proven by using the fact that the corresponding linear transformation $T :\mathbb R^n \to \mathbb R^2$ is onto and hence has a right inverse.
A more elementary solution, which also tells you how to find $D$, is the following:
The systems
$$Ax= \begin{bmatrix} 1 \\ 0 \end{bmatrix} \\
Ay= \begin{bmatrix} 0 \\ 1 \end{bmatrix} \\
$$
are consistent (why?). Let $D$ be the matrix with columns $x$ and $y$ respectively.
Then
$$
AD = \begin{bmatrix} 1 &0 \\ 0&1 \end{bmatrix}
$$

If $AA^T$ is invertible, then $D=A^T(AA^T)^{-1}$. This is the Moore-Penrose inverse; see https://en.wikipedia.org/wiki/Moore%E2%80%93Penrose_inverse.
Below is a series of Python commands to find the pseudo-inverse of a 2 x 4 matrix.
>>> import numpy as np
>>> A = np.array([[2, 4, -6, -10], [4, 8, -8, -8]])
>>> A
array([[  2,   4,  -6, -10],
       [  4,   8,  -8,  -8]])
>>> np.linalg.pinv(A)
array([[-0.06060606,  0.06818182],
       [-0.12121212,  0.13636364],
       [ 0.02272727, -0.05681818],
       [-0.17424242,  0.10227273]])
C = np.linalg.pinv(A)
>>> C.dot(A)
array([[ 1.00000000e+00, -2.22044605e-16],
       [ 8.88178420e-16,  1.00000000e+00]])

Note that since $n < m$, we can only find a left-inverse. Rank-nullity quickly implies that no right-inverse exists.


Applying the Taylor's Theorem, show that if $x > 0$ then $|(1+x)^{1/3} - (1 + \frac x3 - {x^2\over 9})| \le {5x^3\over 81}$ I've been struggling with this problem for weeks now. Seeing as the Taylor's Theorem has up to n terms, I have no idea where to start or stop with it as I wasn't given a n to work with. Can I use the MVT? How would I use the MVT? I'm truly lost, and any help would be appreciated.

Let $f(x)=(1+x)^{1/3}$. It is easy to verify that that the sum of the first three terms of the Taylor  expansion of $f(x)$ about $x=0$ is $1+\frac{x}{3}-\frac{x^2}{9}$.

Note that $f'''(x)=-\frac{10}{27}(1+x)^{-8/3}$.

By the Lagrange form of the remainder, the absolute value of the error when we truncate at the $x^2$ term in the Taylor expansion is
$$\frac{1}{3!}|f'''(c)| |x|^3,$$
where $c$ is between $0$ and $x$. In our case, that shows that the absolute value of the error is
$$\frac{1}{3!}\cdot \frac{10}{27}(1+c)^{-8/3}x^3.\tag{1}$$
Since $x$ is positive, $0\lt (1+c)^{-8/3}\lt 1$. The desired result now follows from (1).

Remark: The Lagrange form of the remainder can be proved using nothing more than the Mean Value Theorem. Probably, however, you are expected to just use the Lagrange form of the remainder, and not to prove it, even for this particular case.

Hint: Check the remainder of alternating series.

Added: In your case $f(x) = (1+x)^{1/3}$

  $$\Bigg| (1+x)^{1/3}-(1-x/3+x^2/9)\Bigg|\leq a_3$$

Where

  $$a_3=\frac{f^{(3)}(0)} {3!}x^3. $$

If calculate the third derivative of the function at $x=0$ you will get

  $$f^{(3)}(0)=\frac{10}{27}. $$


Adjoint Operator to the Derivative Let $V \subset \Bbb R[X]$ be the Vectorspace of all Polynomials of degree $\le 3$.  The inner product on $V$ is defined as follows: $$\langle f,g \rangle:=\int^1_{-1}f(t)g(t)dt$$Let $L:V \to V$ be the derivative $L(p) = p'$

I now have to determine the adjoint Operation $L^*$ so that $$ \langle L(f),g \rangle=\langle f,L^*(g) \rangle$$ Any ideas? ( i tried partial Integration but i don't get anywhere.)

Thanks in advance

Integration by parts $\int_{-1}^1 f'g=[fg]_{-1}^1-\int_{-1}^1 fg'$ is a good idea. But you have to be in a context where the term $[fg]_{-1}^1$ is zero. In order that this condition is always fulfilled,  you must restrict your working space to functions that are zero at both bounds $-1$ and $1$ (it is no longer a vector space). In this framework, we have

$$\int_{-1}^1 f'g=\int_{-1}^1 f(-g)'$$

proving that the adjoint of $L$ is its opposite

$$L^*=-L$$

Remark (a discrete analogy): If you are acustomed to the discrete equivalent of the derivation operator $L$, represented by the infinite matrix:

$$\begin{bmatrix}&\cdots&&&&&\\&-1&1&&&&\\&&-1&1&&&\\&&&-1&1&&\\&&&&-1&1&\\&&&&&\cdots&\\&&&&&&\cdots\\\end{bmatrix}$$
its transpose is as well its opposite.


Orthogonal complex vectors with same magnitude of each entry Given a complex vector $z = (z_1, \ldots, z_n)$, how to find a complex vector $y = (y_1, \ldots, y_n)$ such that $z \cdot y = 0$ (with respect to the complex inner product) and $|y_i| = |z_i|$ for each $i$? Is this always possible? If not, then what conditions does $z$ have to satisfy for this to be possible?

It is not always possible to find $y$ such that $z \cdot y = 0$. For example, if $z=(z_1, 0, \cdots, 0)$, where $|z_1| \neq 0$ then $y = (|z_1|e^{i \theta},0,\cdots,0)$, which implies $$z.y = |z_1|^2 e^{i \phi} \neq 0.$$

The dot product is zero when the 'vectors' are normal. Hence, choose $y=ze^{\pm i\pi/2}=\pm iz$.


Two positive integer with prime number Let $a, b$ be distinct positive integers. Prove that there exists a prime $p$ such that when dividing both $a$ and $b$ by $p$, the remainder of $a$ is less than the remainder of $b$.

How can i solve this?

Just take a prime dividing $a$ but not $b$.


Homology groups of orientable surfaces. Edit: I have a proof here but when I spoke last with my professor, she told me something was close, but not quite. Can someone help me patch this proof? I've been trying to get this down for quite a while and am starting to feel a bit frustrated with myself.

I am trying to show that the second (simplicial) homology group or a compact, connected orientable surface without boundary is ismormophic to $\mathbb Z$.

I can show that this group is non-trivial by triangulating the surface, and considering the chain consisting of all the 2-simplices of the triangulation. Since the surface is orientable, and the simplices are oriented compatibly, each edge of each two simplex will appear twice pointing in opposite directions (a fact which wasn't hard to prove, but I won't type up because LaTeX), and so they cancel. That means the chain consisting of all two-simplices is a cycle. Since there are no three-simplices when dealing with surfaces, there is nothing to be the boundary of. So the two-homology group is non-trivial at least.

Now, to show that $H_2(S)$ for some orientable $S$ is not larger than $\mathbb Z,$ we proceed by contradiction. Suppose that $H_2(S)$ had a second generator. Let's call the above chain of all 2-simplices $K=\sum_{i=1}^n \sigma_i.$ If there is another generator for $H_2(S)$, it is also a cycle, so we can write $L=\sum_{j=1}^k a_j \sigma_j$, with the property that $\partial L=\sum a_j \partial \sigma_j =0.$

Now suppose that some positively oriented 1-simplex $(+)v$ appears in $\partial \sigma_x$, and $(-)v$ appears in $\partial \sigma_y$, then the coefficients $a_x$ and $a_y$ must be equal. Since $v$ was arbitrary, we have $a=a_1=...=a_k.$ So now we can say that $\partial L=\sum_{j=1}^n \sigma_j=0,$ by removing the coefficients and cancelling. This means that $L$ is only a distinct generator for $H_2(S)$ when $k\neq n.$

If this is true then, we can take the difference of the two chains, and produce a third chain so that $K$ is a linear combination of these two chains, producing a basis for $H_2(S)$ isomorphic to $\mathbb Z \bigoplus \mathbb Z$. These two chains are $K-L$ and $L$. However, since $S$ is a connected surface, it is not the disjoint union of two other surfaces, and moreover, since it is connected, it is path-connected. This means that there is a path from any point belonging to a simplex in $K-L$ to any point belonging to a simplex appearing in $L$. However, since there are no 2-simplices between them, the path must go through only 1-simplices of $S$. But then any point in this 1-simplex does not have neighborhoods homeomorphic to 2-disks, so that $S$ is not a surface. A contradiction. So this proves that $k=n,$ so that $L$ is not a distinct generator for $H_2(S).$ Since $H_2(S)$ has only one generator, and $S$ is orientable, $H_2(S)$ is isomorphic to $\mathbb Z$.

What's going wrong here?

Now suppose that some positively oriented 1-simplex $(+)v$ appears in $\partial \sigma_x$, and $(-)v$ appears in $\partial \sigma_y$, then the coefficients $a_x$ and $a_y$ must be equal. Since $v$ was arbitrary, we have $a=a_1=...=a_k.$

This doesn't follow immediately.  You've shown that if $\sigma_x$ and $\sigma_y$ share a boundary edge, then $a_x=a_y$ (to be clear, you should also mention the fact that $\sigma_x$ and $\sigma_y$ are the only two triangles that contain that boundary edge, so that you can't get a $v$ term from $\partial a_j$ for $j\neq x,y$).  But why does this imply that all of the $a_j$ are equal?

The reason is that $S$ is connected.  You know that if $\sigma_x$ and $\sigma_y$ share an edge, then $a_x=a_y$.  But since $S$ is connected, every triangle in $S$ can be reached by starting at $\sigma_x$ and repeatedly "taking the triangle on the other side of a boundary edge".

To be precise, let $\sim$ be the equivalence relation on triangles generated by the relation "shares a boundary edge".  The union of the triangles in any equivalence class is a clopen subset of $S$, and so there can only be one equivalence class by connectedness.  Now the relation $\sim'$ defined by $\sigma_x\sim' \sigma_y$ if $a_x=a_y$ must contain $\sim$, since $a_x=a_y$ whenever $\sigma_x$ and $\sigma_y$ share a boundary edge.  Thus $\sim'$ also has only one equivalence class.

But this actually proves something stronger than you claimed: it proves that you must have $a_1=a_2=\dots=a_n$, not just $a_1=a_2=\dots=a_k$, because nothing in this argument says you have to restrict to $\sigma_x$ such that $a_x\neq 0$.  If you write $a$ for the common value of all the coefficients, you then have $L=aK$.  So every cycle is an integer multiple of $K$.

You wrote already everything down precisely then! You know that a homomorphism between free abelian groups can be represented by a matrix. Write it down and you will see the kernel (i.e. $H_2$) which is $\langle (1,\cdots,1) \rangle$.


Counterintuitive examples in probability I want to teach a short course in probability and I am looking for some counter-intuitive examples in Probability. The results seem to be obviously false but they true or vice versa.
I already found some things. For example these two videos:

Penney's game
How to win a guessing game

In addition, I have found some weird examples of random walks. For example this amazing theorem:

For a simple random walk, the mean number of visits to point $b$ before returning to the origin is equal to $1$ for every $b \neq 0$.

I have also found some advanced examples such as Do Longer Games Favor the Stronger Player?
Could you please do me a favor and share some other examples of such problems? It's very exciting to read yours...

The most famous counter-intuitive probability theory example is the Monty Hall Problem

In a game show, there are three doors behind which there are a car and two goats. However, which door conceals which is unknown to you, the player.
Your aim is to select the door behind which the car is. So, you go and stand in front of a door of your choice.
At this point, regardless of which door you selected, the game show host chooses and opens one of the remaining two doors. If you chose the door with the car, the host selects one of the two remaining doors at random (with equal probability) and opens that door. If you chose a door with a goat, the host selects and opens the other door with a goat.
You are given the option of standing where you are and switching to the other closed door.

Does switching to the other door increase your chances of winning? Or does it not matter?

The answer is that it does matter whether or not you switch. This is initially counter-intuitive for someone seeing this problem for the first time.

If a family has two children, at least one of which is a daughter, what is the probability that both of them are daughters?
If a family has two children, the elder of which is a daughter, what is the probability that both of them are the daughters?

A beginner in probability would expect the answers to both these questions to be the same, which they are not.

Math with Bad Drawings explains this paradox with a great story as a part of a seven-post series in Probability Theory

Nontransitive Dice

Let persons P, Q, R have three distinct dice.

If it is the case that P is more likely to win over Q, and Q is more likely to win over R, is it the case that P is likely to win over R?

The answer, strangely, is no. One such dice configuration is $(\left \{2,2,4,4,9,9 \right\},\left \{ 1,1,6,6,8,8\right \},\left \{ 3,3,5,5,7,7 \right \})$

Sleeping Beauty Paradox

(This is related to philosophy/epistemology and is more related to subjective probability/beliefs than objective interpretations of it.)

Today is Sunday. Sleeping Beauty drinks a powerful sleeping potion and falls asleep.

Her attendant tosses a fair coin and records the result.

The coin lands in Heads. Beauty is awakened only on Monday and interviewed. Her memory is erased and she is again put back to sleep.
The coin lands in Tails. Beauty is awakened and interviewed on Monday. Her memory is erased and she's put back to sleep again. On Tuesday, she is once again awaken, interviewed and finally put back to sleep.

In essence, the awakenings on Mondays and Tuesdays are indistinguishable to her.

The most important question she's asked in the interviews is

  What is your credence (degree of belief) that the coin landed in heads?

Given that Sleeping Beauty is epistemologically rational and is aware of all the rules of the experiment on Sunday, what should be her answer?

This problem seems simple on the surface but there are both arguments for the answer $\frac{1}{2}$ and $\frac{1}{3}$ and there is no common consensus among modern epistemologists on this one.

Ellsberg Paradox

Consider the following situation:

  In an urn, you have 90 balls of 3 colors: red, blue and yellow.
  30 balls are known to be red. All the other balls are either blue or yellow.

  There are two lotteries:

  Lottery A: A random ball is chosen. You win a prize if the ball is red.
  Lottery B: A random ball is chosen. You win a prize if the ball is blue.

Question: In which lottery would you want to participate?

  Lottery X: A random ball is chosen. You win a prize if the ball is either red or yellow.
  Lottery Y: A random ball is chosen. You win a prize if the ball is either blue or yellow.

Question: In which lottery would you want to participate?

If you are an average person, you'd choose Lottery A over Lottery B and Lottery Y over Lottery X.

However, it can be shown that there is no way to assign probabilities in a way that make this look rational. One way to deal with this is to extend the concept of probability to that of imprecise probabilities.

Consider the $d-$dimensional sphere,  then as $d$ goes to infinity the mass concentrates on the equator $x_1=0$


Range of trigonometric functions I would like to know if there is a simple approach to find the range of functions in the form:
$$\sin x\sin2x$$
$$\cos x\cos3x$$
$$\sin 2x\cos 4x$$

For example, finding the range of a function in the form:
$$a\cos\theta + b\sin\theta$$ is simple (the minimum value is $-\sqrt{a^2 + b^2}$ while the maximum value is $\sqrt{a^2 + b^2}$.

It's more interesting when it is not the obvious upper bound $1$. I will take the example $\sin 3x \cos 5 x$. It does not reach the value $1$, so we have some work to do. Let's find an implicit equation for the curve $C \colon \{(\sin(3t), \cos (5t))\ \mid \ t\in [0, 2 \pi]\}$  ( http://en.wikipedia.org/wiki/Lissajous_curve). Skipping some details, it is the curve with equation
$$-1 + 25 x^2 - 200 x^4 + 560 x^6 - 640 x^8 + 256 x^{10} + 9 y^2 -
 24 y^4 + 16 y^6=0$$

It has to do with the Chebyshev polynomials. In fact, a point $(x,y)$ in $\mathbb{R}^2$ is of the form $(\sin (3t), \cos(5t))$ if and only if $1 = P(x) + Q(y)$ where $P(\sin( \alpha))= \sin^2 (5 \alpha)$ and $Q(\cos(\beta)) = \cos^2(3 \beta)$. (the only if is clear, since $\sin^2(15 t) + \cos^2(15 t) = 1$). So it is not that hard to get the implicit form for the curve $C$.

So now we need to find

$$\max x y \ \text{ where }\  -1 + 25 x^2 - 200 x^4 + 560 x^6 - 640 x^8 + 256 x^{10} + 9 y^2 -
 24 y^4 + 16 y^6=0$$

We omit the calculations using Lagrange multipliers. It turns out that the maximum $M$ is the largest root of the equation
$$1073741824\, t^8-1644167168\, t^6+656998400\, t^4-52537500 \,t^2+84375=0$$ $M= 0.96410...$

Well, at least setting the Lagrange multiplier problem in general is not that hard. Solving it is a different thing.

It is not clear to me whether a general easy formula for this maximum exists for general $m$, $n$ for $\max \sin(m t) \cos (n t )$. Maybe a general method, not a general formula that is easy to apply.

Alternatively one writes

$$\sin 3t \cos 5 t = \frac{1}{2}( \sin 8t - \sin 2t)$$

Reduce to an equivalent problem: maximize  $ \sin 4u - \sin u$. Even this one is not straightforward. Certainly the derivative is easy to calculate but the maximal value is again the solution of an equation of degree $8$. Perhaps the advantage is that one can find the solution from the graphs of $\sin u$, $\sin 4u$.

$\bf{Added:}$. If one looks for the maximum value of $\sin mt \cos nt$, it's enough to consider the case $m,n$ relatively prime. Then, even it the maximal value is not $1$, it will get closer to one with the increase of $\max (m,n)$. It is intuitive since the Lissajous curves tend to fill up the square. It would be interesting to investigate how close to $1$ one gets as $\max(m,n) \to \infty$. It appears that number theory, more precisely - rational approximation, appears.

We have always $ a^2\leq a^2+b^2\  \Rightarrow\  |a|\leq\sqrt{a^2+b^2} \Rightarrow \frac{|a|}{\sqrt{a^2+b^2}}\leq 1\Rightarrow-1\leq\frac{a}{\sqrt{a^2+b^2}}\leq1$ so there exists a $\alpha $ that  $ \cos\alpha=\frac{a}{\sqrt{a^2+b^2}}$. Beside  We know that $\sin\alpha=\pm\sqrt{1-\cos^2\alpha}=\pm\sqrt{1-(\frac{a}{\sqrt{a^2+b^2}})^2}=\pm\frac{b}{\sqrt{a^2+b^2}}$. Now let be  $y=a cos \theta+b\sin \theta $ then $\frac{y}{\sqrt{a^2+b^2}}=\frac{a}{\sqrt{a^2+b^2}}\sin \theta+\frac{b}{\sqrt{a^2+b^2}}\cos \theta=\cos\alpha\sin \theta\pm\sin\alpha\cos \theta=\sin (\theta\pm\alpha)$. Because of $-1\leq\sin (\theta\pm\alpha)\leq 1$ then $-1\leq\frac{y}{\sqrt{a^2+b^2}}\leq 1$. Hence $-\sqrt{a^2+b^2}\leq a\cos \theta+b\sin \theta\leq \sqrt{a^2+b^2}$.


Is it possible to construct a strictly monotonic sequence of all rational numbers? I know that the set of all rational numbers is countable, and can be enumerated by a sequence, say $\{a_n\}$.  But can we construct a monotonic $\{a_n\}_{n=1}^{\infty}$, e.g. with $a_k<a_{k+1}$?  It doesn't seem plausible to me, because then $a_1$ would be the smallest rational number, which clearly can't be any finite number.  Am I mistaken?

if $a_k< a_{k+1}$ then $x := \frac{a_{k+1}+a_k}{2}$ is a rational number between those two, so no.

As stated, the answer is no, because the question uses the symbol $<$ which has the implied meaning: The usual ordering of $\mathbb{Q}$ where $\frac{a}{b}<\frac{c}{d}$ iff $ad < bc$ in $\mathbb{Z}$.

But.

As mentioned in another answer, $\mathbb{Q}$ can be well-ordered, i.e. one can define a different order $\prec$ with the property that every nonempty subset of $\mathbb{Q}$ contains a least element with respect to $\prec$. For this ordering, a monotone sequence containing all of the rationals is easy to construct: let $x_1$ be the smallest rational, let $x_2$ be the smallest element of $\mathbb{Q} \setminus \{x_1\}$, etc.


remainder of the division $2^{1990}/1990$ How do we find the remainder of the division $2^{1990}/1990$?
I actually tried it through Fermat's theorem but couldn't arrive at the answer directly.

Well, since 1990 is not prime, you might want to look into Euler's Theorem, which extends Fermat's Little Theorem.


Finding expected value of remaining piece A father has a pie made for his two sons. Eating more than half of the pie will give indigestion to anyone. While he is away, the older son helps himself to a piece of the pie. The younger son then comes and has a piece of what is left by the brother. Assume that the size of each of the two pieces eaten by the sons is random and uniformly distributed over what is currently available.

  What is the expected size of the remaining piece give that no son has indigestion?

After a while of thinking this is what I did.
Let $\theta_1$ be the total angle corresponding to the amount of pie that son $1$ eats. Similarly, let $\theta_2$ be the total angle corresponding to the amount of pie for son $2$.

Let $X$ be the remaining piece. So,
$$
X = 2\pi - \theta_1 - \theta_2 \\
\mathbb{E}(X) = 2\pi - \mathbb{E}(\theta_1) - \mathbb{E}(\theta_2).
$$
So after a while of thinking, given that no son has indigestion, it feels as if son 1 and son 2 are independent, since they eat less than $\pi$...

So $\theta_1 \sim \operatorname{Unif}(0,\pi)$ and also $\theta_2\sim \operatorname{Unif}(0,\pi)$. Then $\mathbb{E}(X) = \pi$.

Is this correct?

$\def\d{\mathrm{d}}\def\peq{\mathrel{\phantom{=}}{} }$As is pointed out by @AndreaL, conditional probability should be used to calculate to desired probability.

Suppose the elder son eats an angle of $X_1$ and the younger son eats an angle of $X_2$. Translating to probabilistic notation, there is $X_1 \sim U(0, 2π)$ and $X_2 \mid X_1 \sim U(0, 2π - X_1)$, and what needs to be found is $E(2π - X_1 - X_2 \mid X_1 \leqslant π, X_2 \leqslant π)$.

The joint density of $(X_1, X_2)$ is$$
f_{X_1, X_2}(x_1, x_2) = f_{X_2 \mid X_1}(x_2 \mid x_1) f_{X_1}(x_1) = \frac{1}{2π(2π - x_1)}, \quad \forall x_1, x_2 > 0, x_1 + x_2 \leqslant 2π
$$
thus\begin{align*}
&\peq P(X_1 \leqslant π, X_2 \leqslant π)\\
&= \iint\limits_{0 < x_1, x_2 \leqslant π} \frac{1}{2π(2π - x_1)} \,\d x_1\d x_2 = \frac{1}{2π} \int_0^π \int_0^π \frac{1}{2π - x_1} \,\d x_2\d x_1\\
&= \frac{1}{2π} \int_0^π \frac{π}{2π - x_1} \,\d x_1 = \frac{\ln 2}{2},
\end{align*}
and\begin{align*}
&\peq E(2π - X_1 - X_2 \mid X_1 \leqslant π, X_2 \leqslant π)\\
&= 2π - E(X_1 + X_2 \mid X_1 \leqslant π, X_2 \leqslant π)\\
&= 2π - \iint\limits_{0 < x_1, x_2 \leqslant π} (x_1 + x_2)·\frac{f_{X_1, X_2}(x_1, x_2)}{P(X_1 \leqslant π, X_2 \leqslant π)} \,\d x_1\d x_2\\
&= 2π - \frac{1}{π\ln 2} \int_0^π \int_0^π \frac{x_1 + x_2}{2π - x_1} \,\d x_2 \d x_1\\
&= 2π - \frac{1}{π\ln 2} \int_0^π \frac{π x_1 + \frac{π^2}{2}}{2π - x_1} \,\d x_1\\
&= 2π - \frac{1}{π\ln 2}·\frac{1}{2} (5 \ln 2 - 2) π^2 = \frac{2 - \ln 2}{2\ln 2} π.
\end{align*}

This is a complicated situation, but the question happens to greatly simply the process of answering it. The question states that no son has indigestion.

Therefore, neither son ate half the pie or more. (amount either son ate: $0 \le eaten \lt .5$)

Therefore, there is always some pie left. (amount left: $0 \lt remaining \le 1$)

Since the portions both sons ate are uniformly random and independent (since neither son eats more than half), in the expected case we just say that they both ate the middle value (halfway between 0 and .5, or .25). And since two sons ate, .25 + .25 = .5, half the pie is gone.

Another way to look at it is to see that the amount of pie left is uniformly random, and deduce that it is .5.

So while it is true that in the situation the amount the second son eats is highly dependent on the amount the first son eats, the amount they eat is independent for this specific question only.

So yes, you are correct.


Prove that $ 3 =\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{...}}}}$ Prove that $ 3 =\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{...}}}}$

According to the source, this can be done easily like
$$(x+1)^2 = 1+2x+x^2$$
$$(x+1)^2 = 1+x(x+2)$$

But how to continue and prove this?

Since $n^2 = 1+(n-1)(n+1)$, then $n = \sqrt{1+(n-1)(n+1)}$.
Repeating the idea with $(n+1)^2 = 1+(n)(n+2)$, yields $n= \sqrt{1+(n-1)\sqrt{1+n(n+2)}}$. Further applications result in the infinite expression $n=\sqrt{1+(n-1)\sqrt{1+n\sqrt{1+(n+1)\sqrt{1+\cdots}}}}$.


Find the area between curve and straight line My first step is to let y of the first equation equal the second:

$2x -1 = x^2 + 6x + 2$

= $x^2 + 4x +3 = 0$

= $(x + 3)(x + 1)$

To find the area, I find the integral of the top most equation and subtract it from the integral of the bottom query, I believe that the top equation is the line so:

$\int_{-3}^{-1}2x-1 dx$

=> $x^2 - x$

=> $(1 -- 1) - (9 -- 3) = -12$

I now find the integral of the curve:

$\int_{-3}^{-1}x^2 + 6x + 2 dx$

=> $\frac{x^3}3 + 3x^2 + 2x$

=> ${(\frac{-1}{3} + 3 - 2}) - (-9 + 27 -6)$

=> $12\frac{1}{3}$

Then I subtract the first integral from the second:

$-12 - 12\frac{1}{3}$

This leads me to a minus number.  Where have I gone wrong?


What is a moving average system? Can someone elaborate on what a moving average system is?

I know that the system is defined as:
$$y[n] = \frac{x[n] + x[n-1] + x[n-2]}{3}$$
How would we draw $y[n]$ given that we have a graph with discrete values for $x[n]$? Can someone actually draw a sample discrete time $x[n]$ graph and show how the corresponding $y[n]$ graph is generated?

Let's consider a few terms of the sequence $y$:

$$
y[0] = \frac{x[0]+x[-1]+x[-2]}{3}
$$

$$
y[1] = \frac{x[1]+x[0]+x[-1]}{3}
$$

$$
y[2] = \frac{x[2]+x[1]+x[0]}{3}
$$

Notice how the values of $y$ are always an average of three values. Also, notice how the indices of $x$ "shift" to the right in the expressions, and the next value gets shifted in. This is as if we have a longer sequence, $\{x[-2],x[-1],x[0],x[1],x[2]\}$, and we have a window of three consecutive values of the sequence, and the window shifts over one place to the right for each term in the $y$ sequence.
$$\{\color{red}{x[-2],x[-1],x[0]},x[1],x[2]\}$$
$$\{x[-2],\color{red}{x[-1],x[0],x[1]},x[2]\}$$
$$\{x[-2],x[-1],\color{red}{x[0],x[1],x[2]}\}$$

This is sometimes called a sliding-window average as well because of this property.

As for a concrete example, let's consider the sequence
$$
x[n] = n^2/10, \;\; n\geq 0 \\
x[n] = 0, \;\; n<0
$$
This is plotted below.

Now, if you do the calculations, you average the first three points, then the second three points, then the third three points, etc. I leave the calculations out, but the result it as follows:

The corresponding numerical values are given in the table below.

n   x       y
-2  0       0
-1  0       0.0333
0   0       0.1667
1   0.1     0.4667
2   0.4     0.9667
3   0.9     1.6667
4   1.6     2.5667
5   2.5     3.6667
6   3.6     4.9667
7   4.9     6.4667
8   6.4     8.1667
9   8.1     10.0667
10  10      12.1667

You need to convolve your data $\mathbf{x}$ with the impluse response of the corresponding FIR filter $\mathbf{h}$. You can learn the details from here.


Minimum of an apparently harmless function of two variables I would like to prove that the minimum of the function

$$
f(x,y):=\frac{(1-\cos(\pi x))(1-\cos (\pi y))\sqrt{x^2+y^2}}{x^2 y^2 \sqrt{(1-\cos(\pi x))(2+\cos(\pi y))+(2+\cos(\pi x))(1-\cos(\pi y))}}
$$

over the domain $[0,1]^2$ is $2\sqrt{2}$. Looking at the 2D plot of the function

one immediately notices that the minimum is $f(1,1) = 2\sqrt{2}$. However, I can't figure out how to prove this in a rigorous way, even if the expression of $f$ seems to have a nice, "quasi-separable" structure...

Any suggestion is welcomed!

Here's an approach that almost works. I haven't been able to find a nice argument to establish the inequality $(\spadesuit)$ below, but verifying the inequality using the code

Reduce[(1 - Cos[Pi*x])^2 - (2 + Cos[Pi*x])*(8 x^4 - 4 x^8) > 0 && 0 < x < 1, x]

in Mathematica gives me the result

0 < x < 1

so I am almost certain the inequality is true. (I get a warning that numeric integration is being used to verify the completeness of the solution set.) If I find an appropriate argument, I'll edit the post.

Since $f$ is symmetric in $x$ and $y$, it suffices to show that $f(x,y)\geq 2\sqrt2$ holds for all $0<x<y<1$. (For points on the boundary, the inequality follows by continuity.)

Define $\phi:(0,1)\to(0,2)$ by $$\phi(x)=1-\cos\pi x.$$ Note that this is an increasing bijective function. For all $x\in(0,1)$ we have $$\phi(x)\geq2x^2,\tag{$\clubsuit$}$$ as can be easily seen using elementary calculus.

To avoid dealing with square roots, we will analyse the square of your function: $$f(x,y)^2=\frac{x^2+y^2}{x^4y^4}\frac{\phi(x)^2\phi(y)^2}{\phi(x)(3-\phi(y))+\phi(y)(3-\phi(x))}.$$ Since $\phi$ is strictly increasing and $x<y$, we have: $$\phi(x)(3-\phi(y))+\phi(y)(3-\phi(x))\leq2\phi(y)(3-\phi(x)).$$ Therefore, $$f(x,y)^2\geq\frac{x^2+y^2}{x^4y^4}\frac{\phi(x)^2\phi(y)^2}{2\phi(y)(3-\phi(x))}\geq\frac{x^2+y^2}{x^4y^2}\frac{\phi(x)^2}{3-\phi(x)}.\tag{$\ast$}$$ where in the second step we have cancelled one of the $\phi(y)$ and used $(\clubsuit)$ to get rid of the other one.

According to Mathematica, the following inequality is true for $x\in(0,1)$: $$\frac{\phi(x)^2}{3-\phi(x)}\geq8x^4-4x^8.\tag{$\spadesuit$}$$
Using this in $(\ast)$ and cancelling $x^4$, we immediately have: $$f(x,y)^2\geq\frac{(8-4x^4)(x^2+y^2)}{y^2}.$$ It remains to show that this last expression is bounded below by $8$, or equivalently: $$(8-4x^4)(x^2+y^2)\geq 8y^2.$$ But this is true, since $$(8-4x^4)(x^2+y^2)-8y^2=4x^2 (2-x^2(x^2+y^2))\geq0$$ obviously holds for $0<x<y<1$.

It is enough to show that both ((x^2 + y^2)^1/2 )/ (x^2 y^2) and the other part

have minimum at (1,1) :

For ((x^2 + y^2)^1/2 )/ (x^2 y^2) the minimum happens at (x,y)=(1,1) and the

minimum is (2^1/2)

For the other part:

Put  1-cosx = A , a-cosx = B

0 < A < = 2 , 0 < B < = 2

Find the minimum point for AB / ((A)(3-B)+(3-A)(B))^1/2  when  0 < A <= 2 , 0 < B < = 2

It happens at (A,B)=(2,2) which means (x,y)=(1,1). The minimum of this is 2.

So the minimum of f(x,y) happens at (x,y)=(1,1) and this is 2(2^1/2)


Geometry with circles. Two circles, with centres O and P respectively, intersect at A and B. The extension of OB intersects the second sircle at C and the extension of PB intersects the first circle at D. A line through B parallel to CD intersects the first circle at Q not equal to B. Prove that AD = BQ.

We will show that $QD=AB$.
Let the extension of $CO$ and $DP$ intersect with their respective circle at $M$ and $N$. It's clear that $M,N,A$ are collinear. So we need to prove that
$$\angle QBD=\angle BMA$$
Since $CD$ is parallel to $QB$, it suffices to show that $\square CDMN$ is cyclic which is equivalent to showing that $\square OPCD$ is cyclic.
By power of points,
$$R(R+BD)=PO^2-r^2$$
, and
$$r(r+BC)=PO^2-R^2$$
, where $R,r$ are denoted by the radii of the first and second circle.
Therefore $R(BD)=r(BC)$, that is $\square OPDC$ is cyclic. The conclusion follows.


Quaternions, Rotations and Real numbers I haven't really formally studied Algebra at anywhere near this level, but I was told about the existence of Quaternions a few years ago and I find them really cool. I also like how pure quaternions are analogous to cross products in $R^3$, and that it gives me a way of doing it algebraically rather than relying on hand rules.

Over the course of using them it's my understanding that multiplying an element by another, both imaginary numbers, rotates it in some manner so as to be orthogonal to both, similar to rotations of $\frac{\pi}{2}$ in the space with one real and one imaginary axis when multiplied by $i$, which is why it's useful for cross products as it will rotate it to be orthogonal to a plane spanned by linear combinations of those two vectors [afaik].

What I can't intuit though is why for any pure quaternion multiplication, there's a negative real part [the scalar product] if it's meant to correspond to a rotation within $R^3$ and the space of pure Quaternions is Isomorphic to $R^3$. Similarly I don't understand why $i^2 = j^2 = k^2 = $ A negative real.

Don't try interpret quaternion multiplication $q_1 q_2$ as immediatedly connected to rotations; it's more indirect.

The operation which performs a rotation is $q u q^*$, where $u$ is a pure vector quaternion (to be rotated), and $q$ is a unit quaternion
$$
q = \cos(\alpha/2) + \sin(\alpha/2) \, n
,
$$
where $n$ is a unit vector giving the axis of rotation, and $\alpha$ is the angle of rotation. (And $q^* = \cos(\alpha/2) - \sin(\alpha/2) \, n$ is the conjugate.)

Then a general quaternion (a constant times a unit quaternion) can be thought of as representing a linear transformation which is a combination of a rotation and a scaling, and the multiplication of two quaternions is an algebraic way of computing the composition of two such linear transformations:
$$
q_1 \bigl( q_2 u q_2^* \bigr) q_1^*
=
(q_1 q_2) u (q_1 q_2)^*
.
$$
For example, a pure vector (unit) quaternion performs a 180 degree rotation around the corresponding axis, but when you compose two 180 degree rotations you don't get a new 180 degree rotation, unless the two axes of rotation happen to be orthogonal. So it makes perfect sense to have a scalar part (nonzero in general) in the quaternion product; it is related to the angle of the composed rotation.

That $i^2 = j^2 = k^2 = -1$ is related to that two rotations à $90^\circ$ give a rotation of $180^\circ$ which reflects (two) axes, i.e. multiply the $x$, $y$ or $z$ values with $-1$.


Let $G = (V, E)$ be a weighted graph in which each edge has a positive integer cost $c_e$ and every cost is... Let $G = (V, E)$ be a weighted graph in which each edge has a positive integer cost $c_e$ and every cost is different from another. Let $G^+ = (V, E)$ be the same graph in which, however, every edge $e$ has cost $c_e+2$.

How can I prove that if $T$ is a minimum spanning tree (MST) for $G$, then $T$ is also a MST for $G^+$?

Let  $G = (V,E,\ell)$ be the graph with the original length function, let  $G^+ = (V,E,\ell^+)$ be the graph where the length of each edge is increased by 2.

Note that, (A) given any spanning tree $T$ in $G$ with total length $\ell(T)$, that same spanning tree $T$ is $G^+$ has length $\ell^+(T)= $ $\ell(T) + 2(n-1)$ in $G^+$, where $n$ is the number of vertices in $G$ and $G^+$, and vice versa (A$'$); given any spanning tree $T$ in $G^+$ with total length $\ell^+(T)$, that same spanning tree in $G$ has length $\ell(T)=$ $\ell^+(T)-2(n-1)$ in $G$. So this implies the following: (B) let $\ell^{\min}$ be the length of the minimum spanning tree in $G$, then the length of the minimum spanning tree in $G^+$ is $\ell^{\min +} =$ $\ell^{\min} + 2(n-1)$. And (B$'$) letting $\ell^{\min +}$ be the minimum spanning tree in $G^+$, the length of the minimum spanning tree in $G$ is $\ell^{\min}=$ $\ell^{\min +} - 2(n-1)$. [So $\ell^{\min +} = \ell^{min} + 2(n-1)$]

So now $T^{\min}$ be a minimum spanning tree of $G = (V,E,\ell)$ [no matter how $T^{\min}$ was found or if it is a unique minimum spanning tree]. Then $T^{\min}$ has length $\ell^{\min}$ in $G$, and so by (A) has length $\ell^{\min}+2(n-1)$ in $G^+$. However, B implies that this is also a minimum spanning tree in $G'$. Likewise, let $T^{\min +}$ be a minimum spanning tree in $G$. Can you show that $T^{\min +}$ is also a minimum spanning tree in $G$, using (A$'$ )and (B$'$)?

Hint:

1) First prove that graphs with distinct edge weights have a unique MST.

2) Suppose $T$ is not a MST in $G^+$. Show how this implies there is another MST in $G$.


Problem in computing of integral by substitution. I want to compute an integral like
$\int_0^{+\infty} \ln(1+x)e^{-x}\,\mathrm dx$.
Then denote $\mu = 1-e^{-x}$, so $x=-\ln(1-\mu)$. Substitute this into the integral, we get

$$\int_0^1 \ln(1-\ln(1-\mu))(1-\mu)\,\mathrm d(-\ln(1-\mu)) = \int_0^1 \ln(1-\ln(1-\mu))\,\mathrm d\mu$$

However, $\ln(1-\ln(1-\mu))$ goes to infinity if $\mu$ goes to 1, which makes the latter integral uncomputable. But the original integral has no such problem. Is there any error in my substitution?

You reduce the integral over the infinite interval to the one over the finite interval. You shall not be surprised that the function under the integral may become unbounded. The point is that you may think of the value of the integral as an area under the graph of the function. Now, what you change a variable, you make a correspondent change of the coordinate system where they area is defined. That shall give you an intuition why the function may (but doesn't have to) become unbounded. However, it still stays integrable.

Your operations with $\mu$ seem fine to me. To support this, I computed both integral in Mathematica (see the image below), they're both well-defined and have the same value.

For some reason, Wolfram Alpha computes only an approximate value of the second integral.

we have
$$\int_{0}^{\infty}e^{\mu x}\ln{(\beta+x)}dx=\dfrac{1}{\mu}[\ln{\beta}-e^{\mu\beta}Ei{(-\beta\mu)}]$$


What price should the units be sold for a revenue of 10450? A company sells 4000 units each for \$2. For every 5 cent increase, 40 fewer are sold. At what price should they sell if they want a revenue of \$10450?

HINT

The demand seems like a linear function of the price, so $D(p) = mp+b$.

Find $m$ from the description of the change in demand when you change prices.
Find $b$ from the point $(2, 4000)$ that you are given.
Revenue is price times units sold, i.e. $R(p) = p D(p)$. Use $D(p)$ you found to express what equation will $R(p)$ obey
Can you solve $R(p) = 10450$?

Model your equation using a quadratic function:

$(4000-40x)(2+0.05x)=y$

This is true because price is given by the second bracket and the units is given by the first bracket.

Now solve for $x$ by plugging in $y=10450$

Then your price is given by $2+0.05x$


Can the product of infinitely many elements from $\mathbb Q$ be irrational? I know there are infinite sums of rational values, which are irrational (for example the Basel Problem). But I was wondering, whether the product of infinitely many rational numbers can be irrational.
Thank you for your answers.

Yes, it can.

Consider any sequence $(a_n)$ of non-zero rational numbers which converges to an irrational number. Then define the sequence $b_n$ by $b_1 = a_1$ and
$$
  b_n = \frac{a_n}{a_{n-1}}
$$
for $n > 1$.

We then have that
$$
  b_1 b_2 \cdots b_n = a_1 \frac{a_2}{a_1} \frac{a_3}{a_2} \cdots \frac{a_n}{a_{n-1}} = a_n.
$$

We thus see that every term of $(b_n)$ is rational, and that the product of the terms of $(b_n)$ is the same as the limit of $a_n$, which is irrational.

There is a simple way to obtain any irrational number as an infinite product:

take any sequence $s_n$ of rational numbers converging to the targeted irrational one (say the approximations of $\pi$ to $n$ decimals);
form the product of the numbers $f_n:=\dfrac{s_{n+1}}{s_n}$, with $f_0=1$.

$$\pi=\prod_{n=0}^\infty f_n=\frac{31}{10}\cdot\frac{314}{310}\cdot\frac{3141}{3140}\cdot\frac{31415}{31410}\cdot\frac{314159}{314150}\cdots$$


Calculate the product distribution Original question was to calculate the $X$, $Y$ and $XY$ distribution, however I have already done the first two, so I am just going to post the question the way I have it.
$$
f_{X,Y}(x,y) = \left\{ \begin{array}{ll}
ye^{-xy-y} & \textrm{when $x>0,y>0$}\\
0 & \textrm{in other case}\\
\end{array} \right.
$$
I calculated both
$$
f_X(x) = \left\{ \begin{array}{ll}
\frac{1}{(x+1)^{2}} & \textrm{when $x>0$}\\
0 & \textrm{in other case}\\
\end{array} \right.
$$
$$
f_Y(y) = \left\{ \begin{array}{ll}
e^{-y} & \textrm{when $y>0$}\\
0 & \textrm{in other case}\\
\end{array} \right.
$$
Since x and  y are dependent, i think i can not use the formula given here https://en.wikipedia.org/wiki/Product_distribution, can someone help?

If we assume that $XY=Z$, then $X=Z/Y$, and $f_Z(z)=0,\ \forall z\le0$.

For $z>0$: $$\begin{align}f_Z(z)&=\int_{-\infty}^{\infty}f_{X,Y}(z/y,y)\frac{1}{|y|}dy\\
&=\int_{0}^{\infty}e^{-z-y}dy\\
&=e^{-z}
\end{align}$$


If a prime ideal has no zero divisors then the ring also has no zero divisors Let $R$ be a commutative ring. Let $P$ be a prime ideal of $R$ such that $P$ has no (non-zero) zero divisors. Prove that $R$ has no (non-zero) zero divisors.

My try:

Let $ab=0$ where $a,b\in R$ .Let us assume that $a\neq 0$ then we prove that $b=0$. Assume that $b \neq 0$.

Now $ab=0\in P$ then either $a\in P $ or $b\in P$ .Now both $a,b$ can't be in $P$ as then $P$ would have zero divisors.

Suppose that $a\notin P$ then $b\in P$.

But how to arrive at a contradiction from here?

Claim: Ring $R$ is an integral domain.

Proof: Suppose $a,b\in R$ with $ab=0$, we must show that $a=0$ or $b=0$. Certainly the images $\overline a,\overline b$ of $a,b$ in $A/P$ satisfy $\overline a\overline b=0$, and since $R/P$ is an integral domain (by the definition of prime ideal) one of $\overline a,\overline b$ is zero, that is $a\in P$ or $b\in P$. Assuming by symmetry the former is the case, then either $b=0$ (in which case we are done), or else $b\neq0$ and the fact that $P$ does not contain zero divisors (of $R$) imply that $a=0$

One more "similar" kind of exercise  from Atiyah Macdonald (If you need  hint to solve this please let me know):

    In a ring $A$, let $\Sigma$ be the set of all ideals in which every element is a zero-divisor. Show that set $\Sigma$ has maximal elements and that every maximal element of $\Sigma$ is a prime ideal. Hence the set of zero-divisors in $A$ is a union of prime ideals.

If you mean by zero divisors in $P$ that for both non zero $x \in P$ and $y \in R$ s.t $xy=0$ then, From your try let $b \in P$ then consider $<b> \subseteq P$ then $ab=0$ in $P$ contradiction.

Otherwise if you say that there are no non-zero $x,y \in P$ s.t. $xy=0$ then the statement is false: take $\Bbb Z_{15}$, and $P=<3>$. Then $P$ is a prime ideal which does not contain $x,y\ne0$ such that $xy=0$, but $\Bbb Z_{15}$ has zero divisors.


Solve $2y''+y'=0$ using power series I got the answer $C_0e^{-x/2}$ by doing
$$\sum_{n=1}^{\infty }\:\left(x^{n-1}\right) \left(2n\left(n+1\right)C_{n+1}\:\right)+\left(nC_n\right) = 0$$
thus giving
$$C_{n+1}= -\dfrac{C_n}{2(n+1)}.$$
This leads me to the answer $y = C_0e^{-x/2}$.

However, the answer in my textbook is $C_1e^{-x/2} + C_0$.  I've found this to be true, but still couldn't figure out a method of solving the equation so that there are two constants.

Edit: what my textbook did was  $$c_{n+2}= -\dfrac{1}{2(n+2)}c_{n+1}.$$  , and then  substitute $$c_{0}=C_0-2c_1$$  and $$c_1=\dfrac{-1}{2}C_1$$    into $y=c_0 +c_1 - \dfrac{1}{2*2!}c_{1} x^2. + \dfrac{1}{2^2*3!}c_{1} x^3........$

Where did the substitutions come from?

$$ \frac{\mathrm{d^2}y }{\mathrm{d} x^2}=-\frac{1}{2} \frac{\mathrm{d} y}{\mathrm{d} x} \\
\frac{\mathrm{d} }{\mathrm{d}x}\left ( \frac{\mathrm{d}y }{\mathrm{d} x} \right ) = -\frac{1}{2} \frac{\mathrm{d} y}{\mathrm{d} x} \\
\ln{\frac{\mathrm{d}y}{\mathrm{d}x}} = \frac{-x}{2}+c \\
\frac{\mathrm{d}y}{\mathrm{d}x} = c_{1}e^{-\frac{x}{2}}
\\ y=c_{1}e^{-\frac{x}{2}}+c_{2}
$$


Prove: $(na_n)$ converges to 0 if $(a_n)_{n=1}^\infty$ is a monotonously falling sequence of non-negative real numbers Prove: Let $(a_n)_{n=1}^\infty$ be a monotonously falling sequence of non-negative real numbers, so that $\sum a_n$ is convergent. Then $(na_n)$ is a sequence converging to 0.

Well, no idea. All help is greatly appreciated!

Because $a_n$ is a non-negative real number, let's say that $c$ is the smallest number of the $a_n$ sequence. That means that $\sum_{n=1}^\infty (n *a_n) \geq \sum_{n=1}^\infty (c)$. Well, if it is that $c>0$ then that series would be divergent. It is not, thus $c=0$, so the series will converge, and may only converge, to $0$.


Calculate $\lim_{x\to0}\frac{\sin ax\cos bx}{\sin cx}$ Calculate: $$\lim_{x\to0} \frac{\sin ax\cos bx}{\sin cx}$$

My attempt:
$$=\lim_{x\to 0} \dfrac {\sin (ax). \cos (bx)}{\sin (cx)}$$
$$=\lim_{x\to 0} \dfrac {\sin (ax)}{ax} \times \dfrac {ax}{\sin (cx)} \times cx \times \dfrac {\cos (bx)}{cx}$$
$$=1\times ax \times 1 \times \dfrac {\cos (bx)}{cx}$$

How do I do further?

A bit of better than your:
$$\lim_{x\rightarrow0}\frac {\sin ax\cos bx}{\sin cx}=\lim_{x\rightarrow0}\left(\cos{bx}\cdot\frac{\sin{ax}}{ax}\cdot\frac{cx}{\sin{cx}}\cdot\frac{a}{c}\right)=\frac{a}{c}$$

$$\lim_{x \rightarrow 0}\frac{\sin \text ax \cos \text bx}{\sin \text cx}$$

Using linear approximation

$$L=f(\text a)+f'(x)(x-\text a)$$

Therefore, at $x=0$
$$\sin \text ax\approx \text ax$$
$$\sin \text cx\approx \text cx$$
$$\cos bx \approx 1$$

$$\lim_{x \rightarrow 0}\frac{\sin \text ax \cos \text bx}{\sin \text cx}=\lim_{x \rightarrow 0}\frac{(\text ax)(1)}{\text cx} \iff \lim_{x \rightarrow 0}\frac{(\text ax)(1)}{\text cx}=\lim_{x \rightarrow 0} \frac{\text a}{\text c}$$

$$\therefore \lim_{x \rightarrow 0} \frac{\text a}{\text c}=\frac{\text a}{\text c}$$


Propositional calculus axiom the other way around I have the following axioms of propositional calculus (as well as modus ponens and the deduction theorem if needed):

$$(a \to (b \to a)) \tag1$$
$$ (((a \to (b \to c)) \to ((a \to b) \to (a \to c))) \tag2 $$
$$ ((\neg b \to \neg a) \to (a \to b)) \tag3$$

I'm trying to deduce the last one the other way around (i.e. $((b \to a) \to (\neg a \to \neg b))$) but never manage to get a point... This isn't something I have to do but I feel as though it could be very convenient in my further proofs.

Any help would be greatly appreciated! Thank you!

We need to assume $\neg a$ is an abbreviation for $a \to \bot$; otherwise, there is no way to introduce negations without having an assumption with a negation in it.

First note that $b \to a, a \to \bot, b \vdash \bot$ by two applications of modus ponens. Then we do three applications of the deduction theorem: first we obtain $b \to a, a \to \bot \vdash \neg b$, then $b \to a \vdash \neg a \to \neg b$, and then finally $\vdash (b\to a) \to (\neg a \to \neg b)$.


Can a matrix satisfy all three of the following properties? Consider an $n \times n$ matrix of the form
$$
A = \begin{bmatrix}
a_1 & a_2 & \ldots & a_{n-1} & a_n \\
1 \\
& 1 \\
& & \ddots \\
& & & 1
\end{bmatrix}
$$
for certain $a_1, \ldots, a_n \in \mathbb{R}$ (and zeroes in all blank spaces). Can this matrix ever (for some $n \in \mathbb{N}$ and some $a_1, \ldots, a_n \in \mathbb{R}$) satisfy all of the following three properties:

$\lvert \lambda \rvert \leq 1$ for all eigenvalues $\lambda$ of $A$
For all eigenvalues $\lambda$ with $\lvert \lambda \rvert = 1$, the algebraic multiplicity is equal to the geometric multiplicity
There exists an eigenvalue $\lambda$ with $\lvert \lambda \rvert = 1$ and such that the multiplicity of $\lambda$ is greater than $1$?

Note that I do not require $A$ to be invertible. I don't immediately see why such a matrix can't exist, but if it could, that would lead to a counterexample to some unproven theorem in my course notes.

I calculated by hand that such a matrix can not exist for $n \leq 3$.

The answer is no. Your matrix $A$ is an avatar of a companion matrix. The geometric multiplicity of an eigenvalue of such a matrix is always $1$;

You can use the following property of a matrix:

Given some n by n matrix, A, the trace of the matrix is equivalent to the sum of its eigenvalues.

The sum of the eigenvalues of the matrix is:

trace (A) =  $a_{1}$ + 0(n-1)

= $a_{1}$ = $\lambda$

Using the above theorem, we can then clearly see the following must hold true:

1.) $\left | \lambda \right | \leq 1 $ iff  $\lambda = 0$ or $\lambda = 1$

2.) The  geometric multiplicity of the matrix is therefore equivalent to the multiplicity of the eigenvalue  $a_{1}$ = 1.

3.) It follows from our first conclusion that the if $\lambda = 1$, the multiplicity must be equal to the number of places in the diagonal that the number 1 occurs. In your example, this can be no more than 1 time. Therefore, the multiplicity of the non-zero value is at most 1.


Prove $(X_n, F_n) $ Martingale $\iff \int_{F} X_{n+1} = \int_{F} X_{n} \forall F \in F_n$ I have some additional questions to this exercise:

Let $(\Omega, F, F_n, P)$ filtered probability space. Let $(X_n)_{n\in\mathbb{N}} \in \mathcal{L}^1(P)$, which is adopted to $(F_n)_{n\in\mathbb{N}}$.

Show $(X_n, F_n) $ Martingale $\iff \int_{F} X_{n+1} = \int_{F} X_{n} \forall F \in F_n$.

I think I have proved the following: "$\Rightarrow$"
$$\int_F X_{n+1} dP = E[1_{F}X_{n+1}] = E[E[1_{F}X_{n+1}|F_n]] = E[1_FE[X_{n+1}|F_n]] = E[1_F X_{n}] = \int_F X_{n} dP $$

Is this correct? Im not sure about this:
$$ E[E[1_{F}X_{n+1}|F_n]] = E[1_FE[X_{n+1}|F_n]] $$

And how should I show the other implication? Any help is appreciated?

The equivalence is just immediate from the definitions of the terms involved. Indeed, with measurability and integrability as background assumptions, we have
$$(X_n, F_n) \  \text{is a martingale}$$ if and only if
$$E(X_{n+1} \mid F_n) = X_n$$ if and only if
$$\int_F X_{n+1} dP = \int_F X_n dP, \ \ \forall F \in F_n,$$
where we used the definition of a martingale first and the definition of conditional expectation second.

As to what you wrote, it's correct, just overkill. The equality you ask about holds because $E(1_FX_{n+1} \mid F_n) = 1_FE(X_{n+1} \mid F_n)$, which is in turn justified by noting that $F \in F_n$ implies that $1_F$ is measurable $F_n$.

The two statements $X_n$ is $F_n$-measurable and $\int_{F} X_{n+1} = \int_{F} X_{n} \forall F \in F_n$

are equivalent to

$$E[X_{n+1} | F_n] = X_n$$

provided $X_n$ is integrable, which it is.

Now for $(X_n)_n$ to be an $((F_n)_n, P)_-$ martingale, we need that

$X_n$ is integrable for each $n$ --> given
$X_n$ is adapted to $F_n$ for each $n$ --> true by definition of $F_n$
$E[X_{n} | F_m] = X_m$ for $m \le n$ and (I guess?) $m, n \in \mathbb N$

Well, $X_m$ is $F_m$-measurable so all that's left to check is $$\int_{F} X_{m} = \int_{F} X_{n} \forall F \in F_m$$

So yeah, we just have to show that

$$\int_{F} X_{m} = \int_{F} X_{n} \forall F \in F_m \iff \int_{F} X_{n+1} = \int_{F} X_{n} \forall F \in F_n$$

For $\leftarrow$, we know that

$$\int_{F} X_{n+1} = \int_{F} X_{n} \forall F \in F_n \tag{1}$$

and

$$\int_{F} X_{n} = \int_{F} X_{n-1} \forall F \in F_{n-1} \tag{2}$$

Actually, $(1)$ implies

$$\int_{F} X_{n+1} = \int_{F} X_{n} \forall F \in F_{n-1} \tag{3}$$

Thus from $(2)$ and $(3)$ we have

$$\int_{F} X_{n+1} = \int_{F} X_{n-1} \forall F \in F_{n-1} \tag{3}$$

Then we work our way down from $n+1$ to $n$ to $n-1$ all the way until $m$. Formally, I guess we take cases. If $m=n-1$, then we're done. If $m=n-2$, then perform another iteration. And so on. I've never seen an iteration kind of proof in maths before. It just seemed like obvious alternative definitions of the martingale property.

Anyhoo for $\rightarrow$, I think you can take it from here. It's similar to proof for $\leftarrow$, perhaps the steps are reversible. I'm hungry now. So, I'll let you try.


How to prove that a function is in fact $L_1$? I'm working on two problems where I need to interchange the limit and integral signs, so I want to evoke Lebesgue's Dominated Convergence Theorem. I now know that the functions I have chosen do indeed dominate, but is there a way to show that the dominating functions are in fact Lebesgue integrable? (I don't want to calculate their integrals...)

One of the functions I am trying to show are Lebesgue integrable are: $\dfrac{1}
{1+x^2}$ over the domain $[0,\infty)$

How would I go about doing this?

Since you asked for more "theoretical" approach, maybe this is something for you.

We have
\begin{equation}
\lim_{x \rightarrow \pm \infty}\frac{x^2}{1+x^2}=1
\end{equation}
and therefore
\begin{equation}
\frac{1}{1+x^2} \leq c \ \frac{1}{x^2},
\end{equation}
for $|x| \geq a$ and suitable constants $a$ and $c$.
On the (compact) interval $ [ -a,a ]$ we can bound the (continuous) function by a constant. And outside of the interval it is just $\frac{1}{x^2}$ scaled by a constant which is known to be integrable outside an interval around zero.

In this case it is quite easy to see what $c$ and $a$ are (even without the limit argument, as pointed out by David C. Ullrich) - but this approach may work in "more complicated" cases: The function considered is continuous and behaves like a integrable function when $x \rightarrow \pm \infty$.

If we're talking about $L^1(\mathbb{R})$, then we should show that the integral is finite (it is).

Also, recall that this neat function is also Riemann integrable - that means that it is Lebesgue integrable. Check this out.