Prove $( \lnot C \implies \lnot B) \implies (B \implies C)$ without the Deduction Theorem The issue is Exercise 1.47 (d) in Elliot Mendelson's "Mathematical Logic". The exercise is to prove $(\lnot C\implies\lnot B)\implies(B\implies C)$ by using the three axioms $(A1,A2,A3)$ without using the Deduction theorem (and without any hypothesis).

The axioms are:

$A1: B\implies(C\implies B)$

$A2: (B\implies(C\implies D))\implies((B\implies C)\implies(B\implies D))$
$A3: (\lnot C\implies\lnot B)\implies((\lnot C\implies B)\implies C)$

The only inference rule is MP.

I have a proof but it is long. My proof is based on the proof of Lemma 1.11 (d) which proves $(\lnot C\implies\lnot B)\implies(B\implies C)$ but uses the Deduction Theorem (Proposition 1.9). Then, like Mendelson suggests in exercise 1.49, I apply the process used to prove the Deduction Theorem to the steps. To be precise, I assume $(\lnot C\implies\lnot B)$ as a Hypothesis $H$. If $C_1,C_2,\dots,C_n$ is a proof of $(B\implies C)$ that uses $H$ then stepwise I prove $H\implies C_i$ for $i=1,\dots,n$. The last step is $H\implies C_n$ which is what we want to prove.

This way requires around 4 uses of Axiom1, 4 of Axiom2, 1 Axiom3, and 9 uses of Modus Ponens.

Do I miss a shorter proof?

I'm able to prove it "independently" from the Deduction Theorem, but the proof is quite longer ...

The axioms are :

  $F \rightarrow (G \rightarrow F)$
  $(F \rightarrow (G \rightarrow H))\rightarrow ((F \rightarrow G) \rightarrow (F \rightarrow H))$
  $(\neg G \rightarrow \neg F) \rightarrow ((\neg G \rightarrow F) \rightarrow G)$

For readibility, I'll organize the proof with some preliminary results :

T1 : $P \rightarrow P$

1) $P \rightarrow ((Q \rightarrow P) \rightarrow P)$ --- Ax.1

2) $P \rightarrow (Q \rightarrow P)$ --- Ax.1

3) $(1) \rightarrow ((2) \rightarrow (P \rightarrow P))$ --- Ax.2

  4) $P \rightarrow P$ --- from 3), 1) and 2) by Modus Ponens twice.

T2 : $(Q \rightarrow R) \rightarrow ((P \rightarrow Q) \rightarrow (P \rightarrow R))$

1) $(P \rightarrow (Q \rightarrow R)) \rightarrow ((P \rightarrow Q) \rightarrow (P \rightarrow R))$ --- Ax.2

2) $(1) \rightarrow ((Q \rightarrow R) \rightarrow (1))$ --- Ax.1

3) $(Q \rightarrow R) \rightarrow (1)$ --- from 1) and 2) by Modus Ponens

4) $(Q \rightarrow R) \rightarrow (P \rightarrow (Q \rightarrow R))$ --- Ax.1

5) $(3) \rightarrow ((4) \rightarrow ((Q \rightarrow R) \rightarrow ((P \rightarrow Q) \rightarrow (P \rightarrow R))))$ --- Ax.2

  6) $(Q \rightarrow R) \rightarrow ((P \rightarrow Q) \rightarrow (P \rightarrow R))$ --- from 5), 3) and 4) by MP twice.

T3 : $P \rightarrow ((P \rightarrow Q) \rightarrow Q)$

1) $(P \rightarrow Q) \rightarrow (P \rightarrow Q)$ --- T1

2) $(1) \rightarrow (((P \rightarrow Q) \rightarrow P) \rightarrow ((P \rightarrow Q) \rightarrow Q))$ --- Ax.2

3) $((P \rightarrow Q) \rightarrow P) \rightarrow ((P \rightarrow Q) \rightarrow Q)$ --- from 1) and 2) by MP

4)  $P \rightarrow ((P \rightarrow Q) \rightarrow P)$ --- Ax.1

5) $(3) \rightarrow ((4) \rightarrow (P \rightarrow ((P \rightarrow Q) \rightarrow Q)))$   --- T2

  6) $P \rightarrow ((P \rightarrow Q) \rightarrow Q)$ --- from 5), 3) and 4) by MP twice.

T4 : $(P \rightarrow (Q \rightarrow R)) \rightarrow (Q \rightarrow (P \rightarrow R))$

1) $((P \rightarrow Q) \rightarrow (P \rightarrow R)) \rightarrow ((Q \rightarrow (P \rightarrow Q)) \rightarrow (Q \rightarrow (P \rightarrow R)))$ --- T2

2) $(P \rightarrow (Q \rightarrow R)) \rightarrow ((P \rightarrow Q) \rightarrow (P \rightarrow R))$ --- Ax.2

3) $Q \rightarrow (P \rightarrow Q)$ --- Ax.1

4) $(1) \rightarrow ((2) \rightarrow ((P \rightarrow (Q \rightarrow R)) \rightarrow ((3) \rightarrow (Q \rightarrow (P \rightarrow R)))))$ --- T2

5) $(P \rightarrow (Q \rightarrow R)) \rightarrow ((3) \rightarrow (Q \rightarrow (P \rightarrow R)))$ --- from 4), 1) and 2) by MP twice

6) $(3) \rightarrow ((3) \rightarrow (Q \rightarrow (P \rightarrow R))) \rightarrow (Q \rightarrow (P \rightarrow R))$ --- T3

7) $((3) \rightarrow (Q \rightarrow (P \rightarrow R))) \rightarrow (Q \rightarrow (P \rightarrow R))$ --- from 6) and 3) by MP

8) $(7) \rightarrow ((5) \rightarrow ((P \rightarrow (Q \rightarrow R)) \rightarrow (Q \rightarrow (P \rightarrow R))))$ --- T2

  9) $(P \rightarrow (Q \rightarrow R)) \rightarrow (Q \rightarrow (P \rightarrow R))$ --- from 8), 7) and 5) by MP twice.

Now for the proof :

1) $(\neg C \rightarrow \neg B) \rightarrow ((\neg C \rightarrow B) \rightarrow C)$ --- Ax.3

2) $B \rightarrow (\neg C \rightarrow B)$ --- Ax.1

3) $(\neg C \rightarrow B) \rightarrow ((\neg C \rightarrow \neg B) \rightarrow C)$ --- from 1) and T4 by Modus Ponens

4) $B \rightarrow ((\neg C \rightarrow \neg B) \rightarrow C)$ --- from T2, 3) and 2) by MP twice

  5) $(\neg C \rightarrow \neg B) \rightarrow (B \rightarrow C)$ --- from T4 and 4) by MP

I don't know how to write those fancy Bs and those fancy Cs that Mendelson uses in that text.  Mendelson has parentheses (deliberately) around all wffs in when talking about the axiom schema, and that makes good sense since it makes substitution for wffs by other wffs just a matter of finding similar symbol patterns.  We don't need to infer a meaning and recognize a pattern, we only need to recognize a pattern when you actually write wffs.

Due the correspondence between Mendelson's semantic statement forms and Mendelson's syntactic wffs, wffs have similar properties to statement forms.  Mendelson indicates in exercise 1.18 on p. 12 of the fifth edition that instead of writing statement forms as he usually does, that Polish notation can get used, which looks like any prefix notation according to him.  The following system has symbols more like the original Polish notation of Lukasiewicz, Meredith, Prior, and other authors whose papers you can read for free via old issues of the Norte Dame Journal of Formal Logic.  Due to how Mendelson writes in his Schaum's Outline of Boolean Algebra, it appears that we can faithfully represent Mendelson's intent using bold capital letters instead of fancy capital letters.  The following puts each axiom schema of the system in Polish notation that I'll use next to the corresponding axiom schema of Mendelson's system.

  Polish               Mendelson

Ax 1  CbCcb .....................  (B -> (C -> B))

Ax 2  CCbCcdCCbcCbd ...   ((B -> (C -> D)) -> ((B -> C) -> (B -> D)))

Ax 3 CCNcNbCCNcbc .....((($\lnot$C) -> ($\lnot$B)) -> ((($\lnot$C) -> B) -> C))

A proof in a system using Polish notation of a certain length implies that a proof in Mendelson's system using Mendelson's notation of the same length can get written.

Axiom Schema                 1  CbCcb
Axiom Schema                 2  CCbCcdCCbcCbd
Axiom Schema                 3  CCNcNbCCNcbc
1 b/CbCcb, c/a               4  CCbCcbCaCbCcb
4 * C1-5                     5  CaCbCcb
2 b/CbCcd, c/Cbc, d/Cbd      6  CCCbCcdCCbcCbdCCCbCcdCbcCCbCcdCbd
6 * C1-7                     7  CCCbCcdCbcCCbCcdCbd
2 b/a, c/b, d/Ccb            8  CCaCbCcbCCabCaCcb
8 * C5-9                     9  CCabCaCcb
5 a/CbCCcbd                  10 CCbCCcbdCbCcb
7 c/Ccb                      11 CCCbCCcbdCbCcbCCbCCcbdCbd
11 * C10-12                  12 CCbCCcbdCbd
9 a/CNcNb, b/CCNcbc, c/d     13 CCCNcNbCCNcbcCCNcNbCdCCNcbc
13 * C3-14                   14 CCNcNbCdCCNcbc
1 b/CCbCCcbdCbd, c/a         15 CCCbCCcbdCbdCaCCbCCcbdCbd
15 * C12-16                  16 CaCCbCCcbdCbd
2 b/a, c/CbCCcbd, d/Cbd      17 CCaCCbCCcbdCbdCCaCbCCcbdCaCbd
17 * C16-18                  18 CCaCbCCcbdCaCbd
18 a/CNcNb, c/Nc, d/c        19 CCCNcNbCbCCNcbcCCNcNbCbc
14 d/b                       20 CCNcNbCbCCNcbc
19 * C20-21                  21 CCNcNbCbc

So, a proof can get written in Mendelson's system in 21 steps if we include the original 3 schema.  If we don't, it can get done in exactly 18 steps.


Principal Minor Theorem for dimension 2 Let $A$ be a real ($2 \times 2$)-matrix such that the map $\mathbb{R}^2 \times \mathbb{R}^2 \to \mathbb{R}$, $\langle x,y \rangle = x^t A y$ is a scalar product. Now I would like to show that the determinant of $A$ must be positive.

Let $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$. I was able to show that $a,d >0$ and $b=c$, so our matrix has the form
$$A = \begin{pmatrix} a & b \\ b & d \end{pmatrix}.$$

The determinant is $ad-b^2$. My idea is it to find a vector $z \in \mathbb{R}^2$ such that $\langle z,z \rangle = \det(A)$ which must be positive due to the positive definiteness of the scalar product.
 If $z = \begin{pmatrix} x \\ y \end{pmatrix}$, then $\langle z,z \rangle = ax^2 + 2bxy + dy^2$. How do I find values for $x,y$ so that this is exactly the determinant of $A$? I think I need to use squares of $a,d$ (which is possible since both are positive) or absolute values, but I am not able to find the solution.

Could you please help me with problem? Thank you!

You must have

$$\langle z,z \rangle = ax^2 + 2bxy + dy^2 >0$$ for all $z\neq 0$. In particular for all $z=(x,1)^T$.

In particular the discriminant of the trinomial

$$a x^2 +2bx +d$$ has to be strictly negative. This is exactly what you’re looking for.


I'm not sure on how to start this higher order diff eq. Solve $$1001y''' + 3.2y'' + \pi y' - \sqrt{4}y = 0$$
Initial Conditions:

$$y(0) = 0, y'(0) = 0, y''(0) = 0$$

I plugged it into the characteristic equation, but I don't know where to go from there...

$$1001r^3 + 3.2r^2 + \pi r - 2 = 0$$

Some help would be appreciated, thanks.

I think the point of this problem is that, by the uniqueness theorem, there is only one solution, and from the initial conditions, it is clear that $y(t)=0$ is that solution.  I don't think you were supposed to mess with the characteristic equation at all.

Case 1: Assuming the last square root applies only to 4 (thus the question becomes trivial). WolframAlpha provides the solution:

$$
y(x) = c_3 e^{0.426569 x} + c_1 e^{-1.81328 x} sin(1.18346 x) + c_2 e^{-1.81328 x} cos(1.18346 x)
$$

Which comes from the characteristic equation you gave. You would have to find the root either numerically or with Cardano's Method after a change in variables. https://en.wikipedia.org/wiki/Cubic_function

The coefficients $c_1,c_2$ and $c_3$ can be found with the initial conditions, but no surprise they're zero.

Case 1: The last square root applies to $y$ as well, so your question becomes a little more convoluted. You can still make a change of variables $u^2=y$, and the equation becomes:

$$
1×(d^3 u(t)^2)/(dt^3) + 3.2×(d^2 u(t)^2)/(dt^2) + π×(du(t)^2)/(dt) - 2 u(t) = 0
$$

$$
(6.4 u(t) + 6 u'(t)) u^(3)(t) = 2 u(t) - 2 π u(t) u'(t) - 6.4 u'(t)^2 - 2 u(t) u^(3)(t)
$$

While I fail to find a straightforward solution for this, you can check that again, no surprise the initial conditions being zero means that $y'''(0)$ is also zero, and thus $y(t)=0$ is a valid solution to your initial value problem.


Does $x\equiv ay \pmod{p^{2m}}$ always have non-trivial solutions $x,y$ with $\lvert x \rvert,\lvert y \rvert \leq p^m$? Given a positive integer $m$, a prime $p$ and an integer $a$, I would like to prove that
$$
x \equiv ay \pmod{p^{2m}} \qquad \lvert x \rvert,\lvert y \rvert \leq p^m
$$
always has at least one solution $(x,y)\in\mathbb{Z^2}$ with $x,y$ not both $0$.

I have no problem if $p\mid a$. Indeed: if $\lvert a \rvert \leq p^m$ then $(a,1)$ is a solution, otherwise $p^m\mid a$ and $(0,p^m)$ will do.

If $p\nmid a$ then $a$ is invertible modulo $p^{2m}$, hence $x \equiv ay \pmod{p^{2m}}$ has many non-trivial solution. However, I don't know how to find one with $\lvert x \rvert,\lvert y \rvert \leq p^m$.

I think this follows from something resembling the pigeon hole principle as follows.

Fix $a$. We can assume that $p^m\nmid a$ as the case $p^m\mid a$ is easy. Consider the set of smallest non-negative remainders modulo $p^{2m}$
$$
S_a=\{ay\bmod p^{2m}\mid 1\le y\le p^m\}.
$$
Observe that there are no repetitions, so $|S_a|=p^m$.

Let us place these $p^m$ pigeons into $p^m$ holes with hole #$j$ consisting of the interval
$I_j:=[jp^m, (j+1)p^m-1]$, $0\le j<p^m$. There are two possibilities. Either two or more pigeons try to enter the same hole, or all the holes are occupied by a single pigeon.

If there is no double occupancy, then one of the pigeons flew into the interval $I_0$. If this pigeon carried the coordinate tag $y_0$, this means that the remainder of $ay_0$ is in the range $I_0$, and we have found our solution: the matching $x_0$ is the remainder of $ay_0$ modulo $p^{2m}$.

The other case is that two pigeons, $ay_1$ and $ay_2$, are in the same hole.
Without loss of generality we can assume that the remainder of $ay_1$ is smaller than that of $ay_2$. Thus the remainder $r$ of
$a(y_2-y_1)$ mod $p^{2m}$ is in the range $0<r<p^m$, and as $0<|y_2-y_1|<p^m$ we have, again, found a solution.

Put $R=\Bbb Z_{p^m}$ then in $R[x,y]$ the polynomial $x-ay$ has non trivial solutions iff $a \not \equiv 0 \pmod {p^m}$. Suppose $a \equiv 0 \pmod {p^n}$ with $n<m$ then take for $y$ a unit in $R$. It is impossible that $ay=0$ because that would imply that $y \equiv 0 \pmod {p^{m-n}}$ and so $y \equiv 0 \pmod {p}$, and this is a property that units just don't have. If $a \equiv 0 \pmod {p^m}$ then clearly $a=ay=0$.


Show that A is compact Let $A = \{(x,y) \in \mathbb{R^2} : x>0, y>0, xy + \frac{1}{x} + \frac{1}{y} \leq \alpha \}$.

I need to prove that  A is compact, $\forall\alpha \in \mathbb{R}.$

My attempt

A is closed since the function $f: (0, \infty)\times(0, \infty) \to \mathbb{R}$,
 $\quad f(x,y) = xy + \frac{1}{x} + \frac{1}{y}  $ is continuous.

I couldn't prove that A is bounded.

How can we do this? Any hints?

Thanks in advance.

If $\alpha\le0$ then $A=\varnothing$. For $\alpha >0$, you can show that $A$ is bounded by finding an upper bound, let $R=\max\{\alpha,1\}$, then  $x\ge R$ or $y\ge R$  implies $(x,y)\notin A$. So, for all $(x,y)\in A$ we have $$x<R\quad\text{and}\quad y<R\qquad\qquad\implies\qquad\qquad \sqrt{x^2+y^2}<\sqrt2R$$

$\displaystyle\alpha \ge xy+\frac{1}{x}+\frac{1}{y} \ge xy $ because $x,y>0$

Now, for fixed $y>0$, $\displaystyle xy\le \alpha \implies x \le \frac{\alpha}{y}$

similarly conclude for fixed $x$


Equation of the projection curve of intersecting spheres How do I find the equation of the projection of the intersection of the spheres $x^{2}+y^{2}+z^{2}=1$ and $ x^{2}+(y-1)^{2}+(z-1)^{2}=1$ onto the $yz$-plane?

All points $(x,y,z)$ on the intersection of the two curves satisfy
$$x^2+y^2+z^2=1\qquad\text{ and }\qquad x^2+(y-1)^2+(z-1)^2=1.$$
In particular, this shows that all such points also satisfy
$$y^2+z^2=(y-1)^2+(z-1)^2,$$
which further simplifies to $z=1-y$. Can you take it from here?

The difference

$$
x^2+y^2+z^2-1-(x^2+(y-1)^2+(z-1)^2-1) = 2(y+z-1) = 0
$$

defines the intersection plane.

Now the intersection plane $y+z-1 = 0$ and the projection plane $O yz$ or $x = 0$ are orthogonal hence, the projection should be a segment. See the picture below when in blue appear the circle from the two spheres intersection and the line as intersection of $Oyz$ and $y+z-1=0$.  This segment limits can be determined thru the intersection of the circles $y^2+z^2=1$ and $(y-1)^2+(z-1)^2=1$


Transition Matrix of M/M/1 Queue We know that for an M/M/1 queue the state space is $S=\{0,1,2,... \}$. Further the probability to go from state $i$ to $i+1$ is $\lambda$ for all $i$ in $S$. Moreover, to go from $i$ to $i-1$ is the probability $\mu$ $\forall i \in S$.

So one can draw the following picture, taken from the wikipedia page on M/M/1 queues:

Now with this, one can establish the transition matrix:

$T=$\begin{pmatrix}
1-\lambda & \lambda & 0 & 0 & 0 & ...\\
\mu & 1-(\mu+ \lambda) & \lambda & 0 & 0 & ... \\
0 & \mu & 1-(\mu+\lambda) & \lambda & 0 & ...\\
0 & 0 &  \mu & 1-(\mu+\lambda) & \lambda & ...\\
&&&&&...
\end{pmatrix}

I know that the transition rate matrix is given by:

$M=$\begin{pmatrix}
-\lambda & \lambda & 0 & 0 & 0 & ...\\
\mu & -(\mu+ \lambda) & \lambda & 0 & 0 & ... \\
0 & \mu & -(\mu+\lambda) & \lambda & 0 & ...\\
0 & 0 &  \mu & -(\mu+\lambda) & \lambda & ...\\
&&&&&...
\end{pmatrix}

How can I get from the transition matrix to the transition rate matrix? Is it even possible?

I appreciate any help! Thank you very much!

More generally... one may want to keep in mind that Markov chains (in discrete time) and Markov processes (in continuous time) are different (although related) objects.

1. A Markov chain $(X_n)$, indexed by $n$ integer, is described by a transition matrix $P$, such that, for every states $(x,y)$ and every $n$, $$\Pr(X_{n+1}=y\mid X_n=x)=P_{xy}.$$ Thus:

  The matrix $P$ has nonnegative entries and every row sum $\sum\limits_yP_{xy}$ is $1$.

The powers of the matrix $P$ yield the transition probabilities of the chain in any number of steps, through the identities

  $$\Pr(X_{n+k}=y\mid X_n=x)=(P^k)_{xy}.$$

2. A Markov process $(Y_t)$, indexed by $t$ real, is described by a transition rate matrix $Q$, often called its generator, such that, informally, for every states $x\ne y$, when $\mathrm dt\to0^+$, $$\Pr(X_{t+\mathrm dt}=y\mid X_t=x)=Q_{xy}\mathrm dt+o(\mathrm dt).$$ Additionally, the diagonal entries are defined as $$Q_{xx}=-\sum_yQ_{xy}.$$
Thus:

  The offdiagonal entries of the matrix $Q$ are nonnegative and every row sum $\sum\limits_yQ_{xy}$ is $0$.

The matrix $Q$ yields the transition probabilities of the process, through the identities

  $$\Pr(X_{t+s}=y\mid X_t=x)=(\mathrm e^{sQ})_{xy},$$

for every nonnegative $(t,s)$, where the exponential is defined by the usual series, always convergent, that is, $$\mathrm e^{sQ}=\sum_{n\geqslant0}\frac{s^n}{n!}Q^n.$$

3. Every Markov process gives rise to some Markov chains, usually called embedded chains, by discretizations of the time. That is, if $(Y_t)$ is a Markov process with generator $Q$ , then for each $\tau\gt0$, $(X_n)$ defined by $$X_n=Y_{n\tau},$$ is a Markov chain with transition matrix $$P=\mathrm e^{\tau Q}.$$ (Exercise: Show that if $Q$ is a generator, that is, with nonnegative offdiagonal entries and every row sum equal to $0$, then $\mathrm e^{\tau Q}$ is a transition matrix, that is, with nonnegative entries and every row sum equal to $1$.)

4. In the other direction, not every Markov chain corresponds to the discretization of a Markov process. The ones that do are called embeddable.

(Exercise: Find some non embeddable transition matrix $P$. Hint: Two states are enough, and watch out for the zero entries of $P$.)

5. Queues in general, and M/M/1 queues in particular, are Markov processes (in continuous time), as explained on the WP page the OP links to. As such, they can be discretized into Markov chains (in discrete time).

Nota: All this can be made rigorous on every finite state space, always, and on countable state spaces, under some technical additional conditions that we left out (and that are satisfied for the queues the OP is interested in).

The matrix is wrong. It should read "1-$\lambda$" instead

Also for the other probabilities it is not $-(\mu + \lambda)$ but rather $1-(\mu + \lambda)$
as probabilities can't be negative, also it's pretty clear by the picture you posted.


Why does the probability of a bivariate Poisson distribution behaves like this? (see example) I am a beginner in working with statistics and I wanted to generate a bivariate Poisson distribution $(X_1, X_2)$ and to do so, I did according to the indication found on wikipedia (bottom of the page, http://en.wikipedia.org/wiki/Poisson_distribution), therefore, I have generated three independent random variables which have a Poisson distribution $(Y_1, Y_2, Y_3)$, with means $\lambda_1 = \lambda_2 = \lambda_3 = 2$.

Provided I have set:

$X_1 = Y_1 + Y_3$

$X_2 = Y_2 + Y_3$

Then,

$X_1$ ~ Poisson($2+2$) = Poisson($4$)

$X_2$ ~ Poisson($2+2$) = Poisson($4$)

In order to compute the bivariate probability I use the formula found on wikipedia (bottom of the page, http://en.wikipedia.org/wiki/Poisson_distribution). When I try to compute the following probabilities (both in Matlab - set the format short - and on the paper), I obtain:

$P(X_1 = 4, X_2 = 4) = 0.0446$

$P(X_1 = 3, X_2 = 3) = 0.0474$

Basically, it indicates that $P(X_1 = 4, X_2 = 4) < P (X_1 = 3, X_2 = 3)$, which seems completely counterintuitive, since the mean for both random variables $X_1$ and $X_2$ is $4$. Initially, I have doubted my Matlab code and my calculation skills, but I tested it thouroughly and I obtain the same results. Is there any potential explanation for this or I am simply wrong in performing my calculations?

Thank you all.

The probability mass of a PoissonDistribution[4] is greater at $x = 3$ than $x = 4$, as you can easily calculate.


How to calculate reliabilty of humans in data-entry tasks? Please help me reason about this:

I have many, simple, data-entry tasks to be executed by humans. Of course they will make mistakes, so the result will be wrong sometimes.

I don't know, before hand, how often a human will make a mistake, but let's assume a person will make a mistake 10% of the times.

If I now let two different persons solve the same task, their results should match. If the results do not match, I could let a third person solve the task and thus find the right answer.

There is also the case where two persons make the same mistake on the same task, giving a wrong but mathcing answer, thus not detecting the failure.

My question is:
How can I calculate the probabilities of an error happening?

Given the answer above, I could then calculate the amount of safety I could get by having 3 or 4 different persons solving the same task.

Thanks a lot!

You need to have a model of the errors, which will depend on the task.  The simplest is where the task is to enter $n$ binary choices, so on each choice there is only one possible error.  Let $p$ be the chance that a single bit is entered incorrectly.  The chance that a given task is completed correctly is then $(1-p)^n$, which corresponds to $90\%$ in your guess.  If you have two people do the same task, the chance they both make a mistake is $0.10^2=0.01=1\%$, but the chance they make an error on the same bit is much smaller.

As the chance of one error is fairly small, there is not much wrong in assuming that nobody makes more than one error.  This lets us say $np=0.10, p=\frac {0.10}n$  Then the chance that two people make the same error is $np^2=\frac {0.01}n$.  The factor $\frac 1n$ comes from the fact that the errors have to match.

If the data is more complicated, like entering names, there are many different errors that can be made, so the chance of matching errors is much lower.  There are probably particular errors that become rather common, however.  If somebody's last name is Jons, you will probably have a relatively high chance it will be entered as Jones.  This is much harder to model.

Since each person works independently, the event "person $1$ makes a mistake performing task $i$" is independent from the event "person $2$ makes a mistake performing task $i$", meaning that the probability of both people making an error for task $i$ is $0.1\cdot 0.1 = 0.01$. This means that if you have $100$ tasks, the average number of errors you can expect (the expected value is $1$.


Let $f(x,y)$ be a function of two variables such that $\displaystyle\lim_{(x,y)\to(0,0)}=5$. Which is true? Let $f(x,y)$ be a function of two variables such that $\displaystyle\lim_{(x,y)\to(0,0)}=5$

Which is true?

(a) We have $f(0,0)=5$

(b) If $(x,y)$ approaches $(0,0)$ but not reaching $(0,0)$, then $f(x,y)$ approaches $5$ but not reaching $5.$

(c) If $(x,y)$ approaches and reaches $(0,0)$, then $f(x,y)$ approaches and reaches $5$

(d) None of them are true.

My solution

  For (b), this seems to be true as this is essentially the definition of limits, isn't it? As $x$ reaches $0,$ y reaches $0,$ but not at $(0,0)$, $f(x,y)$ reaches $5$, but not at $5$ exactly.

But my answer says otherwise..

Please enlighten and thanks!

They are all false.

a) and c) Let f(x) =
$\begin{cases}
5, & \text {if (x,y)} \neq \text{(0,0)} \\
0, & \text{otherwise}
\end{cases}$

b) $f(x,y) = 5$

a) false because there could be a discontinuity

b) false because $f(x,y)$ could be equal to $5$ on the entire interval approaching it.  In that case "but not reaching it" disqualifies this choice.

c) True. The hypothesis says that $f(x,y)$ approaches $5$ as $(x,y)$ approaches $0$.  Since choice c lets $(x,y)$ reach its limit, $f(x,y)$ reaches $5$.


If A and B are symmetric is AB=BA true? Question

if we have matrices $A$ and $B$ which are symmetric. Both of these satisfy:
$$  A=A^T$$
Is $AB=BA$ true ?

Attempt to solve

Now if $A$ and $B$ are symmetric then $AB$ most be symmetric too.

$$ AB=(AB)^T=B^TA^T=BA $$

problem is how do you proof that if $A$ and $B$ are symmetric also $AB$ is symmetric ?

If somone could provide some insight on this that would ge highly appreciated.

Thanks,

Tuki

Let $$A =\begin {bmatrix} a & b\\ b & c \end {bmatrix} \quad\mbox{and}\quad B = \begin {bmatrix} d & e\\ e & f \end {bmatrix}$$
be two symmetric matrices. Then
$$AB-BA =\begin {bmatrix} ad+be & ae + bf\\ bd+ec & be + cf\end {bmatrix} -\begin {bmatrix} ad +be & db + ec\\ ea+bf & eb + fc\end {bmatrix}=\begin {bmatrix} 0 & ?\\ ? & 0\end {bmatrix}.$$
Is the right-hand side necessarily the zero matrix?

If $A$ and $B$ are symmetric matrices then $AB=(BA)^T$, so only if $AB$ is symmetric is it true that $AB=BA$.


rational function limit involving factorials I posted something similar but someone edited the question and added the wrong equation, which gave irrelevant responses.

Lim (2n-1)!/(2n)^n as n approach infinity.

Any method, I would just like a starting, no full answer needed (but will be appreciated)!

Hint. You may recall the Stirling formula
$$
n!\sim\sqrt{2\pi n}\left(\dfrac{n}{e}\right)^n
$$ as $n$ is great, giving
$$
\frac{(2n-1)!}{(2n)^n}\sim\sqrt{2\pi (2n-1)}\left(\dfrac{2n-1}{e}\right)^{2n-1}\frac{1}{(2n)^n} \sim\frac{\sqrt{2\pi}}{e}\left(\dfrac{2n}{e^2}\right)^{n-1/2}
$$
then your limit is $+\infty$ since $\displaystyle \dfrac{2n}{e^2} \longrightarrow +\infty$.


What is the intuition of why convergence in distribution does not imply convergence in probability For me its very counter intuitive (that convergence in Probability and Distribution are not the same), because, conceptually, if two random variables have the same distribution, then they should be considered the same random variable. I know this is wrong. I am not sure why but its wrong (hence the question). Maybe, does it mean they are the same (isomorphic) w.r.t. to their pdf/cdf/pmf? Anyway, I want to understand why I am wrong and fix my mathematical intuition.

The reason I am complaining is because I am completely aware of a counter example. In fact I will provide it in the appendix of this question, however, even after seeing the counter example, I am not sure I could have come up with it. In fact even after seeing a counter example, it still feels wrong. I was wondering if someone could provide a way of how one would come up with such a counter example (or maybe a counter example that is much more obvious), or maybe explain to me why it should be obvious that one can come up with such a counter example.

Appendix:

Recall the definitions of convergence in distribution and convergence in probability:

Convergence in distribution (D):
$$ \lim_{n \rightarrow \infty} | F_{X_n}(x) - F(x)| = 0, \forall x \in X$$

Convergence in probability (P) for sufficiently large $N$ and $\forall \epsilon > 0$:

$$ \lim_{n \rightarrow \infty} Pr[ | X_n - X | \geq \epsilon]= 0$$

i.e. if one goes sufficiently far the sequence of random variables, then, $X_n$ will converge to $X$ (in probability).

I will show the counter example for:

$$D \not\implies P$$

Consider a Bernoulli random variable $X$ with probability of success $\frac{1}{2}$. Now define $X_n$ to be equal to $X$, i.e.

$$X_n = X$$

Therefore, obviously since they are the same r.v. they have the same distribution. Now define:

$$Y = 1 - X$$

Obviously, $Y$ has the same distribution as $X$ and hence, also as $X_n$. So $X_N$ converges in distribution to $Y$.

But, do they converge in probability?

If they do then:

$$ \lim_{n \rightarrow \infty} Pr[ | X_n - Y | \geq \epsilon]= 0$$

i.e. we want their difference to be larger than $\epsilon$ very rarely (in fact we want that event to have zero probability zero as n goes to infinity).

Let's figure out the distance between these two random variables $X_n$ and $Y$:

$$| X_n - Y | = |X_N - (1 - X)| = |X -  1 + X| = |2X - 1| = 1$$

Therefore, their distance is always 1, no matter what. We want their distance to be very far for any $\epsilon$, but notice that for the choice of $\epsilon = \frac{1}{2}$, that the distance between $| X_n - Y | = 1$ is indeed greater than the tolerable threshold (i.e. $1 = | X_n - Y | > \epsilon = \frac{1}{2} $ with probability 1). So for sure their distance is more than $\epsilon$. It doesn't matter what value of $n$ we choose (i.e. it doesn't matter how far we go down the sequence), they are for sure too far apart. More precisely:

$$ \lim_{n \rightarrow \infty}Pr[ | X_n - Y | \geq \epsilon ] = Pr[ 1 \geq \frac{1}{2} ] = 1 $$

which is the opposite of what we need for convergence in probability. End of counter example.

First, in convergence in distribution, the random variables
can be defined on different spaces, whereas convergence in probability requires them all to be defined on the same space (unless the limiting 'random variable' is degenerate).

Second, even artificially requiring that all random variables
are defined on the same space, it is easy to find examples of sequences of random variables that
converge in distribution, but not in probability. Here is
 (perhaps) the most trivial example.

Let the sample space have only two points
$a$ and $b$, each with probability $1/2.$ Let all $X_n$ be
the same with $X_n(a) = 0$ and $X_n(b) = 1$; let $Y$ be
defined as $Y(a) = 1$ and $Y(b) = 0$. (Notice the crucial switch.) Then $F_{X_n}(x)$ has
jumps of $1/2$ at $x = 0$ and $x = 1.$ Also, the CDF of $Y$
is exactly the same function. So $F_{X_n}(x) \equiv F_Y(x)$ and
$|F_{X_n}(x) - F_Y(x)| \equiv 0$, making convergence in
distribution trivial.

By contrast, $|X_n(a) - Y(a)| = 1$ and $|X_n(b) - Y(b)| = 1,$
so $P(|X_n - Y| = 1) = 1,$ and convergence in probability
is impossible.

Finally, for convergence to a constant, the two concepts can
be interpreted as equivalent.

Lim P( Xn(w)-X(w) <= E ) = 1 for all w of the population.
We can induce the same distribution using two different random variables.
Your example shows it good.
If this random variables are different then, p(x=1)=p(y=1) whereas x^-1(1) is a different set of the population from y^-1(1).
This means convergence in distribution is weakening the hypothesis needed for a succession of random variables to converge in probability.
When imagining this, just associate a set of w's to each point of the domain of the distribution of the random variable.


How to prove that the tangent to a circle is perpendicular to the radius drawn to the point of contact? I've tried drawing a parallel chord to the tangent but then how would you prove that the chord is perpendicular to the radius?

Let $O$ be the centre of the circle, let $\ell$ be a tangent line, and let $P$ be the point of tangency.  Suppose that $OP$ is not perpendicular to $\ell$. Draw the line through $O$ which is perpendicular to $\ell$. Then this line meets $\ell$ at a point $Q\ne P$.

Note that $Q$ is outside the circle. Now consider the triangle $OQP$. This is right-angled at $Q$.  So $OP$ is the hypotenuse of this triangle, and is therefore bigger than $OQ$. But this is impossible: since $Q$ is outside the circle, we must have $OP\lt OQ$.

Extend the radius to a line $L$ and use reflection symmetry with respect to this line.


Convert IVP to an equivalent Volterra integral equation Convert the following initial value problem to an equivalent Volterra integral equation:

$
  \begin{cases}
    u'' -u' \sin x + \Bbb e ^x u= x   \\
       u(0)=1\\
   u'(0)=-1\\
  \end{cases}
$

I have solved it and I have got the solution

$$ x- \sin x + x \Bbb e ^x - \Bbb e ^x - \int\limits _{0} ^ x u(t) \Big(\Bbb e ^x (x-t) - \sin x \Big) \Bbb d t ,$$

but our instructor in the university has solved it using another method and he has obtained the solution

$$\frac {x^3} 3 -x +1 + \int \limits _{0} ^x \Big( \sin t -(x-t)(\cos t + \Bbb e ^t) \Big) u(t) \Bbb d t .$$

My question is: why are the two solutions different? Is my solution wrong?

Thanks for help.

As written above, both solutions are wrong. To verify your solution, derive it once; you will get $u'(x) = 1 - \cos x + x \Bbb e ^x + u(x) \sin x$. Deriving it a second time: $u'' (x) = \sin x + \Bbb e ^x + x \Bbb e ^x + u'(x) \sin x - u(x) \cos x$. This clearly does not look like your original equation.

Concerning your instructor's solution, deriving it once gives $u'(x) = x^2 - 1 + u(x) \sin x$; deriving once more gives $u'' (x) = 2x - u(x) \cos x + u' (x) \sin x$. Again, this is very far from the given equation.

Maybe you have made some typing mistakes when writing the above formulae, or maybe you have not taken notes correctly. As you are going to see next, your instructor's formula is very close to the correct one.

Let us start by integrating the given equation; we get

$$u'(y) - u'(0) = \int \limits _0 ^y u' (t) \sin t \Bbb d t - \int \limits _0 ^y \Bbb e ^t u(t) + \int \limits _0 ^y t \Bbb d t .$$

Using that $u'(0) = -1$ and integration by parts in the first integral we get

$$u'(y) = u(y) \sin y - \int \limits _0 ^y u(t) (-\cos t) \Bbb d t - \int \limits _0 ^y \Bbb e ^t u(t) + \frac {y^2} 2 -1 ,$$

or equivalently

$$u'(y) = \frac {y^2} 2 -1 + u(y) \sin y + \int \limits _0 ^y (\cos t - \Bbb e ^t) u(t) \Bbb d t .$$

Now, let us integrate this once more:

$$u(x) - u(0) = \int \limits _0 ^x \frac {y^2} 2 -1 \Bbb d y + \int \limits _0 ^x u(y) \sin y \Bbb d y + \int \limits _0 ^x \int \limits _0 ^y (\cos t - \Bbb e ^t) u(t) \Bbb d t \Bbb d y .$$

We shall now use $u(0) = 1$ and we shall change the order of integration in the last integral in order to get

$$u(x) = \frac {x^3} 6 - x + 1 + \int \limits _0 ^x u(y) \sin y \Bbb d y + \int \limits _0 ^x \int \limits _t ^x (\cos t - \Bbb e ^t) u(t) \Bbb d y \Bbb d t .$$

Since the function in the last integral does not depend on $y$, if we change the letter from $y$ to $t$ in the first integral we get

$$u(x) = \frac {x^3} 6 - x + 1 + \int \limits _0 ^x \Big( \sin t + (x-t) (\cos t - \Bbb e ^t) \Big)u(t) \Bbb d t .$$

As you can see, the only differences between the instructor's formula and the correct one are the denominator ($6$ instead of $3$) and the sign in front of $\cos t$.

$u(x)=x+\sin x-xe^x$-integral from $0$ to $x$ of $(e^x(x-t)-\sin x)u(t)dt$


Is square root of an integer, either an integer or an irrational number, but never (non-integer) rational? Possible Duplicate:
  $\sqrt a$ is either an integer or an irrational number.

$\sqrt{2}$ is irrational number, but $\sqrt{9} = 3$ is an integer.
Are there such integers whose square root is a (non-integer)rational number?

Assume $x^2 =\frac{m}{n}$ where $m \in \mathbb Z$ and $ n \in \mathbb N$

this implies $x = \frac{\sqrt{m}}{\sqrt{n}}$

which means that for an integer to have a rational root it's root must be the ratio of two integer roots. For this ratio to not be an integer m and n must be distinct which implies that x is not an integer quantity.


Question on Probability. What is my Z-Score 20? I have this homework but I am getting a z-score of 20?  Why?  Is my calculation wrong?

The question is

Without assuming that the diameters of apple pies are distributed according to the normal distributions, estimated the probability that the mean diameter is larger than 32 cm.  The sample standard deviation is estimated to be 2.  The sample mean is 28 and the sample size is 100.

When I used CLT (because the sample size is >30).  I am getting a z score of 20?  Is this correct?

Central Limit Theorem indicates that: For samples of size n taken from a population
with mean, $\mu$, and a standard deviation, $\sigma$, $$\mu_{\overline{x}}=\mu, \sigma_{\overline{x}}=\frac{\sigma}{\sqrt{n}}$$
This problem is actually a reversed form of CLT. We are told the sample mean and standard deviation, and we need to compute the population mean and standard deviation.
$$\sigma_{\overline{x}}=2=\frac{\sigma}{\sqrt{100}},\sigma=20$$$$\mu=\mu_{\overline{x}}=28$$
Thus it will just be $P(Z>\frac{32-28}{20})=P(Z>0.2)=1-P(Z<0.2)$

But isn't 2 the standard deviation and not the standard error? isn't the formula 32-28 divided by 2/SQ of 100?  This would indeed have a Z value of 20.


Formula to find the Angle between two slopes I have given two slopes $m_1 = \frac{1}{2}$ and $m_2 = 1$

While finding the angle I made use of the formula $\tan(\theta) = \frac{m_1-m_2}{1+m_1m_2}$

answer is : $\theta = \arctan(\frac{-1}{3})$

But in book the answer is $\theta = \arctan(\frac{1}{3})$.

what would be the right answer?

$$
\frac{1-\frac12}{1+1\cdot\frac 12} = \frac 1 3, \qquad \frac{\frac12-1}{1+\frac12\cdot1} =-\frac13.
$$
Either of these is the angle between those two lines.  Where two lines intersect they form angles with measures adding up to $180^\circ$.  If $\alpha+\beta=180^\circ$ then $\tan\alpha = - \tan\beta$.

The angle is acute. Thus, the answer is the book's answer.


Clear difference between unitary matrix and semi-unitary matrices? What are the clear differences between Unitary and Semi-unitary.

A unitary matrix $U_{n\times n}$ is generally a complex matrix $U$ whose columns (or rows) constitute an orthonormal basis for $\mathbb C^n$. This means $$U^*U=UU^*=I$$
It is also invertible with $U^{−1} = U^∗$.

A semi-unitary matrix $U_{m\times n}$ is a non-square matrix ($m>n$ or $m<n$) where $U^*U=I_n$ or $UU^*=I_m$ which means either the rows or columns of the matrix are orthonormal.

Unitary Matrix: where $U^\dagger U=I$ (the Hermitian of $U$ times $U$ equaling the identity matrix)

Semi-Unitary Matrix: where $U^TU=I$ (the transpose of $U$ times $U$ equaling the identity matrix)

I'll be updating this with resources as I find them. Hope this helps!


How to handle indices with fractional degree? An algebra problem ate my head!!!

  $x$ and $y$ are positive real numbers such that
  $$\sqrt{x^2 + \sqrt[3]{x^4 y^2}} + \sqrt{y^2 + \sqrt[3]{x^2 y^4}} = 512.$$
  Find $x^{2/3} + y^{2/3}$.

It would be a great help if anybody helps me in solving this problem. I tried taking conjugates and all but I didn't get any answer.
thank you

Let,
\begin{align}
& x^{2/3} + y^{2/3} = z \newline
\implies & \left(x^{2/3} + y^{2/3}\right)^{3/2} = z^{3/2}\newline
\implies & z^{3/2} = 512 = 8^3 \tag{1}
\end{align}

To remove the fractional index $3/2$ on the LHS of (1) to obtain the base $z$, we take root $2/3$ (i.e. two-third root) of both sides, i.e.

\begin{align}
& \left(z^{3/2}\right)^{2/3} = \left(8^3\right)^{2/3} \newline
\implies & z^{(3/2) \times (2/3)} = 8^{3 \times 2/3} \newline
\implies & z = 8^2 = 64
\end{align}

Hence,  $x^{2/3} + y^{2/3} = z = 64$.


Proof for the dimension of a subspace of M(n,R) First I will state the exercise and then show what I could get.

Let $A_{1},...,A_{k} \in M(n,R)$ such that for all $ 1 \le i \neq j \le k $:  $ A_{i}^{2}=I$ and $A_{i}A_{j}+A_{j}A_{i}=0$ show that $k \le \frac{n(n+1)}{2}$

Ok, first of all we can see that the condition $ A_{i}^{2}=I$ means that all the matrices in this form are diagonalizable with eigenvalues $\in  \big\{1,-1\big\}  $, so we can state that $A_{i}$ are all invertible matrix and $A_{i}^{-1}=A_{i}$.

Now if I multiply the relation between the matrices for $A_{i}$ i get that $A_{i}A_{j}A_{i}=-A_{j}$ wich means that $A_{j}$ and $-A_{j}$ are similar and because they need to share determinant it has to be 0 if the dimension of the space is odd.
I also know that they commute. I feel like I should be able to find some other relation (maybe about the Ker of the linear application) from $A_{i}A_{j}+A_{j}A_{i}=0$ but I don't know how.

Thanks in advance

I shall work with the assumption that the base field is any field $K$.  Let $V:=\text{Mat}_{n\times n}(K)$.  Suppose that $A_1,A_2,\ldots,A_k\in V$ are such that $A_i^2=1_n$ for every $i\in[k]$, and $A_iA_j+A_jA_i=0_n$ for all $i,j\in[k]$ with $i\neq j$.  Here, $1_r$ and $0_r$ denote, respectively, the $r$-by-$r$ identity matrix and the $r$-by-$r$ zero matrix, and $[m]:=\{1,2,\ldots,m\}$ for all $m\in\mathbb{Z}_{>0}$.

First, we handle the case where $K$ is of characteristic not equal to $2$.  It has been shown that, if $k>1$, then $A_i$ and $-A_i$ are similar matrices for all $i\in[k]$.  That is, if $k>1$, then $$\det(A_i)=\det(-A_i)=(-1)^n\,\det(A_i)\,,$$ making $(-1)^n=1$, whence $n$ is even.  Thus, if $n$ is odd, then $k\leq1$.  From now on, we assume that $k>1$ and $n$ is even.

Consider the linear operator $T_X:V\to V$ defined for each $X\in V$ by
$$T_X(Y):=XY+YX\text{ for any }Y\in V\,.$$
If $X$ is diagonalizable over $K$, then let $x_1,x_2,\ldots,x_n\in K^n$ be eigenvectors of $X$, with respect to eigenvalues $\lambda_1,\lambda_2,\ldots,\lambda_n\in K$, respectively, and let $\chi_1,\chi_2,\ldots,\chi_n\in K^n$ be eigenvectors of $X^\top$, with respect to eigenvalues $\lambda_1,\lambda_2,\ldots,\lambda_n\in K$, respectively.  Then, the $n^2$ linearly independent matrices $x_\mu\,\chi_\nu^\top$ are eigenvectors of $T_X$, with eigenvalues $\lambda_\mu+\lambda_\nu$, for $\mu,\nu\in[n]$.

Without loss of generality, we suppose that
$$A_1=\begin{bmatrix}+1_{\frac{n}{2}}&0_{\frac{n}{2}}
\\0_{\frac{n}{2}}&-1_{\frac{n}{2}}
\end{bmatrix}\,.$$
Then, $A_1$ is diagonalizable with eigenvectors $e_1,e_2,\ldots,e_{\frac{n}{2}}$ corresponding to the eigenvalue $+1$, and with eigenvectors $e_{\frac{n}{2}+1},e_{\frac{n}{2}+2},\ldots,e_n$ corresponding to the eigenvalue $-1$.  Here, $e_1,e_2,\ldots,e_n\in K^n$ are the standard basis vectors of $K^n$.  This shows that the nullspace of $T_{A_1}$ is of dimension $2\left(\dfrac{n}{2}\right)^2=\dfrac{n^2}{2}$, since it is spanned by $e_i\,e_j^\top$ and $e_j\,e_i^\top$ with $i\in\left\{1,2,\ldots,\dfrac{n}{2}\right\}$ and $j\in\left\{\dfrac{n}{2}+1,\dfrac{n}{2}+2,\ldots,n\right\}$.  Thus, from David C. Ullrich's answer, $A_2,A_3,\ldots,A_k$ are linearly independent element of $\ker\big(T_{A_1}\big)$, so we get $$k-1\leq \frac{n^2}{2}\text{ or }k\leq \frac{n^2}{2}+1\leq \frac{n(n+1)}{2}\,.$$
I however believe that the maximum possible value of $k$ is at most $\dfrac{n^2}{4}+1$.

If $K$ is of characteristic $2$, then clearly the matrices $A_1,A_2,\ldots,A_k$ need to only commute.  There is no bound on $k$ (for example, $A_1=A_2=\ldots=A_k=1_n$ works).  However, if we demand that $A_1,A_2,\ldots,A_k$ be linearly independent, then a bound for $k$ can be determined.  First, we may assume that $A_1,A_2,\ldots,A_k$ are upper triangular matrices each of whose diagonal entries is $1$.  Then, $A_2-A_1,A_3-A_1,\ldots,A_k-A_1$ are strictly upper triangular matrices.  The subspace of $V$ composed by strictly upper triangular matrices is of dimension $\dfrac{n(n-1)}{2}$.  That is,
$$k-1\leq \frac{n(n-1)}{2}\text{ or }k\leq \frac{n(n-1)}{2}+1\leq \frac{n(n+1)}{2}\,.$$  I conjecture that, with the extra assumption that the matrices $A_1,A_2,\ldots,A_k$ be linearly independent, $k\leq\left\lfloor\dfrac{n^2}{4}\right\rfloor+1$.

I see the bound is already proved. So i will give a different perspective.
If you add the condition that $A_i^tA_j+A_j A_i^t = 0$,$\forall i,j$ (when $i=j$, this condition applies only when $A_i \neq A_i^t$), then the following is true.
Now consider $A_1,...,A_k$ and now replace $A_i$ by $\frac{1}{\sqrt{2}} (A_i+A^t_i)$,$ \frac{1}{\sqrt{2}} (A_i-A^t_i)$ if $A_i \neq A^t_i$ and remove $A_i^t$ from the set if present or else if $A_i = A^t_i$ keep $A_i$ as it is. Let the replaced matrices be $B_1,...B_k$.

Now $B_1,...,B_{k}$ also satisfies the equations $B^2_i = I$ and $B_i B_j + B_j B_i = 0$.


Proving the number of non-fixed points in a self-inverse function is even I would like some guidance in how to solve this problem. I'm really trying to understand it, so hints and discussion would be preferred over a solution!

Le problem:

  A function $f:S \to S$ is called self-inverting if $f(f(x))=x$ for every $x \in S$. A point $x \in S$ is called a fixed point of $f$ if $f(x)=x$.
  - Prove that, if $f:S \to S$ is a self-inverting function that has $k$ fixed points, then $|S|−k$ is even.

My reasoning so far:
When mapping $f:S \to S$, $f(x)=\frac{1}{x}$ the cardinality of $S$ is equal to $2n$, since every element in $S$ maps to a different element in $S$, except for the case of a fixed point, so that for every fixed point, the cardinality is $2n + k$? I am on the right track? If so, how should I approach proving this?

checkout fixed points theorem.

|S| = |the number of fixed points| + |the number of non-fixed points|

what is |the number of fixed points|?

f(x) = x -> k=1 this means in your set S you have the 1 number that maps to itself. |k| where k=1 is 1.

What is |the number of non-fixed points|?

f'(f(x)) = x .
1 non-fixed point map to 1 'other' point in S.  If there was a number n that described the number of points for x.  Then we would have n non-fixed points mapping to n 'other' points.  To count this number of points I would describe this as 2n points.

This holds true for k=2 (and beyond). You have two numbers that map to themselves and the rest map to another number and back again.(2n)

The cardinality of your set has to be the 2n + k.  If you |S|= 2n + k and you subtract k, then you simply have 2n points.


Cones and partial ordering A positive cone always induces a partial ordering on a vector space. I wonder if the converse holds, so is every partial ordering in a vector space induced by a positive cone, or in other words, is there a bijection between positive cones and partial orderings?

For a partial ordering $\succ$ on a set $\mathcal{S}$, its algebraic dual, the set of $\succ$-monotones $\mathcal{F}$ is a conic partial ordering. Since $f_j \in \mathcal{F}$ admits the conic combination $ \sum_j \mu_j f_j \in \mathcal{F}$ with $\sum_j \mu_j = 1, \mu_j \geq 0$ so it is a cone. The conic partial ordering $\geq_c$ is defined by $f_j \geq_c f_k$ if and only if $f_j - f_k \in \mathcal{F}$. For some conic partial ordering $\mathcal{F}$ over $\mathcal{S}$, in turn, induces the partial ordering $\succ$. So it is true.

I think it is probability not true. Since you are asking for partial order, you can easily define one that does not preserve under scalar multiplication.

e.g.

define a ($\mathbb{R}, \geq$) such that

(1). $x\geq y$ if $(x=1,y=0)$ or $(x=y)$

(2). $x=y$ if $(x=y)$.

It should satisfied the three properties of partial order. But clearly it does not preserved under scalar multiplication. It should be easy to proof that no cone can deduct such partial order.


Definition of divisor associated to a meromorphic function Let $X$ be a complex manifold and let $f$ be a meromorphic function defined on it. Let $Div(X)$ be the group of locally finite sum of analytic irreducible hypersurfaxces of $X$.

One would like to define a divisor associated to $f$ as $$\sum_{i \in I} ord(f)|_{V_i}V_i$$ but I really can't see why this sum should be locally finite.

Let $h$ be analytic and non identically zero on a small disk $U = \{ z \in \Bbb{C}^n,|z|< r\}$ and $Hol(U)$ the ring of meromorphic functions on $U$ that are analytic on some neighborhood of $0$.

Look at the set of principal ideals $S = \{ (\alpha) , \alpha \in Hol(U), \alpha(0) = 0\} $ partially ordered by inclusion. Let $v(\alpha)$ be the degree of the first non-zero term in $\alpha$'s power series. $v(\alpha \beta) = v(\alpha)+v(\beta)$.

If $(\alpha) \supsetneq (\beta)$ then $ \beta =\alpha\gamma$ with $\gamma(0) = 0$ so $v(\gamma) \ge 1$ and $v(\beta) \ge v(\alpha)+1$.

Thus the maximal elements of $S$ are well-defined : $ M = \{ (\phi) \in S, \not \exists (\varphi) \in S, (\varphi) \supsetneq (\phi)\}$.

From there let $h_0 = h$, while $h_j(0) = 0$, pick some $ (\phi_{j+1}) \in M$ such that $(h_j) \subset (\phi_{j+1})$ and let $h_{j+1} = \frac{h_j}{\phi_{j+1}}$. Since $v(h)$ is finite the process terminates obtaining the factorization $$\forall z \in U, \qquad h(z) = h_J(z)\prod_{j=1}^J \phi_j(z)$$ where $J \le v(h)$ and $h_J(0) \ne 0$.

  Todo : show $v(\phi_j) = 1$ so $J = v(h)$.

Find an even smaller disk $V$ where $h_J$ is non-zero and the $\phi_J$ are holomorphic. The local divisor is  defined as $Div(h|_V) = \sum_{j=1}^J Z(\phi_j|_V)$ where $Z(\phi|_V) = \{ z \in V, \phi(z)=0\}$.

If $f$ is meromorphic then $f = \frac{h}{g}$ with $g,h$ holomorphic and $Div(f|_V) = Div(h|_V)-Div(g|_V)$.

The global divisor is obtained by gluing those local divisors.

There is an open set $U$ on which your meromorphic function doesn’t have any zeros or poles, so the only nonzero coefficients come from prime divisors on the complement of $U$ in $X$. This complement is a closed set $Z$. Now for the local finiteness: Take the intersection of $Z$ with a quasicompact open set of $X$, which is Noetherian as a topological space because $X$ is locally Noetherian, as it is a complex manifold and the ring of analytic functions in finitely many variables is Noetherian. (I learned this last fact from Huybrechts’s book Complex Geometry). Every subspace of a Noetherian topological space is Noetherian, so in particular this intersection can have only finitely many irreducible components. This gives you local finiteness.


How to find the values of x and y in equations like ax+by=c? for example this equation: 505x-673y=1 . x=4 and y = 3. but how can I find them with mathematics. What would be the approach here?

This was a comment, but I think deserves to be an answer.

As your question is asking for a method, here it is.

1) First, divide by the common factor $d=GCD(a,b,c)$.
2) Then apply "extended Euclidean algorithm" which will give you  coefficients $x',y'$ called "Bezout coefficients".
3) The final coefficients are : $x'=dx, y'=dy$.

See the details of point 2) here.

This is a simply diophantine equation. To solve this, you have basically to express $673$ in terms of $505$ and $1$ using euclid algorithm. The result, as you have shown is $x=4$ and $y=3$ because:
$$505\cdot4-673\cdot3=1$$


A general equation for the number of rolls to get from N dice to zero dice I am trying to understand how to write a general equation for a math problem that came up during a recent game of Warhammer 40k.

The question is:
Assume you start with a set of 10 regular 6-sided dice. The entire set of dice will be rolled simultaneously. Any of the dice that roll to a 6 will be removed from the set, and the remaining dice will be re-rolled. This will repeat until there are no dice remaining. How many rolls, mathematically, should this process take?

At first thought I was under the assumption that this would be similar to a binary search algorithm, but rather than decreasing the initial set size by $\frac 12$ per iteration, it would decrease only by $\frac 16$, though I keep coming up with answers that are clearly incorrect (huge numbers, negative numbers.)

What would be the most appropriate way to generalize an equation for this problem for a set of $N$ regular 6-sided dice?

Your intuition is reasonable.  The expected number of dice is reduce by a factor $\frac 56$ each roll, so you could ask the number of rolls to get the expected number of dice below $1$.  That gives
$$10\left(\frac 56\right)^n \lt 1\\n\approx 12.6$$
so you should expect to average $13$ rolls but that is very rough.

A more careful approach is to work upwards.  You expect $6$ rolls if you start with one die.  If you start with two dice you have $\frac 1{36}$ chance of getting two sixes and being done.  You have $\frac {10}{36}$ chance of getting one six and being at one die.  You have $\frac {25}{36}$ chance of getting no sixes and being back where you started.  If $x$ is the expected number of rolls starting with two dice you then have
$$x=\frac 1{36} \cdot 1 + \frac {10}{36} \cdot 7 +\frac {25}{36}(1+x)\\
x=\frac {96}{11}\approx 8.73$$
You can keep going with this to get the expected value for ten dice, but the calculations get long.

At each step you have a new geometrically distributed random variable. Each time with a different probability of success.

If you have $k$ dices, then the success ratio is $1 - \left(\frac{5}{6}\right)^k$, i.e. probability of not getting every dice between $1$ and $5$.

In case $k=1$ this is our regular dice.

Now, the expectation of geometrically distributed random variable is $\mathbb{E} = \frac{1}{p}$, where $p$ - is the probability of success.

Denote $Y_i$ - a number of rolls it takes to "get rid" of a next dice, in case you have $i$ at hand.

Then $$\mathbb{E}(Y_i) = \frac{1}{1 - \left(\frac{5}{6}\right)^i}$$

Since the expectation is linear
you have $$\mathbb{E}(Y_{10} + Y_9 + \ldots + Y_1) = \sum\limits_{k=1}^{n = 10} \frac{1}{1 - \left(\frac{5}{6}\right)^k} \approx 21.876$$


Elementary Set Theory: Proof of Distributive Property I'm new to proofs.  While working through a finite math book, I've been asked to prove the distributive properties of set operations.  I'll use the distributive property of union over intersection as the example for my question.

Prove that A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).

The proofs I see for this property aim to show, in two steps, that each side is a subset of the other side.  For example, the first step usually looks something like this:

Let x ∈ A ∪ (B ∩ C). If x ∈ A ∪ (B ∩ C) then x is either in A or in (B and C).
x ∈ A or x ∈ (B and C)
x ∈ A or {x ∈ B and x ∈ C}
{x ∈ A or x ∈ B} and {x ∈ A or x ∈ C}
x ∈ (A or B) and x ∈ (A or C)
x ∈ (A ∪ B) ∩ x ∈ (A ∩ C)
x ∈ (A ∪ B) ∩ (A ∪ C)
x ∈ A ∪ (B ∩ C) => x ∈ (A ∪ B) ∩ (A ∪ C)
Therefore, A ∪ (B ∩ C) ⊂ (A ∪ B) ∩ (A ∪ C)

Then the second step would do something similar to show that (A ∪ B) ∩ (A ∪ C) ⊂ A ∪ (B ∩ C).

My question in this--In jumping from lines 3 to 4, aren't we relying on the very property we're trying to prove?  If so, how is that considered as proof?

I don't have trouble following the steps, and I don't doubt that the property is valid.  But as a newcomer to proofs I'm confused because it looks (to me) like we're using the property to prove itself.  To my non-mathematician mind, drawing a couple of Venn diagrams provides a more intuitive "proof" than manipulating these equations.

Jumping from $3$ to $4$ needs to be justified.

At $3$ you have $x \in A$ or $(x \in B \;\text{and} \; x \in C).$

Then we form two cases. One in which we see what happens when $x \in A$ and second in which $x \in B \cap C$.

Case 1): $x \in A$ -

Since $A \subseteq A \cup B$ and $A \subseteq A \cup C$,

$x \in A \implies x \in A \cup B$ and $x \in A \implies x \in A \cup C$. Thus $x \in A \cup B$ and $x \in A \cup C$. That is $x \in (A \cup B) \cap (A \cup C).$

Case 2):$x \in B \cap C$ -

Since $x \in B$ and $x \in C$, we have $x \in A \cup B$ and $x \in A \cup C$. $(\because B \subseteq A \cup B, C \subseteq A \cup C).$

Thus $x \in (A \cup B) \cap (A \cup C)$ in this case too.

Oops we jumped to $7$ directly. :P

Let's prove that $(A\cup B)\cap C = (A\cup C)\cap(B\cup C)$.

Firstly, let $x \in (A\cup C)\cap(B\cup C)$, then $x \in A\cup C$ and $x \in B\cup C$.

Secondly, there may be two options:

$x \in C$, and therefore $x \in (A\cap B)\cup C$.
$x \notin C$. Then $x \in A$ and $x \in B$. So $x \in A\cap B$, and therefore $x \in (A\cap B)\cup C$.

Finally, $x \in (A\cap B)\cup C$ in both options, and $x \in (A\cup C)\cap (B\cup C)$ is given. Therefore, $(A\cap B)\cup C = (A\cup C)\cap(B\cup C)$.


What is the probability of exactly two out of n persons sharing a birthday? The classical Birthday problem asks for the probability of at least two out of $n$ people sharing the same birthday or sometimes for the least amount $n$ of people required such that with a probability of 50% at least two of them share their birthday. But what about the question how probable it is that also no more than two people share their birthday?

$\mathbb{P}$(no more than two people share their birthday) = $\mathbb{P}$(noone shares their birthday with anyone else) + $\mathbb{P}$(noone shares their birthday with anyone else except one pair)

Now,

$\mathbb{P}$(noone shares their birthday with anyone else) = $\frac{365!}{(365 -n)!}*\frac{1}{365^n}$
$\mathbb{P}$(noone shares their birthday with anyone else except one pair) = $\frac{364!}{(364 - \overline{n-2})!}*\left({n \choose 2} 365\right)*\frac{1}{365^n}$.

n people and only two share a birthday.
Approach: Consider that $n-1$ don't share birthday with anyone. (everyone knows this) $(1)$
The last person will have to share birthday with anyone of the $n-1$ people.
The same has to be done for the initial $n-1$ people who didn't share birthdays.
This will give a sum of $ (n-1) + (n-2) + (n-3) + \dots + \big(n-(n-1)\big) = \frac { n (n-1) } 2 $ $(2)$.
Answer $ = (1) * (2) $


Multivariate Taylor Expansion I am in confidence with Taylor expansion of function $f\colon R \to R$, but I when my professor started to use higher order derivatives and multivariate Taylor expansion of $f\colon R^n \to R$ and $f\colon R^n \to R^m$ I felt lost.

Can somean explain to me from scratch multivariate Taylor?

In particular I don't understand the notation
$$
f(x+h) = \sum_{k=0}^p \frac{1}{k!} f^{(k)}(x)[h,...,h] + O(h^{p+1})
$$
Why we need the k-linear form $ \frac{1}{k!} f^{(k)}(x)[h,...,h]$? This k-linear form is the derivative or the derivative is only $f^{(k)}(x)$?

I'm quite lost. Thank you.

One can think about Taylor's theorem in calculus as applying in the following cases:

Scalar-valued functions of a scalar variable, i.e. $f : \mathbb{R} \rightarrow \mathbb{R}$
Vector-valued functions of a scalar variable, i.e. $f : \mathbb{R} \rightarrow \mathbb{R}^n$
Scalar-valued functions of a vector variable, i.e. $f : \mathbb{R}^n \rightarrow \mathbb{R}$
Vector-valued functions of a vector variable, i.e. $f : \mathbb{R}^n \rightarrow \mathbb{R}^m$

All of these can be derived & proven based on nothing more than integration by parts (the last one needs to be developed in a banach space & the third one is more commonly reduced to the first one which is just a shorthand for re-proving it via integration by parts) if you set things up correctly (as is done in Lang's Undergraduate, Real & Functional Analysis books) & so your main obstacle here is formalism - this is no small obstacle as we'll see below.

Now I'm not sure if your expression for Taylor's formula is map 3 or map 4, one would think it is map 3 since you used the word "linear form" which is standard parlance for maps from vector spaces into a field but you did ask about maps of the form $f : \mathbb{R}^n \rightarrow \mathbb{R}^m$ - are you sure you are differentiating these kinds of maps because they add a whole world of complexity compared to the first 3?

If you're asking about maps of the form in 3 then some intuition is given in this video & some examples & a proof are given in this video. After these you should have enough of a grasp of what's going on & if you focus on developing the formalism properly you should be able to prove it yourself in more general spaces.

If you're actually asking about map 4 then you may be used to the definition of the derivative of the last map as $f : \mathbb{R}^n \rightarrow \mathbb{R}^m$ as something like $\mathcal{f'} : \mathbb{R}^n \rightarrow \mathcal{L}(\mathbb{R}^n,\mathbb{R}^m)$ satisfying all the conditions a differentiable map does, well it's second derivative is defined similarly using a map of the form $f'': \mathbb{R}^n \rightarrow \mathcal{L}(\mathbb{R}^n,\mathcal{L}(\mathbb{R}^n,\mathbb{R}^m))$ & so on, however you see no such entity as a "k-linear form" in any of this & that's because there is a theorem which allows one to think of maps like the second derivative above in terms of multilinear maps & so one can re-cast the theory using multilinear maps which eases the development & allows for nice proofs etc... but without this being explained it might appear odd to randomly start pulling out multilinear maps.

In any case the derivative is a linear map by definition & so that is why you're coming across the word linear, but since you didn't put subscripts on the $[h,...,h]$ I'm not sure how deep I can go, because I see two possibilities here so if the above isn't enough of an explanation as it stands just let me know.

Let's suppose the function we're to investigate is a C(k+1) function within its open domain which includes a as an internal point. The purpose is to somehow find "the best" (polynomial) approximation to the function, at points like x, inside an open sphere around our reference point a, about which we know the value of the function there as well as all the (mixed) partial derivatives up to the kth degree. Now the first (and interstingly the most intuition-based) step to take is to somehow map our orthonormal bases to another orthonormal bases (using orthogonal transformations), such that one base will lie along the vector x-a. Now, with this change of coordinates, it's as though we're changing only one variable to go from a to x. Therefore, the problem turns into that of single variable Taylor expansion of the kth order. The other cool part of the attempt to "construct" the multivariable polynomial with only that for the single-variable at hand is that one will be able to and therefore should find all (mixed) partial derivatives of f at a, "this time in the direction of the vector x-a": (it becomes interesting when one thinks about how what they already have (here: all the (mixed) partials) enables them to obtain what they need to have (here: directional first, second,..., kth derivative of f at a with respect to x-a, if I'm rigth). It may also be illuminating to think about the relationship of the kth derivative in the direction of x-a and that of all the mixed partials at a.) Once the job is completed in thar coordinate system, one, quite clearly, needs to describe all they have obtained until there, in the language of the former coordinate system, thereby having constructed the multivariable polynomial approximation.


How should i prove every finite lattice is bounded? I want to know the outline to how to prove every finite lattice is bounded?

You can get a lower bound on two lattice points by taking their meet (resp. an upper bound by taking their join).  If a lattice has only finitely many points, this "two at a time" approach can be repeated (a finite number of times) to get a common lower bound (resp. common upper bound) on all the points.

Let the lattice $L$ has $n$ elements and the elements of the lattice is $a_1, a_2, \dotso ,a_n$

Now, $$a_1\lor a_2\lor\dotso\lor a_n= (\dotso((a_1\lor a_2)\lor a_3)\dotso a_n-1)\lor a_n)$$ [since $\lor$ is associative].

Since between $L$ is a lattice, between any pair of elements $a_i$ and $a_j$, $(a_i\lor a_j)$ exists,
So, $$(...((a_1\lor a_2)\lor a_3)\dotso a_n-1)\lor a_n)$$ does exist as well.

Similarly we can show if $L$ is a lattice with finite elements then a minimum element does exists.


Find the general values and principal values of $i^{log(i+1)}$? Find the general values of $i^{\log(i+1)}$?

I tried this way...
$\log(i+1)\log(i)=[\log \sqrt2+i(2n\pi+\frac{\pi}{4})]*i(2n\pi+\frac{\pi}{2})$


Laplace transform of $( \sin t)/t$ I have a question about this example from Fundamentals of Differential Equations by Nagle:

Why can we interchange the infinite sum and the Laplace transform $\mathcal{L}$? Specifically, why is it true that
$$\int_0^\infty e^{-st}\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!}t^{2k}dt=
\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!}\int_0^\infty e^{-st}t^{2k}dt$$
I was thinking we must look at the infinite series
$$\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!}e^{-st}t^{2k}$$
(with fixed $s>0$ and as a function of $t$). I was thinking either to a) show the partial sums of the above converge uniformly to swap the integral and infinite sum, but I think uniform convergence is false because the maximal value of each term of the sum occurs at $t=2k$, and we have a $(2k)^{2k}$ term that overpowers everything else.
or b) try to use Dominated convergence, but I need an integrable function that is an upper bound of each partial sum, which I'm also having trouble finding.
Thanks so much in advance for any help.

The interchange of sum and integral is permissible if $\sum \int |\frac {e^{-st}t^{2k}} {(2k+1)!}|\, dt <\infty $ [This is by Fubini's Theorem]. For $t <1$, $\sum \frac {e^{-st}t^{2k}} {(2k+1)!} \leq \sum \frac {e^{-st}} {(2k+1)!} < e^{1-st}$. For $t >1$, $\sum \frac {e^{-st}t^{2k}} {(2k+1)!} \leq \frac 1 t \sum \frac {t^{2k+1}e^{-st}} {(2k+1)!} < e^{-(s-1)t}$. Hence $e^{-(s-1)t}+e^{1-st}$ is an upper bound for the sum for all $t$. Of course $\int_0^{\infty } (e^{-(s-1)t}+e^{1-st}) <\infty $  for $s >1$.


A zero along the imaginary axis of the s-plane If a system has a zero somewhere along the imaginary axis of the s-plane (i.e. it's not on the right half of the plane), does this still make the system nonminimum phase ? Or does the zero have to be strictly on the right half side ?

Also does having a zero on the imaginary axis (especially at the origin) has a significant physical meaning ?

The definition as stated on the Wikipedia page of Minimum phase reads as follows:

  In control theory and signal processing, a linear, time-invariant system is said to be minimum-phase if the system and its inverse are causal and stable.

A causal system means that the system doesn't rely on future inputs for the current state of the system (more zeros than poles to put it really easily for now). And (BIBO) stable meaning that the states are bounded or $\forall t:|x(t)|<\infty$. Note that this doesn't require that the states to converge to zero (which should be the case if you're requiring a-symptotic stability).

Having a system with for example the transfer function $H(s)=\frac{s^2+10}{s^2+1}$ satisfies both these criteria, as well as that the inverse satisfies these criteria, meaning a minimum phase system, but it's easy to see that in both cases the poles of $H(s)$ and $H_{inv}(s)$ have poles on the imaginary axis.

(in case of $H(s)$, $p_{1,2}=+/- i$, and in case of $H_{inv}(s)$, $p_{1,2}=+/- i\sqrt{10}$.)

The physical meaning of this is that the system oscilates around a specific point with a certain frequency.

Always try to think of solutions of systems in exponential powers (since systems are written as differential equations, and the solutions to differential equations normally are in the form of exponential powers, in state-space description $\dot{x}(t)=Ax$ has as a solution $x(t) = e^{At}$). This is also why you want your poles to be in the left half plane, such that you can have solutions of the form $x(t) = e^{p t}$, where if $p<0$ the exponent converges to zero, and if $p$ is complex/purely imaginary you can use Euler's formula to rewrite the solution as sines and cosines.

Which is the case if you have purely imaginary poles you get solutions of the form $e^{ict} = c (\cos(t)+i \sin(t)$.


ratio test on $\sum_{n=1}^{\infty} \frac{n}{5^n}$ $\sum_{n=1}^{\infty} \frac{n}{5^n}$

doing the ratio test i get

$ \frac{n}{5^n} * \frac{5^{n+1}}{n+1} $

which results in

$ L = 5 \frac{n}{n+1} $

as limit reaches infinity on $L$. i get that $L$ is greater than 1 which is undefined behavior. but wolfram confirms it is 5/16. help

You divided thew wrong way around:
$$L=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim_{n\to\infty}\frac{n+1}{5^{n+1}}\frac{5^n}{n}=\lim_{n\to\infty}\frac15\frac{n+1}{n}=\frac15<1$$
So, by the ratio test, the series converges.

Using Cauchy root test, gives us the following:

$$\lim_{n\to\infty}\sqrt[n]{\frac{n}{5^n}}=\frac{1}{5}\lim_{n\to\infty}\sqrt[n]{n}=\frac{1}{5}<1$$

Thus, the given series is converges.

Edit: Yes, I know that the OP asked the ratio test.


Anyway to solve this problem without calculation? The problem is

  Airlines find that each passenger who reserves a seat fails to turn up with probability 0.1 independently from other passengers. So Teeny Weeny Airlines always sell 10 tickets for their 9 seat aero plane while Blockbuster Airways always sell 20 tickets for their 18 seat airplane. Which one is more often overbooked?

The brutal way to solve this is trivial.The probability for Teeny Weeny Airlines to be overbooked is $p_1=0.9^{10}≈0.3487$, and the probability for Blockbuster Airways to be overbooked is $p_2=0.9^{20}+C_{20}^{19}×0.9^{19}×0.1≈0.3917$. So Blockbuster Airways should be more often overbooked.

My question is -- is there any more graceful solution without really calculating the probability? Thank you!

1. Qualitative approach: let the number of passengers not showing up for airline $i$ ($=1,2$) be $$N_i\sim Binom(i\cdot n,\frac 1 {n})$$ with $n=10$ in our case. Note that $N_i$ is centered around $i$ - its mean, mode and median are equal to $i$. When you double the number of passengers, you squish and spread out distribution of $N_1$ into that of $N_2$ - hence the probability of getting the mean/median number of missing passengers exactly goes down (i.e. $P(N_1=1)>P(N_2=2)$). This probability must get distributed both to the right and to the left of this mean/median value - hence $$p_1=P(N_1<1)<P(N_2<2)=p_2$$

2. Poisson approximation to the binomial: $N_i\approx Pois(i)$:
$$p_1=P(N_1=0)\approx e^{-1}$$
$$p_2= P(N_2=0,1)\approx e^{-2}(1+\frac {2^1} {1!})>e^{-1}$$

3. Exact analysis:

$$p_1=(1-\frac 1 n)^n=e^{n\log(1-\frac 1 n)}>e^{-1-\frac 1 {2(n-1)}}>e^{-1}(1-\frac 1 {2(n-1)})$$
$$p_2=p_1p_1+2np_1p_1\frac {1/n}{1-1/n}=p_1p_1(1+\frac 2 {1-1/n})>p_1e^{-1}(1-\frac 1{2(n-1)})(3+\frac 2 {n-1})=p_1e^{-1}(3+\frac 1 {2(n-1)}-\frac 1 {(n-1)^2})\ge\frac 3 ep_1$$
 for all $n\geq 3$.

Check $n=2$ separately to get $$\frac {p_2} {p_1}>\frac 3 e > 1.1$$ for all $n\ge 2$.

For the same amount of tickets sold, odds are in favor of WA.

Consider two successive TWA flights vs. a single WA flight. Given the possible numbers of passengers turning up, the numbers of overbooked flights per $20$ tickets are

$$\begin{align}
10, 10\to 2, 1\\
10, 9\to 2, 1\\
9, 10\to 2, 1\\
10, <9\to 1, 0\\
<9, 10 \to 1, 0\\
<9, <9\to 0, 0\end{align}$$
never in defavor of WA.


Does $\int_0^\infty\frac{x^\theta}{1+x^2}\ \mathrm dx$ converge? I'm wondering
$$\int_0^\infty\frac{x^\theta}{1+x^2}dx$$
equals infinity or just a finite number? Surely when $\theta \geq 1$, it's infinity. But I cannot figure out when less than 1, what will happen.

I think I partially get the answer. Substitute the x with $\tan(t)$, then we get
$$\int_0^{\frac{\pi}{2}}\tan^\theta(t)dt$$, it's infinite for $0<\theta<1$

...actually $$\int_0^{\frac{\pi}{2}}\tan^\theta(t)dt$$ is not infinite. See Rran Gordon's comments.

Just for laughs, let's evaluate this one when $\theta \in (-1,1)$.  The integral may be evaluated by using the residue theorem on the contour integral

$$\oint_C dz \frac{z^{\theta}}{1+z^2}$$

wheere $C$ is a keyhole contour about the positive real axis.  In this case, we have

$$\left ( 1-e^{i 2 \pi \theta}\right ) \int_0^{\infty} dx \frac{x^{\theta}}{1+x^2} = i 2 \pi \left (\frac{e^{i \pi \theta/2}}{2 i} - \frac{e^{i 3 \pi \theta/2}}{2 i}  \right )$$

or

$$\int_0^{\infty} dx \frac{x^{\theta}}{1+x^2} = \frac{\pi}{2 \cos{(\pi \theta/2)}}$$

Note that the integral blows up when $|\theta|=1$.

I am sorry that I am using all the concept through words,I actually I don't know how to write integral and other symbols
anyway coming to question for
Theta<(-1)
 use limit comparisons test with g(x)=gamma(theta+1), where gamma is  the function in gamma function,since limit is infinite implies if intrural(g(x)) diverges then given integral diverges,applying condition for gamma function to divergent,from that we get theta<=(-1)
For proving at -1


Converting a Vector Equation into a Scalar Equation where the Direction Vector is $(1,0)$ I've been working on some textbook problems, where I've had to convert vector equations into scalar equations. I've understood it for the most part, by converting the vector equation into parametric form, and then isolating for the parameter, and then equating them so i have an equation just in terms of x and y.

But, i was stuck on this question,

$[x, y]=[3, -8] + t[1,0]$

As usual, i converted it to parametric form giving me that

$x=3+t$ and $y=-8$

But since $y=-8$ does't have a parameter t, i'm stuck. Thank you!

Hint: what is the equation of a line that is parallel to the x-axis?


UFD and its prime element If $R$ is a UFD and  $a,b \in R$ are relatively prime and $a|bc$, then $a|c$.

we know if a and b relatively prime (a,b)=1 so $a\bot b$.
since $a|bc$ and $a\bot b$ doesn't $a$ has to divide $c$? I dont understand why do we need UFD?

Let $A$ be the ring of polynomials in $4$ variables $x,y,z,w$ with integer coefficients, and let $R$ be the quotient of $A$ by the ideal generated by $xy-zw$. $R$ is an integral domain, but not a UFD. $x$ and $z$ are relatively prime in $R$, and $x$ divides $zw$ in $R$ (because $zw$ is $xy$ in $R$), but $x$ doesn't divide $w$.

A similar example can be produced in any non-UFD. If you have two factorizations into distinct irreducibles, $r=p_1p_2\cdots p_m=q_1q_2\cdots q_n$, you get $p_1$ relatively prime to $q_1$, $p_1$ divides $q_1(q_2\cdots q_n)$, $p_1$ doesn't divide $q_2\cdots q_n$.

Because if you had different factorizations or it weren't a domain, then it wouldn't be true.  Try $\mathbb{Z}_6$.  Then $5 |0 = 3*2$, but it doesn't divide 3 or 2.


How can you compare equality of two vectors of infinite dimension? E.g., say we have 2 vectors of infinite dimension: $$A=<1,0,1,1,...,1>$$$$B=<0,1,1,1,...,1>$$
These are qualitatively not the same vector, though it seems hard to definitively prove that.
If you define equality as $A_i=B_i$ for all elements $i$, then it seems to follow that $A\neq B$.
However, if you define equality between vectors as $\Vert A \Vert =\Vert B \Vert$ and $A$ being co-linear and co-directed with $B$, then these vectors might seem equal, as $\Vert A \Vert = \Vert B \Vert = ∞$, and the angle between the two vectors converges upon $0$ at $∞$ (although by the formula $cos \theta=\frac{<A, B>}{\Vert A \Vert \Vert B \Vert}$, this might technically be indeterminate, I suppose).
Is there a better way to understand and rationalize the relationship between infinite dimension vectors?

Let's use a more concrete example to demonstrate this point, which you are probably already quite familiar with.
Consider all the continuous functions defined on the interval $[0,1]$. This is usually notated as $C[0,1]$. Firstly, is this a vector space? The answer is yes
What is the dimension of this space? It is infinite - as you need to define the value of the function at each point in $[0,1]$ - of which there are infinitely many
Now, when are two vectors (functions) from this space equal? When their pointwise match is equal - or in more mathematical terms
$$f = g \space \text{iff} \space \space ||f-g|| = 0$$
where || is a norm function defined on the vector space


Expected number of frog jumps There is frog jumping forward on a line. Each jump distance is random with a known cumulative distribution function $F$. What is the expected number of jumps to reach (or go beyond) distance $d$ from the origin?

If there is a Wikipedia or other web page describing the formula in terms of $F$, just give a (deep) link. I'd prefer to have the proof as well.

Let $X_i$ denote the distance of jump $i$ of the frog. Each $X_i$ is independent of one-another. The total distance at jump $n$ is then $D_n=X_1+\cdots+X_n$. Define $\tau$ for $D_\tau$, where $\tau$ is the first time such that $D_\tau\geq d$. You are interested in $E(\tau)$. The distribution of $D_n$ is the $n$-fold convolution of $F$'s: $F_n = (F\star)^n:=F\star F\star\cdots \star F$.

$P(\tau> n) = P(D_n<d)$

Now use $E(\tau)=\sum_{n=0}^\infty P(\tau> n)=\sum_{n=0}^\infty P(D_n<d)$

Which we can simplify when $F$ is continuous by noting $P(D_n<d)=P(D_n\leq d)=F_n(d)$. I'm not sure if there's any more simplification you can do without an explicit form for $F$.

The distribution function defines the probability that the random variable is less than or equal to a given x: $P(n\leq x) = F(x)$. In this case, x is the distance of a single jump.

What we need is the expected value V of the distribution function. To get this, we want the derivative of the distribution function, $f(x) = F'(x)$, which produces a function known as the "probability density function" of x. This function will tell us how likely any given jump is to be x units.

Now, given that, the "expected value" of the random variable is equal to:

This "expected value", much like the term states, is the long-term average value of X given a sufficiently large number of trials. For a normal distribution, it is usually the point at which $f(x)$ is at its peak; however many probability functions are not normal (for instance, the probability of an n-sided die rolled once showing a particular face x is 1/n regardless of x). Now, that equation is the general form. We know that the frog always jumps forward, and thus the jump distance is never negative; therefore, we should actually solve the integral for a lower bound of 0:

$$E[x] = \int_0^\infty xf(x)dx$$

$E[x]$ will be the length of the average jump; it has a roughly 50-50 shot to be higher or lower, and thus given enough jumps, the sum of any n jumps should be $E[x] * n$. Now it's arithmetic; solve $E[x] * n >= d$ for n.


Identifyng objects in a category Is there a general way of identifying objects in a category to produce a new category? Something like a quotient by a relation on objects. How would the morphisms behave?

It's not hard to compute a few examples of such quotients. Define $C/\sim$ to be the universal category with a functor from $C$ which is well defined on $~$-equivalence classes. This implies a colimit condition on the morphisms. For instance, if $C$ has two distinct objects, each with endomorphisms $\mathbb{Z}$ and with no morphisms between them, then the universal category identifying the two objects of $C$ is the category with one object and endomorphisms $F_2$, the free group on two letters. If instead $C$ had just one morphism between its two objects, and no endomorphisms but the identity, then identifying the two objects would produce $\mathbf{N}$. To get morphisms between distinct objects $[A],[B]$ of $C/\sim$, you'd take all morphisms $A'\sim B'$ with $A\sim A',B\sim B'$ from $C$ with free composition law, modulo the relations coming from $C$. Obviously this is sketchy, but I don't see any impediment to the construction-nor any hope that you'd be able to compute anything of interest in a significant number of cases.

Your question does not make too much sense, categorically speaking. That's because categories are basically collections of morphisms obeying certain composition rules (see for instance the definition given here, or "arrow only categories" in MacLane'CWM). Objects are an "afterthought" and correspond to identity morphisms. So , in this light, your question is: can we identify/merge identity morphisms? But different identities are just morphisms which have neither tip nor tail in common,  so it is a bit "odd" trying to identify/merge them. What about other morphisms which have neither tip nor tail in common, for example?

What is interesting instead, and is done with congruences - as explained by Hayden in his answer - is to identify/merge morphisms which have both tip and tail in common (ie. they belong to the same Hom set).

Generally speaking, when trying to devise/understand a new construction in category theory, first try to see if it makes sense with morphisms and composition of morphisms, then it will automatically make sense with objects, since objects are nothing but a special kind of morphisms (identity morphisms).


Dealing with log functions I have a function where all variables are linear except for the log function. In one equation I have $\log(x)$ and another equation I have $\log(1-x)$.

How can I linearize $\log(x)$ and $\log(1-x)$?

In general, a function $f:\mathbb{R} \to \mathbb{R}$ can be linearised about a point $x_0$ by taking the first-order Taylor expansion. That is, we define
$$g(x) = f(x_0) + f'(x_0) (x-x_0),$$
and use this function instead.

The linearisation always depends on the linearisation point. I will give the general form where $x_0$ is the linearisation point. For $f_1(x) = \log(x)$,
$$g_1(x) = \log(x_0) + \frac{x-x_0}{x_0},$$
and for $f_2(x) = \log(1-x)$,
$$g_2(x) = \log(1-x_0) + \frac{x-x_0}{x_0-1}.$$


Is $x/x$ equal to $1$ My question is whether $x/x$ is always equal to $1$. I am mostly intersted in real numbers and particularly wonder whether $x/x$ is defined at $x=0$.

On one hand the division should simplify to $1$, on the other hand you should not be allowed to divide by zero.

I have been trying to find whether the simplification 'goes first' or whether the division causes trouble first, but it has proven impossible for me to find usefull search terms.

Note that this question arose after reading this answer and its first comment.

The function $f(x) = \frac{x}{x}$ is defined for all $x \in \mathbb{R}\setminus\{0\}$. Its limits exist from the left and from the right at $0$, but it is not defined at $0$. It doesn't matter if you "simplify" first and then "check" or the other way around, because as you pointed out you're not allowed to divide by $0$. Thus
$$f(x) = \begin{cases} 1 & x \neq 0 \\ \text{undefined} & x = 0\end{cases}.$$
Consider the same question,
$$g(x) = \frac{x^2 - 2x + 1}{x-1}.$$
What is this function?

As linked to in the post you link to, $\frac{0}{0}$ is an indeterminate form - that is, there is no answer at all. It's particularly curious because it's an example of both $\frac{x}{x}$ and $\frac{y}{x}$, the former of which equals $1$ when $x\neq0$ and the latter of which has a limit that tends to either positive or negative infinity when $y\neq0$ as $x$ tends to 0.


A Functional Differential Equation: $f^\prime(x) =\frac{f(2x)}{2f(x)}$ I was having a play with some trig. identities and noticed the following:

$$\cos{x}=\frac{\sin{2x}}{2\sin{x}}.\tag{1}$$

Now, $\cos{x} = \frac{d}{dx}\sin{x}$ so I made the following analogous  differential equation:

  $$f^\prime(x) =\frac{f(2x)}{2f(x)} \tag{2}$$

I have not seen a differential equation which relates a function's derivative to a change in its argument, so I was wondering whether anyone knew what these were called?

Somewhat predictably, $f_1(x)=\sin{x}$ is not the only solution, I found that $f_2(x)=A\sin(\omega x)$ where $a\omega=1$ is also a solution. I then guessed another solution, $e^{\lambda x}$, and found that $f_3(x)=e^{\frac{1}{2}x}$ is also a solution.

My main questions are:

  What are these type of equations called? and are there any other solutions to this one?

Thanks for reading.

CLAIM: The only solutions $f(x)$ to your FDE that are analytic at $x=0$ with $f(0)\ne 0$ are in the form $f(x)=ae^{x/2a}$ with $a\in\mathbb R$.

It is easy to demonstrate that functions of the described form satisfy your FDE, so all that remains is to establish that they are the only solutions.

Let's rewrite your functional equation as
$$2f(x)f'(x)=f(2x)$$
...ignoring the possible issues that this may cause when $f(x)=0$. Now let's assume that $f$ is analytic and that it has the Maclaurin series
$$f(x)=\sum_{n=0}^\infty \frac{x^n a_n}{n!}$$
for some coefficients $a_n$. Then this functional equation is the same as
$$2\bigg(\sum_{m=0}^\infty \frac{x^m a_m}{m!}\bigg)\bigg(\sum_{n=0}^\infty \frac{x^n a_{n+1}}{n!}\bigg)=\sum_{k=0}^\infty \frac{x^k 2^k a_{k}}{k!}$$
If we multiply together the series on the left, we get
$$\begin{align}
2\bigg(\sum_{m=0}^\infty \frac{x^m a_m}{m!}\bigg)\bigg(\sum_{n=0}^\infty \frac{x^n a_{n+1}}{n!}\bigg)
&=\sum_{m,n=0}^\infty \frac{x^{m+n}\cdot 2 a_m a_{n+1}}{m!n!}\\
&=\sum_{k=0}^\infty x^k\sum_{m+n=k}\frac{2a_m a_{n+1}}{m!n!}\\
&=\sum_{k=0}^\infty x^k\sum_{m=0}^k \frac{2a_m a_{k-m+1}}{m!(k-m)!}\\
\end{align}$$
So our initial equality becomes
$$\sum_{k=0}^\infty x^k\sum_{m=0}^k \frac{2a_m a_{k-m+1}}{m!(k-m)!}=\sum_{k=0}^\infty x^k \frac{2^k a_{k}}{k!}$$
Now, by equating coefficients, we have that
$$\sum_{m=0}^k \frac{2a_m a_{k-m+1}}{m!(k-m)!}=\frac{2^k a_{k}}{k!}$$
...for all $k\in \mathbb Z$. This can be interpreted as a recurrence relation for $a_n$, and it demonstrates that given $a_0=f(0)$, all other coefficients, and thus the function $f$, are determined.

Thus, since for any $a_0$, there exists $a\in\mathbb R$ such that $f(x)=ae^{x/2a}$ satisfies $f(0)=a_0$, we have that the value of $f(0)$ determines the function $f(x)$. More specifically,
$$f(x)=f(0)e^{x/2f(0)}$$

NOTE. Because I began this strategy by multiplying both sides of your FDE by $f(x)$, this recurrence messes up if $f(0)=0$. Indeed, if this is the case, then the recurrence shows that $a_n=0$ for all $n$, and $f(x)=0$, even though you have found a function $f_2(x)=\sin(x\omega)/\omega$ that also satisfies $f_2(0)=0$ and is not constant. However, if $f(0)\ne 0$, the recurrence should work out just fine.

To use "Picard's method" might work. You have
$$
y(x)=y_0+\int_{x_0}^x \frac{y(2t)}{y(t)} \, dt
$$

Define
$$
y_{k+1}=y_0+\int_{x_0}^x \frac{y_k(2t)}{y_k(t)}\,dt
$$
and calculate $\lim_{k\rightarrow \infty} y_k$.
I do not  know if the sequence converges.


Solution to $\frac{d^2x}{dt^2}+\frac{kx}{m}=0$ This is the system of differential equations which describes simple harmonic motion:
$$\frac{d^2x}{dt^2}+\frac{kx}{m}=0$$
$$x(0)=A$$
$$\frac{dx}{dt}\bigg|_{t=0}=0$$

By guessing that $x$ is of the form $A\cos\left(\omega t+\phi\right)$, I can arrive at the solution following solution: $$x=A\cos\left(\sqrt{\frac{k}{m}}\text{ }t\right)$$

My question is, how can you derive this solution without assuming that $x$ is a sinusoidal function? I have not yet taken a formal calculus course, so a simple solution would be appreciated.

As noted in the problem statement, the ODE $m\ddot{x}+k x=0$ describes simple harmonic motion. More precisely, it's the equation of motion of such a system (i.e. Newton's second law).

What's more crucial for the present purpose, however, is not Newton's laws but rather energy conservation. To reveal this, suppose we multiply both sides of the equation of motion by $\dot{x}$ to get
$$0=m\dot{x}\ddot{x}+k x\dot{x}=mv\frac{dv}{dt}+kx\frac{dx}{dt}=\frac{d}{dt}\left(\frac12 m v^2+\frac12 k x^2\right)$$
from which we conclude that the energy $E=\frac12 mv^2+\frac12 k x^2$ is conserved. For the initial conditions in the problem, we specifically have

$$\frac12 m\left(\frac{dx}{dt}\right)^2+\frac12 k x^2=\frac12 k A^2.$$

Why is energy conservation useful? Because this ODE is separable:

$$t=\int_{A}^{x(t)}\frac{dx}{dx/dt}=-\sqrt{\frac{m}{k}} \int_{A}^{x(t)}\frac{dx}{\sqrt{A^2-x^2}}.$$

(The minus sign is because, if the object starts from rest with $x(0)=A>0$, it will begin moving with a negative velocity. Other such sign issues can be clarified if one is careful.) Either by trig substitution or by using a table of integrals, one obtains
$$t=\sqrt{\frac m k}\cos^{-1} \frac{x(t)}{A}\implies x(t)=A\cos(t\sqrt{k/m}).$$ So it is indeed a sinusoidal motion.

Solution(Partial): $$\frac{d^2x}{dt^2}=-\frac{kx}{m}$$ $$\implies 2\frac{dx}{dt}\frac{d^2x}{dt^2}=-2\frac{dx}{dt}\frac{kx}{m}$$
On integrating both sides:
$$\implies (\frac{dx}{dt})^2=-\frac{2kx^2}{m}+C$$ Using the initial conditions we get: $$C=\frac{2kA^2}{m}$$
$$\implies \frac{dx}{dt}=\sqrt\frac{2k}{m}\sqrt{(A^2-x^2)}$$ I think you can take it from here.


Functions and Infinitesimals? My textbook says the following:

  $ads = v dv$

where $a$, $s$, $v$ are functions.

I was introduced to integral calculus a semester ago, and Riemann Sums (and integrals) are the only notion of multiplying functions and infinitesimals I'm familiar with.

In the above excerpt, $a$ is not necessarily a function of $s$ yet, you are multiplying it with an infinitesimal of $s$. Could someone give me an intuitive explanation of what it means when you multiply a function with with an infinitesimal of a variable that the function is not of?

IMO, thinking of differentials as infinitesimals is the wrong way to go: if you think of $s$ as a variable quantity, then you should be thinking of $ds$ as a gadget that represents how $s$ varies in different directions. As such, it makes perfect sense to multiply it by other variables, or to add two differential forms.

(don't try to multiply two differential forms: the meaning of that is tricky business, having to do with how things vary in a two-dimensional direction, and has odd properties like $dx dy = -dy dx$. It's not so bad once you're used to it, though; you'll probably be shown the basic idea when you encounter surface integrals)

"In what directions?" you might ask. "Any direction!" is the answer! The really neat thing about doing algebra with differential forms is that they don't really care about how many independent variables you have, or the shape of your parameter space or anything.

If you know your parameter space and have a direction to plug in -- e.g. in $3$-space with coordinates $x,y,z$, if you decide to look at motion in the direction of the vector $(1,2,0)$, you would plug in $dx = 1$, $dy=2$, and $dz=0$ to obtain an actual value for the "rate" at which various quantities are varying. Or if you decide you want to work on the unit sphere rather than all of $3$ space, you add the equation $x^2 + y^2 + z^2 = 1$ to relate your variables... and also the equation $2x dx + 2y dy + 2z dz = 0$ to relate your differentials.

Quick differential geometry aside! If we want to see how things vary in the direction $(1,2,0)$ and we stick on the unit sphere, we would get $2x \cdot 1 + 2y \cdot 2 + 2z \cdot 0 = 0$, or $2x + 4y = 0$. This tells us that the only points on the sphere where we can move in the direction $(1,2,0)$ -- i.e. the points to which $(1,2,0)$ is a tangent vector to the sphere, are the ones where $2x + 4y = 0$.

It is often simpler to use differential forms to relate rates of different quantities. e.g. if $a,v,s$ are all functions of some other variable $t$ -- e.g. $a = \alpha(t)$, $v = \nu(t)$, and $s = \sigma(t)$ -- then $da$, $dv$, and $ds$ are all multiples of $dt$. Specifically,

$$ da = \alpha'(t) dt \qquad dv = \nu'(t) dt \qquad ds = \sigma'(t) dt $$

and so we see how each of the variables $a,v,s$ vary as compared to how $t$ varies. We can even plug these into the original equation to get

$$ a \sigma'(t) dt = v \nu'(t) dt $$

and so everywhere in parameter space, we have either

$$ a \sigma'(t) = v \nu'(t) \qquad \text{or} \qquad dt = 0 $$

The first of these options can be written in the more suggestive form

$$ a \frac{ds}{dt} = v \frac{dv}{dt} $$

In fact, my first understanding of differential forms (long before I ever heard the term) was that things like $ds$ meant "the derivative with respect to some variable, but I haven't decided which, and I'll fill the same one in everywhere once I've decided".

Be careful that you don't try to divide differential forms; doing such a thing only has a chance to be meaningful if you already know one differential form is a multiple of another, which is usually the case when you only have one independent variable. That usually doesn't happen when you have multiple independent variables; e.g. $dx/dy$ is total nonsense in the $xy$-plane.

(the partial derivative $\partial x / \partial y$ makes sense, but I really don't like doing multivariable calculus that way: I usually prefer to stick to differentials, and if I want to see how things vary while holding $x$ constant, I'll simply add in the equation $dx=0$)

Take notice that I said things like "variable" and "variable quantity" rather than "function". IMO, when using this notation style, it's a bad idea to think of your variables as secretly being functions -- doing so gets in the way of being able to use this style of notation fluently, and doubly so when you need to switch back and forth between this style and the style of writing everything as a function.

Even if $a$ is functionally dependent on $s$, it's not a good idea to think of $a$ as being literally a function. Instead, the dependence is expressed by saying that there is a function $f$ such that $a = f(s)$ is true. (and correspondingly, $da = f'(s) ds$)

So good that you asked.  It shows you are thinking, which has been known to improve one's mathematics.

There is not actually such a thing as multiplying by infinitesmals.  When it looks that way, it is always a shortcut for the chain rule.  David H. demonstrates it for this case very nicely in his answer.

You can see that you have to look for the common independent variable, and then for the relationships between the derivatives of the various functions with respect to that variable.

Notational shortcuts such as this one can be very helpful in simplifying computations, or in checking your work.You'll see this shortcut a lot when you take differential equations.

Here's another example: in a physics computation you may have an equation like

$a ft^2 * b (sec)/ ft^2$ = ab (sec). (I wrote "second" here as "sec" rather than the usual s, just for clarity). We know we are not "multiplying" these units of measure, but the notation is very handy.  It allows us to keep the units straight; also if you wind up in units of seconds and you were trying to compute  work, you know you made a mistake.

The main thing is to understand what the shortcut is, so you understand what you are doing.  Notation is to serve you, so you can write things any way that makes them clear to you, but of course you don't want to misuse it because you misunderstand what it means.


Is there a holomorphic function such that $(f(z))^3=z-z^2$ Is there a holomorphic function $f:C-[0,1]$ such that $(f(z))^3=z-z^2$ for all $z\in C-[0,1]$

My intuition tells me that not really, for instance $$(z-z^2)^{\frac{1}{3}}$$

does not have a unique branch on this set, but I do not know how to formally prove it.

The function $z\to z-z^2$ has a double pole at infinity, which means that your $f$ would have a pole of order $2/3$ there!

The zeroes of $f$ must be the same as those of $z-z^2$, namely $0$ and $1$.

What is the winding number of $f(z)$ as $z$ describes a small circle around $0$? Since $z-z^2$ winds around $0$ once, you need an integer that gives $1$ when multiplied by $3$. But that is impossible.


Prove that $\sec^2{\theta}=(4xy)/(x+y)^2$ only when $x=y$ Show that the equation below is only possible when $x=y$
  $$ \sec^2{\theta}=\frac{4xy}{(x+y)^2}$$

The only way I can think of doing this is by rewriting it as

$$ \cos^2{\theta}=\frac{(x+y)^2}{4xy} $$
then using some inequalities to prove it by using:
$$ 0\leq \cos^2{\theta}\leq 1 \;\; \text{ therefore } \;\; 0\leq \frac{(x+y)^2}{4xy}\leq 1 $$
But I have an aversion to using case-based solutions (checking for $x>0$, $y>0$ etc.) since I feel there must be a neater solution to  these kind of problems. So my question is: Is it possible to solve this and these sort of questions using techniques that don't involve checking numerous cases?

We have $\sec^2 \theta\ge 1$ for all $\theta$ at which $\sec\theta$ is defined.  So it is enough to show that $\frac{4xy}{(x+y)^2}\le 1$, with equality only when $x=y$.

To show that $\frac{4xy}{(x+y)^2}\le 1$, we show equivalently that $(x+y)^2\ge 4xy$, or equivalently that $x^2-2xy+y^2\ge 0$. But this is clear, since $x^2-2xy+y^2=(x-y)^2$. And we have equality precisely when $x=y$.

Remark: This is not very different from how you proposed to do things. There are no cases involved. And aversion to cases can be problematic. A consideration of cases (though not in this case) is often a natural approach.


If $\lim_{x\to 1} \frac{f(x)-6}{x-1} = 10$, then what is $\lim_{x\to 1} f(x)$? I was given this question:

  If
  $$\lim_{x\to 1} \frac{f(x)-6}{x-1} = 10$$
  then what is
  $$\lim_{x\to 1} f(x)$$

I am assuming that I will need to use limit laws in reverse, but this doesn't seem to work as the limit of $x-1$ as $x$ approaches $1$ is $0$, which gives an undefined answer.

Let $\alpha: \Bbb{R} \setminus\{1\} \to \Bbb{R}$ be the function defined by
\begin{align}
\alpha(x) = \dfrac{f(x)-6}{x-1}
\end{align}
Then, by assumption, $\lim\limits_{x \to 1}\alpha(x) = 10$. Also, it's easy to see that $\lim \limits_{x \to 1}(x-1) = 0$. Hence, by the rules for products of limits, we know that $\lim \limits_{x \to 1} \big(\alpha(x) \cdot (x-1) \big)$ exists and:
\begin{align}
\lim \limits_{x \to 1} \bigg(\alpha(x) \cdot (x-1) \bigg) &= \bigg(\lim_{x \to 1} \alpha(x) \bigg) \cdot \bigg( \lim_{x \to 1}(x-1)\bigg) \\
&= (10) \cdot 0 \\
&= 0 \tag{$*$}
\end{align}

But, now, let's examine the LHS more carefully. If $x \neq 1$, what is $\alpha(x) \cdot (x-1)$? Well, let's go back to the definition of $\alpha$. It should be clear that if $x \neq 1$, then $\alpha(x) \cdot(x-1) = f(x)-6$. Hence, what $(*)$ shows is that
\begin{align}
\lim_{x\to 1} \bigg( f(x) - 6 \bigg) = 0
\end{align}
or equivalently,
\begin{align}
\lim_{x \to 1} f(x) = 6
\end{align}

The thought process I had when approaching this question is that we are interested in the limit of $f(x)$, and $f(x)$ appears in the numerator. Typically, I would use the rule
\begin{align}
\lim_{x \to 1} \bigg(\dfrac{\phi(x)}{\psi(x)} \bigg) = \dfrac{\lim\limits_{x \to 1} \phi(x)}{\lim\limits_{x \to 1} \psi(x)}
\end{align}
but in this case, we cannot directly apply this rule, because this rule is only valid when $\lim\limits_{x \to 1} \psi(x) \neq 0$ (division by zero is your worst enemy in math!). In this case, the denominator $x-1$ clearly has a limit of $0$ as $x \to 1$.

Therefore, to get around this obstacle, the most natural thing to do is simply "get rid of" the denominator by multiplying the denominator throughout, and as you can see, this was pretty much the idea behind my proof above.

Hopefully this helps you with not only the proof, but more importantly how to think about approaching such problems.

It's possible if $$\lim_{x\rightarrow1}(f(x)-6)=0.$$
Can you end it now?

A full solution:

We have:
 $$\lim_{x\rightarrow1}f(x)=6+\lim\limits_{x\rightarrow1}(f(x)-6)=6+\lim\limits_{x\rightarrow1}\left(\frac{f(x)-6}{x-1}\cdot(x-1)\right)=$$
$$=6+\lim\limits_{x\rightarrow1}\frac{f(x)-6}{x-1}\cdot\lim\limits_{x\rightarrow1}(x-1)=6+10\cdot0=6.$$