Is $(\neg p \lor (p \rightarrow q)) \rightarrow\neg p$ a Tautology? According to the truth table, following is is a Tautology or not?
$(\neg p \lor (p \rightarrow q)) \rightarrow \neg p$

The whole implication is false when $(p \vee (p \rightarrow q))$ is true and $p$ is false. In this case, $p \rightarrow q$ must be true. This happens if $q$ is true or false. So that is not a tautology.


Geometric Sequence Proof Problem:

(a) Determine all nonnegative integers $r$ such that it is possible for an infinite arithmetic sequence to contain exactly $r$ terms that are integers. Prove your answer.
(b) Determine all nonnegative integers $r$ such that it is possible for an infinite geometric sequence to contain exactly $r$ terms that are integers. Prove your answer.

My Solution for $a)$:

$a)$ First, we can analyze the case when $r \geq 2$. In this case, there will be at least $2$ terms in the arithmetic sequence which are integers; $a+md$ and $a+nd$. WLOG, $m>n$. These terms are both integers, so $(m-n)d$ is also an integer. Adding $(m - n)d$ to $a + md$ gives $a + (2m - n)d$, which is also a member of the arithmetic sequence and an integer, so this process can continue, giving the arithmetic sequence an infinite number of terms. Therefore, $r$ can only be $r=1$.

I am currently stuck on $b)$, but I know that we could do a similar approach, but how would I do it?

For part (b):

Take a positive integer $m$.

Construct a GP with first term $m^{r-1}$ and common ratio $\frac 1m$. The GP is

$$\overbrace{ m^{r-1}, m^{r-2},m^{r-3},\cdots\cdots,m^2, m^1, 1}^{r \text{  integer terms}}, \overbrace{\frac 1m, \frac 1{m^2},\cdots\cdots\cdots}^{\text{non-integer terms}}$$

Actually, for (b) any nonnegative integer $r$ works. If you want exactly $r$ terms to be integers, take $2^{r-1}, 2^{r-2}, 2^{r-3},..., 2^2, 2, 1, \frac{1}{2}, ...$
example: $r=3$
$\text{ratio}=\frac{1}{3}$
we get $2^{3-1}, 2^{3-2}, 2^{3-3}$, and those are all the integer terms.
I'm only an 8th grader so there may be a flaw in my thinking, if so please correct me!


positive definite matrix plus positive semi matrix equals positive definite? I have a questions related to the positive definite[PD] matrix and positive semi definite[PSD] matrix

I see and get the property about PD and PSD

1) PD + PD = PD
2) PSD+ PSD = PSD

how about the positive definite[PD] matrix plus positive semi definite matrix ?

(I mean sum of positive definite matrix and positive semi definite matrix : PD + PSD)

Is it right to be positive definite matrix?

For example, If matrix  B is $R \times R$ and it is  sum of identity matrix $I$
and symmetry matrix A

that is, $B=I+A$

1) $I=\det(I)=1>0 $ positive definite
2) $X^{T}AX=X^{T}L^{T}LX=U^{T}U=||U||\geqslant 0 $ positive semidefinite

I think that it would be positive definite, I am not so sure...

So I would like to get some help from you

Thank you very much in advance !

Let us consider $A$ is PD and $B$ is PSD. Then $x^T(A + B)x = x^TAx + x^TBx > 0$ for $0 \neq x \in R^n.$ Hence $A+B$ is PD matrix.

(1) PD+PD=PD is right. Let $A,B\in R^{p\times p}$ are PD. For any nonzero $x\in R^p$,
$$
x^\top(A+B)x=x^\top Ax+x^\top B x>0\text.
$$
(2) PSD+PSD=PSD is not right. One example is that $A=\operatorname{diag}(1,0)$ and $B=\operatorname{diag}(0,1)$. Both of them are PSD, but $A+B=I$ is identity matrix.
(3) PD+PSD=PD should be right. It can be shown as (1).


Need clarification on the way a set former is written I'm being asked to draw a DFA for the language specified by the following set former:

$$\{xabay\, |\, x \in \{a,b\}^*\text{ and }y \in \{a,b\}^*\}$$

Is this any different than the language specified by the set former:

$$\{xabax\, |\, x \in \{a,b\}^*\}$$

They seem equivalent to me, and I just want to make sure I'm not missing some crucial detail related to set former notation.

In the first set former, we can have $x=a$ and $y=b$ whereas in the second one you are restricted to the first and last (group of) letters being the same.

I believe an equivalent statement would be

$$\{xabay\,|\,x,y\in\{a,b\}^*\}$$

however don't quote me on that as my set notation is a bit sketchy and there's probably a small detail that I've forgotten about

Setting $A = \{a,b\}$, the first language can be represented by the regular expression $A^*abaA^*$. Thus it is the set of words having $aba$ as a factor and it is a regular language.

A word $u$ of the second language is also required to have $aba$ as a factor, but a stronger restriction is imposed: one should have $u = pabas$, where the prefix $p$ and the suffix $s$ are equal. This condition makes the second language not regular. Indeed, its intersection with the regular language $a^*abaa^*$ is equal to $\{a^nba^n \mid n > 0\}$ and I let you verify that this language is not regular.


Simple Nonlinear Differential Equation I have tried in vain to search for a closed form solution (involving known functions) to the simple first order equation $\dfrac{dy}{dx}=x^2+y^2, y(0) = 0$. Can anyone help with a solution that does not have the form of an infinite series?

I tried to solve this using maple and here's my code:

dy:=diff(y(x),x)=x^2+y(x)^2;

dsolve({dy,y(0)=0},y(x));

And it gives me a solution in terms of Bessel function, so I don't think there's a solution in terms of elementary functions.

Putting
$$ y = -\dfrac{d(\ln z)}{dx} = -\dfrac{z'}{z}$$
gives
$$y'=\left(\dfrac{z'}{z}\right)^2-\dfrac{z''}{z},$$
and so $$ y' = y^2 - \frac{z''}{z}$$

So in $y' = y^2 + x^2$ the $y^2$ cancels out leaving $-\dfrac{z''}{z} = x^2$, or $z''=-x^2z$, which according to Wolfram has the solutions
$z = c \cdot D[-1/2, u]$, where D is the parabolic cylinder function of fractional order $1/2$ and $u = (1+i)x$
 or (-1+i)x

The relation $y = -\dfrac{d(\ln z)}{dx}$ then leads to
$$(1-i)y = \frac{2D[1/2, u]}{D[-1/2, u]}-u,$$
where $u$ can be $(1+i)x$.  However, Wolfram says $2D[1/2, 1] - D[-1/2, 1] =  1.03133....$ and not $0$ as is required for $y(0)=0$ so there also seems to be a problem with this solution!


"Story" proof of $\sum_{i=j}^n {i \choose j} = {n+1 \choose j+1}$ I am reading a book "Discrete Mathematics for Computer Scientists". One of the exercises asks for a "story" proof of this: $\sum_{i=j}^n {i \choose j} = {n+1 \choose j+1}$.

My question is that:

What is a story proff?
What is the story proof of this identity?

Screenshot of the book:

If you read ${n+1} \choose{j+1}$ as: You have $n+1$ numbered boxes $\{1,...,n+1\}$ and you want to place $j+1$ indistinguishable balls in these boxes (at most one in a box), how many ways is that possible .
Then we look at different cases: To do this look at $x$ defined as the highest number for which the box is occupied. Note $x>j$ since all balls must be placed. This means that the other $j$ balls are distributed in the boxes $\{1,...,x-1\}$. This can be done in ${x-1} \choose {j}$ ways. Now we can sum $x$ over ${j+1,...,n+1}$. Therefor we get $\sum _{i=j}^n {i\choose j}={n+1 \choose j+1}$.

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 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
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 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
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 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,{\rm Li}_{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
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\begin{align}&\color{#66f}{\large\sum_{k\ =\ j}^{n}{k \choose j}}
=\sum_{k\ =\ j}^{n}\oint_{\verts{z}\ =\ 1}
{\pars{1 + z}^{k} \over z^{j + 1}}\,{\dd z \over 2\pi\ic}
=\oint_{\verts{z}\ =\ 1}
{1 \over z^{j + 1}}\sum_{k\ =\ j}^{n}\pars{1 + z}^{k}\,{\dd z \over 2\pi\ic}
\\[5mm]&=\oint_{\verts{z}\ =\ 1}{1 \over z^{j + 1}}\,\pars{1 + z}^{j}\,
{\pars{1 + z}^{n - j + 1} - 1 \over \pars{1 + z} - 1}\,{\dd z \over 2\pi\ic}
\\[5mm]&=\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{n + 1}\over z^{j + 2}}
\,{\dd z \over 2\pi\ic}
-\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{j}\over z^{j + 2}}
\,{\dd z \over 2\pi\ic}
={n + 1 \choose j + 1} -\
\underbrace{j \choose j + 1}_{\ds{=\ \color{#c00000}{0}}}\ = \
\color{#66f}{\large{n + 1 \choose j + 1}}
\end{align}


Connection between Fourier transform and Taylor series Both Fourier transform and Taylor series are means to represent functions in a different form.

What is the connection between these two? Is there a way to get from one to the other (and back again)? Is there an overall, connecting (geometric?) intuition?

Assume that the Taylor expansion $f(x)=\sum_{k=0}^\infty a_k x^k$ is convergent for some $|x|>1$. Then $f$ can be extended in a natural way into the complex domain by writing $f(z)=\sum_{k=0}^\infty a_k z^k$ with $z$ complex and $|z|≤1$. So we may look at $f$ on the unit circle $|z|=1$. Consider $f$ as a function of the polar angle $\phi$ there, i.e., look at the function $F(\phi):=f(e^{i\phi})$. This function $F$ is $2\pi$-periodic, and its Fourier expansion is nothing else but $F(\phi)=\sum_{k=0}^\infty a_k e^{ik\phi}$ where the $a_k$ are the Taylor coefficients of the "real" function $x\mapsto f(x)$ we started with.

I think that the missing link that connects the Fourier transform to the Taylor series expansion is Euler's formula, $e^{\jmath x}=\cos(x) +\jmath \sin(x)$. This celebrated formula establishes a relationship between trigonometric functions of real entities and exponential functions of complex (i.e. imaginary) entities. In doing so, it establishes a fundamental connection with the Fourier transform, which is essentially trigonometric in nature. In fact, the Fourier transform for a function $f(t)$ is, by definition, a Fourier series of $f(t)$ interpreted as $t\rightarrow \infty$, and a Fourier series is, by definition, a linear summation of sin and cos functions. The Taylor Series comes into play in the derivation of Euler's formula.

As previously mentioned, Euler's formula states $e^{\jmath x}=\cos(x) +\jmath \sin(x)$. Hence, it's acceptable to conceptually superimpose the conventional (x, y) unit circle and the real-complex plane, as they both portray the polar Eulerian expression on the continuous interval from $0$ to $2\pi$. We therefore, evaluating Euler's formula at $x=2\pi$, arrive at
$$e^{2\pi*\jmath \theta } = {\cos(}2\pi\theta)+j*{\sin}(2\pi\theta)$$
This expression is intrinsically intertwined with the nature of a Fourier transform because Fourier transforms aim to convert a function from the time domain to the frequency domain.  To do so, we decompose periodic functions into simple, linear sums of sin and cos, and allow this new function to approach infinity.

So, where does the Taylor series fit into all this? We use the Taylor series expansion (or more precisely, the McLauren series expansion) to obtain Euler's formula. Here's the proof (or derivative, depending on your perspective):

Proposition: For any complex number $z$, where $z=x+\jmath y=x+\sqrt{-1}*y$, i.e.  $x=Re \{z \}\ $ and $y=Im \{z \}\ $, it can be said that, for real values $x$,

$$ e^z=e^{j\pi} = \cos(x)+\jmath \sin(x)$$

Lemma 1: The Taylor series expansion  of an infinitely differentiable function $f(x)$ in the neighborhood $a$ is defined
$$f(x)=\sum_{k=0}^{\infty} \frac{f^k(a)}{k!}(x-a)^k$$
We'll say that we're working in the neighborhood $a$ surrounding $\theta$. So, when $\theta = 0$, we get a special case of the Taylor series, called the McLauren series. The expansion of the Taylor (or McLauren) series for $\sin(x)$ and $\cos(x)$ are
$$ \sin (\theta)=\sum_{k=0}^{\infty} \frac{d^{k} \sin(\theta)}{d\theta^k} \mid _{\theta=0} = \sum_{k=0}^{\infty}\frac{\theta^k}{k!} = \theta - \frac{\theta^3}{3!} +  \frac{\theta^5}{5!} - \frac{\theta^7}{7!} ...$$
$$ \cos (\theta)=\sum_{k=0}^{\infty} \frac{d^{k} \cos(\theta)}{d\theta^k} \mid _{\theta=0} = \sum_{k=0}^{\infty}\frac{\theta^k}{k!} = 1 - \frac{\theta^2}{2!} +  \frac{\theta^4}{4!} - \frac{\theta^6}{6!} ...$$
This is not a surprise, as we know that sine and cosine are odd and even functions, respectively. We've now accounted for the trigonometric portions of $e^z$, but have not yet addressed the complex component.

Lemma 2: We apply the previously defined Taylor (more specifically McLauren) series expansion to the function $e^z$ when $z=\jmath x$, and we get
$$e^{\jmath x} =\sum_{k=0}^{\infty} \frac{d^{k} e^{\jmath x}}{d\theta^k} \mid _{\theta=0} = \sum_{k=0}^{\infty}\frac{\theta^k}{k!} = 1 + \jmath x -
$$

Comparing the right-hand terms from the Taylor series expansions of $ \cos (\theta)$, $\sin (\theta)$, and $e^{\jmath \theta}$ performed in Lemmas 1 and 2, we see that summing the expansions from Lemma 1, we arrive, by the procedure I'm too lazy to type out in LaTeX but not too lazy to explain via my good friends Google and krotz, at
$e^{\jmath x}=\cos(x) +\jmath \sin(x)$. [Q.E.D.]

Now that we've proven Euler's formula, we can evaluate it at $x=2\pi$ to acquire

$$e^{2\pi*\jmath \theta } = {\cos(}2\pi\theta)+j*{\sin}(2\pi\theta)$$

Due to the characteristics of the sin and cos functions, it is possible to use simple integration to recover the amplitude of each sin and cos wave represented in a Fourier transform (similar to the reverse of the above proof). In an overwhelming majority of cases, it's highly useful to select Euler's formula as the function to integrate over. Because Euler's formula and the Fourier transform are both (at least, in part) fundamentally trigonometric in nature, the use of Euler's formula greatly simplifies most of the real portions of the Fourier analyses. Additionally, for the complex case, frequencies can be represented inverse-temporally using a combination of Euler's formula and the Fourier series expansion.

So, it's a bit messy and convoluted (etymologically, not integrally), but it really boils down to the fact that the Taylor (or McLauren) series, the Fourier series and transform, and Euler's formula all relate a trigonometrically
The differences between the three arise by nature of application. Taylor series are used to represent functions as infinite sums of their derivatives. Fourier series and transforms are used in linear systems &/or differential equations to convert signals or DEs from the time to frequency domain. Euler's formula is used to relate trigonometric and complex exponential (complexponential?!) funcions, and is also a formula that, when evaluated at $x=\pi$, yields Euler's identity, $e^{\jmath \pi}+1=0$, an equation so austerely eloquent and aesthetically arousing that I'd be down to stare at all day.


Is the following definition of integrals correct? I'd imagine the idea of integrals summing up infinite slices would be formalized something like this:

$$\int_{k = 0}^x f \left( k \right) = \lim_{\omega \to \infty} \sum_{k = 0}^\omega f \left( {k x \over \omega} \right)$$

Is this correct?

No it not correct way to write an integration as this infinite sum rather you can write it like this

∫ˣf(k)dk = lim ω---->∞ (1/ω) Σ(k=0,xω)f(k)

I don't know how to type equations, sorry for that.

EDIT:
$$\int_0^x f(k)dk=\lim_{\omega\to\infty}\frac{1}{\omega}\sum_{k=0}^{x\omega}f(k)$$


Is there a sufficient condition for which derivative of $\sum_{n=0}^{\infty} a_n x^n$ is bounded for all $x \in \mathbb{R}$? Assume that every power series Bounded for $x \in \mathbb{R}$  , I want to know

if there is a sufficient condition for  $(a_n)\neq0$ , for all $n$  so that the derivative of the power series $\sum_{n=0}^{\infty} a_n x^n$ is bounded for all $x \in \mathbb{R}$ ?

Sufficient condition means a condition (on the $a_n$) which implies that there is some constant $c$, such that $\vert\sum_{n=0}^{\infty} a_n x^n\vert\leq c$ for all real $x$.

Any linear combination of functions of the form $x\mapsto\cos(a x+b)$ has the desired property.

So for example the power series defined by the function $\cos(x)+\sin(x)$,
$$\sum_{n=0}^\infty a_n x_n, {\rm  where~} a_{2m}:=(-1)^m/(2m)!, a_{2m+1}:=(-1)^m/(2m+1)!{\rm~for~ all~}m\in\mathbb{N_0}, $$ is bounded by $c=2$ and all coefficients are different from zero!


Models vs. Structures Why are both the terms 'structure' and 'model' used in mathematical logic / model theory?  Are they just holdovers from different subjects or is there a principled reason for having both?

For clarification, I'm not confused about any actual definitions or usages, just why both terms came to be used; I could, after all, survive perfectly well using exclusively one or the other with little chance of confusion.

Models are structures, and structures are models. But when we say "model" we mean that there is a particular theory which holds in the structure, and when we say "structure" we are mainly interested in an arbitrary interpretation of the language.

There is no reason for introducing two different terms. Apparently, somebody introduced one term, and somebody else introduced a different term, either because he didn't like the first guy's term, or he hadn't heard of it. Or maybe it was the same guy and he changed his mind, or forgot what he called it before. How would I know, I'm not a historian (nor a mathematician).

The point is, in mathematics there is no official body with the power to decide what the terminology should be. This is different from other sciences, such as astronomy, where some organization claims the power to decide what's a planet. In mathematics, each writer goes his own way, and anarchy prevails. (If you think "model" vs. "structure" is bad, look at the terminology of graph theory.) Eventually, after a few centuries, a consensus is reached. Obviously, model theory (and graph theory) are too young to have reached that point.


Bilinear Map vs Inner Product What is the difference between a Bilinear Map and a Inner Product?

An inner product is a map from a vector space crossed with itself to $\mathbb{R}$ or $\mathbb{C}$. Also an inner product must be positive definite: $\langle x, x\rangle\geq 0$. A bilinear map, to contrast, is simply a map $A\times B\to C$ for linear spaces $A,B,C$ which is bilinear.

I think positive definiteness is the most important thing that inner products have that bilinear maps do not; in crypto, bilinear maps are typically of the form $G\times G\to G_T$ where $G, G_T$ are groups, neither of can even be totally ordered in a sensible way.

I also like to think about it like this:
If you have column vector, let's just say it's 3D, then the dual space, which is the space of all linear transforms into the scalars, is just the row vectors of the same dimension (3D). And then if you want to get a linear transform into two dimensions instead of just to the scalars, then you'd just add a row below the original row vector of your dual space. Now the space is the 2x3 matrices. And if you want to get all linear transforms into 3D space, then you add another row, and have a 3x3 matrix. Which is why all linear transforms from 3d to 3d are 3x3 matrices.
A similar thing happens with bilinear forms, but with a small twist. In order to be a bilinear map, your starting place for the dual space is not a row vector -- it is a matrix. So if you're working in 3d, and you want a bilinear form into the scalars, your first argument is a row vector, and the second argument is a column vector, and these go on either side of a 3x3 matrix. It's sort of like an inner product (an inner product would be the identity matrix). And now here's where the parallel keeps going: if you want to map into an additional dimension instead of just mapping to the scalars, in the above example, you add a row, but in a bilinear form, you add an additional dimension and put another 3x3 matrix stacked on the other one like two pieces of paper. I imagine it like a 3x3x2 matrix (tensor). And if you wanted to map into 3d, then you'd have a 3x3x3 matrix (tensor).
Cheers!


Counting non-degenerate parallelograms in permutation arrays I am interested in finding out more about the function $P(n)$ that counts the number of distinct non-degenerate parallelograms whose vertices are in distinct rows and columns of an $n$ x $n$ array.  Enumerative C++ code with nested loops shows that $P(4)=6$, $P(5)=62$, and $P(6)=362$ and also produced the values of $P(n)$ for $4 \leq n \leq 129$.  This data (or a heuristic argument) shows that $P(n)$ has $n^6$ growth.  If possible, I would like to find a recurrence relation or generating function for $P(n)$. Curve fitting for large $n$ appears to suggest that there is an asymptotic constant $\alpha \approx 0.0514$ such that $P(n)\sim \alpha n^6$.  I would also like to know if $\alpha$ exists and to know its true value if it does exist.

Theorems 1 and 2 in the new paper

http://www.combinatorics.org/ojs/index.php/eljc/article/view/v23i1p44

that I coauthored solve this problem.


Guessing number between 1-100 always can always be guessed in 7 guess. Why? I'm no good at probability and statistics as this question will soon demonstrate. My question is this, I have an instructor who proved that he could guess a number between 1-100 using the divide and conquer approach. For example, I choose 7.

He: 50

Me: High

He: 25

Me: High

He: 12

Me: High

He: 6

Me: Low

He: 9

Me High

He: 8

Me High

He: 7

See, its not possible for it to take more than 7 guess if your halfway smart. My question is whats the math behind this, and how could I figure the number of guess for like 1-1000, whats the formula?

Thanks!

Hint: $2^7=128>100$.  That is, you can simply guess in the middle and bisect until you get to the solution.

In general, just find the smallest power of $2$ greater than the max number in your set.  For $1000$, you should be able to get it in $10$ guesses, because $2^{10} = 1024$.

8
down vote
accepted
Hint: 27=128>10027=128>100. That is, you can simply guess in the middle and bisect until you get to the solution.

In general, just find the smallest power of 22 greater than the max number in your set. For 10001000, you should be able to get it in 1010 guesses, because 210=1024


Solving for $x$ using $\ln$ I have an equation as follows:

$\ln(a)x + \ln(b) (1 - x) = \ln(c)$

I'd like to solve for $x$, but I think what I've come up with is wrong

$\ln\left(\frac{a}{b}\right)x + \ln(b) = \ln(c)$

$\ln\left(\frac{a}{b}\right)x = ln\left(\frac{c}{b}\right)$

$x = \frac{\ln\left(\frac{c}{b}\right)}{\ln\left(\frac{a}{b}\right)}$


Question about Bishop's "constructive analysis" for people familiar with bishop style constructive analysis only It has been a while since i read Bishop's book "constructive analysis", recently I dug it out of my book shelve and started to read. I came around this observation on the top of page 85.
"A subset Y of $\mathbb{R}$ can be bounded as a
metric space but not bounded as a subset of $\mathbb{R}$."
I have tried to figure out how this can be and came up with nothing so far. Can anybody give me a hint?
PS: As far as I understand Bishop is referring to the metric induced by the absolute value.
PPS: The definition provided:
A metric space $(X, p)$ is called bounded if there exists a real number
$C > 0$, called a bound for $(X, p)$, such that $p(x, y) \leq C \forall x,y \in X$.
A subset $Y$ of a non void metric space $X$ is bounded if, for all
(equivalently, some) $x$ in $X$, the set $Y\cup\{x\}$ with the induced metric p* is a bounded metric space.

Let $Y=\{n\in\mathbb N:n\text{ is the first counterexample to the Goldbach conjecture}\}$.  We don't know whether $Y$ has an element, but if it has one then it has only that one. That is, for any two elements of $Y$, the distance between them is $0$. So (assuming that Bishop's definition of "bounded metric space" is reasonable, i.e., that it's what I would use) I conclude that $Y$ is bounded as a metric space (under any metric). But to be able to say that $Y$ bounded as a subset of $\mathbb R$, we would need (again assuming a reasonable definition) to exhibit real numbers $c$ and $r$ (the center and radius of a metric ball) and prove that all elements of $Y$ are are within $r$ of $c$. And we don't know how to do that.  [You might think we could be sneaky and take $c$ to be the unique element of $Y$ if $Y$ has an element and $0$ otherwise, because then any positive $r$ would work. But that "definition" of $c$ isn't constructive; we don't know which of the two cases holds. Or, as a constructivist would say, we don't know that $Y$ is inhabited or empty.]

Let $Y=[0,\infty[$ and define the distance $d(x,y) = |\tan^{-1}(x) - \tan^{-1}(y)|$ for any $x,y\in Y$. Then $d(x,y)\leq |\tan^{-1}(x)| + |\tan^{-1}(y)| \leq \frac{\pi}{2} + \frac{\pi}{2} = \pi$. So the metric space $Y$ is bounded.
Assume $Y$ is unbounded. Then there is a sequence $\{y_k\}_{k\in\mathbb{N}}$ such that $|y_k|\to\infty$. Assume the metric space $(Y,d)$ is bounded with $d(x,y) = |x-y|$ for any $x,y\in Y$ and with some center $a\in \mathbb{R}$. Take $d_k = d(a,y_k)$. We have $d_k \geq ||a|-|y_k|| \to \infty$. So $d_k\to\infty$.
Some definitions of bounded metric space don't take a center, that's fine too. Since $y_k\to \infty$, we can pick a subsequence $\{y_{k_j}\}_{j\in\mathbb{N}}$ such that $||y_{k_j}| - |y_{k_{j+1}}||\to\infty$ and redefine $d_j = d(y_{k_j},y_{k_{j + 1}})$.


How does $\int \tan x dx$ equal to $ \log( \sec x)$? How can the integration result in $\sec x$? What operations are done here?

Well, let $u=\cos x$ then $du=-\sin x \,dx$ so that we get
$$\int \tan x dx=\int\frac{\sin x}{\cos x}dx=\int-\frac{1}{u}du=-\ln|u|+C=-\ln|\cos x|+C$$

$$=\ln|(\cos x)^{-1}|+C=\ln|\sec x|+C$$

Nothing too complicated. It's just an integral substitution: $$\int \frac{\sin(x)}{\cos(x)} \, dx =- \int \frac{d( \cos(x))}{\cos(x)} = -\log\, \cos(x)+C = \log\, \sec(x)+C$$


Deriving formulas for complex trigonometric functions $\sin(z)$ and $\cos(z)$ I think it is simpler if I just focus on the derivation of $\cos(z)=\frac{e^{iz}+e^{-iz}}{2}$.

I start from the formulae for the complex exponential, assuming $z=x+iy$: $e^{iz}=e^{-y}*(\cos(x)+i\sin(x))$ and $e^{-iz}=e^{y}*(\cos(x)-i\sin(x))$.

And here I start running into problems, I can simply not reproduce the given formula! The furthest I obtain is $$e^{iz}+e^{iz}=(e^{-y}+e^{y})*\cos(x)+i(e^{y}-e^{-y})*\sin(x)$$ and an equivalent formula for the difference.

Please help.

The Taylor series expansions for $e^z$, $\sin{(z)}$ and $\cos{(z)}$ which are valid for all $z\in\mathbb{C}$ are supplied below
$$e^z=1+\frac{z}{1!}+\frac{z^2}{2!}+\dots=\sum_{k=0}^\infty \frac{z^k}{k!}$$
$$\sin{(z)}=\frac{z}{1!}-\frac{z^3}{3!}+\frac{z^5}{5!}-\dots=\sum_{k=0}^\infty \frac{(-1)^kz^{2k+1}}{(2k+1)!}$$
$$\cos{(z)}=1-\frac{z^2}{2!}+\frac{z^4}{4!}-\dots=\sum_{k=0}^\infty \frac{(-1)^kz^{2k}}{(2k)!}$$
If we look at the sum
$$\begin{align}
\cos{(z)}+i\sin{(z)}
&=\sum_{k=0}^\infty \frac{(-1)^kz^{2k}}{(2k)!}+i\sum_{k=0}^\infty \frac{(-1)^kz^{2k+1}}{(2k+1)!}\\
&=\sum_{k=0}^\infty (-1)^k\left(\frac{z^{2k}}{(2k)!}+\frac{i\cdot z^{2k+1}}{(2k+1)!}\right)\\
\end{align}$$
it is in fact equal to
$$\begin{align}
e^{iz}
&=\sum_{k=0}^\infty \frac{(iz)^k}{k!}\\
&=\sum_{k=0}^\infty \frac{(iz)^{2k}}{(2k)!}+\frac{(iz)^{2k+1}}{(2k+1)!}\\
&=\sum_{k=0}^\infty \frac{(-z^2)^{k}}{(2k)!}+\frac{(-z^2)^k(iz)}{(2k+1)!}\\
&=\sum_{k=0}^\infty (-1)^k\left(\frac{z^{2k}}{(2k)!}+\frac{z^{2k}(iz)}{(2k+1)!}\right)\\
&=\sum_{k=0}^\infty (-1)^k\left(\frac{z^{2k}}{(2k)!}+\frac{i\cdot z^{2k+1}}{(2k+1)!}\right)\\
\end{align}$$
for all $z\in\mathbb{C}$. So we can write that
$$e^{iz}=\cos{(z)}+i\sin{(z)}$$
for all $z\in\mathbb{C}$. Using this gives
$$e^{i(-z)}=e^{-iz}=\cos{(-z)}+i\sin{(-z)}=\cos{(z)}-i\sin{(z)}$$
A proof that
$$\sin{(-z)}=-\sin{(z)}$$
$$\cos{(-z)}=\cos{(z)}$$
for all $z\in\mathbb{C}$ comes from looking at the series expansions of $\sin{(-z)}$ and $\cos{(-z)}$ while using the facts that
$$(-z)^{2k}=z^{2k}$$
$$(-z)^{2k+1}=-z^{2k+1}$$
for all $z\in\mathbb{C}$. Finally we have
$$e^{iz}-e^{-iz}=\cos{(z)}+i\sin{(z)}-(\cos{(z)}-i\sin{(z)})=2i\sin{(z)}$$
$$\therefore \sin{(z)}=\frac{e^{iz}-e^{-iz}}{2i}$$
$$e^{iz}+e^{-iz}=\cos{(z)}+i\sin{(z)}+\cos{(z)}-i\sin{(z)}=2\cos{(z)}$$
$$\therefore \cos{(z)}=\frac{e^{iz}+e^{-iz}}{2}$$
with the above identities holding for all $z\in\mathbb{C}$.

Consider $e^{ix}$ and $e^{-ix}$ remember that sine is an odd function and cosine is an even function:

$$e^{ix}=\cos(x)+i\sin(x)$$
$$e^{-ix}=\cos(-x)+i\sin(-x)=\cos(x)-i\sin(x)$$

Sum both sides:

$$e^{ix}+e^{-ix}=2\cos(x)$$

:)


does the infinite series $1 - 1/2 + 2/3 -1/3 + 2/4 - 1/4 + 2/5 -1/5...$ converge? does the infinite series  $1 - 1/2 + 2/3 -1/3 + 2/4 - 1/4 + 2/5 -1/5...$ converge?

I am saying no since it is not nondecreasing. Is this right or should something more be done?

Except for the missing first term, this sequence is equal to the harmonic series, which is divergent. Just group two adjacent terms and you'll see. Making it alternating by splitting the terms doesn't make it more convergent (it rather makes it worse).

1-1/2+2/3-1/3... is converging because if you combine like terms (ie 1-1/2 2/3-1/3 and so on) you get the harmonic series (ie 1+1/2+1/3...) which diverges. Thus 1-1/2+2/3-1/3... diverges.


Partition of a set, definition not clear From wikipedia:

  Equivalently, a set P is a partition of X if, and only if, it does not contain the empty set and:

  The union of the elements of P is equal to X. (The elements of P are said to cover X.)

  The intersection of any two distinct elements of P is empty. (We say the elements of P are pairwise disjoint.)

I clearly understand that the intersection between partition is empty (point 2), but how can the union of a partition can be the all elements in the set?

If it is a partition, shouldnt they be just a part?

I imagine a set divided in 3 and the elements in the first part are not all the elements of the second part.

How do you explain this?

The idea of a partition is that you take a whole (the set $X$) and you divide it to parts.

Now if I cut off an apple into slices (and one core) I have several pairwise disjoint parts of the apple, but if I reassemble the parts I get a whole apple again.

Similarly we require this from a partition of a set. We want that the union of all the parts give us the entire set we partitioned.

A collection of disjoint subsets of a given set. The union of the subsets must equal the entire original set.

For example, one possible partition of {1, 2, 3, 4, 5, 6} is {1, 3}, {2}, {4, 5, 6}.


How to find the absolute extrema of a function on an elliptical cylinder using Lagrange multipliers? Optimize the function $  f(x,y)  =  x^2y  $ on the elliptical cylinder $ \ x^2 \ + \ 2y^2 \ \le \ 6 \ $ using Lagrange Multipliers.

Well, from what I know that I have to find the gradient then to set it equal to zero but I'm not sure about that and about the following steps.

This  optimization problem involving a function of two variables is a nice illustration of how finding the absolute extrema of a one-variable function $ \ f(x) \ $ on a closed interval is extended to more variables.  Within an interval, we look for any values of $ \ x \ $ for which $ \ \frac{df}{dx} \ = \ 0 \ $ , the so-called critical points.  (I am passing over critical points of the type where the first derivative is undefined.)  We then also check the values of $ \ f(x) \ $ at the endpoints of the interval, and finally make a comparison among the values of the function at the critical points within and the endpoints of the interval.

Here, we are looking for the absolute extrema of a function of two variables within an elliptical region (as indicated by the use of an inequality), so we wish to locate any critical points of the function within the ellipse, and compare the values of the function there with values that the function takes on along the elliptical boundary of the region.  So, as with finding extrema on a closed interval, we have two steps to work out.

First, we wish to find the critical points of $ \ f(x,y) \ = \ x^2 y \ $ that may exist; we will then focus on those which lie within the ellipse $ \ x^2 \ + \ 2y^2 \ = \  6 \ $ , which is centered on the origin and has its axes along the $ \ x- \ $ and $ \ y-$ axes.  We calculate the first partial derivatives of the function and set them equal to zero to find

$$ f_x \ = \ 2xy \ = \ 0 \ \ , \ \ f_y \ = \ x^2 \ = 0 \ \ \Rightarrow \  \ x \ = \ 0 \ , \ y \ = \ 0 \ \ . $$

The function then has a single critical point at $ \ (0, \ 0) \ $ .  For all points $ \ (x, \ 0) \ $ and $ \ (0, \ y) \ $ (points on the coordinate axes), the function equals zero.

The second partial derivatives of the function are $ \ f_{xx} \ = \ 2y \ \ , \ \ f_{yy} \ = \ 0 \ \ , \ \ $ and $ f_{xy} \ = \ 2x \ $ .  We can then form the "discriminator index" $ \ D \ = \ f_{xx} \ f_{yy} \ - \ (f_{xy})^2 \ = \ 2y \ \cdot \ 0 \ - \ 4x^2 \ $ , which is intended to aid in classifying critical points.  Unfortunately, $ \ D \ = \ 0 \ $ at the origin, so we learn nothing directly from this; also, we've already seen that the function is flat along the $ \ x- \ $ and $ \ y-$ axes.  If we look at the behavior of the function along the lines $ \ y \ = \ \pm x \ $ , it becomes $ \ f(x, \ \pm x) \ = \ \pm x^3 \ $ .  What we know about the cubic function $ \ y \ = \ x^3 \ $ is that it has odd symmetry about the $ \ y-$ axis and that there is a point of inflection at $ \ x \ = \ 0 \ $ .  Analogously, our function has a saddle point at $ \ (0, \ 0) \ $ , as can be seen in this graph of $ \ z \ = \ x^2 \ y \ $ in the neighborhood of $ \ ( 0, \ 0, \ 0) \ $ .  So the sole critical point of our function is not a relative extremum.

This leaves us to look at the behavior of our function on the surface of the elliptical cylinder.  This is where we may use either "substitution" or Lagrange multipliers.

With "Lagrange", we construct the equation $ \ \nabla f \ = \ \lambda \ \nabla g \ $ , with $ \ g(x,y) \ = \ x^2 \ + \ 2y^2 \ - \ 6 \ $ being the "constraint function" describing the surface of the cylinder.  We obtain from this

$$ f_x \ = \ \lambda \cdot g_x \ \ \Rightarrow \ \ 2xy \ = \ \lambda \cdot 2x \ \ , \ \ f_y \ = \ \lambda \cdot g_y \ \ \Rightarrow \ \ x^2 \ = \ \lambda \cdot 4y \ \ . $$

We may solve these "Lagrange equations" either as

$$ 2xy \ - \ 2 \lambda x \ = \ 0 \ \ \Rightarrow \ \ 2x \ (y \ - \ \lambda) \ = \ 0 \ \ , \ \ x^2 \ = \ 4 \lambda y \ \ ,  $$

or as

$$ \lambda \ = \ \frac{2xy}{2x} \ = \ \frac{x^2}{4y} \ \ , \ \ x, y \ \ne \ 0 \ \ , $$

and consider the cases of $ \ x \ = \ 0 \ $ or $ \ y \ = \ 0 \ $ separately.  If $ \ \lambda \ = \ 0 \ $ , we are forced to have $ \ x \ = \ 0 \ $ , but $ \ y \ $ may be non-zero; in fact, the "constraint ellipse" gives us $ \ 0^2 \ + \ 2y^2 \ = \ 6 \ \Rightarrow \ y \ = \ \pm \sqrt{3} \ $ .  But we already know the function equals zero for $ \ x \ = \ 0 \ $ or $ \ y \ = \ 0 \ $ .  For $ \ \lambda \ \ne \ 0 \ $ , we deduce that

$$ \ \lambda \ = \ y \ \ \Rightarrow \ \ x^2 \ = \ 4y^2 \ \ \Rightarrow \ \ 4y^2 \ + \ 2y^2 \ = \ 6 \ \ \Rightarrow \ \ y \ = \ \pm 1 \ \ , \ \ x \ = \ \pm 2 \ \ . $$

There are four points of interest then on the cylindrical surface:

$ (2, \ 1) \ , \ (-2, \ 1) \ $ -- $ \quad f(\pm 2, \ 1) \ = \ 4 \ \ $ [relative maxima] ;

$ (2, \ -1) \ , \ (-2, \ -1) \ $ -- $ \quad f(\pm 2, \ -1) \ = \ -4 \ \ $ [relative minima] .

These are also the greatest and least values of the function on or within the ellipse, so they are the absolute maximum and minimum values.

[We find the same result by making the substitution $ \ x^2 \ = \ 6 \ - 2y^2 \ $ , implied by the constraint, into the function $ \ f(x,y) \ $ , producing $ \ (6 \ - \ 2y^2 ) \cdot y \ $ .  The critical points are found from

$$ \ \frac{d}{dy} \ [ \ 6y \ - \ 2y^3 \ ] \ = \ 6 \ - \ 6y^2 \ = \ 0 \ \ \Rightarrow \ \  y \ = \ \pm 1 \ \ , \ \ x \ = \ \pm 2 \ \ . ] $$

A graph of the surface $ \ z \ = \ x^2 \ y \ $ is shown below, together with the elliptical cylinder constraint surface.  (The "twins" of the extrema shown are not visible in this view.)

Set $\operatorname{grad} ({\mathbf F}) = \lambda \cdot \operatorname{grad} ({\mathbf G})$ with constraint $X^2+2Y^2 \leq 6$, solve to find max/min.

Edit: Sorry, you can't use Lagrange for the parts inside the boundary of the cylinder...my bad. You must set $\operatorname{grad}({\mathbf F})$ to $0$.


Do constant functions have asymptotes? As far as I have learned a function has a horizontal asymptote $y = k$ if and only if
$$\lim_{x \to \infty} f(x) = k$$ or $$\lim_{x \to -\infty} f(x) = k$$

Now, for a constant function $$f(x) = c$$ we have $$\lim_{x \to \infty} f(x) = \lim_{x \to -\infty} f(x) = c$$

Does this mean that all constant functions have horizontal asymptotes, or is this definition not the one commonly used?

It really depends on how you define asymptotes, and there was much discussion about this matter in this question.  It really comes down to whether or not we should have

$$f(x)=c$$

At only a countably infinite amount of times as $x\to\pm\infty$.  But for all intended purposes, I think stating a constant function is asymptotic to itself, or indeed, that any function is asymptotic to itself, is, though not very useful, acceptably correct by standard definitions.

The function is defined everywhere. It would be an asymptote if the limit exists AND the function is not defined for that y value.


CDF of $X=\min\{X_1, X_1\cdot X_2, X_1\cdot X_2\cdot X_3, \ldots, X_1\cdot X_2 \cdots X_N\}$ $X=\min\{X_1, X_1\cdot  X_2, X_1\cdot  X_2\cdot  X_3, \ldots, X_1\cdot  X_2 \cdots X_N\}$ where $X_i, \forall i\in\{1,2,...,N\}$ are continuous random variables, $X_i\geqslant 0$, e.g., exponential distribution.

I want to find the CDF of $X$ for any general $N$, i.e., $F_X(x)$.

For example, when $N=2, X=\min\{X_1, X_1\cdot X_2\}$, then we can write
$$F_X(x) = \mathsf P(X < x) = \mathsf P(X_1 < x \mid  X_2 \geqslant 1) + \mathsf P(X_1\cdot X_2 < x \mid X_2 < 1).$$

Can any one help me to write $F_X(x)$ for any $N$?

So for $N=3$ we have

$$\begin{align}
F_X(x) & = \mathsf P(X_3\geq 1, X_2\geq 1)\mathsf P(X_1< x\mid X_3\geq 1, X_2\geq 1)\\ & + \mathsf P(X_3\geq 1, X_2< 1)\mathsf P(X_1\cdot X_2 < x\mid X_3\geq 1, X_2< 1) \\ & + \mathsf P(X_3< 1)\mathsf P(X_1\cdot X_2\cdot X_3< x\mid X_3< 1)
\end{align}$$

So in general.
$$\begin{align}
F_X(x) & = \mathsf P(\bigcup_{j=2}^{N} \{X_j\geq 1\})\mathsf P(\{X_1 < x\}\mid \bigcup_{j=2}^{N} \{X_j\geq 1\})
\\ & + \sum_{k=2}^{N}\mathsf P(\{X_k < 1\}\cup \bigcup_{j=k+1}^{N}\{X_j \geq 1\})\mathsf P(\{\prod_{i=1}^k X_i < x\}\mid \{X_k < 1\}\cup \bigcup_{j=k+1}^{N}\{X_j \geq 1\})
\end{align}$$


Adjoints between the category of sets and the category of left G-sets. What functors are there between the category Set and the category of left G sets? Of these functors can you show which are adjoints?

Bonus points for leading this question,  and me, in a  more rigorous direction.

The forgetful functor from $G$-sets to sets has adjoints on both sides.  The left adjoint sends any set $A$ to the product $G\times A$ with the $G$-action defined by $g(h,a)=(gh,a)$ (for all $g,h\in G$ and $a\in A$). This should be called the free $G$-set generated by $A$.  The right adjoint sends $A$ to the set of all functions from $G$ to $A$ (just set-theoretic functions, ignoring the group structure of $G$) with the $G$-action defined by $(gf)(h)=f(hg)$ for all $g,h\in G$ and all functions $f:G\to A$.

Going from G-sets to sets represents a loss of structure. This is the nature behind the forgetful functor designation; it "forgets" what it means to be "G" and only remembers what it means to be "-set".

And in an example of duality, in G-sets this functor has an opposite. And while the conventional way of thinking about opposites is in terms of a forward and backward, positive and negative,  I suggest to you that in category theory it's more fruitful to think of these adjoints as "doing" and "undoing"

As such, if in one direction this adjunction represents a loss of structure (the doing) then going in the other direction must represent a gain in structure (undoing that loss).

The gain in structure that compliments a forgetful loss is known as the FREE FUNCTOR.

The idea behind the free functor is that in going in the opposite direction, from sets to G-sets, you only add the minimal amount of structure to uniquely identify a G-set from any other set in your theory. This is known as a free construction since it is "free" of any superfluous structure except for "G".

(PS: As a comment
 made clear, this is NOT true of all categories, i.e., not all categories with a forgetful functor have a free functor)


Limit, having trouble applying L'Hospital rule I have been asked to solve for this limit:

$$\lim_{x \to 0} \frac{\sin^2(3x)}{1-\cos(2x)}$$

I try to find the derivative of the numerator and denominator:

$$\frac{6\sin(3x)\cos(3x)}{2\sin(2x)}$$ but that still gives me a $\frac{0}{0}$. How do I solve this

Using L'Hôpital's rule, we have
$$\lim\limits_{x \to 0} \frac{\sin^2(3x)}{1-\cos(2x)}$$
$$=\lim\limits_{x \to 0} \frac{\frac{\mathrm d}{\mathrm dx}\sin^2(3x)}{\frac{\mathrm d}{\mathrm dx}\left[1-\cos(2x)\right]}$$
$$=3\lim\limits_{x \to 0} \frac{\sin(3x)\cos(3x)}{\sin(2x)}$$
$$=3\lim\limits_{x \to 0} \frac{\frac{\mathrm d}{\mathrm dx}\left[\sin(3x)\cos(3x)\right]}{\frac{\mathrm d}{\mathrm dx}\sin(2x)}$$
$$=\frac32\lim\limits_{x \to 0} \frac{\cos(3x)\frac{\mathrm d}{\mathrm dx}\sin(3x)+\sin(3x)\frac{\mathrm d}{\mathrm dx}\cos(3x)}{\cos(2x)}$$
$$=\frac92\lim\limits_{x \to 0} \frac{\cos^2(3x)-\sin^2(3x)}{\cos(2x)}$$
$$=\frac92$$

If you'd like to understand how L'Hôpital's rule is applied, see my answer here. We can also solve this limit without L'Hôpital's rule by using the fact that
$$ \lim\limits_{x\to 0}\frac{\sin(x)}{x}=1$$
Therefore
$$\lim\limits_{x \to 0} \frac{\sin^2(3x)}{1-\cos(2x)}$$
$$=\frac12\lim\limits_{x \to 0} \frac{\sin^2(3x)}{\sin^2(x)}$$
$$=\frac12\lim\limits_{x \to 0} \frac{(3x)^2x^2\sin^2(3x)}{(3x)^2x^2\sin^2(x)}$$
$$=\frac92\left(\lim\limits_{x \to 0} \frac{x^2}{x^2}\right)\left(\lim\limits_{x \to 0} \frac{\sin^2(3x)}{(3x)^2}\right)\left(\lim\limits_{x \to 0} \frac{x^2}{\sin^2(x)}\right)$$
$$=\frac92$$


What does $\frac{1}{9801}$ equal to? While playing on my calculator I accidentally calculated $\frac{1}{9801}$ and the value was $0.00010203040506070809101112131415161718192021222324252627282930\ldots$
We observe that there is a nice pattern in the decimal expansion.
but how long will it continue ie. after how many digits will it repeat?

There is a numberphile video related to this here.

Essentially notice that $9801=99^2$. A similar pattern happens with $\frac{1}{9^2}=\frac{1}{81}$. It's value is $0.\dot{0}1234567\dot{9}$, cycling through all the units except $8$ (To see why, try a long division method). The case with $\frac{1}{99^2}$, it cycles through every two digit number, but excludes $98$ from this list... i.e. it is:
$$\frac{1}{99^2}=0.\dot{0}0010203040506\cdots9596979\dot{9}$$ and similar things happen with $\frac{1}{999^2},\frac{1}{9999^2}$, etc.

Every fraction $\frac{a}{b}$ with integers $a$ and $b$ has a pattern which continues endlessly.


Four balls are randomly dropped into four boxes Four balls are randomly dropped into four boxes, where any ball is equally likely to fall into
each box.

For a fixed $k = 0, 1, 2, 3$, let $A_k$ denote the event that exactly $k$ boxes are empty. Compute
$P(A_k)$ for each $k = 0, 1, 2, 3$.

I have computed $P(A_0)$ and $P(A_3)$ which are $\frac{4!}{4^4}=\frac{3}{32}$ and $\frac{4}{4^4}=\frac{1}{64}$ recpectively. But I don't know how to find $P(A_1)$ and $P(A_2)$. Could anybody help me?

Imagine temporarily that each ball had its own color.  We see then that there are $4^4$ equally likely outcomes for where the balls are positioned.  We get as you already did the probabilities $P(A_0)=\dfrac{4!}{4^4}$ and $P(A_3)=\dfrac{4}{4^4}$

Now., for $P(A_1)$, count the numerator using the following multiplication principle argument:

Pick which box was empty: $4$ choices
Pick which box received two balls: $3$ choices
Pick which two colored balls go into the box designated to receive two balls: $\binom{4}{2}$ choices
Pick which colored ball goes into the left-most box designated to receive one ball: $2$ choices
The final ball goes into the final box designated to receive one ball

This gives a total of $4\cdot 3\cdot \binom{4}{2}\cdot 2$ options here for a probability then of

$$P(A_1)=\dfrac{144}{4^4}$$

We can similarly calculate for the other case, taking advantage of the fact that the boxes are arranged from left to right, giving a total of $\binom{4}{2}\binom{4}{2} = 36$ if there were two boxes, each with two balls in them and $4\cdot 3\cdot 4=48$ if there was one box with three and another box with only one, for a total of $84$ outcomes and a probability of:

$$P(A_2)=\dfrac{84}{4^4}$$

Checking, $4!+144+84+4 = 256$ and so we do get our probabilities adding to one as expected.

$P(A_0)=\frac{1}{35}$ because the only possibility is the every has exactly one ball.
$P(A_3)=\frac{4}{35}$ because there are 4 possibility to choose the one who isn't empty.
$P(A_1)=\frac{\binom{4}{1}\cdot3}{35}$ because the first binomial is the way to choose the one without the ball, and the way to give 4 balls to 3 people so that every of them has at least 1 ball.
$P(A_2)=\frac{\binom{4}{2}\cdot\binom{3}{2}}{35}$ because the first binomial is the way to choose the two without the ball, and the second binomial is the way to give 4 balls to 2 people so that every of them has at least 1 ball.


How to use Excels (or LibreOffices) T-test (e.g. is it a Welch test, assymetric input matrix…) I'm using Excels T.TEST and LibreOffices TTEST functions, and their documentation is rather lacking, so I figured I ask some things here. I'm not exactly sure if this is the right site, but I can't find something more fitting and I think I have the basics (what goes in what input field) right…

What exactly do they return?
It's not explicitly stated that it is the p-value, but I would assume so.

What is the difference between Excel and LO, or Excels old TTEST?
The documentation for Excels old function just states that there is a new one that may provide improved accuracy and whose names better reflect their usage. I would assume for for the standard I'm-clueless-about-statistics-I-just-need-significance situation it should not matter, but I'd rather be safe than sorry.

Is a Type 3 T-test a Welch-test?
According to Wikipedia, I should use said Welch-Test, when my samples standard deviations are not identical, and the documentation says Type 3 is for unequal variance (heteroscedastic). Or is that yet another test?

Can I use strangly formated inputs?
My input sheets are formed rather weird:

    0.8 0.7 0.9 0.8        0.6 0.7 0.8
        0.6 0.9 0.7            0.8 0.9
            0.7 0.8                0.7
                0.6

And I just use the whole A1:D4 and G1:I3 parts, with a lot of empty cells (as in T.TEST=(A1:D4;G1:I3;2;3)). Does that matter or do I have to reformat, for example in simple rows without empty cells?

0.8 0.7 0.9 0.8 0.6 0.9 0.7 0.7 0.8 0.6
0.6 0.7 0.8 0.8 0.9 0.7

and thus T.TEST=(A1:J1;A2:F2;2;3)?

Let me see if I understand. You have $N$ locations. For each location $i\in\{1,\ldots,N\}$, you have $n$ observations (about something). You are trying to see, for a pair $(i,j)$ of locations, if the locations are 'different enough' to treat them separately. That is why you calculated correlation between $i$ and $j$, $\rho_{ij}$. And you are trying to decide whether $\rho_{ij}$ is significantly different from zero and want to use $t$-test for that.

If this is so, then $\rho_{ij}\sqrt{\frac{n-2}{1-\rho_{ij}^{2}}}\equiv t'$ is $t$-distributed with $n-2$ degrees of freedom. Wiki on Pearson correlation coefficient is helpful here. From this getting $p$-value should be easy with the Excel's T.INV/T.INV.2T/T.DIST/T.DIST.2T functions. Namely, plugging $t'$ into T.DIST.2T provides $p$-value for the test which has $\rho_{ij}=0$ as a null and $\rho_{ij}\neq0$ as an alternative.


how to prove that the sum of the power two of three reals is less than 5? $a, b$ and $c$ are three real number such that: $(\forall x \in [-1,1]): |ax^2+bx+c|\leq 1$ .

Prove that:

1)   $|c|\leq 1$ .

2) $-1\leq a+c\leq 1$ .

3)Deduce that : $a^2+b^2+c^2\leq 5$.

The first and second questions are easy to prove ( just take $x=0, -1$ or $1$ ... ), but the third one ?!!! I'm sure that the responce must be easy since it's a deduction.

I need your help to solve it, thank you.

$x=0$ gives us $|c| \le 1$.  Similarly set $x = \pm1$ to get $|a \pm b+c| \le 1$.  So we have the inequalities:

$$-1 \le c \le 1 \iff -1 \le -c \le 1\tag{1}$$
$$-1 \le a+b+c \le 1 \iff -1\le -a-b-c \le 1 \tag{2}$$
$$-1 \le a-b+c \le 1 \iff -1 \le -a+b-c \le 1\tag{3}$$

Note the right side of the above three inequalities can be considered the "negatives" of the left sides.  These should be enough for us to prove all sorts of things.

Here,
$$(2)-(1) \implies -2 \le a+b \le 2\implies (a+b)^2 \le 4$$
$$(3)-(1) \implies -2 \le a-b \le 2\implies (a-b)^2 \le 4$$

Now those two give $a^2+b^2=\frac12((a+b)^2+(a-b)^2) \le 4$.
We just need to add to this $|c| \le 1 \implies c^2 \le 1$ to get what we want.

I'm not sure how you're supposed to go straight from 1) and 2) to 3) without some additional steps.

The help I would offer is that it might help to strengthen 2) to the statement $|a+c| + |b| \le 1$ and work from there.


Jumps of independent Lévy processes Suppose I have two independent Lévy processes $(X_t)_t$ and $(Y_t)_t$, both not continuous.

Is anyone familiar and can refer me to a result (or a counterexample) which states that

${\displaystyle \sum_{0\leq s\leq t}}|\Delta X_{s}\cdot \Delta Y_{s}|=0
 $ for all $t\in \mathbb{R}$ a.s?

A different yet equivalent formulation of this is

$\Delta X_{t}=0$  or  $\Delta Y_{t}=0$ a.s. for all $t\in \mathbb{R}$

In words, every two independent Lévy processes have no simultaneous jumps a.s. I know it holds for independent Poisson processes and I'm wondering if it generalizes.

Let $(X_t)_{t \geq 0}$, $(Y_t)_{t \geq0}$ be independent Lévy processes with Lévy triplet $(\ell_1,q_1,\nu_1)$ and $(\ell_2,q_2,\nu_2)$, respectively. By the independence of the processes, we know that $(X_t,Y_t)_{t \geq 0}$ is a Lévy process and using Lévy-Khinchine's formula we find that its Lévy measure equals

$$\nu(dx,dy) := \nu_1(dx) \otimes \delta_{\{0\}}(dy)+ \delta_{\{0\}}(dx)\otimes \nu_2(dy)$$

where $\delta_{\{0\}}$ denotes the Dirac measure centered at $0$ (for a proof see e.g. this question). Roughly speaking, the jumps of the process $(X,Y)$ are concentrated on the coordinate axis. Indeed: Let

$$A(\varepsilon) := \mathbb{R}^2 \backslash \bigg(B[0,\varepsilon] \cup \{(x,y) \in \mathbb{R}^2; x = 0 \vee y=0\} \bigg)$$

for $\varepsilon>0$. Obviously, $A(\varepsilon)$ is open and satisfies $\nu(A(\varepsilon)) = 0$. Consequently, there exists $\Omega(\varepsilon)$, $\mathbb{P}(\Omega(\varepsilon))=1$, such that

$$\forall t \geq 0: \Delta (X_t,Y_t)(\omega) = (\Delta X_t(\omega),\Delta Y_t(\omega)) \notin A(\varepsilon)$$

for all $\omega \in \Omega(\varepsilon)$. Set $\Omega' := \bigcap_{k \in \mathbb{N}} \Omega(1/k)$. Then $\mathbb{P}(\Omega')=1$ and

$$\forall t \geq 0: (\Delta X_t(\omega),\Delta Y_t(\omega))  \notin \mathbb{R}^2 \backslash \{(x,y); x = 0 \vee y=0\}$$

for any $\omega \in \Omega'$. This shows that $X$ and $Y$ do not jump simultaneously a.s.

by Fubini's theorem we know that $\sum_{0\leq s\leq \infty}|\triangle X_{s}\triangle Y_{s}|$ is measurable. We can write it this way $\sum_{i=1}^{\infty}|\triangle X_{s}\triangle Y_{s}|1_{\{i-1\leq s\leq i\}}$. Take expectation, and by independence,the MCT and the continuity in probability of the Levy processes we have

$E\Bigl(\sum_{i=1}^{\infty}|\triangle X_{s}\triangle Y_{s}|1_{\{i-1\leq s\leq i\}}\Bigr)=\sum_{i=1}^{\infty}E\Bigl(|\triangle X_{s}\triangle Y_{s}|1_{\{i-1\leq s\leq i\}}\Bigr)=\sum_{i=1}^{\infty}E\Bigl(|\triangle X_{s}|\Bigr)E\Bigl(|\triangle Y_{s}|\Bigr)1_{\{i-1\leq s\leq i\}}=0$

so $\sum_{0\leq s\leq \infty}|\triangle X_{s}\triangle Y_{s}|=0$ a.s. Since the trajectories of the summand are increasing it implies for all $t>0$.


High School Trigonometry ( Law of cosine and sine) I am preparing for faculty entrance exam and this was the question for which I couldn't find the way to solve (answer is 0). I guess they ask me to solve this by using the rule of sine and cosine:

Let $\alpha$, $\beta$ and $\gamma$  be the angles of arbitrary triangle with sides a, b and c respectively. Then $${b - 2a\cos\gamma \over a\sin\gamma} + {c-2b\cos\alpha \over b\sin\alpha} + {a - 2c\cos\beta \over c\sin\beta}$$ is equal to (answer is zero but I need steps).

The Law of Cosines is equivalent to (and is often proven via) the statements
$$a = b \cos\gamma + c\cos\beta \qquad b = c \cos\alpha + a \cos\gamma \qquad c = a \cos\beta + b \cos\alpha$$
so, your sum becomes
$$\frac{c \cos\alpha - a \cos\gamma}{a\sin\gamma} + \frac{a \cos\beta - b \cos\alpha}{b\sin\alpha} + \frac{b \cos\gamma - c \cos\beta}{c\sin\beta}$$

Further, the Law of Sines allows us to write
$$a = d \sin\alpha \qquad b = d\sin\beta \qquad c = d \sin\gamma$$
where $d$ is the triangle's circumdiameter. This gives

$$\frac{\sin\gamma \cos\alpha - \sin\alpha \cos\gamma}{\sin\alpha\sin\gamma} + \frac{\sin\alpha \cos\beta - \sin\beta \cos\alpha}{\sin\beta\sin\alpha} + \frac{\sin\beta \cos\gamma - \sin\gamma \cos\beta}{\sin\gamma\sin\beta}$$
$$= \left( \cot\alpha - \cot\gamma \right) + \left( \cot\beta - \cot\alpha \right) + \left( \cot\gamma - \cot\beta \right ) = 0 $$

Step 1:

Write cosines relations for a triangle:

$$a^2=b^2+c^2-2bc \cos(A)$$
$$b^2=a^2+c^2-2ac \cos(B)$$
$$c^2=a^2+b^2-2ab \cos(C)$$

Step 2:

Write sinus Area relation for the triangle:

$$Area(ABC)=\frac{ab \sin(C)}{2}=\frac{bc \sin(A)}{2}=\frac{ac \sin(B)}{2}$$

Step 3:

Write cosines relations as you have in your question:

$$a^2-b^2=c(c-2b \cos(A))$$
$$c-2b \cos(A)=\frac{a^2-b^2}{c}$$

$$b^2-c^2=a(a-2c \cos(B))$$
$$a-2c \cos(B)=\frac{b^2-c^2}{a}$$

$$c^2-a^2=b(b-2a \cos(C))$$
$$b-2a \cos(C)=\frac{c^2-a^2}{b}$$

Step 4:

Combine with sinus Area relation

$$c-2b \cos(A)=\frac{a^2-b^2}{c}$$
$$\frac{c-2b \cos(A)}{b \sin(A) }=\frac{a^2-b^2}{c b \sin(A) }$$

$$a-2c \cos(B)=\frac{b^2-c^2}{a}$$
$$\frac{a-2c \cos(B)}{c \sin(B) }=\frac{b^2-c^2}{ac \sin(B)}$$

$$b-2a \cos(C)=\frac{c^2-a^2}{b}$$
$$\frac{b-2a \cos(C)}{a \sin(C) }=\frac{c^2-a^2}{ab \sin(C)}$$

Step 5:

Combine with sinus Area relation

$$\frac{c-2b \cos(A)}{b \sin(A) }=\frac{a^2-b^2}{2.Area(ABC) }$$
$$\frac{a-2c \cos(B)}{c \sin(B) }=\frac{b^2-c^2}{2.Area(ABC)}$$
$$\frac{b-2a \cos(C)}{a \sin(C) }=\frac{c^2-a^2}{2.Area(ABC)}$$

Step 6:

Add them

$$\frac{c-2b \cos(A)}{b \sin(A) }+\frac{a-2c \cos(B)}{c \sin(B) }+\frac{b-2a \cos(C)}{a \sin(C) }=\frac{a^2-b^2}{2.Area(ABC) }+\frac{b^2-c^2}{2.Area(ABC)}+\frac{c^2-a^2}{2.Area(ABC)}$$

$$\frac{c-2b \cos(A)}{b \sin(A) }+\frac{a-2c \cos(B)}{c \sin(B) }+\frac{b-2a \cos(C)}{a \sin(C) }=\frac{a^2-b^2+b^2-c^2+c^2-a^2}{2.Area(ABC) }$$

$$\frac{c-2b \cos(A)}{b \sin(A) }+\frac{a-2c \cos(B)}{c \sin(B) }+\frac{b-2a \cos(C)}{a \sin(C) }=0$$


bijection between prime ideals of $R_p$ and prime ideals of $R$ contained in $P$ Given a ring $R$, I want to show that the localization of $R$ at the prime ideal $P$ of $R$ (denoted as $R_P$) is isomorphic to the set of prime ideals of $R$ contained in $P$. That is:

$$
\text{Spectrum}(R_P)\cong \{I\subseteq P \mid \text{$I$ is an ideal of $R$}\}
$$

From the statment, I can see that $Q\subseteq R_P$ is a prime ideal, then any $x\in Q$ is of the form $x=\frac{a}{b}$, where $a\in R$, but $b\notin P$, from the definition of $R_P$. But how can I show that each such $Q$ relates to an ideal of $R$ contained in $P$.

Sketch:

Let $\,i\colon R\to R_\mathfrak p$, $\,x\mapsto \dfrac x1$ be the canonical morphism. If $\mathfrak q\in\operatorname{Spec}R_{\mathfrak p}$, $\,\mathfrak p'=i^{-1}(\mathfrak q)$ is a prime ideal of $R$  contained in $\mathfrak p$.

It is straightforward to check the inverse  of this mapping is $\,.\mathfrak p'\mapsto \mathfrak p'R_\mathfrak p$

Proof for $Spec(R_P)\cong \{I\in Spec(R)|I\subseteq P \}$:
More generally, The book, "Steps in Commutative Algebra" by Sharp has:

Consider $S=R-P$.

Theorem 5.37 of the book  has similar result for even primary ideals.


Convergence in distribution and calculation of variance I encountered the following argument, and I don't quite understand them.

  The sequence of random variables $(\sqrt n X_n)$ converges in distribution to the random variable $Y \sim \frac{1}{2}\delta_0+\frac{1}{2}N(0, \sigma^2)$. The variance of $Y$ is $\frac{1}{2}\sigma^2$. Half of the time, the sequence converges to $0$ faster than $1/\sqrt n$.

My questions are the followings:

(i) How can I calculate the expectation and variance of $Y$?

(ii) What does it mean by "converges faster"?

Thanks for your help ;)


Statistics Probability of defection of computers produced by a factory Factories $A$ and $B$ produce computers. Factory $A$ produces $4$ times as many computers as factory $B$. The probability that an item produced by factory $A$ is defective is $0.014$, while the probability that an item produced by factory $B$ is defective is $0.048$.
A computer is selected at random and it is found to be defective. What is the probability it came from factory $A$?

I have figured that the fact that factory $A$ produced $4$ times as many computers does not factor into the question.
Let $P(A) = 0.014$
Let $P(B) = 0.048$
$P(A) = P(A ∩ B) + P(A ∩ Bc)$
I rearranged the equation to find the Probability of A and B compliment to find the answer.
$$P(A ∩ Bc) = P(A) - P(A ∩ B) = 0.014 - (0.014 \cdot 0.048) = 0.013328$$
What is the correct answer and solution?

I have figured that the fact that factory $A$ produced $4$ times as many computers does not factor into the question.

No, that tells you the prior probability that a computer comes from factory $A$.   That is $P(A)=4 P(B)$

What you have otherwise been told is the rate of defects conditional on the source.

That is, $~P(D\mid A) = 0.014~$ and also, $~P(D\mid B) = 0.048$

  What is the correct answer and solution?

It's a Bayes' Rule (and Law of Total Probability) questions.

You seek $P(A\mid D)$ the posterior probability that a computer comes from factory#A when given that it is defective.

$$\begin{align}P(A\mid D) ~&=~\dfrac{P(A)P(D\mid A)}{P(A)P(D\mid A)+P(B)P(D\mid B)} \\[1ex] &=~ \dfrac{4\times 0.014}{4\times 0.014~+~0.048} \\[1ex] &=~ \frac 7{13} \\[2ex] &\approx~0.538\end{align}$$

If you select any (might be broken) computer the probability it comes from A is 0.8 or $\frac{4}{5}$ vice versa 0.2 or $\frac{1}{5}$ for B.

If A would't produce 4 times as much computers as B does then the probability of a broken computer coming from A is $\frac{0.014}{0.014+0.042}=\frac{1}{4}$

So the probability that a broken computer comes from A is $\frac{4}{5}\cdot\frac{1}{4}=\frac{1}{5}$


Is it true that at least two of any consecutive $2m$ positive integers cannot be divided by odd prime numbers less than $2m$? Let $m>1$ and $p_2,\cdots,p_k$ are all odd prime numbers less than $2m$.

$q,a_2,\cdots,a_k$ are arbitrarily selected integers(I mean no matter how you choose these numbers).

$$G_1=\lbrace n\in \mathbb Z|q < n \leq q+2m \rbrace,$$

$$G_i=\lbrace n\in \mathbb Z|n\not\equiv a_i \mod {p_i}\rbrace,$$

for $i=2,3,\cdots,k$.

$$G=\bigcap_{i=1}^k{G_i}.$$

  Is it always true that #$|G|\geq 2$?

Especially,is it true that at least two of any consecutive $2m$ positive integers cannot be divided by odd prime numbers less than $2m$?

In other words, if you write $2m$ consecutive positive integers and cross out the multiples of $3, 5,\cdots,p_k$, then at least two of these $2m$ numbers will not be crossed out.

(1) When $m=2,k=2,p_2=3,$ four consecutive integers $q+1,q+2,q+3,q+4$,no matter what is $a_2$,we know that $G$ has at least two elements.

(2) For every $m>1$, we can choose $q$ and $a_i$ so that #$|G|=2$,for example, if

$$q=\frac{R-1}{2}-m,R=\prod_{i=2}^k{p_i},$$
$$a_i=0,i=2,\cdots,k,$$

then $G=\lbrace \frac{R-1}{2},\frac{R+1}{2} \rbrace.$

(3) The case $m=3,k=2,p_2=3,p_3=5.$

#$|G_1 \setminus G_2|=6/3=2,$ #$|G_1 \setminus G_3|\leq [6/5]+1=2,$
hence #$|G|\geq 2.$

(4) When $m\geq 36$, as $$\sum_{2<p<2m}{[\frac{2m}{p}]}>2m-2$$ maybe hold, I don't know if there's a counterexample.

OP asks about every possible choice of $q$. Counterexamples are to be found for low values of $q$, such as sequences $2,3,\dots,2m+1$ and $3,4,\dots,2m+2$, where every member of the sequence must be composed entirely of prime factors $<2m$, and where no two consecutive numbers can each be a power of $2$, i.e. numbers entirely lacking odd prime factors.

In general, if an odd member of a suitable pair of consecutive integers has no (necessarily odd) prime factors smaller than $2m$, then that odd number must either be a prime number larger than $2m$ or a composite number larger than $4m^2$. This suggests that there will be some minimum value of $q$ (a function of $m$) below which OP's conjecture cannot hold.


False Proof that $\sqrt{4}$ is Irrational Everyone with any basic knowledge of number theory knows the classic proof of the irrationality of $\sqrt{2}$. Curious about generalizations using elementary methods,  I looked up the irrationality of $\sqrt{3}$, and found the following:

Say $ \sqrt{3} $ is rational. Then $\sqrt{3}$ can be represented as $\frac{a}{b}$, where a and b have no common factors.
So $3 = \frac{a^2}{b^2}$ and $3b^2 = a^2$. Now $a^2$ must be divisible by $3$, but then so must $a $ (fundamental theorem of arithmetic). So we have $3b^2 = (3k)^2$ and $3b^2 = 9k^2$ or even $b^2 = 3k^2 $ and now we have a contradiction.

Such a proof follows the same basic logic as the proof for $\sqrt{2}$, except for using the fundamental theorem of arithmetic to replace and generalize the trivial fact that $n$ is even if $n^2$ is even.  However,  when I apply this proof format to $\sqrt{4} $ (which is clearly an integer and thus rational)  I get the following:
Say $ \sqrt{4} $ is rational. Then $\sqrt{4}$ can be represented as $\frac{a}{b}$, where a and b have no common factors.
So $4 = \frac{a^2}{b^2}$ and $4b^2 = a^2$. Now $a^2$ must be divisible by $4$, but then so must $a $ (fundamental theorem of arithmetic). So we have $4b^2 = (4k)^2$ and $4b^2 = 16k^2$ or even $b^2 = 4k^2 $, which implies that $b=4n$ by the fundamental theorem. Now we have a contradiction (since can note that both $a$ and $b$ are divisible by $4$ and we assumed they were coprime)
This proof is clearly false,  yet I fail to see where it differs. Where does it do so?

Just because $a^2$ is divisible by $4$, that doesn't mean $a$ is.

Now If you want to prove that square root of a non perfect square is irrational.

Then you can do this what I wrote below as a proof.

Let $p$ is a non perfect square. We are required to prove $\sqrt{p}$ is irrational. Let $\sqrt{p}$

is rational and of the form $\frac{a}{b}$ where $a\;\;and\;\;b$ are coprime and $b\neq{0}$.
Now, we have

$$\sqrt{p}=\frac{a}{b}$$
By squaring both sides we have
$$p=\frac{a^2}{b^2}$$
$$\implies{a^2}=p{b^2}$$
We know that $a\;and\;b$ are coprime to each other and $g.c.d(a,b)=1$.
So we have $$ax+by=1$$ for some non zero integers $x$ and $y$.Then multiply $\sqrt{p}$ on both sides.Then we get
$$\sqrt{p}ax+\sqrt{p}by=\sqrt{p}$$See that $a=\sqrt{p}b$. So we have
$$pbx+ay=\sqrt{p}$$
Now see that the L.H.S belongs to integer set whereas the R.H.S belongs to
$\mathbb{R}\backslash \mathbb{Z}$.

Hence it is a contradiction.So $\sqrt{p}$ is irrational. May be it will help you.


Has lack of mathematical rigour killed anybody before? One of my friends was asking me about tertiary level mathematics as opposed to high school mathematics, and naturally the topic of rigour came up.

To provide him with a brief glimpse as to the difference, I said the following.

  In high school, you were taught that the area of a rectangle is $ab$
  where $a$ is the breadth and $b$ is the height. You can physically see
  this by constructing an $a \times b$ grid and counting the squares it
  forms, provided $a$ and $b$ are integers.

He had agreed and said that it was "obvious" that the area of a rectangle was $ab$. I then responded with:

  What is the area of a rectangle with dimensions $\pi$ by $\sqrt 2$?

He immediately just said $\pi \sqrt 2$, and then I responded with one of the most common questions in mathematics:

  How do you know that for sure?

I had said that it intuitively works for integer values of $a$ and $b$, but how do we KNOW for sure that it works for irrational values of $a$ and $b$? Then I used that as a gateway to explain that in tertiary level mathematics we don't assume such things. There is no "It is clearly true for these easy-to-understand integers, so therefore it is true for all real values" and that everything must be proven.

He then asked me something that I had no answer to:

  I get that we cannot assume these kinds of things, but has there ever been an occasion where an assumption or a lack of rigour has killed someone before?

I am sure that there may exist an example floating somewhere in history, but I cannot think of any.

Do you know of one?

EDIT: Cheers to starsplusplus

A lot of really great responses! However, the majority of them don't quite fit the definition of 'rigour' in the mathematical sense, which is vastly different to the common English term. See this. Many of the answers provided so far have been accidents/deaths caused by a lack of what I feel to be more like procedural rigour as opposed to mathematical rigour.

EDIT 2:

It seems like further clarification is needed regarding what I'm looking for in a response. I was looking for an example of where an individual(s) did something that was mathematically incorrect (not a trivial computational error though) that had a consequence which led to the death of one or more people. So, what do I mean by something mathematically incorrect that isn't a trivial computational error?

Example of something I'm looking for:

Say somebody is part of programming or working out the math behind a missile firing mechanism. In part of their computations, they did one of the following which yielded an incorrect value. This incorrect value caused the missile to fly out of control and cause the death of one or more person(s).

Exchanged a summation with an integral unjustifiably
Needed to use two sequence of numbers that always yielded relatively prime numbers. They used a computer but it didn't find any counter examples, so the programmer assumed that the formula always yields relatively prime integers. However, the counter example lies at $n=99999999999999999999999999$, beyond reasonable computational time.
The limit of a series was to be used at some point in the computations. To calculate it, the person re-arranged terms however they liked and then found a limit. But the series didn't converge absolutely so they could have gotten any value.

Particularly, lack of knowledge of Bayes' theorem, and intuitive use of probability, lead to many misdiagnosed patients all of the time. In fact, some studies suggest that as many as 85%(!) of medical professionals get these type of questions wrong.

A famous example is the following. Given that:

1% of women have breast cancer.
80% of mammograms detect breast cancer when it is there.
10% of mammograms detect breast cancer when it’s not there.

Now say that a woman is diagnosed with breast cancer using a mammogram. What are the chances she actually has cancer?

Ask your friends (including medical students) what their intuition regarding the the answer is, and I'm willing to bet most will say around 80%. The mathematical reasoning people give for this answer is simple: Since the test is right 80% of the time, and the test was positive, the patient has a 80% chance of being sick. Sound correct?

   Would you be surprised to learn that the actual percentage is closer to 10%?

This perhaps surprising result is a consequence of Bayes' theorem: The overall probability of the event (breast cancer in all women) has a crucial role in determining the conditional probability (breast cancer given a mammography).

I hope it's obvious why such a misdiagnoses can be fatal, especially if treatment increases the risk of other forms of cancer, or in reversed scenarios where patients are not given care when tests give negative results.

I think i might be deviating from original question , but let me bring down my opinion , if maths would somehow kill a body , it maybecause excessive rigor not lack of it , like someone who wastes more than half of his days and non slept nights in scientific researches without any result , that would distract him from serious matters in his life and retard him very backward in age , he would miss many things and wake up on a horible deadly truth cause him to commit suicide or self destruction , btw .... this question seems silly for me to care i prefer the first one bein debated

   ok was joking :D time for serious answers

considering an infinite euclidean plan isometry

this graph shows how is a product of two irrational amounts is abstractly feasible (but not perfectly calculable since the human brain conception is bounded to decimal values)

the value shadowed in blue equals to $\int_{0}^{\pi} \sqrt{3}$ ~= $5.44$

the last rounded value is not yet authentic because when we draw a $3.14$ segment perpendicular to $1.73$ long segment we end up with a result nearby $5.43$ this difference of $0.01$ would be gradually and progressivly leading to a killing mistake.

This is an example of deadly inaccuate rounded up error percent.

I will show you now the steps you need to get an irrational product value on your graph paper , take a compass , and a pen , and a ruler with measures.

first draw a circle of radius $r=1Dm = 10cm$ ,  and draw bissectrice of $\pi/2$ with same compass ...

Take same compass again , draw the bissectrice of the angle $\pi/4$ , now you have an angle $\theta=\pi/8$

Draw the projective point on $x$ axis , then take the same compass , and draw a half circle of radius = the segment from the center to the projection of the angle $\theta$

length of new radius is $cos \theta$=$cos \pi/8$

$cos \pi/4 = cos(2(\pi/8))=2cos²(\pi/8)-1$

$cos \pi/8 = \sqrt{\frac{cos (\pi/4) + 1}{2}}= \frac{\sqrt{\sqrt{2}+2}}{2}$

the surface of the blue shadowed segment = $\pi*r^2/8=\frac{\sqrt{2}+2}{24}\pi$

Now be happy :D you have just drawn an irrational product of class $ \pi*\mathbb{N}^{1/2}$ by your own bare hands !!! and it does never equals the approximate decimal calculation result whic is nearby $44.7 mm$


What is the sum of all complex integers? In line with

$$\zeta(-1)=-1/12$$
Could we, by considering

$$f(s)=\sum_{a,b\in\mathbb Z,\;(a,b)\neq(0,0)}\frac{1}{(a+bi)^{s}}$$
Evaluate the sum of all complex integers?

You might want to restrict to positive integers only since...

$$f(s) = \sum_{a>0,b>0} \frac{1}{(a+ib)^s} + \sum_{a>0,b<0} \frac{1}{(a+ib)^s} + \sum_{a<0,b>0} \frac{1}{(a+ib)^s} \\+ \sum_{a<0,b<0} \frac{1}{(a+ib)^s} + \sum_{a=0,b\in Z}\frac{1}{(ib)^s} + \sum_{a\in Z,b=0} \frac{1}{a^s}$$

giving

$$f(s) = (1 + (-1)^s)\left[\left(\sum_{a>0,b>0} \frac{1}{(a+ib)^s} + \frac{1}{(a-ib)^s}\right) + \zeta(s)(1+i^s)\right]$$

So $f(-1)$ (meaning the analytical continuation) will be $0$ (unless the analytical continuation of $\sum_{a>0,b>0} \frac{1}{(a+ib)^s} + ...$ has a simple pole at $s=-1$).

The same happens if one consideres the sum of all integers:

$$g(s) = \sum_{n\in Z, n\not= 0} \frac{1}{n^s} = (1 + (-1)^s)\sum_{n=1} \frac{1}{n^s} = \zeta(s)(1 + (-1)^s)$$

so the 'sum of all integers' are zero.

I'm not sure how to go about proving it in any form of notation, but if we consider the real number axis from -x to +x, and the imaginary axis from -iy to +iy, then all the complex integers will be plotted as iy against x.  Since there are no limits to the size of either real or imaginary parts and given that they may only be integers.  I would have thought that the sum of all the complex integers would be zero.  Mainly because the sum of all imaginary numbers from -iy to +iy is zero, and all the real numbers from -x to +x is zero.  Am I missing something?

I see what you are saying, so 1+(1-1-1)+(1-1-1)...=-∞
But 1-(1-(1-1-1))-(1-(1-1-1))...=+∞

I'm not sure that eventually if all the combinations of number series is included in the argument a situation would arise where the net result is asymmetric.

Thank you Winther, I don't know if everyone will agree with you, but I do. But I'm beggining to see now that with complex numbers the problem may well be different.


Find $\text{Var}(N)$ where $P(N = n|Y = y)$ is $\text{Possion}(y)$; $Y$ is a gamma with parameters $(r,\lambda)$ The question is as follows:

  Suppose that the conditional distribution of $N$, given that $Y = y$, is Poisson with
  mean $y$. Further suppose that $Y$ is a gamma random variable with parameters
  $(r, λ)$, where $r$ is a positive integer.

I want to get Var(N)

I know that $$\text{Var}(N) = E(N^2) - (E(N))^2 = E(\text{Var}(N|Y)) + \text{Var}(E(N|Y))$$
I got stuck on how to calculate $E(N^2)$.
I know that $E(N^2) = E(E(N^2|Y = y))$, but I got stuck onward.

Would greatly appreciate any help!

Ignore the formula $$\operatorname{Var}[N] = \operatorname{E}[N^2] - \operatorname{E}[N]^2.$$  It is the second formula that you want:  $$\operatorname{Var}[N] = \operatorname{E}[\operatorname{Var}[N \mid Y]] + \operatorname{Var}[\operatorname{E}[N \mid Y]].$$

How do you use it?  What is $\operatorname{E}[N \mid Y]$?  It is the expectation of the Poisson distribution with parameter $Y$.  What is $\operatorname{Var}[N \mid Y]$?  This is the variance of the Poisson distribution with parameter $Y$.  So both of these are functions of the random variable $Y$.

Next, compute the expectation and variance of these with respect to the gamma distribution $Y$.

So for example, if $\operatorname{E}[N \mid Y] = Y$, then $$\operatorname{Var}[\operatorname{E}[N \mid Y]] = \operatorname{Var}[Y] = r\lambda^2.$$

Hint: Use the Law of Total Variance. $$\mathsf {Var}(N) = \mathsf E(\mathsf{Var}(N\mid Y))+\mathsf {Var}(\mathsf{E}(N\mid Y))$$

Given that $N\mid Y\sim \mathcal{Pois}(Y) \\ Y\sim\mathcal{Gamma}(r,\lambda)$


Why is an orthogonal matrix called orthogonal? I know a square matrix is called orthogonal if its rows (and columns) are pairwise orthonormal

But is there a deeper reason for this, or is it only an historical reason? I find it is very confusing and the term would let me assume, that a matrix is called orthogonal if its rows (and columns) are orthogonal and that it is called orthonormal if its rows (and columns) are orthonormal but apparently that's not conventional.

I know that square matrices with orthogonal columns have no special interest, but thats not the point. If I read the term orthogonal matrix my first assumption is, that its rows (and columns) are orthogonal what is correct of course, but the more important property is that they are also orthonormal

So, Question:
Why do you call an orthogonal matrix orthogonal and not orthonormal?
Wouldn't this be more precisely and clearly?

A affine transformation which preserves the dot-product on $\mathbb{R}^n$ is called an isometry of Euclidean $n$-space. In fact, one can begin without the assumption of an affine map and derive it as a necessary consequence of dot-product preservation. See the Mazur Ulam Theorem which shows this result holds for maps between finite dimensional normed spaces where the notion of isometry is that the map preserves the norm. In particular, an isometry of $\mathbb{R}^n$ can be expressed as $T(v)=Rv+b$ where $R^TR=I$. The significance of such a transformation is that it provides rigid motions of Euclidean $n$-space. Two objects are congruent in the sense of highschool geometry if and only if some rigid motion carries one object to the other.

My Point? this is the context from which orthogonal matrices gain their name. They correspond to orthogonal transformations. Of course, these break into reflections and rotations according to $\text{det}(R)= -1,1$ respective.

Likely thought: we should just call such transformations orthonormal transformations. I suppose that would be a choice of vernacular. However, we don't, so... they're not orthonormal matrices. But, I totally agree, this is just a choice of terminologies. Here's another: since $R^TR=I$ implies $R$ is either a rotation or reflection let's call the set of such matrices rotoreflective matrices. In any event, I would advocate the change of terminology you advocate, but, the current terminology is pretty well set at this point, so, good luck if you wish to change this culture.

A matrix $V$ with mutually orthogonal columns is called orthogonal because it maps each of the standard orthogonal coordinate directions to a new set of mutually orthogonal coordinate directions. For example, spherical coordinates are orthogonal because
$$
             x=r\sin\phi\cos\theta,\;\; y=r\sin\phi\sin\theta,\;\; z = r\cos\phi
$$
maps the coordinate lines where $r$ alone varies, $\phi$ alone varies, and $\theta$ alone varies into mutually orthogonal curves in $\mathbb{R}^{3}$. This is evidenced in the Jacobian matrix whose columns are mutually orthogonal:
$$
             \frac{\partial(x,y,z)}{\partial(r,\phi,\theta)}
                       = \left[\begin{array}{ccc}
    \sin\phi\cos\theta & r\cos\phi\cos\theta & -r\sin\phi\sin\theta \\
    \sin\phi\sin\theta & r\cos\phi\sin\theta & r\sin\phi\cos\theta \\
    \cos\phi & -r\sin\phi & 0
                               \end{array}\right]
$$
It is well known that the spherical coordinate system is orthogonal because the three different coordinate lines always intersection orthogonally. The determinant of the Jacobian matrix for this orthogonal transformation is easy to determine: it is the product of the lengths of the three columns of the Jacobian matrix, which is the product of the standard spherical distance scale factors $\frac{dl}{dr}=1$, $\frac{dl}{d\phi}=r$, $\frac{dl}{d\theta}=r\sin\phi$. The standard spherical coordinate volume element is the the product of these $dV = r^{2}\sin\phi\,dr\,d\phi\,d\theta$.

A non-zero unitary matrix goes one step further: any set of orthogonal vectors is mapped to another, which is much stronger than the standard coordinate directions being mapped to mutually orthogonal directions. This condition implies that there is a constant $C$ such that the unitary matrix maps all unit vectors to vectors of length $C$. After renormalizing by dividing by $C$, the unitary matrix is distance preserving and preserve angles: $(Vx,Vy)/\|Vx\|\|Vy\|=(x,y)/\|x\|\|y\|$.


What does it mean to integrate a complex function over a real domain? In complex form, we know that the $n$-th Fourier coefficient of a function $f$ is given by $$\int_{-\pi}^{\pi} f(\theta)e^{-in\theta} d\theta.$$

My question: What exactly does it mean to integrate the complex function $f(\theta)e^{-in\theta}$ over a real domain? How can I visualize this—in general and in this particular case?

Let $f:[a,b]\to \mathbb{C}$. Since $f$ is a complex-valued function, then there exist real valued functions $f_1,f_2$ such that $$f(x)=f_1(x)+i_2f(x),$$
($f_1,f_2$ are nothing but the real and imaginary part). You can simply define the integral of $f$ by
$$\int_a^b f(x)dx:=\int_a^b f_1(x)dx+i\int_a^b f_2(x)dx,$$ where $\int_a^b f_1(x)dx$ and $\int_a^b f_2(x)dx$ are the usual integrals of real valued functions.

You can even define integration of a function $f:\mathbb{C}\to \mathbb{C}$ over a countour in the complex plane.

On the positive real line, $\theta= 0$ and on the negative real line, $\theta= \pi$ .  Also $e^{-in\theta}= r cos(\theta)+ irsin(\theta)= r$ on the real line. So that complex integral can be written $\int_{-\pi}^0 f(\pi)x dx+ \int_0^{\pi} f(0)ydy$.


How to show $\zeta (1+\frac{1}{n})\sim n$ How to show $\zeta (1+\frac{1}{n})\sim n$ as $n\rightarrow \infty$ where $\zeta$ is the Riemann zeta function.

And can we say $\lceil \zeta (1+\frac{1}{n}) \rceil=n$ for any positive integer $n\geq 1$. How to prove it?

HINT:

Note that we have

$$\int_2^\infty \frac{1}{x^{1+1/n}}\,dx\le \sum_{k=2}^\infty\frac{1}{k^{1+1/n}}\le \int_1^\infty \frac{1}{x^{1+1/n}}\,dx \tag 1$$

SPOILER ALERT:  Scroll over the highlighted area to reveal the soluiton

  Evaluating the right-hand side of $(1)$, we see that $$\int_1^\infty \frac{1}{x^{1+1/n}}\,dx=n \tag 2$$ while evaluating the left-hand side of $(1)$, we see that $$\int_2^\infty \frac{1}{x^{1+1/n}}\,dx=n2^{-1/n} \tag 3$$Then, note that $$n2^{-1/n}\ge n-\log(2) \tag 4$$Using $(2)-(4)$ in $(1)$, we find that $$n+(1-\log(2)) \le \sum_{k=1}^\infty\frac{1}{k^{1+1/n}}\le n+1$$from which we conclude that $$\sum_{k=1}^\infty\frac{1}{k^{1+1/n}}\sim n$$

Is $\displaystyle \left.\frac d {ds} \, \frac 1 {\zeta(s)}\right|_{s=1}$ equal to $1$?  If so, then $$ 1 = \lim_{s=1} \frac{\dfrac 1 {\zeta(s)} - \dfrac 1 {\zeta(1)}}{s-1} = \lim_{s=1} \frac{\left(\dfrac 1 {\zeta(s)}\right)}{s-1}, $$
so
$$
\zeta(s) \sim \frac 1 {s-1} \text{ as }s\to 1.
$$
Apply this when $s=1 + \dfrac 1 n$.


Cohesive picture of groups, rings, fields, modules and vector spaces. If I understand my algebra correctly every field is a ring and every ring is a group, so when we define modules over rings and vector spaces over fields, we then have that every vector space is a module?

A linear algebra is defined in Hoffman's book as follows.

  Let $F$ be a field. A linear algebra over the field $F$ is a vector space $\mathcal{A}$ over $F$ with an additional operation called multiplication of vectors which assosciates with each pair of vectors $\alpha, \beta \in \mathcal{A}$ a vector $\alpha \beta \in \mathcal{A}$ called the product in such a way that,

  Multiplication is associative: $\alpha (\beta \gamma) = (\alpha \beta) \gamma$
  Multiplication is distributive with respect to addition: $\alpha(\beta + \gamma) = \alpha \beta + \alpha \gamma$ and $(\alpha + \beta)\gamma = \alpha \gamma + \beta \gamma$
  For each scalar c in $F$: $c(\alpha \beta) = (c \alpha)\beta = \alpha(c \beta)$

Is there a more recent term for what Hoffman means by a linear algebra and how does it fit into the whole group-ring-field hierarchy? Lastly, if you define a vector space over a field and a module over a ring, what is defined over a group in this same way?

To reiterate, I have three questions,

Is every vector space a module?
What relationship does a linear algebra to have to vector spaces and modules?
Is there an algebraic structure defined over groups the same way that vector spaces are defined over fields and modules are defined over rings?

Yes: if $F$ is a field, then "module over $F$" is the same thing as "vector space over $F$". Said another way: vector spaces are modules in which the ring of scalars is a field.
A "linear algebra" (or more generally, an "$F$-algebra") is both a ring and an $F$-vector space, in such a way that the ring multiplication is compatible with the $F$-vector space structure.
You have "group actions" if all you have is a set and a map $G\times X\to X$ which is compatible with the operations of $G$. If $X$ has an algebraic structure of its own, e.g., if $X$ is an abelian group, then we talk about $G$-modules (which amounts to having a group homomorphism $G\to\mathrm{Aut}(X)$, where "Aut" are the appropriate structure automorphisms).

All of these can be further generalized to the concept of "general/universal algebra" (fields are not universal algebras, but they can be obtained by weakening the conditions to obtain 'partial algebras'). A great introduction to that is George Bergman's An Invitation to General Algebra and Universal Constructions.

See Arturo's excellent answer above (below?). To this, I wish to add the following:

The definition of an algebra given by Hoffman and clarified by Arturo above is actually that of an associative algebra. Many of the algebras we encounter in classroom mathematics are associative,but it should be noted there are several important types of nonassociative algebras that appear in both algebra and particle physics; algebras where Hoffman's first axiom of the definition fails. The most famous examples are probably the octonions, alternative algebras and the Jordan algebras. And believe it or not,one of the most famous examples is usually studied in high school calculus and is almost never described this way: Three dimensional Euclidean space is a nonassociative algebra under ordinary vector addition and the cross product.In fact, it is both noncommutative and nonassociative under the cross product. (Try and prove it-good exercise for a beginning algebra student!) For much more on nonassociative algebras, the standard text is Richard Schafer's An Introduction To Nonassociative Algebras. A more recent and comprehensive source is Kevin McCrimmon's A Taste Of Jordan Algebras .


No. of pathways from one point to another in a square grid There are 4 by 4 square grid pathways connecting southwest end and northeast end. How many different ways do exist?

My Anwser: $8C4$ I took it that there are 8 right's, RRRRRRRR. However, we need 4 of these R's to be Ups or U's, which implies we need $8C4$ such combinations. Does this make sense?

***YEAH THAT'S RIGHT.

In a 4x4 grid, to end go southwest to northeast end, we have 8 choices of which 4 are right and the other up; then the total number of minimum paths (→ ↑) is 8C4.***


182-Day T-Bill vs. 91-Day T-Bill I am trying to understand how T-Bills work and it would be great if someone could explain me using the following question

  At $t=0$ Smith buys a 182-Day T-Bill with a simple annual discount rate of $10\%$. At $t=91$ Smith sells his T-Bill.  At that moment, the "prevailing quoted annual discount rate of a 91-Day T-Bill is also $10%$."  Find the actual rate of return (91-Day interest rate) that Smith earned during the time he held the T-Bill.

I am thinking that if $d=10\%$ then $i=11.111 \dots \%$ .

So at $t=91$ the interest earned is approximately

$$\frac{1}{4}(11.111 \dots \%)$$

So, I want to say that Smith earned that much, but the answer seems to be about $2.66\%$.

How do we get this?

I have no idea whether this is correct but $$\frac{\left(100-10\times\frac{91}{365}\right)-\left(100-10\times\frac{182}{365}\right)}{\left(100-10\times\frac{182}{365}\right)}\times 100\approx 2.623991$$ while if you replaced $365$ by $360$ it would give about $2.662376$.

This looks like a stupid way to work out interest rates, but who said bankers were clever?


find new average if removing one element from current average Ok, so say that I have a current rating average: 3.3/5

Now I want to say to remove a rating of 4. How do  I find the new average? Or is this even possible?

To fix the problem of thirdender, and put it on a programmatic way.

Substract a value:

average = ((average * nbValues) - value) / (nbValues - 1);

Add a value:

average = average + ((value - average) / nbValues);

For any computer programmers finding this question, the code is very simple.

function removeFromAverage(value, average, samples) {
  if (samples <= 1) {
    return FALSE;
  }
  return ((average - samples) - value) / (samples - 1);
}

value is the value you want to remove from the average, and samples is the number of observations included in the previous average. The if statement is only there in case one sample is used in the previous average, in which case removing that single value would result in zero samples and no calculable average.


Usage of the term $\arg(z)$ Consider the complex number $z = -1 - i$. Is it mathematically correct to say that $\arg(z) = 5\pi/4$? Sure, $5\pi/4$ is not the principle argument of $z$, but it is an element of the set $\arg(z)$. That being said, is $\arg(z) = 5\pi/4$ actually correct, or is it an incorrect use of the term?

Usually I define $\arg(z)=-3 \pi/4$ as the principal value, so that the function is well defined (single valued), but really you can define it as any branch. Priestley's notation is useful:

$$[\arg(z)]=\{\arg(z)+2 n \pi : n \in \mathbb{Z} \}$$


Can someone give me a deeper understanding of implicit differentiation? I'm doing calculus and I want to be an engineer so I would like to understand the essence of the logic of implicit differentials rather than just memorizing the algorithm. Yes, I could probably memorize it and get a 100% on a test, but it means nothing unless I understand it and can acquire a practical understanding of it. I would really appreciate if someone can enlighten me.

What I understand already:

I understand how the derivatives of normal functions are found the long way... i.e.

$ f'(x)=\frac{f(x+h)-f(x))}{(x+h)-x} $

I understand that the derivative of the above equation is found when we find the limit as h, representing the distance between the two points, approaches 0.

I also understand that the chain rule, quotient rule, etc. are just algorithms that speed up the process of finding the derivative. I don't need a proof for that

An example problem

When we do implicit differential equations such as this one:

A ladder is 8.5 m long leaning against a wall, the bottom part of the ladder is 6.0 m from the wall and is sliding away from the wall at a rate of 2.5m/s.

$x^2 + y^2 = h^2$ (Pythagorean theorem) (x is x value, y is y value, h is hypot)
We can find that y = 6.2m.

The derivative is
$2x\frac{\mathrm{d} x}{\mathrm{d} t} + 2y\frac{\mathrm{d} y}{\mathrm{d} t} = 2h\frac{\mathrm{d} h}{\mathrm{d} t}$

What I don't understand: (although I can do them by memorizing the algorithm)

Is the implicit differential... How do we relate all the terms to the change in time? I mean how do we know we can make that equation's derivative with respect to t? Would it be possible to make them all with respect to the change in x? If so please demonstrate, I think that would help a lot as my biggest lack of understanding is how to know what the bottom term should be for each derivative term.

I really appreciate any help, thanks.

Don't treat implicit differentiation as an idea that is distinct from "regular" differentiation.  Just as the chain rule is involved in every derivative, and is not a separate rule that governs only certain situations, e.g.
$$y=x^2=(x)^2$$
$$
\begin{align}
\frac{d}{dx}(y)&=\frac{d}{dx}(x)^2\\
&=2(x)^1\frac{d}{dx}(x)\\
&=2x\cdot 1\\
&=2x
\end{align}
$$
implicit differentiation is something that you're doing all the time, you just don't see it as such.  For example, when you differentiate $y=3x^2+2$, view it as differentiating both sides with respect to the symbol $x$:
$$
\begin{align}
\frac{d}{dx}(y)&=\frac{d}{dx}(3x^2+2)\\
&=\frac{d}{dx}(3x^2)+\frac{d}{dx}(2)\\
&=3\frac{d}{dx}(x^2)+0\\
&=3\cdot 2(x)^1\frac{d}{dx}(x)\quad\text{ (via chain rule)}\\
&=6\cdot x\cdot 1\\
&=6x
\end{align}
$$
I think that there are two things that aid in understanding, both of which treat differentiation in a symbolic manner.

The first is to realize that nothing changes if the equation is rearranged, for example if we change it to $y-3x^2=2$, the same process works.  Differentiating both sides with respect to the symbol $x$ gives:
$$
\begin{align}
\frac{d}{dx}(y-3x^2)&=\frac{d}{dx}(2)\\
\frac{d}{dx}(y)-\frac{d}{dx}(3x^2)&=\frac{d}{dx}(2)\\
\frac{d}{dx}(y)-6x\frac{d}{dx}(x)&=0\\
\frac{dy}{dx}&=6x
\end{align}
$$
The second is to realize (or treat) $x$ as simply being an arbitrary symbol. That is, think of $\dfrac{d}{dx}$ as $\dfrac{d}{d\square}$.  The operations above work just the same.  We would end up with:
$$
\frac{d}{d\square}(y)-6x\frac{d}{d\square}(x)=0
$$
When, if $\square =x$ we get (as above)
$$
\frac{dy}{dx}=6x
$$
while if $\square=y$ we get
$$
\begin{align}
\frac{d}{dy}(y)-6x\frac{d}{dy}(x)&=0\\
1-6x\frac{dx}{dy}&=0\\
\frac{dx}{dy}&=\frac{1}{6x}
\end{align}
$$
Finally, if $\square=t$ then no "cancellation" occurs and we get
$$
\frac{d}{dt}(y)-6x\frac{d}{dt}(x)=0
$$

The only other thing to remember is that the chain rule needs to be applied to all symbols as in:
$$
\frac{d}{d\square}(y)^2=2(y)^1\frac{d}{d\square}(y)
$$

Given: dx/dt = 2.5 m/s ; h = 8.5 m ; x = 6.0 m

Because the ladder is assumed to be constrained to the surfaces - the Pythagorean relationship holds at any given instance in time or space.

Assume ladder doesn't stretch and is rigid so its length (hypotenuse) doesn't change with time. dh/dt = 0

Using d/dt:

Step (1) x*dx/dt = -y*dy/dt

Step (2) find y when x = 6.0 m and h = 8.5 m;

Step (3) plug x, y and dx/dt into (1) and solve dy/dt = -2.4 m/s

Now to answer your question. Assume dh/dx = 0 since it's not stretching

d/dx of our geometric constraint is: 2*x + 2*y*dy/dx = 0

Using d/dx:

Step (1) x  = -y*dy/dx ;

Step (2) again find y when x = 6.0 m and h = 8.5 m and plug x, y into above equation to get dy/dx = -6/6.2

Step (3) dy/dt = (dy/dx)*(dx/dt) ;// relationship defining dy/dt (chain rule again)

Step (4) plug in dy/dx into (3) and plug in dx/dt(given)

Step (5) dy/dt = -6/6.2 * 2.5 = -2.4 m/s

Since we were given a rate which is defined as m/s it only makes sense here to take derivative w/ respective to time b/c as you see it involves less steps but you can solve it either way.


Inverse of a factorial I'm trying to solve hard combinatorics that involve complicated factorials with large values.

In a simple case such as $8Pr = 336$, find the value of $r$, it is easy to say it equals to this: $$\frac{8!}{(8-r)!} = 336.$$

Then $(8-r)! = 336$ and by inspection, clearly $8-r = 5$ and $r = 3$.

Now this is all and good and I know an inverse function to a factorial doesn't exist as there is for functions like sin, cos and tan etc. but how would you possibly solve an equation that involves very large values compared to the above problem without the tedious guess and checking for right values.

Edit: For e.g. if you wanted to calculate a problem like this (it's simple I know but a good starting out problem)
Let's say 10 colored marbles are placed in a row, what is the minimum number of colors needed to guarantee at least $10000$ different patterns? WITHOUT GUESS AND CHECKING

Any method or explanation is appreciated!

I just wrote this answer to an old question. Using $a=1$, we get a close inverse for the factorial function:
$$
n\sim e\exp\left(\operatorname{W}\left(\frac1{e}\log\left(\frac{n!}{\sqrt{2\pi}}\right)\right)\right)-\frac12\tag{1}
$$

The inverse function of $y = x!$ means getting x in terms of $y$ , i.e $x =$ the largest number in factorisation of y as a factorial.(Where factorising as a factorial means you divide $y$ by $2$, then $3$ and so on. You stop when you get $1$) For example let $5040 = x! , x = ?$

Factoring $5040$ as a factorial $5040= 7\times 6\times 5\times 4\times 3\times 2\times 1$ , and $7$ is the largest number of that factorial $\implies x = 7$
In your problem $8!/336 = (8 – r)! , r = ?$

$8!/336 = 120$ , let $(8 – r) = x$ , hence $120 = x! , x = ?$

$120 = 5\times 4\times 3\times 2\times 1$, and the largest number of that factorial $ = x = 5 = (8 – r) \implies r = 3.$


Impossibility of constructing certain regular polygons with straight-edge and compass My question originally rose from a HW exercise in an abstract algebra course, in which I was asked to show that it was impossible to construct a regular heptagon or nonagon.

We have proved in class that it is possible to construct a point $(x,y)\in \mathbb{R}^2$ if and only if the degree of the field extension $[\mathbb{Q}(x,y):\mathbb{Q}]$ is a power of 2. However, I can't manage to use this directly to show that we cannot construct certain polygons. I can show, for example, that the point $(\cos(2\pi/7),\sin(2\pi/7)$ cannot be constructed because of the theorem, but I can't explain why this rules out building some other heptagon which doesn't contain this point.

This link, for example, states that "attempting to construct any polygon is equivalent to finding its internal angles". How can we show that? And how can we show that not being able to construct some point $(\cos \theta,\sin \theta)$ implies that we cannot construct the angle $\theta$ at all?

Perhaps you know that the 60 degree angle cannot be triseceted. Using the interior angle formula for the regular ngon (n-2)180/n for n=3 we have 60. For n=9 we have 7*20=7*(60/3).

Interesting trivia: Of all the mathematical achievements of the mighty Gauss, he personally considered his 1796 proof of the construction of a regular 17-gon as his most beautiful theorem. Gauss found an unexpected link between contructible polygons and prime numbers. Gauss was so pleased with this result that he requested that a regular 17-gon be inscribed on his tombstone. The stonemason declined, stating that the difficult construction would essentially look like a circle. The discovery ultimately led Gauss to choose mathematics instead of philology as a career

  Gauss–Wantzel theorem: A regular $n$-gon (that is, a polygon with n sides) can be constructed with compass and straightedge if and
  only if

  $$ n = 2^a p_1 p_2 p_3 \ldots p_i $$

  where $a$ is a positive integers and each $p_i$ is a prime number of
  the form $2^{2^k} + 1$ (Fermat Prime).

You can find the proofs in this link.


Where can I download Discrete Mathematics lecture videos? Good morning,

I'm doing a course in Discrete Mathematics (so far: Four Colour Theorem, Intro Graph Theory, Intro Logic Theory, Intro Set Theory and Intro Proofs) at University, but unfortunately they don't record the lectures.

Where can I download Discrete Mathematics lecture videos? - If from multiple places, which would you recommend?

Thanks for all suggestions,

Alec Taylor

PS: Notes and worksheets recommendations would be great too :]

Recently I've been studying bioinformatics, and in an effort to get my biology up to scratch, I've been watching online lectures via iTunes U.  There's an astonishing number of free recorded lectures available, including discrete mathematics.

Typically the video and lecture quality is very high, although I have encountered a few lower quality videos.  I'd recommend trying them out before downloading a huge amount (some of the downloads can be quite large).

This is a late response, but MIT's OCW has materials for discrete mathematics.

http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-042j-mathematics-for-computer-science-fall-2010/