Natural deduction proof of $(c ∧ n) → t, h ∧ ¬s, h ∧ ¬(s ∨ c) → p \vdash (n ∧ ¬t) → p$ I'm trying to do a question from Huth and Ryan's book Logic in Computer Science and I am stuck on the following natural deduction proof:

Prove by natural deduction that the sequent

$$(c ∧ n) → t, h ∧ ¬s, h ∧ ¬(s ∨ c) → p \vdash (n ∧ ¬t) → p$$

is valid.

I have tried assuming $n ∧ ¬t$ on line 4 after the premises, and using Modus Tollens from there to get $¬(c ∧ n)$, but I don't know what I can do after that. Any help would be appreciated.

$\begin{align}n, \neg (c\wedge n)\vdash&~ \neg c \\ h\wedge \neg s\vdash&~ \neg s \\ h\wedge \neg s\vdash&~ h \\ \neg c, \neg s\vdash&~ \neg (s\vee c)\\ h, \neg(s\vee c), h\wedge \neg (s\vee c)\to p\vdash & ~ p\end{align}$

More or less.  Format and provide justifications .

Here's a proof that might help:

Here is the OP's attempt:

  I have tried assuming n∧¬t on line 4 after the premises, and using Modus Tollens from there to get $¬(c∧n)$, but I don't know what I can do after that.

The OP reached line 6 of this proof.

After that I used De Morgan's rules (DeM) on line 7. However, doing that left me with a disjunction. I considered the two cases of the disjunction on lines 10-14 and 15-18. They both led to $P$ which allowed me to use disjunction elimination (vE) on line 19. Line 20 used a conditional introduction to conclude the proof.

For more information about these rules or to try the proof checker see the links below.

Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/

P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Winter 2018. http://forallx.openlogicproject.org/


draw line which is perpindicular to another line with a wanted length can anyone help me? I have a line $Ax+By+C=0$ (let's call it Line $1$); I want to draw a perpicidular line (Line $2$) from the center of line $1$ with a length of $10$.
 what I did is get the slope and intercept of line $1$ as well as the its middle point (point $m$).
and I used the length equation $L=\sqrt{(\text{line}\space 2\cdot x-m\cdot x)^2+(\text{line}\space 2\cdot y-m\cdot y)^2}$, but I dont get the right result.
Thanks in advance

Thanks guys for your support. here's what i did ( and yaay it works)
1-As i have the line equation (or points) i got dy and dx.
2- i got the normal vector to the line(-dy,dx) and (dy,-dx) let's call it NormalVec1 and NormalVec2.
3- i got the magnitude of NormalVec
4- i got the UNIT normalvector by dividing NormalVec by its Magnitude.
5- assume that i want a line with length L.
   so it gonna be L/2 on both sides.
6- assume that i want to draw this perpindicular line crosing the original line at its center point M.
7-so the perpindicular line's points are:
endpoint=(M.X+unitNormalVec1*L/2 ,M.X+unitNormalVec1*L/2) ; for normalVec1 is (-dy,dx)
startpoint=(M.X+unitNormalVec2*L/2 ,M.X+unitNormalVec2*L/2) ; for normalVec1 is (dy,-dx)
then you can draw a line segment from these points.
Thanks,

Your code calculates the length of half the existing line. There are many ways to solve your problem. Get start and end coords, sx, sy, ex, ey. Calculate the slope angle with angle=atan(dy/dx), where dx=ex-sx, dy=ey-sy. Add or subtract pi/2 to angle. Calculate midpoint with mx=(ex-sx)/2 and my=(ey-sy)/2. This is the start point of the perpendicular line. Get the end point with mex=mx+10*cos(angle), mey=my+10*sin(angle). Easy as pi/2. :)


Simple question about probability I'm a bit confused about something.

Let's say I want to cross a road to my friend's house. The probability of the road to be opened is $p$, and $1-p$ to be closed. So I know the probability for one direction at a time. What would it be if I would like to go and return on the same day?
note: Let's assume that the road was opened/closed for the whole day.

My answer is $p^2$ (or $(1-p)^2$), depends on the status of the road.

If it is given that the status of the road is the same for the entire day, then the answers of "$p^2$ or $(1-p)^2$" don't make sense. With this interpretation, only the status of the road matters, but it does not matter at all how many times you want to cross the road. With probability $p$ the road is open for the entire day — so this is the probability that you can visit your friend, or visit him and come back home, or visit him and come back home and visit him again, or … (you get the idea). And the probability that you can't cross the road is just $1-p$, regardless of how many times you want to cross it.

I think answer is p,because who comes back and who went are the same,so he knows the road is open or closed,when the probabality of being open is P when he wants to come back anything changes,so thats just one job,so answer is P


Please help me understand the proof Doubt

In third line of the proof, why is Q $\rightarrow$ R ?

Thanks

The proof runs as follows :

1) $P$ --- assumed [a]

2) $\lnot R$ --- assumed [b]

3) $\lnot R \to (P \to \lnot Q)$ --- premise

4) $P \to \lnot Q$ --- from 2) and 3) by modus ponens

5) $\lnot Q$ --- from 1) and 4) by modus ponens

6) $\lnot R \to \lnot Q$ --- from 2) and 5) by Deduction Th (or Conditional Proof), discharging [b]

7) $Q \to R$ --- from 6) by Contraposition

  8)  $P \to (Q \to R)$ --- from 1) and 7) by Deduction Th (or Conditional Proof), discharging [a].

$ \neg R \implies \neg Q \Leftrightarrow Q \implies R$


Is it true that $0.999999999\ldots=1$? I'm told by smart people that
$$0.999999999\ldots=1$$
and I believe them, but is there a proof that explains why this is?

What does it mean when you refer to $.99999\ldots$?  Symbols don't mean anything in particular until you've defined what you mean by them.

In this case the definition is that you are taking the limit of $.9$, $.99$, $.999$, $.9999$, etc.  What does it mean to say that limit is $1$?  Well, it means that no matter how small a number $x$ you pick, I can show you a point in that sequence such that all further numbers in the sequence are within distance $x$ of $1$.  But certainly whatever number you choose your number is bigger than $10^{-k}$ for some $k$.  So I can just pick my point to be the $k$th spot in the sequence.

A more intuitive way of explaining the above argument is that the reason $.99999\ldots = 1$ is that their difference is zero.  So let's subtract $1.0000\ldots -.99999\ldots = .00000\ldots = 0$.  That is,

$1.0 -.9 = .1$

$1.00-.99 = .01$

$1.000-.999=.001$,

$\ldots$

$1.000\ldots -.99999\ldots = .000\ldots = 0$

The more 9's there are in the sequence, the smaller the margin between 1 and that number becomes, and after every 9, the margin become 10 time smaller. For example, for the number 0.9, it is ${1 \over \mathrm{10}^{1}}$ away from reaching 1, 0.99 is ${1 \over \mathrm{10}^{2}}$ away from reaching 1, and 0.999 is ${1 \over \mathrm{10}^{3}}$ away from reaching 1. So if $0.99999999 \dots$ has an $n$ number of 9's in it, the 'spacing' between that number and 1 is ${1 \over \mathrm{10}^{n}}$. Now, since $0.99999999 \dots$ has an infinite number of 9's, it makes sense that the 'space' between $0.9999999 \dots$ and 1 is     ${1 \over \mathrm{10}^{\infty}}$, or $\mathrm{10}^{-\infty}$, which is 0. Now, to prove that $\mathrm{10}^{-\infty}$ is 0. Simply take the log base 10 of both sides, and you'll find that $ \log_{10} 0 = -\infty$. This means that the 'spacing' between $ 0.9999\dots $ and 1 is 0, meaning that $0.99999\dots + 0 = 1$.

Therefore, $0.9999\dots = 1$


Is it possible to have multiple Conjugate Priors? In Bayesian probability theory, can a probability distribution have more than one conjugate prior for the same model parameters?

I know that the Normal distribution has another Normal distribution or the Normal-Inverse-Gamma distribution has conjugate priors, depending on what the model parameters are.

If multiple priors exists, then kindly cite some. If multiple priors do not exist, what is the proof that it cannot exist?

I am not 100% sure I understand your question, but consider a parameter space $\Theta$ with prior probability $f(\theta)$ and a probability distribution $p(x|\theta)$ over a space $X$ indexed by this parameter. The conjugate prior is simply obtained by computing Bayes rule:

\begin{equation*}
p(\theta|x) = \dfrac{f(\theta) p(x|\theta)}{\int_{\theta' \in \Theta}{f(\theta') p(x'|\theta)}}
\end{equation*}

Therefore, if $\theta$ (the model parameter) is given, the posterior distribution $p(\theta|\cdot)$ is unique and given this formula. Or, in other words, if $p(\theta|x)$ and $q(\theta|x)$ are two conjugate priors for $p(x|\theta)$, it is obvious that $p(\theta|x)=q(\theta|x)$ for any $x$.


How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$? How can one prove the statement
$$\lim_{x\to 0}\frac{\sin x}x=1$$
without using the Taylor series of $\sin$, $\cos$ and $\tan$? Best would be a geometrical solution.

This is homework. In my math class, we are about to prove that $\sin$ is continuous. We found out, that proving the above statement is enough for proving the continuity of $\sin$, but I can't find out how. Any help is appreciated.

The area of $\triangle ABC$ is $\frac{1}{2}\sin(x)$.  The area of the colored wedge is $\frac{1}{2}x$, and the area of $\triangle ABD$ is $\frac{1}{2}\tan(x)$.  By inclusion, we get
$$
\frac{1}{2}\tan(x)\ge\frac{1}{2}x\ge\frac{1}{2}\sin(x)\tag{1}
$$
Dividing $(1)$ by $\frac{1}{2}\sin(x)$ and taking reciprocals, we get
$$
\cos(x)\le\frac{\sin(x)}{x}\le1\tag{2}
$$
Since $\frac{\sin(x)}{x}$ and $\cos(x)$ are even functions, $(2)$ is valid for any non-zero $x$ between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$.  Furthermore, since $\cos(x)$ is continuous near $0$ and $\cos(0) = 1$, we get that
$$
\lim_{x\to0}\frac{\sin(x)}{x}=1\tag{3}
$$
Also, dividing $(2)$ by $\cos(x)$, we get that
$$
1\le\frac{\tan(x)}{x}\le\sec(x)\tag{4}
$$
Since $\sec(x)$ is continuous near $0$ and $\sec(0) = 1$, we get that
$$
\lim_{x\to0}\frac{\tan(x)}{x}=1\tag{5}
$$

This is not a rigorous proof, but is instead an intuitive argument. Consider the graph of the sine function and in particular consider the origin $(0,0)$ and some arbitrary point $(x,\sin(x))$ a little bit to the right of it. Connect a secant line between the two points like so:
Now consider the average rate of change of the sine function on that interval $[0,x]$, or alternatively the slope $m$ of that secant line, namely:
$$m(x)=\frac{\sin(x)-\sin(0)}{x-0}$$
But clearly this is just
$$m(x)=\frac{\sin(x)}{x}$$
But then "clearly" (again, this is just for intuitive purposes), as $x\to{0}$, the slope of this secant line approaches the slope of the tangent line to the graph at $x=0$ (i.e. it approaches whatever $\sin'(0)$ should be). Looking at the graph then, it shouldn't feel too unreasonable that this limiting procedure results in a $45$-degree diagonal tangent, hence a slope of $1$.


Other method other than integration by parts I would like to integrate this
$$\int x \ln x\sin x\mathrm dx$$

can I do it with another quick method than using integration by part?

Here is a fun but not very useful trick: if you know the antiderivative already, any integral can be computed using only $u$-substitution. Proof:

$$\int f(x) \, dx$$
Let $F'(x) = f(x)$.
$$u = F(x), du = f(x) \, dx$$
$$\int du = u + C = F(x) + C.$$

So for your integral, we see that it can be solved using the magic $u$-substitution

$$u = -Si(x) + \sin x + \ln x (\sin x - x \cos x).$$

We have

$$du = x \ln x \sin x \, dx$$
$$\int du = u + C = -Si(x) + \sin x + \ln x (\sin x - x \cos x),$$

as desired.


How can i prove that the finite extension field of real number is itself or the field which is isomorphic to complex number? How can i prove that the finite extension field of real number is itself or the field which is isomorphic to complex number ? In deed, this example is included in Fraleght . Abstract Algebra text.
I did try the followings:
 $\mathbb{R}$ is real number. Then $\mathbb{C}$ is explassd as the smallest extension field including $ \mathbb{R} \cup ${$i$}

How about considering this set . Let set $\mathbb{H}$ is the smallest field including
$\mathbb{R} \cup${$i,j,k$} where $i, j, k$ are called Hamilton number or quaternion
their square are equal to $-1$. Firstly, I do know that this set is a ring. But i check that this set is a field.

Of course, $\mathbb{H}$ may be not a field. Becasue, if that is true, then The Fraleght text book is wrong. However, I would like to know the specific reasons and Example's solution .
Please help me to get this.

$\mathbb{H}$ is not a field, because it is not commutative.  Fraleigh is correct that $\mathbb{R}$ and $\mathbb{C}$ are the only finite field extensions of the reals. (Hint: an odd degree polynomial over $\mathbb{R}$ has a real root.)

If you consider division algebras instead of fields, however, there is exactly one more possibility: the one you found, $\mathbb{H}$.

In deed, i did solve that simply.
complex number field is algebraic closure of real number field. So If $M$ is a finite extension field of real number field including complex number field. Then $M$ is also a finite extension field of complex number field. It is a contradiction that complex number field algebraic clusure(it does not have any its algebraic extension field)
I would like to consider another opinion


Fourier Series for $|\cos(x)|$ I'm having trouble figuring out the Fourier series of $|\cos(x)|$ from $-\pi$ to $\pi$.

I understand its an even function, so all the $b_n$s are $0$

$$a_0 = \frac 2 \pi \int_0^\pi |\cos(x)|\,dx = 0$$

$$a_n = \frac 2 \pi \int _0^\pi  |\cos(x)| \cos(nx) \, dx = \frac 2 \pi \int_0^\pi \cos^2(x)\,dx.$$

since for all $j,k$ not equal the integral is zero.

so only $a_1$ remains. is this correct?

How would I evaluate $\sum_{n=1}^\infty (-1)^{n-1} /(4n^2 - 1)\  {}$?

Although $ \int_0^\pi \cos(x)\,dx = 0$, $a_0\ne 0$ because $$\int_0^{\pi/2} |\cos(x)|\,dx=\int_{\pi/2}^{\pi} |\cos(x)|\,dx. $$

We can evaluate it as follows, as can be seen in the plot below

$$a_0 = \frac 1 \pi \int_{-\pi}^\pi |\cos(x)|\,dx=\frac 2 \pi \int_0^\pi |\cos(x)|\,dx=\frac 4 \pi \int_0^{\pi/2} |\cos(x)|\,dx = \frac 4 \pi \int_0^{\pi/2} \cos(x)\,dx=\frac 4 \pi.$$
$$\tag{1}$$

Plot of $\cos x$ (doted line) and $|\cos x|$ (solid line) in the interval $[-\pi,\pi]$.

The coefficients $b_n=0$ as you concluded. As for the $a_n$ coefficients only the odd ones are equal to $0$ (see below). The functions $\cos(x)$ and $\cos(nx)$ are orthogonal in the interval $[-\pi,\pi]$, but $|\cos(x)|$ and $\cos(nx)$ are not. Since

\begin{equation*}
\left\vert \cos (x)\right\vert =\left\{
\begin{array}{c}
\cos (x) \\
-\cos (x)
\end{array}
\begin{array}{c}
\text{if} \\
\text{if}
\end{array}
\begin{array}{c}
0\leq x\leq \pi /2 \\
\pi /2\leq x\leq \pi,
\end{array}
\right. \tag{2}
\end{equation*}

we have that

\begin{eqnarray*}
a_{n} &=&\frac{1}{\pi }\int_{-\pi }^{\pi }\left\vert \cos (x)\right\vert\cos (nx)\,dx=\frac{2}{\pi }\int_{0}^{\pi }\left\vert \cos (x)\right\vert
\cos (nx)\,dx \\
&=&\frac{2}{\pi }\int_{0}^{\pi /2}\left\vert \cos (x)\right\vert \cos
(nx)\,dx+\frac{2}{\pi }\int_{\pi /2}^{\pi }\left\vert \cos (x)\right\vert
\cos (nx)\,dx \\
&=&\frac{2}{\pi }\int_{0}^{\pi /2}\cos (x)\cos (nx)\,dx-\frac{2}{\pi }
\int_{\pi /2}^{\pi }\cos (x)\cos (nx)\,dx. \\
a_{1} &=&\frac{2}{\pi }\int_{0}^{\pi /2}\cos ^{2}(x)\,dx-\frac{2}{\pi }\int_{\pi /2}^{\pi }\cos ^{2}(x)\,dx=0.
\end{eqnarray*}

Using the following trigonometric identity, with  $a=x,b=nx$,
\begin{equation*}
\cos (a)\cos (b)=\frac{\cos (a+b)+\cos (a-b)}{2},\tag{3}
\end{equation*}

we find

\begin{eqnarray*}
a_{2m} &=&\frac{4}{\pi \left( 1-4m^{2}\right) }\cos (\frac{2m\pi }{2})=\frac{
4}{\pi \left( 1-4m^{2}\right) }(-1)^{m} \\
a_{2m+1} &=&\frac{4}{\pi ( 1-4(2m+1)^{2}) }\cos (\frac{(2m+1)\pi
}{2})=0,\qquad  m=1,2,3,\ldots.\tag{4}
\end{eqnarray*}

The expansion of $\left\vert \cos (x)\right\vert $ into a trigonometric
Fourier series in the interval $[-\pi ,\pi ]$ is thus

\begin{equation*}
\left\vert \cos x\right\vert =\frac{a_{0}}{2}+\sum_{n=1}^{\infty }\left(
a_{n}\cos (nx)+b_{n}\sin (nx)\right) =\frac{2}{\pi }+\frac{4}{\pi }
\sum_{m=1}^{\infty }\frac{(-1)^{m}}{1-4m^{2}}\cos (2mx)\tag{5}
\end{equation*}

$$|\sin(x)|\ \text{(blue) and the partial sum  }\frac{2}{\pi }+\frac{4}{\pi }
\sum_{m=1}^{5 }\frac{(-1)^{m}}{1-4m^{2}}\cos (2mx) \ \text{(red) in }[-\pi,\pi]$$

Setting $x=0$ in $(5)$, we obtain

\begin{equation*}
1=\frac{2}{\pi }+\frac{4}{\pi }\sum_{m=1}^{\infty }\frac{(-1)^{m}}{1-4m^{2}}=\frac{2}{\pi }-\frac{4}{\pi }\sum_{n=1}^{\infty }\frac{(-1)^{n-1}}{1-4n^{2}}.\tag{6}
\end{equation*}

Hence

\begin{equation*}
\sum_{n=1}^{\infty }\frac{(-1)^{n-1}}{1-4n^{2}}=\frac{1}{2}-\frac{\pi }{4}.\tag{7}
\end{equation*}

Bro, here you have to understand that, $\cos(x)$ is positive in first quadrant and negative in 2nd quadrant, so according to the rule of modulus, we will need to break the limits again to $0$ to $\frac{π}{2}$ and $\frac{π}{2}$ to $π$, in order to calculate term $a_0$ of the Fourier series. To make it more clear and understand it better, let's say we have $|\sin(x)|$ here the $\sin$ is positive in both the 1 and 2 quadrant. So we don't need to break the limits again like we have to in $\cos(x)$.

All trigonometry are positive in first quadrant
Sin and cosine is positive in 2nd (all other are negative in this quadrant)
Tan and cot is positive in 3rd. all other are negative in this quadrant
Cos and sec is positive in 4th. all other are negative in this quadrant


I don't think W/O replacement events are dependent. Let say there're cards numbered 1 to 10. P(A) is the probability you pick a card whose number is the multiple of 2. P(B) is the probability you pick a card whose number is the multiple of 5. So now you draw two cards without replacement. You don't place the first card you just drew back in the original deck. Then, you draw again from that 9 remaining cards.
In the picture, P(A) is shown as the ordered pairs colored red, and P(B) is shown as the ordered pairs shaded with light blue.
Then I counted them all.

P(A) = (9+9+9+9+9) / (100-10) = 45/90 = 1/2
P(B) = (9+9) / (100-10) = 18/90 = 1/5
P(A∩B) = 9 / 90 = 1/10

So here, I found out that P(A∩B) is actually the multiple of P(A) and P(B), which shouldn't be if they are dependent, meaning they are independent. I've also found out, in a different model, that the probabilities could be dependent each other in the case of W/O replacement such as this one.
I know events with replacements are independent. But it feels like events without replacements doesn't mean dependency. It's more like the case of contraposition of the first one. Events with replacements mean independent, which means, by the contraposition rule, dependent events are the ones done without replacements.
But all the other internet pages say like "if they are done without replacement, then they are dependent". This claim is so prevalent. But one of us, they or I, must be wrong. What am I missing here?

first, "with\without replacement" always appears with "sampling" 。I think the event A and event B you described is not sampling itself.
Normally, if P(A=i) is the probability you pick a card whose number is i,P(B=j) is the probability you pick a card whose number is j. apperantly,  P(B|A) != P(B) without replacement.

A=the first card is even;
B=the second card is 5 or 10.
$P(A)=\frac{1}{2}$ as you said, but $P(B)=P(B\cap A)+P(B\cap A^C)=P(A)P(B|A)+P(b\cap A^C)=\frac{1}{2}\frac{1}{5}+\frac{4}{5} \frac{2}{9}+\frac{1}{5} \frac{1}{9} =\frac{1}{10}+\frac{1}{5}=\frac{3}{10}$
I hope no errors were made.


Find the radius of the circle? Two Circle of an equal of an radii are drawn  , without any overlap , in a semicircle of radius 2 cm.
If these are the largest possible circles that the semicircle can accomodate , then

what is the radius of each of the circles?

Thanks in advance.

This is a difficult problem if you are expected to give a full formal argument. However, it is intuitively quite reasonable to assume that the "best" circles will be tangent to the diameter of the semicircle, tangent to each other, and tangent to the circle of radius $2$.  Under the assumption that these are indeed the best circles, we will find the answer.  (Alternately, we could look at the problem with a full circle instead of a semicircle, and four small circles, and assume that the symmetrical configuration is best.)

Draw a picture, and concentrate on the small circle $\mathcal{K}$ on the right. Let $O$ be the centre of the semicircle, and let $C$ be the centre of $\mathcal{K}$. Let $T$ be the point where $\mathcal{K}$ is tangent to the semicircle.

Note that the line $OT$ is perpendicular to the common tangent line at $T$. Note also that the line $CT$ is perpendicular to the common tangent line at $T$. It follows that the line through $O$ and $T$ is the same line as the line through $C$ and $T$.  In other words, the radius $OT$ of the semicircle passes through $C$.

Let $r$ be the radius of $\mathcal{K}$. Drop a perpendicular from $C$ to the point $P$ on the diameter of the semicircle.  Note that $OP=r$. By the Pythagorean Theorem, $(OC)^2=(PO)^2+(PC)^2=2r^2$, so $OC=\sqrt{2}r$. It follows that
$$2=OT=OC+CT=\sqrt{2}r +r.$$
Now we are finished. We have $r=\frac{2}{\sqrt{2}+1}$. It may be that in the answer given, the denominator was rationalized by multiplying top and bottom by $\sqrt{2}-1$. That gives the answer $r=2(\sqrt{2}-1)$.

Added: Thanks to David Mitra for the diagram!

By construction, the maximum radius of two circles of equal size in a circle of radius r is 1/2r.  The maximum radius of 4 circles of equal size in a circle of radius x is 1/4r.


Find probability that the sum of two dice is not 6 and not 5? When rolling two fair dice, what is P(sum of two dice is not 6 and not 5)?

Calculation:
First, I found the probability of two numbers that would roll a sum of 6:

(1,5) (2,4) (3,3) (4,2) (5,1)  = 5/36 (each probability is 1/36)

Then, I found the probability of two numbers that would roll a sum of 5:

(1,4) (2,3) (3,2) (4,1) = 4/36

My first method was to add the probabilities and then subtract from one to give the probability of NOT rolling a sum of 6 or 5:

(5/36) + (4/36) = 9/36

1 - 9/36 = 27/36 = 0.75

This turned out to be the accepted answer of my online homework. But then, I realized that the original question asked for P(not sum of 6 AND not sum of 5).

So, I recalculated:

1 - 5/36 = 31/36

1 - 4/36 = 32/36

(31/36)(32/36) = 992/1296 ~ 0.7654 (using multiplication rule)

When I entered this fraction in, it was incorrect. But since the questions asks for "AND" and not "or", wouldn't the second probability be the actual correct answer?

By De Morgan's Laws:

not sum of 6 and not sum of 5 is the same to not (sum of 6 or sum of 5)

The second answer is not correct because "not sum of 6" and "not sum of 5" are not independent. So you cannot write the probability of their intersection as product of the individual probabilities (or as you called it "multiplication rule")

The first answer is correct because "sum of 6" and "sum of 5" are disjoint, and the probability of their union is the sum of the probabilities.

The probability that the sum of two dice is 6 and 5 is 9/36.
So the probability that the sum of two dice is not 6 and 5 is 1- 9/36=27/36.


How to prove that the sum of the areas of triangles $ABR$ and $ CDR$ triangle is equal to the $ADR$? In the convex quadrilateral $ABCD$, which is not a parallelogram, the line passing through the centers of the diagonals $AC$ and $BD$ intersects the segment $BC$ at $R$. How to prove that the sum of the areas of triangles $ABR$ and $CDR$ is equal to the area of triangle $ADR$? I have no idea how to do this. Can this be proved with simple geometry?

This is tricky.

We shall use an easy fact that if we are given fixed points $Y,Z$, and a variable point $X$ that changes linearly, then $[XYZ]$ changes linearly, where $[\mathcal{F}]$ denotes the oriented area of $\mathcal{F}$.

Let $M,N$ be midpoints of $AC, BD$. Using the fact we know that the function $MN \ni X \mapsto [ABX]+[CDX]$ is linear. However $$[ABN]+[CDN]=\frac 12 [ABD] + \frac 12 [CDB] = \frac 12 [ABCD]$$ and $$[ABM]+[CDM]=\frac12 [ABC] + \frac 12 [CDA] = \frac 12 [ABCD]$$ so this function is actually constant. In particular $$[ABR]+[CDR]=\frac 12 [ABCD].$$ This implies that $$[DAR] = [ABCD]-([ABR]+[CDR])=[ABCD] - \frac 12 [ABCD]=\frac 12[ABCD] = [ABR]+[CDR].$$

This link gives a great visual representation of this proof.
https://www.mathsisfun.com/geometry/quadrilaterals-interactive.html.

Analytically, I created a complicated proof by using the equation base*height/2 = area_of_a_triangle and cos(angle)*height = base for each of ADR & CDR & ADR and solved for a common relationship.  I recommend finding another way.


I'm a teenager hoping to become a mathematician, but math isn't my forte. Is it possible? I am a sophomore in high school and have discovered the joys of mathematics.  However,due to sheer irony, at times I find the subject difficult. I am what one could label "lopsided" in my faculties, excelling in languages but always floundering with numbers. This may be biological, for I was psychologically tested and found that my linguistic skills were abnormally high but my spatial reasoning...wasn't (I was diagnosed with Asperger's Syndrome). I do not want this to deter me from pursuing a degree in physics/math and finding work in the field. On the other hand, what if I just don't have the ability and fail? I see the mathematics geniuses at my school (or other students who are just better than me) and I feel horribly discouraged. Recently, I took a math test and felt completely lost, even though I studied diligently and received high scores on other assignments, which may be due to the anxiety I felt as I received the exam. The mere thought of that test kills me, as if it is a reminder that I shall never succeed in the subject no matter how hard I try (I was given a B last semester).  I was wondering if those in advanced mathematics have had the same experiences, and somehow overcame their struggles. Is it optimal I should just study a humanities subject even though this is what I like, or is there some way I could find a path in the field? Also, as an aside, what gives you your passion for math? What is the best thing you find about this subject? Perhaps that can force my motivation to the point where I become proficient...

P.S. If you could point me in the direction of some math/physics related extra-curricular/summer activities and programs, it'd be much appreciated.

The title of your question presents an interesting dichotomy.  It is of course true that a necessary condition to succeed as a mathematician is that one be "strong at mathematics"...but exactly what that means is less clear-cut.  And whatever "strong at mathematics" means, it is something that can be developed (or not) over time: it is not something that either inheres in one's soul or doesn't like some sort of Calvinist election.

  This may be biological, for I was psychologically tested and found that my linguistic skills were abnormally high but my spatial reasoning...wasn't (I was diagnosed with Asperger's Syndrome).

Wondering as a teenager what one may or may not be "biologically capable of" is not something that I would recommend: you are placing too much stock in biology and testing.  (Frankly, I am skeptical even of the Asperger's diagnosis.  I am by no means professionally trained in psychology, but in my opinion there is a faddishness in the diagnosis of such "syndromes".  I believe that Asperger's diagnoses are based on measurable cognitive differences, but worry that the implications of these differences may be exaggerated.  To quote from the wikipedia article on Asperger's: "Some researchers have argued that AS can be viewed as a different cognitive style, not a disorder or a disability, and that it should be removed from the standard Diagnostic and Statistical Manual, much as homosexuality was removed.")  Also there is much, much more to mathematics than "spatial reasoning": I am a research mathematician with some kind of international reputation, but my inherent spatial reasoning abilities are no better than average.  I bought a bookcase a few months ago, but delayed assembling it and seem to have lost the instructions: without them, I think I would rather buy a new bookcase than worry about how to assemble the one I already have!  When it comes to applying spatial reasoning to mathematics, I can see that I have to work harder than many of my peers but I can still do it: for instance, when I discuss finding volumes of revolution in calculus I can always identify a few students who can visualize the regions more easily than I can...but overall I am still better at these problems than almost any calculus student I have ever taught because (i) I have so much more experience with these problems than they do and (ii) there are aspects of solving these problems other than spatial reasoning.  I think that if you want to work in certain aspects of low-dimensional topology then it is important to have strong spatial reasoning skills -- sometimes I have seen talks or read papers in which proofs in these areas are done with the aid of pictures that look very confusing to me -- but even here I think that other skills are just as important or more as naked spatial reasoning.  (In fact I did undergraduate research in low-dimensional topology, really enjoyed it, and would be happy to do more of it some day.)

  I do not want this to deter me from pursuing a degree in physics/math and finding work in the field.

It is good to keep an eye on the future, but if you try to deal with the future as if it were the present then you're bringing a lot of unnecessary stress on yourself....we simply don't know so well what the future will bring.  Having a goal of getting an undergraduate degree in physics or math is a good level of future planning for a high school sophomore.  You can also start looking into the various careers that make use of this degree: there are many, and most of them would not go under the name "mathematician".  The best way to stay on track towards this goal is simply to take challenging math classes, do your best to get what you can out of them, and also feed your interest in mathematics by reading up on whatever outside of class mathematical material interests you.

  On the other hand, what if I just don't have the ability and fail?

We all worry about that.  I still worry about that (maybe my definitions of "ability" and "failure" have been adjusted, but the emotional effect is much the same).  I feel very confident that the world's leading mathematicians have these thoughts as well.  They're helpful up to a point -- the point where they drive us to improve ourselves and have lofty goals -- and then they stop being helpful (at the point where we uselessly worry about our innate abilities or future success rather than try to figure out what to do next).

  Recently, I took a math test and felt completely lost, even though I studied diligently and received high scores on other assignments, which may be due to the anxiety I felt as I received the exam. The mere thought of that test kills me, as if it is a reminder that I shall never succeed in the subject no matter how hard I try (I was given a B last semester).

It happens -- none of this is proof positive that you are not cut out for a mathematical career.  When I was in school I usually did very well in my math classes but not always: I remember one "honors precalculus" course I took: the teacher was just very intense and had a way of making things tricky and complicated.  I think he was probably a very good teacher, but his high ambitions and intensity did not translate well for me: I remember when we started a unit on linear equations, a subject which I had seen in several previous courses and knew was easy to understand.  But somehow he made that material complicated as well: he discussed at least four (!!) different forms of equations of lines, and the exam managed to contain questions on linear equations that were really difficult!  For one of the four quarters I got a $B$ in this course, and it was a bit discouraging.  The next year I took calculus and it was both easier and more interesting than the precalculus course I had taken the year before: but I remember that though I found that most everything in the calculus course came very quickly, I also spent more than two hours a night solving calculus exercises...which must have been much more time than I had put into the "harder" precalculus course.

  I was wondering if those in advanced mathematics have had the same experiences, and somehow overcame their struggles

In the first graduate level mathematics course I took, I studied "the wrong version of the Radon-Nikodym Theorem" for the final exam and was unable to do anything with one out of the three questions.  I got a $B$ in that course and later found out that there had been some real question as to whether I should be allowed to continue with the graduate analysis sequence.  In the end I got an award for being one of the best undergraduate math majors.  Being "strong at mathematics" doesn't mean that you are inhumanly perfect: virtually everyone has at least a few lack-of-success stories to tell.

  Is it optimal I should just study a humanities subject even though this is what I like, or is there some way I could find a path in the field? Also, as an aside, what gives you your passion for math? What is the best thing you find about this subject? Perhaps that can force my motivation to the point where I become proficient...

What I find interesting about your question is that while you sound very confident about your skills in language and the humanities, you seem very intent on pursuing a mathematical field....but you don't really say why you like mathematics better than humanities.  (I'm not blaming you: such things can be hard to explain.)  I can identify with this, because all throughout my pre-collegiate education I had skills in the humanities which were as strong as those I had in mathematics and more consistent: I never had a day where e.g. I tried to read Shakespeare and failed!  Perhaps I had the sense that mathematics was deeper and more challenging than the humanities and was drawn to the idea that there was more clear room for improvement.  As a high school student I learned a bit about calculus and saw that there were interesting things that lay far beyond -- like number theory, which I have been interested in since my sophomore year of high school -- and I'm not sure what is the equivalent in the humanities of "gunning for number theory".  On the other hand I took creative writing as a high school sophomore, enjoyed that immensely and felt approximately the same infinite challenge as I did (and still do) with mathematics.  I am somewhat regretful that I gave that up...although, given that I have more than 2000 pages of "extra" mathematical writing on my webpage, perhaps I have not really given it up so completely.  So I wonder:

What is the right path for a high school student who is extremely talented and ambitious in the humanities?

I didn't know the answer to that question then, and I still don't now...but I wish I did.  You might want to look into it.  If you feel much more confident in your abilities in the humanities, shouldn't you at least think about going into the humanities?

Let me also say that the phrase about "forcing my motivation until I become proficient" worries me a bit.  You also ask what gives mathematicians their passion for math and what they like about it.  To me this sounds a little like someone who is thinking of becoming engaged asking an older married friend exactly what it was about her spouse that made her decide to get married.  If you have to ask, then maybe you should be dating someone else!  My passion for mathematics comes from the fact that I love it...it is not really something that can be further analyzed or explained.  If you love mathematics, spend more time with it and develop your knowledge and skills.  Otherwise keep taking mathematics courses at least until you get to college, but keep your mind open to finding the true object of your affections.  It will be out there somewhere...

Been there, done that. Let me explain...

At high school (I'm from Australia so I don't know what all this sophomore/freshman stuff is.. Senior years here) I just barely scraped through my maths classes. At uni I did bachelors of education and science majoring in mathematics. I did not do very well in the math classes. Most of them I just managed to pull passes or the odd credit ( 4 or 5 from a max of 7 with 4 being the pass). However, the education classes I was getting 6 and 7's for. What was different? The assessment. For the education classes most assessment was essay based; the maths was exam based. I have a very bad short term memory. I cannot multiply or often even add two 2 digit numbers in my head. Not because I don't know how, but because by the time I get to the second number I've already forgotten the first. This problem showed in my work at uni; I understood the content and could do it, but when it came to the exams I really struggled. Especially since most of the course's assessment was 80 - 90% in one final 3 hour exam at the end of the semester.

To cut a long story short, I decided that I liked the maths better than teaching and so started looking for a job in that field after I finished. Now I work as a developer/mathematician on an agricultural simulator for a large research organisation. Now, I'll be the first to admit I'm not a pure mathematician. A lot of what I do is actually data analysis, but as you probably know that includes a lot of mathematical and statistical work. I am quite fluent in the language of mathematics but I still have a desk reference of the meaning of math symbols because I can't always remember them. I can do a fourier or solve a system of differential equations but I need a reference manual to help me. Mostly because I do them so rarely I need to jog my memory.

And therein lies my main point. Many jobs in mathematics (excluding academia of course) don't actually require you to do maths. They require you to understand it. All those algebra equations and integrals and matrices they have you do? After you graduate you will probably never do one by hand again. We have computers to do the heavy lifting now. But you need to understand how the equations and integrals, etc work. You need to be able to set up the problem for the computer and understand the output that comes back. You need to be able to communicate your results to others. That is primarily what (in an ideal world) your classes are setting you up for. The exams where you have to do the calculations yourself from memory are to show that you have that understanding.

Now, having said all that. Do your best to do as well in your classes as possible. While it sucks that two people with the same ability can get wildly differing grades simply due to the form of assessment, that's the world we live in. And up to a point, especially when you're a recent graduate, employers will judge you on the basis of those grades simply because you don't have any work experience behind you yet that you can point to. But in the end, realise that your grades do not have to reflect your ability and as long you understand the content, don't let lower grades discourage you from a career in mathematics. The field is much larger and diverse than most people realise, which is probably another topic in and of itself, really.

So the tl;dr version: I sucked pretty hard at maths through school and university. I didn't expect to get a career in it but I did and I love it. Final grades are not always an indicator of mathematical success. If you understand what you're being taught but have trouble when it comes to assessment that can be for reasons other than a lack of ability. If you enjoy, keep going and you will be successful.

Best of luck,
Justin


Integral $\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right) \mathrm dx$ I need help with this integral:
$$I=\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right)\ \mathrm dx.$$
The integrand graph looks like this:
$\hspace{1in}$
The approximate numeric value of the integral:
$$I\approx8.372211626601275661625747121...$$
Neither Mathematica nor Maple could find a closed form for this integral, and lookups of the approximate numeric value in WolframAlpha and ISC+ did not return plausible closed form candidates either. But I still hope there might be a closed form for it.
I am also interested in cases when only numerator or only denominator is present under the logarithm.

I will transform the integral via a substitution, break it up into two pieces and recombine, perform an integration by parts, and perform another substitution to get an integral to which I know a closed form exists.  From there, I use a method I know to attack the integral, but in an unusual way because of the 8th degree polynomial in the denominator of the integrand.

First sub $t=(1-x)/(1+x)$, $dt=-2/(1+x)^2 dx$ to get

$$2 \int_0^{\infty} dt \frac{t^{-1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} $$

Now use the symmetry from the map $t \mapsto 1/t$.  Break the integral up into two as follows:

\begin{align}
& 2 \int_0^{1} dt \frac{t^{-1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} + 2 \int_1^{\infty} dt \frac{t^{-1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} \\
&= 2 \int_0^{1} dt \frac{t^{-1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} + 2 \int_0^{1} dt \frac{t^{1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} \\
&= 2 \int_0^{1} dt \frac{t^{-1/2}}{1-t} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )}
\end{align}

Sub $t=u^2$ to get

$$4 \int_0^{1} \frac{du}{1-u^2} \log{\left (\frac{5-2 u^2+u^4}{1-2 u^2 +5 u^4} \right )}$$

Integrate by parts:

$$\left [2 \log{\left (\frac{1+u}{1-u} \right )} \log{\left (\frac{5-2 u^2+u^4}{1-2 u^2 +5 u^4} \right )}\right ]_0^1 \\- 32 \int_0^1 du \frac{\left(u^5-6 u^3+u\right)}{\left(u^4-2 u^2+5\right) \left(5 u^4-2 u^2+1\right)} \log{\left (\frac{1+u}{1-u} \right )}$$

One last sub: $u=(v-1)/(v+1)$ $du=2/(v+1)^2 dv$, and finally get

$$8 \int_0^{\infty} dv \frac{(v^2-1)(v^4-6 v^2+1)}{v^8+4 v^6+70v^4+4 v^2+1} \log{v}$$

With this form, we may finally conclude that a closed form exists and apply the residue theorem to obtain it.  To wit, consider the following contour integral:

$$\oint_C dz \frac{8 (z^2-1)(z^4-6 z^2+1)}{z^8+4 z^6+70z^4+4 z^2+1} \log^2{z}$$

where $C$ is a keyhole contour about the positive real axis.  This contour integral is equal to (I omit the steps where I show the integral vanishes about the circular arcs)

$$-i 4 \pi \int_0^{\infty} dv \frac{8 (v^2-1)(v^4-6 v^2+1)}{v^8+4 v^6+70v^4+4 v^2+1} \log{v} + 4 \pi^2 \int_0^{\infty} dv \frac{8 (v^2-1)(v^4-6 v^2+1)}{v^8+4 v^6+70v^4+4 v^2+1}$$

It should be noted that the second integral vanishes; this may be easily seen by exploiting the symmetry about $v \mapsto 1/v$.

On the other hand, the contour integral is $i 2 \pi$ times the sum of the residues about the poles of the integrand.  In general, this requires us to find the zeroes of the eight degree polynomial, which may not be possible analytically.  Here, on the other hand, we have many symmetries to exploit, e.g., if $a$ is a root, then $1/a$ is a root, $-a$ is a root, and $\bar{a}$ is a root.  For example, we may deduce that

$$z^8+4 z^6+70z^4+4 z^2+1 = (z^4+4 z^3+10 z^2+4 z+1)  (z^4-4 z^3+10 z^2-4 z+1)$$

which exploits the $a \mapsto -a$ symmetry.  Now write

$$z^4+4 z^3+10 z^2+4 z+1 = (z-a)(z-\bar{a})\left (z-\frac{1}{a}\right )\left (z-\frac{1}{\bar{a}}\right )$$

Write $a=r e^{i \theta}$ and get the following equations:

$$\left ( r+\frac{1}{r}\right ) \cos{\theta}=-2$$
$$\left (r^2+\frac{1}{r^2}\right) + 4 \cos^2{\theta}=10$$

From these equations, one may deduce that a solution is $r=\phi+\sqrt{\phi}$ and $\cos{\theta}=1/\phi$, where $\phi=(1+\sqrt{5})/2$ is the golden ratio.  Thus the poles take the form

$$z_k = \pm \left (\phi\pm\sqrt{\phi}\right) e^{\pm i \arctan{\sqrt{\phi}}}$$

Now we have to find the residues of the integrand at these 8 poles.  We can break this task up by computing:

$$\sum_{k=1}^8 \operatorname*{Res}_{z=z_k} \left [\frac{8 (z^2-1)(z^4-6 z^2+1)  \log^2{z}}{z^8+4 z^6+70z^4+4 z^2+1}\right ]=\sum_{k=1}^8 \operatorname*{Res}_{z=z_k} \left [\frac{8 (z^2-1)(z^4-6 z^2+1)}{z^8+4 z^6+70z^4+4 z^2+1}\right ] \log^2{z_k}$$

Here things got very messy, but the result is rather unbelievably simple:

$$\operatorname*{Res}_{z=z_k} \left [\frac{8 (z^2-1)(z^4-6 z^2+1)}{z^8+4 z^6+70z^4+4 z^2+1}\right ] = \text{sgn}[\cos{(\arg{z_k})}]$$

EDIT

Actually, this is a very simple computation.  Inspired by @sos440, one may express the rational function of $z$ in a very simple form:

$$\frac{8 (z^2-1)(z^4-6 z^2+1)}{z^8+4 z^6+70z^4+4 z^2+1} = -\left [\frac{p'(z)}{p(z)} + \frac{p'(-z)}{p(-z)} \right ]$$

where

$$p(z)=z^4+4 z^3+10 z^2+4 z+1$$

The residue of this function at the poles are then easily seen to be $\pm 1$ according to whether the pole is a zero of $p(z)$ or $p(-z)$.

END EDIT

That is, if the pole has a positive real part, the residue of the fraction is $+1$; if it has a negative real part, the residue is $-1$.

Now consider the log piece.  Expanding the square, we get 3 terms:

$$\log^2{|z_k|} - (\arg{z_k})^2 + i 2 \log{|z_k|} \arg{z_k}$$

Summing over the residues, we find that because of the $\pm1$ contributions above, that the first and third terms sum to zero.  This leaves the second term.  For this, it is crucial that we get the arguments right, as $\arg{z_k} \in [0,2 \pi)$.  Thus, we have

$$\begin{align}I= \int_0^{\infty} dv \frac{8 (v^2-1)(v^4-6 v^2+1)}{v^8+4 v^6+70v^4+4 v^2+1} \log{v} &= \frac12 \sum_{k=1}^8 \text{sgn}[\cos{(\arg{z_k})}] (\arg{z_k})^2 \\ &= \frac12 [2 (\arctan{\sqrt{\phi}})^2 + 2 (2 \pi - \arctan{\sqrt{\phi}})^2 \\ &- 2 (\pi - \arctan{\sqrt{\phi}})^2 - 2 (\pi + \arctan{\sqrt{\phi}})^2]\\ &= 2 \pi^2 -4 \pi \arctan{\sqrt{\phi}} \\ &= 4 \pi \, \text{arccot}{\sqrt{\phi}}\\\end{align}$$

This is not really an answer, but grossly too long for an comment. I didn't know how to simplify it beyond the final solution.

$$I=\int_{-1}^1 \frac{1}{x}\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2x^2+2x+1}{2x^2-2x+1}\right)\text{d}{x}$$

Begin with the substitution of $x=-\cos2a$
$$I=\int_{-1}^1 \frac{1}{-\cos2a}\sqrt{\frac{1-\cos2a}{1+\cos2a}}\ln\left(\frac{2\cos^2 2a-2\cos 2a+1}{2\cos^2 2a-2\cos2a+1}\right)\text{d}{x}$$

By the tangent and cos double angle properties

$$I=\int_{-1}^1 -\sec2a|\tan a|\ln\left(\frac{-2\cos^22a+\cos 4a+2}{2\cos2a+\cos4a+2}\right)\text{d}{a}$$

Were just getting started. Now replace $a=\frac{1}{2}\text{gd}(b)$ where $\text{gd}$ is the Gudermannian function.

$$I=\int_{-1}^1 -\sec(\text{gd}(b))|\tan(\text{gd}(\frac{b}{2}))|\ln\left(\frac{-2\cos^2(\text{gd}(b))+\cos (2\text{gd}(b))+2}{2\cos^2(\text{gd}(b))+\cos (2\text{gd}(b))+2}\right)\text{d}{a}$$

Hehe. Now we get to simplify a bit. This is under the definition of Gudermannian properties.

$$I=\int_{-1}^1 -\text{cosh}\space b|\sinh\frac{b}{2}|\ln\left(\frac{-2\text{sech}^2 b+(\text{sech}^2b+\tanh^2b)+2}{2\text{sech}^2 b+(\text{sech}^2b+\tanh^2b)+2}\right)$$

Now, use properties of $\tanh$ and $\text{sech} $ to simplify even further

$$I=\int_{-1}^1 -\text{cosh}\space b|\sinh\frac{b}{2}|\ln\left(\frac{(1-\text{sech}^2 b)+2}{(1+\text{sech}^2 b)+2}\right)$$

Our goal is to create an $\text{arctanh}$ function, but that will obviously take some serious effort. Factor out a $3$ to generate that $1$ needed even if it makes an ugly factoring.

$$I=\int_{-1}^1 -\text{cosh}\space b|\sinh\frac{b}{2}|\ln\left(\frac{3(1-\frac{\text{sech}^2 b}{3})}{3(1+\frac{\text{sech}^2 b}{3})}\right)$$

And now cut out all of the 3's. After this cut, use a property of $\ln$'s to reciprocate the argument of $\ln$. And multiply 2 and 1/2

$$I=\int_{-1}^1 2\text{cosh}\space b|\sinh\frac{b}{2}|\frac{1}{2}\ln\left(\frac{(1+\frac{\text{sech}^2 b}{3})}{(1-\frac{\text{sech}^2 b}{3})}\right)$$

And what do you know! You're there! Use a property of $\ln$ and $\text{arctanh}$ to generate a much CLEANER form (also by throwing the 2 in front).

$$I=2\int_{-1}^1 \text{cosh}\space b|\sinh\frac{b}{2}|\text{arctanh}(\frac{\text{sech}^2b}{3})$$

This function is even, and we can know that because all parts of what is above, $\cosh b,|\sinh b|, $ etc. all even. So we can do the following.

$$I=4\int_{0}^1 \text{cosh}\space b|\sinh\frac{b}{2}|\text{arctanh}(\frac{\text{sech}^2b}{3})$$

This is just an idea, and like I said not a real solution. I have no idea where to continue beyond this, but I thought it may help to come up with a new idea to solve.


Probability that at least one of 2 balls taken randomly from a pile of 2 red and 3 black is red I get 2 different answers, depending how I approach this, and I need help to see why the error arises.

One solution is to calculate unfavorable combinations probability and substract from 1:
$$1-\frac{C_3^2}{C_5^2}=\frac{7}{10}$$
The other solution is when calculating favorable combinations, to first choose one of the 2 reds, and then choose one of the remaining 4:
$$\frac{C_2^1 \times C_4^1}{C_5^2} = \frac{8}{10}$$
Which is obviously $\neq \frac{7}{10}$.

The first solution is correct. The second solution makes the mistake of counting twice the scenario where both red balls are picked, which has a $\frac1{10}$ chance of occurring. Subtracting this from $\frac8{10}$ yields the correct answer of $\frac7{10}$.

Pick first red ball R1 and you have 4 possible combinations with the other balls: R1R2, R1B1, R1B2, R1B3
Now pick the second red ball R2 and you have 3 possible combinations left without repeating previous combinations (R2R1): R2B1, R2B2, R2B3
Continue with the first black ball B1: B1B2, B1B3
And the second black ball B2: B2B3
That's it, now you have 7 combinations with at least 1 red ball and 3 combinations with only black balls: 7/10


Math terminology: What are rules regarding hyphens? (Nonzero vs. non-zero) This question is geared toward clarifying terminology in writing math.

Which terms are correct and why?

A set $E$ is non-empty.
A set $E$ is nonempty.

The number $x$ is non-negative.
The number $x$ is nonnegative.

The number $y$ is non-positive.
The number $y$ is nonpositive.

The number $z$ is non-zero.
The number $z$ is nonzero.

As a personal preference, I like the way nonzero looks but I prefer to use non-empty so I don't know what the consistency with hyphens is when prefixing "non". I also like the way nonnegative reads but I feel like non-positive looks better than without a hyphen. I don't have any rhyme or reason why I have these preferences and that's why I'm curious to learn if there is a correct version of each term listed above.

Since many (or a nonnegligible amount) of authors just do not care, I guess it's up to personal preference. However, there is a semantic code, like pointed out here.
Thereafter, the prefix non is not hyphenated. And I really think this should be consistent. I had nonzero, nonempty and nonholonomic occuring in my thesis and it was "corrected" by different supervisors into different versions, mostly where non-holonomic was hyphenated and the rest not. I really think this is bad practice, since consistency is lost by this "to highlight" mentality. If on tries to highlight something, why not explain it and treat it consistently in semantics.

To underline my point consider using $X$ for vectors in one direction. For some other vector, to hightlight that it's not in this direction you use $\alpha$. But why not $Y$ with an explanation?! Therefore, semantics is consistent and the highlights are where they belong, in the "comments".

What got me to enforce this was that every title I cited contained the nonhyphenated versions of all those words. Thus, (from a far too small sample, but of good works) I conclude that this is common practice in this field (between robotics and automation).

Oxford Dictionary also advises to use hyphens for prefixes, which lead to vowel collisions, like pre-advised, although nothing further is stated there.

It seems that positive is better word usage than "nonnegative" or "non-negative"? Also should it be not equal to zero as opposed to "nonzero" or 'non-zero"?


every pdf can be regarded as a marginal distribution of a joint pdf suppose we have functions $g\ge0,h\ge0$ that $\int g \, dx=1 , \int h \, dy = 1$. it means $g$ is pdf for random variable $X$ and $h$ is pdf for $Y$.

now how we can prove that there is function $f(x,y)$ that $f$ is joint pdf of $X,Y$ and its marginals are $g,h$?
i know that $f$ is not unique in general,but how can i say there is one $f$?

thanks

Assume $X,Y$ are independent, then what is the joint density and marginals?

Define $F(x,y)=g(x|y)*h(y)$  where we know both functions $g,h$ exist. Then $F(x,y) = f(x,y)$ by Bayes Rule/


Solving $t^2y''-t(t+2)y'+(t+2)y=2t^3$ knowing $y_1 = t, y_2 = te^t$ are solutions to the homogeneous equation $$t^2y''-t(t+2)y'+(t+2)y=2t^3\implies\\y''-\frac{t+2}{t}+\frac{t+2}{t^2}y = 2t$$

My book says that $y_1 = t, y_2 = te^t$ are solutions to the homogeneous equation.

I'm trying a solution of the form $Y = u_1t+u_2te^t$, so the wronskian is

$$W(t, te^t) = t^2e^t$$

Then, the solutions are given by:

$$u_1 = \int\frac{-te^t\cdot 2t}{t^2e^t} \ dt= -2t$$
$$u_2 =  \int\frac{t\cdot 2t}{t^2e^t} \ dt=-2e^{-t}$$

Then, the solution will be:

$$Y = u_1y_1+u_2y_2 = -2t\cdot t + -2e^{-t}\cdot te^t = -2t^2-2t$$

but my book gives $-2t^2$ as the answer. What am I doing wrong?

t is belong to the homogenous equation : y = c1t + c2te^t
the book only give the particular solution part. http://tutorial.math.lamar.edu/Classes/DE/HOVariationOfParam.aspx

maybe... :)


Help proving half the integers in a complete residue system are quadratic residues The following theorem is what I need to prove:

Let $p$ be an odd prime. Then half the numbers not congruent to $0$ in any complete residue system modulo p are perfect squares modulo $p$ and half are not.

I have done examples that help demonstrate this property:

Consider the complete residue system $B=\{1,2,3,…12\}$ where for each value $B$ in we perform $f(b)=b^2 \space (\text{mod} \space 13)$
$f(1)=(1)^2 \space (\text{mod} \space  13)=1$,   $f(2)=(2)^2 \space (\text{mod} \space 13)=4$, $f(3)=(3)^2 \space (\text{mod} \space 13)=4$,  $f(5)=12$, $f(6)=10$,  $f(7)=10$, $f(8)=12$,   $f(9)=3$, $f(10)=9$,   $f(11)=4$,   $f(12)=1$
We see that $f$ maps all the values in the set $B$ to $F(B)=\{1,4,9,12,10\}$. This set $F(B)$ is our quadratic residues $a_i$. Now observe that only half the values in B exist in $F(B)$. This demonstrates what the theorem was discussing; that only half the integers in a complete residue system represent valid quadratic residues.

I can recognize the pattern but I'm at a complete loss for how to prove this theorem. Before anyone tries to close my question, yes, I have read other answer such as: Show that exactly half of the integers in the set are quadratic residues. But I don't understand the logic of the hint. I don't know how $x^2 \equiv (p-x)^2 \pmod p$ was formulated nor do I see how to use it to prove the theorem. Could someone help guide me through a full proof that deliberatly explains the logic of each step?
Please note I have absolutely no knowledge of Group Theory and I'd prefer to focus on a proof that doesn't use Primitive Roots (because I suck at primitive roots lol). Thank you for any help!

I think that the best explanatory approach is to consider my comments, following your question, against the specific example of $(p=11).$
First consider the values $\{1^2, 2^2, 3^2, 4^2, 5^2\}.$  Is it possible that two of those 5 values have the same residue, $\pmod{11}$?
Suppose that $a,b \in \{1,2,3,4,5\} ~: ~a \neq b.$
Further suppose that $a^2 \equiv_{11} b^2.$
Then, $(a-b)(a+b) = (a^2 - b^2) \equiv_{11} 0.$
This implies that $11 ~| ~(a^2 - b^2) = (a-b)(a+b).$
However, by assumption, $a\neq b \implies 11 ~\not| ~(a-b).$
Further, by assumption, $1 \leq a,b \leq 5 \implies (a+b) < 11 \implies 11 ~\not| ~(a+b).$
Therefore, given that 11 is prime, the assumption that there exists $a,b \in \{1,2,3,4,5\} ~: ~a \neq b$ such that $a^2 \equiv_{11} b^2$ has led to a contradiction.
That is, you have that $11 ~| ~(a-b)(a+b)$, yet at the same time, it is impossible for $11$ to divide either $(a-b)$ or $(a+b)$.  Because $11$ is a prime, you know that whenever $11$ divides $(r \times s)$, then either $11 | r$ or $11 | s.$
As a result of this contradiction, I am free to conclude that $\{1^2, 2^2, 3^2, 4^2, 5^2\}$ represent $5$ distinct residues, $\pmod{11}.$

Addendum
The intent of the above analysis is to show that in general, $\{1^2, 2^2, \cdots, [(p-1)/2]^2\}$ must represent $[(p-1)/2]$ distinct residues, $\pmod{11}$.
The next step is to show, continuing to use $(p=11)$ as an example, that the exact number of quadratic residues must be $[(p-1)/2].$
Consider $6^2 = (11 - 5)^2 = 11^2 - [(2) \times (11) \times (5)] + 5^2.$
Clearly, from the above equation, you have that $6^2 \equiv_{11} 5^2.$
This implies that $6^2$ and $5^2$ must represent the exact same quadratic residue, $\pmod{11}.$
Now consider $7^2 = (11 - 4)^2 = 11^2 - [(2) \times (11) \times (4)] + 4^2.$
From the above equation, you have that $7^2 \equiv_{11} 4^2.$
This implies that $7^2$ and $4^2$ must represent the exact same quadratic residue, $\pmod{11}.$
Now consider $8^2 = (11 - 3)^2 = 11^2 - [(2) \times (11) \times (3)] + 3^2.$
From the above equation, you have that $8^2 \equiv_{11} 3^2.$
This implies that $8^2$ and $3^2$ must represent the exact same quadratic residue, $\pmod{11}.$
Now consider $9^2 = (11 - 2)^2 = 11^2 - [(2) \times (11) \times (2)] + 2^2.$
From the above equation, you have that $9^2 \equiv_{11} 2^2.$
This implies that $9^2$ and $2^2$ must represent the exact same quadratic residue, $\pmod{11}.$
Now consider $(10)^2 = (11 - 1)^2 = 11^2 - [(2) \times (11) \times (1)] + 1^2.$
From the above equation, you have that $(10)^2 \equiv_{11} 1^2.$
This implies that $(10)^2$ and $1^2$ must represent the exact same quadratic residue, $\pmod{11}.$
Therefore, you have that the quadratic residues represented by $\{6^2, 7^2, 8^2, 9^2, (10)^2\}$ must be the exact same $5$ quadratic residues as the residues represented by $\{1^2, 2^2, 3^2, 4^2, 5^2\}.$
Therefore, for $(p=11)$ there are exactly $5$ quadratic residues, among $\{1^2, 2^2, \cdots, (10)^2\}.$
This analysis easily generalizes to any odd prime $p$.
The foundation of this analysis is that if $a$ is any element in $\{1,2,\cdots, [(p-1)/2]\}$, then the following two things will be true.

$(p-a)$ must be an element in $\{[(p-1)/2] + 1, \cdots, (p-1)\}.$

$(p-a)^2 = p^2 - (2)(p)(a) + a^2 \equiv_{p} a^2.$

Therefore the $[(p-1)/2]$ quadratic residues represented by
$\{\langle[(p-1)/2] + 1\rangle^2, \cdots, (p-1)^2\}$
must be the exact same quadratic residues as
$\{1^2, 2^2, \cdots, [(p-1)/2]^2\}.$
Therefore, for any odd prime $p$, there will be exactly $[(p-1)/2]$ distinct quadratic residues.

Note
Focusing on $(p=11)$, the quadratic residues are not represented by $\{1,2,3,4,5\}.$
Instead, the quadratic residues are represented by
$\{1^2, 2^2, 3^2, 4^2, 5^2\}.$

Let me answer your queries in different manner.
$S_p = \{ 0, 1, \dots p - 1 \}$ set of complete residues.

$p - 1 \equiv - 1 \; (\text{mod } p)$
$p - 2 \equiv -2 \; (\text{mod } p)$
$\vdots$
$p - \frac{p -1}{2} \equiv - \frac{p-1}{2} \; (\text{mod } p)$

So we can rewrite our set of complete residues:
$- \frac{p-1}{2}, \dots, -1, 0, 1, \dots, \frac{p-1}{2}$

Check that the new set also contains $p$ many elements.

Now, any square number can be thought as a bucket. Only $0$ goes to a bucket alone and others goes in pair $x$ and $-x$.


Function whose graph is dense in the plane Is there a function $f:\mathbb R\to\mathbb R$ such that for every disc in $\mathbb R^2$ the graph of that function has at least one point that lies inside that disc? I searched for something similar but don't know which phrases to use. The general/another problem is that: does there exist a function $f:\mathbb R\to\mathbb R$ such that its graph in some sense has a positive area or even an infinite area?

As mentioned earlier in the comment, the graph of a discontinuous linear map is dense so this answer your first question.

However, a graph of a function $f:\Bbb R\to\Bbb R$ cannot have positive area. This can be deduced from Fubini's theorem, for example, if we interpret the word "area" to mean the (two-dimensional) Lebesgue measure of $\text{Gr}(f)=\{(x,f(x)) : x\in\Bbb R\}$. Indeed, let $E=\text{Gr}(f)$ then
$$\begin{align}
\mu(E) &= \int_{\Bbb R^2} \chi_{E} d\mu \\
&= \int_{\Bbb R}\int_{\Bbb R} \chi_E(x,y) \,dy\,dx
\end{align}$$
but for each $x\in\Bbb R$ we know that
$$
\int_{\Bbb R} \chi_E(x,y) \,dy = 0
$$
because $\chi_E(x,y)=0$ almost everywhere except the point $y=f(x)$, which doesn't effect the value of the integral. Hence we conclude that $\mu(E)=0 $ so $\text{Gr}(f)$ has zero area.

Remark: The situation is different though if you consider from the start a function $f:\Bbb R\to \Bbb R^2$. The meaning of "graph of $f$" would also be different here: it would merely be the image of $\Bbb R$ under $f$, i.e.
$$
G=\{p\in\Bbb R^2: p=f(x) \text{ for some }x\in \Bbb R  \}.
$$
In this case there are many known space-filling curves that have positive area in $\Bbb R^2$. The most well-known is probably Peano's curve.

I do not know if $g$ as you wish exists, it is very odd if there exists, but there is a curve called "Piano's curve" described as such: $\gamma :[0,1]\to [0,1]^2$ s.t $\forall p\in [0,1]^2\, \,\exists t\in[0,1] : \gamma (t)=p$, i.e, $\gamma$ is space filling curve.
you can obviously extend it to any closed square by shifting and rescaling.
I think it answers your last question, this curve has area of 1.


Mathematical Difference between "there is one" and "there is EXACTLY one" I know that I can say ∃x(P(x)) which means there is at least one x for P(x), but how do I express for exactly one?

Here's the questions:

(a) Not everyone in your class has an internet connection.
(b) Everyone except one student in your class has an internet connection.

So for the first one I wrote:

(a) ∀x∃x(¬I(x))

"For all x there exists an x (or more) such that an x does not have an internet connection" (where I is the state of having an internet connection)

(b) Don't know how to express

I could be wrong please correct me since i'm pretty new to expressing this all mathematically

Thanks for help

You are correct that "There exists ..." means that there exists at least one. To say that there is exactly one you need to say the following:

$$\exists x(\varphi(x)\land\forall z(\varphi(z)\rightarrow z=x)).$$

Namely, there exists $x$ satisfying whatever, and whenever $z$ satisfies whatever, $z$ has to be equal to $x$. Also note the scope of the existential quantifier is over the entire statement.

(As a mathematical example, There exists a natural number which is larger than $1$; but there exists exactly one natural number which is smaller than $1$ (here we take $0$ to be a natural number))

Umm... when I studied math logic back there in my uni, we had exact quantification, expressed as $∃!$

So, let $I(x)$ === "x has internet connection", $P$ === "pupils". Then, a) is represented as

$$
\exists x(x \in P \land \neg I(x))
$$

and b) is represented as
$$
\exists!x(x \in P \land \neg I(x))
$$

Difference between them is only in exactness of second statement: it's not just "not everyone", it's "exactly one" who does not have internet.

We, however, used slightly different notation:

$$
x \in P
$$
$$
\exists ! x : I(x) = false
$$

which, arguably, reads more naturally.


Rotating a circle around the y-axis What is the volume obtained when a circle of radius 2 with center (2,0) is rotated about the y-axis? I keep getting 0, but that can't be right!

The points on the right half of the circle are at distance $x=f(y)=2+\sqrt{4-y^2}$ from the $y$-axis, and the points on the left half are at distance $x=g(y)=2-\sqrt{4-y^2}$ from the $y$-axis.

The volume you are trying to calculate is that of the body you get by revolving $x=f(y)$, $y\in[-2,2]$, about the $y$-axis of which you need to subtract the volume of the body gotten by revolving $x=g(y)$,$y\in[-2,2]$ about the $y$-axis. Do you see why?

This can very easily solved by using simple geometry as follows

In general, if a circle, having a radius $r$, is rotated about one of its tangents then the solid obtained is a $\color{blue}{\text{horn torus}}$ with radius $r$ & mean radius $r_m=r$ measured from the center of circle.

The volume of the horn torus is given as $$V_{\text{horn torus}}=(\pi r^2)(2\pi r)=2\pi^2r^3$$

In this case, the circle of  radius $r=2$ is rotated about the y-axis (i.e. tangent) hence the volume of the solid obtained i.e. horn torus is $$2\pi^2(2^3)=16\pi^2$$$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{volume}=16\pi^2\approx 157.9136704\ \text{cubic units}}}$$


What's wrong with this proof of the infinity of primes? While reviewing an online textbook in abstract algebra for my website—which I'm hoping will go live by the end of the month—one of the exercises in the book inspired me to produce a simple, set theoretic proof of the infinity of primes. But it looks wrong! I can't say why exactly—but something looks off about it! If anyone can spot what's wrong with it, let me know, I'm too  tired to think of it now. If it's NOT wrong, well, you get to publish that for free, go ahead, I won't fight you for credit. But boy, I'd be shocked if no one ever thought of this and it's correct!

  There are countably infinite positive integers by definition. Decompose the positive integers into the following partition: the set of all primes and the set of all composite positive integers. Assume there are finitely many primes. Then there must be infinitely many composite positive integers because a union of finitely many finite sets is finite. By the fundamental theorem of arithmetic, each composite must be a unique product of primes. Since there is a finite number of primes, let's say $n$ primes, then there are at most $n!$ products of primes. Therefore, there must be at most n! composite positive integers. But that means the positive integers are a union of 2 finite sets and must be finite and this is a contradiction!

There has to be something wrong with this proof, but for the life of me I can't see what it is right now. I'm probably going to kick myself when someone points it out—it's probably something really trivial.

Any takers?

There is something wrong with your proof. You claim that since there are only $n$ primes, there are only $n!$ composite numbers, which is not true. Even a single prime, $2$ for example, produces infinitely many composites:

$$2,4,8,16,32,\dots, 2^n, \dots$$

That proof appears almost deliberately confusing.

The logical proof is very simple: There are an infinite number of numbers, therefore there are an infinite number of subsets. Any unbounded subset has an infinite quantity of members.

On the other hand, I can paraphrase an old anti-proof:

There are an infinite number of numbers. But not all numbers are primes, therefore there must be a finite number of primes. Any number divided by infinity is zero, therefore there are no prime numbers.


Integral with spherical symmetry over cube Is it possible to calculate the integral

$$I = \int_{-1}^1 \mathrm dx \int_{-1}^1 \mathrm dy \int_{-1}^1 \mathrm dz \frac{1}{x^2 + y^2 + z^2}$$

analytically? I tried using spherical coordinates

$$I = \int_0^{r(\vartheta,\varphi)} \mathrm dr \int_0^\pi \mathrm d\vartheta \int_0^{2\pi} \mathrm d\varphi \sin(\vartheta) \;,$$

but I couldn't come up with a proper expression for $r(\vartheta,\varphi)$, which is the radial component for points on the boundary of the cube $[0,1]^3$.

I. Spherical coordinates

Let's try to do this in spherical coordinates by brute force and see what happens.
$$I = 16\int_R d \Omega \ d r,$$
where $R$ is the region for which $0\leq \theta\leq \pi/2$ and $0\leq\phi\leq \pi/4$.
This region splits into two parts.

In region 1, $0\leq\theta \leq \theta'$ and we integrate up to $z=1$, so $0\leq r \leq 1/\cos\theta$.

In region 2, $\theta' \leq\theta \leq \pi/2$ and we integrate to $x=1$, so $0\leq r \leq 1/(\sin\theta\cos\phi)$.

Here $\theta'$ is a function of $\phi$, $\tan\theta' = \sqrt{1+\tan^2\phi}$.
Notice that $\cos\theta' = 1/\sqrt{2+\tan^2\phi}$.

The integrals over region 1 and 2 are not elementary,
$$\begin{eqnarray*}
I_1 &=& 16 \int_0^{\pi/4} d\phi  \int_0^{\theta'} d\theta \ \sin\theta \int_0^{1/\cos\theta} dr \\
&=& 8  \int_0^{\pi/4} d\phi \ \ln(2+\tan^2\phi) \\
%%%
I_2 &=& 16 \int_0^{\pi/4} d\phi  \int_{\theta'}^{\pi/2} d\theta \ \sin\theta \int_0^{1/(\sin\theta \cos\phi)} dr \\
&=& 16 \int_0^{\pi/4} d\phi \ \sec\phi \ \left(\frac{\pi}{2} - \theta'\right) \\
&=& 8\pi\ln(1+\sqrt2) - 16 \int_0^{\pi/4} d\phi \ \sec\phi \ \tan^{-1}\sqrt{1+\tan^2\phi}.
\end{eqnarray*}$$
It is possible to go further with these integrals, but they are pretty ugly.
Numerically they give $15.3482\cdots$.
Let's try another approach.

II. Divergence theorem

Let's put together the steps in the comments and make it obvious our final answer is real.

Using the divergence theorem for ${\bf F} = \hat r/r$ we find
$$I = 24\int_0^1 d x \int_0^1 d y \frac{1}{x^2+y^2+1},$$
and so, going to polar coordinates,
$$\begin{eqnarray*}
I &=& 48\int_0^{\pi/4} d \phi \int_0^{1/\cos\phi} d r \ \frac{r}{r^2+1} \\
&=& 24\int_0^{\pi/4} d\phi \ \ln(1+\sec^2\phi).
\end{eqnarray*}$$
This integral is nontrivial.

Let us try a series approach and expand in small $\phi$.
We find
$$\begin{eqnarray*}
I &=& 6\pi \ln 2 + 24\int_0^{\pi/4}d \phi \ \left[\ln\left(1-\frac{1}{2}\sin^2\phi\right) - \ln(1-\sin^2\phi)\right] \\
&=& 6\pi \ln 2 + 12\sum_{k=1}^\infty \frac{1}{k}\left(1-\frac{1}{2^k}\right) B_{\frac{1}{2}} \left(k+\frac{1}{2},\frac{1}{2}\right)
\end{eqnarray*}$$
where $B_x(a,b)$ is the incomplete beta function.
The $k$th term of the sum goes like $1/k^{3/2}$.
Notice that $6\pi \ln 2 \approx 13$ so the ``zeroeth'' term is already a pretty good approximation.

Mathematica gives a result that doesn't appear explicitly real, but it can be massaged into
$$I = 24 \mathrm{Ti}_2(3-2\sqrt2) + 6\pi \tanh^{-1}\frac{2\sqrt2}{3} - 24 C,$$
where $\mathrm{Ti}_2(x)$ is the inverse tangent integral, with the series
$$\mathrm{Ti}_2(x) = \sum_{k=1}^\infty (-1)^{k-1} \frac{x^{2k-1}}{(2k-1)^2},$$
and $C$ is the Catalan constant.

The set of limits corresponds to a sphere of radius $1$ ($x$ ranges from $-1$ to $+1$; $y$ ranges from $-1$ to $+1$; and $z$ ranges from $-1$ to $+1$). Therefore we successively integrate:
w.r.t.theta between zero and $\pi$;
 w.r.t. phi between zero and $2\pi$;
and w.r.t. $r$ bet. zero and $1$ (radius vector extends from the origin to $1$).
Thus we get $4\pi$ for the answer.


rewrite logical connectives in terms of not and and so im supposed to rewrite $P\lor Q$, and , $P \Rightarrow Q , P \Leftrightarrow Q$ in terms of $\lnot$ and $\land$

this is what i got

$P \lor Q = ¬ P \land ¬ Q$

$P \land Q = P~Q$ (same thing?)

$P ⇒ Q = Q \lor ¬P$

$P ⇔ Q = ¬ P \land ¬Q$

those loook right?

The first three are right, the fourth should be: $$P \iff Q \equiv (P \land Q) \lor (\lnot P \land \lnot Q) \equiv \lnot (\lnot (P \land Q) \land \lnot (\lnot P \land \lnot Q))$$

almost correct, it is just the opposite
yes, the same thing ($P\land Q$)
go further, use 1. to express it by $\land,\ \lnot$
not correct. use 3. and $P\Leftrightarrow Q = (P\Rightarrow Q)\land(Q\Rightarrow P)$  (and probably simplify if possible)


Let $r$ be a root of the polynomial $p(x) = (\sqrt{5} - 2\sqrt{3})x^3 + \sqrt{3}x - \sqrt{5} + 1$. Find another polynomial $q(x)$ with integer coefficients such that $q(r) = 0$.

I have no clue how to do this question. Can't use rational root theorem and I see no feasible way to get the roots of $p(x)$. Any help would be appreciated.

Start by collecting all the like-terms with a particular radical and moving them to the left-hand-side; I chose $\sqrt{3}$ first. Move all terms without $\sqrt{3}$ to the right-hand-side. Then, factor out the chosen radical. This is shown in $(1)$. Next, square both sides, as shown in $(2)$. Since all terms with $\sqrt{3}$ have been collected, we've taken $\sqrt{3}$ out of the equation. Although, we're still left with all other radicals ($\sqrt{5}$ in this case).

Next, collect all terms with the next radical ($\sqrt{5}$) on one side of the equation. As before, factor out that radical, as shown in $(3)$. And once again, square both sides, as shown in $(4)$. You're left with a polynomial with coefficients in $\mathbb{Z}$. I'll leave the simplification up to you.

$$\begin{aligned}
&(\sqrt{5}-2\sqrt{3})x^3+\sqrt{3}x-\sqrt{5}+1=0
\\
\\&\bigg(-2x^3+x\bigg)\sqrt{3}=\bigg(1-x^3\bigg)\sqrt{5}-1&(1)
\\
\\&3\bigg(-2x^3+x\bigg)^2=5\left(1-x^3\right)^2-2\sqrt{5}\left(1-x^3\right)+1&(2)
\\
\\&3\bigg(-2x^3+x\bigg)^2-5\left(1-x^3\right)^2-1=-2\sqrt{5}\left(1-x^3\right)&(3)
\\
\\&\left(3\bigg(-2x^3+x\bigg)^2-5\left(1-x^3\right)^2-1\right)^2=20\left(1-x^3\right)^2&(4)\end{aligned}$$


Compute definite integral. I am trying to compute the integral:
$$\int_0^{2\pi}\frac{d\theta}{r-cos(\theta)}$$
I tried to set $u = \tan(\frac{\theta}{2})$ but is it impossible because $\tan(\frac{2\pi}{2}) = \tan(\pi) = 0$

I'll appreciate some guidance.

HINT:

Inasmuch as the cosine function is even and $2\pi$-periodic, we have

$$\int_0^{2\pi }\frac{1}{r-\cos(\theta)}\,d\theta=\int_{-\pi}^\pi\frac{1}{r-\cos(\theta)}\,d\theta=2\int_0^\pi \frac{1}{r-\cos(\theta)}\,d\theta$$

For $r>1$ there is no problem that $\tan \theta$ is ever $0$, and then the solution is:

$$\frac{2 \pi}{\sqrt{r^2 - 1}}$$


Why is the number of bee routes between $5$ flowers not equal to $5!$ On a BBC program about algorithms, they talked about solving the travelling salesman problem. They used the analogy of a bee hive with five flowers to visit before returning to the hive? They said that there were 60 possible routes. Am I missing something or should it not be 5! = 120?

You have not missed anything.

Including the beehive, there are $6$ nodes, and the total number of paths for $n$ nodes is given by $(n-1)!$, so it has to be $120$ for this case.

See under "Difficulty" here

Actually, the real difficulty of this type of problem is finding the shortest route, and trying out all possible routes becomes prohibitive as $n$ increases.

PS:

We have to count number of paths from hive (H) to flowers and back
For 3 flowers, say, [ mind it, $n = 4,$ possible routes are

$H-1-2-3-H,\; H-1-3-2-H,\; H-2-1-3-H,$

$H-2-3-1-H, \;H-3-1-2-H,\; H-3-2-1-H,$

$6$ or $3! = (n-1)!$

PPS:

So the number of routes with $5$ flowers = $120$,

but the number of routes of the same length (assuming reverse route doesn't change distance ) = $60$

but the question as given here didn't ask that.


Why are these two expressions equivalent? Let $X$ be a topological space and let $A\subseteq X$. Can someone please tell me
1) what the difference between these two expressions is
2) if they are equivalent (if they aren't can someone give me two counterexamples in which in turn each expression holds and the other doesn't?)
3) how one would formulate the first expression in words ?

$ \forall x \in X: \ \ \forall (x_n)_n$ in $A:\ x_n \rightarrow x \ \Rightarrow x \in A$

$ \forall x \in X: \ \ \exists (x_n)_n$ in $A:\ x_n \rightarrow x \ \Rightarrow x \in A$

(The motivation behind this question is that in a course in topology which I took, the first expression was used in the following theorem that stated that $A$ was closed iff the first expresion would hold:

"Let $X$ be a first countable topological space, $A\subseteq X$ and $x \in X$.
Then we have: $A$ is closed $\Leftrightarrow$ $\forall (x_n)_n$ in $A,\ x_n \rightarrow x \ \Rightarrow x \in A$ "

And the second expression was used in the following theorem that stated that what the closure of $A$, $c(A)$, would look like:

"Let $X$ be a first countable topological space, $A\subseteq X$ and $x \in X$. Then we have: $c(A)= { x\in X |\  \exists (x_n)_n \textrm{ in }A:\ x_n \rightarrow x } $"

So my guess was, that the above expressions should also be equivalent formally (if we take the st theoretic notation from the second theorem away), since they express the same thing - so I think the proof of the equivalence should not use the fact that our topological space is actually first countable and one of the expression is identical to the fact that $A$ is closed.

Please take note, that this is how the theorems were written down on the blackboard; that is all I have (even if it may be incorrect).

Since I'm not very good with formal notation I may have written the above expression falsely.)

The punctuation of the formulas in the question is very odd, and probably not enough to give a unique way of reading the formula.  But here is what the formulas mean if they are read with the usual conventions for quantifiers.

The formula
$$
(\forall x \in X)(\forall (x_n) \in A)[ (x_n) \to x \Rightarrow x \in A]
$$
says that $A$ contains all its sequential adherent points.

The formula
$$
(\forall x \in X)(\exists (x_n) \in A)[ (x_n) \to x \Rightarrow x \in A]
$$
says that for all $x \in X$ there exists a sequence $(x_n)$ of points of $A$ such that if $(x_n) \to x$ then $x \in A$. With the $T_1$ axiom this is simply equivalent to $A$ being nonempty:

If $A$ is nonempty, suppose I am given an $x \in X$. If $x \in A$ let $x_n = x$ for all $n$. Then $(x_n)\to x$ and $x \in A$, so the formula is true. Otherwise, $x \not \in A$. Fix some point $a \in A$ and let $x_n = a$ for all $n$. Then $(x_n) \to a$, so $(x_n) \not \to x$ by $T_1$, which means the formula is true. Either way, if $A$ is nonempty there is a sequence $(x_n)$ that works.
If $A$ is empty, let $x$ be a point of the space. Then there is no sequence $(x_n)$ of points in $A$, so the formula is false.

It is true that the first formula,
$$
(\forall x \in X)(\forall (x_n) \in A)[ (x_n) \to x \Rightarrow x \in A]
$$
is equivalent to the following formula
$$
(\forall x \in X)[((\exists (x_n) \in A)[ (x_n) \to x]) \Rightarrow x \in A]
$$

This is an instance of a general scheme:
$$
((\exists z)\Psi) \Rightarrow \Phi) \Leftrightarrow (\forall x)(\Psi \to \Phi)
$$
which holds if $z$ is not mentioned in $\Phi$ (in this case $z$ is $(x_n)$).

Perhaps the point I am making is that the parentheses are very important, because it makes a difference whether the quantifier is in the hypothesis of a conditional statement, or whether the entire conditional statement is in the scope of the quantifier.

If I'm not misunderstanding the correct meaning of your questions it seems to me quite obvious that they are not the same sequence in general but you have to clarify what you really mean with them.

In particular the second sentence, as I read it, means that every time you have a sequence of elements in $A$ converging at some point $x$ then this point is in $A$. This is true if you have a closed set but in general is not (take $1/n$ in $(0,1)$).

The first sentence (again as I read it) means that if every sequence in $A$ converges to the same point $x$ then this point is in $A$. This seems a little bit confusing because you can always find in a set which as more than one element two different sequence (the constant ones) converging to different points and so the hypotesis of your impications is almost always false.

---Edit---
From the new edit point of view is seems that the the sequences should be red as:

characterize closed set among others via convergent sequences: every time you've got a convergent sequence the limit is a point of the set in which the sequence lays.
characterize the closure of a set $A$. In particular you get the definition of closure (as you have said) taking the second $A$ in the sentence being $c(A)$.

Now, turning to the equivalence if you are asking about logic equivalence clearly there is a gap between the two sentences (not only referring to the $\forall$ and $\exists$ sa my previous answer was about): in particular you are comparing the second part of an equivalence (as in the characterization of a closed set) and a part of the definition of a set. So imo strictly logically speaking you cannot get an equivalence.

Topologically you have to care about separatedness axioms like said above and so on...

I hope I've undestood (better now) your question and that my answer is a little claryfing for you.


In a $\triangle ABC,$ If $\cot A+\cot B+\cot C =\sqrt{3},$ Then prove that $\triangle$ is equilateral. In a $\triangle ABC,$ If $\cot A+\cot B+\cot C =\sqrt{3},$ Then prove that $\triangle$ is equilateral.

$\bf{My\; Try::}$ Using Jensen's Inequality,

Let $f(x)=\cot x\;,$ Where $x\in (0,\pi),$ Then $f'(x) = -\csc^2 x$ and $f''(x) = 2\csc^2 x\cdot \cot x$

So we get $\displaystyle f''(x) = \frac{2\cos x}{\sin^3 x}>0$ in $\displaystyle x\in (0,\pi)-\left\{\frac{\pi}{2}\right\}$

So $$\displaystyle\frac{\cot A+\cot B+\cot C}{3}\geq \cot\left(\frac{A+B+C}{3}\right)=\frac{1}{\sqrt{3}}$$

So we get $\cot A+\cot B+\cot C\geq \sqrt{3}$

But i did not understand how can we prove that $\triangle$ are equilateral

Thanks

You have completed the sum almost all by yourself. That's a really good try using Jensen's inequality.

For a real convex function φ, numbers $x_1$, $x_2$, ...,$ x_n$ in its domain, and positive weights $ai$, Jensen's inequality can be stated as:
$$\varphi\left(\frac{\sum a_i x_i}{\sum a_i}\right) \le \frac{\sum a_i \varphi (x_i)}{\sum a_i} \qquad\qquad(1)$$
and the inequality is reversed if φ is concave, which is
$$\varphi\left(\frac{\sum a_i x_i}{\sum a_i}\right) \geq \frac{\sum a_i \varphi (x_i)}{\sum a_i}.\qquad\qquad(2) $$
Equality holds if and only if $x_1=x_2=\cdots =x_n$ or $φ$ is linear.

It is given here that $$\cot A+\cot B+\cot C=\sqrt3$$

Now, observe that $\cot x$ is not linear.
So the equality in Jensen's inequality holds only if
$$\cot A=\cot B=\cot C$$ or in other words, $$A=B=C$$
It is trivial to conclude now that $A=B=C=60^\circ$ and hence the triangle is equilateral.


Evaluating the definite integral $\int_0^\infty \mathrm{e}^{\sum_ia_ix^{n_i}}\mathrm{d}x $ As a follow-up to a question on evaluating the definite integral $\int_0^\infty \mathrm{e}^{-x^n}\,\mathrm{d}x$, I wish to know if there is a general analytic solution to the related integral where $-x^n$ is replaced by a polynomial of arbitrary degree, namely $$ \int_0^\infty \mathrm{e}^{\sum_ia_ix^{n_i}}\mathrm{d}x $$ for $n_i\in\mathbb{Z}$ and where the individual coefficients $a_i\in\mathbb{R}$ can be positive or negative, but in a manner such that the argument of the exponent (i.e,. the polynomial) is negative.

ALL the $ a_{i} $ are negative aren't they ??.. otherwise the integral would be DIVERGENT

from the property of the exponential $ f(a+b)=f(a)f(b) $ and the fact that

$ \int_{0}^{\infty}dxexp(-x^{n}) = \int_{0}^{\infty}duu^{1/n-1}exp(-u)= \frac{1}{n}\Gamma(1/n)$

so your itnegral will be equal to the product $ \prod _{i}\frac{1}{a_{i}^{1/n_{i}}}\Gamma(1/n_{i}) $ or similar.. :)


Complex number and exponent If $z = -1+ i\sqrt{3}$

Is it possible that to prove by using induction  $z^{2n}+2^n\cdot z^n+2^{2n}=0$ if $n$ is not multiple of $3$.

I know other way of proving it.

Hint: $z^{3n}-2^{3n}=0$ by using Demoivre's theorem.

Notice, $$z=-1+i\sqrt{3}=2\left(-\frac{1}{2}+i\frac{\sqrt 3}{2}\right)$$
$$=2\left(\cos\frac{2\pi}{3}+i\sin \frac{2\pi}{3}\right)$$

Now, we have $$z^{2n}+2^nz^n+2^{2n}=\left(2\left(\cos\frac{2\pi}{3}+i\sin \frac{2\pi}{3}\right)\right)^{2n}+2^n\left(2\left(\cos\frac{2\pi}{3}+i\sin \frac{2\pi}{3}\right)\right)^n+2^{2n}$$ $$=2^{2n}\left(\cos\frac{4n\pi}{3}+i\sin \frac{4n\pi}{3}\right)+2^{2n}\left(\cos\frac{2n\pi}{3}+i\sin \frac{2n\pi}{3}\right)+2^{2n}$$ $$=2^{2n}\left[\cos\frac{4n\pi}{3}+i\sin \frac{4n\pi}{3}+\cos\frac{2n\pi}{3}+i\sin \frac{2n\pi}{3}+1\right]$$
$$=2^{2n}\left[\left(\cos\frac{4n\pi}{3}+\cos\frac{2n\pi}{3}\right)+i\left(\sin \frac{4n\pi}{3}+\sin \frac{2n\pi}{3}\right)+1\right]$$

$$=2^{2n}\left[2\cos(n\pi)\cos\left(\frac{n\pi}{3}\right)+2i\sin(n\pi)\cos\left(\frac{n\pi}{3}\right)+1\right]$$
$$=2^{2n}\left[2\cos(n\pi)\cos\left(\frac{n\pi}{3}\right)+1\right]$$

$\forall\quad n\neq 3m$ where $m$ is an integer.

step 1: setting $n=1$ in the above expression, we get $$2^{2}\left[2\cos(\pi)\cos\left(\frac{\pi}{3}\right)+1\right] $$ $$=4\left[2(-1)\frac{1}{2}+1\right]=0$$ Hence, the given equality is true for $n=1$

Let's assume that it holds for $n=k$ then we get

$$2^{2k}\left[2\cos(k\pi)\cos\left(\frac{k\pi}{3}\right)+1\right]=0$$
$$\iff 2\cos(k\pi)\cos\left(\frac{k\pi}{3}\right)=-\frac{1}{2}$$
3. Substituting $n=k+1$ in the above equality, we get
$$=2^{2(k+1)}\left[2\cos((k+1)\pi)\cos\left(\frac{(k+1)\pi}{3}\right)+1\right]$$
$$=4\cdot2^{2k}\left[\cos\left(k\pi+\pi\right)\cos \left(\frac{k\pi}{3}+\frac{\pi}{3}\right)+1\right]$$

$$=4\cdot2^{2k}\left[-\cos(k\pi)\left\{\cos\frac{k\pi}{3}\cos\frac{\pi}{3}-\sin\frac{k\pi}{3}\sin\frac{\pi}{3}\right\}+1\right]$$
$$=4\cdot2^{2k}\left[-\cos(k\pi)\left\{\cos\frac{k\pi}{3}\frac{1}{2}-\sin\frac{k\pi}{3}\frac{\sqrt 3}{2}\right\}+1\right]$$
$$=4\cdot2^{2k}\left[-\frac{1}{2}\cos(k\pi)\cos\frac{k\pi}{3}+\frac{\sqrt 3}{2}\cos (k\pi)\sin\frac{k\pi}{3}+1\right]$$ substituting value from (2)
$$=4\cdot2^{2k}\left[-\frac{1}{2}\left(-\frac{1}{2}\right)+\frac{\sqrt 3}{2}\cos (k\pi)\sin\frac{k\pi}{3}+1\right]$$  $$=4\cdot2^{2k}\left[\frac{\sqrt 3}{2}\cos (k\pi)\sin\frac{k\pi}{3}+\frac{5}{4}\right]\neq 0$$ Hence, we find from (1), (2) & (3) that the given equality doesn't hold for all positive integers $n\geq 2$


Population mean and expected value = µ? What is the difference between µ when being the population mean, and µ when being the mean or the expected value?

What confuses me is that the same letter is being used to describe two different metrics- or are they? I am solely basing this question on their equation which isn't the same, so where am I missing the point?

There is a difference between the population mean and the sample mean.

Suppose you have a population of a hundred birds. The weight of the birds is normally distributed about a mean of 10kg with a standard deviation of 1kg.

Now suppose you take a random sample of these hundred birds, and you select ten. Unfortunately, your random sample was not very representative of the population, and you ended up picking the lighter ones. The weight of the sampled birds is normally distributed about a mean of 8kg with a standard deviation of 1kg.

Thus, the population of birds has a mean of 10kg. But the sample we have taken from the population of birds has a mean of 8kg.

In terms of notation, we usually denote a population mean with $\mu$, and a sample mean with $\bar{x}$.

The expected value is the population mean. See the Wikipedia article, which is quite informative. The idea here is that when you randomly select a sample, you "expect" the mean of the sample to approximate the population mean as the size of the sample becomes arbitrarily large.

The question you ask is not a trivial one but in fact it is very profound. The fact that you're confused means that you are the kind of person who thinks deeply, which is very commendable.

The source of your confusion is that the population mean,µ, can be understood in two very distinct ways. The simple definition of it is based on an elementary calculation:

formula for population mean,

where N is the size of the population and Xi are it's elements (or observations).

The idea of expected value, however, is a bit more subtle. Imagine drawing an observation from this population repeatedly with replacement i.e. you draw an observation, then place it back in the population, randomize and draw a second observation, and so on. If you make infinitely many draws, then the mean of these infinitely many draws is defined as their expected value. And it can be shown that this expected value exactly equals µ, the population mean.

Note that in the discussion so far I haven't used the idea of the sample mean at all, which shows that there is no need to use the concept of the sample mean to understand the concept of mathematical expectation.

However, it can additionally be shown that the expectation of infinitely many means, each based on a fixed sample size n, is also equal to µ. In other words, you can show that in a repeated sampling sense the sample mean, on average, equals the population mean, which is why it is considered to be an unbiased estimator of the population mean.

Please let me know if you have any follow up questions.


Show that the boundary of a set equals the boundary of its complement $\newcommand{\bdy}{\operatorname{bdy}}$

I'm trying to show that $\bdy(A) = \bdy(A^c)$.

I know that $\bdy(A) = \operatorname{closure} A \setminus \operatorname{int}(A) = (\operatorname{int}(A^c))^c \setminus \operatorname{int}(A)$, but I don't know where to go from there.

Any help or hints would be very much appreciated.

If you know that $\mathrm{closure}(A)=(\mathrm{int}(A^c))^c$, then you also have  $\mathrm{closure}(A^c)=(\mathrm{int}(A))^c$ because $(A^c)^c=A$.  Therefore

$$\mathrm{boundary}(A)=\mathrm{closure}(A)\cap(\mathrm{int}(A))^c = \mathrm{closure}(A)\cap \mathrm{closure}(A^c).$$

The last expression is symmetric in $A$ and $A^c$.

To show that the boundary of a set A is equal to the boundary of it complement , it suffices to show that there exist an element x in between the boundaries of $A$ and $A^c$ .
So if $x$ belongs to boundary of $A$,
This implies there exist $y$ and $z$ such that $y$ belongs to $A$ and $z$ belongs to $A^c$ .
This implies bd $(A)=x=bd (A^c)$. Hence $bd (A)=bd(A^c)$ .


Building $A_k$ such that $\displaystyle\bigcup_{k\in \mathbb{N}}A_{k} = \mathbb{R}$ and $\displaystyle\bigcap_{k\in \mathbb{N}}A_{k} = \mathbb{N}$ I wanted to know how I can find a sequence of sets according to a particular union and intersection of these sets? For example:

Build a sequence of sets such that
$$\bigcup_{k\in \mathbb{N}}A_{k} = \mathbb{R}$$
and
$$\bigcap_{k\in \mathbb{N}}A_{k} = \mathbb{N}$$

Is there a special method or just guess and try?
Thank you in advance.

Example 1

Take $A_1=\mathbb R$, $A_2=A_3=...=\mathbb N$, and we are done !

Example 2

A probably more interesting example, take for all $k$, $$A_k=\bigcup_{i=-\infty }^\infty \left[i-\frac{1}{k},i+\frac{1}{k}\right].$$

The case with $\mathbb R$/$\mathbb N$ is just an example, but the general case is just as easy.

The union consists of any element that is in just one of the $A_k$'s, so set $A_1$ to be that union.

All your $A_k$'s have to contain the intersection, so set the rest of the $A_k$'s to be the intersection. - And you're done.


How to show $\lim_{n\to\infty}\sum_{k=1}^n(1+(k)/n)^3(1/n)=15/4$? How would I show $$\lim_{n\to\infty}\sum_{k=1}^n(1+(k)/n)^3(1/n)=15/4?$$

My attempt is using the Riemann Sum technique. We know $(1+(k)/n)^2=f(\zeta_k)$ and $(1/n)=\Delta x$. So the definite integral goes from $a$ to $b$ where $b-a=1$.

I think I should find the formula for $$\sum_{k=1}^nk^3.$$ This would be $1+8+27+\cdots+n^3$... but then I get stuck.

Hint:

$$S_n:=\sum_{k=1}^n k^3$$ must be a quartic polynomial in $n$ because $S_{n}-S_{n-1}=n^3$ is a cubic polynomial.

Now

$$S_n-S_{n-1}=an^4+bn^3+cn^2+\cdots-a(n-1)^4-b(n-1)^3-c(n-1)^2-\cdots
\\=4an^3+bn^3-bn^3+\cdots$$ and lower order terms. You conclude that

$$S_n\sim\frac{n^4}4$$ and this is enough for you to evaluate the limit.


Can I use my powers for good? I hesitate to ask this question, but I read a lot of the career advice from MathOverflow and math.stackexchange, and I couldn't find anything similar.

Four years after the PhD, I am pretty sure that I am going to leave academia soon. I do enjoy teaching and research, but the alpha-maleness, massive egos and pressure to publish are really unappealing to me, and I have never felt that I had the mathematical power to prove interesting results. However, I am really having trouble thinking of anything else to do. Most people seem to think that the main careers open to mathematicians are in banking and finance. I really want to work in some field where I can use mathematics, but it is also important to me to feel like I am contributing something positive or at least not actively doing harm. For this reason, financial speculation is very unappealing to me, although I do find the underlying mathematics quite fascinating.

Here is my question: what careers which make a positive contribution to society
might be open to academic mathematicians who want to change careers?

If you are in the US, there are several thousand institutions of higher learning, and at many of them there is very little "pressure to publish". At others, the "pressure to publish" can be met by publishing a textbook or some work of scholarship that does not require proofs of interesting (original) results. High schools also need qualified Mathematics teachers. Consider staying in academia, just moving to a different part of it, as an option for using your powers to do good.

I suspect, but cannot be sure, that much of what I've written applies outside the US as well.

The question you seem to be asking is "Can I be paid to do research, without any formal education, if I am very good?" The likely answer is no.  I know many serious researchers across countless fields, and the only ones I am aware of who can make a living without any formal training are inventors who got quite lucky in terms of their ideas and their marketability. Many great artists, mathematicians, and scientists did other things to survive while doing the work they were truly passionate for, and we romanticise many very inspirational people. Research is probably much more a function of diligence and persistence, even when things are depressingly boring, than it is of intelligence or genius. I would suggest reading some of the good bits of advice many modern prodigies have given on the subject. Terence Tao in particular has several good pieces on accelerated education, working hard, and being a genius.

Yes, school can be terribly dull if you are quick. But being able to succeed in that environment is an invaluable human skill, which will prove useful in a research career when looking for funding, or when setting up research groups and similar things, as well as really showing that you have a sense of humility. We can complain about how slow and tedious bureaucracy can be, but if you really are quite good, it is worth your time to learn and understand how to survive in it. Mathematics is not simply something devoid of social interaction or human involvement. Pure and applied mathematics stem from human ideas and are inherently connected to social and cultural concepts, and these connections are often a part of what traditional formal education gives us. I would absolutely advise you to go further than what you see in a classroom, do independent and guided research as early as possible, and test out of whatever classes you can. But skipping the process entirely will handicap your ability to contribute meaningfully to mathematical research.


Integrals on unlimited sets How do you evaluate this expression
$$
\left| \int_{1}^{\infty} 1 \; dx - \int_{1}^{\infty} 1 \; dx \right| \quad ?
$$
Using improper integral definition, this should be an indeterminate $\infty - \infty$ form.

In general, if we integrate on a generic set $A$, when evaluating
$$
\left| \int_{A} f \; dx - \int_{A} g \; dx \right|,
$$
should we make sure that it is not an indeterminate form?

ADDED: probably, in the first case, rules of limits allow to affirm that that difference is $0$. In the second, if improper integrals appears, we can do the same having a well defined result.

You cannot evaluate that expression, for $\infty - \infty$ makes no sense.

Improper integrals of these types are defined in the following way.
$\int_a^{\infty}f(x)dx=\lim\ _{b\rightarrow+\infty}\int_a^bf(x)dx $ assuming $f$ in integrate on $[a,+\infty)$ In your case the answer is zero.if we agree that $\int f+g=\int f+\int g$. Basically infinities cancel each other out.For more about improper integrals check the link out. http://en.wikipedia.org/wiki/Improper_integral
If we wanted to talk generic case weed would have to know about measure theory . Please see the awesome Wikipedia link


Condition for derivative sequence to converge? Let $E_n(R)$ be a sequence of function that converge to $E(R)$, When we can said that $\dot{E}_n(R)$ converge to $\dot{E}(R)$.

I can assume:

  uniform converge of $E_n$ to $E$.
  $E(R)$ convex.
  Continuety of $\ddot{E}$
  Monotonicity of $E_n$ os that the example of $\frac{1}{n}\sin(nx)$ does not work.

I need a positive result so if other condition needed for converge, its OK to point them.

Thanks !

My smart book says:

Uniform convergence of series $\dot{E_n},\,\,x\in \langle a, b\rangle$
Convergence of series $E_n$ at least one point $\in \langle a, b\rangle $(from points 1 and 2 $\Rightarrow$ series $E_n$ convergent uniformly in the interval $\langle a, b\rangle $)

No additional conditions.


finding the order of the polynomials in a polynomial groups under addition My question is:   if  G:= Group of polynomials under addition with elements from $\{0,1,2,3,4,5,6,7,8,9\}$
$f(x)=7x^2+5^x+4$ and $g(x)=4x^2+8^x+6$  ,$f(x)+g(x)=x^2+3x$
what is the order of $f(x)$,$g(x)$,$f(x)+g(x)$
i find that the answers are $10$,$5$ and $10$ . Am i correct??
(i noticed that zero polynomial is the identity here)
then if $h(x)=a_nx^n+...+a_0$ belongs to $G$ what is the order of it? given that

$gcd(a_1,a_2,,,a_n)=1$,

$gcd(a_1,a_2,,,,a_n)=2$,

$gcd(a_1,a_2,,,a_n)=5$,

$gcd(a_1,a_2,,,a_n)=10$

Here i do not understand how to proceeded ?

Find the solution in the attached picture