Pointwise limit of the indicator function: $\lim_{c \to \infty}1_{\{X<kc\}}$ I want to find the pointwise limit of this indicator function:

$$\lim_{c \to \infty}1_{\{X<kc\}}$$
where $k>0$ is a parameter and $1$ is an indicator function and $X$ is finite random variable.

One of my friends says it would be $\lim_{c \to \infty}1_{\{X<kc\}}=1_{\{x<\infty\}}=1$. Please provide me correct answer with explanation. Thank you.

Given $t$ in the domain of $X$, take $M$ so that $M>\frac{X(t)}{k}$. This is possible because $X$ is finite and $k$ is fixed. Then, for all $c\geq M$, we have
$1_{X<kc}(t) = 1$ because $X(t)<Mk\leq ck$. Therefore $\lim_{c\to\infty}1_{X<kc}(t)=1$ for each $t$, which means the pointwise limit $\lim_{c\to\infty}1_{\{X<kc\}}$ is $1$.

$$\lim 1_{A_n} = 1_{\lim A_n}$$

So, convince yourself that $$\lim_{c \to \infty} \{X < kc\} = \{X < \infty\}$$ for $k>0$


$a,b\in A$ are selfadjoint elements of $C^*$-algebras, such that $a\le b$, why is $\|a\|\le \|b\|$ Let $A$ be a unital $C^\ast$-algebra with unit $1_A$.

a) Why is $a\le \|a\|1_A$ for all selfadjoint $a\in A$ and

b) If $a,b\in A$ are selfadjoint such that $a\le b$, why is $\|a\|\le \|b\|$?

I would try it with the continuous functional calculus $$\phi:C(\sigma(a))\to A,\; f\mapsto f(a),$$ for a) with $f(x)=|x|-x$, which is a positive function on $\sigma(a)\subseteq \mathbb{R}$, is it correct?

for b) It is $a\le \|a\|$ and $b\le \|b\|$, therefore $a\le \|b\|$. How can I continue?

Regards

a) If $A=C(X)$, where $X$ is a compact Hausdorff space, for any $a,b\in A_{sa}$,
$$
a\leq b \Leftrightarrow a(x) \leq b(x) \quad\forall x\in X
$$
So if $f\in A_{sa}$ then , $f\leq \|f\|_{\infty}1_A$.

Now, if $a$ is self-adjoint, then $C^{\ast}(a,1_A)$ is abelian and unital, so by Gelfand's theorem, it is isomorphic to $C(X)$ for some compact Hausdorff $X$.

b) Now if $a \leq b$, then by (a), $a\leq \|b\|1_A$. Again look at the abelian $C^{\ast}$-algebra $C^{\ast}(a,1_A)$, and conclude that $\|a\| \leq \|b\|$

Edit: As @TrialandError points out below, part (b) only holds if $0 \leq a$ [it fails trivially even in a commutative $C^{\ast}$-algebra otherwise]

let A be a C*-algebra and a,x in A with 0≤xx*≤a.
questions:
1-  Put bn=(((1/n)+a)^(-1/2))(a^(1/4))x. show that (bn) is a cauchy sequence in A.
2-  Show that the limit b in A satisfies x=(a^(1/4))b and ∥b∥≤∥a∥^(1/4)

for question 1- it is easy but for question 2- how i show that ∥b∥≤∥a∥^(1/4)?


Let $f(x)=\delta(x) $ if $ x\in (-\infty,0] $ and $f(x)=0$ if $x\in (0,\infty)$. Then $I=\int_{-\infty}^{\infty}f(x)dx=?$ Let $\delta$ be the Dirac delta function.

Then, let $f(x)=\delta(x) $ if $ x\in (-\infty,0]
$ and $f(x)=0$ if $x\in (0,\infty)$.Then $I=\int_{-\infty}^{\infty}f(x)dx=?$

Is $I=1/2$? We know $\int_{-\infty}^{\infty}\delta(x)dx=1$. But we can not say that $\delta$ is symmetric about $y-axis$. So we can not conclude that $I=1/2$. Then what will be $I$?

OP is essentially describing an ill-defined product of two distributions$^1$
$$ u~=~\theta \delta, $$
where $\theta$ denotes the Heaviside step function and $\delta$ is the Dirac delta distribution. Products of distributions is a well-known delicate topic, see e.g. this mathoverflow post and Colombeau algebra.

To see what slippery slope lies ahead, consider the following ill-defined chains of formal manipulations:
$$u~=~\theta \delta~=~\theta \theta^{\prime}~=~\frac{1}{2}(\theta^2)^{\prime}
~=~\frac{1}{2}\theta^{\prime}~=~\color{red}{\frac{1}{2}} \delta $$
versus
$$u~=~\theta \delta~=~\theta^2 \delta~=~\theta^2 \theta^{\prime}~=~\frac{1}{3}(\theta^3)^{\prime}
~=~\frac{1}{3}\theta^{\prime}~=~\color{red}{\frac{1}{3}} \delta~! $$
Later OP asks to evaluate the distribution $u[\varphi]$ for the constant function $\varphi=1$. Technically speaking, a constant function $\varphi=1$ is not a valid test function.

--

$^1$ We consider in this answer the "mirror" distribution $\theta(x) \delta(x)$ rather than OP's distribution $\theta(-x) \delta(x)$ to simplify notation. The conceptional challenges will be the same.

Since Delta is defined with $\int_{-\infty}^{\infty}$ you cannot change those limits. What you can do is construct test function as $\phi(x)=1$ if $x\in(-\infty,0]$ and $\phi(x)=0$ if $x\in(0,\infty)$ and ask what is $\int_{-\infty}^{\infty}\phi(x)\delta(x)dx$. The answer is obviously $\phi(0)=1$.


probability that $2$ perticular persons $A$ and $B$ are not next to each other. $n$ persons sit a round table randomly. Find the probability that $2$ perticular
persons $A$ and $B$ are not next to each other.

Attempt: Total ways of arranging $n$ persons sit around a table is $(n-1)!$
probability that $2$ perticular persons are always together is $\displaystyle \binom{n}{2}\cdot (n-1)$
So desired probability is $\displaystyle \frac{1}{(n-1)!}\times \binom{n}{2}\cdot (n-1)$
But anser mention is $\displaystyle \frac{n-3}{n-1}$, could some help me

Due to rotational symmetry, place $A$ at "12 o'clock."

There are $n-3$ places $B$ can be placed, out of a total of $n-1$ positions.

First, let other $(n-2)$ persons sit around the table in $(n-3)!$ ways, then A and B can sit anywhere in the $(n-2)$ gaps between $(n-2)$ persons, but they cannot sit in the same gap. So, first choose two gaps in $^{n-2}C_2$ ways and then you may choose in which of these two gaps A has to sit in $2$ ways. So the final answer is $(n-3)!* ^{n-2}C_2 *2=\dfrac{(n-3)!(n-2)!}{2!(n-4)!}*2=(n-3)*(n-2)!$

Alternative answer:
Total number of ways to make $n$ people sit around the table is $(n-1)!$ ways. But we have to subtract those cases where A and B sit together, but then as they are always together we can think of them as a single person so that it is same as $(n-1)$ people sitting around the table that can happen in $(n-2)!$ ways. But the order of A and B while sitting together can be chosen in 2 ways. So we should actually subtract $2*(n-2)!$. Therefore the final answer is $(n-1)!-2*(n-2)!=(n-3)*(n-2)!$


How to solve 4 variables I received the below puzzle today (via whatsapp):

We tried to solve this, but we can't solve.
We think that this puzzle is wrong.
Can this be solved? Or is this a wrong puzzle?

You can write the unknowns like this:

$a$ ;       $a-9$
$12-a$ ;    $-2-a$

This helps you keep the number of unknowns minimal (just $a$).

Now the horizontal ones are satisfied and so is the 1st vertical.
Then (from the 2nd vertical) you get  $(a-9) + (-2-a) = 2$ which can never be true.
So this puzzle has no solution indeed.

Considering the top two squares as $a$ and $b$, the bottom squares as $c$ and $d$, then
$$a = 4.5 \\
b = -4.5 \\
c = 7.5 \\
d = 6.5$$


Find the value of $(-1)^{1/3}$. Evaluate $(-1)^{\frac{1}{3}}$.

I've tried to answer it by letting it be $x$ so that $x^3+1=0$.
But by this way, I'll get $3$ roots, how do I get the actual answer of $(-1)^{\frac{1}{3}}$??

It depends whether you are trying to solve it as an equation over the reals ($\mathbb{R}$) or over the complex plane ($\mathbb{C}$).
The polynomial $x^3+1$ factors as $x^3+1=(x+1)(x^2-x+1)$. So in the first case, you have exactly one solution: $x=-1$, since the second polynomial has no real roots. If you're looking for roots over $\mathbb{C}$, then you'll have three roots, since the $\sqrt[3]t$ is not a function over $\mathbb{C}$, hence all the 3 roots have equal rights to be called "the root".

Just put it like complex numbers: We know that $z=\sqrt[k]{m_\theta}$, so $z=\sqrt[3]{-1}$ $-1=1_{\pi}$ $\alpha_n=\dfrac{\theta+k\pi}{n}$ $\alpha_0=60$ $\alpha_1=180$ $\alpha_2=300$
So the answers are:
$z_1=1_{\pi/3}$
$z_2=1_{\pi}$
$z_3=1_{5\pi/3}$


cyclic groups homomorphism I have the following task:
"Determine the homomorphism between two cyclic groups. Which are injective, surjective or bijective?"
I already found this for the cyclic group of integers: http://users.math.yale.edu/~auel/courses/370f06/docs/solutions3.pdf page 2, 4.4.
But what about the cyclic groups of Integers modulo n?

We can suppose the cyclic groups are $\mathbf Z/m\mathbf Z$ and $\mathbf Z/n\mathbf Z$ respectively. A homomorphism from the first to the second is determined by the choice of the image $\bar x$ of $\bar 1$, subject to the condition $m \bar x=0$, i.e.
$$\DeclareMathOperator\Hom{Hom}\Hom(\mathbf Z/m\mathbf Z,\mathbf Z/n\mathbf Z) ≃ \operatorname{Ann}_{\mathbf Z/n\mathbf Z}(m).$$
Let $d=\gcd(m,n)$, $m'=\dfrac md$, $n'=\dfrac nd$. Since $m'$ and $n'$ are coprime,
$$\operatorname{Ann}_{\mathbf Z/n\mathbf Z}(m)=n'\mathbf Z/n\mathbf Z ≃ \mathbf Z/d\mathbf Z.$$
Furthermore,

the surjectivity of such a homomorphism means its image, which is contained in $n'\mathbf Z/n\mathbf Z$, is equal to $\mathbf Z/n\mathbf Z$ . This can happen only if $n'=1$, i.e. $n∣m$.
If $\gcd(m,n)=1$, the only homomorphism is the zero homomorphism.
Injectivity means the image of the homomorphism is isomorphic to $\mathbf Z/m\mathbf Z$. Hence, by Lagrange’s theorem, $ m $ has to be a divisor of $ n $. We’ll suppose this is indeed the case. Then the image $ \bar x $ of $\bar 1$ in $ \mathbf Z/n\mathbf Z $ has to be or order $ m $. As the order of $ \bar x$ is $ \dfrac n{\gcd(n, x)} $,  this means $ \gcd(n, x) = \dfrac nm$.
Finally, if a homomorphism is an isomorphism, the above considerations show it implies $ m = n $. The image of $\bar 1 $ is another generator of $\mathbf Z/m\mathbf Z$, i. e. an element $\bar x,\; x < m$, coprime to $ m $.

Property: let $f: G \rightarrow H$ a finite group morphism then
for all $ x\in G$  the order of $f(x)$ divides both, order of $ x$
and order of $H$.

So, if $n$ and $m$ are  coprime, then there is no nonzero morphism
groups from $\Bbb{Z}/n\Bbb{Z}$ to $\Bbb{Z}/m\Bbb{Z}$. else $n=1$ or $m=1$ this last three cases are evidente

If $n$ divides $m$, the number of  choosing the injection from
$\Bbb{Z}/n\Bbb{Z}$  into $\Bbb{Z}/m\Bbb{Z}$ is the number of pairs
$(i, j)$ where $order (i) = oorder(j) = n$. so  in total there are
$\phi(n)^2$;  these injections are surjections therefore
bijections if and only if $n = m$.

If $d = gcd (n, m)$ with $n \neq d \neq m$, there is  no
injections, no surjections, no bijections.


Can the $0$-norm represent determinism? In Scott Aaronson's Quantum Computing since Democritus, he presents classical probability theory as based on the $1$-norm, and QM as based on the $2$-norm.

  Call $\{v_1,\ldots,v_N\}$ a unit vector in the $p$-norm if $|v_1|^{\ p}+\cdots+|v_N|^{\ p}=1$

The slide below is from a presentation of his.

Now, he seems to claim that the $1$-norm and the $2$-norm are the only options here, although earlier in the book he allowed for three choices:

  (1) determinism, (2) classical probabilities, or (3) quantum mechanics.

My question is: Would it be natural to let determinism be simply based on the $0$-norm? Is this problematic? Wouldn't the definition $0^0=0$ (for this purpose) and $p_i\in\{0,1\}$ suffice?

If so, what would be the equivalent to "probability vector" and "amplitude vector" be called?
("Indicator vector"??)

Yes, all of this works out. You just get finite sets and functions between them.

Incidentally, if you want an even more abstract approach to probability that unifies classical and quantum probability into one framework, check out noncommutative probability. I've written about this here and the specialization to the cases Aaronson describes is given here.

I do not think that 'classical probability'  as commonly (stress non-) understood, for instance works at all, if that sentence has any sense. Perhaps, quantum mechanics is trying to tell us this.  With its stronger tie with logic, if anything, quantum probability is less paradoxical.**

To permit me to make an analogy with another 'probabilistic concept'.

Much as with Simpsons paradox, there is just a lot more of this damn concept (of probability) involved in quantum mechanics, and that is perhaps why, on the whole, quantum mechanics is deemed harder to make sense of, then say classical mechanics was.

'Probability' got us into this mess (QM), or the great 'Quantum muddle', as Karl Popper called it.And somehow, making sense of 'probability' is going to have get us out of it. Whether that is, by making sense of probability, or coming to the final realization that the entire concept, was non-sensical to begin with. And the concept of probability be thrown out of the history books, dictionary and academy see S Saunders similar comment in [][1]

History will say which.. Perhaps?.

By this I mean just what it means for an event to be more/less or equal in probability value to another without any use of any other, non logical,non justified/justifiable, if not irrational,  posits. Such as any form of the principal of indifference/non contextuality, which amount to circularity. Asserting that, which one wishes to prove.

Robin Giles had an idea. One needs to real all of his obscure works to get his  at it, its scattered throughout his work.See for example, the work of Robin Giles for instance his [<1992 paper, "A non classical theory of subjective expected utility theory>."][2]

His odd theory involved a decision theory, a theory of probability and quantum probability in general, that in some sense analyed equi-probability and logic as of  the liar paradox re-ified arguably (not just a probabilistic solution). But rather, what it means for two events, to be equi-probable in general.He even tried to give an analysis of twice as with this, on the single case,
that he only alludes to.

Using monotonic-ity alone, or some other such thing.

And I think that is how probability must be analyzed, if is to make any coherent non circular sense at all. Despite this idea being incoherent in itself, if at possible in any sense!

The chance values, do not have to be what they are, as analytic, or 'a priori'. But that does not make the concept, non-a priori.

It is just that the values may not be. Presumably, if at all it would be a form of Nomic logic such as in certain physical conceptions of modal logic/conditional logic or QM which subscribe to physical laws and has nothing to with the so called logical or classical or objective Bayesian view of probability which have no, or little to no, connection to Logic as ordinarily conceived at all.

What other non-circular, justifiable grounds, is there but necessarily truth value equivalence?
 See Frank Arntzenius's draft manuscript on this and otherwise,
See [][7] and
-see <[Chance, Credence and Circles F Cariani,https://philpapers.org/rec/CARCCA-14>][8].

See https://philpapers.org/s/Robin%20Giles See [https://philpapers.org/s/Robin%20Giles][3]

See also R Giles his 1970 paper "Foundations of Quantum Mechanics", and his earlier works: and

In fact Robin Giles, alludes to this as well see "a Logic of Subjective Belief" below.

If interested, One needs to real all of his obscure works to get the general idea. Which he only alludes to here and there  .https://philpapers.org/s/Robin%20Giles See [https://philpapers.org/s/Robin%20Giles][3]

See the links to Robin Giles work on non-classical probability,. non-classical logic, quantum statistic, fuzzy logic and decision theorey, as well as quantum mechanics . See the references, below if interested:

See [<"Introduction to a Logic of Assertions.Robin Giles - 1990 - In Kyburg Henry E., Loui Ronald P. & Carlson Greg N. (eds.), Knowledge Representation and Defeasible Reasoning. Kluwer Academic Publishers. pp. 361--385">][4]

[5]
Giles, Robin, **A generalization of the theory of subjective probability and expected utility, Synthese 90, No.2, 301-343 (1992). ZBL0759.60005.

Giles, Robin, A logic for subjective belief, Found. Probab. Theor., stat. Inference, stat. Theor. Sci., Vol. I, Proc. int. Res. Colloq., London/Canada 1973, 41-72 (1976). ZBL0323.02043.
_Ferm

Fermller, Christian G.; Roschger, Christoph, From games to truth functions: a generalization of Giles's game, Stud. Log. 102, No. 2, 389-410 (2014). ZBL1329.03060.

Fermuller, Christian G., Revisiting Giles's game. Reconciling fuzzy logic and supervaluation, Majer, Ondrej (ed.) et al., Games: Unifying logic, language, and philosophy. Berlin: Springer (ISBN 978-1-4020-9373-9/hbk; 978-1-4020-9374-6/e-book). Logic, Epistemology, and the Unity of Science 15, 209-227 (2009). ZBL1167.03019.

Fermuller, Christian G.; Metcalfe, George_, Giles's game and the proof theory of {\L}ukasiewicz logic, Stud. Log. 92, No. 1, 27-61 (2009). ZBL1185.03041.

See D Luce, 'in Particular a multiplicative Power relation amongst Three Variables
 https://www.google.com.au/search?q=Duncan+Luce+multiplicative+power+relations&ie=utf-8&oe=utf-8&client=firefox-b&gfe_rd=cr&ei=mCkNWaqlKajr8Af4irq4BA

Aerts, Diederik, Quantum axiomatics, Engesser, Kurt (ed.) et al., Handbook of quantum logic and quantum structures. Quantum logic. With a foreword by Anatolij Dvure\v censkij. Amsterdam: Elsevier/North-Holland (ISBN 978-0-444-52869-8/hbk). 79-126 (2009). ZBL1273.81015.

Aerts, Diederik; Coecke, Bob, The creation-discovery-view: Towards a possible explanation of quantum reality, Dalla Chiara, Maria Luisa (ed.) et al., Language, quantum, music. Papers of the 10th international congress of logic, methodology and philosophy of science, Florence, Italy, August 1995. Dordrecht: Kluwer Academic Publishers. Synth. Libr. 281, 105-116 (1999).

ZBL0988.81009.Adamson, Alan; Giles, Robin, A game-based formal system for \L$_\infty$, Stud. Log. 38, 49-73 (1979). ZBL0417.03008.

*[*A Generalization of the Theory of Subjective Probability and Expected Utility.Robin Giles - 1992 - Synthese 90 (2):301 - 343.**

Introduction to a Logic of Assertions.Robin Giles - 1990 - In Kyburg Henry E., Loui Ronald P. & Carlson Greg N. (eds.), Knowledge Representation and Defeasible Reasoning. Kluwer Academic Publishers. pp. 361--385

A Non-Classical Logic for Physics.Robin Giles - 1974 - Studia Logica 33 (4):397 - 415"
*
*-The Concept of a Proposition in Classical and Quantum Physics.Robin Giles - 1979 Studia Logica 38 (4):337 - 353**.][6]

/e-book).%20Logic,%20Epistemology,%20and%20the%20Unity%20of%20Science%2015,%20209-227%20(2009).%20ZBL1167.03019.%20%20%3C/cite%3EFermuller,%20Christian%20G.;%20Metcalfe,%20George_,%20Giles's%20game%20and%20the%20proof%20theory%20of%20%7B%5CL%7Dukasiewicz%20logic,%20Stud.%20Log.%2092,%20No.%201,%2027-61%20(2009).%20ZBL1185.03041.%3C/cite%3E


Maximum area of a isosceles triangle in a circle with a radius r As said in the title, I'm looking for the maximum area of a isosceles triangle in a circle with a radius $r$.

I've split the isosceles triangle in two, and I solve for the area $A=\frac{bh}{2}$*. I have made my base $x$, and solve for the height by using the Pythagorean theorem of the smaller triangle (seen in picture).

$h=r+\sqrt{r^2-x^2}$

So my the formula, I think, for both triangles should be $A=x(r+\sqrt{r^2-x^2})$

But after I solved for the derivative, when put "$= 0$", and checked on my calculator, I got the maximum to be about $4.3301r$, which differs a lot from my book's answer of $\frac{3\sqrt{3}}{4}r^2$*. Is my formula for the area right? Am I going about this the wrong way, or is it just my derivative that is wrong? Thanks in advance

*
 Edited from original post

$A=rx+x\sqrt{r^2-x^2}$

$A'=r+x(\frac{1}{2})(r^2-x^2)^{-\frac{1}{2}}(-2x)+\sqrt{r^2-x^2}$

$r+\sqrt{r^2-x^2}=\frac{x^2}{\sqrt{r^2-x^2}}$

$r\sqrt{r^2-x^2}+(r^2-x^2)=x^2$

$r^2+r\sqrt{r^2-x^2}=2x^2$

$r^2(r^2-x^2)=(2x^2-r^2)^2$

$r^4-r^2x^2=4x^2-4x^2r^2+r^4$

$4x^4=3x^2r^2$

$x=\frac{\sqrt{3}}{2}r$

Let $\theta$ be one-half of the vertex angle (less than a right angle) of the isosceles triangle.

  Exercise: Show that the area of the inscribed triangle is
  $A(\theta) = \dfrac{h b}{2} = \dfrac{(r + r \cos\theta)}{2} (2 r \sin\theta) = r^2 \,(1+\cos\theta) \, \sin\theta $

Differentiating $A$ and setting it to $0$, you will be left with the problem of solving
$\cos\theta = -\cos2 \theta.$

  Exercise: Show that the only $\theta$ in $\left(0, \dfrac{\pi}{2}\right)$ that works is $\dfrac{\pi}{3}$.

Plugging these into $A$ we get the answer: $\dfrac{3\sqrt 3 r^2}{4}.$

Could it be possible that the 2 rs in the drawing are not equal and the other r (diagonal) could be longer than the vertical r in the drawing (or vice versa). In such case, the pythagorean equation h=r + sqrt(r^2 -x^2) is not applicable.


Characteristics of parallel parabola (offset curve) and the formula to find the equation What are the characteristics of parallel parabola? And is there any formula to find the equation of parallel parabola if we know the equation of one parabola?

(see graphics below obtained with Geogebra, using formulas in the text).
Preliminary remark: The keyword "parallel curve" is ambiguous ; you see it replaced by "offset curve" in many works.
If a curve is defined parametrically by $X=\binom{x=x(t)}{y=y(t)}$, a tangent vector is $T=\binom{x'(t)}{y'(t)}$ (speed vector); we can take as normal vector $N=\binom{ \ \ y'(t)}{-x'(t)}$ (check that the dot product $N \bullet T$ is 0), and the unit norm normal vector is $N/\|N\|=\binom{y'(t)/\sqrt{x'^2(t)+y'^2(t)}}{-x'(t)/\sqrt{x'^2(t)+y'^2(t)}},$ with a certain orientation (outwards or inwards with respect to the initial curve.)
Thus, if you want to be at distance $a$ from the initial curve, you proceed from the generic point by making a step of length $a$ in the direction of vector  $N/\|N\|$ giving the parametric representation that can found as well in the Wikipedia article on parallel and offset curves:
$$\tag{1}\left\{\begin{aligned}X&=x+a{\frac {y'}{\sqrt {{x'}^{2}+{y'}^{2}}}}\\Y&=y-a{\frac {x'}{\sqrt {{x'}^{2}+{y'}^{2}}}}.\end{aligned}\right.$$
Let us take the case of the standard parabola $y=x^2$ with parametrization $x=t, \ y=t^2$; we get: $x'=1, \ y'=2t$, yielding
$$\tag{2}\left\{\begin{aligned}X&=t+\frac {2at}{\sqrt {1+4t^2}}\\Y&=t^2-{\frac {a}{\sqrt {1+4t^2}}}.\end{aligned}\right.$$
which is a parametric representation that allows to draw any parallel curve to the initial parabola. According to the sign of $a$, these curves will be external or internal to the parabola (I understand you are interested in external curves).
One can, using a CAS (Mathematica in my case), eliminate $t$ between the 2 expressions above ; one gets a 12th degree implicit equation that can be expressed as the product of two 6th degree factors in variables $X,Y$:
$$\left( 9\,a^2 - 24\,a^4 + 16\,a^6 - 92\,a^2\,X^2 -
    48\,a^4\,X^2 - 9\,X^4 + 48\,a^2\,X^4 - 16\,X^6 + 24\,a^2\,Y -
    32\,a^4\,Y + 18\,X^2\,Y + 24\,a^2\,X^2\,Y + 8\,X^4\,Y -
    9\,Y^2 + 40\,a^2\,Y^2 - 16\,a^4\,Y^2 + 32\,X^2\,Y^2 +
    32\,a^2\,X^2\,Y^2 - 16\,X^4\,Y^2 - 24\,Y^3 + 32\,a^2\,Y^3 +
    32\,X^2\,Y^3 - 16\,Y^4 \right) \,
  \left( a^2 + 8\,a^4 + 16\,a^6 + 20\,a^2\,X^2 - 48\,a^4\,X^2 -
    X^4 + 48\,a^2\,X^4 - 16\,X^6 - 8\,a^2\,Y - 32\,a^4\,Y +
    2\,X^2\,Y - 8\,a^2\,X^2\,Y + 40\,X^4\,Y - Y^2 + 8\,a^2\,Y^2 -
    16\,a^4\,Y^2 - 32\,X^2\,Y^2 + 32\,a^2\,X^2\,Y^2 -
    16\,X^4\,Y^2 + 8\,Y^3 + 32\,a^2\,Y^3 + 32\,X^2\,Y^3 - 16\,Y^4
    \right)=0$$
(please note that $a$ is always present with an even exponent).
One of the factors accounts for the external curve at distance $a$, the other for the internal curve at distance $a$. These expressions are untractable for practical applications.
These offset curves being polynomials with degree 6, they aren't conic curves (2nd degree polynomials), even if they look for example like parabolas when they are close to the "mother curve" (but with an increasingly circular aspect at the bottom).
It is to be noted that internal parallel curves can have a spike.
The second figure gives a different view: the offset curves are implicitly realized as "envelopes" of circles (think to the parabola as an atoll and the inside of the circles as "territorial waters" within a same distance).

Say you have a parabola of the form $ax^2+bx+c$.

Another parabola is parallel if it is of the form $ax^2+bx+k$, where $k$ is a constant that is not equal to $c$.

If $k=c$, then the parabolas would be  identical.


Determine all graphs with matching number = 1 Determine all graphs G, without isolated vertices, such that the
matching number=1.

Could anyone help me this question?
I am really confused about it.

How to determine all possible graphs with this condition?

If your graph is to have matching number 1, you need at least one edge $e_1$, between vertices, say, $v_1$ and $v_2$. This certainly works.

Can we add more things? If we add a new vertex $v_3$, it is required that $v_3$ is not isolated, so, it is connected to something. If it is a new vertex, then our matching number is at least 2. So we can add an extra $v_3$ and assume WLOG that it is connected to $v_1$. (This gives us the path on 3 vertices).

Edit: Here I missed that you can join $v_3$ to BOTH $v_1$ and $v_2$, so, we have $P_3$ (the path on 3 vertices) and $K_3$ (the complete graph on 3 vertices). We can't add any new edges to $K_3$, so the next paragraph is assuming we are dealing with $P_3$.

Can we add more? If we want to add $v_4$, for the same reason as above, it can't be isolated, and can't be connected to something new, so it has to be connected to one of our $v_1,v_2,v_3$. If it is connected to $v_2$ or $v_3$, then our graph contains a path on 4 vertices, which has matching number 2. So, any new edge must be connected to $v_1$.

Continuing in this fashion, you obtain that larger graphs must have a central vertex $v_1$, so that every other vertex is connected to it, and there are no other edges. I'm not sure if such graphs have a name, but I would guess "star-shaped".

You should be able to turn the above intuition into a proof.

Edit: I am told that these are called Star-graphs http://en.wikipedia.org/wiki/Star_(graph_theory)

There is only one family of bipartite graphs for which the matching number is 1. Its commonly denoted as K_{1,r}, r \geq 1. In this graph, one partite set has exactly one vertex, say 'v'; and the other partite set has 'r' vertices, all of which are connected to the one vertex 'v'.


find the solution set of the following inequality with so many radical I get lost

$\sqrt[4]{\frac{\sqrt{x^{2}-3x-4}}{\sqrt{21}-\sqrt{x^{2}-4}}}\geqslant x-5$

Edit
I get online with wolfram
 -5 < x <= -2 || x == -1 || 4 <= x < 5

First let us check where the above formula is defined:

$ x^2-3x-4 \geq 0  \; \; \Leftrightarrow  \; \; x\in ] -\infty , -1] \cup [4,+\infty[$.

$x^2-4 \geq  0  \; \; \Leftrightarrow  \; \; x\in ] -\infty , -2] \cup [2,+\infty[ $.

$ \sqrt{21 } - \sqrt{ x^2-4} > 0  \; \; \Leftrightarrow  \; \;  x\in   ]-5,-2[ \cup ]2,5[.$

Thus  You set of solution must  be subset of  $]-5, -2] \cup [4,5[ $. Note that whenever $x  \in   ]-5, -2] \cup [4,5[ $ the inequality holds true as  $x-5 < 0 $ and the other side  of the inequality is surely positive.

Well, this is a very messy problem.
We start off with:
$\sqrt[4]{\frac{\sqrt{x^{2}-3x-4}}{\sqrt{21}-\sqrt{x^{2}-4}}}\geqslant x-5$

Now we can raise each side to the power $4$.

${\frac{\sqrt{x^{2}-3x-4}}{\sqrt{21}-\sqrt{x^{2}-4}}}\geqslant (x-5)^4$

We can then multiply the denominator to the right side:

$\sqrt{x^{2}-3x-4}\geqslant(x-5)^4 * (\sqrt{21}-\sqrt{x^{2}-4})$

From here, again we can square both sides, receiving:

${x^{2}-3x-4}\geqslant(x-5)^8 * (\sqrt{21}-\sqrt{x^{2}-4})²$

Afterwards, you can expand these.


Where is the flaw in this "proof" that 1=2? (Derivative of repeated addition) Consider the following:

$1 = 1^2$
$2 + 2 = 2^2$
$3 + 3 + 3 = 3^2$

Therefore,

$\underbrace{x + x + x + \ldots + x}_{x \textrm{ times}}= x^2$

Take the derivative of lhs and rhs and we get:

$\underbrace{1 + 1 + 1 + \ldots + 1}_{x \textrm{ times}} = 2x$

Which simplifies to:

$x = 2x$

and hence

$1 = 2$.

Clearly something is wrong but I am unable pinpoint my mistake.

I think the discrete/continuous issue is sort of a red herring.  To me, the problem is forgetting to use the chain rule!

To un-discretize, think of the function $F(u,v) = uv$, which we could think of as $u + \dots + u$, $v$ times.  Then $x^2 = F(x,x)$.  Differentiating both sides gives $2x = F_u(x,x) + F_v(x,x)$, which is perfectly true.  In the fallacious example, the problem is essentially that the $F_v$ term has been omitted.  In some sense, one has forgotten to differentiate the operation "$x$ times" with respect to $x$!  Of course, the notation makes this easier to do.

Even if x were to be an integer, once you have x=2x, a possible value for is 0 which would make that equation true. Thus if x can be zero, then you are simply dividing by a variable that equals 0. And you cannot divide by zero!!!!


Why does this matrix give the derivative of a function? I happened to stumble upon the following matrix:
$$ A = \begin{bmatrix}
    a       & 1  \\
    0       & a
\end{bmatrix}
$$

And after trying a bunch of different examples, I noticed the following remarkable pattern. If $P$ is a polynomial, then:
$$ P(A)=\begin{bmatrix}
    P(a)       & P'(a)  \\
    0       & P(a)
\end{bmatrix}$$

Where $P'(a)$ is the derivative evaluated at $a$.

Futhermore, I tried extending this to other matrix functions, for example the matrix exponential, and wolfram alpha tells me:
$$ \exp(A)=\begin{bmatrix}
    e^a       & e^a  \\
    0       & e^a
\end{bmatrix}$$
and this does in fact follow the pattern since the derivative of $e^x$ is itself!

Furthermore, I decided to look at the function $P(x)=\frac{1}{x}$. If we interpret the reciprocal of a matrix to be its inverse, then we get:
$$ P(A)=\begin{bmatrix}
    \frac{1}{a}       & -\frac{1}{a^2}  \\
    0       & \frac{1}{a}
\end{bmatrix}$$
And since $f'(a)=-\frac{1}{a^2}$, the pattern still holds!

After trying a couple more examples, it seems that this pattern holds whenever $P$ is any rational function.

I have two questions:

Why is this happening?
Are there any other known matrix functions (which can also be applied to real numbers) for which this property holds?

If $$ A = \begin{bmatrix}
    a       & 1  \\
    0       & a
\end{bmatrix}
$$
then by induction you can prove that
$$ A^n = \begin{bmatrix}
    a^n       & n a^{n-1}  \\
    0       & a^n
\end{bmatrix} \tag 1
$$
for $n \ge 1 $. If $f$ can be developed into a power series
$$
  f(z) = \sum_{n=0}^\infty c_n z^n
$$
then
$$
  f'(z) = \sum_{n=1}^\infty n c_n z^{n-1}
$$
and it follows that
$$
 f(A) = \sum_{n=0}^\infty c_n A^n = I + \sum_{n=1}^\infty c_n
\begin{bmatrix}
    a^n       & n a^{n-1}  \\
    0       & a^n
\end{bmatrix} = \begin{bmatrix}
    f(a)       & f'(a)  \\
    0       & f(a)
\end{bmatrix} \tag 2
$$
From $(1)$ and
$$
A^{-1} = \begin{bmatrix}
    a^{-1}       & -a^{-2}  \\
    0       & a^{-1}
\end{bmatrix}
$$
one gets
$$
A^{-n} = \begin{bmatrix}
    a^{-1}       & -a^{-2}  \\
    0       & a^{-1}
\end{bmatrix}^n =
(-a^{-2})^{n} \begin{bmatrix}
    -a       & 1  \\
    0       & -a
\end{bmatrix}^n  \\ =
(-1)^n a^{-2n} \begin{bmatrix}
    (-a)^n       & n (-a)^{n-1}  \\
    0       & (-a)^n
\end{bmatrix} =
\begin{bmatrix}
    a^{-n}       & -n a^{-n-1}  \\
    0       & a^{-n}
\end{bmatrix}
$$
which means that $(1)$ holds for negative exponents as well.
As a consequence, $(2)$ can be generalized to functions
admitting a Laurent series representation:
$$
  f(z) = \sum_{n=-\infty}^\infty c_n z^n
$$

If $e$ satisfies $Xe=eX$ and $e^k=0$, then $f(X+e)=f(X) + ef'(X)+\frac{e^2}{2}f''(X) + \dots + f^{(k-1}(X)\frac{e^{k-1}}{(k-1)!}$ (a finite sum with $k$ terms).

This is true for polynomials and thus for power series that converge in a neighborhood of $X$.


$E = A^C \cap B^C$, is my answer good? Could anyone tell me if my answer is correct regarding this set(complement) question. As to confirm myself that I understand it or not.
$U = \{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15\}$.
$A = \{2,4,6,8,10,12,14\}$,
$B = \{1,2,3,4,5\}$.
Answer i got :
$E = A^C \cap B^C$.
$E = \{1,3,5,7,9,11,13,15\} \cap \{6,7,8,9,10,11,12,13,14,15\}$.
$E = \{9,11,13,15\}.$
Can anyone tell me if my answer is good or not?


Find the inverse of $f(x,y) = (x+3y,3x+y)$ Given the function $f : \mathbb{R}^2 \to \mathbb{R}^2$ as $f(x,y) = (x+3y,3x+y)$. Find $f^{-1}$ .( Assume $f$ is a bijection)

I know how to find $f^{-1} (x) = (3x+2)$ or anything with one variable. But I am not sure how to proceed with 2 variables. Can anyone provide me with a step by step explanation?

Thanks

You need to solve $(u,v) = (x+3y, 3x+y)$ for $x,y$.
In this case we get $3u-v = 8y$, hence $y = \frac{1}{8}(3u-v)$
and the same way we get $x = \frac{1}{8}(3v-u)$.

So $f^{-1}(u,v) = \frac{1}{8}(3v-u, 3u-v)$. You can check that this is true by computing $f\circ f^{-1}$ and $f^{-1}\circ f$.

We want a function that takes us from $(x+3y, 3x+y)$ to $(x,y)$. Let us assume that we can write this as $g(x,y) = (ax+by, cx+dy)$, then $$a(x+3y) + b(3x+y) = x+0y$$ and $$c(x+3y)+d(3x+y) = 0x+y$$

Hence the first equation gives us: $$a+3b = 1 \text{ and } 3a+b = 0$$ The second gives $$c+3d = 0 \text{ and } 3c+d =1.$$

Hence $-b = 3a$ and $a-9a = 1$ which gives $a=-1/8$ and $b=3/8$. Also $c=-3d$ and $-8d = 1$. Hence $d=-1/8$ and $c=3/8$.

This gives $g(x,y) = ( -1/8 x + 3/8 y, 3/8 x - 1/8 y )$


Standard Form for a Parabola What is the standard form for the following problem? I already know that it is a horizontal parabola. I just can't seem to be able to change it into the standard format.

$8y² +96y-12x+240 = 0$

I have gotten it to $x = 2/3 (y + 6)² - 4$, but that's not right.

Edit: How do you find the focus of this parabola?

Edit: I think the problem might be that it is not a vertical parabola.  The $y$ is squared, so it must be a sideways parabola. (Also called a horizontal parabola).


Example of not so simple group ?? Help me with an example of a group having subgroups but it doesn't admit a normal subgroup.. ??

The alternate definition of simple groups using non trivial homomorphic image. Searching for a map that doesn't have a non trivial kernal ??

One classic example is $A_5$, which has plenty of non-trivial subgroups, none of which are normal.  In fact, this is the smallest suitable example.

Hint: Any subgroup of a commutative group is normal. What noncommutative groups do you know.

Bigger Hint: Think about subgroups of $S_5$.

Even Bigger Hint: What about the subgroup of the simple group $A_5$ generated by a product transpositions?

SUPER OMEGA HINT: What is $(12)(45)(12)(34)(12)(45)$?


Proving that $\tan \theta=\cot(90^\circ-\theta)$ when $\theta>90^\circ$ I'm asked to prove $\tan \theta=\cot (90-\theta)$

So $\tan \theta =\frac{a}{b}=\cot(90^\circ-\theta)$

But what if $\theta>90?$

Use the identities
$$
\sin(x-y)=\sin x\cos y-\cos x\sin y
$$
and
$$
\cos(x-y)=\cos x\cos y+\sin x\sin y
$$
to deduce that
$$
\sin(90^{\circ}-\theta)=\cos\theta;\quad \cos(90^{\circ}-\theta)=\sin\theta.
$$
Hence
$$
\cot(90^{\circ}-\theta)
=\frac{\cos(90^{\circ}-\theta)}{\sin(90^{\circ}-\theta)}
=\frac{\sin\theta}{\cos\theta}
=\tan\theta.
$$

At a point the angle of elevation of the top of a tower is such that its tangent is 5/12 . On walking 80m towards the tower, the cotangent of the angle of elevation of the top of the tower is 4/3. The height of the tower will be how much m?


A probability theory question about independent coin tosses by two players Say Bob tosses his $n+1$ fair coins and Alice tosses her $n$ fair coins. Lets assume independent coin tosses. Now after all the $2n+1$ coin tosses one wants to know the probability that Bob has gotten more heads than Alice.

The way I thought of it is this : if Bob gets $0$ heads then there is no way he can get more heads than Alice. Otherwise the number of heads Bob can get which allows him to win is anything in the set $\{1,2,\dots,n+1\}$. And if Bob gets $x$ heads then the number of heads that Alice can get is anything in the set $\{0,1,2,..,x-1\}$. So\begin{align}P(\text{Bob gets more heads than Alice})&= \sum_{x=1}^{n+1} \sum_{y=0}^{x-1} P( \text{Bob gets x heads }\cap \text{Alice gets y heads }) \\[0.2cm]&= \sum_{x=1}^{n+1} \sum_{y=0}^{x-1} \left(C^{n+1}_x \frac{1}{2}^{x} \frac{1}{2}^{n+1-x}\right)\left( C^n_y \frac{1}{2}^y \frac {1}{2}^{n-y}\right)\\[0.2cm]& = \sum_{x=1}^{n+1} \sum_{y=0}^{x-1} \frac{C^{n+1}_x C^n_y}{2^{2n+1}}\end{align}

How does one simplify this?

Apparently the answer is $\frac{1}{2}$ by an argument which looks like this, Since Bob tosses one more coin that Alice, it is impossible that they toss both the same number of heads and the same number of tails.
So Bob tosses either more heads than Alice or more tails than Alice (but not both).
Since the coins are fair, these events are equally likely by symmetry, so both events
have probability 1/2.

the answer is indeed $\frac 12$ .

As an alternative way to see that:  let's pause just before Bob tosses his final (extra) toss.  At this point, there are three possible states:  either Bob is ahead, Alice is ahead, or they are tied.  Let $p$ be the probability that Bob is ahead.  By symmetry, $p$ is also the probability that Alice is ahead (so the probability of a tie is $1-2p$).  Note that symmetry does clearly apply here since they have thrown the same number of tosses.   Bob has exactly two ways to win:  either he is ahead before the last toss, or they are tied and Bob gets $H$ on the last throw.  Thus the probability that Bob eventually wins is $$p+\frac 12 \times (1-2p)=p+\frac 12 -p =\frac 12$$


Dividing by Zero Using Exponents I am sorry if I am missing some fundamental rule disproving my question, but I am very confused here.

So $0^0 = 1$

and

$x/x = x^0 = 1$

so if $x = 0$

$0/0 = 1$

The problem here is people have told me 0 divided by 0 equals undefined, but by using exponents I have proved this wrong. I have to be doing something wrong here. So, what is wrong with what I just did?

Your reasoning is wrong beginning with the line "so if $x=0$", because division by $0$ is undefined. It doesn't matters whether you use a variable or a composite expression as the denominator, as long as it equals $0$, the argument is incorrect.

As is almost always the case with this type of questions, you seem to ignore the fact that mathematics is a deductive science. We begin from axioms and definitions, then demonstrate that some results logically follow from them. The properties of mathematical objects are not known due to observation of the physical world and are not laws of nature, but consequence of their definitions and the logical framework in which they are made (search for "construction of the real numbers".

It makes sense, however, to ask why for all $x \in R$, $x/0$ is undefined. The answer is in short, because it can't be defined meaningfully (in a way that doesn't contradicts the axioms of the real numbers).

Also, it is incorrect that "0 divided by 0 equals undefined". In ordinary treatments of mathematics, it doesn't equals anything. "is" has more than one meaning, not just the same as "equals". However, in formal treatments of mathematics like Metamath (link is to relevant theorem) the value of a function outside its domain (that is it, where it is not defined) equals an arbitrary object.

0^0 is not equal to 1. It is not defined.
however:
(Tending to 0+)^(tending to 0) is 1. This is a standard limit. Thus what you are using is wrong.


In what sense is quantum field theory mathematically incomplete? Is the Yang-Mills existence and mass gap (Millenium Prize problem) essentially what is required?

Or are there more problems in putting QFT on strong mathematical foundations?

For example, the exsitence of measure in doing functional integration in quantum field theory and string theory.

I think that quantum field theory is on a strong mathematical foundations as long as the mathematics behind it, I mean the mathematics on which it relies, is itself on the strong mathematical foundations and if that mathematics is on a strong logical foundations.

So what you practically wanted to ask is will the solution of the Yang-Mills millenium prize problem put quantum field theory on a strong physical foundations, I mean, will that solution complete quantum field theory in any of the meaningful definitions of completeness of physical theory.

But that is rather a strong philosophical issue because what mathematics do in physics is that it serves as a means of describing the phenomena which occur in the world, and only describing, the question of whether physical theory really explains something or not is not (for me) the problem of mathematics but the problem of interpreting the phenomena which occur in a natural world in a good way, and in a way that is in accordance with the experimental data.

So, basically, what you are asking is, in some sense, related to one of the Hilbert´s problems which, roughly stated, is:

Can we build physics so that it is based on a strong axiomatic system and in a way that such an axiomatic system can be described by the help of mathematics?

But that one, of course, is not solved, and I do not see the way in which it could be solved.


There is a number divisible by all integers from 1 to 200, except for two consecutive numbers. What are the two? To reiterate the question, basically there is some number, $n$ that exists that is divisible all the integers $1, \dots, 200$, except for two consecutive numbers in that range. The goal is to find what those two consecutive integers are. The answer isn't trivial though, since $n$ needs to be divisible by all those numbers, it is difficult to find two numbers next to each other such the multiples of those numbers aren't less than $200$ and such that those cannot be prime factorized into numbers that are in the prime factorization of $n$.

I have tried doing this computationally, but the LCM of all the numbers in the range (less two of them) is ginormous and checking the divisibility condition doesn't seem to work on my computer. The problem would be simple if the two numbers didn't have to be consecutive, since we could just select two prime numbers, but since one must be even, this is not possible.

I am trying to think of properties of divisibility that could help, but haven't found anything that worked yet. For example, I was looking for numbers that a prime such that a number before or after it is the square of a prime number. This way, we could say that the prime number itself is omitted from $n$ and that there is only one factor of the square root of the other number in $n$. I am not sure if that would definitely work, but regardless I couldn't find those numbers. I tried another perfect square and a prime number, $196$ and $197$, but there must be enough factors to make two $14$s in $n$, so that doesn't work either.

I am not experienced at all in number theory or discrete math, this is just a brainteaser I have heard. (Also for reference, I do not know the answer to reverse engineer something from). Any help would be appreciated!

Thanks!

Excellent question! The answer is $127$ and $128$... but why?  If you wanted to find a number divisible by $1,2,3,4$ you might first multiply these numbers and say $24$. However, you soon realize $4$ is already a multiple of $2$; you can use just $3\times4$ to get $12$.  Therefore, you need only multiply the largest powers of the primes that factor all of the digits from $2$ to $200$ to get a number that is divisible by all of the integers from $1$ to $200$.

If you do this; you will find the number is $2^7\cdot3^4\cdot5^3\cdot7^2\cdot11^2\cdot13^2\cdot17\cdot19\cdot23\cdot29\cdot\ldots$(the rest of the primes up to $199$) = a very large number.

Next we need to find a restriction to eliminate two consecutive numbers. One of the two numbers must be even.  The only way to remove an even number from the above calculation without modifying any of the other primes is to reduce the power of $2^7$ to $2^6$; this removes the number $128$ from the list.  Since $127$ is also a prime number, it can also be removed from the list without affecting any of the other primes in the list...

I hope this helps.

One of the two consecutive numbers is even lets call it $m$ if $m\neq 128$ then both $m/2$ and $128$ divide the number hence $lcm(m/2,128)$ divides the number and $m$ divides $lcm(m/2,128)$ so by transitivity $m$ divides the number; contradiction so $m=128$ and you're left with two choices for the other one.


Order of the error for the Trapezoidal and Simpson's method of numerical integration What are the order of the error for the Trapezoidal and Simpson's  method of numerical integration? What is the definition of order of the error of a quadrature formula? Is it true that order of error is one higher than that of degree of precision (I know DOP)

Attempt:

I know that the error term of basic Trapezoidal rule for the interval $[x_0, x_1]$ is $$E_T=-\frac{h^3}{12}f''(\xi), ~x_0<\xi<x_1$$ and that of the composite Trapezoidal rule for the interval $[x_0, x_n]$ is $$E_{TC}=-\frac{h^3n}{12}f''(\xi)=-\frac{h^2(b-a)}{12}f''(\xi), ~x_0<\xi<x_n$$

Then what can we say about the order of the error for the Trapezoidal rule? It is $O(h^3)$ or $O(h^2)$? Or can we say that order of error for basic Trapezoidal rule and composite Trapezoidal rule are $O(h^3)$ and $O(h^2)$, respectively? Which one is true? and Please explore the idea of it.

The order of error only makes sense for the composite rule, since for the simple rule the step size $h=x_f-x_0=b-a$ is constant. Only for the composite rule do you get a variable $h=(b-a)/n$ that allows to consider the asymptotic error behavior. Thus $O(h^2)$ for the composite rule, and no asymptotic error for the simple rule.

Since the (composite) Simpson rule can be seen as Richardson extrapolation (first step of the Romberg method) of the symmetric trapezoidal rule, its error order is automatically $O(h^4)$.

We know that trapezoidal method should not be give any error up to linear polynomial and constant. So error will be starting from second order polynomial thus order of error is $o(h^2)$ for trapezoidal.


Is the reverse of the first derivative test statement true? The first derivative test says that if the derivative of a function is positive/negative in the open interval $(x-a,x)$ and negative/positive in the open interval $(x,x+b)$ where both $a$ and $b$ are greater than zero, then there is a local maxima/minima at $x$. Is the reverse of this statement true ?

For example, if $x$ is a local maxima of the function $f(x)$ and if the function $f(x)$ is continuous and differentiable in $(a,b)$ where $a < x < b$, then there exists an interval to the left of $x$, howsoever small, in which the function has a positive slope and an interval to the right of $x$, howsoever small, in which the function has a negative slope.

TL;DR: The answer is no, and I provide a counterexample. See md2perpe’s comment on this answer for more details.

Consider the function:

$f(x) = x^2(\sin(1/x)-1.5)$ for $x \neq 0$ and $f(0) = 0.$

This is what the graph looks like (from WolframAlpha)

￼$f(x) = x^2(\sin(1/x)-1.5)$ ">

$$$$
Then:

$f(x) < 0 $ for all $ x \neq 0$ because $ x^2 > 0 \ \forall x \neq 0 $ and $\sin(1/x)-1.5 < 0 \ \forall x \neq 0$.
Hence the origin is a local (in fact, global) maximum.

Using the same method as the main answer to this question, we see that $f’(0) = 0.$

Furthermore, for $x \neq 0,\ f’(x) = 2x(\sin(1/x) - 1.5) - \cos(1/x),\ $ which is defined for all values $x \neq 0.$

Therefore $f’(x)$ is differentiable for all x in some interval containing the maximum (the origin). (In fact, we see that $f(x)$ is differentiable for all x).

However, we just found what the derivative of $f(x)$ is for $x \neq 0$: $f’(x) = 2x(\sin(1/x) - 1.5) - \cos(1/x)$, which, as $x {\to} 0,\ $ tends to $-\cos(1/x)$, and so oscillates between approximately -1 and 1 (as $x {\to} 0$).

If you want to know where my idea for this came from, I am somewhat familiar with the graph of $g(x) = x^2\sin(1/x)\ $ for $ x \neq 0 $ and $f(0) = 0.$

I considered “pushing the points” (other than the origin) downwards to just below $y=0$. I noted that $x^2\sin(1/x)$ just touches the graph of $ y=x^2$ at various points (as $x {\to} 0$), so I first considered $x^2\sin(1/x) - x^2,\ $ which, by the way, is the same as $x^2(\sin(1/x) - 1),\ $ and this would have been sufficient if the question didn’t require strict maximum. So for a strict maximum, I tried 1.5 instead of 1 and voila!.

Yes, this is true.

By the definition of a local maxima, if $f(k-h)<f(k)>f(k+h)$ for $h \to 0$, $f(k)$ is a point of local maxima. It is evident from this inequality that $\frac{f(k)-f(k-h)}h  = f'(k^-)> 0 $ and $\frac{f(k+h)-f(k)}h  = f'(k^+) < 0$, which proves that the converse is true. A similar proof can also be derived for a local minima.


If $\lim\limits_{h\to 0^+}\frac{f(x+h)-f(x)}{h}=0$ for all $x\in I$, then $f$ is constant. So suppose $I\subset \mathbb{R}$ is an open interval and $f: I\to \mathbb{R}$ a continuous function such that $$\lim\limits_{h\to 0^+}\frac{f(x+h)-f(x)}{h}=0$$ for all $x\in I$. Show that $f$ is constant.

My ideas:

Proof by contraposition. Suppose $f$ is not constant; thus there are $a,b\in I$ with $a<b$ such that $f(a)\neq f(b)$. I am given the hint that I should check the $\sup$ of $$M:=\left \{ x\in [a,b] : \frac{f(x)-f(a)}{x-a} < \frac{f(b)-f(a)}{2(b-a)}\right\}.$$ I tried wrapping my head around the geometric meaning of $\frac{f(x)-f(a)}{x-a} $ and $\frac{f(b)-f(a)}{2(b-a)}$. I came to the conclusion that $\frac{f(x)-f(a)}{x-a} $ approximates the slope of the curve at $a$ for sufficient $x$. $\frac{f(b)-f(a)}{2(b-a)}$ is half of the slope of the secant line connecting $a$ with $b$.

But what does finding the $\sup$ of $M$ mean and how does it contradict the fact that $f$ is not constant?

Following your idea, without loss of generality, and towards a contradiction, $f(b)>f(a)$. Define $g(x)=\frac{f(x)-f(a)}{x-a}$ if $x\neq a$ and $g(a)=0.$ Then $g$ is continuous on $[a,b]$.

Now, set $g(b)=2\epsilon$ and $c=\inf\{x\in [a,b]:g(x)>\epsilon\}.$ Then, continuity of $g$ implies that $c < b$ and $g(c)=\epsilon.$ (Drawing a picture here will help).

But there is a $d\in (c,b]$ such that $|f(x) – f(c)| \le \epsilon(x – c)$  for all $x \in (c,d]$ (why?) and so now we have for all $x\in(c,d],$

$|f(x)-f(a)|\leq |f(x)-f(c)|+|f(c)-f(a)|\leq$

$ \epsilon (x-c)+\epsilon(c-a)= \epsilon(x-a)\Rightarrow g(x)\le \epsilon,$

which contradicts the choice of $c$.

The most simple answer to this problem is to use mean-value theorem.

https://en.wikipedia.org/wiki/Mean_value_theorem#A_simple_application

Look contents 4 : A simple application.


About $f(x)$ satisfying $f(0) = 0, f(1) =1, f(2) = 2, f(3) = 4$, and $f'(x)$ is differentiable on $\mathbb{R}$. let $f(x)$ be a function such that

$$f(0) = 0$$

$$f(1) =1$$ $$f(2) = 2$$ $$f(3) = 4$$ $$f'(x) \text{is differentiable on } \mathbb{R}$$ Prove that there is a number in the interval $(0,3)$ such
that $0 < f''(x)<1$

I'm really stuck.
thanks

Hint: Show that a some point $x \in (0,1)$, $f'(x) = 1$. Then for some $y \in (2,3)$, $f'(y) = 2$. Now use that to deduce the result by applying the MVT again.

Added:

Having found such $x$ and $y$, there must be a $z \in (x,y)$ such that

$$f''(z) = \frac{f'(y) - f'(x)}{y - x}$$

Now think about how you can bound that expression above and below.

I would suggest as a first attempt, curve fitting a polynomial. Let's fit to a cubic, since the general cubic equation has four parameters (and you have four known points).
$$f(x) = ax^3+bx^2+cx+d$$
$$f(0) = a*0^3+b*0^2+c*0+d = 0 \rightarrow d = 0$$
$$f(1) = a*1^3+b*1^2+c*1+d = 1 \rightarrow a+b+c=1$$
$$f(2) = a*2^3+b*2^2+c*2+d = 2 \rightarrow 8a+4b+2c = 2$$
$$f(3) = a*3^3+b*3^2+c*3+d = 4 \rightarrow 27a+9b+3c = 4$$
Solving the last three equations simultaneously gives us:
$$a=-\frac{1}{2}, b=\frac{3}{2}, c = 0$$
Therefore, the actual cubic equation that will pass through all these points is:
$$f(x)=\frac{-x^3}{2}+\frac{3x^2}{2}$$
Since we have the actual function, let's take the derivative:
$$f'(x)=\frac{-3x^2}{2}+3x$$
$$f''(x)=-3x+3$$
Since we have the second derivative being linear, we can figure out where the point $f(x)=0.5$ is. If that point lies inside $(0,3)$, then we have proven the function.
$$f''(x)=3-3x=0.5 \rightarrow x=\frac{-2.5}{-3}=\frac{5}{6}$$
This is inside the domain of the function, so we are good.

Alternatively, we could use the MVT like others have done and say:
$$f''(0)=3$$
$$f''(3)=-6$$
Since the second derivative is continuous (a straight line), by the MVT there exists a value in $(0,3)$ such that $0<f''(x)<1$.


Examples for subspace of a normal space which is not normal Are there any simple examples of subspaces of a normal space which are not normal?

I know closed subspace of a normal space is normal, but open subspace in most cases which I can think of are also normal.

One well-known example is $X=(\omega_1+1)\times(\omega+1)$, where $\omega_1$ is the first uncountable ordinal with the order topology, and $\omega$ is the first infinite ordinal. $X$ is the product of two compact Hausdorff spaces, so $X$ is compact and Hausdorff and therefore normal, but $X\setminus\{\langle\omega_1,\omega\rangle\}$ is not normal: the closed sets $\omega_1\times\{\omega\}$ and $\{\omega_1\}\times\omega$ cannot be separated by disjoint open sets. $X$ is often called the Tikhonov plank, though the name is also sometimes applied to $X\setminus\{\langle\omega_1,\omega\rangle\}$.

Added: We can also appeal to the theorem that a space $X$ is Tikhonov (i.e., $T_1$ and completely regular) iff it is homeomorphic to a subspace of some product of closed unit intervals. Every product of closed unit intervals is compact and Hausdorff, hence normal, but there are many examples of Tikhonov spaces that are not normal, and every non-normal Tikhonov space is an example of a non-normal subspace of a normal space. A couple of the better-known examples of non-normal Tikhonov spaces are the Moore plane and the Sorgenfrey plane. Another example is the space described in this answer, if $\mathscr{D}$ is taken to have cardinality $2^\omega=\mathfrak c$, which the addendum to the answer shows to be possible; non-normality of the space follows from Jones’s lemma.

The topology $T=\{\emptyset , X,\{a\},\{a,b\},\{a,c\}\}$ on $X=\{a,b,c,d\}$.

Take the subspace $Y=\{a,b,c\}$ with Topology $S=\{\emptyset,Y,\{a\},\{a,b\},\{a,c\}\}$. Then Y is not normal but $(X,T)$ is normal.


Working Out Easy Equations does anyone know how to do this equation? I know it's easy but I can't work out what the question means.

When I expanded the first equation:

$(y+4)-(y-3)$

$y^2 -3y +4y - 12$

$y^2-1y-12$

Not sure what I should do after.

Can someone explain how would you work it out in easy terms?

Thank you

When you made the jump from $(y+4)-(y-3)$ to $y^2-3y-4y-12$, you were applying the expansion rules for quantities multiplied together, not subtracted.

multiplication: $(y+4)(y-3)=y^2-3y+4y-12$

Subtraction: $(y+4)-(y-3)=y+4-y+3  = 7$

This is an identity, not an equation. It is true for any real number y. Hence you can't solve the equation to get a single number.


Calculate Exact Value of $\sin\theta, \cos\theta$ and $\tan\theta$ Having trouble getting a start on this problem, any help would be appreciated!

Given point $P = (-3,5)$ is on the terminal arm of angle $\theta$ in standard position. Calculate the exact value of $\sin\theta, \cos\theta$, and $\tan\theta$.

Hints:

What quadrant is the point $\;P = ({\bf x, y}) = (-3, 5)$ located?

Draw the right triangle that point $P$ makes with the $x$ axis - the length of the hypotenuse of the right triangle will equal $\;{\bf h} = \sqrt{(-3)^2 + 5^2} = \sqrt{34}$

Use SOH CAH TOA  to unpack the definitions of $\tan \theta, \;\sin\theta,\;\cos\theta$:

$$\tan\theta = \frac{\text{opposite}}{\text{adjacent}} = \frac yx = \quad?\;$$
$$\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac yh = \quad?\;$$
$$\cos \theta= \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{x}{h} = \quad?$$

$\sin \theta = p/h$ means perpendicular/hypotenuse
$\cos \theta = b/h$ means base/hypotenuse
$\tan \theta = p/b$ means perpendicular/base

This all means, if given two values in a right-angle triangle, we can find another third side of triangle.  Especially for 10 class, this formula helps.


Finding roots of a complex number I am asked to find $2^{1/5} $. Using the formula, and noticing $2 = 2e^{0 i} $, we have

$$ 2^{1/5} = 2^{1/5} \bigg( \cos( \frac{ 0}{5} + \frac{ 2 \pi k}{5} ) + i \sin( \frac{ 0}{n} + \frac{ 2 \pi k}{n} ) \bigg) $$

My confusing. I notice we can also write $2 = 2e^{2 \pi k} = 2e^{4 \pi k} = .... $ Do I need to consider every case, or would it be enough to just consider the case when $2 = 2e^{0i } $ as above?

One way to attack this baby is to remember that  the number of $n$-th complex roots of a number is always $n$, and that those $n$ roots form the vertices of an $n$-sided regular polygon on the Complex plane. So there are $3$ cube roots, forming a triangle, $4$ cube roots, forming a square, $5$ fifth roots, forming a pentagon, and so forth.

Int the above diagram, the pentagon is superimposed on the Complex plane. It has been broken broken up into five segments. The angle between each segment is $1/5$ that of a full circle. That is, each angle is $2\pi/5$. Each of the points, $A$ to $E$ is located by two numbers: One number is the distance from the centre of the figure. The other number is the angle that around the circle that it's located away from point $A$.

The angles are marked on the diagram. $A$ is $0$, (but it's marked on the diagram as $10\pi/5=2\pi$ because travelling $2\pi$ radians from any point on a circle will take you back to where you started).

Your angles are as follows:

$A. 0$
$B. 4\pi/5$
$C. 2\pi/5$
$D. 6\pi/5$
$E. 8\pi/5$

The number you also want to know the argument which is the Real number value of the length of each of those spikes from the centre to a point. That number is the Real value of $2^{1/5}$.

How do you arrange all this information is your final stretch. If we designate $z$ as your Complex root, $r$ as your argument (which is $2^{1/5}$ in each case) and $\theta$ for your angles, your general forms are:

$$z = r(\cos\theta + i\sin\theta)$$  and  $$z = e^{i\theta}$$

(Why these are equivalent is another story.)

So, your five roots are:

$$A.\; 2^{\frac{1}{5}}e^{0i}=2^{\frac{1}{5}}$$
$$B.\; 2^{\frac{1}{5}}e^{2\pi i/5}$$
$$C.\; 2^{\frac{1}{5}}e^{4\pi i/5}$$
$$D.\; 2^{\frac{1}{5}}e^{6\pi i/5}$$
$$E.\; 2^{\frac{1}{5}}e^{8\pi i/5}$$

Alternatively, these can be expressed as:

$$A.\; 2^{\frac{1}{5}}(\cos 0 + i\sin 0)= 2^{\frac{1}{5}}$$
$$B.\; 2^{\frac{1}{5}}(\cos \frac{2\pi}{5} + i\sin \frac{2\pi}{5})$$
$$C.\; 2^{\frac{1}{5}}(\cos \frac{4\pi}{5} + i\sin \frac{4\pi}{5})$$
$$D.\; 2^{\frac{1}{5}}(\cos \frac{6\pi}{5} + i\sin \frac{6\pi}{5})$$
$$E.\; 2^{\frac{1}{5}}(\cos \frac{8\pi}{5} + i\sin \frac{8\pi}{5})$$

These are your five roots, each expressed in two forms. The first form is  polar and the second is rectangular.

Any of these numbers, multiplied by itself five times will yield $2$. Multiplying a rectangular form by itself five times is tedious. Multiplying a polar form is amazingly simple. I'll end off with one example.

$$[2^{\frac{1}{5}}e^{4\pi i/5}]^5=[2^{1/5}]^5[e^{4\pi i/5}]^5=
2e^{20\pi i/5}=2e^{4\pi i}$$

Since $4\pi$ is two full rotations about the circle, $e^{4\pi i}=e^{0 i}=1$
So the equation above finishes:

$$2e^{4\pi i}= 2(1) = 2$$

which is what we wanted to show.

EDIT:
To address the OP directly, you can always add $2\pi$ to any angle without changing the value of $z$ just as you can go around a circle exactly once and end up where you started.

So, you need only show the solution for $\theta=0$ and $\theta=2\pi$ takes care of itself.

Hint: Put $2=x^5$;$x^5=1.2$ Now take 5th roots of both sides. And then use all $5$ 5th roots of unity.


Is the subbasis of a topological space preserved under an open continuous map? Let $X$ and $Y$ be topological spaces, and $f:X\to Y$ a continuous, surjective, and open function. If $\{S_\alpha\}_\alpha$ is a subbasis of $X$, is $\{f(S_\alpha)\}_\alpha$ a subbasis of $Y$ ? That is, if the topology on $X$ is the smallest one containing $\{S_\alpha\}_\alpha$, is the topology on $Y$ the smallest one containing $\{f(S_\alpha)\}_\alpha$ ? Note that the image of a basis of $X$ is a basis of the topology on $Y$.

The answer is NO: Let
$$f:\mathbb{Z}\to\{0,1\}$$
be a function defined by $f(\mbox{even})=0$ and $f(\mbox{odd})=1$.
Give $\mathbb{Z}$ and $\{0,1\}$ the discrete topologies. Then $f$ is open and continuous. The rays in $\mathbb{Z}$  of the form
$$(n,\infty)\qquad\mbox{and}\qquad(-\infty,n)$$
form a subbasis of $\mathbb{Z}$. Their images in $\{0,1\}$ are all the entire set $\{0,1\}$, so they do not form a subbasis of the discrete topology on $\{0,1\}$.

I think yes. Let's explore!

  Let $f \colon X \to Y$ be an open continuous map, where $X$ and $Y$ are topological spaces. Let $\mathscr{S}$ be a subbasis for $X$. Then the collection
  $$ \big\{ \, f(S) \, \colon \, S \in \mathscr{S} \, \big\} \tag{1} $$
  is a subbasis for the subspace $F(X)$ of $Y$. [Does this conclusion hold?]

In what follows, I'll be using the approach used by Munkres in defining subbasis for a topological spaces.

Proof:

  First, we note that, as
  $$ \bigcup_{S \in \mathscr{S}} S = X, $$
  so we must also have
  $$ \bigcup_{S \in \mathscr{S}} f(S) = f \big( \bigcup_{S \in \mathscr{S}} S  \big) = f(X). $$
  Thus the collection (1) is a subbasis for some topology on $f(X)$.

  For any $n \in \mathbb{N}$ and for any sets $S_1, \ldots, S_n \in \mathscr{S}$, the set
  $S_1 \cap \cdots \cap S_n$ is a basis set for the topology of $f(X)$, and conversely any basis set for the topology of $X$ is of the form $S_1 \cap \cdots \cap S_n$, where $n \in \mathbb{N}$ and $S_1, \ldots, S_n \in \mathscr{S}$.

  Let $V$ be any open set in $f(X)$. Then $V = f(X) \cap V^\prime$ for some open set $V^\prime$ in $Y$. Then we find that
  $$
\begin{align} &= f^{-1} \big( f(X) \cap V^\prime \big) \\
&= f^{-1} \big( f(X) \big) \cap f^{-1} \big( V^\prime \big) \\
&= X \cap f^{-1} \big( V^\prime \big) \\
&= f^{-1} \big( V^\prime \big),
\end{align}
$$
  which is an open set in $X$, being the inverse image under the continuous map $f$ of an open set $V^\prime$ in $Y$.

  In short, for any open set $V$ in $f(X)$, the inverse image $f^{-1}(V)$ is again an open set in $X$ and thus expressable as a union of sets of the form $S_1 \cap \cdots \cap S_n$, where $n \in \mathbb{N}$ and $S_1, \ldots, S_n \in \mathscr{S}$.

  However, there is one problem: Unless $f$ is injective, the equality
  $$ f \big( S_1 \cap \cdots \cap S_n \big) = f \big( S_1 \big) \cap \cdots \cap f \big( S_n \big) $$
  need not hold.

  Moreover, although
  $$ f\big( f^{-1}(V) \big) \subset V $$
  holds, the equality
  $$ f\big( f^{-1}(V) \big) = V $$
  does not hold in general if $f$ is not surjective.

Please refer to Sec. 2 in Munkres's Topology, 2nd edition.

Can you figure out the answer to your question from this?


Definition of a strictly increasing function What is the definition of strictly increasing for a function? Why does $f' >0$ imply $f$ is strictly increasing?
Much obliged if anyone can help me .

A function $f:X\to \mathbb R$ defined on a set $X\subset \mathbb R$ is said to be increasing if $f(x)\leq f(y)$ whenever $x<y$ in $X$. If the inequality is strict, i.e., $f(x)<f(y)$ whenever $x<y$ in $X$, then $f$ is said to be strictly increasing.

A similar definition for decreasing and strictly decreasing applies (simply reverse all inequalities).

If $X$ is connected (so, an interval), then the result "$f'>0$ implies $f$ strictly increasing" is a consequence of the mean value theorem (essentially, if $x<y$ and $f(x)=f(y)$, then there must exist a point $z$ with $x\leq z\leq y$ with $f'(z)=0$ since the slope of the secant between $(x,f(x))$ and $(y,f(y))$ would be $0$).

Let $A$ be the open subset where $f$ is differentiable. We know that $f'(x)>0$ for every $x \in A$. $f$ satisfies the criteria of the mean value theorem in $[x_1,x_2] \subset A$, therefore there exists $z \in (x_1, x_2)$ so that $f'(z)= \dfrac {f(x_1)-f(x_2)} {x_1-x_2}$. Since $f'(z)>0$, it follows that $\dfrac {f(x_1)-f(x_2)} {x_1-x_2}$, so $f(x_1)-f(x_2)>0$, i.e. $f(x_1)<f(x_2)$
(because $x_1<x_2$). Therefore $f$ is strictly increasing in $A$.


Geometry problem on the incentre and circumcenter of a triangle I have the following problem:

  In a triangle $ABC$ the line joining incentre and circumcentre is parallel to side $BC$. Prove that $\cos B + \cos C=1$.

Could someone help me solve it?

This question provides simple proofs of the identity: $$\frac{r}{R}+1=\cos A+\cos B+\cos C$$

Referring to Derek's diagram, $\angle CKB = 2\angle A = 2\angle CKR$. Since, the angle made by the chord $BC$ at the point $A$ on the circumcircle of $\Delta ABC$ is half the angle made at the circumcentre $K$.

Also, $\cos A = \cos \angle CKR = \dfrac{KR}{CK}=\dfrac{r}{R}$ (since, IK is parallel to $BC$, we have $KR = r$).

So, $\cos B + \cos C = 1$

Given Line joining Circumcentre and Incentre are parallel
$$r=4R\sin A/2\sin B/2\sin C/2=R\cos A$$
$$4\sin A/2\sin B/2\sin C/2=\cos A$$
$$\cos A+\cos B+\cos C-1=\cos A$$
$$\cos B+\cos C=1$$


A new imaginary number? $x^c = -x$ Being young, I don't have much experience with imaginary numbers outside of the basic usages of $i$. As I was sitting in my high school math class doing logs, I had an idea of something that would allow solving for logs with negative bases or with negative arguments. Taking a similar idea to i, this number, when used as an exponent, would result in a negative number. In other words, as the title says, $x^c = -x$.

For starters, I just quickly mentally ran through simple examples of how it could be used in these cases, (ex. $\log(-100) = 2c$) but I soon began to wonder if it could have other applications and or just if it could be anything else.

I'm also not 100% sure what to tag it, so I kinda guessed.

Robert's answer tells you correctly how to find, given an $x$, a complex $c$ such that $x^c=-x$.

However, it is worth pointing out that you cannot find a single $c$ such $x^c=-x$ will hold for every $x$. There's no existing number that achieves this, and if you try to extend the number system by simply postulating that $c$ exists (which I think was the point of the question), you run into this:
$$ 0 = \frac08 = \frac{-4+4}{8} = \frac{4^c+4}8 = \frac{2^c2^c+4}8 = \frac{(-2)(-2)+4}8 = \frac{4+4}8 = 1 $$
So if you want your $c$ to exist, but don't want to collapse everything into $0=1$, then you have to discard the exponentiation rule $(ab)^c=a^cb^c$. And without that rule, the result isn't really worthy of being called exponentiation in the first place.

I think you have been fooled by a common introduction to complex numbers, which goes like this: "There is no real number whose square is $-1$, but assume we had such a number anyway and call it $i$. Then we can calculate with the assumption $i^2=-1$, and, ta-dah! we have invented a new kind of numbers". This presentation, however, sweeps an important point under the carpet, namely that we had better be sure that assuming $i^2=-1$ doesn't allow us to prove falsehoods such as $0=1$ about the real numbers that we already know and love -- because if it does that means that we haven't invented a new kind of numbers but simply deceived ourselves.

For $i^2=-1$ it turns out that it's not too difficult to prove that it doesn't lead to nonsense. But even when this proof is shown, something is still swept under the carpet here, namely that if we assume $i^2=-1$ and the familiar rule that every number is either 0 or positive or negative (and the usual rules for positive and negative numbers), then we can still derive nonsense. So what's really true is that we can get a $\sqrt{-1}$ if and only if we're willing to drop what we know about positive and negative numbers (or "less than"/"greater than") when we're working with complex numbers.

Extending the concept of number is almost always connected to a loss of some rules that used to hold before the extension. How far we can push the extension depends on how many rules we're willing to let go of. So you can have your $c$, but the cost you must pay is the rules for how exponentiation works. And if you pay that cost, having $c$ becomes pretty useless anyway, so the whole expedition turns out to be pointless.

(But of course we couldn't have known that before we went and checked. The idea was good enough; it just didn't work. Perhaps your next one will.)


How prove $\dfrac{|AB|}{|AC|}\le\sqrt{2}+1$ in $\Delta ABC$ if the $AD\perp BC$,$D\in BC$,and such $$|BC|=2|AD|$$
show that
$$\dfrac{|AB|}{|AC|}\le\sqrt{2}+1$$

since
$$\cot{B}+\cot{C}=\dfrac{BD}{AD}+\dfrac{CD}{AD}=2$$
so
$$\dfrac{AB}{AC}=\dfrac{\sin{C}}{\sin{B}}$$

This is more of an algebra problem in fact, and I misled myself into thinking of Stewart's theorem at first.  In fact if we let the length of $AD$ be 1 and $BC$ be 2 it becomes equivalent to trying to maximise an expression like $\frac{x^2+1}{(2-x)^2+1}$ where $x$ varies between $0$ and $2$  and since the numerator increases while the denominator decreases as $x$ increases it suffices to let $x=2$ giving a bound of $\sqrt{5}$ which is smaller than the given bound.  I do not know if there is another method that gives $\sqrt{2}+1$ though.

I think you have only to check the worst of the cases, so is, prove that $\sqrt{5}\le\sqrt 2 +1$


Even moments = 0 iff $X=0$? Kind of a dumb question. I wasn't able to find this in probability texts, either elementary (Larsen and Marx) or advanced (David Williams), so it's probably wrong.
Is this false?

Let $X$ be a random variable s.t. $E[X^n] < \infty \ \forall n \in \mathbb N$.
$$(\exists n \in \mathbb N, E[X^{2n}] = 0) \iff X=0 \ \text{a.s.}$$

Pf:
'if' $\leftarrow$
is obvious, and the conclusion can be strengthened from $\exists$ to $\forall$.
'only if' $\rightarrow$
$X^{2n} \ge 0$. By Markov's inequality, $\forall a > 0$,
$$P(X^{2n} \ge a) \le \frac{E[X^{2n}]}{a} = 0$$
$$ \iff P(X^{2n} < a) = 1 \ \forall a > 0$$
$$ \iff P(X^{2n} = 0) = 1$$
$$ \iff P(X = 0) = 1$$
QED
Actually the assumption strengthens itself again from $\exists$ to $\forall$.
Is that wrong?
Also, what probability texts, either elementary or advanced, mention anything like this please?

Advanced probability stuff:
$X$ is a random variable in $\mathcal (\Omega, \mathscr F, \mathbb P)$.

Another way to say what you showed: the L2 norm of Xn is zero, so Xn is zero almost surely, so the same is true for X. So the 'well known' fact at the center is the the L^2 norm is a norm (which is more than what you showed). Maybe look at any book about lp spaces? Nice proof though, it's different from the usual one if I recall correctly. – Lorenzo May 12 at 21:56


Tangent map of homomorphism of Lie groups preserves commutator I am trying to prove the following proposition:

  Let $\phi:G\rightarrow H$  be a homomorphism of Lie groups, and $X,Y\in T_eG$. Then, $$  T_e\phi([X,Y]_G) = [T_e\phi X,T_e\phi Y]_H  $$

So we want to prove $ T_e\phi\circ ad^G(X)(Y) = ad^{H}(T_e\phi X,T_e\phi Y) $.
It is clear that $\phi\circ C^G_x = C^H_{\phi(x)} \circ \phi$. Applying $T_e$ to both sides and applying the chain rule yields
$$
       T_e\phi\circ \text{Ad}^G(x) = \text{Ad}^H(\phi(x)) \circ T_e\phi
$$
I now want to apply $T_e$ again and apply the chain rule to get the wanted result. But in order to do that I have to write both sides as functions of $G$. The LHS is already in that form. Applying $T_e$ to the $T_e\phi\circ \text{Ad}^G$ yields
$  T_e\phi\circ \text{ad}^G $. I don't know how to proceed with the RHS: how to write it as a composition in order to apply the tangent operation and the chain rule.

So finally I have an answer that satisfies me:

$$Ad_H(\phi(x))\circ T_e\phi = (R_{T_e\phi}\circ Ad_H\circ\phi)(x)$$
Hence $T_e\phi\circ Ad_G = R_{T_e\phi}\circ Ad_H\circ\phi$. The LHS was taken care of in the question.
Note: $R_{T_e\phi}$ is right composition with the linear map $T_e\phi$ (which is itself linear).
Taking the tangent map of $R_{T_e\phi}\circ Ad_H\circ\phi$ at the identity and applying the chain rule:
$$
      R_{T_e\phi}\circ ad_H\circ T_e\phi
$$
Hence $T_e\phi\circ ad_G=R_{T_e\phi}\circ ad_H\circ T_e\phi $. Applying both sides to $X$ and then to $Y$ we get
$$T_e\phi[X,Y]_G=[T_e\phi X,T_e\phi Y]_H$$

You need to use that $ad : \mathfrak{gl}(\mathfrak{g}) \to \mathfrak{gl}(\mathfrak{g})$ is the derivative of the map $Ad : GL(\mathfrak{g}) \to GL (\mathfrak{g})$.  This will involve chasing around lots of diagrams involving induced maps from $\phi$.


Solve $u_{xx}-3u_{xt}-4u_{tt}=0$ where $u(x,0)=x^2$ and $u_t(x,0)=e^x$ Solve
$$u_{xx}-3u_{xt}-4u_{tt}=0$$
where $u(x,0)=x^2$ and $u_t(x,0)=e^x$.

My workings so far: I have factored the differential equation in the following way:
$$(\delta_x-4\delta_t)(\delta_x+\delta_t)=0$$
where $\delta_x=\frac{\delta}{\delta x}$ etc. Now if we let $v$ be the solution to $(\delta_x+\delta_t)u$ then we have the following two equations:
\begin{eqnarray*}
(\delta_x+\delta_t)u=u_x+u_t=v\\
(\delta_x-4\delta_t)v=v_x-4v_t=0
\end{eqnarray*}
Now for $v$ we simply find
$$v=h(t+4x)$$
where $h$ is an arbitrary function of one variable. Now what remains is find $u$ such that
$$u_x+u_t=h(t+4x)$$
I am stuck here, I thought about making a change of variables $\zeta=x+t$ and $\eta=x-t$ and thus using the product rule to show that $u_x=u_{\zeta}+u_{\eta}$ and $u_t=u_{\zeta}-u_{\eta}$ and thus $u_x+u_t=2u_{\zeta}$ and we need to solve
$$u_{\zeta}=h(t+4x)$$
(I left out the factor 2 because $h$ is an arbitrary function). Do I simply integrate now and conclude
$$u=f(\zeta)h(t+4x)+g(\eta)=f(x+t)h(t+4x)+g(x-t)$$
This seems wrong to me...

Some help would be greatly appreciated!

Note that, since the pde is with constant coefficients, then we proceed as

$$u_{xx}-3u_{xt}-4u_{tt}=0\implies(\delta_x-4\delta_t)(\delta_x+\delta_t)u=0$$

$$\implies(\delta_xu-4\delta_tu=0)\cup(\delta_xu+\delta_tu=0)$$

$$\implies\left\{u(x,t)=F(4x+t)\right\}\cup\left\{u(x,t)=G(t-x)\right\}$$

$$\implies u(x,t)=F(4x+t)+G(t-x)$$

$$\implies u(x,t)=f\left(x+\dfrac{t}{4}\right)+g(x-t)$$

$$u(x,0)=x^2\implies f(x)+g(x)=x^2$$

$$u_t(x,0)=e^x\implies\dfrac{f'(x)}{4}-g'(x)=e^x\implies\dfrac{f(x)}{4}-g(x)=e^x+c$$

$$\therefore f(x)=\dfrac{4x^2+4e^x+4c}{5},g(x)=\dfrac{x^2-4e^x-4c}{5}$$

$$\therefore u(x,t)=\dfrac{4\left(x+\dfrac{t}{4}\right)^2+4e^{x+\frac{t}{4}}+(x-t)^2-4e^{x-t}}{5}$$

I solve with free CAS Maxima. We use Laplace transform method.

Answer:

$$u=\frac{4}{5}e^{x+\frac{t}{4}}-\frac{4}{5}e^{x-t}+x^2+\frac{t^2}{4}$$


Dominated convergence theorem calculating integral How do you evaluate $$\lim_{n \to \infty} \int_{0}^{1}\frac{n\cos(x)}{1+n^2x^{3/2}}\, dx\;\;?$$
I know to use the Dominated convergence theorem, however I cannot find a function that will dominate $$\frac{n \cos(x)}{1+n^2x^{3/2}}.$$
Any help will be much appreciated.

$|\cos(x)| \leq 1$, and $|1 + n^2 x^{3/2}| \geq |2nx^{3/4}|$, so
$$\begin{aligned}
\left|\frac{n \cos(x)}{1+n^2x^{3/2}}\right|
&\leq \left|\frac{n}{2nx^{3/4}}\right| \\
&= \left|\frac{1}{2x^{3/4}}\right|\\
\end{aligned}$$
which is integrable over $[0,1]$. To prove the key inequality $|1 + n^2 x^{3/2}| \geq |2nx^{3/4}|$, let $y = x^{3/4}$, so $y^2 = x^{3/4}$. Then
$$0 \leq (1 - ny)^2 = 1 + n^2 y^2 - 2ny$$
so
$$2ny \leq 1 + n^2 y^2$$
and therefore
$$2nx^{3/4} \leq 1 + n^2 x^{3/2}$$
Both sides are nonnegative since $x \geq 0$, so the absolute values change nothing:
$$|2nx^{3/4}| \leq |1 + n^2 x^{3/2}|$$

Consider a change of variables $y=nx$ (and so we have $\mathop{dy}= n \mathop{dx})$. This allows us to alter $$\frac{n\cos(x)}{1+n^2x^{3/2}}\mathop{dx}$$ into:

$$\dfrac{n\cos \left(\frac{y}{n}\right)}{1+n^2\left(\frac{y}{n}\right)^{3/2}}\mathop{dx}=\dfrac{\cos \left(\frac{y}{n}\right)}{1+n^{1/2}y^{3/2}}\mathop{dy}$$

Consider the dominating function $f(y)=\dfrac{1}{1+y^{3/2}}$

The result then follows from DCT.


How to write $b$ between $a$ and $c$ formally? How to write $b$ between $a$ and $c$ formally ? I mean it could be

1) $a<b<c$

or

2) $a>b>c$

but I want to leave it in the middle which one it is.

If I use the sandwich theorem for instance I think a notation for this might be usefull.

$a<b<c$ and $a>b>c$ are not the same (I assume that was a typo).

But, $a<b<c$ and $c>b>a$ are the same things. In this case, both notations work. The first one looks cooler and easier to understand (to me). By the way, Also, you need to make sure that the inequalities are strict, e.g. $1<1<1$ is false, but $1 \leq 1 \leq 1$ is true.


If I want to show (A and B) imply C with a contrapositive proof, can I assume (B and ~C) and show ~A? If I want to show (A and B) imply C with a contrapositive proof, can I assume (B and ~C) and show ~A?
Is showing (~C and B) implies ~A the same as showing ~C implies ~(A and B)?
Thanks in advance.

If you assume $\lnot C$, then $\lnot C \to [\lnot(A \land B)] \equiv (\lnot A \lor \lnot B)$, by the contrapositve.
If you also know that $B$ holds, then you can conclude that this in turn implies that $\lnot A$ holds, by disjunction elimination, hence you can conclude "If $(\lnot C \land B) \to \lnot A$".  But that only proves
$(1) \; [(A\land B) \to C]\quad \text{premise}$
$(2) B \;\quad \text{premise}$
$(3)\qquad \lnot C \quad \text{Assumption}$
$(4)\qquad\quad \lnot (A\land B) \quad \text{Modus Tollens, 1, 3}$
$(5)\qquad\quad \lnot A \lor \lnot B\quad \text{DeMorgan's, 4}$
$(6) \qquad\quad \lnot\lnot B\quad \text{Double Negation, 2}$
$(7) \qquad\quad \lnot A \quad \text{Disjunctive Syllogism, 5, 6}$
$(8) \qquad\quad\lnot A \lor \lnot B \quad \lor\text{-Introduction}$
$(9) \lnot C \to (\lnot A\lor \lnot B) \quad \text{conditional introduction}$
$\therefore (10) \lnot C \to \lnot (A \land B)\quad \text{DeMorgan's}$
And $$ \lnot C \to \lnot (A \land B) \equiv (A\land B)\to C$$ as its contrapositive, hence the proof is complete.

Perhaps more straight forward for your title question:  Yes.
This requires you understand the equivalence: $$(p \land q) \to r \equiv p \to (q\to r).$$
In your example, that means $(A\land B) \to C \equiv A \to (B\to C).\tag{1}$$
Now (1) is equivalent to its contrapostive:
$$\lnot(B \to C) \to \lnot A \equiv \lnot (\lnot B \lor C)\to \lnot A \equiv (B\land \lnot C)\to \lnot A\tag{2}$$
So with (2), assuming $B \land \lnot C$, then indeed, by Modus Ponens, $\lnot A$ is implied.
By contrapositive equivalence, you hav


Two children paradox: where is my reasoning wrong? I hope here is the good place to be asking this. Apologies otherwise.

The statement is as follows:

"Ms Michu has two children. We know one of the two is a girl, we call that girl Ludivine. What is the probability that Ludivine has a brother, rather than a sister?"

This is with the usual assumptions that there is an equal chance for one person to be a boy or a girl, no cis or gemels, etc.

I read that the answer should be 2/3. Demonstration is :
For any given set of 2 children, there are 4 different equiprobable combinations:
GG BB BG GB
We get rid of the bb combination because there is no girl. That leaves us with 3 possibilities, of which 2 match the criteria of the statement, hence the probability of Ludivine having a brother is of 2/3.

But, if I follow the following reasoning, I obtain a probability of 1/2:
Ludivine(L) is either the elder(e) or the youngest(y), with an equal probability of 1/2.
If she is the elder(e), it gives 2 combinations, with again a probability of 1/2 for each:
B(y) G(L)(e)
G(y) G(L)(e)
If she is the youngest we again have 2 equiprobable combination :
G(L)(y) B(e)
G(L)(y) G(e)
By combining those probabilities, we have 2 cases out of 4 where she has brother. Therefore she has a probability of 2/4 or 1/2 to have a brother.

Now I feel like my reasoning is wrong somewhere, but I can't see where or why, even though i went through quite a few topics on the subject; hence why I'm re-discussing that old topic...
Thanks in advance!

The problem is not really with your reasoning as much as with your interpretation of the question.

In the first interpretation, we know that there is a girl, and then we pick one of them, and then ask what is the probability that the other child is a boy. In this case, since we know that there is a girl, there are as you say exactly 3 equally probable possibilities, GG,GB,BG. And hence we can check that the answer is $\frac{2}{3}$.

In the second interpretation, we pick a child at random, and then it turns out to be a girl, and then we ask what is the probability that the other child is a boy. In this case, there are exactly 2 equally probable possibilities, GG and GB, where the first letter denotes the gender of the picked child, and the second letter denotes the gender of the other child. Clearly the answer is $\frac{1}{2}$. (Your added "reasoning" about the picked child being elder or younger isn't important.)

So the underlying issue is that when specifying a probability question, it is absolutely crucial to clearly state the order in which actions are performed.

How exactly do we know one of the two is a girl? We don't just abstractly know it. There is some story how we received the information. If we meet a person in hospital visiting a sick child, and that is how we gained the information, there is an equal chance that we would have received the information "Ms Michu has at least one boy" or "Ms Michu has at least one girl". If we meet Ms Michu picking up her child at a girls' school, we couldn't have received the information "Ms Michu has at least one boy" because it's a girls' school and she had to pick up a girl.

People with two boys turn up at hospitals visiting a sick child. People with two boys don't turn up at girls' schools to pick up their child. So how we got the information influences the result.


How does population growth relate to $e$? Say that there is a population of $10,000$. From year $0$ to year $1$, we know that the population has grown to $20,000$. How might I model this growth?

It clearly isn't sensible to say that the population suddenly 'jumped' from $10,000$ to $20,000$ as the year drew to a close. A better model might be to split the year up into $365$ segments, where on each day the population is multiplied by $\sqrt[365]{2}$ or approximately $1.0019$. However, even this feels unsatisfying, as the population would have grown during that day, which would have (very slightly) affected the rate of growth. The discrete model used in this example does not take into account the fact that the population would be slightly higher in the afternoon compared to the morning, which (at least in theory) would have an effect on population growth. If I use a continuous model, then I expect to see $e$ crop up. However, I can't seem to find it. Why is this? And, more generally, how does continuous modelling work—is it premised on splitting the year up into $n$ segments, where $n \to \infty$? (An answer that does not use university-level maths would be appreciated, but I understand if this is not possible.)

Your expectation is not unfounded.
We would model it as such: let $f(t)$ be the population at time $t$.
Then $f(0) = 10000$ and $f(1) = 20000$.

We implicitly assume that the growth of the population is proportional to to the population, and moreover that this proportion is constant in time.
This would translate symbolically into

$$f'(t) = cf(t) \tag{$*$},$$

where $c$ is the constant of proportionality, to be determined.
The solution to the differential equation $(*)$ is

$$f(t) = ae^{ct},$$

where $a$ is a constant to be determined.
From $f(0) = a$, we find that $a = 10000$.
For $c$, we compute

$$f(1) = 10000e^c = 20000 \implies e^c = 2 \implies c = \log 2.$$

Wrapping up, we find that the function that continuously describes the growth of of the population through time is

$${f(t) = 10000\cdot 2^t}.$$

$e$ need not show up in the final answer, but it's intrinsically tied to the solution of $(*)$ which is the heart of our problem's modeling.

EDIT: This is only an answer to how continuous compound growth works. In this problem, you would want a different approach.

There is a beautiful answer using calculus so I'll try to put it in not calculus terms.

Let the time period units be small (say, milliseconds). Then every time period, the population (assuming the simplistic compounding model that doesn't account for limit resources, etc.) growths by a factor of $(1+\frac{r}{n})$, where $r$ is the nominal growth rate for the year and $n$ is the number of time periods that fit into a year. In my example about milliseconds, $n$ will obviously be huge.

Imagine this compounding every time period. Since there are $n$ time periods per year, by the end of time $t$ years, the population will be $A_0 (1+\frac{r}{n})^{nt}$, where $A_0$ is the initial value.

What happens if instead of using milliseconds, we decide that we want a finer measurement? The finer the measurement, the larger $n$ becomes. It turns out in the "limit" (just means for $n$ really big), $(1+\frac{r}{n})^{nt} \approx e^{rt}$ and the population is then $\approx A_0 e^{rt}$.

Technically speaking, "compounding continuously" just means the limit of a bunch of small discrete compounds. It's often used when compounding occurs so often that it's not agreed upon how one would stop at discrete steps.


Group $G$ is cyclic $\iff$ every subgroup of $G$ has the form $G^k$, where $G^k=\{g^k\mid g\in G\}$ Let $G$ be a group (finite or infinite).

Given $k \in \Bbb N_+$, define $G^k=\{g^k\mid g\in G\}$

I hope to prove that $G$ is cyclic if and only if every subgroup of $G$ has the form of $G^k$.

It's easy to see that if $G=\langle g\rangle$ is cyclic, then every subgroup $H$ of $G$ must have the form of $\langle g^k \rangle$. Therefore $H=\langle g^k \rangle=\langle g \rangle^k=G^k$.

...excluding the trivial subgroup {e} of an infinite group (complement)

But I get into trouble proving the other side.

I have no idea how to find a counterexample, i.e. there exists a subgroup of $G$ not having the form of $G^k$ if $G$ is not cyclic.

Meanwhile, I tried to prove it directly. But I cannot even prove $G$ is Abelian, considering $G^2$ as a normal(proved) subgroup of $G$.

Could you please help me solve the problem or disprove this proposition?

Thanks a lot! And sorry for my poor English..

Anyway, I'd appreciate for everyone's help. I have got the proof by myself (aha!), as following:

For the trivial subgroup $\{e\}$, there exists some positive integer $n$ so that $\{e\}=G^n$. Then

$\forall a\in G, a^n=e$, which means $o(a)$ (the order of $a$) is a factor of $n$.

As $A=\{m\in\mathbb{N^*}\mid\exists a\in G,o(a)=m\}$ is bounded (namely, $n$ is an upper bound), thus finite, we can get a finite set $B=\{p\in\mathbb{N^*}\mid \exists a\in G\text{ so that }p\text{ is a prime factor of }o(a)\}=\{p_1,\cdots,p_r\}$.

Next, every subgroup $H=G^k$ of $G$ is normal. Notice that

$\forall g\in G,\forall a^k\in H,ga^kg^{-1}=(gag^{-1})^k\in H\implies gHg^{-1}=H$.

According to Sylow's first theorem, $\forall p_i\in B,\exists g_i\in G,s.t. o(g_i)=p_i$. I'd show that $g_i$ is in the center of $G$.

$\forall a\in G,ag_ia^{-1}\in\langle g_i\rangle\implies\exists t\in\mathbb{N},s.t.ag_ia^{-1}=g_i^t$ (if $p_i|t$, then $g_i=a^{-1}ea=e$.Thus without loss of generality, $0<t<p_i$)

Meanwhile, $g_i^{-1}ag_i\in\langle a\rangle\implies\exists s\in\mathbb{N},s.t. g_i^{-1}ag_i=a^s$

We can get $g_ia^s=ag_i=g_i^ta\implies a^{s-1}=g_i^{t-1}\in\langle g_i\rangle$.

For $\gcd(t-1,p_i)=1$, $\langle g_i^{t-1}\rangle=\langle g_i\rangle\implies g_i\in\langle g_i^{t-1}\rangle=\langle a^{s-1}\rangle\implies g_i\in\langle a\rangle\implies g_i$ commutes with $a$. As $a$ is arbitrary, $g_i \in Z(G)$.

Let $g_0=g_1\cdots g_r\in Z(G)$, it's obvious that $o(g_0)=p_1\cdots p_r$.

Then $C=\{a\in G\mid p_1\cdots p_r|o(a)\}\neq\varnothing$, $D=\{m\in\mathbb{N^*}\mid\exists a\in C,o(a)=m\}$ is a bounded non-empty subset of $\mathbb{N}$. We can get $M=\max D$ and select some $h\in C$ so that $o(h)=M$.

There exists some $l$ so that $\langle h\rangle=G^l$. Particularly, $\exists b\in G,s.t. h=b^l$, and $o(h)=o(b^l)=\dfrac{o(b)}{\gcd(l,o(b))}|o(b)$

On one hand, $o(h)\leqslant o(b)$; on the other hand, $b\in C\implies o(b)\leqslant M=o(h)$. Thus $o(h)=o(b)\implies\gcd(o(b),l)=1\implies\gcd(p_1\cdots p_r,l)=1$.

Then $\forall a\in G, \gcd(o(a),l)=1$ ($o(a)$ must have the form of  $\prod_{j=1}^rp_j^{\alpha_j}$). $\exists u,v\in\mathbb{Z},s.t. uo(a)+vl=1\implies a=a^{uo(a)+vl}=(a^v)^l\in G^l=\langle h\rangle$.

Therefore we got the consequence, $G=\langle h\rangle$ is cyclic.

: )

Consider $S_3$ ;

Every subgroup has the form $G^k$ but $S_3$ is still not cyclic.

The subgroups are $\langle (12)\rangle,\langle (23)\rangle,\langle (31)\rangle,\langle (123)\rangle$


Conditional expectation on Bernoulli Process If we have a Bernoulli Process with success probability p, and if X equals the number of successes in first trial, Y the number of successes in first two trials, and Z the number of successes in the first three trials,
how can I find E(XZ | Y)?

I know that X and Z must be conditionally independent given Y but I'm not sure how I can use it here. I couldn't just take E(X | Y)E(Z | Y) can I?

Thanks everyone in advance.

Let $\{X_n:n=1,2,\ldots\}$ be a sequence of i.i.d. $\operatorname{Ber}(p)$ random variables and $\{S_n:n=0,1,2,\ldots\}$ the corresponding Bernoulli process (i.e. $S_0=0$ and $S_n=\sum_{j=1}^n X_j$ for $j>0$). Let $\{\mathcal F_n:n=0,1,\ldots\}$ be the natural filtration of $\{S_n\}$. Then for any $n\geqslant0$ and bounded measurable $f$ we have we have
\begin{align}
\mathbb E\left[f(S_{n+1})\mid \mathcal F_n\right] &= \mathbb E\left[f(S_{n+1})\mathsf 1_{\{X_{n+1}=0\}} \mid\mathcal F_n\right] + \mathbb E\left[f(S_{n+1}\mathsf 1_{\{X_{n+1}=1\}}\mid\mathcal F_n\right]\\
&= \mathbb E\left[f(S_n)\mid\mathcal F_n\right]\mathbb P(X_{n+1}=0) + \mathbb =E\left[f(S_n+1)\mid\mathcal F_n\right]\mathbb P(X_{n+1}=1)\\
&= p\left(f(S_n)+f(S_n+1) \right)\\
&= g(S_n),
\end{align}
where $g$ is the (measurable) map $x\mapsto p\left(f(x)+f(x+1)\right)$. Therefore $\{S_n\}$ is a Markov process, and in particular,
$$
\mathbb E[S_1S_3\mid \mathcal F_2] = S_1\mathbb E[S_3\mid\mathcal F_2] = S_1\left(S_2+p\right).
$$
If we are conditioning on (the $\sigma$-algebra generated by) $S_2$ as opposed to $\mathcal F_2$, then the computation is a bit different:
\begin{align}
\mathbb E[S_1S_3\mid S_2] &= \mathbb E[X_1S_2 + X_1X_3\mid S_2]\\
&= S_2\mathbb E[X_1\mid S_2] + \mathbb E[X_3]\mathbb E[X_1\mid S_2]\\
&= S_2\cdot \frac12 S_2 + p\cdot \frac12 S_2\\
&= \frac12 S_2\left(p+S_2\right).
\end{align}

I know that X and Z must be conditionally independent given Y

You don't know that, because they are not.   Consider the probability of having one success in the fist trial given that you have two successes in the first two trials.$$\mathsf P(X=1\mid Y=2) =1\quad\neq\quad p=\mathsf P(X=1)$$

Likewise the probability of having some successes in the first three trials is dependent on how many successes are given in the first two trials.

Let $A,B,C$ be the indicators for success in the first three trials.  These are independent and identically distributed.

Then $X=A\,, Y=A+B\,, Z=A+B+C$

$\mathsf E(XZ\mid Y) = \mathsf E(A^2+AB+AC\mid A+B) \\ = \mathsf E(A^2\mid A+B)+\mathsf E(AB\mid A+B)+\mathsf E(A\mid A+B)\mathsf E(C)$

Now, clearly $\mathsf E(C)=p$

And $\mathsf E(A\mid A+B) =\frac{A+B}{2}$ due to symmetry. Because $\mathsf E(A+B\mid A+B) = \underbrace{A+B}_Y$

Simmilarly $\mathsf E((A+B)^2\mid A+B)=(A+B)^2$, so...


How to calculate weight of positive and negative values. We have used formula formula to calculate weight as,

$$ w_1 = \frac{s_1}{s_1 + s_2 + s_3};$$
$$ w_2 = \frac{s_2}{s_1 + s_2 + s_3};$$
$$ w_3 = \frac{s_3}{s_1 + s_2 + s_3};$$

However, their is possibility of negative and positive numbers. Even all can be negative or positive. How to calculate weight in this type of situation.

For us -/+ are according to strict algebraic rules. i.e bigger the negative number smaller will be its value.

Thanks.

Let the smallest of the numbers $s_1$, $s_2$, $s_3$ be $m$. Let $t_i=s_i-m+1$ for each $i$. Then the $t_i$ are all positive, and satisfy the same order relatins as the $s_i$, for example, if $s_1\lt s_2$, then $t_1\lt t_2$. So you can use your formula to assign weights to the $t_i$, and then assign those same weights to the $s_i$.

Whether you will like the weights you get, I can't say. Experiment a bit and you may find some other constraint you want your weights to satisfy, then we can refine the technique.

I'm a little confused by your question. When you say the weight can be positive or negative, do you mean just the value can be negative or positive because of the measurement technique, or is it actually a negative weight? I would assume the first (for example, if you slow down really fast in an elevator, and measure your weight in that frame, you'll actually have 'negative' weight).

With that said,  I think what you're after is the RMS value (root mean squared). It's a common technique used to measure velocity since velocity is a vector and can have negative components, but often we care only about it's magnitude. If that is the case for your weight, then do the following.

$w1 = (s1^2/(s1^2 + s2^2 + s3^2))^{1/2}$

$w2 = (s2^2/(s1^2 + s2^2 + s3^2))^{1/2}$

$w3 = (s2^2/(s1^2 + s2^2 + s3^2))^{1/2}$

If you indeed just want the average and weight can be negative, do exactly as what the formulas you provided for us tell you to do.


Too old to start math I'm sorry if this question goes against the meta for posting questions - I attached all the "beware, this is a soft-question" tags I could.

This is a question I've been asking myself now for some time. In most areas, there's a "cut off age" to be good at something. For example, you're not going to make the NHL if you start playing hockey at 20. It just won't happen.

So my question then, how late is too late to start studying math and make a career out of it? By "start studying math" I mean, to really try to understand and comprehend the material (as opposed to just being able to do well in a formal, intuitional environment).

I don't mean this from a "do what you love, its not too late" motivational perspective. I mean this from a purely biological perspective; at approximately what age has your brain's capacity to learn effectively and be influenced by your learning stop? When does the biological clock for learning new math run out?

My reasoning for asking this question is (for those who care): I love math. Really I do. But , having spent the first 21 years of my life in sports/video games/obtaining a degree in a scientific field which I care nothing of/etc, despite all my best attempts at trying to learn math, am I just too late starting to ever actually be good enough at it to make it a career? I've almost completed my second degree (in Math), but find that in many cases, despite how I look at a problem, I lack the intuition to comprehend it. I'm going to single him out (sorry), only as an example, but Qiaochu Yuan is my age.

Note 1: If this question isn't a suitable post, I won't be offended at all if you vote to close - I know this question borders what's acceptable to ask.

Note 2: Thanks to everyone for reading and taking the time for the great responses. Really appreciate it!

Karl Weierstrass was in his 40's when he got his PHD. There are a dozen other counterexamples, a number fairly recent. A good set of examples can be found in the thread on MO here.This myth of "science is a game for the young" is one of the falsest and most destructive canards in modern society. Don't listen to it. You only get one life and when it's over, that's it. When you're dead a hundred million years, you'll be dead the tiniest most infinitesimal fraction of all the time you'll ever be dead. So stop listening to career advice from teenagers, grab a calculus book and get to work. That's my advice.

While I think you are never too old to start anything, I really think Math sports needs to be learned and cultivated from a young age to be truly effective.


Show that $\lim_{n \to \infty} \frac{1}{n!} \frac{n^{n+1}}{e^n} = \infty$ Show that
$$\lim_{n \to \infty} \frac{1}{n!} \frac{n^{n+1}}{e^n} = \infty$$

Wihout using stirlings aproximation to n!

I've tried to compare this to a divergent sequence but didnt work. Also, I dont see how to use L'Hospital or slmething like this.

Thanks in advance!

Lemmas:

This one could easily be proven by induction: $\displaystyle \prod_{k=1}^{n}\left(1+\frac{1}{k} \right) = n+1$
You can try to prove these inequalities yourself since it's not difficult: $\displaystyle \left (1+\frac{1}{k} \right )^k\leq e \leq \left (1+\frac{1}{k}  \right)^{k+{1 \over 2}}\\$ . We will only need the second inequality, but both are easily provable by taking the logarithm of both sides.

Now,

  We first write $n^n/n!$ in a better way: $\displaystyle \frac{n^n}{n!}=\prod_{k=1}^{n-1}\left(1+\frac{1}{k} \right)^n \cdot \prod_{i=1}^{n-1}\prod_{k=1}^{i}  \left(1+\frac{1}{k} \right)^{-1}=\prod_{k=1}^{n-1}\left(1+\frac{1}{k} \right)^n\cdot \prod_{i=1}^{n-1}  \left(1+\frac{1}{i} \right)^{-(n-i)}=\prod_{i=1}^{n-1}  \left(1+\frac{1}{i} \right)^{i}$
  Then, a lower bound: $\displaystyle \prod_{i=1}^{n-1}  \left(1+\frac{1}{i} \right)^{i} \geq \prod_{i=1}^{n-1}\left (e^{1} \cdot  \left(1+\frac{1}{i}\right)^{-1/2} \right )=\frac{e^{n-1}}{\sqrt[2]{n}}$
  Finally, $\displaystyle {n^n \over e^n n!}\geq n^{-1/2}e^{-1} \Rightarrow {n^{n+1} \over e^n n!}\geq {n^{1/2} \over e}$ and we are done.


Re-arranging exponential equation Not quite applicable to many situations, but I'm extending this as more of a "I need help/tips" on re-arranging an equation.

I have something like this:

$$
0.02=\frac{1}{(1+c)}\left(1-e^{-\frac{Z}{y/(1+c))}}\right)
$$
I want to re-arranging this equation for $c$. How would this be done? I know that this is perhaps not the sort of question that should be asked here, but it seems to me a little tricky than I first thought, and I'm all out of ideas!

Am I looking at having to perform a numerical solution?

These equations need the Lambert W function, defined by $W(xe^x)=x$.
Equivalently, if $y=x+e^x$, then $e^y=e^xe^{e^x}$, so $e^x=W(e^y)$, and $x=\log W(e^y)$.
Let $v=\frac{-z(1+c)}y$, then the equation becomes
$$\frac {-yv}{50z}+e^v=1\\
v+\log(-50z/y)+e^{v+\log(-50z/y)}=-50(z/y)+\log(-50z/y)\\
\text{Let }u=v+\log(-50z/y)\\
u+e^u=-50(z/y)+\log(-50z/y)\\
u=\log W(-50(z/y)e^{-50z/y})$$
from which you can untangle $c$.