Number of Paths on a Grid From S to M I was reading about a the derivation of the formula for the number of paths from one corner to another corner of a H by W grid here and I wondered whether it is possible to apply the result: $\binom{(H-1)(W-1)}{H-1}$ to find the number of paths from a given square on the top row of the grid to another selected square in the bottom row.

For example the number of paths from B to J.
I thought of reducing the grid to just the columns B to D, counting the paths there and then adding the possible paths from every other square outside the reduced grid. However I had trouble in finding a formula for the possible paths from outside of the reduced grid.
In a path you cant repeat a square, and you can move to any adjacent square.

A possible (but involved) approach
You could represent moves by complex numbers:
$1$ is move right, $-1$ is move left, $i$ is move up, $-i$ is move down.
Then a path is a sequence such as $1,-i, -1, ...$
The conditions for a valid path then have simple arithmetical equivalences.
To move from B to J The total sum of the numbers must be $2-5i$.
To stay in the grid For any natural number $n$, the sum $S_n$ of the first $n$ numbers must  satisfy $4\ge \Re (S_n)\ge -1, 0\ge \Im (S_n)\ge -5. $
To not visit any square twice No sum of successive numbers must have sum $0$.
This looks suitable to be programmed if that is of interest.

In the original source, note that each step in the path was Right or Down.  With those rules, any path from B to J requires 2 Rights and 5 Downs, in any order.  There are $\binom{5+2}{2} = 21$ such paths.  But to go from B to G, say, you would need Left and Down moves.  If you allow Left and Right both, then the number of paths is infinite, as you could move Right Left Right Left... as long as you like.
Edit: With the restriction about not visiting a position more than once, the number of paths with every direction allowed is finite but large (compared to the board size).  For instance, the number of paths between opposite corners of a $2 \times n$ grid is $2^{n-1}$, already exponential.


Apex of an Exponential Function Is there a way of calculating where the apex of an exponential lies? There's probably a deeper / more mathematical way of explaining what this is exactly. The image hopefully demonstrates what I mean. I'm not sure if derivatives help and perhaps numerical methods need to be used?

You can compute the maximum of the curvature of the curve $y=e^x$. This can be done by the formula
$$\kappa=\frac{y''}{(1+(y')^2)^{3/2}}=\frac{e^x}{(1+e^{2x})^{3/2}}.$$
A simple computation shows that this is maximal for $x=-\frac{\ln 2}{2}$ (after edit).

However, for the function $A\,e^{Bx}$ where $A,B>0$, the result may be different and the computed quantity is not invariant with respect to the "scaling" of your graph. Note that you have a different scale in the $x$- and $y$-axis in your picture.

By a "measurement on the screen" it seems to me that your 4 units on the $x$-axis correspond to $100$ on the $y$-axis in your graph. So you really draw the function $\frac{1}{25} e^x$ (using an equi-scaled graph) which leads to the maximal curvature in $\frac{-\ln 2}{2}+\ln 25\simeq 2.87$ which is already approximately 3 as you want.

I guess exponential is too crazy for that kind of stuff - it's growing - well, exponentially - so basically, always growing more than ever before - not just getting bigger - but getting more bigger than it ever got bigger before. All I'm saying is that the derivative always grows, i.e. that the second derivative is positive. Problem is, all derivatives are positive anyway ... So it's also getting more more bigger than it ever got more bigger before, and so on ...

At the end of the day, such apexes as searched are everywhere, given that the slope is always steeper after any point than it was before, by a margin that was never so great before, ...


When is the union of $\sigma$-algebras atomless? Suppose that we are given a probability space $(\Omega, \mathcal{F}, \mathsf P)$ and an increasing sequence of $$\mathcal{F}_1\subset \ldots \subset\mathcal{F}_n\subset \mathcal{F}_{n+1} \subset \ldots \subset \mathcal{F}$$
of $\sigma$-algebras. Assume that $\mathsf P|_{\mathcal{F}_n}$ is atomless for each $n$. Is so $\mathsf P$ restricted to the $\sigma$-algebra generated by the union of $\mathcal{F}_n$ ($n\in \mathbb{N}$)?

Actually, something much stronger is true: If $P$ is a measure on a $\sigma$-algebra $\mathcal{F}$ and $\mathcal{G}\subseteq\mathcal{F}$ is a $\sigma$-subalgebra on which $P$ is atomless and $\sigma$-finite, then $P$ is atomless on all of $\mathcal{F}$.  To show this, suppose $A\in \mathcal{F}$ is an atom.  Then for any $B\in \mathcal{G}$, either $P(A\cap B)=0$ or $P(A\cap B)=P(A)$.  Let $\mathcal{I}$ be the collection of all $B\in\mathcal{G}$ such that $P(A\cap B)=0$.  Then $\mathcal{I}$ is a closed under countable unions, and must contain all $B$ such that $P(B)<P(A)$.  But since $P$ is atomless and $\sigma$-finite on $\mathcal{G}$, every element of $\mathcal{G}$ can be written as a union of countably many sets $B\in\mathcal{G}$ such that $P(B)<P(A)$.  It follows that $\mathcal{I}$ is all of $\mathcal{G}$.  But clearly $\Omega\not\in \mathcal{I}$, so this is a contradiction.

An atom is an indivisible set in the sigma algebra with positive measure.

Consider $\mathcal{F}_n$ a sigma algebra of subsets of the space $[0,1]^2$.

The construction is a bit artificial. so I will divide it into steps.

1) consider a the sets $[0,1] \cup (n,n+1)\backslash E_n$ as the basic sets of your sigma algebra $\mathcal{F}_n$ ($E_n  \subset [n-1,n+1]$ is a denumerable set)

Note that there are no atoms in $\mathcal{F}_n$

2) set P([0,1]) = 1

Note that $[0,1]\in \sigma(\mathcal{F}_1 \cup \mathcal{F}_2)$ $[0,1]$ is indivisible and has positive measure. Therefore it is an Atom.


Differential Equation @ 2nd Order How to solve this? $$\dfrac{d^2x}{dt^2}=-kx^2$$

I need it in physics problem.https://physics.stackexchange.com/questions/60202/non-shm-oscillatory-motion/     Though solved without it but still it would be better to have this.

The differential equation is not easy to solve. The usual trick is to let $v=\frac{dx}{dt}$. Multiply both sides by $2\frac{dx}{dt}$.

On the left-hand side we have the derivative of $v^2$. On the right, we have the derivative of $-\frac{2k}{3}x^3$. So $v^2=-\frac{2k}{3}x^3+C$. We end up with a separable equation. Unfortunately, unless we are very lucky and $C=0$, we end up with something that cannot be integrated in elementary terms.

I think this might be part of the solution

let x(t) be a Taylor series

the co-efficients of the taylor series contain x(0), x'(0), x''(0) ... etc
I think these can be derived from d2xdt2=−kx2 and some initial conditions for x(0) = a, and x'(0) = b

http://i.stack.imgur.com/nRi4k.jpg

The final recurrence relation can be used to compute the x(0), x'(0), x''(0) co-efficients which can then be used to construct the taylor series for x(t)

It may not be nice but I believe it will work ...


Determine if it is a tautology, contradiction, or contingency I'm trying to figure out is it a tautology, contradiction, or contingency?
$(∀x.∀y.p(x,y))↔(∀y.∀x.p(x,y))$
I think $(∀x.∀y.p(x,y))$ and $(∀y.∀x.p(x,y))$ both true, so the answer is tautology, but my teacher told me I'm wrong.
Who can tell me why?

The truth of the individual statements $\forall x \forall y \ p(x,y)$ and $\forall y \forall x \ p(x,y)$ depends on what $p(x,y)$ is saying and what the domain is over which you are quantifying. For example, if $p(x,y)$ means that $x=y$, then $\forall x \forall y P(x,y)$ is saying that everything is identical to everything ... which would be false for most domains though if your domain contains exactly one object, it would actually be true.

So no, you should not try to evaluate the truth-values of the individual statements, since you cannot know those.

Also, even if you would know both statements to be true, then that would only tell you that for that one possible interpretation, the biconditional is true, but in order for the biconditional to be a tautology, the biconditional needs to be true in every possible interpretation, not just for one possible interpretation.

So, if you think the biconditional is a tautology, you need to provide a general argument that shows that no matter how we interpret $p(x,y)$, and no matter what domain we use, the left and right side of the biconditional will always have the same truth-value.

If, on the other hand, you think the statement is a contradiction, you need to show how no matter how we interpret $p(x,y)$, and no matter what domain we use, the left and right side of the biconditional will always have the opposite truth-value.

And finally, if you think the statement is neither, then you need to come up with some interpretation in which the two statements have the same truth-value, as well as an interpretation where they have opposite truth-values.

It's the three ones under disjunction, if sets with x, y are given and for all the elements of those sets p(x,y) is true, then ∀x.∀y.p(x,y) is true and implies ∀x.∀y.p(x,y) so your statement (∀x.∀y.p(x,y))↔(∀y.∀x.p(x,y)) is a tautology, if the given sets can't make true p(x,y) for some x, y, then ~(∀x.∀y.p(x,y)) that implies ~(∀x.∀y.p(x,y)) (since it's iff and by modus tollens) so your statament is not a tautology but it's not necessarily a contradiction too, thus what I can say about it it's that your statement it's a contradiction xor your statement it's a tautology xor your statement it's a contigency, in fact nothing special to find.


Derivative of a function when it is squared. Was wondering when you are for example finding $dw/dt$ but you are given a function like $w^2(t)=r^2-2\cos(t)$, when r is some constant, how you are supposed to solve it? Are you supposed to incorporate chain rule?

thanks

You've got $$w^2=r^2-2\cos(t).$$ Now differentiate everything with respect to $t$ to give $$\underbrace{2w \cdot\frac{\mathrm{d}w}{\mathrm{d}t}}_{\text{chain rule}}=0+2\sin(t).$$

Now re-arrange this to find $\frac{\mathrm{d}w}{\mathrm{d}t}.$

Note that $r$ is constant, so $r^2$ is constant, so the derivative of $r^2$ wrt $\ t\ $ is zero.

A derivative, conceptually, is the rate of change.  Constants do not change over the dependent variables.

r is a constant, there for r^2 is a constant.  The derivative of a constant is 0, thus can be effectively ignored.


Trigonometry: Law of Sines How can this be solved using law of sines? I get a different answer then when I solve it using law of cosines..

Let $\theta$ be the angle opposite $4$. Then we have
$$\frac{\sin\theta}{4}=\frac{\sin(65^\circ)}{5}.$$
Calculation shows that $\theta$ is about $46.47$ degrees, or $180^\circ$ minus that. The second is impossible, it would make the angle sum greater than $180^\circ$.

Now we can find the third angle by subtraction, and use the Sine Law again to find $a$.

I'm new to using latex so please excuse the displayed equations. The underlying mathematics should be accurate. Feel free to correct it if you know how.

The interior sum of angles of a triangle equal $180^\circ$. Therefore we can use [Law of Sines][1] to find $\angle A$ opposite a by deduction.

Let B be the angle opposite 4 and C=$65^\circ$. Then we have c=5.

$$\frac a{\sin A^ \circ}=\frac b{\sin B^\circ}=\frac c{\sin C^\circ}$$
$$\frac b{\sin B^\circ}=\frac c{\sin C^\circ}$$
$$\frac 4{\sin B^\circ}=\frac 5{\sin 65^\circ}$$
$$\frac 1{\sin B^\circ}=\frac 5{4\sin 65^\circ}$$
$${\sin B^\circ}=\frac {4\sin 65^\circ}5$$
$$B^\circ=\sin^{-1} ({\frac {4\sin 65^\circ}5})$$
$B=46.47269373^\circ$

Now we can find the third angle by [Sum of Angles of triangle][1] = $180^\circ$

$$180^\circ=A+B+C$$
$$180^\circ-(A+C)=B$$
$$B=68.53^\circ$$

Now you can use either [law of cosines] or [law of sines] to find $a$

Law of Cosines states:
$$a^2=b^2+c^2-2bc\cos A^\circ$$
$$a=\sqrt{b^2+c^2-2bc\cos A^\circ}$$
$$a=\sqrt{5^2+4^2-2(5x4)\cos 68.53^\circ}$$
$$a=4.91$$


Pure geometry proof for triangle problem Here is the orthologic triangles theorem:

  $ABC$ and $A'B'C'$ are triangles in the same plane. If the perpendiculars from $A$, $B$ and $C$ to $B'C'$, $C'A'$ and $A'B'$ respectively are concurrent, prove that so are the perpendiculars from $A'$, $B'$ and $C'$ to $BC$, $CA$ and $AB$ respectively.

In my book two solutions have been given using coordinate geometry and another using vectors. But I was wondering if you can solve this using pure geometry. Any help would be appreciated.

Recall well-known lemma:

Lemma 1. Let $D,E,F$ be points on lines $BC, CA, AB$. Then lines perpendicular to $BC, CA, AB$ through $D,E,F$ are concurrent if and only if $BD^2+CE^2+AF^2=DC^2+EA^2+FB^2$.

Recall another known lemma:

Lemma 2. For any points $A,B,C,D$ the following equivalence holds: $AC \perp BD \iff AB^2+CD^2=BC^2+DA^2$.

We can easily generalize lemma 1 using lemma 2 to the following

Lemma 3. Let $A,B,C,D,E,F$ be any points. Then lines perpendicular to $BC, CA, AB$ through $D,E,F$ are concurrent if and only if $BD^2+CE^2+AF^2=DC^2+EA^2+FB^2$.

Now we are ready to prove the theorem.

First, we put $(A,B,C,D,E,F):=(A',B',C',A,B,C)$ in lemma 3 and we get the equality $$B'A^2+C'B^2+A'C^2=AC'^2+BA'^2+CB'^2.$$ Now we put $(A,B,C,D,E,F)=(A,B,C,A',B',C')$ in lemma 3 and conclude that the equality above implies that perpendiculars to $BC, CA, AB$ through $A',B',C'$ are concurrent. $\square$

There is a rotation and scaling (without translation) with respect to  common  orthocenter  of ABC and A'B'C'. Thus the two triangles are duals, or, alternately belong to a set of dialation/rotation transforms.

So its proof is easier ( I believe) with complex number sets.


Do the real numbers and the complex numbers have the same cardinality? So it's easy to show that the rationals and the integers have the same size, using everyone's favorite spiral-around-the-grid.

Can the approach be extended to say that the set of complex numbers has the same cardinality as the reals?

Yes.

$$|\mathbb R|=2^{\aleph_0}; |\mathbb C|=|\mathbb{R\times R}|=|\mathbb R|^2.$$

We have if so:

$$|\mathbb C|=|\mathbb R|^2 =(2^{\aleph_0})^2 = 2^{\aleph_0\cdot 2}=2^{\aleph_0}=|\mathbb R|$$

If one wishes to write down an explicit function, one can use a function of $\mathbb{N\times 2\to N}$, and combine it with a bijection between $2^\mathbb N$ and $\mathbb R$.

One particularly nice class of bijections from $\Bbb R$ to $\Bbb C = \Bbb R^2$, which is in my opinion a little bit similar to the spiral around the grid, is given by the space-filling curves.


Describe the non-trivial periodic orbit of a system of ode's Consider the system

$\frac{dx}{dt}=y+ax-x(x^2+y^2),\;x(0)=x_0$

$\frac{dy}{dt}=-x+ay-y(x^2+y^2),\;y(0)=y_0$

The question is describe for $a> 0$ the non-trivial periodic orbit and determine its period.

My ideas:

Where, the fixed point is $(x,y)=(0,0)$ and using the polar coordinates, I obtained the following system of ode's

$\frac{dr}{dt}=ar-r^3$

$\frac{d\theta}{dt}=-1$.

If $\frac{dr}{dt}<0$ then $r\geq \sqrt{a}$ and if $\frac{dr}{dt}>0$ then $0<r\leq \sqrt{a}$.

From here I don't know how to continue to find the orbit.

Thanks for your help.

Hint: You almost have it.  $dr/dt = 0$ if $r = \sqrt{a}$.

From

$$
\cases{
\frac{dr}{dt}=ar-r^3\\
\frac{d\theta}{dt}=-1
}
$$

with $\theta = t_0-t$ we have after integration

$$
\frac{\ln \left(r_1^2-a\right)-\ln \left(r_2^2-a\right)-2 \ln (r_1)+2 \ln (r_2)}{2
   a} = t_2-t_1 = -2\pi
$$

and solving for $r_1$

$$
r_1 = \frac{\sqrt{a} e^{2 \pi  a} r_2}{\sqrt{e^{4 \pi  a} r_2^2+a-r_2^2}}
$$

and $r_1 = r_2 = \sqrt a$

NOTE

From

$$
\cases{
x = r(t)\cos(\theta(t))\\
y = r(t)\sin(\theta(t))\\
\theta(t) = t_0 -t
}
$$

we have $T = -2\pi$ so along any orbit we have

$$
\frac{\ln \left(r_1^2-a\right)-\ln \left(r_2^2-a\right)-2 \ln (r_1)+2 \ln (r_2)}{2
   a} = t_2-t_1 = T
$$

and solving for $r_1$

$$
r_1 = \frac{\sqrt{a} r_2}{\sqrt{\left(a-r_2^2\right) e^{2 a T}+r_2^2}}
$$

so along a limit cycle, after a period $T$ we have $r_1 = r_2\Rightarrow r_1 = r_2 = \sqrt{a}$

characterizing a circle with radius $\sqrt a$ and period $T$


Derivation of the integral Evaluate
  $$\large\frac{d}{dx}\int_{0}^{\large\int_0^{e^x}{\cos (s)\,\mathrm  ds}}\sec(t^2)\,\mathrm dt$$

I got the answer to be $$e^x\cdot\sec(\sin^2(e^x))\cdot \cos(e^x)$$ but do not know if this is correct and if not some suggestions?

It is

$$ [ \tan(\int_0^{e^{x^2}}\cos sds)]' =  [ \tan(\sin e^{x^2})]'$$

   $$\sec^2(\sin e^{x^2})  \cdot \cos e^{x^2}  \cdot e^{x^2} \cdot  (2x)  $$

Can you proceed from here?


If a spherical triangle has two equal sides then then it has two equal angles Suppose we have a spherical triangle $ABC$ in which two sides are equal, say $AC=BC$. I want to show that this necessarily means that the angles opposite these sides are equal.

I've tried playing around with both the spherical cosine law and the spherical sine law, but neither of these seem to be getting me very far.

Spherical cosine law: $$\sin(a)\sin(b)\cos(\gamma)=\cos(c)-\cos(a)\cos(b)$$

How is best to do this?

To prove such statement does not really require the fifth postulate, hence it is true both in Euclidean, spherical and hyperbolic geometry. Let $d$ be the geodetic distance on the sphere. $d(A,C)=d(B,C)$ implies that $A$ and $B$ belong to a geodetic circle centered at $C$. Let $M_C$ be the midpoint of the $AB$ arc on the sphere: by the definition of perpendicular bisector as a geometric locus, both $C$ and $M_C$ belong to the perpendicular bisector of $AB$ and the spherical triangles $AM_C C$, $BM_C C$ are congruent by SAS.

The figure is symmetrical about the angle bisector of the vertex angle.


Exercise 11, chapter 2 in Lie Groups, Lie Algebras, and Representations: An Elementary Introduction I am reading the book: Lie Groups, Lie Algebras, and Representations: An Elementary Introduction  by Brian C. Hall. I am stuck at the following exercise: exercise 11, chapter 2 . Can you help me?

Suppose $G$ is a matrix lie group in $GL(n, \mathbb{C})$ with the lie algebra $\mathfrak{g}$. suppose that $A$ is  in G and that $||A-I||<1$,so the power series for $\log A$ is convergent. Is it necessarily the case that $\log A$ is in $\mathfrak{g}$？ prove or give a counterexample.

In this book, $\log A$ is defined as a power series:
$$ log A = \sum _{n=1}^{\infty} (-1)^{n+1}(A-I)^n/n$$
whenever the series converges.

I know that if $A$ is small enough,  then we can prove that it is true.

I reposted a closely related question here and got a good answer.

To summarize: No, $\log(A)$ may not be in $\mathfrak{g}$. Counterexamples can be found in the group of $n$-th roots of unity for large $n$, or in the line on the torus
$$G_a = \left\{\begin{pmatrix}
e^{2\pi it} & 0 \\
0 & e^{2\pi ita}
\end{pmatrix}
: t \in \mathbb{R} \right\}$$
for $a \in \mathbb{Q}$ with large denominator.

In the cases of ordinary numbers, under the definition
$$ \log x = \sum _{n=1}^{\infty} (-1)^{n+1}(x-1)^n/n, $$
we can expand
$$ e^{\log x} = C_0 + C_1 (x-1)+ C_2 (x-1)^2 + \cdots .$$
Because $ e^{\log x} =x $, we should get
$$ C_0 =1, C_1 =1 ,C_i=0 \, (i\geq 2) . $$
In the cases of matricies, under the definition
$$ \log A = \sum _{n=1}^{\infty} (-1)^{n+1}(A-I)^n/n, $$
we can also expand $ e^{\log x} . $
Because $ \left[ (A-I)^k,(A-I)^l \right]=0, $ the coefficients of this series are the same as in the cases of ordinary numbers. So
$$ e^{\log A} = I + (A-I) = A $$
Thus it is necessarily the case that $\log A$ is in $\mathfrak{g}$.


If $\int_a^b f(x) \ \mathrm{d}x = \int_a^b g(x) \ \mathrm{d}x$ then $\exists x \in [a,b]$ with $f(x) = g(x).$ I am trying to prove the following:

Take $f, g:[a,b] \to \mathbb{R}$ such that $f$ and $g$ are continuous. If
$$\int_a^b f(x) \ \mathrm{d}x = \int_a^b g(x) \ \mathrm{d}x,$$

then there exists some $c \in [a,b]$ such that $f(c) = g(c).$

Here's my current proof. I'd welcome any feedback regarding correctness and clarity.

Current Proof

Assume that there exists no such $c$. There are then three possibilities. First, it is possible that $f(x) > g(x)$ $\forall$ $x \in [a,b]$. However, this cannot be, since then we would have
$$\int_a^b f(x) \ \mathrm{d}x > \int_a^b g(x) \ \mathrm{d}.x$$

Similarly, we cannot have $g(x) > f(x)$ $\forall$ $x \in [a,b]$, since then
$$\int_a^b f(x) \ \mathrm{d}x < \int_a^b g(x) \ \mathrm{d}x.$$
Thus, there exists some $x \in [a,b]$ such that $f(x) > g(x)$ and some $y \neq x$ such that $g(y) > f(y).$ Assume without loss of generality that $x < y.$ Consider a new function $h:[a,b] \to \mathbb{R}$ defined by
$$h(x) = f(x) - g(x).$$
Clearly, $h$ is continuous, as the difference of two continuous functions. From the above, we have that $h(x) > 0$ and $h(y) < 0.$ Apply the Intermediate Value Theorem to $h$ on the interval $(x,y).$ Thus, there exists some $c \in (x, y)$ such that $f(c) = 0$. Since $(x, y) \subset [a, b]$, we have found an element of $[a,b]$ such that $h(c) = 0 \implies f(x) = g(x).$

While your proof is correct, there are ways you could simplify it.  For example, as noted in the comments, you could apply Rolle's theorem to $\int_a^x f(x') - g(x') dx'$ since you know it will be zero at the endpoints $a$ and $b$.

It can be proved:
    ∫{a}^{b}f(x)dx=∫{a}^{b}g(x)dx
    and
    f(x)≥g(x), for all x∈[a,b],
    then there is c∈[a,b] such that f(c)=g(c).


Finding tangents to a circle with a straightedge There is a geometric construction that I heard years ago and I still haven't figured out why it works despite several attempts.
Playing with pen, paper and GeoGebra makes me confident that it does indeed work.
Could someone explain it to me?

Task:
Given a circle and a point outside it, construct the two tangents to the circle through the point, using only a straightedge.

Solution:
Draw any three different lines through the given point $P$ that intersect the circle twice.
Let $A_1,A_2,B_1,B_2,C_1,C_2$ be the six intersection points, with the same letter corresponding to the same line and the index 1 corresponding to the point closer to $P$.
Let $D$ be the point where the lines $A_1B_2$ and $A_2B_1$ intersect, and similarly $E$ for the lines $B_1C_2$ and $B_2C_1$.
Draw a line through $D$ and $E$.
This line meets the circle at two points, $F$ and $G$.
The tangents are $PF$ and $PG$.

It is a consequence of Pascal's theorem in Projective geometry valid for any conic, the circle is a special case. When separation tends to zero, the secant points  $(OA_1, OA_2)$ where $(A_1, A_2) $ are extremities of chord on circle  which multiply to tangent point $F$ as

$ OA_1 OA_2= OB_1 OB_2=...= OD_1 OD_2= PF^2$ as constant product, a property of the circle.

The locus,chord  $ FG $, is a straight line. As a property of circle it is the harmonic mean of segments $ (OA_1, OA_2) $ as,for all chords e.g.,

$$  2/PF = 1/OA_1 + 1/OA_2. $$

Conclusion:

Pascal line is the chord of harmonic means of a pencil/circle intersection

Pascal Line valid on ellipse also, but I do not know if that is the harmonic locus. (posted Sept 2, 2016).

The answer is because when you fix one line through $P$ that intersects the circle at $A_1$ and $A_2$ and vary a second line and hence its intersection points $B_1$ and $B_2$, the point $E$ defined by you varies on a line.

This should be fairly easy and boring to check with analytic geometry.

It bugs me too that I cannot see the reason in some ratios or some such though.


German and French words for "basic sequence" Wikipedia says that a sequence $(x_n)$ is a basic sequence iff it is a Schauder basis of its closed linear span. I was wondering whether there is a French or German word for 'basic sequence'? How would you translate this into these two languages?

If you just know how to translate it into one of these languages, this would be extremely helpful too.

The German translation is simply "Basisfolge", see e.g. these lectures notes (Definition 1.7).

In French, the word "sequence" is translated by "suite" and "basic sequence" by "suite basique".


Finding $f$ such that $f(xy) = xf(y)+yf(x)-2xy$ given $f'(1)=3$ Let $f$ be a differentiable function satisfying the relation                $$f(xy) = xf(y)+yf(x)-2xy$$ where $x, y>0$ and $f'(1)=3$ then prove that the equation f(x) = k has two solutions in
 $k\in(-e^{-3}, 0)$

I tried differentiating this function but couldn't get anything from it. How to proceed here?

Hint: you can compute $f$ explicitly. Let $g(x)=\frac {f(x)} x -2$ and verify that $g(xy)=g(x)+g(y)$. Do you know how to find all continuous functions satisfying this equation?. [$f(x)=x(c\log\, x+2)$].

Hint: you can compute f explicitly. Let g(x)=f(x)x−2 and verify that g(xy)=g(x)+g(y). Do you know how to find all continuous functions satisfying this equation?. [f(x)=x(clogx+2)].


quadratic and bilinear forms Does a quadratic form always come from symmetric bilinear form ?
We know when $q(x)=b(x,x)$ where $q$ is a quadratic form and $b$ is a symmetric bilinear form.
But when we just take a bilinear form and $b(x,y)$ and write $x$ instead of $y$,does it give us a quadratic form ?

Yes, every quadratic form (over a finite-dimensional $\mathbb{R}$-space) can be expressed in terms of a symmetric bilinear form, because if your quadratic form $Q(x)$ (for $x \in \mathbb{R}^n$) is written as
$$ Q(x) = \sum_{i\le j} c_{ij} x_i x_j $$
then $Q(x) = x^T Ax$, where $A$ is a symmetric matrix given by $A = (a_{ij})$ with
$$a_{ij} = a_{ji} = \frac{c_{ij}}{2}, i < j, \quad a_{ii} = c_{ii}$$

If we have b symmetric bilinear form we can get q quadratic form
$q\colon V \to \mathbb{R}$
q(v)=b(v,v)

conversely if q is a quadratic form
$q\colon V \to \mathbb{R}$
we can define
$\frac 12$(q(v+w)-q(v)-q(w)):=b(v,w)

the vital answer is you just get a bilinear form not always a symmetric bilinear
form.
because the definition $\frac 12$(q(v+w)-q(v)-q(w)) leads us to $\frac 12$(b(v,w)+b(w,v))


Limit of Riemannian integrals with Lebesgue Show: $\lim\limits_{n\rightarrow\infty}\int_0^nx^i(1-\frac{x}{n})^ndx=i!$ for $i\geq 0$.

Hello there. Question is how do this? Which relation between Lebesgue and Riemann integrals must be used? Appreciate any hints. Thank you.

It should be
$$
0\lt\int_0^n x^i\left(1-\frac{x}{n}\right)^ndx\le
\int_0^n x^ie^{-x}dx\le\int_0^\infty x^ie^{-x}dx=\Gamma(i+1)=i!.
$$


Can an open set be covered by proper open subsets? Let $(X, \tau)$ be a topological space, let $U\in\tau$. An open cover of $U$ is a set $\{U_i \ |\ i\in I\}$ (of open sets $U_i$) whose union $\bigcup U_i$ contains $U$.

If $U\subsetneq X$, then $U$ admits an open covering by open sets $\{U_i \ |\ i\in I\}$, where an arbitrary open set $U_i$ may not be a subset of  $U$. (I think.)

Does $U$ admit an open covering by open sets $\{U_i \ |\ i\in I\}$ where every open set $U_i$ is a proper subset of $U$? Less formally: given an arbitrary open set $U$, can we find an open cover of $U$ made up only of open sets "inside" of $U$?

If the answer is no, then what are the open sets $U$ that admit such "internal" coverings?

Is the answer any different if we replace $U$ by an arbitrary closed set?

It depends. Consider $\mathbb{R}$ with the usual topology and $U = (0,1)$, then we can write $(0,1) = \bigcup_{n=2}^\infty\left(\frac1n,1\right)$, for instance. Since every open set $U\subset\mathbb{R}$ is the union of intervals, it means that every open subset of $\mathbb{R}$ can be written as the union of proper open subsets of itself. It is easy to see that the same argument can be applied to general normed vector spaces.

However, if a singleton $\{x\}$ is an open set of your space then there are no proper open subsets of $\{x\}$ to unite.

Show that $K = [0,1]$ cannot be covered by open sets that are contained by $K$.

If $U$ is open, then $U$ can be covered by $\{U\}$. If $U$ is an open singleton, then there is no cover by proper subsets.

When $U$ is open, then for all $x$ in $U$, there is an open base set $U_x$ with $x$ in $U_x$ subset $U$. Often the base sets can be chosen to be proper subsets.


Simple proof of integration in polar coordinates? In every example I saw of integration in polar coordinates the
Jacobian determinant is used, not that I have a problem with the Jacobian,
but I wondered if there's a simpler way to show this which will also give
me some more intuition about the Jacobian.

If I try to simply write the differentials:

\begin{align}
x & = r \cos \theta\\
y & = r \sin \theta\\
dx & = dr \cos \theta - r \sin \theta\ d\theta\\
dy & = dr \sin \theta + r \cos \theta\ d\theta\\
\end{align}

In a double integral you integrate $dxdy$, so if I try to plug in the results I'll get something which is not $r d\theta dr$
\begin{align}
dxdy & = \left(dr \cos \theta - r \sin \theta\ d\theta \right) \left( dr \sin \theta + r \cos \theta\ d\theta\right)\\
& = dr^2 \cos \theta \sin \theta  - r^2 d\theta^2 \cos \theta\ \sin\ \theta + r\ dr\ d\theta\ (\cos^2 \theta\ - \sin^2\theta )
\end{align}

I don't think I can go anywhere from here,
I'm not sure if it's just a calculation mistake or the entire logic is bad.

How do I get this right?

Thanx :)

Of course, if you break $\mathbb{R}^2$ into a polar grid

$\hspace{3.5cm}$

the small slightly curved rectangles have area $r\,\mathrm{d}\theta\,\mathrm{d}r$.

However, it seems that you are interested in looking at
$$
\begin{align}
\mathrm{d}y\,\mathrm{d}x
&=(\sin(\theta)\,\mathrm{d}r+r\cos(\theta)\,\mathrm{d}\theta)(\cos(\theta)\,\mathrm{d}r-r\sin(\theta)\,\mathrm{d}\theta)\\
&=r\,\mathrm{d}\theta\,\mathrm{d}r
\end{align}
$$
and why the $\mathrm{d}r^2$ and $\mathrm{d}\theta^2$ terms disappear and the $\mathrm{d}r\,\mathrm{d}\theta$ and $\mathrm{d}\theta\,\mathrm{d}r$ have different signs.

Let's start with
$$
\begin{align}
\mathrm{d}x&=\cos(\theta)\,\mathrm{d}r-r\sin(\theta)\,\mathrm{d}\theta\\
\mathrm{d}y&=\sin(\theta)\,\mathrm{d}r+r\cos(\theta)\,\mathrm{d}\theta
\end{align}
$$
rewritten as
$$
\begin{bmatrix}\mathrm{d}x\\\mathrm{d}y\end{bmatrix}
=\begin{bmatrix}\cos(\theta)\\\sin(\theta)\end{bmatrix}\mathrm{d}r
+\begin{bmatrix}-r\sin(\theta)\\r\cos(\theta)\end{bmatrix}\mathrm{d}\theta
$$
Therefore, the displacements $\color{green}{\mathrm{d}r}$ and $\color{red}{\mathrm{d}\theta}$ get mapped to  $\color{green}{\begin{bmatrix}\cos(\theta)\\\sin(\theta)\end{bmatrix}\mathrm{d}r}$ and $\color{red}{\begin{bmatrix}-r\sin(\theta)\\r\cos(\theta)\end{bmatrix}\mathrm{d}\theta}$ in $\mathbb{R}^2$:

$\hspace{3cm}$

where the area in gray is given by $\color{green}{\begin{bmatrix}\cos(\theta)\\\sin(\theta)\end{bmatrix}\mathrm{d}r}\times\color{red}{\begin{bmatrix}-r\sin(\theta)\\r\cos(\theta)\end{bmatrix}\mathrm{d}\theta}=r\,\mathrm{d}r\,\mathrm{d}\theta$.

The fact that the cross product is involved is the reason that the $\mathrm{d}r^2$ and $\mathrm{d}\theta^2$ terms disappear and the $\mathrm{d}r\,\mathrm{d}\theta$ and $\mathrm{d}\theta\,\mathrm{d}r$ have different signs. This, and its $n$-dimensional analogs, are why we use wedge products and differential forms when changing variables.

We know that
$$x=r\cos\theta,\quad y= r\sin\theta$$
Differentiate b.s respectively, we get
$$dx= -r\sin\theta\, d\theta+ \cos\theta\, dr,\quad dy= r\cos\theta\, d\theta + \sin\theta\, dr$$

Then
  $$\begin{align}dx\cdot dy &= (-r\sin\theta\, d\theta+ \cos\theta\, dr).(r\cos\theta\, d\theta + \sin\theta\, dr)\\ &= - r\cdot r \sin\theta\cdot\cos\theta\, d\theta\cdot d\theta + r\cos\theta\cdot\cos\theta\, dr\cdot d\theta - r \sin\theta\sin\theta\, dr\,d\theta +\sin\theta\cdot\cos\theta\,dr\,dr\end{align}$$


If $X$ is an integrable random variable and that $A$ is an event, how to show $E(X|A) = E(X|\mathbb{1}_A)$? Suppose that $X$ is an integrable random variable and that $A$ is an event. How can I show that $E(X|A) = E(X|\mathbb{1}_A)$? Is this something that is taken by definition?

Let's assume that $A$ is an event with $0<P(A)<1$ so that
$$
\mathrm{E}[X\mid A]:=\mathrm{E}[X\mathbf{1}_A]/P(A)
$$
as well as $\mathrm{E}[X\mid A^c]$ are well-defined.

The identity in your post is in fact wrong but instead we have the following
$$
\mathrm{E}[X\mid \mathbf{1}_A]=\mathrm{E}[X\mid A]\mathbf{1}_A+\mathrm{E}[X\mid A^c]\mathbf{1}_{A^c}.
$$
To show this identity you have to show that the right-hand side (let's call it $Y$) satisfy the criteria for being a conditional expectation of $X$ given the random variable $\mathbf{1}_A$. That is, we need to show that $Y$

is $\sigma(\mathbf{1}_A)$-measurable,
is integrable,
satisfies $\int_B Y\,\mathrm dP=\int_BX\mathrm dP$ for all $B\in\sigma(\mathbf{1}_A)$.

This should be more or less straightforward once you've realized what $\sigma(\mathbf{1}_A)$ looks like.

It is not exactly true.   However, yes, everything is basically definitional.

$\mathsf E[X\mid A]$ is the (conditional) expected value of $X$ when event $A$ occurs.   This is a constant; a real value (assuming $X$ is a real valued random variable).

$\mathsf E[X\mid \mathbf 1_A]$ is the conditional expectation of $X$ measured against the indicator random variable of $A$.   This is a random variable.   That means it is a function of the support of the indicator random variable.  $$\mathsf E[X\mid \mathbf 1_A]~:~\{0, 1\}\mapsto \Bbb R$$

Thus the first expected value is actually that function evaluated with an argument of $1$.   Also the evaluation at $0$ has similar meaning.

$$\mathsf E[X\mid A] = \mathsf E[X\mid \mathbf 1_A](1) \quad = \mathsf E[X\mid\mathbf 1_A{=}1]\\ \mathsf E[X\mid A^\complement] = \mathsf E[X\mid \mathbf 1_A](0) \quad = \mathsf E[X\mid \mathbf 1_A{=}0]~~$$


What are the differences between Hilbert's axioms and Euclid's axioms? Euclid had his axioms.

Why would we need Hilbert's modern axiomatization of Euclidean geometry?

What are key differences between the two sets of axioms?

The axioms of Euclid are insufficient, by a fair distance, for proving the theorems in Elements. The first mistake is in the first proposition, which is essentially about drawing an equilateral triangle.

The procedure involves drawing certain circles. Then the equilateral triangle has vertices at certain places where the circles meet. But Euclid does not show that the circles indeed do meet.

One major omission was noticed by Pasch, in the nineteeth century. Suppose that $ABC$ is a triangle, and let $\ell$ be a line that goes through a point $P$ on an edge of $\triangle ABC$. Then there is a point $Q$ (possibly equal to $P$,) on another edge of the triangle, such that $\ell$ passes through $Q$. The assumption that this is true is used tacitly in Elements. It is neither derived nor derivable from Euclid's axioms.

Around $1900$, Hilbert did a thoroughgoing axiomatization, with all details filled in. The result is vastly more complicated than the partial axiomatization by Euclid. One feature of the Hilbert axiomatization is that it is second-order. A benefit is that one can then prove that, for example, the Euclidean plane can be coordinatized using the real numbers.

Later, in the $1930$'s, Tarski produced an axiomatization that is first-order. Inevitably, we lose the result that the models are isomorphic to the coordinate plane $\mathbb{R}^2$. But there are bonuses, such as the later result (again by Tarski) that there is a decision procedure for elementary geometry.

Because Euclid's axioms do not represent a formal and rigorous axiomatic system. Since some postulates can be proven by others so they aren't independent. Also there are some theorems such pasch's theorem in Euclidean geometry but not provable using Euclid's axioms. There are other methodological reasons that makes it necessary to build a new more complet axiomatic system.


Totally ordered abelian group Let $\Gamma$ be a totally ordered abelian group (written additively), and let $K$ be a field. A valuation of $K$ with values in $\Gamma$ is a mapping $v:K^* \to \Gamma$ such that

$1)$ $v(xy)=v(x)+v(y)$
$2)$ $v(x+y) \ge \min (v(x),v(y))$,
for all $x,y \in K^*$.

Show that the set $A=\{x\in K^*:v(x) \ge 0\} \cup \{0\}$    is a valuation ring of $K$.

I want to show that $-1 \in A$. From $v(1)=v(1)+v(1)$, $v(1)=0$ and $v(1)=v(-1)+v(-1)$ so that $2v(-1)=0$. Is it possible to conclude that $v(-1)=0$?

In general for any totally ordered abelian group, does $2x=0$ imply $x=0$?

Yes.  If $x$ is positive then so is $nx$ for any $n \in \mathbb{Z}^+$ and if $x$ is negative then $-x$ is positive and $nx = 0$ iff $n(-x) = 0$.

In other words, a totally ordered abelian group is necessarily torsionfree.  More interestingly, the converse also holds: any torsionfree abelian group can be totally ordered (in at least one way). See Section 17.2 of these notes for the proof.

It does follow that $\nu(-1)=0$.

Here is an elementary argument. Suppose firstly that $\nu(-1)>0$. Then $2\nu(-1)=\nu(-1)+\nu(-1)>0$; now $2\nu(-1)=0$ implies $0>0$, contradiction.

The same argument, \textit{mutatis mutandis}, shows that you cannot have $\nu(-1)<0$.


Let $A,B$ be compact subsets of $X$. Prove that $A \cap B$ is compact. Let $A,B$ be compact subsets of $X$. Prove that $A \cap B$ is compact.

Attempt: Suppose by contrapositive, that  $A \cup B$ is compact. Then
let $V$ be an open cover of $A \cup B$. Then let $A$ be compact, then $ V$ has a finite subcover of $A$ . But suppose $B$ is not compact. Then $V$ does not contain a finite subcover of $B$. Thus their union will not be have a finite subcover of $ V$ that converse $A \cup B$.

Can someone please help me? I already proved if $A, B$ are compact , then their union is compact. So I was trying to use a similar argument.
Any feedback/help would really help. Thanks

There is a problem with your contrapositive.

Hint: show that a closed part of a compact set is compact.

In any topological space if you suppose that A and B are compact then it holds that A can be written as a finite cover of open sets and so can B (definition of compactness).
So if you intersect open sets you still get open sets therefore that should be a finite cover of open sets of = (A intersection B)
and again according to defenition the intersection is compact.

ok,  this happens only in hauusdorf spaces as I have seen and it's incorrect for every space. I've seen a counter example: (intersection of two compacts isn't compact) Y-with the discrete topology Y is infinite and X is taken to be X=Y uninon {c1} union {c2}, where {c1} and {c2} are two arbitary points. The topology on X is defined to be all the open sets in Y. Now can anyone understand this counter example? It doesn't make sense...


Extremal values of $(x+3)^\frac{1}{3} - x^\frac{1}{3}$ $f(x) = (x+3)^\frac{1}{3} - x^\frac{1}{3}$

I am trying to find the extremal values of $f$. I start by differentiating, equating the derivative to 0, and solving that for x:

$f'(x)= \frac{1}{3} ((x+3)^\frac{-2}{3} - x^\frac{-2}{3}) = 0$

$(x+3)^2=x^2$

$x^2 + 6x + 9 = x^2$

$x = \frac{-3}{2}$

The extremal value of $f$ is at $f(\frac{-3}{2})$.

$f(\frac{-3}{2})= 2.289428485$ (decimal approximation)

However, when I try to verify this result with a CAS, WolframAlpha says something else, and SymboLab talks about the saddle of some interval, which I am not familiar with.

Is my working correct?

Your calculus is correct and $x=-\frac{3}{2}$ is a maximum as $f''(-\frac{3}{2})<0$ as noted in the answer of @dewypeters.

But note that the first derivative $f'$ is not defined for $x=-3$ and $x=0$ that are points where the graph of the function has a vertical tangent.

The result of WolframAlpha may depend on this fact and, maybe, by the particular way in which you are write the question (that I cannot see, but you can see the answers to my question about the roots of the first derivative : A WolframAlpha error?). Anyway you can see here the graph of your function by WA.

I've got it.

$f''(\frac{-3}{2}) < 0$, therefore the extremal is a maximum point.


Prove that a continuous function has given property The problem: Suppose $f$ is a continuous function on [0,2] and $f(0) = f(2)$. Prove that there exists $x,y \in [0,2]$ such that $|x-y| = 1$ and $f(x) = f(y)$.

Intuitively this makes sense after thinking about it for some time, but I do not know have to get going.

Consider the function $g(x) = f(x+1)-f(x)$ on $[0,1]$. We have $g(0) = f(1) - f(0) = - (f(2)-f(1)) = -g(1)$, hence $g$ has a zero on $[0,1]$ by the intermediate value theorem.

Function $g(x)=f(x+1)-f(x)$ is continuous on $[0,1]$ and  $g(0)=f(1)-f(0)=f(1)-f(2)=-g(1)$. Then $g(x)=0$ for some $x$ (intermediate value) and $y=x+1$ then $f(y)=f(x)$ and $|y-x|=1$


Why are randomly drawn vectors nearly perpendicular in high dimensions I am struggling understanding this finding. Can somebody explain intuitively why randomly drawn high-dimensional vectors will tend to be mutually orthogonal? I realize that intuition in high dimensions might be too much to ask for, still, an explanation without having to integrate over several pages of symbols would be preferred.

A random uniform unit vector is $X/\|X\|$ where $X$ is standard normal, thus the scalar product of two independent unit vectors $U$ and $V$ is $\langle U,V\rangle=\langle X,Y\rangle/(\|X\|\cdot\|Y\|)$ where $X$ and $Y$ are independent and standard normal. When $n\to\infty$, by the law of large numbers, $\|X\|/\sqrt{n}\to1$ almost surely and $\|Y\|/\sqrt{n}\to1$ almost surely, and by the central limit theorem, $\langle X,Y\rangle/\sqrt{n}$ converges in distribution to a standard one-dimensional normal random variable $Z$.

Thus, $\sqrt{n}\cdot\langle U,V\rangle\to Z$ in distribution, in particular, for every $\varepsilon\gt0$, $P(|\langle U,V\rangle|\geqslant\varepsilon)\to0$. In this sense, when $n\to\infty$, the probability that $U$ and $V$ are nearly orthogonal goes to $1$.

Likewise, $k$ independent uniform unit vectors are nearly orthogonal with very high probability when $n\to\infty$, for every fixed $k$.

As the brief proof showed below, it does not matter to suppose that the vector c is the diagonal vector of the n vector from the standard basis in the n-dimensional vector space with inner-product. It means that the projection vector of c onto vi is the same length(norm) with each other (irrelevant to i). Any randomly drawn vector is the same as c.


Is true that if $V \cong W$ and $Z \cong Y$ then $V\times Z \cong W \times Y?$ Let $V,W,Z,Y$ finite dimensional vector spaces. Is true that if $V \cong W$ and $Z \cong Y$ then $V\times Z \cong W \times Y?$

My attempt:

Suppose it is true: Then there is $f : V \to W$ isomorphism and $g : Z\to Y$ isomorphism.

Them $\dim V = \dim W$ and $\dim Z = \dim Y.$

So define $$h : V\times Z \to W\times Y,$$
$$h(v,z) = (f(v),g(w)).$$

Note that $\dim V \times Z = \dim W\times Y.$

Even more, suppose that $h(v,z) = (0,0).$ Then $f(v) = 0$ and $g(w) = 0$. So since $f$ and $g$ are injectors $v = 0$ and $w = 0.$

So the claim is proved.

There are two ways I would like to think about this problem, which you have essentially highlighted. One is that the function you define is clearly injective and linear, and hence an isomorphism.

The other way is that it's really easy to know whether two f.d. vector spaces over the same field are isomorphic; you simply have to check whether they have the same dimension (this view is similar to picking a basis). Then, isomorphism follows from the easy-to-prove fact that $$\dim X\times Y=\dim X + \dim Y $$

Yes, this works. If you take the trouble to show surjectivity this will work in arbitrary dimensions as well.


Significance of Inflection Points in Biological Application This problem appears in our calculus book. It asks to compute the maximum concentration of drug in the blood and the time when the maximum occurs as well as to find the number of inflection points and then it asks for the time at which the inflection points occur and finally the "significance" of the inflection points. The model given is
$$C(t)=0.5t^2e^{-0.6t}$$
where $C(t)$ is the concentration. It was very easy to find that the maximum concentration of $\frac{50}{9}e^{-2}$ occurs at $t=10/3$ and that there are two inflection points, occurring at $t=\frac{10\pm 5\sqrt{2}}{3}$. But when the author asks for the "significance" of the inflection points I'm not sure what he means. An inflection point is a zero of the second derivative occurs. It is where the graph is "flat" (i.e. the curvature is zero) but in this case I believe the author wants the "biological" significance, and there I am stuck. Is it when the drug is being absorbed and excreted? Perhaps there is something about drug metabolism that I need to know and do not? Thank you in advance for any help!

When "the rate of variation of $C(t)$ is maximal" you would expect maximum effect of the drug at that time $t$.  For example: one has headache and takes the pain killer drug, at $t=23$ min, he will have maximum effect of the drug.

A better example: an antibiotic for infection. The guy should take tablets regularly every 8 hours and the maximum effect of the drug is at $t=4$ hours.  In real research, these times should be computed in trials to be able to say how often a tablet should be taken (dosage and many other things of the tablet are important as well).

Thanks Did for answer.


Demonstrating if a transformation is linear by using linear independence/dependence Can examining the linear dependence/independence of the vectors in the domain and codomain demonstrate that a transformation is linear?

I think that if there are linearly independent vectors in the domain that have images in the codomain which are also linearly independent, then the transformation is linear. Is this correct?

What about if there are vectors that are found to be linearly dependent in the domain and codomain?

Not true. For example, let $e_1,\ldots,e_n$ be a basis for $\mathbb{R}^n$ and define $T:\mathbb{R}^n\to\mathbb{R}^n$ by
$$T(e_1)=e_1,\quad\ldots,\quad T(e_n)=e_n,\quad\text{and}\quad T(v)=0 \text{ for all }v\notin\{e_1,\ldots,e_n\}.$$
This satisfies your description, although it is clearly not linear.

Linear independence is a more fundamental property then transformation linearity-a linear transformation cannot transform a linearly dependent set into one which is linearly dependent. Indeed, a linear transformation is linear largely because the basis of the dual space for a vector space is linearly independent. Spenser above gave a good counterexample-others aren't hard to come up with.


Conditional Probability: Distinction between $P(Fri|Defective)$ and $P(Defective|Fri)$ Question:

The factory quality control department discovers that the conditional probability of making a manufacturing mistake in its precision ball bearing production is 4% on Tuesday, 4% on Wednesday, 4% on Thursday, 8% on Monday, and 12% on Friday.

The Company manufactures an equal amount of ball bearings (20%) on each weekday.
What is the probability that a defective ball bearing was manufactured on a Friday?

My solution:
Given, P(defective|Monday)=0.08
P(defective|Tuesday)=0.04
P(defective|Wednesday)=0.04
P(defective|Thursday)=0.04
P(defective|Friday)=0.12

Now, the question asks for P(Friday|Defective)? Is my logic correct? Could someone help in how to finish the problem?

So far you have properly identified $P(D|M), \dots, P(D|F).$ But you need to look at Bayes' Theorem and figure out
how to get the 'inverse' or 'posterior' probability $P(F|D).$ Here is an
outline; you should match the steps with Bayes' Theorem.

Notice that $P(D \cap F),\, P(D|F)$ and $P(F|D)$ are all probabilities
involving events $D$ and $F.$ But they refer to different populations:
$P(D \cap F)$ refers to the entire weekly production of ball bearings;
$P(D|F)$ refers to the population of ball bearings made on Friday; and $P(F|D)$ refers to the population of all defective ball bearings, asking what
proportion of them were made on Friday.

$P(D \cap M) = P(M)P(D|M) = .2(.08) = .16.$ This uses what is sometimes
called the 'general multiplication rule'.

$P(D) = P(D \cap M) + P(D \cap Tu) + \cdots + P(D \cap F) = 0.064.$ This uses
what is sometimes called the 'law of total probability'.

Using R statisical software as a calculator:

p.d = sum(.2*c(.08, .04, .04, .04, .12));  p.d
## 0.064

$P(F|D) = P(D \cap F)/P(D) = .024/.064 = 0.375.$ This uses Bayes' Theorem.

While only 20% of all ball bearings are made on Friday, 37.5% of defective ball bearings were made on Friday.

This problem's just a simple application of Bayes's theorems. We want to know $ P(Friday | defective)$ and we already have the inverse conditional probabilities e.g. $P(defective| Monday)$, $P(defective| Tuesday)$, etc.  The prior probability is 0.2 (this is written as:  "20% of production on each weekday", 0.2 * 5 = 1): there are five days on the work week and we are indifferent to which day we choose.
Bayes Theorem:
$P(A|B) = \frac{P(B|A)*P(A)} {P(B)}$
$P(Defective| Friday) = P(B|A) = .12 $,  makes the first term multiplied by :
$P(A) = P(Friday) = P(any other day) = 0.2$, which is the prior probability.
In the denominator, $P(B)$ is the sum of the joint probabilities each one multiplied by the prior probability :
$ P(defective | Monday)P(Monday) + P(defective|tuesday)P(Tuesday)...+ etc., (0.08 + 0.04 + 0.04 ...) * 0.2.$
Since the prior probabilities are equal:
$$ P(Friday | defective) = \frac {0.12 * 0.2}{0.32 * 0.2} = 0.375 $$


Free-fall according to Newton's gravitation law Most analysis of free-fall assume that bodies fall with constant acceleration. If however one analyses  free-fall according to Newton's gravitation law, one is lead to a differential equation which I don't know how to solve.

The differential equation one is lead to[1] is $\frac{d^2x}{dt^2}=-\frac{1}{x^2}$. Does this equation have a closed form solution with given initial conditions $x_0,v_0$?

[1] If one picks the units so that $MG=1$, fixes one mass at $x=0$ and places the falling mass at some $x>0$.

Multiply the equation by $dx/dt$ and integrate once to get
$$
\frac{1}{2}\Bigl(\frac{dx}{dt}\Bigr)^2=\frac{1}{x}+\frac{v_0^2}{2}-\frac{1}{x_0}.
$$
Then
$$
\frac{dx}{dt}=\pm\sqrt{\frac{2}{x}+v_0^2-\frac{2}{x_0}}\,,
$$
a first order equation in separated variables whose solution is
$$
\int_{x_0}^x\frac{dz}{\sqrt{\frac{2}{z}+v_0^2-\frac{2}{x_0}}}=\pm t.
$$
The integral can be computed explicitly, although I do not think you can find a closed form for $x$ in terms of $t$.

I managed to do it

NewtonGravitySolution

The final solution I had was:

$t=\sqrt{\frac {R}{2Gm} } \biggl( x \sqrt{\frac {R}{x} -1} +tan^{-1}\sqrt{\frac{R} {x}-1 }  \biggr)$

$R$ is the initial position where $v=0$

If you plug in values for Earth with Mass = Earth's Mass you can plot it.


Let $f\in L^p$. Can we say $\|f\|_{L^{p}} \leq \epsilon$ on $|x|\geq R$ for large $R$? Let $f\in L^{p}(\mathbb R), (1\leq p <\infty)$ and $\epsilon>0.$

  My Question: Can we expect  to find $R>1$ (may be large) so  that
   $\|f\|_{L^{p}(B_R)} \leq \epsilon$  on $B_{R}=\{x:|x|\geq R\}$?

The function $ g_n = \chi_{[-n,n]} |f|^p$ satisfies

$$0\le g_1 \le g_2\le \cdots \le |f|^p$$

and $g_n \to |f|^p$ pointwisely on $\mathbb R$. Thus the monotone convergence theorem implies that

$$\left( \int_{-n}^n |f|^p\right)^{\frac{1}{p}} \to \|f\|_p.$$

If $\int_\mathbb{R} |f|^p$ is finite, then for all $\epsilon >0$ there is $n$ so that

$$ \|f\|^p_p -\epsilon^p \le \int_{-n}^n |f|^p \le \|f\|^p_p,$$

which implies

$$\left(\int_{|x|\ge n} |f|^p\right)^{\frac 1p} <\epsilon.$$

Let $f$ consist of small non-overlapping rectangular bumps of height 1
centered on 1,2,3,$\cdots $,n,$\cdots $ with width $w_{n}=1/n^{k}$ with
sufficiently large $k$. Then $f\in L^{p}$ but will never become smaller than
$\varepsilon $ as $x$ increases.


Are there any nicely factoring cubic equations with the same first three coefficients? For example is there a factorable equation of the form $(x+1)^3+c$ where $c\neq 0$, more generally is there a factorable equation of the form $(x+a)^{n}+c$ again with the condition that $c\neq0$, but also $n>2$?

For $c \ne 0$ the equation $(x+a)^n + c = 0$ has exactly one real root if $n$ is odd, and two or no roots if $n$ is even and $c \lt 0\,$, $c \gt 0$ respectively. Therefore, there is no factorization possible in linear (real) factors for $n \ge 3$.

Here's a quick answer — just showing some possibilities, but certainly far from complete. Since $a^3+b^3=(a+b)(a^2-ab+b^2)$, pick $c$ to be a perfect cube, and you'll have a nicely factorable expression. The same trick works for all odd powers $n$.


Can a sphere be considered the limit of a regular polyhedron with infinite faces? I was trying to approximate the surface area of the sphere.
I remember that Archimede approximated the circumference of a circle by calcualting the perimeter of its inscribed and circumscribed regular polygons while n ( number of sides of the polygon) increased.
Anyway when I looked for a similiar procedure for the sphere, I discovered that there are only 5 convex regualar polyhedrons (platonic solids), so it doesn't seem to be possible to approximate the surface of the sphere with the same procedure.
I then looked to Goldberg and Geodesic polyhedrons but I didn't understand really how they are constructed so I'm not convinced "they fit for the procedure."
So I'm asking, if there are only 5 regular polyhedrons can be the sphere considered the limit of a regular polyhedron with infinite faces?
If not, are Goldberg and Geodesic Polyhedrons the alternative to approximate the surface of a sphere?

As for your first part of the question, since regular polyhedra, it is evident that one cannot talk about approximation proccedures. As for your second part, you can approximate the surface of a sphere with Goldberg Polyhedra but they are a little bit "out of context" if you are looking to mimic the "Method of Exhaustion" used in ancient times (for more about Goldberg Polyhedra, see here, for more about the "Method of Exhaustion" on two-dimensional shapes, see here).

It could be "better" to think as follows - with better I mean "historically more accurate":

Take a sphere and a maximal circle $C$ of it - this is a circle that occurs as the intersection of the sphere with a plane passing through its centre. Then consider two regular polygons $A_1A_2\dots A_n$ and $B_1B_2\dots B_n$ such that $A_1A_2\dots A_n$ is inscribed in $C$ and $B_1B_2\dots B_n$ is circumscribed to $C$, as follows (it is a regular dodecagon - 12-sided regular polygon):

Here, the circle $C$ is in black and the two polygons are red and blue. Now, consider any diametre of the above circle, let $d=A_1A_7$, and rotate around $d$ this circle - this is a 3D rotation. So, the rotation of $C$ would give us the initial sphere and the rotation of the two polygons would give us two solids that are not, of course regular polyhedra, not even polyhedra, but they consist of parts of cones with generating line being each side of the initial polygon.

Despite the fact that such solids seem to be quite complex, their volume and their surface area can be calculated relatively easily. What is needed is the formula for the surface area of a part of a cone between its base and a plane that its parallel to its base, which can be found using the formula for the cone's surface area:
$$S=\pi\rho l$$
where $\rho$ is the cone's radius and $l$ is the length of its side.

A theoretical solution could be this. One of the definitions of a polyhedra is that it is an intersection of half-spaces. Now, the sphere $S$ separable, so there is a sequence $(e_j)_{i=1}^{\infty} \subset S$ so that the points $e_j$ are dense in $S.$ Consider the unique half-spaces $H_j$ so that $e_j$ belongs to the boundary of $H_j$ and $S \subset H_j$. Consider the intersections $\bigcap_{j=1}^k H_j$ and let $k \to \infty$. Now, the boundaries of those polyhedra get closer and closer to sphere.


Prove that the straight line joining the middle point of the hypotenuse of a right angled triangle to the right angle is equal to half the hypotenuse. I am supposed to use the following 8 theorems only to prove the above prepositions:

Theorem 1: If a ray stands on a line , then the sum of the adjacent angles formed is $180 $deg.

Theorem 2: If two lines intersect , then the vertically opposite angles are equal.

Theorem 3: If a transversal cuts two parallel lines, then each pair of alternate angles are equal, and the interior angles on the same side of the transversal are supplementary.

Theorem 4: Lines which are parallel to the same line are parallel to each other.

Theorem 5: The sum of the three angles of a triangle is $180$deg.

Theorem 6: If one side of a triangle is produced , the exterior angle so formed is equal to the sum of the interior opposite angles.

Theorem 7: The angles opposite to equal sides of a triangle are equal in an isosceles triangle.

Theorem 8: The bisector of the vertical angle of an isosceles triangle bisects the base and is perpendicular to the base.

I tried to get the solution but could not apply theorems 7 and 8 since it is a right angle triangle. I don't use concept of rotation. I use ASA, SAS, SSS and RHS Postulates and converse of theorem 1,3,7 and 8. can anyone tell me how the proof looks like? Thanks in advance...Srikanth

in $ \triangle ABC$, $\angle B=90^\circ$, $AD=DC$

we draw $DO \perp BC$. So $BO=OC= \frac{BC}{2}$ and $DO= \frac{AB}{2} $ [midpoint theorem]
so $ BD^2=DO^2+BO^2=\frac{BC^2+AB^2}{4}=\frac{AC^2}{4}$
so $BD= \frac{AC}{2}$ (proved)

I didn't read Theorems 1–8, but offer the following proof: The midpoint $M$ of the hypotenuse $AB$ is the center of a Thales circle with  diameter $AB$, and $MC$ is a radius of this circle.


The ultimate formula to factor them all. Context

I am working on Integer factorization problem, I found a formula for factoring numbers, and I need your help to simplify it. First I will explain how I get there and then I  present the problem (This is actually an improvement of my previously posted idea).

Background

I want to find the factors of $d$ or in other words I want to know for what values of $x$ do you need to divide $d$ to get an integer solutions?

$$\frac{d}{x} = \left\lfloor\frac{d}{x}\right\rfloor$$

I found that in small ranges $\left\lfloor\frac{d}{x}\right\rfloor$ act as a linear line, and if you look on $x$ values higher then $\sqrt{d}$ you will see that the slope is equal to $1$. So in general you can write it like this:

$$\frac{d}{x} = \left\lfloor\frac{d}{\lfloor\sqrt{ d }\rfloor}\right\rfloor +y - 1*(x-\lfloor\sqrt{ d }\rfloor)$$

$\left\lfloor\frac{d}{\lfloor\sqrt{ d }\rfloor}\right\rfloor$ is the starting point for $x=\lfloor\sqrt{ d }\rfloor$
$1*(x-\lfloor\sqrt{ d }\rfloor)$ is the predicted change for the next $x-\lfloor\sqrt{ d }\rfloor$ values, assuming that it is a straight line
$y$ is the error, as we don't really sure in what of those straight line $\frac{d}{x} = \left\lfloor\frac{d}{x}\right\rfloor$

So if I extract $x$ I will get the next thing:

$$x = \frac{a+b + y+ \sqrt{(a+b+y)^2 - 4d}}{2}$$

$a = \left\lfloor\frac{d}{\lfloor\sqrt{ d }\rfloor}\right\rfloor$
$b  = \lfloor\sqrt{ d }\rfloor$
$y$ positive integer

Question

When the next equation is an integer:

$$x = \frac{a+b + y+ \sqrt{(a+b+y)^2 - 4d}}{2}$$

$a,b,d$ are positive known integers
$y$ must be positive integer

Right now I am finding the solution by checking all the options. So it takes $546$ attempts to factor $652450640273$ to it's two dividers $787021$, $829013$

What can I do to reduce the number of options that I have for $y$?

Notice that, for any $k\in [\sqrt{d},d]$, there is a $y$ such that that choice of $y$ forces $x$ to equal $k$ - that is, our choice of $y$ completely determines our choice of $x$. Furthermore, given that it straightforwards to compute $y$ given $x$ and $x$ given $y$, it is clear that the problem of determining $y$ is, in terms of computation, equivalent to factoring $d$. There is a great deal of research into that subject, if you are really really intent on finding $y$.

The fact that your calculations don't reduce the problem of factoring isn't too surprising - you have come up with what is, roughly, the linear approximation of $\frac{d}x$ around $x=\sqrt{d}$ - so you can approximate the term $\frac{d}x$ with that. However, linear approximations are good at telling you "Well $\frac{d}x$ lies somewhere near this value on the number line" - but what you need to know is not the general area where $\frac{d}x$ lives (a sort of global property) but a local property: Is $\frac{d}x$ an integer? It is impossible to determine this with approximations - the error term completely determines the answer to that, so asking what the error term is is, in essence, rephrasing the original question.

ST1 Factorisation by example, massively better than Fermat Factorisation. This overall method works for all numbers, but the example is 1 of a group of 8 within 36 possibles and is at a limit requiring only 2 pieces of logic and 1 guess. The other 28 require another term.

N = 3,911,290,831 (p x q)

3911290831/30 leaves remainder 1 therefore choices for B and C in (N-(BxC))/30 from (1 1) (7 13) (11 11) (19 19) (17 23) or (29 29)

(N - (BxC))/30

Let's try (11 11): (N - 121)/30 = (3911290831-121)/30 = 130376357(N2)

Sqrt((3911290831/900) = 2187(ignore remainder)

1300376357(N2) ends in 7. 2187 plus 2187+1 does not produce a number ending in 7, a requirement, therefore add 1. 2188 plus 2188+1 does produce a number ending in 7. Because using (11 11), (2188 + 2189 just the digits)/3 must leave remainder 1 , it doesn't. 2193 + 2194 ends in 7 and divided by 3 leaves remainder 1. Bingo!

(2193-n)30) +11 x (2194+n)30) +11 = 3,911,290,831 (p x q)

n = 684

p = 45311 q = 86321 45311 x 86321 = 3911290831 One number is almost twice the other. 1000 times better than Fermat? If I've written this wrong, sorry, it does work, l'm just losing the will!


How to prove b is positive? If A * B is positive and I know that A is positive, how do I prove that B is positive? I have to prove it with just basic laws of numbers, e.g. associative property and stuff like that.

If we know that $ab>0$ and that $a>0$, then we can divide both sides of the inequality by $a$ to find that $b>0$. Thus, $b$ is positive. Remember that dividing both sides of an inequality does not change the inequality as long as what we're dividing by is positive.

If $b$ were $0$ then $ab$ would be $0$ too, which is not positive.

If $b$ were negative, then $ab$ would be negative (since positive times negative makes negative).

The only remaining possibility is that $b$ is positive.


Why is $\;n^2-\frac{n^2}{2} =\frac{n^2}{2}\;$? Could someone please expand on how to get from $\;\displaystyle\left( n^2-\frac{n^2}{2}\right)\;$ to $\;\left(\dfrac{n^2}{2}\right)\;?\;$

I can't seem to wrap my head around that.

This reduces to proving that $1-\frac 12 = \frac 12$. In order to do this, expand $1$ into the geometric series $$1=\sum_{i=1}^{\infty}\frac 1{2^i},$$ and divide both sides by $2$ to get $$\frac 12=\sum_{i=1}^{\infty}\frac 1{2^{i+1}}=\sum_{i=2}^{\infty}\frac 1{2^i}.$$ It follows that $$1-\frac 12 = \sum_{i=1}^{\infty}\frac 1{2^i}-\frac 1{2^1}=\sum_{i=2}^{\infty}\frac 1{2^{i}}=\frac 12.$$

Actually, if you consider an integer division (so called Euclidian division) the proposition becomes:
$$
n^2 - \frac{n^2}{2} = \frac{n^2}{2} + \text{mod}{(n^2,2)}
$$

Indeed, by definition of the Euclidian division of the integer $n^2$ by the integer $2$, one has:
$$
n^2 = 2\times\frac{n^2}{2} + \text{mod}{(n^2,2)}
$$
where $\text{mod}{(a,b)}$ denotes the remainder of the euclidian division of an integer $a$ by an integer $b$.

Now, by (cleverly enough) remarking that the term $2\times \frac{n^2}{2}$ equals to $\frac{n^2}{2} + \frac{n^2}{2}$ one can write:
$$
\begin{aligned}
n^2 &= 2\times\frac{n^2}{2} + \text{mod}{(n^2,2)} \\
n^2 &= \frac{n^2}{2} + \frac{n^2}{2} + \text{mod}{(n^2,2)} \\
\Leftrightarrow n^2 - \frac{n^2}{2}&= \frac{n^2}{2} + \text{mod}{(n^2,2)} \\
\end{aligned}
$$
that ends the proof ;)


coplanar lines in 3d plane If we are given two lines

Coplaner then how can we find the value of k.

I think if they are coplaner then their cross product should be zero .

In the solution it is given as

But I could not understand what they have done.

Hint: Examine both lines in parametric form. If their vectors are parallel then they are certainly coplanar. If their vectors are not parallel, two lines are coplanar if and only iff they intersect; otherwise, they are skew.  Hope it helps.

I think if they are coplaner then their cross product should be zero.

This statement is at least unclear, and not quite meaningful or correct. First of all, what do "they" and "their" refer to? If you mean the lines, then this statement is meaningless because cross product is an operation on vectors, not lines. It would be meaningful when talking about vectors, but then it's not true — the cross product of two (nonzero) vectors is zero means that they are collinear, not coplanar.

Let $\vec{a}$ and $\vec{b}$ be the directional vectors of two lines, and $A$ and $B$ some points on the same lines, respectively. (All of these can be seen from the given equations.) The condition given in the answers requires that the three vectors $\vec{a}$, $\vec{b}$, and $\vec{AB}$ be coplanar to guarantee that the two lines are. To check whether they are coplanar, we can set up the triple scalar product of these vectors. Effectively it finds the volume of the parallelepiped (up to the sign) built on these three vectors as sides. If the volume is zero, the parallelepiped is "flat" due to the three vectors lying in the same plane.

Why do we want $\vec{a}$, $\vec{b}$, and $\vec{AB}$ to be coplanar? Consider two cases. (1) If $\vec{a}\parallel\vec{b}$, then the lines are parallel and therefore coplanar. Note that in this case the three vectors are also coplanar regardless of the third vector. (2) Otherwise, we need to distinguish between intersecting lines (coplanar) and  skew lines (not coplanar). If the lines are intersecting, then all their points lie in the same plane as $\vec{a}$ and $\vec{b}$, therefore $\vec{AB}$ must lie in that same plane.


why do we use cosine as the expression of vector dot product? When we do vector products, we use two different methods.
One is the vector dot product, another is vector cross product.
The equation of the vector dot product is
$$\textbf A \cdot \textbf B =|\textbf A| | \textbf B| \cos\theta,$$
where $\theta$ is the angle between the vectors $\textbf A$ and $\textbf B$.

Why do we use cosine as the expression?

The dot product of two vectors $A$ and $B$ is just the product of the magnitude of one vector with the scalar projection of the other one on itself. Hence the $cos$ term. Also, note that the $cos$ function is greater for smaller angles, and lesser for larger ones, just like the length of the projection. (both are the same thing, actually.)

It is actually the definition of the dot product of two vectors.


Solving a cubic system of equations I have the following equation system:

$A_1 x + B_1 y + C_1 z + D_1 xy + E_1 xz + F_1 yz + G_1 xyz = M_1$
$A_2 x + B_2 y + C_2 z + D_2 xy + E_2 xz + F_2 yz + G_2 xyz = M_2$
$A_3 x + B_3 y + C_3 z + D_3 xy + E_3 xz + F_3 yz + G_3 xyz = M_3$

$A_1$, $B_1$, ..., $M_1$, $A_2$, $B_2$, ..., $M_2$, $A_3$, $B_3$, ..., $M_3$ are known.

Trying to get $x$ based on $y$ and $z$ from the first equation, then substituting it in the second equation, then getting $y$ based on $z$ and substituting it in the third equasion seems a nightmare. How to solve this equation system?

A general cubic system of three equations in three unknown has at most 27 solutions. This can be proved by reducing it to a 27 degree polynomial equation in one variable and apply Gauss theorem. Finding the polynomial give you full control of the number of solutions. However, there is a problem with this scheme since it the polynomial of such high degree might be numerically unstable. Another method is Newtons Raphsons method which will give you a numerical solution of the problem.

A possible workaround can be the following:

Let's multiply the first equation with with $-G_2$, the second equation with $G_1$, and add them.
Now multiply the second equation with $-G_3$, the third one with
$G_2$, and add them.
Finally multiply the third equation with $-G_1$, and the first one with $G_3$ and add them.

Theoretically we have now tree new equations, but we escaped from the $xyz$ part.

In the same way let's escape from xy, xz, xz.

We have now a simple equation system with 3 equations and 3 unknowns:

$AA_1x + BB_1y + CC_1z = MM_1$

$AA_2x + BB_2y + CC_2z = MM_2$

$AA_3x + BB_3y + CC_3z = MM_3$

that's easy to solve.

If you think this wouldn't always work please let me know, or if you have a better solution it is welcomed!


Well-Ordering Principle Let S be a nonempty subset of $\{n \in \mathbb{Z}: n \le 0\}$ Let $S$ be a nonempty subset of $\{n \in  \mathbb{Z}: n \le 0\}$. Use the Well-Ordering Principle to show that $S$ has a
greatest element; in other words, prove there exists $s \in  S$ such that $s \ge  x$ for all $x \in S$.

I can clearly tell that the greatest element here is $0$ just by looking at the set. My problem comes by way of what this question is asking me to do. Am I supposed to simply identify $0$ as the greatest element? If so, here is my attempt:

let $S$ be a nonempty subset of  $\{n \in  \mathbb{Z}: n \le 0\}$, if $s \in  S$, then $s \in \{n \in  \mathbb{Z}: n \le 0\}$ The greatest element of $S$ is $0$. Thus $s \ge x$ for all $x \in S$.

$S$ can be any subset of the non-positive integers!

For example, if $S=\{-1,-2,-3 \dots\}$, then in fact $-1$ is the maximal element.

but there is still hope to use your idea.

Indeed, if $S$ were all of $- \mathbb N$, then $0$ would be the maximal element. But since it is not, assume for contradiction, that it has no  maximal element. Then $0 \notin S$, since if it were, then since $S \subset -\mathbb N$, all of its elements would be less than zero as well. Then, $S \subset - \mathbb N\setminus \{0\}$. In this case, what is the maximal element of $-\mathbb N\setminus \{0\}$ can $-1 \in S$ if $S$ has no maximal element?

Hint: What can you say about the set $-S$?

Notice that, as others have pointed out, $0$ need not belong to $S$.


Is a sequence of all the same numbers monotonic? I'm wondering based on the definition of monotonicity:

  A sequence where $a_n\geq a_{n+1}$ for all $n\in\mathbb{N}$ is monotonic.

So given that the sequence $a_n = 3$ is all the same numbers and is neither increasing or decreasing, is it monotonic?

Yes, a constant sequence (a number repeated indefinitely) is inceed monotonic: it is both monotonic non-decreasing, and monotonic non-increasing.

Hence, one can require that a sequence be strictly monotonic increasing or strictly monotonic decreasing. Under such a restriction, a constant sequence is neither strictly increasing nor strictly decreasing monotonically.

yes, because constant sequence is both increasing and decreasing sequence. so that it is monotonic.


$dX_t=\alpha X_t \,dt + \sqrt{X_t} \,dW_t, $ with $X_0=x_0,\,\alpha,\sigma>0.$ Compute $E[X_t] $ and $E[Y]$ for $Y=\lim_{t\to\infty}e^{-\alpha t}X_t$ First of all i suppose that there is no easy analytical solution to the SDE? I got the hint to use Gronwall to show that there is a constant $C$ , s.t. $E[X_{t\land\tau_n}]\le C$ for all $n\in \mathbb{N}$, with $\tau_n=\inf(t\ge0: |X_t|\ge n)\land n.$

It is unclear to me how the hint is helping. If I could show it, I still would only have an upper bound for $E[X_t]$.

I think Y will have expectation $x_0$, since the drift gets eliminated but I don`t know how to prove the existence of Y and my guess.

Edit: $\alpha$ and $ \sigma$ are constants

Assume for the moment being that we already know that $$\sup_{t \in [0,T]} \mathbb{E}(|X_t|) \leq C < \infty \tag{1}$$ for any $T>0$. Follow the following steps to solve the problem:

Recall the following statement: If $f$ is a progressively mapping such that $\mathbb{E}\left( \int_0^T f(s)^2 \, ds \right)<\infty$ for any $T>0$, then $$M_t := \int_0^t f(s) \, dW_s$$ is a martingale; in particular $$\mathbb{E}(M_t)=\mathbb{E}(M_0)=0.$$
Use Step 1 and Fubini's theorem to show that $$\mathbb{E}(X_t)-x_0 =\alpha \int_0^t \mathbb{E}(X_s) \, ds. \tag{2}$$
By $(2)$, the mapping $u(t) := \mathbb{E}(X_t)$ solves the ordinary differential equation $$\frac{d}{dt} u(t) = \alpha u(t), \qquad u(0)=x_0.$$ Solve it.
Apply Itô's formula to show that $$e^{-\alpha t} X_t - x_0 = \int_0^t e^{-\alpha s} \sqrt{X_s} \, W_s.$$
Conclude that $$Y=\lim_{t \to \infty} e^{-\alpha t} X_t$$ almost surely and in $L^1$. (Hint: Show that $e^{-\alpha t}X_t$ is an $L^2$-bounded martingale.)
Because of the $L^1$-convergence, it follows from Step 4 that $$\mathbb{E}(Y) = \lim_{t \to \infty} \mathbb{E}(e^{-\alpha t} X_t) = x_0.$$

It remains to prove $(1)$:

Show that the process $$M_t^{(n)} := \int_0^{t \wedge \tau_n} \sqrt{X_s} \, dW_s$$ is a martingale for any $n \in \mathbb{N}$; here $$\tau_n := \inf\{t \geq 0; |X_t| \geq n\} \wedge n.$$ In particular, $M_t^{(n)}$ has expectation zero.
Use Itô's formula and step 1 to show that $$\mathbb{E}(X_{t \wedge \tau_n})-x_0 = \alpha \mathbb{E}\left( \int_0^{t \wedge \tau_n} X_s \, ds \right) \leq \alpha \int_0^t \mathbb{E}(X_{s \wedge \tau_n}) \, ds.$$
Apply the Gronwall lemma to prove that there exists a constant $C=C(T,x_0)>0$ such that $$\sup_{t \leq T} \mathbb{E}(X_{t \wedge \tau_n}) \leq C \qquad \text{for all $n \geq 1$}.$$ Use Fatous lemma to conclude that $$\sup_{t \leq T} \mathbb{E} \left(  X_t \right) \leq C.$$

You did not specify what kind of function $\alpha$ is. Assume it is a constant. Take expectation of the SDE, as $\mathbf E[dW_t]=0$,
$$dx = \alpha xdt \Longleftrightarrow x(t)=x(0)e^{\alpha t}\Longleftrightarrow y(t)=e^{-\alpha t}x(t)=x(0),$$
where $x(t):=\mathbf E[X_t]$ and $y(t):=\mathbf E[e^{-\alpha t}X_t]$.

I will add the existence of $\mathbf E[X_t]$ later, but soon.


Mechanical vibration: single degree of freedom model of wheel mounted on a spring I think the author of my book has this solution wrong and would like some feedback on my thoughts (see Example 1.4.1).

The solution states that the total energy of the system is represented by the spring, rotation of the wheel, and translation of the wheel, therefore constant total energy E is given by

\begin{align} E = (1/2)mv^2 + (1/2)J\omega^2 + (1/2)kx^2 \end{align}

where $J$ is the inertia of the wheel, equal to $mr^2$, and omega is the angular velocity equal to $v/r$.

I think in this scenario the translational energy and rotational energy are the same quantity -- that is, there cannot be rotation without translation (since he defined "no slipping occurs and no energy is lost at contact"), and vice-versa; or, any time the object is rotating the energy of rotation will be consumed to produce translation, and vice-versa. Thus the equation should be written

\begin{align} E = (1/2)mv^2 + (1/2)kx^2 \end{align}

And the maximum kinetic energy will be

\begin{align} T_{max} = (1/2)m(\omega_n)^2A^2 \end{align}

yielding

\begin{align} \omega_n^2 = k/m \end{align}

Thank you in advance for all responses. This is my first time posting on stack exchange.

This is more a physics problem then math. But here are some points that you should consider:

The moment of inertia of a filled disk is not $J=mr^2$, but $J=\frac 12 mr^2$, so the kinetic energy of rotation would be half of the kinetic energy of translation. $$K.E.R.=\frac 12 \frac12 mr^2\frac{v^2}{r^2}=\frac 14 mv^2$$
You have $J=mr^2$ only if the mass of the wheel is uniformly distributed on the periphery of the disk. So bodies with the same mass can have different moments of inertia. If the kinetic energy of translation is the same as the kinetic energy of rotation, which formula would you use? Why is the other formula wrong?
Let's look at a simpler case, a wheel rolling with constant velocity $v$. The mass of the disk is concentrated in two points, on opposite sides of the disk. We have at each of these points a mass $m/2$. The kinetic energy of the center of the mass is $$K.E._{CM}=\frac12(m/2+m/2)v^2=\frac{mv^2}2$$
But this must be equal to the sum of kinetic energies of the two particles. When one of the particle is at the bottom, the velocity of that particle is $v_b=0$. At the same time, the velocity of the particle at the top is $v_t=2v$. The sum of kinetic energies is then $$\frac 12 \frac m2 0^2+\frac 12\frac m2 (2v)^2=mv^2\ne \frac{mv^2}2$$
So if the sum is $mv^2$ but the kinetic energy of the center of mass is only half, where did the other half go? The answer is in the kinetic energy of rotation.

The energy of the system at a configuration $(x,\dot{x}, \dot{\theta})$ is given by
$E(x,\dot{x}, \dot{\theta}) = {1 \over 2} m \dot{x}^2 + {1 \over 2} J \dot{\theta}^2 + {1 \over 2} k x$.

In this particular example, there is an additional constraint which is that
$\dot{x} = r \dot{\theta}$.

When the object rolls, it will 'see' an effective mass of $m+ {J \over r^2}$. That is, it will behave as a single point mass with no inertial component of mass $m+ {J \over r^2}$.

The formula for the resonant frequency follows from the usual formula.

Aside:

Consider two systems of mass $m$, (i) is just a frictionless sliding mass and (ii) is an ideal, non slipping hoop with the mass at any positive radius.

The kinetic energies are (i) ${1 \over 2} m \dot{x}^2$ and (ii) $m \dot{x}^2$ (regardless of radius).


Euler's Identity in Degrees Since we have a simple conversion method for converting from radians to degrees, $\frac{180}{\pi}$ or vice versa, could we apply this to Euler's Identity, $e^{i\pi}=-1$ and traditionally in radians, to produce an equation that is in degrees; one that may or may not be as simplistic or beautiful as the radian version, but yet is still quite mathematically true?

If not, why?
If so, how did you derive it?

-K

I suppose the explanation here made it sound more difficult than just replacing pi with 180.

If you define $n^{\circ} = n \pi/180$, then you could write
$$ e^{i\pi} = e^{180^\circ i} = -1.$$

I don't think this is a good idea, since the exponential is usually defined as the power series

$$\exp(x)=\sum_{k=0}^\infty \frac{x^k}{k!}.$$

I would say degrees $°$ is some kind of unit like meters. In the power series, you would get something like $(y°)^k$ and sums of (all) different powers of the unit $°$. This makes as much sense as adding meters to square meters.

Of course you can just write $\exp(i180°) = -1$, but in my opinion it is not well-defined nor useful.


Unique critical point does not imply global maximum/global minimum I have actually two questions here, but both are very much related so I decided to put them both in this question.

From Wikipedia I found the following example of a function that has a single critical point which is a local minimum, yet is not a global minimum:

$f(x,y) = x^2 + y^2(1-x)^3$

Indeed the only critical point of $f$ is $(0,0)$ which is a local minimum by the second derivative test, but $(0,0)$ is not a global minimum because $f(4,1) = -11$. Why does this happen? I am having some trouble understanding intuitively why this is true. I know that $f$ has a minimum when restricted to a compact set. Then $(0,0)$ is a global minimum in any closed ball around $(0,0)$, and thus $(0,0)$ is a global maximum?? This is not true because $f$ is a counterexample, but I can't quite understand why.

In some cases we do have a global minimum at the unique critical point. For example, let

$f(x,y) = (-x+y)^2 + (x-1)^2 + (x+y-1)^2$

Then $f$ has a unique critical point at $(\frac{2}{3}, \frac{1}{6})$. The critical point turns out to be a global minimum (according to the graph and wolframalpha). How would I prove that the point is a global minimum? And how does one in general decide that a function has a global minimum/maximum on a critical point when the function is defined on all of $\mathbb{R}^2$, particularly when there is only one critical point?

Your intuitive picture is probably as follows: Imaging filling water into the hollow at the origin. As the water rises, since there are points lower than the origin, you expect the water to start spilling over into the lower lying areas at some point. And that point would need to be a critical point! The way to resolve the paradox is to note that before any spilling over occurs, the lake will in fact extend off to infinity. Look at a cross section of your function graph for fixed $y$, as $y$ starts at $0$ and then grows. In other words, study the function $x\mapsto x^2+y^2(1-x)^3$ as $y^2$ grows. You will see a local minimum, starting at the origin and approaching $x=1$ from below, and a local maximum, starting at infinity and approaching $x=1$ from above. As $y^2$ grows, the value at the local minimum grows, while the value at the local maximum decreases. So the aforementioned lake will extend out to $(1,\pm\infty)$ and then start to spill over the edge (the local maximum) at infinity before it spills over anywhere else.

A picture is tremendously helpful. If you have access to maple, try this:

plots[animate](plot,
  [x^2+y^2*(1-x)^3, x = 0 .. 2, view = [0 .. 2, -1 .. 2]], y = 0 .. 10);

Or you can take a look at this static plot of $f(x,y)$. It is easy to see how $(0,0)$ is a local, but not a global, maximum:

A critical point can be either a minimum, a maximum, a saddle point or none of these three, i.e. undecidable. For a two-variable function f(x,y), if the determinant of its Hessian is positive at the critical point (x*,y*), and fxx(x*,y*) is positive, then the critical point(x*, y*) is a local minimum. Refer to the list of tests in: https://en.wikipedia.org/wiki/Second_partial_derivative_test


Is there a name for a 2-dimensional dumbbell like shape? Is there a mathematical name for a 2-dimensional shape in the general form of a dumbbell?  That is two circular nodes connected by a center beam such as shown in this image from this answer.  It could also be thought of as a circle that has been pinched together.  I believe topology does not help us here since there are no holes, and thus such as shape is the same as a circle, but perhaps geometry gives us a name for it?

The shape reminds of a Cassini oval (http://mathworld.wolfram.com/CassiniOvals.html) but without a more precise equation, one cannot be more specific.


Inclusion-Exclusion Principle for basic combinatorics problem... How many ways are there to pick five people for a committee if there are six
  (different) men and eight (different) women and the selection must include at
  least one man and one woman?

I know to solve this with very basic combinatorial analysis (though I know the principle is also basic). We find the total people we can choose for the committee and proceed with cases: 1 where we have all men on the committee, 1 where we have all women on the committee and then we subtract the two cases from our total. This gives us
$${14 \choose 5} - {8 \choose 5}-{6 \choose 5}.$$

How would I compute for Inclusion-Exclusion Principle, though? I've no idea where to begin with it other drawing a venn diagram and labeling set M and W...

You have more-or-less done it already.

For $i \in \{0,1,\ldots\}$, let $a_i$ be the number of committees with $\geq i$ absent genders.

We include $a_0=\binom{14}{5}$.

We exclude $a_1=\binom{8}{5}+\binom{6}{5}$.

We include $a_2=0$.

We exclude $a_3=0$.

And so on.

In my opinion, this is the answer:

1m and 4 w : 6 X 8C4
2m and 3 w : 6C2 X 8C3
3m and 2 w : 6C3 X 8C2
4m and 1 w : 6C4 X 8

Since these are all mutually exclusive, they can all be added.


If you have 100 spheres packed into a sphere shape, how many will be on the surface? My question is more about ratios. I'm wondering is there a calculator or formula I can input an X number of Spheres. If the spheres are packed in a spherical shape what is the ratio of The interior spheres compared to the spheres on the surface?
Thanks

This is an unsolved problem. Here is a description for $X\le 12$.

Also, follow this other thread for more information on the subject

The maximum density of an infinite number of congruent identical spheres packed within a larger sphere is 0.74048... or Pi/(sqrt(18)), as stated in the Kepler Conjecture, partially proven by Gauss and not formally proven by Thomas C. Hale until 2014.

The Kepler Conjecture states that “no arrangement of equally sized spheres filling space has a greater average density than that of the cubic close packing (face-centered cubic) and hexagonal close packing arrangements.”

I actually worked this out myself (but didn’t prove it) by calculating how many spheres could fit into a regular tetrahedral container and a square pyramidal container.

When the number of spheres got well into the millions, the values converged at .074048...

I still have the Excel spreadsheets that I used to do the calculations.


Integrate $\sin^{-1}\frac{2x}{1+x^2}$ Integrate $\sin^{-1}\frac{2x}{1+x^2}$

The solution is given in my reference as: $2x\tan^{-1}x-\log(1+x^2)+C$.

But, is it a complete solution ?

My Attempt
$$
\int 2\tan^{-1}x \, dx=\int \tan^{-1}x \cdot 2\,dx=\tan^{-1}x\int2\,dx-\int\frac{1}{1+x^2}\int2\,dx\cdot dx\\
=\tan^{-1}x \cdot2x-\int\frac{2x}{1+x^2}\,dx=2x\tan^{-1}x-\log(1+x^2)+C
$$
$$
2\tan^{-1}x=\begin{cases}\sin^{-1}\frac{2x}{1+x^2}\text{ if }|x|\leq{1}\\
\pi-\sin^{-1}\frac{2x}{1+x^2}\text{ if }|x|>{1}\text{ and }x>0\\
-\pi-\sin^{-1}\frac{2x}{1+x^2}\text{ if }|x|>{1}\text{ and }x<0
\end{cases}\\
\sin^{-1}\frac{2x}{1+x^2}=\begin{cases}2\tan^{-1}x\text{ if }|x|\leq{1}\\
\pi-2\tan^{-1}x\text{ if }|x|>{1}\text{ and }x>0\\
-\pi-2\tan^{-1}x\text{ if }|x|>{1}\text{ and }x<0
\end{cases}
$$
$$
\int\sin^{-1}\frac{2x}{1+x^2}\,dx=\begin{cases}\int2\tan^{-1}x\,dx&\text{ if } |x|\leq{1}\\\int\pi\, dx-\int2\tan^{-1}x\,dx&\text{ if }|x|>{1}&\text{ and } x>0\\-\int\pi \,dx-\int2\tan^{-1}x\,dx&\text{ if }|x|>{1}&\text{ and } x<0\end{cases}=\begin{cases}\color{red}{2x\tan^{-1}x-\log(1+x^2)+C\text{ if } |x|\leq{1}}\\\pi x-2x\tan^{-1}x+\log(1+x^2)+C&\text{ if }|x|>{1}&\text{ and } x>0\\-\pi x-2x\tan^{-1}x+\log(1+x^2)+C&\text{ if }|x|>{1}&\text{ and } x<0\end{cases}$$

So don't we have two more cases for our solution rather than that is given in my reference, right ?

You're right; but it depends on what the full question is; if one is asked to compute
$$
\int_{0}^{1}\arcsin\frac{2x}{1+x^2}\,dx
$$
then just the antiderivative $2\arctan x$ is sufficient. Not if one wants to compute an integral involving points outside the interval $[-1,1]$.

Here's a shorter way to get at your result.

Consider
$$
f(x)=\arcsin\frac{2x}{1+x^2}
$$
Then
$$
f'(x)=\frac{1}{\sqrt{1-\dfrac{4x^2}{(1+x^2)^2}}}\frac{2(1+x^2)-4x^2}{(1+x^2)^2}
=\frac{1+x^2}{|1-x^2|}\frac{2(1-x^2)}{(1+x^2)^2}
$$
hence
$$
f'(x)=\begin{cases}
\dfrac{2}{1+x^2} & x\in(-1,1) \\[4px]
-\dfrac{2}{1+x^2} & x\in(-\infty,-1)\cup(1,\infty)
\end{cases}
$$
Therefore, knowing that $f(0)=0$,
$$
f(x)=\begin{cases}
-\pi-2\arctan x & x<-1 \\[4px]
2\arctan x & -1\le x\le 1 \\[4px]
\pi-2\arctan x & x>1
\end{cases}
$$

In order to find an antiderivative we can consider
$$
\int\arctan x\,dx=x\arctan x-\int\frac{x}{1+x^2}\,dx
=x\arctan x-\frac{1}{2}\log(1+x^2)
$$
Thus an antiderivative of $f$ has the form
$$
F(x)=\begin{cases}
c_- -\pi x-2x\arctan x+\log(1+x^2) &\qquad x<-1 \\[4px]
2x\arctan x-\log(1+x^2) &\qquad -1\le x\le 1 \\[4px]
c_+ +\pi x-2x\arctan x+\log(1+x^2) &\qquad x>1
\end{cases}
$$
and you just need to determine $c_-$ and $c_+$ to ensure continuity at $-1$ and $1$.

The other antiderivatives differ from $F$ by a constant.

We know that $\tan(\theta/2) = \frac{\sin(\theta)}{1+\cos(\theta)}$ and $\cos(\theta) = \sqrt{1-\sin^2(\theta)}$.
So, letting $x = \sin(\theta)$ $(\implies \theta = \arcsin(x))$ we have:

$\tan(\theta/2) = \frac{x}{1+\sqrt{1-x^2}} \implies \frac{\theta}{2} = \arctan\left(\frac{x}{1+\sqrt{1-x^2}}\right) \implies \arcsin(x) = 2\arctan\left(\frac{x}{1+\sqrt{1-x^2}}\right)$.

With that in hands, we just have to calculate $\arcsin\left(\frac{2x}{1+x^2}\right) = 2\arctan\left(\frac{\frac{2x}{1+x^2}}{1+\sqrt{1-\frac{4x^2}{1+2x^2+x^4}}}\right)$.

Now observe that $\frac{\frac{2x}{1+x^2}}{1+\sqrt{1-\frac{4x^2}{1+2x^2+x^4}}} = \frac{\frac{2x}{1+x^2}}{1+\sqrt{\frac{1-2x^2+x^4}{1+2x^2+x^4}}} = \frac{\frac{2x}{1+x^2}}{1+\frac{1-x^2}{1+x^2}} = \frac{\frac{2x}{1+x^2}}{\frac{2}{1+x²}} = x$. I thik this solve the equality you were struggling with.


Find the equation of a parabola (in general form) Find the equation of the parabola with axis parallel to the $y$-axis, passing through $(1/2,-5/2),(3/2,-9/4)$ and $(-7/2,3/2)$.

$\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
\begin{align}
{\rm y}\pars{x}&=
A\pars{x - \half}\pars{x - {3 \over 2}}
+B\pars{x - {3 \over 2}}\pars{x + {7 \over 2}}
+C\pars{x - \half}\pars{x + {7 \over 2}}
\end{align}

  In order to determine $\ds{A, B , C}$ use
  $$
{\rm y}\pars{-\,{7 \over 2}} = {3 \over 2}\,,\qquad
{\rm y}\pars{\half} = -\,{5 \over 2}\,,\qquad
{\rm y}\pars{3 \over 2} = -\,{9 \over 4}
$$


Hypothesis testing, chance To find out whether the inhabitants of two South Pacific islands may be regarded as having the same racial ancestry, an anthropologist determines the cephalie indices of six adult
males from each island, getting $x_1$ = 77.4 and $x_2$ = 72.2 and the corresponding standard
deviation $s_1 = 3.3$ and $s_2 = 2.1$. Test at $0.01$ level of significance whether the difference between the two sample means can reasonably be attributed to chance. Assume that the
populations sampled are normal and have equal variances.

I am confused on what the hypotheses should be: should it be $H_0: \mu_1 - \mu_2 = 0$ and $H_1: \mu_1-\mu_2 \neq 0$?

Pooled 2-sample t test. Minitab is one of the relatively few statistical packages
that will perform a pooled 2-sample t test on summarized data
(that is, without seeing the 12 actual data values). You
have correctly used 6 + 6 -2 = 10 degrees of freedom to get
your critical value 3.169. Also, Minitab gets $T = 3.26$ for the test
statistic, which agrees with your result.
Here is Minitab output.

Two-Sample T-Test and CI

Sample  N   Mean  StDev  SE Mean
1       6  77.40   3.30      1.3
2       6  72.20   2.10     0.86

Difference = μ (1) - μ (2)
Estimate for difference:  5.20
95% CI for difference:  (1.64, 8.76)
T-Test of difference = 0 (vs ≠):
  T-Value = 3.26  P-Value = 0.009  DF = 10
Both use Pooled StDev = 2.7659

Welch 2-sample t-test. However, as a matter of statistical practice, I would not be willing
to assume the two population variances are equal. (If the population
means might be different, why not also the population variances?)
The Welch version of the 2-sample t test does not assume equal
population variances. Standard statistical practice is not to assume
equal variances unless you have strong prior evidence of equality. (In this
situation, that might come from measuring cephalic indexes for many other
Pacific islands and having found that means vary, but that variances don't.)

Typically the Welch 2-sample t test uses a reduced degrees of freedom
to compensate for possibly unequal variances. A formula involving sample
sizes and sample standard deviations is used to find the adjusted degrees
of freedom. Here is Minitab output for the Welch 2-sample t test.

Two-Sample T-Test and CI

Sample  N   Mean  StDev  SE Mean
1       6  77.40   3.30      1.3
2       6  72.20   2.10     0.86

Difference = μ (1) - μ (2)
Estimate for difference:  5.20
95% CI for difference:  (1.52, 8.88)
T-Test of difference = 0 (vs ≠):
  T-Value = 3.26  P-Value = 0.012  DF = 8

Notice that df = 8 here instead of 10 for the pooled test.
Also, notice that there is no note at the end about using
a pooled standard deviation. The P-value is a little larger
for the Welch test, but is still below 0.05, so $H_0$ is
rejected here as well.


Railway track and Cyclist crossing, Motion Problem. A railway track runs parallel to a road until a turn brings the road to railway crossing. A cyclist rides along the road everyday at a constant speed 20 km/hr. He normally meets a train that travels in same direction at the crossing. One day he was late by 25 minutes and met the train 10 km before the railway crossing.Find the speed of the train.

My Solution:

I suppose the cyclist starts from a fixed point everyday markes by the red dot. If the situation was A, and if the cyclist wakes up late then train would have crossed the crossing and went straight and there seems no possibility that cyclist would be able to catch it 10 km before the crossing. So situation B would be correct as even if he goes late, train comes to him on his self, so he can catch the train 10 km before.
[contradiction: "a train that travels in same direction"]

As he is 25 minutes late, the train gets 25 minutes extra to cover some distance and cyclist 25 minutes less so he will cover less distance.

[contradiction: if cyclist gets 25 minutes less he would cover $25/60\;.\;20=8.\bar 3$ km before the track but where does 10 km come from]

[contradiction2: if we suppose cyclist takes t min daily to reach crossing, then taking $v_t$ the speed of train $v_t(t+25/60)+v(t-25/60)=v_tt+vt\implies v_t=v=20??$]

Suppose train daily starts $v_tt$ distance before the crossing and the cyclist starts from his house and covers $20t$ distance daily, the difference in starting points thus is $v_tt-20t$. Now since he is late the train travels in time $t_2$: $v_tt_2+10=v_tt$ and cyclist travels in time $t_2$: $20t_2+10=20t$ a distance $v_t(25/60)+v_tt_2=v_tt-10$ where $t_2$ is the time cyclist travels after waking up. The cylist would travel a distance $20t_2=20t-10$, this on slving gives $v_t=120$!!

Time taken by the cyclist to cover the distance of 10 km=(10×1000)/(20×1000/60)=30minutes. It means cyclist was 30 minutes away from the crossing the moment he met the train. So the man will reach the crossing in 30-25=5minutes. Therefore the train had to move 10 km in 5 minutes to meet the cyclist on a usual day at the crossing.
Train speed=10 km/5 minutes= 120km per hour.


$n^4 + 4^n$ is a not a prime Prove that $n^4 + 4^n$ is not a prime for all $n > 1$ and $n \in \mathbb{N}$.

This question appeared in the undergrad entrance exam of  the Indian Statistical institute.

When $n$ is even the proof is simple.
For $n = 2m+1$ I am utterly stuck.

Let $n=2k+1$ with $k\geq 1$, then
$$n^4+4^n=n^4+4 \cdot 4^{2k}=n^4+4\cdot (2^k)^4=(n^2+2\cdot 2^{2k}+2^{k+1}n)(n^2+2\cdot 2^{2k}-2^{k+1}n).$$
Thus $n^4+4^n$ can be factored into non-trivial factors, when $n$ is odd.

Remember doing this as an Exercise from Ivan Niven's Number Theory book. I think this follows from an identity of Sophie Germain, namely $$a^{4} + 4b^{4} = (a^{2}+2b^{2}-2ab)(a^{2}+2b^{2}+2ab).$$

See also this article on the identity.


Showing that $\sum_{k\mathop=n}^{\infty} \frac 1 {k^2} < \frac 2 {2n-1}$ I need to show that

$$\sum_{k\mathop=n}^{\infty} \frac 1 {k^2} < \frac 2 {2n-1}$$

Could I have a small hint?

I have tried many things, like bounding it above with the corresponding integral but this is too weak. I have tried subtracting $1$ from each of the denominators (excluding the first term) and then evaluating the resulting telescoping sum, but even that turns out to be too weak.

Hint: Compare with the integral of $1/x^2$, by comparison with the integral, you obtain that $\sum_{k\geq n}{1\over k^2}<\int_{n-{1\over 2}}^{+\infty}{1\over x^2}dx={2\over{2n-1}}$.


is $|x|^3$ differentiable at [-1, 2] I first defined the function for $x$>0 and $x$<0.

For $x$>0 the function is $x^3$

For $x$<0 function is =-$x^3$

I differentiated it and both cases turned out to be 0.

so left hand limit=RHL=0.

So the function must be differentiable.

am I correct?

Yes, you have
$$\lim_{h \to 0^+} \frac{|h|^3-|0|^3}{h}=\lim_{h \to 0^+} \frac{h^3}{h}=\lim_{h \to 0^+} h^2 =0$$
And
$$\lim_{h \to 0^-} \frac{|h|^3-|0|^3}{h}=\lim_{h \to 0^-}\frac{-h^3}{h}=\lim_{h \to 0^+} -h^2 =0$$
So $|x|^3$ is differentiable at $x=0$ and its derivative is $0$.

In any other $x \in [-1,2] \setminus \{0\}$ you can use the usual derivative rules since there aren't problems with the derivative.

Edit:
If there's function $g(x)=|f(x)|\\$
$g'(x) = sgn(f(x)).f'(x)$


How do you solve $\lim_{n\to \infty}{n[(1+\frac{c}{n})^{n}-(1-\frac{c}{n})^{-n}]}$ without L’Hopital's Rule? I know that solution is: $-c^2e^c$

$$\lim\nolimits_{n\to \infty}{n\left[\left(1+\frac{c}{n}\right)^{n}-\left(1-\frac{c}{n}\right)^{-n}\right]}$$

Hint: common factor to $\left(1+\frac{c}{n}\right)^{n}$ and then to use approaches

As a big hint, assuming WLOG that $c > 0$, we have as $n \to \infty$

$$\left(1+\frac{c}{n}\right)^{n}\uparrow e^c  \downarrow\left(1-\frac{c}{n}\right)^{-n},$$

and for all $n > c$

$$0 \leqslant n\left[\left(1-\frac{c}{n}\right)^{-n}-\left(1+\frac{c}{n}\right)^{n}\right] = n\left(1-\frac{c}{n}\right)^{-n}\left[1 -  \left(1-\frac{c^2}{n^2}\right)^{n}\right] \leqslant \ldots$$

See if you can finish by first applying Bernoulli's inequality to the last term on the RHS and then applying the squeeze theorem.

$$
\lim\limits_{n\to\infty} \left[ n\left( 1 + \frac{c}{n} \right)^n - n\left( 1 - \frac{c}{n} \right)^{-n}\right] =
\lim\limits_{n\to\infty} \left[ (n + c)\left( 1 + \frac{c}{n} \right)^{n-1} - (n - c)\left( 1 - \frac{c}{n} \right)^{-n - 1}\right] = \\
\lim\limits_{n\to\infty} \left[ (n + c)\underbrace{\left( 1 + \frac{c}{n} \right)^{-1}}_{1}\underbrace{\left( 1 + \frac{c}{n} \right)^{n}}_{e^c} - (n - c)\underbrace{\left( 1 - \frac{c}{n} \right)^{-1}}_{1}\underbrace{\left( 1 - \frac{c}{n} \right)^{-n}}_{e^c}\right] = 2\cdot c \cdot e^c\\
$$