Solve $f(x+f(2y))=f(x)+f(y)+y$ Find all $f:\mathbb{R}^+\to \mathbb{R}^+$ such that for each $x$ and $y$ in $\mathbb{R}^+$, $$f(x+f(2y))=f(x)+f(y)+y$$

Note:
$f(x)=x+b$ is a solution for all  $b\in\mathbb{R}^+$ but I can not prove it.

EDIT: I now have a complete solution.

$$f:\mathbb{R}^+\to\mathbb{R}^+$$
$$f(x+f(2y))=f(x)+f(y)+y$$

Note that $f(a+f(2x_1)+f(2x_2)+...+f(2x_n))=f(a)+f(x_1)+x_1+f(x_2)+x_2+...+f(x_n)+x_n$

Let us calculate the expression $f(x+f(2(y+f(2z)))$ in two ways:
$$f(x+f(2(y+f(2z)))=f(x)+f(y+f(2z))+y+f(2z)=f(x)+f(y)+f(z)+z+y+f(2z)$$
$$f(x+f(2(y+f(2z)))=f(x+f(2y+f(2z)+f(2z)))=f(x+f(2y)+2f(z)+2z)=f(x+2z)+f(y)+y+2f(\frac{z}{2})+2\cdot{}\frac{z}{2}$$
Thus, cancelling $f(y)+y+z$ we get
$$f(x)+f(z)+f(2z)=f(x+2z)+2f(\frac{z}{2})$$
i.e. (for $y=2z$)
$$f(x+y)=f(x)+f(y)+f(\frac{y}{2})-2f(\frac{y}{4})$$

Since the expression $f(x+y)-f(x)-f(y)$ is symmetric in $x$ and $y$, we get that $f(\frac{y}{2})-2f(\frac{y}{4})=f(\frac{x}{2})-2f(\frac{x}{4})$, i.e. $f(2y)-2f(y)=f(2x)-2f(x)$ for all $x,y$. so $f(2x)-2f(x)$ is some constant $-c$. So we obtained:
$$f(x+y)=f(x)+f(y)-c$$
Define $i=inf_{\mathbb{R}^+}(f(x))$. Clearly, $0\le{}i\le{}\infty$. as $x$ and $y$ go through all positive real values, the infimum of LHS is $i$ and the infimum of RHS is $2i-c$, therefore $c=i$. since $f(y)-i\ge{}0$, we get $f(x+y)\ge{}f(x)$, so $f$ is monotonically increasing.

Define $g:\mathbb{R}^+\to{}\mathbb{R}^+$ as $g(x)=f(x)-i$. from $f(x+y)=f(x)+f(y)-c$ we get that $g$ is additive, i.e. $g(x+y)=g(x)+g(y)$. Thus, by induction, $g(k)=kg(1)$ for all $k\in{}\mathbb{N}$. This argument can be expanded for all $k\in\mathbb{Q}^+$ (why?), so $g(q)=qg(1)$ for all $q\in{}\mathbb{Q}^+$. but since $f$ is monotonic we get that $g$ is monotonic, so $g(x)=xg(1)$ for all $x>0$. Thus $g$ is linear, so $f$ is linear. Substituting $f(x)=ax+b$ in the original equation yields the solutions:
$$a=1$$
$$a=-\frac{1}{2}, b=0$$
We are interested in solutions that give only positive values in the image of $f$. Thus $f(x)=x+b$ for some constant $b\ge{}0$. This function satisfies the equation for all $b\ge{}0$.

Another solution is
$f(x) =-x/2$.

Here,
at the end of my ramblings,
is how I found it.

Inspired by the comment
that a solution is
$x+b$,
let
$f(x) = x+g(x)$.

Then
$x+2y+g(2y)+g(x+2y+g(2y))
=x+g(x)+y+g(y)+y
$
or
$$g(2y)+g(x+2y+g(2y))
=g(x)+g(y).
$$

Letting $y=0$,
this is
$g(0)+g(x+g(0))
=g(x)+g(0)
$
or
$g(x) = g(x+g(0))$.

If $g(0) = 0$,
this tells nothing.
If $g(0) \ne 0$,
$g$ is periodic
with period $g(0)$.
In particular,
$g(0)
= g(g(0))
= g(g(g(0)))
=...
$.

Setting $x=0$,
this is
$g(2y)+g(2y+g(2y))
=g(0)+g(y)
$.
This doesn't seem to help much.

If $x=2y$,
this is
$g(2y)+g(4y+g(2y))
=g(2y)+g(y)
$
or
$g(y)
=g(4y+g(2y))
$,
so
$g(y+g(0))
=g(4y+g(2y))
$.

If
$y+g(0)
=4y+g(2y)$,
then
$g(2y)
=g(0)-3y
$
or
$g(y)
=g(0)-3y/2
$.

This suggests trying
$f(x) = ax+b$.
Then
$a(x+2ay+b)+b
=a(x+y)+2b+y
$
or
$2a^2y+ab
=(a+1)y+b
$
or
$0
=(2a^2-a-1)y+b(a-1)
$.
Therefore
$0=(2a^2-a-1$
or
$a
=\dfrac{1\pm\sqrt{1+8}}{4}
=\dfrac{1\pm 3}{4}
=1, -\frac12
$.

If $a=1$
then any $b$ works.

If
$a=-\frac12$,
then
$b=0$,
so
$-x/2$
is a solution.
To check,
$f(x+f(2y))
=f(x-y)
=(y-x)/2
$
and
$f(x)+f(y)+y
=-x/2-y/2+y
=(y-x)/2
$.

That's all I have.


Let X be a random variable with the following p.m.f. f(x) = cx(4 − x), x = 1, 2, 3 Let $X$ be a random variable with the following p.m.f. $f(x) = cx(4 − x), x = 1, 2, 3$
and zero, otherwise.

(i) Find c

(ii) Find $E(X)$ and $Var(X).$

(iii) Find the moment generating function.

(iv) Draw a graph of the probability histogram of the pmf of $X$

I was able to find the value for c by doing $1=c(1(4-1) + 2(4-2) + 3(4-3))$ which gives $c = 1/10.$

So what you do is we know $E(x)=x(f(x))$
And V$ar(x)=x^2(f(x))-(E(x))^2$

All you do is plug the numbers in and its very easy from then on


Why are "join" and "meet" named as they are? In the context of partially ordered sets, why are the words for supremum and infimum "join" and "meet"? I find the nomenclature puzzling, especially since the English words "join" and "meet" are synonyms, but denote opposite concepts when talking about posets.

Does anyone know how these concepts got these names?

(Cross-posted on HSM)

"Join" and "meet" for partially ordered sets could be based on elementary set theory, where you have the union ($\cup$) or intersection ($\cap$) of two or more sets; join $(\vee)$ and meet $(\wedge)$ are their analogues for posets.


How to put 9 pigs into 4 pens so that there are an odd number of pigs in each pen? So I'm tutoring at the library and an elementary or pre K student shows me a sheet with one problem on it:

Put 9 pigs into 4 pens so that there are an odd number of pigs in each pen.

I tried to solve it and failed! Does anybody know how to solve this? This question seems ridiculously difficult and impossible IMO.

Since this has been exposed, some claim, to be a riddle and not a bona fide math question, why not completely drive the stake through its heart:

Pen 1: 7 pigs

Pen 2: 1 pig

Pen 3: 3 pigs

Pen 4: -2 pigs

Now the astute reader will note that -2 pigs is a pretty darn odd number of pigs to be in a pen!

I assume in elementary school they are not expected to know the parity of zero, so why not this:

3,3,3,0

Thats 9, and all "not even" (again, forget about the parity of zero).


Prove that $(a,b)\times(c,d)\subset \mathbb{R}^2$ is an open set. Let $\|\cdot\|$ be a norm of $\mathbb{R}^2$.
Prove that
$$\forall x\in (a,b)\times(c,d)\subset\mathbb{R}^2,\; \exists \varepsilon>0,\; B(x,\varepsilon)\subseteq (a,b)\times (c,d)$$
where $\displaystyle B(x,\varepsilon)=\{v\in \mathbb{R}^2: \|v - x\|<\varepsilon \}$.

My attempt:
Let $x=(x_1,x_2)\in (a,b)\times (c,d)$. So $a<x_1<b\land c<x_2<d$.

Figure suggests that the radius of the open ball (blue) ought be smaller than any of the green lines (or equal to the shortest one).
Let $\displaystyle\varepsilon =\min\{\|(x_1,d)-x\|,\|(x_1,c)-x\|,\|(a,x_2)-x\|,\|(b,x_2)-x\|\}=\|\alpha-x\|$.
And let $v\in B(x,\varepsilon)$, that is
$$\|v-x\|<\|\alpha-x\|.\tag{*}$$
I can't find a way to prove that $(*)$ implies $v\in (a,b)\times (c,d)$.
How could I continue the proof? If I'm not going in the right direction a hint would also help.

On a finite product,  $\mathbb R\times\mathbb R $, in this case,  the product and box topology coincide.  So the cross product of two open sets is open, by definition.
Put differently,  if $\pi_1,\pi_2$ are the projections onto the first and second factor, then $(a,b)×(c,d)=\pi_1^{-1}(a,b)\cap\pi_2^{-1}(c,d) $.


I've solved it by taking modulus of each side. Is there any other possible way to solve this question? If $\left(a,b,c,d\right) \in \mathbb{R}^4$ and if $$a+bi=\frac{c}{d+\cos(x)+i\sin(x)}$$

Show that $$(d^2-1)(a^2+b^2) +c^2 = 2acd$$

cross multiplication leads to the equation$$ad+a\cos(x)-b\sin(x)+i(bd+b\cos(x)+a\sin(x))=c$$
your equation is true
writing
$$a\cos(x)-b\sin(x)=-ad$$
and
$$a\sin(x)+b\cos(x)=-bd$$
squaring both equations and adding we get
$$a^2(\sin^2(x)+\cos^2(x))+b^2(\sin^2(x)+\cos^2(x))=a^2d^2+b^2d^2$$
can you finish?
from this equation we get
$$a^2+b^2=d^2(a^2+b^2)$$ thus $$d^2=1$$ if $$a^2+b^2\neq 0$$


Evaluate $\lim_{x\to\infty}\left[x - \sqrt[n]{(x - a_1)(x - a_2)\ldots(x - a_n)}\right]$ Evaluate the following the limit:
$$\lim_{x\to\infty}\left[x - \sqrt[n]{(x - a_1)(x - a_2)\ldots(x - a_n)}\right]$$

I tried expressing the limit in the form $f(x)g(x)\left[\frac{1}{f(x)} - \frac{1}{g(x)}\right]$ but it did not help.

\begin{align}
& \lim_{x\to\infty} x - \sqrt[n]{(x - a_1)(x - a_2)\ldots(x - a_n)} \\
= {} & \lim_{x\to\infty} x\left(1-\sqrt[n]{\left(1 - \frac{a_1}{x}\right)\left(1 - \frac{a_2}{x}\right)\ldots\left(1 - \frac{a_n}{x}\right)} \right) \\
= {} & \lim_{x\to\infty} x\left(1-1 +\frac{1}{n}\frac{\sum a_i}{x}\right)\\
= {} & \frac{1}{n}{\sum a_i}
\end{align}

You know that the product $(x - a_1)(a - a_2)\cdots(x - a_n)$ is equal to $x^n + \cdots + a_1a_2\cdots a_n$, where the $\cdots$ between $x^n$ and the product of the $a_i$'s is made of powers of $x$ with an exponent that is smaller than $n$. After you've seen this, it's easy to see that the only term that matters inside that root is $x^n$, because all the other ones can be left out when $x$ goes to $\infty$.

So your limit is:

$$\lim_{x\to\infty} (x - \sqrt[n]{x^n}),$$

and now it will all depend on whether $n$ is even or odd. If it's even, then $\sqrt[n]{x^n} = \lvert x\rvert$, otherwise $\sqrt[n]{x^n} = x$. So, if $x$ goes to $-\infty$, for example:

$$\lim_{x\to-\infty} (x - \sqrt[n]{x^n}) = \begin{equation}
   \begin{cases}
   -\infty, & \text{if $n$ is even;} \\0, & \text{if $n$ is odd.}
   \end{cases}
\end{equation}$$

Otherwise, if $x$ goes to $+\infty$, it's easy to see that the limit is $0$ in both cases.


Precise value of $\coth(x)$ for large values of $x$? We know that for large values of $x$, $\cosh (x)\approx \sinh (x)$, and then $\lim_{x\to \infty } \, \coth (x)=1$. My question is that for example is the result of $\coth (100)$ exactly one? Or it is an estimated value?

${\cosh(100)}$ will actually be a very large number. Did you mean ${\coth(100)}$? If so, no, it will not be exactly $1$, it's estimated. But it will be very, very close to it. The fact the limit is $1$ as ${x\rightarrow\infty}$ just means that they behave "asymptotically" the same as ${x}$ get's larger.

Large $x$, or $x\gg1$, means $e^{-x}\ll1$. Then, $\coth x $ can be expressed as
$$\coth x =\frac{e^x+e^{-x}}{e^x-e^{-x}}
= \frac{1+e^{-2x}}{1-e^{-2x}}\approx(1+e^{-2x})^2\approx1+2e^{-2x}
$$
For example, $\coth 100 \approx 1+2e^{-200}$.


Law of composition in a group Most textbooks on group theory define the law of composition, say $\star$, as a function $\star: S \times S \to S$. The next part of the definition is sending $(a,b) \mapsto a \star b$. This confuses me somewhat. $\star$ is the function, but we're in effect treating it  as thought it were an operation and writing it between $a$ and $b$ instead of $\star(a,b)$. I'm ok with the abuse of the notation, but the definition is circular, defining $\star$ in terms of itself. One definition I say defines an operation $m: S \times S \to S$ by $(a,b) \mapsto a \cdot b$, but then it doesn't define the symbol $\cdot$, which is really the function/operation.
Is this all just an abuse of notation?

As long as your map $\star$ (or $\cdot$, or etc.) has no property to fulfill, feel free to denote the image of a pair $(a,b)$ of the domain by any between the two equivalent symbols $\star(a,b)$ and $a\star b$. But, as commented by @CharlesHudgins, as soon as you have to deal with properties for your map to fulfill (e.g. associativity), infix notation brings undoubtable practical benefits.


applying logarithm law question Here is my equation (below) on which I am applying log

$X=\frac{a}{b}\left ( c-d \right )$

so far I applied it as

$\log X=\log(a)-\log(b)+\left [ \log\left ( c \right )-\log\left ( d \right ) \right ]$

The equation you have given will on taking log give-

logX=log(a/b)+log(c-d)

Now as karolis commented

log(c−d)≠logc−logd
log(c+d)≠logc+logd

But log(a/b) =log(a) - log(b)
And loga*b=log(a) +log(b)

So we get

logX=[log(a)−log(b)]*log(c-d)


Prove the following relation between side lengths Let $ABC $ be any triangle , right-angled at $A$ , with $D$ any point on the side $AB$. The line $DE$ is drawn parallel to $BC$ to meet the side AC at the point E. F is the foot of the perpendicular drawn from E to BC. If $AD=x_1$ , $DB=x_2$ , $BF=x_3$, $EF=x_4$ and $AE=x_5$.  Then show that

$$\frac{x_1}{x_5} + \frac{x_2}{x_5} = \frac {x_1x_3  + x_4x_5}{x_3x_5 - x_1x_4}$$

My attempt was finding similar triangles but there are a lot of them and I really can't seem to make any progress

Note that $\tan \angle AEB=\frac{x_1+x_2}{x_5}$
and $\angle AEB = \angle AED +\color{blue}{\angle BED}=\angle AED +\color{blue}{\angle EBF}$ given that lines $DE$ and $BF$ are parallel.

Using the sum of angles formula for the tangent function we have
$$\tan \angle AEB=\frac{\tan\angle AED +\tan\angle EBF}{1-\tan\angle AED\tan\angle EBF}$$
where $\tan \angle AED=\frac{x_1}{x_5}$ and $\tan \angle EBF=\frac{x_4}{x_3}$.

Substituting results in
$$\frac{x_1+x_2}{x_5}=\frac{\frac{x_1}{x_5}+\frac{x_4}{x_3}}{1-\frac{x_1}{x_5}\frac{x_4}{x_3}}=\frac{x_1x_3+x_4x_5}{x_3x_5-x_1x_4}$$

I think it's easy to do this using coordinates, as there are many right angles.

So, fix triangle ABC in such a way that A = origin, C = (1, 0), B = (0, b) with b > 0. Let D = (0, d) for some 0 <= d <= b.

Then you can calculate the coordinates of points E and F to get that
E = (d/b, 0), F = ( (b^3 + d)/(b(b^2 + 1)), (b - d) / (b^2 + 1) ).

And then calculate all the lengths to get the required result.

It involves a bit of nasty algebra, so perhaps it's not a shortest solution, but at least a straightforward way to do without thinking too much, just performing mechanical calculations.


Find a formula for $g'''(a)$ Suppose $f$ has third derivation on $\mathbb R$ and for a real number $a$, $f(a)=a$ and $f'(a)=-1$.
Suppose that $g(x)=f(f(x)).$
 Find the value of $g'(a)$ and $g''(a)$.
Find a simple formula for $g'''(a)$.

Note :  According to the formula for derivation of combination of two functions, I calculated $g'(a)=1$.

We have, by the chain rule: $g'(x)=f'(f(x))f'(x)$. Hence

$$ g'(a)=f'(f(a))f'(a)=f'(a)^2=1.$$

Now its your turn to compute $g''(a)$ and $g'''(a)$.


Is a Normed Vector Space Necessary to Prove Path Connectedness? Path connectedness seems to be defined in a topological space, but can the existence of a path be proven without using the functions of vector addition, scalar multiplication and norm ?

For example, $R^n$ is a metric space (with the Euclidean metric) and has a corresponding induced topology. But to prove that a ball (open or closed) is path connected I would construct a mapping from $ t \in [0, 1]$ to  $R^n$ defining a path between $a, b \in R^n$  by $y = a + t (b - a)$ and then use the norm to show that this is continuous and within the ball. In this case (and in general) can one prove path connectedness using only metric or topological concepts ?

After comments and a couple of answers, perhaps I understand my question better. It was prompted particularly in consideration of $R^n$. It seems to be confirmed (by comments) that one cannot prove the existence of a continuous path in $R^n$ without defining a function which relies on vector addition, scalar multiplication and norm.

Path connectedness is defined for a topological space and $R^n$ induces a topology and has additional metric properties, but requires yet more properties (normed vector space) in order to prove path connectedness. This strikes me as odd because normally when a concept is defined in a mathematical structure one would expect to be able to prove various results within that structure. In this case I would have expected some, perhaps non-constructive proof, of existence of a path: apparently not.

I imagine what you are asking is this:

Let $V$ be a real vector space with a topology that is compatible with addition and scalar multiplication. This means that addition $a: V \times V \to V, (x,y) \mapsto a(x,y)=x+y$ and scalar multiplication $m: \mathbb R \times V \to V, (\lambda,x) \mapsto m(\lambda,x)=\lambda \cdot x$ are continuous functions. Does this imply that $V$ is path connected?

The answer is yes.

Before the thing itself is done some simple statements:

First, let $f: X_1 \to Y_1$ and $g: X_2 \to Y_2$ be continuos functions. Define $\langle f, g \rangle :X_1 \times X_2 \to Y_1 \times Y_2$, $(a,b) \mapsto (f(a),g(b))$. Then $\langle f ,g \rangle$ is also continuous.

Second, any map from the singleton space to another topological space is continuous.

Thirdly if $\{p\}$ is the singleton space, $X$ a topological space, $X \times \{p\}$ is homeomorphic to $X$.

To see the first statement: if $U$ is open on $Y_1 \times Y_2$ then $U$ is of the form $U=\bigcup_{i \in I}U_{i,1} \times U_{i,2}$ with $U_{i,1}, U_{i,2}$ open on $Y_1, Y_2$ because sets of that form are a basis of the product topology. Note that $$\langle f, g\rangle^{-1}(\bigcup_{i \in I}U_{i,1} \times U_{i,2})=\bigcup_{i \in I}\langle f, g\rangle^{-1}(U_{i,1} \times U_{i,2})=\bigcup_{i \in I}f^{-1}(U_{i,1})\times g^{-1}(U_{i,2})$$

Since $f,g$ are continuous $f^{-1}(U_{i,1})\times g^{-1}(U_{i,2})$ is open in $X_1 \times X_2$ and $\langle f, g\rangle^{-1}(U)$ is then open for any open $U$.

The second and third statements are trivial.

Now define for any $x \in V$ $s_x: \{p\} \to V, p \mapsto x$. Then define $\gamma_x :=m(\langle id, s_x \rangle) : [0,1]\times\{p\} \to V$. Because $[0,1] \times \{p\}$ is the same as $[0,1]$ I will just throw away the dependence of construction on the singleton set from now on. As a composition of various continuous maps $\gamma_x$ is also continuous, the specific form is $\gamma_x(t)= t \cdot x$, so $\gamma_x$ is a continuous path connecting the points $0$ and $x$.

You can either use that two points being connected by paths is an equivalence relation, or define $\gamma_{x,y}:=a(\langle \gamma_x, \gamma_y \circ i \rangle$ with $i:[0,1] \to [0,1], t \mapsto 1-t$ to find $\gamma_{x,y}(t) = t\cdot x + (1-t) \cdot y$ is a continuous path connecting $x$ and $y$.

This is then a general statement that requires only a compatibility of the vector space structure with the topology. If no such compatibility is required, the property does not need to hold, the topology has now no connection with the vector space structure apart the set $V$ being limited in what cardinality it may have (basically no finite sets apart from $\{0\}$ and no countable sets are allowed).

A metric space might not be connected (and therefore path connected). Consider for example the space $(0,1) \cup (2,3)$ equipped with the standard real distance.

One important property of a Normed Vector Space is that it is convex, hence path connected.

I don't know if it precisely answer what you had in mind...


Is it wrong to tell children that $1/0 =$ NaN is incorrect, and should be $∞$? I was on the tube and overheard a dad questioning his kids about maths.  The children were probably about 11 or 12 years old.

After several more mundane questions he asked his daughter what $1/0$ evaluated to. She stated that it had no answer. He asked who told her that and she said her teacher. He then stated that her teacher had "taught it wrong" and it was actually $∞$.

I thought the Dad's statement was a little irresponsible.  Does that seem like reasonable attitude? I suppose this question is partly about morality.

The usual meaning of $a/b=c$ is that $a=b\cdot c$. Since for $b=0$ we have $0\cdot x=0$ for any $x$, there simply isn't any $c$ such that $1=0\cdot c$, unless we throw the properties of arithmetic to the garbage (i.e. adding new elements which do not respect laws like $a(x+y)=ax+ay$).

So "undefined" or "not a number" is the most correct answer possible.

However:

It is sometimes useful to break the laws of arithmetic by adding new elements such as "$\infty$" and even defining $1/0=\infty$. It is very context-dependent and assumes everyone understands what's going on. This is certainly not something to be stated to kids as some general law of Mathematics.

Also:

I believe that the common misconception of "$1/0=\infty$" comes from elementary Calculus, where the following equality holds: $\lim_{x\to 0^+}\frac{1}{x} = \infty$. This cannot be simplified to a statement like $\frac{1}{0}=\infty$ because of two problems:

$\lim_{x\to 0^-}\frac{1}{x} = -\infty$, so the "direction" of the limit matters; moreover, because of this, $\lim_{x\to 0}\frac{1}{x}$ is undefined.
By writing $\lim f(x)=\infty$ we don't really mean that something gets the value "$\infty$" - in Calculus $\infty$ is what we call "potential infinity" - it describes a property of a function (namely, that for every $N>0$ we can find $x_N$ such that $f(x_N)>N$ and $x_N$ is in some specific neighborhood).

As @hardmath indicated there is indeed a floating point representation for infinity.

added:

To address the non-math side of this question. IMHO, it was bad form for the father to contradict authority to his child and sets a bad example without setting a better attitude such as "there may be another answer". I think... it's infinity, what do you think? Here he shows his child he can be vulnerable have a different opinion, even if it technically wrong, and that can be OK if you consider the possibility that someone could be wrong and find a way to verify supposition of facts and confirm or deny them in a rational non-confrontational way. That also demonstrates tolerance and rational objectivity to determine truth by asking if she knows the reason in a non-confrontational way and learn to discuss differences of opinions without fear of criticism as overheard in this example.

" If 'a' is zero and the payload is zero, then it represents infinity."  Is the technical reason why the teacher was correct.

However for teaching purposes such as the concept of the distance of the universe, the Big Bang and the weight of an atom, it may be useful to consider teaching in "metaphors" so that approaching a zero numerator or denominator, which may become a Singularity when you approach infinity with a real number.

It is also useful to define the order of rules when the paradox of;

"anything multiplied by zero is zero" which contradicts...
"anything divided by zero is infinity"

...Actually (2) really bad boo boo ;) (2.) is not true, as it is only a conditionally true if the numerator is also zero, otherwise, it is NaN.

So explain to the child that 0/0 is a special case with religion. ;) or something of

greater than human intelligence.

Fun Fact: Thanks to John Wallis who introduced the infinity symbol to mathematical literature with his works on the collisions of heavenly bodies at university in the women's dorm. Consider how fortunate we are that we dont have to call it a Wallis instead of ∞ infinity. However John eventually got his immortality with an asteroid in 2000 and his name assigned to it.


explain why one can write $\hat{f}(\xi)=\lim_{n\to\infty}\frac{1}{\sqrt{2\pi}}\int_{-n}^{n}e^{-i\xi x}f(x)dx$ when $f\in L^2(\mathbb{R})$ Let $f\in L^1(\mathbb{R})$ where the measure is taken to be the Lebesgue measure. The Fourier transform of $f$ is the function $\hat{f}$ defined as
$$\hat{f}(\xi)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-i\xi x}f(x)dx \qquad ,\xi\in \mathbb{R}$$

(Plancherel Forumla) If $f\in L^1(\mathbb{R})\cap L^2(\mathbb{R})$ then $\hat{f}\in L^2(\mathbb{R})$ and $\|f\|_2=\|\hat{f}\|_2$

(*) Assuming this result, we can extend the Fourier transform to an isometric operator $L^2(\mathbb{R})\to L^2(\mathbb{R})$

Let $f\in L^2(\mathbb{R})$ be arbitrary. Using the idea (*), explain why one can write $$\hat{f}(\xi)=\lim_{n\to\infty}\frac{1}{\sqrt{2\pi}}\int_{-n}^{n}e^{-i\xi x}f(x)dx$$

In this question,

Is the equality $$\lim_{n\to\infty}\frac{1}{\sqrt{2\pi}}\int_{-n}^{n}e^{-i\xi x}f(x)dx=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-i\xi x}f(x)dx$$ correct?
If the above equality is correct, do we need to show that $\int_{-\infty}^{\infty}e^{-i\xi x}f(x)dx<\infty$  i.e.  $\int_{-\infty}^{\infty}|f(x)|dx<\infty$?

How can we solve the question? Thanks!

Update

Let $f_n=\chi_{[-n,n]}f$. Is it enough to show the following steps?

$f_n\in L^1(\mathbb{R})\cap L^2(\mathbb{R})$
$f_n \to f$ in $L^2$ and $L^1$ as $n\to\infty$
$\hat{f_n}=\frac{1}{\sqrt{2\pi}}\int_{-n}^{n} e^{-i\xi x}f(x)dx$

If the answer yes, my attempt is the following:

Note that $|f_n|<|f|$ implies $\int_{\mathbb{R}}|f_n|^2<\int_{\mathbb{R}}|f|^2<\infty$ . So $f_n\in L^2(\mathbb{R})$. On the other hand we have
$$\int_{\mathbb{R}}|f_n|=\int_{[-n,n]}|f_n|\leq \sqrt{\int_{[-n,n]}|f_n|^2}\sqrt{\int_{[-n,n]}\textbf{1}^2}=\sqrt{\int_{[-n,n]}|f_n|^2}.\mathcal{L}([-n,n])<\infty$$ So $f_n\in L^1(\mathbb{R})$ as well.
$|f_n-f|^2= \left\{
  \begin{array}{lr}
    0 & \text{on $[-n,n]$}\\
    |f|^2 &  \text{otherwise}
  \end{array}
\right.
$

So $$\|f_n-f\|_2^2=\int |f_n-f|^2=\int_{[-n,n]}|f_n-f|^2+\int_{\mathbb{R}-[-n,n]}|f_n-f|^2=\int_{\mathbb{R}-[-n,n]}|f|^2$$
So $\|f_n-f\|_2\to 0$ as $n\to \infty$. Similarly $\|f_n-f\|_1 \to 0$ as $n\to\infty$

By definition, it is obvious.

The following is typically how one would go about equating the $L^2(\mathbb{R})$ extension of the Fourier transform to the classical integral transform.

Theorem: Let $f \in L^2(\mathbb{R})$, and suppose that the following exists for almost every $\xi \in \mathbb{R}$:
$$
           g(\xi) = \lim_{n\rightarrow\infty}\frac{1}{\sqrt{2\pi}}\int_{-n}^{n} f(t)e^{-it\xi}dt.
$$
Then $g = \hat{f}$ a.e., where $\hat{f}$ is $L^2(\mathbb{R})$ limit of the integral on the right.

Proof: Suppose $f \in L^2(\mathbb{R})$ and define
$$
             g_n(\xi)=\frac{1}{\sqrt{2\pi}}\int_{-n}^{n}f(t)e^{-i\xi t}dt.
$$
The integral on the right is absolutely convergent for every $n$ because  $\chi_{[-n,n]}f \in L^1$. By Parseval's identity,
$$
          \lim_n \|g_n-\hat{f}\|_{L^2}=0.
$$
By standard results of $L^2$, there is a subsequence $\{ g_{n_k} \}$ that converges pointwise a.e.. to $\hat{f}$. By assumption, $\{ g_n(\xi) \}$ converges pointwise a.e. to $g$. Hence, $g=\hat{f}$ must hold a.e.. $\blacksquare$

$f \in L^2$

$\hat{f}= \lim\limits_{u,v\mapsto \infty}\frac{1}{\sqrt{2\pi}}\int_{-u}^v fe^{-isx}dx$

multiply $f$ by a bump function $g_n$ with compact support in $[-u,v]$

such that $ \lim\limits_{n\mapsto \infty}g_nf=f$ on $[-u,v]$

let $h_n=g_nf$$\quad$ $h_n \in L^1 \bigcap L^2$ and $\hat{h}_n \in L^2 $(Plancherel theorem)

Also $\lim\limits_{n\mapsto \infty}\hat{h}_n=\frac{1}{\sqrt{2\pi}}\int_{-u}^v fe^{-isx}dx$

The good part is that the sequence {$h_n$} is uniformly integrable over $[-u,v]$: this follows from property of bump function.
These imply $\hat{f} \in L^2$

Fourier Inversion theorem:$$
     \lim_{u,v\rightarrow\infty}\left\|\frac{1}{\sqrt{2\pi}}\int_{-u}^{v}\hat{h}_n(s)e^{isx}ds-h_n\right\|_{L^1(\mathbb{R})} = 0. \;\;\;
$$

Plancherel theorem:$$
     \lim_{u,v\rightarrow\infty}\left\|\frac{1}{\sqrt{2\pi}}\int_{-u}^{v}\hat{h}_n(s)e^{isx}ds-h_n\right\|_{L^2(\mathbb{R})} = 0. \;\;\;
$$

for sufficiently large $n,u,v$:$$\int_{-\infty}^{\infty}(\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}\hat{h}_n(s)e^{isx}ds-\int_{-u}^{v}\frac{1}{\sqrt{2\pi}}\hat{f}_n(s)e^{isx}ds)^2dx \le \delta$$

$$\int_{-\infty}^{\infty}({h}_n-f)^2dx=\int_{-\infty}^{\infty}(\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}\hat{h}_n(s)e^{isx}ds-f)^2dx \le \epsilon$$

writing $(a-b)^2=(a-c+c-b)^2=(a-c)^2+(c-b)^2+2(a-c)(c-b)$ and by Cauchy–Schwarz inequality:

$$\int_{-\infty}^{\infty}(f-\int_{-u}^{v}\frac{1}{\sqrt{2\pi}}\hat{f}(s)e^{isx}ds)^2 dx\le \int_{-\infty}^{\infty}((\int_{-u}^{v}\frac{1}{\sqrt{2\pi}}\hat{f}(s)e^{isx}ds-\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}\hat{h}_n(s)e^{isx}ds)^2+(\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}\hat{h}_n(s)e^{isx}ds-f)^2)dx +(2\delta \epsilon)^{1/2}\le\delta+\epsilon+2(\delta \epsilon)^{1/2}$$


If $a$|$c$ and $b$|$c$, then $\frac {ab}{(a,b)}$|$c$ How can  I prove that for $a,b,c \in ℕ^*$, if $a$|$c$ and $b$|$c$, then $\frac {ab}{(a,b)}$|$c$?

This is what I've tried:

$a$|$c$ and  $b$|$c$ implies that  $ba$|$bc$ and  $ab$|$ac$, so  $ab$|$bcx + acy$ and  $ab$|$c(bx+ay)$. We know that for some $x$ and $y$, $bx+ay$$=(a,b)$, so $\frac {ab}{(a,b)}$|$c$. Is this right?

Thanks.

Yes, $\ a,b\mid c\,\Rightarrow\, ab\mid ac,bc\,\Rightarrow\, ab\mid(ac,bc)=(a,b)c\,\Rightarrow\,ab/(a,b)\mid c,\ $ where, for variety, I have replaced your use of the Bezout Identity by the GCD Distributive Law.

Remark $\ $ The reverse implications are also true, which shows $\,{\rm lcm}(a,b) = ab/\gcd(a,b).$

Write a = k(a,b), and b = s(a,b) where (k,s) = 1. So a divides c ==> c = ma =mk(a,b),

and b divides c ==> c = nb = ns(a,b) ==> c/(a,b) = mk = ns. This means k divides c/(a,b), and

s also divides c/(a,b). Since (k,s) = 1, ks divides c/(a,b) . This means:

[a/(a,b)]*[b/(a,b)] divides c/(a,b) , and cancel out (a,b) we get the answer.


What is the difference between $\{1, 2\}\cup\{3\}$ and $\{1, 2, 3\}$? I have a simple question related to set theory.
Is there any difference between the following two sets?

What is the difference between $\{1, 2\}\cup\{3\}$ and $\{1, 2, 3\}$?

There is no difference; they are the same set.

The union of $\{1,2\}$ and $\{3\}$ is equal to $\{1,2,3\}$.  The intersect of $\{1,2\}$ and $\{3\}$ would be equal to $0$.


If $(a,b)=1$ and $n$ is a prime, prove that $(a^n+b^n)/(a+b)$ and $(a+b)$ have no common factor, unless $a+b$ is a multiple of $n$ I have no hints to get started. This is a really tough problem involving prime powers and it's the first time I have seen such question.

I need some good hints to get started. Thanks!

Please do not use congruence, limits or derivatives because they are out of scope of the chapter which has this question.

Assume $a+b$ is not a multiple of $n$.

For $n=2$, a common prime factor of $a+b$ and $\frac{a^2+b^2}{a+b}=a+b-\frac{2ab}{a+b}$ must be $2(=n)$, which is excluded, or must divide one of $a,b$ and hence both.
Thus we may assume that $n$ is odd.

Let $q$ be a prime with $q^k\| a+b$, $k\ge 1$. So $q\nmid n$ and we have $b=-a+rq^k$ with $q\nmid r$.
We need to show that $q^{k+1}\nmid a^n+b^n$:
$$a^n+b^n=a^n+(-a)^n+n(-a)^{n-1}rq^k+{n\choose 2}(-a)^{n-2}r^2q^{2k}+\ldots\equiv na^{n-1}rq^k\pmod{q^{k+1}}$$
So we want to show that $q\nmid na^{n-1}r$. We already know $q\nmid r$ and $q\nmid n$. Remains the possibility that  $q\mid a$, but then also  $q\mid (a+b)-a=b$, contradicting $\gcd(a,b)=1$. Hence indees $q\nmid na^{n-1}r$, $q^{k+1}\nmid a^n+b^n$ and ultimately $q\nmid \frac{a^n+b^n}{a+b}$.

well , the problem may seem confusing but it can be easily approached using fermat theorem
    as, a^n==a mod n .....(1)
 and b^n== b mod n ....(2)
 now, adding both
    a^n+b^n==a+b mod n
 now if (a+b) is a multiple of n,
   a^n+b^n==0 mod n,
  dividing by (a+b) on both the sides we can imply that ,
 a^n+b^n/a+b and a+b have no common factors unless a+b is a multiple of n
 thanks...waiting for response!


Trace of the matrix power Say I have matrix $A = \begin{bmatrix}
a & 0 & -c\\
0 & b & 0\\
-c & 0 & a
\end{bmatrix}$.

What is matrix trace
tr(A^200)

Thanks much!

For this kind of problems, first you have to get eigenvalues and their corresponding eigenvectors. For this problem,
$$ \lambda_1=b,\lambda_2=a-c,\lambda_3=a+c, v_1=(0,1,0)^T,v_2=(\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}})^T,v_3=(-\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}})^T. $$
Let
$$ P=(v_1,v_2,v_3)=\left(\begin{matrix}0&\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}\\
1&0&0\\
-\frac{1}{\sqrt{2}}&0&\frac{1}{\sqrt{2}}
\end{matrix}\right). $$
Then it is easy to check $P^T=P^{-1},P^TAP=\text{diag}(\lambda_1,\lambda_2,\lambda_3)$. Hence
\begin{eqnarray*}
\text{Tr}(A^{200})&=&\text{Tr}(P^TA^{200}P)\\
&=&\text{Tr}((P^TAP)^{200})\\
&=&\lambda_1^{200}+\lambda_2^{200}+\lambda_3^{200}\\
&=&b^{200}+(a-c)^{200}+(a+c)^{200}.
\end{eqnarray*}

You just need to calculate Eigen values bcoz sum of eigen values =$tr(A)$ and if "a" is the eigen value of matrix $A$ then "$a^n$" is the eigen value of $A^n$.
thus,
$$Tr(A^{200})=b^{200}+(a−c)^{200}+(a+c)^{200}$$


linear system straight line solution For a matrix

$
A = \begin{bmatrix}
-1 & 3 \\
0 & -2
\end{bmatrix}
$

I have obtained the Eigenvalues $\lambda = -2$ and $\lambda = -1$.

Now, after trying to find the straight line solution for $\lambda = -1$, I get:

$
\begin{bmatrix}
0 & 3 \\
0 & -1
\end{bmatrix}
\begin{bmatrix}
x \\
y
\end{bmatrix} =
\begin{bmatrix}
0 \\
0
\end{bmatrix}
$

...

Now what does it mean?
$ 3y = 0, -y = 0$, but what is the eigenvector ? ? ?

Can it be a vector (0,1) ?

Thanks.

You showed that $A\begin{pmatrix}x\\ y\end{pmatrix}=-\begin{pmatrix}x\\ y\end{pmatrix}$ if and only if $y=0$.

Thus the set of eigenvectors for the eigenvalue $-1$ is $\{(x,0)\mid x\ne0\}$ and the eigenspace for the eigenvalue $-1$ is $E_{-1}=\{(x,0)\mid x\in\mathbb R\}$. Note that $E_{-1}$ is the subspace generated by the vector $(1,0)$.

$3y=0$ and $-y=0$ implies $y=0$. For a linear transformation $T$ representing the matrix, $\IM T = \{(0,0)\}$ for every $x$.
So for every $x,y$ you get $(0,0)$.
Therefore you get as the $\KER T=\{(1,0)\}$.
$\DeclareMathOperator{\IM}{Im}\DeclareMathOperator{\KER}{Ker}$


Measure theory convention that $\infty \cdot 0 = 0$ In the preface of Terry Tao's notes on measure theory he states that in the extended real number setting we adopt the convention that $\infty \cdot 0 = 0 \cdot \infty = 0.$

He explains that it's a useful convention which makes it natural to define integration from below (e.g. integrals as supremums of nonnegative simple functions). He seems to be saying, "Let's define it this way, because it makes our lives simpler." Is more justification not needed?

This 'fact' helped me during my midterm today in showing that $\mu(E \times \mathbb{R})=0$ for any measure-zero set $E$, but I felt sleazy using it.

Any thoughts (philosophical or otherwise) on why or how we can do this and still have self respect as mathematicians?

You need to distinguish between facts and convention.  A convention is useful if it helps to organize a number of facts.  In the context of measure theory, there are several facts that have the appearance of "$0\times \infty =0$" that are proven from the definitions, and thus the convention allows for an easier way to state facts.

A couple of examples:

If $\mu(X)=0$ and $f(x)=+\infty$ for all $x\in X$, then the definition of $\int_X f\,d\mu$ leads to a value of $0$.  On the other hand, this could be thought of as "$\mu(X)\times \infty=0\times \infty$", integrating a constant function.
If $f(x)=0$ for all $x\in X$ and $\mu(X)=\infty$, then the definition of $\int_X f\,d\mu$ leads to a value of $0$.  On the other hand, this could be thought of as "$\mu(X)\times 0=\infty\times 0$", integrating a constant function.

As a consequence of 1. and 2. above, adopting the convention $0\times \infty=0$ (along with the more intuitive $a\times \infty=\infty$ if $a>0$) allows the following result to be stated:

$$\int_{X} C\,d\mu=\mu(X)\times C,$$

for all constants $C\in[0,\infty]$ regardless of the size of $\mu(X)$.  The convention is used only in stating and using the result in condensed form, not in proving it.

Another example that you highlight: If $E$ has measure zero, so does $E\times \mathbb R$.  This is not proved using the convention.  I would think of it as $$\mu(E\times \mathbb R)=\mu\left(\bigcup_n E\times [-n,n]\right)\leq\sum_n\mu(E\times[-n,n])=\sum_n 0=0,$$ where the equation $\mu(E\times[-n,n])=0$ requires justification, but nothing with infinite measure is involved.

Somehow it feels like the way to be "in the same side" of Continuum Hypotesis: http://en.m.wikipedia.org/wiki/Continuum_hypothesis

The sum of countable many sets is also countable... so $\infty \cdot 0$ should be 0.


In a triangle ABC, (b + c) cos A + (c + a) cos B + (a + b) cos C is equal to In a triangle $ABC$

$$(b + c)\cos A + (c + a)\cos B + (a + b)\cos C=?$$

we know

$$a = b \cos C + c \cos B$$
$$b= a \cos C + c \cos A$$
$$c= b \cos A + a \cos B$$
adding together we get
$$(a+b) \cos C + (b+c) \cos A + (c+a) \cos B =a +b+c $$

If we write:

$b+c=2S-a$
$c+a = 2S-b$
$a+b= 2S -c$

Then by expanding the bracket and taking $2S$ as common factor we can write $2S(\cos a +\cos b +\cos c)-(a\cos a+b\cos b+c\cos c)$
Now since $a+b+c=180 \cos a +\cos b +\cos c = 1+$ 4\sin a/2.sinb/2.sinc/2 and acosa+bcosb +ccosc is $S$.
Therefore solving further we get $2S(1+4 \sin a/2\sin b/2\sin c/2)-S = S + 4 r/R$  where $r$ is inradius and $R$ is circumradius as $4 \sin a/2 \sin b/2\ sin c/2 is r/R$ .

I'm stuck up after here.

Help me if I'm doing something wrong


Induction principle: $n^2-7n+12≥0$ for every $n≥3$ How can I prove that $n^2-7n+12≥0$ for every $n≥3$?

I know that for $n=3$ I have $0≥0$ so the inductive Hypothesis is true.

Now for $n+1$ I have $(n+1)^2-7(n+1)+12=n^2-5n+6$ and now I don't know how to go on...


Help with Rudin rank theorem proof! I am struggling through Rudin's proof of the rank theorem (9.32) in the baby Rudin book. There is a part in the proof where he claims that for a finite-dimensional linear operator A, if the set V is open, then A(V) is an open subset of the range of A. I have seem things about the open mapping theorem involving Banach spaces, but I am not on that level yet and I don't see why the justification of this statement could possibly involve Banach spaces, considering this book does not talk about those. How does Rudin justify this statement, at the level of this book?

Thanks!

Pick any $x_0 \in V$. We will show that $Ax_0 $ is an interior point of $A(V)$.

By translating (i.e. consider $V - x_0$ instead of $V$), we can assume $x_0 = 0$.

Let $y_1, \dots, y_n$ be a basis of $\rm{Range}(A)$ and choose $x_1, \dots, x_n$ with $y_i = Ax_i$ for each $i$.

As $V$ is open with $0 \in V$, there is some $\varepsilon > 0$ such that $\sum_i \alpha_i x_i \in V$ holds for all $\alpha_1, \dots, \alpha_n$ with $|\alpha_i| < \varepsilon$ for all $i$.

This implies that $A(V)$ contains the set

$$
\bigg\{ \sum_i \alpha_i y_i \mid |\alpha_1|, \dots, |\alpha_n| < \varepsilon\bigg\}.
$$

Why does that imply your claim?

This appears to be true only when range of A have dimension larger than 0. Otherwise, $A(V)=0$, which is not open.


Neighbor Interaction in a Random List Assuming a random arbitrarily long list where each element has a $50\%$ chance of being a $0$ or a $1$, such as:

$0001101101$

What is the chance of having a neighbor that isn't the same?  For example, in the above list the first and second item have no different neighbor but the third and fourth do.

I'm working on this for a project dealing with AB testing in a stream of items and I realized there might be interaction between the two sample sets but I'm not sure how much on average (and math class was a very long time ago so I'm rusty).  My instinct says $50\%$ but I don't trust my instincts.

Consider the following scenarios for an interior digit $d$ (i.e. not the first or last digit):

$D_1$: Exactly one neighboring digit is different from $d$
$D_2$: Exactly two neighboring digits are different from $d$
$D_{\ge1}$: At least one neighboring digit is different from $d$ (this is what you're looking for)

We can use some simple counting arguments to find that $P(D_1)=\frac{1}{2}$ and $P(D_2)=\frac{1}{4}$. From here, just note that $P(D_{\ge1})=P(D_1)+P(D_2)$ since $D_1$ or $D_2$ would qualify as a success condition, and we get our answer:

$P(D_{\ge1})=P(D_1)+P(D_2)=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}$.

For exterior digits, $D_1$ is the only possible success condition, so we have $P(D_1)=\frac{1}{2}$.

Edit: Alternatively, for an interior digit $d$, consider the scenario $S_2$: Exactly two neighboring digits are the same as $d$. A simple counting argument shows that $P(S_2)=\frac{1}{4}$. Since $D_{\ge1}=\lnot S_2$, we have

$P(D_{\ge1})=1-P(S_2)=1-\frac{1}{4}=\frac{3}{4}$

To have a neighbor that isn't the same, assuming you pick beyond the first digit and before the last digit, is given by,

(Probability of a neighbor being different)=(probability of two digits not being the same)+(probability of one digit not being the same)=(probability of a digit not being the same)*(probability of a digit not being the same)+(probability of digit not being the same)

In math symbols,

$$P=(1-{1 \over 2})^2+(1-{1 \over 2})={3 \over 4}$$

So there is a ${3 \over 4}$ chance of you getting a neighbor that is different.

If you look at the first or last digit, the probability is indeed ${1 \over 2}$. Can you think of why?


Does chess have more Nash equilibria than you can find through backwards induction? All equilibria found with backwards induction on a tree of a perfect information game are Nash equilibria, but in general the reverse is not true:

        y
(1)---+---(0, 20)             ← Also a Nash equilibrium when (2) announces
      | n         y             he will play y: (0, 20) > (-10, -2)
      +---(2)---+---(-10, -2)
                | n
                +---( +5, -1) ← Solution through backwards induction

(In class, we've called this the "icecream game", (1) is the mom who needs to decide whether to buy his son an ice cream and (2) is the son who needs to decide whether to cry or not about it.)

However, Chess (or Checkers, or Tic Tac Toe) are different from the "icecream game" because the payoffs are either (1, −1) (white win), (−1, 1) (black win) or (0,0) (draw).

Do these games still allow Nash equilibria that can't be found through backward induction?

Chess is a zero-sum game, so all Nash equilibria will lead to the same of three outcomes: White wins. Black wins. Draw. By Zermelo's theorem, in the first case white can force a win. In the second case, black can force a win. In the third case, both can force a draw.

It is not known which case holds, but in the first two cases, there will be many Nash equilibria (a whole continuum). If one player plays a strategy that guarantees a win for her, her strategy combined with any strategy of the other player together will constitute a Nash equilibrium. Even if both players can force a draw, there may be several ways to do so.

So in conclusion, chess has probably more than one Nash equilibrium. If there is only one Nash equilibrium, it will end in a draw.

Remark: Zermelo did not use backward induction and his proof was not based on chess having a stopping rule.

Generally speaking chess theoretically always should end in a draw if the best moves are always played by both sides.  The reason, being that a +4 advantage which is rare to achieve is very hard to win with, ie mating with a bishop is not possible nor is it with a horse alone in a king end game.  Having a pawn is good, but the opponent can generally sacrifice that piece for the pawn to prevent a queen.  Mating with a knight + bishop is a 48 move process, and you need to play it perfect to avoid the 50 move draw rule.  This is why so many games end in draws.  The only trump card is to force a draw, taking risks in other lines, can simply be refuted if the opponent can work it out OTB.


Finding conditional expectation and conditional probability Let $\xi $ be a random variable in $(\Omega, \mathcal{F}, \mathbb{P})$ distributed according to $\mathrm{Unif}\{-2,...,3\}$. Hence,
$$\mathbb{P}(\xi = -2) = \mathbb{P}(\xi = -1) =\mathbb{P}(\xi = 0) =\mathbb{P}(\xi = 1) =\mathbb{P}(\xi = 2) =\mathbb{P}(\xi = 3) =1/6$$
Now let $\mathcal{D}=\{D_1,D_2,D_3\}$ where $D_1=\{\xi<0\}$ $D_2=\{\xi \in [0,2]\}$ $D_3=\{ \xi >2 \}$.

I need to find:

1) $\mathbb{E}(\xi^2|\mathcal{D})$ (The conditional expectation).

2) $\mathbb{P}(\xi<1|\mathcal{D})$.

I am not sure, but I think that for 1) I should use formula $\mathbb{E}(\xi^2 |\mathcal{D})(\omega)= \sum_{i} \frac{\mathbb{E}(\xi^2 \mathbf{1}_{D_i)}}{\mathbb{P}(D_i)} \mathbf{1}_{D_i} (\omega) $. But how should I use it?

I assume that $\mathbb E[\xi^2\mid\mathcal D]$ denotes the same as $\mathbb E[\xi^2\mid\mathcal\sigma(\mathcal D)]$

$\mathbb{E}\left[\xi^{2}\mid\mathcal{D}\right]$ is measurable wrt
$\sigma\left(\mathcal{D}\right)$ implying here the existence of coefficients
$a,b,c$ with $$\mathbb{E}\left[\xi^{2}\mid\mathcal{D}\right]=a\mathbf{1}_{\xi<0}+b\mathbf{1}_{0\leq\xi\leq2}+c\mathbf{1}_{\xi>2}\tag1$$

Secondly we have the equalities: $$\mathbb{E}\left[\xi^{2}\mathbf{1}_{D}\right]=\mathbb{E}\left[\mathbb{E}\left[\xi^{2}\mid\mathcal{D}\right]\mathbf{1}_{D}\right]\text{ for every }D\in\sigma\left(\mathcal{D}\right)\tag2$$

Substituting $\left(1\right)$ in $(2)$ we find for  $D=\{\xi<0\}$, $\{0\leq\xi\leq2\}$ and $\{\xi>2\}$:

$\mathbb{E}\left[\xi^{2}\mathbf{1}_{\xi<0}\right]=aP\left(\xi<0\right)$
$\mathbb{E}\left[\xi^{2}\mathbf{1}_{0\leq\xi\leq2}\right]=bP\left(0\leq\xi\leq2\right)$
$\mathbb{E}\left[\xi^{2}\mathbf{1}_{\xi>2}\right]=cP\left(\xi>2\right)$

Or equivalently

$a=\mathbb{E}\left[\xi^{2}\mid\xi<0\right]=\frac{1}{2}\left(-2\right)^{2}+\frac{1}{2}\left(-1\right)^{2}=\frac{5}{2}$
$b=\mathbb{E}\left[\xi^{2}\mid0\leq\xi\leq2\right]=\frac{1}{3}0^{2}+\frac{1}{3}1^{2}+\frac{1}{3}2^{2}=\frac{5}{3}$
$c=\mathbb{E}\left[\xi^{2}\mid\xi>2\right]=3^{2}=9$.

So we end up with: $$\mathbb{E}\left[\xi^{2}\mid\mathcal{D}\right]=\frac{5}{2}\mathbf{1}_{\xi<0}+\frac{5}{3}\mathbf{1}_{0\leq\xi\leq2}+9\mathbf{1}_{\xi>2}$$

edit:

Observe that $\mathbb P(\xi<1\mid\mathcal D)=\mathbb E(\mathbf1_{\xi<1}\mid\mathcal D)$.

We can find it exactly as above if we replace $\xi^2$ there by $\mathbf1_{\xi<1}$:

$\mathbb{E}\left[\mathbf1_{\xi<1}\mathbf{1}_{\xi<0}\right]=aP\left(\xi<0\right)$
$\mathbb{E}\left[\mathbf1_{\xi<1}\mathbf{1}_{0\leq\xi\leq2}\right]=bP\left(0\leq\xi\leq2\right)$
$\mathbb{E}\left[\mathbf1_{\xi<1}\mathbf{1}_{\xi>2}\right]=cP\left(\xi>2\right)$

Or equivalently:

$a=\mathbb P(\xi<1\mid\xi<0)=1$
$b=\mathbb P(\xi<1\mid0\leq\xi\leq2)=\frac13$
$c=\mathbb P(\xi<1\mid \xi>2)=0$

So we end up with: $$\mathbb{P}\left[\xi<1\mid\mathcal{D}\right]=\mathbf{1}_{\xi<0}+\frac{1}{3}\mathbf{1}_{0\leq\xi\leq2}$$

First, we define the underlying probability space $(\Omega, \mathcal{A}, \mathbb{P}) := (\{-2,...,3\}, 2^{\{-2,...,3\}}, \mathrm{Unif}\{-2,...,3\})$. $\xi$ is a random variable from $\Omega$ into itself, say $\xi \equiv \mathrm{id}_\Omega$. Hence, $\xi(\omega) = \omega$.
Note that $\mathbb{E}[\xi^2|\mathcal{D}]$ is also a random variable, from $\Omega$ into itself. Hence, we need to compute its value (in principal) for every $\omega \in \Omega$. Also note that the outcome really depends on the definition of $\xi$ above. Only the distribution of $\mathbb{E}[\xi^2|\mathcal{D}]$ is unique.
We can now just apply the formula:
\begin{align}
\mathbb{E}[\xi^2|\mathcal{D}](-2) &= \frac{(-2)^2 \cdot 1/6 + (-1)^2 \cdot 1/6}{1/3} = 2.5\\
\mathbb{E}[\xi^2|\mathcal{D}](-1) &= \frac{(-2)^2 \cdot 1/6 + (-1)^2 \cdot 1/6}{1/3} = 2.5\\
\mathbb{E}[\xi^2|\mathcal{D}](0) &= \frac{0^2 \cdot 1/6 + 1^2 \cdot 1/6 + 2^2 \cdot 1/6}{1/2} = 5/3\\
\mathbb{E}[\xi^2|\mathcal{D}](1) &= \frac{0^2 \cdot 1/6 + 1^2 \cdot 1/6 + 2^2 \cdot 1/6}{1/2} = 5/3\\
\mathbb{E}[\xi^2|\mathcal{D}](2) &= \frac{0^2 \cdot 1/6 + 1^2 \cdot 1/6 + 2^2 \cdot 1/6}{1/2} = 5/3\\
\mathbb{E}[\xi^2|\mathcal{D}](3) &= \frac{3^2 \cdot 1/6}{1/6} = 9.
\end{align}
If I didn't do any mistakes, this should be the correct conditional expectation, given the modelling of $\xi$.
The second example you can probably solve yourself, remember that $\mathbb{P}(\xi < 1|\mathcal{D} ) = \mathbb{E}[\mathbf{1}_{\xi < 1}|\mathcal{D}]$.


3 trams are coming every 10, 15 and 15 minutes. On average, how long do I have to wait for any tram to come? 3 trams are coming to the stop every 10, 15 and 15 minutes.
 On average, how long do I have to wait for any tram to come?

It's a practical problem, not some kind of a riddle for which I have a surprising magic trick or an answer. I really don't know. I was waiting for a tram when this question come to my mind. So, if you ask me for example "how the trams are driving?" my answer will be I don't know, I have the same (or lesser) tram knowledge as you. Assume some accurate (probably probabilistic;) model and present the answer, for example "5 minutes" + showing how you obtain this result. Perfect answer will generalize the problem, answering how long do we have to wait when the trams come every $x_1, x_2, x_3...$ minutes. But even the basic problem is not as easy as it is looking, so feel warned.

There are (at least) three reasonable probability models for this problem: (a) the Poisson process model (see heropup's answer), (b) the assumption that the trams arrive on time according to a scheme known to the user (for  examples see PhiNotPi's and user1008646's answers), and (c) the assumption that the trams arrive on time with unknown but equidistributed phases. In the following I shall treat model (c).

Model (c) is equivalent to the following: A random point $P=(X,Y,Z)$ is chosen in the block $$B:=\{(x,y,z)\>|\>0\leq x\leq15,\ 0\leq y\leq 15,\ 0\leq z\leq 10\}\ .$$
The waiting time $T$ is then given by $T=\min\{X,Y,Z\}$. The points $P$ with waiting time between $t$ and $t+dt$ are lying in the union of three rectangular panels of thickness $dt$ and having a distance $t$ from the planes $x=0$, $y=0$, and $z=0$ respectively. The area of the two vertical panels is $(10-t)(15-t)$, and the area of the horizontal one is $(15-t)(15-t)$. It follows that the probability distribution function $f_T$ of the waiting time is given by
$$f_T(t)={1\over2250}(525-80t+3t^2)\qquad(0\leq t\leq10)\ .$$
From this we obtain the expected waiting time as
$$E(T)=\int_0^{10} t\>f_T(t)\ dt={85\over27}\ .$$

Suppose that none of the trains come simultaneously. Wait significant amount of time, say $N >> 1$ hours. Then the interval of $N$ hours will be punctured by $6N$ arrivals of the 10-min train, $4N$ arrivals of the 1st 15-min train, and $4N$ arrivals of the 2nd 15-min train. This breaks the $N$-hour interval into $14N$ waits between arrivals. Therefore the average wait is $60/14$ minutes.

EDIT: as pointed out in the comment, $60/14$ is the average interval length between the trains; however, the average wait should be $1/2$ of the interval length. Therefore the correct answer for the wait is $60/28$.

EDIT 2: as pointed out by the commenter, the answer is, generally speaking, incorrect. On the, eh, 3rd thought I realized the cause of my error: the average wait is not proportional to the average size of the interval; it's proportional to the 2nd moment of the interval.

Let me elaborate on the above a bit: suppose that the interval lengths are $X_i,1\le i
\le n$. Let $\Sigma_{i=1}^n X_i=1$ for simplicity. The probability of landing in the interval $X_i$ is $X_i$. The average wait on the assumption that you landed in $X_i$, is $X_i/2$. Therefore the average wait among all intervals is $\Sigma_{i=1}^n X_i\cdot X_i/2=1/2\cdot M_2$, where $M_2$ is the 2nd moment of the distribution $X_i$. The prior computations applied to the 1st moment $M_1$, also known as average.

So the correct answer should make further assumption on the distribution of the intervals.


Set-Theoretic Omniscience and Set Definition It occurred to me the other day that a set can, in some sense, know more than I do.  This is because a set does not contain duplicate objects, and yet if the same object is defined in two different ways it may be very difficult for me to realize that they are the same (it could even be NP-hard).  That's what I'm calling "Omniscience."

I realize that there is no paradox here, and I'm not trying to create one.  I'm just interested in various philosophies of set theory, in constructivism, alternate axioms, and related issues so I'm wondering if anyone has taken issue with the "omniscience" of sets and tried to go about set theory in a different way because of it.

Another thing about sets is that two sets defined differently may be equal in extensionality, but completely different as far as how I actually go about determining their membership.  In fact there may be theorems from Complexity Theory about how hard it is for me to identify two sets as the same.  Is there any approach to math which takes this into account, perhaps treating sets as processes rather than totalities?

The distinction you're seeing is a very real one, and is treated seriously in both philosophical and mathematical logic; the prevalance of the extensional approach should be taken as a sign of its strength rather than the weakness/uninterestingness of intensional approaches.

You're getting at precisely the extensionality versus intensionality issue. The extensional perspective (as per the axiom) says that a set is determined entirely by its elements; the intensional view, by contrast, says that there is additional relevant data, and usually this data is taken to be some meaning or process assigned to the set.

It's important to note that this issue persists outside set theory. For a natural language example, "morningstar" and "eveningstar" are extensionally equivalent but are a priori different notions; so we can imagine an "intensional" universe of names sitting above an "extensional" universe of referents. This is a huge motivation in philosophical logic, and in particular modal logic; see e.g. here. It's also a natural aspect of intuitionism, or any philosophy of mathematics which views the mathematical universe as a "product of the mind" (or similar).

To be a bit more specific:

Your particular focus on how hard it is to "know" that two things are equal takes us to a very precise notion on logic, namely provable equality (or more generally "provable ---ness"). This is generally a hugely important idea, even when we're only interested in a semantic and extensional picture of things. I think the place this is most obvious is in Henkin's proof of the completeness theorem. Given a consistent theory $\Gamma$, Henkin (after perhaps modifying $\Gamma$ in a technical way) looks at the structure $\mathcal{T}(\Gamma)$ assigned to $\Gamma$ as follows:

Elements of $\mathcal{T}(\Gamma)$ are equivalence classes of terms in the language of $\Gamma$, with the equivalence relation being "$\Gamma$-provable equality."
The language of $\Gamma$ is then interpreted on this set of equivalence classes by saying that relations hold precisely when they provably hold: e.g. if $U$ is a unary relation symbol in the language of $\Gamma$ and $t$ is a term in the language of $\Gamma$, then $U$ holds of the equivalence class of $t$ in $\mathcal{T}(\Gamma)$ iff $\Gamma$ proves $U(t)$.

So the objects of $\mathcal{T}(\Gamma)$ are "really" just names for objects. Note that this is all taking place inside classical logic: the intensional perspective is useful even within extensional mathematics.

Conversely, one could always argue that "intensionality is extensionality in disguise" - whenever we move from an "extensional world of objects" to an "intensional world of objects-via-definitions," what we've really done is move to an "extensional world of definitions" - our new objects are definitions, and they're treated extensionally! This is of similar spirit to how many-valued logic can be situated inside first-order logic, or how classical logic can be embedded into intuitionistic logic, or ... Basically each perspective is so broad that it can simulate the other. This is a good thing for each (in my opinion).

Another example of this phenomenon is the idea of "provable totality" in proof theory. Roughly speaking, a function is provably total relative to a given theory if it has some definition which the theory proves defines a total function: that is, the theory should prove $\forall x\exists !y\varphi(x,y)$. The actual notions of provable totality considered arise by restricting attention to definitions of certain forms, e.g. via Turing machines, and understanding what functions a theory proves are total (in a given sense) gives us lots of information about the theory as a whole. This is closely connected with a similar but more technical notion in proof theory: proof-theoretic ordinals.

It occurred to me the other day that a set can, in some sense, know more than I do. This is because a set does not contain duplicate objects...

Duplicate elements are only an issue in set-builder notation, e.g. $X=\{0, 1, 1\}$

Translating this set-builder notation into the language of standard set theory, we would have:

$\forall a: [a\in X \iff a=0 \lor a=1 \lor a=1]$

Dropping the second disjunction, it would then be trivial to prove that:

$\forall a: [a\in X \iff a=0 \lor a=1]$

Translating this back to set-builder notation, we would have:

$X=\{0, 1\}$


Exact match of assets and liabilities. Liabilities of $1$ each are due at then ends of periods $1$ and $2$. There are three securities available to produce asset income to cover these liabilities, as follows:

(i) A bond due at the end of period 1 with coupon at rate .01 per period, valued at a periodic yield of 14%

(ii) A bond due at the end of period 2 with coupon at rate .02 per period, valued at a periodic yield of 15%

(iii) A bond due at the end of period 2 with coupon at rate .20 per period, valued at a periodic yield of 14.95%

Determine the cost of the portfolio that exactly-matches asset income to liabilities due using

(a) bonds (i) and (ii) only.

(b) bonds (i) and (iii) only

(c) Show that the combination of securities in (b) minimizes the cost of all exact-marching portfolios made up of a combination of the three securities. Note that the minimum cost exact-matching portfolio does not use the highest yielding security in this case.

My attempt was to match the assets with liabilities for t=1 and t=2 here is how I did it:

$L_1$ at $t=1$ is $1(1.14^{-1})=0.877$

$L_2$ at $t=2$ is $1(1.15^{-2})=0.75614$

$X(1.01)(1.14^{-1})+Y(0.02)(1.15^{-1})=0.877$

$Y(1.02)(1.15^{-2})=0.75614$

Solving these leads to $X=0.97$ and $Y=0.98$

$PV=0.97(1.01)(1.14^{-1})+0.98(0.02(1.15^{-1})+1.02(1.15^{-2})=1.633$

Granted that review this afterward I realize I am just turning around and not actually solving the problem. Could anyone help me out and explain what are the proper steps and why?

Myy solution is to solve for bond 2 first, because to reach the exact matching, we need to ensure that A=L at all time, and the 1st bond couldn't cover the liability at the end of period 2. Then substitute the result of bond 2 into the equation of period 1.
Hope this will be helpful.


How the sample mean changes when you add a new observation I don't know how to solve in my statistics homework problem

Sample

In a class of 6 students, their average age is 21. When the teacher joins the class, the avg age becomes 26. What's the age of the teacher ?

Is there a formula to do this or a series of substitutions ?

If you have the sample average based on $n$ samples, $\overline{x}_n$, and add a new observation, $x$, then the new sample average is

$$ \overline{x}_{n+1} = \frac{ n \overline{x}_n + x }{n + 1} $$

using the data you provided, $\overline{x}_n = 21$, $n = 6$ and $\overline{x}_{n+1} =  26$, so you know 3 out of 4 unknowns:

$$ 26 = \frac{ 6 \cdot 21 + x }{6 + 1} $$

and you can solve for the last unknown, $x$, which I'll leave to you. I hope this helps!!

Given: $n_1=6,n_2=7$ $\text{Mean}_1=21\text{ Mean}_2=26$ Age of the Teacher$=(n2\times \text{ Mean}_2)-(n1\times \text{Mean}_1)$
Age of the Teacher$=(7\times 26)-(6\times 21)$
Age of the Teacher$=56$


When to use these fractions properly? I'm self-taught in mathematics, however, I have missed many basic ideas of arithmetic, where I'll usually use symbolab to help me get through some simplification of equations. although, I'm starting to get annoyed by this, as its impeding my learning. I would really like your support in helping me understand the distinction in fractions likeso:
$$[a] \space \cfrac{x}{1-x\over x}$$ when this could be simplified as either:
$$[1] \space\frac{x}{(1-x)x} $$
or $$[2] \space \frac{x^2}{1-x}$$
Though, I've also used symbolab and it's given me fractions likeso:
$$[b] \space \cfrac{x \over1+x}{1-x\over x}$$
Where it will be split like the first fraction into either (1) or (2). Are there rules that help me decide if its either (1) or (2) given equation (a), and when is it acceptable to use these for equation (b).
Or does it not matter for (b), and that the result should ensue likeso:
$$[b] \space \cfrac{x \over1+x}{1-x\over x} = \frac{x}{1+x}\frac{1-x}{x}$$
Thank you for the support! this should help me from relying on symbolab all the time for this.

$$\frac{x}{\frac{y}{z}}=\frac{(x)}{\left(\frac{y}{z}\right)}=x\left/\frac{y}{z}\right.=\frac{xz}{y}$$
$$\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{\left(\frac{a}{b}\right)}{\left(\frac{c}{d}\right)}=\frac{a}{b}\left/\frac{c}{d}\right.=\frac{ad}{bc}$$
In both cases, the middle fraction line is the "longest", so there are implicit brackets around the numerators and the denominators.

2 is right I think. This is just a pure notation question. The situation tells for sure. Use common sense lol.


A challenging question on Riemann integration ( Reference- Real Analysis/A.J.White) If $f:[a,b]→R$ is Riemann integrable over [a,b] and $f≥0$, then given  $ϵ>0$ there exists $δ>0$ such that if $P=(y_1,...,y_m)$ (which is a partition of [a.b]) and max{$y_{s+1}$-$y_{s}$:s=1,...,m-1}<$δ$, then $U(P,f)-\int_{a}^{b}f<ϵ$.
$$$$I would much appreciate if someone could help me out with this question. Thanks

Complete answer: $[L\leqslant I\leqslant U\land U-L\leqslant\varepsilon]\implies U-I\leqslant\varepsilon$.

Here is the answer I figured out and I am putting it here.  Let ε>0 be arbitrary. Then there exists a partition $P_1$={$x_0$,...,$x_p$} in  $[a,b]$ such that $U(P_1,f)<\int_{a}^{b}f+ϵ/2$
 Choose $δ=ϵ/4K(p-1)$ where K is the upper bound of $f(x)$.
Let P be any partition of $[a,b]$ with $||P||<δ$.

Let $P'$ be the union of $P$ and $P_1$ .

$P'$ has at most $(p-1)$ points that $P$. $$U(P,f)-2K(p-1)δ⩽U(P',f)⩽U(P_1,f)<\int_{a}^{b}f+ϵ/2$$ Thus,

$$ U(P,f)<\int_{a}^{b}f+ϵ$$ for any partition with $||P||<δ$.


Proof that $(1+\epsilon)^n = O(1+n\epsilon)$ How to prove that $(1+\epsilon)^n = O(1+n\epsilon)$ ?

So far I proved the following:

By the binomial, $(1+\epsilon)^n > 1+n\epsilon$

Also $\epsilon^n$ = 0 when n-> infinity.

Edit: n constant. $\epsilon$ -> 0.

$(1+\epsilon)^n=e^{n \times log(1+\epsilon)}$

$=1+n \times log(1+\epsilon)+\frac{(n \times log(1+\epsilon))^2}{2!}+...,$ Taylor series

$=1+n\epsilon+...   ,$  Please develop this serie well to see what is happening,  since $\epsilon^n\simeq0$ and $log(1+\epsilon)=\sum_{n=0}^{\infty}(-1)^n\frac{\epsilon^{n+1}}{n+1}$, the result then follow:

$(1+\epsilon)^n=\circ(1+n\epsilon)$


Pigeonhole principle and finding the total ways for this The question is:

Given any $7$ positive integers whose sum is $100$, show there are $3$ of the numbers whose sum is at least $43$.

So I'm having trouble identifying the pigeonholes. However, I think I know the pigeons: the total number of solutions $\{x_1,x_2,\ldots,x_7\}$ of the equation
$$x_1 + x_2 + \ldots + x_7 = 100$$
where $x_k\geq 1$ for all $k=1,2,\ldots,7$.

This looks like a partitioning problem (since it says "any $7$ positive integers", so $1 + 1 + 1 + 1 + 1 + 1 + 94 = 100$ is the same as
$94 + 1 + 1 + 1 + 1 + 1 + 1 = 100$),
so I'm not sure if I interpreted it correctly.

Divide the given values into 3 sets, two of three elements and one with a single element. Note that at least one of the given elements is less than or equal to 14. Suppose the set with only one member contains this value. Thus, the sum of the two sets of 3 members is greater than or equal to 86. In fact, at least one of the two groups had a sum of their members greater or equal (in the case of both have the same sum) than 43.

Reworded to work with pigeons and pigeonholes, this question is asking: "Say you have 100 pigeons and you wished to stuff them in $7$ [pigeon]holes. Show that there must be some set of $3$ holes such that there are at least $43$ pigeons in them."

We can start by attempting to evenly distribute the pigeons into the holes, with $14$ per hole and $2$ left over. Since all possible sets of $3$ pigeonholes have only $42$ pigeons in them, when we add the $2$ leftover pigeons, there must be at least one set of $3$ pigeonholes that have at least $43$ pigeons (at least one pigeonhole will have $15$, plus $2$ others of $14$).


Distinguishing sets according to more fine-grained notions than cardinality. I'm interested in distinguishing sets according to more fine-grained notions than cardinality. Now I don't know a thing about computability theory, but it seems to me that considering sets up to "computable bijection" is an inherently sensible way to achieve this.

To this end, I was thinking:

By an enumerated set, let us mean a pair $(X,x)$ such that $X$ is equipped with a bijection to $x : \mathbb{N} \rightarrow X$.
Study the category whose objects are enumerated sets, such that an arrow $f : (X,x) \rightarrow (Y,y)$ is just a function $U(f) : X \rightarrow Y$ with the property that $y^{-1} \circ U(f) \circ x$ is a computable function. Note that we're not requiring the diagram involving $x,y$ and $f$ to commute.

Are people talking about this sort of thing? If so, where can I learn more?

For subsets of the natural numbers, and for sets naturally connected with such subsets via indexing, the subject has been studied for a very long time. The Wikipedia article on the Arithmetical Hierarchy will give a start. There has been particularly fine-grained analysis of r.e. degrees, huge literature.

You may wish to extend your ordering range to the ordinals, which are in some senses more fine-grained than cardinals.

You may also wish to read about order-embeddings and the constellation of related ideas that are linked and referenced there.


Of all the possible combinations of positive numbers that sum to 10, which has the largest multiplication? Of all the possible combinations of positive numbers that sum to 10, which has the largest multiplication?

I had also got a clue: it's related to e.

Please help! (I need explanation aswell)

Notice: i said positive not natural so you can use fractions.

Suppose we have $a_1$, $a_2$, $\dotsc$, $a_n$ all positive and summing to $10$. Then by the AM–GM inequality their product is maximized when $a_1 = a_2 = \dotsb = a_n$.  Since for $n \geq 10$ you'd have a product of numbers less than or equal to $1$, you only have to compute $(10/n)^n$ for $1 \leq n \leq 9$ and you find that the maximum occurs for $n=4$ with $2.5^4 = 39.0625$.

You were all wrong actually... i checked with my teacher and the asnwer is e 3.67879441 (10/e) times so: e^(10/e) = 39.5986256


How to proceed for this problem on analysis Find all differentiable functions $f\colon [0,\infty)\to [0,\infty)$ for which $f(0)=0$ and $f^{\prime}(x^2)=f(x)$ for any $x\in [0,\infty)$.

I have tried to reduce to the form $f(x)=f'(x^{\frac{1}{2^n}})$. but it is not coming. Is there any other way?

First note that $f$ and $f'$ are both nondecreasing.

By MVT, there is a $0 < \lambda < 1$ such that $f(1)=f'(\lambda)$. But $f'(\lambda)=f(\sqrt{ \lambda })$ and since $f$ is nondecreasing, it should be constant on $[\sqrt{ \lambda }, 1]$. Therefore, if $\sqrt{ \lambda } < x < 1$, we have $f'(x)=0=f(\sqrt{x})$. By continuity, $f(1)=0$.

Take $0 < \delta < 1$ and suppose that $f(n)=0$. Then
$$f(n + \delta) = \int_n^{n + \delta} f'(x)dx $$
$$ \le \delta f'(n + \delta) $$
$$ = \delta f(\sqrt{n + \delta}) $$
$$ \le \delta f(n + \delta) $$

The last inequality follows because for $n \ge 1$, we have $1 < \sqrt{n + \delta} < n + \delta$. So $f(n+\delta) \le 0$ and then $f(n + \delta)=0$. By continuity, $f(n+1)=0$ and the induction is complete.

We prove this in two steps. First, I prove $f$ must be identically $0$ on $[0,1]$, then I prove the result for $[0, \infty)$. We prove the more general result where $f$ is assumed $f:[0, \infty) \to \mathbb{R}$

  $f \equiv 0$ on $[0,1]$.

Proceeding by contradiction, suppose $f$ is not identically $0$ on $[0,1]$. By the extreme value theorem, $|f|$ attains a maximum value on $[0,1]$. Let $c$ be an arbitrary maximizer of $|f|$ on $[0,1]$ and note $|f(c)|>0$. Note that this would also imply that $c^2$ is a maximizer of $|f'|$ on $[0,1]$, since $|f(c)| = |f'(c^2)|$.

We have that $$|f(c)| = \left|\int_{0}^{c} f'(t) \ dt \right|\leq \int_{0}^{c} |f'(t)| dt \leq \int_{0}^{c} |f'(c^2)| dt = c|f'(c^2)| = c|f(c)|$$  In particular, $$|f(c)| \leq c|f(c)|$$

Since $|f(c)|>0$, we can divide through by $|f(c)|$ to simplify the inequality to $$1 \leq c$$

The above argument implies that any maximizer $c$ of $f$ on $[0,1]$ must satisfy $1 \leq c$. Since $c \in [0,1]$, this implies that $c=1$ is the unique maximizer of $f$ on $[0,1]$.

Note by the mean value theorem $\frac{f(1) - f(0)}{1-0} = f(1) = f'(\delta)$ for some $\delta \in (0,1)$. Hence $f(1) = f'(\delta) = f(\sqrt{\delta})$ and $\sqrt{\delta} \neq 1$ since $\delta \neq 1$. But this means that $\sqrt{\delta}$ is a maximizer of $|f|$ on $[0,1]$ not equal to $1$, which contradicts that $c=1$ is the unique maximizer.

  $f \equiv 0$ on $[0, \infty)$.

By contradiction. Suppose $f$ satisfies the given conditions and yet is not identically $0$. Put $n_0 = \text{max}\{n \in \mathbb{N}: f \ \text{is identically zero on} \ [0,n]\}$. We have that $f(x) \neq 0$ for some $ x \in [n_0, n_0+1]$. Note that $n_0 \geq 1$ since we've already done $[0,1]$. Choose $c$ which maximizes $|f|$ on $[n_0, n_0 + 1]$ .  We can assume without loss of generality that $f(c) > 0$, since otherwise define $g=-f$, noting that $g$ satisfies the hypotheses for the problem, and use that to obtain the contradiction instead. Since $f$ is identically $0$ on $[0, n_0]$, it actually follows that $f(c) = \text{max}_{x \in [0, n_0+1]} f(x)$

We have that $f(c)  = f'(\delta)(c-n_0) = f(\sqrt{\delta})(c-n_0)$ where $\delta \in (n_0, c)$ by the mean value theorem. Since $c-n_0 \leq 1$, this implies $f(\sqrt{\delta}) \geq f(c)$. Certainly it's impossible that $f(\sqrt{\delta}) > f(c)$, since $\sqrt{\delta} \in [0, n_0 + 1]$ and $c$ maximizes $f$ on $[0, n_0 + 1]$. Hence we have that $f(\sqrt{\delta}) = f(c)$, so, in fact, $\sqrt{\delta}$ also maximizes $f$ on $[0, n_0 + 1]$.

While it should be clear that $\sqrt{\delta} \in [0, n_0 + 1]$, one may also note more precisely that we must have $\sqrt{\delta} \in (n_0, n_0 + 1]$. This is because $f$ is identically $0$ in $[0, n_0]$ but $f(\sqrt{\delta})>0$. Moreover, note $\sqrt{\delta} < \delta$ since $\delta>n_0 \geq 1$. With this in mind, we have that $n_0 < \sqrt{\delta} < \delta < n_0 + 1$, so in particular $\sqrt{\delta} - n_0 < 1$.

We use the mean value theorem one more time to finish off the proof. Note that $f(\sqrt{\delta}) - f(n_0) = f(\sqrt{\delta}) = f'(\gamma)(\sqrt{\delta} - n_0) = f(\sqrt{\gamma})(\sqrt{\delta} - n_0)$ where $\gamma \in (n_0, \sqrt{\delta})$. Since $\sqrt{\delta} - n_0<1$, we may conclude that $f(\sqrt{\gamma}) > f(\sqrt{\delta})$, which contradicts that $\sqrt{\delta}$ maximizes $f$ on $[0,n_0 + 1]$. QED.


Linear algebra - Find a quadratic function so that it fits these data Fit a quadratic function of the form $f(t)=c_0+c_1t+c_2t^2$ to the data points (0, 0), (1, -9), (2, -2), (3, -19), using least squares.

I'd like to know how to use least squares to do this. Thanks.

Xβ=y
design matrix X=
1 0 0
1 1 1
1 2 4
1 3 9
observation vector y =
0
-9
-2
-19
parametric vector β =
β0
β1
β2
Using least squares:  X^{T}*Xβ=X^{T}y
Solve for β to find the function $f(t)=β0+β1t+β2t^2$

You want to fit the function $f(t)=c_0+c_1t+c_2t^2$ to the data points $(0,0)$, $(1,-9)$, $(2,-2)$, and $(3,19)$, where we with a data point $(x,y)$ mean a combination $(t_0,f(t_0))$ for some real number $t_0$.

Consider the first point $(0,0)$. It says that $f(0)=0$, i.e., that $c_0+0c_1+0^2c_2=0$. This can also be written as $(1,0,0^2)(c_0,c_1,c_2)'=0$ (the apostrophe denotes transpose). Similarly we get that $(1,1,1^2)(c_0,c_1,c_2)'=-9$. Now do the same exercise for the points $(2,-2)$ and $(3,19)$. Summarize the equations in $c=(c_0,c_1,c_2)'$ as $Ac=b$ with design matrix $A$ and column vector $b=(0,-9,-2,19)'$. The least squares method gives the least squares estimator $\hat{c}=(A'A)^{-1}A'b$ of $c$. This now gives you the least squares fit $f(t)=\hat{c}_0+\hat{c}_1t+\hat{c}_2t^2$.


GRE Subject Test - Past Papers, Books, Advice This is not for the Maths part of the General GRE. This is for the GRE Subject Test in Maths. Feel free to add or comment.

How do I know the definition of rings or of anything on the GRE given that definitions can vary?
What does the subject GRE measure?

  I think trying to relearn an entire undergraduate degree's worth of mathematics purely for this test would be an enormous waste of precious time and energy. You can't make this test your life,you have other priorities. I'd begin talking to graduate students who have been through the test-see what they'd recommend. - Mathemagician1234

http://www.mathematicsgre.com/

http://www.physicsgre.com/viewtopic.php?f=13&t=1078

Top web resource:
DM Ashura aka Bill Shillito: https://www.mathsub.com/

Official resources:
(Official) 0568
http://www.math.ucla.edu/~cmarshak/GRE1.pdf
http://web.archive.org/web/20150701191646/http://rambotutoring.com/GR0568-solutions.pdf
http://online.fliphtml5.com/ebvz/nytg/
https://web.archive.org/web/20161023164437/http://online.fliphtml5.com/ebvz/nytg/#p=1
http://www.math.ucla.edu/~iacoley/gre/Practice%202.pdf
http://www.math.ucla.edu/~iacoley/gre/Practice%202%20solutions.pdf

(Official) 1268 aka 1768
http://www.math.ucla.edu/~iacoley/gre/Practice%201.pdf
http://www.math.ucla.edu/~iacoley/gre/Practice%201%20solutions.pdf
http://seansmathematicsblog.blogspot.hk/2016/08/solutions-to-math-subject-gre-sample.html
https://drive.google.com/file/d/0B4qQg_AuKUglUnd6YWIyYWNudWM/view?usp=drive_web
http://webpages.sou.edu/~stonelakb/math/testprep
http://www.rambotutoring.com/GR1268-solutions.pdf
https://www.scribd.com/doc/315056948/GR1268-Solutions
https://www.scribd.com/user/285425100/RG
http://www.mathematicsgre.com/viewtopic.php?t=3494
https://drive.google.com/file/d/0B-uVGGkZosoPcGVRcDNPN0d2ZVU/view

(Official) 9367
http://www.math.ucla.edu/~cmarshak/GRE2.pdf
http://www.math.ucla.edu/~iacoley/gre/Practice%203.pdf
http://www.math.ucla.edu/~iacoley/gre/Practice%203%20solutions.pdf
http://seansmathematicsblog.blogspot.hk/2016/08/solutions-to-mathematics-subject-gre.html
https://drive.google.com/file/d/0B4qQg_AuKUglY0I1bkZ5azl6NDQ/view

(Official) 8767
http://www.math.ucla.edu/~cmarshak/GRE3.pdf
http://www.math.ucla.edu/~iacoley/gre/Practice%204.pdf
http://www.math.ucla.edu/~iacoley/gre/Practice%204%20solutions.pdf
http://seansmathematicsblog.blogspot.hk/2016/08/solutions-to-mathematics-subject-gre_7.html
https://drive.google.com/open?id=0B4qQg_AuKUgld2hoV3h0OFBaZHM

(Official) 9768
http://www.math.ucla.edu/~cmarshak/GRE4.pdf
http://www.math.ucla.edu/~iacoley/gre/Practice%205.pdf
http://www.math.ucla.edu/~iacoley/gre/Practice%205%20solutions.pdf

(Official) 35-item practice test
http://www.math.ucla.edu/~iacoley/gre/othertests/math97-99.pdf
https://www.geneseo.edu/~johannes/GRE%20Practice%20Problems.pdf

(Official) ETS Major Field Test in Mathematics Sample Questions
https://www.ets.org/Media/Tests/MFT/pdf/mft_mathII.pdf
http://www.wou.edu/~beaverc/404/S14/major%20field%20test.pdf

Web resources:
Rambotutoring.com practice test and answers
http://www.rambotutoring.com/GREpractice.pdf
http://www.rambotutoring.com/GREpracticeanswers.pdf

More practice tests (rea i think)
http://www.math.ucla.edu/~iacoley/gre/othertests/math01.pdf
http://www.math.ucla.edu/~iacoley/gre/othertests/math02.pdf
http://www.math.ucla.edu/~iacoley/gre/othertests/math03.pdf
http://www.math.ucla.edu/~iacoley/gre/othertests/math04.pdf
http://www.math.ucla.edu/~iacoley/gre/othertests/math05.pdf
http://www.math.ucla.edu/~iacoley/gre/othertests/math06.pdf
http://www.math.ucla.edu/~iacoley/gre/othertests/math01e.pdf
http://www.math.ucla.edu/~iacoley/gre/othertests/math02e.pdf
http://www.math.ucla.edu/~iacoley/gre/othertests/math03e.pdf
http://www.math.ucla.edu/~iacoley/gre/othertests/math04e.pdf
http://www.math.ucla.edu/~iacoley/gre/othertests/math05e.pdf
http://www.math.ucla.edu/~iacoley/gre/othertests/math06e.pdf

GRE Subject Test Math Practice (App)
https://play.google.com/store/apps/details?id=com.varsitytutors.learningtools.gresubjecttestmath&hl=en

Sheir.Org GRE Mathematics Sample Questions
http://www.sheir.org/gre_mathematics_sample_questions.html
http://www.sheir.org/gre-mathematics-sample-test.html

Sheir.Org GRE Mathematics Practice Test
http://www.sheir.org/gre_math_practice_questions.html
http://www.sheir.org/gre-math-practice-test.html
http://www.sheir.org/gre-math-subject-test-practice.html
http://www.sheir.org/gre-mathematics-subject-test-practice.html

FPSC Mathematics Lecturer Test Sample Questions
http://www.sheir.org/fpsc-math-sample-questions.html

FPSC Mathematics Lecturer Test Past Papers
http://www.sheir.org/fpsc-mathematics-past-papers.html
http://www.sheir.org/fpsc-mathematics-past-test.html
http://www.sheir.org/fpsc-mathematics-past-test-papers.html
http://www.sheir.org/math-lecturer-past-papers.html
http://www.sheir.org/mathematics-lecturer-test-past-papers.html

Sample Questions (1-10) FPSC Mathematics Lecturer test BS-17.pdf
Mathematics Subject test sample questions MCQs (21-30).pdf
Mathematics Subject test sample questions MCQs (31-40).pdf
Mathematics Subject test sample questions MCQs (41-50).pdf

UCLA Christian Parkinson
http://www.math.ucla.edu/~chparkin/gre.html

UCLA Charlie Marshak
http://www.math.ucla.edu/~cmarshak/GREWorkshop.html
https://web.archive.org/web/20160911233702/http://www.math.ucla.edu/~cmarshak/GREworkshopProblems.pdf
https://web.archive.org/web/20160911233717/http://www.math.ucla.edu/~cmarshak/GREworkshopSolutions.pdf
https://www.math.ucla.edu/~cmarshak/GRE/Problems/GREBootcampProblems.pdf
https://www.math.ucla.edu/~cmarshak/GRE/Solutions/GREBootcampSolutions.pdf
https://www.math.ucla.edu/~cmarshak/GRE/GREProb.pdf
https://www.math.ucla.edu/~cmarshak/GRE/GRESol.pdf

UCSB
http://web.math.ucsb.edu/~padraic/ucsb_2014_15/math_gre_w2015/
http://web.math.ucsb.edu/~padraic/ucsb_2014_15/math_gre_w2015/math_gre_w2015.html
http://web.math.ucsb.edu/~padraic/ucsb_2014_15/math_gre_w2015/math_gre_w2015_syllabus.pdf
http://web.math.ucsb.edu/~padraic/ucsb_2014_15/math_gre_w2015/math_gre_w2015_lecture1.pdf
http://web.math.ucsb.edu/~padraic/ucsb_2014_15/math_gre_w2015/math_gre_w2015_lecture2.pdf
http://web.math.ucsb.edu/~padraic/ucsb_2014_15/math_gre_w2015/math_gre_w2015_lecture3.pdf
http://web.math.ucsb.edu/~padraic/ucsb_2014_15/math_gre_w2015/math_gre_w2015_lecture4.pdf
http://web.math.ucsb.edu/~padraic/ucsb_2014_15/math_gre_w2015/math_gre_w2015_lecture5.pdf
http://web.math.ucsb.edu/~padraic/ucsb_2014_15/math_gre_w2015/lecture%206.pdf
http://web.math.ucsb.edu/~padraic/ucsb_2014_15/math_gre_w2015/math_gre_w2015_lecture7.pdf
http://web.math.ucsb.edu/~padraic/ucsb_2014_15/math_gre_w2015/math_gre_w2015_lecture8.pdf
http://web.math.ucsb.edu/~padraic/ucsb_2014_15/math_gre_w2015/math_gre_w2015_lecture10.pdf
http://web.math.ucsb.edu/~padraic/ucsb_2014_15/math_gre_w2015/math_gre_w2015_practice_test.pdf

Clark University Study Group for the GRE Subject Test in Mathematics
http://aleph0.clarku.edu/~djoyce/GREmath/
http://aleph0.clarku.edu/~djoyce/GREmath/linalg.pdf
http://aleph0.clarku.edu/~djoyce/GREmath/linalgDiscuss.pdf
http://aleph0.clarku.edu/~djoyce/GREmath/highcalc.pdf

SubjectMath.Com
http://media.wix.com/ugd/57c670_fb82db77a2134bd6836923eda969cda6.pdf
http://media.wix.com/ugd/57c670_a4c54c95e72c44aa8944ef3172ec6f25.pdf
http://media.wix.com/ugd/57c670_2d7c1c187d394b8aa39d14cc3bf4665e.pdf
http://media.wix.com/ugd/57c670_5deab4f8898941cf8eed1f07bb339fb9.pdf
https://www.youtube.com/watch?v=qXoZ8n72v5M
https://www.youtube.com/watch?v=-LYugb1LO9g
https://www.youtube.com/watch?v=EPPMAVwz59Y
https://www.youtube.com/watch?v=C2xJWoD95_k
https://www.youtube.com/watch?v=84nZMwr7saU

GRE Subject Calculus
https://www.coursehero.com/file/10329376/GRE-Subject-Calculus/
GRE Subject Additional Topics
https://www.coursehero.com/file/10329377/GRE-Subject-Additional-topics

UChicago has a very large list of unique questions that I don't see linked here (if it is, sorry I just missed it!). See: math.uchicago.edu/~min/GRE – Jonathan Rayner

Just found a YouTube channel called ProfOmarMath - has a selection of good videos discussing content and dissecting GRE questions. See youtube.com/playlist?list=PLBiVnG9A5gceFbk7oTBHALETI2SUmplIL - Jordan Mitchell Barrett

Book resources:

Morris Bramson --> I think this is better than REA because Bramson tests more on understanding while REA tests more on knowledge. Idk. Disagreements welcome.

https://www.coursehero.com/file/14751549/Morris-Bramson-Mathematics-Subject-Test-AdvanceBookFi-org/

REA (rated low by DM Ashura aka Bill Shillito)

https://books.google.com/books?id=Njk8Ea2IAU0C
https://www.amazon.com/GRE-Mathematics-Test-Preparation/dp/0738608386
REA_GRE_MATH_SUB.pdf

Princeton

https://www.amazon.com/Cracking-Mathematics-Subject-Test-Edition/dp/0375429727
Quote: This book is for GRE Math Subject Test, NOT for GRE General Test.
Cracking The Gre Math Sunject Test.pdf
Cracking+the+GRE+Mathematics+Subject+Test(2010).pdf
Typos
http://www.mathematicsgre.com/viewtopic.php?t=38
http://www.mathematicsgre.com/viewtopic.php?t=1432
http://www.mathematicsgre.com/viewtopic.php?f=1&t=234
http://www.mathematicsgre.com/viewtopic.php?t=234
-
Calculus Practice
https://www.coursehero.com/file/p2an7lo/P-A-21-B-16-C-15-D-12-E-10-78-%D0%9D%D0%B0%D0%BC-%D0%B2%D0%B0%D0%B6%D0%B5/
-
UCI
http://sites.uci.edu/gremath/files/2013/08/First-Day-Test-.pdf
http://sites.uci.edu/gremath/files/2011/09/multivariable_calculus.pdf
http://sites.uci.edu/gremath/files/2011/09/more_calculus_from_test2.pdf
http://sites.uci.edu/gremath/files/2011/09/calculus_practice_test3.pdf
http://sites.uci.edu/gremath/files/2011/09/ALGEBRA_REVIEW1.pdf
http://sites.uci.edu/gremath/files/2011/09/linear_algebra_from_test2.docx
http://sites.uci.edu/gremath/files/2013/08/chapter-1.pdf
http://sites.uci.edu/gremath/files/2013/08/chapter-2.pdf
http://sites.uci.edu/gremath/files/2013/08/chapter-3.pdf
http://sites.uci.edu/gremath/files/2013/08/chapter-4.pdf
http://sites.uci.edu/gremath/files/2013/08/chapter-5.pdf
http://sites.uci.edu/gremath/files/2013/08/chapter-6.pdf
http://sites.uci.edu/gremath/files/2013/08/Chapter-7.pdf

Web resources (Advice)
GRE advice&strategy - be realistic
Post by j0equ1nn » Sat Apr 04, 2009 4:08 pm
http://www.mathematicsgre.com/viewtopic.php?f=1&t=217
Maths SE questions
recommending books for GRE math subject test
https://math.stackexchange.com/questions/16514/math-gre-subject-exam
Multivariable Calculus for GRE
Books to study for Math GRE, self-study, have some time.
GRE Math Subject Test
Ultimate GRE Prep

Maths SE Tags:
https://math.stackexchange.com/questions/tagged/gre-exam
http://academia.stackexchange.com/questions/tagged/gre
https://math.stackexchange.com/search?q=gre

Resources suggested to me:

http://academicpublishers.in/site/book_details/competitive-mathematics/mathematics-for-competitive-examinations

https://mualphatheta.org/past_tests#tests2018

http://scherk.pbworks.com/w/page/14864231/Quiz%3A%20Linear%20algebra

Schaum's outlines "Advanced Mathematics for Engineers and Scientists" has some excellent ODE, linear algebra, and complex problems that mimic the ones I have seen on practice exams.


Prove log(log(n)) is Big-O (log(n)) I need to show that:

$g(n) = \log(\log(n))= O(\log(n))$

This is what I have so far:

Choose $k = 1$

Suppose $n > 1$ then:

$\log(n)  <  n$

$\log(\log(n)) <  \log(n)$

But I can't figure out what my C should be if this is the correct answer?

If $0<f(n)<g(n)$ then $c=1$ works to show $f(n)=O(g(n))$.


Say $f\ge 0$ and $f \in R(\alpha)$ over a finite interval, is $f^p$ integrable in the same interval? The title pretty much says it, $p$ is some real $>0$, $a$ and $b$ both real, $f$ bounded and $\int d\alpha$ is the Riemann-Stieltjes integral.

It  is trivially true if $f>0$ since then $f^p = e^{p \log f}$ over the whole interval but I'm not sure how to start if we allow points with $f(x)=0$. In this case $f^p$ is not the composition of a continuous function with $f$ so the trivial argument valid before is useless.

My analysis background is Rudin's PoMA ch1-6 so a answer without measure theory would be welcome.

The theorem you want is Theorem 6.11 in Rudin POMA. The function $f^p$ is given by the composition of f with the continuous function $h(t)=|t|^p$. I won't reproduce the proof of the theorem here, but basically, you first use uniform continuity of the function $h$ in the interval $[0,M]$, where $M=\sup f$, to find for any given $\varepsilon>0$ a $\delta>0$ (that you can always take less than $\varepsilon$) such that $|a^p-b^p|\le \varepsilon$ for all $a,b\in [0,M]$ with $|a-b|\le \delta$. Then you apply the definition of Riemann-Stieltjes integral to find a lower sum and an upper sum for $f$ which differ by $\delta^2$. Then you study the sup and inf of $f^p$ in each of the partitions considering the cases in which $\sup_{I_i} f-\inf_{I_i}\le \delta$ or $\ge \delta$.

Consider the function $f(x) =\frac{1}{\sqrt{x}}$ and $f^2(x)$ with $\alpha(x) =x$ on the interval $(0,1) $.


How to prove that $x^{1-x}+(1-x)^{x}\leq x^{1/2}+(1-x)^{1/2}$? Let $x\in [0,1]$,try to prove that:
$$x^{1-x}+(1-x)^{x}\leq x^{1/2}+(1-x)^{1/2}$$
My try:

let $x=\sin ^{2}t$,and it is equal to show that  $$\sin^{2}t^{\cos^2{t}}+\cos^{2}t^{\sin^2{t}}\leq \sin t+\cos t$$

but still nothing.

Thanks :-)

I use https://eudml.org/doc/223938 , pdf-file
page $135$, section $7$ with $\enspace a:=1-x\enspace $ and $\enspace b:=x$ .

Because of the symmetry of the inequality $\enspace x^{1-x}+(1-x)^x\leq \sqrt{x}+\sqrt{1-x}\enspace $ at $\enspace x=0.5$ ,

see $\enspace x\to 1-x$, it’s enough to check the mentioned inequality for $\enspace 0\leq x\leq 0.5$

instead of  $\enspace 0\leq x\leq 1$ .

We have to proof for $\enspace 0\leq x\leq 0.5$ :

$(1.1) \hspace{1cm}\displaystyle x^{1-x}\leq\sqrt{4x^2(1-x)+2x(1-x)(1-2x)\ln(1-x)}$

$(1.2) \hspace{1cm}\displaystyle \sqrt{4x^2(1-x)+2x(1-x)(1-2x)\ln(1-x)}\leq\sqrt{x}(1-(1-x)(1-2x)\ln 2)$

$(1.3) \hspace{1cm}\displaystyle -4+8x^2+(8x-6)(1-x)(1-2x)\ln 2<4x-2+(8x-6)\ln(2(1-x))$

$(1.4) \hspace{1cm}\displaystyle x(3-2x)(3-4x)\ln 2<1$

$(2.1) \hspace{1cm}\displaystyle (1-x)^x\leq\sqrt{1-4x^2(1-x)-2x(1-x)(1-2x)\ln(1-x)}$

$(2.2) \hspace{1cm}\displaystyle \sqrt{1-4x^2(1-x)-2x(1-x)(1-2x)\ln(1-x)}\leq\sqrt{1-x}(1+x(1-2x)\ln 2)$

$(2.3) \hspace{1cm}\displaystyle (1-4x)(\ln 2)^2-\frac{2}{1-x}<-1\leq\frac{3-4x}{(1-x)^2(1-2x)^2}-\frac{4}{1-2x}$

$(3) \hspace{1.3cm}\displaystyle \sqrt{x}(1-(1-x)(1-2x)\ln 2)+\sqrt{1-x}(1+x(1-2x)\ln 2)\leq \sqrt{x}+\sqrt{1-x}$

The proofs are carried out from the bottom upwards.

To $(3)$ :

$0\leq x\leq 0.5\enspace $ => $\enspace 0\leq x\leq 1-x\enspace $ => $\enspace \sqrt{x}\leq\sqrt{1-x}\enspace $ => $\enspace x\sqrt{1-x}\leq (1-x)\sqrt{x}$

=> $\enspace x\sqrt{1-x}(1-2x)\ln 2\leq (1-x)\sqrt{x}(1-2x)\ln 2$

=> $\enspace \sqrt{x}(-(1-x)(1-2x)\ln 2)+\sqrt{1-x}(x(1-2x)\ln 2)\leq 0$

=> $\enspace \sqrt{x}(1-(1-x)(1-2x)\ln 2)+\sqrt{1-x}(1+x(1-2x)\ln 2)\leq \sqrt{x}+\sqrt{1-x}$

To $(2.3)$ :

On the one hand it’s $\enspace (1-x)(1-4x)(\ln 2)^2\leq (\ln 2)^2<1\leq 1+x\enspace $ and therefore

$\displaystyle (1-4x)(\ln 2)^2-
\frac{2}{1-x}<-1\enspace $ and on the other hand because of $\enspace x\leq 0.5\enspace $ we get

$(a)\hspace{1cm}x^2(-1+2x)\leq 0\enspace$ => $\enspace (1-x)^2(3+2x)\leq 3-4x\enspace\enspace$ and

$(b)\hspace{1cm}(1-x)^2(1-2x)(3+2x)\leq (1-x)^2(3+2x)$ .

With $\enspace (a) \enspace $ and $\enspace (b) \enspace $ follows $\enspace (1-x)^2(1-2x)(3+2x)\leq 3-4x$

and therefore $\enspace\displaystyle -1\leq \frac{3-4x}{(1-x)^2(1-2x)^2}-\frac{4}{1-2x}$ .

To $(2.2)$ based on $(2.3)$ :

It’s $\enspace\displaystyle \frac{d}{dx}(-2+x(1-2x)(\ln 2)^2+2\ln(2(1-x)))=(1-4x)(\ln 2)^2-\frac{2}{1-x}\enspace$ and

$\displaystyle \frac{d}{dx}(\frac{1}{(1-x)(1-2x)}-\frac{2}{1-2x})=\frac{3-4x}{(1-x)^2(1-2x)^2}-\frac{4}{1-2x}\enspace$ and it follows

$\displaystyle \frac{d}{dx}(-2+x(1-2x)(\ln 2)^2+2\ln(2(1-x)))<\frac{d}{dx}(\frac{1}{(1-x)(1-2x)}-\frac{2}{1-2x})$ .

Together with the common point

$\displaystyle (-2+x(1-2x)(\ln 2)^2+2\ln(2(1-x)))|_{x=0.5}=-2=(\frac{1}{(1-x)(1-2x)}-\frac{2}{1-2x})|_{x=0.5}\enspace $ does it mean that

$-2+x(1-2x)(\ln 2)^2+2\ln(2(1-x)) \enspace $ and $\enspace\displaystyle \frac{1}{(1-x)(1-2x)}-\frac{2}{1-2x}\enspace $ don’t touch each other for $\enspace 0\leq x<0.5$ .

Taking a value of this value range, e.g. $x=0$, we get

$\displaystyle (\frac{1}{(1-x)(1-2x)}-\frac{2}{1-2x})|_{x=0}=-1<$

$\displaystyle <-2+2\ln 2=(-2+x(1-2x)(\ln 2)^2+2\ln(2(1-x)))|_{x=0}$

and therefore $\enspace\displaystyle \frac{1}{(1-x)(1-2x)}-\frac{2}{1-2x}\leq -2+x(1-2x)(\ln 2)^2+2\ln(2(1-x))$ .

Elementary transformations lead to

$1-4x^2(1-x)-2x(1-x)(1-2x)\ln(1-x)\leq (1-x)(1+x(1-2x)\ln 2)^2$ .

To $(2.1)$ : $\enspace$ That’s formula $(7.2)$, see the link above.

To $(1.4)$ :

With $\enspace\displaystyle \frac{d}{dx}(x(3-2x)(3-4x)\ln 2)=((x-\frac{3}{4})^2-\frac{3}{16})24\ln 2:=0\enspace $ follows

$\displaystyle x_{1,2}=\frac{3\pm\sqrt{3}}{4}\enspace $ and with $\enspace\displaystyle \frac{d^2}{dx^2}(x(3-2x)(3-4x)\ln 2)=(2x-\frac{3}{2})24\ln 2<0$

for $\enspace 0\leq x\leq 0.5\enspace $ follows concavity of $\enspace x(3-2x)(3-4x)\ln 2$ .

Therefore we get

$ x(3-2x)(3-4x)\ln 2\leq\max(x(3-2x)(3-4x)\ln 2)=$

$=(x(3-2x)(3-4x)\ln 2)|_{x=(3-\sqrt{3})/4}=0.75\sqrt{3}\ln 2<1$ .

To $(1.3)$ based on $(1.4)$ :

It’s $\enspace 2x<2x+1-x(3-2x)(3-4x)\ln 2\enspace $ and with

$4x^2\leq 2x\enspace $ we get $\enspace 4x^2<2x+1-x(3-2x)(3-4x)\ln 2$ .

Elementary transformations lead to

$4x^2+(4x-3)(1-x)(1-2x)\ln 2<2x+1+(4x-3)\ln 2$ .

Because of $\enspace 4x-3<0\enspace $ and $\enspace \ln(2(1-x))\leq \ln 2$

we have $(4x-3)\ln 2\leq (4x-3)\ln(2(1-x))\enspace$ and it follows

$4x^2+(4x-3)(1-x)(1-2x)\ln 2<2x+1+(4x-3)\ln 2\leq$

$\leq 2x+1+(4x-3)\ln(2(1-x))$

and with elementary transformations using the left and right side we get

$-4+8x^2+(8x-6)(1-x)(1-2x)\ln 2<4x-2+(8x-6)\ln(2(1-x))$ .

To $(1.2)$ based on $(1.3)$ :

It’s

$\displaystyle \frac{d}{dx}(1-4x(1-x)+((1-x)(1-2x)\ln 2)^2)=-4+8x^2+(8x-6)(1-x)(1-2x)\ln 2$

and

$\displaystyle \frac{d}{dx}(2(1-x)(1-2x)\ln(2(1-x)))=4x-2+(8x-6)\ln(2(1-x))$

so that we get

$\displaystyle \frac{d}{dx}(1-4x(1-x)+((1-x)(1-2x)\ln 2)^2)<\frac{d}{dx}(2(1-x)(1-2x)\ln(2(1-x)))$ .

Together with the common point

$(1-4x(1-x)+((1-x)(1-2x)\ln 2)^2)|_{x=0.5}=0=(2(1-x)(1-2x)\ln(2(1-x)))|_{x=0.5}\enspace $

does it mean that

$1-4x(1-x)+((1-x)(1-2x)\ln 2)^2\enspace $ and $\enspace 2(1-x)(1-2x)\ln(2(1-x))$

don’t touch each other for $\enspace 0\leq x<0.5$ .

Taking a value of this value range, e.g.  $x=0$, we get

$(1-4x(1-x)+((1-x)(1-2x)\ln 2)^2)|_{x=0}=1+(\ln 2)^2>$

$>2\ln 2=(2(1-x)(1-2x)\ln(2(1-x)))|_{x=0}$

and therefore $\enspace 2(1-x)(1-2x)\ln(2(1-x))\leq 1-4x(1-x)+((1-x)(1-2x)\ln 2)^2$.

Elementary transformations lead to

$4x^2(1-x)+2x(1-x)(1-2x)\ln(1-x)\leq x(1-(1-x)(1-2x)\ln 2)^2$ .

To $(1.1)$: $\enspace$ That’s formula $(7.1)$, see the link above.

Now we have

$x^{1-x}+(1-x)^x$

$\leq\sqrt{4x^2(1-x)+2x(1-x)(1-2x)\ln(1-x)}$

$\hspace{0.5cm}+\sqrt{1-4x^2(1-x)-2x(1-x)(1-2x)\ln(1-x)}$

$\leq\sqrt{x}(1-(1-x)(1-2x)\ln 2)+\sqrt{1-x}(1+x(1-2x)\ln 2)$

$\leq\sqrt{x}+\sqrt{1-x}$ .

Hint:as we see $x+(1-x)=1$ take $a=x ,b=1-x \to a+b=1 , 0< a,b <1$
you have to prove $$a^b+b^a \leq \sqrt a+\sqrt b$$


Is the Lie bracket of two vector fields well defined? I want to understand what exactly means to ask the question if the Lie bracket $[X,Y]$ of two vector fields $X,Y\in \mathcal{XM}$, where $\mathcal{M}$ is a differentiable manifold, is well defined.

What does it mean conceptually that the Lie bracket is well defined?
How could I prove that indeed it is well defined?
Can I do it use it a specific basis, say, $e_x = \partial/\partial x$ and $e_y = \partial / \partial y$?

The standard definition of the Lie bracket is: For a differentiable function $f$ on $M$ and two vector fields $X, Y$ on $M$,
$$
[X,Y](f)= XY(f) - YX(f).
$$
You need to check that this defines a derivation on the space of functions on $M$. Doing so is a very nice exercise in the definitions (the more interesting part is verifying the Leibnitz rule).

If $X$ and $Y$ are two vector fields, $[X,Y]=DY(X)-DX(Y)$. Here $DX$ is the differential of $X$.

You can see this https://en.wikipedia.org/wiki/Lie_bracket_of_vector_fields


Power sequence in matrices I read the following result:
If $A$ is $n\times n$ matrix and $\rho(A)$ denotes the spectral radius of $A$. Then if $\rho(A)<1$, then $\lim\limits_{m\rightarrow \infty} A^m\rightarrow \boldsymbol{0}$, where $\boldsymbol{0}$ is zero matrix.
Now result is easily proved to be true for diagonalizable matrices because if $PAP^{-1}$ is diagonal matrix $D$, then $D=diag(d_1, d_2, \dots, d_n)$ where $|d_i|<1$. Then $PA^mP^{-1}=diag(d_1^m, d_2^m, \dots, d_n^m)\rightarrow \boldsymbol{0}$. So $\lim\limits_{m\rightarrow \infty} A^m\rightarrow \boldsymbol{0}$.
Now we know that $\lim\limits_{m\rightarrow \infty} A^m\rightarrow \boldsymbol{0}$ is same as $\lim\limits_{m\rightarrow \infty} \|A^m\|\rightarrow \boldsymbol{0}$ for any matrix norm $\|\cdot\|$ on $\mathbb M_n(\mathbb C)$.
And in any matrix norm, diagonalizable matrices are dense in $\mathbb M_n(\mathbb C)$. Can we use some continuity argument to prove the result for any matrix? (we also know eigenvalues are continuous function on $\mathbb M_n(\mathbb C)$.

Can we use some continuity argument to prove the result for any matrix?

Yes, I have such an argument, albeit a roundabout one.
Note that Gelfand's formula $\rho(A)=\lim_{m\to\infty}\|A^m\|^{1/m}$ can be proved without using the statement that $\rho(A)<1$ iff $\lim_{m\to\infty}A^m=0$. Suppose $\rho(A)<1$. Pick any submultiplicative matrix norm. By Gelfand's formula, $\lim_{m\to\infty}\|A^m\|^{1/m}<1$. It follows that $\|A^M\|<1$ for some positive integer $M$. Now, for all $m\ge 0$, write $m=qM+r$ by Euclidean division. Then $\|A^m\|\le\|A^M\|^q\max_{0\le r<M}\|A^r\|$. When $m\to\infty$, we have $q\to\infty$ and hence $\|A^m\|\to0$. Since all norms are equivalent on a finite-dimensional vector space, we conclude that $\lim_{m\to\infty}A^m=0$ with respect to every (submultiplicative or non-submultiplicative) matrix norm.
But where is continuity used? It's in the proof of Gelfand's formula.


Find $\lim_{n \to \infty} n(\frac{1+i}{2})^n$ Find $\lim_{n \to \infty} n(\frac{1+i}{2})^n$.

I don't know how to solve this limit. Should I use the fact that $\lim_{n \to \infty} n(\sqrt{2}/2)^n\cos(n \pi / 4)$ and $\lim_{n \to \infty} n(\sqrt{2}/2)^n\sin(n \pi / 4)$ for the real et imaginary part of $n(\frac{1+i}{2})^n$.

Can anyone give me a hint to solve the problem?

One may use the fact that, for any $|q|<1$, we have
$$
\lim_{n \to \infty}n|q|^n=0.
$$ Apply it here, observing that
$$
\left|n\left(\frac{1+i}{2}\right)^n\right|\leq n\left|\frac1{\sqrt{2}}\right|^n.
$$

Consider that :

$$(1+i)^4 = ((1+i)^2)^2=(1+2i-1)^2=4i^2=-4$$

So for any $n=4(n\,div\,4)+(n\,mod\,4)$ we have :

$$\left|n\left(\frac{1+i}{2}\right)^n\right|=n2^{-n+2(n\,div\,4)+(n\,mod\,4)}|(1+i)^{n\,mod\,4}|$$

Now the magnitude of this cannot ever be $>1$ so this converges to zero ( converging along distinct sets of values, but converging all the same to zero ).