Irrationality of $\sqrt{2\sqrt{3\sqrt{4\cdots}}}$ In this question it is stated that Somos' quadratic recurrence constant
$$\alpha=\sqrt{2\sqrt{3\sqrt{4\sqrt{\cdots}}}}$$
is an irrational number. [update: the author of that question is no longer claiming to have a proof of this]

This fact seems by no means trivial to me. The algebraic numbers $\sqrt{2}$, $\sqrt{2\sqrt{3}}$, $\sqrt{2\sqrt{3\sqrt{4}}}$, $\dots$ do not converge quickly enough to $\alpha$, so one cannot reuse the proof of Liouville's theorem in this case.

Approximation arguments do not seem a good way, since
$$ \sqrt{2\sqrt{2\sqrt{2\sqrt{\cdots}}}}=2$$
is rational instead!

What am I missing?

Your original post implies you have thought of, or know of, a good way of using Liouville's theorem to solve the problem. Here's the start of some ideas on the issue that perhaps you will be able to finish off.

We can write $\alpha$ in the following way:
$$\alpha=\prod_{n=1}^\infty\left(1+\frac{1}{n}\right)^{2^{-(n-1)}}.$$

We will also define the partial product

$$\alpha_N=\prod_{n=1}^N\left(1+\frac{1}{n}\right)^{2^{-(n-1)}}.$$

Then, for some integers $p$, $q$:

$$\left|\alpha-\frac{p}{q}\right|\leq\left|\alpha-\alpha_N\right|+\left|\alpha_N-\frac{p}{q}\right|$$.

Now, the first of these values can be quickly bounded:

\begin{eqnarray}\alpha-\alpha_N&=&\alpha_N\left(\prod_{n=N+1}^\infty\left(1+\frac{1}{n}\right)^{2^{-(n-1)}}-1\right)\\
&\leq& \alpha_N\left(\prod_{n=N+1}^\infty\left(2\right)^{2^{-(n-1)}}-1\right)\\
&\leq& \alpha_N\left(2^{2^{-(N-1)}}-1\right) \\
&\leq& 4\alpha_N\cdot 2^{-N} \\
&\leq& 4\alpha\cdot 2^{-N}
\end{eqnarray}

However, I can't see how to bound the other expression, $\left|\alpha_N-\frac{p}{q}\right|$. Because the $\alpha_N$ are algebraic, we can't instantly point to the existence of some rationals that approximate them ``well enough'' for our purposes here, as far as I can see.

I may be completely out of line here and my thinking doesn't seem to be at the level of most of the other proofs presented, but I wonder if the following could lead to a proof. Please do not consider this a rigorous proof by any stretch.

First, use the distributive property of roots to separate $\sqrt{2}$ from $\sqrt{\sqrt{3\sqrt{4\sqrt{...}}}}$ which gives us $\sqrt{2}*\sqrt{\sqrt{3\sqrt{4\sqrt{...}}}}$ Now, most have seen the proof from contradiction that $\sqrt{2}$ is irrational. Would it not then just be an issue of showing that no matter the remaining sequence, when multiplied by $\sqrt{2}$, it can never produce a rational number? Essentially, an irrational number cannot be multiplied by any such number as to produce a rational number $\frac{p}{q}$. Am I completely off base here?


Field extension of odd degree whose Galois group is trivial Let $n\geq 1$ be an odd integer. Show there exists a field extension $L$ over $\Bbb{Q}$ of degree $n$ such that $|\mathrm{Gal}(L/\Bbb{Q})|= 1$.

Recall the following theorem:

  Let $F/K$ be a finite separable extension of degree $n$. Then $L/K$ is normal if and only if the group of $K$-automorphisms of $L$ has $n$ elements.

In particular, this means that your extension cannot be normal (except for the trivial case $n =1$). Hence, I will use the notation $\mathrm{Gal}(L/K)$ for non-normal extensions, denoting the group of $K$-automorphisms of $L$.

For all odd $n \ge 3$, the extension $L= \Bbb{Q}(\sqrt[n]{2})$ does the job. This is because any $\Bbb{Q}$-automorphism of $L$ must fix $\Bbb{Q}$ (by definition), and must fix $\sqrt[n]{2}$ (it's the unique root of $X^n-2$ belonging to $L$); in particular, identity is the unique $\Bbb{Q}$-automorphism of $L$.

For $n=2$ this is impossible, since any quadratic extension of $\Bbb{Q}$ is normal, and its Galois group has two elements.

The answer of Crostul is right only for odd $n$. For an even $n$ it doesn't work, because all the roots are $ \sqrt[n]{2}e^{2 \pi i k/n}$ for $k$ from $0$ to $ n-1.$
So if n is even, in the exponent we get $\frac{ 2 \pi i K}{n}= \frac{ \pi i k}{n/2}$ for $K=0,...,n-1$.
At this moment  for $K_1 = 0,... n/2-1$ and $K_2= n/2 ,...,n$  we realize that $e^{\frac{\pi i k_1}{n/2}}=-e^{\frac{\pi i k_2}{n/2}}$ due to the fact that $e^{\frac{n/2 \pi i}{n/2}}=e^{\pi i}=-1$

Hence $ \sqrt[n]{2}, -\sqrt[n]{2} \in \mathbb{Q}(\sqrt[n]{2})$
and there is no longer a unique root in $\mathbb{Q}(\sqrt[n]{2})$


Proving that a group is closed under multiplication by scalars I need to prove that the group $U = \{A \in M_n(\mathbb{R})\mid AB=BA\}$ is closed under multiplication by scalars.

So I let $\alpha \in \mathbb{R}$ and let $A \in U$, which means I need to prove that $\alpha A \in U$.

$(\alpha A)B = \alpha(AB) = \alpha(BA)$

But that does not prove what I need, I need to prove that $B(\alpha A) \in U$.

Where do I go from here?

It seems that the $U$ is a subspace of $V=M_{n\times n}(\mathbb R)$. So you are on a right path. In fact, if $A,C\in U$ then $$(aA+bC)=(aA)B+(bC)B=a(AB)+b(CB)=...=B(aA)+B(bC)=B(aA+bC)$$. Note that I amasuuming $B$ is a certain fixed matrix in vector space $V$

No, you are not asked to prove that $B(\alpha A)\in U$. What you should prove is the following:

  if $A\in U$, then $\alpha A\in U$ for any $\alpha\in\mathbb{R}$,

or equivalently, if $AB=BA$, then $(\alpha A)B=B(\alpha A)$ for any scalar $\alpha$. This should be trivial because both sides of the last equality are equal to $\alpha(AB)$.

Alternatively, if we define $f:A\mapsto AB-BA$, then $f$ is a linear map defined on $M_n(\mathbb{R})$ and $U$ is its nullspace. Since every vector subspace is closed under scalar multiplication, the assertion follows.


Does the homoemorphism preserve the following: Let $F$ be a homomorphism between $U$ and $V$ where $U \subset X$ and  $V$ is open in the space $Y$.

Then is $U$ open in the space $X$?

If so how can we show this?

thanks

Let $Y=\{a,b\}$ with the discrete topology and let $X=\mathbb{R}$. Letting $U=\{0\}$ and $V=\{a\}$, it is clear that $U$ is not open in $X$ even though $F:U\to V$ defined by $F(0)=a$ is a homeomorphism.

Take $X=\Bbb R, Y=[0,1),V=[0,\frac{1}{2})$.

Consider $F(x)=x$ then $F:U\to V$ is a homeomorphism but $U=[0,\frac{1}{2})$ is not open in $\Bbb R$


Solve $\cos\theta-3\cos2\theta+\cos3\theta=\sin\theta-3\sin2\theta+\sin3\theta$ My attempt:
\begin{align*}
\cos\theta-3\cos2\theta+\cos3\theta&=\sin\theta-3\sin2\theta+\sin3\theta\\
\cos\theta-3\cos2\theta+4\cos^3\theta-3\cos\theta&=\sin\theta-3\sin2\theta+3\sin\theta-4\sin^3\theta\\
-2\cos\theta-3\cos2\theta+4\cos^3\theta&=4\sin\theta-3\sin2\theta-4\sin^3\theta
\end{align*}
I have faced several symmetric trigonometry problems most of them need to use product to sum identities, but this one I can't continue.

It's $$2\cos2\theta\cos\theta-3\cos2\theta=2\sin2\theta\cos\theta-3\sin2\theta$$ or
$$(2\cos\theta-3)(\cos2\theta-\sin2\theta)=0$$ or
$$\tan2\theta=1.$$
Can you end it now?

It is a false identity. Click here to see the graphs.


How does this limit with a square root behave as it approaches infinity? $$
\lim_{x\to \infty}\frac{2x}{\sqrt{4x^2+1}}
$$

Can I substitute $\sqrt{4x^2}$ for $2x$ in the denominator to show that this limit approaches $1$?

Yes, dividing both the numerator and the denominator by $2x$ is fine, and converting $2x$ into $\sqrt{4x^2}$ is fine. The reason that this is fine is that as $x\to \infty$, we know that (eventually) $x$ and therefore $2x$ is strictly positive, and you can always do those things with positive numbers.

So it is not the case that $\frac{2x}{\sqrt{4x^2 + 1}} = \frac{1}{\sqrt{1+\frac{1}{4x^2}}}$. It is, however, the case that they are equal once $x$ is sufficiently close to $\infty$, and for our limit that's all we care about.

Yes it is completely correct in this case,since for $x>0$ $\sqrt{4x^2}=2x$ , and equivalent to get out $2x$ from the denominator.


Information-theoretic proof of Gödel's Theorem? I'm looking for an information-theoretic proof of Gödel's Theorem that goes something like this, without any reference to diagonalization:

Every axiom system in the scope of Gödel's Theorem has a finite number of bits.
It requires an infinite number of bits to specify the all the truths of number theory.
By the Soundness theorem, no new bits can be introduced by deduction.
So no such axiom system as specified in part 1 above can fully axiomatize number theory.

Does such a proof exist? Is it even feasible? Please include references with your answer. Thanks

I think that parts (1) and (2) would already prove the theorem. The heart is part (2), which phrased another way says "no finite number of bits can encode all truths of number theory".   The difficult thing with the proof would be making formal sense of "finite number of bits".

This makes no sense to me.  Yes, some axiom systems can be written down using a finite number of bits.  From those axioms infinitely many theorems may be deduced.  Does it not require infinitely many bits to specify them?  How does your "proof" distinguish between systems that are complete, such as Presburger arithmetic, and systems that are not?


Efficiently find the y on a line at x when only having the start and end point Hi Mathematics StackExchange,

Click this image for a graph(I'm not allowed to post images yet on the Math StackExchange)

Please bare with me here :). So, I have a line. By a line, I mean I have a start point of the line, and the end point. (Also I have the degrees of the angle if that matters).

Now let's say I want to get the the point on the line where it intersects x=10, as you can see in the image.

What would be the most efficient way to do this, with the least calculations? The only way I could think of is making a formula from the two points, and plugging in 10 as x. The reason I'm asking this is because I'm programming something where this calculation has to be done many times per second.

Thanks, I appreciate all responses.

The most efficient way to solve this problem would be to consider where you expect this intersection to happen, then have your code search in a neighbourhood around that point, instead of traverse the whole space looking for an intersection. i.e. if your slope of the line is always near 1, you want to look for intersections in a neighbourhood of $x=10$.

Your question is vague, though. I'm not sure this is something that can be optimized with such sparse detail.


What is the integral $\int\limits_1^3\frac{e^{\frac{1}{x}}}{x^{2}}dx$? What is

$$\int_1^3 \frac{e^{\frac{1}{x}}}{x^{2}} dx ?$$

How do you approach solving this?

Let $u = \frac{1}{x}$. Then $du = -\frac{1}{x^2} \ dx$. Hence we now have

$$- \int_{x=1}^{x=3} e^u \ du = -e^u = -e^\frac{1}{x}$$

Plug in the bounds now to get your answer of $e-e^\frac{1}{3}$.

The following integral can be computed by substitution, $t=\frac{1}{x}$
$$\int_1^3 \frac{e^{\frac{1}{x}}}{x^2}dx=-\int_1^\frac{1}{3} e^tdt=e-e^\frac{1}{3}$$


How do I evaluate $\sum_{n=1}^{\infty}\frac{n}{2n-1} - \frac{n+2}{2n+3}$? $$\sum_{n=1}^{\infty}\frac{n}{2n-1} - \frac{n+2}{2n+3}$$
I've tried combining the sum, telescoping series and even trying to make an Nth partial sum but nothing seems to budge. I'm not sure where to go...

It becomes easier if you generalise the problem: You are calculating the sum of f(n) - f(n+2), where f(n) = n / (2n-1).
The partial sums for the sum up to k are f(1)-f(3) when k=1, f(1)+f(2)-f(3)-f(4) = f(1)+f(2)-f(k+1)-f(k+2) when k=2, and from then on the partial sum is f(1)+f(2)-f(k+1)-f(k+2) for all k.
In this case, since f(k+1) and f(k+2) both converge towards 1/2, the limit is f(1)+f(2)-1 = 1+2/3-1 = 2/3. The same method will always work if f(k+1)+f(k+2) has a limit, and otherwise the sum doesn’t converge.


PDE : Mixture of Wave and Heat equations Today I was given the following equation : $$\frac{1}{c^2}u_{tt} + \frac{1}{D}u_t = u_{xx}$$ with initial conditions : $u(x,0) = 1$ if $|x|<L$ and $0$ otherwise, $u_t(x,0) = 0$. So fairly simple initial conditions.

I can see that there is a bit of wave and heat equation so I first solved each case but I couldn't "glue" the answers together. If $c$ gets large, then the equation will behave like a heat equation and similarly, if $D$ is large then it will behave like a wave equation.

Using dimensional analysis I deduced that if $\frac{c^2}{LD}$ is the criterion to say if $c$ and $D$ are "large".

I know that I can solve the equation using separation of variable but what would be a way to be able see how the solution behaves without solving it? Like being able to sketch a solution for varying $t$ would be really nice.

Cheers

A substitution of the form $u=e^{-at}v$ with $a=c^2/(2D)$ transforms the equation into the telegraph equation $$\frac1{c^2}v_{tt}-v_{xx}=bv$$ with $b=a/(2D)$. The telegraph equation is a much studied equation.


Let $f : X \rightarrow f(X)$ be continuous and X connected. Then $f(X)$ is connected. Let $f : X \rightarrow f(X)$ be continuous and X connected. Then $f(X)$ is connected.

I'm coming across a major problem with the following proof:

Suppose $f(X)$ is not connected. Then $\exists$ non-empty sets $U,V$ open in $f(X)$ such that $f(X) = U \cup V$, and $U,V$ are disjoint.
Since $f$ is continuous, $f^{-1}(U)$ and $f^{-1}(V)$ are open in $X$ (where $f^{-1}$ is the pre-image). We want to show $f^{-1}(U)$ and $f^{-1}(V)$ disconnect X for a contradiction.

It is now easy to show that X is the union of $f^{-1}(U)$ and $f^{-1}(V)$ and that $f^{-1}(U)$ and $f^{-1}(V)$ are non-empty. However, it seems to me that $f^{-1}(U)$ and $f^{-1}(V)$ are disjoint if and only if $f$ is 1-1. Am I mistaken?

$f^{-1}(U)\cap f^{-1}(V)=f^{-1}(U\cap V)=$
$f^{-1}(\emptyset)=\emptyset$
because it is always true that the intersection of the inverse image with respect to a function is the inverse image of the intersection

In general for connected proofs, it is much easier to argue by contradiction. So we shall assume for the sake of contradiction that $f(X)$ is not connected. Hence, we see there exists $B_1, B_2$ where both are non-empty such that $\overline{B_1} \cap B_2 = \emptyset$ and $\overline{B_2} \cap B_1 = \emptyset$ such that $f(X) =B_{1} \cup B_{2}$.

Now we want this to show that $X$ is not connected. So how do we go from our function back to $X$? Well we shall consider the pre-image.

Let $X_1$ = $f^{-1}(B_1) \cap X$, and let $X_2$ = $f^{-1}(B_2) \cap X$. It should be intuitively that these sets make $X$ not connected.

Now we claim that $X = X_1 \cup X_2$. Indeed, $X_1 \cup X_2 = (f^{-1}(B_1) \cap X) \cup (f^{-1}(B_2) \cap X) = f^{-1}(B_1 \cup B_2) \cap X = f^{-1}(f(X)) \cap X = X \cap X = X. $

Now we consider $\overline{X_1} \cap X_2 = \overline{(f^{-1}(B_1) \cap A)} \cap (f^{-1}(B_2) \cap A) \subset \overline{f^{-1}(B_1)} \cap f^{-1}(B_2)$

Now here's the crucial need of continuity: $f: X \rightarrow Y$ is continuous if and only if for every $B \subset Y$ $\overline{f^{-1}(B)} \subset f^{-1}(\overline{B})$ (for the proof of this, it should be in the beginning of Rudin's chapter on continuity). Hence, we see

$\overline{f^{-1}(B_1)} \cap f^{-1}(B_2) \subset f^{-1}(\overline{B_1}) \cap f^{-1}(B_2)$ =$ f^{-1}(\overline{B_1} \cap B_2)$  = $f^{-1}(\emptyset)$ = $\emptyset$, so we have $\overline{B_1} \cap B_2 \subset \emptyset$. S we see $\overline{B_1} \cap B_2 = \emptyset$.

Now repeat the above proof for $\overline{B_2} \cap B_1$. Now we have shown $X = B_1 \cap B_2$ where $\overline{B_2} \cap B_1 = \overline{B_1} \cap B_2  = \emptyset$, so $X$ is disconnected, but this is our desired contradiction.


Which is greater as $n$ gets larger, $f(n)=2^{2^{2^n}}$ or $g(n)=100^{100^n}$? It is the first time I met such a question:

  Which is greater as $n$ gets larger, $f(n)=2^{2^{2^n}}$ or $g(n)=100^{100^n}$?

Intuitively I think $f(n)$ would gradually become larger as $n$ gets larger, but I find it hard to produce an argument.
Is there any trick to use for this type of question?

Take logs

$$\log\log\log(f)=\log\log\log(2^{2^{2^n}})=n\log(2\log(2\log(2)))$$
$$\log\log(g)=\log\log(100^{100^n})=n\log (100(\log (100)))$$

Which are both linear and grow at the same rate, so to say. Now exponentiate to retrieve $f,g$.

Try taking the logs. Log is a monotonic transformation.


How Can The Following Language Possibly Be Regular? $L =$ {$(01)^a$$x(10)^b$ | $a=b, b > 0, x∈${$0,1$}*}

This is a question where according to the key, yes, the language is regular, but no explanation is given. However, if $a=b$, then this can be rewritten as:

$L =$ {$(01)^n$$x(10)^n$ | $n > 0, x∈${$0,1$}*}

And $L =$ {$(01)^n$$(10)^n$ | $n > 0$} is a famous example of an expression which is NOT regular, so how on earth could inserting the $x$ in between the first two exponents allow it to be regular? I'm just not seeing any regular expression that could describe this, which is necessary for a language to be regular. What am I missing?

Since $x$ can be anything, it can swallow most of the 01s and 10s.
So it's the same as $\{01x10:x\in\{0,1\}^*\}$.

Hint: The intermediate symbol $x$ has no impact on the non-regularity of the language. The general method to show that a language is non-regular is the pumping lemma and this language is similar to $\{a^nb^n\mid n\geq 1\}$ or better $\{a^nxb^n\mid n\geq 1\}$.


Inverse of ratio function $$ f(x) =\frac{4x^3}{x^2+1}$$

Question is
$$ \frac d{dx} (f^{-1}(2))=? $$

Now i know how to invert exponential function, rational function and etc but i don't understand how can i invert this function

Try to exchange x and y

So
$$ x=\frac{4y^3}{y^2+1} $$

And then couldn't find out how to do

In General, because an inverse is the reflection of the function in the line $y=x$, the slope of the tangent to the inverse at the point $(x,f^{-1}(x))$ must be the reciprocal of the slope of the tangent to the function at $(f^{-1}(x) ,x)$
$$
\frac d{dx}f^{-1}(x)=\frac 1 {f'( f^{-1}(x))}
$$

In your case $f(1)=2$ so $f^{-1}(2)=1$

So
$$
\frac d{dx}f^{-1}(2)=\frac 1 {f'( 1)}
$$

The inverse function is:

$$\frac{1}{12} \left(\frac{x^2}{\sqrt[3]{x^3+12 \sqrt{3} \sqrt{x^2
   \left(x^2+108\right)}+216 x}}+\sqrt[3]{x^3+12 \sqrt{3} \sqrt{x^2
   \left(x^2+108\right)}+216 x}+x\right)$$

and its derivative is:

$$\frac{1}{12} \left(-\frac{\left(3 x^2+\frac{6 \sqrt{3} \left(2 x^3+2
   \left(x^2+108\right) x\right)}{\sqrt{x^2 \left(x^2+108\right)}}+216\right) x^2}{3
   \left(x^3+12 \sqrt{3} \sqrt{x^2 \left(x^2+108\right)}+216 x\right)^{4/3}}+\frac{2
   x}{\sqrt[3]{x^3+12 \sqrt{3} \sqrt{x^2 \left(x^2+108\right)}+216 x}}+\frac{3
   x^2+\frac{6 \sqrt{3} \left(2 x^3+2 \left(x^2+108\right) x\right)}{\sqrt{x^2
   \left(x^2+108\right)}}+216}{3 \left(x^3+12 \sqrt{3} \sqrt{x^2
   \left(x^2+108\right)}+216 x\right)^{2/3}}+1\right)$$

Here is the function (red), and its inverse (green) and the reflection line y = x (dashed blue):


Distribution functions: differentials in the numerator or denominator One paper I'm looking at says,

  $n(M, z) \, dM \, dz$, the number of sources with mass $M$ at a
  redshift $z$, in the mass interval $dM$ occurring in the redshift
  interval $dz$.

While another says that,

  $\frac{d^2n}{dM dz}$, the number of sources with mass and redshift in the range $[M, M+dM]$ and $[z, z+dz]$ respectively.

What is the difference (conceptually and technically, please) between these formalisms -- and the (different?) meaning(s) of the distribution $n$?

Perhaps unrelated: sometimes when integrating numerically (e.g. to find the total number of sources over all parameters) we have to include the differential itself, i.e. multiply by $\sum n_i \Delta z_i = n_i (z_{i+1} - z_i)$, whereas other times we just sum the distribution $\sum n_i$ --- is this because the latter version implicitly includes the differential-elements within it?

Both representations are of the same thing, depending upon the choice and preference of the author.

These two similar representation, account to the fact that in the function, $n(M, z)$ is may well be possible, that, $M := M(z)$, that is, it could be in itself, a function of $z$, please note that this fact is implicit and may not be always be true. This is the supposed technicality behind these representations, otherwise these two are the different visuals pointing to the same thing.

Let's consider an example which illustrates this point:
consider the integral, $$\int_{\pi/4}^{\pi/2}(1 + e^{tan(\theta)}) sec^{2}\theta d\theta.$$

Please note that this integral can also be written as $$\int_{\pi/4}^{\pi/2}(1 + e^{tan(\theta)}) d({tan(\theta)}),$$
since $$\frac{d(tan\theta)}{d\theta} = sec^{2}\theta$$
so, $$d(tan\theta) = sec^{2}\theta d\theta.$$

So, whether you say $d(tan\theta)$ or $sec^{2}\theta d\theta$, it represents the same thing.


Formulas for calculating pythagorean triples I'm looking for formulas or methods to find pythagorean triples. I only know one formula for calculating a pythagorean triple and that is euclid's which is:

$$\begin{align}
&a = m^2-n^2 \\
&b = 2mn\\
&c = m^2+n^2
\end{align}$$

With numerous parameters.

So are there other formulas/methods?

It doesn't seem cumbersome to me.  Just loop on $m$ starting at $2$.  Loop on $n$ starting at $1$ or $2$ depending on the parity of $m$ and going up to $m-1$.  Check the GCD using Euclid's algorithm.  If it is $1$ you have a primitive set, so calculate $a,b,c$.  If you want all sets up to some $N$, multiply $a,b,c$ by all values up to $\lfloor \frac Nc \rfloor$.  Stop the $m$ loop at $\sqrt N$

I can tell formula for any odd number but this formula always works.

take any odd number=(x) . Solve for $(a+a+1 = x^2)$.

$(a)$ and $(a+1)$ will be the two numbers that will satisfy pythagoran triplet with $x$.

$(x)$ should be a natural number.

$(a+1)$ will be hypotenuse, $(a)$ will be the longer side ( if you consider perpendicular is greater, then $(a)$ is the perpendicular) and $(x)$ will be the shortest side.

This is the new trick which is invented by me shreyas.


Estimating $P(A,B|X)$ given $P(A|X)$, $P(B|X)$ and $P(B|A)$ I have to deal with a situation where I have the following knowledge:

$P(A)$
$P(B)$
$P(A|X)$
$P(B|X)$
$P(B|A)$

I do now want to compute $P(A,B|X)$.
Firstly, I observed that $P(A,B|X) = P(B|A,X) \cdot P(A|X)$. So what I need is a function which takes as input $P(B), P(B|X)$ and $P(B|A)$ and estimates $P(B|A,X)$.

My intuition is that I don't have enough information to logically deduce a solution. So what I want is an estimation which is as sensible as possible.
Therefore, I made a list of necessary conditions for this function.

Firstly, in order to be probability distribution, it is necessary that

P: $0 \leq P(B|A,X) \leq 1$

Then, I checked some special cases, for the values of $P(B|A)$:

A1: $~P(B|A,X) \rightarrow 1~$ for $~P(B|A) \rightarrow 1, ~P(B|X) \not = 0$
A2: $~P(B|A,X) = P(B|X)~$ if $~P(B|A) = P(B)$
A3: $~P(B|A,X) = 0~$ if $~P(B|A) = 0$

And then also for the values of $P(B|X)$:

X1: $~P(B|A,X) \rightarrow 1~$ for $~P(B|X) \rightarrow 1, ~P(B|A) \not = 0$
X2: $~P(B|A,X) = P(B|A)~$ if $~P(B|X) = P(B)$
X3: $~P(B|A,X) = 0~$ if $~P(B|X) = 0$

Furthermore it should hold that

M: $~P(B|A,X)$ is strictly monotonically increasing in $P(B|A)$ and in $P(B|X)$.

I considered the function $P(B|A,X) = \min(1, \frac{P(B|X) \cdot P(B|A)}{P(B)})$ which satisfies P, A2, A3, X2, X3, M but not A1, X1.

Can you come up with a function that satisfies all of these constraints?

Thanks in advance for ideas :)

I found a function which satisfies all of the above constraints and is not too complicated:

$P(B|A,X) = \min \left(1, P(B|X) \cdot P(B|A) \cdot \left[ \frac{(P(B|X)-P(B))^2 \cdot (P(B|A)-P(B))^2}{(1-P(B|X)) \cdot (1-P(B|A))} + \frac{1}{P(B)}  \right] \right)$

This is what the function looks like in a 3D-plot:

x := P(B|X), y := P(B|A), P(B) := 0.8

I cannot give any reasons for its structure, apart from the fact that it fulfils the above necessary constraints. My intuition is that I don't have enough information to logically deduce a function, which is why I only asked for an estimation which is as sensible as possible.

So, before you downvote me, give me suggestions on how this function should be different!


Uniformly continuous functions on $\mathbb{R}$. Possible Duplicate:
  Is a uniformly continuous function vanishing at $0$ bounded by $a|x|+c$?

On the wikipedia page for Modulus of Continuity, it states that for $f$ uniformly continuous on $\mathbb{R}$, $\exists a,b>0 \in \mathbb{R}$ such that $|f(x)| \leq a|x|+b$ for all $x \in \mathbb{R}$. Can someone please explain why this is?

Regards, MM.

$f$ is uniformly continuous, thus given $\epsilon>0$ there exists $\delta>0$ such that $|x-y|\leq \delta$ yields $|f(x)-f(y)|\leq \epsilon$. By chosing any sequence $(x_k)_{k=0}^N$ such that $x_0=0$, $x_N=x$ and $|x_{k+1}-x_k|\leq \delta$, we have
$$
|f(x)-f(0)|\leq \sum_{k=0}^{N-1} |f(x_{k+1})-f(x_k)|\leq N\epsilon.
$$
But since $|x|/\delta\geq N$, you indeed obtain

$$
|f(x)|\leq (\epsilon/\delta)|x|+|f(0)|.$$


Why $\sqrt{-1 \cdot {-1}} \neq \sqrt{-1}^2$? I know there must be something unmathematical in the following but I don't know where it is:
\begin{align}
\sqrt{-1} &= i \\\\\
\frac1{\sqrt{-1}} &= \frac1i \\\\
\frac{\sqrt1}{\sqrt{-1}} &= \frac1i \\\\
\sqrt{\frac1{-1}} &= \frac1i \\\\
\sqrt{\frac{-1}1} &= \frac1i \\\\
\sqrt{-1} &= \frac1i \\\\
i &= \frac1i \\\\
i^2 &= 1 \\\\
-1 &= 1 \quad !!?
\end{align}

Between your third and fourth lines, you use $\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}$.  This is only (guaranteed to be) true when $a\ge 0$ and $b>0$.

edit:  As pointed out in the comments, what I meant was that the identity $\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}$ has domain $a\ge 0$ and $b>0$.  Outside that domain, applying the identity is inappropriate, whether or not it "works."

In general (and this is the crux of most "fake" proofs involving square roots of negative numbers), $\sqrt{x}$ where $x$ is a negative real number ($x<0$) must first be rewritten as $i\sqrt{|x|}$ before any other algebraic manipulations can be applied (because the identities relating to manipulation of square roots [perhaps exponentiation with non-integer exponents in general] require nonnegative numbers).

This similar question, focused on $-1=i^2=(\sqrt{-1})^2=\sqrt{-1}\sqrt{-1}\overset{!}{=}\sqrt{-1\cdot-1}=\sqrt{1}=1$, is using the similar identity $\sqrt{a}\sqrt{b}=\sqrt{ab}$, which has domain $a\ge 0$ and $b\ge 0$, so applying it when $a=b=-1$ is invalid.

what you are saying is similar to $1^2=(-1)^2$ so $1=-1$, math does not work that way


Formally, why does a logical contradiction have probability zero? In terms of formal probability theory, why does an event representing a logical contradiction (such as $A \wedge \neg A$) always have probability zero?

I understand intuitively why this is the case, but I don't see anything in the axioms of probability that directly explains why $A \wedge \neg A$ must have probability zero (or, equivalently, why a logical tautology such as $A \vee \neg A$ must have probability 1).

We see that in probability, we represent the event $A$ as a set of elements in our sample space, and $\neg A$ as the complement of $A$ in our sample space. Thus, in probabilistic terms,

$$P(A \wedge \neg A) = P\left(A \cap \overline{A}\right) = P(\emptyset)$$

by the definition of the complement. And by the Kolmogorov Axioms, we see that

$$P(\emptyset) = 0.$$

A logical contradiction can never be true.   Something that surely will never happen (an impossibility) has probability measure of zero.

A tautology will always be true.   Something that surely will happen (a certainty) has a probability measure of one.

That is all.


Show that operator is continuous and selfadjoint (or not) In this thread
Show compactness/ noncompactness of an operator by approximation
I came to the conclusion that the operator
$$
T\colon\ell^2\to\ell^2, (a_n)_{n\in\mathbb{N}}\mapsto\left(\frac{a_n+a_{n+1}}{2}\right)_{n\in\mathbb{N}}
$$
is not compact.
Now my task is to show whether this operator is (i) continuous (or not) and (ii) selfadjoint (or not).

(i) Concerning continuity I wrote down
$$
\lVert Tx\rVert_{\ell^2}^2=\frac{1}{4}\sum\limits_{n\geq 1}\lvert x_n+x_{n+1}\rvert^2\leq\sum\limits_{n\geq 1}\left(\lvert x_n\rvert+\lvert x_{n+1}\rvert\right)^2\leq\sum\limits_{n\geq 1}(2\lvert x_n\rvert)^2=4\lVert x\rVert_{\ell^2}^2,
$$
so
$$
\lVert Tx\rVert_{\ell^2}\leq 2\lVert x\rVert_{\ell^2}
$$
and so $T$ is continuous, because it is linear and bounded.

(ii) But I do not know exactly how to show if $T$ is selfadjoint. I try to put it more precisely: $\ell^2$ is a Hilbert space with scalar product
$$
\langle (x_n),(y_n)\rangle=\sum\limits_{n\geq 1}x_n\overline{y_n}.
$$
Therefore I started with
$$
\langle Tx,y\rangle=\sum\limits_{n\geq 1}\frac{x_n+x_{n+1}}{2}\overline{y_n}.
$$
Now I have to work with this in order to get an expression with is $\langle x,T^*y\rangle$. But I do not know how to continue.

Could you please say me if (i) is correct and help me with (ii)?

With regards

math12

The operator is not self-adjoint. Consider $x=(1,2,0,\dots,0,\dots)$ and $y=(3,4,0,\dots,0,\dots)$.
It is linear and bounded, and you argued correctly.

This is not an answer for the question but I want to comment on part (i). It seems to me that your estimate
$$\lVert Tx\rVert_{\ell^2}\leqslant 2\lVert x\rVert_{\ell^2}$$is too rough. My calculation shows that
$$\lVert Tx\rVert_{\ell^2}\leqslant \lVert x\rVert_{\ell^2}$$and it turns out that $\|T\|=1$ since $\|T\| \geqslant \sqrt{1-\frac 3{4n}}$ for any $n$ if you choose the point $(1,\dots,1,0,\dots)$ to test $\|T\|$.


$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\cdots}}}}}$ approximation Is there any trick to evaluate this or this is an approximation, I mean I am not allowed to use calculator.

  $$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\cdots}}}}}$$

$$x = \sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\cdots}}}}}$$

$$x = \sqrt{7x}$$

$$x^2 - 7x = 0$$

$$x(x - 7) =0 \implies x = 7$$

Because $\sqrt{7} > 0$ we reject the $x=0$ solution.


Limits in Indeterminate form. Why is the   $\lim_{x  \rightarrow \ 0}$$x^x$ indeterminate and not $1$ ?.

I first approached this as the zero polynomial, but still couldn't wrap my head around why this would still be an indeterminate form.

Use the fact that $e^{lnx}=x$ to rewrite the problem as $$\lim_{x\to 0} e^{lnx^x}=\lim_{x\to 0} e^{xlnx}=e^{\lim_{x\to 0}xlnx}=e^0=1$$

Giving it $1$ as result would assume giving a meaning to $\frac{0}{0}$.

If $0^0 = n$, where $n$ is a number, then $n=0^0 =0^{1-1} = \frac{0}{0}$.

And this would be in contrast with other math rules.


Simple probability calculation for dice game "Schock" I have a problem to calculate a probability within my programm I am currently writing on. I thought about it the whole day yesterday but I don't get it. Maybe it is quite simple for you guys.

It is a dice game without entire information. I have three or more player. I can calculate the possibility to win for each player against each other player. Using these numbers, I want to calculate the general possibility to win for each player.

Simple example: 3 player. Every player has 50% against each other. I used a tree to calculate the overall possibility, but that seems to fail. The possibility should be 33% for each player to have the best roll?! But using the tree would calculate 1 * 0.5 (beating villain 1) * 0.5 (beating villain 2) = 0.25.

Can someone please point me to the right direction?

Update:
The game is called "Schock", a german tavern-game. Every player has 3 dices and up to 3 rolls, which will take place consecutively. When all player finished their roll, the player with the lowest rating roll gets minus points equal to the highest ranking rolls' value.

Player may choose to let some of their dices unrevealed.

Update 2, to make more clear what my problem is:
The numbers above are just example numbers to keep things simple. I already have a routine which is able to calculate odds between two players' dicecups, so lets take it as given that I can predict in any situation the odds between two players' cups even if they did not finish their rolls yet and have any number of dices unrevealed.
The core of my problem is using these numbers (1vs1) to calculate odds between more than two player, which is a pure mathematical problem.
My first approach was a tree, but this fails. As it was stated in the comments, this may be a complete false approach as players' rolls rankings are not independent of each other.
So the core question is: Is it possible to use all 1vs1 odds to calculate 1vsAll odds and how.

Maybe with a concrete situation:
Lets assume 3 players playing. Which all have made 3 rolls, so the round is finished. Before all cups are revealed, this is the situation:

Player 1: Has two 1s and one dice unrevealed.
Player 2: Has one 1 and two dices unrevealed.
Player 3: Has all dices unrevealed.

Given all 1 vs 1 odds to have the better roll:

P1 vs P2: 0.88426 (88%)
P1 vs P3: 0.97299 (97%)
P2 vs P3: 0.50759 (51%)

Without even knowing the rules of the game, is it possible to calculate the total chance for Player 1 to win or Player 3 not to loose using the given numbers?

The issue at hand is that the probabilities aren't independent - indeed, one result is necessarily defined by the other two - if A beats B and B beats C, then A necessarily beats C. And just knowing the probabilities for each of the three two-player matches isn't enough to define the system.

Allow me to demonstrate with a simple example of a similar game, but with easily-analysed properties. A bag holds four balls, each labelled between 1 and 4. Each of the three picks a ball out of the bag, and the one with the highest number wins. This is used to avoid issues of equal scores causing rematches, etc.

Ahead of time, it's easy to determine that the probability that each player wins is $\frac13$, and that the probability that any specific player beats any other player is $\frac12$. We can easily write that
$$
Pr(``\text{A wins''}) = Pr(A>B) - Pr(C>A>B)
$$
Now, as we noted, $Pr(A>B)=\frac12$ and $Pr(``\text{A wins''})=\frac13$, so we have that $Pr(C>A>B)=\frac16$. But this means that $Pr(C>A|A>B)=\frac13\neq Pr(C>A)$, and thus the events aren't independent. To see why this is, let's look at the situation where A>B. In this case, there are six possible combinations.

--- NOTE: Below here, until noted otherwise, probabilities are for case in which $A>B$, left off for notational simplicity ---

Specifically, if we denote the combinations as (A,B), then we have (2,1), (3,1), (3,2), (4,1), (4,2), and (4,3). Of these situations, $Pr(C>A=4)=0$, $Pr(C>A=3)=\frac12$, and $Pr(C>A=2)=1$, where $Pr(C>A=n)$ is shorthand for $Pr(C>A|A=n)$. And given the probabilities that arise, we have
$$
Pr(C>A) = \sum_{n=2}^4 Pr(C>A=n)Pr(A=n)\\
= 0\cdot \frac12+\frac12\cdot \frac13+1\cdot\frac16 = \frac13
$$
Now suppose that we allow partial knowledge. We know that $B\neq2$ and $B\neq 3$. Now the set of combinations becomes (2,1), (3,1), (4,1). This means that we now have
$$
Pr(C>A) = 0\cdot\frac13 + \frac12\cdot\frac13 + 1\cdot\frac13 = \frac12
$$

--- NOTE: Assumption that $A>B$ is removed from here ---

When we have $B\not\in\{2,3\}$, we can see that $Pr(A>B)=\frac12$. We also have that $Pr(C>A|A>B) = \frac12$, and so
$$
Pr(``\text{A wins''}) = \frac12 - \frac12\cdot\frac12 = \frac14
$$
Notice that $Pr(A>B)$, $Pr(A>C)$, and $Pr(B>C)$ didn't change, but the final probabilities did? Notice that, in the latter case, $Pr(C>A)=Pr(C>A|A>B)$?

So it's not enough to know the probabilities for each pair, you need to know how they interact, and that depends on the specifics of what each player shows (not just how many dice are shown, but number of dice+score on those dice).

If you make a probability tree for each player:
for Player 1 $P_1$
$$P_{win,win}=0.5*0.5=0.25$$
$$P_{win,loss}=0.5*0.5=0.25$$
$$P_{loss,win}=0.5*0.5=0.25$$
$$P_{loss,loss}=0.5*0.5=0.25$$

Summing all the probabilitites equals 1.
This is the same for all the other players ($2$ and $3$)


maximum radius of a circle inscribed in an ellipse Consider an ellipse with major and minor axes of length $10$ and $8$ respectively. The radius of the largest circle that can be inscribed in this ellipse, given that the centre of this circle is one of the focus of the ellipse.

I attempted to solve this by using a very simple concept:

All points of the circle should either lie inside or on the circle. Hence assuming any point on the circle to be $x=ae+r\cos \theta$ and $y=r\sin \theta$  which satisfies:

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}<1$$

I got a quadratic in $\cos \theta$ and I made the equation to be true independent of theta which gave $r\in [a-ae,a+ae]$ what is wrong to solve it by this method.

I would also like to know any other methods to solve this problem.

Hint:

Parametrize the ellipse as $(x,y) = (5 \cos t, 4 \sin t)$.
Note that a focus is located at $(3,0)$.
Obtain an expression for the distance of a point on the ellipse to this focus.
Find a value of $t$ for which this distance is minimized.
Compute the minimum distance. This is the radius of the largest circle centered at the focus.

Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} =1$.
Parametric point on the ellipse is $(a\cos t, b\sin t)$ . Distance of centre $(0,0)$ from this point is given by $(a-ae \cos t)$. (Substitute in distance formula and simplify to obtain that expression)

$(a-ae \cos t)$ in this case is $(5-3\cos t)$.

It is minimum for $\cos t=1$ and its value is $2$. Hence , the radius of the largest circle that can be inscribed is $2$ units.


Is it obvious that $f\in \mathcal C^m([a,r])$ for all $r<b$ implies that $f\in \mathcal C^m([a,b))$? In a book I'm reading, to prove that $f\in \mathcal C^m([a,b))$ they proved that $f\in \mathcal C^m([a,r])$ for all $r\in [a,b)$.

Q1) Is it that obvious that $f\in \mathcal C^m([a,b))$ when  $f\in \mathcal C^m([a,r])$ for all $r\in [a,b)$ ?

Q2) If $a<b-1$, can we "obviously" say that $$\mathcal C^m([a,b))=\bigcup_{n\in\mathbb N^*}\mathcal C^m\left(\left[a,b-\frac{1}{n}\right]\right) \ \ ?$$

Because unfortunately it's not that obvious for me, and I'm not really sure it's correct to be honnest.

Q1) It is not true that if $f$ is once continuously differentiable, then it is $m$ times so. What their proof is doing is saying that you can prove the result with $m$ derivatives by repeatedly applying the one derivative case or that the general case is similar to the $m = 1$ case.

Q2) It is an intersection, not a union. (Remember, a union is a "or" and an intersection is an "and", so that the union of those sets is the set of functions that is continuous;y $m$ times differentiable on AT LEAST ONE $[a,b-1/n]$, but an intersection of those sets would require that it is on ALL those subintervals.) Second, you shouldn't require $a-b > 1$, but instead require that $n > 1/(b-a)$ so that $1/n < b-a$.

Ask yourself, if a function is $m$ times continuously differentiable on $[a,b)$, then what does that mean? It means that they derivatives exist and are continuous at each point in $[a,b)$, because being continuously $m$ times differentiable is a local property: it only needs to be checked on some small interval containing the point. So, if a function is continuously $m$ times differentiable on $[a,b)$ then it is clearly continuously $m$ times differentiable on the sub intervals $[a,b-1/n]$ for $n$ large enough. Likewise, if it is differentiable on each of these subintervals, then for any point $x \in [a,b)$ there is an $n$ such that $x \in [a, b-1/n)$ on which $f$ is continuously $m$ times differentiable and so $f$ is continuously $m$ times differentiable near $x$. This means that $f$ is continuously $m$ times differentiable on the entirety of $[a,b)$.

Is it that obvious that $f\in C^m([a,b))$ when $f\in C^1([a,r])$ for all $r\in [a,b)$?

Yes, it is correct.

$f\in C^m([a,b))$ can be written as $f\in C^m([a,b-\varepsilon])$ for all $\varepsilon >0$ and $\varepsilon\leq b-a$ (to make sure right point of interval is bigger than left one).

Let $\varepsilon > 0$ and $\varepsilon\leq b-a$ be arbitrary but fixed. Then $a\leq b-\varepsilon<b$. Hence, we can set $r:=b-\varepsilon$. Since $f\in C^m([a,r])$ by assumption and $\varepsilon$ was arbitrary, it follows $f\in C^m([a,b))$.

Intuitively you can always choose $r$ as near to $b$ that it fits.

  If $a<b-1$ can we “obviously” say that
  $$C^m([a,b)) = \bigcup_{n\in \mathbb N^*}C^m\bigg(\bigg[a,b-\frac1 n\bigg]\bigg)\mathrm?$$

No, this is not correct. As @4-ier pointed out in his answer it has to be the intersection.
So
$$C^m([a,b)) = \bigcap_{n\in \mathbb N^*}C^m\bigg(\bigg[a,b-\frac1 n\bigg]\bigg) $$ is correct.

You need the assumption $a<b-1$ because if you set $n=1$ you get the interval $[a,b-1]$ where the right point has to be greater then the left one.

You can show the equality similar as for the above question. $\frac1 n\to 0$ for $n\to\infty$ but $\frac1 n > 0$ for all $n\in \mathbb N^*$. (Intuitively, this means you can approach $b$ as far as you want but you cannot reach it). Hence, you get the above equality.


Computing the integral of $\log(\sin x)$ How to compute the following integral?
$$\int\log(\sin x)\,dx$$

Motivation: Since $\log(\sin x)'=\cot x$, the antiderivative $\int\log(\sin x)\,dx$ has the nice property $F''(x)=\cot x$. Can we find $F$ explicitly? Failing that, can we find the definite integral over one of intervals where $\log (\sin x)$ is defined?

You can calculate
$$
\int_0^\pi\log(\sin x)\,dx = -\pi\log2
$$
and integrating up to $\pi/2$ would give half of this.

Note that integrating $\log(\sin x)$ from 0 to $\pi/2$ is the same as integrating $\log(\cos x)$ so that
$$
\begin{align}
\int_0^{\pi/2}\log(\sin x)\,dx &= \frac12\int_0^{\pi/2}\log(\sin x\cos x)\,dx\\
&= \frac12\int_0^{\pi/2}\log(\sin 2x)\,dx - \frac{\pi}{4}\log 2.
\end{align}
$$
After a change of variables, this can be rearranged to get the result.

For the indefinite integral, you have this closed form:

  $$ \frac{i{x}^{2}}{2}+x\ln  \left( \cos \left( x \right)  \right) -x\ln
 \left( 1+{{\rm e}^{2\,ix}} \right) +\frac{i}{2} Li_2 ( -{
{\rm e}^{2\,ix}} ), $$

where $Li_2$ is a polylogarithm.


Does $xe^{1/\log(x)}$ have an oblique asymptote when x tends to infinity? Does $xe^{1/\log(x)}$ have an oblique asymptote when $x$ tends to infinity?
 If so, what is the equation of this asymptote and how can we find it?

If $f(x) = xe^{1/\ln x}$ has an oblique asymptote $y=Ax+B$, then $A = \lim_{x\to\infty} f(x)/x$, but the existence of this limit is not enough to conclude that there is an oblique asymptote.

We also need that $B = \lim_{x\to\infty} f(x)-Ax$ exists.

In your example, we have
$$A = \lim_{x\to\infty} \frac{xe^{1/\ln x}}{x} = \lim_{x\to\infty} e^{1/\ln x} = 1$$
but
\begin{align}
B &= \lim_{x\to\infty} xe^{1/\ln x} - x \\
  &= \lim_{x\to\infty} x(e^{1/\ln x}-1) \qquad [t=1/\ln x] \\
  &= \lim_{t\to 0^+} e^{1/t}(e^t-1) = +\infty,
\end{align}
so your $f$ doesn't have an oblique asymptote when $x\to\infty$.

Using Ruslan's definition,
we want to see if
$\lim_{x \to \infty} \dfrac{x e^{1/\ln x}}{x}
$
exists.

$\dfrac{x e^{1/\ln x}}{x}
=e^{1/\ln x}
$
and,
since
$\lim_{x \to \infty} 1/\ln x
= 0
$,
$\lim_{x \to \infty} e^{1/\ln x}
= e^0
=1
$.

So it does have a
oblique asymptote
with slope $1$.


What are some counter-intuitive results in mathematics that involve only finite objects? There are many counter-intuitive results in mathematics, some of which are listed here. However, most of these theorems involve infinite objects and one can argue that the reason these results seem counter-intuitive is our intuition not working properly for infinite objects.

I am looking for examples of counter-intuitive theorems which involve only finite objects. Let me be clear about what I mean by "involving finite objects". The objects involved in the proposed examples should not contain an infinite amount of information. For example, a singleton consisting of a real number is a finite object, however, a real number simply encodes a sequence of natural numbers and hence contains an infinite amount of information. Thus the proposed examples should not mention any real numbers.

I would prefer to have statements which do not mention infinite sets at all. An example of such a counter-intuitive theorem would be the existence of non-transitive dice. On the other hand, allowing examples of the form $\forall n\ P(n)$ or $\exists n\ P(n)$ where $n$ ranges over some countable set and $P$ does not mention infinite sets would provide more flexibility to get nice answers.

What are some examples of such counter-intuitive theorems?

100 prisoners problem.

Citing Sedgewick and Flajolet, the problem reads as follows:

The director of a prison offers 100 death row prisoners, who are numbered from 1 to 100, a last chance. A room contains a cupboard with 100 drawers. The director randomly puts one prisoner's number in each closed drawer. The prisoners enter the room, one after another. Each prisoner may open and look into 50 drawers in any order. The drawers are closed again afterwards. If, during this search, every prisoner finds his number in one of the drawers, all prisoners are pardoned. If just one prisoner does not find his number, all prisoners die. Before the first prisoner enters the room, the prisoners may discuss strategy—but may not communicate once the first prisoner enters to look in the drawers. What is the prisoners' best strategy?

Surprisingly, there exists a strategy with surviving probability more than 30%. It is connected to the fact---also non-intuitive---that a big random permutation is quite likely to contain "small" cycles only.

Maybe not right on the money, but worth mentioning: (subtly) faulty dissections. For example, as shown here, it is seemingly possible to dissect an 8 x 8 square into a 5 x 13 rectangle.


Prove that the minimum of a functional doesn't exist Prove that there is no smooth solution ho the minimization problem:
$$\mathcal{L} (u)= \int_{0}^1 e^{-u'}+u^2 dx$$
Where the admissible space is $X =\{ u \in \mathcal{c}^2 [0,1] | u(0)=0,  u(1)=1
       \} $
UPDATE:
GOAL: I am trying to define a sequence of functions $u_n(x)$ in $X$ s.t their integrals $\mathcal{L}(u_n) \to 0$
What I have done so far:
Define $u_n(x)$ in the following manner:
$$u_n (x)= \begin{cases}
0, \quad 0 \leq x\leq 1-\frac{1}{n} \\[2ex]
ax^2 +bx+ c, \quad 1-\frac{1}{n}<x \leq 1
\end{cases}$$
where $n=1,2,3,..$. Require that $u_n$ satisfy:
\begin{align}
 u_n(1-\frac{1}{n})=0\\
 u_n(1)=1\\
 u_n'(1-\frac{1}{n})=0\\
\end{align}
We then have:
\begin{align}
a+b+c=1\\
a(1-\frac{1}{n})^2+b(1-\frac{1}{n})+c=0\\
2a(1-\frac{1}{n})+b=0
\end{align}
Find $a,b,c$ then:
$$u_n (x)= \begin{cases}
0, \quad 0 \leq x\leq 1-\frac{1}{n} \\[2ex]
n^2x^2 -2n(n-1)x+ (n-1)^2, \quad 1-\frac{1}{n}<x \leq 1
\end{cases}$$
\begin{align}
\int_0^1 e^{-u_n(x)'} + u_n^2(x) dx &= \int_0^{1-\frac{1}{n}} e^{0} + 0 dx +  \int_{1-\frac{1}{n}}^1 e^{-(2n^2 x -2n(n-1))} + (n^2x^2 -2n(n-1)x+ (n-1)^2)^2 dx\\& = 1- \dfrac{1}{n}+ \dfrac{\mathrm{e}^{-2n}\left(\left(2n+5\right)\mathrm{e}^{2n}-5\right)}{10n^2}
\end{align}
$$\dfrac{\mathrm{e}^{-2n}\left(\left(2n+5\right)\mathrm{e}^{2n}-5\right)}{10n^2} \to 0$$ and $$1-\dfrac{1}{n} \to 1$$

We can more or less use the same proof technique as in my Math.SE answer here.

Define function $$f(a)~:=~(a+1)e^{-a}.\tag{1}$$
Now it is a fact that
$$ \exists ! a_0 <0:~~ f(a_0)~=~0. \tag{2}$$
$$ \forall a> a_0: ~~ 0~<~f(a)~\leq~1.\tag{3}$$
$$ \forall a< a_0: ~~ f(a)~<~0.\tag{4}$$
Here $a_0=-1$. There is a first integral to OP's EL equation. Together with OP's boundary conditions, it becomes $$  f(u^{\prime}(0))~=~ f(u^{\prime}) + u^2
~=~f(u^{\prime}(1))+1 .\tag{5} $$
In light of eqs. (2)-(4) we see that eq. (5) is impossible to satisfy for $u^{\prime}(1)>a_0$ because then $f(u^{\prime}(1))>0$ while $f(u^{\prime}(0))\leq 1$.

The EL equation reads
$$ \left(-e^{-u^{\prime}}\right)^{\prime}~=~2u.\tag{6} $$
If in the interval $[x_0,1]$ the function $u > 0$ is positive, then $-e^{-u^{\prime}}$ (and thereby $u^{\prime}$) is increasing, and hence $u^{\prime}\geq 0$. Then
$$u^{\prime}(1)~\geq~ 0,\tag{7}$$
which according to section 1 leads to a contradiction. $\Box$

Observe that you cannot have $\mathcal{L}(u)=0$, since that would imply $u=0$ a.e., which contradicts the fact that $u(1)=1$ (+continuity). Therefore, $\mathcal{L}(u)>0$ for all $u\in X$.
As someone already mentioned, you can consider the sequence $u_n(x)=x^n$, which satisfies $\mathcal{L}(u_n)\to 0$ (by the dominated convergence theorem).
This implies that $\mathcal{L}$ does not attain its minimum on $X$.


Relation between bounded first derivative and behaviour of a function Let me begin with an example first:

Consider $\;f:\mathbb R \rightarrow \mathbb R\;$ such that $\;f(x)=\tanh x\;$. Then its plot is given below:

Now, the first derivative of $\;f\;$ is equal to $\;f'(x)=\frac{2}{e^x + e^{-x}} \equiv \frac{1}{\cosh x} \;$ which is a bounded function. Its plot is also given below:

I observe that $\;f\;$ goes from the value $-1$ to the value $+1$ in a "smooth" ,kind of, way.(I'm sorry I lack of the best word to describe the behaviour of $\;f\;$)

I wonder if the above behaviour of $\;f\;$ is related to the fact that its first derivative is bounded.

If my assumption is true, why does this hold?
If I had a function which first derivative wasn't bounded, then what could I claim about the behaviour of $\;f\;$ in this case?

I'm searching for some examples but I can't find any...

I hope I made my question clear enough. Any help would be valuable.

Thanks in advance!

Take your function to be

$$
 f (x) =
  \begin{cases}
 0 & \text{if } x<0 \\
 x+1 & \text{if } x \geq 0 \\
  \end{cases}
$$

Then $f$ has a bounded derivative but it is not "smooth" because it is discontinuous at $0$.

If your derivative is not bounded, then you can't claim much about boundness of the original function. For example, $f=\sqrt{x}$ on $[0,1]$ is a bounded function, but it's derivative is not bounded. However, $g(x)=e^x$ is a function that is not bounded and whose derivative is not bounded.

If your 1st derivative is not bounded then either your function is not continuous from both side of domain or it is discontinuous
e.g.
consider these function
\begin{cases}
 \sqrt{x}, & \text{x ∈ [ 0 , 1 ]}\end{cases}
as it has derivative unbounded near zero hence it has not left hand side continuity
and,
\begin{cases}
 1 / x, & \text{x ∈ R}\end{cases}
but in this case the function is not continuous at x = 0
therefore unbounded discontinuity does tells us so much about continuity and much more stuff


One question regarding a subring of $Q$ which will contain 1. Let $R$ be a subring of $Q$ containing 1. Then which of the following is/are true?

$R$ is  a PID.

$R$ contains infinitely many prime ideals.

$R$ contains a prime ideal which is not a maximal ideal.

For every maximal ideal $m$ , $R/m$ will be finite.

My Try : 2. This is not true. $Q$ itself is the counter example. $Q$ has no ideal.

This is correct. Because {0} is in every $R$.

If we take the subring [$a/3^k$: where k is non negative integer]. Then $Z$ is a maximal ideal of this subring but $R/Z$ is not finite. so False.
I have no idea about option 1.

Have I gone wrong anywhere? Please correct me if I have and tell me what will happen for the option 1.

2, 3, and 4 are obviously false, since $\Bbb{Q}$ is an obvious counterexample in each case.

1 is true since the subrings are various localizations of $\Bbb{Z}$, and localizations of PIDs are PIDs.

2 is false, let $p$ be a prime in $\mathbb Z$ and let $R$ be the ring generated by $\mathbb Z$ and $\{n^{-1}:n\in\mathbb Z, p\nmid n$}.  In, other words every member of $\mathbb Z$ is a unit unless it is in the prime ideal $(p)$.  It's not hard to show that the only non-trivial ideal is $(p)$. So, $R$ has only one prime ideal.


How would I use root/ratio test on $\sum_{n=1}^\infty\left(\frac{n}{n+1}\right)^{n^2}$? WolframAlpha tells me that the sum $$\displaystyle \sum_{n=1}^\infty\left(\frac{n}{n+1}\right)^{n^2}\tag1$$ converges by the ratio test, but I have absolutely no idea how I would use the ratio test on this sum. Also, the root test is inconclusive, because I'm just gonna get $\lim\limits_{n\to 0} \left(\frac{n}{n+1}\right)^{2}$, which is $1$.

(the limit symbol isn't showing up properly but I don't know how to do it better)

It's a good idea to use the root test, but you simplified the expression incorrectly. Instead, you should get
$$ \lim_{n\to\infty}\Big(\frac{n}{n+1}\Big)^{\frac{n^2}{n}}=\lim_{n\to\infty}\Big(\frac{n}{n+1}\Big)^{n}=\frac{1}{e}$$
and since $\frac{1}{e}<1$ it follows that the series converges.

I wouldn't recommend it, but the ratio test works as well:

$$\frac{\left(\frac{n+1}{n+2}\right)^{(n+1)^2}}{\left(\frac{n}{n+1}\right)^{n^2}} = \frac{\left(\frac{n+1}{n+2}\right)^{n^2}}{\left(\frac{n}{n+1}\right)^{n^2}}\cdot \left(\frac{n+1}{n+2}\right)^{2n+1} = \left(\frac{(n+1)^2}{n(n+2)}\right)^{n^2}\cdot \left(\frac{n+1}{n+2}\right)^{2n+1} = \left(1+\frac{1}{n^2+2n}\right)^{n^2}\cdot\left(1-\frac 1{n+2}\right)^{2n+1}\longrightarrow e\cdot e^{-2}=\frac 1 e$$


Ratios and Fractions I am trying to assist my daughter with holiday work, please help me with these questions:

Q1: Find the ratio $a:b$, when  $\frac{a+b}{a-b}=\frac{11}{5}$

Q2: Find the the ratio $a:b:c$ when   $\frac{1}{c}=\frac{2}{3b}=\frac{3}{4a}$

Q3: Evaluate $\frac{3a-b}{a+2b}$ if $\frac{a}{b}=\frac{4}{5}$

For the first, try to get things in terms of the quantity $x=a/b$. Use the fact that $r+s$ can be written as $s(\frac{r}{s}+1)$, which you can verify by multiplying it out.
$$\frac{11}{5}=\frac{b\left(\frac{a}{b}+1\right)}{b\left(\frac{a}{b}-1\right)}=\frac{x+1}{x-1}.$$
Solve this for $x$.

For the second, take the reciprocal:
$$c=\frac{3}{2}b=\frac{4}{3}a.$$

For the third, again, try to get things in terms of $x=a/b=4/5$.
$$\frac{3a-b}{a+2b}=\frac{b\left(3\frac{a}{b}-1\right)}{b\left(\frac{a}{b}+2\right)}=\frac{3\cdot\frac{4}{5}-1}{\frac{4}{5}+2}.$$

Note that all three of Ragnar's answers are wrong at the time of this post, so don't be alarmed if your answers don't match his.

Q1: You can rewrite this to $5(a-b)=11(a+b)$, and thus $-6a=16 b$. It follows that $\frac ab=-\frac 38$.

Q2: Take the inverse: $c=\frac 32 b=\frac 43 a$, thus $a:b:c=\frac 34:\frac 23:1$.

Q3: $\frac{3a-b}{a+2b}=\frac{2\frac ab-1}{\frac ab+2}=\frac{2\frac 45-1}{\frac 45+2}=\frac{\frac 35}{\frac {14}5}=\frac 3{14}$


Expectation of function of random variable? How to calculate the expectation of function of random variable without using probability density function? Note:- only cumulative distribution function is available.
For example $E[g(X)]$=? where X is  nonnegative r.v. with CDF $F_{X}(x)$.

If $X\geqslant0$ almost surely and if $g$ is regular,
$$
\mathrm E(g(X))=g(0)+\int_0^{+\infty}g'(x)\cdot(1-F_X(x))\cdot\mathrm dx.
$$
Proof: integrate with respect to $\mathrm P$ both sides of the almost sure relation
$$
g(X)=g(0)+\int_0^{+\infty}g'(x)\cdot[x\lt X]\cdot\mathrm dx,
$$
where $[\ \ ]$ denotes Iverson bracket.

At least formally, the probability density function is $f(X) = dF/dX$, so

  $ E[g(X)] = \int dX f(X) g(X) = \int dX g(X) dF/dX = Fg(X_f) - Fg(X_i) - \int dX F(X) dg/dX $.

So if $g(X)$ is differentiable, and $g$ is finite at the endpoints of the domain of $X$ ($X_f$ and $X_i$), then you can try to evaluate the integral above.


Are there any memorization techniques that exist for math students? I just watched this video on Ted.com entitled:

Joshua Foer: Feats of memory anyone can do

and it got me thinking about memory from a programmers perspective, and since programming and mathematics are so similar I figured I post here as well. There are so many abstract concepts and syntactic nuances that are constantly encountered, and yet we still manage to retain that information.

The memory palace may help in remembering someone's name, a sequence of numbers, or a random story, but are there any memorization techniques that can better aid those learning new math concepts?

You shouldn't try to learn mathematics through memorization at all. It will get you nowhere: anything that can be memorized can be looked up these days. What you should try to learn is the underlying concepts and the way they relate to each other. If you understand those well enough, you won't need to memorize anything.

Think of learning mathematics as being like learning, say, chess. Would you learn how to play chess by memorizing openings? Well, maybe that could work, but it's probably a better idea to learn how to play chess by, y'know, playing a lot of chess.

While some of the comments above have some validity, I can see you just needing to memorize some quick formulas. I do agree practice makes perfect. But what many of the memorizers AND Joshua Foer said is yes, they are tricks but they FORCE you to memorize things. And Foer goes on to say that they aren't even really tricks. They are actually harder to do that just doing it, because you're kind of setting up a safety net in your mind, and it often involved preparatory work.

Anthony Metivier is another memory "guru" who.. I guess.. throws out all forms of memorization stuff EXCEPT the "memory palace" (google it if need be). Though, while he focuses only on memory palaces, other systems like the Major Memory System, (where you associate the phonetic alphabet with numbers) make guest appearances.

http://www.magneticmemorymethod.com/mmmpodcast-episode-003-memorizing-mathematical-formulas/

https://litemind.com/major-system/

^^^^^More info here^^^^^

So, in Metivier's mind, if you have this..

notice vowels and w, h, y are ignored
..you can associate formulas with words like:

C2 = CaN (You keep the c, ignore the a, use the n)

'=' becomes an image of something meaningful to you, he chooses to use a FLAG

A2 = yawN (again, since you're ignoring the letters y, a, w, you can place them anywhere to make words)

'+' = this becomes another image of your choice- try using a cross, or hospital symbol

B2 = Maybe this is BuN, BaNe, etc.

Now, he went a slightly different (I believe more complicated) route. I simplified this for you if you're new.

You now have images. You throw them into a story. And he would say "USE A MEMORY PALACE TOO!"

The idea would be you would walk through.. say.. your house and in your mind, you're placing these items in your house, in order.

Anyone into memory techniques will say to use your own images, your own words, your own palaces. Do not expect to find any Website on memory palaces and have things mapped out for you. Things like the Major System and Peg System will suggest things for you but if they aren't "sticky" TOSS THEM OUT.

Read more on building memory palaces and rules for doing so.

Other sources:

The Bible of Memory folks-
Google "The Memory Book" by Harry Lorayne

This one focuses on academics-
Super Memory Super Student by Harry Lorayne


Given a die, what is the probability that the second roll of a die will be less than the first roll? If you are given a die and asked to roll it twice. What is the probability that the value of the second roll will be less than the value of the first roll?

There are various ways to answer this.  Here is one:

There is clearly a $1$ out of $6$ chance that the two rolls will be the same, hence a $5$ out of $6$ chance that they will be different.  Further, the chance that the first roll is greater than the second must be equal to the chance that the second roll is greater than the first (e.g. switch the two dice!), so both chances must be $2.5$ out of $6$ or $5$ out of $12$.

Let's look at it: $$<6: 1, 2, 3, 4, 5$$$$<5: 1, 2, 3, 4$$$$<4: 1, 2, 3$$$$<3: 1, 2$$$$<2: 1$$$$<1:$$ From there, there are $15$ possibilities and there are $36$ outcomes of any two numbers on a $6$-sided die rolled (one after the other), which simplifies to $5\over 12$.


Axiomatic characterization of the rational numbers We have the well-known Peano axioms for the natural numbers and the real numbers can be characterized by demanding them to be a Dedekind-complete, totally ordered field (or some variation of this).

But I never saw any axiomatic characterization of the rational numbers! Either they are constructed out of the natural numbers or are found as a subset of the reals.

I know that the rational numbers are unique in the sense that they are the smallest totally ordered field. But it is somewhat unsatisfactory for me to define them to be a totally ordered field which has an order-preserving embedding into any other totally ordered field. This would be like defining the real numbers to be an Archimedean totally ordered field such that every other Archimedean totally ordered field can be order-preservingly embedded into it - this is some ugly definition (for me) and I find the usual one much better.

  So what is a nice axiomatic characterization of the rational numbers?

Preamble

I believe that the OP is seeking a characterization of $ \mathbb{Q} $ using only the first-order language of fields, $ \mathcal{L}_{\text{Field}} $. Restricting ourselves to this language, we can try to uncover new axioms, in addition to the usual field axioms (i.e., those that relate to the associativity and commutativity of addition and multiplication, the distributivity of multiplication over addition, the behavior of the zero and identity elements, and the existence of a multiplicative inverse for each non-zero element), that describe $ \mathbb{Q} $ uniquely.

Any attempt to describe the smallest field satisfying a given property must prescribe a method of comparing one field with another (namely using field homomorphisms, which are injective if not trivial), but such a method clearly cannot be formalized using $ \mathcal{L}_{\text{Field}} $.

1. There Exists No First-Order Characterization of $ \mathbb{Q} $

The answer is ‘no’, if one is seeking a first-order characterization of $ \mathbb{Q} $. This follows from the Upward Löwenheim-Skolem Theorem, which is a classical tool in logic and model theory.

Observe that $ \mathbb{Q} $ is an infinite $ \mathcal{L}_{\text{Field}} $-structure of cardinality $ \aleph_{0} $. The Upward Löwenheim-Skolem Theorem then says that there exists an $ \mathcal{L}_{\text{Field}} $-structure (i.e., a field) $ \mathbb{F} $ of cardinality $ \aleph_{1} $ that is an elementary extension of $ \mathbb{Q} $. By definition, this means that $ \mathbb{Q} $ and $ \mathbb{F} $ satisfy the same set of $ \mathcal{L}_{\text{Field}} $-sentences, so we cannot use first-order logic to distinguish $ \mathbb{Q} $ and $ \mathbb{F} $. In other words, as far as first-order logic can tell, these two fields are identical (an analogy may be found in point-set topology, where two distinct points of a non-$ T_{0} $ topological space can be topologically indistinguishable). However, $ \mathbb{Q} $ and $ \mathbb{F} $ have different cardinalities, so they are not isomorphic. This phenomenon is ultimately due to the fact that the notion of cardinality cannot be formalized using $ \mathcal{L}_{\text{Field}} $. Therefore, any difference between the two fields can only be seen externally, outside of first-order logic.

2. Finding a Second-Order Characterization of $ \mathbb{Q} $

This part is inspired by lhf's answer below, which I believe deserves more credit. We start by formalizing the notion of proper subfield using second-order logic.

Let $ P $ be a variable for unary predicates. Consider the following six formulas:
\begin{align}
\Phi^{P}_{1} &\stackrel{\text{def}}{\equiv} (\exists x) \neg P(x); \\
\Phi^{P}_{2} &\stackrel{\text{def}}{\equiv} P(0); \\
\Phi^{P}_{3} &\stackrel{\text{def}}{\equiv} P(1); \\
\Phi^{P}_{4} &\stackrel{\text{def}}{\equiv} (\forall x)(\forall y)((P(x) \land P(y)) \rightarrow P(x + y)); \\
\Phi^{P}_{5} &\stackrel{\text{def}}{\equiv} (\forall x)(\forall y)((P(x) \land P(y)) \rightarrow P(x \cdot y)); \\
\Phi^{P}_{6} &\stackrel{\text{def}}{\equiv} (\forall x)((P(x) \land \neg (x = 0)) \rightarrow (\exists y)(P(y) \land (x \cdot y = 1))).
\end{align}
What $ \Phi^{P}_{1},\ldots,\Phi^{P}_{6} $ are saying is that the set of all elements of the domain of discourse that satisfy the predicate $ P $ forms a proper subfield of the domain. The domain itself will be a field if we impose upon it the first-order field axioms. Hence,
$$
\{ \text{First-order field axioms} \} \cup \{ \text{First-order axioms defining characteristic $ 0 $} \} \cup \{ \neg (\exists P)(\Phi^{P}_{1} ~ \land ~ \Phi^{P}_{2} ~ \land ~ \Phi^{P}_{3} ~ \land ~ \Phi^{P}_{4} ~ \land ~ \Phi^{P}_{5} ~ \land ~ \Phi^{P}_{6}) \}
$$
is a set of first- and second-order axioms that characterizes $ \mathbb{Q} $ uniquely because of the following two reasons:

Up to isomorphism, $ \mathbb{Q} $ is the only field with characteristic $ 0 $ that contains no proper subfield.
If $ \mathbb{F} \ncong \mathbb{Q} $ is a field with characteristic $ 0 $, then $ \mathbb{F} $ does not model this set of axioms. Otherwise, interpreting “$ P(x) $” as “$ x \in \mathbb{Q}_{\mathbb{F}} $” yields a contradiction, where $ \mathbb{Q}_{\mathbb{F}} $ is the copy of $ \mathbb{Q} $ sitting inside $ \mathbb{F} $.

Here we give an axiomatic characterization for the rational numbers under addition.

Let $(\Bbb Z, 0, 1, +)$ be given.

Let $(M,0,1,+)$ be a monoid (not a singleton) satisfying the following:

P-1: For every integer $d \gt 0$ there exist an injective homomorphism (necessarily unique)

$\tag 1 \mu_d:\Bbb Z \to M \quad \text{with } \mu_d(d) = 1$

P-2: For every $x,y \in M$ there exists an integer  $d \gt 0$ such that

$\tag 2  x,y \in \mu_d(\Bbb Z)$

Theorem: The monoid $M$ in unique up to isomorphism and $(M,0,1,+) \equiv (\Bbb Q,0,1,+)$.


Ways to put numbered balls in boxes no box being empty I know the formula for putting $n$ identical balls in $r$ different boxes such that each box has at least 1 ball, but what is the formula for putting $n$ different balls in $r$ different boxes, no box being empty?

Thanks!

You can do this by inclusion/exclusion. There are $\binom r0r^n$ ways of putting the $n$ balls into the $r$ boxes. From this we have to subtract the $\binom r1(r-1)^n$ ways of putting the $n$ balls into just $r-1$ of the boxes. To this we have to add the $\binom r2(r-2)^n$ ways of putting the $n$ balls into just $r-2$ of the boxes, and so on, so

$$a_n=\sum_{k=0}^r\binom rk(-1)^{r-k}k^n=\left.\left(q\frac{\partial}{\partial q}\right)^n(q-1)^r\right|_{q=1}\;.$$

Distribute n out of k balls in the n numbered boxes in n! ways (no box could have more than one ball). Distribute the remaining k-n balls in n^(k-n) ways (no restriction). Total arrangement =n!* n^(k-n) ways.


How to prove $L(f, P_{1}) \leq U(f, P_{2})$ under given conditions. I would like to show that under the assumptions of the following Theorem, if $P_{1}$ and $P_{2}$ are partitions of $[a, b]$ then $L(f, P_{1}) \leq U(f, P_{2})$, and I would like to use this result to directly prove the theorem.

  Theorem: Suppose that the functions $f : [a, b] \rightarrow \mathbb{R}$ and $g : [a, b] \rightarrow \mathbb{R}$ are integrable and
  that

  $$f(x) \leq g(x) $$

  for all $x$ in $[a, b]$. Then,

  $$\int_{a}^{b} f \leq \int_{a}^{b} g.$$

So, I recognize the above Theorem as the monotonicity of the integral. Also, since $f$ is integrable, I tried doing some things with the Archimedes-Riemann theorem, but I didn't get anywhere. I've tried for a couple hours now, but still cannot progress. Maybe the refinement lemma can be used here to construct a second partition and proceed from there. I think the ultimate goal here is to relate the lower and upper Darboux sums to the integral, which I cannot seem to do. Any help is appreciated.

My attempt:

Since $f$ is integrable, the value of $\int_{a}^{b} f$ exists. Also, for any partition $P$, $L(f, P) \leq \int_{a}^{b} \leq U(f, P)$, which means that $L(f, P_{1}) \leq U(f, P_{2})$.

Now, there are a sequence of partitions $\{P_{n}\}$ for $f$ on $[a, b]$ such that $\lim_{n\to \infty} L(f, P_{n}) = \int_{a}^{b} f$, and there is a sequence of partitions $\{Q_{n}\}$ for $g$ on $[a, b]$ such that $\lim_{n\to \infty} U(g, Q_{n}) = \int_{a}^{b} g.$

But since $L(f, P_{1}) \leq U(f, P_{2})$, we have $L(f, P_{n}) \leq U(f, Q_{n})$ for every $n$. So,  $\lim_{n \to \infty} L(f, P_{n}) \leq \lim_{n\to\infty} U(f, Q_{n})$.

Finally $\int_{a}^{b} f = \lim_{n\to\infty} L(f, P_{n}) \leq \lim_{n\to \infty} U(f, Q_{n}) \leq \lim_{n\to\infty} U(g, Q_{n}) = \int_{a}^{b} g$, as desired.

The comparison of upper and lower Darboux sums for a given function over two partitions can be done by using the union of two partitions and you should try to prove it.

The monotone nature of the integral on the other hand is equivalent to the fact that if a function is non-negative then so is its integral. Why? Because any lower Darboux sum is non-negative and the integral is supremum of all lower sums so that it must also be non-negative. I don't think you need to compare upper and lower Darboux sums here.

Your proof is unnecessarily long. Just take any sequence $P_n$ of partitions such that $\operatorname {gap} P_n\to 0$ as $n\to\infty$ and then $$\int_{a} ^{b} f(x) \, dx=\lim_{n\to\infty} L(f, P_n) \leq \lim_{n\to\infty} L(g, P_n) =\int_{a} ^{b} g(x) \, dx$$ A Riemann integral (if it exists) is the limit of upper as well as lower Darboux sums and hence it is sufficient here to consider only one type of sum either upper or lower but no need to deal with both.

Consider the union $P:=P_1\cup P_2$ of the two partitions. Then $P$ is finer than both $P_1$ and $P_2$. It follows that
$$L(f,P_1)\leq L(f,P)\leq U(f,P)\leq U(f,P_2)\>,$$
as desired in the first sentence of your question.


Pure Mathematics proof for $(-a)b$ =$-(ab)$ How can I prove this without using the Multiplicative Property of zero? $(a*0=0)$

$$\begin{array}{rll}
(-a)b
&= (-a)b + 0 &0 \text{ is identity for addition}\\
&= (-a)b + (ab + (-ab))&-ab \text{ is additive inverse of } ab\\
&= ((-a)b+ab) + (-ab)&\text{addition is associative}\\
&= ((-a+a)b) + (-ab)&\text{multiplication is distributive}\\
&= \color{blue}{(0b)} + (-ab)&-a \text{ is additive inverse of } a\\
&= \color{blue}{(0b + 0)} + (-ab)&0 \text{ is identity for addition}\\
&= \color{blue}{(0b + (0b + (-0b)))} + (-ab)& -0b \text{ is addition inverse of } 0b\\
&= \color{blue}{((0b + 0b) + (-0b))} + (-ab)& \text{addition is associative}\\
&= \color{blue}{((0+0)b + (-0b))} + (-ab)&\text{multiplication is distributive}\\
&= \color{blue}{(0b + (-0b))} + (-ab)&0 \text{ is identity for addition}\\
&= \color{blue}{0} + (-ab)&-0b \text{ is additive inverse of } 0b\\
&= -ab &0 \text{ is identity for addition}
\end{array}$$

Please note that the part in $\color{blue}{\text{blue}}$ is a sub-proof of the statement $0b = 0$.

by Munna shigri

a(-b)=-(ab)

proof:

0=0

b+(-b)=0.   {b+(-b)=0}

a{b+(-b)}=a0.      {multiplying by (a )on both side}

ab+a(-b)=a0.       {by distributive property}

ab+a(-b)=0.         {a0=0}

{ab+a(-b)}+(-(ab)=0+(-ab). { by adding (-ab) on both side}

{a(-b)+ab}+(-ab)=(-ab)+0.    {by comutative property}

a(-b)+{ab+(-ab)}=(-ab)+0.    {by associative property}

a(-b)+0=(-ab)+0.                    {a+(-a)=0}

a(-b)=-ab.     proved                             {a+0=a}


$0^0$ is undefined, but sometimes defined as $1$? When typing a calculation to Wolfram Alpha it always takes $0^0$ as undefined, but for some calculations I need to preform, I specifically need it to be equal $1$.

Is there a way to make it understand $0^0$ as $1$, or is there at least a workaround for this? (cases of $0^n$ where $n>0$ should still be $0$)

To me $0^0$ is very nicely defined as the cardinality of the set of maps from the empty set to the empty set, hence equals $1$. However, in the context of limits one must be aware that the binary operation of exponentiation is not continuous at $(0,0)$ (and not even defined in an open neighbourhood of $(0,0)$), which implies that $a_n\to 0$, $b_n\to 0$ gives us no idea what might happen to $a_n^{b_n}$ as $n\to\infty$. Therefore $0^0$ is called an indeterminate form, just like $\frac 00$; however the latter is not only an indeterminate form but also undefined. As this is the most-encountered specimen of indeterminate form, it is not surprising that confusing "indeterminate" and "undefined" is wide-spread.

Note that "indeterminate form" is really about the unevaluated expression $0^0$. Normally, if two things are equal, they have the same properties, so if $0^0$ is indeterminate and $0^0=1$, we conclude that $1$ is indeterminate - which it is of course not. The important detail is that we are not talking about the value of $0^0$ as being indeterminate, but rather the "syntactic" expression (and that's why those things are called indeterminate forms, not indeterminate values)

To clarify, we say that the ("syntactic") form $a\circ b$ is an indeterminate form if $a_n\to a$ and $b_n\to b$ does not allow conclusions about the existence or value of $\lim_{n\to\infty}a_n\circ b_n$.

Thus writing something like
$$ \lim_{n\to\infty}a_n^{b_n}=0^0=1$$
is probably wrong because $a_n\to 0$, $b_n\to 0$ does not warrant the first equality. In contrast, writing
$$ \lim_{n\to\infty}\frac{a_n}{b_n}=\frac00=?$$
is definitely wrong because you cannot equate something with an undefined expression.

Similarly to $0^0=1$, it is customary in some branches of math to define $0\cdot \infty=0$. But being defined does not mend the fact that $0\cdot \infty$ is an indeterminate form in the above sense.

As they probably introduce exponentiation on the natural numbers in most schools, for any natural number $n$, we have:

$n^2 = n\times n$
$n^3 = n\times n \times n$
$n^4 =  n\times n \times n\times n$
$\vdots$
$n^{m+1} = n^m\times n\space\space\space$ for $m\ge 2$
$\vdots$

Now, if we drop the restriction that $m\ge 2$, it turns out that this scheme can be used to determine $n^m$ for all natural numbers $n$ and $m$, except for $n=m=0$. In fact, any value will work for $0^0$ in this scheme.
You can pick any convenient value for $0^0$. The values $0$ or $1$ result in some nice outcomes. Or you can leave it undefined as they probably taught you in school.


The sequence $A_n=\prod_{k=1}^n\left(1+\frac{k}{n^2}\right)$ is decreasing Let $A$ be the sequence of real numbers defined by :
$$\forall n\in\mathbb{N}^\star,\,A_n=\prod_{k=1}^n\left(1+\frac{k}{n^2}\right)$$
I know how to prove that this sequence converges to $\sqrt e$, using the following inequalities :
$$\forall t>0,\,t-\frac{t^2}2\leqslant\ln(1+t)\leqslant t$$
I found numerical evidence that $(A_n)$ is decreasing, but wasn't able to prove it. Any help will be appreciated.

A few observations to begin with

Pr1. $1+x\leq e^x$ and $1-x\leq e^{-x}$ for $\forall x$.

These inequalities are well known.

Pr2. $$\sum\limits_{k=1}^n \frac{1}{n^2+k} < \int\limits_{n^2+1}^{n^2+n+1}\frac{1}{x}dx=\log{\left(1+\frac{n}{n^2+1}\right)}$$

Easy to show using Riemann_sum.

Pr3. Function $f(x)=\log{\left(1+\frac{x}{x^2+1}\right)}-\frac{2x^2-1}{x^2(2x+1)} \leq 0$ for $\forall x\geq2$.

Easy to check from $f(2)\approx -0.013527763 <0$ and
$\lim\limits_{x\to\ +\infty}f(x)=0$. Then
$$f'(x)= \frac{(5x^5-9x^3-10x^2-8x-2)}{x^3(2x+1)^2(x^2+1)(x^2+x+1)} >0, \forall x\geq2$$
which means $f(x)$ is ascending for $\forall x\geq2$, i.e. it ascends to $0$ from $f(2)<0$. If we assume $\exists x_0>2: f(x_0)>0$, then (because $f$ is ascending) $f(x)\geq f(x_0)>0$ for $\forall x\geq x_0>2$. This contradicts the fact that $\lim\limits_{x\to\ +\infty}f(x)=0$.

Pr4. $\frac{n^2}{(n+1)^2}+\frac{n^2}{(n+1)^2}\cdot\frac{2n+1}{n^2+k} = 1-\frac{k(2n+1)}{(n^2+k)(n+1)^2}$

From
$$\frac{n^2}{(n+1)^2}+\frac{n^2}{(n+1)^2}\cdot\frac{2n+1}{n^2+k} - 1 =\\
\frac{n^2(n^2+k)+n^2(2n+1)-(n^2+k)(n+1)^2}{(n^2+k)(n+1)^2}=
-\frac{k(2n+1)}{(n^2+k)(n+1)^2}$$

Now, let's rewrite $A_n$ as
$$A_n=\prod\limits_{k=1}^n\left(1+\frac{k}{n^2}\right)=
\frac{1}{n^{2n}}\prod\limits_{k=1}^n(n^2+k)$$
And check
$$\frac{A_{n+1}}{A_n}=\frac{\frac{1}{(n+1)^{2(n+1)}}\prod\limits_{k=1}^{n+1}((n+1)^2+k)}{\frac{1}{n^{2n}}\prod\limits_{k=1}^n(n^2+k)}=\\
\frac{n^{2n}}{(n+1)^{2(n+1)}}\cdot((n+1)^2+n+1)\cdot\prod\limits_{k=1}^n\frac{(n+1)^2+k}{n^2+k}=\\
\left(\frac{n}{n+1}\right)^{2n}\cdot\left(1+\frac{1}{n+1}\right)\cdot\prod\limits_{k=1}^n\left(1+\frac{2n+1}{n^2+k}\right)=\\
\left(1+\frac{1}{n+1}\right)\cdot\prod\limits_{k=1}^n\left(\frac{n^2}{(n+1)^2}+\frac{n^2}{(n+1)^2}\cdot\frac{2n+1}{n^2+k}\right)\overset{\color{red}{Pr4}}{=}\\
\left(1+\frac{1}{n+1}\right)\cdot\prod\limits_{k=1}^n\left(1-\frac{k(2n+1)}{(n^2+k)(n+1)^2}\right) \overset{\color{red}{Pr1}}{\leq} $$
$$e^{\frac{1}{n+1}} \cdot e^{-\sum\limits_{k=1}^n \frac{k(2n+1)}{(n^2+k)(n+1)^2}}=
e^{\color{blue}{\frac{1}{n+1}-\frac{2n+1}{(n+1)^2}\left(\sum\limits_{k=1}^n \frac{k}{n^2+k}\right)}} \tag{1}$$
Let's look at
$$\color{blue}{\frac{1}{n+1} - \frac{2n+1}{(n+1)^2}\left(\sum\limits_{k=1}^n \frac{k}{n^2+k}\right)} < 0 \iff \\
n+1 - (2n+1)\left(\sum\limits_{k=1}^n \frac{k}{n^2+k}\right) <0 \iff \\
\sum\limits_{k=1}^n \frac{k}{n^2+k} > \frac{n+1}{2n+1}=1-\frac{n}{2n+1} \iff \\
\sum\limits_{k=1}^n \left(1-\frac{n^2}{n^2+k}\right) > 1-\frac{n}{2n+1} \iff \\
-\sum\limits_{k=1}^n \frac{n^2}{n^2+k} > 1-\frac{n}{2n+1}-n=-\frac{2n^2-1}{2n+1} \iff $$
$$\sum\limits_{k=1}^n \frac{1}{n^2+k} < \frac{2n^2-1}{n^2(2n+1)} \tag{2}$$
which is true for large enough $n$ since
$$\sum\limits_{k=1}^n \frac{1}{n^2+k} \overset{\color{red}{Pr2}}{<} \log{\left(1+\frac{n}{n^2+1}\right)} \overset{\color{red}{Pr3}}{\leq} \frac{2n^2-1}{n^2(2n+1)}$$
Returning back to $(1)$
$$\frac{A_{n+1}}{A_n} < e^0=1$$
for large enough $n$.


Krull dimension of the ring $K[t^{-1},t]$ Can anyone explain me why the Krull dimension of the ring $K[t^{-1},t]$, where $K$ is a field, is $1$?

I know I should show some efforts here, in MSE, but I actually couldn't find any clue about that.

Any hints or answer would be highly appreciated.

$K[t^{-1},t] = K[t]_t$, meaning $K[t]$ localized at $t$. Hence the prime ideals in $K[t^{-1},t]$ are in one-to-one correspondence with the prime ideals in $K[t]$ which avoid the multiplicative set $\{t^n; n \in \mathbb{N}\}$. It now follows easily that $\dim(K[t^{-1},t]) \leq \dim(K[t]) = 1$.

Also $0 \subset (t-1)$ is a chain of prime ideals in $K[t^{-1},t]$, so that $\dim(K[t^{-1},t]) \geq 1$. The conclusion follows.

For the irreducible polynomial $f(y,t) = yt-1$$\in$ $K[y,t]$ , Krull dimension of the quotient ring $K[y,t]/(f(y,t)$ is $1$ so from the isomorphism $ K[t^{-1},t] \rightarrow K[y,t]/(f(y,t)) $ , Krull dimension of the ring $ K[t^{-1},t]$ is $1$.


Simpson's rule to estimate distance traveled given velocity at certain points Problem:

A boat drives a steady course with a variable speed for 4 hours. The speed is registered at regular intervals in meters per second.

The registration shows $2.4, 4.4, 7.6, 8.4, 8.6, 7.9, 8.3, 8.7, 7.7, 6.5, 7.1, 6.7, 1.4$ (sorry, but I'm making a point out of this later).

Use Simpson's Rule to estimate how far the boat has traveled during the four hours, and its average speed.

Additional question; why can Simpson's Rule be used to estimate the boat's distance traveled?

My progress:

There are 13 recordings, which is an odd number, so that should be fine.

Using the Composite Simpson's Rule (with coefficients 1,4,2,4,2...,2,4,1) , I get $$\frac19\left[ 2.4 + 4(4.4) + 2(7.6) + 4(8.4) + \ldots + 4(6.7) + 1.4 \right] \approx 28.09$$ (unless I've made some careless mistake).

Here $\frac19$ comes from $\frac h3$ (from formula) where $h = \frac13$.

Now, as I understand it, this number would - since we're estimating the area under the "curve" (i.e. using Simpson to estimate a definite integral given points) - be a number for the distance traveled.

But since the numbers are given in $\frac ms$, wouldn't this $28.09m$ be an oddly low number? What am I missing here?

Any help appreciated!

I believe your $\frac{1}{9}$ is obtained by $\frac{1}{3}\cdot\frac{1}{3}$ where one of the $\frac{1}{3}$ is in hours. So instead you should multiply by $3600$, which is the number of seconds in an hour.

I have calculated the answer using simpson's one-third rule(the distance travelled by boat in 4 hours) 93.56307km.


If $f$ continuous at $[1,\infty)$ and $\int _1^\infty\,f\left(x\right)dx$ converge, then $\int _1^\infty\frac{f\left(x\right)}{x}dx$ also converge? I need to prove or disprove this statement.
I didn't find any counter example. So I tried to prove it.
I know that $\int _1^\infty\,f\left(x\right)dx$ converge so the limit

$\lim _{t\to \infty }\int _1^tf\left(x\right)dx$ exists

I also know that $f$ is coninuous , so there's a function $F\left(x\right)\:=\:\int _1^xf\left(s\right)ds$

Such that $F'(x) = f(x)$

That's all I know and I dont know how to put this together.

If $f$ is non-negative, then
$$
\int_1^x\frac{f(t)\,dt}{t}\le\int_1^x f(t)\,dt,
$$
and hence $\int_1^\infty\frac{f(t)\,dt}{t}$ converges.

In general, let $\,F(x)=\int_1^x f(t)\,dt$. Since $\lim_{x\to\infty} F(x)=a\in\mathbb R$, then $F$ is bounded. Say $\vert F(x)\rvert\le M$. Then
$$
\int_1^x\frac{f(t)\,dt}{t}=\frac{F(t)}{t}\,\Big|_1^x+\int_1^x \frac{F(t)\,dt}{t^2} \\
=\frac{F(x)}{x}+\int_1^x \frac{F(t)\,dt}{t^2}.
$$
Observe that
$$
\lim_{x\to\infty}\frac{F(x)}{x}=0
$$
while the limit
$$
\lim_{x\to\infty}\int_1^x \frac{F(t)\,dt}{t^2},
$$
exists, since $F$ is bounded.

Note. The fact that $\lim_{x\to\infty}\int_1^x f(t)\,dt$ exists does NOT imply that $f$ is integrable over $[1,\infty)$, i.e., $f\in L^1[1,\infty)$.

So you can use a comparison theorem which says that if $g(x) \leq h(x)$ and $\int_1^\infty h(x) \,dx$ converges then $\int_1^\infty g(x) \,dx$ also converges.  Since in your problem $x >1$ then $\frac{f(x)}{x} \leq f(x)$, you know that $\int_1^\infty \frac{f(x)}{x} \,dx$ converges if $\int_1^\infty f(x) \, dx$ converges.


different domains for each variable in predicate logic Before I begin, note that this is significantly outside what I've studied, so if it's a load of crap just let me know.
Up till now, everything I've covered assumes the same domain for all variables.
I'm trying to define an arbitrary one-to-one relationship between elements in two different sets, and want each of the variables x and y in the formula below to be in the domain of its particular set:

X = {X1, X2, X3}
Y = {Y1, Y2, Y3}

That is, I want:

x ∈ X, y ∈ Y

If possible, how would I go about specifying this?
Also, I've phrased the formula

∀x ∃y [ A(x, y) ∧ ¬A(¬x, y) ]

Where A(x, y) indicates an relationship between x and y. Is this statement well formed? Particularly the ¬x in ¬A(¬x, y)? And does it mean what I think it means - That A(x, y) for all x ∈ X except the x in question, and for the y in question, must be false in order for the statement ot be true? That is, does it define a one-to-one relationship?
Thanks a lot,
Wyatt

No, I don't think it makes sense to write $\neg x$ in this context.

The simplest way to specify that $x\in X$ and $y\in Y$ is to make them clauses in your statement. For example, to write that for every $x\in X$ there should exist a $y\in Y$ such that $A(x,y)$ holds, you would usually write:

  $$\forall x \left(x\in X\rightarrow \exists y(y\in Y\wedge A(x,y)\right).$$

For the second part: to say that no other $z\in X$ will also be associated to $y$, the easiest why is to say that anyone who is associated to $y$ is going to be equal to $x$; so you would want a clause $A(z,y)\rightarrow x=z$.

To put it all together:

  $$\forall x\left( x\in X\rightarrow \exists y(y\in Y\wedge A(x,y)\wedge \forall z(z\in X\wedge A(z,y)\rightarrow x=z))\right).$$

It is also possible to use common (but perhaps slightly abusive) notation and write

  $$\forall x\in X\Bigl(\exists y\in Y\Bigl(\forall z\in X\Bigl(A(x,y)\wedge (A(z,y)\rightarrow x=z)\Bigr)\Bigr).$$

Added: As noted in several comments, the above does not completely describe a one-to-one function; it only describes a relation with domain $X$, codomain $Y$, such that for each $y\in Y$ there is at most one $x\in X$ such that $A(x,y)$ holds; this does not fully describe a function (unless you happen to know some other things about $X$ and $Y$, e.g., that they are both finite and of equal size, in which case the other condition will follow), nor a bijective function. In general, the clause you want for "function-ness" is the dual of the condition you have: for all $x\in X$, and for all $y\in Y$, $A(x,y)\wedge A(x,y')\rightarrow y=y'$ (so each input has at most one output). For bijectivity (so you have a true correspondence) you also need that for all $y\in Y$ there exists $x\in X$ such that $A(x,y)$.

There are lots of ways to write a formula A(x,y) that defines a bijection between two sets. If you're trying to keep the depth of quantifiers to a minimum, you could use
$$
\begin{align}
&  (\forall x)(\exists y) [ x \in X \to y \in Y \land A(x,y)] \\
& {} \land (\forall y)(\exists x) [ y \in Y \to x \in X \land A(x,y)]\\
& {} \land (\forall x,x',y,y') ([x \in X \land y \in Y) \\
& \qquad {} \to [(A(x,y) \land A(x',y) \to x=x']
 \land [A(x,y) \land A(x,y') \to y = y'])]\\
\end{align}
$$

This can be written more elegantly with the abbreviations $(\forall x \in X) $ and $(\exists x \in X)$:
$$
\begin{align}
(\forall x \in X)\Phi &\equiv (\forall x) [ x \in X \to \Phi] \\
(\exists x \in X)\Phi &\equiv (\exists x) [ x \in X \land \Phi] \\
\end{align}
$$
like so:
$$
\begin{align}
&  (\forall x \in X)(\exists y\in Y) A(x,y) \land (\forall y \in Y )(\exists x \in X) A(x,y)\\
& {} \land (\forall x \in X)(\forall x' \in X)(\forall y \in Y)(\forall y' \in Y)([(A(x,y) \land A(x',y) \to x=x'] \\
& \qquad \qquad {} \land [A(x,y) \land A(x,y') \to y = y'])\\
\end{align}
$$


Confusion in fraction notation $$a_n = n\dfrac{n^2 + 5}{4}$$

In the above fraction series, for $n=3$ I think the answer should be $26/4$, while the answer in the answer book is $21/2$ (or $42/4$). I think the difference stems from how we treat the first $n$. In my understanding, the first number is a complete part and should be added to fraction, while the book treats it as part of fraction itself, thus multiplying it with $n^2+5$.
So, I just want to understand which convention is correct.

This is from problem 6 in exercise 9.1 on page 180 of the book Sequences and Series.

Here is the answer sheet from the book (answer 6, 3rd element):

$3,8,15,24,35$
$\dfrac{1}{2},\dfrac{2}{3},\dfrac{3}{4},\dfrac{4}{5},\dfrac{5}{6}$
$2, 4, 8, 16 \text{ and } 32$
$-\dfrac{1}{6},\dfrac{1}{6},\dfrac{1}{2},\dfrac{5}{6},\dfrac{7}{6}$
$25,-125,625,-3125,15625$
$\dfrac{3}{2},\dfrac{9}{2},\dfrac{21}{2},21,\dfrac{75}{2}$
$65, 93$
$\dfrac{49}{128}$
$729$
$\dfrac{360}{23}$
$3, 11, 35, 107, 323$; $3+11+35+107+323+...$
$-1,\dfrac{-1}{2},\dfrac{-1}{6},\dfrac{-1}{24},\dfrac{-1}{120}$; $-1+(\dfrac{-1}{2})+(\dfrac{-1}{6})+(\dfrac{-1}{24})+(d\frac{-1}{120})+...$
$2, 2, 1, 0, -1$; $2+2+1+0+(-1)+...$
$1,2,\dfrac{3}{5},\dfrac{8}{5}$

in elementary school math the fraction $x\frac{y}{z}$ usually means $x+\frac{y}{z}$ and is called a mixed fraction.

However these are almost never used after junior high.

Most of the time when you see $x\frac{y}{z}$ the two terms should be multiplied, so it is equal to $\frac{xy}{z}$.

Here's the Cliff Notes version.  Actually cut and pasted from Cliff Notes:
"•Two variables (letters) next to each other: ab means a times b"

This is always the case, even if the variable is identified as an integer.  If a=3 and b=4 then ab=12, NOT 34.  To understand why 3 1/4 is not 3 x 1/4 it is actually simple - it follows a different rule of notation.  when a number is in front of a fraction they are considered parts of a total number that consists of a whole and a fractional part.
One final example of notation.  Intuitively we know that 34 is thirty-four.  it's not 7, ie 3+4, and it's not 12, ie 3x4.  The notation is our common base 10 system: so 34 is just an abbreviation for (3x10) + (4x1).  Since we do it so often we don't think about it, it's intuitive and we don't look for exceptions.  That's how you should treat ab:  it's a times b always.


Prove that for all integers $n > 3$, $y_{n+1} = 2 x_n$ Let $x_n$ be the number of 0/1 strings of length $n$, not including the sequence 010. Let $y_n$ be the number of 0/1 strings of length $n$, not including 0011 or 1100. Prove that for all integers $n > 3$, $y_{n+1} = 2x_n$.

I don't know where to start with this one, especially with the exclusion of certain sequences.

Any help getting me on the right track would be appreciated. Thanks.

EDIT:It is difficult to write this formally, so I'm just going to give you a hint:

For $n\geq 4$, you can write $0/1$ strings of length $n$ in $2^n$ ways. To calculate $x_n$, we should calculate how many of those strings include $010$. You can show by induction that this number is number is $2^{n-1}-n$, so $x_n=2^n-(2^{n-1}-n)=2^{n-1}+n$ (alternatively, you can show this formula for $x_n$ by induction. Also, if you write all sequences of length $n$ for $n=3,4,5$ in a matrix, for example, in the lexicographical order, and then mark the sequences containing $010$, you get a nice pattern, which may help you understand where did this number come from:$$\begin{bmatrix}n=3 & n=4
\\000&0000
\\001&0001
\\010(X)&0010(X)
\\011&0011
\\100&0100(X)
\\101&0101(X)
\\110&0110
\\111&0111
\\&1000
\\&1001
\\&1010(X)
\\&1011
\\&1100
\\&1101
\\&1110
\\&1111
\end{bmatrix}$$
for $n=5$ the marks make something like $(X)-(X)(X)-(X)(X)(X)(X)-(X)-(X)(X)-(X)$, and so on).
Assuming the result of the exercise, we should have $y_n=2x_{n-1}=2^{n-1}+2n-2$, which you can also showto be valid by induction (making a similar table may help you again).


Continuous bijection from $\mathbb R-\{0\}$ to $\mathbb R$ I know such a function exists if continuity is not required.

How do you prove it doesn't exist when continuity is required?

For the sake of completeness,  note that a continuous bijection from $\mathbb R$ to $\mathbb R-{0}$ cannot exist either (the IVT applies trivially).

Assume $f: \mathbb{R}\setminus \{0\} \to \mathbb{R}$ is a continuous bijection. Denote $\mathbb{R}_{>0} = \{x \in \mathbb{R} \mid x > 0\}$ and $\mathbb{R}_{<0} = \{x \in \mathbb{R} \mid x < 0\}$.

Spaces $\mathbb{R}_{>0}$ and $\mathbb{R}_{<0}$ are connected and $f$ is continuous, therefore $f(\mathbb{R}_{>0})$ and $f(\mathbb{R}_{>0})$ are also connected. Since $f$ is a bijection, we see that $\mathbb{R}$ is a union of two disjoint non-empty connected subsets: $\mathbb{R} = f(\mathbb{R}_{<0}) \cup f(\mathbb{R}_{>0})$.
Connected subsets in $\mathbb{R}$ are precisely the same thing as intervals. It's easy to see then that one of the two sets $f(\mathbb{R}_{<0})$ and $f(\mathbb{R}_{>0})$ must be unbounded on one side and closed on the other side. For instance, let's assume that $f(\mathbb{R}_{>0}) = [y, +\infty)$, where $y \in \mathbb{R}$. (The other possible cases are analogous).
Let $x = f^{-1}(y)$. The restricted function $f|_{\mathbb{R}_{>0}}$ reaches its minimum at point $x > 0$. It follows that there are $x_1$ and $x_2$ somewhere close to $x$ such that $0 < x_1 < x < x_2$, $f(x_1) > f(x)$ and $f(x_2) > f(x)$. Now it's easy to reach a contradiction with the injectivity of $f$ using the intermediate value theorem. Done.

The topology is different. $\mathbb{R}=(-\infty,\infty)$ has two open ends and $\mathbb{R}-\{0\}=(-\infty,0)\cup(0,\infty)$ has four. Continuous functions map boundaries to boundaries, which cannot happen bijectively. For instance, $\log |x|$ is an example that maps two-to-one.

Continuous bijection only exists between spaces with equal topology.


Prove that $\lim_{n\to \infty} x_n = 0$ if $\lvert f(x) - f(y) \rvert \leq \frac{1}{2} \lvert x - y \rvert$ , $f(0) = 0$ and $x_n = f(x_{n-1})$ Suppose $f: \mathbb{R} \to \mathbb{R}$ satisfies $\lvert f(x) - f(y) \rvert \leq \frac{1}{2} \lvert x - y \rvert$ for all $x,y \in \mathbb{R}$, and $f(0) = 0$. Let $x_0 \in \mathbb{R} $ be arbitrary. Define $x_1 = f(x_0)$, $x_2 = f(x_1)$, etc. so that $x_n = f(x_{n-1})$ for all $n\in \mathbb{Z^+}$. Prove that $\lim_{n\to \infty} x_n = 0$.

I am aware that one of this means that the function $f$ is Lipschitz which implies that it is uniformly continuous.

This is my attempt at the proof:
Suppose $\varepsilon > 0$.
We have
$$\lvert x_n - 0 \rvert = \lvert f(x_{n-1}) - f(0) \rvert \leq \frac{1}{2} \lvert x_{n-1} \rvert.
$$

From uniform continuity, we can choose $\delta >0$ such that $\lvert x - y \rvert < \delta$ implies that $\lvert f(x) - f(y) \rvert < \varepsilon$.
I want to somehow use this information to find an $N\in \mathbb{Z^+}$ such that $\vert x_n - 0 \rvert < \varepsilon$ whenever $n\geq N$ but I'm stuck on how to finish.

Why not iterate what you have? $$|x_n| \leq \frac{1}{2}|x_{n-1}| \leq \frac{1}{2^2} |x_{n-2}| \leq \dots \leq \frac{1}{2^n} |x_0|$$  Note $\frac{1}{2^n} \to 0$ as $n \to \infty.$

This is more or less the proof of the Banach Fixed Point Theorem, but I'll include the argument here. Notice that $$\lvert x_{n+1} - x_{n} \rvert = \lvert f(x_{n}) - f(x_{n-1}) \rvert \le \frac 1 2 \lvert x_{n} - x_{n-1} \rvert $$ and since this holds for all $n \in \mathbb Z^+$, we can use induction to see that $$\lvert x_{n+1} - x_{n} \rvert \le \frac{1}{2^n} \lvert f(x_1) - f(x_0) \rvert = \frac C {2^n}$$ where $C = \lvert f(x_1) - f(x_0) \rvert$ is a constant. Now fix $N \in \mathbb Z^+$, And note that for $m > n > N$, \begin{align*} \lvert x_m - x_n\rvert &= \lvert (x_m - x_{m-1}) - (x_{m-1} - x_{m-2}) - \cdots - (x_{n+1} - x_n) \rvert \\ &\le \sum^{m-1}_{k=n} \lvert x_{k+1} - x_{k} \rvert\\
&\le \sum^{m-1}_{k=n} \frac{C}{2^k} \le \sum^\infty_{k=N} \frac{C}{2^k} = \frac{C(1/2)^N}{1-(1/2)} \to 0\,\,\,\, \text{ as } N\to \infty.
\end{align*}  This shows that $\{x_n\}$ is a Cauchy sequence and thus converges to some $x^* \in \mathbb R$. Now taking the limit on both sides of the equation and using continuity of $f$, we see that $$x_{n+1} = f(x_n) \,\,\,\,\, \implies \,\,\,\,\, x^* = f(x^*).$$ That is $x^*$ is a fixed point of $f$.

Now suppose that for $x,y \in \mathbb R$, $$f(x) = x \,\,\,\,\,\, \text{ and } \,\,\,\,\,\, f(y) = y.$$ Then $$0 \le \lvert x - y \rvert = \lvert f(x) - f(y) \rvert \le \frac 1 2 \lvert x - y \rvert$$ which shows that $\lvert x - y \rvert = 0$ so $x=y$. Thus $f$ has only one fixed point and so $f(x^*) = x^*$ implies that $x^* = 0$. Thus $x_{n}\to 0.$


Does $2 \log(-1) = 0$? It occurred to me, following the rule of logarithms, that $a\log(b)=\log(b^a)$

Thus $2 \log(-1) = \log((-1)^2) = \log(1) = 0$

In complex analysis the answer is no.  You do not have $\ln(a^b)=b\ln(a)$, it simply isn't an identity when we move beyond positive real arguments.

Things are different, however, in $p$-adics.  Select base 2 and consider the Maclaurin series

$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}...$

Plug in $x=-2$ and convert all terms to $2$-adics; the series converges $2$-adically ... to $...000000 = 0$!  We indeed have $2\ln(-1)=2×0=0$ in $2$-adics.

It's an example of how $p$-adics logarithms differ from what we are used to seeing in real or complex analysis.  In the case of $2$-adics, if we want to have the one-to-one and "inverse of exponential" properties that often make logarithms useful we have to limit the domain to arguments $\equiv 1 \bmod 4$.

As far as I know you can assign as input to any lnx function only numbers greater than 0, meaning that the following must happen: x>0. This is very obvious, cause there cannot exist a number such that when it is raised to the e to be equal to -1. So this 2ln(-1) cannot be defined. The same happens to any function of the form logax.


Is my six sided dice weighted? I was playing a game with my son, and it seemed that we rolled two or five about half the time. He was wondering if the die was weighted.

So I re-rolled it a bunch of times and jotted down the results:

1: 4 times
2: 12 times
3: 13 times
4: 4 times
5: 13 times
6: 8 times

I realise that this is a small sample (only 54 datapoints), but to my very untrained eye, the 1 and 4 seem very low. I was wondering how I would go about determining the probability that the die is fair. I was expecting about 9 each. I did a ChiTest in excel, which I am not sure how to calculate manually and got 0.069

Does this mean that there's a 6.9% chance that the die is unweighted, and therefore a 93.1% chance it's weighted, or have I misinterpreted the results entirely?

This is not the right interpretation of the $0.069$ number. Indeed, this wrong interpretation is extremely common, and people from doctors to judges to scientists to policy makers routinely get this wrong (and go on to make poor decisions as a result).

The $0.069$ number means this: if your die is fair, then the probability of getting a result at least as extreme as the one you observed is $0.069$. In other words, if you do sets of 54 trials over and over again, you'll see distribution as extreme or more in 6.9% of the trials.

You can use this informtaion however you want to make deductions about whether the die is fair or unfair. However, it makes no sense to talk about "the probability that the die is fair" unless we have an underlying space of all possible events as reference (all dice in the world, for example, or all ways of manufacturing dice). The result from a chi-square test simply does not provide that type of information.

I would not consider your die as weighted, we can see this as a binomial distribution problem
$$P(X=r)={{n}\choose{r}}P^{r}(1-P)^{n-r}$$

This basically models your case, $r$ is the number of succeses, $n$ would be the number of trials and $P$ the probability of succes. Replacing and computing for rolling number 1 four times in 54 trials

$$P(X=4)\approx0.0268$$

Which would be 2.68%, since it is not incredibly low, considering all the posibilities (one 1, two 1s,..., 54 ones) I would guess it is not. But you could always apply the large number rule, meaning the more tries, the more the real possibility asimilates to the theorital possibility.