TITLE: Variable mass systems
QUESTION [0 upvotes]: I'm studying this article:
https://ocw.mit.edu/courses/aeronautics-and-astronautics/16-07-dynamics-fall-2009/lecture-notes/MIT16_07F09_Lec14.pdf
and I have the following doubt:
On page 3, when he derives the equation of motion (11), he says that the external forces $\mathbf{F}$ acts only on $m$, and not also on $dm$ (see equation (9)). Why? If we think that $\mathbf{F}$ is the gravity force (as he does in the following pages in the Rocket example), then why $dm$ does not feel gravity?

REPLY [0 votes]: The reaction force on the rocket engine depends on the change in momentum of dm which is caused by the engine, but not that caused by gravity.


TITLE: Writing an iterated double integral in two forms
QUESTION [2 upvotes]: Let $f\left(x,\:y\right)\:=\:x^2e^{x^2}$ and let $R$ be the triangle bounded by the lines $x=5$, $x=y/2$, and $y=x$ in the $xy$-plane.
Express $\int _RfdA$ in two different ways.
After sketching the region, I got that the first way to write the integral would simply be:
$\int _0^5\int _x^{2x}\:x^2e^{x^2}dydx$
However, I was stuck on how to write it in the other way. With the region, there is no obvious way to write the integral as one in terms of $y$. I feel that this may mean that I need to have the sum of two different double integrals but I am not entirely sure how that would work in this case.
Would I have to subtract the higher $x=(1/2)y$ line from the $x=y$ line or is there some other way?
Any help would be highly appreciated!

REPLY [3 votes]: You are correct that the other expression has to be written as a sum of two integrals.  Sketching the region, we see that we can split the region into an upper and lower section.  The lower section is from $y = 0$ to $y = 5$, and the upper is from $y = 5$ to $y = 10$.  In this lower section, $x$ will vary from $y/2$ to $y$, and in the upper section, $x$ will vary from $y/2$ to $5$, since here $y > 5$ and the boundary becomes the vertical line $x = 5$.   Writing out the integral, we have
$$
\int_{0}^{5}\int_{y/2}^{y} x^2 e^{x^2} dx dy + \int_{5}^{10} \int_{y/2}^{5} x^2 e^{x^2} dx dy .
$$


TITLE: What is the reason behind the name 3n-display?
QUESTION [4 upvotes]: In the paper "The display of a formal $p$-divisible group" Zink defines some objects and calls them $3n$-display. A $3n$-display over $R$ is a quadruple $P$, $Q$, $F$, $F^1$ such that $P$ is a projective $W(R)$ module, $Q$ is a submodule and $F:P\to P$, $F^1 :Q\to P$ are functions satisfying some conditions. I don't know why these things are called $3n$-display and I don't see any relation to $3$ or $n$ in the definition. Does someone know the reason behind this name?

REPLY [6 votes]: I have not found a source where Thomas Zink explains the name, however arXiv:1906.00899 explains it as an abbreviation of “not-necessarily-nilpotent” (or $3n$-) displays. See also Travaux de Zink, page 343.


TITLE: Was it necessary to have $g(x)\neq4$ and $x\neq4$ in the assumptions of this problem?
QUESTION [1 upvotes]: Suppose $f'(4)=3 , g'(4)=7,g(4)=4,g(x)=4,g(x)\neq4$ for $x\neq4$ find:
$1)$$\lim_{x\to4}\frac{f(x)-f(4)}{g(x)-4}\quad\quad\quad\quad\quad\quad\quad\quad2)\lim_{x\to4}\frac{f(g(x))-f(4)}{x-4}$

I know how to solve this problem ( we can do it by rewriting first one as $\lim_{x\to4}\dfrac{{\frac{f(x)-f(4)}{x-4}} }{\frac{g(x)-g(4)}{x-4}}$ and second one as $\lim_{x\to4}\dfrac{f(g(x))-f(g(4))}{g(x)-g(4)}\times\dfrac{g(x)-g(4)}{x-4}$ and using defenition of derivative.)
My question : Was it necessary to say $g(x)\neq4$ and finding the values for $x\neq4$? I also don't understand how $g(x)\neq4$ if we have $g(x)=4$ in the assumptions.

REPLY [1 votes]: For the first limit to exist, it's necessary that there exists a small interval $(4-\delta,4+\delta)$ in which $g(x)\ne4$ except for $g(4)=4$ itself. (If there is no such interval, then the quotient is undefined at a sequence of points tending to $x=4$, which means the limit cannot be defined.)
Rather than bother saying the bit about the interval $(4-\delta,4+\delta)$, I believe the writer chose a simpler hypothesis, namely that $g(x)\ne4$ for $x\ne4$. This is stronger than is really required (as discussed above), but simpler to say and to deal with. But the real point is to just ensure that the quotient itself is defined enough to make the limit make sense in the first place.


TITLE: Is there an operational isomorphism from $(\mathbb{Z},+)$ to $(\mathbb{Q}^{+},\cdot)$?
QUESTION [3 upvotes]: Let $\left(\mathbb{Z},+\right)$ and $\left(\mathbb{Q}^{+},\cdot\right)$ be groups (let the integers be a group with respect to addition and the positive rationals be a group with respect to multiplication). Is there a function $\phi\colon\mathbb{Z}\mapsto\mathbb{Q}^{+}$ such that:

$\phi(a)=\phi(b) \implies a=b$ (injection)
$\forall p\in\mathbb{Q}^{+} : \exists a\in\mathbb{Z} : \phi(a)=p$ (surjection)
$\phi(a+b) = \phi(a)\cdot\phi(b)$ (homomorphism)

? If so, provide an example. If not, disprove.

REPLY [3 votes]: Every nontrivial subgroup of $\mathbb{Z}$ is infinite cyclic. But, $\mathbb{Q}^+$ contains a rank 2 abelian subgroup, namely all rational numbers of the form $2^m 3^n$ for integer values of $m,n$.


TITLE: Proof that chromatic number is $< 9$
QUESTION [1 upvotes]: Let $G= (V,E)$ be a graph such that:
$E = E_1 \cup E_2$
$G_1 = (V, E_1)$ is planar
$G_2 = (V, E_2)$ is forest
 Proof that chromatic number is $< 9$
My observations
Each planar graph has chromatic number $\le 4$. Also each forest has chromatic number $2$ if $|V| > 1$
But how can I connect that and have convince about chromatic number $<9$?

REPLY [1 votes]: Given colorings $C_1$ of $G_1$ and $C_2$ of $G_2$, construct a new coloring $C_1 \times C_2$ where  $C_1 \times C_2(v) = ( C_1(v),C_2(v))$. This is a valid coloring of the graph (check this!). Since there are at most 4 colors in $C_1$ and at most 2 in $C_2$, there are at most 8 in  $C_1 \times C_2$.


TITLE: The probability for inner product for two unit sphere uniformly distributed random vectors $|u\cdot v|<\alpha$
QUESTION [1 upvotes]: Following this question Why do we have that $u\cdot v$ converges weakly to a standard Gaussian random variables as $n\to \infty$?.
Let $u$ and $v$ be two random vectors on $R^n$ that are independent and uniformly distributed on the unit sphere. That means we can represent it as Gaussian random vectors $g\sim N(0, I_n)$, $$u=\frac{g}{\|g\|_2}.
$$
Here we know that $u\cdot v$ is a Gaussian with mean 0 and width $n^{-1/2}$.
Let $X=u\cdot v$ which is a Gaussian with mean $0$ and width $n^{-1/2}$.
Why do we have $$P(|X|\le \alpha)=\alpha \sqrt{n}$$ for $0<\alpha<n^{-1/2}$? Does this result coincide with the concentration inequality of the Gaussian $X$?

I can just get the asymptotic bound that means $u\cdot v=O_p(n^{-1/2})$. For non-asymptotic upper bound, we have this answer: Can we get the concentration inequality of the inner product of two unit vectors distributed on the sphere?
$$
P(|u\cdot v-E[u\cdot v]|\ge t)\le e^{-t^2/2}
$$
Since we have $E[u]=-E[u]$, then $E[u\cdot v]=E[(-u)\cdot v]$. Hence, $E[u\cdot v]=0$.
So $P(|u\cdot v|\ge t)\le e^{-t^2/2}$.
But how to show that $P(|u\cdot v|\le t)=t\sqrt{n}$?

REPLY [1 votes]: As discussed, the question here relates to this question.
Fix $n\in\mathbb Z_{>0}$. Let $X_1,X_2\sim\mathcal N(0,I_n)$ be independent standard normal, let $Z_i=\|X_i\|_2$ and $X^\circ_i=\frac{1}{Z_i}X_i$ for $i\in\{1,2\}$. We know the following distributions.

The $X_{1,1},\dots,X_{1,n},X_{2,1},\dots,X_{2,n}$ are i.i.d. standard normal.
This suggests that $Z_i^2=\sum_{j=1}^nX_{i,j}^2$ is Chi-squared with $n$ degrees of freedom.
The inner product $X_1\cdot X_2=\sum_{j=1}^nX_{1,j}X_{2,j}=X^{\mathrm{T}}MX$ follows a generalized Chi-squared distribution, where $X=(X_{1,1},\dots,X_{1,n},X_{2,1},\dots,X_{2,n})\sim\mathcal N(0,I_{2n})$ (due to independence) and $M=\begin{pmatrix}0 & I_n\\ 0 & 0\end{pmatrix}$.
The normalized vectors $X^\circ_1,X^\circ_2$ are i.i.d. uniform on the unit sphere.

What we do not know, is the distribution of $X^\circ_1\cdot X^\circ_2$, in particular it is not Gaussian (remember that $|X^\circ_1\cdot X^\circ_2|\le 1$ almost surely, this is certainly not normally distributed).
So, what do we know about $X^\circ_1\cdot X^\circ_2$?

This post suggests that $\mathbb P(|X^\circ_1\cdot X^\circ_2|\ge\sqrt{\ln(n)/n})\le a\exp(-b\ln(n))=an^{-b}=o(1)$, meaning that $|X^\circ_1\cdot X^\circ_2|$ converges to $0$ in probability. Let $V=\sqrt{n}X^\circ_1\cdot X^\circ_2\in[-\sqrt{n},\sqrt{n}]$, and $\delta(n)=o(1)$. Then this result also suggests that $\lim_{n\rightarrow 0}\mathbb P(\delta(n)V\ge\varepsilon)=0$ for all $\varepsilon$, so $\delta(n)V$ converges to $0$ in probability. One would say that $c(n)=\sqrt{n}$ is the appropriate scaling for $X^\circ_1\cdot X^\circ_2$. If you choose a scaling constant $c(n)=o(\sqrt{n})$, then the limiting distribution is trivial, meaning a one-point mass on $0$. If you choose $c(n)=\omega(\sqrt{n})$, then the limiting distribution is not defined (one could show that the limit of any bounded set would have probability $0$ in the limit).
This post suggests that $V$ converges to $W\sim\mathcal N(0,1)$ in distribution. So, what we do not know is the distribution of $V$. What we do know, is that for every measurable $\mathcal E\subseteq\mathbb R$ we have $\lim_{n\rightarrow\infty}\mathbb P(V\in\mathcal E)=\mathbb P(W\in\mathcal E)$.

Since we do not know the distribution of $V$ and $X^\circ_1\cdot X^\circ_2$, there is little hope to obtain probabilities of events, say $\mathbb P(|X^\circ_1\cdot X^\circ_2|\le\alpha)$. What we have understood, however, are the asymptotics. I mean, we have just said that we know that $\lim_{n\rightarrow\infty}\mathbb P(|V|\le\alpha)=\mathbb P(|W|\le\alpha)$, for all $\alpha\in\mathbb R_{\ge 0}$.
At this point, we really have to pay attention.

Taking limits: We know that if we fix $\alpha$ and then take the limit, we have $\lim_{n\rightarrow\infty}\mathbb P(|V|\le\alpha)=\mathbb P(|W|\le\alpha)$. This means that $\alpha$ cannot depend on $n$.
We have to remember the scaling. This means we have $\lim_{n\rightarrow\infty}\mathbb P(|X^\circ_1\cdot X^\circ_2|\le\frac{\alpha}{\sqrt{n}})=\mathbb P(|W|\le\alpha)$.

The answer in this post addresses a random variable $Y\sim\mathcal N(0,1/n)$. How does this fit here?

We have $\sqrt{n}Y\sim\mathcal N(0,1)$ and hence $\sqrt{n}Y$ and $W$ have the same law.
One might be tempted to compare $X^\circ_1\cdot X^\circ_2$ and $Y$, since $V=\sqrt{n}X^\circ_1\cdot X^\circ_2$ converges in distribution to $\sqrt{n}Y$. However, convergence in distribution is very weak (hence the name weak convergence), and we need to achieve significantly more to justify a comparison of $X^\circ_1\cdot X^\circ_2$ and $Y$.

This explains that we cannot (yet reliably) say that $X=u\cdot v$ has the law $\mathcal N(0,1/n)$, not even in an approximate sense. If you want to take a closer look at the finite size behavior of $X^\circ_1\cdot X^\circ_2$, I'd be happy to, but maybe in a new question. In the remainder of this answer I will focus on $Y\sim\mathcal N(0,1/n)$ and $\mathbb P(|Y|\le\alpha)$, since this was the actual question.
So, let $\alpha\in\mathbb R$. Recall from above that we have $\sqrt{n}Y\sim\mathcal N(0,1)$, so $\sqrt{n}Y$ and $W$ have the same law and we can write $\mathbb P(|Y|\le\alpha)=\mathbb P(|W|\le\sqrt{n}\alpha)$.
The question is why we have $\mathbb P(|W|\le\sqrt{n}\alpha)=\sqrt{n}\alpha$, the answer is: we don't.
We have
$$\mathbb P(|W|\le\sqrt{n}\alpha)=\int_{-\sqrt{n}\alpha}^{\sqrt{n}\alpha}\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}\mathrm{d}x.$$
Some people would write $\mathbb P(|W|\le\sqrt{n}\alpha)=\Phi(\sqrt{n}\alpha)-\Phi(-\sqrt{n}\alpha)=1-2\Phi(-\sqrt{n}\alpha)$, where $\Phi$ denotes the cumulative distribution function of $\mathcal N(0,1)$. There's no better way to write this.
Now, let's turn to the claim. In fact, we can get somewhat decent approximations. Let's use the shorthand $\phi(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}$. As explained above, we want to know how $\mathbb P(|W|\le\sqrt{n}\alpha)$ and $\sqrt{n}\alpha$ are related for $0\le\alpha\le 1/\sqrt{n}$. So, let's do ourselves a favor first and just consider
$\mathbb P(|W|\le a)$ and $a$ for $0\le a\le 1$, which is equivalent (to give some background, we look at the probability to be within an interval of at most one standard deviation from the mean, that's how it was intended in the answer in the first place). Using that $\phi$ has its unique maximum at $0$, we obtain the bounds $e^{-1/2}\sqrt{\frac{2}{\pi}}a=\frac{2e^{-1/2}a}{\sqrt{2\pi}}\le 2a\phi(1)\le 2a\phi(a)\le\mathbb P(|W|\le a)\le 2a\phi(0)=\frac{2a}{\sqrt{2\pi}}=\sqrt{\frac{2}{\pi}}a$.
What does this tell us? It tells us that, strictly speaking, we have $\mathbb P(|W|\le a)\le \sqrt{\frac{2}{\pi}}a<a$, so the assumption $\mathbb P(|W|\le a)=a$ is wrong. But, not so strictly speaking, we have $\mathbb P(|W|\le a)=c(a)a$, where $e^{-1/2}\sqrt{\frac{2}{\pi}}\le c(a)\le \sqrt{\frac{2}{\pi}}$ is a coefficient depending on $a$, somewhere between $0.4839$ and $0.7979$, so the error is not catastrophic using this approximation. Since the argument in the answer was not formalized (and doesn't depend on constants, only on orders), this approximation is reasonable.
As I mentioned before, if you want to understand the error made by approximating $X^\circ_1\cdot X^\circ_2$ with $Y$ (respectively $\frac{1}{\sqrt{n}}W$), then please post a new question, I'd be happy to look into it.


TITLE: Simplifying Boolean Expression (De Morgan)
QUESTION [1 upvotes]: my question is basically, have I done the simplification correct so far? Is the answer correct? What is that simplification step I'm missing? Could anyone possibly outline the simplification steps for me in this question as I'm slightly confused?

REPLY [0 votes]: @amWhy already pointed out the mistake in your first step, and correctly indicated that on line 3 you should have:
$(A \lor B) \land \neg B$
What I want to add to this is that this last line is such a common pattern that there is an equivalence rule for it:
Reduction
$P \land (\neg P \lor Q) \Leftrightarrow P \land Q$
(this makes sense, since for $\neg P \lor Q$ to be true in addition to $P$ being true means that $Q$ has to be true. In other words, the term $P$ makes the term $\neg P \lor Q$ reduce to just $Q$)
And of course, for any equivalence, we always have a dual equivalence by swapping the $\land$'s and $\lor$'s, so we also have:
$P \lor (\neg P \land Q) \Leftrightarrow P \lor Q$
Anyway, given this Reduction principle, you last step is really just one step:
$(A \lor B) \land \neg B$ (Reduction!)
$A \land \neg B$
So yes, remember this Reduction rule: it's real handy to have when doing Boolean Algebra!


TITLE: Existence of probability measure, $\lim_{N\to\infty}\frac{1}{N}\sum_{j=1}^N f(x_j)$
QUESTION [2 upvotes]: Problem statement:

Let $x_j$ be a sequence in $[0,1]$.  Consider  $$\lim_{N\to\infty}\frac{1}{N}\sum_{j=1}^N f(x_j)\hspace 2cm (\star).$$  If ($\star$) exists for all $f\in C[0,1]$ then there is a positive Borel measure $\mu$ on [0,1] with $\mu([0,1])=1$ such that ($\star)=\int_{[0,1]} f d\mu$ for all $f\in C[0,1]$.  Furthermore, if $x_j\to x$ then ($\star$) exists for all $f$, and the above applies.  Which $\mu$ do we get?

Note:
There are other similar questions like this out there, but they don't ask you to show that $\mu([0,1])=1$ and/or ask about a subsequence instead of just a convergent subsequence.
Attempt:
Given that ($\star$) holds for all $f$ I can show that the the map that sends $f$ to ($\star$) is a positive linear functional.  Applying (one version) of the Riez Representation Theorem gives a measure $\mu$ as needed.  However, how do I know $\mu([0,1])=1$?
As for what happens when $x_j\to x$, I suspect since $f(x_j)\to f(x)$ and we are averaging that $(\star)=f(x)$.

REPLY [1 votes]: As for $\mu([0,1])=1$, consider the constant function $f(x) = 1$ on $x\in [0,1]$.
For the second statement, if $f\in C[0,1]$ then $f(x_j) \rightarrow f(x)$.
In this case, the limit will converge to $f(x)$.
Thus, the measure is the Dirac measure at the point $x$.


TITLE: Primes Represented By Quadratic Forms With Discriminant $-35$
QUESTION [2 upvotes]: I have been working through an exam question about which primes are represented by quadratic forms of discriminant $-35$.
So far I have been able to show that the only two reduced forms with discriminant $-35$ are $x^2 + xy + 9y^2$ and $3x^2 + xy + 3y^2$.
Next I showed that only primes $p$ with $\left(\frac{p}{5}\right)\left(\frac{p}{7}\right) = 1$ are represented by a form with discriminant $-35$.
However now the question asks me to show that if $\left(\frac{p}{5}\right) = \left(\frac{p}{7}\right) = 1$, $p$ is represented by $x^2 + xy + 9y^2$ and if $\left(\frac{p}{5}\right) = \left(\frac{p}{7}\right) = -1$, $p$ is represented by $3x^2 + xy + 3y^2$ and I can't see how to do it.
Please can someone show me how to do this?

REPLY [2 votes]: ADDED: it is worth your time to calculate exactly how the possible steps in Gauss reduction of a positive binary form match perfectly with matrix updates, given $Z$ take an "elementary" matrix $E$ and calculate $E^T ZE$
In the two by two case, we usually demand positive determinant, so the legal elementary matrices are
$$
E =
\left(
\begin{array}{cc}
1 & s \\
0 & 1
\end{array}
\right)
$$
$$
E =
\left(
\begin{array}{cc}
1 & 0 \\
s & 1
\end{array}
\right)
$$
$$
E =
\left(
\begin{array}{cc}
0 & -1 \\
1 & 0
\end{array}
\right)
$$
the way you wrote it, you need quadratic reciprocity. We begin with a (positive) prime  $p \neq 2,5,7$ such that
$$  (-35|p) = 1.$$ That means we can find some
$$  w^2 \equiv -35 \pmod p,  $$
$$  w^2 + 35 \equiv 0 \pmod p .  $$
Now if $w$ is even, we replace it by $p - w \equiv -w \pmod p.$ Now $w^2 + 35 \equiv 0 \pmod 4,$ together
$$ w^2 + 35 \equiv 0 \pmod {4p}. $$ This means there is an integer $t$ with
$$  w^2 + 35 = 4pt.  $$ Finally,
$$  w^2 - 4pt = -35. $$ We have constructed the binary form with discriminant $-35$ andcoefficients
$$  \langle p , \; w , \; t \rangle \; .  $$ Let
$H$ be the two by two matrix
$$
H =
\left(
\begin{array}{cc}
2p & w \\
w & 2t
\end{array}
\right)
$$
Reduce this, meaning construct the matrix $P$ with integer elements and determinant $1,$ such that $G =P^T HP$ is reduced; either
$$
G =
\left(
\begin{array}{cc}
2 & 1 \\
1 & 18
\end{array}
\right)
$$
OR
$$
G =
\left(
\begin{array}{cc}
6 & 1 \\
1 & 6
\end{array}
\right)
$$
Whichever one happens, take
$$  Q = P^{-1}, $$
and $Q^T G Q = H$
shows how to represent $p$ with integers.


TITLE: Ιnitial Value Problem $y'=(x-y)^{1/2}$, $y(2)=2$
QUESTION [1 upvotes]: Let the initial value problem :
  $$y'=(x-y)^{1/2}, y(2)=2$$
  Study the existence of solutions in a domain around the initial value and study its uniqueness too.

Attempt :
In order to show the uniqueness, I'll work around the function $f(x,y) = (x-y)^{1/2}$ and it's Lipschitz condition with respect to $y$ :
$$|f(x,y_2) - f(x,y_1)| = |(x-y_2)^{1/2} - (x-y_1)^{1/2}|$$
For the function $g(y) = (x-y)^{1/2}$, we can use the Mean Value Theorem since it's continuous and differentiable for $x-y \geq 0 \Leftrightarrow x \geq y$, for some $ξ \in (y_1,y_2) $ :
$$\big((x-y)^{1/2}\big)'|_{y=ξ} = \frac{(x-y_2)^{1/2} - (x-y_1)^{1/2}}{y_2 - y_1}$$
$$\Leftrightarrow$$
$$-\frac{1}{2\sqrt{x-ξ}}(y_2-y_1) = (x-y_2)^{1/2}-(x-y_1)^{1/2}$$
$$\Leftrightarrow$$
$$\bigg|-\frac{1}{2\sqrt{x-ξ}}\bigg|\bigg|y_2-y_1\bigg| = \bigg| (x-y_2)^{1/2}-(x-y_1)^{1/2}\bigg|$$
Now, I am not sure if I can make an assumption about a constant $L$ with respect to $y$. I observe that $k=\bigg|-\frac{1}{2\sqrt{x-ξ}}\bigg|$ is independent of $y_2,y_1$ but I cannot see how to make an inequality out of this, since it's value would change for any different $x$. Can you give me some help regarding how to show that there exists a Lipschitz constant or not ?
Also, regarding the solution, how would I study the existence of solutions in a domain around $y(2) = 2$ ? Can you please explain thoroughly with a complete solution regarding this, because it's an issue I have in a lot of similar problems.

REPLY [2 votes]: Let $z(x)=x-y(x)$, then $z'(x)=1-y'(x)$ hence $$z'(x)=F(z(x))\qquad z(2)=0$$ with $$F:u\mapsto1-\sqrt{u}$$
This is an autonomous differential equation with $F(z(2))\ne0$ hence, for every $x\geqslant2$, $$\int_0^{z(x)}\frac{du}{1-\sqrt{u}}=x-2$$ that is, using the change of variable $v=\sqrt u$, $$x-2=\int_0^{\sqrt{z(x)}}\frac{2vdv}{1-v}=\left.-2\log(1-v)-2v\right|_0^{\sqrt{z(x)}}=-2\log(1-\sqrt{z(x)})-2\sqrt{z(x)}$$ or, equivalently, $$(1-\sqrt{z(x)})\,e^{\sqrt{z(x)}}=e^{1-x/2}$$ One sees that $z(x)$ increases from $z(2)=0$ to $z(+\infty)=1$, and that, for every $x\geqslant2$, $y(x)$ is the unique solution of the identity $$(1-\sqrt{x-y(x)})\,e^{\sqrt{x-y(x)}}=e^{1-x/2}$$
On the other hand, again using $F(z(2))>0$, one sees that the solution is not defined on $x<2$.
Nota: The fact that $F(z(2))>0$ implies that every solution $z(x)$ is instantaneously positive, that is, that $z(x)>0$ for every $x>2$, and then Cauchy-Lipschitz uniqueness and existence criterion applies. However, a rigorous argument showing uniqueness and existence starting at $z(2)=0$ is lacking (Cauchy-Peano-Arzelà provides existence, confirmed by the explicit computations above, but the hypotheses of Osgood's criterion, for example, are not met).


TITLE: Force of interest problem from Exam FM
QUESTION [1 upvotes]: Tawny makes a deposit into a bank account which credits interest at a nominal interest rate of $10\%$ per annum, convertible semiannually.  At the same time, Fabio deposits $1000$ into a different bank account, whic is credited with simple interest.  At the end of $5$ years, the forces of interest on the two account are equal, and Fabio's account has accumulated to $Z$.  Determine $Z$.

The following is the way I tried.
Tawny has compound interest, semiannual, $10\%$ per annum.  So her force of interest is
$$\delta_T = \ln {1.1}$$
Fabio's simple interest has a force dependent on time $t$, so
$$\delta_F = \frac{d}{dt}(1+it) = \frac{i}{1+it}$$
At the end of $5$ years, the forces are equal to each other, so
$$\ln{1.1}=\frac{i}{1+5i}$$
implies
$$i\approx 0.1821$$
So I get the value of $Z$ as
$$Z = 1000(1+5i) \approx 1910.40$$
However, the answer is supposedly $1953$.  Can someone explain to me how they got that or what I am doing wrong?

REPLY [2 votes]: A nominal rate of 10% per annum, compound semiannually, means a rate of 5% per 6 month period.  Therefore, the annual effective rate is $1.05^2 - 1 = 10.25\%$, not 10%.


TITLE: Find all Pythagorean triples $x^2+y^2=z^2$ where $x=21$
QUESTION [2 upvotes]: Consider the following theorem:

If $(x,y,z)$ are the lengths of a Primitive Pythagorean triangle, then  $$x = r^2-s^2$$ $$y = 2rs$$ $$z = r^2+z^2$$ where $\gcd(r,s) = 1$ and $r,s$ are of opposite
  parity.

According to the previous theorem,My try is the following:
since $x = r^2-s^2$, $x$ is difference of two squares implying that $x \equiv 0 \pmod 4$. But $x=21 \not \equiv 0 \pmod 4$. Hence, there are no triangles having such $x$.
Is that right?
Added:
My argument is false here. Please refer to the appropriate answer.

REPLY [0 votes]: Since $A$ may be any odd number $\ge3$ and we can find one or more triples for any odd leg $A\ge 3$ using a function of $(m,A).$
We know that $A=m^2-n^2\implies n=\sqrt{m^2-A}.$
$$\text{We can let }n=\sqrt{m^2-A}\text{ where }\lceil\sqrt{A}\space\rceil\le m\le \frac{A+1}{2}$$
$\text{Note: }n\in \mathbb{R}\implies \sqrt{A}\lt m\land n<m\implies m\le\frac{a+1}{2}$
$$\text{For A=21}\quad m_{min}=\lceil\sqrt{21}\space\rceil=5\qquad \qquad m_{max}\frac{22}{2}=11$$
Testing for $5\le m\le 11$, we find $\mathbf{integers}$ for $(m,n)=(5,2)\text{ and }(11,10)$
$$f(5,2)=(21,20,29)\qquad f(11,10)=(21,220,221$$
But $21=3*7$
$$\text{For A=3}\quad m_{min}=\lceil\sqrt{3}\space\rceil=2\quad m_{max}=\frac{4}{2}=2\quad \quad 7f(2,1)=7(3,4,5)=21,28,35$$
$$\text{For A=7}\quad m_{min}=\lceil\sqrt{7}\space\rceil=3\quad m_{max}=\frac{8}{2}=4\quad \quad 3f(4,3)=3(7,24,25)=21,72,75$$