TITLE: harold domar growth model: solve the differential equation by integration
QUESTION [0 upvotes]: harold domar growth model: solve the differential equation (by integration)
When arranging the differential equation like this:
$$ \frac{dY}{dt} \cdot \frac{1}{Y(t)} = (\frac{s}{v})$$ and integrating both parts, I get the solution: $Y(t) = e^((\frac{s}{v})\cdot)\cdot C$, which is correct apparently
$C$ is the intergration constant, $s$ and $v$ are some parameters of the Harold Domar model
However, whenever I try to use the same technique of integrating right and left hand side when rearranging the equation like this:
$\frac{dY}{dt} - \frac{s}{v}\cdot Y(t) = 0$ , I get nowhere, well, I get something like $Y(1-\frac{s}{v} \cdot t) = c$.
How come it doesn't seem to work with different rearrangements of the equation? Does it matter? Or is it just my lack of integration skills? Many other rearrangements are possible, but I only find the solution with this particular one and I wonder why.

REPLY [1 votes]: $$Y'-\frac1\tau Y=0$$ can be integrated by noting that
$$\left(Y'-\frac1\tau Y\right)e^{-t/\tau}=Y'e^{-t\tau}+(e^{-t/\tau})'Y=(Ye^{-t/\tau})'.$$
Then
$$Ye^{-t/\tau}=C$$ and $$Y=Ce^{t/\tau}.$$
This technique brings no benefit as regards an homogeneous equation, but is useful to find non-homogeneous solutions.
$$Y'-\frac1\tau Y=Z$$ yields
$$Ye^{-t/\tau}=\int Z\,e^{-t/\tau}\,dt$$
and
$$Y=e^{t/\tau}\int Z\,e^{-t/\tau}\,dt.$$


TITLE: Prove that $6|2n^3+3n^2+n$
QUESTION [4 upvotes]: My attempt at it: $\displaystyle 2n^3+3n^2+n= n(n+1)(2n+1) = 6\sum_nn^2$
This however reduces to proving the summation result by induction, which I am trying to avoid as it provides little insight.

REPLY [0 votes]: to check for divisibility by 6 a number must be divisible by both 2 and 3 so we will prove that
$2n^3 + 3n^2 + n
= n (2n^2 + 3n +1)
= n (n+1) (2n+1)$
If $n$ is even then 2 divides $n$ and $n+1$ will be odd so $n+1$ can be $3k+2$ or $3k$ where $k$ is some integer.
If $3k+1 = n+1$ as it would make $n$ itself a multiple of 6 as our assumption that $n$ is even
So if $n+1$ is $3k$ it can be divided by 3 so no problem.
But if $n+1 = 3k+2$ then $2n+1$ will be $2(3k+1) +1 = 6k+3$ which is divisible by 3
and hence by 6.
Case 2: If $n$ is odd then $n+1$ is even and thus divisible by 2.

if $n$ is $3k$ then it is divisible by 3 so no issues
if $n$ is $3k+2$ as it makes $n+1=3k+3$ which is itself a multiple of 6 as our earlier assumption is that $n+1$ is even
if $n$ is $3k+1$ then $2n+1$ becomes $2(3k+1)+1=6k+3$ which is divisible by 3
and hence proved


TITLE: Combinatorics. Please explain me how to do it.
QUESTION [2 upvotes]: In how many different ways three persons A, B, C having 6, 7 and 8 one rupee coins respectively can donate Rs.10 collectively? This isn't a homework question. Please explain me the steps. Thank you!

REPLY [0 votes]: Let A,B,C donate $x_1,x_2,x_3$ coins with $x_i\geq 0$.
Then A/Q $\sum x_i=10$                   .....(1)
At first lets find all the solutions of this equation in integers.
The no. of such solution is $12C2$
Now we will find the no. of solution in which $x_1\geq 7$,(these solutions cant be considered),
To find this lets replace $x_1$ by $x+6$ where $x\geq 1$ putting this into 1 we have $x+x_2+x_3=4$ we will find no. of such solns. $5C2$
In this way we will find the other cases which are not possible.
Namely when $x_2\geq8$ in this case we have $4C2$ solutions , and the last one when $x_3\geq 9$ then we have $3C2$ solutions.
All the cases which cant be posiible are disjoint implying the total no. of solution =(total no. of cases)-(cases not possible).
$12C2-5C2-4C2-3C2$


TITLE: Solution methods for system of ODE's
QUESTION [0 upvotes]: I'm reading Coddington's "Introudction to ordinary differential equations". In chapter 6, the author discusses systems of differential equations. One of the exercises is to find the solution $\hat{\phi}$ the system $$y_1'=y_1,$$ $$y_2'=y_1 + y_2,$$
which satisfies $\hat{\phi}(0)=(1,2).$ Yet, there is no discussion about a systematic way to solve these kinds of systems. I know how to solve it by first writing the system in matrix form $$\hat{y}=\bigg(\begin{matrix}  1&0 \\1 &1 \end{matrix}\bigg)\ \bigg(\begin{matrix} y_1 \\y_2 \end{matrix}\bigg)$$
and then finding the general solutions with the eigenvalues/vectors of this matrix. This method is not mentioned in the book, so I think there might be other ways to solve this system. So are the other solution methods? I think its useful to have other tools besides the eigenvalue/vector type, since it is quite tedious to apply that to large systems.

REPLY [1 votes]: $$y'_1-y_1=0$$
$$ (y_1e^{-t})'=0$$
Integrate:
$$ y_1e^{-t}=C_1$$
$$\implies y_1=e^{t}C_1$$
Then solve the second DE:
$$y'_2-y_2=y_1$$
$$(y_2e^{-t})'=y_1e^{-t}$$
$$(y_2e^{-t})'=C_1$$
Integrate:
$$ y_2e^{-t}=C_1t+C_2$$
$$\implies y_2=e^{t}(C_1t+C_2)$$


TITLE: Prove $\int_{0}^{1}{x\arcsin(x) dx} \leq \frac{\pi}{4}$
QUESTION [0 upvotes]: I know how you can show this geometrically, but is there any way to prove this algebraically?

REPLY [0 votes]: You can use the following two known inequalities
$$x \leq |x|$$ and $$\left|\int_a^b f(x)dx\right|\leq \int_a^b|f(x)|dx.$$
We have
\begin{align*}
\int_0^1x\sin^{-1}xdx &\leq \left|\int_0^1x\sin^{-1}xdx\right|\\
&\leq \int_0^1x|\sin^{-1}x|dx\\
&\leq \frac{\pi}{2}\int_0^1xdx=\frac{\pi}{4}.
\end{align*}


TITLE: How to find the limit:$\lim_{n\to \infty}\left(\sum_{k=n+1}^{2n}\left(2(2k)^{\frac{1}{2k}}-k^{\frac{1}{k}}\right)-n\right)$
QUESTION [3 upvotes]: How to find the limit:$$\lim_{n\to \infty}\left(\sum_{k=n+1}^{2n}\left(2(2k)^{\frac{1}{2k}}-k^{\frac{1}{k}}\right)-n\right)$$
I can't think of any way of this problem
Can someone to evaluate this?
Thank you.

REPLY [2 votes]: $$k^{\frac1k}=e^{\frac{\log k}{k}}\sim1+\frac{\log k}{k}$$
$$2k^{\frac{1}{2k}}=e^{\frac{\log 2k}{2k}}\sim1+\frac{\log 2k}{2k}$$
$$ 2k^{\frac{1}{2k}}-k^{\frac1k}\sim1+\frac{\log 2}{k}$$
$$\sum_{k=n+1}^{2n}\left(2(2k)^{\frac{1}{2k}}-k^{\frac{1}{k}}\right)\sim n+\log 2\sum_{k=n+1}^{2n}\frac1k\sim n+\log 2 \, \log \left(\frac{2n}{n+1}\right)$$
$$\sum_{k=n+1}^{2n}\left(2(2k)^{\frac{1}{2k}}-k^{\frac{1}{k}}\right)-n\sim \log 2 \, \log \left(\frac{2n}{n+1}\right)\to(\log2)^2$$


TITLE: Making a neat orthonormal basis with one vector being: $\frac{1}{\sqrt{77}}\begin{pmatrix} 4 & 6 & 5 \end{pmatrix}^T$
QUESTION [0 upvotes]: I'm trying to make an orthonormal basis, where one vector is:
$\frac{1}{\sqrt{77}}\begin{pmatrix}
4 \\
6 \\
5
\end{pmatrix}$
After that I need to do a lot of other calculations (by hand ... ). This is part of a Schur decomposition. I'm trying to get a little neat orthonormal basis here as otherwise I'm making thousands of mistakes.
By just using gram-schmidt I get a very ugly basis. Is there a way to do this a little bit more sophisticated ? For example, I would like to call $t=\frac{1}{\sqrt {77}}$. And having all basisvectors of the form $$u=
t\begin{pmatrix}
a \\
b \\
c
\end{pmatrix}$$
where $a,b,c$ is not some kind of exotic $-\frac{\sqrt{61}}{\sqrt{4697}}$ kind of thing.

REPLY [3 votes]: You can do a variant of the Euclidean algorithm on $\{4,6,5\}$ to produce $0$ to get a nice second vector. A quick ad-hoc calculation gives:

$1 = 1 \cdot 6 - 1 \cdot 5$
$1 = 1 \cdot 5 - 1 \cdot 4$
$0 = 1 \cdot 1 - 1 \cdot 1 = 1 \cdot 4 - 2 \cdot 5 + 1 \cdot 6$

So we know $(4,6,5)$ is orthogonal to $(1,1,-2)$.
A more systematic method would be to use lattice reduction on the orthogonal complement to $(4,6,5)$. First do integer row reduction:
$$
\left( \begin{array}{ccc|c}
1 & 0 & 0 & 4
\\ 0 & 1 & 0 & 6
\\ 0 & 0 & 1 & 5
\end{array} \right) \mapsto \left( \begin{array}{ccc|c}
-1 & 0 & 1 & 1
\\ 6 & 1 & -6 & 0
\\ 5 & 0 & -4 & 0
\end{array} \right)$$
(you may, of course, get a different matrix) Then take the two rows that give zero and lattice reduction gives
$$\left( \begin{array}{ccc}
-1 & -1 & 2
\\ 3 & -2 & 0
\end{array} \right)$$
(I used the LLL command in sage) So $(-1,-1,2)$ is the shortest integer vector perpendicular to $(4,6,5)$. Then you could take the cross product to get a third.
The nice thing about the lattice reduction approach is that it generalizes to higher dimensions. If you already have $m$ orthogonal vectors in an $n$-dimensional space, you can do row reduction to get a basis for the integer orthogonal complement to your $m$ vectors, and do LLL to get a short one, and take that as your $(m+1)$-th vector.
Of course, the later vectors will probably be relatively large. Trying to make them all short is a much harder problem, I think, and I'm not sure how to go about doing that. My first instinct would be to take any orthogonal vectors, then try some sort of simulated annealing optimization procedure to find smaller sets of orthogonal vectors.


TITLE: Computing pullbacks of certain function
QUESTION [0 upvotes]: Let $I^3 : [0, 1]^3 \to \mathbb{R}^3, I^3(x) = x$. Define $I^3 _{(i, \alpha)} : [0, 1]^2 \to [0, 1]^3, \alpha \in \{0, 1\}$ like that : $$I^3_{(1, a)}(x^1, x^2) = I^3(\alpha, x^1, x^2), I^3_{(2, a)}(x^1, x^2) = I^3(x^1, \alpha, x^2), I^3_{(3, a)}(x^1, x^2) = I^3( x^1, x^2, \alpha).$$
I would like to compute pullbacks $(I^3_{i, \alpha})^*(dx^1 \wedge dx^2)$, however I am lost in what is happening here. How could I start computing something simpler eg. $(I^3_{1, \alpha})^*(dx^2)$?

REPLY [0 votes]: $(I^3_{(1,\alpha)})^*dx^1 = 0 \\
(I^3_{(1,\alpha)})^*dx^2 = dx^1 \\
(I^3_{(1,\alpha)})^*dx^3 = dx^2 \\
(I^3_{(2,\alpha)})^*dx^1 = dx^1 \\
(I^3_{(2,\alpha)})^*dx^2 = 0 \\
(I^3_{(2,\alpha)})^*dx^3 = dx^2 \\
(I^3_{(3,\alpha)})^*dx^1 = dx^1 \\
(I^3_{(3,\alpha)})^*dx^2 = dx^2 \\
(I^3_{(3,\alpha)})^*dx^3 = 0 \\
(I^3_{(1,\alpha)})^∗(dx^1∧dx^2)=0∧dx^1=0 \\
(I^3_{(2,\alpha)})^*(dx^1 \wedge dx^2) = dx^1 \wedge 0 = 0 \\
(I^3_{(3,\alpha)})^∗(dx^1∧dx^2)=dx^1∧dx^2
$


TITLE: Double Slit Experiment with detector on the double slit screen itself
QUESTION [0 upvotes]: I’ve looked for a version of the double slit experiment in which the detector (of photo, electron, etc.) is also placed on the slit screen itself, the side facing the emitter. Thinking: As we know, the probability wave of a single emitted photon passes through both slits (assuming no measurement there) and then the photon is measured in a specific spot on the screen on the other side (as photons collect, an interference pattern emerges). The space between the slits (and above, etc.) must also be included in the probability set of the photon’s journey or impact point—ie is included in the wave function. In other words, presumably, not all photons register on the detector screen, some must be lost to areas outside the detector. Would measuring a case where the photon is measured hitting the screen instead of passing through the slit(s) disrupt the interference pattern of photons on the final detector screen (the one on the other side of the slits)?  Thanks in advance…

REPLY [1 votes]: No, in principle detecting photons that would otherwise have hit the barrier, rather than passing through the slits in the barrier, should not affect the interference pattern.
Quantum mechanical explanations of the two slits experiment typically treat the barrier as something that blocks the propagation of the wave function everywhere apart from through the slits. Effectively it means that the barrier is assumed to be opaque, so it will either reflect or absorb particles that hit it.
The barrier could, for example, be a photographic film with two slits cut into it, which would show where particles had impinged upon it.


TITLE: Differential Geometry-Hodge Star
QUESTION [2 upvotes]: The Hodge star is given by $$*(dx^{i_1}\wedge dx^{i_2}\wedge....\wedge dx_{i_p})=\frac{1}{(n-p)!}e_{i_1 i_2....i_p i_{p+1}...i_n}dx^{i_{p+1}}\wedge dx^{i_{p+2}}\wedge....\wedge dx^{i_n}$$
The question is to prove $$**\omega_p=(-1)^{p(n-p)}\omega_p$$
by repeating the * operator on a p-form $\omega_p$. Now, how shall I prove it?
Here, $e_{ijk....}$ is antisymmetric tensor in n-dimension

REPLY [3 votes]: First of all, note that in the formula you give us,
$$*(dx^{i_1}\wedge dx^{i_2}\wedge\dots\wedge dx^{i_p})=\frac{1}{(n-p)!}e_{i_1 i_2\dots i_p i_{p+1}...i_n}dx^{i_{p+1}}\wedge dx^{i_{p+2}}\wedge....\wedge dx^{i_n},
$$
there is a summation involved! In fact, there are $n-p$ sums, since the indices are being contracted. I won't use Einstein's convention in this post.
Explicitly we have:
$$*(dx^{i_1}\wedge dx^{i_2}\wedge\dots\wedge dx^{i_p})=\frac{1}{(n-p)!}\sum_{j_{p+1},\dots j_{n}=1}^ne_{i_1 i_2\dots i_p j_{p+1}...j_n}dx^{j_{p+1}}\wedge dx^{j_{p+2}}\wedge....\wedge dx^{j_n},
$$
where $e_{\cdots}$ is the Levi-Civita tensor.
In fact you're summing the same $(n-p)$-form $(n-p)!$ times! This follows from the definition of the Levi-Civita tensor and that $dx_i\wedge dx_j = -dx_j\wedge dx_i$.
To find how many times you're summing the same term observe that whenever an index is repeated, the term vanishes. In this case, the indices in the sum can take only $n-p$ distinct values and must all be different. So we get $(n-p)!$ possible combinations.
From this we get that (no sum involved!):
$$
*(dx^{i_1}\wedge dx^{i_2}\wedge\dots\wedge dx^{i_p})=e_{i_1 \dots i_n}dx^{i_{p+1}}\wedge dx^{i_{p+2}}\wedge....\wedge dx^{i_n}.
$$
Now, applying again the star operator we get:
\begin{align}
**(dx^{i_1}\wedge dx^{i_2}\wedge\dots\wedge dx^{i_p})
&=e_{i_1 \dots i_n}*dx^{i_{p+1}}\wedge dx^{i_{p+2}}\wedge....\wedge dx^{i_n}\\
&=e_{i_1\dots i_p i_{p+1}\dots i_n} e_{i_{p+1}\dots i_{n} i_1\dots i_{p}} dx^{i_1}\wedge dx^{i_2}\wedge\dots\wedge dx^{i_p}.
\end{align}
To conclude, observe that
$$
e_{i_1\dots i_p i_{p+1}\dots i_n} e_{i_{p+1}\dots i_{n} i_1\dots i_{p}} = (-1)^{p(n-p)}.
$$
(To show this count how many permutations you must do on the second term to obtain the first.)